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1 MECHANICS SECTION 1 QUESTIONS 1.

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Two vectors have magnitudes in the ratio 1:2. Their vector sum is perpendicular to the smaller vector. The angle between the vectors is a) 1500 b) between 1500 and 1200 0 c) between 1200 and 900 c) 120 Two satellites A and B revolve round earth in circular orbits of radius r and 1.01r. The percentage increase in the period of B with respect to A is a) 0.5 b) 0.75 c) 1 d) 1.5 Which of the following pair has not same dimension? a) Curie and angular velocity b) velocity gradient and frequency of light c) Gravitational field and latent heat d) potential gradient and force per unit charge Three particles of same mass rest on the corners of an equilateral triangle. The gravitational force between any two of them has a magnitude F. Then gravitational force on any of them has magnitude a) F b) 2F c) √3 F d) √2 F The momentum of a car is increased by 10% and then decreased by 10%. In the process the kinetic energy of car a) increases by 2% b) decreases by 2% c) increases by 1% d) decreases by 1% The position vector of a particle is r = (a cos ωt ) ˆi + a (sin ωt ) ˆj . The velocity vector is a) perpendicular to position vector b) parallel to position vector c)always directed towards origin d) always directed away from the origin Read the following statements A and B: A: If a quantity has units, it will have dimensions. B: If a quantity has dimensions, it will have units Of these statements a) Both A and B are true b) Both A and B are false c) A is true but B is false d) B is true but A is false The equation of the motion of a body initially at rest is given by

dv = 8-4v, where v is dt

velocity of the body at the time. Check the wrong statement a) terminal velocity of the body is 2m/s b) initial acceleration of the body is 8 m/s2 c) the velocity of the body when acceleration is half the initial value is 1 m/s d) velocity is related to time by the equation v = at + bt2 where a and b are constants.

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A set of N cubical blocks rest on a smooth horizontal table such that distance between 2 near surfaces of the block is L. The block at one end is given a speed v to the next at the time t = 0. All collisions are perfectly inelastic. then the last block starts to move after a time ( N − 1) L ( N − 1)L N ( N − 1)L N ( N − 1)L a) b) c) d) v 2v 2v v 10. In the previous question, centre of mass of the system will have a final speed v v c) v d) a) zero b) 2 N 11. A rocket moves with a speed v where air resistance is proportional to the cube of its speed. If the speed of the rocket is to be increased by 1%, the increase in the power will be a) 1% b) 2% c) 3% d) 4% 3 12. Water issuing out as a horizontal jet with a speed v1 at a rate V m per second strikes a plate normally moving towards it with a speed v2. If density of water is p and water splashes down after hitting the plate, the force on the plate in newton is 9.

a) pv1 V

b) p(v1 +v2) V

c) p(v2-v1)V

d)

pv 1 2 V v1 + v 2

13. Read the following statements carefully.

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Statements x: A vector quantity has magnitude and direction. Statement y: All quantities, which have magnitude and direction, are vectors. Here a) both statements are correct b) both statements are not correct c) x is correct and y is not d) y is correct and x is not If R is the radius of earth’s orbit round sun, the angular momentum of earth round sun is proportional to a) R2 b) R3/2 c) R d) R1/2 Two particles P and Q move towards each other due to mutual gravitational force. At a certain time the speed of P is v and speed of Q is 2v. The speed of centre of mass of the system is a) 0 b) v/2 c) 3v/2 d) v Two masses 0.4 and 0.6 kg are suspended from the zeroeth and hundredth cm division of a meter scale of negligible mass. It is required to rotate the meter scale with constant angular velocity with minimum work. The axis of rotation should pass through at a) 40 cm b) 50 cm c) 60 cm d) 70 cm An infinite number of point masses m kg each are placed at a distances x = 1cm, x = 2cm, x = 8 cm.....etc from a point 0 on the x-axis. The gravitational potential at the point 0 in J/kg is a) infinity b) zero c) -100 Gm d) -200 Gm A simple pendulum of length L has a bob of mass m. It sis displaced by an angle θ to the vertical. If T is a tension of a string, v is the speed of the bob then, check the correct statement a) T = mg cos θ

b) T = mg cos θ -

mv 2 L

mv 2 d) T cos θ = mg L n 19. The gravitational force between a planet and a satellite varies by a distance law f αr . The period of the satellite will be independent of distance if n is equal to a) 1 b) 2 c) -1 d) 3/2 c) T = mg cosθ +

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3

20. Two boys A and B hold two ends of a light rope carrying a 1 kg weight at the centre. If the

length of the rope is 1 metre the work done by the boys to straighten the rope is a) 0 b) infinity c) 4.9 J d) 9.8 J 21. Two particles of mass m each move in a circle of radius r due to their mutual gravitational force. The period of their motion is

a)

22.

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2π

r3 Gm

b ) 4π

r3 Gm

c) π

r3 Gm

d ) 2π

Gm

r3 A car, a truck and a lorry of masses in the ratio 1:2:3 have equal kinetic energy. Their braking forces are also same. When brakes are applied at the same time, they are brought to rest in distances s1, s2 and s3 respectively. Then a) s1 > s2 > s3 b) s1 < s2 < s3 c) s1 = s2 = s3 d) s1 + s2 = s3 A shell is fired with an initial velocity u at an angle α to the horizontal. At highest point on its path, it explodes into two equal fragments. One falls vertically down. The distance between the fragments when they reach the ground is b) u2 sin 2α/ 2g a) u2 sin 2α/g 2 d) 3u2 sin 2α/ g c) 2u sin 2α/ g The time of a flight of a projectile is 4 s. One second after projection it moves at an angle 450. Its angle of projection is b) tan-1 (4/3) c) tan-1 (3/2) d) tan-1 (2) a) 600 The velocity vector of the projectile in the above question is parallel to a ) ˆi + 2 ˆi b) 2 ˆi + ˆi c) 2 ˆi + 3 ˆi d ) 3 ˆi + 2 ˆi

26. A fly wheel starting from rest and rotating with uniform angular acceleration turns through

100 rad during third second. During fifth second it will turn through an angle (in rad) a) 100 b) 166.7 c) 180 d) 200 G G G 27. When three forces f 1 = 3ˆi + 4ˆj − 5kˆ , f 2 = 3ˆi + 2ˆj − k and f 3 = x (ˆi + ˆj − kˆ ) act on a body, the body moves with uniform velocity. The value of x here is a) -6 b) 6 c) -9 d) 9 28. A satellite is moving in a circular orbit of radius x around the earth of radius R. The acceleration of the satellite is gR R R2 c) g a) g b) d) g 2 x x x 29. Check the correct statement for a body in uniform circular motion

a) the body has constant centripetal force c) the body has constant velocity

b) the body has varying centripetal force d ) the body obeys T2 α R3 law 30. A body of mass 2 kg moves with a velocity (3 ˆi + 4ˆj) t. The magnitude of force acting on it at this instant is a) 6N b) 8N c) 10N d) 7N 2 31. A particle moves in a parabolic path whose equation is y = 4ax , with a constant x component of velocity equal to A. The acceleration of the particle is a ) 4aAkˆ b) 8aA 2 ˆj c) 4aA 2 kˆ d) 8aA 2 kˆ 32. For an earth satellite longer orbits will correspond to

a) longer periods and larger velocity

b) longer periods and smaller velocity

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c) smaller periods and smaller velocity

d) smaller periods and larger velocity

33. The velocity-time graph of an aero plane moving along a run way is a parabola. This means

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aero plane has a) constant but large acceleration b) acceleration proportional to time c) acceleration proportional to square of time d) acceleration proportional to cube of time The period of a simple pendulum with an iron bob is T. It will be less than T, when the bob is subjected to a) a magnetic field directed upwards b) an electric field directed upwards c) a magnetic field directed downwards d) an electric field directed downwards A body falls from a height h and rebounds from a surface of coefficient of restitution e = 1/√2. The number of collisions it will make before it stops is equal to a) 1 b) 2 c) 4 d) infinite The total distance travelled by the body before it stops to rebound is a) 2h b) 3h c) 4h d) 3√2h A stone is thrown vertically up. It reaches a maximum height of 50 m. The height at which kinetic energy becomes 60% of the initial value is (in m) a) 30

b) 50 0.6

c) 50 0.4

d) 20

38. A trolley of mass 100 kg moves on a smooth horizontal rails with a monkey of mass 20 kg

inside it, with a speed of 10 m/s. The monkey at some point jumps vertically up and catches overhanging branch of a tree. The speed of the trolley after this would be a) 12 m/s b) 10 m/s c) 15 m/s d) 8.33 m/s G ˆ ˆ 39. F is a force represented by the vector F = 2i + 4 j and s is displacement vector given by G ˆ ˆ s = i + j. The component of force in the direction of displacement is a)

20

b)

2

c) 3 2

d) 6

40. A body of mass 1 kg is dropped from the top of a tower on a windy day. The wind exerts a

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constant horizontal force of 20N on the body. The path of the body will be a) a parabola b) a straight line c) an ellipse d) an irregular curve A stone A is dropped from the top of a tower of height 40 m. At the same time another stone B is projected from the bottom of the tower with an initial velocity such that the stones collide midway. The initial speed of second stone B is (Take g = 10 m/s2) a) 5 m/s b) 10 m/s c) 20 m/s d) 15 m/s A TV signal sent from a station A to Insat 1-C is received back in a neighboring station B in a time t seconds. The value of t here is nearest to a) 1 s b) 0.1 s c) 0.125 s d) 0.24 s A person is standing at a distance of 8m from a train. The train starts with acceleration 1 m/s2. The person also starts at the same time with an acceleration 2m/s2. The time in which he catches the train is a) 2 s b) 4s c) 8s d) 16s A particle moves with a constant velocity parallel to the x-axis . Its angular momentum with respect to the origin a) increases b) decreases

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c) first increases and then decreases

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d) remains the same 0

45. The escape velocity of a body projected at an angle 5 with the horizontal from the earth in (in

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km/s) is a) 11.2 b) 11.2 cos 50 c) 11.2 sin 50 d) infinite Two uniform gold spheres of radius R each exert a gravitational force of F when in contact. The gravitational force of two uniform gold spheres of radius 2R in contact will be a) F/16 b) F/4 c) F d) 16F The altitude of earth satellite in terms of radius of earth R, when it has a velocity 4km/s is a) R b) 2R c) 3R d) 4R Two forces 100N and 150N act on a body of mass 100 kg. The minimum acceleration this body can have under their combined action is (in m/s2) a) 3.5 b) 1.5 c) 1.25 d) 0.5 A passenger in a car taking a curve feels a centrifugal force on him. This force is a) exerted by car on him b) exerted by ground on him through the car c) the reaction to the centripetal force d) due to inertia The gravitational field at a point at a distance of r from the centre of a uniform sphere of radius R for r < R is proportional to a) 1/r2 b) 1/r c) r d) r2 A particle moving in a horizontal circle of radius r has centripetal force F = -k/r2, where k is a constant. The kinetic energy of the particle is c) 2k/r d) k/2r a) k/r b) k2/r2 The minimum acceleration with which a fireman can climb down a rope of breaking tension 60% of his weight is a) 0.4 g b) 0.6 g c) 0.2 g d) 0.3 g ABC is a triangular plate of uniform thickness with sides AB: BC: CA in the ratio 1:3:5. If IAB is the moment of inertia about AB, IBC that about BC and ICA that about CA, which one of the following is correct ? b) IAB + IBC = ICA c) ICA is minimum d) IAB < IBC a) IBC = IAB A block takes t seconds to slide down a smooth inclined plane of angle 450. The same block takes 2t seconds to slide down a rough inclined plane of same angle. The coefficient of friction of the second plane is a) 0.5 b) 1/√2 c) 0.75 d) 1/2√2 A heavy nucleus at rest emits can α particle. Then a) momentum of α is greater than that of nucleus b) momentum of nucleus is greater than that of α c) kinetic energy of nucleus is greater than that of α d) kinetic energy of α is greater than that of nucleus A thin uniform needle has a moment of inertia I1 about centre of mass axis. It is bent into a ring. The moment of inertia is I2 about c.m. then I1 / I2 will be b) π2/4 c) π2/12 d) 3/π2 a) π2/3 Assuming earth to be a uniform sphere of density ρ, period of a satellite close to earth is related to the density by the equation ‘t’ is proportional to a) ρ b) √ρ c) 1/√ρ d) 1/ρ Three forces each of magnitude 1 N act from one corner towards the other three corners of a square. Their sum has a magnitude nearest to a) 3 N b) 1 N c) 1.4 N d) 2.4 N

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59. A laboratory on earth can be taken as

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a) an inertial frame strictly b) an inertial frame approximately c) a non-inertial frame strictly d) either depending on where we are looking from A particle is falling freely along an inclined plane. It travels the last 1/3 of the distance with 1/nth of the total time of fall. Here n is nearest to a) 2 b) 3 c) 4 d) 5 A car is travelling on a level road of coefficient of static friction 0.8. The car cannot have an acceleration more than a) 0.8 m/s2 b) 8 m/s2 c) 4 m/s2 d) 10 m/s2 A mixture of sea water of specific gravity 1.1 is mixed with fresh water of specific gravity 1. If the mixture has specific gravity 1.04 the ratio of volume of sea water to volume of fresh water is a) 1:1.04 b) 5:4 c) 5:2 d) 3:2 A disc with a small hole is rotating through centre of mass axis with a constant period. If the centre of the disc is heated the period of rotation will a) increase b) decrease c) remain same d) may increase or decrease A uniform cylinder of length L and mass M having cross-sectional area A is suspended, with its length vertical, from a fixed point by a mass less spring of force constant k, such that it is half-submerged in a liquid of density ρ at equilibrium position. When the cylinder is given a small downward push and released it oscillates with small amplitude and with a frequency

a)

1 k − Aρg 2π M

1/ 2

1 k − ρgL2 c) 2π M

b)

1 k + Aρg 2π M

d)

1 k − Aρg 2π Aρg

1/ 2

1/ 2

1/ 2

65. Imagine a light planet revolving around a very massive star in a circular orbit of radius R with

a period of revolution T. If the gravitational force of attraction between the planet and the star is proportional to R-5/2, then a) T2 is proportional to R2 b) T2 is proportional to R7/2 2 3/2 d) T2 is proportional to R3.75 c) T is proportional to R 66. *A particle is acted upon by a force is constant magnitude which is always perpendicular to the velocity of the particle. The motion of the particle takes place in a plane. It follows that a) its velocity is constant b) its acceleration is constant c) its kinetic energy is constant d) it moves in a circular path 67. 1 kg of cotton is balanced by 1 kg of a brass weight in a balance covered with a bell jar. If the bell jar is evacuated, a) the cotton pan will go down b) the brass pan will go down c) both will remain in equilibrium d) cotton will first go up and then come down 68. *A diametrical tunnel is dug along the earth. A particle is dropped from a point at a height h directly above the tunnel. Assuming earth’s density to be uniform, a frictional forces are negligible, a) the particle reaches the same height on the other side b) the particle will have zero acceleration when it passes through the centre of the earth. c) the particle has maximum acceleration at the point of release

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d) the particle will oscillate simple harmonically 69. When a running athlete stops after a race

a) only his momentum is conserved by himself b) only his kinetic energy is conserved by himself c) both momentum and kinetic energy are conserved by him d) neither momentum nor kinetic energy is conserved by him 70. A block of mass 0.1 kg is held against a wall by applying a horizontal force of 5N on the block. If the coefficient of friction between the block and the wall is 0.5 the magnitude of the frictional force acting on the block is a) 2.5 N b) 0.98 N c) 4.9 N d) 0.49 N 71. A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration ac is varying with time t as ac = k2rt2 where k is a constant. The power delivered to the particle by the forces acting on it is a) 2πmk2r2t b) mk2r2t c) (mk2r2t5)/3 d) zero 72. One end of a long metallic wire of length L is tied to the ceiling. The other end is tied to a massless spring of spring constant K. A mass m hangs freely from the free end of the spring. The area of cross section and the Young’s modulus of the wire are A and Y respectively. If the mass is slightly pulled down and released, it will oscillate with a time period T equal to

a ) 2π

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b) 2π

m(YA + KL) YAK

mL mYA d ) 2π YA KL Among the forces in nature, friction can be classified into a) gravitational b)electro magnetic c) weak nuclear d) strong nuclear When a body covers one quadrant of a circle of radius 0.1 metre, the ratio of distance travelled by it to the magnitude of displacement is a) 1 b) π/√2 c) π/2 d) π/2√2 When a person walks along east, the force of friction is along a) east b) west c) any where from east to west d) anywhere from north to south If distance of earth from sun reduces to half the present value, the number of days in an year would be nearly a) 730 b) 229 c) 182 d) 130 Two bodies of same mass but of different material are released from same height in a medium where air resistance cannot be neglected. Then a) the denser body reaches ground first b) the less dense body reaches ground first c) both fall through in the same time d) cannot be answered from the data The level of water in a trough when a train is at rest is horizontal. The train starts to move with uniform acceleration towards right. Then water level will a) still be horizontal b) incline up towards rear of the train c) incline up towards front of the train d) become concave shaped. c) 2 π

73.

m K

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79. (Fig) Three equal weights of mass 2 kg each are hanging on a string passing over a fixed

pulley as shown in the fig. What is the tension of the string connecting weights B and C ? a) 1.3 N b) 1.3 kgwt c) 3.3 N d) 19.6 N 80. If a body moves under constant power, the distance s travelled by it will be related to the time t by the equation a) s α t3/2 b) s α t2/3 B 2 d) s α t1/2 c) s α t 81. Two bodies M and N of equal mass are suspended from two separate massless springs of spring constant k1 and k2 C A respectively. If the two bodies oscillate vertically such that their maximum velocities are equal, the ratio of amplitude of vibration of M to that of N is equal to a)

k1 k2

b)

k2 k1

c)

k2 k1

d)

1

k1 k2

82. If in the above question, the bodies oscillate so that they have equal maximum acceleration,

the amplitude ratio of M to N is a)

k1 k2

b)

k1 k2

c)

k2 k1

d)

1

k2 k1

6

83. *A linear harmonic oscillator of force constant 2 x 10 N/m and amplitude 0.01 m has a total

mechanical energy of 160 J. Its a) maximum potential energy is 100 J b) maximum kinetic energy is 100 J c) maximum potential energy is 160 J d) minimum potential energy is zero 84. A boat which has a speed of 5 km/h in still water crosses a river width 1 km along the shortest possible path in 15 minutes. The velocity of the river water in km/h is a) 1

b) 3

c) 4

d)

41

85. A particle executes simple harmonic motion of frequency, f. The frequency with which its

kinetic energy oscillates is

f d) 4f 2 86. A shell is fired from a cannon with a velocity V(m/s) at an angle θ with the horizontal direction. At the highest point in its path it explodes into two pieces of equal mass. One of the pieces retraces its path to the cannon and the speed (in m/s) of the other piece immediately after the explosion is 3 3 a) 2 v cosθ b) 3v cosθ c) v cosθ d) v cosθ 2 2 a) 2f

b) f

c)

V0 αt (1-e ), where V0 is a α

87. The position of a particle at time t, is given by the relation: x(t) =

constant and α > 0. The dimensions of V0 and α are, respectively, a) M0L1T0 and T1 b) M0L1T1 and T c) M0L1T-1 and T2 d) M0L1T-1 and T-1 88. In a simple harmonic motion, when the displacement is half the amplitude, what fraction of the total energy is kinetic?

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3 1 1 b) zero c) d) 4 4 2 th 89. If radius of earth contracts to (1/n) , with its mass remaining constant, the duration of the day in hour will be 24 24 a) 24 b) c) 24 n d) 2 n n 90. A uniform chain of mass M and length L rests on a smooth table with one fourth of its length hanging. the work required to pull the hanging portion is MgL MgL MgL MgL a) d) b) c) 2 4 16 32 91. When n vectors of different magnitudes are added the sum is zero. Then n cannot be a) 11 b) 6 c) 4 d) 2 92. *In figure, .the spring balance A reads 2 kg with a block m suspended form it. A balance B reads 5 kg when a beaker A with liquid is put on the pan of the balance. The two balances are now so arranged that the hanging mass is inside the liquid in the beaker as shown in fig. In this situation a) the balance A will read more than 2 kg b) the balance B will read more than 5 kg c) the balance A will read less than 2 kg and B will read more than 5 kg B d) the balances A and B will read 2 kg and 5 kg respectively 93. The moment of inertia of a thin square plate ABCD, (Fig) of uniform thickness about an axis passing through the centre 0 and perpendicular to the plane of the plate is (where I1, I2, I3, I4 are MI about the axis in its I I plane) a) I1 + I2 b) I3 + I4 c) I1 + I3 d) I1 + I2 + I3 + I4 I 94. A tube of length L is filled completely with an incompressible liquid of mass M and closed at both the ends. The tube is then rotated in a horizontal plane about I one of its ends with a uniform angular velocity ω. The force exerted by the liquid at the other end is a) Mω2 L/2 b) Mω2L c) Mω2L/4 d) Mω2L2/2 95. A car is moving in a circular horizontal track of radius 10m with a constant speed of 10 m/s. A plumb bob is suspended from the roof of the car by a light rigid rod of length 1.00m. The angle made by the rod with the track is a) zero b) 300 c) 450 d) 600 96. During a rainstorm, raindrops are observed to be striking the ground at an angle of θ with the vertical. A wind is blowing horizontally at the speed of 5.0 m s-1. The speed of raindrop is (in m/s) 5 5 a) 5 sinθ b) c) 5 cosθ d) sin θ cos θ 97. A 2 kg stone at the end of a string 1 m long is whirled in a vertical circle at a constant speed of 4 m s-1. The tension in the string will be 52 N when the stone is a) at the top of the circle b) halfway down a)

1

3

2

4

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c) at the bottom of the circle

d) anywhere on the circle

98. The volume of an air bubble is doubled when it rises from the bottom to the surface of a lake.

The depth of the lake is approximately a) 10 m b) 15 m c) 25 m d) 30 m 99. A body of certain material floats with 9/10 of volume submerged in water in earth. In a planet, where acceleration due to gravity is five times that of earth, the submerged fraction will be

9 5 9 9 9 b) c) d) 10 50 10 250 100. If the change in the value of g at a height h above the surface of the earth (radius R) is the same as that at a depth x below it (assuming h > T0), at a distance r from the source. If the temperature and distance are doubled, the power P1 received by the surface becomes approximately (P original power) a) P b) 2P c) 4P d) 16 P Two rods of same length and area but made of different materials are 1 welded together as shown in the fig. If λ1 and λ2 are the thermal conductivity of each of them respectively, the thermal conductivity 2 of the system will be a) λ1 + λ2 b) λ1 λ2 / λ1 + λ2 c) 2λ1 λ2 / λ1 +λ2 d) λ1 + λ2/2

78. A bullet is stopped by a target the temperature rise of the bullet is ∆T, without loss of energy.

If the bullet’s velocity is doubled and it loses 50% of the energy, the rise in temperature of the bullet will be a) ∆T b) 2∆T c) 3 ∆T d) 4 ∆T 0 0 79. One mol of monatomic gas is heated from 0 C to 100 C at constant pressure. The change in its internal energy is a) 2.3 J b) 12.4 J c) 120 J d) 1250 J 80. If a temperature of source and sink of a Carnot engine is increased by 10% its efficiency a) increases by 10% b) decreases by 10% c) increases by 1% d) remains the same 0 81. A test tube of water at 4 C is immersed inside a large block of ice. Which of the following will happen? a) Water in the test tube will freeze b) Water in the test tube will not cool c) Water in the test tube will reach freezing point but will not freeze

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d) water in the test tube will reach between freezing point and 40 C 82. Which of the following cases, the work done by a gas is minimum ? P Y

Y

Y

X V (1 )

a) 1

Y

X (2 )

b) 2

X (3 )

c) 3

X (4 )

d) 4

83. If distance between sun and earth is reduced to half the present value, solar constant

84.

a) increases to 4 times b) increases to 2 times c) decreases to 4 times d) remains the same A cylinder of radius R made of a material of thermal conductivity K1 is surrounded by a cylindrical shell of inner radius R and outer radius 2R R 1 2 made of a material of thermal conductivity K2. The K two ends of the combined system are maintained at two different temperatures. there is no loss of heat K across the cylindrical surface and the system is in a steady state. The effective thermal conductivity of the system is 2R a) K1 + K2 b) K1 K2 / (K1 +K2) c) K1 + 3K2)/4 d) (3K1 + K2)/4 A cube of side a is heated through 1oC. If α is the linear expansivity of the material of the cube, its increase in area is a) a2α b) 4a2α c) 6a2α d) 12 a2α The maximum temperature of 1g hot water which can be mixed with 1g ice at 0o C so that the temperature remains at 0o C, is b) 200 C c) 50o C d) 80o C a) 10o C 3.12 kJ of heat is supplied to a mass of monatomic gas produces a temperature rise. If the same amount of heat is supplied to a diatomic gas of same mass, the temperature rise will be a) same b) 7/5 times c) 3/5 times d) 5/7 times o 16 g of oxygen at STP is mixed with 14 g of nitrogen at 10 C . The resulting temperature of the mixture will be a) 0oC b) 5oC o c) slightly more than 5 C d) slightly less than 5oC A substance of mass M requires an input power P to remain in the molten state at its melting point. When the power source is turned off the sample completely solidifies in a time t. The latent heat of fusion of the substance is a) Pt/M b) P/Mt c) Mt/T d) MP/t o 2 g of He is heated through 1 C at constant pressure. . The external work done in the process is a) R/2 b) R c) 3R/2 d) 2R The RMS speed of hydrogen molecules at a temperature T is x m/s. At a temperature 4T, hydrogen dissociates. At this temperature the RMS speed of hydrogen will be a) 2x b) 4x c) (2√2) x d) 4√2 x Two metals A and B are used to make a compensated pendulum. If the linear expansivity of A to that of B is in the ratio 3:1, the ratio of length of A to that of B will be 2

1

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a) 3:1

b) 1:3

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c) 1.5:1

d) 1:1.5

93. 1 mol of hydrogen is mixed with 1 mol of helium at the same temperature. The ratio of

specific heats γ of the mixture is a) 1.67 b) 1.4 c) 1.5 d) 1.53 94. A thin steel wire of length `L’ increases in length by 1% when heated through a certain range of temperature. If a thin steel plate of area 2LxL is heated through same range of temperature, the percentage increase in area will be a) 1% b) 2% c) 3% d) 4% 95. A gas of γ = 4/3 is heated at constant pressure what percentage of total heat supplied is used

as external work ? a) 25% b) 20% c) 10% d) 5% o 96. An electric heater of constant power takes 50 seconds to turn x g of ice at 0 C into water at o o 100 C. The time it will take to turn this water into steam at 100 C is a) 25 s b) 50 s c) 100 s d) 150 s 97. A mixture is formed with one mol of monatomic gas, 1 mol of diatomic gas and 1 mol of triatomic gas. The external work done when this mixture is heated through 1oC is a) R b) 6R c) 1.5R d) 3R 98. The ratio of diameter of molecule A to that of B is 1:2. The mean free path of A to B when they have same concentration, will be in the ratio a) 1:2 b) 1:4 c) 2:1 d) 4:1 o o 99. 22.2 g of ice at 0 C is mixed with 22.2g of water at 22.2 C. The resulting temperature of the mixture is a) 22.2o C b) 11.1o C c) 0o C d) 5.6o C 100. An ideal monatomic gas is heated at constant pressure . What fraction of heat energy supplied to it is used as increase in internal energy? a) 2/5 b) 3/5 c) 5/7 d) 3/7 101. If R/Cp for a gas is 0.28, the gas could be a) hydrogen b) helium c) carbon dioxide d) mixture of helium and hydrogen 102. If big ice cube falls from a height 3.36 m. Nearly what fraction of it will melt? a) 0.1% b) 1 % c) 10% d) 0.01% 103. *A thermos flask containing hot coffee is vigorously shaken . Check the correct statements a) The temperature of the coffee will increase b) The internal energy of system will increase c) Heat is given to the system d) Work is done on the system 104. The temperature of a hot source is 1000 K. If we use a powerful convex lens to converge heat radiation from the source to a point, at this point a) we can produce a temperature well above 1000 K b) we can produce a temperature of 1000 K c) we can produce a temperature only below 1000 K d) we can produce a temperature above 1000 K if the source has sufficient energy. 105. During winter the snow does not melt at once when sun’s rays fall on it, because it a) reflects sun’s rays b) has high specific heat c) has high latent heat of fusion d) is due to all the three given above

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106. In an adiabatic change the pressure P and temperature T of a monatomic gas is related by the

equation P is proportional to Tc. Here c is equal to a) 5/3 b) 2/5 c) 3/5 d) 5/2 3/2 107. An ideal gas is found to obey the equation PV = constant. If such a gas at a temperature T is compressed adiabatically to half its volume the final temperature would be a) 2 T b) 4 T c) 1.4 T d) 2.8 T 108. The RMS speed of hydrogen molecules is v. The RMS speed of a mixture of hydrogen and oxygen of volumes in the ratio 5:1 will be 21 6 21 c) d) 6 21 3 109. The temperature of an ideal gas is increased from 120 K to 480 K. The rms speed of its molecules increases by a) 100% b) 200% c) 300% d) 400% 3 110. A container of volume 1m is equally divided by a partition . One part contains an ideal gas at 300 K and the other part is vacuum . The whole system is thermally insulated from the surroundings. When the partition is removed the gas expands to occupy the whole volume. The temperature of the gas will be a) 150 K b) 300 K c) 600 K d) 75 K 111. An ideal gas with a pressure P, volume V and temperature T is expanded isothermally to a volume 2V and pressure Pi . If the same gas is expanded adiabatically to a volume 2V, the final pressure is Pa. The ratio specific heat of the gas is 1.67. The ratio Pa /Pi is equal to b) 2-0.67 c) 20.67 d) 2 a) 21.67 112. The ratio of speed of sound in hydrogen to rms speed of hydrogen molecules at STP is equal to 7 15 5 9 a) b) c) d) 15 7 9 5 a) √5 v

b)

113. The pressure of a gas contained in a vessel is P. If mass of each molecule is reduced to half

and rms speed doubled, the pressure will be a) P b) P/2 c) P/4 d) 2P 114. A given mass of gas of volume V is heated at a constant pressure so that the rms velocity is doubled. The new volume it occupies is a) V/2 b) 2V c) √2V d) 4V 115. The mean free path of molecules λ is related to the temperature T by λ proportional to

a) T b) 1/T c) T2 d) 1/T2 116. If heat supplied to a gas only increases internal energy of the gas, the process is a) isothermal b) adiabatic c) isobaric d) isochoric -2/5 117. The equation to the adiabatic change of a gas is given by T is proportional to V . The heat o required to raise the internal energy of 1 mol of this gas through 1 C is a) (5/2)R b) (3/2)R c) (7/2)R d) 3R 118. If A is the dimensions of specific heat and B is the dimensions of gravitational potential, the dimensions of the ratio A/B will be that of a) temperature b) Wein’s constant c) mass x temperature d) 1/temperature 119. *The internal energy of a gas depends on a) pressure b) volume c) temperature d) molecular separation

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51

120. Which of the following remains constant when water boils to steam ?

a) potential energy b) kinetic energy c) internal energy d) entropy 121. The rms speed of certain gas at a temperature is v. If the temperature of the gas is increased by 10% and then decreased by 10% , the rms speed a) remains the same as v b) increases by 1% of v c) decreases by 1% of v d) decreases by 0.5% of v 122. The pressure and temperature of a given mass of gas is increased to 3 times each. In this process the rms velocity a) increases to 3 times b) increases to 9 times c) increases to 3√3 times d) increases to√3 times 123. The pressure of a gas is decided by a) total random collision of molecules b) random collision of molecules per second c) average collision of molecules per second d) average collision per unit area per second 124. *Cooking vessels made with stainless steel are provided with extra copper bottom. This is because a) copper has more density b) copper has more thermal conductivity c) copper has more specific heat d) copper has less specific heat o 125. A body A having 1000 J of heat energy at 20 C is mixed with a body B having 500 J of heat o energy at 40 C. Check the correct statement: a) Heat will flow from A to B b) Heat will flow from B to A c) Heat will not flow between the two bodies d) The final temperature of the mixture will be 30o C 126. A refrigerator has a coefficient of performance of 10. When it works at a power of 700 W, the heat removed by it from the substance to be cooled per second is a) 7000 J b) 70 J c) 700 J d) 4200 J 127. x mol of monatomic gas is mixed with x mol of diatomic gas. The heat in calorie required to raise the temperature of the mixture through 10o C is a) 40 x b) 80x c) 20 x d) 60 x 128. The efficiency of a Carnot’s engine is y ( y is a fraction). When the temperature of the sink is increased by 200 C, the efficiency becomes 0.8 y. The temperature of the source is a) 50/y b) 40/0.2y c) 100/y d) 100/0.2 y 129. The efficiency of a Carnot engine can be increased by a) increasing both the temperature of source and sink b) decreasing both the temperature of source and sink c) decreasing the temperature of source and increasing in the temperature of sink d) increasing the temperature of source and decreasing in the temperature of sink o 130. By what height should water fall so that it will be warmer by 1 C? Assume 30% of energy of water is wasted a) 600 m b) 300 m c) 240 m d) 420 m 131. If the radius and surface temperature of a black body decrease by 1% each, the power radiated by it

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a) decreases by 2% b) decreases by 4% c) decreases by 5% d) decreases by 6% 132. At upper atmosphere, where kinetic temperature is 1000 K, an astronaut feels a) very cold b) cold c) very hot d) warm 6 133. In an atomic bomb explosion, the temperature produced is about 10 K. In which region of electro- magnetic spectrum does the emitted radiation belong to? a) ultra violet region b) infra red region c) X-ray region d) visible region 134. The unit of thermal resistance is a) KW-1 b) mW-1K c) WK-1 d) Wm-1

SECTION 2: ANSWERS 1

(b)

2

(c)

3

(d)

4

(a)

5

(c)

6

(d)

7

(a)

8

(d)

9

(d)

10

(d)

11

(a)

12

(c)

13

(c)

14

(d)

15

(a),(d)

16

(a)

17

(d)

18

(c)

19

(b)

20

(a)

21

(c)

22

(c)

23

(d)

24

(a)

25

(b)

26

(d)

27

(b)

28

(d)

29

(d)

30

(b)

31

(d)

32

(b)

33

(d)

34

(b)

35

(a)

36

(c)

37

(c)

38

(d)

39

(b)

40

(d)

41

(a)

42

(a)

43

(c)

44

(d)

45

(b)

46

(c)

47

(a)

48

(a)

49

(c)

50

(b)

51

(d)

52

(a)

53

(b)

54

(c)

55

(d)

56

(d)

57

(b)

58

(c)

59

(b)

60

(c)

61

(b)

62

(b)

63

(c)

64

(c)

65

(b)

66

(b)

67

(c)

68

(d)

69

(d)

70

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71

(d)

72

(a)

73

(a)

74

(a)

75

(c)

76

(c)

77

(d)

78

(b)

79

(d)

80

(d)

81

(c)

82

(a)

83

(a)

84

(c)

85

(d)

86

(d)

87

(c)

88

(b)

89

(a)

90

(a)

91

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92

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93

(c)

94

(b)

95

(a)

96

(d)

97

(d)

98

(d)

99

(c)

100

(b)

101

(a)

102

(d)

103

(a,b,d)

104

(c)

105

(c)

106

(d)

107

(c)

108

(c)

109

(a)

110

(b)

111

(b)

112

(a)

113

(d)

114

(d)

115

(b)

116

(d)

117

(a)

118

(d)

119

(c,d)

120

(b)

121

(d)

122

(d)

123

(d)

124

(b,d)

125

(b)

126

(a)

127

(b)

128

(c)

129

(d)

130

(a)

131

(d)

132

(a)

133

(c)

134

(a)

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53

SECTION 3 SOLUTIONS 1.

When the balloon sinks below the surface, hydrostatic pressure increases, decreasing the volume (Boyle’s law). This reduces up thrust (buoyant force), which makes it to sink more until it sinks to the bottom. (Note: the answer is not surface tension because it will oppose sinking) Ans.b

2.

The terminal velocity is proportional to a2 (a radius) .v = velocity of the smaller drop, v2 that of resulting drop,

2a 2 (ρ − σ)g If v1 is the terminal 9η

v1 a 1 2 = . a2 = 21/3 x a1. This gives v2 a 22

4.

v2 = 22/3 x v1 = 22/3 x 1 m/s. Ans.c When coal-tar is heated viscosity decreases with heating. Then it is laid. After it cools down viscosity increases and it settles. ( Indirect theory notes). Ans.d Ans.a

5.

r h ρ g = 2 S cosθ. That is rxh = constant. A = πr2. Thus we find √Axh = constant. When

6.

Using the formula r h ρ g = 2S, when warm water is taken surface tension decreases. So the capillary height will decrease. However density is less for warm water. So the question can be answered only if the relative decrease of the two are known. Ans.d

7.

Once again we use the relation r h ρ g = 2S. Capillary rise h will be less if g is more. This happens only in the case of lift accelerating upward. Here effectively g is equal to g+a. Therefore h is less. Ans.a

3.

area is increased to twice, height decreases to

2 times. Ans.c

The depression at centre 1 t of a metre scale is given by δ =

4Mgl 3

. Thus δ is proportional to bd 3 Y 13. When metre scale is cut into 2 equal halves, 1 becomes 1/2 δ becomes (1/2)3 = 1/8 of x. Ans.d 9. Heating reduces surface tension. Hence B is correct. But only insoluble and semi soluble impurities reduce surface tension. Highly soluble impurity increases surface tension of the solvent. Hence A is wrong. Ans.d 10. When a body floats in a liquid, the immersed fraction is=density of body/density of liquid. When water is heated its density increases up to 40C, and then decreases . The immersed fraction therefore first decreases and then increases. Ans.d 8.

11. If e is increase in length, using α =

∆L e = we have = α . Thus e = α. Young’s modulus L∆T l× l

FL F ×1 , i.e. = E. F = Ee = Eα. Ans.a Ae 1× e 12. Heat lost = mcθ. For a given quality of heat we require only the minimum mass of water for cooling because of a high value of specific heat. Ans.c 13. During melting, the temperature is constant. Hence the molecules do not gain kinetic energy. The heat supplied is used to gain potential energy, decreasing the separation of molecules. The volume of ice decreases on melting. Ans.c E=

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14. The coefficient of thermal conductivity is a constant for a given material and will not depend

on any of these. (The heat conducted will depend on these factors). Ans.d 15. Using L1 = L0 (1+ α ∆T) where α is linear expansivity.

We have L10 = L0 (1+ 10α). (a) is correct. Since areal expansion β = 2α, A10 = A0 (1+βt) = L2 (1+20α). But the total area is that of six faces. Hence (b) is wrong. Since density at 00C, d0 is related to density at t0C by d d0 d0 = the equation d0 = dt(1+γt), we have d t = 0 = (d) is correct and (c) is 1 + γt 1 + (3α ×10) 1 + 30α wrong. Ans.a,d 0

16. Use the information supplied in indirect theory notes Here θ < 80 C. Ans.a 17. Cv for helium(monatomic gas) is (3/2) R and for hydrogen (diatomic) is (5/2)R. Hence Cv

for mixture is 4R/2 joule/mole K = 2R J/mol K = 16.6. (Use table given in theory notes) Ans.d 18. According to Kirchoff’ law, a good absorber of a certain wavelength is also a good emitter. If sodium emits two bright yellow lines, it will absorb radiation of the same wavelength when a more intense source is kept behind it. An arc source is more intense than sodium. Therefore, sodium absorbs these wavelengths and produces two dark lines under the bright background of the arc source. Ans.c 0

0

19. We have mCp ∆T = 70 cal, m = 2 mol, ∆T = 5 C. This gives Cp = 7 cal / mol C. Cp – Cv =

20.

21. 22.

23.

R where R = 2 cal/mol0 C. Therefore Cv = 7-2 = 5. Thus the heat required at constant volume to raise through 50C = mCv∆T = 2 x 5 x 5 = 50 cal. Ans.b We have to use here Wein’s displacement law, λmT = constant, where λm is the wavelength of radiation carrying maximum energy and T the temperature of the star. For a very hot star T is high. λm will be low. In the visible region lower wave length corresponds to the violet Ans.a By Boltzmann’s law, thermal energy is of the order of kBT where kB is Boltzmann’s constant Taking room temperature T = 300 K, E = 1.38 x 10-23 x 300/(1.6 x 10-19) ≈ 0.02 eV Ans.c When heated at constant pressure, by Charles’ law temperature should increase to 4 times. The rms velocity is proportional to square root of temperature. So the rms velocity becomes √4 = 2 times. Ans.c A given mass of water has minimum volume at 40C. When heated or cooled from this temperature water expands and overflows. Hence temperature of the room is 40C. Ans.d

4 πr 3 ρ rρ = which is proportional to r. 3 4πr 2 3 Heat radiated by S1 r1 4 4 = . Mass of S1 = 3 x mass of S2. Therefore πr13ρ = 3 x πr23ρ. Heat radiated by S 2 r2 3 3 1/3 This gives (r1/r2) = (3) . Ans.a 25. According to Maxwellian distribution of velocities, average velocity of a molecule depends only on the temperature. (Similar to one postulate of kinetic theory). Since the temperatures are same in the three vessels, the average speed of molecules will also be same. Ans.b 26. If Cv is molar heat capacity at constant volume, Cp molar heat capacity at constant pressure, heat required to increase internal energy of n mol of gas in n Cv ∆T. Total heat supplied to the gas i.e. to increase internal energy and to do external work, is nCp∆T. The ratio of 24. Heat energy radiated per unit area per second =

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increase in internal energy to total energy is

55 nC v ∆T C v 1 5 = = = for a diatomic gas. This nC p ∆T C p γ 7

can also be done from Indirect theory notes short-cuts. Ans.d aT 2 = AV-c-BV-1 (1) At constant temperature, RT + b = constant V Vc A and aT2 = constant B. It is given in the question here P = AVm –BVn. (2). Comparing (1) and (2) we get m = -c and n = -1. Ans.b 28. 22 gm of CO2 is half mol of CO2 16 gm of O2 is half mol of O2. Cv for CO2 is 3R. Cv for O2 is 5R/2. Let t be the final temperature of the mixture. Then by law of mixtures, 5 0.5 x 3R x (t-27) = 0.5 x R x (37-t). This gives t = 31.54 ≈ 320C. Ans.d 2 29. If Q1 and Q2 are the rate of flow of heat through the rods by the law of heat conduction Q = 27. Re-writing P =

RT + b

−

2

Q=

A(T1 − T2 ) t Q1 A 1 L 2 πr1 2 L 2 1 1 1 , = × = = × = . Ans.d L Q 2 A 2 L 1 πr2 2 L1 2 2 8

30. We are mixing here one mol of hydrogen i.e. monatomic gas with one mol of He that is

(3R/2) + (5R/2) (5R/2) + (7R/2) =2R. Cp (mixture) = = 3R. 2 2 Thus the required ratio is γ is 3R/2R = 1.5. This information is also given in indirect theory notes. Ans.b 31. Since specific heat varies with temperature here, this should be done by integration. Heat diatomic gas. Cv (mixture) =

required to raise from 10 to 200C is

34.

35. 36.

37.

38.

20

10

10

∫ mC V dT = ∫ 2 x 0.5T dT = 150 J. Ans d

3RT .When temperature is doubled T becomes 2T. When oxygen M dissociates, its molecular mass M becomes atomic mass M/2. Substituting these values the new rms speed = 2v. Ans.b If earth has no atmosphere, there will be no molecules to communicate heat energy by collision. Kinetic energy of molecules transferred to us by collision is the main reason for our feeling of heat. Hence it reduces considerably. Ans.d A cooking vessel should have low specific heat capacity so that smaller amount of heat will produce larger rise in temperature. Also it should have high thermal conductivity to conduct maximum heat. Ans.b The external work = Cp –Cv. For 1 mol Cp-Cv = R. 1g of hydrogen is (1/2) mol. Therefore external work = R/2. R = 2 cal/g. R/2 = 1 cal. Ans.a When the two tanks balance mass of air should be equal to mass of helium. Volume of two tanks are same. Hence density of helium should be made nearly 7.5 times so that its density becomes equal to that of air. Density is proportional to pressure. Hence pressure has to be increased to 7.5 times. Ans.c Clearly the vessel has 1/4 mol of gas with 1/8 mol of oxygen and 1/8 mol of nitrogen. When all oxygen is removed we have 1/8 mol of gas left. Hence pressure will fall to 1/2 of the 8 atmosphere. Ans.c Ans.d

32. v(rms speed) =

33.

20

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39. For an ideal gas, Cp – Cv = R, for 1 mol. Cp-Cv is the external work done per mol. Here we

have two mol of ideal gas. Hence the external work will be 2 R. Ans.b

40. According to one of adiabatic relations, TV γ −1 = constant.

Comparing this with given equation, γ-1 = 2/5. γ = 7/5. The gas is diatomic. The heat required to increase internal energy through 10C = CV = 5/2 R. Ans.d 41. Heat required to melt = mL, where m is mass, and is latent heat of fusion. Here m is same. L depends on material, hence the same. Ans.a Jm −1s −1 K −1 λ = = m2s-1. Ans.a dc (kg m −3 )( J kg −1 K −1 )

42. Substituting the units of 43. x = 2π

dx 1 dl dx ∆T 1 = = α ∆T ⇒ . = k l . Taking logarithms and differentiating = x 2 l x 2α g

dx = (1/2)x α ∆T. Ans.c 2

44. For an ideal gas PV = RT (1). It is given VP = constant K (2) . Squaring equation 1 and

P2V2

R 2T 2

. This gives V = constant x T2. That is T is proportional VP K2 to √V. When the volume is doubled, temperature becomes √2 times. Ans.d

dividing by 2, we get

2

=

45. The maximum efficiency of a heat engine is =

efficiency is 40%. i.e. =

T1 − T2 2100 − 700 2 = = . The actual T1 2100 3

2 . The ratio of actual efficiency of maximum efficiency 5

(2 / 5) = 0.6 = 60% Ans.b (2 / 3)

46. Since expansion is isothermal i.e. at constant temperature, only volume and pressure will

change. Change of pressure at constant temperature will not affect rms velocity. Ans.c 47. Using first law of thermo dynamics dQ = dU + dW, we have here dW = -dU (decrease of internal energy). dQ=0, means change is adiabatic. Decrease of internal energy decreases temperature, that is the gas is expanding. . The gas is undergoing adiabatic expansion. Ans.a 48. According to the first law of thermodynamics dQ = dU + dW. Here dQ = 100 cal = 420 J. External work dW = 20 J. Substituting in the above equation, 420 = dU + 20, dU = 400 J. Since dU is positive, internal energy increases. Ans.a 4

49. We use here Stefan’s law of radiation E = σT . If E1 is the new energy after increasing the 4

temperature to (3/2)T, E1 = σ [(3/2)T] = σ (81/16)T4 = nearly 5 σ T4. Energy increases from σT4 to 5 σT4 i.e. by four times or 400%. Ans.c Stefan’s law, heat energy radiated eper second E = σ AT4. 50. By 2

2 4 2 4 E r1 T1 1 2 E 1 σ4πr1 T1 = = = ⇒ 1 = 1 . Ans.b E 2 σ4πr2 2 T2 4 r2 T2 4 1 E2

51. According to the law of heat conduction heat conducted is directly proportional to area, and

inversely proportional to length at thermal equilibrium. In the second case, heat conducted doubles due to increase in area, and doubles due to decrease in length. Thus heat conducted becomes four times, i.e. 16 W. Ans.d

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57

52. Using Wien’s displacement law λmT = constant, T1/T2 = λ2/λ1 = f1/f2. (Since frequency is

inversely proportional to wave length). This gives T1/T2 = 2/2.5 = 4/5. Ans.a

4

53. Assuring the star to be a black body and applying Stefan’s law, (E αT ) the ratio of 1/4

1/4

temperature of the stars should be T1/T2 = (E1/E2) =(1/16) = 1/2. By Wein’s displacement law, λmT = constant. The ratio of wavelengths carrying maximum energy of the stars λ1/λ2 should be T2/T1 = 2/1. Ans.b

[x

2

2

]

− x 1 2 ρL , where t is the time to 2λθ increase the thickness of ice from x1 to x2, ρ density, L latent heat and λ thermal conductivity of ice. θ the temperature difference other things being constants, t is proportional to x22 – x12. t 1 12 − 0 2 1 = = , where t1 is a time to increase thickness from 0 to 1 and t2 from 1 to 2. t 2 2 2 − 12 3 Here t1 = 10 hour, t2 ,therefore, is 30 h. Ans.c 55. A system is thermodynamic equilibrium if no force acts, (mechanical equilibrium), no chemical reaction takes place ( chemical equilibrium), and no heat exchange takes place (thermal equilibrium). Ans.d 56. This is adiabatic expansion. Since no external heat is supplied internal energy should decrease. Hence (b) is correct. During adiabatic change entropy remains constant. Hence (c) is correct. Ans.d 57. Using the first law of thermodynamics, the increase in internal energy dU = dQ-dW. dU is the difference between external energy supplied (20x12 = 240 W) and heat transferred (10W). i.e. 230 W. Ans.b 54. Thickness of formation of ice in a pond is given by t =

4

58. By Stefan’s law energy radiated E = σT . When current is doubled heat energy becomes 4 2

4

times. (H α i ) .That is T becomes 4 times. T increases to 41/4 = √2 times. Ans.c 59. When liquids of different mass, specific heat and temperature are mixed final temperature is m c t + m 2 c 2 t 2 + m 3c 3 t 3 .. given by Σ(mc ∆T) / Σ(mc). For three liquids this will be 1 1 1 m1c1 + m 2 c 2 + m 3 c 3 Here m1 = m2 = m3. Therefore final temperature will be (c1t1 + c2t2 + c3t3) /c1 + c2 + c3. Ans.b 60. According to Boltzmann’s law average kinetic energy = (3/2)kT, per molecule. When this energy = 1 electron volt = 1.6 x 10-19J, we equate the two.Assuming k = 1.38 x 10-23, we get T = 7700 K. Ans.c 61. The thermal resistance (similar to electrical resistance) is directly proportional to the length (1) and inversely proportional to the area (A). Therefore, it can be written as R (thermal) = ρ1/A, where ρ is thermal resistivity . Thermal conductivity λ is reciprocal of the thermal resistivity i.e. λ = 1/ρ. Thus we get R (thermal) = 1/λA. Ans.b 62. In thermal equilibrium heat flowing (fig.) across any portion should be same. Hence, if the temperature of the intervening layer is T, 2λ (T-O) = λ(36-T), B A This gives T = 120C. Ans.b 0 T 36 63. The rate of loss of heat is proportional to area of the surface if other factors affecting radiation like mass, temperature difference remain the same. Since all have same mass, same density, they have same volume. For a given volume sphere has least surface area and circular plate maximum surface area. Hence circular plate cools fastest. Ans.c

58

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64. By Newton’s law of cooling, rate of cooling is proportional to the mean excess temperature of

65. 66. 67.

68.

the body over the surroundings. In the first case, the mean temperature difference is 65-300C. In the second case, the temperature difference is 55-30 = 250C. Hence rate of cooling will be in the ratio 35/25 = 7/5. So the time taken for cooling from 60 to 50 = 10 x (7/5) = 14 minutes. Ans.c Density of a liquid increases with fall in temperature. Density of benzene will be more in winter. Hence a given volume of benzene will weigh more in winter. Ans.b From kinetic theory we can deduce Boyle’s law. Boyle’s law is violated at high pressure and low temperature. Hence most of the postulates break down at this stage. Ans.b When temperature is increased at constant volume, the average kinetic energy and hence rms speed of molecules increase. Hence they make more number of collisions on the wall with higher velocity. Ans.c Kinetic energy of one mol of monatomic gas at STP (3/2)RT, which by calculation (3/2) x 8.3 x 273 = 3400 J. 1 gm of helium is (1/4) mol. Hence its kinetic energy is 3400/4 = 850 J. (Note: kinetic energy of monatomic gas at STP is one of the constants to be remembered). Ans.d

69. Heat supplied H = nCv ∆T, where n is number of mole, Cv specific heat at constant volume,

∆T rise in temperature. Here heat supplied is same. n is same. Hence C1 ∆T1 = C2 ∆T2. This gives ∆T1/∆T2 = C2/C1 = Cv of monoatomic gas/Cv of diatomic gas. (3R/2)/ (5R/2) = 3/5. Ans.d 70. For a solid cubical expansivity γ = 3α nearly. A crystal is anisotropic and hence does not

expand in the same way in all directions. Therefore γ = α1 + α2 + α3 = (13 + 231 + 231) 10-7 = 475 x 10-7. Ans.c 71. Here the expansion is free. That is not external work is done on the system or by the system. No heat is supplied to the system either. Therefore there will be no change in the internal energy of the system. Ans.d 72. According to Boyle’s law PV = constant at constant temperature. Therefore if P1 is the new pressure PV + PV = P1V, which gives P1 = 2P. Ans.a 73. By Wein’s law λ1T1 = λ2T2, where λ1, λ2 are wavelengths carrying maximum energy. This

gives T1/T2 = 4800/3600 = 4/3. Ratio of total power radiated = T14/T24 = (4/3)4 = 256/81. Ans.a 74. Using the first law of thermodynamics, dQ = dU + dW, dQ = 0, for adiabatic process. (d) is correct, Also dW= dU.. (b) is correct). For isothermal process there is no change in temperature and hence dU is zero. (c is correct). The increase in internal energy whether at constant pressure or constant volume = nCv ∆T. (a is wrong). Ans.a 75. The adiabatic elasticity of any gas is γP. Helium is a monatomic gas. For helium γ = 1.66

and P = nearly 105 Pa. Ans.c 76. When temperature T is doubled, heat energy radiated per second becomes 16 times (Stefan’s law E αT4). When r is doubled heat received becomes 1/4. (inverse square law I α 1/r2). Hence heat energy received becomes 16 x 1/4 = 4 times. Ans.c 77. If T1 and T2 are the temperatures at the two ends of the system, at thermal equilibrium, the heat conducted through the rod respectively are Q1 = λ1A (T1 –T2)/x, Q2 = λ2A (T1-T2)/x, per second. The total heat conducted Q = Q1 + Q2 (1) . If λ is the conductivity of the composite rod, then heat conducted per second Q = 2Aλ (T1-T2/x. Substituting these values of Q, Q1, Q2 in (1) we get λ = λ1 + λ2/2. Ans.d

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59

78. Here kinetic energy of the bullet is converted into heat energy. If the velocity of the bullet is

doubled kinetic energy of bullet (1/2)mv2 becomes four times and when 50% of it is lost. Kinetic energy remains two times. The rise in temperature will be two times. Ans.b

79. The increase in the internal energy = m Cv ∆T. For a monatomic gas Cv = (3/2)R. ∆T =

1000C or 100K (Temperature intervals will be same in both scales). Hence dU = (3/2)x 8.3 x 100 = 1245 J. Ans.d

80. η = T1 – T2/T1. When T1 and T2 are increased by 10% all temperatures increase by 10%.

Hence (T1-T2)/T1 remains the same. Ans.d 0

81. Water will reach 0 C but will not freeze, because it cannot give out latent heat. Latent heat

will not be conducted as there is no temperature difference at this stage. Ans.c 82. The work done by a gas is the area of P-V graph. This area is minimum for graph (1) where

there is no increase in volume i.e. dV = 0. (Alternatively dQ = dU + PdV. Here dV = 0. Therefore PdV = 0. Ans.a 83. Solar constant is amount of heat energy received per unit area in one second on earth. If P is power radiated by sun ,. solar constant S0 = P/4πd2, where d is the distance between sun and earth. When d is reduced to half, heat received per unit area per second i.e. S0 becomes 4 times. Ans.a K 1 πR 2 (θ1 − θ 2 ) , x where θ1 and θ2 are the temperatures at the two ends (θ1 > θ2) and x is the distance between K π(2R 2 − πR 2 )θ1 − θ 2 . If K is them. Heat flowing across the cylindrical shell is Q2 = 2 x the equivalent thermal conductivity of the system then, the total heat flowing = Kπ(2R)2 (θ1-θ2)/x = Q. Equating Q1 + Q2 = Q, we get K = (K1 + 3K2)/4. Ans.c 85. Increase in area ∆A = A β ∆T, where ∆T is increase in temperature, A original area, β areal expansivity . A cube has 6 surfaces. So the total area A = 6 a2. Areal expansivity β = 2α. ∆T = 1oC = 1K. This gives ∆A = 12a2α. Ans.d o 86. Latent heat of fusion of ice is 80 cal/gm. So temperature will remain at 0 C if we supply a maximum heat 80 cal. Since specific heat of water in CGS system is 1, 1 g of water when cools from 80 to 0 can give 80 cal of heat. Use also indirect theory notes for answering this quickly. Ans.d 84. Heat flowing per second across the cylinder of radius R is given by Q 1 =

[

]

87. Heat supplied = mCv∆T, where Cv is specific heat at constant volume.

If C1 and C2 are the specific heat of monatomic and diatomic gases, ∆T1 and ∆T2, temperature rise, we have C1∆T1 C (3 / 2)R 3 = C2∆T2, for the same mass . 1 = This gives ∆T2 = ∆T1. Ans.c C 2 (5 / 2)R 5

88. 16 g of oxygen is half a mole. 14 g of nitrogen is also (1/2) a mole. Both of them are

diatomic. Hence their Cv will be same. Using heat lost by nitrogen = heat gained by oxygen formula , we have m CV (10-T) = m CV (T-0). Hence the temperature will be 5oC. Ans.b 89. A power P keeps the substance in the molten state i.e in equilibrium . This means the substance is supplied with an energy Pt to keep in the state. So latent heat of fusion = heat energy supplied / mass = Pt/M. Ans.a o 90. The external work done when 1 mol of gas is heated through 1 C or 1 K is equal to Cp-Cv = R. 2 g of helium is (1/2) mole. Hence external work will be (R/2). Ans. a

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3RT where M is molecular mass. When temperature increases to 4T, M hydrogen dissociates, its molecular mass M becomes atomic mass M/2. Thus the RMS speed 3R x 4T 8 x 3RT = = 8 x = 2 2 x Ans.c becomes M/2 M

91. The RMS speed =

92. Use information supplied in indirect theory notes. . For compensated pendulum, we should

have lengths in the ratio inverse of expansivity i.e. 1:3. Ans b 93. Use information given in indirect theory notes. This is for practice. Hydrogen is diatomic and helium is monatomic. Ans.c 2 94. We have A = 2LxL = 2L . Taking logarithms and differentiating dA/A = 2(dL/L). Here dL/L = 1%. Therefore dA/A = 2%. Ans.b 95. The external work done = R = Cp- CV, for 1 mol. The heat supplied at constant pressure = Cp Cp − CV C 1 1 for 1 mol for 1o rise in temperature. The required ratio is = 1− V = 1− = . Cp CP γ 4 Use indirect theory short-cuts. Ans.a 96. Use information supplied in theory notes. To melt x g of ice we need 80 x calorie of heat. To raise the temperature of x g of water from 0 to 100o C we need 100 x calorie heat. Total 180x. To turn x g of water at 100o C into steam, we need 540x calorie of heat, which is 3 times 180x. So the required time will be 3x50 = 150 s. Ans.d 97. Here we have a total of 3 mol of gas. Whatever may be a gas the external work done for 1 mol Cp-CV = R. For 3 mole the external work will be 3R. Ans.d 98. The mean free path λ is inversely proportional to the square of the diameter. ( formula λ =

1/√2πσ2n). Therefore mean free path will be in the ratio 22:12.= 4:1 Ans.d 99. Use information supplied in in indirect theory notes. Ans.c 100. The ratio of increase in internal energy to the total heat energy supplied is clearly Cv/Cp, which in turn is equal to 1/γ. For a monatomic gas this value is 3/5. Ans.b 101. Since it is given that R/Cp =0.28, we have Cp = R/0.28, which is nearly 3.5 R, or (7/2) R. This is the value of Cp of a diatomic gas. The only diatomic gas listed is hydrogen. Ans a 102. Here the potential energy of ice is converted into heat energy for melting. Mgh = mL,where m is the mass of ice melting, L latent heat of fusion. With g nearly 10 and h = 3.36m, we get m/M = gh/L= 10x3.36/336000 = 0.0001= 0.01% Ans d 103. Since mechanical work is done on the system during shaking , (d) is correct. Due to this the temperature of the coffee will increase. (a) is correct. Heat is not directly added to the system. Hence (c) is wrong. By first law of thermo dynamics ∆Q = ∆U+ ∆W. Here ∆Q = 0. Therefore ∆U = - ∆W ( The negative sign here is due to work done on the system. That is internal energy of the system increases. (b) is correct. Ans. a,b,d 104. Whatever may be the energy of the source, we cannot produce a temperature above 1000 K at the point. If the temperature of the point becomes above 1000 K, it would mean that we have transferred heat from cold body to hot body. This will be in violation of second law of thermodynamics. Ans.c 105. The latent heat of fusion of ice is 80 cal/g, which is equal to 336 J/g. This is a large value. So it takes some time for ice to get this much of heat. Specific heat has nothing to do with melting, because there is no temperature change during melting. Ans.c

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61

T γ P 1− γ = = constant. This means P = constant Thus we have c =(γ/γ-1) For a monatomic gas γ = 5/3. Therefore c = 5/2. Ans.d

106. According to one of the adiabatic relations (γ/γ-1)

x T

.

107. Comparing with the standard adiabatic equation PVγ = constant, we find γ = 3/2. The

relation between temperature and volume is TVγ-1 = constant. That is TV(1/2) = constant . That is T is proportional to V-1/2. When V is halved T becomes (1/2)-(1/2) = 2(1/2) T = √2 T= 1.4 T. Ans.c

108. The RMS speed is proportional to

1 ρ

Density of the mixture is equal to

Since ρo = 16 times ρ H , The RMS velocity is equal to

6 v 21

5ρ H + ρ O 21 = ρH 6 6

Ans.c

109. We know Crms is proportional to √T. So when the temperature increases from 120 to 480 K

that is four times the rms speed should increase to √4 = 2 times. The rms speed thus increases to 200% and hence it increases by 100%. Ans a 110. Here the gas is undergoing free expansion i.e. no energy is supplied from outside and no energy is taken from inside. Therefore the temperature will remain the same. Ans.b γ

111. For isothermal expansion the equation is PV = constant and for adiabatic equation PV is

constant. PV = Pix2V. PVγ = Pa (2V)γ . This will give 112. The speed of sound is given by

Pa 2 2 = γ = 1.67 = 2 −0.67 Ans.b Pi 2 2

γRT and RMS velocity M

3RT The required ratio is M

γ = 3

7 7 , where we have assumed γ for hydrogen as . Ans.a 15 5 2

2

113. P = (1/3)nmc , where c is mean square velocity. When m is made half , c is doubled, the

pressure becomes [4(1/2)] = 2 times. Ans.d 114. To double the rms velocity the temperature has to increased to 4 times because Crms is

proportional to √T. When temperature increases to 4 times at constant pressure volume increases to 4 times. Ans.d 115. When a gas is heated molecules move faster making more number of collisions per second decreasing distance between collisions. Ans.b 116. From the first law thermodynamics ∆Q = ∆U+∆W.

Here only internal energy changes, which means ∆W = 0. But ∆W = PdV, where dV is increase in volume. Thus we have dV = 0. That is volume is constant. The process is isochoric. Ans.d γ-1

1-γ

117. Comparing with standard adiabatic relation TV = constant, we have T proportional to V

. Here 1-γ = -2/5. This gives γ = 7/5. The gas is diatomic. The heat required to raise internal energy by 1oC (which is Cv) = (5/2)R. Ans a 118. A = energy/(mass x temperature), while Gravitational potential (B) is energy/mass. This makes A/B equal to 1/temperature. Ans d 119. The internal energy of a gas molecule is the sum of potential and kinetic energy. The kinetic energy depends on temperature. (c) is correct. The potential energy depends on distance between molecules. (d) is correct. Ans.c & d

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120. When water boils molecular separation changes and hence potential energy changes.

So internal energy also changes. Since temperature is constant during boiling, kinetic energy remains constant. Entropy increases. Ans.b

121. If T is the temperature, Crms is proportional to √T.

When the temperature is increased by 10%, it becomes (110%)T and when decreased 10% from there, it becomes (110x90/100)T =

99%T. Thus the temperature decreases by 1% and the rms velocity being proportional to T decreases by 0.5%. Ans.d. 122. When pressure and temperature are increased at the same time, consider first increase of pressure. The rms velocity does not change. Now consider increase of temperature. When temperature is increased to 3 times rms velocity increases to √3 times. Ans.d 123. Since the number of molecules is large and they are moving at random, we cannot calculate exact number of collisions per second. So we take average number colliding per second. This gives force. Since pressure = force/area we take number of collision per second per unit area. Ans. d 124. Cooking vessels should have high conductivity to conduct heat. That is (b) is correct. They should have low specific heat so that smaller amount of heat will produce larger temperature rise. (d) is correct. Recall the properties of cooking vessels given in one of the previous questions. Density has nothing to do with conduction. . Ans. b & d. 125. Heat flows from a body of higher temperature to the body of lower temperature and not from the body of higher heat capacity to that of lower heat capacity. Hence (b) is correct and (a) is wrong. The temperature attained at thermal equilibrium T is such that heat lost by B is equal to the heat gained by A. This is calculated using the formula 1000(T-20) = 500(40-T). We have used 1000 and 500 as the product of mass and specific heat. This gives T not equal to 30oC. Ans.b 126. If ω is the coefficient of performance of the refrigerator, Q2 heat removed from the sink, W

external energy supplied, then Q2/W = ω. This gives Q2 = ωW = 10x700 = 7000 J/s. Ans a. 127. Use information supplied in theory notes. The specific heat at constant volume of the mixture is equal to [(3/2)Rx + (5/2)Rx] /2x = 2R. The heat required to raise the temperature of the mixture through 10oC = (2x)(2R)10 = 40Rx. In CGS system R = 2 cal/moloC. So the rrequired heat will be is 80x cal. Ans.b 128. Using the formula (T1-T2)/T1 = y, when T2 is increased by 20, this equation becomes [T1-(T2+20)]/T1 = 0.8y. Solving these equations we get T1 = 100/y. Ans.c

T1 − T2 T T = 1 − 2 .To get maximum efficiency 2 should be T1 T1 T1 minimum. That is T2 should be decreased and T1 should be increased. Ans.d 130. The potential energy of water is converted into heat. So if h is the height mgh x 70/100 = mc∆T. Here ∆T is 1o C and therefore h = cx100/g x 70. c = 4200 J/kg K, g = 10 m/s2, h = 600 m. If g is taken as 9.8 m/s2 the nearest answer will be 600 m. Ans.a 129. The efficiency is given by

2 4

131. The heat radiated by a black body per second is equal to E = σ 4πR T . Taking logarithms

dE dR dT =2 +4 = 2x1% + 4x1% = 6%. Ans.d E R T 132. The kinetic temperature 1000 K means each molecule will have a kinetic energy (1/2)kT, with T = 1000 K. This will be large compared to the energy of a molecule at our place. However, the number of molecules per unit volume at upper atmosphere is very very small. Hence total and differentiating,

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63

energy of the molecules per unit volume will be small and energy communicated to the astronaut by collision will also be very small. Ans.a 133. Here also we use Wein’s law. In this question you have to guess the value of Wein’s constant approximately. Wein’s constant = 2x10-3 mK. So using λ mT = this constant, we have λ m = 2x10-3/106 = 2x10-9 m = 2nm. This wavelength is in the X-ray region.. Ans.c 134. Thermal resistance is proportional to length (L),and inversely proportional to area (A). So we write R(thermal)= R = L/λA. where λ is thermal conductivity i.e. reciprocal of thermal = KW-1. Ans.a resistivity.⇒ R = (m)/Wm-1K-1m2

3 WAVE MOTION AND SHM SECTION 1 QUESTIONS 1.

If ω is angular frequency and k wave vector of a sound wave, its velocity is ω k c) ωk d) ωk b) a) k ω

2.

Which of the following does not represent a wave ? a) y = f(x-vt) b) y = ymsin k(x+vt) c) y = ymlog(x-vt) d) y = f(x2-v2t2) When a sound wave reflects from boundary seperating two media, which of the following could change ? a) frequency b) wavelength c) speed d) phase Two strings A and B made of same material are stretched by same tension. The radius of string A is double the radius of B. A transverse wave travels in A with speed VA and in B with speed VB. The ratio VA / VB is a) 1/2 b) 2 c) 1/4 d) 4 The speed of sound through a certain medium at 270C and 105 Nm-2 pressure is 200 ms-1. If the temperature increases to 1270C and pressure falls to 0.5 x 105 Nm-2, the speed of sound (in ms-1) is 400 50 2 100 2 c) 100 2 a) b) d) 3 3 3

3.

4.

5.

6.

7.

8.

9.

When temperature increases, the frequency of a tuning fork a) remains the same b) decreases c) increases d) may increase or decrease depending on the material A sound wave of frequency 500 hertz has a velocity 350 m/s. The minimum distance between the two particles having a phase difference of 600 is nearest to a) 0.7 cm b) 12 cm c) 70 cm d) 120 cm The speed of sound through oxygen at a temperature T is v ms-1. If the temperature increases to 2T and oxygen gas dissociates into atomic oxygen, the speed of sound a) remains the same b) becomes √2 v c) becomes 2v d) becomes nearly 9v/5 A supersonic jet produces waves in air. The wave front is a) spherical b) paraboloidal c) ellipsoidal d) conical

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65

10. A vibrating tuning fork is rotated about a vertical axis through its stem. An observer standing near will hear, in one revolution a) 4 maxima and 4 minima b) 2 maxima and 2 minima c) 8 maxima and 8 minima d) uniform sound 11. A rope of length L and mass m hangs freely from a ceiling. If v is the velocity of transverse wave produced in it and x is the distance from the free end, then v is proportional to a) x0 b) x-1/2 c) x1/2 d) x 12. A wire has frequency f, under a certain tension. Its length is doubled by uniformly stretching. Its frequency under the same tension will be a) 2f b) f c) 1.4 f d) 0.7 f 13. The ratio of velocity of sound in hydrogen to that in helium at STP is a)

42 25

b)

25 42

c)

2

d)

1 2

14. A hollow metallic tube open at both ends is in resonance with a tuning fork of frequency 256 Hz. If the tube now is immersed half inside water it will resonante to a frequency (in Hz) a) 1024 b) 512 c) 256 d) 128 15. The quality of a tone a) decreases with loudness b) depends on the overtones present c) is proportional to the pitch d) is inversely proportional to amplitude 16. When a motor boat sails in water, the waves produced are a) longitudinal b) transverse c) neither longitudinal nor transverse d) both longitudinal and transverse 17. The waves produced in a sonometer wire are a) transverse, stationary but not polarised b) transverse, progressive and polarised c) transverse, progressive and unpolarised d) longitudinal 0 18. When two waves of equal amplitude A and equal frequency at a phase difference of 60 superpose, the amplitude of the resulting wave is a) 2A

b) √2 A

c) 2√2 A

d)

3A

19. A long steel pipe is tapped at one end. A listener at the other end hears two sounds

a) b) c) d)

of the same intensity, at the same time less intense sound first, more intense later more intense sound first, less intense later of the same intensity, but at different times

20. A wave represented by the equation y = a cos(kx-ωt) is superposed on another wave to form a

stationary wave such that the point x = 0 is a node. The equation for the other wave is a) a sin (kx+ωt ) b) -a cos(kx-ωt) c) -a cos(kx+ωt) d) -a sin(kx-ωt) 21. A hospital uses an ultrasonic scanner to locate tumours in a tissue. The operating frequency of the scanner is 4.2 MHz. The speed of sound in a tissue is 1.7 km s-1. The wavelength of sound in the tissue is close to a) 4 x 10-3 m b) 8 x 104 m

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d)

8 x 10-3 m

22. A sound wave of wavelength λ travels towards the right horizontally with a velocity V. It

strikes and reflects from a vertical plane surface, travelling at a speed v towards the left. The number of positive crests striking in a time interval of five seconds on the wall is 5(V + v) 5(V − v) ( V + v) (V − v) b) c) d) a) λ λ 5λ 5λ 23. Two waves represented by equations y1 = 10 sin (2000 πt) and y2 = 10 sin [2000 πt +

π ] are 2

superposed at a point at an instant t. The amplitude of the superposed wave is a) 10 units b) 20 units c) 14 units d) zero 24. Which of the following gives a pure sine wave ? a) a note from a piano b) vibrating veena wire c) a vibrating tuning fork d) heart beats of a normal person x ]. The maximum particle λ velocity is equal to four times the wave velocity if λ is equal to (wave length) πY0 πY0/4 c) πY0 b) d) 2πY0 2 26. A wave equation which gives displacement along the y direction can be written as y = 10-4 sin(60t+2x) where x and y are in metres and t in seconds. This represents a wave a) of velocity v = 30 ms-1 in the negative x-direction b) of wavelength = πm c) of frequency f = 30/πHz d) of amplitude = 10-4 travelling in negative x-direction 27. Which of the following is not a travelling wave equation ? x a) Y = Ymsin k(x-vt) b) Y = Ym cos 2π − vt λ 25. A transverse wave is described by the equation Y = Y0sin 2π[ft -

x t c) Y = Ym sin 2π − λ T

d) Y = 2Ym sin kx cosωt

28. Spherical waves are emitted from a 5.0 W source in an isotropic non-absorbing medium. The

intensity of wave at a distance of 2.0 m from the source is a) 0.020 W/m2 b) 0.20 W/m2 c) 0.10 W/m2

d) 0.01 W/m2

29. The equation to a wave travelling in a spring can be written as y = cos π (100t-x), where

distances are in m. Its wavelength is a) 2π b) 1m

c) 0.05m

d) 2m

2πx sin 100πt where x and y are in cm, t in s, the node appears at x 30. In the equation y = 4 cos 50 equal to (in cm) a) 12.5 b) 50 c) 20 d) 100/2π 31. Two sound producing bodies produce progressive waves given by Y1 = 4 sin(400 πt),

3 sin(404 πt). A person situated nearby will hear a) two beats per second of intensity ratio 4;3 b) two beats per second of intensity ratio 49:1

y2 =

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67

c) four beats per second of intensity ratio 49:1 d) four beats per second of intensity ratio 4:3 32. Two waves having intensity in the ratio 9:1 produce superposition. The ratio of the maximum to minimum intensity of the superposed wave is a) 10/8 b) 4/1 c) 9/1 d) 2/1 33. The displacement of particles in a string in x-direction is represented by y. Among the following expression for y, those describing wave motion are a) coskx sinωt b) k2x2-ω2t2 c) cos(k2x2-ω2t2) d) cos2(kx+ωt) 34. A plane wave of sound travelling in air is incident on a plane water surface. The angle of incidence is 600. Assuming Snell’s law to be valid for sound waves, it follows, a) the wave will be refracted into water away from normal b) the wave will be refracted into water towards normal c) the wave will graze the surface of separation d) the wave will be reflected into air 35. The equation y = sin 4πt + cos 5πt is

a) periodic but not harmonic b) periodic but not simple harmonic c) periodic and simple harmonic d) neither periodic nor harmonic 36. The amplitude of a wave disturbance propagating in the positive x-direction is given by y = 1/(1+x2) at time t = 0 and by y = 1/[1+(x-1)2] at t = 2 seconds, where x and y are in metres. the shape of the wave disturbance does not change during the propagation. The velocity of the wave is b) 0.5 ms-1 c) 1.5 ms-1 d) 2 ms-1 a) 1 ms-1 37. A wave is represented by the equation y = A sin [10πx + 15πt +

π ] where x is in metres and t 3

in seconds. The expression represents a) a wave travelling in the positive x-direction with a velocity 1.5 ms-1 b) a wave travelling in the negative x-direction with a velocity 0.66 ms-1 c) a wave travelling in the negative x-direction having a wavelength 0.2 m d) a wave travelling in the positive x-direction having a wavelength 0.2 m 38. The displacement y of a particle executing periodic motion is given by y = t sin 1000 t. This expression may be considered to be a resultant of the superposition 4 cos2 2 of a) three waves b) two waves c) four waves d) five waves 39. A wave is represented by the equation y = A sin 2(10t-20x). Its wavelength is a) 20m b) π/20)m c) 10m d) (π/10)m 40. A wave disturbance in a medium is described by y (x,t) = 0.02 cos (50πt +

where x and y are in metres and in seconds. For this wave a) a node occurs at x = 0.15 m b) an antinode occurs at x = 0.3m c) the speed of the component wave is 5.0 m/s d) all the above are correct

π ) cos (10πx), 2

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41. If λ1, λ2, λ3 are the wavelengths of the wave giving resonance in the fundamental, second and

third harmonic modes respectively of a pipe closed at one end, the ratio of the wavelengths, λ2, λ2, λ3 in the order is a) 1:2:3 b) 1:3:5 c) 6:3:2 d) 15:5:3

42. The equation to the motion of a wave can be written as x = 2 sin

which the maximum velocity are attained by this wave will be a) 0, 6, 12 s, etc b) 0, 4, 8 s, etc c) 0, 3,6,9 s, etc

π t. The instants of time at 6 d) 2, 4,6 s, etc

43. An object of specific gravity ρ is hung from a thin steel wire. The fundamental frequency

form transverse standing waves in the wire is 300 Hz. The object is immersed in water so that one half of its volume is submerged. The new fundamental frequency, in Hz, is 2ρ − 1 2ρ − 1 a) 300 b) 300 2ρ 2ρ 2ρ 2ρ c) 300 d) 300 2 ρ − 1 2 ρ −1 44. The velocity of sound in air is 348 m/s. Two sources producing waves of wavelength 2m and 3m are sounded together. Which of the following is not correct ? a) They will produce 58 beats per second b) They will produce beats which cannot be detected the human ear c) The beats can be heard at the given rate in (a) d) They will produce frequencies whose difference will increase with increasing temperature 45. *The power transmitted in a vibrating string by a wave of frequency f and amplitude A is proportional to a) frequency of the wave b) amplitude of wave c) square of frequency of the wave d) square of the amplitude of the wave 46. Oxygen is 16 times heavier relative to hydrogen. Equal volumes of hydrogen and oxygen are mixed. The ratio of the velocity of sound in the mixture to that in oxygen is a)

1 8

b)

32 17

c)

17 . 25

d)

8

47. The maximum velocity of simple harmonic motion which produces a wave is 4 m/s.

Its maximum acceleration is 16 m/s2. Its amplitude of the motion is a) 1m b) 2m c) 4m d) 8m 48. A source of sound x gives five beats per second, when sounded with another source of frequency 100 /s. The second harmonic of the source x together with a source of frequency 205/s gives five beats per second. The frequency of the source x is a) 100/s b) 105/s c) 205/s d) 95/s 49. If the diameter of resonance tube of constant length is increased, the frequency emitted by it will a) remain the same b) increase slightly c) decrease slightly d) decrease considerably 50. If pressure increases by 1 atmosphere and temperature increases by 1 K, the velocity of sound a) decreases by 0.6 ms-1 b) increases by 0.6 ms-1 -1 c) increases by 60 ms d) decreases by 60 ms-1

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69

51. A stretched wire is divided into three segments whose frequencies are in the ratio 2:3:4. Their

52.

53.

54.

55.

56.

57.

58.

59.

60.

61.

lengths are in the ratio a) 2:3:4 b) 4:3:2 c) 3:2:1 d) 6:4;3 -4 -1 The linear density of a wire is 1.3 x 10 kgm . A transverse wave represented by the equation y = 0.02 sin(x+30t) is propagating along the wire, where x and y are in m and t in s. The tension of the wire is a) 0.48 N b) 0.12 N c) 1.2 N d) 4.8 N When beats are produced by two progressive waves of the same amplitude and of nearly the same frequency, the ratio of maximum loudness produced to loudness of one of the waves will be n, where n is a) 1 b) 2 c) 2.5 d) 4 A man standing between two cliffs claps his hands and hears a series of echoes such that one echo is heard at intervals of 1 s and two echoes in every alternate second. If speed of sound in air is 340 ms-1, the distance between the cliffs should be a) 340 m b) 680 m c) 510 m d) 170 m In a good tuning fork a) only the fundamental is excited b) the first overtone and the fundamental mode are excited c) only the first overtone is excited d) the fundamental and odd harmonics are excited A man facing a wall holds a tuning fork of frequency 256 between himself and a vertical wall. He moves the tuning fork towards the wall with a velocity 1/100 of the velocity of sound. The number of beats heard per minute would be nearly a) 75 b) 150 c) zero d) 300 A tuning fork of frequency 256 Hz is excited and held at the mouth of a closed pipe of frequency 250 Hz. Pick out the correct statement a) 6 beats per second will be heard b) 12 beats per second will be heard c) 3 beat per second will be heard d) No beats will be heard A uniform rope of mass 6 kg hangs vertically from a rigid support. A block of mass 2 kg is attached to the free end of the rope. A transverse pulse of wavelength 0.06 m is produced at the lower end of the rope. The wavelength of the pulse when it reaches the top is ( in m) a) 0.06 b) 0.12 c) 0.03 d) 0.24 *Two identical straight wires are stretched so as to produce 6 beats per second when vibrating simultaneously. On changing the tension slightly in one of them, the beat frequency still remains unchanged. Denoting by T1, T2 the higher and the lower initial tensions in the strings, it could be said that while making the above changes in tension a) T2 was decreased b) T2 was increased d) T1 was decreased c) T1 was increased *The prongs of a vibrating tuning fork are immersed in water. Then a) velocity of the waves will decrease b) amplitude of the waves will decrease c) frequency of the waves will decrease d) wavelength of the waves will increase Which of the following cannot produce super position? a) Light waves b) Micro waves c) Radio waves d) Laser beam

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62. A wave has an angular frequency of 200 rad/s and a wave vector 2 m . The wave velocity is

(m/s) a) 100 b) 100/π c) 100π d) 200/π o 63. The speed of sound in oxygen at a pressure of 1 atmosphere and a temperature of 27 C is x o m/s. The speed of sound in hydrogen at a pressure of two atmospheres and at 27 C is a) 2x b) 4x c) √2x d) 2 √2x 64. Two waves of equal amplitude and equal wavelength arrive at a point at a path difference of λ/6. The ratio of resultant intensity at the point to the maximum intensity will be a) 1/4 b) 1/2 c) 3/8 d) 3/4. 65. A wave equation is represented by y =2 cos2x sin3t. The phase difference between two adjacent particles will be ( in degrees) a) 0 b) 30 c) 60 d) 90 66. Two waves are represented by the equation y1= 10 sin πt. y2 = 10 sin [πt+

67.

68.

69.

70.

71.

π ]. The amplitude 3

of the resulting wave at a given point is a) 10 b) 20 c) 10√2 d) 10√3 If a graph is drawn between amplitude of a wave from a source arriving at certain point with distance from the source and if the medium absorbs no energy, the graph will be a) a straight line having a negative slope b) an irregular curve c) a rectangular hyperbola d) an inverse square law graph The speed of transverse waves through a wire stretched by a tension T is v. If the wire is doubled in itself and stretched by the same tension, the speed of transverse waves will be a) same as v b) (√2)v c) v/√2 d) 2v *When stationary waves are produced in a medium which of the following characteristics change at the antinodes ? a) density b) pressure c) phase d) temperature A particle on the trough of a wave of period T at a given instant will come to mean position after a time a) T b) 3T/4 c) T/4 d) T/2 At the same temperature speed of sound in 50% humid air is v1 and that in 70% humid air is v2. Then a) v1 > v2 b) v1 = v2 c) v2 > v1 d) v1 is greater than or less than v2 depending on the climate

72. A progressive wave has an equation y = A sin2π(10t-20x) where the distances are in metre

and time in seconds. The distance between a crest and the nearest trough of this wave in cm is a) 10 b) 2.5 c) 5 d) 5/π 73. A wave incident on a medium has an amplitude 1 m. The reflected wave has an amplitude 0.5 m. The percentage of intensity transmitted into the medium is equal to a) 75 b) 50 c) 25 d) 12.5 74. The amplitude of the transmitted wave in the previous question (in m) is a) 0.75 b) 0.5 c) 0.25 d) 0.866

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71

75. A war plane is flying at a speed Mac 1.5. The waves produced by it will be

a) plane

b) spherical

c) cylindrical

d) conical

76. A wave is represented by the equation y = A sin (ωt-kx). The ratio of maximum particle

velocity to wave velocity is a) A/k b) k/A c) kA d) ωA 77. If the tension of a sonometer wire is increased by 21%, the frequency of the note emitted by it increases by a) 10% b) 21% c) 32.5% d) 42% 78. If a source of sound of frequency f and a listener approach each other with a velocity equal to 1/20 of velocity of sound, the apparent frequency heard by the listener will be 19 20 21 21 f f a) d) f b) f c) 21 21 20 19 79. A tuning fork gives 5 beats when vibrated with 40 cm length of a sonometer wire. If the

length of the wire is shortened by 1 cm, the number of beats is still the same. The frequency of the fork is : a) 385 b) 320 c) 395 d) 400 80. A tuning fork of frequency f is immersed in water and vibrated. Compared to that in air, a) its period will decrease b) the wavelength of sound emitted by it will decrease c) the velocity of sound waves will decrease d) its amplitude will decrease 81. An observer moves a tuning fork of frequency 350 Hz towards a wall with a speed (1/100) th of speed of sound. The number of beats heard by him per second will be a) 4 b) 5 c) 7 d) 14 82. If temperature increases by 2%, the velocity of sound will increase by a) 1% b) √2 % c) 2% d) 2√2 % 83. Two progressive waves represented by the equations y1 = Asin(100πt) and y2 = A sin(104πt)

produce beats. An observer standing near the sources hears a maximum and the next minimum separated by an interval of time a) 1/2 s b) 1/4 s c) 3/4 s d) 1 s 84. Imagine a hypothetical gas in which sound travels with a speed equal to the rms speed of molecules of the gas. The heat required to increase the internal energy of this gas through 1 K will be a) 2R b) 3R c) (3/2)R` d) (1/2) R π x (4 t + ) , where distances are 4 16 in centimetre and time in second. The phase difference of the same particle at a time interval of 0.8 second is ( in degrees) a) 36 b) 72 c) 102 d) 144 π x (2 t + ) , where distances 86. The equation to a plane wave is given by the equation y = 6 sin 4 8 are in centimetre and time in seconds. Which of the following statement(s) is/are correct for this plane wave? a) The wave is travelling along the positive x-direction. b) The wave has a wavelength of 64 cm. 85. A plane wave train is represented by the equation y = 5 sin

72

87.

88.

89.

90.

91.

92.

93.

94.

95.

96.

97.

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c) The wave has a wavelength of 32 cm. d) The phase difference between two points separated by a distance of 0.2 metre is 112.5o. .Read the following statements carefully: A) Two sine waves of same frequency and amplitude travelling in the opposite direction superpose, giving stationary waves. B) A sine wave can superpose only with a sine wave. Of these statements, a) B is true but A need not be true. b) A is true, but B need not be true c) Both A and B are true d) Both A and B are not true Assuming that it is always possible to see light from a distant star accelerating away from earth, which of the following is likely to happen? a) The star will turn yellow and then red and remain so b) The star will turn blue. c) The star will turn violet and remain so d) The star will turn red and then invisible. Two simple harmonic motions of same period and different amplitudes and no phase difference are combined at right angles. The path of the resultant motion will be a) a circle b) a straight line c) an ellipse d) a parabola If the phase difference between the two motions in previous question is 90o the resultant motion will be a) a circle b) a straight line c) an ellipse d) a parabola If the phase difference between the two motions given in the previous question is 90o and the amplitude of the two motions are same, then the path will be a) a circle b) a straight line c) an ellipse d) a parabola A spring of force constant k and mass m attached with it, has a period T. If the mass is fully immersed in a liquid of density half its density, the period will be a) T/2 b) 2T c) T/√2 d) √2 T If a graph is plotted with period of a pendulum and its length, the graph will be a) a parabola b) a rectangular hyperbola c) a straight line d) an irregular curve The acceleration due to gravity in moon is (1/5)th that of earth. The period of a seconds pendulum in moon will be a) 2√6 s b) 2/√6 s c) 12 s d) 2 s The period of a pendulum on the surface of the earth is T. At 3 quarters down earth its period will be a) T/4 b) T/2 c) (√3/2)T d) 2T *A simple pendulum will have infinite period a) inside an earth satellite b) at the centre of earth c) inside a lift under a free fall state d) inside a lift accelerating up with a value ‘g’ A spring mass system makes 12 oscillations per minute. The spring is cut into two halves and they are suspended from the same point with the other ends carrying the same mass. The number of oscillations made by the system will be ( in one minute) a ) 12 b) 24 c) 6 d) 12√2

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73

98. A uniform cubic plank of side L is floating in a liquid of density twice its density. When it

is immersed slightly and released it will oscillate simple harmonically with a period a ) 2π

L g

b) 2π

2L g

c) 2 π

L 2g

d ) 4π

L g

99. A body executing SHM has a velocity 3cm/s when the displacement is 4 cm .

It has a velocity of 4 cm/s when the displacement is 3 cm. The amplitude of motion of this body is a) 7 b) 6 c) 6.2 d) 5 100. A toy gun uses to fire a spring of force constant K. The spring is compressed by x metre before a shot is fired . If the mass of the shot is M, the height reached by the shot when projected vertically is b) Kx2/Mg c) 2Kx2/Mg d) K2x2/Mg a) Kx2/2Mg 101. A particle oscillating simple harmonically, has an equation x= 5 cos (4πt+π/3), where t is in

seconds and x in metre. Its acceleration at a time t = 4 s, in m/s2 is c) 40π d) 40π2 a) 20π b) 20π2 102. A simple pendulm oscillates with an angular frequency 4 rad/s. The ratio of maximum acceleration to maximum velocity of this pendulum will be a) 4 b) 2 c) 1 d) 0.5 103. A listener and a source are at rest. A strong wind blows from source towards the listener. The apparent frequency heard by the listener will be a) equal to the original frequency b) less than the original frequency c) more than the original frequency d) cannot be said from the data 104. A whistle is rotating in a horizontal circle. An observer A is standing at the centre of the circle and another observer B is standing outside the circle. If f is the true frequency of the whistle, then the frequency heard by A and B will be a) changing frequency by both A and B b) same frequency by B and changing frequency by A c) same frequency by A and B d) changing frequency by B and same frequency by A 105. In which of the following cases Doppler effect cannot be observed ? a) The source and the listener moving with same velocity in opposite direction. b) The source moving and wind blowing with the same speed c) The listener moving and wind blowing with the same velocity d) The source and the listener moving with same velocity in the same direction.

SECTION 2: ANSWERS 1

(a)

2

(c)

3

(d)

4

(a)

5

(b)

6

(b)

7

(b)

8

(d)

9

(d)

10

(a)

11

(c)

12

(d)

13

(a)

14

(c)

15

(b)

16

(d)

17

(a)

18

(d)

19

(c)

20

(c)

21

(c)

22

(a)

23

(c)

24

(c)

25

(b)

26

(a,b,c,d)

27

(d)

28

(c)

29

(d)

30

(a)

31

(b)

32

(b)

33

(a),(d)

34

(d)

35

(b)

36

(b)

37

(c)

38

(a)

39

(b)

40

(d)

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41

(d)

42

(a)

43

(a)

44

(c)

45

(c),(d)

46

(b)

47

(a)

48

(b)

49

(c)

50

(b)

51

(d)

52

(b)

53

(d)

54

(c)

55

(a)

56

(d)

57

(d)

58

(b)

59

(b),(d)

60

(b),(d)

61

(d)

62

(a)

63

(b)

64

(d)

65

(a)

66

(d)

67

(c)

68

(b)

69

(a),(b),(d) 70

(c)

71

(c)

72

(b)

73

(a)

74

(d)

75

(d)

76

(c)

77

(a)

78

(b)

79

(c)

80

(d)

81

(c)

82

(a)

83

(b)

84

(d)

85

(d)

86

(a),(b),(d) 87

(b)

88

(d)

89

(b)

90

(c)

91

(a)

92

(c)

93

(a)

94

(d)

95

(d)

96

(a),(b),(c) 97

(b)

98

(c)

99

(d)

100

(a)

101

(d)

(a)

103

(a)

104

(d)

105

(d)

102

SECTON 3 SOLUTIONS 2πf ω 2π = , because 2πf = ω, = k . Ans.a 2π / λ k λ

1.

V = fλ=

2.

While (a) and (b) are travelling waves, (d) is the superposition of two travelling waves, f(xvt) and f(x+vt). A logarithmic function such as (c) cannot represent a wave motion, because it has no regular period. Ans.c Frequency is the property of the source and hence does not change. Since wave returns to the medium, speed does not change. v = fλ. Hence wavelength λ also does not change. The only quantity that could change due to reflection is the phase. Ans.d

3.

4.

Velocity of transverse waves through string = Here T is same. vA = vB

5.

6.

7.

T , where T is tension, µ is linear density. µ

µ = πr2d, where d is density rA = 2rB.

i.e. µA = 4µB .

Therefore

µB 1 1 = = . Ans.a 4 2 µA

The speed of sound does not change with pressure. V127 400 400 . Ans.b = . From this V127 = V27 300 3

It is directly proportional to √T.

v E . v= When temperature increases, E (Young’s modulus) decreases and λ ρ so does density. But the rate of decrease in Young’s modulus is more than the rate of decrease of density. Therefore v decreases and f also decreases. Ans.b Frequency f =

λ= v/f = 350/500 = 0.7 m. A path difference of λ = phase difference of 2π radian or 3600. Hence a phase difference of 600 = a path difference of λ/6 = 0.7/6 m = nearly 12 cm. Ans.b

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75

γRT (1). Temperature T becomes 2T. When oxygen dissociates, molecular mass M M becomes atomic mass M/2. γ (diatomic) = 7/5, becomes γ(monoatomic) = 5/3. Substituting these values in (1), we find nearest answer is d. Ans.d 9. A supersonic jet has a speed more than the speed of sound. Hence wave front coming from it will be conical. Ans.d 10. When the tuning fork is vibrating condensations travel along the X axis producing two maxima. Rarefactions travel along the Y axis producing two maxima. Thus we have four maxima. In between two maxima there will be minimum. Thus, the observer will hear 4 maxima and 4 minima in one revolution. Ans.a 8.

v=

T . If free lower end of the rope is taken as origin, as µ x increases tension also increases because tension is provided by hanging portion of the rope. Since velocity v is proportional to √T, and T α x, v α √x. Ans.c 12. By the law of transverse vibration, frequency is inversely proportional to length and inversely proportional to square root of linear density. When the wire is stretched, its length becomes twice and area decreases to half, because the volume is constant. The frequency decreases to half due to the increase in length and increases to √2 times due to the decrease in area. Thus frequency becomes (1/2) x √2 = 1/√2 of the original value = 0.7 f. Ans.d 11. The velocity of transverse waves v =

13. v =

v γ M γRT . When T is constant, 1 = 1 2 , where subscript 1 is for hydrogen and 2 for M v2 γ 2 M1

helium. γ1 = 7/5, γ2 = 5/3. M1 = 2g/mol. M2 = 4g/mol. This gives v1/v2 = 42 / 25 . Ans.a 14. The fundamental frequency of an open pipe of length L is v/2L. When it is immersed half inside water, its length becomes half i.e. L/2, and also it becomes a closed pipe. The frequency of a closed pipe is v/4L1. Substituting in this equation L1 = L/2, we get the new v v = . Ans.c frequency as (4 × L / 2) 2L 15. The tonal quality depends on the number of overtones present, and their relative intensity.

Ans.b 16. When a motor boat travels in water its propeller cuts the water surface lateraly at the same

time pushes backward. This produces both longitudinal and transverse waves. Ans.d 17. The waves produced in sonometer are transverse (vibration perpendicular to the wave

propagation) and stationary. Sound waves cannot be polarised. Ans.a 18. When two waves of amplitudes A1 and A2 at a phase difference of φ superpose, the resulting

amplitude is given by

A 1 2 + A 2 2 + 2A 1 A 2 cos φ . Here A1 = A2 = A. φ = 600. This gives

the resulting amplitude as √3 A. Ans.d 19. The types of waves reaching the listener are those propagating through air and those through

steel. Steel is more elastic. Hence the amplitude and the intensity of waves through steel will be more. The velocity of sound waves through steel is greater and hence they reach quicker. Ans.c 20. The given wave is a cosine wave and one travelling along the positive x-direction. To form stationary waves, the two waves should travel in the opposite direction. That is the required wave should have a term (kx+ωt). Also a cosine wave can be combined only with a cosine

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wave. The only cosine wave travelling along the negative x-direction in the given list is – a cos(kx+ωt). We can check combining the two, i.e. a cos (kx-ωt) and –a cos(kx+ωt), we get 2a sin(kx) sin (ωt), for which x = 0 is a node. Ans.c 21. Using the equation v = fλ, we have -λ =

v 1.7 × 1000 = = 4x10 − 4 m. Ans.c f 4.2 × 10 6

22. The relative velocity of sound waves with respect to the wall is V+v. Hence the apparent

frequency of the waves striking the wall will be (V+v)/λ. Each wave has one positive crest. Hence the number of positive crests will be same as the frequency. For 5 seconds, this number will be [5(V+v)]/ λ. Ans.a 23. The resultant amplitude A of two waqves of amplitudes a1 and a2 at a phase difference φ is

(a12+a22+2a1a2 cosφ)1/2. Substituting a1 = 10, a2 = 10 and φ = 900, we get A = 14.1. Ans.c 24. A tuning fork emits a pure sine wave and also gives a single frequency. That is, it has no harmonics. Ans.c 25. The maximum particle velocity of a SHM of amplitude Y0 and frequency f is 2πfY0. The

πY0 . Ans.b 2 26. Comparing with the standard equation of a transverse progressive wave, Y = A sin(ωt + kx), we find A = 10-4, ω = 2πf = 60, which gives f = (30/π) Hz. k = 2π/λ = 2. That is λ= π.m. Speed of the wave v = ω/k = 60/2 = 30 m/s. Hence all the answers are correct. Ans.a,b,c,d 27. The first 3 are progressive waves, while the last one is super position of two progressive waves, i.e. stationary wave. Ans.d 28. Intensity at a point is the energy received per unit area per second or power received per unit area. Here power is distributed at the point of reception on the surface of a sphere of radius P 5 2m. Therefore I = = 0.1 Wm-2 . Ans.c = 4πr 2 4π × 2 2 wave velocity is fλ. For 2πfY0 to be equal to 4fλ, λ has to be

29. Comparing with the standard equation, Y = A cos(ωt+kx) with Y = 3 cos(100πt-πx), we find

ω= 100π, k = π ; 2π/λ = π, λ = 2m. Ans.d

30. Comparing with the standard equation for a stationary wave, y = 2A cos(kx) sin(ωt) we find

the amplitude term is 2A cos(kx).

A node appears when amplitude is 0, i.e. kx =

π . 2

2πx π = . This gives x = 12.5 cm. Ans.a 50 2 31. Comparing with standard equation, Y = A sin (2πft), we find the frequency of waves are 200

Hz and 202 Hz. So, the number of beats will be 2 per second. The intensity ratio will be the ratio of the square of the maximum amplitude (4+3) and the minimum amplitude (4-3) which is 72:12 = 49:1. Ans.b 32. Intensity is proportional to the square of amplitude. Therefore the amplitudes are in the ratio 3:1. The ratio of the maximum and the minimum amplitudes is 4:2. So the ratio of the maximum intensity to the minimum intensity is 16:4 = 4:1. You can also use indirect theory short-cut formula. Ans.b 33. (b) and (c) cannot represent a wave motion because they involve square of x and t and therefore will not have a regular period. (d) involves cos2, hence cannot be negative but cos2θ

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77

can be written as (1+cos2θ)/2 and hence could be periodic. (a) represents superposition of two waves. Ans.a,d 34. Since Snell’s law is valid, the refractive index n of the medium is given by velocity of sound in air/velocity of sound in medium = 330/1400 = 0.235, where we have taken velocity of sound in air and water roughly as 330 and 1400 m/s. If C is the critical angle, then the refractive index n is given by = sin C/sin90 = n = 0.235. (note the difference with the optics equation n = 1/sin C). We get C = nearly 140. (We need not actually calculate but only need know that it is less than 600, incident angle). Thus the angle of incidence is greater than this critical angle. Hence the wave will totally reflect into air. Ans.d 35. According to Fourier theorem, a period function f(t) can be expanded as a number of harmonic functions. Here y = f(t) is periodic since RHS is harmonic. But for a simple harmonic function we should have same period for harmonic functions. Here periods of cos 4πt and sin 5πt are not same. Hence it is not simple harmonic. Ans.b 36. Writing the general expression for y in terms of x as y =

At t = 2s, y =

1 1 + [ x − v(2)] 2

1 1 + ( x − vt )

2

, at t = 0, y =

1 (1 + x 2 )

. Comparing with the given equation we get 2v= 1 and v =

.

0.5

m/s. Ans.b 37. Comparing with the standard equation, y = A sin (ωt-kx+φ) we ω is 15π and k = 10π.

ω 2π = − 1.5 ms −1 . Also = k = 10π. λ = 0.2 m. The wave is travelling along the k λ negative x direction with a velocity 1.5 ms-1 and has a wavelength 0.2 m. Ans.c

Velocity =

38. If we have an equation of the form cos kx sin ωt, we know, it is due to the superposition of

two wave motions. If we have an equation of the form cos2 at sin bt, it can be shown to be superposition of three sine waves. Ans.a

39. Since the equation has no π, comparing the standard form y = a sin (ωt-kx) with the given

equation y = a sin(20t-40x), we find k = 40. 2π/λ = 40, λ = π/20m. Ans.b 40. The given equation is that of a stationary wave and can be written as y = 0.02 cos (10 πx) sin

(50πt). Comparing with standard equation y = A cos(kx) sin (ωt) we find first node appears when cos 10 πx = 0. That is 10πx = π/2. x = 0.05 m. Next nodes will be at 0.15 m, 0.25 m, 0.35 m etc (choice a) Anti-node appears when cos 10πx = 1. That is 10πx = nπ. So the antinodes will be at 0, 0.1 m, 0,2m, 0.3 m (choice b). Velocity of the wave v = ω/k = 50π/10π = 5m/s. (choice c). Since all are correct the choice is d. Ans.d 41. In a closed pipe of length L resonance appears for wavelengths L = λ1/4, L = 3λ2/4 and L =

5λ3/4, neglecting end correction. 1:(1/3):(1/5) = 15:5:3. Ans.d

This will give λ2 : λ2 : λ3 = 4: (4/3): (4/5) that is

π π π π cos ( t). This will be maximum when cos t is maximum. i.e. t = 6 6 6 6 nπ, where n is an integer. This gives t = 0, 6,12,18 etc. [Alternate way of doing: Comparing with the standard equation, cos(2πt/T), we find period of wave motion T is 2π/T = π/6. T = 12 s. In a simple harmonic motion maximum velocity is attained during half the period, because the particle will cross equilibrium position twice in a period. i.e. 6,12,18 s etc. Ans.a

42. Velocity dx/dt = 2

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43. By law of transverse vibration f α T .

f1 = f2

T1 = T2

M1 where M1 g is weight in air and M2

M2g is weight under water. M2 = M1 - ∆M1, where ∆M1 is loss of weight in water. M1 = Vρ. M2 = Vρ-(V/2)x1. M1/M2 = 2ρ/2ρ-1, where we have taken specific gravity of water as 1. This gives

f2 M2 2ρ − 1 2ρ − 1 = = ⇒ f 2 = 300 × Ans.a f1 M1 2ρ 2ρ

44. Frequency difference = f1-f2 = (V/λ1– (V/λ2)) = (348/2)-(348/3) = 174-116 = 58. The beat

frequency will therefore be 58 Hz. (a) is correct. Since human ear cannot detect more than 8 beats/s, (b) is correct. (c) is wrong. When temperature increases, velocity of sound will increase, frequency will increase and number of beats will also increase (d) is correct. Ans.c 45. The power transmitted by the wave is the energy of simple harmonic motion per second. This we know as (1/2)mω2A2. Thus the energy will be proportional to the square of frequency f and square of amplitude A. Ans.c & d 46. The density of the mixture is ρm = (v1ρ1 + v2ρ2)/(v1+v2) = [(v/2)16 + (v/2)1]/v = 17/2.

Velocity of sound = same v(mixure) = v(oxygen )

value ρ(oxy )

γP / ρ . i.e. velocity of sound is proportional to 1/ρ, since both have for

γ.

Therefore

v(mixture)/v(oxygen)

16 32 = .Ans.b = 17 / 2 17 ρ(mix ) 2

47. v (maximum) = ωA......(1) and a (maximum)= ω A.....(2) for simple harmonic motion of

amplitude A and angular frequency ω. Squaring equation (1) and dividing by (2) we get ω2A2/ω2A = 16/16 = 1. That is A = 1m. Ans.a

48. From the first statement frequency of x = 100 ± 5 = 105 or 95. From the second statement,

second harmonic of the source x has to be 205 ± 5 = 210 or 200. This means x is either 105 or 100. The common value of both conclusions is 105 per second. Ans.b 49. According to an empirical formula the end correction of the resonance tube is given by 0.3 d, where d is the diameter of the tube. The corrected equation will therefore be L1 + e = λ/4, for first resonance. If d is increased, e increases, λ increases and f decreases slightly. Ans.c 50. The increase in pressure does not change the velocity of sound. When the temperature increases by 10C or 1 K, the velocity of sound increases roughly by 0.6 ms-1. (Recall formula Vt = V0 + 0.6 t). Ans.b 51. According to one of the laws of transverse vibration, frequency is inversely proportional to length. So the ratio of length will be 1/2 : 1/3 : 1/4, which is 6:4:3. Ans.d 52. We compare with the standard equation for a travelling wave, Y = A sin (ωt + kx). We find ω

= 30, k = 1, velocity ω/k = 30 ms-1. The velocity of transverse waves in a string is given by V

=

T / µ , where T is the stretching tension µ is the linear density. µ here is given as 1.3 x 10-

. T = V2µ = 0.12 N. Ans.b 2 53. Let a be the amplitude of each wave . Then intensity of one wave is proportional to a . The maximum amplitude is the sum of the two, i.e. 2a. Hence maximum intensity is proportional to 4a2. The required ratio n here is 4a2/a2 = 4. Ans.d 54. The man should be standing 1/3 from one of the cliffs and 2/3 from the other cliff. The sound takes 1/2 second to travel to the nearest cliff and 1/2 second to return. Similarly it takes 1 4

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second to travel to farther cliff and 1 second to return. Distances between man and nearest cliff = distance travelled by sound in 1/2 s = 170 m. This gives distance between cliffs 170 x 3 = 510 m. Ans.c 55. A tuning fork gives only the fundamental frequency and it gives a pure sine wave. That means it has no harmonics. Ans.a 56. Here the tuning fork is a source moving away from the observer, while the sound reflected from the wall is moving towards the observer. Thus we have two sources of the same frequency, one moving away and the other moving towards the observer with the same velocity. If V is the velocity of sound Us the velocity of source, the apparent frequency 2fU s Vf Vf 1 − = difference V , we get. number if Us T2 , f1 > f2 : f1-f2

= 6 Hz. By increasing the lower tension T2, we can increase f2 such that f2-f1 = 6Hz. Siimilar we can decrease the higher tension T1 such that f decreases making f2-f1 = 6 Hz. Ans.b & d 60. The amplitude of vibration will decrease due to the resistance produced by the water medium. Hence (b) is correct. The velocity of sound in water (1400 m/s) is greater than its velocity in air (340 m/s). Hence (a) is wrong. The frequency is the property of the source and hence does not change. Thus (c) is wrong. v = fλ. As v increase, λ also should increase. Hence (d) is correct. Ans.b & d 61. The principle of super position is common for all wave motion. But in the case of the laser beam it is already an intense beam got by a number of photons in phase. So it has a very large amplitude and hence difficult to superpose. Ans.d 62. The wave velocity in terms of angular frequency ω and wave vector k is v = ω/k = 200/2

=100 m/s Ans.a 63. Here the temperature is same. The variation of pressure does not change velocity of sound.

Using the formula v = vH = 4x. Ans.b

γ RT / M , we find vH/vO =

M(O 2 ) / M (H 2 ) = 32 / 2 =4 . This gives

64. Use the information supplied already in indirect theory notes. A path difference of λ/6 means

a phase difference of (2π/λ)x(λ/6) = π/3 radian. Intensity at the point I = Iocos2 (π/6). This gives I/Io = cos230 = 3/4 . Ans.d 65. Use the information supplied already in theory notes. to recognise that the given equation is that of a stationary wave. In a stationary wave all particles vibrate in phase. The phase difference between any two adjacent particles will be zero. Ans.a

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66. Use the information supplied already in theory notes. The resultant amplitude is got by

substituting a = 10, φ = 60o, which gives 10x√3. Ans.d 67. The intensity at a point is inversely proportional to square of the distance from the source. Hence the amplitude (a2 = I) is inversely proportional to the distance from the point to the source. The graph between them will be a rectangular hyperbola. Ans.c 68. The speed of transverse waves through a wire is given by

T / µ , where T is stretching

tension and µ linear density. When the wire is doubled in length the linear density which is mass/length becomes half the original value. Thus µ becomes µ/2. The velocity becomes √2 times. Ans.b 69. In a stationary wave formation at the antinode pressure is minimum, which means density also is minimum. Pressure is maximum and density maximum at nodes. Phase does not change in stationary wave. The temperature changes because when sound wave propagates through a medium the medium undergoes adiabatic change. Ans.a,b,d 70. Trough is a point of minimum (negative maximum) displacement. From here to mean position ( point of zero displacement) the time taken will be one quarter of a period. Ans.c γP / ρ . When air is humid, more water vapour is present. This will decrease the mass of air in a given volume and hence the density of air. So the velocity of sound will increase. Ans.c

71. The velocity of sound in air at a given temperature =

72. Comparing with standard equation y = Asin(ωt-kx), we find k is equal to 2πx20. But k is

equal to 2π /λ. This gives λ = (1/20) m = 5 cm. Distance between crest and trough = λ /2 = 2.5 cm. Ans.b 73. Since amplitude reduces to half after reflection, the reflected intensity which is a square of the amplitude, reduces to 1/4th i.e. 25%. The transmitted intensity will be 75%. Ans.a 3 / 4 = √3/2 = 0.866. Ans.d 75. Use the information supplied already in theory notes.. Speed Mac 1.5 means it travels with a speed of 1.5 times speed of sound. Hence the waves produced will be conical. Ans.d 74. Since the transmitted intensity is 75% i.e. 3/4 , the transmitted amplitude is

76. Use the information supplied already in theory notes. Maximum particle velocity is ωA.

Wave velocity = ω/k. Dividing the two equations, the required ratio is kA. Ans.c

77. By one of the laws of transverse vibration, frequency f is proportional to √T or = k√T. When

121T = 1.10 k√T. This means the frequency 100 increases to 1.1 f i.e. increases by 10%. Ans.a 78. The apparent frequency heard when a source and listener approach each other with a velocity V + UL Us and UL respectively, is given by the Doppler effect formula f ' ' = f . Substituting V − Us tension is increased by 21% f becomes k

here UL= Us=(1/20) V, where V is the velocity of sound in air, we get f’’ = (21/19) f. Ans.b 79. If f is the frequency of the tuning fork then f1 = f-5. f2 = f+5. Using one of the laws of

transverse vibration, we have f1l1 = f2l2. That is (f-5)40 = (f+5)39. Solving we get f = 395 Hz. Ans.c 80. Due to high damping (resistance) exerted by water, the amplitude and the intensity will decrease. So (d) is correct. The period is a property of the source and hence will remain

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81

constant. (a) is wrong. The velocity of sound in water is greater than that in air. Hence (c) is wrong. v= fλ. When v increases and f remains constant, λ should also increase. Hence (b) is wrong. Ans.d. 81. Use the equation supplied in theory notes. The number of beats heard will be 2fUs/V. Here Us/V = 1/100. f= 350 Hz. Substituting these values, we get the number of beats per second as 7. ( Note: The velocity of sound in air is not given in the question. Hence you cannot assume it for calculation. Our formula has only the ratio of velocity of tuning fork to the velocity of sound. This is the advantage of using it) Ans c. 82. The velocity of sound is directly proportional to the square root of temperature . i. e. v= k√T.

To find change of speed for small variation of temperature, we should differentiate this equation. This is the shortest way to do it. Differentiating, dv/v = (1/2) dT/T. Since it is given here dT/T = 2%, we get dv/v = 1%. Ans a 83. Comparing with the standard equation to a progressive wave y = A sin ωt, we have the

frequencies of the waves producing beats are 2πf1 = 100π, and 2π f2 = 104πt. This gives f1 = 50 Hz and f2 = 52 Hz. The number of beats produced per second, that is difference in frequency = 2. The time interval between two beats is, therefore, 1/2 a second. The time interval between one maximum and next minimum is 1/4 a second. Ans b. 84. We have already seen in answering a similar question in one of previous section, the ratio of velocity of sound to rms velocity of molecules of the gas is

γ / 3 . If these two are same, the

gas must have γ = 3. That is Cp/Cv = 3. This gives Cp = 3 Cv. What is required is Cv. Cp-Cv = R. ⇒ 3Cv - Cv = R. This gives Cv = R/2. Ans. d 85. We have to find the wave period first and then use information given in theory notes . For this we compare with any standard equation , but it is always good to rely on one standard equation, which is given in the theory notes. The comparison between y =A sin (ωt+kx) gives ω as equal to π. That is π = 2πf. This gives f = 1/2 and the wave period T as 2 seconds. Now use the information in theory notes . We get the time difference of 0.8 second corresponds to a phase difference of 360x0.8/2 = 144o. Ans d 86. Comparing with the standard equation as in the previous question, we find k = π/32 = 2π/λ.

This gives λ = 64 cm. This makes (b) correct and (c) wrong. A difference in the path of 64 cm is equal to a phase difference of 2π radian or 360o. Thus 20 cm should produce a phase difference of 360x20/64 = 112.5o. This makes (d) correct Since there is a positive sign between t and x terms, the wave is travelling along the negative x direction. Thus (a) is correct. Ans a,b,d 87. The statement (A) is correct. But sine wave can superpose with a cosine wave, because cosine wave is a sine wave with a phase difference of 90o. Hence (B) need not always be true. Ans b 88. As the star is moving away from the observer, it should show red shift. As its velocity is increasing , the light from it will shift towards red, will become red and then infrared. The star will become invisible since then, as infrared is invisible. Ans d 89. If two motions are represented by equations x = a sinωt and y = b sinωt, we have x/a = sinωt,

y/b = sinωt. So x/a = y/b. i.e. y = (b/a)x. This is an equation to a straight line passing through the origin inclined at an angle tan-1(b/a). Ans.b 90. Here the two motions can be represented by equations x = a sinωt and y = b sin(ωt+π/2) =

b cosωt. Squaring and adding, we get (x2/a2)+(y2/b2) = 1. This is equation to an ellipse, whose axes coincide with coordinate axes. Ans.c

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91. Here the two motions can be represented by equations x = a sinωt and y = a sin(ωt+π/2) =

a cosωt Squaring and adding, we get x2+y2 =a2. This is equation to a circle whose centre is origin and whose radius is a. Ans.a

92. The period T = 2π m / k . When the mass is fully immersed in the liquid, the buoyant force

will be (m/2)g. Hence the apparent weight will be (m/2)g. 2 π m / 2 k = T/√2. Ans.c

The new period will be 2

93. The period of a pendulum T is directly proportional to √L. That is T = k√L. Squaring T =

constant x L. So the graph between T and L will be a parabola. (Note: The graph between T2and L will be a straight line which usually is drawn in a school laboratory). Ans.a 94. A seconds pendulum by definition is one which has a period of two seconds, wherever it is . Ans.d. 95. T = 2π L / g . At 3 quarters down earth the value of acceleration due to gravity is equal to g1

= [R-(3/4)R]g/R. ⇒g1 = (1/4)g. The period will be equal to 2π L / g 1 = 2T. Ans. d 96. A simple pendulum will have infinite period when acceleration due to gravity is zero or

effectively zero. This happens inside a satellite, at the centre of earth and a lift in free fall state. This makes first three answers correct. In a lift accelerating up ‘g’ will be effectively more than the normal value. This makes the last answer wrong. Ans a,b,c 97. The frequency of oscillation of a spring is given by n = 1/( 2π m / k ).

When the spring is cut into two halves, each half will have a force constant of 2k. When they are suspended from the same point with a common mass, they are in parallel. Hence the force constants add to 4k. The new frequency n1 will be equal to

1 2π

k = 2n = 24. Ans.b. m

98. If x is the distance through which plank is immersed, A cross sectional area, then weight of

liquid displaced = -Ax2dg = -kx. Here this force provides restoring force. Force constant k = A2dg. The period of oscillation will be = 2π m / k =2π ALd / A 2dg = 2π L / 2g . (Can also be answered from indirect theory short-cuts.) Ans.c 99. The velocity of SHM having an amplitude A at a displacement x is given by v= ω

A 2 − x 2 . Here we have v= 3 when x= 4 . This gives 3 = ω A 2 − 4 2 (1). 2

Also v= 4

2

when x= 3. This gives 4 = ω A − 3 . (2). Dividing (1) by (2), squaring and simplifying we get A = 5 cm. Ans d 2 100. The potential energy when the spring is compressed is (1/2)Kx . When the shot is fired upward, this energy is converted into gravitational potential energy of the shot. This gives (1/2)Kx2 = Mgh. ⇒ h= Kx2/2Mg. Ans a 101. The acceleration of SHM is -ω2x. Here ω = 4π. Hence the acceleration will be -16π2[5 cos

(4πt+π/3)]. When we substitute t = 4, this reduces to 80π2cos(π/3) = 40π2. Ans d.

102. The maximum velocity of SHM in terms of ω is ωA, where A is the amplitude and the

maximum acceleration will be ω2A, in magnitude(because acceleration has negative sign). Dividing the two, we get the ratio of maximum acceleration to maximum .velocity as ω2A/ωA =ω = 4. Ans. a 103. The apparent change in the frequency due to Doppler effect depends on the relative velocity between the source and observer. The motion of medium alone cannot produce Doppler

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effect. Here the relative velocity between the source and the observer is zero and hence no change in frequency. . Ans.a 104. For an observer standing outside the circle maximum and minimum (i.e. more than and less than f) will be heard due to Doppler effect. For an observer at the centre of the circle, the velocity of the source is always perpendicular to the line joining the observer and the source. Hence no Doppler change in frequency. Ans.d 105. Except (d), in all other cases Doppler shift in frequency will take place. In (d) the relative velocity between the source and the observer is zero. Hence no Doppler effect. Ans.d

4 ELECTROSTATICS SECTION 1 QUESTIONS 1.

2.

3.

4.

5.

6.

7.

8.

9.

When a body is negatively charged by electostatic conduction, its mass will a) slightly increase b) slightly decrease c) remain same d) increase or decrease depending on material Two charges q1, q2 are separated by a distance r. If a metal plate of thickness x is introduced between the charges the force between the charges F will now be a) F = 0 b) F = A[q1q2/(r-x)2] 2 c) F = A(q1q2/r ) d) F = A[q1q2/(r+x)2] Two point charges +4 mC and –1mC are separated by a distance of d. The ratio force acting on them will be a) 1:4 b) 1:16 c) 1:1 d) 1:-1 *Two insulated copper spheres of radius 0.1 m and 0.15 m are given equal charges. They are connected by a wire. Then a) both spheres attain the same charge b) both spheres attain the same potential c) smaller sphere attains greater charge d) bigger sphere attains greater charge Three point charges +q each are placed on the vertices of an equilateral triangle, which lie on the circumference of a circle of radius r. The electric field at the centre of the circle is a) Aq/r2 b) zero c) 3Aq/r2 d) 3q/r2 Inside a uniformly charged solid sphere,the electric field intensity at any point distant r from the centre of the sphere varies as a) 1/r2 b) r2 c) r2 d) r A charge q is located at the centre of a cube. The electric flux through any one face of the cube is a) q/ε0 b) zero c) q/4εo d) q/6εo When a soap bubble is negatively charged a) it collapses b) it increases in size c) it decreases in size d) it remains as such Two metallic spheres of radii 1 cm and 2 cm are given charges 10-2 C and 5 x 10-2 C, respectively. If they are connected by a conducting wire, the final charge on the smaller sphere is a) 2 x 10-2 C b) 4 x 10-2 C c) 1 x 10-2 C d) 3 x 10-2 C

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10. There is an electric field E along x-direction. If the work done in moving a charge 0.2 C

through a distance 2 metres along a line making anangle 600 with the x-axis is 4 J, the value of E is a) 4 N/C b) 8 N/c c) √3 N/C d) 20 N/C 11. A particle of mass 1 kg and charge 1 C falls through a potential of 1 V. Its velocity is a) 1.4 ms-1 b) 2 ms-1 c) 0.5 ms-1 d) 1 ms-1 12. The dimensions of εv is the same as that of

a) capacitance b) capacitance/unit length c) capacitance/area d) potential/length -11 13. The radius of the hydrogen atom is 5 x 10 m and the charge on electron or proton is of magnitude 1.6 x 10-19C. At the ground state, the dipole moment of the atom ( in C m)is b) 16 x 10-30 c) 0 d) 4x10-30 a) 8 x 10-30 14. Two conductors of capacitances 1 µF and 2 µF are charged to 200 V and 100 V respectively,

15.

16.

17.

18.

19.

20.

21.

22.

23.

and then connected by a wire. The final potential of the connected system is a) 150 V b) 106 V c) 133 V d) 187 V The dielectric strength of a certain material is 106 Vm-1. The breakdown voltage to be applied across a 1 mm thick specimen is a) 106 V b) 109 V c) 103 V d) 2 x 106 V A charge q is placed at the centre of the line joining two equal charges Q. The system of the three charges will be in equilibrium if q is equal to a) -Q/2 b ) -Q/4 c) +Q/4 d) +Q/2 *A parallel plate capacitor is charged and the charging battery is then disconnected. If the plates of the capacitor are moved farther apart by means of insulating handles, a) the charge on the capacitor increases b) the voltage across the plates increases c) the capacitance increases d) the electrostatic energy stored in the capacitor increases A certain charge is divided in the ratio m: n so that when kept at a given distance, the force between them is maximum. The ratio m:n is a) 2;1 b) 4:1 c) 1:1 d) 3:1 A condenser is connected to a battery . If a dielectric slab is introduced fully in the intervening space between the condensor plates, the electric field between the plates will a) decrease b) increase c) remain the same d) cannot be answered from the data An electron falls between two parallel plates of separation 2m, in which a uniform electric field 2.5 V/m exists. The energy gained by electron is equal to a) 2.5 eV b) 1 eV c) 4 x 10-19 J d) 8 x 10-19J A condenser of capacity C is connected to a battery of potential difference V. If q is the charge given by the battery, the energy given by the battery and energy stored by the condenser are respectively a) (1/2)qV, (1/2) qV b) qV, (1/2) QV c) (1/2)qV, qV d) qV, qV The potential due to an electric dipole on the axial line at a point r from the dipole is V. The potential at a distance 2r from the dipole will be a) V/2 b) V/4 c) V/8 d) V/16 *Which of the following statement(s) is/are true? a) If electric field at a point is zero, electric potential will always be zero

86

24. 25.

26.

27.

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29.

30.

31.

32.

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b) If electric potential at a point is zero, electric field will always be zero c) Electric field can exist at point even when the potential is zero at the point d) Electric potential can exist even when the field is zero at the point The number of lines of force coming out from +1C is a) 1.11 x 1011 b) 1.11 x 10-10 c) 1.1 x 1019 d) 1 Two identical capacitors are joined in parallel and charged to a potential V. Then these are separated and connected in series with the positive plate of one connected to the negative of the other. Then a) the charges on the plates connected together are destroyed b) the charges on the free plates are enhanced c) the energy stored in the system increases d) the potential difference between the plates becomes 2V If the distance between the plates of a parallel plates condenser is increased, its potential will a) remain the same b) decrease c) increase d) decrease expontentially The insulating property of air breaks down at the intensity of electric field 3 x 106 Vm-1. The maximum charge that can be given to a sphere of diameter 5m is about a) 2 x 10-9 C b) 2 x 10-3 C c) 2 x 10-5 C d) 2 x 10-6 C A hollow metallic sphere of radius 5 cm is charged such that the potential on its surface is 10V. The potential at the centre of the sphere is a) 10 V b) zero c) same as at a point 10 cm away from the surface d) same as at a point 25 cm away from the surface Two identical small metal balls are given charges equal to 10 units and –20 units. They are allowed to touch each other and are again separated to the same distance as before. The ratio of the force between the two balls in the two cases respectively is a) -8:1 b) 8:1 c) 1:8 d) 1:8 Two concentric thin metallic spheres of radii R1 and R2 (R1 > R2) bear charges Q1 and Q2 respectively. The potential at radius r between R1 and R2 is a) 1/4 πε0 [ (Q1/R1) + (Q2/R2)] b) 1/4 πε0 [(Q1/r ) + (Q2/r)] c) 1/4 πε0 (Q1/R1 ) + (Q2/r) d) 1/4 πε0 (Q1 Q2 / R1 R2) A parallel plate capacitor is charged. With charging battery connected, the plates of the capacitor are moved further apart by means of insulating handles a) the electric field between the plates increases b) the electric field between the plates decreases c) the electric field between the plates remains the same d) cannot be said from the data. K (Fig) A parallel plate capacitor with plate area A and t separation is filled with dielectric as shown. The dielectric t K constants are k1 and k2. The capacitance will be b) ε0A (k1+ k2)/k1 k2t a) ε0A (k1 +k2)/t d) 2ε0A (k1 +k2)/k1 k2 t c) 2ε0A k1k2/(k1+k2)t 1

1

2

2

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33. Three protons A,B and C are located between the plates of a parallel plate condenser such that

34.

35.

36.

37. 38.

39. 40.

41.

42.

B is midway between the plates, A at a distance from one plate and C is at an equal distance from the other plate. (Fig.) Check the correct statement(s) a) The force on all the protons will be same b) The force on proton B will be zero A C c) The force on A and C will be same in magnitude and direction d) The force on the proton A and C will be equal and opposite B There is a non-uniform electric field along x-axis as shown in fig.. The field increases at a uniform rate along +ve x-axis. A dipole is kept inside the field as shown. Which one of the following statements is correct for dipole ? a) dipole moves along positive x-axis and rotates clockwise b) dipole moves along negative x-axis and rotates clockwise -q c) dipole moves along positive x-axis and rotates anticlockwise d) dipole moves along negative x-axis and rotates anti-clockwise Two spheres A and B of radius a and b respectively are at the same potential. The ratio of the surface density of charge of A to that of B x-a xis is a) b/a b) a/b c) a2/b2 +q 2 2 d) b /a In a certain region of space there exists a uniform electric field of 2 000 k Vm-1. A rectangular coil of sides 10 cm x 20 cm is kept in XY plane. The electric flux through the coil in SI will be a) zero b) 40 c) 4 x 105 d) 4 An lisolated metal sphere of radius R is given a charge q. Its potential energy will be b) q/4πεoR c) q/8πεoR d) q2/8πεoR a) q2/4πεoR The electric field at a certain point is 10 NC-1. The electric lines of force crossing unit area around the point at right angles to it is a) εo b) 1/εo c) 5 d) 10 0 The number of lines crossing at an angle 30 with the surface in previous question is a) εo b) 1/εo c) 5 d) 10 2 A rectangular coil of area 200 cm is placed in an electrostatic field 200 kˆ , in Y-Z plane. The electric flux through the coil will be a) zero b) 4 x 104 c) 40 d) 4 Two small balls are given equal positive charge Q coulomb each and are suspended by two insulating strings of length L each (metre) from a hook fixed to a stand. If the arrangement is taken to a satellite orbiting round earth, the angle θ between the two strings will be a) 00 b) 900 c) 1800 d) 00 < θ < 1800 The tension of each string in this position as given in previous question will be a) 0 c)

b) Q2

8πε 0 L2

d)

Q2 16πε 0 L2 Q2 4πε 0 L2

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43. An electron kept in an electric field of strength E experiences a force equal to its weight. If m

44.

45.

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49.

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51.

52.

53.

is the mass of electron, the value of E is equal to a) mg b) mg/e c) mg/e2 d) e/mg A hollow charged metal sphere has a radius r. If the potential difference between its surface and a point at distance 3 r from the centre is V, then the electric field intensity at a distance 3r from the centre is a) V/6r b) V/4r c) V/3r d) V/2r If one penetrates into a uniformly charged sphere, the electric field strength E will a) decrease b) increase c) remain the same as that at the surface d) be zero at all points If all the electrons are removed from 1 g of hydrogen atom, the charge it will have would be nearly a) 106 C b) 104 C c) 1.6 x 10-19C d) 105C The electric potential V at a certain point distant x (in metre) is given by V(x) = 5x2+10x-9 volt. The electric field at x = 1 m will be a) 20 V/m b) -10 V/m c)- 23 V/m d) -20 V/m At any point on the right bisector of an electric dipole a) the electric field is zero b) the electric potential is zero c) the electric field is perpendicular to the dipole d) the electric field is parallel to the dipole A parallel plate capacitor of plate area A and plate separation d is charged to potential difference V and then the battery is disconnected. A slab of dielectric constant K is then inserted between the plates of the capacitor so as to fill the space between the plates. The work done on the system in inserting the slab is a) ε0 AV2/d b) ε0 KAV2/2d 2 d) ε0 AKV2/2d(K-1) c) ε0 (K-1)AV /2Kd *A proton moving with a constant velocity passes through a region of space without any change in its velocity. If E and B represent the electric and magnetic fields respectively, this region of space may have a) E = 0, B = 0 b) E = 0, B not equal to 0 c) E not equal to 0, B = 0 d) E not equal 0, B not equal to 0 Two equal negative charge –q are placed at points (0,a) and (0,-a) . A positive charge Q is released from rest from a point 2a, 0 on the x-axis. The charge Q will a) execute S.H.M about the origin b) move to origin and remain at rest c) move to infinity d) execute oscillations but not simple harmonic 1000 small water drops each of radius and charge q coalesce to form a single bigger drop. The ratio of potential of the bigger drop to that of the smaller one is a) 1 b) 10 c) 100 d) 1000 An infinite number of electric charge each of magnitude +e are arranged along the X-axis at distance x= 1m, x = 2m, x = 4m, x = 8m ....etc. The electrostatic potential at the origin is a) ∞ b) e/2πεo c) e/3πεo d) e/5πεo

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54. If the charges in the above question are alternately + and – beginning with a positive charge,

the potential at the origin will be a) zero b) e/2πεo c) e/3πεo d) e/6πεo 55. In the arrangement as given in question (53), the electric field at the origin will be a) zero b) e/2πεo c) e/3πεo d) e/5πεo e) e/6πεo 56. In the arrangement as given in question (54), the electric field at the origin will be a) zero b) e/2πεo c) e/3πεo d) e/5πεo 57. Two condensers of capacity 0.3 µ F and 0.6 µF respectively are connected in series. the

combination is connected across a potential of 6 volts. The ratio of energies stored by the first condensor to that of second will be a) 1/2 b) 4 c) 1/4 d) 2 58. A solid conducting sphere having a charge Q is surrounded by an uncharged concentric conducting hollow spherical shell. Let the potential difference between the surface of the solid sphere and that of the outer surface of the solid sphere and that of the outer surface of the hollow shell be V. If the shell is now given a charge of –3Q, the new potential difference between the same two surface is a) V b) 2V c) 4V d) -2V 59. A ball carrying a positive charge hangs from a silk thread. If we keep a positive test charge q0 at a point and measure the force F on the test charge, then it can be said that the electric field strength E at the point will be a) > F/q0 b) = F/q0 c) < F/q0 d) F/2q0 60. A condenser of capacitance 2 µF is charged to 200 V. It is now discharged through a resistor

61.

62.

63.

64.

65.

the heat produced in the resistor is a) 400 J b) 0.02 J c) 0.04 J d) 0.08 J The capacity of a parallel plate f condenser is 5 µF. When a glass plate is placed between the plates of the condenser, its potential difference reduces to 1/8 of the original value. the value of the dielectric constant of glass is a) 1.6 b) 8 c) 5 d) 40 The force acting on a charged particle kept between the plates of a charged parallel plate condenser is F. If one of the plates of the condenser is removed, then the force acting on the same particle will be a) 0 b) F/2 c) F d) 2F Two identical capacitors are connected first in series and then in parallel to the same source. The ratio of energy of the system will be a) 1:4 b) 4:1 c) 1:2 d) 2:1 *Two identical charged spheres are suspended by strings of equal length making an angle of 400 with each other. If they are immersed in a liquid of density less than the density of the material of the spheres, then a) the electrostatic force between them will increase b) the electrostatic force between them will decrease c) the net downward force on the spheres will decrease d) the net downward force will remain the same Vandergraff generator can be used for a) charging of batteries b) checking voltmeter markings

90

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c) producing very high magnetic fields d) accelerating charged particles Three point charges +q, -q, -q are placed on the corners of an equilateral triangle of side r. The potential energy of the system will be (A = 1/4πεo) a) 0 b) 3Aq2/r c) -2Aq2/r d) Aq2/r An uncharged metallic sphere A suspended between positively and negatively charged metal plates is given a small push towards the +ve plate. Which one of the following statements is correct? a) A touches +ve plate and remains there b) A touches +ve plate and then moves towards –ve plate and remains there c) A oscillates between the two plates with a constant time period d) A oscillates between the two plates with an increasing time-period A particle A has charge +q and another particle B has charge +4q. Each of them has a mass of m. They are allowed to fall from rest through the same electrical potential difference. The ratio of their speeds vA/vB will be a) 2:1 b) 1:2 c) 1:4 d) 4:1 The radius hydrogen atom is 0.53 x 10-10 metre. The electrostatic potential produced by the proton at the side of the electron will be a) 27.2 V b) 13.6 V c)-27.2V d) -13.6 V The effective capacitance between the points A and D in adjacent figure is a) 3 µF b) 21 µF c) 1/3 µF d) 1 µF A charged particle of mass 8 g remains stationary on a vertical electrical field 1000 kN/C. The number of fundamental quantum of charges carried by it is b) 5 x 1013 a) 5 x 1011 3 14 c) 5 x 10 d) 5 x 1015 3 3 3 3 3 Two identical copper spheres of mass 1 g each kept at C D a distance of 1m carry equal unbalanced negative A B charge. They repel with a force of 2.56 nN. The number of extra electrons carried by either of them is 3 a) 3.3 x 109 b) 3.3 x 1010 11 c) 3.3 x 10 d) 3.3 x 1012 Voltx coulomb has a dimensions of a) force b) acceleration c) velocity d) torque An electron moves along x-direction with a uniform speed. An electric field is applied along the z-direction. the path of electron will be a) a parabola in x-y plane b) a circle in x-y plane c) a parabola in x-z plane d) a straight line along z axis If the applied field is uniform magnetic in the z-direction, path of electron will be a) a helix in x-z plane b) a helix in x-y plane c) a circle in x-z plane d) a circle in x-y plane A conducting plate carries a charge +q. Another identical plate carrying zero charge is brought from infinity and kept at a distance r from the first plate. The second plate is earthed. The electrostatic energy between the plates is (A = 1/4 πεo) a) 0 b) -Aq2/r2 c) Aq2/r d) -Aq2/r

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77. The electric field at a distance 2 m from a charged plane sheet is E. When electric field at

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86.

infinity will be a) E b) E/2 c) infinity d) 0 A simple pendulum having a bob of mass m carries a charge q. When the bob is kept in a uniform electric field E in the horizontal direction, the inclination of the string with vertical is b) tan-1 (qE/mg) a) cos-1(qE/mg) -1 c) sin (qE/mg) d) cot-1(qE/mg) If electric field at a point from a charged needle is plotted with distance which of the following will be the correct graph ? a) a straight line of negative slope b) a straight line with positive slope c) a rectangular hyperbola d) an exponential graph A charged particle of mass m charge q is released from rest in an electric field of strength E. After a time t seconds the kinetic energy of the particle will be a) 2E2t2/mq b) E2mq2/2t2 c) E2q2t2/2m d) qE Two charges exert a force F when kept at a distance r between them in air. The distance at which they exert same force in a medium of relative permittivity 4 is a) r/4 b) r/2 c) 4r d) 2r Two conducting spheres of radii 5 cm and 3 cm are equally charged. The ratio of their potentials is a) 3:5 b) 5:3 c) 9:25 d) 25:9 Two conducting spheres of radii 5 cm and 3 cm charged to the same potential. The ratio of their charges will be a) 3:5 b) 5:3 c) 9:25 d) 25:9 Two conducting spheres of radii 5 cm and 3 cm are equally charged. The ratio of electric fields on their surfaces will be a) 3:5 b) 5:3 c) 9:25 d) 25:9 Two conducting spheres of radii 5 cm and 3 cm are charged so that the electric field on their surfaces is same. The ratio of the charges will be a) 3:5 b) 5:3 c) 9:25 d) 25:9 A charge q1 exerts some force on another charge q2. If a third charge q3 is brought near q1, the force exerted by q1 on q2 a) decreases b) remains unchanged c) increases d) decrease for like and increase for unlike

87. Twelve identical condenser plates are given. Two such plates will give a capacitance of 1 µF.

The maximum capacitance which we can have from these plates is (in µF) a) 1 b) 6 c) 10 d) 11 88. A point charge q is placed at the corner of a square of side a. The potential difference between one adjacent corner and opposite corner is [A = 1/4πεo] a) Aq/a b) 10.7 Aq/a c) 0.35 Aq/a d) 0.29 Aq/a 89. Two conducting spheres A and B are charged to +10 µC and +20 µC. They have radii 3 cm

and 6 cm respectively. If they are connected by a wire a) charge flows from A to B and heat is produced in the wire b) charge flows from B to A and heat is produced in the wire c) no charge flows between A and B and no heat is produced in the wire

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d) charge flows from B to A and no heat is produced in the wire 90. The Coulomb’s law will not operate at a distance less than a) 10-8m b) 10-10m c) 10-12m d) 10-15 m 91. A charged particle will move along an electric line of force if a) it is free b) it is accelerated c) it is moving with a uniform velocity d) it is moving with a non uniform velocity 92. Two given point charges are placed at a fixed distance in the following media. In which of the medium the Coulomb force between them will be minimum? a) water b) oil c) mica d) paper 93. Three charges q1, q2, q3 are kept at the corners of an equilateral triangle. The force between q1 and q2 is F. If q3 is removed from the third corner force between q1 and q2 a) increases b) decreases c) remains the same d) cannot be said from the data 94. Three charges +2q, -q, -q are placed at the corners of an equilateral triangle. If V is electric potential and E the electric field at the centre of the triangle, then a) E = 0, V ≠ 0 b) E ≠ 0, V = 0 c) E ≠ 0, V ≠ 0 d) E = 0, V = 0 95. An electric line of force in x-y plane is given by the equation x2+y2 = 1. A particle with a unit positive charge initially at rest at a point x = 1, y = 0 in the x-y plane a) will move opposite the line of force b) will move at an angle to the line of force c) will continue to be at rest d) will move along the line of force 96. A charge of one coulomb is placed at the centre of a cube of side a. The number of lines of force coming out from one face of the cube is a) 1.11x 1011/a2 b) 1.11x 1011a2 c) 6x1.11x 1011/a2 d) 1/6) 1.11x 1011 97. Two charges +1C and -1C are placed at points (0,0) and (1,0) respectively. A non-uniform electric field exists along the X axis which increases at a uniform rate of 1V/m2. If the force on +1C is 5 N, the force on -1C is a) 6 N b) -6 N c) 5 N d) -5 N 98. The electric lines of force coming out a charged body ‘A’ is parallel and equidistant up to infinity. This body is a) a solid sphere b) a hollow sphere c) a needle d) a plane sheet 99. The electric field due to a dipole at a distance 2 m from it is E. The electric field at a distance 4 m from it will be a) E/2 b) E/4 c) E/8 d) E/16 100. A hollow metallic sphere P is charged to a potential V volt . Another sphere Q is charged to a potential V/2. If Q is placed inside P and they are connected by a wire, then a) charge will flow from P to Q until the potential becomes 3V/4 b) whole charge will flow from P to Q c) the whole charge will flow from Q to P d) no charge flows 101. The potential due a point charge at a distance 2 m from it is 6 V. The field at the same point due to the same charge will be a) 3 N/C b) 6 N/C c) 12 N/C d) 0

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93

102. An electric dipole kept in a uniform electric field is disturbed slightly. Which of the following will happen? a) It returns immediately to the equilibrium state b) It moves away from the equilibrium state c) It oscillates and returns to the equilibrium state d) It oscillates for ever. 103. An electric dipole is subjected to two mutually perpendicular uniform fields F along the X axis and 2F along the Y axis. The dipole will a) oscillate in XY plane b) settle at an angle 45owith X axis c) settle at an angle tan-1 (2) with X axis d) settle at an angle tan-1(1/2) with X axis 104. An charged particle is moving with a velocity 10 k m/s. It is subjected to an electric field 50 i N/C. The path of the particle will be a) a straight line b) a parabola in XZ plane c) a parabola in YZ plane d) a parabola in XY plane 105. If the applied field in the above question is magnetic, the path of the particle will be a) a straight line b) a circle in Y-Z plane c) a circle in X-Y plane d) a circle in X-Z plane 106. Three point charges each 1/3 µC are brought from infinity and are placed at the corners of an equilateral triangle of side 1 m. The work done for this will be a) 9 mJ b) 3 mJ c) 0 d) 4.5 mJ 107. Two spheres of radii 1 cm and 2 cm have equal charge density. The ratio of the electric field on their surfaces in the given order will be a) 1 b) 2/1 c) 4/1 d) 1/2 108. A cube of side a is placed in an electric field Eo i . The net number of flux lines passing through the cube is b) 2a2 Eo c) 4 a2 Eo d) 0 a) 6a2 Eo 109. The relative permittivity of a medium is x and the dielectric strength of the medium is y. For the medium to be a good insulator, it should have a) high x and low y b) high y and low x c) low y and low x d) high x and high y 110. A point charge q is kept at the centre O of a circle of radius r. A and B are two points in the circumference of the circle. ∠AOB = 60o. The work done in taking a unit positive charge from A to B along the smaller arc will be q 5q q a) b) c) d) zero. 4πε o r 6 x 4πε o r 6 x 4πε o r

111. A solid sphere of radius R has a uniform charge density ρ per unit volume. The electric field at a point inside at a distance r ( r < R) from the centre of the sphere is given by rρ Rρ 2 rρ a) zero b) c) d) 3ε o 3ε o 3ε o 112. Two soap bubbles P and Q are charged with same surface density of charge σ, but P positively and Q negatively. Then a) both the bubbles expand b) P contracts while Q expands

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c) Q contracts while P expands

d) both the bubbles contract

113. Two capacitors 1 pF and 2 pF are charged to 200Vand 100 V respectively. They are then connected so that the positive plate of the one is connected to the negative plate of the other. The final potential of the connected system will be a) 150 V b) 106 V c) 133 V d) 0 V 114. The mass of an α particle is nearly 8000 times the mass of an electron. An α particle is accelerated through 1 volt. The energy of this particle will be a) 1 eV b) 2 eV c) 8000 eV d) 16000 eV 115. *An electric dipole is formed with charges +q and -q points (-1,0) and (+1,0) respectively on the x axis. Check the correct statements. a) The y-axis will be an equipotential line. b) The electric field at all points on the y-axis will have same magnitude and direction. c) The electric field at all points on the y-axis will be along the positive x-axis. d) The electric field at all points in the y-axis will have same magnitude but different direc tions. 116. The magnitude of torque on a dipole is doubled, when the angle made by the dipole with the field is increased to three times. The initial angle of the dipole with the field is a) 15o b) 45o c) 60o d) 30o 117. Three identical capacitors are connected in parallel. Then they connected in series. The difference between the effective capacitance is 16 µF. Then capacity of each in µF is a) 3 b) 6 c) 8/3 d) 16/3 118. A condenser of capacitance 2 µF is connected to a battery through a resistor 3 ohm. The battery spends energy of 4 m J to charge the condenser fully. The heat produced across the resistor during charging is equal to (in mJ) a) 1 b) 3/2 c) 2 d) 2/3 119. n identical capacitors are connected in parallel to a potential difference V. These capacitors are reconnected in series, their charges being left undisturbed. The potential difference obtained is a) V/n b) (n-1)V c) n2V d) nV 120. An isolated metallic object is charged in vacuum to a potential V, its electrostatic energy being E. It is then disconnected from the source of potential, its charge being left unchanged. It is immersed in a large volume of dielectric with dielectric constant K. Its electrostatic energy becomes a) E/K b) KE c) EK/V d) E2K/V 121. A small metal ball is suspended in a uniform electric field using an insulated string. If high energy X-ray beam falls on the ball a) the ball will be deflected towards positive plate b) the ball will be deflected towards negative plate c) the ball will continue to remain in equilibrium d) the ball will oscillate between the plates 122. Two metal spheres of radii R1 and R2 are charged with Q1 and Q2 respectively. Then they are connected by a wire. The charge flows a) until Q1 = Q2 b) until Q1R1 = Q2R2 c) until Q1R2 = Q2R1 d) never

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95

123. A capacitor of capacitance 5 µF is charged to a potential difference of 100 V. The battery is disconnected, and then it is connected in parallel to an uncharged capacitor of capacitance C. The potential difference measured across this combination is 25V. The capacitance C is a) 20 µF b) 10 µF c) 5 µF d) 15 µF 124. In the previous question , the ratio of electrostatic energy of the system before to that after the connection of C is a) 4/1 b) 1/2 c) 2/1 d) 1/4 125. A parallel plate condenser has an energy U. The plates of the condenser are pulled using an insulating handle until their separation is 3 times the original value. The work done in the process is a) 3U b) 2U c) U d) 3/2 U 126. The minimum radius of a body which will hold a charge of 1 C so that the surrounding air will not become ionised is nearest to a) 1 cm b) 1 m c) 50 m d) the radius of earth 127. One thousand identical drops each of capacity 2µF are charged to a potential 100 V each. If they coalesce into a single drop, the capacity of single drop will be a) 20µF b) 200µF c) 2000µF d) 4000µF 128. The potential of the resulting drop given in the previous is equal to a) 100 V b) 1000 V c) 10000 V d) 105 V 129. Check which one of the following statements is not correct. a) The SI unit of charge is greater than CGS (esu). b) The SI unit of capacity is greater than CGS(esu).. c) The SI unit of energy is greater than CGS unit d) The SI unit of potential is greater than CGS (esu) 130. The capacity of a parallel plate condenser with air as dielectric is C. If the condenser is filled with 3 dielectrics of equal thickness and dielectric constants 2,3,4 the capacity will be a) (36/13)C b) (13/12)C c) (18/13)C d) 9C 131. Two condensers of capacitance C each are connected in series to a battery of potential difference V. The same condensers are later connected in parallel to the same battery. The ratio of energy of the system in the two cases in the given order is a) 1/2 b) 1/4 c) 2 d) 4 132. A parallel plate condenser contains a dielectric of relative permittivity 2. The condenser is charged by a battery. The battery is disconnected and the dielectric slab is removed. In the process the energy of the condenser a) decreases by 50% b) increases by 50% c) decreases by 100% d) increases by 100% 133. An external electric field E0 is applied across a dielectric. At any point inside the dielectric the electric field a) will be greater than E0 b) will be less than E0 d) will be greater or less depending on direction. c) will be equal to E0 134. * If the negatively charged plate of a parallel plate condenser is removed to infinity a) the potential of the positive plate increases b) the potential of the positive plate decreases c) the capacity of the positive plate increases d) the capacity of the positive plate decreases

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135. Ten identical charged drops each having an energy E, coalesce to form a single drop. The energy of the resulting drop is c) 105/3E d) 102/3 E a) 10 E b) 103/2E 136. What is a capacitance between A and B in theadjacent diagram ?( all capacitances are in µF)

a) 2

b) 4

c) 6

d) 8

137. The effective capacitance between the points A and B in the adjacent diagram (all in µF) is a) 2 b) 4 c) 6 d) 0.5 138. In the adjacent figure, each edge of the cube contains a capacitor of value C. The total capacitance of the circuit when a battery is connected between A and B will be, a) (5/6) C b) (6/5) C c) zero d) infinite 139. An infinite network of capacitors, each 2 µF, is made as shown in the figure below. The capacitance between A and B (in µF) is a) 2 b) 2.6 c) 3.2 d) 0 140. A parallel plate capacitor has a separation d and a capacitance of 100 pF. If a metal foil of thickness d/3 is introduced between the plates, the new capacitance will be (in pF ) a) 300 b) 150 c) 100 d) 67 141. Two conductors have equal volume and carry equal charge. Then a) they have same potential b) they have same capacity c) they have same energy d) all the above three will be different. Solve the following two problems within a maximum time of 2 minutes using only one equation. 142. A capacitor of capacitance 2µF is charged to 200 V. Another capacitor of capacitance of 2µF is charged to 100 V. They are connected in parallel. What is the heat energy produced in the connection wires? 143. Three capacitors of capacitance 1µF, 2µF and 3µF are connected to a source of potential difference 110 V. What is the potential difference across each of them? 144. If E is the electric field, E2εoεr has the dimensions of

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97

a) energy b) energy per unit area c) energy per unit length d) energy per unit volume 145. Two point charges q and -3q are placed at a given distance. The electric field at the site of q is E. Then the electric field at the site of -3q is a) -E b) -E/3 c) -3E d) +E/3 146. A ring of radius R carries a uniformly distributed charge +Q. A point charge -q is placed on the axis of the ring at a distance 2R from the centre of the ring and released from rest. The charge a) executes oscillatory motion but not simple harmonic. b) moves to the centre of the ring and remains at rest there c) remains in equilibrium at the point d) executes simple harmonic motion along the axis 147. Three identical spheres of masses m1, m2, m3 are charged positively and negatively and no charge respectively. Which of the following is correct ? a) m1 > m2 > m3 b) m2 > m3 > m1 c) m2 > m3 < m1 d) m2 > m3 = m1 148. A parallel plate air capacitor has a capacitance of 100 pF. The plates are at a distance apart. If a metallic wire of very small thickness is introduced parallel to plates between them, the new capacitance will be a) 100 pF b) < 100 pF c) >100 pF d) 0 149. * A capacitor C1 of capacitance 1 microfarad and another capacitor C2 of capacitance 2 microfarad are separately charged fully by a common battery. The two capacitors are then separately allowed to discharge through equal resistors at time t = 0. Then a) the current in each of the two discharging circuits is zero at t = 0 b) the currents in the two discharging circuits at t = 0 are equal but not zero c) the currents in the two discharing circuits at t = 0 are unequal d) capacitor C1 loses 50% of its initial charge sooner than C2 loses 50% of its initial charge 150. The plates of a parallel plate capacitor of capacitance C are connected to a battery of emf 12V. A dielectric of relative permittivity K is introduced in between the plates of the capacitor. The capacitance C and potential V a) both increase to K times b) both remain the same c) C increases K times and V decreases to K times d) C increases to K times and V remains the same 151. Two point charges are kept at a distance in a medium. Each charge feels the presence of the other a) instantaneously b) in about a second c) after a small but finite time d) after a time of the order of a few minutes 152. The maximum amount of electrostatic energy density which air can have so that it will not break its insulating property is a) 900 J/m3 b) 90 J/m3 c) 40 J/m3 d) 1 J/m3 153. Two point charges -2 µC and +4 µC are placed at points A and B as shown in figure below. Which of the points marked in the figure is a possible null point (point of zero electric field) for the arrangement ?

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a) P

b) R

c) Q

d) all the three points

154. In the figure given along with previous question, which point will possibly a point of zero

potential ? a) P

b) R

c) Q

d) P and R

155. An infinite number of capacitors each of 2 µF are connected as shown in the figure below.

What is the capacitance between the points A and B? a) 1.6 µF

b) o.66 µF

c) 0.8 µF

d) 1.2 µF

156. Two rings each of radius R are kept perpendicular to the plane of the paper so that their

centres P and Q are at a distance R. They carry uniform charge q each. The potential due to one ring at its centre is V. Then the potential due to both rings at the centre P will be a) 2V b) √2V c) 0 d) (√2+1)V/√2 157. Four identical condensers connected in series have an effective capacitance 1 µF. One of them is removed from the combination and connected in parallel with the rest. The effective capacitance will now be (in µF) a) 7/4 b) 7/3 c) 16/3 d) 3/4 Solve the following problems within a maximum time of 2 minutes. 158. 12 identical capacitors each of 1 µF are given to you. How will you get a capacitance of 3/4 µF using all of them? Draw the circuit diagram. 159. How will you obtain a capacitance of 1/3 µF using all of them?

SECTION 2:ANSWERS 1

(a)

2

(a)

3

(d)

4

(b),(d)

5

(b)

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(d)

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(d)

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(b)

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(b),(d)

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(c),(d)

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(a)

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(c)

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(a),(c)

34

(d)

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(d)

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(d)

39

(c)

40

(a)

41

(c)

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(b)

43

(b)

44

(d)

45

(a)

46

(d)

47

(d)

48

(b),(d)

49

(c)

50

(a,b,d)

51

(d)

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(c)

53

(b)

54

(d)

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(c)

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58

(a)

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61

(b)

62

(b)

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(a)

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(b),(c)

65

(d)

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(d)

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(a)

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(b)

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(c)

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(c)

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(b)

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(a)

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(c)

85

(d)

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(b)

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(d)

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(d)

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(a)

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94

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95

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(d)

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98

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99

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(a)

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(a),(d)

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1/200 J

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60,30,20

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3/4

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SECTION 3 SOLUTIONS 1.

2.

3.

4.

5.

q Charging by conduction is addition of electrons when negatively charged) or removing electrons when positively charged). Here electrons are added. The mass slightly E E increases. Ans.a A metal plate has infinite dielectric constant. The force between charges separated partly or fully by it will be E q (1/4πε0εr) (q1q2/r2) with εr = α. Thus the force between them q will be zero. Ans.a The force between two point charges q1 and q2 kept at a distance F12 and F21 will have same magnitude (since the force depends on the product of the charges) and opposite direction. i.e. F21 = -F12. Hence the ratio will 1: -1. Ans.d The potential on the surface of a charged sphere is Aq/r. When same charges are given sphere of smaller radius will have greater r potential. When connected by wire charge flows from higher to lower potential making the potential same (choice b). This R reduces charge of smaller sphere and increases charge of bigger sphere (choice d) Ans.b & d Refer to the adjacent figure. The electric field at the centre of the

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MORE PRACTICE PAPERS FOR IIT-JEE circle 0 has a magnitude Aq/r2 due to each charge. Their directions are shown. Three such vectors form the sides of an equilateral triangle and by rules of vector addition the sum will be zero. Ans.b

6.

Refer to the adjacent figure. If ρ is the charge per unit volume, ρ = q/4πR3. Charge on the smaller sphere will be q(4/3)πr3/(4/3)πR3 = qr3/R3 where r is the radius of the smaller sphere. The electric field at the point is A (charge inside the sphere)/r2. This gives field = Aqr3/R3r2 i.e. α to r. Ans.d

7.

According to Gauss’s theorem total flux crossing normally through the whole cube = q/ε0. This is flux coming out of all the six faces of the cube. Hence flux through one face = q/6ε0. Ans.d When a conductor is charged, there is an outward mechanical force given by (σ2/2ε0)N/m2, where σ is charge radius of the bubble. It goes on expanding until surface tension of soap solution balances this force and brings about an equilibrium. Ans.b

8.

9.

The capacitance of a spherical conductor is 4πε0r. When connected by a wire the final potential = V = q1`+q2/c1+c2 = (1 x 10-2 + 5 x 10-2)/4πε0x3x10-2 = 2/4πε0 volt. Charge of smaller sphere = potential x capacity = (2/4πε0) x 4πε0 (1 x 10-2 )= 2 x 10-2 C. Ans.a

10. Work W = Fscosθ, where F is force, s is distance θ is angle between force and displacement. Here F = qE, where q is charge E is electric field. Thus W = qE cosθ. That is 4 = 0.2 x E x 2 x cos600. This gives E = 20 N/C. Ans.d 11. If v is the velocity of a particle of mass m, falling under a potential difference of V volt, 12. 13.

14. 15. 16.

17.

18.

electrostatic potential energy eV = kinetic energy (1/2) mv2 ⇒ v = 2 m/s. Ans.a Unit of ε0 is farad/metre, capacitance per unit length. This can be easily checked from the equation of capacitance of a parallel plate condenser, C = ε0A/d, where a is area of the plates and d distance between the plates. ε0 = Cd/A. Ans.b At ground state of hydrogen atom, the centre of mass of the revolving electron and that of the nucleus will be same point that is centre of atom. Hence the separation will be zero. The dipole moment is the product of charge and separation. Dipole moment P = qa = qx0 =0. Ans.c The resulting potential is total charge/total capacitance (q1+q2)/(C1+C2) =(C1V1+C2V2)/(C1+C2) = 133 V. Ans.c Dielectric strength is the minimum value of the electric field to break down the specimen. The corresponding potential difference is field x distance = 106 Vm-1 x 10-3 m = 1000 V. Ans.c Charge q is already in equilibrium. Consider the equilibrium of one of the Qs.This will be in equilibrium if force between Q and Q is equal and opposite to force between Q and q. If we take distance between Q and Q as 2a, distance between Q and q is a. Then, we have AQq/a2 = -A(QxQ)/(2a)2. This will solve to q = -Q/4. Ans.b When the charging battery is disconnected, the charge on the capacitor remains constant. Hence answer (a) is wrong. When plates are separated more the capacitance of the condenser ε0εrA/d decreases as d increases. So answer (c) is wrong. Q = V/C. Q is constant. C decreases. Hence V increases (b) is correct. The electrostatic energy of a condenser is Q2/2C. Q is constant C decreases and therefore energy increases. (d) is correct. Ans.b & d Let Q be the charge and q be its fraction. Then the charges are q and Q-q. The force between charges, F = [A(Q-q)q]/r2. The force is maximum when dF/dq = 0. This gives q = Q/2. Hence force is maximum between two charges when kept at a given distance if they are divided in the ratio 1:1. Ans.c

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101

19. Since battery connection exists, potential difference V is constant. Distance between the plates d is constant . Electric field V/d, therefore, is constant. Ans.c 20. Energy gained by the electron = q x V, where q is charge and V is potential difference. Since electric field E is uniform, V = Exd, d separation between the plates. Therefore energy = qEd = 1.6 x 10-19 x 2.5 x 2 = 8 x 10-19 J. Ans.d 21. The work done by the battery to charge the condenser is qV J. However, the energy stored by the condenser is only (1/2)qV. The factor 1/2 appears here because during charging potential increases uniformly from 0 to V. Hence we take the average value. Ans.b 22. The field due to a dipole on the axial line at a distance is Ax 2p/r3. Since field is potential gradient, potential should be proportional to 1/r2. Hence when r is doubled, V becomes V/4. Ans.b 23. At a point electric potential or field can exist even when the other is zero. Examples are given in indirect theory notes. Ans.c, d. 24. The number of lines of force coming out from a unit charge in CGS system is 4π and iun SI = 1/ε0. Value of ε0 = 8.85 x 1012. Hence 1/ε0 = 1/(9 x 10-12) = 1012/9 = 1.11 x 1011. This is also given in indirect theory notes. Ans.a 25. Two plates get equal potential V and charge Q = CV. They are joined in series. No charge flows as they carry the same potential, and in a series connection the charge is same. Potential difference = 2V. Ans.d 26. When the distance is increased capacitance = ε0ε0A/d decreases. Since charge is the same, the potential increases by the equation, Q = CV. Ans.c 27. If q is the maximum charge it can hold, q/4πε0r2 is the electric field on the surface. q = 4πε0r2 E. Here r = 5/2 m. 4πε0 = (9 x 109)-1. We have from this q = 2 x 10-3 C. Ans.b 28. The field inside a charge hollow sphere is zero. Therefore the potential at all points will be same inside. Ans.a 29. Force of attraction (F1) = A(10) x (-20)/r2 = -200A/r2 where A is the constant appearing in Coulomb’s law. When the balls touch each other, the radii being the same, the capacitance will also be the same. So the charge will distribute equally. The charge q1 = q2 = (10-20)/2 = -5 units. (F2) = A(-5) x (-5)/r2 = 25 A/r2 , F1:F2 = -8:1. Ans.a 30. Taking 1/4πε0= A, the potential at the surface of larger sphere AQ1/R1 should be = potential inside at a distance r from the centre because then only the electric field inside larger sphere will be zero. The potential due to smaller sphere will be AQ2/r, assuming charge concentrated at centre. So the total potential = (AQ1/R1) + (AQ2/r). Ans.c 31. When charging battery is connected V remains constant. When distance is increased, electric field V/d dcreases.. Ans.b. 32. Using the formula C = ε0Ak1k2 / (k1t2 + k2t1), for a parallel plate condenser filled with 2 dielectrics, we have here t1 = t2 = t/2, where is total distance between the plates. This gives C = 2ε0Ak1k2/(k1+k2)t. Ans.c 33. The electric field between parallel plate condenser is uniform and is given by E = σ/ε0, from Gauss’s law. Force = charge x field. Hence the force on all the protons will be same in magnitude and direction. Ans.a,c 34. Since the field increasing along x-axis force on –q is greater than force on +q. Hence the dipole will move along direction of greater force i.e. negative x-axis. The force on charge will produce a torque which rotates the dipole anti-clockwise. Ans.d 35. Taking 1/4πε0 = A, we have Aq1/a = Aq2/b. That is q1/q2 = a/b ....(1) If σ1 and σ2 are ratio of surface density of the two spheres, σ1 / σ2 = (q1 /4πa2)/(q24πb2) = (q1/q2)(b2/a2). Substituting from (1) for q1/q2, we get σ1/σ2 = b/a. Ans.a

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G G 36. E = 200 kˆ. The area vector S is perpendicular to the plane of the coil that is along the z G G G axis. Therefore S = 10 x 20 x 10-4 kˆ. The flux φ = E. S = 2000 x 200 x 10-4 ( kˆ. kˆ ) = 40 units. Ans.b 37. The energy of a charged conductor is given by (1/2) qV or (1/2) CV2. For a spherical conductor V = q/4πεoR. Therefore energy will be q2/8πεoR. Ans.d 38. The intensity of electric field at a point is the number of lines of force crossing unit area around the point normally. Here it is 10. Ans.d G G 39. φ = E. dS E dS cosθ, where θ is the angle between field vector and area vector. Here the lines make an angle 300 with the surface, which means they make an angle 600 with the area vector. (Recall area vector is normal to the surface). Thus φ = 10 x 1 x cos600 = 5. Ans.c 40. Since the coil is Y-Z plane, its area vector will be 200 ˆi cm2. φ = E. dS = (200 kˆ ). (200 x 41. 42.

43. 44. 45. 46. 47. 48.

49.

50.

51.

10-4 ˆi ) =0, because kˆ. ˆi = 0. Ans.a Inside a satellite there is no other force except electrostatic force (apparent weight of the balls is zero). Therefore due to the electrostatic force of repulsion the balls settle at maximum distance. Both the strings will be horizontal. Angle between them will be 1800 . Ans.c The tension of each string will be equal to the electrostatic force of repulsion experienced by a 1 QxQ ball. The distance between the balls is 2L. Therefore the force = Ans.b 4πε 0 (2L) 2 Force on electron due to electric field = eE. If this is equal to weight of electron, we have mg = eE. E = mg/e. Ans.b Taking A = 1/4πεo, potential on this surface = Aq/r. Potential at a point 3r from the centre Aq/3r. It is given (Aq/r)-(Aq/3r) = V. That is V = (2/3)Aq/r. field at 3r = -d/dr = (Aq/3r) = Aq/3r2 = V/2r. Ans.d Refer theory notes and previous similar question, in which it has been shown that the electric field inside a charged conducting sphere is proportional to distance from the centre. Therefore field decreases as we penetrate from the surface towards the centre. Ans.a 1 g of hydrogen 6 x 1023 atoms nearly. If all electrons are removed it will have 6 x 1023 protons left. The total charge of all these protons will be 1.6 x 10-19 x 6 x 1023 = 9.6 x 104 ≈ 105 C. Ans.d Electric field E = -dV/dx. That is E = -(10x+10). At the point x = 1m, E will be equal to – (10x1+10) = -20 V/m. Ans. d At a point on the perpendicular bisector of a dipole, the charges are at equal distance. Hence the electric potential will be A(q/r) + A(-q/r) = 0. That is (b) is correct. The electric field at this point (on the equitorial line) will be Ap/r3, and this always be parallel to the axis of the dipole. (d) is correct. Ans.b & d. (Note A = 1/4πεo) Since the battery is disconnected, charge Q on the condenser remains the same. The capacitance of the condenser when the slab is inserted increases to K times. Work done = difference in the electrostatic energy = (Q2/2C)-(Q2/2CK) = Q2(K-1)/2KC. Since Q = CV and C = εoA/d, this reduces to εoAV2(K-1)/2dK. Ans.c If both E and B are zero, the proton will certainly continue with constant velocity. (a) is correct. Even when B is not equal to zero, if it is parallel to proton motion, it will go undflected. Hence (b) is correct. If E is not equal to zero proton will be accelerated (c) is wrong. If both E and B exist, and the force due to the fields are equal and opposite i.e. eE = evB, the proton would move with constant velocity v = E/B. (d) is correct. Ans. a,b & d When the positive charge is released, the x-components 2Fx of coulomb force F will provide restoring force for oscillatory motion, while y components cancel. However, coulomb force

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52. 53. 54. 55. 56.

57. 58. 59.

60. 61.

62.

63. 64. 65. 66. 67. 68.

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obeys inverse squarelaw, while we need a force proportional to displacement for SHM. Hence the charge will oscillate but not simple harmonically. Ans.d If n drops of same radius and charge coalesce, the potential of resulting drop V1 can be found by the formula V1 = n2/3 V ( Indirect theory notes), where V is the potential of smaller drop. The required ratio, therefore, V1 /V is (1000)2/3 = 100. Ans.c Taking A = 1/4πεo, the potential at the origin will be V(0) = (Ae/1)+(Ae/2)+(Ae/4)+(Ae/8) +.....to infinity. This is equal to Ae[1+(1/2)+(1/4)+(1/8)....]= 2e/4πεo, where we observe the sum of geometric series in the bracket is 2. Ans.b In this case V(0) = (Ae/1) – (Ae/2) + (Ae/4)- (Ae/8)+.....= Ae[(1)-(1/2)+(1/4)-(1/8)....]= Ae(2/3) = e/6πεo, where we observe the sum of geometric series in the bracket is 2/3. Ans.d The electric field at the origin E(0) = (Ae/12)+(Ae/22) + (Ae/42) + (Ae/82) ...= Ae[1+(1/22) + (1/32) + (Ae/42)-(Ae/82)+......= Ae[1+(1/22) + (1/42) +.......] = 3/4πεo, where we observe the sum of geometric series in the bracket is 4/3. Ans.c The electric field E(0) = (Ae/12) – (Ae/22) + (Ae/42)-(Ae/82) +....= Ae[1-(1/22)+ (1/42).....] = Ae(4/5), where we observe the sum of the geometric series in the bracket is 4/5. Hence E(0) = e/5πεo. Ans.d Since charge is same we use equation for energy (1/2)q2/C. This means energy U α 1/C. The ratio of energy will be U1/U2 = C2/C1 = 0.6/0.3 = 2. Ans.d The field inside a charged hollow conductor is zero. This will be satisfied only if the potential difference remains the same value V. Ans.a In order to measure electric field, we place a test charge q0. The test charge will have to be as small as possible because otherwise it will produce its own field at the point and thus will disturb the electric field to be measured. Hence the measured field will always be less than the actual field. Ans.c The electrostatic energy lost is the heat energy produced. Hence (1/2) CV2 is the heat energy produced. Hence C = 2 x 10-6 F. V = 200 volt. We get energy = 0.04 J. Ans.c When glass plate is fully inserted, capacitance increases to εr times, where εr is dielectric constant of glass. Since charge is constant potential will decrease to 1/εr times. Here 1/εr is 1/8. Therefore εr = 8. Ans.b Electric field between parallel plate condenser is uniform and is σ/ε0, where σ is charge per unit area. When one plate is removed, the field becomes that due to one charged plate. The field will be either σ/2ε0 or - σ/2ε0. Force on a charged particle = field x charge. Hence the force will reduce to half. Ans.b Since they are connected to the same source, V is constant. Use the equation for energy (1/2)CV2 . In the first case capacitance is C/2 and in the second case 2C. So the ratio of energy will be 1: 4. Ans.a When the spheres are immersed in liquid due to buoyant force their apparent weight will be less. Hence net downward force will decrease (c) is correct. Electrostatic force is balanced by the horizontal component of tension. Hence (b) is correct. Ans.b & c Vandegraff generator produces very high electric potential of the order of one million volt so that charged particles can be accelerated using its potential. Ans.d Potential energy of a system of three given point charges = (-Aq2/r) + (-Aq2/r) + (Aq2/r) = Aq2/r. Ans.d A uniform electric field exists between the plates. A metal sphere suspended as shown when slightly displaced will move towards one plate. Due to restoring force produced by the string it moves in the opposite direction. Thus it will oscillate with a constant period. Ans.c The electrostatic energy gained by A and B will be respectively qV, 4qV. This is equal to respective mechanical kinetic energies (1/2) mvA2, (1/2)mvB2 . From this we get vA2 / vB2 = 1/4. vA / vB is 1:2. Ans.b

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69. At the ground state, potential energy of hydrogen atom is –27.2 eV. Its kinetic energy (half of potential energy) is 13.6 eV, which makes the total energy as –13.6 eV. Since electrostatic potential energy of electron is –27.2 eV, electrostatic potential at the site of electron due to proton is +27.2 V (+ as its is due to proton). Ans.a 70. The effective capacitance between B and C is 3µF. (Balanced Wheatstone’s bridge of capacitors). This is in series with 3 µF between A and B and 3 µF between C and D. These three 3 µF capacitors in series add to 1 µF by series addition rule. Ans.d 71. Here weight mg = qE. Therefore q = mg/E = 8 x 10-3 x 10/1000 x 103 = 8 x 10-9 C. The number of quantum charges = q/charge of electron = 8 x 109 / 1.6 x 10-19 = 5 x 1011. Ans.a 72. If n is the number of electrons carried by each and e is the electron charge, we have A(ne)2/I2 = 2.56x10-9 N by Coloumb’s law, where A = 1/4πεo = 9 x 109 . This simplifies to n2 = 1020/9. n = 1010/3 = 3.33 x 109. Ans.a. (Note: The numbers given here are easy for simplification) 73. Volt x coulomb = potential x charge = energy, which has the same dimension as torque. Ans.d 74. When electric field is perpendicular to the direction of motion of electron with uniform speed, electron will have velocity and acceleration perpendicular at the beginning. Hence its path will be a parabola in the plane of velocity (X) and acceleration (Z). Ans.c 75. When applied magnetic field is along Z direction and velocity along X-direction by Lorentz force equation q v x B the force will be along Y direction. So electron will move in a circle containing velocity vector and force vector. That is X-Y plane. Ans.d 76. Here we have two plates forming a parallel plate condenser. The electrostatic energy between the plates = Aq(-q)/r = -Aq2/r. Ans.d 77. The field due to charged plane sheet = σ/2εo, which is uniform and does not depend on distance. So it will be same at infinity. Ans.a 78. If θ is angle made by the string with vertical, the forces acting on the bob are mg vertically downward and qE horizontal. The string will settle at an angle θ, such that tanθ = horizontal force/vertical force = qE/mg. (Recall tangent law in dynamics). Ans.b 79. The electric field due to a cylinder = λ/2πε0r, where λ is charge per unit length. The field is proportional to 1/r. The graph between two quantities inversely proportional to each other will be a rectangular hyperbola. Ans.c 80. Acceleration = qE/m. Velocity after t seconds v = vo+at = 0 + (qE/m)t. Kinetic energy = (1/2) mv2 = (1/2)m(qEt/m)2 = (1/2)q2E2t2/m. Ans.c 81. According to Columb’s law if r is distance in air and r1 is a distance in the medium for the forces to be equal r2 should be equal to εrr12. Equating we get r1 = r/√εr = r/2. Ans.b 82. Potential V = Aq/r, where A = 1/4πεo. Here q is the same. Hence V α 1/r. V1/V2 = r2/r1 = 3/5. Ans.a 83. Here V is same. Using V = Aq/r, q α r. q1 /q2 = r1 / r2 = 5/3. Ans.b 84. Electric field at the surface = Aq/r2, where A = 1/4πεo. Here q is same. Hence E α 1/r2. E1/E2 = r22 /r12 = 9/25. Ans.c 85. Since electric field in the surfaces are same , Aq/r12 = Aq2/r22. This gives q1/q2 =25/9. Ans.d 86. The force between q1 and q2 depends only on q1 and q2 and their separation according to Coloumb’s law. Any intervening charge q3 will not change the force between these two. Ans.b 87. With twelve condenser plates we can form eleven condensers. If the plates are arranged parallel at equal distance and alternate plates are connected to positive of a battery, other

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105

alternate plates to negative, these condensers will be parallel. Its capacity will be 11 µF. Ans.d 88. Taking A = 1/4πεo, potential due to charge at the adjacent corner = Aq/a. Potential due to charge in the opposite corner = Aq/√2a. Potential difference = Aq/a [1-(1/√2)]= Aq/a[1-0.71] = 0.29 x Aq/a, where we have taken 1/√2 = 0.71. Ans.d 89. Potential on the surface of A is = potential on the surface of B, because the ratio 10/3 = 20/6 (potential = 1/4 πεo(q/r). Since there is no potential difference charge will not flow. Ans.c 90. Use the information supplied already in theory notes.. This law fails inside the nucleus. The inter-nuclear distance is of the order of 10-15m. The nuclear forces follow a different set of laws. Ans.d 91. An electric charge will move along the line of force only if it is free, i.e. it is not under the action of any other force. Hence it cannot move along the line of force if it is accelerated. If its velocity is uniform, still it need not follow the line of force if its velocity is inclined to the field. Ans.a 92. The dielectric constant of pure water is 81. Among the given media water has maximum dielectric constant. Since the coulomb force is inversely proportional to dielectric constant εr , it is minimum for water. Ans a 93. Use the information supplied already in theory notes.. The force between q1 and q2 will not change due to the presence or absence of the third charge q3.. Ans.c 2q q q 1 ( − − ), where r is 94. At the centre the electric potential due to the charges will be r r 4 πε o r the distance from the corner to the centre of the triangle. The electric field at the centre will not be zero, because the fields are in different directions due to charges and hence will not cancel. Ans.b 95. In an electric field the line of force is the path followed by an isolated (or free) charge. Initially the charge is in the line of force because x = 1, y= 0 is a point in the circle. Hence when it is free, it will move along the line of force. Ans.d. 96. Use the information supplied in theory notes.. The total number of lines of force coming from the charge is 1.11x 1011. This is distributed to six faces. The number of lines of force coming from one face = (1/6) 1.11x 1011. Ans.d. 97. The force on 1C is 5 N means electric field at the origin is (force / charge) = 5/1 = 5 volt/m. Since the field increases at a rate 1 V/m for every metre, the field at x = 1 should be 5+1 = 6 V/m. The force on -1 C = charge x field = -1x6 = -6 N. Ans.b ( 98. Use information supplied in theory notes. The electric field due to a charged plane sheet σ/2εo ) is same at all points. The field is uniform throughout. Ans.d.. 99. Use the information supplied already in theory notes. When the distance increases to twice the field decreases to 1/23 of E. i.e. E/8. Ans.c 100. The field inside a charged hollow conductor is zero. In order to make the field zero, the charges will have to flow from inner hollow sphere Q to the outer one. Ans.c 101. The potential due to a point charge q is given by V = q/ 4πεor. The field due to a point charge

at the same point will have a magnitude E = q/4πεor2. From these two equations, we find E = V/r = 6/2 = 3 N/C. Ans.a 102. When a dipole is disturbed slightly, there will be a restoring torque trying it bring back to equilibrium. Due to the action of the torque, it returns. But it would have acquired kinetic energy of rotation, when it returns to equilibrium. Therefore, it continues in the opposite

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direction. Thus it oscillates. When all its energy is spent against dissipative forces, it returns to equilibrium. Ans.c 103. The two torques acting on the dipole are pFsinθ and p2F sin(90-θ), where θ is angle made by

the dipole with X axis at equilibrium. When the two torques are equal, the dipole will remain in equilibrium. Equating, we get tanθ = 2. Ans.c 104. The particle here is subjected to an electric field perpendicular to its velocity, because velocity is along Z-axis and the field along X axis. Hence the path will be a parabola in the plane of velocity and the field vectors. Ans.b 105. According to Lorentz’s force equation the force on the particle will be F = qv z xB x . Since the

vector product of velocity and field will be perpendicular to them, the force vector will be along the Y axis. This will provide centripetal force for circular motion. Therefore, the particle will have a velocity along Z-axis and centripetal acceleration along the Y-axis. The particle will move in a circle in Y-Z plane. Ans.b 106. The work done is the total potential energy of the system. If each charge is q, then the total energy will be 3Aq2/r. Here A = 1/4πεo = 9x109. q = (1/3 µC) = (10-6 / 3) C. This gives work as equal to 3 m J. Ans.b 2

107. The electric field on the surface of a sphere charged with q is q/4πεor , where r is radius of the 2

sphere. This can be written as σ/εo, because σ = q/4πr . Since σ is same, the electric field will also be same. Ans.a 108. The cube is placed so that the electric field is parallel to one of the edges along the X-axis. . The number of lines of force entering the face perpendicular to the edge is equal to the number of lines of force leaving. Hence the net flux will be zero. Ans.d 109. It is essential that a good insulator should have high value for dielectric constant or relative permittivity. The dielectric strength is the electric field to produce break down in the substance. This also will have to be high. Ans.d 110. The potential at the points A and B due to the point charge kept at the centre are same. This means potential difference is zero. Hence work done when a charge is moved between the points will be zero. You can also use theory note short cuts given. Ans.d 111. If we imagine a sphere of radius r inside bigger sphere, the question asks to find the field at the surface of this smaller sphere. The electric field at the surface of this smaller sphere is produced by the charge inside it. Since volume distribution of charge is uniform, the charge inside the smaller sphere will be (4/3)πr3ρ. Let this charge be q1. The electric field will be q1/ rρ 4πεor2. Substituting for q1, we get the field at the surface of smaller sphere as . Ans.b 3ε o 112. If the surface density of charge is σ, there will be an outward force on the soap bubble =

(σ2/2εo) N/m2 due to charging. This force will always expand the soap bubble, whether σ is positive or negative. So the soap bubble will always expand when charged. Ans.a. 113. The resulting potential is total charge/total capacitance. Since the positive plate of one is connected to the negative plate of the other, the charges cancel, i.e. total charge will be q1-q2. q − q 2 C1V1 − C 2V 2 1x 200 − 2 x100 = = 0 . Ans.d = Thus V= 1 1+ 2 C1 + C 2 C1 + C 2 114. The electrostatic potential energy = charge x potential = qV. This will be in joule.

If we write it in terms of electron charge e, then the charge carried by the α particle q = 2e. The energy will be 2eV joule = 2V electron volt = 2 x1 electron volt. Ans.b

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107

115. Any point on the y-axis is at equal distance from both the charges. Hence the potential due to

them will be equal in value and opposite in sign, which makes (a) correct. Any point on the y-axis is a point on the equitorial line of the dipole. Hence the field will depend on 1/y3. (Look up Theory notes supplied). So (b) and (d) are wrong. The electric field at all points on the y-axis will be in a direction parallel to the axis of the dipole, in this case +x axis. This makes (c) correct. Ans.a,c 116. The magnitude of the torque is given by pE sinθ, where θ is the angle made by the dipole

with the field. When the angle is made three times the torque becomes twice only if θ= 30o, since sin 30o = 0.5 and sin 90o = 1. Ans.d 117. Use the information supplied in theory notes.. When three identical condensers are joined in parallel the effective value is 3C and in series the effective value is C/3. It is given here 3C(C/3) = 16. Solving C = 6µF. Ans.b 118. Use the information supplied in theory notes. The half of the energy spent by the battery is used to charge the condenser, while the other half goes as heat. Therefore, heat produced is 2 mJ. Ans.c 119. When capacitors are in parallel, the charge is nCV. When they are in series, the charge is (C/n)V1, where V1 is the new potential. Since charge is constant, we have nCV = CV1/n. This gives V1 = n2V. You can also use theory notes to get this straight. Ans.d 120. The electrostatic energy of a charged body E = (1/2)QV. When this body is immersed in a dielectric, its capacity becomes KC and potential becomes V/K. [Use information given in theory notes tables]. The new energy will be E1 = (1/2)QV/K = E/K. Ans.a 121. When X-rays fall on the metal ball, some of the electrons are removed from the metal due to ionisation. This produces a net positive charge on the metal ball. Due to this the ball will deflect towards negative plate. Ans.b 122. Use the information supplied in theory notes. Charge flows until the potentials are equal i.e. Q1/4πεoR1 = Q2/4πεoR2. That is Q1/R1 = Q2/R2 or Q1R2 = Q2R1. Ans.c 123. Use the information supplied in theory notes. Here the potential is reduced from 100 to 25 V. This means the capacitance has been increased to 4 times i.e. from 5 to 20 due to the connection. So increase in capacitance is 20-5 = 15 µF. Ans.d 124. Since charge is conserved, use equation for electrostatic energy E = QV/2. V decreases from 100 to 25 V.i.e. to 1/4 th. So energy also decreases to 1/4. Ans.a 2 125. Use the equation for energy U = Q /2C. When plates are separated to 3 times capacitance becomes C/3. The energy becomes 3Q2/2C = 3U. Thus the energy increases from U to 3U. The work done = increase in energy = 2U. Ans.b 126. Use the information supplied in theory notes. If a sphere of radius R holds 1 C of charge the electric field on its surface should not exceed 3x106 N/C. This means Q/ 4πεoR2 = 3x106V/m. That is, 9x109/R2=3x106, which gives R = 3000 , which is nearly 50 m. Ans. c 127. Use the information supplied already in theory notes. The resulting capacity is equal to

10001/3x 2 = 10x2 = 20µF. Ans.a 128. Use the information supplied in indirect theory notes. The potential of resulting drop is equal to 10002/3x100 = 100x100 = 10000 V. Ans.c 129. Usually in international system, units are bigger than those in in CGS system. The units in CGS system are stat coulomb for charge, stat volt for potential and stat farad for capacitance and erg for energy. 1C = 3x109stat -coulomb. 1F = 9x1011 stat farad. 1 J = 107 erg. 1V =

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(1/300) stat volt. From these we find the CGS unit of potential stat volt = 300 V i.e. bigger than volt. Ans.d ε 0A ε A . Here . With air C 0 (t 1 / k 1 ) + (t 2 / k 2 ) + (t 3 / k 3 ) t t1=t2=t3=t/3. k1=2. k2=3 k3=4. Substituting these values we get C1= (36/13)C. Ans.a 2 131. Since the potential is same we use equation for energy (1/2) CV . So E1/E2 =C1/C2. C1/C2.= 1/4 from indirect theory notes. Thus the ratio E1/E2 = 1/4. Ans.b 132. When the battery is disconnected the charge remains constant. Hence use the equation for energy as Q2/2KC with dielectric and Q2/2C without the dielectric . Thus energy increases in the ratio 1/K:1 i.e.1/2:1 That is,the energy increases by 100%. Ans.d. 133. When an electric field is applied to a dielectric it becomes polarised. Polarisation produces dipoles. These dipoles produce a field at any point inside which opposes the external field. Hence the field at any point will be E0 - the field produced by the dipoles i.e. it will be less than E0. Ans.b 134. When an uncharged plate is brought from infinity near to a positively charged plate, the potential of positive plate decreases and its capacity increases. (Principle of a condenser). So when this plate is removed capacity decreases and potential increases. So (a) and (d) are correct, (b) and (c) are wrong. Ans.a,d. 135. Use the information supplied already in theory notes. If Q is the charge of each drop and V is the potential, when drops coalesce the charge becomes 10 Q and potential becomes 102/3V. Since energy is proportional to the product of Q and V, it will become 10x102/3 = 105/3 of that of one drop. You can also use theory notes short-cuts. Ans.c total capaci tance C is equal to 136. Since all the capacitors are in parallel 1 1 1 1 1 1 1 4 + 2 + 1 + + + + ....to ∞ = 4 1 + + + .... = 8µF 2 4 8 16 2 4 8 . Ans.d where we have used the information that geometric series in the bracket has a sum 2. Ans.d 137. The arrangement in the triangle part ABC is a balanced Wheatstone’s bridge of capacitors. The effective capacitance between the points A and B of this bridge is 1µF. This is parallel with 1µF across DE. Therefore the total capacitance 1+1 =2µF. Ans.a 138. Let a charge of 6q flow from the battery. It divides into three equal values 2q each as shown in figure. Each of the 2q divides to two condensers as q each. Then these charges rejoin as 6q. Taking any closed path and adding potential difference across each capacitor, we have 2q/C + q/C+2q/C = 6q/C1, where C1 is the equivalent capacitance of A and B.⇒C1= (6/5) C. Ans.b 139. Let X be the capacitance between A and B. (fig below). In this infinite network we have repeating units. Here one unit is a parallel an series capacitor. Break the circuit after one such unit. We find the circuit will have the same structure as the original network. Hence the capacitance between E and F will also be X itself. X across EF is in series with 2 . This gives 2X/2+X. This is in parallel with 130. Using the equationc C1 =

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109

2 across CD. This will give (2X/2+X)+2. This should be equal to X, i.e. capacitance across 2X + 2 = X . Solving this equation we gets X = 3.2µF. Ans.c AB. Therefore, we have 2+X 140. Before introducing the metal, let the capacitance be C. The C = εoA/d. If a dielectric of

thickness t and dielectric constant K is introduced the new capacitance C1 is given by 3ε A 3 ε0A 3 C1 = Here K =∞ and t =d / 3 . C1 = 0 = C = x100 = 150 pF. Ans.b 1 2d 2 2 d − t (1 − ) K 141. The potential, the capacity and energy depend on the shape and the charge of the conductor. Hence they will not be same even if they have same volume. Ans.d 142. Use the information supplied already in theory notes. The electrostatic energy lost is equal to heat energy gained. Therefore the heat energy produced = 1 2 x2 1 2 −6 x10 ( 200 − 100) = J. 2 2+2 200 143. In a series connection capacitors get same charge. Using the equation Q =VC, we find V is proportional to 1/C. V1:V2:V3 = (1/1): (1/2): (1/3) = 6:3:2. So all we have do is to divide 110 volt in this ratio. The potentials will be 60V, 30V and 20V respectively. 2

144. The equation for energy density i.e. energy per unit volume is given by (1/2) εoεrE . Ans.d 145. When two charges are placed at a distance, the force on them will be the same in magnitude

and opposite in direction. Force on q = field x charge = Eq. Force on -3q = Eq. Field at the site of -3q = force/charge = = qE/-3q = -E/3. Ans.b 146. When the charge is released there will be force of attraction towards the ring. The horizontal component of the force along the axis will provide the restoring force. This force will always be directed towards the centre of the ring. The particle, thus, will have oscillatory motion. But the coloumb forces follow inverse square law. For simple harmonic motion we need force proportional to displacement. Hence the motion will not be simple harmonic. Ans.a 147. A negatively charged body gains electrons. So its mass increases. A positively charged body loses electrons and its mass decreases. Hence negatively charged sphere has more mass and positively charged one has less mass. Ans.b 148. Use the equation for capacitance of a parallel plate condenser with a dielectric of thickness t introduced . If C is the capacitance when medium is air we have C = εoA/d. ε0A C1 = . Here K = ∞ and t =0. This gives C1 = C. Ans.a 1 d − t( 1 − ) K 149. A condenser discharges according to the equation q = qoe

-t/RC

, where R is the value of the resistor in series with C. Current i = dq/dt = -qoe (1/RC). Since q0/C is constant equal to charging potential V, we find, the current i is not zero at time t = 0,but is constant equal to q0/CR. Therefore (a) and (c) are wrong but (b) is correct. Also dq/dt is more when C is less, from the above equation. This means condesner of smaller capacitance loses charge more quickly. Hence (d) is correct. Ans.,b, d 150. Since the condenser is connected to the battery, whatever may happen, the potential remains the same. When a dielectric is introduced, the capacity will increase to K times. Ans. d. 151. The electromagnetic force is transferred through photons. A charge feels the presence of the other when photons are exchanged between them. Hence the time taken will be the time for -t/RC

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photons to travel the distance between them which will be about 10-8s. This time is small but finite. Ans.c 152. Use the information supplied in theory notes . The insulating property of air breaks down The corresponding energy density is when the electric field ≈ 3x106V/m. 1 1 = ε o E 2 = x9 x10 −12 x(3x10 6 ) 2 = 40 J / m 3 . Ans.c 2 2 153. Since the charges are of opposite sign no null point will be got between the charges. So R is not a null point. Since the field will have to be equal in magnitude, the point will have to be nearer to the smaller charge. Hence a possible null point is P. Ans.a 154. The potential at any point is produced by positive and negative charges. Hence it may cancel at a point between them, or at a point nearer to the smaller charge. Since potential is a scalar, we need not look for the direction. Hence the two points P and R could have zero potential. Note in these questions we find only location of points and not the distance. Ans.d 155. The solution to this question is similar to the one given in indirect theory notes for resistors. Here one repeating unit is a series and a parallel capacitor. Break the circuit after one unit and replace it with X, as shown in adjacent figure. We get 2+X and 2 in series which is equal to 2(2+X)/2+2+X. Thus we have 2(2+X)/2+2+X = X. Solving we get X= 1.2 µF. Ans.d 156. At centre of one ring, say P, the potential V due to the charge on it is

V = q/4πεoR. At the same point the elements of the other ring are at a distance √2R. Therefore the potential at P Total potential is equal to V+V/√2 = due to the other ring is q/(4πεo√2R) = V/√2. (√2+1)V/√2. Ans.d. 157. Use the information supplied in theory notes. Each condenser should have a capacitance 4 µF. When one of them is removed the other three are in series. They will have a capacitance 4/3 µF. This in parallel with 4 µF will give a net capacitance 16/3 µF. Ans.c 158. Connect 4 condensers in series. It will have a capacity of 1/4 µF. Connect such three groups in parallel. Total capacity will be 3x(1/4) = 3/4 µF. This is shown in the figure below. 159. Connect six of them in series. The effective capacitance will be 1/6 µF. Make two such

groups and connect both groups in parallel. This will give 2x1/6= 1/3 µF

5 ELECTRICITY SECTION 1 QUESTIONS 1.

The current through 60 ohm resistor in A is (Fig) 120

60

60

120

6 V

2.

a) 0.025 b) 1/30 c) 2/30 d) 1 The current through 60 ohm resistor in the following network in A is (Fig) 30

6V

70

60 30

40 50

Fig. 3

3.

4.

5.

a) 0.08 b) 0.06 c) 0.04 d) 0.02 A uniform wire of resistance 4 ohm is bent into a square loop. The resistance between the midpoints of its opposite sides would be a) 4Ω b) 2Ω c) 1Ω d) 0.5Ω It is required to get largest amount of heat energy from a resistance wire of length 0.5 m connected with a battery of negligible resistance. This can be done a) by joining the wire directly to battery b) by cutting the wires in two equal parts and then joining all of them in parallel to the battery c) by cutting the wire in two four equal pieces and then connecting all of them in parallel to the battery d) by joining only half the wire A silver voltameter and a zinc voltameter are connected in series and current i is passed for a time t liberating w kg of zinc. The mass of the silver deposited is nearly a) W b) 3.5 W c) 2.4 W d) 1.7 W

112

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9.

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An electric immersion heater of 1.08 kW is immersed in water. After the water has reached a temperature of 1000C, how much time will be required to produce 100 g of steam ? a) 50 s b) 420 s c) 105 s d) 210 s Two resistors in parallel have their resistances in the ratio 1:3. A source is connected to the combination. The ratio of heats produced in a given time in the two resistors is a) 3:1 b) 9:1 c) 1:3 d) 1:9 The terminals of battery of emf 12 V and internal resistance 1 Ω are connected to a circular coil of resistance 16 Ω at two points distant a quarter of the circumference of the coil. The current flowing through the smaller arc of the circle in A is a) 3.0 b) 2.25 c) 0.75 d) 0.5 A uniform wire of length 5 cm is carrying a steady current. The electric field inside it is 0.2 V/m. The potential difference across the ends of the wire is a) 0.1 V b) 0.5 V c) 5 V d) 10-2 V Two wires A and B are in series. When heated, the combination has same resistance at all temperatures. A is of germanium. Then B should be of a) carbon b) aluminium c) silicon d) alloy of carbon and silicon A cell of emf X is connected across a resistor R. The potential difference across the wire is measured as Y. The internal resistance of the cell should be a) X-Y/(R-X) b) (X-Y)R c) (X-Y)R/X d) (X-Y)R/Y A battery of 2 volt emf and internal resistance 1 ohm sends a current of 1 A through an external load. If two such batteries are connected in series, the current through the same load would be a) 1A b) 2 A c) 1.5 A d) 1.33 A A rectangular loop carrying a current i is situated near a long straight wire such that the wire is parallel to one of the sides of the loop and is in the plane of the loop. If a steady current i is established in the wire as shown in the fig. 1, the loop will a) rotate about an axis parallel to the wire’ b) move away from the wire c) move towards the wire d) remain stationary A wire of given length is first bent in one loop and the next it is bent in three loops. If the same current is passed in both the cases, the ratio of magnetic induction at their centres will be a) 1:4 b) 1:9 c) 9:1 d) 1:3 A circular loop of mass m and radius r is in a horizontal (x-y plane) table as shown in fig 2. A uniform magnetic field B is applied parallel tro x-axis. The current in the loop, so that its one edge just lifts from the table is a) mg/πr2B b) mg/πrB c) mg/2πrB d) πrB/mg The magnetic field at the centre of the cube of edge of length a carrying a current i is a) zero b) (8µoi)/(√2 a) c) (8µoi)/2√2 a) d) (2µoi)/2√2 a)

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17. An electron enters along east in a region in magnetic equator. It will be deflected a) vertically up b) horizontally along east c) horizontally along west d) vertically down 18. A current of i ampere is flowing through a loop of a circle of radius r metre which subtends an angle θ as shown. The magnetic field at the centre of the loop is a) µoiθ/4πr b) µoi sinθ/πr2 c) µoi sinθ/2r d) µoiθ/4r 19. If a graph is plotted between magnetic field due to a long straight conductor and distance, the graph will be a) straight line with positive slope b) straight line with negative slope c) parabola d) rectangular hyperbola 20. A conducting circular loop of radius r carries a constant current i. It is placed in a uniform magnetic field Bo such that Bo is perpendicular to the plane of the loop. The magnetic force acting on the loop is a) irBo b) 2πriBo c) zero d) π riBo 21. A proton and electron of equal momentum enter into a uniform field at right angles to the velocity. The radius of their tracks will be in the ratio a) 1:1840 b) 1840:1 c) 1:43 d) 1:1 22. A charged particle is undergoing a circular motion in a uniform magentic field. The time period is independent of a) speed b) mass c) charge d) intensity of magnetic field 23. A one MeV energy alpha particle describes a circle of radius x in a uniform magnetic field. A proton of the same energy will describe in the same field a circle of radius a) x b) x/2 c) 2x d) 1.4 x 24. A circular coil of wire having 100 turns and radius 1m is arranged in plane perpendicular to the magnetic meridian. A current of 1 A is passed through the coil. A horizontal magnetic needle at the centre will show a deflection a) tan-1 (1.57) b) tan-1 (3.14) c) 0 -1 d) tan (0.85) 25. In the Lorentz force equation F = qv x B,. which of the following is correct ? a) v and B are always perpendicular b) v and F are always perpendicular c) v and B cannot be parallel d) F and B be parallel 26. A long straight conduct or is bent into the shape as shown in Fig.6 It carries a current i A and the radius of the circular loop is a m, The magnetic field at the centre of the loop is a) 0 b) µo i/2π a c) µo i(π+1)2π a d) µo i(π-1)/2π a

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27. A thin wire is bent to the form of a square loop ABCD.(fig.7) A battery is connected between A and C as shown in fig. the magnetic induction due to the current in the loop at the centre O a) points into the plane of the paper b) points along OP, the bisector of angle BOC c) zero d) points out of the plane of the paper 28. The magnetic field produced by any side at the centre of the square loop in the adjacent figure 8 is of magnitude B. The magnetic field at the centre due to the whole loop is a) 0 b) 4B c) 2B d) B 29. The magnetic induction at a point P distant 4 cm from a long current carrying wire is X tesla. The induction at a distance 12 cm from the same wire would be (in tesla) a) X b) X/3 c) 9X d) X/9 30. A circular loop carrying a current of i A is bent into a circular coil of 4 turns.The new magnetic moment of the coil will be a) 4 times the original b) 2 times the original c) 1/4 of the original d) half of the original 31. Two particles X and Y having equal charges, after being accelerated through the same potential difference, enter a region of uniform magnetic field and describe circular paths of radii R1 and R2 respectively. The ratio of mass of X that of Y is a) (R1 /R2)1/2 b) (R2/R1) c) (R1/R2) d) (R1 / R2)2 32. Am2 has the same dimension of a) IT b) NT c) JT-1 d) NT-1 33. The torque acting in a magnetic needle when it make 300 with a uniform field is x. The torque acting when it is perpendicular to the field is a) x b) 3x c) 2x d) (3/2) x 34. A thin uniform magnetic needle of period T is broken into n parts of equal length. The period of one part will be a) nT b) T/n c) n2T d) T/n2 35. Magnetic shielding to an instrument can be provided by covering with a) soft iron b) plastic c) copper d) aluminium 36. Isogonic lines are those which join places of a) zero dip b) zero declination c) equal dip d) equal declination 37. If a dip needle stands horizontal at a place, the place is a) magnetic meridian b) perpendicular to magnetic meridian c) magnetic pole d) magnetic equator 38. Two magnets placed one over the other oscillate with a period of 16 s. When one of them is reversed, the period is 8 s. The ratio of their magnetic moments is a) 3;1 b) 1:3 c) 5:3 d) 3:5

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39. The susceptibility of a substance is negative. This substance could be a) bismuth b) iron c) nickel d) aluminium 40. Two magnets of same pole strength p but of lengths 3L and 4L respectively are arranged to form a right angle so that the north pole of one touches the south pole of the other. The moment of the combination is a) pL b) 7 pL c) 5 pL d) 6 pL 41. The correct dip at a place is δ. If the dip is measured in a plane making and angle θ with the magnetic meridian as δ1 , then δ1 will be b) tan-1 (tanδ sec θ) a) tan-1 (tanδ cosθ) d) tan-1 (tanδ cosec θ) c) tan-1 (tanδ sinθ) 42. The total intensity of earth’s magnetic field at the magnetic equator is 0.4 cgs units. At a place where the dip needle reads 900, the vertical intensity will be (in cgs units) a) 0.1 b) 0.2 c) 0,.4 d) 0 43. Which of the following substance has magnetism independent of temperature? a) copper b) aluminium c) iron d) manganese 44. Which of the following material shows the property of hysterisis ? a) copper b) cobalt c) aluminium d) silver 45. Which of the following is most suitable for making the beam of a chemical balance ? a) brass b) iron c) aluminium d) nickel 46. The field due to a short magnet on the axis at distance x is n times the field due to the magnet at a distance 2x. Here n is a) 2 b) 4 c) 8 d) 16 47. Soft iron is used for electromagnets because of a) high limit of magnetic saturation b) large area for hysterisis curve c) large coercivity d) large retentivity 48. Which of the following is a paramagnetic substance ? a) bismuth b) antimony c) water d) chromium 49. A steel wire of length L has a magnetic moment M. It is bent into a semi circular arc. The new magnetic moment is a) M/ 2π b) ML/2π c) M/L d) 2 M/π 50. A magnet of moment m is kept in stable equilibrium in a uniform magnetic field of intensity B. If it is rotated through an angle of 1800, the work done is a) mB b) 2mB c) mB/2 d) zero 51. A line joining all places on the earth having zero magnetic dip is a) magnetic line of force b) magnetic equator c) magnetic meridian d) isoclinic line 52. A magnet falls vertically through a horizontal copper ring. Its acceleration will be a) always less than g b) always greater than g c) always equal to g d) less than g when approaches and greater than g when recedes 53. The phase difference between the flux linkage and gthe induced e.m.f in a rotating coil in a uniform magnetic field is

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a) π b) π/2 c) π/4 d) zero 54. The induced emf in a coil of wire produced by motion of a magnet will not depend on a) number of turns in the coil b) area of the coil c) resistance of the coil d) speed of the magnet 55. The energy stored in an inductor is in the form of a) electric b) mechanical c) thermal d) magnetic 56. An aeroplane with wing span of 50 m flies at 540 km/h horizontally along north south. The vertical component of earth’s magnetic field is 0.2 gauss. The induced emf across the wing tips is a) 0.15 b) 15 V c) 1500 V d) 150 V 57. A square loop PQRS of resistance 1 Ω and side 10 cm is moved perpendicular to a uniform magnetic field of 1 T. The ends of the loop are connected to a net work of resistors as shown in fig. The speed of the loop so that a current of 1 mA flows through it is (in m/s) a) 10-2 b) 2 x 10-2 c) 20 d) 10 58. A circuit has a coil of self inductance 20 mH and a resistance of 3 Ω. The induced emf in the circuit when current changes at a rate 5 A/s is (in volt) a) 0.1 b) 0.1/3 c) 0.3 d) 0.2 59. Lenz’s law is a consequence of the law of conservation of a) charge b) momentum c) mass d) energy 60. An electron moves along the line AB which lies in the same plane as the circular loop of conducting wire as a shown in the fig.2 What will be the direction of the current induced if any in the loop? a) no current will be induced b) the current will be clockwise c) the current will be anti-clockwise d) there will be an emf but no current 61. In what way should be conductor AB be moved in a magnetic field such that the current flows as shown in fig. 3. a) vertically up ward b) towards left c) towards right d) vertically downward 62. J A-2 is the unit of a) permeability b) permittvity c) self indction d) energy density 63. A fan leaf of length L rotates in a uniform magnetic field of

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64.

65.

66.

67.

68.

69. 70.

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intensity B, with a constant angular velocity ω . The induced emf between the centre and end of the leaf will be a) B12 ω b) 2 B12 ω c) (1/2)B12 ω d) B 1ω A coil has a self inductance L. If the length and breadth of the coil are doubled keeping number of turns per unit length constant the self inductance will a) remain same b) become 2 times c) become 4 times d) become 8 times The coefficient of mutual induction between the two coils will not depend on a) number of turns of the secondary b) distance between primary and secondary c) cross sectional area of the primary d) the rate of change of current in the primary The equivalent inductance between points P and Q in fig.4 is a) 2H b) 6H c) 8/3H d) 4/9H Two coils have self inductances L1 = 8 mH and L2 = 2mH. The current in the two coils are increased at the same constant rate. At a certain instant the power given to the two coils are same. At that instant if currents in the coils are I1 and I2 respectively, then I1/I2 = a) 4/1 b) 1/4 c) 2/1 d) 1 A conducting square loop of side L and resistance R moves in its plane with a uniform velocity v perpendicular to one of its sides. A magnetic induction B constant in time and space pointing perpendicular to and into the plane of the loop exits everywhere. The current induced in the loop is a) BLv/R clockwise b) BLv/R anti-clockwise c) BLv/R clockwise d) 2BLv/R anti-clockwise Which of the following has the dimension of time ? a) LR b) L/R c) R/L d) 1/CR Which of the following is not correct ? a) ohm = henry per second b) farad x henry = second2 c) farad = joule/volt d) ohm x farad = second

71. The flux through a closed circuit of resistance 10 Ω varies with time according to the equation φ = 12t2 + 4t + 1. The induce current in the circuit 1/4 second after start will be (in ampere) a) 10 b) 2.75 c) 0.275 d) 1 72. The emf in an AC circuit is represented by the equation e = 5 sin ωt and current by i = 2 cos ωt. The average power consumed in the circuit is (in W) a) 10 b) 10√2 c) 0 d) 5√2 73. An alternating voltage of rms value V is applied to a circuit containing a resistor, inductor, and a capacitor. Four voltmeters are connected across resistor, inductor, apacitor and supply. If their readings are V1, V2, V3 and V respectively, then a) V1 + V2 + V3 = V b) V12 + V22 + V32 = V2 2 2 2 c) V1 + (V2 – V3) = V d) (V1 – V2)2 + V32 = V2 74. If V1 , V2, V3 are instantaneous voltage across the components and V = Vo sinωt supply voltage a) V1 + V2 + V3 = V

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b) V12 + V22 + V32 = V2 c) V12 + (V2 – V3)2 = V2 d) (V1 – V2)2 + V32 = V2 75. In figure2 a 6 volt battery is connected to a Neon lamp which needs 10 volts to glow. An inductor L is connected in parallel. When the key K is switched on which of the following happens ? a) bulb glows immediately but dimly b) bulb first does not glow but glows dim later c) bulb never glow d) bulb glows bright 76. When the current is switched off in the above circuit in previous a) bulb goes off immediately b) bulb goes off slowly c) bulb keeps on glowing d) bulb turns brighter and goes off 77. An A.C voltage of rms value 200 V is applied to a circuit containing a resistor of 60 Ω, inductor of reactance 180 Ω and a capacitor of reactance 100 Ω. The rms value of current in the circuit is (in A) a) 2 b) 2.8 c) 10/7 d) 10/19 78. The phase difference between current and voltage in the above circuit given in previous question is a) tan-1 (3/4) b) tan-1 (4/3) c) tan-1 (5/3) d) tan-1 ((3/5) 79. The power factor of circuit given in previous question is a) 0.44 b) 0.8 c) 0.6 d) 1 80. The average power consumed by the circuit in watt is a) 400 b) 200 c) 282 d) 240 81. An electro magnetic wave can be produced by a) positive and negative charge b) steady current in a wire c) varying electric current d) a strong magnet 82. An alternating voltage 230 V, 50 Hz is applied to a circuit containing an inductance, resistance and a capacitance. At any moment for rms voltages, a) the total voltage across each when added will be equal to 230 V b) the voltage across each will be 230 V c) the voltage when added across the three should be less than 230 V d) the voltage across the inductance can be greater than 230 V 83. We have a source of red light and a blue light of same power. Then the red light gives photons per second a) same in number, but of less energy b) more in number, but of less energy c) less in number and of less energy d) more in number and of more energy 84. The instantaneous values of current and voltage in an A.C circuit are I = 4 sin ωt and E = 100 cos[ωt + (π/3)] respectively. The phase difference between voltage and current is

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a) 7π/6

b)

5π/6

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c)

2π/3

d) π/3

85. An A.C given by e(t) = 282 sin (120π t) is applied across a resistance of 20 Ω. An A.C ammeter connected in series records a current of a) 14.1 A b) 7.05 A c) 10 A d) 5 A 86. Two coils of inductance L1, L2 are connected in series. The value of inductance is equal to L1 + L2 if a) distance between them is large b) distance between them is small c) they are would one over the other d) same always 87. Which of the following is least scattered by a medium a) X-rays b) visible light c) infra red d) radiowaves 88. Take wave length of blue light roughly as 440 nm and that of red light as 660 nm. The ratio of intensity of red light scattered to that of blue light by earth’s atmosphere is nearly equal to a) 2/3 b) 3/2 c) 4/9 d) 1/5 89. Assuming the radius of earth to be 6400 km, the maximum distance to which a TV signal can be received on earth using an antenna of height 500 m is equal to a) 80 km b) 80 x √2 km c) 40 x √2 km d) 160 km 90. If we plot a graph between the reactance of a capacitor and frequency of A.C passing through it , the graph will be a) a straight line having a positive slope b) a straight line having a negative slope c) a parabola d) a rectangular hyperbola 91. Two point charges +q and –q are held fixed at (-d,0) and (d,0) respectively, of (X,Y) coordinate system. Then a) E at all points on the Y-axis is along ˆi G b) the electric field E at all points on the X-axis, has the same direction. c) dipole moment is 2 pd directed along ˆi d) work has to be done in bringing a test charge from infinity to the origin. 92. A battery is connected between two points A and B on the circumference of a uniform conducting ring of radius r and resistance R. One of the arcs AB of the ring subtends an angle θ at the centre. The value of the magnetic induction at the centre due to the current in the ring is (fig) a) proportional to 2 (1800-θ) b) inversely proportional to r3 c) zero only if θ = 1800 d) zero for all values of θ 93. The wire loop PQRSP formed by joining two semicircular wire of radii R1 and R2 carries a current i as shown in the fig. 3. The magnetic field B at the point C is µ I 1 1 − a) 0 2π R 1 R 2

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b)

µ0I 1 1 − 4π R 1 R 2

c)

µ0I 1 1 − 4 R1 R 2

d)

µ0I 1 1 − 2 R1 R 2

94. A proton of mass 1.67 x 10-27 kg and charge 1.6 x 10-19 C is projected with a speed of 2x106 m/s at an angle of 600 to the Xaxis . If a uniform magnetic field of 0.104 Tesla is applied along the Y-axis, the path of the proton is a) a circle of time eperiod π x 10-7 s b) a circle of time period 2π x 10-7 s c) a helix of time period 2π x 10-7 s d) a helix of time period 4 π x 10-7 s 95. H+ , He+ and O2+ all having the same kinetic energy pass through a region in which there is a uniform magnetic field perpendicular to their velocity. The mass of H+. He+ and O2+ are 1 amu, 4 amu and 16 amu respectively. Then b) O2+ will be deflected most a) H+ will be deflected most + 2+ c) He and O will be deflected equally d) all will be deflected equally 96. A microammeter has a resistance of 100 Ω and a full scale range of 50 µA. It can be used as a voltmeter or as a higher range ammeter provided a resistance is added to it. Pick the correct range and resistance combination(s) a) 50 V range with 10 kΩ resistance in series b) 10 V range with 200 kΩ resistance in series c) 5 mA range 1 Ω resistance in parallel d) 10 mA range with 1Ω resistance in parallel 97. A piece of copper and another of germanium are cooled from room temperature to 80 K. The resistance of a) each of them increases b) each of them decreases c) copper increases and germanium decreases d) copper decreases and germanium increases 98. A battery of internal resistance 4 Ω is connected to the network of resistance as shown. In order to give the maximum power to the network, the value of R in Ω should be a) 4/9 b) 8/9 c) 2 d) 18 99. The current through a wire changes with time as given by the equation i = t . the correct value of the rms current within the time interval t = 2 to t = 4 s will be a)

3A

b) 3A

c) 3 3 A

d) 3 2 A

100. Two heater wires of equal resistance are first connected in series and then in parallel, to the same source. The ratio of heat produced in them are in that order a) 2:1 b) 1:2 c) 4:1 d) 1:4

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101. The electric flux through a hemispherical surface of radius R placed in a uniform electric field of intensity E parallel to the axis of its circular plane is b) 2πR2E c) R2E d) 4/3πR3E a) 2RE2 102. To get maximum current in a resistance of 3Ω, one uses n rows of m cells each, connected in parallel. If the total number of cells is 24 and internal resistance of a cell 0.5 Ω, then a) m = 12 n = 2 b) m = 8, n = 3 c) m = 2 n = 12 d) m = 6, n = 4 103. A particle of mass m and charge q is placed at rest in a uniform electric field E and released along the y-axis. The kinetic energy it attains after moving a distance y is a) 1/2 qEy b) qE2y c) qEy d) 1/2 m(qEy) 104. An electric dipole in a uniform field is slightly disturbed from equilibrium position. Its period of oscillation is a ) 2π I / pE

b) 2π pE / I

c) 2π p / IE

d ) 2π IE / p

105. A house is served by a 220 V supply line. In a circuit protected by a fuse marked 9 A, the maximum number of 60 W lamps in parallel which can be turned on at the same time is a) 44 b) 20 c) 22 d) 33 106. The unit of thermo electric power is a) W b) VK-1 c) V d) Vs-1 107. A uniform wire of resistance 2 Ω and length 4 m is melted and reformed into a uniform wire of length 2m. Its resistance now will be (in Ω) a) 2 b) 0.5 c) 4 d) 0.25 108. A particle carrying a charge of 1 C enters into a uniform magnetic field 3 j with a velocity 2j+k. The force on the particle has magnitude in N and direction respectively a) √7, +x axis b) √7, -x axis c) 3, + x axisi d) 3, -x axis 109. ML2T-3 I-2 is the dimension of a) potential difference b) electric field c) permitivity d) resistance 110. Which of the following is correct ? a) Resistivity = electric field/current density b) Resistivity = current density/electric field c) Resistivity = 1/current density d) Resistivity = current density/potential 111. Four bulbs connected in series consume a total power of P watt. One of the bulbs is fused and the other three are connected in series to the same source. The total power consumed (in W) is a) P b) 3P/4 c) 8P/3 d) 4P/3 112. The best combination of properties of a metal to make a standard resistor are a) high resistivity and low temperature coefficient b) high temperature coefficient and high linear expansivity. c) low linear expansivity and high temperature coefficient d) low temperature coefficient and high melting point . 113. Two wires A and B of same material have same mass but radii in the ratio 1:2. Their resistance ratio RA/RB will be a) 1:4 b) 1:8 c) 1:16 d) 16:1 114. A primary and a secondary cell have same emf. Then

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a) both give same current b) primary gives more current c) secondary gives more current d) cannot be said from the data 115. You are supplied with two resistors. Using them it is possible to get resistance value 9,27,6 and 18 ohms. The resistance of resistors are a) 9 and 18 b) 27 and 18 c) 6 and 18 d) 27 and 6 116. Four cells each of same emf and same internal resistance are connected in series. The combination gives a current i. If one of the cells is reversed the current decreases by 2 1 a) 25% b) 50% c) 66 % d)33 % 3 3 117. An ammeter of resistance G reads 1 ampere per division. It is to be converted into an ammeter reading n amperes per division. The resistance to be connected in parallel is a) nG b) G/(n-1) c) G/n d) (n-1)G 118. *Two resistors having resistances R1 and R2 at 0oC and temperature coefficient α1 and α2 respectively are joined in series. Check the correct statement(s): a) This combination will have same resistance at all temperatures if R1α2 = R2α1 b) This combination will have same resistance at all temperatures if R1α1 = R2α2 c) Both resistors should be made of metals d) One resistor should be made of metal and the other of semi conductor. Solve the following problems within a maximum time of 2 minutes 119. What is the reading of the ammeter (A) in figure1? 120. What is a current through 2 ohm resistor in the network shown in fig 2

Fig.1 Fig.2 121. What is the reading of the ammeter A in the fig 3?

Fig.4

Fig.3

Fig.5

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123

122. What is the reading of ammeter A in the figure 4? 123. A uniform circular loop has a resistance of 2 ohm. A straight conductor of resistance 1 ohm is connected diametrically in this conductor across AB, as shown in figure 5. What is the number of electrons flowing through the section AB (straight) per second ? 124. A battery of emf E and internal resistance 1 ohm is connected as driving cell in a potentiometer. The balancing length of a Daniel cell is found to be L m. If another battery of same emf and internal resistance 2 ohm is connected as driving cell, the balancing length of the same Daniel cell will be a) more than L b) less than L c) equal to L d) cannot be said from the data. 125. Four wires of same material, labelled as P, Q, R, S, lengths in the ratio 1: √2 : √3: 2 and radii also in the same ratio are available. Which of them would be most suitable for making a fuse wire? a) P b) Q c) R d) S 126. *Which of the following statements are correct? a) the terminal voltage of a cell will always be less than its emf b) the terminal voltage of a cell can exceed its emf under certain conditions c) the terminal voltage of a cell can be less than or greater than its emf d) the terminal voltage of a cell will always be greater than its emf 127. In the adjacent figure two cells of emf E1 and E2 are connected as shown with a resistor 9 ohm. The terminal voltage across E1 is (in volt) a) 7.66 b) 8.33 c) 8 d) 0 128. In the same fig. terminal voltage across E2 is (in V) a) 12 b) 12.66 c) 11.34 d) 8 129. In the same figure the potential difference across the points P and Q is ( in volt) a) 11.33 b) 11 c) 8.33 d) 3 . 130. In the adjacent figure, the heat developed per second across the 3 ohm resistor is H. The heat developed across 1 ohm resistor per second is a) H/3 b) 2H/11 c) 11H/2 d) 3H 131. In the adjacent figure the lamp of resistance 4 ohm shines with maximum brightness. The internal resistance of the cell is (in ohm) a) 4 b) 2

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c) 8 d )nearly zero 132. The equivalent resistance between the points A and B in the figure below is (in ohm) a) 1 b) 3/4 c) 4/3 d) 10 133. A heater has three coils in parallel having resistances 100, 100 and 200 ohm with the option of using any one or all of them. When 200 ohm coil alone is used, the heat required to boil certain mass of water is 5 minutes. When all the three coils are used, the heat required to boil the same amount of water will be ( in minutes) a) 25 b) 1 c) 5/4 d) 10/3 134. Charge flows at a steady rate through a conductor of non-uniform cross sectional area. Which of the following remains constant? a) drift speed of electrons b) current density c) current d) all the three 135. *The resistivity of a wire depends on a) its length b) its cross sectional area c) its material d) temperature 136. An electric bulb rated as 500 W at 100 V is to be used in a circuit of supply voltage 150 V. Then for the bulb to deliver 500 W, a resistance R a) equal to 20 ohm is to be connected in series b) equal to 20 ohm is to be connected in parallel c) equal to 20/3 ohm is to be connected in series d) equal to 10 ohm is to be connected in series 137. A wire is uniformly stretched to two times its length. The conductance of the wire a) remains the same b) becomes 1/2 c) becomes 1/4 d) becomes 4 times 138. Three cells each of emf E are connected in series to an external resistor of value 10 ohm. The current through the external resistor is i . When the same cells are connected in parallel to the same resistor, the current is still i. The internal resistance of each cell (in ohm) is a) 30 b) 10/3 c) 10 d) 3 139. A conductor has n electrons per unit volume. It has current density j. If specific charge of electron is K, then the total momentum of n electrons is equal to a) jK b) K/j c) j/K d) j2K 140. Two electrons of speeds v and 2v enter into a uniform transverse magnetic field. The ratio of their frequencies of circular motion will be a) 1:1 b) 1:2 c) 2:1 d) 1:4 141. A charged particle of mass m moving with a velocity is subjected to a magnetic field of 20ˆi and an electricfield 1000ˆj. If this particle continues with the same velocity in the same direction, its momentum is ( all units in SI) a) 100m b) 50m c) 200m 142. A cyclotron cannot be used to accelerate a) α particles b) protons c) electrons

d) (1/50)m d) deuterons

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143. A circular loop is placed so that its plane is parallel to a uniform magnetic field. In this magnetic field, it experiences a) a torque b) a force c) both a torque and a force d) neither a torque nor a force 144. A circular loop carrying current is placed perpendicular to a uniform magnetic field. It experiences a) a torque b) a force c) both a torque and a force d) neither a torque or a force 145. Two circular loops are made from wires of lengths in the ratio 1:2 . If they carry same currents, the ratio of their magnetic moments is a) 1:2 b) 2:1 c) 1:4 d) 4:1 146. In a uniform transverse magnetic field, a charged particle in motion has a) constant momentum and kinetic energy b) varying momentum but constant kinetic energy c) constant momentum but varying kinetic energy d) varying momentum and varying kinetic energy 147. The ratio of magnetic field at the centre of a current carrying circular coil of radius R to the field at a distance R from the centre along its axis is a) 1:8 b) 1:4 c) 1:2√2 d) √8:1 148. An electron describes a circular orbit of radius r, with a speed v. The magnetic field produced at the site of nucleus is µ ev µ ev µ e2v µ ev a) o 2 b) o c) o d ) o 2 e) 0 2πr 2πr 4πr 2πr 149. The final energy of a charged particle coming out of a cyclotron is independent of a) mass of the particle b) charge of the particle c) the magnetic field d) speed of the particle 150. A proton and a deuteron of same kinetic energy enter into a uniform magnetic field. The ratio of radius of the track of deuteron to that of proton will be a) 1:1 b) √2:1 c) 1:√2 d) 2:1 151. An electron enters into a uniform magnetic field with equal components of velocity parallel and perpendicular to the field . The path of electron will be a) a circle b) an ellipse c) a helix d) a straight line making an angle 45o with field 152. In a hollow copper pipe carrying DC, the magnetic field produced will be a) only outside b) only inside c) both inside and outside d) neither inside nor outside 153. A charged particle accelerated by a potential V enters into a uniform magnetic field. The radius of its path will be proportional to a) V b) 1/V c) V2 d) √V 154. Two long parallel straight conductors at a distance ‘a’ carry equal currents i. If the force between them is F, which of the following graphs will be a parabola? a) between F and i b) between F and i2 c) between F and a d) between F and 1/a 155. A beam of electrons move from north to south. A straight conductor is held horizontally with current flowing from north to south below the electron beam. The electron beam will a) be pulled down b) remain unaffected

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c) be pulled up d) be pulled horizontally 156. Three identical conductors AB, BC, CA are joined to form the sides of an equilateral triangle and carry same current as shown in the adjacent figure. The magnetic field due to each conductor at the centroid of the triangle of side ‘a’is X. The total magnetic field at the centroid due to all the conductors will be a) 3X towards the reader b) 3X into the paper c) be zero d) 3X cos 60o towards the reader 157. A helium nucleus makes a full rotation in a circle of radius 0.8 m in 2 s. The value of magnetic induction at the centre of the circle (in T) is a) 10-19/µo b) 10-19µo c) 2 x 10-19/µo d) 2 x 10-10 µo 158. A steel needle of length L has a magnetic moment M. The needle is bent at the mid point so that two parts are at right angles. The magnetic moment now will be a) M b) M/2 c) 2M d) M/√2 159. A uniform circular loop has a resistance of 4 ohm. It is connected to a battery as shown in fig. The ratio of resistance of smaller and larger arc is 1:3. When current is passed as shown, the magnetic field at the centre due to the longer arc ABC to the smaller arc AC a) will be in the ratio 1:3 in the same direction b) will be in the ratio 3:1 in the same direction c) will be in the ratio 1:3 in the opposite direction d) will be zero 160. A tangent galvanometer has a deflection of 45o, for a given current i when the coil is in magnetic meridian. If the coil is inclined at an angle 10o with the meridian, the deflection for the same current will be a) more than 45o b) less than 45o o d) cannot be said from the data c) 45 161. The period of oscillation of a magnet in a uniform field is T. If two such identical magnets are placed one over the other so that the like poles point in the same direction, the period in the same field will be a) T/2 b) T c) √2T d) T/√2 162. The magnetic field at a point (P) distant r from one end of a long straight conductor carrying a current is µ i µ i a) o b) o 2πr 8πr µ oi µ i d) o 4πr πr 163. *Check which of the following statements are correct regarding diamagnetism: a) Diamagnetism is present in all the substances b) Diamagnetism can be explained from Lenz’s law. c)

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c) Diamagnetic effect increases with temperature d) Diamagnetic effect of an atom depends on the radius of electron orbits 164. Two long straight conductors P and Q carry currents i and 2i respectively in the same direction. A third conductor carrying a current i is kept midway between P and Q. Then R a) will remain in equilibrium b) will move towards P c) will move towards Q d) will oscillate between P and Q 165. Which of the following is out of place here with respect to conservation of energy principle ? a) Lenz’s law b) Kirchoff’s loop theorem c) Recoil of a gun d) Bernoulli’s theorem 166. A war plane is flying in the region of magnetic equator. In which of the following cases will there be no induced emf on the wings of the plane? a) when flying vertically b) when flying horizontally c) when climbing up at an angle d) when coming down at an angle 167. A constant current is flowing through a straight conductor from left to right due to a source of emf. When the source is switched off a) an induced current will flow from left to right b) an induced current will flow from right to left c) no induced current will flow d) cannot be said from the data 168. .The flux linking with a circuit is given by φ= t3+3t-7. The graph between induced emf and time will be a) a straight line through origin b) a straight line but not through origin c) a parabola through origin d) a parabola but not through origin 169. A straight wire is held by an insulating handle and moved across a uniform magnetic field . Then a) an emf is induced in the wire b) a current is induced in the wire c) a charge is induced in the wire d) all the three are induced in the wire 170. If the induced emf and the inducing emf are in the same direction, it will violate a) the law of conservation of momentum b) the law of conservation of charge c) the law of conservation of current d) the law of conservation of energy 171. In order to move a conductor in a non-uniform magnetic field with a velocity 2 m/s, energy is spent at a rate 20 W. The force acting on the conductor is a) 10 N b) 5 N c) 20 N d) 40 N 172. An A.C of frequency 50 Hz and rms value √2 A flows through the primary of a transformer. If mutual inductance between primary and the secondary is 1 H, the maximum induced emf in the secondary is (in volt) a) 280 b) 400 c) 560 d) 140 173. Two similar circular coaxial loops carrying equal currents in the same direction approach each other along their common axis. Then a) the current in each loop will increase

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b) the current in each loop will decrease c) the current in both the loops will remain the same d) current in one loop will increase and that in the other loop will decrease 174. A current is sent through a vertical spring carrying a weight, from top to bottom. The spring a) will stretch more b) will remain as such c) will contract d) will break 175. The self inductance of a coil depends on a) change of flux through the coil b) rate of change of flux through a coil c) cross sectional area of the coil d) number of turns in the coil 176. When a DC is passed through an inductor of self inductance L, the energy stored in it is a) 1/2 Li2 b) Li2 c) 2Li2 d) 0 177. A circuit has a self inductance of 1 H and carries a current of 2A. To prevent sparking when the circuit is broken, a capacitor which can withstand 400 volts is used. The least capacitance of the capacitor is a) 12.5 µF b) 50 µF c) 25 µF d) 100 µF 178. The maximum mutual inductance between two coils is 6 H and the difference in their self inductance is 5 H. Then, the sum of their self inductance is (in H) a) 11 b) 13 c) 20 d) 22 179. *If L,Q,R represent inductance, charge and resistance respectively then the units of a) QR/L will be that of current b) Q2R3/L2 will be that of power c) QL/R will be that of current d) Q3R2/L will be that of power 180. A coil of metal wire is stationary in a strong non-uniform magnetic field . Then a) no e.m.f. is induced in the coil b) a varying e.m.f. is induced in the coil but no current c) a constant e.m.f. and a current are induced in the coil d) a constant e.m.f is induced in the coil, but no current. 181. The e.m.f. across the secondary of a transformer will not depend on a) voltage across the primary b) spacing of the primary turns c) core of the primary d) resistance of the secondary 182. Ohm per henry has the dimensions of a) time b) current c) angular velocity d) magnetic flux 183. An AC voltage E = 200√2 sin (100t) is applied to a 1µF capacitor through an AC ammeter. The reading of the ammeter will be (in mA) a) 10 b) 20 c) 40 d) 80 184. The phase angle between current and voltage in an AC circuit is tan-1 (4/3). The power factor of the circuit is a) 3/4 b) 3/5 c) 4/5 d) 0 185. A coil and a magnet move along a common axis as shown in the adjacent figure. In which of the following cases will there be no induced emf in the coil? a) When both move with same velocity in same

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direction as shown in figure b) When both move with the same velocity in the opposite direction c) When the coil moves with a velocity v while the magnet is at rest d) When the magnet moves with a velocity v while coil is at rest. 186. An AC represented by the equation e = 282 sin (100πt) is applied to the primary of a step up transformer whose turns are in the ratio 1:2. If an AC voltmeter is connected across the secondary, the reading of the voltmeter will be a) 350 V b) 282 V c) 564 V d) 400 V 187. An AC voltage represented by the equation e = 200 sin(100πt) is applied to a circuit which contains a pure inductor. The time lag between voltage and current in the inductor will be ( in s) a) 1/200 b) 1/100 c) 1/50 d) 1/25 188. The equation for energy density of magnetic field is a) (1/2) µoB2 b) B2/2µo c) B2/µo d) µoB2 189. A conductor of length L and resistance R kept in a uniform magnetic field B is removed from the field with a velocity v. The power spent in this process is a) B2L2v2/R b) BL2v2/R c)B2L2v2/2R d) BLv/R 190. Three identical coils A,B,C are placed with their planes parallel as shown in the adjacent figure. The coils A and C carry currents as shown in figure. The coils B and C are fixed while the coil A moves with uniform velocity towards B. Then a) no current is induced in B by the two coils. b) a clockwise current is induced in B. c) equal currents are induced in B by the two coils which cancels d) an anti-clockwise current is induced in B 191. *An AC voltage of angular frequency ω is applied to a circuit which consists of an inductor of inductance L and a capacitor of capacitance C in parallel. Then across the inductance a) current is maximum when ω2 = 1/LC b) voltage is maximum when ω2 = 1/LC c) current is minimum when ω2 = 1/LC d) voltage is minimum when ω2 = 1/LC 192. In an LCR circuit given below an AC voltage 200V, 50 Hz is applied. V1, V2, V3 are three voltmeters connected across the resistor, the inductor and the capacitor. A is an ammeter. At resonance the reading of V2 is 300 V. Then the reading of V1 and V3 are respectively a) 200 V, 100 V b) 300 V, 100 V c) 200 V, 0 V d) 200 V, 300 V 193. If the resistance R = 200 ohm in the above circuit, the reading of the ammeter A at resonance will be equal to (in ampere) a) 1 b) 1/4 c) 1.4 d) 2.8

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194. If in the above circuit when not in resonance, the volt meter V1 reads 40 V, V2 reads 50 V and V3 reads 20 V. The source voltage is a) 110 V b) 10 V c) 50 V d) 70 V 195. An AC circuit has voltage and current represented respectively by E = 100 sin(100t) V and I = 100 sin (100t+π/3) mA . The average power dissipated in the circuit is a) 0 W b) 5000 W c) 5 W d) 2.5 W 196. A 1.0 µF condenser is charged to 50 volts. The charging battery is then disconnected and a 10 mH coil is connected across the capacitor so that it oscillates . What is the maximum current in the coil ? (Assume that the circuit contains no resistance) : a) 0.25 A b) 0.50 A c) 0.75 A d) 1.00 A 197. When an AC voltage of rms value 240 V is applied to a circuit having a pure inductor, the current through the circuit is 3 A. When the same voltage is applied to a circuit having a pure resistor, the current is 4 A. If the same voltage is applied to a circuit having this resistor and inductor in series, the current through the circuit will be ( in A) a) 3.5 b) 7 c) 1 d) 2.4 198. A capacitor, an inductor are connected to an electric bulb with a source of AC. If the frequency of AC supplied is increased from a small value, then the brightness of the bulb a) increases b) decreases c) remains the same d) first increases and then decreases 199. If x is the relative permeability and y is the relative permitivity of a medium, the refractive index of the medium is equal to a)

x y

b)

1 xy

c) xy

d)

y x

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SECTION 2: ANSWERS 1

(b)

2

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3

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(a),(c)

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(b),(d)

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1A

120

0.3 A

121

6/11A

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0

123

1.25x1019 124

(a)

125

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(b),(c)

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(a),(b),(d) 164

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(c)

SECTION 3 SOLUTIONS 1. 2.

3. 4.

5.

6.

7.

Take the current from the battery as 2i. It divides i each through 2 branches. Applying Krchoff’s law to any closed network containing battery, we have 120i + 60i = 6. i = 1/30 A. Ans.b No current will pass throgh 40 Ω resistor because of the short-circuited path. Add 70 and 50 in series gives 120 Ω. Hence current 6V has to divide between two parallel paths 60 Ω and 120 Ω. Take the current from the battery as 3i for convenience of division. Of this, ‘2i ‘ flows through 60 Ω and ‘i ‘ through 120 Ω. By Kirchoff’s law for a closed path (30x3i) + (60x2i) + (30x3i) = 6. i = 1/50 A. Current through 60 Ω resistor = 2i = 0.04 A. Ans.c Since the wire is uniform, the two sections between mid points A and B of the sides have resistance 2 Ω each. They are in parallel. The effective resistance would be 1Ω . Ans.c When the wire is cut into four equal parts, each part will have a resistance R/4. When these four are joined in parallel the resistance will be R/16. Heat produced = E2/(R/16) = 16 E2/R which is more than E2/R and more than any of the other alternatives. Ans.c By Faraday’s second law of electrolysis, the masses of silver and zinc deposited will be proportional to the respective equivalent weights. Equivalent weight of silver is 108 and that of zinc 32. These are nearly in the ratio 3.5:1. Ans.b L is the latent heat of vapourisation of water, the heat required for producing 1 g of steam. L = 540 cal = 540 x 4.2 J. Energy supplied = 1080 J/s. Time to boil 100 g of water = (540 x 4.2 x 100)/1080 = 210 s. Ans.d Let E be the voltage of the source. The heat produced in the two wires is E2/R, E2/3R respectively. The ratio of heat produced is 3:1. Ans.a

Here the two parts of the coil having resistance 12Ω and 4 Ω are in parallel. The effectrive resistance is 12x4/16 = 3Ω. Current = emf/total resistance = 12/(3+1) = 3 A. When this current branches into two parts of the coil having resistance 12 Ω and 4 Ω, the current passing through smaller resistance part of the cloil is = total current x 3/4 = 3x(3/4) = 2.25 A. Ans.b 9. Electric field strength = 0.2 Vm-1. Length of the wire = 5 x 10-2 m. Therefore required potential difference across the ends of the wire = 0.2 V x 5 x 10-2 = 10-2 V. Ans.d 10. If the combination has the same resistance at all temperatures, the rate of increase of one should be equal to the rate of decrease of the resistance of other. Germanium is a semiconductor. So its resistance should decrease on heating.. Therefore the other should be aluminium. Ans.b 8.

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11. Let B be the internal resistance of the cell. The current through the circuit is X/(R+B). The potential difference across the resistance = XR/(R+B). This is given as Y. Equating XR/(R+B) = Y, we get B = (X-Y)R/Y). This can also be answered directly from theory notes short-cuts. Ans.d 12. Let R be the external load. Current = 2/(1+R) = 1 A. Thus the load = 1 Ω. When two batteries are in series, emf becomes 4 V and total resistance is 2 (internal) + 1(external) = 3Ω. Current = 4/3 = 1.33 . Ans.d 13. There will be attractive force between wire and nearer side of the loop, repulsive force between wire and farther side of the loop. There will be no force between wire and perpendicular sides. Since attractive force is greater, this side being nearer, net force will be attractive. Ans.c 14. Magnetic field at the centre = µoni/2a. When wire is bent into three loops, field becomes 3 times (Bα n). Radius becomes 1/3. Due to this field becomes 3 times (Bα 1/a). Thus field becomes 9 times. Ans.b 15. In a uniform magnetic field torque due to the field = iAB. When this torque is equal and opposite to that produced by weight of the loop about edge, the edge will start to lift. Therefore mgr = iAB = i πr2 B. i = mg/π rB. Ans.b 16. The magnetic field produced by current in the four edges of one face of the loop at centre will cancel with the magnetic field produced by the currents in the four edges of opposite face. Thus the net field will be zero. Ans.a 17. Applying Fleming’s left hand rule with forefinger from south to north (earth’s field), middle finger from east to west (opposte to electron motion-current), the thumb will point down. Ans.d 18. According to Biot-Savart law, magnetic field = µo i dl sin 900 /4π r2. Here dl = length of the loop = r x θ. Substituting this value, we get field = µoirθ/4πr2 = µoiθ/4πr. Ans.a 19. The magnetic field due to a long straight conductor is given by B = µ 0 I / 2πa That is field B is proportional to 1/a. The graph between two inversely proportional quantities will be a rectangular hyperbola. Ans.d 20. When the magnetic field is perpendicular to the plane of the loop, area vector is parallel to the field. (Recall area has direction along the normal to the loop). The force depends on AxB and will be zero. Ans.c 21. The radius of a charged particle in a uniform field is given by r = mv/qB, where p is momentum. Here p, q and B are same. Hence radius will also be same. Or you can use indirect theory notes short-cuts. Ans.d 22. Use theory notes short-cuts. Ans.a 23. The radius of a charged particle in a uniform magnetic field is r = mv/qB = 2mE /qB, where p is momentum, K kinetic energy. The mass of a particle is 4 times that of proton and its charge is 2 times that of proton. So the radius will be same as that of proton. Ans.a 24. The magnetic field produced by a circular loop at centre = µ 0 ni/2a. The field will be along north-south in the horizontal plane when the coil is perpendicular of magnetic meridian. The horizontal component of earth’s magnetic field Bo will be parallel to it. Thus the two field add. There will be no deflection of magnet in compass box. Ans.c

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25. v x B is a vector perpendicular to the plane of v and B. Thus F is always perpendicular to v and B. Any angle is possible between v and B. The angle between v and B decides the magnitude of the force, not the direction. Ans.b 26. The field due to the straight portion of the conductor at the centre = µoi / 2πa. The field due to circular portion of the wire is µoi / 2a. These two fields are in the opposite direction. µ i µ i µ i Hence net field = 0 − 0 = 0 [π − 1] . Ans.d 2a 2πa 2πa 27. Magnetic field produced by AB and CD are equal and opposite in direction at the centre. Similarly magnetic field produced by BC and AD are equal and opposite. Hence net field will be zero. Ans.c 28. pplying dl x r rule of Biot-Savart law to find the direction of magnetic field due to a straight conductor, we find the field due to all four conductors at the centre will be pointing into the paper and hence they add. Ans.b 29. The magnetic induction due to a straight wire is inversely proportional to the distance (B = µ0 i / 2πa). When the distance is increased from 4 to 12 cm, magnetic induction decreases to 1/3 of X. Ans.b 30. The magnetic moment of a current loop = inA = inπr2. When the loop is bent as 4 turns, the radius reduces to 1/4. The area becomes 1/16. Area of n loops (n = 4) will be 4 x 1/16 = 1/4 times. Magnetic moment becomes 1/4 of the original. Ans.c 31. The radius of the particles r = mv/qB = p/qB = 2mE / qB . Here q is same. Since the accelerating potential is same, energy E is same. Hence r is proportional to √m. R1 /R2 = m1 / m 2 . m1 /m2 = R12/ R22. Ans.d 32. Am2 is the unit of magnetic moment. Similarly joule/tesla is also a unit of magnetic moment. This can be seen from the equation potential energy = mB cosθ. m has the unit of energy/B = JT-1 Ans.c 33. Torque = mB sin θ. So T1 / T2 = sin 30/sin 90 = 1/2. Here T1 = T. Therefore T2 = 2T. =2x. Ans.c 34. When magnet is cut into n pieces, each piece will have a mass 1/n and length 1/n, so that moment of inertia ( mass x 12/12) becomes 1/n3. Moment of magnet (length x polestrength) becomes 1/n. Period becomes

( 1 / n3 )x n = (1/n) of T. You can also use theory notes short-

cuts. Ans.b 35. Magnetic shielding is protection from external magnetic field. By using a material such as soft iron of high permeability, the lines of force of external field can be confined to the cover. Ans.a 36. Isogonic lines join places of equal declination, while agonic lines join places of zero declination. Look up indirect theory notes. Ans.d 37. A dip needle stands in the direction of earth’s resultant magnetic field. At magnetic equator, earth’s field has only horizontal component, that is, vertical component is zero. Hence it stands horizontal. Ans.d 38. Using the formula, m1/m2 = (T12 + T22) / (T12 – T22), we get m1/m2 = 5/3. Ans.c 39. The susceptibility of a diamagnetic substance is negative. In the given list the only diamagnetic substance is bismuth. Ans.a

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40. Magnetic moment is a vector. The moment of smaller magnet = 3Lp. The moment of larger magnet 4Lp. Adding them as vectors, the sum=

(3pL) 2 + (4pL) 2 = 5pL. Ans.c

41. If V is the vertical component of earth’s field. H horizontal component, the angle of dip δ at the place is given by tan δ = V/H. If the dip is measured when the instrument makes an angle θ with magnetic meridian, it is called apparent dip. Here we have to substitute H cos θ in the place of H. If δ1 is the apparent dip, tanδ1 = V/Hcosθ = tanδ sec θ. δ1 = tan-1 (tan δ sec θ) Ans.b 42. A dip needle reads 900 (or stands vertical) at magnetic poles. Here only vertical component of earth’s field exists, i.e. horizontal component is zero. Hence the total intensity will also be 0.4 cgs units. You can also answer this question from the table given in indirct theory notes. Ans.c 43. The magnetism of a diamagnetic substance is independent of temperature. So we look for a diamagnetic substance, the only one in the list being copper. Ans.a 44. Hysterisis is the property shown by only ferro magnets. The only ferro magnet in the given list is cobalt. Ans.b 45. If the beam of a chemical balance is made of a magnetic material, earth’s vertical component will exert a torque on it and it will deflect losing accuracy. Hence a non-magnetic material such as brass can be used. Ans.a 46. The field due to a short magnet at a distance on the axial line is given by [µo/4π] 2m/d3. The field is proportional to 1/d3. The ratio of field at distance x and 2x will be 23:13 = 8:1. Therefore n, here is equal to 8. Ans.c 47. Electromagnets are temporary magnets. So they need high degree of retention of magnetism. Large area would mean more loss of energy while high coercivity would mean larger field to destroy magnetism, which are not desirable for temporary magnets. Look up indirect theory notes too. Ans.d 48. Ans.d 49. If p is the pole strength, M moment, then M = pL. When bent into a semicircle of radius r, L = πr. r = L/π. The new magnetic moment = pole strength x distance between poles = p x 2r = p x 2L/π = 2M/π. Ans.d 50. When a magnet is in stable equilibrium (parallel to the field) its potential energy –mB. When rotated through 1800, it becomes anti-parallel to the field. Its potential energy then = mBcos1800 = + mB. The work done in the process = difference in the potential energy = 2mB. Ans.b 51. At all places in magnetic equator, angle of dip will be zero. (Note: isoclinic line is the line joining places of equal dip not zero dip.) Ans.b 52. By Lenz’s law induced emf in the ring will oppose relative motion between the ring and the magnet. Hence the magnet will always have acceleration less than free fall acceleration ‘g’. Ans.a 53. Flux linkage in the coil φ = NAB cos θ. Induced emf = -dφ / dt = NABω sin θ. The phase difference therefore will be 900. (i.e.that between cosθ and sin θ) Ans.b 54. The induced emf will depend on all the three except the resistance of the coil. Induced current will depend on the resistance, but not the induced emf. Ans.c

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55. The energy stored in the inductor = (1/2) Li2 is due to magnetic field produced inside the conductor coil when a current passes through it. Hence the energy is stored as magnetic field. Ans.d 56. Induced emf = Blv sin θ. Since it flies along the north-south, it will intercept only vertical lines of force of earth. B = 0.2 x 10-4 T, 1 = 50 m, v = 540 x 5/18 ms-1. θ = 900 gives e = 0.15 V. Ans.a 57. The network is a balanced wheatstone’s bridge of effective resistance 1Ω between points connected to the loop. Induced emf = BLv sin 900 , Induced current = BLv/R , which given as 1 mA . With B = 1 T, L = 0.1 m,R = resistance of the loop (1Ω) + resistance of the network (1Ω) = 2Ω, we get V = 2 x 10-2 m/s. Ans.b 58. e = LdI / dt = 20 x 10-3 x 5 /1 = 0.1V. (The resistance of the coil is not necessary here because we are finding emf. ) Ans.a 59. Both induced emf and inducing emf are energies. Only if they are in the opposite direction the total energy will be conserved. Hence it is the consequency of law of conservation of energy. (Also given in indirect theory notes). Ans.d 60. Magnetic lines of force produced by flow of electron (current) are not intercepted by the loop, because they are in the same plane. Hence d φ/dt = 0. Ans.a 61. Applying Fleming’s right hand rule fore finger in direction of the field, middle finger in the direction of the current, the thumb will point into the paper, that is vertically downward. Ans.d 62. Using the formula energy of an inductor E = (1/2) Li2 we can see the unit of self induction can also be swritten as E/i2, J/A2. Ans.c 63. The equation to induced emf of a fan leaf is similar to equation to induced emf between the two ends of a conducting rod in a uniform magnetic field. It is (1/2) B12ω. Ans.c 64. Self inductance L = µoµrN2 A/l. Since N/l is constant, we rewrite the equation as L = N2 µ 0 µ r 2 Al . When length and breadth are doubled A becomes 4 times, length l becomes l twice. So, self inductance L becomes 8 times, because N2/l2 is constant. Ans.d 65. The coefficient of mutual induction does not depend on the current or rate of change of current in the primary. (Note the equation of M = µoµr NpNs/L). Only the induced emf in the secondary depends on the rate of change of current in the primary. Ans.d 66. Inductances add according to the rule of addition of resistors. 2 and 4 in parallel will give 2x4/6 = (4/3) H. This in series with 2/3 gives (4/3) + (2/3) = 2H. Ans.a dI1 dI and L 2 2 dt dt dI 2 dI1 dI 2 = I2 L 2 . It given = . dt dt dt

67. Let induced emf in the two coils at the instant of time t be respectively L 1 Power = voltage x current is same. That is I1 L1 Hence I1 L1 = I2 L2 . Therefore,

dI1 dt

I1 L 2 2 = = . Ans.b I 2 L1 8

68. Since magnetic field perpendicular to the plane and the loop moves in its plane, the flux intercepted by the loop changes. Using palm rule induced emf will be BLv and induced current BLv/R anticlockwise.. Ans.b 69. Use indirect theory notes. Ans.b

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70. Farad = coulomb / volt and not joule/volt. Hence (c) is wrong. All others can be checked and found to be correct remembering L/R , CR and LC have dimensions of time. Ans.c 71. e = -dφ / dt = -(24t +4). At the time t = 1/4 s induced emf will have a magnitude (24 x 1/4) + 4 = 10 V. The induced current will be induced emf / resistance = 10/10 = 1 A. Ans.d 72. The average power in an A C circuit = Erms Irms cos φ, where φ is the phase difference between current and voltage. Here φ = 900. (Phase difference between cos ωt and sin ωt) therefore cos φ = 0. Ans.c 73. Voltmeters read rms value of A.C voltages. In AC circuit voltages add according to vector rules. There is no phase difference between current and voltage in a resistor. Current lags 900 in an inductor and leads 900 in a capacitor. The vector sum is of voltages will be

V1 2 + (V2 − V3 ) 2 = V2. Ans.c 74. In an AC circuit instantaneous voltages follow Ohm’s law. So we can write just as for DC circuit, V1 + V2 + V3 = V. Ans.a 75. When the current is switched on the induced emf in the coil will oppose the growth of the current due to self induction. So the bulb will not glow immediately. When current becomes steady it glows dim because it gets only 6 V when it needs 10. Ans.b 76. During decay of current the induced emf in the inductor will oppose decay of current by Lenz’s law and momentarily shoots up current, and then goes off. Ans.d (Note:These two questions are drawn from experimental demonstration of of self induction). R 2 + ( X L − X c ) 2 where XL , Xc are inductive and capacitative reactance. Here R = 60, XL Erms/Z = 200/100 = 2 A. Ans.a

77. Impedance Z =

78. Refer to theory note tabales.. Tan φ = XL – Xc/R = (180-100)/60 = 4/3. Ans.b 79. Power factor = cos φ = R/Z = 60/100 = 0.6. Ans.c 80. Average power Prms = Erms Irms cos φ = 200 x 2 x 0.6 = 240 W. Ans.d 81. An electro magnetic wave is produced by a tuning circuit which consists of an inductor and a cpacitor. Here is exchange of inductor’s magnetic energy 1/2 Li2 and condensor’s electrostatic energy 1/2 CV2 produces electro-magnetic waves. Ans.c 82. In an AC circuit, voltages add according to a vector rule. Voltages across inductance and capacitance oppose. Hence all that we can say is VR2 + (VL – Vc)2 = V2 where VR . VL, Vc and V are the voltages across the resistor, inductor, capacitor and the supply voltage respectively. Hence it is possible for VL or Vc to exceed V. Ans.d 83. The energy of a photon is given by Planck’s law E = hv = hc/λ. The blue light has less wave length and hence gives photons of more energy. If P is the power of the source, n is no. of photons emitted per second, we have nhc/λ= P. Since nhc/λ is constant, the number of photons is more for light of larger wave length, i.e. red. Hence red light gives photons of less energy but more in number. Ans.b 84. To compare the phase difference between voltage and current, we should have both voltage and current either as sine functions or cosine function. So we rewrite E = 100 sin [ωt + π/2 + π/3] = 100 sin {ωt + 5π/6]. Comparing this with current I = 4 sin ωt, we find phase difference is 5π/6. Ans.b 85. The peak values of AC, E0 = 282 V. Rms value of AC Rrms = Eo/√2 = 282/1.41 = 200 V. Irms = Erms / resistance = 200/20 = 10 A. AC instruments show rms value. Ans.c

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86. The effective inductance between them is given by L = L1 + L2 ± 2M, where M is the mutual inductance between them. When the coils are far apart, mutual inductance between them becomes zero. Then only the self inductance becomes L1 + L2 . Ans.a 87. Scattered intensity is proportional to 1/λ4. Of the given electro-magnetic radiation radio waves have longest wavelength. Hence they are scattered least. Ans.d 88. Using the Rayleigh’s scattering law Iα 1/λ4 , λ (blue)/ λ (red) = 440/660 = 2/3. I(red)/I(blue) = [λ(blue)/ λ(red)]4 = (2/3)4 = 16/81 = 1/5 . Ans.d 89. The maximum distance ground waves can reach by an antenna of height h = 2Rh , where R is radius of earth. Here h = 0.5 km. This gives the maximum distance as 80 km. Ans.a 90. The capacitative reactance Xc = 1/ωC = 1/2πf C. Thus Xc is inversely proportional to f. The graph between them will be rectangular hyperbola. Ans.d 91. Any point on the Y-axis is on the equitorial line of the dipole. Hence field will be parallel to the axis. (a is correct). Electric field at a point between the charges will be along positive Xaxis, but outside along negative X-axis (b is wrong). Dipole moment = -2pd i, as it is negative to positive charge. (c is wrong) work = potential difference x charge. Since potential at infinity and origin are zero, work will be zero. (d is wrong). Ans.a

µ 0 Idl

by Biot-Savart law. Since resistance is proportional to 4πr 2 length, flux density = constant x I1R1 /r2 for one arc for the other arc constant x I2R2/r2. As the potential difference across the two paths are same I1R1 = I2R2, The fields will be same in magnitude. Since currents are in the opposite direction, the fields ancel. Ans.d

92. Flux density the centre =

93. We have here two semi circular coils of radii R1 and R2. The field due to one µoi/4R1 ( half of a coil of one turn) and due to the other µoi/4R2. Since the currents are in the opposite directions field oppose. Ans.c 94. Since the proton is projected at an angle (not parallel or perpendicular) to the field its path will be a helix. Hence choices a and b are wrong. Time period (independent of velocity) = 2πm/qB = 2πx1.67x10-27 / 1.6 x 10-19 x 0.104 = 6.3 x 10-7 s or 2πx10-7s. Ans.c 95. The radius of a charged particle in a uniform magnetic field r = p/qB = is the kinetic energy and p momentum. Thus we find r α to 1/r and hence +

q/m

2mE / qB , where E

m / q Deflection is proportional

+

Deflection of H is proportional to q/√1. (choice a) Deflection of

He is proportional to q/√4 = q/2. Deflection of O2+ is proportional to 2q/√16 = q/2. choice c). Ans.a & c 96. The shunt resistance S to make a galvanometer an ammeter is given by S = Ig /I-Ig, where Ig is the current through galvanometer I is current through the main circuit, G galvanometer resistance. To convert into an ammeter we have to add a resistor in parallel. Since the answer carries 1 Ω we substitute S = 1Ω, Ig = 50 µA, G = 100 Ω, we get I = 5 mA. (choice c). To convert into voltmeter we have to connect a high resistor in series. If R is the value of the resist or in series R = (V/Ig) – G. Here substituting V = 10 volt, Ig = 50 µA, G = 100 Ω, we get R = 200000 Ω, i.e. 200 kΩ. (choice. b) Ans. b & c 97. Copper is a metal and its resistance decreases on cooling. Germanium is a semi conductor. Its resistance increases on cooling. Ans.d 98. Redrawing thediagram we find the arrangement is a balanced Wheasone’s bridge. The effective value of resistance between A and B is RAB = 3R x 6R/9R = 2R. Maximum power is

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139

transfered when internal resistance is equal to external resistance . That is 2R = 4, R = 2 Ω. Ans.c 99. Irms =

1 4 2 ∫ I dt 2 2

= 3 A. Ans.a

100. Use the information supplied in theory notes. Ratio of two equal resistors when connected in series and in parallel is n2. Here n = 2. Hence ratio of resistances in series and parallel will be 4. Heat produced = E2/R. The required ratio will be 1:4. Ans.d 101. Electric flux by Gauss’s theorem φ = E ds cosθ. Here the area intercepting the lines of forces which is parallel to the axis of the circular plane is the surface area of the hemisphere. Therefore φ = 2πR2E. Ans.b 102. The current through the arrangement, which consists of n cells each of emf E and internal mnE resistance r connected in series with m rows in parallel is , where R is external mr + nR resistance. The current will be maximum if mr = nR. Here r = 0.5 Ω and R = 3Ω. This gives m = 6n. Since mn = 24, total no. of cells we have m = 12 and n = 2.. Ans.a 103. Kinetic energy of a charged particle in an electric field = Charge x potential difference = charge x electric field x distance if the field is uniform. Here it is qEy. Ans.c 104. The torque on a dipole when it makes an angle θ = pEθ, when θ is small. This restoring torque should be equal to 1α. Thus 1α = -pEθ. α = -(pE/I)θ. The angular acceleration is proportional to angular displacement. The motion is S.H.M. The period is 2π I / pE This answer can also be found from similarity with period of magnet in a uniform field 2π I / mB . In the place of a magnetic field B, electrif field E will come. Ans.a 105. Current through one bulb = 60/220 A. If n is maximum number of bulbs that can be used, then n x 60/220 = 9. This gives n = 33. Ans.d 106. Thermo electric power = dE/dT = emf/temperature difference. Ans b 107. Use the information supplied in indirect theory notes. When the wire is halved in length the resistance becomes 1/4. Ans.b 108. Using the equation F = q vxB with q = 1 C, we have the force as –3i. Ans.d 109. ML2T3 is the dimension of power. Hence the given quantity is power/current2 = resistance. Ans.d 110. We start from Ohm’s law j = σE. Resistivity = 1/σ = E/j = electric field / current density.

Ans.a 111. Let the voltage across the source be E and resistance of each bulb be R.

Then power across the bulbs will be E2/4R. This is given as P. When one of them is removed, the power consumed will be E2/3R. Since E2/3R = ( E2/4R) x (4/3), it will be equal to (4/3)P. Ans.d 112. A standard resistor should not change resistance. So it should have low temperature coefficient. It should not melt quickly due to the heat produced in it. Hence it should have high melting point. Ans.d 2 2 113. Since the masses and hence the volumes are equal πr1 L1 = πr2 L2.

2

2

2

2

r R ρL ρL L r r r L 1 r2 = 2 . The ratio 1 = 21 / 22 = 1 x 2 2 = 2 2 x 2 2 = 2 R 2 πr1 πr2 L 2 r1 L2 r1 r1 r1 r1 = 16. Ans.d

4

From this we get

2 .This is equal to 1

4

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114. A secondary cell has less internal resistance compared to a primary cell. Hence it gives more

current. Ans.c 115. Using two resistors we can get 4 resistance values. They are each of their value, both of them

in series and both of them in parallel. In a series connection the resistances add . So you should choose two values given in the question whose sum is the third value given there. Hence the resistances should be 18 and 9. 18 and 9 in series gives 27 and in parallel gives 6 Ans.a 116. If E is emf of each cell and r its internal resistance, when connected in series current i = 4E/4r = E/r. When one of the cells is reversed the emf is equal to E+E+E-E = 2E. However internal resistances still add to 4r. The new current i1 = 2E/4r = (1/2)i. The current reduces by 50%. Ans.b 117. When n ampere enters the combination, 1 ampere should flow the ammeter while n-1 ampere should flow through the shunt. If S is the resistance of the shunt, then (n-1)S = 1xG. ⇒ S= G/n-1. Ans.b 118. The increase in resistance when it is heated through ∆T is given by R1α1 ∆T. For the other

resistor it should be R2α2 ∆T. These should be equal for all ∆T. This makes (b) correct and (a) wrong. If the combination in series should have same resistance at all temperatures, the resistance of one should increase by the same amount as that of the decrease of the other. So one should be a metal and the other a semi-conductor. This can also be answered from indirect theory notes. Ans. b & d 119. The network here is repeated network . Reduce it starting from left end. The value will be outer resistance 2 ohm. Current is equal to emf / resistance = 2/2= 1 A. 120. Use the information given in theory notes. Current through parallel network divides in the ratio of reciprocal of resistance. So divide 1.1 A in the ratio (1/1): (1/2): 1/3) = 6:3:2. Current through 2 ohm resistor equals 1.1 x (3/11) = 0.3 A 121. Use the information current divides in the reciprocal ratio of resistance. Take the current from 6V battery as 3i. Of these 2i passes through 1 ohm and i through 2 ohm. By Kirchoff’s law 1x2i+3x3i = 6. Solving, i = 6/11 A 122. The arrangement between P and Q (that is points connected to battery) is a balanced Wheatstone’s bridge. 3 ohm here is in the place of galvanometer, while 2 ohm resistors are 4 arms of the bridge. Since the current through galvanometer is zero in a balanced bridge, i =0. 123. The arrangement is three one ohm resistors in parallel. They are each half of the loop and diametrical wire AB. Take the current as 3i. Applying Kirchoff’s law to a closed loop containing i = 2A. Use the information supplied in theory notes. The number of electrons will be 2x6.25x1018 = 1.25x1019. 124. The balancing length L of a cell of emf E1 is given by E1 = irL, where r is the resistance per unit length of potentiometer wire and i current through it. When driving cell has more internal resistance, current i decreases and for a given E1, L increases. Ans.a 125. Use the information supplied in theory notes. We look for wire of maximum resistance. Resistance R = ρL/πr2. Resistance is maximum for the wire which has L/r2 maximum. Their values are 1/1, √2/2, √3/3, 2/4. Hence it is maximum for the wire P. Ans.a 126. Usually in a circuit consisting of one cell the terminal voltage cannot exceed emf. However if the circuits has two or more cells connected in opposition, the terminal voltage of a cell of smaller emf can exceed its emf. Hence statements b,c are correct (a) and (d) are wrong. [There is a question which follows to illustrate this ]. Ans.b,c.

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141

127. The current through the circuit is equal to net emf / total resistance = (12-8)/(1+2+9) = (1/3)

A. The drop across E1 = current x internal resistance = (1/3)V. So terminal voltage is equal to 8+(1/3) = 8.33V. Ans.b (Note:-The drop is added with the emf because current flows from positive to negative inside the cell. This is due to the presence of a cell of higher emf 12V) 128. The drop across E2 = current x internal resistance = 2/3 V. Terminal voltage = 12-(2/3) = 11.34V. Ans.c 129. The potential difference across the points P and Q is the same as the potential difference across the resistor 9 ohm, because the point P is one end of the resistor and Q the other end. The required voltage = currentxresistance = (1/3)x9 = 3 V. Ans.d 130. Whatever may be the value of R, the potential difference across the 3 resistors 1,2,3 ohm are same. Let this potential difference be V. It is given V2/3= H. We are asked V2/1. This is 3H. Ans.d 131. Since the lamp lights with maximum brightness, maximum power is transferred from the source to the lamp. This happens when resistance of the lamp is equal to internal resistance of the source (Maximum power transfer theorem)i.e. 4 ohm Ans.a 132. (Fig) .Looking carefully into the network, we can find that it has symmetry. The upper and lower parts are two balanced Wheatstone’s bridges. Since the current through galvanometer is zero, remove those 2 ohm resistors at centre and redraw the figure as shown in figure above. This reduces as shown in adjacent figure. Here three resistors 4,2,4 are in parallel. The effective resistance R is given by 1/R=(1/4)+(1/2)+(1/4), which gives R = 1 ohm Ans.a 2 133. Since all the resistors are in parallel, use the equation for heat produced V /R per second. Let 2 V /200 = H. When all the resisitors are connected in parallel, the effective resistance (adding them) is 40 ohm. The heat produced now will be V2/40 per second which is 5H. So time to boil same amount of water will be (1/5) of 5 minutes. Ans.b 134. Unless you read carefully, you will miss the correct answer. Since the rate of flow of charge is constant, current is constant. Ans c 135. The resistivity of a wire depends on its material. Note that the resistivity does not depend on the length or cross sectional area It is the resistance which depends on them. So (c) is correct while (a) and (b) are wrong. The resistivity also depends on the temperature, which makes (d) correct. Ans. c,d 2 2 136. The resistance of the bulb R = V /P = 100 /500 = 20 ohm. When 150 V is supplied 100 V should be across the bulb and 50 V should be across a series resistor. In a series connection potential difference across a resistor is proportional to the resistance. Hence the value of the resistor should be half of that of the bulb = 10 ohm. Ans.d 2 137. Use information supplied in theory notes The resistance of the wire becomes 2 = 4 times. The conductance of the wire which is reciprocal of resistance becomes 1/4. Ans.c 138. If E is the emf of each cell and r internal resistance i =

3E . When in parallel i will be 10 + 3r

E . Equating the two and solving r = 10 ohm. Ans.c 10 + ( r / 3)

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139. If v is the drift speed of electron, current i = nev. Multiplying both sides by m, mi = mnev =

e(nmv). Here nmv is a total momentum p of all electrons. This gives p= mi/e = i/(e/m)= i/K, where K is the specific charge e/m of the electron. Ans.c 140. Use the information supplied in theory notes. Both electrons will have same period and hence the same frequency. Ans.a 141. Use the information supplied in theory notes The speed of the particle = E/B = 1000/20 = 50. Its momentum is equal to mass x velocity = 50m. Ans.b 142. When a particle is accelerated, its velocity increases. According to the theory of relativity, its mass also increases. The mass increase is more for lighter particle electron as it can be accelerated to close to the speed of light. The period T = 2πm/qB does not remain constant and hence the resonance condition will not be satisfied. Ans c. 143. A current loop is like a magnet. In a uniform magnetic field a magnet experiences only a G G G torque. No force Torque τ = i A x B . When plane of the loop is parallel to the field, the area vector is perpendicular to the field. Hence τ = iAB sin 90o. Ans.a 144. Here angle between area vector and field vector is 0. Hence

τ = iAB sin0o = 0. No torque,

no force. Ans.d 145. Since the lengths are in the ratio 1:2, the radii of loops will be also in the ratio 1:2. Areas will be in the ratio 1:4. Magnetic moment = current x area of loop. ⇒ Ratio of moments 1:4. Ans.c 146. Use the information supplied in theory notes. The charged particle here has uniform cirular motion. In a uniform circular motion speed is constant. Hence kinetic energy = (1/2) mv2 = constant. But the velocity varies. Hence momentum also varies. Ans.b µ o nR 2 i µ ni . F(centre) = o Dividing, we get F (centre)/ F(axis) = 147. F(axis) = 2 2 3/ 2 2R 2( R + R ) √8R3/R3 = √8/1. Ans.d 148. When an electron moves with a speed v , current i = charge x frequency =

e x (v/2πr) µ o i µ o ev µ o ev ev = Ans.a . Field B = = = 2 πr 2r 2 r 2 πr 4 πr 2 149. Use information supplied in theory notes The final momentum p = qBR. The final kinetic energy = p2/2m = q2B2R2/2m which is independent of the speed. Ans.d 150. Use the information supplied already in theory notes Since E is same, radius of the tracks will Radius (deuteron ) m (d ) q p 2m e x = 2 .Ans.b = = be proportional to the ratio (√m)/q. x m e Radius(proton ) m(p) q d o

151. Since the velocity components are equal the electron should be entering at an angle 45 with

the field. Use the information supplied in theory notes. The path will be a helix. Ans.c G 152. If B is the magnetic field, then by Ampere’s circuital theorem, B. dl = µ o i , where i is the

∫

current enclosed by the path along which integral is taken. If the point is outside , path will enclose a current, and hence the field B will not be zero. If the point is inside, the integral path will not enclose a current. The current i and hence the field B will be zero. Ans.a

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143

153. Use the information supplied

in theory notes. The energy E = qV, where V is the accelerating potential. Since r is proportional to √E, it will be proportional to √V as well. Ans.d

µ 0i 2 . The graph between F and i 2πa 2 will be a parabola, since F is proportional to i . The graph between F and i2 will be a straight line. The graph between F and a will be a rectangular hyperbola while the graph between F and 1/a will be a straight line. Ans.a 155. Since the electron is moving from north to south, it is equal to a current flowing from south to north. Thus the electron current and the current through the conductor are in opposite direction. Hence they repel. The electron beam will be pulled up. Ans.c 156. The magnetic field due to each conductor X has same magnitude. Its direction is given by the vector i dl x r and will be pointing into the plane of the paper . All the fields add in magnitude and direction Ans.b 157. Current i = charge / time = 2e/T, where 2e is charge of helium nucleus. The magnetic field at µ i µ 2e µ x 2x1.6x10 −19 centre B = o = o = o =µ o 10 −19 . Ans.b 2r 2r T 2x 0.8x 2 154. The equation for force between the parallel wires is F =

158. Let p be the pole strength. Then the magnetic moment M is equal to px L. When the needle is

bent the poles are at A and B as shown in the adjacent figure. The length of the magnet is equal to the distance between poles = L/√2. So the new magnetic moment = pL/√2 = M√2. Ans.d o

2

159. The magnetic field at the centre is µoiLsin90 /4πr . = constant x iL, where L is the length of

the arc. Since the loop is uniform its length is proportional to the resistance. So the field becomes constant x iR. For a parallel connection i1R1=i2R2 i.e. iR =constant. So the magnetic field will be same due to the two arcs. But the field will be the opposite direction as the currents are in opposite directions. They cancel. Ans.d 160. For a tangent galvanometer i = K tanθ, where K is 2rBH/µon. When the coil is arranged at an angle 10o with the meridian, the horizontal component of earth’s field BH will only be BH cos 10o. For a constant current i the deflection θ will increase. Ans.a 161. The period of oscillation of a magnet is given by 2π I / MB , where I is moment of inertia of

the magnet. For two identical magnets, one placed over the other as given, I becomes 2I and M becomes 2M. Thus the period remains the same. Ans.b 162. The field at one end point along the perpendicular bisector of a straight conductor is got µ i µ i from the equation o (sinθ1+sinθ2) = o as here θ1 = 0 and θ2 = 90. Ans. c 4 πr 4 πr 163. Diamagnetism is produced due to induced effect when the atom is placed in a magnetic field Hence it is present in all substances. But it shows only in substances where para and ferro effects are absent. (a) is correct. Since it is due to electromagnetic induction it can be explained by Lenz’s law. (b) is correct. It does not depend on temperature. (c) is wrong. Induced magnetic moment depends on induced current and on the radius of the orbit. Hence (d) is correct. Ans.a,b,d 164. The force between parallel conductors carrying current in the

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same direction will be attractive . Since the distance of R is same from P and Q, the force will be proportional to the product i1i2. i1i2 = i2, for P and R and 2i2 for R and Q. Therefore the attractive force of Q will be greater. Thus R will move towards Q . Ans.c. 165. Except recoil of gun all others are one or other form of law conservation of energy. In recoil of gun momentum is conserved . But kinetic energy of the gun is less than that of a shot. So kinetic energy is not conserved. Ans.c 166. In the region of magnetic equator earth’s magnetic field has only horizontal component. So when the plane flies horizontally it cannot intercept magnetic lines of force of earth. Hence no induced e.m.f. In all other cases the magnetic lines of force are intercepted . Ans.b 167. When the source is switched off, the current decreases. By Lenz’s law, induced current should oppose the decrease and hence should flow in the same direction, i,e, from left to right. Ans.a 2

168. The induced emf |e| = dφ/dt = 3t + 3. This is equation to a parabola . When t = 0, e = 3.

Hence the graph will not pass through the origin. Ans.d 169. When a conductor moves across (perpendicular to) the field , an emf is induced in the conductor. Since the circuit is not closed, no current flows through the wire and hence no charge flows through the wire. Ans.a 170. Use the supplied in theory notes . If the induced and inducing emf are in the same direction, it does not obey Lenz’s law. Lenz’s law follows from law of conservation of energy. Ans.d 171. Using the formula force x velocity = power, we have power = 20 W, v = 2 m/s. This gives F = 10 N..Ans.a 172. The peak value of AC = Irms √2 = 2A. The current changes from 2 A (positive maximum) to -

2 A (negative maximum) in half of a cycle, i.e. (1/100) s. So induced emf |e | = M di/dt = 1[2-(-2)]100 = 400 V. Ans.b 173. When loops come nearer, the flux linking with each loop due to the current in the other will increase. That is dφ /dt is positive. The induced emf, therefore -dφ /dt will be negative. That is current in each loop will decrease due to this induced emf. This is in accordance with the Lenz’s law. Ans.b 174. Each turn of the spring can be considered as a loop. So each turn becomes a magnet. It will have opposite poles at top and bottom. One turn will be attracted by the next turn because of the opposite polarity near. Thus the spring will contract. Ans.c

µoµ r N 2A . Thus self inductance depends on A and N/l. l This makes (c,d) correct. The self inductance will not depend on flux or rate of change of flux. It is the induced emf which depends on rate of change of flux. Hence (a) and (b) are wrong. Ans.c,d 176. An inductor stores magnetic energy when current is changing and an emf is induced. A steady current does not produce induced emf. Ans.d 177. Here the magnetic energy of the inductor (1/2)Li2 is converted into electrostatic energy of the capacitor (1/2)CV2. When V is maximum C is minimum. Equating the two C = 1x22/4002 = 25x10-6 F. Ans.c 175. The self inductance L is given by

178. If L1 and L2 are self inductance, maximum mutual inductance M =

L1L2 = 36, L1-L2= 5. L1 +L2 =

( L1 − L 2 )

2

+ 4 L 1 L 2 = 13 H. Ans.b

L 1 L 2 . Here M = 6, so

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145

179. Use the information supplied in theory notes. We know L/R has the dimensions of time .

QR/L will be Q/t = charge / time = current. Q2R3/L2 = (Q2R2/ L2) x R = (Q2/t2 )xR = I2xR, because Q/t = current. I2xR is power. Ans .a ,b 180. To have induced e.m.f there should be a change in the flux linking with the wire, i.e. there should be relative motion. Here the coil is at rest. Ans.a 181. The induced e.m.f. in the secondary of a transformer will depend on the mutual inductance which, in turn, depends on spacing of primary (number of turns per metre), core (µr) of the transformer. The induced e.m.f. will not depend on the resistance of the secondary. Only the induced current will depend on the resistance of the secondary. Ans.d 182. Use the information supplied in theory notes. ohm/henry = R/L, which has the dimensions of reciprocal of time i.e .dimension of angular velocity T-1. Ans.c 183. Comparing with standard equation E = Eo sinωt, Eo = 200√2. Erms = 200 V, ω = 100. Irms =

Erms/Z. Here Z = reactance = 1/ Cω ⇒ Irms = ErmsCω = 200x1x10-6x100 = 20x10-3A = mA. Ans.b

20

184. If φ is the phase angle between current and voltage, it is given here tanφ = (4/3). Power factor

of a circuit = cosφ. cosφ will be (3/5). Ans.b 185. To have an induced emf in the coil, there should be a relative velocity between the magnet

and the coil. When both move with same velocity in the same direction , the relative velocity is zero. There is no change in flux. Hence the induced emf is zero. Ans.a 186. Comparing the given equation with the standard equation e = Eo sinωt, we find Eo

= 282 V. Erms=282/√2 = 200 V. Since the transformer is step-up, the secondary voltage will be 200x2= 400 V. (Recall that the AC voltmeter will show only rms value). Ans.d

187. Here again, comparing with the standard equation Eo sinωt, we find ω = 100π. 2πf = 100π,

which gives the frequency of AC f= 50 Hz. Period of AC = 1/50 s. The time lag of 1/50 second corresponds to a phase lag of 360o. In a purely inductive circuit, the phase lag between current and voltage is 90o. This corresponds to a time lag of (1/4) of 1/50 = 1/200 s. Ans. a

188. The energy density ( energy per unit volume) of magnetic field is given by B2/2µo. Note the

difference in the equation for energy density of electric field (1/2) εoE2. Ans.b 189. The induced emf in the conductor e = BLv. The power when removed from the field = e2/R = B2L2v2/R. Ans. a 190. Since there is a relative motion between A and B a current will be induced in B by the motion of A. By Lenz’s law the current in B will be in an opposite direction to that in A, i.e. anticlockwise. There will be no current due to C, because there is no relative motion (no change of flux) between B and C. Ans.d 2

191. In a parallel resonant circuit, current is minimum at resonance. ω = 1/LC is the condition for

resonance. Hence (c) is correct. Across L the voltage and the current differ by a phase angle 90o. Hence when current is minimum voltage will be maximum. Thus (b) is correct. Ans. b, c 192. At resonance the voltage across L and that across C should be equal and opposite. So the reading of V3 should be 300 V. The supply voltage 200 V should be across the resistor. So V1 will be 200 V. Ans.d 193. Use the information supplied already in theory notes. At resonance inductive reactance and capacitative reactance cancel. Hence the net impedance Z = R = 200 ohm. Current = Erms / Z = 200/200 = 1 A. Ans.a

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194. Use the information supplied in theory notes. The voltages add according to vector rule. i.e.

V=

VR 2 + ( VL − Vc ) 2 = 40 2 + (50 − 20) 2 = 50V. Ans.c

100 100x10 −3 π cos = 2.5 W (Note the current is in mA) Ans.d. x 3 2 2

195. P = Erms Irms cosφ =

2

196. The electrostatic energy of the condenser (1/2) CV is converted into magnetic energy (1/2)

Li2 during discharge. Equating the two we get i = CV 2 / L = 1x10 −6 x50 2 / 10x10 −3 = 0.5 A. Ans.b 197. The reactance X = 240/3 = 80 ohm. Resistance R = 240/4 = 60 ohm. When the two are connected in series, we have an LR circuit. The impedance Z = R 2 + X 2 = 60 2 + 80 2 = 100 ohm. Current i = Erms/Z = 240/100 = 2.4 A. Ans.d 198. This is an LCR circuit. Here R is the resistance of the bulb. When the frequency of AC is increased, the current increases. When the frequency becomes equal to the resonant frequency (Lω = 1/Cω), the current reaches a maximum value. Then it decreases. Since brightness of the bulb depends on the current, it also increases and then decreases. Ans.d. 1 1 c c 199. The speed of e.m waves through a medium c1 = , = = = µε µ0ε0 µ r εr µrεr xy where c is the speed of e.m.waves through free space. From Optics, we know speed through a medium c1 = c/n, (2)where n is the refractive index of the medium. Comparing the two equations(1)and (2) we find n = xy Ans.c

6 OPTICS SECTION 1 QUESTIONS 1.

2.

3.

4.

A diverging beam of light falls on a plane mirror. The image formed by the mirror is a) real, erect b) virtual , inverted c) virtual, erect d) real, inverted In a pond water is 10 m deep. A candle flame is held 15 m above the surface of water. If the refractive index of water is 4/3, the image of the candle flame is formed at a) 15 m deep b) (45/4)m deep c) 10 m deep d) (30/4)m deep Which of the following colours will have minimum critical angle when light passes from glass to air ? a) green b) blue c) yellow d) orange *A ray of light is incident at an angle 450 in an equilateral prism of refractive index √2. The ray a) will totally reflect from the second face. b) will just graze the second surface. c) will run parallel to the incident ray. d) will undergo minimum deviation.

Light of wavelength λ and frequency f is incident from air to glass of refractive index µ . Inside glass the frequency and the wavelength of the light are a) f, µ λ b) µ f, µ λ c) f, λ/µ d) µ f, λ/µ 6. Two convex lenses of equal focal length f are kept at a distance coaxially. A parallel beam of light incident on the first lens emerges from the second lens as a parallel beam. The distance between the lenses is a) f b) 2f c) f/2 d) f/4 7. Imagine a hypothetical convex lens which can pass all the electromagnetic radiations through it. For which of the following radiations will the lens have maximum focal length? a) X-rays b) γ rays c) micro waves d) infra red rays 8. If a is the distance of an object from the focus of a concave mirror and b is the distance of its real image from the focus, the graph between a and b will be a) a straight line of positive slope b) a rectangular hyperbola c) a straight line with negative slope d) a parabola 9. A thin rod of length f/3 is placed along the optical axis of a concave mirror of focal length f such that its image which is real and elongated, just touches the rod. Its magnification is a) 3/2 b) 5/2 c) 1/2 d) 7/2 10. A concave mirror has a focal length of f cm. An object is placed such that it forms a two fold magnified real image. The distance between the object and the image will be ( in cm) 5.

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a) f b) 1.5f c) 2f d) 2.5f 11. Two plane mirrors are inclined to each other at a certain angle. A ray of light first incident on one of them at an angle of 100 with the mirror retraces its path after five reflections. The angle between the mirror is a) 12o b) 22o c) 30o d) 20o 12. If µ 21 represents

13. 14.

15.

16.

17.

the refractive index of the

second medium with respect to the first

medium, then µ 21 µ 32 µ 43 will be equal to b) µ 31 c) µ 14 d) µ 13 a) µ 41 The intensity of light at a distance r from the axis of long cylindrical source is proportional to a) 1/r2 b) 1/r c) 1/r1/2 d) 1/r3 A ray of light incident on an equilateral prism of refractive index 1.41 suffers minimum deviation. The angle of incidence on the first face is a) 60o b) 30o c) 40o d) 45o A ray of light is incident on two plane mirrors inclined at an angle deviates through 240o. The number of images of an object formed by the mirrors is equal to a) 3 b) 2 c) 5 d) 6 A fish is at a depth of 4 m from the water surface. A bird is at a height 3 m vertically above from the water surface. The distance at which the fish sees the bird is a) 7 m b) 9 m c) 8 m d) 10 m A fish under water at a depth h sees the outside world within a cone of radius r at water surface. If µ is refractive index of water, the value of r here is h h a) h µ 2 −1 b) h µ − 1 c) d) 2 µ −1 µ −1

18. A plano-convex lens of focal length f and radius of curvature R is silvered on the plane surface as shown in the adjacent figure The focal length of this will be a) f b) f/2 c) R d) R/2 19. The above plano-convex lens is silvered on the convex surface, as shown in the adjacent figure, so that it becomes a concave mirror. The focal length of this concave mirror will be a) R/2 b) R c) (R/2µ) d) 2R/µ 20. A ray of light passes through three media of refractive indices as shown in the adjacent figure. Then a) µ1 = µ2 , µ3 > µ2 b) µ1 = µ2, µ3 < µ2 c) µ1 = µ2 = µ3 d) µ1 < µ2, µ2 = µ3 21. A plano–convex lens fits exactly with a plano-concave lens as shown in the adjacent figure. Their refractive indices and focal lengths are shown in the figure. The focal length of the combination is

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a)

R 2(µ 2 − µ1 )

c) R

b)

2R (µ 2 − µ1 )

d)

R µ 2 − µ1

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22. Light energy falls at a rate E Wm-2 on a perfectly reflecting plane mirror. The pressure exerted by the light on the mirror is (c speed of light) a) E/c b) 2E/c c) E/3600 c d) E/7200 c 23. A ray of light is incident at an angle 45o into a medium of µ=√2 . A part of light is reflected and a part refracted. The angle between reflected and refracted rays in degrees will be a) 45 b) 75 c) 90 d) 105 24. An object is kept symmetrically between two plane mirrors inclined at an angle 90o. The images formed by the mirror will lie in a) a circle b) a parabola c) a straight line d) an irregular curve 25. A source of light is kept in an isotropic medium. If the amplitude of wave reaching a point is plotted with distance from the source to the point, the graph will be a) a straight line b) an inverse square law graph c) a parabola d) rectangular hyperbola 26. A lamp is suspended vertically above a table at a height. When the height of suspension is reduced by 2%, the intensity at the centre of the table a) increases by 4% b) increases by 3% c) increases by 2% d) increases by 8% 27. Two lamps A and B are placed at a distance 2 m and 3 m from a screen respectively. The intensity of illumination on the screen due to the two lamps are equal. If PA and PB are the powers of two lamps, then PA/PB = a) 2/3 b) 4/9 c) 3/2 d) 9/4 28. The refractive index of a prism depends on a) angle of the prism A b) angle of minimum deviation D c) both A and D d) neither A nor D Solve the following two problems within a maximum time of 2 minute. Use only one equation. 29. In a displacement method the distance between object and the screen is 90 cm. The focal length of the lens is 20 cm. What is the distance through which lens is displaced between magnified and diminished image ? 30. In a displacement method the lens is displaced through a distance 45 cm. The magnification of one of the images = 4. What is the focal length of the lens? 31. Young’s experiment is performed with white light, which of the following colours will have minimum band width ? a) green b) yellow c) blue d) red 32. If Young’s experiment were performed so that interfering beams travel in water. The ratio of width of band in air to that in water i.e. β(air) / β(water) = a) 1 b) 1/2 c) 3/4 d) 4/3 33. Two light beams have same wave vector k. One travels through a distance L1 in a medium of refractive index µ1 and the other L2 in a medium of refractive index µ2. On emerging their phase difference will be b) k(µ1L2-µ2 L1) a) k (µ1L1-µ2L2)

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d) (µ1L2-µ2 L1)/k c) (µ1L1-µ2L2)/k If one of the slits is twice as wide as the other in Young’s experiment, the ratio of intensity of bright band to dark band will be a) 4:1 b) 2:1 c) 3:1 d) 9:1 *.Which of the following cannot produce interference? a) Two narrow sources of blue light and red light b) One real narrow source and one narrow virtual source of same colour c) Two virtual monochromatic sources d) Two different parts of the same source *If the width of both the coherent slits in Young’s experiment is progressively increased a) the width of bands will increase b) the width of bands will decrease c) the contrast between dark and bright bands will decrease d) the bands will disappear after some time The relation between critical angle 'C' and angle of polarisation of a medium 'i' is a) i = tan-1(cosec C) b) i = cot-1 (cosec C) -1 c) i = sin (tan C ) d) i = tan-1(sin C)

38. A laser beam of wave length λ passing through a slit of width ‘a’ is sent to moon. The linear spread of the beam when it reaches to moon is a) Da/λ b) Dλ/a c) Da/2λ d) Dλ/2a 39. When the laser beam given in previous question reaches moon, the area to which it spreads is equal to a) D2a/λ b) D2λ2/a2 c) ( Da/2λ)2 d) Dλ2/a2 40. Which of the following colours will have minimum polarising angle in a given medium? a) violet b) blue c) green d) indigo 2 2 2 41. If D10 = x, D15 = y, D20 = z, where D10, D15 and D20 are the diameters of 10th, 15th and 20th dark ring of a Newton’s ring system, then a) y-x = 2(z-y) b) 2(y-x) = z-y c) z-x = 2y d) y-x = z-y 42. For a certain experiment it is required to diffract Hα line of Balmer series. The wave length of this line is about 660 nm. Which of the following grating would be most suitable for this? a) A grating having 106 lines /m b) A grating having 104 lines/m c) A grating having 6000 lines/m d) A grating having 3x108 lines/m 43. *. Which of the following statements are correct about optic axis? a) It is a line b) It is a direction c) The crystal is symmetric about it d) The speed of O and E rays will be same along it. 44. If the acute angle of a biprism is increased, the width of the fringes produced by it a) decreases b) increases c) remains the same d) cannot be said from the data 45. The polarising angle of ordinary glass is equal 57o. If this glass is immersed in water, the polarising angle will be a) more than 570 b) equal to 570 0 c) less than 57 d) exactly equal to 450

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46. In the earth’s atmosphere the ratio of amplitude of blue light scattered to that of red light , if their wave lengths are 400 nm and 600 nm respectively, is a) 9/4 b) 4/9 c) 81/16 d) 16/81 47. In the figure below unpolarised light of intensity I0 = 64 W/m2 is incident on the first

48.

49. 50.

51.

52.

53.

polaroid N1. The intensity of light coming from the second polaroid N2 is I2, = 2 W/m2. The angle between the axis of N1 and that of N2 is equal to a) 30o b) 600 c) cos-1 (1/4) d) cos-1(1/8) Suppose the polaroid N2 in the previous question is rotated keeping N1 fixed. During one rotation, an observer will see a) a maximum and a minimum b) four maxima and four minima c) two maxima and two minima d) light of varying intensity but no minimum and maximum Laser beam can be used to standardise a) distance b) time c) angle d) temperature 14 A laser beam of frequency 3x10 Hz passes through an aperture of diameter 1 mm. The angular spread of the beam in radian is equal to a) 10-2 b) 10-4 -3 c) 5x10 d) 10-3 Which of the following statement(s) are true for photons ? a) Their rest mass is zero b) Their rest mass is infinite c) They cannot produce interference d) They travel with a speed equal to that of light waves If the width of source slit in Young’s experiment is increased, a) the fringe width will increase b) the fringe width will decrease c) the bright bands will become more bright d) the visibility of the system will become less A ray of light of intensity I is incident on a parallel glass-slab at a point A as shown in adjacent figure. It undergoes partial reflection and refraction. At each reflection 25% of incident energy is reflected. The rays AB and A′B′ undergo interference. The ratio Imax/Imin is: a) 4:1 b) 8:1 c) 7:1 d) 49:1

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54. The position of the direct image obtained at O, when a monochromatic beam of light is passed through a plane transmission grating at normal incidence is shown in the adjacent figure. The diffracted images A,B and C correspond to the first, second and third order diffraction. When the source is replaced by another source of shorter wavelength a) all the four images shift in the direction C to O b) the images C,B and A will shift towards O c) all the four images shift in the direction O to C d) the images C, B and A will shift away from O 55. If the coherent slits in Young’s experiment are narrowed so that their width becomes comparable with the wavelength of light, a) the fringes will become broad b) the fringes will become narrower c) the fringes will become unequally spaced d) the fringes will disappear and uniform illumination will set in 56. A mixture of red light of λ = 6000 Ao and blue light of λ = 4400 Ao is incident normally on an air film. The minimum thickness of the film for which the reflected light will appear blue is a) 3x10-7 m b) 6 x 10-7m c) 2.2 x 10-7 m d) 4.4 x 10-7 m 57. Which of the following gives maximum field of view ? a) plane mirror b) concave mirror c) convex mirror d) cylindrical mirror 58. A ray of light falls on a plane mirror at angle of incidence i. The deviation produced on the ray by the mirror is a) 2i b) (180-2i) c) (180 + 2i) d) (360-2i) 59. When a convergent beam of light is incident on a plane mirror, the image formed is a) erect and real b) erect and virtual c) inverted and real d) inverted and virtual 60. When the moon is near horizon, it appears bigger due to a) refraction b) scattering of light c) diffraction d) all the three 61. How many images of himself does an observer see on the mirror surfaced two adjacent walls and the ceiling of a rectangular room? a) 3 b) 5 c) 6 d) 9 62. A glass slab is placed in the path of converging beam of light. The point of convergence of light a) moves towards the glass slab b) moves away from the glass slab c) remains at the same point d) undergoes a lateral shift 63. A plane glass slab is placed on the letters of the following colours. The letters of which colour is raised maximum ? a) red b) yellow c) green d) blue 64. A ray of light passes from vacuum into a medium at an angle of incidence twice the angle of refraction. The angle of incidence is a) cos-1 (µ/2) b) 2 cos-1 (µ/2) d) 2 sin-1 (1/µ) d) 2 sin-1 (µ/2) 14 65. A light wave of frequency 5 x 10 Hz enters a medium of refractive index 1.5. In the medium the velocity of the light wave is a) 3 x 108 m/s b) 2 x 108 m/s c) (5/1014/1.5) m/s d) 1.5 x 5 x 1014 m/s

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66. A small plane mirror is rotating with a constant frequency of n rotations per second. With what linear velocity (in m/s) a spot of light reflected from the mirror moves along a spherical screen of radius of curvature R, if the mirror is at the centre of curvature of screen ? a) 2πnR b) 2nR c) 4nR d) 4πnR 67. When a source of light is 1 m away, a photosensitive meter gives a delfection of 8 divisions, if the source is moved 1m more, the deflection will be a) 8 divisions b) 4 divisions c) 2 divisions d) 9 divisions 68. The critical angle of a prism is 360. The maximum angle of the prism for which an emergent ray is possible is a) 720 b) 540 c) 360 d) 160 0 69. A thin prism P1 of angle 4 and ,made from glass of refractive index 1.54 is combined with another thin prism P2 made from glass of refractive index 1.72 so that ray undergoes no deviation. The angle of prism P2 is b) 40 c) 30 d) 2.60 a) 5.330 70. The distance between an object and a divergent lens is x times the focal length of the lens. The lateral magnification m produced by the lens will be a) x b) 1/x c) x+1 d) 1/x+1 71. The colour of light which travels with minium speed in glass is a) red b) yellow c) green d) blue 72. A convex lens is placed between an object and screen which are at a fixed distance apart. For one position of the lens the magnification of the image obtained on the screen is m1. When the lens is moved by a distance d, the magnification of the image obtained on the same screen is m2. The focal length of the lens is (m1 > m2) a) d/(m1-m2) b) d/(m1 + m2) c) dm1/m2 d) m2/m1 73. In a displacement method using convex lens we obtain two images for separation of the lens d. One image is magnified as much as the other is diminished. If m is the magnification of one image, the focal length of the lens is a) d/(m-1) b) md(m2-1) c) d/(m2-m) d) dm2/m1 74. If the earth had no atmosphere, the duration of the day would have been a) longer b) the same c) shorter d) shorter or longer depending on the season 75. A short linear object of length b lies along the axis of a concave mirror of focal length f at a distance u from the pole of the mirror. The size of the image is approximately equal to f a) b u − f

1/ 2

f b) b u −f

2

u − f c) b f

2

u − f d) b f

76. Two convex lenses separated by a distance are brought into contact. The focal power of the combination a) decreases b) increases c) remains the same d) cannot be said from the data 77. A transparent cube of 12 cm edge contains a small air bubble. Its apparent depth is 5 cm when viewed through one face and 3 cm when viewed through the opposite face. The refractive index of the material of the cube is a) 1.5 b) 12/7 c) 12/5 d) 5/3 78. A source of light located at a point A where AB = BC = 1 m as shown in fig. The intensity of illumination at B is P. The intensity of illumination at C is

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80.

81.

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a) 0.36 P b) 0.5 P c) 0.7 P d) 0.44 P A diver from inside water looks at an outside object whose natural colour is green. He sees the object as a) green b) red c) violet d) yellow An ant is approaching a convex lens with a constant speed up to the first focus. The speed of the image of ant formed by the lens. a) remains the same b) increases uniformly c) first decreases and then increases d) first increases and then decreases A thin lens of focal length f and aperture diameter d forms an image of intensity I. The central part of the aperture up to diameter d/2 is blocked by an opaque paper. The focal length and image intensity would respectively be a) f, I/4 b) f/2, I/2 c) 3f/4, I/2 d) f, I/2 Two thin lenses of focal lengths f1 and f2 are placed at a distance d between them. For power of the combination to be zero, the separation of the lenses must be f f +f a) f1 + f2 d) f1 f 2 b) 1 2 c) 1 2 f2

83. A blue cross illuminated by white light under white background is observed through a red filter. The observer will see a) a red cross on a black background b) a red cross on a blue background c) a black cross on a red background d) a black cross on a blue background 84. The angle of a prism is A and the angle of minimum deviation is 180-2A. the refractive index of the prism is a) sin (A/2) b) cos (A/2) c) tan (A/2) d) cot (A/2) 85. A glass prism has a refractive index of 1.5 and refracting angle of 900. A ray of light is incident on it at an angle of incidence of 300. The angle of emergence will be in degrees a) 60 b) 30 c) 45 d) the ray will not emerge 86. A ray of light is incident at an angle of 600 on one face of a prism of refracting angle 300. The emergent ray makes an angle 300 with the incident ray. The refractive index of the prism is a) 1.41 b) 1.73 B c) 1.5 d) 1.6 87. A ray of monochromatic light is incident on one face of a prism of angle 750. It is incident on the second face at critical angle. If µ = √2, the angle of incidence on the first face of the prism is (in degrees) a) 30 b) 45 c) 60 d) 90 88. A beam of light consisting of red, green blue colours is incident on right angled prism normally on the face AB. The refractive indices of the prism for red, green and blue are

45° A

C

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89.

90. 91.

92.

93.

94.

95.

96.

97.

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1.39, 1.44 and 1.47 respectively. The prism will a) separate red colour from blue and green b) separate part of blue colour from red and green c) separate all the three colours from one another d) not separate even partially any colour from the others A prism of critical angle 420 is immersed in water of critical angle 490. The critical angle of the prism inside water will be a) between 420 and 490 b) less than 420 0 c) greater than 49 d) equal to 490 The angle with horizontal at which an under-water swimmer will see the setting sun is a) 410 b) 320 c) 490 d) 600 A spherical air bubble in water will act as a) a convex lens b) a concave lens c) plane glass plate d) plano-concave lens When a lens is moved towards the object from a distance 20 cm to 15 cm, the magnification of the image remains the same. The focal length of the lens is a) 16.8 cm b) 15.5 cm c) 18.2 cm d) 17.5 cm Which of the following is an achromatic combination pair of a telescope objective? a) Lenses of of power +3D and - 5D b) Lenses of f = 40 cm and power –3 D c) Lenses of f = -50 cm and power +2D d) Lenses of f = +20 cm and power –4.5 D A convex lens A of focal length 20 cm and a concave lens B of focal length 5 cm are kept along the same axis with a d distance d between them.. If a parallel beam of light falling on a leaves B as a [parallel beam of light falling on A leaves B as a parallel beam, then d is equal to (in cm) a) 25 b) 22.5 c) 20 d) 15 A parallel beam of white light falls on a combination of a concave and a convex lens, both of the same material. Their focal lengths are 15 and 30 cam respectively for the mean wavelength in white light. On the other side of the lens system, one sees a) white pattern b) no pattern at all c) coloured pattern with violet at inner side d) coloured pattern with violet at outer side Consider Fraunhofer diffraction pattern obtained with a single slit, illuminated at normal incidence. At the angular position of the first diffraction minimum, the phase difference (in radians) between the wavelets from the opposite edges of the slit is a) π b) 2π c) π/4 d) π/2 *A planet is observed by an astronomical refracting telescope having objective of focal length 16 m and eyepiece of focal length 2 cm. Then a) The distance between objective and eyepiece is 16.02 m b) The angular magnification of the planet is –800 c) The image of the planet is inverted

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d) The objective is larger than the eye-piece 98. Interference fringes are formed with light of wavelength 400 nm in Young’s experiment with separation of slit 1 mm and screen distance 1 m. The angular width of fringe (in radian) is a) 0.4 b) 0.04 c) 0.004 d) 0.0004 99. If Young’s experiment with one source of light and two slits be performed so that interfering beam traverse in water instead of air, a) the fringes will be less in number b) the fringes will be broader c) the fringes will be narrower d) no fringes will be obtained 100. In Young’s interference experiment with one source and two slits, one slit is covered with a cellophane sheet so that only half of the intensity is transmitted. Then a) no fringes will be obtained b) bright fringes will be brighter and dark fringes will be darker c) all fringes will be darker d) bright fringes will be less bright and dark fringes will be less dark 101. In a Young’s double slit experiment, the interference pattern is found to have an intensity ratio of 9 between bright and dark fringes. This implies that a) the intensities at the screen due to the two slits are 5 units and 4 units respectively b) the intensities all the screen due to the two slits are 4 units and 1 unit respectively c) the amplitude ratio is 3/2 d) the amplitude ratio is 3 102. In Young’s double slit experiment the central fringe is white. What colour of fringe will you see nearest to the central fringe ? a) violet b) green c) red d) yellow 103. *White light is used to illuminate the two slits in a Young’s double slit experiment. The separation between the slits is b and the screen is at a distance d ( >> b) from the slits. At a point on the screen directly in front of one of the slits, certain wave lengths are missing. Some of these missing wavelengths are a) λ = b2/d b) λ = 2b2/d c) λ = b2/3d d) λ = 2b2/3d 104. In Young’s double slit experiment using light of wavelength λ, the intensity of one bright fringe is I units. The intensity of light at a point where the path difference is λ/4 is a) I/2 b) I/4 c) I/3 d) I/9 105. The polarising angle of diamond is 670. The critical angle of diamond is nearest to b) 340 c) 450 d) 600 a) 220 106. A ray of light enters from a rarer medium to a denser medium at an angle of incidence i. The reflected and refracted rays make an angle of 900 with each other. The angles of reflection and refraction are r and r’ respectively. The critical angle of denser medium is a) sin-1 (tan r’)-1 b) sin-1 (tan r’1)-1 -1 -1 c) sin (tan i) d) tan-1(sin i) 107. Two nicol prisms (polariser and analyser) have their axes an angle 300 in between. If I is the intensity of light falling on first nicol,. Then the intensity of emerging light from second nicol is a) (1/8) I b) (3/8)I c) (1/4)I d) (1/2)I

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108. A point source of light is to be suspended above the centre of a circular table of radius R. In order to produce maximum illumination at the edge of the table, the height of the light source should be a) R b) b) 2 R c) R / 2 d) R/2 109. If a star emitting yellow light starts accelerating towards the earth, its colour as seen from earth will a) turn gradually red b) turn gradually yellow c) remain unchanged d) turn gradually blue 110. An astronomical telescope has large aperture to a) reduce spherical aberration b) have high resolution c) have low dispersion d) increase the span observation 111. The distance between virtual sources in a biprism of angle α and refractive index µ, if the source is placed at distance a from it, is a) 2 (µ-1) α b) (µ-1)α a/2 c) (µ-1) α d) 2(µ-1)α a 112. In Newton’s rings the radius of the first dark ring 2 mm. The radius of the third dark ring will be a) 3 mm b) 2 mm c) 1.7 mm d) 3.4 mm 113. If Young’s experiment could be performed with the following radiations, which of them will give bands of minimum width? a) red light b) radio waves c) violet light d) micro waves 114. Which of the following is inevitable for a Fraunhoffer diffraction experiment? a) White light source b) a colour filter c) a telescope d) a convex lens 115. Which of the following can be polarised ? a) α rays b) β rays c) γ rays d) all the three 116. Energies of photon of four different electromagnetic radiations are given below. Which energy value corresponds to a visible photon ? a) 1 eV b) 2.5 eV c) 5 eV d) 1000 eV 117. The first crucial experiment if in favour of wave theory was carried out by a) Huygens b) Foucault c) Newton d) Thomas Young 118. An interference pattern is observed by Young’s double slit experiment. If now the separation between coherent sources is halved and the distance of screen from coherent sources is doubled ; the now fringe width a) becomes double b) becomes one-fourth c) remains the same d) becomes four times 119. The two slits in Young’s double slit experiment are illuminated by two different parts of same sodium lamp. Then a) uniform illumination will be seen b) interference pattern with dark and bright fringes is observed c) interference pattern with circular yellow fringes is observed d) complete darkness will be seen 120. Young’s experiment is performed with while light. The wavelength of red light = 660 nm. The wavelength of blue light = 440 nm. The minimum order of the red light fringe which coincides with a blue light fringe is equal to a) 2 b) 3 c) 4 d) 5

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121. In Young’s experiment monochromatic light is used to illuminate the slits A and B. Interference fringes are observed on a screen placed in front of the slits. Now if a thin glass plate is placed in the path of the beam coming from A, then a) the fringes will disappear b) the fringe width will increase c) there will be no change in the fringe width d) the fringes will shift 122. The wavelength of light plays no role in a) interference b) diffraction c) polarisation d) resolving power -1 123. The critical angle of a certain medium is sin (3/5). The polarising angle of the medium is a) sin-1 (4/5) b) tan-1 (5/3) c) tan-1 (3/4) d) tan-1 (4/3) 124. Double refraction of light is shown by a) quartz and calcite only b) calcite only c) calcite and ice only d) calcite, ice and quartz 125. A diffraction grating has 5000 lines/cm. The maximum order visible with wavelength 6000A0 is a) 2 b) 3 c) 4 d) 50 126. If a drop of water is introduced between the glass plate and a convex lens in a Newton’s ring system, the ring system a) contracts b) expands c) remains the same d) first expands, then contracts 127. A fly is sitting on the objective of a telescope pointed towards the moon. What effect is expected in a photograph of the moon taken through the telescope? a) the entire field of view will be blocked b) there will be image of the fly on the photograph c) there will be no effect at all d) there ill be reduction in the intensity of the image 128. A lens is placed between a source of light and a wall. It forms images of area A1 and A2 on the wall for its two different positions. The area of the source of light is a)

b) ( A 1 + A 2 ) / 2

A1 A 2

c) [( A 1 + A 2 / 2]

2

d ) [(1 / A 1 ) + (1 / A 2 )] −1

129. The diameter of the moon is 3.5 x 103 km and its distance from the earth is 3.8 x 105 km. It is seen by a telescope having the focal lengths of the objective and the eye piece 4 m and 10 cm respectively. The diameter of the image of moon would be approximately a) 20 b) 200 c) 400 d) 500 130. An eye piece consists of two lenses of focal lengths f1 and f2. In order that the combination is achromatic, the lenses are to be separated by a distance a) 2f1 f2/f1 + f2 b) 2f1 f2 / f1 – f2 c) f1 + f2 /2 d) f1 – f2 131. A person can see clearly up to 3 m. What is the power of the lens he should use so that he can see upto 12 m ? a) +0.5 D b) -0.5 D c) +0.25 D d) -0.25 D 132. If the accelerating potential increases to 4 times in an electron microscope, its resolving power R would change to a) R/4 b) 4R c) 2R d) R/2 133. Which of the following is true for a quarter wave plate ?

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a) It is made of glass and produces a phase difference π/2 b) It is made of glass and produces a phase difference π/4 c) It is made of quartz and produces a phase difference π/2 d) It is made of quartz and produces a phase difference π/4 134. Which of the following can diffract sound waves from a tuning fork of the frequency 384 Hz? a) a sphere of diameter 10 m b) a sphere of diameter 1 m c) a sphere of diameter 1 cm d) a sphere of diameter 1 mm 135. The exposure time of a camera lens at the f/2.8 setting is 1/200 second. The correct time of exposure at f/5.6 setting is a) 0.02 sec b) 0.04 sec c) 0.20 sec d) 0.40 sec

SECTION 2: ANSWERS 1

(c)

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30 cm

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(b),(c),(d) 44

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SECTION 3 SOLUTIONS 1.

Use the information supplied in theory notes. The diverging beam is coming from a real object. Hence the plane mirror forms a virtual image. Virtual images are erect. Ans.c

2.

Here we are getting reflected image of the flame. So its distance will be equal to the distance of the object. Ans.a Use the information supplied in the table in indirect theory notes. Critical angle is minimum for light which has minimum wave length in the given list. Here it is blue. Ans.b

3. 4.

5.

6.

7.

8.

9.

By Snell’s law of refraction sin i1/ sin r1 = √2. This will give r1 = 30o , as i1 = 45o. Since r1 + r2 = A = 600, r2 will also be 30o. This gives i2 = 45o. Thus the ray will emerge from the second face at an angle of 45o. So (a), (b),(c) are wrong . Since i1 = i2 and r1 = r2, the ray undergoes minimum deviation. So (d) is correct. Ans. d Use the information supplied in indirect theory notes wave motion part. The frequency of a wave remains constant. Therefore we eliminate answers (b) and (d). The wavelength decreases when light passes from rarer to denser medium. Ans.c Use the formula for focal length of combination of two lenses or power. Since parallel beam emerges as a parallel beam, the focal power of the system should be zero. i.e. 1 1 1 a 1 = + − . =P=0.⇒a = 2f. Ans.b F f f f xf F Use the information supplied in the table given in indirect theory notes in Optics. The focal length of a lens increases with wavelength. In the given list of electromagnetic radiations, micro waves have the maximum wave length. Ans.c We use Newton’s equation for the relation between distance of an object and real image from the focus, ab = f2. Since ab is equal to a constant, the graph between a and b will be a rectangular hyperbola. Ans.b The image of an object at a distance of 2f is formed at 2f itself. Hence one end of a rod should be at 2f. The image of this end at 2f is formed at 2f itself. The other end will be at a distance 2f-(f/3)=5f/3. The image of this end at a distance 5f/3 is formed at distance v given by the equation 1/u + 1/v = 1/f, with u = 5f/3. This gives v = 5f/2. The length of the image of the rod is thus 5f/2-2f = f/2. The length of the rod is f/3. Length of image f / 2 3 Magnification = = = . Ans.a Length of object f / 3 2

10. Use the formula given for magnification. m= f/u-f. Here m=+2 (real image). This gives u = 1.5 f. Since v/u=2, v= 3f. The distance between object and image for real image = v-u =3f 1.5 f = 1.5 f. Ans.b 11. Use information supplied in theory notes. Let θ be the angle between the mirrors. After the first reflection, the ray is deviated through 180-2i, where i is the angle of incidence. It finds the second mirror at an angle θ. So it deviates through 2θ. ( In direct theory notes) During third, fourth and fifth reflections it undergoes a deviation of 2θ each. The total deviation of the ray after five reflections is therefore 180-2i+ 8θ. Since the ray retraces its path, we have 180-2i+8θ =180o. The angle i=100 . This gives θ = 20o. Ans.d

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12. Using

the

relation

µ21=

µ2 , µ1

we

161

can

rewrite

the

given

quantity

as

=

µ2 µ3 µ4 µ4 x x = = µ 41 . Ans a µ1 µ 2 µ 3 µ1

13. If E is the total light energy given by the source per second , the light energy received per unit area per second of the curved surface of the cylinder = E/2πr per unit length. Thus intensity is proportional to 1/r. Ans.b ( A + D) A / sin . At minimum deviation, (A+D)/2 = angle of 2 2 incidence i. Substituting A= 60o , we get sin i/ sin30 = µ =1.41 or √2. This gives i = 45o. Ans.d

14. Refractive index µ = sin

15. Use the information supplied in theory notes. The angle between the mirrors θ is given by 360-2θ = angle of deviation. This gives 360-2θ = 240. θ = 60o. Number of images = (360/θ)-1 = 5. Ans.c 16. The apparent height of the bird from the surface of water as seen by the fish = refractive index of water x 3 = (4/3) x 3 = 4. The total distance between the fish and the bird as seen by the fish is equal to 4+4 = 8 m. Ans c 17. Use the information supplied theory notes . tanC = r/h. …(1) To find tan C, sinC = 1/µ . cosC =

µ 2 − 1 / µ. . tanC =1/ µ 2 − 1 ⇒ r = h tanC = h/ µ 2 − 1 . Ans.c

18. Since plane surface is silvered, when a ray of light is incident it reflects normally on this surface. So it refracts twice on the spherical surface. Therefore, this is equal to a convex lens. Use the information in indirect theory notes, which gives the focal length of convex lens made of two plano-convex lenses is half of that of plano-convex lens. Ans. b 19. Here refraction takes place on the two plane surfaces and reflection the spherical surface. If F is the focal length of the system, 1/F = (1/f) of mirror + 2(1/f) of lens. Focal length of mirror is equal to R/2. (1/f) of lens is equal to ( µ-1)/R. Substituting these values in the above 1 2 2(µ −1) 1 2µ R ⇒ = . F= equation ,we get = + ..Ans.c F R R F R 2µ 20. Light is passing from first medium to second medium goes straight. Hence µ1 = µ2. When it passes from second medium to third medium it diverges. That means the third medium is denser compared to the second. That is µ3 > µ2. Ans.a 21. Using lens makers formula P = (µ -1)/R, we have for the first lens P1 =-(µ1 – 1)/R and P2 =+(µ2 –1)/R. (negative sign for concave). The power of the combination P = P1 + P2 = (µ2-µ1) / R. Focal length of the combination = 1/P = R/(µ2-µ1). Ans.d 22. Momentum of a photon p = h/λ. Energy E = hc/λ = pc. So p = E/c. Force = change in momentum per second = p-(-p) = 2p = 2E/c. Since energy is falling on unit area, this force will be equal to pressure. Ans.b 23. Refer to the adjacent fig. Angle of refraction r1 is got from the equation sin45/sinr1 = √2. This gives r1 = 30o.. Angle between reflected and refracted rays = 180-75= 1050. Ans.d 24. There will be three images at equal distance from the object. They will lie in a circle. Ans.a

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25. By inverse square law , the intensity of light at a point is proportional to 1/r2. The amplitude (square root of intensity) is proportional to 1/r. The graph will be rectangular hyperbola. Ans.d 26. The intensity I = P cosθ/r2. Taking logarithm and differentiating, dI/I = -2(dr/r). Here dr/r = 2%. Therefore dI/I = -4%. The minus sign shows intensity increases when r decreases. Ans. a 27. By inverse square law I = P cosθ/r2. Here θ = 0o. So PA/22 = PB/32. PA/PB = 4/9. Ans.b 28. This is tricky question. The refractive index depends only on the material of the prism and the colour of light. It does not depend on A and D. You should not be carried off by the (A + D) A equation µ = sin / sin . This is only a relation between µ, A, D. That is you can 2 2 calculate µ, if you know A and D. Ans.d 29. Use the simplified formula for finding d. d = D 2 − 4fD = 90 2 − (4 x 20x 90) = 30 cm . .Ans. 30 cm 30. Use the simplified formula connecting focal length f, distance of displacement d and md 4 x 45 magnification m. f = 2 = = 12 cm . .Ans.12 cm m −1 4 2 −1 31. Use the information supplied in theory notes. β αλ. Of the given colours blue has minimum wavelength and hence will have minimum band width. Ans.c 32. Use the information supplied already in two sections of earlier theory notes The band width velocity of light in air β(air ) λ(air ) 4 = = = decreases according to the formula β( water ) λ ( water ) velocity of light in water 3 (refractive index of water). Ans.d 33. When the beams travel through different media, their path difference can be compared if we find their equivalent path in air. If a beam travels L1 in a medium of refractive index µ1, its equivalent path in air is µ1L1. Similarly for the second beam, equivalent path in air is µ2L2. Therefore their path difference is (µ1L1-µ2L2). Use the information supplied in wave motion theory notes to convert path difference to phase difference. The corresponding phase difference is equal to

2π (µ1L1-µ2L2)= k(µ1L1-µ2L2). Ans.a λ

34. Use the information supplied in theory notes in wave motion. The amplitude of wave transmitted is proportional to the slit width. The maximum amplitude is 3 units and the minimum amplitude 1 unit. The ratio Imax/Imin = A max2 / Amin2 = 9/1. Ans.d 35. This question is to test whether all the necessary conditions of interference are known to you. The necessary condition for interference is that the sources should be coherent. That is, they should emit light of (1)same wavelength, (2) same amplitude and (3) zero or constant phase difference. (a) violates the first condition while (d) violates the third condition. (Note that virtual sources can produce interference. For example two virtual coherent sources are used in Biprism, while one real source and one virtual source is used in Lloyd’s mirror.) Ans. a, d. 36. There will be no change in the fringe width because the fringe-width does not depend on width of slits but depends on the distance between slits. [Theory notes]. Hence (a) and (b) are wrong. The contrast between bright and dark bands will decrease, bright bands becoming less bright and dark bands becoming less dark. When the width becomes very large, interference

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condition will be violated and bands will disappear, Ans. c,d

163

uniform illumination will set in .

37. We have tan i = µ. Also µ = 1/sin C = cosec C. This gives tan i = cosec C. i = tan-1(cosec C). Ans.a 38. The spread of a laser beam is only due to diffraction when it passes through the slit. The angular spread is equal to the angle of diffraction θ = λ/a. If D is the distance to moon, the linear spread = angular spread x distance = Dλ/a. Ans.b 39. The areal spread = square of linear spread = (Dθ)2 = (Dλ/a)2 Ans.b 40. Use the information given in table in indirect theory notes. The refractive index is minimum for green. Since tan i =µ, we have ‘i’ is minimum for green. Ans.c 41. Use the information supplied in theory notes. Dn2-Dm2 α (n-m). Hence D152- D102 = D202- D152. That is y-x = z-y. Ans.d 42. For diffraction the spacing of the grating should be of the order of wave length of light. Number of lines = 1/ λ = 1/6.6x10-7 m = 1.5x106/m. (Note: 10/6.6 = 3/2), which is nearest to answer a. Ans.a 43. Optic axis is a direction and not a line. So (b) is correct. (c) and (d) are the other two properties of optic axis. Ans. b,c,d 44. Use the information supplied in theory notes . When α is increased, the distance between virtual sources d increases. The band width β = Dλ/d (same as Young’s experiment )decreases. Ans.a 45. Using the equation tani = µ, when glass is immersed in water, we have to replace µ by µgw, where µgw is refractive index of glass with respect to water. µgw = µg/µw, which is less than µ. Hence tan i decreases, when immersed in water. Ans.c. 46. Use Rayleigh’s law of scattering. The intensity of scattered light is proportional to 1/λ4. Since intensity is proportional to square of amplitude, the amplitude will be proportional to 1/λ2. The ratio of required amplitudes will be A(blue)/A(red) = 6002/4002 = 9/4. Ans.a 47. Use the information supplied in indirect theory notes I = cos2 θ=

I0 cos 2 θ . Here Io = 64, I2 =2.⇒ 2

1 1 ⇒ cosθ = . Ans.c 16 4

48. Here again use the same theory note as in previous question . When angle of rotation of N2 changes from 0 to 360o, I2 reaches two maxima (θ = 0 and 180) and two minima (when θ = 90 and 270). Ans.c 49. Laser beam is highly coherent, monochromatic and unidirectional. Hence it can be used to standardise distance. Ans.a 50. Use the information and formula found in a similar question in this section. The spread of laser beam is due to diffraction at the aperture . Angle of diffraction θ=λ/a. λ = velocity / frequency = 3x108/ 3x1014 = 10-6m. This gives θ= 10-6/10-3= 10-3 radian. Ans.d 51. Statements (a),(d) are true for photons. Since photons have zero rest mass, (b) is wrong. They can produce interference, which we explain due to their wave nature . Hence (c) is not true. Ans.a,d 52. Let x be the width of the source slit and y the distance between the sources slit and the plane of the two coherent slits. For a distinct visibility of interference fringes, the condition to be satisfied is x/y < λ/d. So if x is increased or y is decreased, visibility will be poor. Ans.d

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53. The percentage of light reflected from A is 25% = 0.25. If ‘a’ is the amplitude of the incident light, the amplitude reflected at AB = 0.25 a= 0.5a = (1/2)a Intensity reflected from C = 0.25 of 0.75I. Light transmitted at A′ = 0.75x0.25x0.75 of the total I. The amplitude of A′B′ = 0.75 x 0.25 x 0.75 a = 0.375a = (3/8)a. Thus the amplitude of two components are AB and A’B’ are (1/2)a and (3/8)a. Imax/Imin = [(1/2)a+(3/8)a]2 / [(1/2)a-(3/8)a]2 = 49/1. Ans.d 54. The equation for diffraction grating sinθ= Nnλ, where n is the order, N no. of lines /m of grating and λ the wavelength of light used. When λ decreases, θ also decreases. Hence A,B,C move towards O. However there will be no change in O because it is centre of image i.e. direct and not diffracted . Ans.b 55. When the width of slits become comparable with the wavelength of light used, two slit diffraction pattern results. The diffraction bands are of unequal width. Hence we shall see small number of unequally spaced bands. Ans.c 56. The condition for a film to appear dark by reflection is 2µt cosr = nλ. For minimum thickness n = 1, for normal vision r = 0, for air µ = 1. This reduces the above equation to 2t = λ. The film will appear blue if red is missing . So here we substitute the value of λ for red= 6000Ao =6x10-7m. This gives t=3x10-7m..Ans.a 57. All the objects from the convex mirror to infinity are brought to focus between the pole and the focus of the mirror. Hence large span can be covered. Also the image is erect (This is the reason for using convex mirror as rear – view mirror in a motor car). Ans.c 58. The angle of the deviation is the angle between incident ray and reflected ray. Producing the incident ray we find it is 180-2i here. Ans.b 59. A beam of light converging to a reflecting or a refracting surface is coming from a virtual object. The image of a virtual object formed by a plane mirror is real. Also plane mirror images are erect. Ans.a 60. Ans.a 61. No of images between two adjacent perpendicular mirror = (360/90)-1 = 3. We have here three pairs of perpendicular mirrors. The number of images seen will be 9. Three of them will be common. Thus the number of images will be 9-3 = 6. 62. The glass slab produces an extra path difference of (µ-1)t, where µ is a refractive index, ‘t’ thickness of the slab. This increased path moves the converging point farther. Ans.b 63. Refractive index µ = actual depth / apparent depth. Thus apparent depth is proportional to 1/µ. The letter is raised maximum, when apparent depth is minimum. That is µ should be maximum. For the given colours it happens for blue. Ans.d 64. sin i/sin r = µ. Here i = 2r. That is sin 2r/sin r = µ.. 2 sinr cosr/sin r = µ. This gives cos r = µ/2. r = cos-1 ( µ /2). i = 2r =2cos-1 (µ/2). Ans.b 65. Speed of light in the medium = speed of light in air / µ i.e. 3 x 108 / 1.5 = 2x 108 m/s (Frequency of light is not necessary here) Ans.b 66. When the mirror rotates through an angle θ, the reflected ray rotates through an angle 2θ. The angle turned by the reflected ray in one second ω = 2 x 2πn = 4πn rad/s. The linear velocity of the spot of light will be equal to v = ωR = 4πnR m/s. Ans.d 67. According to the inverse square law, the intensity will be inversely proportional to the square of the distance from the source. When the distance is doubled, the intensity becomes onefourth. The deflection is proportional to the intensity and hence will be 2 divisions. Ans.c

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165

68. The maximum angle for which an emergent ray is possible is 2C. This can be shown as follows: Angle of prism Amax = (r1)max + (r2)max. The maximum values of i1 and i2 will be 900 each, Hence the maximum values of r1 and r2 each will be equal to C. Amax = 2C. If the angle of the prism increases beyound 2C, the ray will totally reflect from the second face. You can also answer this question from theory notes shortcuts. Ans.a 69. If µ1 and µ2 are the refractive indices of the two prisms the conducrion for non deviation is (µ1 – 1) A1 = (µ2-1) A2, where we have used the formula for deviation of small angle prisms, d = (µ -1) A. Here A1 = 40, µ1 = 1.54, µ2 = 1.72 we get A2 = 30. Ans. c 70. Using the formula for magnification in terms of object distance u, m = f/ (u-f). Here u = xf. Therefore m = f/(xf-f). i.e. m = -f/[xf-(-f) ] = -1/(x+1), applying signs. The negative sng in magnification shows that the image is virtual. Ans.d 71. The speed of light is inversely proportional to the refractive index. Blue has the maximum refractive index in the given set and hence has the minimum velocity. This question can be answered from the theory notes table. Ans.d 72. This is displacement method. If D is the distance between the object and the screen, ‘d’ the separation of the two positions of the lens throwing two images on the screen, then m1 = (D+d)/(D-d), and m2 = (D-d) / (D+d), m1 – m2 = d/f, f = d/(m1 – m2). This also can be answered from the indirect theory notes. Ans.a 73. f = d/(m1-m2). Here if m1 is taken as m, m2 = 1/m. Then f becomes md/ (m2-1). Ans.b 74. The molecules of the earth’s atmosphere refract the sunlight so that the sun is visible before sunrise and after sunset. Thus presence of atmosphere makes the day longer. In the absence of atmosphere, sunlight can been seen only after sunrise and before sunset. The day will be shorter. Ans.c 75. The linear magnification of an object is magnification when the object is placed parallel to the principal axis. This is given by dv/du. Differentiating mirror formula 1/u+ 1/v = 1/f, −

du u2

−

dv v2

2

=0⇒

dv v2 f =− 2 = . du u u − f

2

Length of image f . = Length of object u − f

This gives the

2

f length of the image as b .This formula is given in indirect theory notes. Ans.d u − f 76. The power of the combination P = P1 + P2 – dP1P2, where d is their separation. When they are brought into contact, d = 0 and hence P increases. Ans.b 77. Let x be the actual depth of air bubble. When viewed from one face µ = x/5. When viewed from the opposite face µ = (12-x)/3. This gives x = 7.5 and µ = 1.5. Ans.a 78. If S is the power of the source, θ angle of incidence, intensity of illumination is P = Scosθ/r2. In the first case (θ = 0) P = S cosθ/12 and in the second case power P1 = Scos450/(√2)2. This gives P1 = P/2√2 = 0.35 P. Ans.a 79. The speed of light and hence wavelength of light in water is less than that in air . λ (water ) = λ (air) / µ. Taking µ = 4/3, we find wavelength of green (nearly 540 nm,) could reduce to 540 x 3/4 = that of nearly violet. (Note: This question can be answered even without the above calculation, because the only wavelength less than of green given here is violet.) Ans.c 80. Differentiating lens formula 1/u + 1/v = 1/f, with respect to time we get, dv/dt = (v2/u2) du/dt. Since v/u = m, magnification and object speed du/dt is constant, dv/dt is proportional to –m2. When the object approaches from infinity to focus the image is real and hence m is positive, dv/dt is negative i.e. it decreases. When the object is at the focus the image moves to infinity

166

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on the other side and hence is virtual m is –ve and dv/dt is positve. Hence dv/dt first decreases and then increases. Ans.c 81. There will be no change in focal length. However, reduction of diameter to half reduces the area covered to 1/4 and hence intensity reduces to 1/4. Ans.a 82. Power of the combination of two lenses separated by a distanc d is 1/f = p = (1/f1) + (1/f2)-d/f1 f2). When p = 0, we have (1/f1) +(1/f2) = (d/f1 f2.) This gives d = f1 + f2. Ans.a 83. The red filter allows only red light from the white background. The blue cross will appear dark under red light. Hence one sees black cross under red back ground. Ans.c 84. Here D = 180 – 2A. µ = sin [(A+D) / 2 ] / sin [A/2] becomes cot (A/2). Ans.d 85. sini1 / sinr1 = 1.5, i1 = 300. Hence r1 < 300. r2 = 90-r1 should be greater than 600. Thus the ray is incident in the second face at an angle greater than critical angle which is nearly 420 for a prism of µ = 1.5. It will totally reflect from the second face. Ans.d 86. Here i1 = 600 , d = 300. i1 + i2 = A+d. With A = 300, we get i2 = 0. Hence r2 = 0. r1 + r2 = A. This gives r1 = 300. µ = sin1 / sinr2 = sin 60/sin 30 = √3. Ans.b 87. Here r2 = critical angle C. Since µ = 1/sin C = √2, C = 450 r1 +r2 = 750. Thus r1 = 300. Using sini1 / sin r1 =√2, we get i1 = 450. Ans.b 88. All radiation pass normally the face AB and is incident on the face AC at an angle 450. 1/sinC = µ. Also 1/sin 45 = √2 = 1.41. for red µ = 1.39, hence C is greater than 450 .Thus red will refract. But for the other two colours, µ > 1.41, hence C is less than 450. Thus they will totally reflect from the face AC. So the prism will separate red from the other two colours. Ans.a 89. The relation between critical angle C and refractive index µ is µ = 1/sin C. When a prism is µg is immersed in water, the refractive index of the prism with respect to water µ gw = µw less than that of water or glass. Hence the critical angle will be more than that of water and glass. (Claculation of critical angle is not necessary here, but it can be shown to be 620 ). Ans.c 90. This is given in indirect theory notes. An underwater swimmer sees the setting sun at an angle with vertical equal to critical angle of water 490 The angle with the horizontal will be 900 – 490 = 410. Ans.a 91. A lens may be convex or concave depending on where we are looking from. The usual convention is to look from the rarer medium. An air bubble inside water will act as a concave lens when we look from inside air, which rarer medium here...Ans. b 92. It is clear that one image is real and the other image virtual. Using the equation for magnification in terms of object distance f/u –f, we have, f/(20-f) = -f/(15-f), which gives f = 17.5 cm. Ans.d 93. An achromatic combination of lenses should be made from one concave and one convex lens, with the convex lens having a greater focal power (or shorter length). This is to make the rays forming the final image to converge. These two conditions are satisfied only for the combination given as d, where focal powers are +5D and –4.5 D. Ans.d 94. Since the parallel beam incident leaves as such, the system here acts as a plane glass i.e. power P = 0. Refer to the solution of question (26) the distance between the lenses should be d = f1 + f2 . Here f1 = 20 cm and f2 = -5 cm. Therefore d = 15 cm. Ans.d

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167

95. The given combination has focal length –30 cm, using the combination formula (1/f1) + (1/f2) = 1/F. Hence light falling on the system will produce divergence of the rays. Since violet has more deviation that red, violet will appear to then outer edge. Ans.d 96. The diffraction minimum appears for a path difference equal to n λ ( a sin θ = nλ). A path difference of λ = a phase difference of 2 π radian. Hence the minimum appears for phase difference 2 π, 4π, 6π etc. Ans.b 97. The distance between objective and eye-piece is f0 + fe = 16 + 0.02 = 16.02 m (a is correct). Since the final image is virtual the magnificatrion is negative and equal to f0/fe = -800 (b is correct). An astronomical telescope produces inverted image and hence (c) is correct. The objective is larger than the eye-piece in aperture, (d) is correct. Ans.a,b,c,d Dλ , where D is the distance between the source and the d microscope. d separation of slits and λ wavelength of light. Angular width of the fringe β/D = λ/d = 400 x 10-9 /1 x 10-3. = 4 x 10-4 rad = 0.0004 rad. Ans.d 99. The speed of light inwater is less, and hence the wavelength of light in water is less than that in air. From the equation β = Dλ/d, we understand β will decrease, other values remaining the same. Ans.c 100. Here the amplitude of the waves producing interference is not the same. If the amplitude of a wave coming from the full slit is ‘a’ that coming from the other is only a/ 2 . The maximum amplitude is less than 2a. Hence bright fringes will be less bright. The minimum amplitude is not zero. Hence dark fringes will not be perfectly dark. Ans.d 101. I (maximum )/I(minimum) = [a1 + a2 / a1 – a2]2 where a1, a2 are amplitude of the two waves. Here (a1 + a2)2 = 1. Thus a1 + a2 = 3, a1 – a2 = 1. The amplitudes a1 and a2 are 2 and 1 unit respectively. The intensity of the two waves will be 4 and 1 unit respectively. It should be noticed that intensity cannot be subtracted, as it is a scalar quantity. Amplitude is a vector quantity. This can also be answered using intensity formula given in indirect theory notes. Ans.b 98. The width of a fings β =

102. Because the central fringe is white, the source is white light. The band width β is proportional to wave length λ. The band width is smaller for shorter wavelength. Violet has the shortest wavelength in the visible region. Hence violet fringe will be seen next white. Ans.a 103. When the waves from the two slits reach a point directly in front of a slit as shown in fig. distance travelled by one wave is d and the other d 2 + b 2 . Since b b is qbB q (b − a )B qaB q (b + a )B a) b) c) d) m m m m Two equal point charges Q each are fixed at x = -a and x = +a on the x-axis. Another point charge q is placed at the origin. The change in the electrical potential energy of q, when it is displaced by a small distance x along the x-axis, the approximately proportional to a) x b) x2 c) x3 d) 1/x A long straight wire along the z-axis carries a current I in the negative z direction. The G magnetic vector field B at a point having coordinates (x,y) in the z = 0 plane is µ 0 I ( yˆi − xˆj) µ 0 I ( xˆi + yˆj) a) b) 2 2 2π ( x + y ) 2π ( x 2 + y 2 ) µ 0 I ( xˆj − yˆi ) µ 0 I ( xˆi − yˆj) c) d) 2 2 2π ( x + y ) 2π ( x 2 + y 2 ) A 100 W bulb B1 and two 60 W bulbs B2 and B3 are connected to a 250 V source, as shown in the figure. Now W1, W2 and W3 are the output powers of the bulbs B1, B2 and B3 respectively, then a) W1 > W2 = W3 b) W1 > W2 > W3 c) W1 < W2 = W3 d) W1 < W2 < W3 As shown in the figure 2 P and Q are two coaxial conducting loops separated by some distance. When the switch S is closed, a clockwise current Ip flows in P (as seen by E) and an induced current IQ1 flows in Q. The switch remains closed for a long time. When S is opened, a current IQ2 flows in Q. Then the direction of IQ1 and IQ2 (as seen by E) are a) respectively clockwise and anti-clokwise b) both clockwise c) both anti-clockwise d) respectively anti-clockwise and clockwise A short-circuited coil is placed in a time-varying magnetic field. Electrical power is dissipated due to the current induced in the coil. If the number of turns were to be quadrupled and the wire radius halved, the electrical power dissipated would be

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7.

a) halved b) the same c) doubled d) quadrupled Two identical capacitors have the same capacitance C. One of them is charged to potential V1 and the other to V2. The negative ends of the capacitors are connected together. When the positive ends are also connected, the decrease in energy of the combined system is a)

8.

1 C(V12 − V2 2 ) 4

b)

1 C( V12 + V2 2 ) 4

c)

1 C(V1 − V2 ) 2 4

d)

1 C( V1 + V2 ) 2 a 4

The magnetic field lines due to a bar

magnet are correctly shown in fig.3 An ideal gas is taken through the cycle A →B→C→A, as shown in the figure 4. If the net heat supplied to the gas in the cycle is 5 J, the work done by the gas in the process C →A is a) -5 J b) -10 J c) -15 J d) -20 J 10. Which of the following graphs correctly represents the variation of β = -(dV/dP)/V with P for an ideal gas at constant temperature

9.

11. A Hydrogen atom and a Li++ ion are both in the second excited state. If lH and lLi are their respective electronic angular momenta, and EH and ELi their respective energies, then a) lH > lLi and |EH| > |ELi| b) lH = lLi and |EH| < |ELi| c) lH = lLi and |EH| > |ELi| d) lH < lLi and |EH| < |ELi| 12. An ideal black-body at room temperature is thrown into a furnace. It is observed that a) initially it is the darkest body and at later times the brightest b) it is the darkest body at all times c) it cannot be distinguished at all times d) initially it is the darkest body and at later times it cannot be distinguished 13. The potential difference applied to an X-rays tube is 5 kV and the current through it is 3.2 mA. Then the number of electrons striking the target per second is a) 2 x 1016 b) 5 x 1016 17 c) 1 x 10 d) 4 x 1015

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14. Which of the following processes represents a gamma-decay ? a) AXZ + γ→ AXZ-1 + a + b b) AXZ + 1n0 → A-3XZ-2+c c) AXZ → AXZ + f d) A XZ + e-1 → AXZ-1 +g 15. The half-life of 215At is 100 µs. The time taken for the radioactivity of a sample of decay to 1/16th of its initial value is a) 400 µs b) 6.3 µs c) 40 µs d) 300 µs 16. An observer can see through a pin-hole the top end of a thin rod of height h, placed as shown in the figure. The beaker height is 3h and its radius h. When the beaker is filled with a liquid up to a height 2h, he can see the lower end of the rod. Then the refractive index of the liquid is (fig.6 below) a) 5/2 b) (5 / 2)

215

At to

c) 3 / 2 d) 3/2 17. Which one of the following spherical lenses does not exhibit dispersion? The radii of curvature of the surfaces of the lenses are as given in the diagrams.(fig 6 next column)?

18. In the ideal double-slit experiment, when a glass-plate (refractive index 1.5) of thickness t is introduced in the path of one of the interfering beams (wave-length λ), the intensity at the position where the central maximum occurred previously remains unchanged. The minimum thickness of the glass-plate is a) 2λ b) 2λ/3 c) λ/3 d) λ 19. Two plane mirrors A and B are aligned parallel to each other, as shown in the figure. A light ray is incident at an angle of 300 at a point just inside one end of A. The plane of incidence coincides with the plane of the figure. The maximum number of times the ray undergoes reflections (including the first one) before it emerges out is a) 28 b) 30 c) 32 d) 34 20. A wooden block, with a coin placed on its top, floats in water as shown in fig. The distance l and h are shown there. After some time the coin falls into the water (fig.8). Then a) l decreases and h increases b) l increases and h decreases c) both l and h increase d) both l and h decrease

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197

21. A simple pendulum is oscillating without damping. When the displacement of the bob is less than maximum, its acceleration vector a is correctly shown in (fig.9)

22. A cylinder rolls up an inclined plane, reaches some height, and then rolls down (without slipping throughout these motions). The directions of the frictional force acting on the cylinder are a) up the incline while ascending and down the incline while descending b) up the incline while ascending as well as descending c) down the incline while ascending and up the incline while descending d) down the incline while ascending as well as descending 23. A circular platform is free to rotate in a horizontal plane about a vertical axis passing through

its center. A tortoise is sitting at the edge of the platform. Now, the platform is given an angular velocity ω0. When the tortoise moves along a chord of the platform with a constant velocity (with respect to the platform), the angular velocity of the platform ω(t) will vary with time t as (fig.10)

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24. A particle, which is constrained to move along the x-axis, is subjected to a force in the same direction which varies with the distance x of the particle from the origin as F(x) = -kx + ax3. Here k and a are positive constants. For x ≥ 0, the functional form of the potential energy U(x) of the particle is (fig.11)

25. Two blocks of masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse gives a velocity of 14 m/s to the heavier block in the direction of the lighter block. The velocity of the center of mass is a) 30 m/s b) 20 m/s c) 10 m/s d) 5 m/s 26. A siren placed at a railway platform is emitting sound of frequency 5 kHz. A passenger sitting in a moving train A records a frequency of 5.5 kHz while the train approaches the siren. during his return journey in a different train B he records a frequency of 6.0 kHz while approaching the same siren. The ratio of the velocity of train B to that of train A is a) 242/252 b) 2 c) 5/6 d) 11/6 27. A sonometer wire resonates with a given tuning fork forming standing waves with five antinodes between the two bridges when a mass of 9 kg is suspended from the wire. when this mass is replaced by a mass M, the wire resonates with the same tuning fork forming three antinodes for the same positions of the bridges. The value of M is a) 25 kg b) 5 kg c) 12.5 kg d) 1/25 kg 28. An ideal spring with spring-constant k is hung from the ceiling and a block of mass M is attached to its lower end. the mass is released with the spring initially unstretched. Then the maximum extension in the spring is a) 4 Mg/k b) 2 Mg/k c) Mg/k d) Mg/2k 29. A geo-stationary satellite orbits around the earch in a circular orbit of radius 36000 km. Then, the time period of a spy satellite orbiting a few hundred kilometers above the earth’s surface (REarth = 6400 km) will approximately be a) 1/2 h b) 1 h c) 2 h d) 4 h

30. The effective resistance between points P and Q of the electrical circuit shown in the figure is a)

2Rr R+r

b)

8R ( R + r ) ) 3R + r

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c) 2r+ 4 R 1.

2.

d)

5R 2 + 2r

SOLUTIONS The radius of a charged particle in a uniform magnetic field is r = mv/qB. If the particle is not to enter the region x > b, the radius should have only a maximum value b-a. So b-a = mv/qB ⇒ b = qB (b-a) /m. Ans.B The potential energy between the two charges q,q and the charge Q at the origin is Qq 1 Qq + when charge is displaced by a distance x. Before displacement the 4πε 0 a − x a + x potential energy is

1 2Qq . 4πε 0 a

The change in energy can be found by finding the

difference between the two terms. This will be ∆U = 3.

will be

5. 6.

7.

2πε 0 a 3

, which is proportional to x2.

Ans.B The radius vector joining the conductor to the point is r = ix+jy. The magnitude of the field µ0 I µ I . (1) The direction of the field will be unit vector perpendicular will be 0 = 2πr 2π x 2 + y 2 to the radius vector r. This will be

4.

Qqx 2

ˆi y − ˆjx x2 + y2

(2). From (1) and (2) the magnetic field vector

µ 0 I ( yˆi − xˆj) . Ans.A 2π ( x 2 + y 2 )

In a parallel connection power is same. So power across B3 is the same as power across B1 and B2 together. So W3 is maximum. In a series connection power is directly proportional to resistance. 60 W lamp has greater resistance. So it has greater power. i.e. W2 > W1. Thus W1 < W2 < W3. Ans.D When the current is increasing in P clockwise induced current in Q is anti-clockwise by Lenz’s law. When current is decreasing in P, the induced current in Q is clockwise. Ans.D Induced emf e is proportional to number of turns N. Power is proportional to (emf)2. When the number of turns is made for four times, power increases to 16 times. When radius of wire is halved, resistance increases to four times. Since length increases to four times, resistance increases to 4 x 4 = 16 times. Thus power e2/r remains constant. Ans.B Use the short cut equation for loss of electrostatic energy ∆U = here C1 = C2 ⇒ ∆U =

C1C2 (V1 − V2 )2 . We have 2(C1 + C2 )

1 C( V1 − V 2 ) 2 Ans.C 4

8.

Magnetic lines of force start from north pole and end at south pole outside and continue from south pole to north pole inside the magnet. Ans.D

9.

Work done along AB = 10 (2-1) = 10 J. Work done along BC = 0 (no change in volume). Work done along AC be equal to W. Then 10 + 0 + W = 5, that is heat supplied. This gives W = -5J. Ans.A

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10. Bulk modulus of an ideal gas = K =

P 1 ∆V / V . Compressibility β = . Thus we = ( ∆V / V) K P

find β is inversely proportional to P. hyperbola. Ans.A

The graph between them will be a rectangular

11. By Bohr’s postulate, the angular momentum in nth orbit = n(h/2π). So it is same for hydrogen

and helium atom. The energy of hydrogen atom is -13.6/n2 eV for nth state and -13.6 Z2 / n2, for helium atom. So it is more for helium atom. Ans.B

12. A black body is a good emitter and a good absorber. When it is thrown into the furnace, it

first becomes dark. Then it reemits all radiations. It becomes brightest. Ans.A 13. If n is the number of electrons striking the target, e its charge, current I = ne. n = I/e = 3.2 x

2 x 1016. Potential will not decide the number of electrons. It decides 10-3/ 1.6 x 10-19 = only the energy of electrons. Ans.A

14. Only in a γ decay there will be no change in mass number and atomic number. Ans.C 15. Use the short cut equation

N0 = 2t / T N

Here N0/N = 16 ⇒ t/T = 4 ⇒ t is four times T.=

400 µs. Ans.A 0

16. From fig. when the lower end of the rod is seen, the angle of incidence is 45

.

sin i sin i 1 . r = 450. This gives sin i = 1/ 5 ⇒ µ = 5 / 2 . Ans.B = µ ag ⇒ = sin r sin r µ ga 17. We use here Lens maker’s formula

1 1 1 . For no dispersion, d (1/f) should be = (µ − 1) + f R R 2 1

zero ⇒ R1 = -R2. i.e. one surface must be convex and the other concave. Ans.C. 18. If µ is the refractive index of glass, n is a number of fringes shifting, for central maximum to

remain unchanged (µ-1)t = nλ ⇒ t = 2λ. Ans.A 0

19. During each reflection, there is a lateral displacement of x = d tan 30 , where d is the distance

between the glass plates. If n is the number of reflections, then n d tan 30 = length of the plate L. n x 0.2(1/ 3 ) = 2 3 ⇒ n=30. Total number = 30+initial+final = 32. Ans.C 20. By law of flotation, weight of floating body = weight of liquid displaced. When the coin is

removed, the weight of body decreases. Hence it rises and ‘l’ decreases. When the coin falls into water, it displaces water equal to its volume. The volume of coin is numerically less than its mass. Hence water level falls ‘h’ decreases. Ans.D 21. A body in motion along a curve has two accelerations. They are tangential acceleration at along the tangent and centripetal acceleration ar along the radius towards centre. Its vector G G G sum a = a t + a r . Ans.C 22. On an inclined plane the force of gravity is always downwards, even if the cylinder moves up or down. Hence frictional force is always upwards. Ans.B 23. The initial distance of the tortoise from centre of circle O is equal to radius of platform r. As it moves from A its distance from O decreases, reaches minimum at O, then increases. Its moment of inertia I = mr2 ecreases and then increases. As the platform is a closed system Iω = Crms . So ω first increases and then decreases. Ans.B

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24. If F is the force and U potential energy, F= −

dU Kx 2 ax 4 ⇒ U = − ∫ F dx = − + . For U = 0, the dX 2 4

equation is quadratic in x. Hence the graph of U against x will touch x axis at two points. This will eliminate answers B and C From the equation we find F = 0 when x = 0 . dU = 0 ⇒ The slope of graph U against X should be 0 at the origin. This eliminates ⇒ dX x =0

answer A. Ans.D 25. The spring mass system has only internal force velocity of centre of mass v(c.m) = m1v1 + m 2 v 2 (10x14) + (4 x 0) = 10 m/s Ans.C = m1 + m 2 14 26. If VA is velocity of A, VB that of B, V velocity of sound, the frequencies heard by the

passenger are f1 =

V f (V + VA ) f (V + VB ) ⇒ B = 2 . Given f = 5, f1 = 5.5 f2 = 6, we get Ans.B , f2 = V V VA th

27. The frequency of sonometer wire when it vibrates in n n 2L

T n = µ 2L

mode (in ‘n’ loops) is fn =

Mg ⇒ n M = constant, n is the number of nodes. µ

M1 n = 2 ⇒ M1= 9 kg. M2 n1

⇒ M2 = 25 kg Ans. A 28. If A is maximum extension, the work done in stretching = kinetic energy of the opring. Work 2Mg 1 = force x distance. the force here is constant = Mg ⇒ MgA = kA 2 ⇒ A = Ans.B k 2 3/2 29. By Kepler’s third T α R if T is the period of geostationary satellites Ts that of spy satellite T 36000 + 6400 = Ts 6400

3/ 2

= 7 7 ⇒ Ts =

24 7 7

nearly 2h. One can also answer the question with

the theory information that the period of a satellite very close to earth is 84 minutes. The distance of spy satellite should be slightly more as it is orbiting a few hundred km above earth. The nearest answer is 2h. Ans.C 30. By symmetry remove 2R from the centre of the net work. (Note this is similar to Wheatsone’s bridge.) This leaves three resistances 2R + 2R, 2R+2R, r+ r, in parallel. This gives

2Rr Ans.A R+r

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IIT-SCREENING TEST QUESTIONS 2001 1.

2.

3.

4.

5.

A quantity X is given by ε 0 L

∆V , where ε0 is the permittivity of free space, L is a length, ∆V ∆t

is a potential difference and ∆t is a time interval. The dimensional formula for X is the same as that of A) resistance B) charge C) voltage D) current The pulleys and strings shown in the figure are smooth and of negligible mass. For the system to remain in equilibrium, the angle θ should be A) 00 B) 300 0 C) 45 D) 600 A string of negligible mass going over a clamped pulley of mass m supports a block of mass M as shown in the figure. The force on the pulley by the clamp is given by A)

2 Mg

B)

2 mg

C)

(M + m) 2 + m 2 g

D)

( M + m) 2 + M 2 g

2m m

m m

An insect crawls up a hemispherical surface very slowly (see Fig). The coefficient of friction between the insect and the surface is 1/3. If the line joining the centre of the hemispherical surface to the insect makes an angle α with the vertical, the maximum possible value of α is given by A) cot α = 3 B) tanα = 3 C) sectα = 3 D) cosecα = 3 Two particles of masses m1 and m2 in projectile motion have velocities v1 and v2 respectively at time t = 0. They collide at time t0 . Their velocities become v1’ and v2’ at time 2t0 while still moving in air. The value of m1v1’ + m2v2’ )-(m1v1 + m2v2) is A) zero

B)

(m1 + m2) gt0

C) 2(m1 + m2) gt0

D)

1 (m1 + m2) gt0 2

M

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6.

One quarter sector is cut from a uniform circular disc of radius R. This sector has mass M. It is made to rotate about a line perpendicular to its plane and passing through the centre of the original disc. Its moment of inertia about the axis of rotation is 1 MR 2 2 1 C) MR 2 8 A)

7.

B)

1 MR 2 4

D)

2 MR 2

A simple pendulum has a time period T1 when on the earth’s surface and T2 when taken to a height R above the earth’s surface, where R is the radius of the earth. The value of T2/T1 is B) √2

A) 1 8.

C) 4

D) 2

A small block of shot into each of the four tracks as shown below. Each of the tracks rises to the same height. The speed with which the block enters the track is the same in all cases. At the highest point of the track the normal reaction is maximum in

v

9.

v v v (A) (B) (C) (D) A hemispherical portion of radius R is removed from the bottom of a cylinder of radius R. The volume of the remaining cylinder is V and its mass is M. It is suspended by a string in a liquid of density ρ, where it stays vertical. The upper surface of the cylinder is at a depth h below the liquid surface. The force on the bottom of the cylinder by the liquid is A) Mg

h

B) Mg-V ρg

ρ

C) Mg- πR2 h ρg D) ρg (V+πR2h)

2R

10. The ends of a stretched wire of length L are fixed at X = 0

and X = L. In one experiment, the displacement of the wire is Y1 = A sin (πx/L) sin ωt and energy is E1 and in another experiment its displacement is Y2 = A sin (2πx/L) sin 2ωt and energy is E2 . Then A) E2 = E1 B) E2 = 2E1 C) E2 = 4E1 D) E2 = 16 E1 11. Two pulses in a stretched string whose centres are initially 8 cm apart are moving towards each other as shown in the Fig. The speed of each pulse is 2 cm/s. After 2 seconds, the total energy of the pulses will be A) zero C) purely potential 12. P-V plots for two shown in the Fig.

B) purely kinetic D) partly kinetic and partly potential gases during adiabatic processes are Plots 1 and 2 should correspond

8 cm

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respectively to A) He and O2

P

B) O2 and He C) He and Ar D) O2 and N2 1 2

13. Three rods made of the same material and having

the same cross section have been joined as shown in the Fig. Each rod is of the same length. The left and right ends are kept at 00C and 900C respectively. The temperature of the junction of the three rods will be

V 90°C

0

A) 45 C B) 600C 0°C C) 300C D) 200C 0 14. When a block of iron floats in mercury at 0 C, a fraction k1 of its volume is submerged, while at the temperature 600C, a fraction k2 is seen to be submerged. If the coefficient of volume expansion of iron is γFe, and that of mercury γHg then the ratio k1/k2 can be expressed as A)

1 + 60 γ Fe 1 + 60 γ Hg

B)

D)

1 + 60 γ Fe 1 − 60 γ Hg

D)

90°C

1 − 60 γ Fe 1 + 60 γ Hg 1 + 60 γ Hg 1 + 60 γ Fe

15. In a given process on an ideal gas, dW = 0 and dQ < 0. Then for the gas

A) the temperature will decrease B) the volume will increase C) the pressure will remain constant D) the temperature will increase 16. Three positive charges of equal value q are placed at the vertices of an equilateral triangle. The resulting lines of force should be sketched as in

(A)

(B)

(C)

(D)

17. A uniform electric field pointing in positive x-direction exists in a region. Let A be the origin,

B the point on the x-axis at x = +1 cm and C be the point on the y-axis at y = +1 cm. Then the potentials at the points A,B and C satisfy A) VA < VB C) VA < VC

B) VA > VB D) VA > VC

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18. Consider the situation shown in the Fig. The capacitor A has a charge q on it whereas B is

uncharged. The charge appearing on the capacitor B a long time after the switch is closed is A) zero

B) q/2

C) q

D) 2q

q _ +

19. A coil having N turns is wound tightly in the form of a

spiral with inner and outer radii a and b respectively. When a current passes through the coil the magnetic field at the centre is µ NI A) 0 b C)

b µ0 NI In 2(b − a ) a

2µ0 NI B) a D)

_

+

_

+

S

_

+ A

B

b µ0 NI N In 2( b − a ) a

20. Two particles A and B of masses mA and mB respectively

and having the same charge are moving in a plane. A uniform magnetic field exists perpendicular to this plane. The speeds of the particles are vA and vB respectively and trajectories of particles are as shown in the Fig. Then

A B

A) mAvA < mB vB B) mAvA > mB vB C) mA < mB and vA < vB D) mA = mB and vA = vB 21. Two circular coils can be arranged in any of the three situations shown in the Fig. Their

mutual inductance will be

(A) A) maximum in situation (A)

C) maximum in situation (C)

(B)

(C) B) maximum in situation (B)

D) the same in all situations

22. A wire of length L and 3 identical cells of negligible internal resistances are connected in

series. Due to the current the temperature of the wire is raised by ∆T in a time t. A number N of similar cells is now connected in series with a wire of the same material and cross-section but of length 2L. The temperature of the wire is raised by the same amount ∆T in the same time t. The value of N is A) 4

B) 6

C) 8

D) 9

23. A ray of light passes through four transparent media with refractive indices µ1, µ2, µ3 and µ4

as shown in the Fig. The surfaces of all media are parallel. If the emergent ray CD is parallel to the incident ray AB, we must have

206

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B) µ2 = µ3

C) µ3 = µ4

D) µ4 = µ1

D

µ1 µ2 µ3

24. A given ray of light suffers minimum deviation in an

equilateral prism P. Additional prisms Q and R of identical shape and of the same material as P are now added as shown in the Fig. The ray will now suffer

C

B

µ4

A

A) greater deviation B) no deviation C) same deviation as before

Q

D) total internal reflection

P

R

25. Two beams of light having intensities 1 and 4I interfere

to produce a fringe pattern on a screen. The phase difference between the beams is π/2 at point A and π at point B. Then the difference between the resultant intensities at A and B is A) 21

B) 41

C) 51

D) 71

26. The transition from the state n = 4 to n = 3 in a hydrogen-like atom results in ultraviolet

radiation. Infra-red radiation will be obtained in the transition A) 2 → 1

B)

3→2

C) 4 → 2

27. The intensity of X-rays from a Coolidge tube is

plotted against wavelength λ as shown in the Fig. The minimum wavelength found is λC and the wavelength of the ⇒ Kα line is λK . As the accelerating voltage is increased

D) 5 → 4

I

A) λK - λC increases B) λK - λC decreases C) λK increases

C

K

D) λK decreases 28. The electron emitted in beta radiation originates from

A) inner orbits of atoms

V

R

V

C

B) free electrons existing in nuclei C) decay of a neutron in a nucleus D) photon escaping from the nucleus 29. In the given circuit with steady current, the potential

drop across the capacitor must be A) V

B)

V/2

C) V/3

D)

2V/3

2V

2R

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30. A metallic square loop ABCD is moving in its own

plane with velocity v in a uniform magnetic field perpendicular to its plane as shown in the Fig. An electric field is induced

A

B

D

C

A) in AD, but not in BC B) in BC but not in AD C) neither in AD nor in BC D) in both AD and BC 31. A radioactive sample consists of two distinct species having equal number of atoms initially.

The mean life time of one species is τ and that of the other is 5τ. The decay products in both cases are stable. A plot is made of the total number of radioactive nuclei as a function of time. Which of the following Fig.s best represents the form of this plot ? N

N

N

N

t t t t (A) (B) (C) (D) 32. In the given circuit, it is observed that the current I is independent of the value of the resistance R6. Then the resistance values must satisfy R5

A) R1 R2 R5 = R3 R4 R6 B)

1 1 1 1 + = + R 5 R 6 R1 + R 2 R 3 + R 4

R1

I

C) R1 R4 = R2 R3

R6

R2

D) R1 R3 = R2 R4 = R5 R6

R3 R4

33. In a Young’s double slit experiment, 12 fringes are

observed to be formed in a certain segment of the screen when light of wavelength 600 nm is used. If the wavelength of light is changed to 400 nm, number of fringes observed in the same segment of the screen is given by A) 12

B) 18

C) 24

34. A non-planar loop of conducting wire carrying a

current I is placed as shown in the Fig. Each of the straight sections of the loop is of length 2a. The magnetic field due to this loop at the point P(a,0,) points in the direction A) C)

1 2 1 3

( −ˆj + kˆ )

B)

( −ˆi + ˆj + kˆ )

D)

D) 30 z I

1

( −ˆj + kˆ + ˆi ) 3 1 ˆ ˆ (i + k ) 2

y x

2a

35. A particle executes simple harmonic motion between x = -A and x = +A. The time taken for

it to go from 0 to A/2 is T1 and to go from A/2 to A is T2. Then

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A) T1 < T2

B) T1 > T2

C) T1 = T2

D) T1 = 2T2

SOLUTIONS 1.

Substituting the units of the given quantity we have εo L

F V FxV C dV xmx = = = A. . = m t t t dt

Thus the given quantity is current. Ans.D 2.

Consider the free body diagram of mass m. We find mg = T (1). Next consider the free body diagram of mass√2 mg. We find that the vertical component of the tension should be equal to weight √2mg⇒√2mg = 2Tcosθ = 2mgcosθ ,since T = mg from equation (1). ⇒ θ = 45o. Ans.C 2T cos T

T

T m

T sin

T sin

mg

3.

2 mg

Consider free body diagram of the block as shown in fig.1. We can write Mg = T (1). Now consider figure 2 which is the free body diagram of the pulley. If F is the net force on the F T

T

Mg F1

mg

T

pulley, it should be equal to the vector sum of the two forces mg+T and T.⇒ 2

F= ( mg + T ) 2 + T 2 = (mg + Mg ) 2 + ( Mg ) 2 = (m + M ) + M

co s

5.

g. Ans.D

Let P be the point where insect leaves the sphere. This happens when the centripetal force vanishes. i.e. N-mgcosα = mv2/r should become zero. ⇒ N= mg cosα (1). At this point the frictional force along the tangent µN should be equal to mgsinα i.e. µN = mgsinα (2). Dividing N equation (2) by (1), we get tan α = µ ⇒ cotα = P 1/µ = 3. Ans.A The given quantity (m1v1'+m2v2')- (m1v1+m2v2) is the magnitude of change in momentum of the mg mg sin projectiles. Change in momentum = Impulse I. Impulse I =force x time. The only force acting on the system here is the force of gravity. i.e. weight for the two bodies (m1+m2)g. The time for which the force acts is 2to. So the impulse = (m1+m2)g x 2to. Ans. C. m g

4.

2

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6.

Moment of inertia is a scalar quantity. It can be added , if the axis of rotation is the same. The moment of inertia of the quarter sector is one-fourth of the moment of inertia of the disc as a whole. The moment of inertia of the disc about an axis passing through centre of mass perpendicular to the plane is 4MR2/2, since 4M is the mass of the disc. The moment of inertia of the sector = (1/4) 4MR2/2 = MR2/2. Ans. A

7.

If g1 is acceleration due to gravity at the surface of earth and g2 its value at an altitude h=R, we have g 1 =

GM R2

(1) and g 2 =

GM

=

(R + h ) 2

GM (R + R ) 2

=

GM 4R 2

=

g1 4

(2).

Dividing the two

equations, g1/g2 = 4. The periods of the pendulum at the surface of earth and at the altitude R are given T1= 2π 8.

T l l , T2= 2π ⇒ 1 = g1 g2 T2

g2 = g1

1 1 = ⇒ T2 = 2T1.. Ans.D 4 2

If N is the normal reaction at the highest point, mg weight, R radius of the track, then the centripetal force is given by mg − N =

mV 2 mV 2 ⇒ N = mg − , where V is velocity at highest R R

point..Here reaction N is maximum, when radius of the track R is minimum because v is constant. In the given choices R is minimum in fig. (A). Ans.A 9.

The force on the bottom of the cylinder F' - Force on the top (F) = Buoyant force (B). This gives force at the bottom F' = F+B. F = Thrust on the top surface πr 2 hρg . Buoyant force B = Weight of liquid displaced = Vρg. ⇒ F' = πr 2 hρg +Vρg = ρg( πr 2 h +V). Ans.D. This question can also be answered by simple reasoning, once we know the thrust at the bottom is sum of thrust at the top and buoyant force. The thrust at the top depends only on the height, density of liquid and g. The buoyant force will depend only on the volume of cylinder and density of the liquid. So the required equation should not contain the mass of the cylinder. This eliminates all the first three choices. 2

2

10. The energy of SHM is given by (1/2)mω A , where A is the amplitude. In both the given

equations maximum displacement (amplitude) is A. But the angular frequency of the first is ω, while that of the second is 2ω. This gives

E 2 ( 2ω) 2 = = 4 ⇒ E 2 = 4E1 .. Ans. C. E1 ω2

11. After two seconds each pulse would have travelled 4 cm and would arrive at the same point so

that their maximum will be at the same points. That is, each particle will be at the respective mean position i.e position of zero displacement. At mean position the energy of a simple harmonic particle is purely kinetic. Ans.B 12. The slope of an adiabatic curve is equal to the ratio of specific heats γxP. The slope of curve

2 is greater than that of 1. γ of monatomic gas is greater than that of diatomic gas. This would mean that curve 2 corresponds to monatomic and curve 1 diatomic. Oxygen is diatomic and He monatomic. Ans. B. 13. Let T be the temperature of the common junction of three rods at thermal equilibrium. Then

heat conducted/s between the left end and common junction= heat conducted per s between common junctions and the two right ends. i.e.

λA(T − 0) λA(90 − T ) λA(90 − T) , where λ = + x x x

is the thermal conductivity of the material of the rod, x length of each rod, A cross sectional area. This gives T = 600. Ans. B.

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14. Fraction of the immersed volume = density of body/density of liquid. This gives k1 =

density of Fe at 0o C o

density of Hg at 0 C

=

d (Fe ) density of Fe at 60o C d 0 ( Fe) = 60 . We know , k2 = o d 0 ( Hg) d density of Hg at 60 C 60 ( Hg )

d 0 ( Fe) = d 60 (1 + γ Fe 60) , d 0 ( Hg ) = d 60 (1 + γ Hg 60) . This gives

1 + 60 γ Fe k1 , . Ans.A. = k 2 1 + 60 γ Hg

15. We use the first law of thermodynamics dQ = dU+dW. It is given dQ is negative and dW= 0.

This would mean dU is negative. The internal energy decreases. Temperature of system decreases. Ans.A. 16. In choice (A), lines of force starting from a positive charge end at another positive charge.

Hence (A) is wrong. In choice (B) and choice (D) lines of force form closed loops. This cannot happen for positive charges. Therefore, choice (C) is correct. Ans.C. 17. Since a uniform electric field exists along the positive X-axis, a free positive charge will

move from A to B. We also know a positive charge moves from higher to a lower potential. This means potential at VA is greater than VB. Ans.B. 18. The right plate of the capacitor B is isolated. For charges to flow from A to B, and B to work

as a capacitor, the right plate will have to be earthed or connected by a wire to some other plate. Under the given condition B cannot attain any charge. Ans.A. 19. The equation for magnetic field due to a spiral of inner radius a and outer radius b will act as a

coil of width b-a and should contain both a and b. This rules out the choices (A) and (B). In choice (D) the power of current is IN. This will make the equation for magnetic field dimensionally incorrect. The correct choice, therefore, is (C). Ans.C. 20. The radius of a charged particle in a uniform magnetic field is given by r =

are same for both the particles. From figure rA is greater than rB.⇒ >mBvB.

mv . Here q and B qB

mA v A m B v B > ⇒ mAvA qB qB

Ans.B.

21. The mutual inductance between the two coils is maximum when the flux produced by one coil

links maximum with the other. The magnetic field produced by a coil is along the axis. So flux lines will come along the axis. The second coil will intercept maximum flux when it presents maximum cross sectional area to the flux. This happens in a situation shown in fig.(A). Ans.A. 22. We shall assume that the wires are uniform so that the length is proportional to mass and

proportional to resistance. If E is the emf of each cell, c the specific heat of the wire, then the heat

produced

[ NE ] 2 = 2mc∆T 2R

in

the

two

cases

respectively

will

be

[3E] 2 = mc∆T R

(1)

(2) Dividing the two equations we get N2 = 36 ⇒ N=6. Ans.B.

23. By Snell's law of refraction for media 1 and 4, µ1 sin θ1 = µ 4 sin θ4 ⇒

sin θ1 µ 4 = ⇒ µ 4 = µ1 sin θ4 µ1

since θ1 = θ 4 .Ans.D. 24. If we consider the prisms Q and R joined together, they form a parallel-sided glass. A ray of

light entering from P to this combination should go parallel after refraction through the

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combination Q and R. So the deviation of the ray after passing through P will remain the same even after passing through Q and R. Ans.C 25. The resultant intensity when two waves of intensity I1 and I2 superpose at a phase difference

θ

is

given

by

I = I1 + I 2 + 2 I1 I 2 cos θ .

and I B = I + 4I + 2 I 4I cos π = I

Using

this

I A = I + 4I + 2 I 4I cos π / 2 = 5I

⇒ I A -IB= 4I. Ans.B

26. The energy difference between quantum states goes on decreasing as En is proportional to

1/n2. The energy difference for infrared should be less than that for for ultraviolet radiation. If transition from 4 to 3 produces ultra-violet radiation, then the energy difference will be less only for transition from 5 to 4. Ans. D

27. The wavelength of Kα radiation λK is a constant for a given target material. The wavelength hc of radiation is related to accelerating voltage by the equation λC = . That is λ is inversely eV

proportional to V. increases. Ans.A.

When the accelerating voltage is increased, λC decreases. ⇒λK-λC

28. Electrons do not exist in the nucleus. Decay of neutron produces electron and proton. Ans.C. 29. At the steady state no current flows through the branch containing condenser. Considering 2V − V V the rest of the circuit, the current i = = . If we now consider the upper part of the R + 2R 3R VR loop and the part containing condenser, the charging potential of the condenser is V+ -V = 3R V . Ans.C. 3 30. The induced e.m.f when a conductor of length l moves in a magnetic field B with a velocity v

is Blv sinθ. Here the sides AD and BC are perpendicular to the field θ= 900 and hence will have induced e.m.f.s. The sides AB and CD are parallel to the field and hence the induced e.m.f will be Blv sin0 =0. The induced e.m.fs produce potential difference in the side AD and BC and hence an electric field. Ans.D

31. What ever may be the value of mean life or half life, the number of atoms remaining to

undergo disintegration should only decrease with time according to the exponential law N = N 0 e −λt .This rules out answers (A),(B) and (C). Ans.D. 32. If the current through R6 is independent of the value of R6, then the branches R1, R3, R4, R2

should form a balanced Wheat stone's bridge, with R6 taking the place of galvanometer resistance. For balance, we should have

R1 R 2 = ⇒ R 1 R 4 = R 2 R 3 Ans.C. R3 R 4

33. The bandwidth in Young's experiment is given by β =

Dλ i.e. β is proportion al to λ. When d

λ

decreases from 600 to 400 nm, the bandwidth decreases to 2/3. The number of bands in a given distance x = x/ λ increases to 3/2 times i.e. 12x3/2 = 18. Ans.B 34. Treat the given loops as two separate loops one in the YZ plane and the other in the XY plane.

The magnetic field produced by the loop is perpendicular to the plane of the loop. So it should be respectively along +X axis and +Z axis from the current direction. Our answer

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should, therefore, contain unit vectors along the X-axis ˆi and Z-axis kˆ .The unit vector in the direction of ˆi + kˆ is the unit vector in the direction of magnetic field. This is

1 2

(ˆi + kˆ ) .

Ans.D. 35. The equation to displacement of S.H.M is given by A sinθ or A sin ωt, where A is amplitude.

We know sinθ increases fast first and then increases slowly. ( For example sin 0 = 0, sin 300= 1/2 and sin 900 =1). So the displacement increases from 0 to A/2 quicker than that from A/2 to A. So time necessary to cover the first half of the amplitude is less than that for the second half. ( If T is the period of S.H.M, we can easily prove that T1 = T/12 and T2 = T/6.) Ans.A

View more...
2.

3.

4.

5.

6.

7.

8.

Two vectors have magnitudes in the ratio 1:2. Their vector sum is perpendicular to the smaller vector. The angle between the vectors is a) 1500 b) between 1500 and 1200 0 c) between 1200 and 900 c) 120 Two satellites A and B revolve round earth in circular orbits of radius r and 1.01r. The percentage increase in the period of B with respect to A is a) 0.5 b) 0.75 c) 1 d) 1.5 Which of the following pair has not same dimension? a) Curie and angular velocity b) velocity gradient and frequency of light c) Gravitational field and latent heat d) potential gradient and force per unit charge Three particles of same mass rest on the corners of an equilateral triangle. The gravitational force between any two of them has a magnitude F. Then gravitational force on any of them has magnitude a) F b) 2F c) √3 F d) √2 F The momentum of a car is increased by 10% and then decreased by 10%. In the process the kinetic energy of car a) increases by 2% b) decreases by 2% c) increases by 1% d) decreases by 1% The position vector of a particle is r = (a cos ωt ) ˆi + a (sin ωt ) ˆj . The velocity vector is a) perpendicular to position vector b) parallel to position vector c)always directed towards origin d) always directed away from the origin Read the following statements A and B: A: If a quantity has units, it will have dimensions. B: If a quantity has dimensions, it will have units Of these statements a) Both A and B are true b) Both A and B are false c) A is true but B is false d) B is true but A is false The equation of the motion of a body initially at rest is given by

dv = 8-4v, where v is dt

velocity of the body at the time. Check the wrong statement a) terminal velocity of the body is 2m/s b) initial acceleration of the body is 8 m/s2 c) the velocity of the body when acceleration is half the initial value is 1 m/s d) velocity is related to time by the equation v = at + bt2 where a and b are constants.

2

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A set of N cubical blocks rest on a smooth horizontal table such that distance between 2 near surfaces of the block is L. The block at one end is given a speed v to the next at the time t = 0. All collisions are perfectly inelastic. then the last block starts to move after a time ( N − 1) L ( N − 1)L N ( N − 1)L N ( N − 1)L a) b) c) d) v 2v 2v v 10. In the previous question, centre of mass of the system will have a final speed v v c) v d) a) zero b) 2 N 11. A rocket moves with a speed v where air resistance is proportional to the cube of its speed. If the speed of the rocket is to be increased by 1%, the increase in the power will be a) 1% b) 2% c) 3% d) 4% 3 12. Water issuing out as a horizontal jet with a speed v1 at a rate V m per second strikes a plate normally moving towards it with a speed v2. If density of water is p and water splashes down after hitting the plate, the force on the plate in newton is 9.

a) pv1 V

b) p(v1 +v2) V

c) p(v2-v1)V

d)

pv 1 2 V v1 + v 2

13. Read the following statements carefully.

14.

15.

16.

17.

18.

Statements x: A vector quantity has magnitude and direction. Statement y: All quantities, which have magnitude and direction, are vectors. Here a) both statements are correct b) both statements are not correct c) x is correct and y is not d) y is correct and x is not If R is the radius of earth’s orbit round sun, the angular momentum of earth round sun is proportional to a) R2 b) R3/2 c) R d) R1/2 Two particles P and Q move towards each other due to mutual gravitational force. At a certain time the speed of P is v and speed of Q is 2v. The speed of centre of mass of the system is a) 0 b) v/2 c) 3v/2 d) v Two masses 0.4 and 0.6 kg are suspended from the zeroeth and hundredth cm division of a meter scale of negligible mass. It is required to rotate the meter scale with constant angular velocity with minimum work. The axis of rotation should pass through at a) 40 cm b) 50 cm c) 60 cm d) 70 cm An infinite number of point masses m kg each are placed at a distances x = 1cm, x = 2cm, x = 8 cm.....etc from a point 0 on the x-axis. The gravitational potential at the point 0 in J/kg is a) infinity b) zero c) -100 Gm d) -200 Gm A simple pendulum of length L has a bob of mass m. It sis displaced by an angle θ to the vertical. If T is a tension of a string, v is the speed of the bob then, check the correct statement a) T = mg cos θ

b) T = mg cos θ -

mv 2 L

mv 2 d) T cos θ = mg L n 19. The gravitational force between a planet and a satellite varies by a distance law f αr . The period of the satellite will be independent of distance if n is equal to a) 1 b) 2 c) -1 d) 3/2 c) T = mg cosθ +

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3

20. Two boys A and B hold two ends of a light rope carrying a 1 kg weight at the centre. If the

length of the rope is 1 metre the work done by the boys to straighten the rope is a) 0 b) infinity c) 4.9 J d) 9.8 J 21. Two particles of mass m each move in a circle of radius r due to their mutual gravitational force. The period of their motion is

a)

22.

23.

24.

25.

2π

r3 Gm

b ) 4π

r3 Gm

c) π

r3 Gm

d ) 2π

Gm

r3 A car, a truck and a lorry of masses in the ratio 1:2:3 have equal kinetic energy. Their braking forces are also same. When brakes are applied at the same time, they are brought to rest in distances s1, s2 and s3 respectively. Then a) s1 > s2 > s3 b) s1 < s2 < s3 c) s1 = s2 = s3 d) s1 + s2 = s3 A shell is fired with an initial velocity u at an angle α to the horizontal. At highest point on its path, it explodes into two equal fragments. One falls vertically down. The distance between the fragments when they reach the ground is b) u2 sin 2α/ 2g a) u2 sin 2α/g 2 d) 3u2 sin 2α/ g c) 2u sin 2α/ g The time of a flight of a projectile is 4 s. One second after projection it moves at an angle 450. Its angle of projection is b) tan-1 (4/3) c) tan-1 (3/2) d) tan-1 (2) a) 600 The velocity vector of the projectile in the above question is parallel to a ) ˆi + 2 ˆi b) 2 ˆi + ˆi c) 2 ˆi + 3 ˆi d ) 3 ˆi + 2 ˆi

26. A fly wheel starting from rest and rotating with uniform angular acceleration turns through

100 rad during third second. During fifth second it will turn through an angle (in rad) a) 100 b) 166.7 c) 180 d) 200 G G G 27. When three forces f 1 = 3ˆi + 4ˆj − 5kˆ , f 2 = 3ˆi + 2ˆj − k and f 3 = x (ˆi + ˆj − kˆ ) act on a body, the body moves with uniform velocity. The value of x here is a) -6 b) 6 c) -9 d) 9 28. A satellite is moving in a circular orbit of radius x around the earth of radius R. The acceleration of the satellite is gR R R2 c) g a) g b) d) g 2 x x x 29. Check the correct statement for a body in uniform circular motion

a) the body has constant centripetal force c) the body has constant velocity

b) the body has varying centripetal force d ) the body obeys T2 α R3 law 30. A body of mass 2 kg moves with a velocity (3 ˆi + 4ˆj) t. The magnitude of force acting on it at this instant is a) 6N b) 8N c) 10N d) 7N 2 31. A particle moves in a parabolic path whose equation is y = 4ax , with a constant x component of velocity equal to A. The acceleration of the particle is a ) 4aAkˆ b) 8aA 2 ˆj c) 4aA 2 kˆ d) 8aA 2 kˆ 32. For an earth satellite longer orbits will correspond to

a) longer periods and larger velocity

b) longer periods and smaller velocity

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c) smaller periods and smaller velocity

d) smaller periods and larger velocity

33. The velocity-time graph of an aero plane moving along a run way is a parabola. This means

34.

35.

36. 37.

aero plane has a) constant but large acceleration b) acceleration proportional to time c) acceleration proportional to square of time d) acceleration proportional to cube of time The period of a simple pendulum with an iron bob is T. It will be less than T, when the bob is subjected to a) a magnetic field directed upwards b) an electric field directed upwards c) a magnetic field directed downwards d) an electric field directed downwards A body falls from a height h and rebounds from a surface of coefficient of restitution e = 1/√2. The number of collisions it will make before it stops is equal to a) 1 b) 2 c) 4 d) infinite The total distance travelled by the body before it stops to rebound is a) 2h b) 3h c) 4h d) 3√2h A stone is thrown vertically up. It reaches a maximum height of 50 m. The height at which kinetic energy becomes 60% of the initial value is (in m) a) 30

b) 50 0.6

c) 50 0.4

d) 20

38. A trolley of mass 100 kg moves on a smooth horizontal rails with a monkey of mass 20 kg

inside it, with a speed of 10 m/s. The monkey at some point jumps vertically up and catches overhanging branch of a tree. The speed of the trolley after this would be a) 12 m/s b) 10 m/s c) 15 m/s d) 8.33 m/s G ˆ ˆ 39. F is a force represented by the vector F = 2i + 4 j and s is displacement vector given by G ˆ ˆ s = i + j. The component of force in the direction of displacement is a)

20

b)

2

c) 3 2

d) 6

40. A body of mass 1 kg is dropped from the top of a tower on a windy day. The wind exerts a

41.

42.

43.

44.

constant horizontal force of 20N on the body. The path of the body will be a) a parabola b) a straight line c) an ellipse d) an irregular curve A stone A is dropped from the top of a tower of height 40 m. At the same time another stone B is projected from the bottom of the tower with an initial velocity such that the stones collide midway. The initial speed of second stone B is (Take g = 10 m/s2) a) 5 m/s b) 10 m/s c) 20 m/s d) 15 m/s A TV signal sent from a station A to Insat 1-C is received back in a neighboring station B in a time t seconds. The value of t here is nearest to a) 1 s b) 0.1 s c) 0.125 s d) 0.24 s A person is standing at a distance of 8m from a train. The train starts with acceleration 1 m/s2. The person also starts at the same time with an acceleration 2m/s2. The time in which he catches the train is a) 2 s b) 4s c) 8s d) 16s A particle moves with a constant velocity parallel to the x-axis . Its angular momentum with respect to the origin a) increases b) decreases

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c) first increases and then decreases

5

d) remains the same 0

45. The escape velocity of a body projected at an angle 5 with the horizontal from the earth in (in

46.

47. 48.

49.

50.

51.

52.

53.

54.

55.

56.

57.

58.

km/s) is a) 11.2 b) 11.2 cos 50 c) 11.2 sin 50 d) infinite Two uniform gold spheres of radius R each exert a gravitational force of F when in contact. The gravitational force of two uniform gold spheres of radius 2R in contact will be a) F/16 b) F/4 c) F d) 16F The altitude of earth satellite in terms of radius of earth R, when it has a velocity 4km/s is a) R b) 2R c) 3R d) 4R Two forces 100N and 150N act on a body of mass 100 kg. The minimum acceleration this body can have under their combined action is (in m/s2) a) 3.5 b) 1.5 c) 1.25 d) 0.5 A passenger in a car taking a curve feels a centrifugal force on him. This force is a) exerted by car on him b) exerted by ground on him through the car c) the reaction to the centripetal force d) due to inertia The gravitational field at a point at a distance of r from the centre of a uniform sphere of radius R for r < R is proportional to a) 1/r2 b) 1/r c) r d) r2 A particle moving in a horizontal circle of radius r has centripetal force F = -k/r2, where k is a constant. The kinetic energy of the particle is c) 2k/r d) k/2r a) k/r b) k2/r2 The minimum acceleration with which a fireman can climb down a rope of breaking tension 60% of his weight is a) 0.4 g b) 0.6 g c) 0.2 g d) 0.3 g ABC is a triangular plate of uniform thickness with sides AB: BC: CA in the ratio 1:3:5. If IAB is the moment of inertia about AB, IBC that about BC and ICA that about CA, which one of the following is correct ? b) IAB + IBC = ICA c) ICA is minimum d) IAB < IBC a) IBC = IAB A block takes t seconds to slide down a smooth inclined plane of angle 450. The same block takes 2t seconds to slide down a rough inclined plane of same angle. The coefficient of friction of the second plane is a) 0.5 b) 1/√2 c) 0.75 d) 1/2√2 A heavy nucleus at rest emits can α particle. Then a) momentum of α is greater than that of nucleus b) momentum of nucleus is greater than that of α c) kinetic energy of nucleus is greater than that of α d) kinetic energy of α is greater than that of nucleus A thin uniform needle has a moment of inertia I1 about centre of mass axis. It is bent into a ring. The moment of inertia is I2 about c.m. then I1 / I2 will be b) π2/4 c) π2/12 d) 3/π2 a) π2/3 Assuming earth to be a uniform sphere of density ρ, period of a satellite close to earth is related to the density by the equation ‘t’ is proportional to a) ρ b) √ρ c) 1/√ρ d) 1/ρ Three forces each of magnitude 1 N act from one corner towards the other three corners of a square. Their sum has a magnitude nearest to a) 3 N b) 1 N c) 1.4 N d) 2.4 N

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59. A laboratory on earth can be taken as

60.

61.

62.

63.

64.

a) an inertial frame strictly b) an inertial frame approximately c) a non-inertial frame strictly d) either depending on where we are looking from A particle is falling freely along an inclined plane. It travels the last 1/3 of the distance with 1/nth of the total time of fall. Here n is nearest to a) 2 b) 3 c) 4 d) 5 A car is travelling on a level road of coefficient of static friction 0.8. The car cannot have an acceleration more than a) 0.8 m/s2 b) 8 m/s2 c) 4 m/s2 d) 10 m/s2 A mixture of sea water of specific gravity 1.1 is mixed with fresh water of specific gravity 1. If the mixture has specific gravity 1.04 the ratio of volume of sea water to volume of fresh water is a) 1:1.04 b) 5:4 c) 5:2 d) 3:2 A disc with a small hole is rotating through centre of mass axis with a constant period. If the centre of the disc is heated the period of rotation will a) increase b) decrease c) remain same d) may increase or decrease A uniform cylinder of length L and mass M having cross-sectional area A is suspended, with its length vertical, from a fixed point by a mass less spring of force constant k, such that it is half-submerged in a liquid of density ρ at equilibrium position. When the cylinder is given a small downward push and released it oscillates with small amplitude and with a frequency

a)

1 k − Aρg 2π M

1/ 2

1 k − ρgL2 c) 2π M

b)

1 k + Aρg 2π M

d)

1 k − Aρg 2π Aρg

1/ 2

1/ 2

1/ 2

65. Imagine a light planet revolving around a very massive star in a circular orbit of radius R with

a period of revolution T. If the gravitational force of attraction between the planet and the star is proportional to R-5/2, then a) T2 is proportional to R2 b) T2 is proportional to R7/2 2 3/2 d) T2 is proportional to R3.75 c) T is proportional to R 66. *A particle is acted upon by a force is constant magnitude which is always perpendicular to the velocity of the particle. The motion of the particle takes place in a plane. It follows that a) its velocity is constant b) its acceleration is constant c) its kinetic energy is constant d) it moves in a circular path 67. 1 kg of cotton is balanced by 1 kg of a brass weight in a balance covered with a bell jar. If the bell jar is evacuated, a) the cotton pan will go down b) the brass pan will go down c) both will remain in equilibrium d) cotton will first go up and then come down 68. *A diametrical tunnel is dug along the earth. A particle is dropped from a point at a height h directly above the tunnel. Assuming earth’s density to be uniform, a frictional forces are negligible, a) the particle reaches the same height on the other side b) the particle will have zero acceleration when it passes through the centre of the earth. c) the particle has maximum acceleration at the point of release

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7

d) the particle will oscillate simple harmonically 69. When a running athlete stops after a race

a) only his momentum is conserved by himself b) only his kinetic energy is conserved by himself c) both momentum and kinetic energy are conserved by him d) neither momentum nor kinetic energy is conserved by him 70. A block of mass 0.1 kg is held against a wall by applying a horizontal force of 5N on the block. If the coefficient of friction between the block and the wall is 0.5 the magnitude of the frictional force acting on the block is a) 2.5 N b) 0.98 N c) 4.9 N d) 0.49 N 71. A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration ac is varying with time t as ac = k2rt2 where k is a constant. The power delivered to the particle by the forces acting on it is a) 2πmk2r2t b) mk2r2t c) (mk2r2t5)/3 d) zero 72. One end of a long metallic wire of length L is tied to the ceiling. The other end is tied to a massless spring of spring constant K. A mass m hangs freely from the free end of the spring. The area of cross section and the Young’s modulus of the wire are A and Y respectively. If the mass is slightly pulled down and released, it will oscillate with a time period T equal to

a ) 2π

74.

75.

76.

77.

78.

b) 2π

m(YA + KL) YAK

mL mYA d ) 2π YA KL Among the forces in nature, friction can be classified into a) gravitational b)electro magnetic c) weak nuclear d) strong nuclear When a body covers one quadrant of a circle of radius 0.1 metre, the ratio of distance travelled by it to the magnitude of displacement is a) 1 b) π/√2 c) π/2 d) π/2√2 When a person walks along east, the force of friction is along a) east b) west c) any where from east to west d) anywhere from north to south If distance of earth from sun reduces to half the present value, the number of days in an year would be nearly a) 730 b) 229 c) 182 d) 130 Two bodies of same mass but of different material are released from same height in a medium where air resistance cannot be neglected. Then a) the denser body reaches ground first b) the less dense body reaches ground first c) both fall through in the same time d) cannot be answered from the data The level of water in a trough when a train is at rest is horizontal. The train starts to move with uniform acceleration towards right. Then water level will a) still be horizontal b) incline up towards rear of the train c) incline up towards front of the train d) become concave shaped. c) 2 π

73.

m K

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79. (Fig) Three equal weights of mass 2 kg each are hanging on a string passing over a fixed

pulley as shown in the fig. What is the tension of the string connecting weights B and C ? a) 1.3 N b) 1.3 kgwt c) 3.3 N d) 19.6 N 80. If a body moves under constant power, the distance s travelled by it will be related to the time t by the equation a) s α t3/2 b) s α t2/3 B 2 d) s α t1/2 c) s α t 81. Two bodies M and N of equal mass are suspended from two separate massless springs of spring constant k1 and k2 C A respectively. If the two bodies oscillate vertically such that their maximum velocities are equal, the ratio of amplitude of vibration of M to that of N is equal to a)

k1 k2

b)

k2 k1

c)

k2 k1

d)

1

k1 k2

82. If in the above question, the bodies oscillate so that they have equal maximum acceleration,

the amplitude ratio of M to N is a)

k1 k2

b)

k1 k2

c)

k2 k1

d)

1

k2 k1

6

83. *A linear harmonic oscillator of force constant 2 x 10 N/m and amplitude 0.01 m has a total

mechanical energy of 160 J. Its a) maximum potential energy is 100 J b) maximum kinetic energy is 100 J c) maximum potential energy is 160 J d) minimum potential energy is zero 84. A boat which has a speed of 5 km/h in still water crosses a river width 1 km along the shortest possible path in 15 minutes. The velocity of the river water in km/h is a) 1

b) 3

c) 4

d)

41

85. A particle executes simple harmonic motion of frequency, f. The frequency with which its

kinetic energy oscillates is

f d) 4f 2 86. A shell is fired from a cannon with a velocity V(m/s) at an angle θ with the horizontal direction. At the highest point in its path it explodes into two pieces of equal mass. One of the pieces retraces its path to the cannon and the speed (in m/s) of the other piece immediately after the explosion is 3 3 a) 2 v cosθ b) 3v cosθ c) v cosθ d) v cosθ 2 2 a) 2f

b) f

c)

V0 αt (1-e ), where V0 is a α

87. The position of a particle at time t, is given by the relation: x(t) =

constant and α > 0. The dimensions of V0 and α are, respectively, a) M0L1T0 and T1 b) M0L1T1 and T c) M0L1T-1 and T2 d) M0L1T-1 and T-1 88. In a simple harmonic motion, when the displacement is half the amplitude, what fraction of the total energy is kinetic?

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9

3 1 1 b) zero c) d) 4 4 2 th 89. If radius of earth contracts to (1/n) , with its mass remaining constant, the duration of the day in hour will be 24 24 a) 24 b) c) 24 n d) 2 n n 90. A uniform chain of mass M and length L rests on a smooth table with one fourth of its length hanging. the work required to pull the hanging portion is MgL MgL MgL MgL a) d) b) c) 2 4 16 32 91. When n vectors of different magnitudes are added the sum is zero. Then n cannot be a) 11 b) 6 c) 4 d) 2 92. *In figure, .the spring balance A reads 2 kg with a block m suspended form it. A balance B reads 5 kg when a beaker A with liquid is put on the pan of the balance. The two balances are now so arranged that the hanging mass is inside the liquid in the beaker as shown in fig. In this situation a) the balance A will read more than 2 kg b) the balance B will read more than 5 kg c) the balance A will read less than 2 kg and B will read more than 5 kg B d) the balances A and B will read 2 kg and 5 kg respectively 93. The moment of inertia of a thin square plate ABCD, (Fig) of uniform thickness about an axis passing through the centre 0 and perpendicular to the plane of the plate is (where I1, I2, I3, I4 are MI about the axis in its I I plane) a) I1 + I2 b) I3 + I4 c) I1 + I3 d) I1 + I2 + I3 + I4 I 94. A tube of length L is filled completely with an incompressible liquid of mass M and closed at both the ends. The tube is then rotated in a horizontal plane about I one of its ends with a uniform angular velocity ω. The force exerted by the liquid at the other end is a) Mω2 L/2 b) Mω2L c) Mω2L/4 d) Mω2L2/2 95. A car is moving in a circular horizontal track of radius 10m with a constant speed of 10 m/s. A plumb bob is suspended from the roof of the car by a light rigid rod of length 1.00m. The angle made by the rod with the track is a) zero b) 300 c) 450 d) 600 96. During a rainstorm, raindrops are observed to be striking the ground at an angle of θ with the vertical. A wind is blowing horizontally at the speed of 5.0 m s-1. The speed of raindrop is (in m/s) 5 5 a) 5 sinθ b) c) 5 cosθ d) sin θ cos θ 97. A 2 kg stone at the end of a string 1 m long is whirled in a vertical circle at a constant speed of 4 m s-1. The tension in the string will be 52 N when the stone is a) at the top of the circle b) halfway down a)

1

3

2

4

10

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c) at the bottom of the circle

d) anywhere on the circle

98. The volume of an air bubble is doubled when it rises from the bottom to the surface of a lake.

The depth of the lake is approximately a) 10 m b) 15 m c) 25 m d) 30 m 99. A body of certain material floats with 9/10 of volume submerged in water in earth. In a planet, where acceleration due to gravity is five times that of earth, the submerged fraction will be

9 5 9 9 9 b) c) d) 10 50 10 250 100. If the change in the value of g at a height h above the surface of the earth (radius R) is the same as that at a depth x below it (assuming h > T0), at a distance r from the source. If the temperature and distance are doubled, the power P1 received by the surface becomes approximately (P original power) a) P b) 2P c) 4P d) 16 P Two rods of same length and area but made of different materials are 1 welded together as shown in the fig. If λ1 and λ2 are the thermal conductivity of each of them respectively, the thermal conductivity 2 of the system will be a) λ1 + λ2 b) λ1 λ2 / λ1 + λ2 c) 2λ1 λ2 / λ1 +λ2 d) λ1 + λ2/2

78. A bullet is stopped by a target the temperature rise of the bullet is ∆T, without loss of energy.

If the bullet’s velocity is doubled and it loses 50% of the energy, the rise in temperature of the bullet will be a) ∆T b) 2∆T c) 3 ∆T d) 4 ∆T 0 0 79. One mol of monatomic gas is heated from 0 C to 100 C at constant pressure. The change in its internal energy is a) 2.3 J b) 12.4 J c) 120 J d) 1250 J 80. If a temperature of source and sink of a Carnot engine is increased by 10% its efficiency a) increases by 10% b) decreases by 10% c) increases by 1% d) remains the same 0 81. A test tube of water at 4 C is immersed inside a large block of ice. Which of the following will happen? a) Water in the test tube will freeze b) Water in the test tube will not cool c) Water in the test tube will reach freezing point but will not freeze

48

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d) water in the test tube will reach between freezing point and 40 C 82. Which of the following cases, the work done by a gas is minimum ? P Y

Y

Y

X V (1 )

a) 1

Y

X (2 )

b) 2

X (3 )

c) 3

X (4 )

d) 4

83. If distance between sun and earth is reduced to half the present value, solar constant

84.

a) increases to 4 times b) increases to 2 times c) decreases to 4 times d) remains the same A cylinder of radius R made of a material of thermal conductivity K1 is surrounded by a cylindrical shell of inner radius R and outer radius 2R R 1 2 made of a material of thermal conductivity K2. The K two ends of the combined system are maintained at two different temperatures. there is no loss of heat K across the cylindrical surface and the system is in a steady state. The effective thermal conductivity of the system is 2R a) K1 + K2 b) K1 K2 / (K1 +K2) c) K1 + 3K2)/4 d) (3K1 + K2)/4 A cube of side a is heated through 1oC. If α is the linear expansivity of the material of the cube, its increase in area is a) a2α b) 4a2α c) 6a2α d) 12 a2α The maximum temperature of 1g hot water which can be mixed with 1g ice at 0o C so that the temperature remains at 0o C, is b) 200 C c) 50o C d) 80o C a) 10o C 3.12 kJ of heat is supplied to a mass of monatomic gas produces a temperature rise. If the same amount of heat is supplied to a diatomic gas of same mass, the temperature rise will be a) same b) 7/5 times c) 3/5 times d) 5/7 times o 16 g of oxygen at STP is mixed with 14 g of nitrogen at 10 C . The resulting temperature of the mixture will be a) 0oC b) 5oC o c) slightly more than 5 C d) slightly less than 5oC A substance of mass M requires an input power P to remain in the molten state at its melting point. When the power source is turned off the sample completely solidifies in a time t. The latent heat of fusion of the substance is a) Pt/M b) P/Mt c) Mt/T d) MP/t o 2 g of He is heated through 1 C at constant pressure. . The external work done in the process is a) R/2 b) R c) 3R/2 d) 2R The RMS speed of hydrogen molecules at a temperature T is x m/s. At a temperature 4T, hydrogen dissociates. At this temperature the RMS speed of hydrogen will be a) 2x b) 4x c) (2√2) x d) 4√2 x Two metals A and B are used to make a compensated pendulum. If the linear expansivity of A to that of B is in the ratio 3:1, the ratio of length of A to that of B will be 2

1

85.

86.

87.

88.

89.

90.

91.

92.

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a) 3:1

b) 1:3

49

c) 1.5:1

d) 1:1.5

93. 1 mol of hydrogen is mixed with 1 mol of helium at the same temperature. The ratio of

specific heats γ of the mixture is a) 1.67 b) 1.4 c) 1.5 d) 1.53 94. A thin steel wire of length `L’ increases in length by 1% when heated through a certain range of temperature. If a thin steel plate of area 2LxL is heated through same range of temperature, the percentage increase in area will be a) 1% b) 2% c) 3% d) 4% 95. A gas of γ = 4/3 is heated at constant pressure what percentage of total heat supplied is used

as external work ? a) 25% b) 20% c) 10% d) 5% o 96. An electric heater of constant power takes 50 seconds to turn x g of ice at 0 C into water at o o 100 C. The time it will take to turn this water into steam at 100 C is a) 25 s b) 50 s c) 100 s d) 150 s 97. A mixture is formed with one mol of monatomic gas, 1 mol of diatomic gas and 1 mol of triatomic gas. The external work done when this mixture is heated through 1oC is a) R b) 6R c) 1.5R d) 3R 98. The ratio of diameter of molecule A to that of B is 1:2. The mean free path of A to B when they have same concentration, will be in the ratio a) 1:2 b) 1:4 c) 2:1 d) 4:1 o o 99. 22.2 g of ice at 0 C is mixed with 22.2g of water at 22.2 C. The resulting temperature of the mixture is a) 22.2o C b) 11.1o C c) 0o C d) 5.6o C 100. An ideal monatomic gas is heated at constant pressure . What fraction of heat energy supplied to it is used as increase in internal energy? a) 2/5 b) 3/5 c) 5/7 d) 3/7 101. If R/Cp for a gas is 0.28, the gas could be a) hydrogen b) helium c) carbon dioxide d) mixture of helium and hydrogen 102. If big ice cube falls from a height 3.36 m. Nearly what fraction of it will melt? a) 0.1% b) 1 % c) 10% d) 0.01% 103. *A thermos flask containing hot coffee is vigorously shaken . Check the correct statements a) The temperature of the coffee will increase b) The internal energy of system will increase c) Heat is given to the system d) Work is done on the system 104. The temperature of a hot source is 1000 K. If we use a powerful convex lens to converge heat radiation from the source to a point, at this point a) we can produce a temperature well above 1000 K b) we can produce a temperature of 1000 K c) we can produce a temperature only below 1000 K d) we can produce a temperature above 1000 K if the source has sufficient energy. 105. During winter the snow does not melt at once when sun’s rays fall on it, because it a) reflects sun’s rays b) has high specific heat c) has high latent heat of fusion d) is due to all the three given above

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106. In an adiabatic change the pressure P and temperature T of a monatomic gas is related by the

equation P is proportional to Tc. Here c is equal to a) 5/3 b) 2/5 c) 3/5 d) 5/2 3/2 107. An ideal gas is found to obey the equation PV = constant. If such a gas at a temperature T is compressed adiabatically to half its volume the final temperature would be a) 2 T b) 4 T c) 1.4 T d) 2.8 T 108. The RMS speed of hydrogen molecules is v. The RMS speed of a mixture of hydrogen and oxygen of volumes in the ratio 5:1 will be 21 6 21 c) d) 6 21 3 109. The temperature of an ideal gas is increased from 120 K to 480 K. The rms speed of its molecules increases by a) 100% b) 200% c) 300% d) 400% 3 110. A container of volume 1m is equally divided by a partition . One part contains an ideal gas at 300 K and the other part is vacuum . The whole system is thermally insulated from the surroundings. When the partition is removed the gas expands to occupy the whole volume. The temperature of the gas will be a) 150 K b) 300 K c) 600 K d) 75 K 111. An ideal gas with a pressure P, volume V and temperature T is expanded isothermally to a volume 2V and pressure Pi . If the same gas is expanded adiabatically to a volume 2V, the final pressure is Pa. The ratio specific heat of the gas is 1.67. The ratio Pa /Pi is equal to b) 2-0.67 c) 20.67 d) 2 a) 21.67 112. The ratio of speed of sound in hydrogen to rms speed of hydrogen molecules at STP is equal to 7 15 5 9 a) b) c) d) 15 7 9 5 a) √5 v

b)

113. The pressure of a gas contained in a vessel is P. If mass of each molecule is reduced to half

and rms speed doubled, the pressure will be a) P b) P/2 c) P/4 d) 2P 114. A given mass of gas of volume V is heated at a constant pressure so that the rms velocity is doubled. The new volume it occupies is a) V/2 b) 2V c) √2V d) 4V 115. The mean free path of molecules λ is related to the temperature T by λ proportional to

a) T b) 1/T c) T2 d) 1/T2 116. If heat supplied to a gas only increases internal energy of the gas, the process is a) isothermal b) adiabatic c) isobaric d) isochoric -2/5 117. The equation to the adiabatic change of a gas is given by T is proportional to V . The heat o required to raise the internal energy of 1 mol of this gas through 1 C is a) (5/2)R b) (3/2)R c) (7/2)R d) 3R 118. If A is the dimensions of specific heat and B is the dimensions of gravitational potential, the dimensions of the ratio A/B will be that of a) temperature b) Wein’s constant c) mass x temperature d) 1/temperature 119. *The internal energy of a gas depends on a) pressure b) volume c) temperature d) molecular separation

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51

120. Which of the following remains constant when water boils to steam ?

a) potential energy b) kinetic energy c) internal energy d) entropy 121. The rms speed of certain gas at a temperature is v. If the temperature of the gas is increased by 10% and then decreased by 10% , the rms speed a) remains the same as v b) increases by 1% of v c) decreases by 1% of v d) decreases by 0.5% of v 122. The pressure and temperature of a given mass of gas is increased to 3 times each. In this process the rms velocity a) increases to 3 times b) increases to 9 times c) increases to 3√3 times d) increases to√3 times 123. The pressure of a gas is decided by a) total random collision of molecules b) random collision of molecules per second c) average collision of molecules per second d) average collision per unit area per second 124. *Cooking vessels made with stainless steel are provided with extra copper bottom. This is because a) copper has more density b) copper has more thermal conductivity c) copper has more specific heat d) copper has less specific heat o 125. A body A having 1000 J of heat energy at 20 C is mixed with a body B having 500 J of heat o energy at 40 C. Check the correct statement: a) Heat will flow from A to B b) Heat will flow from B to A c) Heat will not flow between the two bodies d) The final temperature of the mixture will be 30o C 126. A refrigerator has a coefficient of performance of 10. When it works at a power of 700 W, the heat removed by it from the substance to be cooled per second is a) 7000 J b) 70 J c) 700 J d) 4200 J 127. x mol of monatomic gas is mixed with x mol of diatomic gas. The heat in calorie required to raise the temperature of the mixture through 10o C is a) 40 x b) 80x c) 20 x d) 60 x 128. The efficiency of a Carnot’s engine is y ( y is a fraction). When the temperature of the sink is increased by 200 C, the efficiency becomes 0.8 y. The temperature of the source is a) 50/y b) 40/0.2y c) 100/y d) 100/0.2 y 129. The efficiency of a Carnot engine can be increased by a) increasing both the temperature of source and sink b) decreasing both the temperature of source and sink c) decreasing the temperature of source and increasing in the temperature of sink d) increasing the temperature of source and decreasing in the temperature of sink o 130. By what height should water fall so that it will be warmer by 1 C? Assume 30% of energy of water is wasted a) 600 m b) 300 m c) 240 m d) 420 m 131. If the radius and surface temperature of a black body decrease by 1% each, the power radiated by it

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a) decreases by 2% b) decreases by 4% c) decreases by 5% d) decreases by 6% 132. At upper atmosphere, where kinetic temperature is 1000 K, an astronaut feels a) very cold b) cold c) very hot d) warm 6 133. In an atomic bomb explosion, the temperature produced is about 10 K. In which region of electro- magnetic spectrum does the emitted radiation belong to? a) ultra violet region b) infra red region c) X-ray region d) visible region 134. The unit of thermal resistance is a) KW-1 b) mW-1K c) WK-1 d) Wm-1

SECTION 2: ANSWERS 1

(b)

2

(c)

3

(d)

4

(a)

5

(c)

6

(d)

7

(a)

8

(d)

9

(d)

10

(d)

11

(a)

12

(c)

13

(c)

14

(d)

15

(a),(d)

16

(a)

17

(d)

18

(c)

19

(b)

20

(a)

21

(c)

22

(c)

23

(d)

24

(a)

25

(b)

26

(d)

27

(b)

28

(d)

29

(d)

30

(b)

31

(d)

32

(b)

33

(d)

34

(b)

35

(a)

36

(c)

37

(c)

38

(d)

39

(b)

40

(d)

41

(a)

42

(a)

43

(c)

44

(d)

45

(b)

46

(c)

47

(a)

48

(a)

49

(c)

50

(b)

51

(d)

52

(a)

53

(b)

54

(c)

55

(d)

56

(d)

57

(b)

58

(c)

59

(b)

60

(c)

61

(b)

62

(b)

63

(c)

64

(c)

65

(b)

66

(b)

67

(c)

68

(d)

69

(d)

70

(c)

71

(d)

72

(a)

73

(a)

74

(a)

75

(c)

76

(c)

77

(d)

78

(b)

79

(d)

80

(d)

81

(c)

82

(a)

83

(a)

84

(c)

85

(d)

86

(d)

87

(c)

88

(b)

89

(a)

90

(a)

91

(c)

92

(b)

93

(c)

94

(b)

95

(a)

96

(d)

97

(d)

98

(d)

99

(c)

100

(b)

101

(a)

102

(d)

103

(a,b,d)

104

(c)

105

(c)

106

(d)

107

(c)

108

(c)

109

(a)

110

(b)

111

(b)

112

(a)

113

(d)

114

(d)

115

(b)

116

(d)

117

(a)

118

(d)

119

(c,d)

120

(b)

121

(d)

122

(d)

123

(d)

124

(b,d)

125

(b)

126

(a)

127

(b)

128

(c)

129

(d)

130

(a)

131

(d)

132

(a)

133

(c)

134

(a)

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53

SECTION 3 SOLUTIONS 1.

When the balloon sinks below the surface, hydrostatic pressure increases, decreasing the volume (Boyle’s law). This reduces up thrust (buoyant force), which makes it to sink more until it sinks to the bottom. (Note: the answer is not surface tension because it will oppose sinking) Ans.b

2.

The terminal velocity is proportional to a2 (a radius) .v = velocity of the smaller drop, v2 that of resulting drop,

2a 2 (ρ − σ)g If v1 is the terminal 9η

v1 a 1 2 = . a2 = 21/3 x a1. This gives v2 a 22

4.

v2 = 22/3 x v1 = 22/3 x 1 m/s. Ans.c When coal-tar is heated viscosity decreases with heating. Then it is laid. After it cools down viscosity increases and it settles. ( Indirect theory notes). Ans.d Ans.a

5.

r h ρ g = 2 S cosθ. That is rxh = constant. A = πr2. Thus we find √Axh = constant. When

6.

Using the formula r h ρ g = 2S, when warm water is taken surface tension decreases. So the capillary height will decrease. However density is less for warm water. So the question can be answered only if the relative decrease of the two are known. Ans.d

7.

Once again we use the relation r h ρ g = 2S. Capillary rise h will be less if g is more. This happens only in the case of lift accelerating upward. Here effectively g is equal to g+a. Therefore h is less. Ans.a

3.

area is increased to twice, height decreases to

2 times. Ans.c

The depression at centre 1 t of a metre scale is given by δ =

4Mgl 3

. Thus δ is proportional to bd 3 Y 13. When metre scale is cut into 2 equal halves, 1 becomes 1/2 δ becomes (1/2)3 = 1/8 of x. Ans.d 9. Heating reduces surface tension. Hence B is correct. But only insoluble and semi soluble impurities reduce surface tension. Highly soluble impurity increases surface tension of the solvent. Hence A is wrong. Ans.d 10. When a body floats in a liquid, the immersed fraction is=density of body/density of liquid. When water is heated its density increases up to 40C, and then decreases . The immersed fraction therefore first decreases and then increases. Ans.d 8.

11. If e is increase in length, using α =

∆L e = we have = α . Thus e = α. Young’s modulus L∆T l× l

FL F ×1 , i.e. = E. F = Ee = Eα. Ans.a Ae 1× e 12. Heat lost = mcθ. For a given quality of heat we require only the minimum mass of water for cooling because of a high value of specific heat. Ans.c 13. During melting, the temperature is constant. Hence the molecules do not gain kinetic energy. The heat supplied is used to gain potential energy, decreasing the separation of molecules. The volume of ice decreases on melting. Ans.c E=

54

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14. The coefficient of thermal conductivity is a constant for a given material and will not depend

on any of these. (The heat conducted will depend on these factors). Ans.d 15. Using L1 = L0 (1+ α ∆T) where α is linear expansivity.

We have L10 = L0 (1+ 10α). (a) is correct. Since areal expansion β = 2α, A10 = A0 (1+βt) = L2 (1+20α). But the total area is that of six faces. Hence (b) is wrong. Since density at 00C, d0 is related to density at t0C by d d0 d0 = the equation d0 = dt(1+γt), we have d t = 0 = (d) is correct and (c) is 1 + γt 1 + (3α ×10) 1 + 30α wrong. Ans.a,d 0

16. Use the information supplied in indirect theory notes Here θ < 80 C. Ans.a 17. Cv for helium(monatomic gas) is (3/2) R and for hydrogen (diatomic) is (5/2)R. Hence Cv

for mixture is 4R/2 joule/mole K = 2R J/mol K = 16.6. (Use table given in theory notes) Ans.d 18. According to Kirchoff’ law, a good absorber of a certain wavelength is also a good emitter. If sodium emits two bright yellow lines, it will absorb radiation of the same wavelength when a more intense source is kept behind it. An arc source is more intense than sodium. Therefore, sodium absorbs these wavelengths and produces two dark lines under the bright background of the arc source. Ans.c 0

0

19. We have mCp ∆T = 70 cal, m = 2 mol, ∆T = 5 C. This gives Cp = 7 cal / mol C. Cp – Cv =

20.

21. 22.

23.

R where R = 2 cal/mol0 C. Therefore Cv = 7-2 = 5. Thus the heat required at constant volume to raise through 50C = mCv∆T = 2 x 5 x 5 = 50 cal. Ans.b We have to use here Wein’s displacement law, λmT = constant, where λm is the wavelength of radiation carrying maximum energy and T the temperature of the star. For a very hot star T is high. λm will be low. In the visible region lower wave length corresponds to the violet Ans.a By Boltzmann’s law, thermal energy is of the order of kBT where kB is Boltzmann’s constant Taking room temperature T = 300 K, E = 1.38 x 10-23 x 300/(1.6 x 10-19) ≈ 0.02 eV Ans.c When heated at constant pressure, by Charles’ law temperature should increase to 4 times. The rms velocity is proportional to square root of temperature. So the rms velocity becomes √4 = 2 times. Ans.c A given mass of water has minimum volume at 40C. When heated or cooled from this temperature water expands and overflows. Hence temperature of the room is 40C. Ans.d

4 πr 3 ρ rρ = which is proportional to r. 3 4πr 2 3 Heat radiated by S1 r1 4 4 = . Mass of S1 = 3 x mass of S2. Therefore πr13ρ = 3 x πr23ρ. Heat radiated by S 2 r2 3 3 1/3 This gives (r1/r2) = (3) . Ans.a 25. According to Maxwellian distribution of velocities, average velocity of a molecule depends only on the temperature. (Similar to one postulate of kinetic theory). Since the temperatures are same in the three vessels, the average speed of molecules will also be same. Ans.b 26. If Cv is molar heat capacity at constant volume, Cp molar heat capacity at constant pressure, heat required to increase internal energy of n mol of gas in n Cv ∆T. Total heat supplied to the gas i.e. to increase internal energy and to do external work, is nCp∆T. The ratio of 24. Heat energy radiated per unit area per second =

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increase in internal energy to total energy is

55 nC v ∆T C v 1 5 = = = for a diatomic gas. This nC p ∆T C p γ 7

can also be done from Indirect theory notes short-cuts. Ans.d aT 2 = AV-c-BV-1 (1) At constant temperature, RT + b = constant V Vc A and aT2 = constant B. It is given in the question here P = AVm –BVn. (2). Comparing (1) and (2) we get m = -c and n = -1. Ans.b 28. 22 gm of CO2 is half mol of CO2 16 gm of O2 is half mol of O2. Cv for CO2 is 3R. Cv for O2 is 5R/2. Let t be the final temperature of the mixture. Then by law of mixtures, 5 0.5 x 3R x (t-27) = 0.5 x R x (37-t). This gives t = 31.54 ≈ 320C. Ans.d 2 29. If Q1 and Q2 are the rate of flow of heat through the rods by the law of heat conduction Q = 27. Re-writing P =

RT + b

−

2

Q=

A(T1 − T2 ) t Q1 A 1 L 2 πr1 2 L 2 1 1 1 , = × = = × = . Ans.d L Q 2 A 2 L 1 πr2 2 L1 2 2 8

30. We are mixing here one mol of hydrogen i.e. monatomic gas with one mol of He that is

(3R/2) + (5R/2) (5R/2) + (7R/2) =2R. Cp (mixture) = = 3R. 2 2 Thus the required ratio is γ is 3R/2R = 1.5. This information is also given in indirect theory notes. Ans.b 31. Since specific heat varies with temperature here, this should be done by integration. Heat diatomic gas. Cv (mixture) =

required to raise from 10 to 200C is

34.

35. 36.

37.

38.

20

10

10

∫ mC V dT = ∫ 2 x 0.5T dT = 150 J. Ans d

3RT .When temperature is doubled T becomes 2T. When oxygen M dissociates, its molecular mass M becomes atomic mass M/2. Substituting these values the new rms speed = 2v. Ans.b If earth has no atmosphere, there will be no molecules to communicate heat energy by collision. Kinetic energy of molecules transferred to us by collision is the main reason for our feeling of heat. Hence it reduces considerably. Ans.d A cooking vessel should have low specific heat capacity so that smaller amount of heat will produce larger rise in temperature. Also it should have high thermal conductivity to conduct maximum heat. Ans.b The external work = Cp –Cv. For 1 mol Cp-Cv = R. 1g of hydrogen is (1/2) mol. Therefore external work = R/2. R = 2 cal/g. R/2 = 1 cal. Ans.a When the two tanks balance mass of air should be equal to mass of helium. Volume of two tanks are same. Hence density of helium should be made nearly 7.5 times so that its density becomes equal to that of air. Density is proportional to pressure. Hence pressure has to be increased to 7.5 times. Ans.c Clearly the vessel has 1/4 mol of gas with 1/8 mol of oxygen and 1/8 mol of nitrogen. When all oxygen is removed we have 1/8 mol of gas left. Hence pressure will fall to 1/2 of the 8 atmosphere. Ans.c Ans.d

32. v(rms speed) =

33.

20

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39. For an ideal gas, Cp – Cv = R, for 1 mol. Cp-Cv is the external work done per mol. Here we

have two mol of ideal gas. Hence the external work will be 2 R. Ans.b

40. According to one of adiabatic relations, TV γ −1 = constant.

Comparing this with given equation, γ-1 = 2/5. γ = 7/5. The gas is diatomic. The heat required to increase internal energy through 10C = CV = 5/2 R. Ans.d 41. Heat required to melt = mL, where m is mass, and is latent heat of fusion. Here m is same. L depends on material, hence the same. Ans.a Jm −1s −1 K −1 λ = = m2s-1. Ans.a dc (kg m −3 )( J kg −1 K −1 )

42. Substituting the units of 43. x = 2π

dx 1 dl dx ∆T 1 = = α ∆T ⇒ . = k l . Taking logarithms and differentiating = x 2 l x 2α g

dx = (1/2)x α ∆T. Ans.c 2

44. For an ideal gas PV = RT (1). It is given VP = constant K (2) . Squaring equation 1 and

P2V2

R 2T 2

. This gives V = constant x T2. That is T is proportional VP K2 to √V. When the volume is doubled, temperature becomes √2 times. Ans.d

dividing by 2, we get

2

=

45. The maximum efficiency of a heat engine is =

efficiency is 40%. i.e. =

T1 − T2 2100 − 700 2 = = . The actual T1 2100 3

2 . The ratio of actual efficiency of maximum efficiency 5

(2 / 5) = 0.6 = 60% Ans.b (2 / 3)

46. Since expansion is isothermal i.e. at constant temperature, only volume and pressure will

change. Change of pressure at constant temperature will not affect rms velocity. Ans.c 47. Using first law of thermo dynamics dQ = dU + dW, we have here dW = -dU (decrease of internal energy). dQ=0, means change is adiabatic. Decrease of internal energy decreases temperature, that is the gas is expanding. . The gas is undergoing adiabatic expansion. Ans.a 48. According to the first law of thermodynamics dQ = dU + dW. Here dQ = 100 cal = 420 J. External work dW = 20 J. Substituting in the above equation, 420 = dU + 20, dU = 400 J. Since dU is positive, internal energy increases. Ans.a 4

49. We use here Stefan’s law of radiation E = σT . If E1 is the new energy after increasing the 4

temperature to (3/2)T, E1 = σ [(3/2)T] = σ (81/16)T4 = nearly 5 σ T4. Energy increases from σT4 to 5 σT4 i.e. by four times or 400%. Ans.c Stefan’s law, heat energy radiated eper second E = σ AT4. 50. By 2

2 4 2 4 E r1 T1 1 2 E 1 σ4πr1 T1 = = = ⇒ 1 = 1 . Ans.b E 2 σ4πr2 2 T2 4 r2 T2 4 1 E2

51. According to the law of heat conduction heat conducted is directly proportional to area, and

inversely proportional to length at thermal equilibrium. In the second case, heat conducted doubles due to increase in area, and doubles due to decrease in length. Thus heat conducted becomes four times, i.e. 16 W. Ans.d

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57

52. Using Wien’s displacement law λmT = constant, T1/T2 = λ2/λ1 = f1/f2. (Since frequency is

inversely proportional to wave length). This gives T1/T2 = 2/2.5 = 4/5. Ans.a

4

53. Assuring the star to be a black body and applying Stefan’s law, (E αT ) the ratio of 1/4

1/4

temperature of the stars should be T1/T2 = (E1/E2) =(1/16) = 1/2. By Wein’s displacement law, λmT = constant. The ratio of wavelengths carrying maximum energy of the stars λ1/λ2 should be T2/T1 = 2/1. Ans.b

[x

2

2

]

− x 1 2 ρL , where t is the time to 2λθ increase the thickness of ice from x1 to x2, ρ density, L latent heat and λ thermal conductivity of ice. θ the temperature difference other things being constants, t is proportional to x22 – x12. t 1 12 − 0 2 1 = = , where t1 is a time to increase thickness from 0 to 1 and t2 from 1 to 2. t 2 2 2 − 12 3 Here t1 = 10 hour, t2 ,therefore, is 30 h. Ans.c 55. A system is thermodynamic equilibrium if no force acts, (mechanical equilibrium), no chemical reaction takes place ( chemical equilibrium), and no heat exchange takes place (thermal equilibrium). Ans.d 56. This is adiabatic expansion. Since no external heat is supplied internal energy should decrease. Hence (b) is correct. During adiabatic change entropy remains constant. Hence (c) is correct. Ans.d 57. Using the first law of thermodynamics, the increase in internal energy dU = dQ-dW. dU is the difference between external energy supplied (20x12 = 240 W) and heat transferred (10W). i.e. 230 W. Ans.b 54. Thickness of formation of ice in a pond is given by t =

4

58. By Stefan’s law energy radiated E = σT . When current is doubled heat energy becomes 4 2

4

times. (H α i ) .That is T becomes 4 times. T increases to 41/4 = √2 times. Ans.c 59. When liquids of different mass, specific heat and temperature are mixed final temperature is m c t + m 2 c 2 t 2 + m 3c 3 t 3 .. given by Σ(mc ∆T) / Σ(mc). For three liquids this will be 1 1 1 m1c1 + m 2 c 2 + m 3 c 3 Here m1 = m2 = m3. Therefore final temperature will be (c1t1 + c2t2 + c3t3) /c1 + c2 + c3. Ans.b 60. According to Boltzmann’s law average kinetic energy = (3/2)kT, per molecule. When this energy = 1 electron volt = 1.6 x 10-19J, we equate the two.Assuming k = 1.38 x 10-23, we get T = 7700 K. Ans.c 61. The thermal resistance (similar to electrical resistance) is directly proportional to the length (1) and inversely proportional to the area (A). Therefore, it can be written as R (thermal) = ρ1/A, where ρ is thermal resistivity . Thermal conductivity λ is reciprocal of the thermal resistivity i.e. λ = 1/ρ. Thus we get R (thermal) = 1/λA. Ans.b 62. In thermal equilibrium heat flowing (fig.) across any portion should be same. Hence, if the temperature of the intervening layer is T, 2λ (T-O) = λ(36-T), B A This gives T = 120C. Ans.b 0 T 36 63. The rate of loss of heat is proportional to area of the surface if other factors affecting radiation like mass, temperature difference remain the same. Since all have same mass, same density, they have same volume. For a given volume sphere has least surface area and circular plate maximum surface area. Hence circular plate cools fastest. Ans.c

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64. By Newton’s law of cooling, rate of cooling is proportional to the mean excess temperature of

65. 66. 67.

68.

the body over the surroundings. In the first case, the mean temperature difference is 65-300C. In the second case, the temperature difference is 55-30 = 250C. Hence rate of cooling will be in the ratio 35/25 = 7/5. So the time taken for cooling from 60 to 50 = 10 x (7/5) = 14 minutes. Ans.c Density of a liquid increases with fall in temperature. Density of benzene will be more in winter. Hence a given volume of benzene will weigh more in winter. Ans.b From kinetic theory we can deduce Boyle’s law. Boyle’s law is violated at high pressure and low temperature. Hence most of the postulates break down at this stage. Ans.b When temperature is increased at constant volume, the average kinetic energy and hence rms speed of molecules increase. Hence they make more number of collisions on the wall with higher velocity. Ans.c Kinetic energy of one mol of monatomic gas at STP (3/2)RT, which by calculation (3/2) x 8.3 x 273 = 3400 J. 1 gm of helium is (1/4) mol. Hence its kinetic energy is 3400/4 = 850 J. (Note: kinetic energy of monatomic gas at STP is one of the constants to be remembered). Ans.d

69. Heat supplied H = nCv ∆T, where n is number of mole, Cv specific heat at constant volume,

∆T rise in temperature. Here heat supplied is same. n is same. Hence C1 ∆T1 = C2 ∆T2. This gives ∆T1/∆T2 = C2/C1 = Cv of monoatomic gas/Cv of diatomic gas. (3R/2)/ (5R/2) = 3/5. Ans.d 70. For a solid cubical expansivity γ = 3α nearly. A crystal is anisotropic and hence does not

expand in the same way in all directions. Therefore γ = α1 + α2 + α3 = (13 + 231 + 231) 10-7 = 475 x 10-7. Ans.c 71. Here the expansion is free. That is not external work is done on the system or by the system. No heat is supplied to the system either. Therefore there will be no change in the internal energy of the system. Ans.d 72. According to Boyle’s law PV = constant at constant temperature. Therefore if P1 is the new pressure PV + PV = P1V, which gives P1 = 2P. Ans.a 73. By Wein’s law λ1T1 = λ2T2, where λ1, λ2 are wavelengths carrying maximum energy. This

gives T1/T2 = 4800/3600 = 4/3. Ratio of total power radiated = T14/T24 = (4/3)4 = 256/81. Ans.a 74. Using the first law of thermodynamics, dQ = dU + dW, dQ = 0, for adiabatic process. (d) is correct, Also dW= dU.. (b) is correct). For isothermal process there is no change in temperature and hence dU is zero. (c is correct). The increase in internal energy whether at constant pressure or constant volume = nCv ∆T. (a is wrong). Ans.a 75. The adiabatic elasticity of any gas is γP. Helium is a monatomic gas. For helium γ = 1.66

and P = nearly 105 Pa. Ans.c 76. When temperature T is doubled, heat energy radiated per second becomes 16 times (Stefan’s law E αT4). When r is doubled heat received becomes 1/4. (inverse square law I α 1/r2). Hence heat energy received becomes 16 x 1/4 = 4 times. Ans.c 77. If T1 and T2 are the temperatures at the two ends of the system, at thermal equilibrium, the heat conducted through the rod respectively are Q1 = λ1A (T1 –T2)/x, Q2 = λ2A (T1-T2)/x, per second. The total heat conducted Q = Q1 + Q2 (1) . If λ is the conductivity of the composite rod, then heat conducted per second Q = 2Aλ (T1-T2/x. Substituting these values of Q, Q1, Q2 in (1) we get λ = λ1 + λ2/2. Ans.d

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59

78. Here kinetic energy of the bullet is converted into heat energy. If the velocity of the bullet is

doubled kinetic energy of bullet (1/2)mv2 becomes four times and when 50% of it is lost. Kinetic energy remains two times. The rise in temperature will be two times. Ans.b

79. The increase in the internal energy = m Cv ∆T. For a monatomic gas Cv = (3/2)R. ∆T =

1000C or 100K (Temperature intervals will be same in both scales). Hence dU = (3/2)x 8.3 x 100 = 1245 J. Ans.d

80. η = T1 – T2/T1. When T1 and T2 are increased by 10% all temperatures increase by 10%.

Hence (T1-T2)/T1 remains the same. Ans.d 0

81. Water will reach 0 C but will not freeze, because it cannot give out latent heat. Latent heat

will not be conducted as there is no temperature difference at this stage. Ans.c 82. The work done by a gas is the area of P-V graph. This area is minimum for graph (1) where

there is no increase in volume i.e. dV = 0. (Alternatively dQ = dU + PdV. Here dV = 0. Therefore PdV = 0. Ans.a 83. Solar constant is amount of heat energy received per unit area in one second on earth. If P is power radiated by sun ,. solar constant S0 = P/4πd2, where d is the distance between sun and earth. When d is reduced to half, heat received per unit area per second i.e. S0 becomes 4 times. Ans.a K 1 πR 2 (θ1 − θ 2 ) , x where θ1 and θ2 are the temperatures at the two ends (θ1 > θ2) and x is the distance between K π(2R 2 − πR 2 )θ1 − θ 2 . If K is them. Heat flowing across the cylindrical shell is Q2 = 2 x the equivalent thermal conductivity of the system then, the total heat flowing = Kπ(2R)2 (θ1-θ2)/x = Q. Equating Q1 + Q2 = Q, we get K = (K1 + 3K2)/4. Ans.c 85. Increase in area ∆A = A β ∆T, where ∆T is increase in temperature, A original area, β areal expansivity . A cube has 6 surfaces. So the total area A = 6 a2. Areal expansivity β = 2α. ∆T = 1oC = 1K. This gives ∆A = 12a2α. Ans.d o 86. Latent heat of fusion of ice is 80 cal/gm. So temperature will remain at 0 C if we supply a maximum heat 80 cal. Since specific heat of water in CGS system is 1, 1 g of water when cools from 80 to 0 can give 80 cal of heat. Use also indirect theory notes for answering this quickly. Ans.d 84. Heat flowing per second across the cylinder of radius R is given by Q 1 =

[

]

87. Heat supplied = mCv∆T, where Cv is specific heat at constant volume.

If C1 and C2 are the specific heat of monatomic and diatomic gases, ∆T1 and ∆T2, temperature rise, we have C1∆T1 C (3 / 2)R 3 = C2∆T2, for the same mass . 1 = This gives ∆T2 = ∆T1. Ans.c C 2 (5 / 2)R 5

88. 16 g of oxygen is half a mole. 14 g of nitrogen is also (1/2) a mole. Both of them are

diatomic. Hence their Cv will be same. Using heat lost by nitrogen = heat gained by oxygen formula , we have m CV (10-T) = m CV (T-0). Hence the temperature will be 5oC. Ans.b 89. A power P keeps the substance in the molten state i.e in equilibrium . This means the substance is supplied with an energy Pt to keep in the state. So latent heat of fusion = heat energy supplied / mass = Pt/M. Ans.a o 90. The external work done when 1 mol of gas is heated through 1 C or 1 K is equal to Cp-Cv = R. 2 g of helium is (1/2) mole. Hence external work will be (R/2). Ans. a

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3RT where M is molecular mass. When temperature increases to 4T, M hydrogen dissociates, its molecular mass M becomes atomic mass M/2. Thus the RMS speed 3R x 4T 8 x 3RT = = 8 x = 2 2 x Ans.c becomes M/2 M

91. The RMS speed =

92. Use information supplied in indirect theory notes. . For compensated pendulum, we should

have lengths in the ratio inverse of expansivity i.e. 1:3. Ans b 93. Use information given in indirect theory notes. This is for practice. Hydrogen is diatomic and helium is monatomic. Ans.c 2 94. We have A = 2LxL = 2L . Taking logarithms and differentiating dA/A = 2(dL/L). Here dL/L = 1%. Therefore dA/A = 2%. Ans.b 95. The external work done = R = Cp- CV, for 1 mol. The heat supplied at constant pressure = Cp Cp − CV C 1 1 for 1 mol for 1o rise in temperature. The required ratio is = 1− V = 1− = . Cp CP γ 4 Use indirect theory short-cuts. Ans.a 96. Use information supplied in theory notes. To melt x g of ice we need 80 x calorie of heat. To raise the temperature of x g of water from 0 to 100o C we need 100 x calorie heat. Total 180x. To turn x g of water at 100o C into steam, we need 540x calorie of heat, which is 3 times 180x. So the required time will be 3x50 = 150 s. Ans.d 97. Here we have a total of 3 mol of gas. Whatever may be a gas the external work done for 1 mol Cp-CV = R. For 3 mole the external work will be 3R. Ans.d 98. The mean free path λ is inversely proportional to the square of the diameter. ( formula λ =

1/√2πσ2n). Therefore mean free path will be in the ratio 22:12.= 4:1 Ans.d 99. Use information supplied in in indirect theory notes. Ans.c 100. The ratio of increase in internal energy to the total heat energy supplied is clearly Cv/Cp, which in turn is equal to 1/γ. For a monatomic gas this value is 3/5. Ans.b 101. Since it is given that R/Cp =0.28, we have Cp = R/0.28, which is nearly 3.5 R, or (7/2) R. This is the value of Cp of a diatomic gas. The only diatomic gas listed is hydrogen. Ans a 102. Here the potential energy of ice is converted into heat energy for melting. Mgh = mL,where m is the mass of ice melting, L latent heat of fusion. With g nearly 10 and h = 3.36m, we get m/M = gh/L= 10x3.36/336000 = 0.0001= 0.01% Ans d 103. Since mechanical work is done on the system during shaking , (d) is correct. Due to this the temperature of the coffee will increase. (a) is correct. Heat is not directly added to the system. Hence (c) is wrong. By first law of thermo dynamics ∆Q = ∆U+ ∆W. Here ∆Q = 0. Therefore ∆U = - ∆W ( The negative sign here is due to work done on the system. That is internal energy of the system increases. (b) is correct. Ans. a,b,d 104. Whatever may be the energy of the source, we cannot produce a temperature above 1000 K at the point. If the temperature of the point becomes above 1000 K, it would mean that we have transferred heat from cold body to hot body. This will be in violation of second law of thermodynamics. Ans.c 105. The latent heat of fusion of ice is 80 cal/g, which is equal to 336 J/g. This is a large value. So it takes some time for ice to get this much of heat. Specific heat has nothing to do with melting, because there is no temperature change during melting. Ans.c

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61

T γ P 1− γ = = constant. This means P = constant Thus we have c =(γ/γ-1) For a monatomic gas γ = 5/3. Therefore c = 5/2. Ans.d

106. According to one of the adiabatic relations (γ/γ-1)

x T

.

107. Comparing with the standard adiabatic equation PVγ = constant, we find γ = 3/2. The

relation between temperature and volume is TVγ-1 = constant. That is TV(1/2) = constant . That is T is proportional to V-1/2. When V is halved T becomes (1/2)-(1/2) = 2(1/2) T = √2 T= 1.4 T. Ans.c

108. The RMS speed is proportional to

1 ρ

Density of the mixture is equal to

Since ρo = 16 times ρ H , The RMS velocity is equal to

6 v 21

5ρ H + ρ O 21 = ρH 6 6

Ans.c

109. We know Crms is proportional to √T. So when the temperature increases from 120 to 480 K

that is four times the rms speed should increase to √4 = 2 times. The rms speed thus increases to 200% and hence it increases by 100%. Ans a 110. Here the gas is undergoing free expansion i.e. no energy is supplied from outside and no energy is taken from inside. Therefore the temperature will remain the same. Ans.b γ

111. For isothermal expansion the equation is PV = constant and for adiabatic equation PV is

constant. PV = Pix2V. PVγ = Pa (2V)γ . This will give 112. The speed of sound is given by

Pa 2 2 = γ = 1.67 = 2 −0.67 Ans.b Pi 2 2

γRT and RMS velocity M

3RT The required ratio is M

γ = 3

7 7 , where we have assumed γ for hydrogen as . Ans.a 15 5 2

2

113. P = (1/3)nmc , where c is mean square velocity. When m is made half , c is doubled, the

pressure becomes [4(1/2)] = 2 times. Ans.d 114. To double the rms velocity the temperature has to increased to 4 times because Crms is

proportional to √T. When temperature increases to 4 times at constant pressure volume increases to 4 times. Ans.d 115. When a gas is heated molecules move faster making more number of collisions per second decreasing distance between collisions. Ans.b 116. From the first law thermodynamics ∆Q = ∆U+∆W.

Here only internal energy changes, which means ∆W = 0. But ∆W = PdV, where dV is increase in volume. Thus we have dV = 0. That is volume is constant. The process is isochoric. Ans.d γ-1

1-γ

117. Comparing with standard adiabatic relation TV = constant, we have T proportional to V

. Here 1-γ = -2/5. This gives γ = 7/5. The gas is diatomic. The heat required to raise internal energy by 1oC (which is Cv) = (5/2)R. Ans a 118. A = energy/(mass x temperature), while Gravitational potential (B) is energy/mass. This makes A/B equal to 1/temperature. Ans d 119. The internal energy of a gas molecule is the sum of potential and kinetic energy. The kinetic energy depends on temperature. (c) is correct. The potential energy depends on distance between molecules. (d) is correct. Ans.c & d

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120. When water boils molecular separation changes and hence potential energy changes.

So internal energy also changes. Since temperature is constant during boiling, kinetic energy remains constant. Entropy increases. Ans.b

121. If T is the temperature, Crms is proportional to √T.

When the temperature is increased by 10%, it becomes (110%)T and when decreased 10% from there, it becomes (110x90/100)T =

99%T. Thus the temperature decreases by 1% and the rms velocity being proportional to T decreases by 0.5%. Ans.d. 122. When pressure and temperature are increased at the same time, consider first increase of pressure. The rms velocity does not change. Now consider increase of temperature. When temperature is increased to 3 times rms velocity increases to √3 times. Ans.d 123. Since the number of molecules is large and they are moving at random, we cannot calculate exact number of collisions per second. So we take average number colliding per second. This gives force. Since pressure = force/area we take number of collision per second per unit area. Ans. d 124. Cooking vessels should have high conductivity to conduct heat. That is (b) is correct. They should have low specific heat so that smaller amount of heat will produce larger temperature rise. (d) is correct. Recall the properties of cooking vessels given in one of the previous questions. Density has nothing to do with conduction. . Ans. b & d. 125. Heat flows from a body of higher temperature to the body of lower temperature and not from the body of higher heat capacity to that of lower heat capacity. Hence (b) is correct and (a) is wrong. The temperature attained at thermal equilibrium T is such that heat lost by B is equal to the heat gained by A. This is calculated using the formula 1000(T-20) = 500(40-T). We have used 1000 and 500 as the product of mass and specific heat. This gives T not equal to 30oC. Ans.b 126. If ω is the coefficient of performance of the refrigerator, Q2 heat removed from the sink, W

external energy supplied, then Q2/W = ω. This gives Q2 = ωW = 10x700 = 7000 J/s. Ans a. 127. Use information supplied in theory notes. The specific heat at constant volume of the mixture is equal to [(3/2)Rx + (5/2)Rx] /2x = 2R. The heat required to raise the temperature of the mixture through 10oC = (2x)(2R)10 = 40Rx. In CGS system R = 2 cal/moloC. So the rrequired heat will be is 80x cal. Ans.b 128. Using the formula (T1-T2)/T1 = y, when T2 is increased by 20, this equation becomes [T1-(T2+20)]/T1 = 0.8y. Solving these equations we get T1 = 100/y. Ans.c

T1 − T2 T T = 1 − 2 .To get maximum efficiency 2 should be T1 T1 T1 minimum. That is T2 should be decreased and T1 should be increased. Ans.d 130. The potential energy of water is converted into heat. So if h is the height mgh x 70/100 = mc∆T. Here ∆T is 1o C and therefore h = cx100/g x 70. c = 4200 J/kg K, g = 10 m/s2, h = 600 m. If g is taken as 9.8 m/s2 the nearest answer will be 600 m. Ans.a 129. The efficiency is given by

2 4

131. The heat radiated by a black body per second is equal to E = σ 4πR T . Taking logarithms

dE dR dT =2 +4 = 2x1% + 4x1% = 6%. Ans.d E R T 132. The kinetic temperature 1000 K means each molecule will have a kinetic energy (1/2)kT, with T = 1000 K. This will be large compared to the energy of a molecule at our place. However, the number of molecules per unit volume at upper atmosphere is very very small. Hence total and differentiating,

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63

energy of the molecules per unit volume will be small and energy communicated to the astronaut by collision will also be very small. Ans.a 133. Here also we use Wein’s law. In this question you have to guess the value of Wein’s constant approximately. Wein’s constant = 2x10-3 mK. So using λ mT = this constant, we have λ m = 2x10-3/106 = 2x10-9 m = 2nm. This wavelength is in the X-ray region.. Ans.c 134. Thermal resistance is proportional to length (L),and inversely proportional to area (A). So we write R(thermal)= R = L/λA. where λ is thermal conductivity i.e. reciprocal of thermal = KW-1. Ans.a resistivity.⇒ R = (m)/Wm-1K-1m2

3 WAVE MOTION AND SHM SECTION 1 QUESTIONS 1.

If ω is angular frequency and k wave vector of a sound wave, its velocity is ω k c) ωk d) ωk b) a) k ω

2.

Which of the following does not represent a wave ? a) y = f(x-vt) b) y = ymsin k(x+vt) c) y = ymlog(x-vt) d) y = f(x2-v2t2) When a sound wave reflects from boundary seperating two media, which of the following could change ? a) frequency b) wavelength c) speed d) phase Two strings A and B made of same material are stretched by same tension. The radius of string A is double the radius of B. A transverse wave travels in A with speed VA and in B with speed VB. The ratio VA / VB is a) 1/2 b) 2 c) 1/4 d) 4 The speed of sound through a certain medium at 270C and 105 Nm-2 pressure is 200 ms-1. If the temperature increases to 1270C and pressure falls to 0.5 x 105 Nm-2, the speed of sound (in ms-1) is 400 50 2 100 2 c) 100 2 a) b) d) 3 3 3

3.

4.

5.

6.

7.

8.

9.

When temperature increases, the frequency of a tuning fork a) remains the same b) decreases c) increases d) may increase or decrease depending on the material A sound wave of frequency 500 hertz has a velocity 350 m/s. The minimum distance between the two particles having a phase difference of 600 is nearest to a) 0.7 cm b) 12 cm c) 70 cm d) 120 cm The speed of sound through oxygen at a temperature T is v ms-1. If the temperature increases to 2T and oxygen gas dissociates into atomic oxygen, the speed of sound a) remains the same b) becomes √2 v c) becomes 2v d) becomes nearly 9v/5 A supersonic jet produces waves in air. The wave front is a) spherical b) paraboloidal c) ellipsoidal d) conical

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65

10. A vibrating tuning fork is rotated about a vertical axis through its stem. An observer standing near will hear, in one revolution a) 4 maxima and 4 minima b) 2 maxima and 2 minima c) 8 maxima and 8 minima d) uniform sound 11. A rope of length L and mass m hangs freely from a ceiling. If v is the velocity of transverse wave produced in it and x is the distance from the free end, then v is proportional to a) x0 b) x-1/2 c) x1/2 d) x 12. A wire has frequency f, under a certain tension. Its length is doubled by uniformly stretching. Its frequency under the same tension will be a) 2f b) f c) 1.4 f d) 0.7 f 13. The ratio of velocity of sound in hydrogen to that in helium at STP is a)

42 25

b)

25 42

c)

2

d)

1 2

14. A hollow metallic tube open at both ends is in resonance with a tuning fork of frequency 256 Hz. If the tube now is immersed half inside water it will resonante to a frequency (in Hz) a) 1024 b) 512 c) 256 d) 128 15. The quality of a tone a) decreases with loudness b) depends on the overtones present c) is proportional to the pitch d) is inversely proportional to amplitude 16. When a motor boat sails in water, the waves produced are a) longitudinal b) transverse c) neither longitudinal nor transverse d) both longitudinal and transverse 17. The waves produced in a sonometer wire are a) transverse, stationary but not polarised b) transverse, progressive and polarised c) transverse, progressive and unpolarised d) longitudinal 0 18. When two waves of equal amplitude A and equal frequency at a phase difference of 60 superpose, the amplitude of the resulting wave is a) 2A

b) √2 A

c) 2√2 A

d)

3A

19. A long steel pipe is tapped at one end. A listener at the other end hears two sounds

a) b) c) d)

of the same intensity, at the same time less intense sound first, more intense later more intense sound first, less intense later of the same intensity, but at different times

20. A wave represented by the equation y = a cos(kx-ωt) is superposed on another wave to form a

stationary wave such that the point x = 0 is a node. The equation for the other wave is a) a sin (kx+ωt ) b) -a cos(kx-ωt) c) -a cos(kx+ωt) d) -a sin(kx-ωt) 21. A hospital uses an ultrasonic scanner to locate tumours in a tissue. The operating frequency of the scanner is 4.2 MHz. The speed of sound in a tissue is 1.7 km s-1. The wavelength of sound in the tissue is close to a) 4 x 10-3 m b) 8 x 104 m

66

MORE PRACTICE PAPERS FOR IIT-JEE c) 4 x 10-4 m

d)

8 x 10-3 m

22. A sound wave of wavelength λ travels towards the right horizontally with a velocity V. It

strikes and reflects from a vertical plane surface, travelling at a speed v towards the left. The number of positive crests striking in a time interval of five seconds on the wall is 5(V + v) 5(V − v) ( V + v) (V − v) b) c) d) a) λ λ 5λ 5λ 23. Two waves represented by equations y1 = 10 sin (2000 πt) and y2 = 10 sin [2000 πt +

π ] are 2

superposed at a point at an instant t. The amplitude of the superposed wave is a) 10 units b) 20 units c) 14 units d) zero 24. Which of the following gives a pure sine wave ? a) a note from a piano b) vibrating veena wire c) a vibrating tuning fork d) heart beats of a normal person x ]. The maximum particle λ velocity is equal to four times the wave velocity if λ is equal to (wave length) πY0 πY0/4 c) πY0 b) d) 2πY0 2 26. A wave equation which gives displacement along the y direction can be written as y = 10-4 sin(60t+2x) where x and y are in metres and t in seconds. This represents a wave a) of velocity v = 30 ms-1 in the negative x-direction b) of wavelength = πm c) of frequency f = 30/πHz d) of amplitude = 10-4 travelling in negative x-direction 27. Which of the following is not a travelling wave equation ? x a) Y = Ymsin k(x-vt) b) Y = Ym cos 2π − vt λ 25. A transverse wave is described by the equation Y = Y0sin 2π[ft -

x t c) Y = Ym sin 2π − λ T

d) Y = 2Ym sin kx cosωt

28. Spherical waves are emitted from a 5.0 W source in an isotropic non-absorbing medium. The

intensity of wave at a distance of 2.0 m from the source is a) 0.020 W/m2 b) 0.20 W/m2 c) 0.10 W/m2

d) 0.01 W/m2

29. The equation to a wave travelling in a spring can be written as y = cos π (100t-x), where

distances are in m. Its wavelength is a) 2π b) 1m

c) 0.05m

d) 2m

2πx sin 100πt where x and y are in cm, t in s, the node appears at x 30. In the equation y = 4 cos 50 equal to (in cm) a) 12.5 b) 50 c) 20 d) 100/2π 31. Two sound producing bodies produce progressive waves given by Y1 = 4 sin(400 πt),

3 sin(404 πt). A person situated nearby will hear a) two beats per second of intensity ratio 4;3 b) two beats per second of intensity ratio 49:1

y2 =

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67

c) four beats per second of intensity ratio 49:1 d) four beats per second of intensity ratio 4:3 32. Two waves having intensity in the ratio 9:1 produce superposition. The ratio of the maximum to minimum intensity of the superposed wave is a) 10/8 b) 4/1 c) 9/1 d) 2/1 33. The displacement of particles in a string in x-direction is represented by y. Among the following expression for y, those describing wave motion are a) coskx sinωt b) k2x2-ω2t2 c) cos(k2x2-ω2t2) d) cos2(kx+ωt) 34. A plane wave of sound travelling in air is incident on a plane water surface. The angle of incidence is 600. Assuming Snell’s law to be valid for sound waves, it follows, a) the wave will be refracted into water away from normal b) the wave will be refracted into water towards normal c) the wave will graze the surface of separation d) the wave will be reflected into air 35. The equation y = sin 4πt + cos 5πt is

a) periodic but not harmonic b) periodic but not simple harmonic c) periodic and simple harmonic d) neither periodic nor harmonic 36. The amplitude of a wave disturbance propagating in the positive x-direction is given by y = 1/(1+x2) at time t = 0 and by y = 1/[1+(x-1)2] at t = 2 seconds, where x and y are in metres. the shape of the wave disturbance does not change during the propagation. The velocity of the wave is b) 0.5 ms-1 c) 1.5 ms-1 d) 2 ms-1 a) 1 ms-1 37. A wave is represented by the equation y = A sin [10πx + 15πt +

π ] where x is in metres and t 3

in seconds. The expression represents a) a wave travelling in the positive x-direction with a velocity 1.5 ms-1 b) a wave travelling in the negative x-direction with a velocity 0.66 ms-1 c) a wave travelling in the negative x-direction having a wavelength 0.2 m d) a wave travelling in the positive x-direction having a wavelength 0.2 m 38. The displacement y of a particle executing periodic motion is given by y = t sin 1000 t. This expression may be considered to be a resultant of the superposition 4 cos2 2 of a) three waves b) two waves c) four waves d) five waves 39. A wave is represented by the equation y = A sin 2(10t-20x). Its wavelength is a) 20m b) π/20)m c) 10m d) (π/10)m 40. A wave disturbance in a medium is described by y (x,t) = 0.02 cos (50πt +

where x and y are in metres and in seconds. For this wave a) a node occurs at x = 0.15 m b) an antinode occurs at x = 0.3m c) the speed of the component wave is 5.0 m/s d) all the above are correct

π ) cos (10πx), 2

68

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41. If λ1, λ2, λ3 are the wavelengths of the wave giving resonance in the fundamental, second and

third harmonic modes respectively of a pipe closed at one end, the ratio of the wavelengths, λ2, λ2, λ3 in the order is a) 1:2:3 b) 1:3:5 c) 6:3:2 d) 15:5:3

42. The equation to the motion of a wave can be written as x = 2 sin

which the maximum velocity are attained by this wave will be a) 0, 6, 12 s, etc b) 0, 4, 8 s, etc c) 0, 3,6,9 s, etc

π t. The instants of time at 6 d) 2, 4,6 s, etc

43. An object of specific gravity ρ is hung from a thin steel wire. The fundamental frequency

form transverse standing waves in the wire is 300 Hz. The object is immersed in water so that one half of its volume is submerged. The new fundamental frequency, in Hz, is 2ρ − 1 2ρ − 1 a) 300 b) 300 2ρ 2ρ 2ρ 2ρ c) 300 d) 300 2 ρ − 1 2 ρ −1 44. The velocity of sound in air is 348 m/s. Two sources producing waves of wavelength 2m and 3m are sounded together. Which of the following is not correct ? a) They will produce 58 beats per second b) They will produce beats which cannot be detected the human ear c) The beats can be heard at the given rate in (a) d) They will produce frequencies whose difference will increase with increasing temperature 45. *The power transmitted in a vibrating string by a wave of frequency f and amplitude A is proportional to a) frequency of the wave b) amplitude of wave c) square of frequency of the wave d) square of the amplitude of the wave 46. Oxygen is 16 times heavier relative to hydrogen. Equal volumes of hydrogen and oxygen are mixed. The ratio of the velocity of sound in the mixture to that in oxygen is a)

1 8

b)

32 17

c)

17 . 25

d)

8

47. The maximum velocity of simple harmonic motion which produces a wave is 4 m/s.

Its maximum acceleration is 16 m/s2. Its amplitude of the motion is a) 1m b) 2m c) 4m d) 8m 48. A source of sound x gives five beats per second, when sounded with another source of frequency 100 /s. The second harmonic of the source x together with a source of frequency 205/s gives five beats per second. The frequency of the source x is a) 100/s b) 105/s c) 205/s d) 95/s 49. If the diameter of resonance tube of constant length is increased, the frequency emitted by it will a) remain the same b) increase slightly c) decrease slightly d) decrease considerably 50. If pressure increases by 1 atmosphere and temperature increases by 1 K, the velocity of sound a) decreases by 0.6 ms-1 b) increases by 0.6 ms-1 -1 c) increases by 60 ms d) decreases by 60 ms-1

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69

51. A stretched wire is divided into three segments whose frequencies are in the ratio 2:3:4. Their

52.

53.

54.

55.

56.

57.

58.

59.

60.

61.

lengths are in the ratio a) 2:3:4 b) 4:3:2 c) 3:2:1 d) 6:4;3 -4 -1 The linear density of a wire is 1.3 x 10 kgm . A transverse wave represented by the equation y = 0.02 sin(x+30t) is propagating along the wire, where x and y are in m and t in s. The tension of the wire is a) 0.48 N b) 0.12 N c) 1.2 N d) 4.8 N When beats are produced by two progressive waves of the same amplitude and of nearly the same frequency, the ratio of maximum loudness produced to loudness of one of the waves will be n, where n is a) 1 b) 2 c) 2.5 d) 4 A man standing between two cliffs claps his hands and hears a series of echoes such that one echo is heard at intervals of 1 s and two echoes in every alternate second. If speed of sound in air is 340 ms-1, the distance between the cliffs should be a) 340 m b) 680 m c) 510 m d) 170 m In a good tuning fork a) only the fundamental is excited b) the first overtone and the fundamental mode are excited c) only the first overtone is excited d) the fundamental and odd harmonics are excited A man facing a wall holds a tuning fork of frequency 256 between himself and a vertical wall. He moves the tuning fork towards the wall with a velocity 1/100 of the velocity of sound. The number of beats heard per minute would be nearly a) 75 b) 150 c) zero d) 300 A tuning fork of frequency 256 Hz is excited and held at the mouth of a closed pipe of frequency 250 Hz. Pick out the correct statement a) 6 beats per second will be heard b) 12 beats per second will be heard c) 3 beat per second will be heard d) No beats will be heard A uniform rope of mass 6 kg hangs vertically from a rigid support. A block of mass 2 kg is attached to the free end of the rope. A transverse pulse of wavelength 0.06 m is produced at the lower end of the rope. The wavelength of the pulse when it reaches the top is ( in m) a) 0.06 b) 0.12 c) 0.03 d) 0.24 *Two identical straight wires are stretched so as to produce 6 beats per second when vibrating simultaneously. On changing the tension slightly in one of them, the beat frequency still remains unchanged. Denoting by T1, T2 the higher and the lower initial tensions in the strings, it could be said that while making the above changes in tension a) T2 was decreased b) T2 was increased d) T1 was decreased c) T1 was increased *The prongs of a vibrating tuning fork are immersed in water. Then a) velocity of the waves will decrease b) amplitude of the waves will decrease c) frequency of the waves will decrease d) wavelength of the waves will increase Which of the following cannot produce super position? a) Light waves b) Micro waves c) Radio waves d) Laser beam

70

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62. A wave has an angular frequency of 200 rad/s and a wave vector 2 m . The wave velocity is

(m/s) a) 100 b) 100/π c) 100π d) 200/π o 63. The speed of sound in oxygen at a pressure of 1 atmosphere and a temperature of 27 C is x o m/s. The speed of sound in hydrogen at a pressure of two atmospheres and at 27 C is a) 2x b) 4x c) √2x d) 2 √2x 64. Two waves of equal amplitude and equal wavelength arrive at a point at a path difference of λ/6. The ratio of resultant intensity at the point to the maximum intensity will be a) 1/4 b) 1/2 c) 3/8 d) 3/4. 65. A wave equation is represented by y =2 cos2x sin3t. The phase difference between two adjacent particles will be ( in degrees) a) 0 b) 30 c) 60 d) 90 66. Two waves are represented by the equation y1= 10 sin πt. y2 = 10 sin [πt+

67.

68.

69.

70.

71.

π ]. The amplitude 3

of the resulting wave at a given point is a) 10 b) 20 c) 10√2 d) 10√3 If a graph is drawn between amplitude of a wave from a source arriving at certain point with distance from the source and if the medium absorbs no energy, the graph will be a) a straight line having a negative slope b) an irregular curve c) a rectangular hyperbola d) an inverse square law graph The speed of transverse waves through a wire stretched by a tension T is v. If the wire is doubled in itself and stretched by the same tension, the speed of transverse waves will be a) same as v b) (√2)v c) v/√2 d) 2v *When stationary waves are produced in a medium which of the following characteristics change at the antinodes ? a) density b) pressure c) phase d) temperature A particle on the trough of a wave of period T at a given instant will come to mean position after a time a) T b) 3T/4 c) T/4 d) T/2 At the same temperature speed of sound in 50% humid air is v1 and that in 70% humid air is v2. Then a) v1 > v2 b) v1 = v2 c) v2 > v1 d) v1 is greater than or less than v2 depending on the climate

72. A progressive wave has an equation y = A sin2π(10t-20x) where the distances are in metre

and time in seconds. The distance between a crest and the nearest trough of this wave in cm is a) 10 b) 2.5 c) 5 d) 5/π 73. A wave incident on a medium has an amplitude 1 m. The reflected wave has an amplitude 0.5 m. The percentage of intensity transmitted into the medium is equal to a) 75 b) 50 c) 25 d) 12.5 74. The amplitude of the transmitted wave in the previous question (in m) is a) 0.75 b) 0.5 c) 0.25 d) 0.866

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71

75. A war plane is flying at a speed Mac 1.5. The waves produced by it will be

a) plane

b) spherical

c) cylindrical

d) conical

76. A wave is represented by the equation y = A sin (ωt-kx). The ratio of maximum particle

velocity to wave velocity is a) A/k b) k/A c) kA d) ωA 77. If the tension of a sonometer wire is increased by 21%, the frequency of the note emitted by it increases by a) 10% b) 21% c) 32.5% d) 42% 78. If a source of sound of frequency f and a listener approach each other with a velocity equal to 1/20 of velocity of sound, the apparent frequency heard by the listener will be 19 20 21 21 f f a) d) f b) f c) 21 21 20 19 79. A tuning fork gives 5 beats when vibrated with 40 cm length of a sonometer wire. If the

length of the wire is shortened by 1 cm, the number of beats is still the same. The frequency of the fork is : a) 385 b) 320 c) 395 d) 400 80. A tuning fork of frequency f is immersed in water and vibrated. Compared to that in air, a) its period will decrease b) the wavelength of sound emitted by it will decrease c) the velocity of sound waves will decrease d) its amplitude will decrease 81. An observer moves a tuning fork of frequency 350 Hz towards a wall with a speed (1/100) th of speed of sound. The number of beats heard by him per second will be a) 4 b) 5 c) 7 d) 14 82. If temperature increases by 2%, the velocity of sound will increase by a) 1% b) √2 % c) 2% d) 2√2 % 83. Two progressive waves represented by the equations y1 = Asin(100πt) and y2 = A sin(104πt)

produce beats. An observer standing near the sources hears a maximum and the next minimum separated by an interval of time a) 1/2 s b) 1/4 s c) 3/4 s d) 1 s 84. Imagine a hypothetical gas in which sound travels with a speed equal to the rms speed of molecules of the gas. The heat required to increase the internal energy of this gas through 1 K will be a) 2R b) 3R c) (3/2)R` d) (1/2) R π x (4 t + ) , where distances are 4 16 in centimetre and time in second. The phase difference of the same particle at a time interval of 0.8 second is ( in degrees) a) 36 b) 72 c) 102 d) 144 π x (2 t + ) , where distances 86. The equation to a plane wave is given by the equation y = 6 sin 4 8 are in centimetre and time in seconds. Which of the following statement(s) is/are correct for this plane wave? a) The wave is travelling along the positive x-direction. b) The wave has a wavelength of 64 cm. 85. A plane wave train is represented by the equation y = 5 sin

72

87.

88.

89.

90.

91.

92.

93.

94.

95.

96.

97.

MORE PRACTICE PAPERS FOR IIT-JEE

c) The wave has a wavelength of 32 cm. d) The phase difference between two points separated by a distance of 0.2 metre is 112.5o. .Read the following statements carefully: A) Two sine waves of same frequency and amplitude travelling in the opposite direction superpose, giving stationary waves. B) A sine wave can superpose only with a sine wave. Of these statements, a) B is true but A need not be true. b) A is true, but B need not be true c) Both A and B are true d) Both A and B are not true Assuming that it is always possible to see light from a distant star accelerating away from earth, which of the following is likely to happen? a) The star will turn yellow and then red and remain so b) The star will turn blue. c) The star will turn violet and remain so d) The star will turn red and then invisible. Two simple harmonic motions of same period and different amplitudes and no phase difference are combined at right angles. The path of the resultant motion will be a) a circle b) a straight line c) an ellipse d) a parabola If the phase difference between the two motions in previous question is 90o the resultant motion will be a) a circle b) a straight line c) an ellipse d) a parabola If the phase difference between the two motions given in the previous question is 90o and the amplitude of the two motions are same, then the path will be a) a circle b) a straight line c) an ellipse d) a parabola A spring of force constant k and mass m attached with it, has a period T. If the mass is fully immersed in a liquid of density half its density, the period will be a) T/2 b) 2T c) T/√2 d) √2 T If a graph is plotted with period of a pendulum and its length, the graph will be a) a parabola b) a rectangular hyperbola c) a straight line d) an irregular curve The acceleration due to gravity in moon is (1/5)th that of earth. The period of a seconds pendulum in moon will be a) 2√6 s b) 2/√6 s c) 12 s d) 2 s The period of a pendulum on the surface of the earth is T. At 3 quarters down earth its period will be a) T/4 b) T/2 c) (√3/2)T d) 2T *A simple pendulum will have infinite period a) inside an earth satellite b) at the centre of earth c) inside a lift under a free fall state d) inside a lift accelerating up with a value ‘g’ A spring mass system makes 12 oscillations per minute. The spring is cut into two halves and they are suspended from the same point with the other ends carrying the same mass. The number of oscillations made by the system will be ( in one minute) a ) 12 b) 24 c) 6 d) 12√2

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73

98. A uniform cubic plank of side L is floating in a liquid of density twice its density. When it

is immersed slightly and released it will oscillate simple harmonically with a period a ) 2π

L g

b) 2π

2L g

c) 2 π

L 2g

d ) 4π

L g

99. A body executing SHM has a velocity 3cm/s when the displacement is 4 cm .

It has a velocity of 4 cm/s when the displacement is 3 cm. The amplitude of motion of this body is a) 7 b) 6 c) 6.2 d) 5 100. A toy gun uses to fire a spring of force constant K. The spring is compressed by x metre before a shot is fired . If the mass of the shot is M, the height reached by the shot when projected vertically is b) Kx2/Mg c) 2Kx2/Mg d) K2x2/Mg a) Kx2/2Mg 101. A particle oscillating simple harmonically, has an equation x= 5 cos (4πt+π/3), where t is in

seconds and x in metre. Its acceleration at a time t = 4 s, in m/s2 is c) 40π d) 40π2 a) 20π b) 20π2 102. A simple pendulm oscillates with an angular frequency 4 rad/s. The ratio of maximum acceleration to maximum velocity of this pendulum will be a) 4 b) 2 c) 1 d) 0.5 103. A listener and a source are at rest. A strong wind blows from source towards the listener. The apparent frequency heard by the listener will be a) equal to the original frequency b) less than the original frequency c) more than the original frequency d) cannot be said from the data 104. A whistle is rotating in a horizontal circle. An observer A is standing at the centre of the circle and another observer B is standing outside the circle. If f is the true frequency of the whistle, then the frequency heard by A and B will be a) changing frequency by both A and B b) same frequency by B and changing frequency by A c) same frequency by A and B d) changing frequency by B and same frequency by A 105. In which of the following cases Doppler effect cannot be observed ? a) The source and the listener moving with same velocity in opposite direction. b) The source moving and wind blowing with the same speed c) The listener moving and wind blowing with the same velocity d) The source and the listener moving with same velocity in the same direction.

SECTION 2: ANSWERS 1

(a)

2

(c)

3

(d)

4

(a)

5

(b)

6

(b)

7

(b)

8

(d)

9

(d)

10

(a)

11

(c)

12

(d)

13

(a)

14

(c)

15

(b)

16

(d)

17

(a)

18

(d)

19

(c)

20

(c)

21

(c)

22

(a)

23

(c)

24

(c)

25

(b)

26

(a,b,c,d)

27

(d)

28

(c)

29

(d)

30

(a)

31

(b)

32

(b)

33

(a),(d)

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(d)

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(b)

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(b)

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(c)

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(a)

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(d)

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(a)

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(c),(d)

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(a)

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(b),(d)

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(a)

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(c)

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(b)

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(a),(b),(d) 70

(c)

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(c)

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(a)

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(d)

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(d)

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(a)

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(b)

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(c)

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(d)

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(c)

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(a)

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(b)

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(d)

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(d)

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(a),(b),(d) 87

(b)

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(d)

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(b)

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(c)

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(a)

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(c)

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(a)

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(d)

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(d)

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(b)

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(c)

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(d)

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(d)

(a)

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(a)

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(d)

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(d)

102

SECTON 3 SOLUTIONS 2πf ω 2π = , because 2πf = ω, = k . Ans.a 2π / λ k λ

1.

V = fλ=

2.

While (a) and (b) are travelling waves, (d) is the superposition of two travelling waves, f(xvt) and f(x+vt). A logarithmic function such as (c) cannot represent a wave motion, because it has no regular period. Ans.c Frequency is the property of the source and hence does not change. Since wave returns to the medium, speed does not change. v = fλ. Hence wavelength λ also does not change. The only quantity that could change due to reflection is the phase. Ans.d

3.

4.

Velocity of transverse waves through string = Here T is same. vA = vB

5.

6.

7.

T , where T is tension, µ is linear density. µ

µ = πr2d, where d is density rA = 2rB.

i.e. µA = 4µB .

Therefore

µB 1 1 = = . Ans.a 4 2 µA

The speed of sound does not change with pressure. V127 400 400 . Ans.b = . From this V127 = V27 300 3

It is directly proportional to √T.

v E . v= When temperature increases, E (Young’s modulus) decreases and λ ρ so does density. But the rate of decrease in Young’s modulus is more than the rate of decrease of density. Therefore v decreases and f also decreases. Ans.b Frequency f =

λ= v/f = 350/500 = 0.7 m. A path difference of λ = phase difference of 2π radian or 3600. Hence a phase difference of 600 = a path difference of λ/6 = 0.7/6 m = nearly 12 cm. Ans.b

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γRT (1). Temperature T becomes 2T. When oxygen dissociates, molecular mass M M becomes atomic mass M/2. γ (diatomic) = 7/5, becomes γ(monoatomic) = 5/3. Substituting these values in (1), we find nearest answer is d. Ans.d 9. A supersonic jet has a speed more than the speed of sound. Hence wave front coming from it will be conical. Ans.d 10. When the tuning fork is vibrating condensations travel along the X axis producing two maxima. Rarefactions travel along the Y axis producing two maxima. Thus we have four maxima. In between two maxima there will be minimum. Thus, the observer will hear 4 maxima and 4 minima in one revolution. Ans.a 8.

v=

T . If free lower end of the rope is taken as origin, as µ x increases tension also increases because tension is provided by hanging portion of the rope. Since velocity v is proportional to √T, and T α x, v α √x. Ans.c 12. By the law of transverse vibration, frequency is inversely proportional to length and inversely proportional to square root of linear density. When the wire is stretched, its length becomes twice and area decreases to half, because the volume is constant. The frequency decreases to half due to the increase in length and increases to √2 times due to the decrease in area. Thus frequency becomes (1/2) x √2 = 1/√2 of the original value = 0.7 f. Ans.d 11. The velocity of transverse waves v =

13. v =

v γ M γRT . When T is constant, 1 = 1 2 , where subscript 1 is for hydrogen and 2 for M v2 γ 2 M1

helium. γ1 = 7/5, γ2 = 5/3. M1 = 2g/mol. M2 = 4g/mol. This gives v1/v2 = 42 / 25 . Ans.a 14. The fundamental frequency of an open pipe of length L is v/2L. When it is immersed half inside water, its length becomes half i.e. L/2, and also it becomes a closed pipe. The frequency of a closed pipe is v/4L1. Substituting in this equation L1 = L/2, we get the new v v = . Ans.c frequency as (4 × L / 2) 2L 15. The tonal quality depends on the number of overtones present, and their relative intensity.

Ans.b 16. When a motor boat travels in water its propeller cuts the water surface lateraly at the same

time pushes backward. This produces both longitudinal and transverse waves. Ans.d 17. The waves produced in sonometer are transverse (vibration perpendicular to the wave

propagation) and stationary. Sound waves cannot be polarised. Ans.a 18. When two waves of amplitudes A1 and A2 at a phase difference of φ superpose, the resulting

amplitude is given by

A 1 2 + A 2 2 + 2A 1 A 2 cos φ . Here A1 = A2 = A. φ = 600. This gives

the resulting amplitude as √3 A. Ans.d 19. The types of waves reaching the listener are those propagating through air and those through

steel. Steel is more elastic. Hence the amplitude and the intensity of waves through steel will be more. The velocity of sound waves through steel is greater and hence they reach quicker. Ans.c 20. The given wave is a cosine wave and one travelling along the positive x-direction. To form stationary waves, the two waves should travel in the opposite direction. That is the required wave should have a term (kx+ωt). Also a cosine wave can be combined only with a cosine

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wave. The only cosine wave travelling along the negative x-direction in the given list is – a cos(kx+ωt). We can check combining the two, i.e. a cos (kx-ωt) and –a cos(kx+ωt), we get 2a sin(kx) sin (ωt), for which x = 0 is a node. Ans.c 21. Using the equation v = fλ, we have -λ =

v 1.7 × 1000 = = 4x10 − 4 m. Ans.c f 4.2 × 10 6

22. The relative velocity of sound waves with respect to the wall is V+v. Hence the apparent

frequency of the waves striking the wall will be (V+v)/λ. Each wave has one positive crest. Hence the number of positive crests will be same as the frequency. For 5 seconds, this number will be [5(V+v)]/ λ. Ans.a 23. The resultant amplitude A of two waqves of amplitudes a1 and a2 at a phase difference φ is

(a12+a22+2a1a2 cosφ)1/2. Substituting a1 = 10, a2 = 10 and φ = 900, we get A = 14.1. Ans.c 24. A tuning fork emits a pure sine wave and also gives a single frequency. That is, it has no harmonics. Ans.c 25. The maximum particle velocity of a SHM of amplitude Y0 and frequency f is 2πfY0. The

πY0 . Ans.b 2 26. Comparing with the standard equation of a transverse progressive wave, Y = A sin(ωt + kx), we find A = 10-4, ω = 2πf = 60, which gives f = (30/π) Hz. k = 2π/λ = 2. That is λ= π.m. Speed of the wave v = ω/k = 60/2 = 30 m/s. Hence all the answers are correct. Ans.a,b,c,d 27. The first 3 are progressive waves, while the last one is super position of two progressive waves, i.e. stationary wave. Ans.d 28. Intensity at a point is the energy received per unit area per second or power received per unit area. Here power is distributed at the point of reception on the surface of a sphere of radius P 5 2m. Therefore I = = 0.1 Wm-2 . Ans.c = 4πr 2 4π × 2 2 wave velocity is fλ. For 2πfY0 to be equal to 4fλ, λ has to be

29. Comparing with the standard equation, Y = A cos(ωt+kx) with Y = 3 cos(100πt-πx), we find

ω= 100π, k = π ; 2π/λ = π, λ = 2m. Ans.d

30. Comparing with the standard equation for a stationary wave, y = 2A cos(kx) sin(ωt) we find

the amplitude term is 2A cos(kx).

A node appears when amplitude is 0, i.e. kx =

π . 2

2πx π = . This gives x = 12.5 cm. Ans.a 50 2 31. Comparing with standard equation, Y = A sin (2πft), we find the frequency of waves are 200

Hz and 202 Hz. So, the number of beats will be 2 per second. The intensity ratio will be the ratio of the square of the maximum amplitude (4+3) and the minimum amplitude (4-3) which is 72:12 = 49:1. Ans.b 32. Intensity is proportional to the square of amplitude. Therefore the amplitudes are in the ratio 3:1. The ratio of the maximum and the minimum amplitudes is 4:2. So the ratio of the maximum intensity to the minimum intensity is 16:4 = 4:1. You can also use indirect theory short-cut formula. Ans.b 33. (b) and (c) cannot represent a wave motion because they involve square of x and t and therefore will not have a regular period. (d) involves cos2, hence cannot be negative but cos2θ

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can be written as (1+cos2θ)/2 and hence could be periodic. (a) represents superposition of two waves. Ans.a,d 34. Since Snell’s law is valid, the refractive index n of the medium is given by velocity of sound in air/velocity of sound in medium = 330/1400 = 0.235, where we have taken velocity of sound in air and water roughly as 330 and 1400 m/s. If C is the critical angle, then the refractive index n is given by = sin C/sin90 = n = 0.235. (note the difference with the optics equation n = 1/sin C). We get C = nearly 140. (We need not actually calculate but only need know that it is less than 600, incident angle). Thus the angle of incidence is greater than this critical angle. Hence the wave will totally reflect into air. Ans.d 35. According to Fourier theorem, a period function f(t) can be expanded as a number of harmonic functions. Here y = f(t) is periodic since RHS is harmonic. But for a simple harmonic function we should have same period for harmonic functions. Here periods of cos 4πt and sin 5πt are not same. Hence it is not simple harmonic. Ans.b 36. Writing the general expression for y in terms of x as y =

At t = 2s, y =

1 1 + [ x − v(2)] 2

1 1 + ( x − vt )

2

, at t = 0, y =

1 (1 + x 2 )

. Comparing with the given equation we get 2v= 1 and v =

.

0.5

m/s. Ans.b 37. Comparing with the standard equation, y = A sin (ωt-kx+φ) we ω is 15π and k = 10π.

ω 2π = − 1.5 ms −1 . Also = k = 10π. λ = 0.2 m. The wave is travelling along the k λ negative x direction with a velocity 1.5 ms-1 and has a wavelength 0.2 m. Ans.c

Velocity =

38. If we have an equation of the form cos kx sin ωt, we know, it is due to the superposition of

two wave motions. If we have an equation of the form cos2 at sin bt, it can be shown to be superposition of three sine waves. Ans.a

39. Since the equation has no π, comparing the standard form y = a sin (ωt-kx) with the given

equation y = a sin(20t-40x), we find k = 40. 2π/λ = 40, λ = π/20m. Ans.b 40. The given equation is that of a stationary wave and can be written as y = 0.02 cos (10 πx) sin

(50πt). Comparing with standard equation y = A cos(kx) sin (ωt) we find first node appears when cos 10 πx = 0. That is 10πx = π/2. x = 0.05 m. Next nodes will be at 0.15 m, 0.25 m, 0.35 m etc (choice a) Anti-node appears when cos 10πx = 1. That is 10πx = nπ. So the antinodes will be at 0, 0.1 m, 0,2m, 0.3 m (choice b). Velocity of the wave v = ω/k = 50π/10π = 5m/s. (choice c). Since all are correct the choice is d. Ans.d 41. In a closed pipe of length L resonance appears for wavelengths L = λ1/4, L = 3λ2/4 and L =

5λ3/4, neglecting end correction. 1:(1/3):(1/5) = 15:5:3. Ans.d

This will give λ2 : λ2 : λ3 = 4: (4/3): (4/5) that is

π π π π cos ( t). This will be maximum when cos t is maximum. i.e. t = 6 6 6 6 nπ, where n is an integer. This gives t = 0, 6,12,18 etc. [Alternate way of doing: Comparing with the standard equation, cos(2πt/T), we find period of wave motion T is 2π/T = π/6. T = 12 s. In a simple harmonic motion maximum velocity is attained during half the period, because the particle will cross equilibrium position twice in a period. i.e. 6,12,18 s etc. Ans.a

42. Velocity dx/dt = 2

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43. By law of transverse vibration f α T .

f1 = f2

T1 = T2

M1 where M1 g is weight in air and M2

M2g is weight under water. M2 = M1 - ∆M1, where ∆M1 is loss of weight in water. M1 = Vρ. M2 = Vρ-(V/2)x1. M1/M2 = 2ρ/2ρ-1, where we have taken specific gravity of water as 1. This gives

f2 M2 2ρ − 1 2ρ − 1 = = ⇒ f 2 = 300 × Ans.a f1 M1 2ρ 2ρ

44. Frequency difference = f1-f2 = (V/λ1– (V/λ2)) = (348/2)-(348/3) = 174-116 = 58. The beat

frequency will therefore be 58 Hz. (a) is correct. Since human ear cannot detect more than 8 beats/s, (b) is correct. (c) is wrong. When temperature increases, velocity of sound will increase, frequency will increase and number of beats will also increase (d) is correct. Ans.c 45. The power transmitted by the wave is the energy of simple harmonic motion per second. This we know as (1/2)mω2A2. Thus the energy will be proportional to the square of frequency f and square of amplitude A. Ans.c & d 46. The density of the mixture is ρm = (v1ρ1 + v2ρ2)/(v1+v2) = [(v/2)16 + (v/2)1]/v = 17/2.

Velocity of sound = same v(mixure) = v(oxygen )

value ρ(oxy )

γP / ρ . i.e. velocity of sound is proportional to 1/ρ, since both have for

γ.

Therefore

v(mixture)/v(oxygen)

16 32 = .Ans.b = 17 / 2 17 ρ(mix ) 2

47. v (maximum) = ωA......(1) and a (maximum)= ω A.....(2) for simple harmonic motion of

amplitude A and angular frequency ω. Squaring equation (1) and dividing by (2) we get ω2A2/ω2A = 16/16 = 1. That is A = 1m. Ans.a

48. From the first statement frequency of x = 100 ± 5 = 105 or 95. From the second statement,

second harmonic of the source x has to be 205 ± 5 = 210 or 200. This means x is either 105 or 100. The common value of both conclusions is 105 per second. Ans.b 49. According to an empirical formula the end correction of the resonance tube is given by 0.3 d, where d is the diameter of the tube. The corrected equation will therefore be L1 + e = λ/4, for first resonance. If d is increased, e increases, λ increases and f decreases slightly. Ans.c 50. The increase in pressure does not change the velocity of sound. When the temperature increases by 10C or 1 K, the velocity of sound increases roughly by 0.6 ms-1. (Recall formula Vt = V0 + 0.6 t). Ans.b 51. According to one of the laws of transverse vibration, frequency is inversely proportional to length. So the ratio of length will be 1/2 : 1/3 : 1/4, which is 6:4:3. Ans.d 52. We compare with the standard equation for a travelling wave, Y = A sin (ωt + kx). We find ω

= 30, k = 1, velocity ω/k = 30 ms-1. The velocity of transverse waves in a string is given by V

=

T / µ , where T is the stretching tension µ is the linear density. µ here is given as 1.3 x 10-

. T = V2µ = 0.12 N. Ans.b 2 53. Let a be the amplitude of each wave . Then intensity of one wave is proportional to a . The maximum amplitude is the sum of the two, i.e. 2a. Hence maximum intensity is proportional to 4a2. The required ratio n here is 4a2/a2 = 4. Ans.d 54. The man should be standing 1/3 from one of the cliffs and 2/3 from the other cliff. The sound takes 1/2 second to travel to the nearest cliff and 1/2 second to return. Similarly it takes 1 4

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second to travel to farther cliff and 1 second to return. Distances between man and nearest cliff = distance travelled by sound in 1/2 s = 170 m. This gives distance between cliffs 170 x 3 = 510 m. Ans.c 55. A tuning fork gives only the fundamental frequency and it gives a pure sine wave. That means it has no harmonics. Ans.a 56. Here the tuning fork is a source moving away from the observer, while the sound reflected from the wall is moving towards the observer. Thus we have two sources of the same frequency, one moving away and the other moving towards the observer with the same velocity. If V is the velocity of sound Us the velocity of source, the apparent frequency 2fU s Vf Vf 1 − = difference V , we get. number if Us T2 , f1 > f2 : f1-f2

= 6 Hz. By increasing the lower tension T2, we can increase f2 such that f2-f1 = 6Hz. Siimilar we can decrease the higher tension T1 such that f decreases making f2-f1 = 6 Hz. Ans.b & d 60. The amplitude of vibration will decrease due to the resistance produced by the water medium. Hence (b) is correct. The velocity of sound in water (1400 m/s) is greater than its velocity in air (340 m/s). Hence (a) is wrong. The frequency is the property of the source and hence does not change. Thus (c) is wrong. v = fλ. As v increase, λ also should increase. Hence (d) is correct. Ans.b & d 61. The principle of super position is common for all wave motion. But in the case of the laser beam it is already an intense beam got by a number of photons in phase. So it has a very large amplitude and hence difficult to superpose. Ans.d 62. The wave velocity in terms of angular frequency ω and wave vector k is v = ω/k = 200/2

=100 m/s Ans.a 63. Here the temperature is same. The variation of pressure does not change velocity of sound.

Using the formula v = vH = 4x. Ans.b

γ RT / M , we find vH/vO =

M(O 2 ) / M (H 2 ) = 32 / 2 =4 . This gives

64. Use the information supplied already in indirect theory notes. A path difference of λ/6 means

a phase difference of (2π/λ)x(λ/6) = π/3 radian. Intensity at the point I = Iocos2 (π/6). This gives I/Io = cos230 = 3/4 . Ans.d 65. Use the information supplied already in theory notes. to recognise that the given equation is that of a stationary wave. In a stationary wave all particles vibrate in phase. The phase difference between any two adjacent particles will be zero. Ans.a

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66. Use the information supplied already in theory notes. The resultant amplitude is got by

substituting a = 10, φ = 60o, which gives 10x√3. Ans.d 67. The intensity at a point is inversely proportional to square of the distance from the source. Hence the amplitude (a2 = I) is inversely proportional to the distance from the point to the source. The graph between them will be a rectangular hyperbola. Ans.c 68. The speed of transverse waves through a wire is given by

T / µ , where T is stretching

tension and µ linear density. When the wire is doubled in length the linear density which is mass/length becomes half the original value. Thus µ becomes µ/2. The velocity becomes √2 times. Ans.b 69. In a stationary wave formation at the antinode pressure is minimum, which means density also is minimum. Pressure is maximum and density maximum at nodes. Phase does not change in stationary wave. The temperature changes because when sound wave propagates through a medium the medium undergoes adiabatic change. Ans.a,b,d 70. Trough is a point of minimum (negative maximum) displacement. From here to mean position ( point of zero displacement) the time taken will be one quarter of a period. Ans.c γP / ρ . When air is humid, more water vapour is present. This will decrease the mass of air in a given volume and hence the density of air. So the velocity of sound will increase. Ans.c

71. The velocity of sound in air at a given temperature =

72. Comparing with standard equation y = Asin(ωt-kx), we find k is equal to 2πx20. But k is

equal to 2π /λ. This gives λ = (1/20) m = 5 cm. Distance between crest and trough = λ /2 = 2.5 cm. Ans.b 73. Since amplitude reduces to half after reflection, the reflected intensity which is a square of the amplitude, reduces to 1/4th i.e. 25%. The transmitted intensity will be 75%. Ans.a 3 / 4 = √3/2 = 0.866. Ans.d 75. Use the information supplied already in theory notes.. Speed Mac 1.5 means it travels with a speed of 1.5 times speed of sound. Hence the waves produced will be conical. Ans.d 74. Since the transmitted intensity is 75% i.e. 3/4 , the transmitted amplitude is

76. Use the information supplied already in theory notes. Maximum particle velocity is ωA.

Wave velocity = ω/k. Dividing the two equations, the required ratio is kA. Ans.c

77. By one of the laws of transverse vibration, frequency f is proportional to √T or = k√T. When

121T = 1.10 k√T. This means the frequency 100 increases to 1.1 f i.e. increases by 10%. Ans.a 78. The apparent frequency heard when a source and listener approach each other with a velocity V + UL Us and UL respectively, is given by the Doppler effect formula f ' ' = f . Substituting V − Us tension is increased by 21% f becomes k

here UL= Us=(1/20) V, where V is the velocity of sound in air, we get f’’ = (21/19) f. Ans.b 79. If f is the frequency of the tuning fork then f1 = f-5. f2 = f+5. Using one of the laws of

transverse vibration, we have f1l1 = f2l2. That is (f-5)40 = (f+5)39. Solving we get f = 395 Hz. Ans.c 80. Due to high damping (resistance) exerted by water, the amplitude and the intensity will decrease. So (d) is correct. The period is a property of the source and hence will remain

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81

constant. (a) is wrong. The velocity of sound in water is greater than that in air. Hence (c) is wrong. v= fλ. When v increases and f remains constant, λ should also increase. Hence (b) is wrong. Ans.d. 81. Use the equation supplied in theory notes. The number of beats heard will be 2fUs/V. Here Us/V = 1/100. f= 350 Hz. Substituting these values, we get the number of beats per second as 7. ( Note: The velocity of sound in air is not given in the question. Hence you cannot assume it for calculation. Our formula has only the ratio of velocity of tuning fork to the velocity of sound. This is the advantage of using it) Ans c. 82. The velocity of sound is directly proportional to the square root of temperature . i. e. v= k√T.

To find change of speed for small variation of temperature, we should differentiate this equation. This is the shortest way to do it. Differentiating, dv/v = (1/2) dT/T. Since it is given here dT/T = 2%, we get dv/v = 1%. Ans a 83. Comparing with the standard equation to a progressive wave y = A sin ωt, we have the

frequencies of the waves producing beats are 2πf1 = 100π, and 2π f2 = 104πt. This gives f1 = 50 Hz and f2 = 52 Hz. The number of beats produced per second, that is difference in frequency = 2. The time interval between two beats is, therefore, 1/2 a second. The time interval between one maximum and next minimum is 1/4 a second. Ans b. 84. We have already seen in answering a similar question in one of previous section, the ratio of velocity of sound to rms velocity of molecules of the gas is

γ / 3 . If these two are same, the

gas must have γ = 3. That is Cp/Cv = 3. This gives Cp = 3 Cv. What is required is Cv. Cp-Cv = R. ⇒ 3Cv - Cv = R. This gives Cv = R/2. Ans. d 85. We have to find the wave period first and then use information given in theory notes . For this we compare with any standard equation , but it is always good to rely on one standard equation, which is given in the theory notes. The comparison between y =A sin (ωt+kx) gives ω as equal to π. That is π = 2πf. This gives f = 1/2 and the wave period T as 2 seconds. Now use the information in theory notes . We get the time difference of 0.8 second corresponds to a phase difference of 360x0.8/2 = 144o. Ans d 86. Comparing with the standard equation as in the previous question, we find k = π/32 = 2π/λ.

This gives λ = 64 cm. This makes (b) correct and (c) wrong. A difference in the path of 64 cm is equal to a phase difference of 2π radian or 360o. Thus 20 cm should produce a phase difference of 360x20/64 = 112.5o. This makes (d) correct Since there is a positive sign between t and x terms, the wave is travelling along the negative x direction. Thus (a) is correct. Ans a,b,d 87. The statement (A) is correct. But sine wave can superpose with a cosine wave, because cosine wave is a sine wave with a phase difference of 90o. Hence (B) need not always be true. Ans b 88. As the star is moving away from the observer, it should show red shift. As its velocity is increasing , the light from it will shift towards red, will become red and then infrared. The star will become invisible since then, as infrared is invisible. Ans d 89. If two motions are represented by equations x = a sinωt and y = b sinωt, we have x/a = sinωt,

y/b = sinωt. So x/a = y/b. i.e. y = (b/a)x. This is an equation to a straight line passing through the origin inclined at an angle tan-1(b/a). Ans.b 90. Here the two motions can be represented by equations x = a sinωt and y = b sin(ωt+π/2) =

b cosωt. Squaring and adding, we get (x2/a2)+(y2/b2) = 1. This is equation to an ellipse, whose axes coincide with coordinate axes. Ans.c

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91. Here the two motions can be represented by equations x = a sinωt and y = a sin(ωt+π/2) =

a cosωt Squaring and adding, we get x2+y2 =a2. This is equation to a circle whose centre is origin and whose radius is a. Ans.a

92. The period T = 2π m / k . When the mass is fully immersed in the liquid, the buoyant force

will be (m/2)g. Hence the apparent weight will be (m/2)g. 2 π m / 2 k = T/√2. Ans.c

The new period will be 2

93. The period of a pendulum T is directly proportional to √L. That is T = k√L. Squaring T =

constant x L. So the graph between T and L will be a parabola. (Note: The graph between T2and L will be a straight line which usually is drawn in a school laboratory). Ans.a 94. A seconds pendulum by definition is one which has a period of two seconds, wherever it is . Ans.d. 95. T = 2π L / g . At 3 quarters down earth the value of acceleration due to gravity is equal to g1

= [R-(3/4)R]g/R. ⇒g1 = (1/4)g. The period will be equal to 2π L / g 1 = 2T. Ans. d 96. A simple pendulum will have infinite period when acceleration due to gravity is zero or

effectively zero. This happens inside a satellite, at the centre of earth and a lift in free fall state. This makes first three answers correct. In a lift accelerating up ‘g’ will be effectively more than the normal value. This makes the last answer wrong. Ans a,b,c 97. The frequency of oscillation of a spring is given by n = 1/( 2π m / k ).

When the spring is cut into two halves, each half will have a force constant of 2k. When they are suspended from the same point with a common mass, they are in parallel. Hence the force constants add to 4k. The new frequency n1 will be equal to

1 2π

k = 2n = 24. Ans.b. m

98. If x is the distance through which plank is immersed, A cross sectional area, then weight of

liquid displaced = -Ax2dg = -kx. Here this force provides restoring force. Force constant k = A2dg. The period of oscillation will be = 2π m / k =2π ALd / A 2dg = 2π L / 2g . (Can also be answered from indirect theory short-cuts.) Ans.c 99. The velocity of SHM having an amplitude A at a displacement x is given by v= ω

A 2 − x 2 . Here we have v= 3 when x= 4 . This gives 3 = ω A 2 − 4 2 (1). 2

Also v= 4

2

when x= 3. This gives 4 = ω A − 3 . (2). Dividing (1) by (2), squaring and simplifying we get A = 5 cm. Ans d 2 100. The potential energy when the spring is compressed is (1/2)Kx . When the shot is fired upward, this energy is converted into gravitational potential energy of the shot. This gives (1/2)Kx2 = Mgh. ⇒ h= Kx2/2Mg. Ans a 101. The acceleration of SHM is -ω2x. Here ω = 4π. Hence the acceleration will be -16π2[5 cos

(4πt+π/3)]. When we substitute t = 4, this reduces to 80π2cos(π/3) = 40π2. Ans d.

102. The maximum velocity of SHM in terms of ω is ωA, where A is the amplitude and the

maximum acceleration will be ω2A, in magnitude(because acceleration has negative sign). Dividing the two, we get the ratio of maximum acceleration to maximum .velocity as ω2A/ωA =ω = 4. Ans. a 103. The apparent change in the frequency due to Doppler effect depends on the relative velocity between the source and observer. The motion of medium alone cannot produce Doppler

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effect. Here the relative velocity between the source and the observer is zero and hence no change in frequency. . Ans.a 104. For an observer standing outside the circle maximum and minimum (i.e. more than and less than f) will be heard due to Doppler effect. For an observer at the centre of the circle, the velocity of the source is always perpendicular to the line joining the observer and the source. Hence no Doppler change in frequency. Ans.d 105. Except (d), in all other cases Doppler shift in frequency will take place. In (d) the relative velocity between the source and the observer is zero. Hence no Doppler effect. Ans.d

4 ELECTROSTATICS SECTION 1 QUESTIONS 1.

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9.

When a body is negatively charged by electostatic conduction, its mass will a) slightly increase b) slightly decrease c) remain same d) increase or decrease depending on material Two charges q1, q2 are separated by a distance r. If a metal plate of thickness x is introduced between the charges the force between the charges F will now be a) F = 0 b) F = A[q1q2/(r-x)2] 2 c) F = A(q1q2/r ) d) F = A[q1q2/(r+x)2] Two point charges +4 mC and –1mC are separated by a distance of d. The ratio force acting on them will be a) 1:4 b) 1:16 c) 1:1 d) 1:-1 *Two insulated copper spheres of radius 0.1 m and 0.15 m are given equal charges. They are connected by a wire. Then a) both spheres attain the same charge b) both spheres attain the same potential c) smaller sphere attains greater charge d) bigger sphere attains greater charge Three point charges +q each are placed on the vertices of an equilateral triangle, which lie on the circumference of a circle of radius r. The electric field at the centre of the circle is a) Aq/r2 b) zero c) 3Aq/r2 d) 3q/r2 Inside a uniformly charged solid sphere,the electric field intensity at any point distant r from the centre of the sphere varies as a) 1/r2 b) r2 c) r2 d) r A charge q is located at the centre of a cube. The electric flux through any one face of the cube is a) q/ε0 b) zero c) q/4εo d) q/6εo When a soap bubble is negatively charged a) it collapses b) it increases in size c) it decreases in size d) it remains as such Two metallic spheres of radii 1 cm and 2 cm are given charges 10-2 C and 5 x 10-2 C, respectively. If they are connected by a conducting wire, the final charge on the smaller sphere is a) 2 x 10-2 C b) 4 x 10-2 C c) 1 x 10-2 C d) 3 x 10-2 C

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10. There is an electric field E along x-direction. If the work done in moving a charge 0.2 C

through a distance 2 metres along a line making anangle 600 with the x-axis is 4 J, the value of E is a) 4 N/C b) 8 N/c c) √3 N/C d) 20 N/C 11. A particle of mass 1 kg and charge 1 C falls through a potential of 1 V. Its velocity is a) 1.4 ms-1 b) 2 ms-1 c) 0.5 ms-1 d) 1 ms-1 12. The dimensions of εv is the same as that of

a) capacitance b) capacitance/unit length c) capacitance/area d) potential/length -11 13. The radius of the hydrogen atom is 5 x 10 m and the charge on electron or proton is of magnitude 1.6 x 10-19C. At the ground state, the dipole moment of the atom ( in C m)is b) 16 x 10-30 c) 0 d) 4x10-30 a) 8 x 10-30 14. Two conductors of capacitances 1 µF and 2 µF are charged to 200 V and 100 V respectively,

15.

16.

17.

18.

19.

20.

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and then connected by a wire. The final potential of the connected system is a) 150 V b) 106 V c) 133 V d) 187 V The dielectric strength of a certain material is 106 Vm-1. The breakdown voltage to be applied across a 1 mm thick specimen is a) 106 V b) 109 V c) 103 V d) 2 x 106 V A charge q is placed at the centre of the line joining two equal charges Q. The system of the three charges will be in equilibrium if q is equal to a) -Q/2 b ) -Q/4 c) +Q/4 d) +Q/2 *A parallel plate capacitor is charged and the charging battery is then disconnected. If the plates of the capacitor are moved farther apart by means of insulating handles, a) the charge on the capacitor increases b) the voltage across the plates increases c) the capacitance increases d) the electrostatic energy stored in the capacitor increases A certain charge is divided in the ratio m: n so that when kept at a given distance, the force between them is maximum. The ratio m:n is a) 2;1 b) 4:1 c) 1:1 d) 3:1 A condenser is connected to a battery . If a dielectric slab is introduced fully in the intervening space between the condensor plates, the electric field between the plates will a) decrease b) increase c) remain the same d) cannot be answered from the data An electron falls between two parallel plates of separation 2m, in which a uniform electric field 2.5 V/m exists. The energy gained by electron is equal to a) 2.5 eV b) 1 eV c) 4 x 10-19 J d) 8 x 10-19J A condenser of capacity C is connected to a battery of potential difference V. If q is the charge given by the battery, the energy given by the battery and energy stored by the condenser are respectively a) (1/2)qV, (1/2) qV b) qV, (1/2) QV c) (1/2)qV, qV d) qV, qV The potential due to an electric dipole on the axial line at a point r from the dipole is V. The potential at a distance 2r from the dipole will be a) V/2 b) V/4 c) V/8 d) V/16 *Which of the following statement(s) is/are true? a) If electric field at a point is zero, electric potential will always be zero

86

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b) If electric potential at a point is zero, electric field will always be zero c) Electric field can exist at point even when the potential is zero at the point d) Electric potential can exist even when the field is zero at the point The number of lines of force coming out from +1C is a) 1.11 x 1011 b) 1.11 x 10-10 c) 1.1 x 1019 d) 1 Two identical capacitors are joined in parallel and charged to a potential V. Then these are separated and connected in series with the positive plate of one connected to the negative of the other. Then a) the charges on the plates connected together are destroyed b) the charges on the free plates are enhanced c) the energy stored in the system increases d) the potential difference between the plates becomes 2V If the distance between the plates of a parallel plates condenser is increased, its potential will a) remain the same b) decrease c) increase d) decrease expontentially The insulating property of air breaks down at the intensity of electric field 3 x 106 Vm-1. The maximum charge that can be given to a sphere of diameter 5m is about a) 2 x 10-9 C b) 2 x 10-3 C c) 2 x 10-5 C d) 2 x 10-6 C A hollow metallic sphere of radius 5 cm is charged such that the potential on its surface is 10V. The potential at the centre of the sphere is a) 10 V b) zero c) same as at a point 10 cm away from the surface d) same as at a point 25 cm away from the surface Two identical small metal balls are given charges equal to 10 units and –20 units. They are allowed to touch each other and are again separated to the same distance as before. The ratio of the force between the two balls in the two cases respectively is a) -8:1 b) 8:1 c) 1:8 d) 1:8 Two concentric thin metallic spheres of radii R1 and R2 (R1 > R2) bear charges Q1 and Q2 respectively. The potential at radius r between R1 and R2 is a) 1/4 πε0 [ (Q1/R1) + (Q2/R2)] b) 1/4 πε0 [(Q1/r ) + (Q2/r)] c) 1/4 πε0 (Q1/R1 ) + (Q2/r) d) 1/4 πε0 (Q1 Q2 / R1 R2) A parallel plate capacitor is charged. With charging battery connected, the plates of the capacitor are moved further apart by means of insulating handles a) the electric field between the plates increases b) the electric field between the plates decreases c) the electric field between the plates remains the same d) cannot be said from the data. K (Fig) A parallel plate capacitor with plate area A and t separation is filled with dielectric as shown. The dielectric t K constants are k1 and k2. The capacitance will be b) ε0A (k1+ k2)/k1 k2t a) ε0A (k1 +k2)/t d) 2ε0A (k1 +k2)/k1 k2 t c) 2ε0A k1k2/(k1+k2)t 1

1

2

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33. Three protons A,B and C are located between the plates of a parallel plate condenser such that

34.

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37. 38.

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B is midway between the plates, A at a distance from one plate and C is at an equal distance from the other plate. (Fig.) Check the correct statement(s) a) The force on all the protons will be same b) The force on proton B will be zero A C c) The force on A and C will be same in magnitude and direction d) The force on the proton A and C will be equal and opposite B There is a non-uniform electric field along x-axis as shown in fig.. The field increases at a uniform rate along +ve x-axis. A dipole is kept inside the field as shown. Which one of the following statements is correct for dipole ? a) dipole moves along positive x-axis and rotates clockwise b) dipole moves along negative x-axis and rotates clockwise -q c) dipole moves along positive x-axis and rotates anticlockwise d) dipole moves along negative x-axis and rotates anti-clockwise Two spheres A and B of radius a and b respectively are at the same potential. The ratio of the surface density of charge of A to that of B x-a xis is a) b/a b) a/b c) a2/b2 +q 2 2 d) b /a In a certain region of space there exists a uniform electric field of 2 000 k Vm-1. A rectangular coil of sides 10 cm x 20 cm is kept in XY plane. The electric flux through the coil in SI will be a) zero b) 40 c) 4 x 105 d) 4 An lisolated metal sphere of radius R is given a charge q. Its potential energy will be b) q/4πεoR c) q/8πεoR d) q2/8πεoR a) q2/4πεoR The electric field at a certain point is 10 NC-1. The electric lines of force crossing unit area around the point at right angles to it is a) εo b) 1/εo c) 5 d) 10 0 The number of lines crossing at an angle 30 with the surface in previous question is a) εo b) 1/εo c) 5 d) 10 2 A rectangular coil of area 200 cm is placed in an electrostatic field 200 kˆ , in Y-Z plane. The electric flux through the coil will be a) zero b) 4 x 104 c) 40 d) 4 Two small balls are given equal positive charge Q coulomb each and are suspended by two insulating strings of length L each (metre) from a hook fixed to a stand. If the arrangement is taken to a satellite orbiting round earth, the angle θ between the two strings will be a) 00 b) 900 c) 1800 d) 00 < θ < 1800 The tension of each string in this position as given in previous question will be a) 0 c)

b) Q2

8πε 0 L2

d)

Q2 16πε 0 L2 Q2 4πε 0 L2

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43. An electron kept in an electric field of strength E experiences a force equal to its weight. If m

44.

45.

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48.

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50.

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52.

53.

is the mass of electron, the value of E is equal to a) mg b) mg/e c) mg/e2 d) e/mg A hollow charged metal sphere has a radius r. If the potential difference between its surface and a point at distance 3 r from the centre is V, then the electric field intensity at a distance 3r from the centre is a) V/6r b) V/4r c) V/3r d) V/2r If one penetrates into a uniformly charged sphere, the electric field strength E will a) decrease b) increase c) remain the same as that at the surface d) be zero at all points If all the electrons are removed from 1 g of hydrogen atom, the charge it will have would be nearly a) 106 C b) 104 C c) 1.6 x 10-19C d) 105C The electric potential V at a certain point distant x (in metre) is given by V(x) = 5x2+10x-9 volt. The electric field at x = 1 m will be a) 20 V/m b) -10 V/m c)- 23 V/m d) -20 V/m At any point on the right bisector of an electric dipole a) the electric field is zero b) the electric potential is zero c) the electric field is perpendicular to the dipole d) the electric field is parallel to the dipole A parallel plate capacitor of plate area A and plate separation d is charged to potential difference V and then the battery is disconnected. A slab of dielectric constant K is then inserted between the plates of the capacitor so as to fill the space between the plates. The work done on the system in inserting the slab is a) ε0 AV2/d b) ε0 KAV2/2d 2 d) ε0 AKV2/2d(K-1) c) ε0 (K-1)AV /2Kd *A proton moving with a constant velocity passes through a region of space without any change in its velocity. If E and B represent the electric and magnetic fields respectively, this region of space may have a) E = 0, B = 0 b) E = 0, B not equal to 0 c) E not equal to 0, B = 0 d) E not equal 0, B not equal to 0 Two equal negative charge –q are placed at points (0,a) and (0,-a) . A positive charge Q is released from rest from a point 2a, 0 on the x-axis. The charge Q will a) execute S.H.M about the origin b) move to origin and remain at rest c) move to infinity d) execute oscillations but not simple harmonic 1000 small water drops each of radius and charge q coalesce to form a single bigger drop. The ratio of potential of the bigger drop to that of the smaller one is a) 1 b) 10 c) 100 d) 1000 An infinite number of electric charge each of magnitude +e are arranged along the X-axis at distance x= 1m, x = 2m, x = 4m, x = 8m ....etc. The electrostatic potential at the origin is a) ∞ b) e/2πεo c) e/3πεo d) e/5πεo

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54. If the charges in the above question are alternately + and – beginning with a positive charge,

the potential at the origin will be a) zero b) e/2πεo c) e/3πεo d) e/6πεo 55. In the arrangement as given in question (53), the electric field at the origin will be a) zero b) e/2πεo c) e/3πεo d) e/5πεo e) e/6πεo 56. In the arrangement as given in question (54), the electric field at the origin will be a) zero b) e/2πεo c) e/3πεo d) e/5πεo 57. Two condensers of capacity 0.3 µ F and 0.6 µF respectively are connected in series. the

combination is connected across a potential of 6 volts. The ratio of energies stored by the first condensor to that of second will be a) 1/2 b) 4 c) 1/4 d) 2 58. A solid conducting sphere having a charge Q is surrounded by an uncharged concentric conducting hollow spherical shell. Let the potential difference between the surface of the solid sphere and that of the outer surface of the solid sphere and that of the outer surface of the hollow shell be V. If the shell is now given a charge of –3Q, the new potential difference between the same two surface is a) V b) 2V c) 4V d) -2V 59. A ball carrying a positive charge hangs from a silk thread. If we keep a positive test charge q0 at a point and measure the force F on the test charge, then it can be said that the electric field strength E at the point will be a) > F/q0 b) = F/q0 c) < F/q0 d) F/2q0 60. A condenser of capacitance 2 µF is charged to 200 V. It is now discharged through a resistor

61.

62.

63.

64.

65.

the heat produced in the resistor is a) 400 J b) 0.02 J c) 0.04 J d) 0.08 J The capacity of a parallel plate f condenser is 5 µF. When a glass plate is placed between the plates of the condenser, its potential difference reduces to 1/8 of the original value. the value of the dielectric constant of glass is a) 1.6 b) 8 c) 5 d) 40 The force acting on a charged particle kept between the plates of a charged parallel plate condenser is F. If one of the plates of the condenser is removed, then the force acting on the same particle will be a) 0 b) F/2 c) F d) 2F Two identical capacitors are connected first in series and then in parallel to the same source. The ratio of energy of the system will be a) 1:4 b) 4:1 c) 1:2 d) 2:1 *Two identical charged spheres are suspended by strings of equal length making an angle of 400 with each other. If they are immersed in a liquid of density less than the density of the material of the spheres, then a) the electrostatic force between them will increase b) the electrostatic force between them will decrease c) the net downward force on the spheres will decrease d) the net downward force will remain the same Vandergraff generator can be used for a) charging of batteries b) checking voltmeter markings

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c) producing very high magnetic fields d) accelerating charged particles Three point charges +q, -q, -q are placed on the corners of an equilateral triangle of side r. The potential energy of the system will be (A = 1/4πεo) a) 0 b) 3Aq2/r c) -2Aq2/r d) Aq2/r An uncharged metallic sphere A suspended between positively and negatively charged metal plates is given a small push towards the +ve plate. Which one of the following statements is correct? a) A touches +ve plate and remains there b) A touches +ve plate and then moves towards –ve plate and remains there c) A oscillates between the two plates with a constant time period d) A oscillates between the two plates with an increasing time-period A particle A has charge +q and another particle B has charge +4q. Each of them has a mass of m. They are allowed to fall from rest through the same electrical potential difference. The ratio of their speeds vA/vB will be a) 2:1 b) 1:2 c) 1:4 d) 4:1 The radius hydrogen atom is 0.53 x 10-10 metre. The electrostatic potential produced by the proton at the side of the electron will be a) 27.2 V b) 13.6 V c)-27.2V d) -13.6 V The effective capacitance between the points A and D in adjacent figure is a) 3 µF b) 21 µF c) 1/3 µF d) 1 µF A charged particle of mass 8 g remains stationary on a vertical electrical field 1000 kN/C. The number of fundamental quantum of charges carried by it is b) 5 x 1013 a) 5 x 1011 3 14 c) 5 x 10 d) 5 x 1015 3 3 3 3 3 Two identical copper spheres of mass 1 g each kept at C D a distance of 1m carry equal unbalanced negative A B charge. They repel with a force of 2.56 nN. The number of extra electrons carried by either of them is 3 a) 3.3 x 109 b) 3.3 x 1010 11 c) 3.3 x 10 d) 3.3 x 1012 Voltx coulomb has a dimensions of a) force b) acceleration c) velocity d) torque An electron moves along x-direction with a uniform speed. An electric field is applied along the z-direction. the path of electron will be a) a parabola in x-y plane b) a circle in x-y plane c) a parabola in x-z plane d) a straight line along z axis If the applied field is uniform magnetic in the z-direction, path of electron will be a) a helix in x-z plane b) a helix in x-y plane c) a circle in x-z plane d) a circle in x-y plane A conducting plate carries a charge +q. Another identical plate carrying zero charge is brought from infinity and kept at a distance r from the first plate. The second plate is earthed. The electrostatic energy between the plates is (A = 1/4 πεo) a) 0 b) -Aq2/r2 c) Aq2/r d) -Aq2/r

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77. The electric field at a distance 2 m from a charged plane sheet is E. When electric field at

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infinity will be a) E b) E/2 c) infinity d) 0 A simple pendulum having a bob of mass m carries a charge q. When the bob is kept in a uniform electric field E in the horizontal direction, the inclination of the string with vertical is b) tan-1 (qE/mg) a) cos-1(qE/mg) -1 c) sin (qE/mg) d) cot-1(qE/mg) If electric field at a point from a charged needle is plotted with distance which of the following will be the correct graph ? a) a straight line of negative slope b) a straight line with positive slope c) a rectangular hyperbola d) an exponential graph A charged particle of mass m charge q is released from rest in an electric field of strength E. After a time t seconds the kinetic energy of the particle will be a) 2E2t2/mq b) E2mq2/2t2 c) E2q2t2/2m d) qE Two charges exert a force F when kept at a distance r between them in air. The distance at which they exert same force in a medium of relative permittivity 4 is a) r/4 b) r/2 c) 4r d) 2r Two conducting spheres of radii 5 cm and 3 cm are equally charged. The ratio of their potentials is a) 3:5 b) 5:3 c) 9:25 d) 25:9 Two conducting spheres of radii 5 cm and 3 cm charged to the same potential. The ratio of their charges will be a) 3:5 b) 5:3 c) 9:25 d) 25:9 Two conducting spheres of radii 5 cm and 3 cm are equally charged. The ratio of electric fields on their surfaces will be a) 3:5 b) 5:3 c) 9:25 d) 25:9 Two conducting spheres of radii 5 cm and 3 cm are charged so that the electric field on their surfaces is same. The ratio of the charges will be a) 3:5 b) 5:3 c) 9:25 d) 25:9 A charge q1 exerts some force on another charge q2. If a third charge q3 is brought near q1, the force exerted by q1 on q2 a) decreases b) remains unchanged c) increases d) decrease for like and increase for unlike

87. Twelve identical condenser plates are given. Two such plates will give a capacitance of 1 µF.

The maximum capacitance which we can have from these plates is (in µF) a) 1 b) 6 c) 10 d) 11 88. A point charge q is placed at the corner of a square of side a. The potential difference between one adjacent corner and opposite corner is [A = 1/4πεo] a) Aq/a b) 10.7 Aq/a c) 0.35 Aq/a d) 0.29 Aq/a 89. Two conducting spheres A and B are charged to +10 µC and +20 µC. They have radii 3 cm

and 6 cm respectively. If they are connected by a wire a) charge flows from A to B and heat is produced in the wire b) charge flows from B to A and heat is produced in the wire c) no charge flows between A and B and no heat is produced in the wire

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d) charge flows from B to A and no heat is produced in the wire 90. The Coulomb’s law will not operate at a distance less than a) 10-8m b) 10-10m c) 10-12m d) 10-15 m 91. A charged particle will move along an electric line of force if a) it is free b) it is accelerated c) it is moving with a uniform velocity d) it is moving with a non uniform velocity 92. Two given point charges are placed at a fixed distance in the following media. In which of the medium the Coulomb force between them will be minimum? a) water b) oil c) mica d) paper 93. Three charges q1, q2, q3 are kept at the corners of an equilateral triangle. The force between q1 and q2 is F. If q3 is removed from the third corner force between q1 and q2 a) increases b) decreases c) remains the same d) cannot be said from the data 94. Three charges +2q, -q, -q are placed at the corners of an equilateral triangle. If V is electric potential and E the electric field at the centre of the triangle, then a) E = 0, V ≠ 0 b) E ≠ 0, V = 0 c) E ≠ 0, V ≠ 0 d) E = 0, V = 0 95. An electric line of force in x-y plane is given by the equation x2+y2 = 1. A particle with a unit positive charge initially at rest at a point x = 1, y = 0 in the x-y plane a) will move opposite the line of force b) will move at an angle to the line of force c) will continue to be at rest d) will move along the line of force 96. A charge of one coulomb is placed at the centre of a cube of side a. The number of lines of force coming out from one face of the cube is a) 1.11x 1011/a2 b) 1.11x 1011a2 c) 6x1.11x 1011/a2 d) 1/6) 1.11x 1011 97. Two charges +1C and -1C are placed at points (0,0) and (1,0) respectively. A non-uniform electric field exists along the X axis which increases at a uniform rate of 1V/m2. If the force on +1C is 5 N, the force on -1C is a) 6 N b) -6 N c) 5 N d) -5 N 98. The electric lines of force coming out a charged body ‘A’ is parallel and equidistant up to infinity. This body is a) a solid sphere b) a hollow sphere c) a needle d) a plane sheet 99. The electric field due to a dipole at a distance 2 m from it is E. The electric field at a distance 4 m from it will be a) E/2 b) E/4 c) E/8 d) E/16 100. A hollow metallic sphere P is charged to a potential V volt . Another sphere Q is charged to a potential V/2. If Q is placed inside P and they are connected by a wire, then a) charge will flow from P to Q until the potential becomes 3V/4 b) whole charge will flow from P to Q c) the whole charge will flow from Q to P d) no charge flows 101. The potential due a point charge at a distance 2 m from it is 6 V. The field at the same point due to the same charge will be a) 3 N/C b) 6 N/C c) 12 N/C d) 0

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102. An electric dipole kept in a uniform electric field is disturbed slightly. Which of the following will happen? a) It returns immediately to the equilibrium state b) It moves away from the equilibrium state c) It oscillates and returns to the equilibrium state d) It oscillates for ever. 103. An electric dipole is subjected to two mutually perpendicular uniform fields F along the X axis and 2F along the Y axis. The dipole will a) oscillate in XY plane b) settle at an angle 45owith X axis c) settle at an angle tan-1 (2) with X axis d) settle at an angle tan-1(1/2) with X axis 104. An charged particle is moving with a velocity 10 k m/s. It is subjected to an electric field 50 i N/C. The path of the particle will be a) a straight line b) a parabola in XZ plane c) a parabola in YZ plane d) a parabola in XY plane 105. If the applied field in the above question is magnetic, the path of the particle will be a) a straight line b) a circle in Y-Z plane c) a circle in X-Y plane d) a circle in X-Z plane 106. Three point charges each 1/3 µC are brought from infinity and are placed at the corners of an equilateral triangle of side 1 m. The work done for this will be a) 9 mJ b) 3 mJ c) 0 d) 4.5 mJ 107. Two spheres of radii 1 cm and 2 cm have equal charge density. The ratio of the electric field on their surfaces in the given order will be a) 1 b) 2/1 c) 4/1 d) 1/2 108. A cube of side a is placed in an electric field Eo i . The net number of flux lines passing through the cube is b) 2a2 Eo c) 4 a2 Eo d) 0 a) 6a2 Eo 109. The relative permittivity of a medium is x and the dielectric strength of the medium is y. For the medium to be a good insulator, it should have a) high x and low y b) high y and low x c) low y and low x d) high x and high y 110. A point charge q is kept at the centre O of a circle of radius r. A and B are two points in the circumference of the circle. ∠AOB = 60o. The work done in taking a unit positive charge from A to B along the smaller arc will be q 5q q a) b) c) d) zero. 4πε o r 6 x 4πε o r 6 x 4πε o r

111. A solid sphere of radius R has a uniform charge density ρ per unit volume. The electric field at a point inside at a distance r ( r < R) from the centre of the sphere is given by rρ Rρ 2 rρ a) zero b) c) d) 3ε o 3ε o 3ε o 112. Two soap bubbles P and Q are charged with same surface density of charge σ, but P positively and Q negatively. Then a) both the bubbles expand b) P contracts while Q expands

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c) Q contracts while P expands

d) both the bubbles contract

113. Two capacitors 1 pF and 2 pF are charged to 200Vand 100 V respectively. They are then connected so that the positive plate of the one is connected to the negative plate of the other. The final potential of the connected system will be a) 150 V b) 106 V c) 133 V d) 0 V 114. The mass of an α particle is nearly 8000 times the mass of an electron. An α particle is accelerated through 1 volt. The energy of this particle will be a) 1 eV b) 2 eV c) 8000 eV d) 16000 eV 115. *An electric dipole is formed with charges +q and -q points (-1,0) and (+1,0) respectively on the x axis. Check the correct statements. a) The y-axis will be an equipotential line. b) The electric field at all points on the y-axis will have same magnitude and direction. c) The electric field at all points on the y-axis will be along the positive x-axis. d) The electric field at all points in the y-axis will have same magnitude but different direc tions. 116. The magnitude of torque on a dipole is doubled, when the angle made by the dipole with the field is increased to three times. The initial angle of the dipole with the field is a) 15o b) 45o c) 60o d) 30o 117. Three identical capacitors are connected in parallel. Then they connected in series. The difference between the effective capacitance is 16 µF. Then capacity of each in µF is a) 3 b) 6 c) 8/3 d) 16/3 118. A condenser of capacitance 2 µF is connected to a battery through a resistor 3 ohm. The battery spends energy of 4 m J to charge the condenser fully. The heat produced across the resistor during charging is equal to (in mJ) a) 1 b) 3/2 c) 2 d) 2/3 119. n identical capacitors are connected in parallel to a potential difference V. These capacitors are reconnected in series, their charges being left undisturbed. The potential difference obtained is a) V/n b) (n-1)V c) n2V d) nV 120. An isolated metallic object is charged in vacuum to a potential V, its electrostatic energy being E. It is then disconnected from the source of potential, its charge being left unchanged. It is immersed in a large volume of dielectric with dielectric constant K. Its electrostatic energy becomes a) E/K b) KE c) EK/V d) E2K/V 121. A small metal ball is suspended in a uniform electric field using an insulated string. If high energy X-ray beam falls on the ball a) the ball will be deflected towards positive plate b) the ball will be deflected towards negative plate c) the ball will continue to remain in equilibrium d) the ball will oscillate between the plates 122. Two metal spheres of radii R1 and R2 are charged with Q1 and Q2 respectively. Then they are connected by a wire. The charge flows a) until Q1 = Q2 b) until Q1R1 = Q2R2 c) until Q1R2 = Q2R1 d) never

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95

123. A capacitor of capacitance 5 µF is charged to a potential difference of 100 V. The battery is disconnected, and then it is connected in parallel to an uncharged capacitor of capacitance C. The potential difference measured across this combination is 25V. The capacitance C is a) 20 µF b) 10 µF c) 5 µF d) 15 µF 124. In the previous question , the ratio of electrostatic energy of the system before to that after the connection of C is a) 4/1 b) 1/2 c) 2/1 d) 1/4 125. A parallel plate condenser has an energy U. The plates of the condenser are pulled using an insulating handle until their separation is 3 times the original value. The work done in the process is a) 3U b) 2U c) U d) 3/2 U 126. The minimum radius of a body which will hold a charge of 1 C so that the surrounding air will not become ionised is nearest to a) 1 cm b) 1 m c) 50 m d) the radius of earth 127. One thousand identical drops each of capacity 2µF are charged to a potential 100 V each. If they coalesce into a single drop, the capacity of single drop will be a) 20µF b) 200µF c) 2000µF d) 4000µF 128. The potential of the resulting drop given in the previous is equal to a) 100 V b) 1000 V c) 10000 V d) 105 V 129. Check which one of the following statements is not correct. a) The SI unit of charge is greater than CGS (esu). b) The SI unit of capacity is greater than CGS(esu).. c) The SI unit of energy is greater than CGS unit d) The SI unit of potential is greater than CGS (esu) 130. The capacity of a parallel plate condenser with air as dielectric is C. If the condenser is filled with 3 dielectrics of equal thickness and dielectric constants 2,3,4 the capacity will be a) (36/13)C b) (13/12)C c) (18/13)C d) 9C 131. Two condensers of capacitance C each are connected in series to a battery of potential difference V. The same condensers are later connected in parallel to the same battery. The ratio of energy of the system in the two cases in the given order is a) 1/2 b) 1/4 c) 2 d) 4 132. A parallel plate condenser contains a dielectric of relative permittivity 2. The condenser is charged by a battery. The battery is disconnected and the dielectric slab is removed. In the process the energy of the condenser a) decreases by 50% b) increases by 50% c) decreases by 100% d) increases by 100% 133. An external electric field E0 is applied across a dielectric. At any point inside the dielectric the electric field a) will be greater than E0 b) will be less than E0 d) will be greater or less depending on direction. c) will be equal to E0 134. * If the negatively charged plate of a parallel plate condenser is removed to infinity a) the potential of the positive plate increases b) the potential of the positive plate decreases c) the capacity of the positive plate increases d) the capacity of the positive plate decreases

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135. Ten identical charged drops each having an energy E, coalesce to form a single drop. The energy of the resulting drop is c) 105/3E d) 102/3 E a) 10 E b) 103/2E 136. What is a capacitance between A and B in theadjacent diagram ?( all capacitances are in µF)

a) 2

b) 4

c) 6

d) 8

137. The effective capacitance between the points A and B in the adjacent diagram (all in µF) is a) 2 b) 4 c) 6 d) 0.5 138. In the adjacent figure, each edge of the cube contains a capacitor of value C. The total capacitance of the circuit when a battery is connected between A and B will be, a) (5/6) C b) (6/5) C c) zero d) infinite 139. An infinite network of capacitors, each 2 µF, is made as shown in the figure below. The capacitance between A and B (in µF) is a) 2 b) 2.6 c) 3.2 d) 0 140. A parallel plate capacitor has a separation d and a capacitance of 100 pF. If a metal foil of thickness d/3 is introduced between the plates, the new capacitance will be (in pF ) a) 300 b) 150 c) 100 d) 67 141. Two conductors have equal volume and carry equal charge. Then a) they have same potential b) they have same capacity c) they have same energy d) all the above three will be different. Solve the following two problems within a maximum time of 2 minutes using only one equation. 142. A capacitor of capacitance 2µF is charged to 200 V. Another capacitor of capacitance of 2µF is charged to 100 V. They are connected in parallel. What is the heat energy produced in the connection wires? 143. Three capacitors of capacitance 1µF, 2µF and 3µF are connected to a source of potential difference 110 V. What is the potential difference across each of them? 144. If E is the electric field, E2εoεr has the dimensions of

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97

a) energy b) energy per unit area c) energy per unit length d) energy per unit volume 145. Two point charges q and -3q are placed at a given distance. The electric field at the site of q is E. Then the electric field at the site of -3q is a) -E b) -E/3 c) -3E d) +E/3 146. A ring of radius R carries a uniformly distributed charge +Q. A point charge -q is placed on the axis of the ring at a distance 2R from the centre of the ring and released from rest. The charge a) executes oscillatory motion but not simple harmonic. b) moves to the centre of the ring and remains at rest there c) remains in equilibrium at the point d) executes simple harmonic motion along the axis 147. Three identical spheres of masses m1, m2, m3 are charged positively and negatively and no charge respectively. Which of the following is correct ? a) m1 > m2 > m3 b) m2 > m3 > m1 c) m2 > m3 < m1 d) m2 > m3 = m1 148. A parallel plate air capacitor has a capacitance of 100 pF. The plates are at a distance apart. If a metallic wire of very small thickness is introduced parallel to plates between them, the new capacitance will be a) 100 pF b) < 100 pF c) >100 pF d) 0 149. * A capacitor C1 of capacitance 1 microfarad and another capacitor C2 of capacitance 2 microfarad are separately charged fully by a common battery. The two capacitors are then separately allowed to discharge through equal resistors at time t = 0. Then a) the current in each of the two discharging circuits is zero at t = 0 b) the currents in the two discharging circuits at t = 0 are equal but not zero c) the currents in the two discharing circuits at t = 0 are unequal d) capacitor C1 loses 50% of its initial charge sooner than C2 loses 50% of its initial charge 150. The plates of a parallel plate capacitor of capacitance C are connected to a battery of emf 12V. A dielectric of relative permittivity K is introduced in between the plates of the capacitor. The capacitance C and potential V a) both increase to K times b) both remain the same c) C increases K times and V decreases to K times d) C increases to K times and V remains the same 151. Two point charges are kept at a distance in a medium. Each charge feels the presence of the other a) instantaneously b) in about a second c) after a small but finite time d) after a time of the order of a few minutes 152. The maximum amount of electrostatic energy density which air can have so that it will not break its insulating property is a) 900 J/m3 b) 90 J/m3 c) 40 J/m3 d) 1 J/m3 153. Two point charges -2 µC and +4 µC are placed at points A and B as shown in figure below. Which of the points marked in the figure is a possible null point (point of zero electric field) for the arrangement ?

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a) P

b) R

c) Q

d) all the three points

154. In the figure given along with previous question, which point will possibly a point of zero

potential ? a) P

b) R

c) Q

d) P and R

155. An infinite number of capacitors each of 2 µF are connected as shown in the figure below.

What is the capacitance between the points A and B? a) 1.6 µF

b) o.66 µF

c) 0.8 µF

d) 1.2 µF

156. Two rings each of radius R are kept perpendicular to the plane of the paper so that their

centres P and Q are at a distance R. They carry uniform charge q each. The potential due to one ring at its centre is V. Then the potential due to both rings at the centre P will be a) 2V b) √2V c) 0 d) (√2+1)V/√2 157. Four identical condensers connected in series have an effective capacitance 1 µF. One of them is removed from the combination and connected in parallel with the rest. The effective capacitance will now be (in µF) a) 7/4 b) 7/3 c) 16/3 d) 3/4 Solve the following problems within a maximum time of 2 minutes. 158. 12 identical capacitors each of 1 µF are given to you. How will you get a capacitance of 3/4 µF using all of them? Draw the circuit diagram. 159. How will you obtain a capacitance of 1/3 µF using all of them?

SECTION 2:ANSWERS 1

(a)

2

(a)

3

(d)

4

(b),(d)

5

(b)

6

(d)

7

(d)

8

(b)

9

(a)

10

(d)

11

(a)

12

(b)

13

(c)

14

(c)

15

(c)

16

(b)

17

(b),(d)

18

(c)

19

(c)

20

(d)

21

(b)

22

(b)

23

(c),(d)

24

(a)

25

(d)

26

(c)

27

(b)

28

(a)

29

(a)

30

(c)

31

(b)

32

(c)

33

(a),(c)

34

(d)

35

(a)

36

(b)

37

(d)

38

(d)

39

(c)

40

(a)

41

(c)

42

(b)

43

(b)

44

(d)

45

(a)

46

(d)

47

(d)

48

(b),(d)

49

(c)

50

(a,b,d)

51

(d)

52

(c)

53

(b)

54

(d)

55

(c)

56

(d)

57

(d)

58

(a)

59

(c)

60

(c)

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99

61

(b)

62

(b)

63

(a)

64

(b),(c)

65

(d)

66

(d)

67

(c)

68

(b)

69

(a)

70

(d)

71

(a)

72

(a)

73

(d)

74

(c)

75

(d)

76

(d)

77

(a)

78

(b)

79

(c)

80

(c)

81

(b)

82

(a)

83

(b)

84

(c)

85

(d)

86

(b)

87

(d)

88

(d)

89

(c)

90

(d)

91

(a)

92

(a)

93

(c)

94

(b)

95

(d)

96

(d)

97

(b)

98

(d)

99

(c)

100

(c)

101

(a)

102

(c)

103

(c)

104

(b)

105

(b)

106

(b)

107

(a)

108

(d)

109

(d)

110

(d)

111

(b)

112

(a)

113

(d)

114

(b)

115

(a),(c)

116

(d)

117

(b)

118

(c)

119

(d)

120

(a)

121

(b)

122

(c)

123

(d)

124

(a)

125

(b)

126

(c)

127

(a)

128

(c)

129

(d)

130

(a)

131

(b)

132

(d)

133

(b)

134

(a),(d)

135

(c)

136

(d)

137

(a)

138

(b)

139

(c)

140

(b)

141

(d)

142

1/200 J

143

60,30,20

144

(d)

145

(b)

146

(a)

147

(b)

148

(a)

149

(b),(d)

150

(d)

151

(c)

152

(c)

153

(a)

154

(d)

155

(d)

156

(d)

157

(c)

158

3/4

159

1/3

SECTION 3 SOLUTIONS 1.

2.

3.

4.

5.

q Charging by conduction is addition of electrons when negatively charged) or removing electrons when positively charged). Here electrons are added. The mass slightly E E increases. Ans.a A metal plate has infinite dielectric constant. The force between charges separated partly or fully by it will be E q (1/4πε0εr) (q1q2/r2) with εr = α. Thus the force between them q will be zero. Ans.a The force between two point charges q1 and q2 kept at a distance F12 and F21 will have same magnitude (since the force depends on the product of the charges) and opposite direction. i.e. F21 = -F12. Hence the ratio will 1: -1. Ans.d The potential on the surface of a charged sphere is Aq/r. When same charges are given sphere of smaller radius will have greater r potential. When connected by wire charge flows from higher to lower potential making the potential same (choice b). This R reduces charge of smaller sphere and increases charge of bigger sphere (choice d) Ans.b & d Refer to the adjacent figure. The electric field at the centre of the

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6.

Refer to the adjacent figure. If ρ is the charge per unit volume, ρ = q/4πR3. Charge on the smaller sphere will be q(4/3)πr3/(4/3)πR3 = qr3/R3 where r is the radius of the smaller sphere. The electric field at the point is A (charge inside the sphere)/r2. This gives field = Aqr3/R3r2 i.e. α to r. Ans.d

7.

According to Gauss’s theorem total flux crossing normally through the whole cube = q/ε0. This is flux coming out of all the six faces of the cube. Hence flux through one face = q/6ε0. Ans.d When a conductor is charged, there is an outward mechanical force given by (σ2/2ε0)N/m2, where σ is charge radius of the bubble. It goes on expanding until surface tension of soap solution balances this force and brings about an equilibrium. Ans.b

8.

9.

The capacitance of a spherical conductor is 4πε0r. When connected by a wire the final potential = V = q1`+q2/c1+c2 = (1 x 10-2 + 5 x 10-2)/4πε0x3x10-2 = 2/4πε0 volt. Charge of smaller sphere = potential x capacity = (2/4πε0) x 4πε0 (1 x 10-2 )= 2 x 10-2 C. Ans.a

10. Work W = Fscosθ, where F is force, s is distance θ is angle between force and displacement. Here F = qE, where q is charge E is electric field. Thus W = qE cosθ. That is 4 = 0.2 x E x 2 x cos600. This gives E = 20 N/C. Ans.d 11. If v is the velocity of a particle of mass m, falling under a potential difference of V volt, 12. 13.

14. 15. 16.

17.

18.

electrostatic potential energy eV = kinetic energy (1/2) mv2 ⇒ v = 2 m/s. Ans.a Unit of ε0 is farad/metre, capacitance per unit length. This can be easily checked from the equation of capacitance of a parallel plate condenser, C = ε0A/d, where a is area of the plates and d distance between the plates. ε0 = Cd/A. Ans.b At ground state of hydrogen atom, the centre of mass of the revolving electron and that of the nucleus will be same point that is centre of atom. Hence the separation will be zero. The dipole moment is the product of charge and separation. Dipole moment P = qa = qx0 =0. Ans.c The resulting potential is total charge/total capacitance (q1+q2)/(C1+C2) =(C1V1+C2V2)/(C1+C2) = 133 V. Ans.c Dielectric strength is the minimum value of the electric field to break down the specimen. The corresponding potential difference is field x distance = 106 Vm-1 x 10-3 m = 1000 V. Ans.c Charge q is already in equilibrium. Consider the equilibrium of one of the Qs.This will be in equilibrium if force between Q and Q is equal and opposite to force between Q and q. If we take distance between Q and Q as 2a, distance between Q and q is a. Then, we have AQq/a2 = -A(QxQ)/(2a)2. This will solve to q = -Q/4. Ans.b When the charging battery is disconnected, the charge on the capacitor remains constant. Hence answer (a) is wrong. When plates are separated more the capacitance of the condenser ε0εrA/d decreases as d increases. So answer (c) is wrong. Q = V/C. Q is constant. C decreases. Hence V increases (b) is correct. The electrostatic energy of a condenser is Q2/2C. Q is constant C decreases and therefore energy increases. (d) is correct. Ans.b & d Let Q be the charge and q be its fraction. Then the charges are q and Q-q. The force between charges, F = [A(Q-q)q]/r2. The force is maximum when dF/dq = 0. This gives q = Q/2. Hence force is maximum between two charges when kept at a given distance if they are divided in the ratio 1:1. Ans.c

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101

19. Since battery connection exists, potential difference V is constant. Distance between the plates d is constant . Electric field V/d, therefore, is constant. Ans.c 20. Energy gained by the electron = q x V, where q is charge and V is potential difference. Since electric field E is uniform, V = Exd, d separation between the plates. Therefore energy = qEd = 1.6 x 10-19 x 2.5 x 2 = 8 x 10-19 J. Ans.d 21. The work done by the battery to charge the condenser is qV J. However, the energy stored by the condenser is only (1/2)qV. The factor 1/2 appears here because during charging potential increases uniformly from 0 to V. Hence we take the average value. Ans.b 22. The field due to a dipole on the axial line at a distance is Ax 2p/r3. Since field is potential gradient, potential should be proportional to 1/r2. Hence when r is doubled, V becomes V/4. Ans.b 23. At a point electric potential or field can exist even when the other is zero. Examples are given in indirect theory notes. Ans.c, d. 24. The number of lines of force coming out from a unit charge in CGS system is 4π and iun SI = 1/ε0. Value of ε0 = 8.85 x 1012. Hence 1/ε0 = 1/(9 x 10-12) = 1012/9 = 1.11 x 1011. This is also given in indirect theory notes. Ans.a 25. Two plates get equal potential V and charge Q = CV. They are joined in series. No charge flows as they carry the same potential, and in a series connection the charge is same. Potential difference = 2V. Ans.d 26. When the distance is increased capacitance = ε0ε0A/d decreases. Since charge is the same, the potential increases by the equation, Q = CV. Ans.c 27. If q is the maximum charge it can hold, q/4πε0r2 is the electric field on the surface. q = 4πε0r2 E. Here r = 5/2 m. 4πε0 = (9 x 109)-1. We have from this q = 2 x 10-3 C. Ans.b 28. The field inside a charge hollow sphere is zero. Therefore the potential at all points will be same inside. Ans.a 29. Force of attraction (F1) = A(10) x (-20)/r2 = -200A/r2 where A is the constant appearing in Coulomb’s law. When the balls touch each other, the radii being the same, the capacitance will also be the same. So the charge will distribute equally. The charge q1 = q2 = (10-20)/2 = -5 units. (F2) = A(-5) x (-5)/r2 = 25 A/r2 , F1:F2 = -8:1. Ans.a 30. Taking 1/4πε0= A, the potential at the surface of larger sphere AQ1/R1 should be = potential inside at a distance r from the centre because then only the electric field inside larger sphere will be zero. The potential due to smaller sphere will be AQ2/r, assuming charge concentrated at centre. So the total potential = (AQ1/R1) + (AQ2/r). Ans.c 31. When charging battery is connected V remains constant. When distance is increased, electric field V/d dcreases.. Ans.b. 32. Using the formula C = ε0Ak1k2 / (k1t2 + k2t1), for a parallel plate condenser filled with 2 dielectrics, we have here t1 = t2 = t/2, where is total distance between the plates. This gives C = 2ε0Ak1k2/(k1+k2)t. Ans.c 33. The electric field between parallel plate condenser is uniform and is given by E = σ/ε0, from Gauss’s law. Force = charge x field. Hence the force on all the protons will be same in magnitude and direction. Ans.a,c 34. Since the field increasing along x-axis force on –q is greater than force on +q. Hence the dipole will move along direction of greater force i.e. negative x-axis. The force on charge will produce a torque which rotates the dipole anti-clockwise. Ans.d 35. Taking 1/4πε0 = A, we have Aq1/a = Aq2/b. That is q1/q2 = a/b ....(1) If σ1 and σ2 are ratio of surface density of the two spheres, σ1 / σ2 = (q1 /4πa2)/(q24πb2) = (q1/q2)(b2/a2). Substituting from (1) for q1/q2, we get σ1/σ2 = b/a. Ans.a

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G G 36. E = 200 kˆ. The area vector S is perpendicular to the plane of the coil that is along the z G G G axis. Therefore S = 10 x 20 x 10-4 kˆ. The flux φ = E. S = 2000 x 200 x 10-4 ( kˆ. kˆ ) = 40 units. Ans.b 37. The energy of a charged conductor is given by (1/2) qV or (1/2) CV2. For a spherical conductor V = q/4πεoR. Therefore energy will be q2/8πεoR. Ans.d 38. The intensity of electric field at a point is the number of lines of force crossing unit area around the point normally. Here it is 10. Ans.d G G 39. φ = E. dS E dS cosθ, where θ is the angle between field vector and area vector. Here the lines make an angle 300 with the surface, which means they make an angle 600 with the area vector. (Recall area vector is normal to the surface). Thus φ = 10 x 1 x cos600 = 5. Ans.c 40. Since the coil is Y-Z plane, its area vector will be 200 ˆi cm2. φ = E. dS = (200 kˆ ). (200 x 41. 42.

43. 44. 45. 46. 47. 48.

49.

50.

51.

10-4 ˆi ) =0, because kˆ. ˆi = 0. Ans.a Inside a satellite there is no other force except electrostatic force (apparent weight of the balls is zero). Therefore due to the electrostatic force of repulsion the balls settle at maximum distance. Both the strings will be horizontal. Angle between them will be 1800 . Ans.c The tension of each string will be equal to the electrostatic force of repulsion experienced by a 1 QxQ ball. The distance between the balls is 2L. Therefore the force = Ans.b 4πε 0 (2L) 2 Force on electron due to electric field = eE. If this is equal to weight of electron, we have mg = eE. E = mg/e. Ans.b Taking A = 1/4πεo, potential on this surface = Aq/r. Potential at a point 3r from the centre Aq/3r. It is given (Aq/r)-(Aq/3r) = V. That is V = (2/3)Aq/r. field at 3r = -d/dr = (Aq/3r) = Aq/3r2 = V/2r. Ans.d Refer theory notes and previous similar question, in which it has been shown that the electric field inside a charged conducting sphere is proportional to distance from the centre. Therefore field decreases as we penetrate from the surface towards the centre. Ans.a 1 g of hydrogen 6 x 1023 atoms nearly. If all electrons are removed it will have 6 x 1023 protons left. The total charge of all these protons will be 1.6 x 10-19 x 6 x 1023 = 9.6 x 104 ≈ 105 C. Ans.d Electric field E = -dV/dx. That is E = -(10x+10). At the point x = 1m, E will be equal to – (10x1+10) = -20 V/m. Ans. d At a point on the perpendicular bisector of a dipole, the charges are at equal distance. Hence the electric potential will be A(q/r) + A(-q/r) = 0. That is (b) is correct. The electric field at this point (on the equitorial line) will be Ap/r3, and this always be parallel to the axis of the dipole. (d) is correct. Ans.b & d. (Note A = 1/4πεo) Since the battery is disconnected, charge Q on the condenser remains the same. The capacitance of the condenser when the slab is inserted increases to K times. Work done = difference in the electrostatic energy = (Q2/2C)-(Q2/2CK) = Q2(K-1)/2KC. Since Q = CV and C = εoA/d, this reduces to εoAV2(K-1)/2dK. Ans.c If both E and B are zero, the proton will certainly continue with constant velocity. (a) is correct. Even when B is not equal to zero, if it is parallel to proton motion, it will go undflected. Hence (b) is correct. If E is not equal to zero proton will be accelerated (c) is wrong. If both E and B exist, and the force due to the fields are equal and opposite i.e. eE = evB, the proton would move with constant velocity v = E/B. (d) is correct. Ans. a,b & d When the positive charge is released, the x-components 2Fx of coulomb force F will provide restoring force for oscillatory motion, while y components cancel. However, coulomb force

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52. 53. 54. 55. 56.

57. 58. 59.

60. 61.

62.

63. 64. 65. 66. 67. 68.

103

obeys inverse squarelaw, while we need a force proportional to displacement for SHM. Hence the charge will oscillate but not simple harmonically. Ans.d If n drops of same radius and charge coalesce, the potential of resulting drop V1 can be found by the formula V1 = n2/3 V ( Indirect theory notes), where V is the potential of smaller drop. The required ratio, therefore, V1 /V is (1000)2/3 = 100. Ans.c Taking A = 1/4πεo, the potential at the origin will be V(0) = (Ae/1)+(Ae/2)+(Ae/4)+(Ae/8) +.....to infinity. This is equal to Ae[1+(1/2)+(1/4)+(1/8)....]= 2e/4πεo, where we observe the sum of geometric series in the bracket is 2. Ans.b In this case V(0) = (Ae/1) – (Ae/2) + (Ae/4)- (Ae/8)+.....= Ae[(1)-(1/2)+(1/4)-(1/8)....]= Ae(2/3) = e/6πεo, where we observe the sum of geometric series in the bracket is 2/3. Ans.d The electric field at the origin E(0) = (Ae/12)+(Ae/22) + (Ae/42) + (Ae/82) ...= Ae[1+(1/22) + (1/32) + (Ae/42)-(Ae/82)+......= Ae[1+(1/22) + (1/42) +.......] = 3/4πεo, where we observe the sum of geometric series in the bracket is 4/3. Ans.c The electric field E(0) = (Ae/12) – (Ae/22) + (Ae/42)-(Ae/82) +....= Ae[1-(1/22)+ (1/42).....] = Ae(4/5), where we observe the sum of the geometric series in the bracket is 4/5. Hence E(0) = e/5πεo. Ans.d Since charge is same we use equation for energy (1/2)q2/C. This means energy U α 1/C. The ratio of energy will be U1/U2 = C2/C1 = 0.6/0.3 = 2. Ans.d The field inside a charged hollow conductor is zero. This will be satisfied only if the potential difference remains the same value V. Ans.a In order to measure electric field, we place a test charge q0. The test charge will have to be as small as possible because otherwise it will produce its own field at the point and thus will disturb the electric field to be measured. Hence the measured field will always be less than the actual field. Ans.c The electrostatic energy lost is the heat energy produced. Hence (1/2) CV2 is the heat energy produced. Hence C = 2 x 10-6 F. V = 200 volt. We get energy = 0.04 J. Ans.c When glass plate is fully inserted, capacitance increases to εr times, where εr is dielectric constant of glass. Since charge is constant potential will decrease to 1/εr times. Here 1/εr is 1/8. Therefore εr = 8. Ans.b Electric field between parallel plate condenser is uniform and is σ/ε0, where σ is charge per unit area. When one plate is removed, the field becomes that due to one charged plate. The field will be either σ/2ε0 or - σ/2ε0. Force on a charged particle = field x charge. Hence the force will reduce to half. Ans.b Since they are connected to the same source, V is constant. Use the equation for energy (1/2)CV2 . In the first case capacitance is C/2 and in the second case 2C. So the ratio of energy will be 1: 4. Ans.a When the spheres are immersed in liquid due to buoyant force their apparent weight will be less. Hence net downward force will decrease (c) is correct. Electrostatic force is balanced by the horizontal component of tension. Hence (b) is correct. Ans.b & c Vandegraff generator produces very high electric potential of the order of one million volt so that charged particles can be accelerated using its potential. Ans.d Potential energy of a system of three given point charges = (-Aq2/r) + (-Aq2/r) + (Aq2/r) = Aq2/r. Ans.d A uniform electric field exists between the plates. A metal sphere suspended as shown when slightly displaced will move towards one plate. Due to restoring force produced by the string it moves in the opposite direction. Thus it will oscillate with a constant period. Ans.c The electrostatic energy gained by A and B will be respectively qV, 4qV. This is equal to respective mechanical kinetic energies (1/2) mvA2, (1/2)mvB2 . From this we get vA2 / vB2 = 1/4. vA / vB is 1:2. Ans.b

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69. At the ground state, potential energy of hydrogen atom is –27.2 eV. Its kinetic energy (half of potential energy) is 13.6 eV, which makes the total energy as –13.6 eV. Since electrostatic potential energy of electron is –27.2 eV, electrostatic potential at the site of electron due to proton is +27.2 V (+ as its is due to proton). Ans.a 70. The effective capacitance between B and C is 3µF. (Balanced Wheatstone’s bridge of capacitors). This is in series with 3 µF between A and B and 3 µF between C and D. These three 3 µF capacitors in series add to 1 µF by series addition rule. Ans.d 71. Here weight mg = qE. Therefore q = mg/E = 8 x 10-3 x 10/1000 x 103 = 8 x 10-9 C. The number of quantum charges = q/charge of electron = 8 x 109 / 1.6 x 10-19 = 5 x 1011. Ans.a 72. If n is the number of electrons carried by each and e is the electron charge, we have A(ne)2/I2 = 2.56x10-9 N by Coloumb’s law, where A = 1/4πεo = 9 x 109 . This simplifies to n2 = 1020/9. n = 1010/3 = 3.33 x 109. Ans.a. (Note: The numbers given here are easy for simplification) 73. Volt x coulomb = potential x charge = energy, which has the same dimension as torque. Ans.d 74. When electric field is perpendicular to the direction of motion of electron with uniform speed, electron will have velocity and acceleration perpendicular at the beginning. Hence its path will be a parabola in the plane of velocity (X) and acceleration (Z). Ans.c 75. When applied magnetic field is along Z direction and velocity along X-direction by Lorentz force equation q v x B the force will be along Y direction. So electron will move in a circle containing velocity vector and force vector. That is X-Y plane. Ans.d 76. Here we have two plates forming a parallel plate condenser. The electrostatic energy between the plates = Aq(-q)/r = -Aq2/r. Ans.d 77. The field due to charged plane sheet = σ/2εo, which is uniform and does not depend on distance. So it will be same at infinity. Ans.a 78. If θ is angle made by the string with vertical, the forces acting on the bob are mg vertically downward and qE horizontal. The string will settle at an angle θ, such that tanθ = horizontal force/vertical force = qE/mg. (Recall tangent law in dynamics). Ans.b 79. The electric field due to a cylinder = λ/2πε0r, where λ is charge per unit length. The field is proportional to 1/r. The graph between two quantities inversely proportional to each other will be a rectangular hyperbola. Ans.c 80. Acceleration = qE/m. Velocity after t seconds v = vo+at = 0 + (qE/m)t. Kinetic energy = (1/2) mv2 = (1/2)m(qEt/m)2 = (1/2)q2E2t2/m. Ans.c 81. According to Columb’s law if r is distance in air and r1 is a distance in the medium for the forces to be equal r2 should be equal to εrr12. Equating we get r1 = r/√εr = r/2. Ans.b 82. Potential V = Aq/r, where A = 1/4πεo. Here q is the same. Hence V α 1/r. V1/V2 = r2/r1 = 3/5. Ans.a 83. Here V is same. Using V = Aq/r, q α r. q1 /q2 = r1 / r2 = 5/3. Ans.b 84. Electric field at the surface = Aq/r2, where A = 1/4πεo. Here q is same. Hence E α 1/r2. E1/E2 = r22 /r12 = 9/25. Ans.c 85. Since electric field in the surfaces are same , Aq/r12 = Aq2/r22. This gives q1/q2 =25/9. Ans.d 86. The force between q1 and q2 depends only on q1 and q2 and their separation according to Coloumb’s law. Any intervening charge q3 will not change the force between these two. Ans.b 87. With twelve condenser plates we can form eleven condensers. If the plates are arranged parallel at equal distance and alternate plates are connected to positive of a battery, other

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alternate plates to negative, these condensers will be parallel. Its capacity will be 11 µF. Ans.d 88. Taking A = 1/4πεo, potential due to charge at the adjacent corner = Aq/a. Potential due to charge in the opposite corner = Aq/√2a. Potential difference = Aq/a [1-(1/√2)]= Aq/a[1-0.71] = 0.29 x Aq/a, where we have taken 1/√2 = 0.71. Ans.d 89. Potential on the surface of A is = potential on the surface of B, because the ratio 10/3 = 20/6 (potential = 1/4 πεo(q/r). Since there is no potential difference charge will not flow. Ans.c 90. Use the information supplied already in theory notes.. This law fails inside the nucleus. The inter-nuclear distance is of the order of 10-15m. The nuclear forces follow a different set of laws. Ans.d 91. An electric charge will move along the line of force only if it is free, i.e. it is not under the action of any other force. Hence it cannot move along the line of force if it is accelerated. If its velocity is uniform, still it need not follow the line of force if its velocity is inclined to the field. Ans.a 92. The dielectric constant of pure water is 81. Among the given media water has maximum dielectric constant. Since the coulomb force is inversely proportional to dielectric constant εr , it is minimum for water. Ans a 93. Use the information supplied already in theory notes.. The force between q1 and q2 will not change due to the presence or absence of the third charge q3.. Ans.c 2q q q 1 ( − − ), where r is 94. At the centre the electric potential due to the charges will be r r 4 πε o r the distance from the corner to the centre of the triangle. The electric field at the centre will not be zero, because the fields are in different directions due to charges and hence will not cancel. Ans.b 95. In an electric field the line of force is the path followed by an isolated (or free) charge. Initially the charge is in the line of force because x = 1, y= 0 is a point in the circle. Hence when it is free, it will move along the line of force. Ans.d. 96. Use the information supplied in theory notes.. The total number of lines of force coming from the charge is 1.11x 1011. This is distributed to six faces. The number of lines of force coming from one face = (1/6) 1.11x 1011. Ans.d. 97. The force on 1C is 5 N means electric field at the origin is (force / charge) = 5/1 = 5 volt/m. Since the field increases at a rate 1 V/m for every metre, the field at x = 1 should be 5+1 = 6 V/m. The force on -1 C = charge x field = -1x6 = -6 N. Ans.b ( 98. Use information supplied in theory notes. The electric field due to a charged plane sheet σ/2εo ) is same at all points. The field is uniform throughout. Ans.d.. 99. Use the information supplied already in theory notes. When the distance increases to twice the field decreases to 1/23 of E. i.e. E/8. Ans.c 100. The field inside a charged hollow conductor is zero. In order to make the field zero, the charges will have to flow from inner hollow sphere Q to the outer one. Ans.c 101. The potential due to a point charge q is given by V = q/ 4πεor. The field due to a point charge

at the same point will have a magnitude E = q/4πεor2. From these two equations, we find E = V/r = 6/2 = 3 N/C. Ans.a 102. When a dipole is disturbed slightly, there will be a restoring torque trying it bring back to equilibrium. Due to the action of the torque, it returns. But it would have acquired kinetic energy of rotation, when it returns to equilibrium. Therefore, it continues in the opposite

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direction. Thus it oscillates. When all its energy is spent against dissipative forces, it returns to equilibrium. Ans.c 103. The two torques acting on the dipole are pFsinθ and p2F sin(90-θ), where θ is angle made by

the dipole with X axis at equilibrium. When the two torques are equal, the dipole will remain in equilibrium. Equating, we get tanθ = 2. Ans.c 104. The particle here is subjected to an electric field perpendicular to its velocity, because velocity is along Z-axis and the field along X axis. Hence the path will be a parabola in the plane of velocity and the field vectors. Ans.b 105. According to Lorentz’s force equation the force on the particle will be F = qv z xB x . Since the

vector product of velocity and field will be perpendicular to them, the force vector will be along the Y axis. This will provide centripetal force for circular motion. Therefore, the particle will have a velocity along Z-axis and centripetal acceleration along the Y-axis. The particle will move in a circle in Y-Z plane. Ans.b 106. The work done is the total potential energy of the system. If each charge is q, then the total energy will be 3Aq2/r. Here A = 1/4πεo = 9x109. q = (1/3 µC) = (10-6 / 3) C. This gives work as equal to 3 m J. Ans.b 2

107. The electric field on the surface of a sphere charged with q is q/4πεor , where r is radius of the 2

sphere. This can be written as σ/εo, because σ = q/4πr . Since σ is same, the electric field will also be same. Ans.a 108. The cube is placed so that the electric field is parallel to one of the edges along the X-axis. . The number of lines of force entering the face perpendicular to the edge is equal to the number of lines of force leaving. Hence the net flux will be zero. Ans.d 109. It is essential that a good insulator should have high value for dielectric constant or relative permittivity. The dielectric strength is the electric field to produce break down in the substance. This also will have to be high. Ans.d 110. The potential at the points A and B due to the point charge kept at the centre are same. This means potential difference is zero. Hence work done when a charge is moved between the points will be zero. You can also use theory note short cuts given. Ans.d 111. If we imagine a sphere of radius r inside bigger sphere, the question asks to find the field at the surface of this smaller sphere. The electric field at the surface of this smaller sphere is produced by the charge inside it. Since volume distribution of charge is uniform, the charge inside the smaller sphere will be (4/3)πr3ρ. Let this charge be q1. The electric field will be q1/ rρ 4πεor2. Substituting for q1, we get the field at the surface of smaller sphere as . Ans.b 3ε o 112. If the surface density of charge is σ, there will be an outward force on the soap bubble =

(σ2/2εo) N/m2 due to charging. This force will always expand the soap bubble, whether σ is positive or negative. So the soap bubble will always expand when charged. Ans.a. 113. The resulting potential is total charge/total capacitance. Since the positive plate of one is connected to the negative plate of the other, the charges cancel, i.e. total charge will be q1-q2. q − q 2 C1V1 − C 2V 2 1x 200 − 2 x100 = = 0 . Ans.d = Thus V= 1 1+ 2 C1 + C 2 C1 + C 2 114. The electrostatic potential energy = charge x potential = qV. This will be in joule.

If we write it in terms of electron charge e, then the charge carried by the α particle q = 2e. The energy will be 2eV joule = 2V electron volt = 2 x1 electron volt. Ans.b

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115. Any point on the y-axis is at equal distance from both the charges. Hence the potential due to

them will be equal in value and opposite in sign, which makes (a) correct. Any point on the y-axis is a point on the equitorial line of the dipole. Hence the field will depend on 1/y3. (Look up Theory notes supplied). So (b) and (d) are wrong. The electric field at all points on the y-axis will be in a direction parallel to the axis of the dipole, in this case +x axis. This makes (c) correct. Ans.a,c 116. The magnitude of the torque is given by pE sinθ, where θ is the angle made by the dipole

with the field. When the angle is made three times the torque becomes twice only if θ= 30o, since sin 30o = 0.5 and sin 90o = 1. Ans.d 117. Use the information supplied in theory notes.. When three identical condensers are joined in parallel the effective value is 3C and in series the effective value is C/3. It is given here 3C(C/3) = 16. Solving C = 6µF. Ans.b 118. Use the information supplied in theory notes. The half of the energy spent by the battery is used to charge the condenser, while the other half goes as heat. Therefore, heat produced is 2 mJ. Ans.c 119. When capacitors are in parallel, the charge is nCV. When they are in series, the charge is (C/n)V1, where V1 is the new potential. Since charge is constant, we have nCV = CV1/n. This gives V1 = n2V. You can also use theory notes to get this straight. Ans.d 120. The electrostatic energy of a charged body E = (1/2)QV. When this body is immersed in a dielectric, its capacity becomes KC and potential becomes V/K. [Use information given in theory notes tables]. The new energy will be E1 = (1/2)QV/K = E/K. Ans.a 121. When X-rays fall on the metal ball, some of the electrons are removed from the metal due to ionisation. This produces a net positive charge on the metal ball. Due to this the ball will deflect towards negative plate. Ans.b 122. Use the information supplied in theory notes. Charge flows until the potentials are equal i.e. Q1/4πεoR1 = Q2/4πεoR2. That is Q1/R1 = Q2/R2 or Q1R2 = Q2R1. Ans.c 123. Use the information supplied in theory notes. Here the potential is reduced from 100 to 25 V. This means the capacitance has been increased to 4 times i.e. from 5 to 20 due to the connection. So increase in capacitance is 20-5 = 15 µF. Ans.d 124. Since charge is conserved, use equation for electrostatic energy E = QV/2. V decreases from 100 to 25 V.i.e. to 1/4 th. So energy also decreases to 1/4. Ans.a 2 125. Use the equation for energy U = Q /2C. When plates are separated to 3 times capacitance becomes C/3. The energy becomes 3Q2/2C = 3U. Thus the energy increases from U to 3U. The work done = increase in energy = 2U. Ans.b 126. Use the information supplied in theory notes. If a sphere of radius R holds 1 C of charge the electric field on its surface should not exceed 3x106 N/C. This means Q/ 4πεoR2 = 3x106V/m. That is, 9x109/R2=3x106, which gives R = 3000 , which is nearly 50 m. Ans. c 127. Use the information supplied already in theory notes. The resulting capacity is equal to

10001/3x 2 = 10x2 = 20µF. Ans.a 128. Use the information supplied in indirect theory notes. The potential of resulting drop is equal to 10002/3x100 = 100x100 = 10000 V. Ans.c 129. Usually in international system, units are bigger than those in in CGS system. The units in CGS system are stat coulomb for charge, stat volt for potential and stat farad for capacitance and erg for energy. 1C = 3x109stat -coulomb. 1F = 9x1011 stat farad. 1 J = 107 erg. 1V =

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(1/300) stat volt. From these we find the CGS unit of potential stat volt = 300 V i.e. bigger than volt. Ans.d ε 0A ε A . Here . With air C 0 (t 1 / k 1 ) + (t 2 / k 2 ) + (t 3 / k 3 ) t t1=t2=t3=t/3. k1=2. k2=3 k3=4. Substituting these values we get C1= (36/13)C. Ans.a 2 131. Since the potential is same we use equation for energy (1/2) CV . So E1/E2 =C1/C2. C1/C2.= 1/4 from indirect theory notes. Thus the ratio E1/E2 = 1/4. Ans.b 132. When the battery is disconnected the charge remains constant. Hence use the equation for energy as Q2/2KC with dielectric and Q2/2C without the dielectric . Thus energy increases in the ratio 1/K:1 i.e.1/2:1 That is,the energy increases by 100%. Ans.d. 133. When an electric field is applied to a dielectric it becomes polarised. Polarisation produces dipoles. These dipoles produce a field at any point inside which opposes the external field. Hence the field at any point will be E0 - the field produced by the dipoles i.e. it will be less than E0. Ans.b 134. When an uncharged plate is brought from infinity near to a positively charged plate, the potential of positive plate decreases and its capacity increases. (Principle of a condenser). So when this plate is removed capacity decreases and potential increases. So (a) and (d) are correct, (b) and (c) are wrong. Ans.a,d. 135. Use the information supplied already in theory notes. If Q is the charge of each drop and V is the potential, when drops coalesce the charge becomes 10 Q and potential becomes 102/3V. Since energy is proportional to the product of Q and V, it will become 10x102/3 = 105/3 of that of one drop. You can also use theory notes short-cuts. Ans.c total capaci tance C is equal to 136. Since all the capacitors are in parallel 1 1 1 1 1 1 1 4 + 2 + 1 + + + + ....to ∞ = 4 1 + + + .... = 8µF 2 4 8 16 2 4 8 . Ans.d where we have used the information that geometric series in the bracket has a sum 2. Ans.d 137. The arrangement in the triangle part ABC is a balanced Wheatstone’s bridge of capacitors. The effective capacitance between the points A and B of this bridge is 1µF. This is parallel with 1µF across DE. Therefore the total capacitance 1+1 =2µF. Ans.a 138. Let a charge of 6q flow from the battery. It divides into three equal values 2q each as shown in figure. Each of the 2q divides to two condensers as q each. Then these charges rejoin as 6q. Taking any closed path and adding potential difference across each capacitor, we have 2q/C + q/C+2q/C = 6q/C1, where C1 is the equivalent capacitance of A and B.⇒C1= (6/5) C. Ans.b 139. Let X be the capacitance between A and B. (fig below). In this infinite network we have repeating units. Here one unit is a parallel an series capacitor. Break the circuit after one such unit. We find the circuit will have the same structure as the original network. Hence the capacitance between E and F will also be X itself. X across EF is in series with 2 . This gives 2X/2+X. This is in parallel with 130. Using the equationc C1 =

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109

2 across CD. This will give (2X/2+X)+2. This should be equal to X, i.e. capacitance across 2X + 2 = X . Solving this equation we gets X = 3.2µF. Ans.c AB. Therefore, we have 2+X 140. Before introducing the metal, let the capacitance be C. The C = εoA/d. If a dielectric of

thickness t and dielectric constant K is introduced the new capacitance C1 is given by 3ε A 3 ε0A 3 C1 = Here K =∞ and t =d / 3 . C1 = 0 = C = x100 = 150 pF. Ans.b 1 2d 2 2 d − t (1 − ) K 141. The potential, the capacity and energy depend on the shape and the charge of the conductor. Hence they will not be same even if they have same volume. Ans.d 142. Use the information supplied already in theory notes. The electrostatic energy lost is equal to heat energy gained. Therefore the heat energy produced = 1 2 x2 1 2 −6 x10 ( 200 − 100) = J. 2 2+2 200 143. In a series connection capacitors get same charge. Using the equation Q =VC, we find V is proportional to 1/C. V1:V2:V3 = (1/1): (1/2): (1/3) = 6:3:2. So all we have do is to divide 110 volt in this ratio. The potentials will be 60V, 30V and 20V respectively. 2

144. The equation for energy density i.e. energy per unit volume is given by (1/2) εoεrE . Ans.d 145. When two charges are placed at a distance, the force on them will be the same in magnitude

and opposite in direction. Force on q = field x charge = Eq. Force on -3q = Eq. Field at the site of -3q = force/charge = = qE/-3q = -E/3. Ans.b 146. When the charge is released there will be force of attraction towards the ring. The horizontal component of the force along the axis will provide the restoring force. This force will always be directed towards the centre of the ring. The particle, thus, will have oscillatory motion. But the coloumb forces follow inverse square law. For simple harmonic motion we need force proportional to displacement. Hence the motion will not be simple harmonic. Ans.a 147. A negatively charged body gains electrons. So its mass increases. A positively charged body loses electrons and its mass decreases. Hence negatively charged sphere has more mass and positively charged one has less mass. Ans.b 148. Use the equation for capacitance of a parallel plate condenser with a dielectric of thickness t introduced . If C is the capacitance when medium is air we have C = εoA/d. ε0A C1 = . Here K = ∞ and t =0. This gives C1 = C. Ans.a 1 d − t( 1 − ) K 149. A condenser discharges according to the equation q = qoe

-t/RC

, where R is the value of the resistor in series with C. Current i = dq/dt = -qoe (1/RC). Since q0/C is constant equal to charging potential V, we find, the current i is not zero at time t = 0,but is constant equal to q0/CR. Therefore (a) and (c) are wrong but (b) is correct. Also dq/dt is more when C is less, from the above equation. This means condesner of smaller capacitance loses charge more quickly. Hence (d) is correct. Ans.,b, d 150. Since the condenser is connected to the battery, whatever may happen, the potential remains the same. When a dielectric is introduced, the capacity will increase to K times. Ans. d. 151. The electromagnetic force is transferred through photons. A charge feels the presence of the other when photons are exchanged between them. Hence the time taken will be the time for -t/RC

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photons to travel the distance between them which will be about 10-8s. This time is small but finite. Ans.c 152. Use the information supplied in theory notes . The insulating property of air breaks down The corresponding energy density is when the electric field ≈ 3x106V/m. 1 1 = ε o E 2 = x9 x10 −12 x(3x10 6 ) 2 = 40 J / m 3 . Ans.c 2 2 153. Since the charges are of opposite sign no null point will be got between the charges. So R is not a null point. Since the field will have to be equal in magnitude, the point will have to be nearer to the smaller charge. Hence a possible null point is P. Ans.a 154. The potential at any point is produced by positive and negative charges. Hence it may cancel at a point between them, or at a point nearer to the smaller charge. Since potential is a scalar, we need not look for the direction. Hence the two points P and R could have zero potential. Note in these questions we find only location of points and not the distance. Ans.d 155. The solution to this question is similar to the one given in indirect theory notes for resistors. Here one repeating unit is a series and a parallel capacitor. Break the circuit after one unit and replace it with X, as shown in adjacent figure. We get 2+X and 2 in series which is equal to 2(2+X)/2+2+X. Thus we have 2(2+X)/2+2+X = X. Solving we get X= 1.2 µF. Ans.d 156. At centre of one ring, say P, the potential V due to the charge on it is

V = q/4πεoR. At the same point the elements of the other ring are at a distance √2R. Therefore the potential at P Total potential is equal to V+V/√2 = due to the other ring is q/(4πεo√2R) = V/√2. (√2+1)V/√2. Ans.d. 157. Use the information supplied in theory notes. Each condenser should have a capacitance 4 µF. When one of them is removed the other three are in series. They will have a capacitance 4/3 µF. This in parallel with 4 µF will give a net capacitance 16/3 µF. Ans.c 158. Connect 4 condensers in series. It will have a capacity of 1/4 µF. Connect such three groups in parallel. Total capacity will be 3x(1/4) = 3/4 µF. This is shown in the figure below. 159. Connect six of them in series. The effective capacitance will be 1/6 µF. Make two such

groups and connect both groups in parallel. This will give 2x1/6= 1/3 µF

5 ELECTRICITY SECTION 1 QUESTIONS 1.

The current through 60 ohm resistor in A is (Fig) 120

60

60

120

6 V

2.

a) 0.025 b) 1/30 c) 2/30 d) 1 The current through 60 ohm resistor in the following network in A is (Fig) 30

6V

70

60 30

40 50

Fig. 3

3.

4.

5.

a) 0.08 b) 0.06 c) 0.04 d) 0.02 A uniform wire of resistance 4 ohm is bent into a square loop. The resistance between the midpoints of its opposite sides would be a) 4Ω b) 2Ω c) 1Ω d) 0.5Ω It is required to get largest amount of heat energy from a resistance wire of length 0.5 m connected with a battery of negligible resistance. This can be done a) by joining the wire directly to battery b) by cutting the wires in two equal parts and then joining all of them in parallel to the battery c) by cutting the wire in two four equal pieces and then connecting all of them in parallel to the battery d) by joining only half the wire A silver voltameter and a zinc voltameter are connected in series and current i is passed for a time t liberating w kg of zinc. The mass of the silver deposited is nearly a) W b) 3.5 W c) 2.4 W d) 1.7 W

112

6.

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16.

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An electric immersion heater of 1.08 kW is immersed in water. After the water has reached a temperature of 1000C, how much time will be required to produce 100 g of steam ? a) 50 s b) 420 s c) 105 s d) 210 s Two resistors in parallel have their resistances in the ratio 1:3. A source is connected to the combination. The ratio of heats produced in a given time in the two resistors is a) 3:1 b) 9:1 c) 1:3 d) 1:9 The terminals of battery of emf 12 V and internal resistance 1 Ω are connected to a circular coil of resistance 16 Ω at two points distant a quarter of the circumference of the coil. The current flowing through the smaller arc of the circle in A is a) 3.0 b) 2.25 c) 0.75 d) 0.5 A uniform wire of length 5 cm is carrying a steady current. The electric field inside it is 0.2 V/m. The potential difference across the ends of the wire is a) 0.1 V b) 0.5 V c) 5 V d) 10-2 V Two wires A and B are in series. When heated, the combination has same resistance at all temperatures. A is of germanium. Then B should be of a) carbon b) aluminium c) silicon d) alloy of carbon and silicon A cell of emf X is connected across a resistor R. The potential difference across the wire is measured as Y. The internal resistance of the cell should be a) X-Y/(R-X) b) (X-Y)R c) (X-Y)R/X d) (X-Y)R/Y A battery of 2 volt emf and internal resistance 1 ohm sends a current of 1 A through an external load. If two such batteries are connected in series, the current through the same load would be a) 1A b) 2 A c) 1.5 A d) 1.33 A A rectangular loop carrying a current i is situated near a long straight wire such that the wire is parallel to one of the sides of the loop and is in the plane of the loop. If a steady current i is established in the wire as shown in the fig. 1, the loop will a) rotate about an axis parallel to the wire’ b) move away from the wire c) move towards the wire d) remain stationary A wire of given length is first bent in one loop and the next it is bent in three loops. If the same current is passed in both the cases, the ratio of magnetic induction at their centres will be a) 1:4 b) 1:9 c) 9:1 d) 1:3 A circular loop of mass m and radius r is in a horizontal (x-y plane) table as shown in fig 2. A uniform magnetic field B is applied parallel tro x-axis. The current in the loop, so that its one edge just lifts from the table is a) mg/πr2B b) mg/πrB c) mg/2πrB d) πrB/mg The magnetic field at the centre of the cube of edge of length a carrying a current i is a) zero b) (8µoi)/(√2 a) c) (8µoi)/2√2 a) d) (2µoi)/2√2 a)

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113

17. An electron enters along east in a region in magnetic equator. It will be deflected a) vertically up b) horizontally along east c) horizontally along west d) vertically down 18. A current of i ampere is flowing through a loop of a circle of radius r metre which subtends an angle θ as shown. The magnetic field at the centre of the loop is a) µoiθ/4πr b) µoi sinθ/πr2 c) µoi sinθ/2r d) µoiθ/4r 19. If a graph is plotted between magnetic field due to a long straight conductor and distance, the graph will be a) straight line with positive slope b) straight line with negative slope c) parabola d) rectangular hyperbola 20. A conducting circular loop of radius r carries a constant current i. It is placed in a uniform magnetic field Bo such that Bo is perpendicular to the plane of the loop. The magnetic force acting on the loop is a) irBo b) 2πriBo c) zero d) π riBo 21. A proton and electron of equal momentum enter into a uniform field at right angles to the velocity. The radius of their tracks will be in the ratio a) 1:1840 b) 1840:1 c) 1:43 d) 1:1 22. A charged particle is undergoing a circular motion in a uniform magentic field. The time period is independent of a) speed b) mass c) charge d) intensity of magnetic field 23. A one MeV energy alpha particle describes a circle of radius x in a uniform magnetic field. A proton of the same energy will describe in the same field a circle of radius a) x b) x/2 c) 2x d) 1.4 x 24. A circular coil of wire having 100 turns and radius 1m is arranged in plane perpendicular to the magnetic meridian. A current of 1 A is passed through the coil. A horizontal magnetic needle at the centre will show a deflection a) tan-1 (1.57) b) tan-1 (3.14) c) 0 -1 d) tan (0.85) 25. In the Lorentz force equation F = qv x B,. which of the following is correct ? a) v and B are always perpendicular b) v and F are always perpendicular c) v and B cannot be parallel d) F and B be parallel 26. A long straight conduct or is bent into the shape as shown in Fig.6 It carries a current i A and the radius of the circular loop is a m, The magnetic field at the centre of the loop is a) 0 b) µo i/2π a c) µo i(π+1)2π a d) µo i(π-1)/2π a

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27. A thin wire is bent to the form of a square loop ABCD.(fig.7) A battery is connected between A and C as shown in fig. the magnetic induction due to the current in the loop at the centre O a) points into the plane of the paper b) points along OP, the bisector of angle BOC c) zero d) points out of the plane of the paper 28. The magnetic field produced by any side at the centre of the square loop in the adjacent figure 8 is of magnitude B. The magnetic field at the centre due to the whole loop is a) 0 b) 4B c) 2B d) B 29. The magnetic induction at a point P distant 4 cm from a long current carrying wire is X tesla. The induction at a distance 12 cm from the same wire would be (in tesla) a) X b) X/3 c) 9X d) X/9 30. A circular loop carrying a current of i A is bent into a circular coil of 4 turns.The new magnetic moment of the coil will be a) 4 times the original b) 2 times the original c) 1/4 of the original d) half of the original 31. Two particles X and Y having equal charges, after being accelerated through the same potential difference, enter a region of uniform magnetic field and describe circular paths of radii R1 and R2 respectively. The ratio of mass of X that of Y is a) (R1 /R2)1/2 b) (R2/R1) c) (R1/R2) d) (R1 / R2)2 32. Am2 has the same dimension of a) IT b) NT c) JT-1 d) NT-1 33. The torque acting in a magnetic needle when it make 300 with a uniform field is x. The torque acting when it is perpendicular to the field is a) x b) 3x c) 2x d) (3/2) x 34. A thin uniform magnetic needle of period T is broken into n parts of equal length. The period of one part will be a) nT b) T/n c) n2T d) T/n2 35. Magnetic shielding to an instrument can be provided by covering with a) soft iron b) plastic c) copper d) aluminium 36. Isogonic lines are those which join places of a) zero dip b) zero declination c) equal dip d) equal declination 37. If a dip needle stands horizontal at a place, the place is a) magnetic meridian b) perpendicular to magnetic meridian c) magnetic pole d) magnetic equator 38. Two magnets placed one over the other oscillate with a period of 16 s. When one of them is reversed, the period is 8 s. The ratio of their magnetic moments is a) 3;1 b) 1:3 c) 5:3 d) 3:5

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115

39. The susceptibility of a substance is negative. This substance could be a) bismuth b) iron c) nickel d) aluminium 40. Two magnets of same pole strength p but of lengths 3L and 4L respectively are arranged to form a right angle so that the north pole of one touches the south pole of the other. The moment of the combination is a) pL b) 7 pL c) 5 pL d) 6 pL 41. The correct dip at a place is δ. If the dip is measured in a plane making and angle θ with the magnetic meridian as δ1 , then δ1 will be b) tan-1 (tanδ sec θ) a) tan-1 (tanδ cosθ) d) tan-1 (tanδ cosec θ) c) tan-1 (tanδ sinθ) 42. The total intensity of earth’s magnetic field at the magnetic equator is 0.4 cgs units. At a place where the dip needle reads 900, the vertical intensity will be (in cgs units) a) 0.1 b) 0.2 c) 0,.4 d) 0 43. Which of the following substance has magnetism independent of temperature? a) copper b) aluminium c) iron d) manganese 44. Which of the following material shows the property of hysterisis ? a) copper b) cobalt c) aluminium d) silver 45. Which of the following is most suitable for making the beam of a chemical balance ? a) brass b) iron c) aluminium d) nickel 46. The field due to a short magnet on the axis at distance x is n times the field due to the magnet at a distance 2x. Here n is a) 2 b) 4 c) 8 d) 16 47. Soft iron is used for electromagnets because of a) high limit of magnetic saturation b) large area for hysterisis curve c) large coercivity d) large retentivity 48. Which of the following is a paramagnetic substance ? a) bismuth b) antimony c) water d) chromium 49. A steel wire of length L has a magnetic moment M. It is bent into a semi circular arc. The new magnetic moment is a) M/ 2π b) ML/2π c) M/L d) 2 M/π 50. A magnet of moment m is kept in stable equilibrium in a uniform magnetic field of intensity B. If it is rotated through an angle of 1800, the work done is a) mB b) 2mB c) mB/2 d) zero 51. A line joining all places on the earth having zero magnetic dip is a) magnetic line of force b) magnetic equator c) magnetic meridian d) isoclinic line 52. A magnet falls vertically through a horizontal copper ring. Its acceleration will be a) always less than g b) always greater than g c) always equal to g d) less than g when approaches and greater than g when recedes 53. The phase difference between the flux linkage and gthe induced e.m.f in a rotating coil in a uniform magnetic field is

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a) π b) π/2 c) π/4 d) zero 54. The induced emf in a coil of wire produced by motion of a magnet will not depend on a) number of turns in the coil b) area of the coil c) resistance of the coil d) speed of the magnet 55. The energy stored in an inductor is in the form of a) electric b) mechanical c) thermal d) magnetic 56. An aeroplane with wing span of 50 m flies at 540 km/h horizontally along north south. The vertical component of earth’s magnetic field is 0.2 gauss. The induced emf across the wing tips is a) 0.15 b) 15 V c) 1500 V d) 150 V 57. A square loop PQRS of resistance 1 Ω and side 10 cm is moved perpendicular to a uniform magnetic field of 1 T. The ends of the loop are connected to a net work of resistors as shown in fig. The speed of the loop so that a current of 1 mA flows through it is (in m/s) a) 10-2 b) 2 x 10-2 c) 20 d) 10 58. A circuit has a coil of self inductance 20 mH and a resistance of 3 Ω. The induced emf in the circuit when current changes at a rate 5 A/s is (in volt) a) 0.1 b) 0.1/3 c) 0.3 d) 0.2 59. Lenz’s law is a consequence of the law of conservation of a) charge b) momentum c) mass d) energy 60. An electron moves along the line AB which lies in the same plane as the circular loop of conducting wire as a shown in the fig.2 What will be the direction of the current induced if any in the loop? a) no current will be induced b) the current will be clockwise c) the current will be anti-clockwise d) there will be an emf but no current 61. In what way should be conductor AB be moved in a magnetic field such that the current flows as shown in fig. 3. a) vertically up ward b) towards left c) towards right d) vertically downward 62. J A-2 is the unit of a) permeability b) permittvity c) self indction d) energy density 63. A fan leaf of length L rotates in a uniform magnetic field of

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64.

65.

66.

67.

68.

69. 70.

117

intensity B, with a constant angular velocity ω . The induced emf between the centre and end of the leaf will be a) B12 ω b) 2 B12 ω c) (1/2)B12 ω d) B 1ω A coil has a self inductance L. If the length and breadth of the coil are doubled keeping number of turns per unit length constant the self inductance will a) remain same b) become 2 times c) become 4 times d) become 8 times The coefficient of mutual induction between the two coils will not depend on a) number of turns of the secondary b) distance between primary and secondary c) cross sectional area of the primary d) the rate of change of current in the primary The equivalent inductance between points P and Q in fig.4 is a) 2H b) 6H c) 8/3H d) 4/9H Two coils have self inductances L1 = 8 mH and L2 = 2mH. The current in the two coils are increased at the same constant rate. At a certain instant the power given to the two coils are same. At that instant if currents in the coils are I1 and I2 respectively, then I1/I2 = a) 4/1 b) 1/4 c) 2/1 d) 1 A conducting square loop of side L and resistance R moves in its plane with a uniform velocity v perpendicular to one of its sides. A magnetic induction B constant in time and space pointing perpendicular to and into the plane of the loop exits everywhere. The current induced in the loop is a) BLv/R clockwise b) BLv/R anti-clockwise c) BLv/R clockwise d) 2BLv/R anti-clockwise Which of the following has the dimension of time ? a) LR b) L/R c) R/L d) 1/CR Which of the following is not correct ? a) ohm = henry per second b) farad x henry = second2 c) farad = joule/volt d) ohm x farad = second

71. The flux through a closed circuit of resistance 10 Ω varies with time according to the equation φ = 12t2 + 4t + 1. The induce current in the circuit 1/4 second after start will be (in ampere) a) 10 b) 2.75 c) 0.275 d) 1 72. The emf in an AC circuit is represented by the equation e = 5 sin ωt and current by i = 2 cos ωt. The average power consumed in the circuit is (in W) a) 10 b) 10√2 c) 0 d) 5√2 73. An alternating voltage of rms value V is applied to a circuit containing a resistor, inductor, and a capacitor. Four voltmeters are connected across resistor, inductor, apacitor and supply. If their readings are V1, V2, V3 and V respectively, then a) V1 + V2 + V3 = V b) V12 + V22 + V32 = V2 2 2 2 c) V1 + (V2 – V3) = V d) (V1 – V2)2 + V32 = V2 74. If V1 , V2, V3 are instantaneous voltage across the components and V = Vo sinωt supply voltage a) V1 + V2 + V3 = V

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b) V12 + V22 + V32 = V2 c) V12 + (V2 – V3)2 = V2 d) (V1 – V2)2 + V32 = V2 75. In figure2 a 6 volt battery is connected to a Neon lamp which needs 10 volts to glow. An inductor L is connected in parallel. When the key K is switched on which of the following happens ? a) bulb glows immediately but dimly b) bulb first does not glow but glows dim later c) bulb never glow d) bulb glows bright 76. When the current is switched off in the above circuit in previous a) bulb goes off immediately b) bulb goes off slowly c) bulb keeps on glowing d) bulb turns brighter and goes off 77. An A.C voltage of rms value 200 V is applied to a circuit containing a resistor of 60 Ω, inductor of reactance 180 Ω and a capacitor of reactance 100 Ω. The rms value of current in the circuit is (in A) a) 2 b) 2.8 c) 10/7 d) 10/19 78. The phase difference between current and voltage in the above circuit given in previous question is a) tan-1 (3/4) b) tan-1 (4/3) c) tan-1 (5/3) d) tan-1 ((3/5) 79. The power factor of circuit given in previous question is a) 0.44 b) 0.8 c) 0.6 d) 1 80. The average power consumed by the circuit in watt is a) 400 b) 200 c) 282 d) 240 81. An electro magnetic wave can be produced by a) positive and negative charge b) steady current in a wire c) varying electric current d) a strong magnet 82. An alternating voltage 230 V, 50 Hz is applied to a circuit containing an inductance, resistance and a capacitance. At any moment for rms voltages, a) the total voltage across each when added will be equal to 230 V b) the voltage across each will be 230 V c) the voltage when added across the three should be less than 230 V d) the voltage across the inductance can be greater than 230 V 83. We have a source of red light and a blue light of same power. Then the red light gives photons per second a) same in number, but of less energy b) more in number, but of less energy c) less in number and of less energy d) more in number and of more energy 84. The instantaneous values of current and voltage in an A.C circuit are I = 4 sin ωt and E = 100 cos[ωt + (π/3)] respectively. The phase difference between voltage and current is

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a) 7π/6

b)

5π/6

119

c)

2π/3

d) π/3

85. An A.C given by e(t) = 282 sin (120π t) is applied across a resistance of 20 Ω. An A.C ammeter connected in series records a current of a) 14.1 A b) 7.05 A c) 10 A d) 5 A 86. Two coils of inductance L1, L2 are connected in series. The value of inductance is equal to L1 + L2 if a) distance between them is large b) distance between them is small c) they are would one over the other d) same always 87. Which of the following is least scattered by a medium a) X-rays b) visible light c) infra red d) radiowaves 88. Take wave length of blue light roughly as 440 nm and that of red light as 660 nm. The ratio of intensity of red light scattered to that of blue light by earth’s atmosphere is nearly equal to a) 2/3 b) 3/2 c) 4/9 d) 1/5 89. Assuming the radius of earth to be 6400 km, the maximum distance to which a TV signal can be received on earth using an antenna of height 500 m is equal to a) 80 km b) 80 x √2 km c) 40 x √2 km d) 160 km 90. If we plot a graph between the reactance of a capacitor and frequency of A.C passing through it , the graph will be a) a straight line having a positive slope b) a straight line having a negative slope c) a parabola d) a rectangular hyperbola 91. Two point charges +q and –q are held fixed at (-d,0) and (d,0) respectively, of (X,Y) coordinate system. Then a) E at all points on the Y-axis is along ˆi G b) the electric field E at all points on the X-axis, has the same direction. c) dipole moment is 2 pd directed along ˆi d) work has to be done in bringing a test charge from infinity to the origin. 92. A battery is connected between two points A and B on the circumference of a uniform conducting ring of radius r and resistance R. One of the arcs AB of the ring subtends an angle θ at the centre. The value of the magnetic induction at the centre due to the current in the ring is (fig) a) proportional to 2 (1800-θ) b) inversely proportional to r3 c) zero only if θ = 1800 d) zero for all values of θ 93. The wire loop PQRSP formed by joining two semicircular wire of radii R1 and R2 carries a current i as shown in the fig. 3. The magnetic field B at the point C is µ I 1 1 − a) 0 2π R 1 R 2

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b)

µ0I 1 1 − 4π R 1 R 2

c)

µ0I 1 1 − 4 R1 R 2

d)

µ0I 1 1 − 2 R1 R 2

94. A proton of mass 1.67 x 10-27 kg and charge 1.6 x 10-19 C is projected with a speed of 2x106 m/s at an angle of 600 to the Xaxis . If a uniform magnetic field of 0.104 Tesla is applied along the Y-axis, the path of the proton is a) a circle of time eperiod π x 10-7 s b) a circle of time period 2π x 10-7 s c) a helix of time period 2π x 10-7 s d) a helix of time period 4 π x 10-7 s 95. H+ , He+ and O2+ all having the same kinetic energy pass through a region in which there is a uniform magnetic field perpendicular to their velocity. The mass of H+. He+ and O2+ are 1 amu, 4 amu and 16 amu respectively. Then b) O2+ will be deflected most a) H+ will be deflected most + 2+ c) He and O will be deflected equally d) all will be deflected equally 96. A microammeter has a resistance of 100 Ω and a full scale range of 50 µA. It can be used as a voltmeter or as a higher range ammeter provided a resistance is added to it. Pick the correct range and resistance combination(s) a) 50 V range with 10 kΩ resistance in series b) 10 V range with 200 kΩ resistance in series c) 5 mA range 1 Ω resistance in parallel d) 10 mA range with 1Ω resistance in parallel 97. A piece of copper and another of germanium are cooled from room temperature to 80 K. The resistance of a) each of them increases b) each of them decreases c) copper increases and germanium decreases d) copper decreases and germanium increases 98. A battery of internal resistance 4 Ω is connected to the network of resistance as shown. In order to give the maximum power to the network, the value of R in Ω should be a) 4/9 b) 8/9 c) 2 d) 18 99. The current through a wire changes with time as given by the equation i = t . the correct value of the rms current within the time interval t = 2 to t = 4 s will be a)

3A

b) 3A

c) 3 3 A

d) 3 2 A

100. Two heater wires of equal resistance are first connected in series and then in parallel, to the same source. The ratio of heat produced in them are in that order a) 2:1 b) 1:2 c) 4:1 d) 1:4

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121

101. The electric flux through a hemispherical surface of radius R placed in a uniform electric field of intensity E parallel to the axis of its circular plane is b) 2πR2E c) R2E d) 4/3πR3E a) 2RE2 102. To get maximum current in a resistance of 3Ω, one uses n rows of m cells each, connected in parallel. If the total number of cells is 24 and internal resistance of a cell 0.5 Ω, then a) m = 12 n = 2 b) m = 8, n = 3 c) m = 2 n = 12 d) m = 6, n = 4 103. A particle of mass m and charge q is placed at rest in a uniform electric field E and released along the y-axis. The kinetic energy it attains after moving a distance y is a) 1/2 qEy b) qE2y c) qEy d) 1/2 m(qEy) 104. An electric dipole in a uniform field is slightly disturbed from equilibrium position. Its period of oscillation is a ) 2π I / pE

b) 2π pE / I

c) 2π p / IE

d ) 2π IE / p

105. A house is served by a 220 V supply line. In a circuit protected by a fuse marked 9 A, the maximum number of 60 W lamps in parallel which can be turned on at the same time is a) 44 b) 20 c) 22 d) 33 106. The unit of thermo electric power is a) W b) VK-1 c) V d) Vs-1 107. A uniform wire of resistance 2 Ω and length 4 m is melted and reformed into a uniform wire of length 2m. Its resistance now will be (in Ω) a) 2 b) 0.5 c) 4 d) 0.25 108. A particle carrying a charge of 1 C enters into a uniform magnetic field 3 j with a velocity 2j+k. The force on the particle has magnitude in N and direction respectively a) √7, +x axis b) √7, -x axis c) 3, + x axisi d) 3, -x axis 109. ML2T-3 I-2 is the dimension of a) potential difference b) electric field c) permitivity d) resistance 110. Which of the following is correct ? a) Resistivity = electric field/current density b) Resistivity = current density/electric field c) Resistivity = 1/current density d) Resistivity = current density/potential 111. Four bulbs connected in series consume a total power of P watt. One of the bulbs is fused and the other three are connected in series to the same source. The total power consumed (in W) is a) P b) 3P/4 c) 8P/3 d) 4P/3 112. The best combination of properties of a metal to make a standard resistor are a) high resistivity and low temperature coefficient b) high temperature coefficient and high linear expansivity. c) low linear expansivity and high temperature coefficient d) low temperature coefficient and high melting point . 113. Two wires A and B of same material have same mass but radii in the ratio 1:2. Their resistance ratio RA/RB will be a) 1:4 b) 1:8 c) 1:16 d) 16:1 114. A primary and a secondary cell have same emf. Then

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a) both give same current b) primary gives more current c) secondary gives more current d) cannot be said from the data 115. You are supplied with two resistors. Using them it is possible to get resistance value 9,27,6 and 18 ohms. The resistance of resistors are a) 9 and 18 b) 27 and 18 c) 6 and 18 d) 27 and 6 116. Four cells each of same emf and same internal resistance are connected in series. The combination gives a current i. If one of the cells is reversed the current decreases by 2 1 a) 25% b) 50% c) 66 % d)33 % 3 3 117. An ammeter of resistance G reads 1 ampere per division. It is to be converted into an ammeter reading n amperes per division. The resistance to be connected in parallel is a) nG b) G/(n-1) c) G/n d) (n-1)G 118. *Two resistors having resistances R1 and R2 at 0oC and temperature coefficient α1 and α2 respectively are joined in series. Check the correct statement(s): a) This combination will have same resistance at all temperatures if R1α2 = R2α1 b) This combination will have same resistance at all temperatures if R1α1 = R2α2 c) Both resistors should be made of metals d) One resistor should be made of metal and the other of semi conductor. Solve the following problems within a maximum time of 2 minutes 119. What is the reading of the ammeter (A) in figure1? 120. What is a current through 2 ohm resistor in the network shown in fig 2

Fig.1 Fig.2 121. What is the reading of the ammeter A in the fig 3?

Fig.4

Fig.3

Fig.5

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123

122. What is the reading of ammeter A in the figure 4? 123. A uniform circular loop has a resistance of 2 ohm. A straight conductor of resistance 1 ohm is connected diametrically in this conductor across AB, as shown in figure 5. What is the number of electrons flowing through the section AB (straight) per second ? 124. A battery of emf E and internal resistance 1 ohm is connected as driving cell in a potentiometer. The balancing length of a Daniel cell is found to be L m. If another battery of same emf and internal resistance 2 ohm is connected as driving cell, the balancing length of the same Daniel cell will be a) more than L b) less than L c) equal to L d) cannot be said from the data. 125. Four wires of same material, labelled as P, Q, R, S, lengths in the ratio 1: √2 : √3: 2 and radii also in the same ratio are available. Which of them would be most suitable for making a fuse wire? a) P b) Q c) R d) S 126. *Which of the following statements are correct? a) the terminal voltage of a cell will always be less than its emf b) the terminal voltage of a cell can exceed its emf under certain conditions c) the terminal voltage of a cell can be less than or greater than its emf d) the terminal voltage of a cell will always be greater than its emf 127. In the adjacent figure two cells of emf E1 and E2 are connected as shown with a resistor 9 ohm. The terminal voltage across E1 is (in volt) a) 7.66 b) 8.33 c) 8 d) 0 128. In the same fig. terminal voltage across E2 is (in V) a) 12 b) 12.66 c) 11.34 d) 8 129. In the same figure the potential difference across the points P and Q is ( in volt) a) 11.33 b) 11 c) 8.33 d) 3 . 130. In the adjacent figure, the heat developed per second across the 3 ohm resistor is H. The heat developed across 1 ohm resistor per second is a) H/3 b) 2H/11 c) 11H/2 d) 3H 131. In the adjacent figure the lamp of resistance 4 ohm shines with maximum brightness. The internal resistance of the cell is (in ohm) a) 4 b) 2

124

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c) 8 d )nearly zero 132. The equivalent resistance between the points A and B in the figure below is (in ohm) a) 1 b) 3/4 c) 4/3 d) 10 133. A heater has three coils in parallel having resistances 100, 100 and 200 ohm with the option of using any one or all of them. When 200 ohm coil alone is used, the heat required to boil certain mass of water is 5 minutes. When all the three coils are used, the heat required to boil the same amount of water will be ( in minutes) a) 25 b) 1 c) 5/4 d) 10/3 134. Charge flows at a steady rate through a conductor of non-uniform cross sectional area. Which of the following remains constant? a) drift speed of electrons b) current density c) current d) all the three 135. *The resistivity of a wire depends on a) its length b) its cross sectional area c) its material d) temperature 136. An electric bulb rated as 500 W at 100 V is to be used in a circuit of supply voltage 150 V. Then for the bulb to deliver 500 W, a resistance R a) equal to 20 ohm is to be connected in series b) equal to 20 ohm is to be connected in parallel c) equal to 20/3 ohm is to be connected in series d) equal to 10 ohm is to be connected in series 137. A wire is uniformly stretched to two times its length. The conductance of the wire a) remains the same b) becomes 1/2 c) becomes 1/4 d) becomes 4 times 138. Three cells each of emf E are connected in series to an external resistor of value 10 ohm. The current through the external resistor is i . When the same cells are connected in parallel to the same resistor, the current is still i. The internal resistance of each cell (in ohm) is a) 30 b) 10/3 c) 10 d) 3 139. A conductor has n electrons per unit volume. It has current density j. If specific charge of electron is K, then the total momentum of n electrons is equal to a) jK b) K/j c) j/K d) j2K 140. Two electrons of speeds v and 2v enter into a uniform transverse magnetic field. The ratio of their frequencies of circular motion will be a) 1:1 b) 1:2 c) 2:1 d) 1:4 141. A charged particle of mass m moving with a velocity is subjected to a magnetic field of 20ˆi and an electricfield 1000ˆj. If this particle continues with the same velocity in the same direction, its momentum is ( all units in SI) a) 100m b) 50m c) 200m 142. A cyclotron cannot be used to accelerate a) α particles b) protons c) electrons

d) (1/50)m d) deuterons

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143. A circular loop is placed so that its plane is parallel to a uniform magnetic field. In this magnetic field, it experiences a) a torque b) a force c) both a torque and a force d) neither a torque nor a force 144. A circular loop carrying current is placed perpendicular to a uniform magnetic field. It experiences a) a torque b) a force c) both a torque and a force d) neither a torque or a force 145. Two circular loops are made from wires of lengths in the ratio 1:2 . If they carry same currents, the ratio of their magnetic moments is a) 1:2 b) 2:1 c) 1:4 d) 4:1 146. In a uniform transverse magnetic field, a charged particle in motion has a) constant momentum and kinetic energy b) varying momentum but constant kinetic energy c) constant momentum but varying kinetic energy d) varying momentum and varying kinetic energy 147. The ratio of magnetic field at the centre of a current carrying circular coil of radius R to the field at a distance R from the centre along its axis is a) 1:8 b) 1:4 c) 1:2√2 d) √8:1 148. An electron describes a circular orbit of radius r, with a speed v. The magnetic field produced at the site of nucleus is µ ev µ ev µ e2v µ ev a) o 2 b) o c) o d ) o 2 e) 0 2πr 2πr 4πr 2πr 149. The final energy of a charged particle coming out of a cyclotron is independent of a) mass of the particle b) charge of the particle c) the magnetic field d) speed of the particle 150. A proton and a deuteron of same kinetic energy enter into a uniform magnetic field. The ratio of radius of the track of deuteron to that of proton will be a) 1:1 b) √2:1 c) 1:√2 d) 2:1 151. An electron enters into a uniform magnetic field with equal components of velocity parallel and perpendicular to the field . The path of electron will be a) a circle b) an ellipse c) a helix d) a straight line making an angle 45o with field 152. In a hollow copper pipe carrying DC, the magnetic field produced will be a) only outside b) only inside c) both inside and outside d) neither inside nor outside 153. A charged particle accelerated by a potential V enters into a uniform magnetic field. The radius of its path will be proportional to a) V b) 1/V c) V2 d) √V 154. Two long parallel straight conductors at a distance ‘a’ carry equal currents i. If the force between them is F, which of the following graphs will be a parabola? a) between F and i b) between F and i2 c) between F and a d) between F and 1/a 155. A beam of electrons move from north to south. A straight conductor is held horizontally with current flowing from north to south below the electron beam. The electron beam will a) be pulled down b) remain unaffected

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c) be pulled up d) be pulled horizontally 156. Three identical conductors AB, BC, CA are joined to form the sides of an equilateral triangle and carry same current as shown in the adjacent figure. The magnetic field due to each conductor at the centroid of the triangle of side ‘a’is X. The total magnetic field at the centroid due to all the conductors will be a) 3X towards the reader b) 3X into the paper c) be zero d) 3X cos 60o towards the reader 157. A helium nucleus makes a full rotation in a circle of radius 0.8 m in 2 s. The value of magnetic induction at the centre of the circle (in T) is a) 10-19/µo b) 10-19µo c) 2 x 10-19/µo d) 2 x 10-10 µo 158. A steel needle of length L has a magnetic moment M. The needle is bent at the mid point so that two parts are at right angles. The magnetic moment now will be a) M b) M/2 c) 2M d) M/√2 159. A uniform circular loop has a resistance of 4 ohm. It is connected to a battery as shown in fig. The ratio of resistance of smaller and larger arc is 1:3. When current is passed as shown, the magnetic field at the centre due to the longer arc ABC to the smaller arc AC a) will be in the ratio 1:3 in the same direction b) will be in the ratio 3:1 in the same direction c) will be in the ratio 1:3 in the opposite direction d) will be zero 160. A tangent galvanometer has a deflection of 45o, for a given current i when the coil is in magnetic meridian. If the coil is inclined at an angle 10o with the meridian, the deflection for the same current will be a) more than 45o b) less than 45o o d) cannot be said from the data c) 45 161. The period of oscillation of a magnet in a uniform field is T. If two such identical magnets are placed one over the other so that the like poles point in the same direction, the period in the same field will be a) T/2 b) T c) √2T d) T/√2 162. The magnetic field at a point (P) distant r from one end of a long straight conductor carrying a current is µ i µ i a) o b) o 2πr 8πr µ oi µ i d) o 4πr πr 163. *Check which of the following statements are correct regarding diamagnetism: a) Diamagnetism is present in all the substances b) Diamagnetism can be explained from Lenz’s law. c)

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c) Diamagnetic effect increases with temperature d) Diamagnetic effect of an atom depends on the radius of electron orbits 164. Two long straight conductors P and Q carry currents i and 2i respectively in the same direction. A third conductor carrying a current i is kept midway between P and Q. Then R a) will remain in equilibrium b) will move towards P c) will move towards Q d) will oscillate between P and Q 165. Which of the following is out of place here with respect to conservation of energy principle ? a) Lenz’s law b) Kirchoff’s loop theorem c) Recoil of a gun d) Bernoulli’s theorem 166. A war plane is flying in the region of magnetic equator. In which of the following cases will there be no induced emf on the wings of the plane? a) when flying vertically b) when flying horizontally c) when climbing up at an angle d) when coming down at an angle 167. A constant current is flowing through a straight conductor from left to right due to a source of emf. When the source is switched off a) an induced current will flow from left to right b) an induced current will flow from right to left c) no induced current will flow d) cannot be said from the data 168. .The flux linking with a circuit is given by φ= t3+3t-7. The graph between induced emf and time will be a) a straight line through origin b) a straight line but not through origin c) a parabola through origin d) a parabola but not through origin 169. A straight wire is held by an insulating handle and moved across a uniform magnetic field . Then a) an emf is induced in the wire b) a current is induced in the wire c) a charge is induced in the wire d) all the three are induced in the wire 170. If the induced emf and the inducing emf are in the same direction, it will violate a) the law of conservation of momentum b) the law of conservation of charge c) the law of conservation of current d) the law of conservation of energy 171. In order to move a conductor in a non-uniform magnetic field with a velocity 2 m/s, energy is spent at a rate 20 W. The force acting on the conductor is a) 10 N b) 5 N c) 20 N d) 40 N 172. An A.C of frequency 50 Hz and rms value √2 A flows through the primary of a transformer. If mutual inductance between primary and the secondary is 1 H, the maximum induced emf in the secondary is (in volt) a) 280 b) 400 c) 560 d) 140 173. Two similar circular coaxial loops carrying equal currents in the same direction approach each other along their common axis. Then a) the current in each loop will increase

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b) the current in each loop will decrease c) the current in both the loops will remain the same d) current in one loop will increase and that in the other loop will decrease 174. A current is sent through a vertical spring carrying a weight, from top to bottom. The spring a) will stretch more b) will remain as such c) will contract d) will break 175. The self inductance of a coil depends on a) change of flux through the coil b) rate of change of flux through a coil c) cross sectional area of the coil d) number of turns in the coil 176. When a DC is passed through an inductor of self inductance L, the energy stored in it is a) 1/2 Li2 b) Li2 c) 2Li2 d) 0 177. A circuit has a self inductance of 1 H and carries a current of 2A. To prevent sparking when the circuit is broken, a capacitor which can withstand 400 volts is used. The least capacitance of the capacitor is a) 12.5 µF b) 50 µF c) 25 µF d) 100 µF 178. The maximum mutual inductance between two coils is 6 H and the difference in their self inductance is 5 H. Then, the sum of their self inductance is (in H) a) 11 b) 13 c) 20 d) 22 179. *If L,Q,R represent inductance, charge and resistance respectively then the units of a) QR/L will be that of current b) Q2R3/L2 will be that of power c) QL/R will be that of current d) Q3R2/L will be that of power 180. A coil of metal wire is stationary in a strong non-uniform magnetic field . Then a) no e.m.f. is induced in the coil b) a varying e.m.f. is induced in the coil but no current c) a constant e.m.f. and a current are induced in the coil d) a constant e.m.f is induced in the coil, but no current. 181. The e.m.f. across the secondary of a transformer will not depend on a) voltage across the primary b) spacing of the primary turns c) core of the primary d) resistance of the secondary 182. Ohm per henry has the dimensions of a) time b) current c) angular velocity d) magnetic flux 183. An AC voltage E = 200√2 sin (100t) is applied to a 1µF capacitor through an AC ammeter. The reading of the ammeter will be (in mA) a) 10 b) 20 c) 40 d) 80 184. The phase angle between current and voltage in an AC circuit is tan-1 (4/3). The power factor of the circuit is a) 3/4 b) 3/5 c) 4/5 d) 0 185. A coil and a magnet move along a common axis as shown in the adjacent figure. In which of the following cases will there be no induced emf in the coil? a) When both move with same velocity in same

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direction as shown in figure b) When both move with the same velocity in the opposite direction c) When the coil moves with a velocity v while the magnet is at rest d) When the magnet moves with a velocity v while coil is at rest. 186. An AC represented by the equation e = 282 sin (100πt) is applied to the primary of a step up transformer whose turns are in the ratio 1:2. If an AC voltmeter is connected across the secondary, the reading of the voltmeter will be a) 350 V b) 282 V c) 564 V d) 400 V 187. An AC voltage represented by the equation e = 200 sin(100πt) is applied to a circuit which contains a pure inductor. The time lag between voltage and current in the inductor will be ( in s) a) 1/200 b) 1/100 c) 1/50 d) 1/25 188. The equation for energy density of magnetic field is a) (1/2) µoB2 b) B2/2µo c) B2/µo d) µoB2 189. A conductor of length L and resistance R kept in a uniform magnetic field B is removed from the field with a velocity v. The power spent in this process is a) B2L2v2/R b) BL2v2/R c)B2L2v2/2R d) BLv/R 190. Three identical coils A,B,C are placed with their planes parallel as shown in the adjacent figure. The coils A and C carry currents as shown in figure. The coils B and C are fixed while the coil A moves with uniform velocity towards B. Then a) no current is induced in B by the two coils. b) a clockwise current is induced in B. c) equal currents are induced in B by the two coils which cancels d) an anti-clockwise current is induced in B 191. *An AC voltage of angular frequency ω is applied to a circuit which consists of an inductor of inductance L and a capacitor of capacitance C in parallel. Then across the inductance a) current is maximum when ω2 = 1/LC b) voltage is maximum when ω2 = 1/LC c) current is minimum when ω2 = 1/LC d) voltage is minimum when ω2 = 1/LC 192. In an LCR circuit given below an AC voltage 200V, 50 Hz is applied. V1, V2, V3 are three voltmeters connected across the resistor, the inductor and the capacitor. A is an ammeter. At resonance the reading of V2 is 300 V. Then the reading of V1 and V3 are respectively a) 200 V, 100 V b) 300 V, 100 V c) 200 V, 0 V d) 200 V, 300 V 193. If the resistance R = 200 ohm in the above circuit, the reading of the ammeter A at resonance will be equal to (in ampere) a) 1 b) 1/4 c) 1.4 d) 2.8

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194. If in the above circuit when not in resonance, the volt meter V1 reads 40 V, V2 reads 50 V and V3 reads 20 V. The source voltage is a) 110 V b) 10 V c) 50 V d) 70 V 195. An AC circuit has voltage and current represented respectively by E = 100 sin(100t) V and I = 100 sin (100t+π/3) mA . The average power dissipated in the circuit is a) 0 W b) 5000 W c) 5 W d) 2.5 W 196. A 1.0 µF condenser is charged to 50 volts. The charging battery is then disconnected and a 10 mH coil is connected across the capacitor so that it oscillates . What is the maximum current in the coil ? (Assume that the circuit contains no resistance) : a) 0.25 A b) 0.50 A c) 0.75 A d) 1.00 A 197. When an AC voltage of rms value 240 V is applied to a circuit having a pure inductor, the current through the circuit is 3 A. When the same voltage is applied to a circuit having a pure resistor, the current is 4 A. If the same voltage is applied to a circuit having this resistor and inductor in series, the current through the circuit will be ( in A) a) 3.5 b) 7 c) 1 d) 2.4 198. A capacitor, an inductor are connected to an electric bulb with a source of AC. If the frequency of AC supplied is increased from a small value, then the brightness of the bulb a) increases b) decreases c) remains the same d) first increases and then decreases 199. If x is the relative permeability and y is the relative permitivity of a medium, the refractive index of the medium is equal to a)

x y

b)

1 xy

c) xy

d)

y x

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SECTION 2: ANSWERS 1

(b)

2

(c)

3

(c)

4

(c)

5

(b)

6

(d)

7

(a)

8

(b)

9

(d)

10

(b)

11

(d)

12

(d)

13

(c)

14

(b)

15

(b)

16

(a)

17

(d)

18

(a)

19

(d)

20

(c)

21

(d)

22

(a)

23

(a)

24

(c)

25

(b)

26

(d)

27

(c)

28

(b)

29

(b)

30

(c)

31

(d)

32

(c)

33

(c)

34

(b)

35

(a)

36

(d)

37

(d)

38

(c)

39

(a)

40

(c)

41

(b)

42

(c)

43

(a)

44

(b)

45

(a)

46

(c)

47

(d)

48

(d)

49

(d)

50

(b)

51

(b)

52

(a)

53

(b)

54

(c)

55

(d)

56

(a)

57

(b)

58

(a)

59

(d)

60

(a)

61

(d)

62

(c)

63

(c)

64

(d)

65

(d)

66

(a)

67

(b)

68

(b)

69

(b)

70

(c)

71

(d)

72

(c)

73

(c)

74

(a)

75

(b)

76

(d)

77

(a)

78

(b)

79

(c)

80

(d)

81

(c)

82

(d)

83

(b)

84

(b)

85

(c)

86

(a)

87

(d)

88

(d)

89

(a)

90

(d)

91

(a)

92

(d)

93

(c)

94

(c)

95

(a),(c)

96

(b),(c)

97

(d)

98

(c)

99

(a)

100

(d)

101

(b)

102

(a)

103

(c)

104

(a)

105

(d)

106

(b)

107

(b)

108

(d)

109

(d)

110

(a)

111

(d)

112

(d)

113

(d)

114

(c)

115

(a)

116

(b)

117

(b)

118

(b),(d)

119

1A

120

0.3 A

121

6/11A

122

0

123

1.25x1019 124

(a)

125

(a)

126

(b),(c)

127

(b)

128

(c)

129

(d)

130

(d)

131

(a)

132

(a)

133

(b)

134

(c)

135

(c),(d)

136

(d)

137

(c)

138

(c)

139

(c)

140

(a)

141

(b)

142

(c)

143

(a)

144

(d)

145

(c)

146

(b)

147

(d)

148

(a)

149

(d)

150

(b)

151

(c)

152

(a)

153

(d)

154

(a)

155

(c)

156

(b)

157

(b)

158

(d)

159

(d)

160

(a)

132

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161

(b)

162

(c)

163

(a),(b),(d) 164

(c)

165

(c)

166

(b)

167

(a)

168

(d)

169

(a)

170

(d)

171

(a)

172

(b)

173

(b)

174

(c)

175

(c),(d)

176

(d)

177

(c)

178

(b)

179

(a),(b)

180

(a)

181

(d)

182

(c)

183

(b)

184

(b)

185

(a)

186

(d)

187

(a)

188

(b)

189

(a)

190

(d)

191

(b),(c)

192

(d)

193

(a)

194

(c)

195

(d)

196

(b)

197

(d)

198

(d)

199

(c)

SECTION 3 SOLUTIONS 1. 2.

3. 4.

5.

6.

7.

Take the current from the battery as 2i. It divides i each through 2 branches. Applying Krchoff’s law to any closed network containing battery, we have 120i + 60i = 6. i = 1/30 A. Ans.b No current will pass throgh 40 Ω resistor because of the short-circuited path. Add 70 and 50 in series gives 120 Ω. Hence current 6V has to divide between two parallel paths 60 Ω and 120 Ω. Take the current from the battery as 3i for convenience of division. Of this, ‘2i ‘ flows through 60 Ω and ‘i ‘ through 120 Ω. By Kirchoff’s law for a closed path (30x3i) + (60x2i) + (30x3i) = 6. i = 1/50 A. Current through 60 Ω resistor = 2i = 0.04 A. Ans.c Since the wire is uniform, the two sections between mid points A and B of the sides have resistance 2 Ω each. They are in parallel. The effective resistance would be 1Ω . Ans.c When the wire is cut into four equal parts, each part will have a resistance R/4. When these four are joined in parallel the resistance will be R/16. Heat produced = E2/(R/16) = 16 E2/R which is more than E2/R and more than any of the other alternatives. Ans.c By Faraday’s second law of electrolysis, the masses of silver and zinc deposited will be proportional to the respective equivalent weights. Equivalent weight of silver is 108 and that of zinc 32. These are nearly in the ratio 3.5:1. Ans.b L is the latent heat of vapourisation of water, the heat required for producing 1 g of steam. L = 540 cal = 540 x 4.2 J. Energy supplied = 1080 J/s. Time to boil 100 g of water = (540 x 4.2 x 100)/1080 = 210 s. Ans.d Let E be the voltage of the source. The heat produced in the two wires is E2/R, E2/3R respectively. The ratio of heat produced is 3:1. Ans.a

Here the two parts of the coil having resistance 12Ω and 4 Ω are in parallel. The effectrive resistance is 12x4/16 = 3Ω. Current = emf/total resistance = 12/(3+1) = 3 A. When this current branches into two parts of the coil having resistance 12 Ω and 4 Ω, the current passing through smaller resistance part of the cloil is = total current x 3/4 = 3x(3/4) = 2.25 A. Ans.b 9. Electric field strength = 0.2 Vm-1. Length of the wire = 5 x 10-2 m. Therefore required potential difference across the ends of the wire = 0.2 V x 5 x 10-2 = 10-2 V. Ans.d 10. If the combination has the same resistance at all temperatures, the rate of increase of one should be equal to the rate of decrease of the resistance of other. Germanium is a semiconductor. So its resistance should decrease on heating.. Therefore the other should be aluminium. Ans.b 8.

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11. Let B be the internal resistance of the cell. The current through the circuit is X/(R+B). The potential difference across the resistance = XR/(R+B). This is given as Y. Equating XR/(R+B) = Y, we get B = (X-Y)R/Y). This can also be answered directly from theory notes short-cuts. Ans.d 12. Let R be the external load. Current = 2/(1+R) = 1 A. Thus the load = 1 Ω. When two batteries are in series, emf becomes 4 V and total resistance is 2 (internal) + 1(external) = 3Ω. Current = 4/3 = 1.33 . Ans.d 13. There will be attractive force between wire and nearer side of the loop, repulsive force between wire and farther side of the loop. There will be no force between wire and perpendicular sides. Since attractive force is greater, this side being nearer, net force will be attractive. Ans.c 14. Magnetic field at the centre = µoni/2a. When wire is bent into three loops, field becomes 3 times (Bα n). Radius becomes 1/3. Due to this field becomes 3 times (Bα 1/a). Thus field becomes 9 times. Ans.b 15. In a uniform magnetic field torque due to the field = iAB. When this torque is equal and opposite to that produced by weight of the loop about edge, the edge will start to lift. Therefore mgr = iAB = i πr2 B. i = mg/π rB. Ans.b 16. The magnetic field produced by current in the four edges of one face of the loop at centre will cancel with the magnetic field produced by the currents in the four edges of opposite face. Thus the net field will be zero. Ans.a 17. Applying Fleming’s left hand rule with forefinger from south to north (earth’s field), middle finger from east to west (opposte to electron motion-current), the thumb will point down. Ans.d 18. According to Biot-Savart law, magnetic field = µo i dl sin 900 /4π r2. Here dl = length of the loop = r x θ. Substituting this value, we get field = µoirθ/4πr2 = µoiθ/4πr. Ans.a 19. The magnetic field due to a long straight conductor is given by B = µ 0 I / 2πa That is field B is proportional to 1/a. The graph between two inversely proportional quantities will be a rectangular hyperbola. Ans.d 20. When the magnetic field is perpendicular to the plane of the loop, area vector is parallel to the field. (Recall area has direction along the normal to the loop). The force depends on AxB and will be zero. Ans.c 21. The radius of a charged particle in a uniform field is given by r = mv/qB, where p is momentum. Here p, q and B are same. Hence radius will also be same. Or you can use indirect theory notes short-cuts. Ans.d 22. Use theory notes short-cuts. Ans.a 23. The radius of a charged particle in a uniform magnetic field is r = mv/qB = 2mE /qB, where p is momentum, K kinetic energy. The mass of a particle is 4 times that of proton and its charge is 2 times that of proton. So the radius will be same as that of proton. Ans.a 24. The magnetic field produced by a circular loop at centre = µ 0 ni/2a. The field will be along north-south in the horizontal plane when the coil is perpendicular of magnetic meridian. The horizontal component of earth’s magnetic field Bo will be parallel to it. Thus the two field add. There will be no deflection of magnet in compass box. Ans.c

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25. v x B is a vector perpendicular to the plane of v and B. Thus F is always perpendicular to v and B. Any angle is possible between v and B. The angle between v and B decides the magnitude of the force, not the direction. Ans.b 26. The field due to the straight portion of the conductor at the centre = µoi / 2πa. The field due to circular portion of the wire is µoi / 2a. These two fields are in the opposite direction. µ i µ i µ i Hence net field = 0 − 0 = 0 [π − 1] . Ans.d 2a 2πa 2πa 27. Magnetic field produced by AB and CD are equal and opposite in direction at the centre. Similarly magnetic field produced by BC and AD are equal and opposite. Hence net field will be zero. Ans.c 28. pplying dl x r rule of Biot-Savart law to find the direction of magnetic field due to a straight conductor, we find the field due to all four conductors at the centre will be pointing into the paper and hence they add. Ans.b 29. The magnetic induction due to a straight wire is inversely proportional to the distance (B = µ0 i / 2πa). When the distance is increased from 4 to 12 cm, magnetic induction decreases to 1/3 of X. Ans.b 30. The magnetic moment of a current loop = inA = inπr2. When the loop is bent as 4 turns, the radius reduces to 1/4. The area becomes 1/16. Area of n loops (n = 4) will be 4 x 1/16 = 1/4 times. Magnetic moment becomes 1/4 of the original. Ans.c 31. The radius of the particles r = mv/qB = p/qB = 2mE / qB . Here q is same. Since the accelerating potential is same, energy E is same. Hence r is proportional to √m. R1 /R2 = m1 / m 2 . m1 /m2 = R12/ R22. Ans.d 32. Am2 is the unit of magnetic moment. Similarly joule/tesla is also a unit of magnetic moment. This can be seen from the equation potential energy = mB cosθ. m has the unit of energy/B = JT-1 Ans.c 33. Torque = mB sin θ. So T1 / T2 = sin 30/sin 90 = 1/2. Here T1 = T. Therefore T2 = 2T. =2x. Ans.c 34. When magnet is cut into n pieces, each piece will have a mass 1/n and length 1/n, so that moment of inertia ( mass x 12/12) becomes 1/n3. Moment of magnet (length x polestrength) becomes 1/n. Period becomes

( 1 / n3 )x n = (1/n) of T. You can also use theory notes short-

cuts. Ans.b 35. Magnetic shielding is protection from external magnetic field. By using a material such as soft iron of high permeability, the lines of force of external field can be confined to the cover. Ans.a 36. Isogonic lines join places of equal declination, while agonic lines join places of zero declination. Look up indirect theory notes. Ans.d 37. A dip needle stands in the direction of earth’s resultant magnetic field. At magnetic equator, earth’s field has only horizontal component, that is, vertical component is zero. Hence it stands horizontal. Ans.d 38. Using the formula, m1/m2 = (T12 + T22) / (T12 – T22), we get m1/m2 = 5/3. Ans.c 39. The susceptibility of a diamagnetic substance is negative. In the given list the only diamagnetic substance is bismuth. Ans.a

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40. Magnetic moment is a vector. The moment of smaller magnet = 3Lp. The moment of larger magnet 4Lp. Adding them as vectors, the sum=

(3pL) 2 + (4pL) 2 = 5pL. Ans.c

41. If V is the vertical component of earth’s field. H horizontal component, the angle of dip δ at the place is given by tan δ = V/H. If the dip is measured when the instrument makes an angle θ with magnetic meridian, it is called apparent dip. Here we have to substitute H cos θ in the place of H. If δ1 is the apparent dip, tanδ1 = V/Hcosθ = tanδ sec θ. δ1 = tan-1 (tan δ sec θ) Ans.b 42. A dip needle reads 900 (or stands vertical) at magnetic poles. Here only vertical component of earth’s field exists, i.e. horizontal component is zero. Hence the total intensity will also be 0.4 cgs units. You can also answer this question from the table given in indirct theory notes. Ans.c 43. The magnetism of a diamagnetic substance is independent of temperature. So we look for a diamagnetic substance, the only one in the list being copper. Ans.a 44. Hysterisis is the property shown by only ferro magnets. The only ferro magnet in the given list is cobalt. Ans.b 45. If the beam of a chemical balance is made of a magnetic material, earth’s vertical component will exert a torque on it and it will deflect losing accuracy. Hence a non-magnetic material such as brass can be used. Ans.a 46. The field due to a short magnet at a distance on the axial line is given by [µo/4π] 2m/d3. The field is proportional to 1/d3. The ratio of field at distance x and 2x will be 23:13 = 8:1. Therefore n, here is equal to 8. Ans.c 47. Electromagnets are temporary magnets. So they need high degree of retention of magnetism. Large area would mean more loss of energy while high coercivity would mean larger field to destroy magnetism, which are not desirable for temporary magnets. Look up indirect theory notes too. Ans.d 48. Ans.d 49. If p is the pole strength, M moment, then M = pL. When bent into a semicircle of radius r, L = πr. r = L/π. The new magnetic moment = pole strength x distance between poles = p x 2r = p x 2L/π = 2M/π. Ans.d 50. When a magnet is in stable equilibrium (parallel to the field) its potential energy –mB. When rotated through 1800, it becomes anti-parallel to the field. Its potential energy then = mBcos1800 = + mB. The work done in the process = difference in the potential energy = 2mB. Ans.b 51. At all places in magnetic equator, angle of dip will be zero. (Note: isoclinic line is the line joining places of equal dip not zero dip.) Ans.b 52. By Lenz’s law induced emf in the ring will oppose relative motion between the ring and the magnet. Hence the magnet will always have acceleration less than free fall acceleration ‘g’. Ans.a 53. Flux linkage in the coil φ = NAB cos θ. Induced emf = -dφ / dt = NABω sin θ. The phase difference therefore will be 900. (i.e.that between cosθ and sin θ) Ans.b 54. The induced emf will depend on all the three except the resistance of the coil. Induced current will depend on the resistance, but not the induced emf. Ans.c

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55. The energy stored in the inductor = (1/2) Li2 is due to magnetic field produced inside the conductor coil when a current passes through it. Hence the energy is stored as magnetic field. Ans.d 56. Induced emf = Blv sin θ. Since it flies along the north-south, it will intercept only vertical lines of force of earth. B = 0.2 x 10-4 T, 1 = 50 m, v = 540 x 5/18 ms-1. θ = 900 gives e = 0.15 V. Ans.a 57. The network is a balanced wheatstone’s bridge of effective resistance 1Ω between points connected to the loop. Induced emf = BLv sin 900 , Induced current = BLv/R , which given as 1 mA . With B = 1 T, L = 0.1 m,R = resistance of the loop (1Ω) + resistance of the network (1Ω) = 2Ω, we get V = 2 x 10-2 m/s. Ans.b 58. e = LdI / dt = 20 x 10-3 x 5 /1 = 0.1V. (The resistance of the coil is not necessary here because we are finding emf. ) Ans.a 59. Both induced emf and inducing emf are energies. Only if they are in the opposite direction the total energy will be conserved. Hence it is the consequency of law of conservation of energy. (Also given in indirect theory notes). Ans.d 60. Magnetic lines of force produced by flow of electron (current) are not intercepted by the loop, because they are in the same plane. Hence d φ/dt = 0. Ans.a 61. Applying Fleming’s right hand rule fore finger in direction of the field, middle finger in the direction of the current, the thumb will point into the paper, that is vertically downward. Ans.d 62. Using the formula energy of an inductor E = (1/2) Li2 we can see the unit of self induction can also be swritten as E/i2, J/A2. Ans.c 63. The equation to induced emf of a fan leaf is similar to equation to induced emf between the two ends of a conducting rod in a uniform magnetic field. It is (1/2) B12ω. Ans.c 64. Self inductance L = µoµrN2 A/l. Since N/l is constant, we rewrite the equation as L = N2 µ 0 µ r 2 Al . When length and breadth are doubled A becomes 4 times, length l becomes l twice. So, self inductance L becomes 8 times, because N2/l2 is constant. Ans.d 65. The coefficient of mutual induction does not depend on the current or rate of change of current in the primary. (Note the equation of M = µoµr NpNs/L). Only the induced emf in the secondary depends on the rate of change of current in the primary. Ans.d 66. Inductances add according to the rule of addition of resistors. 2 and 4 in parallel will give 2x4/6 = (4/3) H. This in series with 2/3 gives (4/3) + (2/3) = 2H. Ans.a dI1 dI and L 2 2 dt dt dI 2 dI1 dI 2 = I2 L 2 . It given = . dt dt dt

67. Let induced emf in the two coils at the instant of time t be respectively L 1 Power = voltage x current is same. That is I1 L1 Hence I1 L1 = I2 L2 . Therefore,

dI1 dt

I1 L 2 2 = = . Ans.b I 2 L1 8

68. Since magnetic field perpendicular to the plane and the loop moves in its plane, the flux intercepted by the loop changes. Using palm rule induced emf will be BLv and induced current BLv/R anticlockwise.. Ans.b 69. Use indirect theory notes. Ans.b

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70. Farad = coulomb / volt and not joule/volt. Hence (c) is wrong. All others can be checked and found to be correct remembering L/R , CR and LC have dimensions of time. Ans.c 71. e = -dφ / dt = -(24t +4). At the time t = 1/4 s induced emf will have a magnitude (24 x 1/4) + 4 = 10 V. The induced current will be induced emf / resistance = 10/10 = 1 A. Ans.d 72. The average power in an A C circuit = Erms Irms cos φ, where φ is the phase difference between current and voltage. Here φ = 900. (Phase difference between cos ωt and sin ωt) therefore cos φ = 0. Ans.c 73. Voltmeters read rms value of A.C voltages. In AC circuit voltages add according to vector rules. There is no phase difference between current and voltage in a resistor. Current lags 900 in an inductor and leads 900 in a capacitor. The vector sum is of voltages will be

V1 2 + (V2 − V3 ) 2 = V2. Ans.c 74. In an AC circuit instantaneous voltages follow Ohm’s law. So we can write just as for DC circuit, V1 + V2 + V3 = V. Ans.a 75. When the current is switched on the induced emf in the coil will oppose the growth of the current due to self induction. So the bulb will not glow immediately. When current becomes steady it glows dim because it gets only 6 V when it needs 10. Ans.b 76. During decay of current the induced emf in the inductor will oppose decay of current by Lenz’s law and momentarily shoots up current, and then goes off. Ans.d (Note:These two questions are drawn from experimental demonstration of of self induction). R 2 + ( X L − X c ) 2 where XL , Xc are inductive and capacitative reactance. Here R = 60, XL Erms/Z = 200/100 = 2 A. Ans.a

77. Impedance Z =

78. Refer to theory note tabales.. Tan φ = XL – Xc/R = (180-100)/60 = 4/3. Ans.b 79. Power factor = cos φ = R/Z = 60/100 = 0.6. Ans.c 80. Average power Prms = Erms Irms cos φ = 200 x 2 x 0.6 = 240 W. Ans.d 81. An electro magnetic wave is produced by a tuning circuit which consists of an inductor and a cpacitor. Here is exchange of inductor’s magnetic energy 1/2 Li2 and condensor’s electrostatic energy 1/2 CV2 produces electro-magnetic waves. Ans.c 82. In an AC circuit, voltages add according to a vector rule. Voltages across inductance and capacitance oppose. Hence all that we can say is VR2 + (VL – Vc)2 = V2 where VR . VL, Vc and V are the voltages across the resistor, inductor, capacitor and the supply voltage respectively. Hence it is possible for VL or Vc to exceed V. Ans.d 83. The energy of a photon is given by Planck’s law E = hv = hc/λ. The blue light has less wave length and hence gives photons of more energy. If P is the power of the source, n is no. of photons emitted per second, we have nhc/λ= P. Since nhc/λ is constant, the number of photons is more for light of larger wave length, i.e. red. Hence red light gives photons of less energy but more in number. Ans.b 84. To compare the phase difference between voltage and current, we should have both voltage and current either as sine functions or cosine function. So we rewrite E = 100 sin [ωt + π/2 + π/3] = 100 sin {ωt + 5π/6]. Comparing this with current I = 4 sin ωt, we find phase difference is 5π/6. Ans.b 85. The peak values of AC, E0 = 282 V. Rms value of AC Rrms = Eo/√2 = 282/1.41 = 200 V. Irms = Erms / resistance = 200/20 = 10 A. AC instruments show rms value. Ans.c

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86. The effective inductance between them is given by L = L1 + L2 ± 2M, where M is the mutual inductance between them. When the coils are far apart, mutual inductance between them becomes zero. Then only the self inductance becomes L1 + L2 . Ans.a 87. Scattered intensity is proportional to 1/λ4. Of the given electro-magnetic radiation radio waves have longest wavelength. Hence they are scattered least. Ans.d 88. Using the Rayleigh’s scattering law Iα 1/λ4 , λ (blue)/ λ (red) = 440/660 = 2/3. I(red)/I(blue) = [λ(blue)/ λ(red)]4 = (2/3)4 = 16/81 = 1/5 . Ans.d 89. The maximum distance ground waves can reach by an antenna of height h = 2Rh , where R is radius of earth. Here h = 0.5 km. This gives the maximum distance as 80 km. Ans.a 90. The capacitative reactance Xc = 1/ωC = 1/2πf C. Thus Xc is inversely proportional to f. The graph between them will be rectangular hyperbola. Ans.d 91. Any point on the Y-axis is on the equitorial line of the dipole. Hence field will be parallel to the axis. (a is correct). Electric field at a point between the charges will be along positive Xaxis, but outside along negative X-axis (b is wrong). Dipole moment = -2pd i, as it is negative to positive charge. (c is wrong) work = potential difference x charge. Since potential at infinity and origin are zero, work will be zero. (d is wrong). Ans.a

µ 0 Idl

by Biot-Savart law. Since resistance is proportional to 4πr 2 length, flux density = constant x I1R1 /r2 for one arc for the other arc constant x I2R2/r2. As the potential difference across the two paths are same I1R1 = I2R2, The fields will be same in magnitude. Since currents are in the opposite direction, the fields ancel. Ans.d

92. Flux density the centre =

93. We have here two semi circular coils of radii R1 and R2. The field due to one µoi/4R1 ( half of a coil of one turn) and due to the other µoi/4R2. Since the currents are in the opposite directions field oppose. Ans.c 94. Since the proton is projected at an angle (not parallel or perpendicular) to the field its path will be a helix. Hence choices a and b are wrong. Time period (independent of velocity) = 2πm/qB = 2πx1.67x10-27 / 1.6 x 10-19 x 0.104 = 6.3 x 10-7 s or 2πx10-7s. Ans.c 95. The radius of a charged particle in a uniform magnetic field r = p/qB = is the kinetic energy and p momentum. Thus we find r α to 1/r and hence +

q/m

2mE / qB , where E

m / q Deflection is proportional

+

Deflection of H is proportional to q/√1. (choice a) Deflection of

He is proportional to q/√4 = q/2. Deflection of O2+ is proportional to 2q/√16 = q/2. choice c). Ans.a & c 96. The shunt resistance S to make a galvanometer an ammeter is given by S = Ig /I-Ig, where Ig is the current through galvanometer I is current through the main circuit, G galvanometer resistance. To convert into an ammeter we have to add a resistor in parallel. Since the answer carries 1 Ω we substitute S = 1Ω, Ig = 50 µA, G = 100 Ω, we get I = 5 mA. (choice c). To convert into voltmeter we have to connect a high resistor in series. If R is the value of the resist or in series R = (V/Ig) – G. Here substituting V = 10 volt, Ig = 50 µA, G = 100 Ω, we get R = 200000 Ω, i.e. 200 kΩ. (choice. b) Ans. b & c 97. Copper is a metal and its resistance decreases on cooling. Germanium is a semi conductor. Its resistance increases on cooling. Ans.d 98. Redrawing thediagram we find the arrangement is a balanced Wheasone’s bridge. The effective value of resistance between A and B is RAB = 3R x 6R/9R = 2R. Maximum power is

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transfered when internal resistance is equal to external resistance . That is 2R = 4, R = 2 Ω. Ans.c 99. Irms =

1 4 2 ∫ I dt 2 2

= 3 A. Ans.a

100. Use the information supplied in theory notes. Ratio of two equal resistors when connected in series and in parallel is n2. Here n = 2. Hence ratio of resistances in series and parallel will be 4. Heat produced = E2/R. The required ratio will be 1:4. Ans.d 101. Electric flux by Gauss’s theorem φ = E ds cosθ. Here the area intercepting the lines of forces which is parallel to the axis of the circular plane is the surface area of the hemisphere. Therefore φ = 2πR2E. Ans.b 102. The current through the arrangement, which consists of n cells each of emf E and internal mnE resistance r connected in series with m rows in parallel is , where R is external mr + nR resistance. The current will be maximum if mr = nR. Here r = 0.5 Ω and R = 3Ω. This gives m = 6n. Since mn = 24, total no. of cells we have m = 12 and n = 2.. Ans.a 103. Kinetic energy of a charged particle in an electric field = Charge x potential difference = charge x electric field x distance if the field is uniform. Here it is qEy. Ans.c 104. The torque on a dipole when it makes an angle θ = pEθ, when θ is small. This restoring torque should be equal to 1α. Thus 1α = -pEθ. α = -(pE/I)θ. The angular acceleration is proportional to angular displacement. The motion is S.H.M. The period is 2π I / pE This answer can also be found from similarity with period of magnet in a uniform field 2π I / mB . In the place of a magnetic field B, electrif field E will come. Ans.a 105. Current through one bulb = 60/220 A. If n is maximum number of bulbs that can be used, then n x 60/220 = 9. This gives n = 33. Ans.d 106. Thermo electric power = dE/dT = emf/temperature difference. Ans b 107. Use the information supplied in indirect theory notes. When the wire is halved in length the resistance becomes 1/4. Ans.b 108. Using the equation F = q vxB with q = 1 C, we have the force as –3i. Ans.d 109. ML2T3 is the dimension of power. Hence the given quantity is power/current2 = resistance. Ans.d 110. We start from Ohm’s law j = σE. Resistivity = 1/σ = E/j = electric field / current density.

Ans.a 111. Let the voltage across the source be E and resistance of each bulb be R.

Then power across the bulbs will be E2/4R. This is given as P. When one of them is removed, the power consumed will be E2/3R. Since E2/3R = ( E2/4R) x (4/3), it will be equal to (4/3)P. Ans.d 112. A standard resistor should not change resistance. So it should have low temperature coefficient. It should not melt quickly due to the heat produced in it. Hence it should have high melting point. Ans.d 2 2 113. Since the masses and hence the volumes are equal πr1 L1 = πr2 L2.

2

2

2

2

r R ρL ρL L r r r L 1 r2 = 2 . The ratio 1 = 21 / 22 = 1 x 2 2 = 2 2 x 2 2 = 2 R 2 πr1 πr2 L 2 r1 L2 r1 r1 r1 r1 = 16. Ans.d

4

From this we get

2 .This is equal to 1

4

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114. A secondary cell has less internal resistance compared to a primary cell. Hence it gives more

current. Ans.c 115. Using two resistors we can get 4 resistance values. They are each of their value, both of them

in series and both of them in parallel. In a series connection the resistances add . So you should choose two values given in the question whose sum is the third value given there. Hence the resistances should be 18 and 9. 18 and 9 in series gives 27 and in parallel gives 6 Ans.a 116. If E is emf of each cell and r its internal resistance, when connected in series current i = 4E/4r = E/r. When one of the cells is reversed the emf is equal to E+E+E-E = 2E. However internal resistances still add to 4r. The new current i1 = 2E/4r = (1/2)i. The current reduces by 50%. Ans.b 117. When n ampere enters the combination, 1 ampere should flow the ammeter while n-1 ampere should flow through the shunt. If S is the resistance of the shunt, then (n-1)S = 1xG. ⇒ S= G/n-1. Ans.b 118. The increase in resistance when it is heated through ∆T is given by R1α1 ∆T. For the other

resistor it should be R2α2 ∆T. These should be equal for all ∆T. This makes (b) correct and (a) wrong. If the combination in series should have same resistance at all temperatures, the resistance of one should increase by the same amount as that of the decrease of the other. So one should be a metal and the other a semi-conductor. This can also be answered from indirect theory notes. Ans. b & d 119. The network here is repeated network . Reduce it starting from left end. The value will be outer resistance 2 ohm. Current is equal to emf / resistance = 2/2= 1 A. 120. Use the information given in theory notes. Current through parallel network divides in the ratio of reciprocal of resistance. So divide 1.1 A in the ratio (1/1): (1/2): 1/3) = 6:3:2. Current through 2 ohm resistor equals 1.1 x (3/11) = 0.3 A 121. Use the information current divides in the reciprocal ratio of resistance. Take the current from 6V battery as 3i. Of these 2i passes through 1 ohm and i through 2 ohm. By Kirchoff’s law 1x2i+3x3i = 6. Solving, i = 6/11 A 122. The arrangement between P and Q (that is points connected to battery) is a balanced Wheatstone’s bridge. 3 ohm here is in the place of galvanometer, while 2 ohm resistors are 4 arms of the bridge. Since the current through galvanometer is zero in a balanced bridge, i =0. 123. The arrangement is three one ohm resistors in parallel. They are each half of the loop and diametrical wire AB. Take the current as 3i. Applying Kirchoff’s law to a closed loop containing i = 2A. Use the information supplied in theory notes. The number of electrons will be 2x6.25x1018 = 1.25x1019. 124. The balancing length L of a cell of emf E1 is given by E1 = irL, where r is the resistance per unit length of potentiometer wire and i current through it. When driving cell has more internal resistance, current i decreases and for a given E1, L increases. Ans.a 125. Use the information supplied in theory notes. We look for wire of maximum resistance. Resistance R = ρL/πr2. Resistance is maximum for the wire which has L/r2 maximum. Their values are 1/1, √2/2, √3/3, 2/4. Hence it is maximum for the wire P. Ans.a 126. Usually in a circuit consisting of one cell the terminal voltage cannot exceed emf. However if the circuits has two or more cells connected in opposition, the terminal voltage of a cell of smaller emf can exceed its emf. Hence statements b,c are correct (a) and (d) are wrong. [There is a question which follows to illustrate this ]. Ans.b,c.

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127. The current through the circuit is equal to net emf / total resistance = (12-8)/(1+2+9) = (1/3)

A. The drop across E1 = current x internal resistance = (1/3)V. So terminal voltage is equal to 8+(1/3) = 8.33V. Ans.b (Note:-The drop is added with the emf because current flows from positive to negative inside the cell. This is due to the presence of a cell of higher emf 12V) 128. The drop across E2 = current x internal resistance = 2/3 V. Terminal voltage = 12-(2/3) = 11.34V. Ans.c 129. The potential difference across the points P and Q is the same as the potential difference across the resistor 9 ohm, because the point P is one end of the resistor and Q the other end. The required voltage = currentxresistance = (1/3)x9 = 3 V. Ans.d 130. Whatever may be the value of R, the potential difference across the 3 resistors 1,2,3 ohm are same. Let this potential difference be V. It is given V2/3= H. We are asked V2/1. This is 3H. Ans.d 131. Since the lamp lights with maximum brightness, maximum power is transferred from the source to the lamp. This happens when resistance of the lamp is equal to internal resistance of the source (Maximum power transfer theorem)i.e. 4 ohm Ans.a 132. (Fig) .Looking carefully into the network, we can find that it has symmetry. The upper and lower parts are two balanced Wheatstone’s bridges. Since the current through galvanometer is zero, remove those 2 ohm resistors at centre and redraw the figure as shown in figure above. This reduces as shown in adjacent figure. Here three resistors 4,2,4 are in parallel. The effective resistance R is given by 1/R=(1/4)+(1/2)+(1/4), which gives R = 1 ohm Ans.a 2 133. Since all the resistors are in parallel, use the equation for heat produced V /R per second. Let 2 V /200 = H. When all the resisitors are connected in parallel, the effective resistance (adding them) is 40 ohm. The heat produced now will be V2/40 per second which is 5H. So time to boil same amount of water will be (1/5) of 5 minutes. Ans.b 134. Unless you read carefully, you will miss the correct answer. Since the rate of flow of charge is constant, current is constant. Ans c 135. The resistivity of a wire depends on its material. Note that the resistivity does not depend on the length or cross sectional area It is the resistance which depends on them. So (c) is correct while (a) and (b) are wrong. The resistivity also depends on the temperature, which makes (d) correct. Ans. c,d 2 2 136. The resistance of the bulb R = V /P = 100 /500 = 20 ohm. When 150 V is supplied 100 V should be across the bulb and 50 V should be across a series resistor. In a series connection potential difference across a resistor is proportional to the resistance. Hence the value of the resistor should be half of that of the bulb = 10 ohm. Ans.d 2 137. Use information supplied in theory notes The resistance of the wire becomes 2 = 4 times. The conductance of the wire which is reciprocal of resistance becomes 1/4. Ans.c 138. If E is the emf of each cell and r internal resistance i =

3E . When in parallel i will be 10 + 3r

E . Equating the two and solving r = 10 ohm. Ans.c 10 + ( r / 3)

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139. If v is the drift speed of electron, current i = nev. Multiplying both sides by m, mi = mnev =

e(nmv). Here nmv is a total momentum p of all electrons. This gives p= mi/e = i/(e/m)= i/K, where K is the specific charge e/m of the electron. Ans.c 140. Use the information supplied in theory notes. Both electrons will have same period and hence the same frequency. Ans.a 141. Use the information supplied in theory notes The speed of the particle = E/B = 1000/20 = 50. Its momentum is equal to mass x velocity = 50m. Ans.b 142. When a particle is accelerated, its velocity increases. According to the theory of relativity, its mass also increases. The mass increase is more for lighter particle electron as it can be accelerated to close to the speed of light. The period T = 2πm/qB does not remain constant and hence the resonance condition will not be satisfied. Ans c. 143. A current loop is like a magnet. In a uniform magnetic field a magnet experiences only a G G G torque. No force Torque τ = i A x B . When plane of the loop is parallel to the field, the area vector is perpendicular to the field. Hence τ = iAB sin 90o. Ans.a 144. Here angle between area vector and field vector is 0. Hence

τ = iAB sin0o = 0. No torque,

no force. Ans.d 145. Since the lengths are in the ratio 1:2, the radii of loops will be also in the ratio 1:2. Areas will be in the ratio 1:4. Magnetic moment = current x area of loop. ⇒ Ratio of moments 1:4. Ans.c 146. Use the information supplied in theory notes. The charged particle here has uniform cirular motion. In a uniform circular motion speed is constant. Hence kinetic energy = (1/2) mv2 = constant. But the velocity varies. Hence momentum also varies. Ans.b µ o nR 2 i µ ni . F(centre) = o Dividing, we get F (centre)/ F(axis) = 147. F(axis) = 2 2 3/ 2 2R 2( R + R ) √8R3/R3 = √8/1. Ans.d 148. When an electron moves with a speed v , current i = charge x frequency =

e x (v/2πr) µ o i µ o ev µ o ev ev = Ans.a . Field B = = = 2 πr 2r 2 r 2 πr 4 πr 2 149. Use information supplied in theory notes The final momentum p = qBR. The final kinetic energy = p2/2m = q2B2R2/2m which is independent of the speed. Ans.d 150. Use the information supplied already in theory notes Since E is same, radius of the tracks will Radius (deuteron ) m (d ) q p 2m e x = 2 .Ans.b = = be proportional to the ratio (√m)/q. x m e Radius(proton ) m(p) q d o

151. Since the velocity components are equal the electron should be entering at an angle 45 with

the field. Use the information supplied in theory notes. The path will be a helix. Ans.c G 152. If B is the magnetic field, then by Ampere’s circuital theorem, B. dl = µ o i , where i is the

∫

current enclosed by the path along which integral is taken. If the point is outside , path will enclose a current, and hence the field B will not be zero. If the point is inside, the integral path will not enclose a current. The current i and hence the field B will be zero. Ans.a

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143

153. Use the information supplied

in theory notes. The energy E = qV, where V is the accelerating potential. Since r is proportional to √E, it will be proportional to √V as well. Ans.d

µ 0i 2 . The graph between F and i 2πa 2 will be a parabola, since F is proportional to i . The graph between F and i2 will be a straight line. The graph between F and a will be a rectangular hyperbola while the graph between F and 1/a will be a straight line. Ans.a 155. Since the electron is moving from north to south, it is equal to a current flowing from south to north. Thus the electron current and the current through the conductor are in opposite direction. Hence they repel. The electron beam will be pulled up. Ans.c 156. The magnetic field due to each conductor X has same magnitude. Its direction is given by the vector i dl x r and will be pointing into the plane of the paper . All the fields add in magnitude and direction Ans.b 157. Current i = charge / time = 2e/T, where 2e is charge of helium nucleus. The magnetic field at µ i µ 2e µ x 2x1.6x10 −19 centre B = o = o = o =µ o 10 −19 . Ans.b 2r 2r T 2x 0.8x 2 154. The equation for force between the parallel wires is F =

158. Let p be the pole strength. Then the magnetic moment M is equal to px L. When the needle is

bent the poles are at A and B as shown in the adjacent figure. The length of the magnet is equal to the distance between poles = L/√2. So the new magnetic moment = pL/√2 = M√2. Ans.d o

2

159. The magnetic field at the centre is µoiLsin90 /4πr . = constant x iL, where L is the length of

the arc. Since the loop is uniform its length is proportional to the resistance. So the field becomes constant x iR. For a parallel connection i1R1=i2R2 i.e. iR =constant. So the magnetic field will be same due to the two arcs. But the field will be the opposite direction as the currents are in opposite directions. They cancel. Ans.d 160. For a tangent galvanometer i = K tanθ, where K is 2rBH/µon. When the coil is arranged at an angle 10o with the meridian, the horizontal component of earth’s field BH will only be BH cos 10o. For a constant current i the deflection θ will increase. Ans.a 161. The period of oscillation of a magnet is given by 2π I / MB , where I is moment of inertia of

the magnet. For two identical magnets, one placed over the other as given, I becomes 2I and M becomes 2M. Thus the period remains the same. Ans.b 162. The field at one end point along the perpendicular bisector of a straight conductor is got µ i µ i from the equation o (sinθ1+sinθ2) = o as here θ1 = 0 and θ2 = 90. Ans. c 4 πr 4 πr 163. Diamagnetism is produced due to induced effect when the atom is placed in a magnetic field Hence it is present in all substances. But it shows only in substances where para and ferro effects are absent. (a) is correct. Since it is due to electromagnetic induction it can be explained by Lenz’s law. (b) is correct. It does not depend on temperature. (c) is wrong. Induced magnetic moment depends on induced current and on the radius of the orbit. Hence (d) is correct. Ans.a,b,d 164. The force between parallel conductors carrying current in the

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same direction will be attractive . Since the distance of R is same from P and Q, the force will be proportional to the product i1i2. i1i2 = i2, for P and R and 2i2 for R and Q. Therefore the attractive force of Q will be greater. Thus R will move towards Q . Ans.c. 165. Except recoil of gun all others are one or other form of law conservation of energy. In recoil of gun momentum is conserved . But kinetic energy of the gun is less than that of a shot. So kinetic energy is not conserved. Ans.c 166. In the region of magnetic equator earth’s magnetic field has only horizontal component. So when the plane flies horizontally it cannot intercept magnetic lines of force of earth. Hence no induced e.m.f. In all other cases the magnetic lines of force are intercepted . Ans.b 167. When the source is switched off, the current decreases. By Lenz’s law, induced current should oppose the decrease and hence should flow in the same direction, i,e, from left to right. Ans.a 2

168. The induced emf |e| = dφ/dt = 3t + 3. This is equation to a parabola . When t = 0, e = 3.

Hence the graph will not pass through the origin. Ans.d 169. When a conductor moves across (perpendicular to) the field , an emf is induced in the conductor. Since the circuit is not closed, no current flows through the wire and hence no charge flows through the wire. Ans.a 170. Use the supplied in theory notes . If the induced and inducing emf are in the same direction, it does not obey Lenz’s law. Lenz’s law follows from law of conservation of energy. Ans.d 171. Using the formula force x velocity = power, we have power = 20 W, v = 2 m/s. This gives F = 10 N..Ans.a 172. The peak value of AC = Irms √2 = 2A. The current changes from 2 A (positive maximum) to -

2 A (negative maximum) in half of a cycle, i.e. (1/100) s. So induced emf |e | = M di/dt = 1[2-(-2)]100 = 400 V. Ans.b 173. When loops come nearer, the flux linking with each loop due to the current in the other will increase. That is dφ /dt is positive. The induced emf, therefore -dφ /dt will be negative. That is current in each loop will decrease due to this induced emf. This is in accordance with the Lenz’s law. Ans.b 174. Each turn of the spring can be considered as a loop. So each turn becomes a magnet. It will have opposite poles at top and bottom. One turn will be attracted by the next turn because of the opposite polarity near. Thus the spring will contract. Ans.c

µoµ r N 2A . Thus self inductance depends on A and N/l. l This makes (c,d) correct. The self inductance will not depend on flux or rate of change of flux. It is the induced emf which depends on rate of change of flux. Hence (a) and (b) are wrong. Ans.c,d 176. An inductor stores magnetic energy when current is changing and an emf is induced. A steady current does not produce induced emf. Ans.d 177. Here the magnetic energy of the inductor (1/2)Li2 is converted into electrostatic energy of the capacitor (1/2)CV2. When V is maximum C is minimum. Equating the two C = 1x22/4002 = 25x10-6 F. Ans.c 175. The self inductance L is given by

178. If L1 and L2 are self inductance, maximum mutual inductance M =

L1L2 = 36, L1-L2= 5. L1 +L2 =

( L1 − L 2 )

2

+ 4 L 1 L 2 = 13 H. Ans.b

L 1 L 2 . Here M = 6, so

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145

179. Use the information supplied in theory notes. We know L/R has the dimensions of time .

QR/L will be Q/t = charge / time = current. Q2R3/L2 = (Q2R2/ L2) x R = (Q2/t2 )xR = I2xR, because Q/t = current. I2xR is power. Ans .a ,b 180. To have induced e.m.f there should be a change in the flux linking with the wire, i.e. there should be relative motion. Here the coil is at rest. Ans.a 181. The induced e.m.f. in the secondary of a transformer will depend on the mutual inductance which, in turn, depends on spacing of primary (number of turns per metre), core (µr) of the transformer. The induced e.m.f. will not depend on the resistance of the secondary. Only the induced current will depend on the resistance of the secondary. Ans.d 182. Use the information supplied in theory notes. ohm/henry = R/L, which has the dimensions of reciprocal of time i.e .dimension of angular velocity T-1. Ans.c 183. Comparing with standard equation E = Eo sinωt, Eo = 200√2. Erms = 200 V, ω = 100. Irms =

Erms/Z. Here Z = reactance = 1/ Cω ⇒ Irms = ErmsCω = 200x1x10-6x100 = 20x10-3A = mA. Ans.b

20

184. If φ is the phase angle between current and voltage, it is given here tanφ = (4/3). Power factor

of a circuit = cosφ. cosφ will be (3/5). Ans.b 185. To have an induced emf in the coil, there should be a relative velocity between the magnet

and the coil. When both move with same velocity in the same direction , the relative velocity is zero. There is no change in flux. Hence the induced emf is zero. Ans.a 186. Comparing the given equation with the standard equation e = Eo sinωt, we find Eo

= 282 V. Erms=282/√2 = 200 V. Since the transformer is step-up, the secondary voltage will be 200x2= 400 V. (Recall that the AC voltmeter will show only rms value). Ans.d

187. Here again, comparing with the standard equation Eo sinωt, we find ω = 100π. 2πf = 100π,

which gives the frequency of AC f= 50 Hz. Period of AC = 1/50 s. The time lag of 1/50 second corresponds to a phase lag of 360o. In a purely inductive circuit, the phase lag between current and voltage is 90o. This corresponds to a time lag of (1/4) of 1/50 = 1/200 s. Ans. a

188. The energy density ( energy per unit volume) of magnetic field is given by B2/2µo. Note the

difference in the equation for energy density of electric field (1/2) εoE2. Ans.b 189. The induced emf in the conductor e = BLv. The power when removed from the field = e2/R = B2L2v2/R. Ans. a 190. Since there is a relative motion between A and B a current will be induced in B by the motion of A. By Lenz’s law the current in B will be in an opposite direction to that in A, i.e. anticlockwise. There will be no current due to C, because there is no relative motion (no change of flux) between B and C. Ans.d 2

191. In a parallel resonant circuit, current is minimum at resonance. ω = 1/LC is the condition for

resonance. Hence (c) is correct. Across L the voltage and the current differ by a phase angle 90o. Hence when current is minimum voltage will be maximum. Thus (b) is correct. Ans. b, c 192. At resonance the voltage across L and that across C should be equal and opposite. So the reading of V3 should be 300 V. The supply voltage 200 V should be across the resistor. So V1 will be 200 V. Ans.d 193. Use the information supplied already in theory notes. At resonance inductive reactance and capacitative reactance cancel. Hence the net impedance Z = R = 200 ohm. Current = Erms / Z = 200/200 = 1 A. Ans.a

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194. Use the information supplied in theory notes. The voltages add according to vector rule. i.e.

V=

VR 2 + ( VL − Vc ) 2 = 40 2 + (50 − 20) 2 = 50V. Ans.c

100 100x10 −3 π cos = 2.5 W (Note the current is in mA) Ans.d. x 3 2 2

195. P = Erms Irms cosφ =

2

196. The electrostatic energy of the condenser (1/2) CV is converted into magnetic energy (1/2)

Li2 during discharge. Equating the two we get i = CV 2 / L = 1x10 −6 x50 2 / 10x10 −3 = 0.5 A. Ans.b 197. The reactance X = 240/3 = 80 ohm. Resistance R = 240/4 = 60 ohm. When the two are connected in series, we have an LR circuit. The impedance Z = R 2 + X 2 = 60 2 + 80 2 = 100 ohm. Current i = Erms/Z = 240/100 = 2.4 A. Ans.d 198. This is an LCR circuit. Here R is the resistance of the bulb. When the frequency of AC is increased, the current increases. When the frequency becomes equal to the resonant frequency (Lω = 1/Cω), the current reaches a maximum value. Then it decreases. Since brightness of the bulb depends on the current, it also increases and then decreases. Ans.d. 1 1 c c 199. The speed of e.m waves through a medium c1 = , = = = µε µ0ε0 µ r εr µrεr xy where c is the speed of e.m.waves through free space. From Optics, we know speed through a medium c1 = c/n, (2)where n is the refractive index of the medium. Comparing the two equations(1)and (2) we find n = xy Ans.c

6 OPTICS SECTION 1 QUESTIONS 1.

2.

3.

4.

A diverging beam of light falls on a plane mirror. The image formed by the mirror is a) real, erect b) virtual , inverted c) virtual, erect d) real, inverted In a pond water is 10 m deep. A candle flame is held 15 m above the surface of water. If the refractive index of water is 4/3, the image of the candle flame is formed at a) 15 m deep b) (45/4)m deep c) 10 m deep d) (30/4)m deep Which of the following colours will have minimum critical angle when light passes from glass to air ? a) green b) blue c) yellow d) orange *A ray of light is incident at an angle 450 in an equilateral prism of refractive index √2. The ray a) will totally reflect from the second face. b) will just graze the second surface. c) will run parallel to the incident ray. d) will undergo minimum deviation.

Light of wavelength λ and frequency f is incident from air to glass of refractive index µ . Inside glass the frequency and the wavelength of the light are a) f, µ λ b) µ f, µ λ c) f, λ/µ d) µ f, λ/µ 6. Two convex lenses of equal focal length f are kept at a distance coaxially. A parallel beam of light incident on the first lens emerges from the second lens as a parallel beam. The distance between the lenses is a) f b) 2f c) f/2 d) f/4 7. Imagine a hypothetical convex lens which can pass all the electromagnetic radiations through it. For which of the following radiations will the lens have maximum focal length? a) X-rays b) γ rays c) micro waves d) infra red rays 8. If a is the distance of an object from the focus of a concave mirror and b is the distance of its real image from the focus, the graph between a and b will be a) a straight line of positive slope b) a rectangular hyperbola c) a straight line with negative slope d) a parabola 9. A thin rod of length f/3 is placed along the optical axis of a concave mirror of focal length f such that its image which is real and elongated, just touches the rod. Its magnification is a) 3/2 b) 5/2 c) 1/2 d) 7/2 10. A concave mirror has a focal length of f cm. An object is placed such that it forms a two fold magnified real image. The distance between the object and the image will be ( in cm) 5.

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a) f b) 1.5f c) 2f d) 2.5f 11. Two plane mirrors are inclined to each other at a certain angle. A ray of light first incident on one of them at an angle of 100 with the mirror retraces its path after five reflections. The angle between the mirror is a) 12o b) 22o c) 30o d) 20o 12. If µ 21 represents

13. 14.

15.

16.

17.

the refractive index of the

second medium with respect to the first

medium, then µ 21 µ 32 µ 43 will be equal to b) µ 31 c) µ 14 d) µ 13 a) µ 41 The intensity of light at a distance r from the axis of long cylindrical source is proportional to a) 1/r2 b) 1/r c) 1/r1/2 d) 1/r3 A ray of light incident on an equilateral prism of refractive index 1.41 suffers minimum deviation. The angle of incidence on the first face is a) 60o b) 30o c) 40o d) 45o A ray of light is incident on two plane mirrors inclined at an angle deviates through 240o. The number of images of an object formed by the mirrors is equal to a) 3 b) 2 c) 5 d) 6 A fish is at a depth of 4 m from the water surface. A bird is at a height 3 m vertically above from the water surface. The distance at which the fish sees the bird is a) 7 m b) 9 m c) 8 m d) 10 m A fish under water at a depth h sees the outside world within a cone of radius r at water surface. If µ is refractive index of water, the value of r here is h h a) h µ 2 −1 b) h µ − 1 c) d) 2 µ −1 µ −1

18. A plano-convex lens of focal length f and radius of curvature R is silvered on the plane surface as shown in the adjacent figure The focal length of this will be a) f b) f/2 c) R d) R/2 19. The above plano-convex lens is silvered on the convex surface, as shown in the adjacent figure, so that it becomes a concave mirror. The focal length of this concave mirror will be a) R/2 b) R c) (R/2µ) d) 2R/µ 20. A ray of light passes through three media of refractive indices as shown in the adjacent figure. Then a) µ1 = µ2 , µ3 > µ2 b) µ1 = µ2, µ3 < µ2 c) µ1 = µ2 = µ3 d) µ1 < µ2, µ2 = µ3 21. A plano–convex lens fits exactly with a plano-concave lens as shown in the adjacent figure. Their refractive indices and focal lengths are shown in the figure. The focal length of the combination is

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a)

R 2(µ 2 − µ1 )

c) R

b)

2R (µ 2 − µ1 )

d)

R µ 2 − µ1

149

22. Light energy falls at a rate E Wm-2 on a perfectly reflecting plane mirror. The pressure exerted by the light on the mirror is (c speed of light) a) E/c b) 2E/c c) E/3600 c d) E/7200 c 23. A ray of light is incident at an angle 45o into a medium of µ=√2 . A part of light is reflected and a part refracted. The angle between reflected and refracted rays in degrees will be a) 45 b) 75 c) 90 d) 105 24. An object is kept symmetrically between two plane mirrors inclined at an angle 90o. The images formed by the mirror will lie in a) a circle b) a parabola c) a straight line d) an irregular curve 25. A source of light is kept in an isotropic medium. If the amplitude of wave reaching a point is plotted with distance from the source to the point, the graph will be a) a straight line b) an inverse square law graph c) a parabola d) rectangular hyperbola 26. A lamp is suspended vertically above a table at a height. When the height of suspension is reduced by 2%, the intensity at the centre of the table a) increases by 4% b) increases by 3% c) increases by 2% d) increases by 8% 27. Two lamps A and B are placed at a distance 2 m and 3 m from a screen respectively. The intensity of illumination on the screen due to the two lamps are equal. If PA and PB are the powers of two lamps, then PA/PB = a) 2/3 b) 4/9 c) 3/2 d) 9/4 28. The refractive index of a prism depends on a) angle of the prism A b) angle of minimum deviation D c) both A and D d) neither A nor D Solve the following two problems within a maximum time of 2 minute. Use only one equation. 29. In a displacement method the distance between object and the screen is 90 cm. The focal length of the lens is 20 cm. What is the distance through which lens is displaced between magnified and diminished image ? 30. In a displacement method the lens is displaced through a distance 45 cm. The magnification of one of the images = 4. What is the focal length of the lens? 31. Young’s experiment is performed with white light, which of the following colours will have minimum band width ? a) green b) yellow c) blue d) red 32. If Young’s experiment were performed so that interfering beams travel in water. The ratio of width of band in air to that in water i.e. β(air) / β(water) = a) 1 b) 1/2 c) 3/4 d) 4/3 33. Two light beams have same wave vector k. One travels through a distance L1 in a medium of refractive index µ1 and the other L2 in a medium of refractive index µ2. On emerging their phase difference will be b) k(µ1L2-µ2 L1) a) k (µ1L1-µ2L2)

150

34.

35.

36.

37.

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d) (µ1L2-µ2 L1)/k c) (µ1L1-µ2L2)/k If one of the slits is twice as wide as the other in Young’s experiment, the ratio of intensity of bright band to dark band will be a) 4:1 b) 2:1 c) 3:1 d) 9:1 *.Which of the following cannot produce interference? a) Two narrow sources of blue light and red light b) One real narrow source and one narrow virtual source of same colour c) Two virtual monochromatic sources d) Two different parts of the same source *If the width of both the coherent slits in Young’s experiment is progressively increased a) the width of bands will increase b) the width of bands will decrease c) the contrast between dark and bright bands will decrease d) the bands will disappear after some time The relation between critical angle 'C' and angle of polarisation of a medium 'i' is a) i = tan-1(cosec C) b) i = cot-1 (cosec C) -1 c) i = sin (tan C ) d) i = tan-1(sin C)

38. A laser beam of wave length λ passing through a slit of width ‘a’ is sent to moon. The linear spread of the beam when it reaches to moon is a) Da/λ b) Dλ/a c) Da/2λ d) Dλ/2a 39. When the laser beam given in previous question reaches moon, the area to which it spreads is equal to a) D2a/λ b) D2λ2/a2 c) ( Da/2λ)2 d) Dλ2/a2 40. Which of the following colours will have minimum polarising angle in a given medium? a) violet b) blue c) green d) indigo 2 2 2 41. If D10 = x, D15 = y, D20 = z, where D10, D15 and D20 are the diameters of 10th, 15th and 20th dark ring of a Newton’s ring system, then a) y-x = 2(z-y) b) 2(y-x) = z-y c) z-x = 2y d) y-x = z-y 42. For a certain experiment it is required to diffract Hα line of Balmer series. The wave length of this line is about 660 nm. Which of the following grating would be most suitable for this? a) A grating having 106 lines /m b) A grating having 104 lines/m c) A grating having 6000 lines/m d) A grating having 3x108 lines/m 43. *. Which of the following statements are correct about optic axis? a) It is a line b) It is a direction c) The crystal is symmetric about it d) The speed of O and E rays will be same along it. 44. If the acute angle of a biprism is increased, the width of the fringes produced by it a) decreases b) increases c) remains the same d) cannot be said from the data 45. The polarising angle of ordinary glass is equal 57o. If this glass is immersed in water, the polarising angle will be a) more than 570 b) equal to 570 0 c) less than 57 d) exactly equal to 450

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151

46. In the earth’s atmosphere the ratio of amplitude of blue light scattered to that of red light , if their wave lengths are 400 nm and 600 nm respectively, is a) 9/4 b) 4/9 c) 81/16 d) 16/81 47. In the figure below unpolarised light of intensity I0 = 64 W/m2 is incident on the first

48.

49. 50.

51.

52.

53.

polaroid N1. The intensity of light coming from the second polaroid N2 is I2, = 2 W/m2. The angle between the axis of N1 and that of N2 is equal to a) 30o b) 600 c) cos-1 (1/4) d) cos-1(1/8) Suppose the polaroid N2 in the previous question is rotated keeping N1 fixed. During one rotation, an observer will see a) a maximum and a minimum b) four maxima and four minima c) two maxima and two minima d) light of varying intensity but no minimum and maximum Laser beam can be used to standardise a) distance b) time c) angle d) temperature 14 A laser beam of frequency 3x10 Hz passes through an aperture of diameter 1 mm. The angular spread of the beam in radian is equal to a) 10-2 b) 10-4 -3 c) 5x10 d) 10-3 Which of the following statement(s) are true for photons ? a) Their rest mass is zero b) Their rest mass is infinite c) They cannot produce interference d) They travel with a speed equal to that of light waves If the width of source slit in Young’s experiment is increased, a) the fringe width will increase b) the fringe width will decrease c) the bright bands will become more bright d) the visibility of the system will become less A ray of light of intensity I is incident on a parallel glass-slab at a point A as shown in adjacent figure. It undergoes partial reflection and refraction. At each reflection 25% of incident energy is reflected. The rays AB and A′B′ undergo interference. The ratio Imax/Imin is: a) 4:1 b) 8:1 c) 7:1 d) 49:1

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54. The position of the direct image obtained at O, when a monochromatic beam of light is passed through a plane transmission grating at normal incidence is shown in the adjacent figure. The diffracted images A,B and C correspond to the first, second and third order diffraction. When the source is replaced by another source of shorter wavelength a) all the four images shift in the direction C to O b) the images C,B and A will shift towards O c) all the four images shift in the direction O to C d) the images C, B and A will shift away from O 55. If the coherent slits in Young’s experiment are narrowed so that their width becomes comparable with the wavelength of light, a) the fringes will become broad b) the fringes will become narrower c) the fringes will become unequally spaced d) the fringes will disappear and uniform illumination will set in 56. A mixture of red light of λ = 6000 Ao and blue light of λ = 4400 Ao is incident normally on an air film. The minimum thickness of the film for which the reflected light will appear blue is a) 3x10-7 m b) 6 x 10-7m c) 2.2 x 10-7 m d) 4.4 x 10-7 m 57. Which of the following gives maximum field of view ? a) plane mirror b) concave mirror c) convex mirror d) cylindrical mirror 58. A ray of light falls on a plane mirror at angle of incidence i. The deviation produced on the ray by the mirror is a) 2i b) (180-2i) c) (180 + 2i) d) (360-2i) 59. When a convergent beam of light is incident on a plane mirror, the image formed is a) erect and real b) erect and virtual c) inverted and real d) inverted and virtual 60. When the moon is near horizon, it appears bigger due to a) refraction b) scattering of light c) diffraction d) all the three 61. How many images of himself does an observer see on the mirror surfaced two adjacent walls and the ceiling of a rectangular room? a) 3 b) 5 c) 6 d) 9 62. A glass slab is placed in the path of converging beam of light. The point of convergence of light a) moves towards the glass slab b) moves away from the glass slab c) remains at the same point d) undergoes a lateral shift 63. A plane glass slab is placed on the letters of the following colours. The letters of which colour is raised maximum ? a) red b) yellow c) green d) blue 64. A ray of light passes from vacuum into a medium at an angle of incidence twice the angle of refraction. The angle of incidence is a) cos-1 (µ/2) b) 2 cos-1 (µ/2) d) 2 sin-1 (1/µ) d) 2 sin-1 (µ/2) 14 65. A light wave of frequency 5 x 10 Hz enters a medium of refractive index 1.5. In the medium the velocity of the light wave is a) 3 x 108 m/s b) 2 x 108 m/s c) (5/1014/1.5) m/s d) 1.5 x 5 x 1014 m/s

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153

66. A small plane mirror is rotating with a constant frequency of n rotations per second. With what linear velocity (in m/s) a spot of light reflected from the mirror moves along a spherical screen of radius of curvature R, if the mirror is at the centre of curvature of screen ? a) 2πnR b) 2nR c) 4nR d) 4πnR 67. When a source of light is 1 m away, a photosensitive meter gives a delfection of 8 divisions, if the source is moved 1m more, the deflection will be a) 8 divisions b) 4 divisions c) 2 divisions d) 9 divisions 68. The critical angle of a prism is 360. The maximum angle of the prism for which an emergent ray is possible is a) 720 b) 540 c) 360 d) 160 0 69. A thin prism P1 of angle 4 and ,made from glass of refractive index 1.54 is combined with another thin prism P2 made from glass of refractive index 1.72 so that ray undergoes no deviation. The angle of prism P2 is b) 40 c) 30 d) 2.60 a) 5.330 70. The distance between an object and a divergent lens is x times the focal length of the lens. The lateral magnification m produced by the lens will be a) x b) 1/x c) x+1 d) 1/x+1 71. The colour of light which travels with minium speed in glass is a) red b) yellow c) green d) blue 72. A convex lens is placed between an object and screen which are at a fixed distance apart. For one position of the lens the magnification of the image obtained on the screen is m1. When the lens is moved by a distance d, the magnification of the image obtained on the same screen is m2. The focal length of the lens is (m1 > m2) a) d/(m1-m2) b) d/(m1 + m2) c) dm1/m2 d) m2/m1 73. In a displacement method using convex lens we obtain two images for separation of the lens d. One image is magnified as much as the other is diminished. If m is the magnification of one image, the focal length of the lens is a) d/(m-1) b) md(m2-1) c) d/(m2-m) d) dm2/m1 74. If the earth had no atmosphere, the duration of the day would have been a) longer b) the same c) shorter d) shorter or longer depending on the season 75. A short linear object of length b lies along the axis of a concave mirror of focal length f at a distance u from the pole of the mirror. The size of the image is approximately equal to f a) b u − f

1/ 2

f b) b u −f

2

u − f c) b f

2

u − f d) b f

76. Two convex lenses separated by a distance are brought into contact. The focal power of the combination a) decreases b) increases c) remains the same d) cannot be said from the data 77. A transparent cube of 12 cm edge contains a small air bubble. Its apparent depth is 5 cm when viewed through one face and 3 cm when viewed through the opposite face. The refractive index of the material of the cube is a) 1.5 b) 12/7 c) 12/5 d) 5/3 78. A source of light located at a point A where AB = BC = 1 m as shown in fig. The intensity of illumination at B is P. The intensity of illumination at C is

154

79.

80.

81.

82.

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a) 0.36 P b) 0.5 P c) 0.7 P d) 0.44 P A diver from inside water looks at an outside object whose natural colour is green. He sees the object as a) green b) red c) violet d) yellow An ant is approaching a convex lens with a constant speed up to the first focus. The speed of the image of ant formed by the lens. a) remains the same b) increases uniformly c) first decreases and then increases d) first increases and then decreases A thin lens of focal length f and aperture diameter d forms an image of intensity I. The central part of the aperture up to diameter d/2 is blocked by an opaque paper. The focal length and image intensity would respectively be a) f, I/4 b) f/2, I/2 c) 3f/4, I/2 d) f, I/2 Two thin lenses of focal lengths f1 and f2 are placed at a distance d between them. For power of the combination to be zero, the separation of the lenses must be f f +f a) f1 + f2 d) f1 f 2 b) 1 2 c) 1 2 f2

83. A blue cross illuminated by white light under white background is observed through a red filter. The observer will see a) a red cross on a black background b) a red cross on a blue background c) a black cross on a red background d) a black cross on a blue background 84. The angle of a prism is A and the angle of minimum deviation is 180-2A. the refractive index of the prism is a) sin (A/2) b) cos (A/2) c) tan (A/2) d) cot (A/2) 85. A glass prism has a refractive index of 1.5 and refracting angle of 900. A ray of light is incident on it at an angle of incidence of 300. The angle of emergence will be in degrees a) 60 b) 30 c) 45 d) the ray will not emerge 86. A ray of light is incident at an angle of 600 on one face of a prism of refracting angle 300. The emergent ray makes an angle 300 with the incident ray. The refractive index of the prism is a) 1.41 b) 1.73 B c) 1.5 d) 1.6 87. A ray of monochromatic light is incident on one face of a prism of angle 750. It is incident on the second face at critical angle. If µ = √2, the angle of incidence on the first face of the prism is (in degrees) a) 30 b) 45 c) 60 d) 90 88. A beam of light consisting of red, green blue colours is incident on right angled prism normally on the face AB. The refractive indices of the prism for red, green and blue are

45° A

C

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89.

90. 91.

92.

93.

94.

95.

96.

97.

155

1.39, 1.44 and 1.47 respectively. The prism will a) separate red colour from blue and green b) separate part of blue colour from red and green c) separate all the three colours from one another d) not separate even partially any colour from the others A prism of critical angle 420 is immersed in water of critical angle 490. The critical angle of the prism inside water will be a) between 420 and 490 b) less than 420 0 c) greater than 49 d) equal to 490 The angle with horizontal at which an under-water swimmer will see the setting sun is a) 410 b) 320 c) 490 d) 600 A spherical air bubble in water will act as a) a convex lens b) a concave lens c) plane glass plate d) plano-concave lens When a lens is moved towards the object from a distance 20 cm to 15 cm, the magnification of the image remains the same. The focal length of the lens is a) 16.8 cm b) 15.5 cm c) 18.2 cm d) 17.5 cm Which of the following is an achromatic combination pair of a telescope objective? a) Lenses of of power +3D and - 5D b) Lenses of f = 40 cm and power –3 D c) Lenses of f = -50 cm and power +2D d) Lenses of f = +20 cm and power –4.5 D A convex lens A of focal length 20 cm and a concave lens B of focal length 5 cm are kept along the same axis with a d distance d between them.. If a parallel beam of light falling on a leaves B as a [parallel beam of light falling on A leaves B as a parallel beam, then d is equal to (in cm) a) 25 b) 22.5 c) 20 d) 15 A parallel beam of white light falls on a combination of a concave and a convex lens, both of the same material. Their focal lengths are 15 and 30 cam respectively for the mean wavelength in white light. On the other side of the lens system, one sees a) white pattern b) no pattern at all c) coloured pattern with violet at inner side d) coloured pattern with violet at outer side Consider Fraunhofer diffraction pattern obtained with a single slit, illuminated at normal incidence. At the angular position of the first diffraction minimum, the phase difference (in radians) between the wavelets from the opposite edges of the slit is a) π b) 2π c) π/4 d) π/2 *A planet is observed by an astronomical refracting telescope having objective of focal length 16 m and eyepiece of focal length 2 cm. Then a) The distance between objective and eyepiece is 16.02 m b) The angular magnification of the planet is –800 c) The image of the planet is inverted

156

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d) The objective is larger than the eye-piece 98. Interference fringes are formed with light of wavelength 400 nm in Young’s experiment with separation of slit 1 mm and screen distance 1 m. The angular width of fringe (in radian) is a) 0.4 b) 0.04 c) 0.004 d) 0.0004 99. If Young’s experiment with one source of light and two slits be performed so that interfering beam traverse in water instead of air, a) the fringes will be less in number b) the fringes will be broader c) the fringes will be narrower d) no fringes will be obtained 100. In Young’s interference experiment with one source and two slits, one slit is covered with a cellophane sheet so that only half of the intensity is transmitted. Then a) no fringes will be obtained b) bright fringes will be brighter and dark fringes will be darker c) all fringes will be darker d) bright fringes will be less bright and dark fringes will be less dark 101. In a Young’s double slit experiment, the interference pattern is found to have an intensity ratio of 9 between bright and dark fringes. This implies that a) the intensities at the screen due to the two slits are 5 units and 4 units respectively b) the intensities all the screen due to the two slits are 4 units and 1 unit respectively c) the amplitude ratio is 3/2 d) the amplitude ratio is 3 102. In Young’s double slit experiment the central fringe is white. What colour of fringe will you see nearest to the central fringe ? a) violet b) green c) red d) yellow 103. *White light is used to illuminate the two slits in a Young’s double slit experiment. The separation between the slits is b and the screen is at a distance d ( >> b) from the slits. At a point on the screen directly in front of one of the slits, certain wave lengths are missing. Some of these missing wavelengths are a) λ = b2/d b) λ = 2b2/d c) λ = b2/3d d) λ = 2b2/3d 104. In Young’s double slit experiment using light of wavelength λ, the intensity of one bright fringe is I units. The intensity of light at a point where the path difference is λ/4 is a) I/2 b) I/4 c) I/3 d) I/9 105. The polarising angle of diamond is 670. The critical angle of diamond is nearest to b) 340 c) 450 d) 600 a) 220 106. A ray of light enters from a rarer medium to a denser medium at an angle of incidence i. The reflected and refracted rays make an angle of 900 with each other. The angles of reflection and refraction are r and r’ respectively. The critical angle of denser medium is a) sin-1 (tan r’)-1 b) sin-1 (tan r’1)-1 -1 -1 c) sin (tan i) d) tan-1(sin i) 107. Two nicol prisms (polariser and analyser) have their axes an angle 300 in between. If I is the intensity of light falling on first nicol,. Then the intensity of emerging light from second nicol is a) (1/8) I b) (3/8)I c) (1/4)I d) (1/2)I

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157

108. A point source of light is to be suspended above the centre of a circular table of radius R. In order to produce maximum illumination at the edge of the table, the height of the light source should be a) R b) b) 2 R c) R / 2 d) R/2 109. If a star emitting yellow light starts accelerating towards the earth, its colour as seen from earth will a) turn gradually red b) turn gradually yellow c) remain unchanged d) turn gradually blue 110. An astronomical telescope has large aperture to a) reduce spherical aberration b) have high resolution c) have low dispersion d) increase the span observation 111. The distance between virtual sources in a biprism of angle α and refractive index µ, if the source is placed at distance a from it, is a) 2 (µ-1) α b) (µ-1)α a/2 c) (µ-1) α d) 2(µ-1)α a 112. In Newton’s rings the radius of the first dark ring 2 mm. The radius of the third dark ring will be a) 3 mm b) 2 mm c) 1.7 mm d) 3.4 mm 113. If Young’s experiment could be performed with the following radiations, which of them will give bands of minimum width? a) red light b) radio waves c) violet light d) micro waves 114. Which of the following is inevitable for a Fraunhoffer diffraction experiment? a) White light source b) a colour filter c) a telescope d) a convex lens 115. Which of the following can be polarised ? a) α rays b) β rays c) γ rays d) all the three 116. Energies of photon of four different electromagnetic radiations are given below. Which energy value corresponds to a visible photon ? a) 1 eV b) 2.5 eV c) 5 eV d) 1000 eV 117. The first crucial experiment if in favour of wave theory was carried out by a) Huygens b) Foucault c) Newton d) Thomas Young 118. An interference pattern is observed by Young’s double slit experiment. If now the separation between coherent sources is halved and the distance of screen from coherent sources is doubled ; the now fringe width a) becomes double b) becomes one-fourth c) remains the same d) becomes four times 119. The two slits in Young’s double slit experiment are illuminated by two different parts of same sodium lamp. Then a) uniform illumination will be seen b) interference pattern with dark and bright fringes is observed c) interference pattern with circular yellow fringes is observed d) complete darkness will be seen 120. Young’s experiment is performed with while light. The wavelength of red light = 660 nm. The wavelength of blue light = 440 nm. The minimum order of the red light fringe which coincides with a blue light fringe is equal to a) 2 b) 3 c) 4 d) 5

158

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121. In Young’s experiment monochromatic light is used to illuminate the slits A and B. Interference fringes are observed on a screen placed in front of the slits. Now if a thin glass plate is placed in the path of the beam coming from A, then a) the fringes will disappear b) the fringe width will increase c) there will be no change in the fringe width d) the fringes will shift 122. The wavelength of light plays no role in a) interference b) diffraction c) polarisation d) resolving power -1 123. The critical angle of a certain medium is sin (3/5). The polarising angle of the medium is a) sin-1 (4/5) b) tan-1 (5/3) c) tan-1 (3/4) d) tan-1 (4/3) 124. Double refraction of light is shown by a) quartz and calcite only b) calcite only c) calcite and ice only d) calcite, ice and quartz 125. A diffraction grating has 5000 lines/cm. The maximum order visible with wavelength 6000A0 is a) 2 b) 3 c) 4 d) 50 126. If a drop of water is introduced between the glass plate and a convex lens in a Newton’s ring system, the ring system a) contracts b) expands c) remains the same d) first expands, then contracts 127. A fly is sitting on the objective of a telescope pointed towards the moon. What effect is expected in a photograph of the moon taken through the telescope? a) the entire field of view will be blocked b) there will be image of the fly on the photograph c) there will be no effect at all d) there ill be reduction in the intensity of the image 128. A lens is placed between a source of light and a wall. It forms images of area A1 and A2 on the wall for its two different positions. The area of the source of light is a)

b) ( A 1 + A 2 ) / 2

A1 A 2

c) [( A 1 + A 2 / 2]

2

d ) [(1 / A 1 ) + (1 / A 2 )] −1

129. The diameter of the moon is 3.5 x 103 km and its distance from the earth is 3.8 x 105 km. It is seen by a telescope having the focal lengths of the objective and the eye piece 4 m and 10 cm respectively. The diameter of the image of moon would be approximately a) 20 b) 200 c) 400 d) 500 130. An eye piece consists of two lenses of focal lengths f1 and f2. In order that the combination is achromatic, the lenses are to be separated by a distance a) 2f1 f2/f1 + f2 b) 2f1 f2 / f1 – f2 c) f1 + f2 /2 d) f1 – f2 131. A person can see clearly up to 3 m. What is the power of the lens he should use so that he can see upto 12 m ? a) +0.5 D b) -0.5 D c) +0.25 D d) -0.25 D 132. If the accelerating potential increases to 4 times in an electron microscope, its resolving power R would change to a) R/4 b) 4R c) 2R d) R/2 133. Which of the following is true for a quarter wave plate ?

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159

a) It is made of glass and produces a phase difference π/2 b) It is made of glass and produces a phase difference π/4 c) It is made of quartz and produces a phase difference π/2 d) It is made of quartz and produces a phase difference π/4 134. Which of the following can diffract sound waves from a tuning fork of the frequency 384 Hz? a) a sphere of diameter 10 m b) a sphere of diameter 1 m c) a sphere of diameter 1 cm d) a sphere of diameter 1 mm 135. The exposure time of a camera lens at the f/2.8 setting is 1/200 second. The correct time of exposure at f/5.6 setting is a) 0.02 sec b) 0.04 sec c) 0.20 sec d) 0.40 sec

SECTION 2: ANSWERS 1

(c)

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SECTION 3 SOLUTIONS 1.

Use the information supplied in theory notes. The diverging beam is coming from a real object. Hence the plane mirror forms a virtual image. Virtual images are erect. Ans.c

2.

Here we are getting reflected image of the flame. So its distance will be equal to the distance of the object. Ans.a Use the information supplied in the table in indirect theory notes. Critical angle is minimum for light which has minimum wave length in the given list. Here it is blue. Ans.b

3. 4.

5.

6.

7.

8.

9.

By Snell’s law of refraction sin i1/ sin r1 = √2. This will give r1 = 30o , as i1 = 45o. Since r1 + r2 = A = 600, r2 will also be 30o. This gives i2 = 45o. Thus the ray will emerge from the second face at an angle of 45o. So (a), (b),(c) are wrong . Since i1 = i2 and r1 = r2, the ray undergoes minimum deviation. So (d) is correct. Ans. d Use the information supplied in indirect theory notes wave motion part. The frequency of a wave remains constant. Therefore we eliminate answers (b) and (d). The wavelength decreases when light passes from rarer to denser medium. Ans.c Use the formula for focal length of combination of two lenses or power. Since parallel beam emerges as a parallel beam, the focal power of the system should be zero. i.e. 1 1 1 a 1 = + − . =P=0.⇒a = 2f. Ans.b F f f f xf F Use the information supplied in the table given in indirect theory notes in Optics. The focal length of a lens increases with wavelength. In the given list of electromagnetic radiations, micro waves have the maximum wave length. Ans.c We use Newton’s equation for the relation between distance of an object and real image from the focus, ab = f2. Since ab is equal to a constant, the graph between a and b will be a rectangular hyperbola. Ans.b The image of an object at a distance of 2f is formed at 2f itself. Hence one end of a rod should be at 2f. The image of this end at 2f is formed at 2f itself. The other end will be at a distance 2f-(f/3)=5f/3. The image of this end at a distance 5f/3 is formed at distance v given by the equation 1/u + 1/v = 1/f, with u = 5f/3. This gives v = 5f/2. The length of the image of the rod is thus 5f/2-2f = f/2. The length of the rod is f/3. Length of image f / 2 3 Magnification = = = . Ans.a Length of object f / 3 2

10. Use the formula given for magnification. m= f/u-f. Here m=+2 (real image). This gives u = 1.5 f. Since v/u=2, v= 3f. The distance between object and image for real image = v-u =3f 1.5 f = 1.5 f. Ans.b 11. Use information supplied in theory notes. Let θ be the angle between the mirrors. After the first reflection, the ray is deviated through 180-2i, where i is the angle of incidence. It finds the second mirror at an angle θ. So it deviates through 2θ. ( In direct theory notes) During third, fourth and fifth reflections it undergoes a deviation of 2θ each. The total deviation of the ray after five reflections is therefore 180-2i+ 8θ. Since the ray retraces its path, we have 180-2i+8θ =180o. The angle i=100 . This gives θ = 20o. Ans.d

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12. Using

the

relation

µ21=

µ2 , µ1

we

161

can

rewrite

the

given

quantity

as

=

µ2 µ3 µ4 µ4 x x = = µ 41 . Ans a µ1 µ 2 µ 3 µ1

13. If E is the total light energy given by the source per second , the light energy received per unit area per second of the curved surface of the cylinder = E/2πr per unit length. Thus intensity is proportional to 1/r. Ans.b ( A + D) A / sin . At minimum deviation, (A+D)/2 = angle of 2 2 incidence i. Substituting A= 60o , we get sin i/ sin30 = µ =1.41 or √2. This gives i = 45o. Ans.d

14. Refractive index µ = sin

15. Use the information supplied in theory notes. The angle between the mirrors θ is given by 360-2θ = angle of deviation. This gives 360-2θ = 240. θ = 60o. Number of images = (360/θ)-1 = 5. Ans.c 16. The apparent height of the bird from the surface of water as seen by the fish = refractive index of water x 3 = (4/3) x 3 = 4. The total distance between the fish and the bird as seen by the fish is equal to 4+4 = 8 m. Ans c 17. Use the information supplied theory notes . tanC = r/h. …(1) To find tan C, sinC = 1/µ . cosC =

µ 2 − 1 / µ. . tanC =1/ µ 2 − 1 ⇒ r = h tanC = h/ µ 2 − 1 . Ans.c

18. Since plane surface is silvered, when a ray of light is incident it reflects normally on this surface. So it refracts twice on the spherical surface. Therefore, this is equal to a convex lens. Use the information in indirect theory notes, which gives the focal length of convex lens made of two plano-convex lenses is half of that of plano-convex lens. Ans. b 19. Here refraction takes place on the two plane surfaces and reflection the spherical surface. If F is the focal length of the system, 1/F = (1/f) of mirror + 2(1/f) of lens. Focal length of mirror is equal to R/2. (1/f) of lens is equal to ( µ-1)/R. Substituting these values in the above 1 2 2(µ −1) 1 2µ R ⇒ = . F= equation ,we get = + ..Ans.c F R R F R 2µ 20. Light is passing from first medium to second medium goes straight. Hence µ1 = µ2. When it passes from second medium to third medium it diverges. That means the third medium is denser compared to the second. That is µ3 > µ2. Ans.a 21. Using lens makers formula P = (µ -1)/R, we have for the first lens P1 =-(µ1 – 1)/R and P2 =+(µ2 –1)/R. (negative sign for concave). The power of the combination P = P1 + P2 = (µ2-µ1) / R. Focal length of the combination = 1/P = R/(µ2-µ1). Ans.d 22. Momentum of a photon p = h/λ. Energy E = hc/λ = pc. So p = E/c. Force = change in momentum per second = p-(-p) = 2p = 2E/c. Since energy is falling on unit area, this force will be equal to pressure. Ans.b 23. Refer to the adjacent fig. Angle of refraction r1 is got from the equation sin45/sinr1 = √2. This gives r1 = 30o.. Angle between reflected and refracted rays = 180-75= 1050. Ans.d 24. There will be three images at equal distance from the object. They will lie in a circle. Ans.a

162

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25. By inverse square law , the intensity of light at a point is proportional to 1/r2. The amplitude (square root of intensity) is proportional to 1/r. The graph will be rectangular hyperbola. Ans.d 26. The intensity I = P cosθ/r2. Taking logarithm and differentiating, dI/I = -2(dr/r). Here dr/r = 2%. Therefore dI/I = -4%. The minus sign shows intensity increases when r decreases. Ans. a 27. By inverse square law I = P cosθ/r2. Here θ = 0o. So PA/22 = PB/32. PA/PB = 4/9. Ans.b 28. This is tricky question. The refractive index depends only on the material of the prism and the colour of light. It does not depend on A and D. You should not be carried off by the (A + D) A equation µ = sin / sin . This is only a relation between µ, A, D. That is you can 2 2 calculate µ, if you know A and D. Ans.d 29. Use the simplified formula for finding d. d = D 2 − 4fD = 90 2 − (4 x 20x 90) = 30 cm . .Ans. 30 cm 30. Use the simplified formula connecting focal length f, distance of displacement d and md 4 x 45 magnification m. f = 2 = = 12 cm . .Ans.12 cm m −1 4 2 −1 31. Use the information supplied in theory notes. β αλ. Of the given colours blue has minimum wavelength and hence will have minimum band width. Ans.c 32. Use the information supplied already in two sections of earlier theory notes The band width velocity of light in air β(air ) λ(air ) 4 = = = decreases according to the formula β( water ) λ ( water ) velocity of light in water 3 (refractive index of water). Ans.d 33. When the beams travel through different media, their path difference can be compared if we find their equivalent path in air. If a beam travels L1 in a medium of refractive index µ1, its equivalent path in air is µ1L1. Similarly for the second beam, equivalent path in air is µ2L2. Therefore their path difference is (µ1L1-µ2L2). Use the information supplied in wave motion theory notes to convert path difference to phase difference. The corresponding phase difference is equal to

2π (µ1L1-µ2L2)= k(µ1L1-µ2L2). Ans.a λ

34. Use the information supplied in theory notes in wave motion. The amplitude of wave transmitted is proportional to the slit width. The maximum amplitude is 3 units and the minimum amplitude 1 unit. The ratio Imax/Imin = A max2 / Amin2 = 9/1. Ans.d 35. This question is to test whether all the necessary conditions of interference are known to you. The necessary condition for interference is that the sources should be coherent. That is, they should emit light of (1)same wavelength, (2) same amplitude and (3) zero or constant phase difference. (a) violates the first condition while (d) violates the third condition. (Note that virtual sources can produce interference. For example two virtual coherent sources are used in Biprism, while one real source and one virtual source is used in Lloyd’s mirror.) Ans. a, d. 36. There will be no change in the fringe width because the fringe-width does not depend on width of slits but depends on the distance between slits. [Theory notes]. Hence (a) and (b) are wrong. The contrast between bright and dark bands will decrease, bright bands becoming less bright and dark bands becoming less dark. When the width becomes very large, interference

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condition will be violated and bands will disappear, Ans. c,d

163

uniform illumination will set in .

37. We have tan i = µ. Also µ = 1/sin C = cosec C. This gives tan i = cosec C. i = tan-1(cosec C). Ans.a 38. The spread of a laser beam is only due to diffraction when it passes through the slit. The angular spread is equal to the angle of diffraction θ = λ/a. If D is the distance to moon, the linear spread = angular spread x distance = Dλ/a. Ans.b 39. The areal spread = square of linear spread = (Dθ)2 = (Dλ/a)2 Ans.b 40. Use the information given in table in indirect theory notes. The refractive index is minimum for green. Since tan i =µ, we have ‘i’ is minimum for green. Ans.c 41. Use the information supplied in theory notes. Dn2-Dm2 α (n-m). Hence D152- D102 = D202- D152. That is y-x = z-y. Ans.d 42. For diffraction the spacing of the grating should be of the order of wave length of light. Number of lines = 1/ λ = 1/6.6x10-7 m = 1.5x106/m. (Note: 10/6.6 = 3/2), which is nearest to answer a. Ans.a 43. Optic axis is a direction and not a line. So (b) is correct. (c) and (d) are the other two properties of optic axis. Ans. b,c,d 44. Use the information supplied in theory notes . When α is increased, the distance between virtual sources d increases. The band width β = Dλ/d (same as Young’s experiment )decreases. Ans.a 45. Using the equation tani = µ, when glass is immersed in water, we have to replace µ by µgw, where µgw is refractive index of glass with respect to water. µgw = µg/µw, which is less than µ. Hence tan i decreases, when immersed in water. Ans.c. 46. Use Rayleigh’s law of scattering. The intensity of scattered light is proportional to 1/λ4. Since intensity is proportional to square of amplitude, the amplitude will be proportional to 1/λ2. The ratio of required amplitudes will be A(blue)/A(red) = 6002/4002 = 9/4. Ans.a 47. Use the information supplied in indirect theory notes I = cos2 θ=

I0 cos 2 θ . Here Io = 64, I2 =2.⇒ 2

1 1 ⇒ cosθ = . Ans.c 16 4

48. Here again use the same theory note as in previous question . When angle of rotation of N2 changes from 0 to 360o, I2 reaches two maxima (θ = 0 and 180) and two minima (when θ = 90 and 270). Ans.c 49. Laser beam is highly coherent, monochromatic and unidirectional. Hence it can be used to standardise distance. Ans.a 50. Use the information and formula found in a similar question in this section. The spread of laser beam is due to diffraction at the aperture . Angle of diffraction θ=λ/a. λ = velocity / frequency = 3x108/ 3x1014 = 10-6m. This gives θ= 10-6/10-3= 10-3 radian. Ans.d 51. Statements (a),(d) are true for photons. Since photons have zero rest mass, (b) is wrong. They can produce interference, which we explain due to their wave nature . Hence (c) is not true. Ans.a,d 52. Let x be the width of the source slit and y the distance between the sources slit and the plane of the two coherent slits. For a distinct visibility of interference fringes, the condition to be satisfied is x/y < λ/d. So if x is increased or y is decreased, visibility will be poor. Ans.d

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53. The percentage of light reflected from A is 25% = 0.25. If ‘a’ is the amplitude of the incident light, the amplitude reflected at AB = 0.25 a= 0.5a = (1/2)a Intensity reflected from C = 0.25 of 0.75I. Light transmitted at A′ = 0.75x0.25x0.75 of the total I. The amplitude of A′B′ = 0.75 x 0.25 x 0.75 a = 0.375a = (3/8)a. Thus the amplitude of two components are AB and A’B’ are (1/2)a and (3/8)a. Imax/Imin = [(1/2)a+(3/8)a]2 / [(1/2)a-(3/8)a]2 = 49/1. Ans.d 54. The equation for diffraction grating sinθ= Nnλ, where n is the order, N no. of lines /m of grating and λ the wavelength of light used. When λ decreases, θ also decreases. Hence A,B,C move towards O. However there will be no change in O because it is centre of image i.e. direct and not diffracted . Ans.b 55. When the width of slits become comparable with the wavelength of light used, two slit diffraction pattern results. The diffraction bands are of unequal width. Hence we shall see small number of unequally spaced bands. Ans.c 56. The condition for a film to appear dark by reflection is 2µt cosr = nλ. For minimum thickness n = 1, for normal vision r = 0, for air µ = 1. This reduces the above equation to 2t = λ. The film will appear blue if red is missing . So here we substitute the value of λ for red= 6000Ao =6x10-7m. This gives t=3x10-7m..Ans.a 57. All the objects from the convex mirror to infinity are brought to focus between the pole and the focus of the mirror. Hence large span can be covered. Also the image is erect (This is the reason for using convex mirror as rear – view mirror in a motor car). Ans.c 58. The angle of the deviation is the angle between incident ray and reflected ray. Producing the incident ray we find it is 180-2i here. Ans.b 59. A beam of light converging to a reflecting or a refracting surface is coming from a virtual object. The image of a virtual object formed by a plane mirror is real. Also plane mirror images are erect. Ans.a 60. Ans.a 61. No of images between two adjacent perpendicular mirror = (360/90)-1 = 3. We have here three pairs of perpendicular mirrors. The number of images seen will be 9. Three of them will be common. Thus the number of images will be 9-3 = 6. 62. The glass slab produces an extra path difference of (µ-1)t, where µ is a refractive index, ‘t’ thickness of the slab. This increased path moves the converging point farther. Ans.b 63. Refractive index µ = actual depth / apparent depth. Thus apparent depth is proportional to 1/µ. The letter is raised maximum, when apparent depth is minimum. That is µ should be maximum. For the given colours it happens for blue. Ans.d 64. sin i/sin r = µ. Here i = 2r. That is sin 2r/sin r = µ.. 2 sinr cosr/sin r = µ. This gives cos r = µ/2. r = cos-1 ( µ /2). i = 2r =2cos-1 (µ/2). Ans.b 65. Speed of light in the medium = speed of light in air / µ i.e. 3 x 108 / 1.5 = 2x 108 m/s (Frequency of light is not necessary here) Ans.b 66. When the mirror rotates through an angle θ, the reflected ray rotates through an angle 2θ. The angle turned by the reflected ray in one second ω = 2 x 2πn = 4πn rad/s. The linear velocity of the spot of light will be equal to v = ωR = 4πnR m/s. Ans.d 67. According to the inverse square law, the intensity will be inversely proportional to the square of the distance from the source. When the distance is doubled, the intensity becomes onefourth. The deflection is proportional to the intensity and hence will be 2 divisions. Ans.c

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165

68. The maximum angle for which an emergent ray is possible is 2C. This can be shown as follows: Angle of prism Amax = (r1)max + (r2)max. The maximum values of i1 and i2 will be 900 each, Hence the maximum values of r1 and r2 each will be equal to C. Amax = 2C. If the angle of the prism increases beyound 2C, the ray will totally reflect from the second face. You can also answer this question from theory notes shortcuts. Ans.a 69. If µ1 and µ2 are the refractive indices of the two prisms the conducrion for non deviation is (µ1 – 1) A1 = (µ2-1) A2, where we have used the formula for deviation of small angle prisms, d = (µ -1) A. Here A1 = 40, µ1 = 1.54, µ2 = 1.72 we get A2 = 30. Ans. c 70. Using the formula for magnification in terms of object distance u, m = f/ (u-f). Here u = xf. Therefore m = f/(xf-f). i.e. m = -f/[xf-(-f) ] = -1/(x+1), applying signs. The negative sng in magnification shows that the image is virtual. Ans.d 71. The speed of light is inversely proportional to the refractive index. Blue has the maximum refractive index in the given set and hence has the minimum velocity. This question can be answered from the theory notes table. Ans.d 72. This is displacement method. If D is the distance between the object and the screen, ‘d’ the separation of the two positions of the lens throwing two images on the screen, then m1 = (D+d)/(D-d), and m2 = (D-d) / (D+d), m1 – m2 = d/f, f = d/(m1 – m2). This also can be answered from the indirect theory notes. Ans.a 73. f = d/(m1-m2). Here if m1 is taken as m, m2 = 1/m. Then f becomes md/ (m2-1). Ans.b 74. The molecules of the earth’s atmosphere refract the sunlight so that the sun is visible before sunrise and after sunset. Thus presence of atmosphere makes the day longer. In the absence of atmosphere, sunlight can been seen only after sunrise and before sunset. The day will be shorter. Ans.c 75. The linear magnification of an object is magnification when the object is placed parallel to the principal axis. This is given by dv/du. Differentiating mirror formula 1/u+ 1/v = 1/f, −

du u2

−

dv v2

2

=0⇒

dv v2 f =− 2 = . du u u − f

2

Length of image f . = Length of object u − f

This gives the

2

f length of the image as b .This formula is given in indirect theory notes. Ans.d u − f 76. The power of the combination P = P1 + P2 – dP1P2, where d is their separation. When they are brought into contact, d = 0 and hence P increases. Ans.b 77. Let x be the actual depth of air bubble. When viewed from one face µ = x/5. When viewed from the opposite face µ = (12-x)/3. This gives x = 7.5 and µ = 1.5. Ans.a 78. If S is the power of the source, θ angle of incidence, intensity of illumination is P = Scosθ/r2. In the first case (θ = 0) P = S cosθ/12 and in the second case power P1 = Scos450/(√2)2. This gives P1 = P/2√2 = 0.35 P. Ans.a 79. The speed of light and hence wavelength of light in water is less than that in air . λ (water ) = λ (air) / µ. Taking µ = 4/3, we find wavelength of green (nearly 540 nm,) could reduce to 540 x 3/4 = that of nearly violet. (Note: This question can be answered even without the above calculation, because the only wavelength less than of green given here is violet.) Ans.c 80. Differentiating lens formula 1/u + 1/v = 1/f, with respect to time we get, dv/dt = (v2/u2) du/dt. Since v/u = m, magnification and object speed du/dt is constant, dv/dt is proportional to –m2. When the object approaches from infinity to focus the image is real and hence m is positive, dv/dt is negative i.e. it decreases. When the object is at the focus the image moves to infinity

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on the other side and hence is virtual m is –ve and dv/dt is positve. Hence dv/dt first decreases and then increases. Ans.c 81. There will be no change in focal length. However, reduction of diameter to half reduces the area covered to 1/4 and hence intensity reduces to 1/4. Ans.a 82. Power of the combination of two lenses separated by a distanc d is 1/f = p = (1/f1) + (1/f2)-d/f1 f2). When p = 0, we have (1/f1) +(1/f2) = (d/f1 f2.) This gives d = f1 + f2. Ans.a 83. The red filter allows only red light from the white background. The blue cross will appear dark under red light. Hence one sees black cross under red back ground. Ans.c 84. Here D = 180 – 2A. µ = sin [(A+D) / 2 ] / sin [A/2] becomes cot (A/2). Ans.d 85. sini1 / sinr1 = 1.5, i1 = 300. Hence r1 < 300. r2 = 90-r1 should be greater than 600. Thus the ray is incident in the second face at an angle greater than critical angle which is nearly 420 for a prism of µ = 1.5. It will totally reflect from the second face. Ans.d 86. Here i1 = 600 , d = 300. i1 + i2 = A+d. With A = 300, we get i2 = 0. Hence r2 = 0. r1 + r2 = A. This gives r1 = 300. µ = sin1 / sinr2 = sin 60/sin 30 = √3. Ans.b 87. Here r2 = critical angle C. Since µ = 1/sin C = √2, C = 450 r1 +r2 = 750. Thus r1 = 300. Using sini1 / sin r1 =√2, we get i1 = 450. Ans.b 88. All radiation pass normally the face AB and is incident on the face AC at an angle 450. 1/sinC = µ. Also 1/sin 45 = √2 = 1.41. for red µ = 1.39, hence C is greater than 450 .Thus red will refract. But for the other two colours, µ > 1.41, hence C is less than 450. Thus they will totally reflect from the face AC. So the prism will separate red from the other two colours. Ans.a 89. The relation between critical angle C and refractive index µ is µ = 1/sin C. When a prism is µg is immersed in water, the refractive index of the prism with respect to water µ gw = µw less than that of water or glass. Hence the critical angle will be more than that of water and glass. (Claculation of critical angle is not necessary here, but it can be shown to be 620 ). Ans.c 90. This is given in indirect theory notes. An underwater swimmer sees the setting sun at an angle with vertical equal to critical angle of water 490 The angle with the horizontal will be 900 – 490 = 410. Ans.a 91. A lens may be convex or concave depending on where we are looking from. The usual convention is to look from the rarer medium. An air bubble inside water will act as a concave lens when we look from inside air, which rarer medium here...Ans. b 92. It is clear that one image is real and the other image virtual. Using the equation for magnification in terms of object distance f/u –f, we have, f/(20-f) = -f/(15-f), which gives f = 17.5 cm. Ans.d 93. An achromatic combination of lenses should be made from one concave and one convex lens, with the convex lens having a greater focal power (or shorter length). This is to make the rays forming the final image to converge. These two conditions are satisfied only for the combination given as d, where focal powers are +5D and –4.5 D. Ans.d 94. Since the parallel beam incident leaves as such, the system here acts as a plane glass i.e. power P = 0. Refer to the solution of question (26) the distance between the lenses should be d = f1 + f2 . Here f1 = 20 cm and f2 = -5 cm. Therefore d = 15 cm. Ans.d

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167

95. The given combination has focal length –30 cm, using the combination formula (1/f1) + (1/f2) = 1/F. Hence light falling on the system will produce divergence of the rays. Since violet has more deviation that red, violet will appear to then outer edge. Ans.d 96. The diffraction minimum appears for a path difference equal to n λ ( a sin θ = nλ). A path difference of λ = a phase difference of 2 π radian. Hence the minimum appears for phase difference 2 π, 4π, 6π etc. Ans.b 97. The distance between objective and eye-piece is f0 + fe = 16 + 0.02 = 16.02 m (a is correct). Since the final image is virtual the magnificatrion is negative and equal to f0/fe = -800 (b is correct). An astronomical telescope produces inverted image and hence (c) is correct. The objective is larger than the eye-piece in aperture, (d) is correct. Ans.a,b,c,d Dλ , where D is the distance between the source and the d microscope. d separation of slits and λ wavelength of light. Angular width of the fringe β/D = λ/d = 400 x 10-9 /1 x 10-3. = 4 x 10-4 rad = 0.0004 rad. Ans.d 99. The speed of light inwater is less, and hence the wavelength of light in water is less than that in air. From the equation β = Dλ/d, we understand β will decrease, other values remaining the same. Ans.c 100. Here the amplitude of the waves producing interference is not the same. If the amplitude of a wave coming from the full slit is ‘a’ that coming from the other is only a/ 2 . The maximum amplitude is less than 2a. Hence bright fringes will be less bright. The minimum amplitude is not zero. Hence dark fringes will not be perfectly dark. Ans.d 101. I (maximum )/I(minimum) = [a1 + a2 / a1 – a2]2 where a1, a2 are amplitude of the two waves. Here (a1 + a2)2 = 1. Thus a1 + a2 = 3, a1 – a2 = 1. The amplitudes a1 and a2 are 2 and 1 unit respectively. The intensity of the two waves will be 4 and 1 unit respectively. It should be noticed that intensity cannot be subtracted, as it is a scalar quantity. Amplitude is a vector quantity. This can also be answered using intensity formula given in indirect theory notes. Ans.b 98. The width of a fings β =

102. Because the central fringe is white, the source is white light. The band width β is proportional to wave length λ. The band width is smaller for shorter wavelength. Violet has the shortest wavelength in the visible region. Hence violet fringe will be seen next white. Ans.a 103. When the waves from the two slits reach a point directly in front of a slit as shown in fig. distance travelled by one wave is d and the other d 2 + b 2 . Since b b is qbB q (b − a )B qaB q (b + a )B a) b) c) d) m m m m Two equal point charges Q each are fixed at x = -a and x = +a on the x-axis. Another point charge q is placed at the origin. The change in the electrical potential energy of q, when it is displaced by a small distance x along the x-axis, the approximately proportional to a) x b) x2 c) x3 d) 1/x A long straight wire along the z-axis carries a current I in the negative z direction. The G magnetic vector field B at a point having coordinates (x,y) in the z = 0 plane is µ 0 I ( yˆi − xˆj) µ 0 I ( xˆi + yˆj) a) b) 2 2 2π ( x + y ) 2π ( x 2 + y 2 ) µ 0 I ( xˆj − yˆi ) µ 0 I ( xˆi − yˆj) c) d) 2 2 2π ( x + y ) 2π ( x 2 + y 2 ) A 100 W bulb B1 and two 60 W bulbs B2 and B3 are connected to a 250 V source, as shown in the figure. Now W1, W2 and W3 are the output powers of the bulbs B1, B2 and B3 respectively, then a) W1 > W2 = W3 b) W1 > W2 > W3 c) W1 < W2 = W3 d) W1 < W2 < W3 As shown in the figure 2 P and Q are two coaxial conducting loops separated by some distance. When the switch S is closed, a clockwise current Ip flows in P (as seen by E) and an induced current IQ1 flows in Q. The switch remains closed for a long time. When S is opened, a current IQ2 flows in Q. Then the direction of IQ1 and IQ2 (as seen by E) are a) respectively clockwise and anti-clokwise b) both clockwise c) both anti-clockwise d) respectively anti-clockwise and clockwise A short-circuited coil is placed in a time-varying magnetic field. Electrical power is dissipated due to the current induced in the coil. If the number of turns were to be quadrupled and the wire radius halved, the electrical power dissipated would be

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7.

a) halved b) the same c) doubled d) quadrupled Two identical capacitors have the same capacitance C. One of them is charged to potential V1 and the other to V2. The negative ends of the capacitors are connected together. When the positive ends are also connected, the decrease in energy of the combined system is a)

8.

1 C(V12 − V2 2 ) 4

b)

1 C( V12 + V2 2 ) 4

c)

1 C(V1 − V2 ) 2 4

d)

1 C( V1 + V2 ) 2 a 4

The magnetic field lines due to a bar

magnet are correctly shown in fig.3 An ideal gas is taken through the cycle A →B→C→A, as shown in the figure 4. If the net heat supplied to the gas in the cycle is 5 J, the work done by the gas in the process C →A is a) -5 J b) -10 J c) -15 J d) -20 J 10. Which of the following graphs correctly represents the variation of β = -(dV/dP)/V with P for an ideal gas at constant temperature

9.

11. A Hydrogen atom and a Li++ ion are both in the second excited state. If lH and lLi are their respective electronic angular momenta, and EH and ELi their respective energies, then a) lH > lLi and |EH| > |ELi| b) lH = lLi and |EH| < |ELi| c) lH = lLi and |EH| > |ELi| d) lH < lLi and |EH| < |ELi| 12. An ideal black-body at room temperature is thrown into a furnace. It is observed that a) initially it is the darkest body and at later times the brightest b) it is the darkest body at all times c) it cannot be distinguished at all times d) initially it is the darkest body and at later times it cannot be distinguished 13. The potential difference applied to an X-rays tube is 5 kV and the current through it is 3.2 mA. Then the number of electrons striking the target per second is a) 2 x 1016 b) 5 x 1016 17 c) 1 x 10 d) 4 x 1015

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14. Which of the following processes represents a gamma-decay ? a) AXZ + γ→ AXZ-1 + a + b b) AXZ + 1n0 → A-3XZ-2+c c) AXZ → AXZ + f d) A XZ + e-1 → AXZ-1 +g 15. The half-life of 215At is 100 µs. The time taken for the radioactivity of a sample of decay to 1/16th of its initial value is a) 400 µs b) 6.3 µs c) 40 µs d) 300 µs 16. An observer can see through a pin-hole the top end of a thin rod of height h, placed as shown in the figure. The beaker height is 3h and its radius h. When the beaker is filled with a liquid up to a height 2h, he can see the lower end of the rod. Then the refractive index of the liquid is (fig.6 below) a) 5/2 b) (5 / 2)

215

At to

c) 3 / 2 d) 3/2 17. Which one of the following spherical lenses does not exhibit dispersion? The radii of curvature of the surfaces of the lenses are as given in the diagrams.(fig 6 next column)?

18. In the ideal double-slit experiment, when a glass-plate (refractive index 1.5) of thickness t is introduced in the path of one of the interfering beams (wave-length λ), the intensity at the position where the central maximum occurred previously remains unchanged. The minimum thickness of the glass-plate is a) 2λ b) 2λ/3 c) λ/3 d) λ 19. Two plane mirrors A and B are aligned parallel to each other, as shown in the figure. A light ray is incident at an angle of 300 at a point just inside one end of A. The plane of incidence coincides with the plane of the figure. The maximum number of times the ray undergoes reflections (including the first one) before it emerges out is a) 28 b) 30 c) 32 d) 34 20. A wooden block, with a coin placed on its top, floats in water as shown in fig. The distance l and h are shown there. After some time the coin falls into the water (fig.8). Then a) l decreases and h increases b) l increases and h decreases c) both l and h increase d) both l and h decrease

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197

21. A simple pendulum is oscillating without damping. When the displacement of the bob is less than maximum, its acceleration vector a is correctly shown in (fig.9)

22. A cylinder rolls up an inclined plane, reaches some height, and then rolls down (without slipping throughout these motions). The directions of the frictional force acting on the cylinder are a) up the incline while ascending and down the incline while descending b) up the incline while ascending as well as descending c) down the incline while ascending and up the incline while descending d) down the incline while ascending as well as descending 23. A circular platform is free to rotate in a horizontal plane about a vertical axis passing through

its center. A tortoise is sitting at the edge of the platform. Now, the platform is given an angular velocity ω0. When the tortoise moves along a chord of the platform with a constant velocity (with respect to the platform), the angular velocity of the platform ω(t) will vary with time t as (fig.10)

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24. A particle, which is constrained to move along the x-axis, is subjected to a force in the same direction which varies with the distance x of the particle from the origin as F(x) = -kx + ax3. Here k and a are positive constants. For x ≥ 0, the functional form of the potential energy U(x) of the particle is (fig.11)

25. Two blocks of masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse gives a velocity of 14 m/s to the heavier block in the direction of the lighter block. The velocity of the center of mass is a) 30 m/s b) 20 m/s c) 10 m/s d) 5 m/s 26. A siren placed at a railway platform is emitting sound of frequency 5 kHz. A passenger sitting in a moving train A records a frequency of 5.5 kHz while the train approaches the siren. during his return journey in a different train B he records a frequency of 6.0 kHz while approaching the same siren. The ratio of the velocity of train B to that of train A is a) 242/252 b) 2 c) 5/6 d) 11/6 27. A sonometer wire resonates with a given tuning fork forming standing waves with five antinodes between the two bridges when a mass of 9 kg is suspended from the wire. when this mass is replaced by a mass M, the wire resonates with the same tuning fork forming three antinodes for the same positions of the bridges. The value of M is a) 25 kg b) 5 kg c) 12.5 kg d) 1/25 kg 28. An ideal spring with spring-constant k is hung from the ceiling and a block of mass M is attached to its lower end. the mass is released with the spring initially unstretched. Then the maximum extension in the spring is a) 4 Mg/k b) 2 Mg/k c) Mg/k d) Mg/2k 29. A geo-stationary satellite orbits around the earch in a circular orbit of radius 36000 km. Then, the time period of a spy satellite orbiting a few hundred kilometers above the earth’s surface (REarth = 6400 km) will approximately be a) 1/2 h b) 1 h c) 2 h d) 4 h

30. The effective resistance between points P and Q of the electrical circuit shown in the figure is a)

2Rr R+r

b)

8R ( R + r ) ) 3R + r

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c) 2r+ 4 R 1.

2.

d)

5R 2 + 2r

SOLUTIONS The radius of a charged particle in a uniform magnetic field is r = mv/qB. If the particle is not to enter the region x > b, the radius should have only a maximum value b-a. So b-a = mv/qB ⇒ b = qB (b-a) /m. Ans.B The potential energy between the two charges q,q and the charge Q at the origin is Qq 1 Qq + when charge is displaced by a distance x. Before displacement the 4πε 0 a − x a + x potential energy is

1 2Qq . 4πε 0 a

The change in energy can be found by finding the

difference between the two terms. This will be ∆U = 3.

will be

5. 6.

7.

2πε 0 a 3

, which is proportional to x2.

Ans.B The radius vector joining the conductor to the point is r = ix+jy. The magnitude of the field µ0 I µ I . (1) The direction of the field will be unit vector perpendicular will be 0 = 2πr 2π x 2 + y 2 to the radius vector r. This will be

4.

Qqx 2

ˆi y − ˆjx x2 + y2

(2). From (1) and (2) the magnetic field vector

µ 0 I ( yˆi − xˆj) . Ans.A 2π ( x 2 + y 2 )

In a parallel connection power is same. So power across B3 is the same as power across B1 and B2 together. So W3 is maximum. In a series connection power is directly proportional to resistance. 60 W lamp has greater resistance. So it has greater power. i.e. W2 > W1. Thus W1 < W2 < W3. Ans.D When the current is increasing in P clockwise induced current in Q is anti-clockwise by Lenz’s law. When current is decreasing in P, the induced current in Q is clockwise. Ans.D Induced emf e is proportional to number of turns N. Power is proportional to (emf)2. When the number of turns is made for four times, power increases to 16 times. When radius of wire is halved, resistance increases to four times. Since length increases to four times, resistance increases to 4 x 4 = 16 times. Thus power e2/r remains constant. Ans.B Use the short cut equation for loss of electrostatic energy ∆U = here C1 = C2 ⇒ ∆U =

C1C2 (V1 − V2 )2 . We have 2(C1 + C2 )

1 C( V1 − V 2 ) 2 Ans.C 4

8.

Magnetic lines of force start from north pole and end at south pole outside and continue from south pole to north pole inside the magnet. Ans.D

9.

Work done along AB = 10 (2-1) = 10 J. Work done along BC = 0 (no change in volume). Work done along AC be equal to W. Then 10 + 0 + W = 5, that is heat supplied. This gives W = -5J. Ans.A

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10. Bulk modulus of an ideal gas = K =

P 1 ∆V / V . Compressibility β = . Thus we = ( ∆V / V) K P

find β is inversely proportional to P. hyperbola. Ans.A

The graph between them will be a rectangular

11. By Bohr’s postulate, the angular momentum in nth orbit = n(h/2π). So it is same for hydrogen

and helium atom. The energy of hydrogen atom is -13.6/n2 eV for nth state and -13.6 Z2 / n2, for helium atom. So it is more for helium atom. Ans.B

12. A black body is a good emitter and a good absorber. When it is thrown into the furnace, it

first becomes dark. Then it reemits all radiations. It becomes brightest. Ans.A 13. If n is the number of electrons striking the target, e its charge, current I = ne. n = I/e = 3.2 x

2 x 1016. Potential will not decide the number of electrons. It decides 10-3/ 1.6 x 10-19 = only the energy of electrons. Ans.A

14. Only in a γ decay there will be no change in mass number and atomic number. Ans.C 15. Use the short cut equation

N0 = 2t / T N

Here N0/N = 16 ⇒ t/T = 4 ⇒ t is four times T.=

400 µs. Ans.A 0

16. From fig. when the lower end of the rod is seen, the angle of incidence is 45

.

sin i sin i 1 . r = 450. This gives sin i = 1/ 5 ⇒ µ = 5 / 2 . Ans.B = µ ag ⇒ = sin r sin r µ ga 17. We use here Lens maker’s formula

1 1 1 . For no dispersion, d (1/f) should be = (µ − 1) + f R R 2 1

zero ⇒ R1 = -R2. i.e. one surface must be convex and the other concave. Ans.C. 18. If µ is the refractive index of glass, n is a number of fringes shifting, for central maximum to

remain unchanged (µ-1)t = nλ ⇒ t = 2λ. Ans.A 0

19. During each reflection, there is a lateral displacement of x = d tan 30 , where d is the distance

between the glass plates. If n is the number of reflections, then n d tan 30 = length of the plate L. n x 0.2(1/ 3 ) = 2 3 ⇒ n=30. Total number = 30+initial+final = 32. Ans.C 20. By law of flotation, weight of floating body = weight of liquid displaced. When the coin is

removed, the weight of body decreases. Hence it rises and ‘l’ decreases. When the coin falls into water, it displaces water equal to its volume. The volume of coin is numerically less than its mass. Hence water level falls ‘h’ decreases. Ans.D 21. A body in motion along a curve has two accelerations. They are tangential acceleration at along the tangent and centripetal acceleration ar along the radius towards centre. Its vector G G G sum a = a t + a r . Ans.C 22. On an inclined plane the force of gravity is always downwards, even if the cylinder moves up or down. Hence frictional force is always upwards. Ans.B 23. The initial distance of the tortoise from centre of circle O is equal to radius of platform r. As it moves from A its distance from O decreases, reaches minimum at O, then increases. Its moment of inertia I = mr2 ecreases and then increases. As the platform is a closed system Iω = Crms . So ω first increases and then decreases. Ans.B

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24. If F is the force and U potential energy, F= −

dU Kx 2 ax 4 ⇒ U = − ∫ F dx = − + . For U = 0, the dX 2 4

equation is quadratic in x. Hence the graph of U against x will touch x axis at two points. This will eliminate answers B and C From the equation we find F = 0 when x = 0 . dU = 0 ⇒ The slope of graph U against X should be 0 at the origin. This eliminates ⇒ dX x =0

answer A. Ans.D 25. The spring mass system has only internal force velocity of centre of mass v(c.m) = m1v1 + m 2 v 2 (10x14) + (4 x 0) = 10 m/s Ans.C = m1 + m 2 14 26. If VA is velocity of A, VB that of B, V velocity of sound, the frequencies heard by the

passenger are f1 =

V f (V + VA ) f (V + VB ) ⇒ B = 2 . Given f = 5, f1 = 5.5 f2 = 6, we get Ans.B , f2 = V V VA th

27. The frequency of sonometer wire when it vibrates in n n 2L

T n = µ 2L

mode (in ‘n’ loops) is fn =

Mg ⇒ n M = constant, n is the number of nodes. µ

M1 n = 2 ⇒ M1= 9 kg. M2 n1

⇒ M2 = 25 kg Ans. A 28. If A is maximum extension, the work done in stretching = kinetic energy of the opring. Work 2Mg 1 = force x distance. the force here is constant = Mg ⇒ MgA = kA 2 ⇒ A = Ans.B k 2 3/2 29. By Kepler’s third T α R if T is the period of geostationary satellites Ts that of spy satellite T 36000 + 6400 = Ts 6400

3/ 2

= 7 7 ⇒ Ts =

24 7 7

nearly 2h. One can also answer the question with

the theory information that the period of a satellite very close to earth is 84 minutes. The distance of spy satellite should be slightly more as it is orbiting a few hundred km above earth. The nearest answer is 2h. Ans.C 30. By symmetry remove 2R from the centre of the net work. (Note this is similar to Wheatsone’s bridge.) This leaves three resistances 2R + 2R, 2R+2R, r+ r, in parallel. This gives

2Rr Ans.A R+r

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IIT-SCREENING TEST QUESTIONS 2001 1.

2.

3.

4.

5.

A quantity X is given by ε 0 L

∆V , where ε0 is the permittivity of free space, L is a length, ∆V ∆t

is a potential difference and ∆t is a time interval. The dimensional formula for X is the same as that of A) resistance B) charge C) voltage D) current The pulleys and strings shown in the figure are smooth and of negligible mass. For the system to remain in equilibrium, the angle θ should be A) 00 B) 300 0 C) 45 D) 600 A string of negligible mass going over a clamped pulley of mass m supports a block of mass M as shown in the figure. The force on the pulley by the clamp is given by A)

2 Mg

B)

2 mg

C)

(M + m) 2 + m 2 g

D)

( M + m) 2 + M 2 g

2m m

m m

An insect crawls up a hemispherical surface very slowly (see Fig). The coefficient of friction between the insect and the surface is 1/3. If the line joining the centre of the hemispherical surface to the insect makes an angle α with the vertical, the maximum possible value of α is given by A) cot α = 3 B) tanα = 3 C) sectα = 3 D) cosecα = 3 Two particles of masses m1 and m2 in projectile motion have velocities v1 and v2 respectively at time t = 0. They collide at time t0 . Their velocities become v1’ and v2’ at time 2t0 while still moving in air. The value of m1v1’ + m2v2’ )-(m1v1 + m2v2) is A) zero

B)

(m1 + m2) gt0

C) 2(m1 + m2) gt0

D)

1 (m1 + m2) gt0 2

M

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6.

One quarter sector is cut from a uniform circular disc of radius R. This sector has mass M. It is made to rotate about a line perpendicular to its plane and passing through the centre of the original disc. Its moment of inertia about the axis of rotation is 1 MR 2 2 1 C) MR 2 8 A)

7.

B)

1 MR 2 4

D)

2 MR 2

A simple pendulum has a time period T1 when on the earth’s surface and T2 when taken to a height R above the earth’s surface, where R is the radius of the earth. The value of T2/T1 is B) √2

A) 1 8.

C) 4

D) 2

A small block of shot into each of the four tracks as shown below. Each of the tracks rises to the same height. The speed with which the block enters the track is the same in all cases. At the highest point of the track the normal reaction is maximum in

v

9.

v v v (A) (B) (C) (D) A hemispherical portion of radius R is removed from the bottom of a cylinder of radius R. The volume of the remaining cylinder is V and its mass is M. It is suspended by a string in a liquid of density ρ, where it stays vertical. The upper surface of the cylinder is at a depth h below the liquid surface. The force on the bottom of the cylinder by the liquid is A) Mg

h

B) Mg-V ρg

ρ

C) Mg- πR2 h ρg D) ρg (V+πR2h)

2R

10. The ends of a stretched wire of length L are fixed at X = 0

and X = L. In one experiment, the displacement of the wire is Y1 = A sin (πx/L) sin ωt and energy is E1 and in another experiment its displacement is Y2 = A sin (2πx/L) sin 2ωt and energy is E2 . Then A) E2 = E1 B) E2 = 2E1 C) E2 = 4E1 D) E2 = 16 E1 11. Two pulses in a stretched string whose centres are initially 8 cm apart are moving towards each other as shown in the Fig. The speed of each pulse is 2 cm/s. After 2 seconds, the total energy of the pulses will be A) zero C) purely potential 12. P-V plots for two shown in the Fig.

B) purely kinetic D) partly kinetic and partly potential gases during adiabatic processes are Plots 1 and 2 should correspond

8 cm

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respectively to A) He and O2

P

B) O2 and He C) He and Ar D) O2 and N2 1 2

13. Three rods made of the same material and having

the same cross section have been joined as shown in the Fig. Each rod is of the same length. The left and right ends are kept at 00C and 900C respectively. The temperature of the junction of the three rods will be

V 90°C

0

A) 45 C B) 600C 0°C C) 300C D) 200C 0 14. When a block of iron floats in mercury at 0 C, a fraction k1 of its volume is submerged, while at the temperature 600C, a fraction k2 is seen to be submerged. If the coefficient of volume expansion of iron is γFe, and that of mercury γHg then the ratio k1/k2 can be expressed as A)

1 + 60 γ Fe 1 + 60 γ Hg

B)

D)

1 + 60 γ Fe 1 − 60 γ Hg

D)

90°C

1 − 60 γ Fe 1 + 60 γ Hg 1 + 60 γ Hg 1 + 60 γ Fe

15. In a given process on an ideal gas, dW = 0 and dQ < 0. Then for the gas

A) the temperature will decrease B) the volume will increase C) the pressure will remain constant D) the temperature will increase 16. Three positive charges of equal value q are placed at the vertices of an equilateral triangle. The resulting lines of force should be sketched as in

(A)

(B)

(C)

(D)

17. A uniform electric field pointing in positive x-direction exists in a region. Let A be the origin,

B the point on the x-axis at x = +1 cm and C be the point on the y-axis at y = +1 cm. Then the potentials at the points A,B and C satisfy A) VA < VB C) VA < VC

B) VA > VB D) VA > VC

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18. Consider the situation shown in the Fig. The capacitor A has a charge q on it whereas B is

uncharged. The charge appearing on the capacitor B a long time after the switch is closed is A) zero

B) q/2

C) q

D) 2q

q _ +

19. A coil having N turns is wound tightly in the form of a

spiral with inner and outer radii a and b respectively. When a current passes through the coil the magnetic field at the centre is µ NI A) 0 b C)

b µ0 NI In 2(b − a ) a

2µ0 NI B) a D)

_

+

_

+

S

_

+ A

B

b µ0 NI N In 2( b − a ) a

20. Two particles A and B of masses mA and mB respectively

and having the same charge are moving in a plane. A uniform magnetic field exists perpendicular to this plane. The speeds of the particles are vA and vB respectively and trajectories of particles are as shown in the Fig. Then

A B

A) mAvA < mB vB B) mAvA > mB vB C) mA < mB and vA < vB D) mA = mB and vA = vB 21. Two circular coils can be arranged in any of the three situations shown in the Fig. Their

mutual inductance will be

(A) A) maximum in situation (A)

C) maximum in situation (C)

(B)

(C) B) maximum in situation (B)

D) the same in all situations

22. A wire of length L and 3 identical cells of negligible internal resistances are connected in

series. Due to the current the temperature of the wire is raised by ∆T in a time t. A number N of similar cells is now connected in series with a wire of the same material and cross-section but of length 2L. The temperature of the wire is raised by the same amount ∆T in the same time t. The value of N is A) 4

B) 6

C) 8

D) 9

23. A ray of light passes through four transparent media with refractive indices µ1, µ2, µ3 and µ4

as shown in the Fig. The surfaces of all media are parallel. If the emergent ray CD is parallel to the incident ray AB, we must have

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B) µ2 = µ3

C) µ3 = µ4

D) µ4 = µ1

D

µ1 µ2 µ3

24. A given ray of light suffers minimum deviation in an

equilateral prism P. Additional prisms Q and R of identical shape and of the same material as P are now added as shown in the Fig. The ray will now suffer

C

B

µ4

A

A) greater deviation B) no deviation C) same deviation as before

Q

D) total internal reflection

P

R

25. Two beams of light having intensities 1 and 4I interfere

to produce a fringe pattern on a screen. The phase difference between the beams is π/2 at point A and π at point B. Then the difference between the resultant intensities at A and B is A) 21

B) 41

C) 51

D) 71

26. The transition from the state n = 4 to n = 3 in a hydrogen-like atom results in ultraviolet

radiation. Infra-red radiation will be obtained in the transition A) 2 → 1

B)

3→2

C) 4 → 2

27. The intensity of X-rays from a Coolidge tube is

plotted against wavelength λ as shown in the Fig. The minimum wavelength found is λC and the wavelength of the ⇒ Kα line is λK . As the accelerating voltage is increased

D) 5 → 4

I

A) λK - λC increases B) λK - λC decreases C) λK increases

C

K

D) λK decreases 28. The electron emitted in beta radiation originates from

A) inner orbits of atoms

V

R

V

C

B) free electrons existing in nuclei C) decay of a neutron in a nucleus D) photon escaping from the nucleus 29. In the given circuit with steady current, the potential

drop across the capacitor must be A) V

B)

V/2

C) V/3

D)

2V/3

2V

2R

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30. A metallic square loop ABCD is moving in its own

plane with velocity v in a uniform magnetic field perpendicular to its plane as shown in the Fig. An electric field is induced

A

B

D

C

A) in AD, but not in BC B) in BC but not in AD C) neither in AD nor in BC D) in both AD and BC 31. A radioactive sample consists of two distinct species having equal number of atoms initially.

The mean life time of one species is τ and that of the other is 5τ. The decay products in both cases are stable. A plot is made of the total number of radioactive nuclei as a function of time. Which of the following Fig.s best represents the form of this plot ? N

N

N

N

t t t t (A) (B) (C) (D) 32. In the given circuit, it is observed that the current I is independent of the value of the resistance R6. Then the resistance values must satisfy R5

A) R1 R2 R5 = R3 R4 R6 B)

1 1 1 1 + = + R 5 R 6 R1 + R 2 R 3 + R 4

R1

I

C) R1 R4 = R2 R3

R6

R2

D) R1 R3 = R2 R4 = R5 R6

R3 R4

33. In a Young’s double slit experiment, 12 fringes are

observed to be formed in a certain segment of the screen when light of wavelength 600 nm is used. If the wavelength of light is changed to 400 nm, number of fringes observed in the same segment of the screen is given by A) 12

B) 18

C) 24

34. A non-planar loop of conducting wire carrying a

current I is placed as shown in the Fig. Each of the straight sections of the loop is of length 2a. The magnetic field due to this loop at the point P(a,0,) points in the direction A) C)

1 2 1 3

( −ˆj + kˆ )

B)

( −ˆi + ˆj + kˆ )

D)

D) 30 z I

1

( −ˆj + kˆ + ˆi ) 3 1 ˆ ˆ (i + k ) 2

y x

2a

35. A particle executes simple harmonic motion between x = -A and x = +A. The time taken for

it to go from 0 to A/2 is T1 and to go from A/2 to A is T2. Then

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A) T1 < T2

B) T1 > T2

C) T1 = T2

D) T1 = 2T2

SOLUTIONS 1.

Substituting the units of the given quantity we have εo L

F V FxV C dV xmx = = = A. . = m t t t dt

Thus the given quantity is current. Ans.D 2.

Consider the free body diagram of mass m. We find mg = T (1). Next consider the free body diagram of mass√2 mg. We find that the vertical component of the tension should be equal to weight √2mg⇒√2mg = 2Tcosθ = 2mgcosθ ,since T = mg from equation (1). ⇒ θ = 45o. Ans.C 2T cos T

T

T m

T sin

T sin

mg

3.

2 mg

Consider free body diagram of the block as shown in fig.1. We can write Mg = T (1). Now consider figure 2 which is the free body diagram of the pulley. If F is the net force on the F T

T

Mg F1

mg

T

pulley, it should be equal to the vector sum of the two forces mg+T and T.⇒ 2

F= ( mg + T ) 2 + T 2 = (mg + Mg ) 2 + ( Mg ) 2 = (m + M ) + M

co s

5.

g. Ans.D

Let P be the point where insect leaves the sphere. This happens when the centripetal force vanishes. i.e. N-mgcosα = mv2/r should become zero. ⇒ N= mg cosα (1). At this point the frictional force along the tangent µN should be equal to mgsinα i.e. µN = mgsinα (2). Dividing N equation (2) by (1), we get tan α = µ ⇒ cotα = P 1/µ = 3. Ans.A The given quantity (m1v1'+m2v2')- (m1v1+m2v2) is the magnitude of change in momentum of the mg mg sin projectiles. Change in momentum = Impulse I. Impulse I =force x time. The only force acting on the system here is the force of gravity. i.e. weight for the two bodies (m1+m2)g. The time for which the force acts is 2to. So the impulse = (m1+m2)g x 2to. Ans. C. m g

4.

2

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6.

Moment of inertia is a scalar quantity. It can be added , if the axis of rotation is the same. The moment of inertia of the quarter sector is one-fourth of the moment of inertia of the disc as a whole. The moment of inertia of the disc about an axis passing through centre of mass perpendicular to the plane is 4MR2/2, since 4M is the mass of the disc. The moment of inertia of the sector = (1/4) 4MR2/2 = MR2/2. Ans. A

7.

If g1 is acceleration due to gravity at the surface of earth and g2 its value at an altitude h=R, we have g 1 =

GM R2

(1) and g 2 =

GM

=

(R + h ) 2

GM (R + R ) 2

=

GM 4R 2

=

g1 4

(2).

Dividing the two

equations, g1/g2 = 4. The periods of the pendulum at the surface of earth and at the altitude R are given T1= 2π 8.

T l l , T2= 2π ⇒ 1 = g1 g2 T2

g2 = g1

1 1 = ⇒ T2 = 2T1.. Ans.D 4 2

If N is the normal reaction at the highest point, mg weight, R radius of the track, then the centripetal force is given by mg − N =

mV 2 mV 2 ⇒ N = mg − , where V is velocity at highest R R

point..Here reaction N is maximum, when radius of the track R is minimum because v is constant. In the given choices R is minimum in fig. (A). Ans.A 9.

The force on the bottom of the cylinder F' - Force on the top (F) = Buoyant force (B). This gives force at the bottom F' = F+B. F = Thrust on the top surface πr 2 hρg . Buoyant force B = Weight of liquid displaced = Vρg. ⇒ F' = πr 2 hρg +Vρg = ρg( πr 2 h +V). Ans.D. This question can also be answered by simple reasoning, once we know the thrust at the bottom is sum of thrust at the top and buoyant force. The thrust at the top depends only on the height, density of liquid and g. The buoyant force will depend only on the volume of cylinder and density of the liquid. So the required equation should not contain the mass of the cylinder. This eliminates all the first three choices. 2

2

10. The energy of SHM is given by (1/2)mω A , where A is the amplitude. In both the given

equations maximum displacement (amplitude) is A. But the angular frequency of the first is ω, while that of the second is 2ω. This gives

E 2 ( 2ω) 2 = = 4 ⇒ E 2 = 4E1 .. Ans. C. E1 ω2

11. After two seconds each pulse would have travelled 4 cm and would arrive at the same point so

that their maximum will be at the same points. That is, each particle will be at the respective mean position i.e position of zero displacement. At mean position the energy of a simple harmonic particle is purely kinetic. Ans.B 12. The slope of an adiabatic curve is equal to the ratio of specific heats γxP. The slope of curve

2 is greater than that of 1. γ of monatomic gas is greater than that of diatomic gas. This would mean that curve 2 corresponds to monatomic and curve 1 diatomic. Oxygen is diatomic and He monatomic. Ans. B. 13. Let T be the temperature of the common junction of three rods at thermal equilibrium. Then

heat conducted/s between the left end and common junction= heat conducted per s between common junctions and the two right ends. i.e.

λA(T − 0) λA(90 − T ) λA(90 − T) , where λ = + x x x

is the thermal conductivity of the material of the rod, x length of each rod, A cross sectional area. This gives T = 600. Ans. B.

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14. Fraction of the immersed volume = density of body/density of liquid. This gives k1 =

density of Fe at 0o C o

density of Hg at 0 C

=

d (Fe ) density of Fe at 60o C d 0 ( Fe) = 60 . We know , k2 = o d 0 ( Hg) d density of Hg at 60 C 60 ( Hg )

d 0 ( Fe) = d 60 (1 + γ Fe 60) , d 0 ( Hg ) = d 60 (1 + γ Hg 60) . This gives

1 + 60 γ Fe k1 , . Ans.A. = k 2 1 + 60 γ Hg

15. We use the first law of thermodynamics dQ = dU+dW. It is given dQ is negative and dW= 0.

This would mean dU is negative. The internal energy decreases. Temperature of system decreases. Ans.A. 16. In choice (A), lines of force starting from a positive charge end at another positive charge.

Hence (A) is wrong. In choice (B) and choice (D) lines of force form closed loops. This cannot happen for positive charges. Therefore, choice (C) is correct. Ans.C. 17. Since a uniform electric field exists along the positive X-axis, a free positive charge will

move from A to B. We also know a positive charge moves from higher to a lower potential. This means potential at VA is greater than VB. Ans.B. 18. The right plate of the capacitor B is isolated. For charges to flow from A to B, and B to work

as a capacitor, the right plate will have to be earthed or connected by a wire to some other plate. Under the given condition B cannot attain any charge. Ans.A. 19. The equation for magnetic field due to a spiral of inner radius a and outer radius b will act as a

coil of width b-a and should contain both a and b. This rules out the choices (A) and (B). In choice (D) the power of current is IN. This will make the equation for magnetic field dimensionally incorrect. The correct choice, therefore, is (C). Ans.C. 20. The radius of a charged particle in a uniform magnetic field is given by r =

are same for both the particles. From figure rA is greater than rB.⇒ >mBvB.

mv . Here q and B qB

mA v A m B v B > ⇒ mAvA qB qB

Ans.B.

21. The mutual inductance between the two coils is maximum when the flux produced by one coil

links maximum with the other. The magnetic field produced by a coil is along the axis. So flux lines will come along the axis. The second coil will intercept maximum flux when it presents maximum cross sectional area to the flux. This happens in a situation shown in fig.(A). Ans.A. 22. We shall assume that the wires are uniform so that the length is proportional to mass and

proportional to resistance. If E is the emf of each cell, c the specific heat of the wire, then the heat

produced

[ NE ] 2 = 2mc∆T 2R

in

the

two

cases

respectively

will

be

[3E] 2 = mc∆T R

(1)

(2) Dividing the two equations we get N2 = 36 ⇒ N=6. Ans.B.

23. By Snell's law of refraction for media 1 and 4, µ1 sin θ1 = µ 4 sin θ4 ⇒

sin θ1 µ 4 = ⇒ µ 4 = µ1 sin θ4 µ1

since θ1 = θ 4 .Ans.D. 24. If we consider the prisms Q and R joined together, they form a parallel-sided glass. A ray of

light entering from P to this combination should go parallel after refraction through the

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combination Q and R. So the deviation of the ray after passing through P will remain the same even after passing through Q and R. Ans.C 25. The resultant intensity when two waves of intensity I1 and I2 superpose at a phase difference

θ

is

given

by

I = I1 + I 2 + 2 I1 I 2 cos θ .

and I B = I + 4I + 2 I 4I cos π = I

Using

this

I A = I + 4I + 2 I 4I cos π / 2 = 5I

⇒ I A -IB= 4I. Ans.B

26. The energy difference between quantum states goes on decreasing as En is proportional to

1/n2. The energy difference for infrared should be less than that for for ultraviolet radiation. If transition from 4 to 3 produces ultra-violet radiation, then the energy difference will be less only for transition from 5 to 4. Ans. D

27. The wavelength of Kα radiation λK is a constant for a given target material. The wavelength hc of radiation is related to accelerating voltage by the equation λC = . That is λ is inversely eV

proportional to V. increases. Ans.A.

When the accelerating voltage is increased, λC decreases. ⇒λK-λC

28. Electrons do not exist in the nucleus. Decay of neutron produces electron and proton. Ans.C. 29. At the steady state no current flows through the branch containing condenser. Considering 2V − V V the rest of the circuit, the current i = = . If we now consider the upper part of the R + 2R 3R VR loop and the part containing condenser, the charging potential of the condenser is V+ -V = 3R V . Ans.C. 3 30. The induced e.m.f when a conductor of length l moves in a magnetic field B with a velocity v

is Blv sinθ. Here the sides AD and BC are perpendicular to the field θ= 900 and hence will have induced e.m.f.s. The sides AB and CD are parallel to the field and hence the induced e.m.f will be Blv sin0 =0. The induced e.m.fs produce potential difference in the side AD and BC and hence an electric field. Ans.D

31. What ever may be the value of mean life or half life, the number of atoms remaining to

undergo disintegration should only decrease with time according to the exponential law N = N 0 e −λt .This rules out answers (A),(B) and (C). Ans.D. 32. If the current through R6 is independent of the value of R6, then the branches R1, R3, R4, R2

should form a balanced Wheat stone's bridge, with R6 taking the place of galvanometer resistance. For balance, we should have

R1 R 2 = ⇒ R 1 R 4 = R 2 R 3 Ans.C. R3 R 4

33. The bandwidth in Young's experiment is given by β =

Dλ i.e. β is proportion al to λ. When d

λ

decreases from 600 to 400 nm, the bandwidth decreases to 2/3. The number of bands in a given distance x = x/ λ increases to 3/2 times i.e. 12x3/2 = 18. Ans.B 34. Treat the given loops as two separate loops one in the YZ plane and the other in the XY plane.

The magnetic field produced by the loop is perpendicular to the plane of the loop. So it should be respectively along +X axis and +Z axis from the current direction. Our answer

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should, therefore, contain unit vectors along the X-axis ˆi and Z-axis kˆ .The unit vector in the direction of ˆi + kˆ is the unit vector in the direction of magnetic field. This is

1 2

(ˆi + kˆ ) .

Ans.D. 35. The equation to displacement of S.H.M is given by A sinθ or A sin ωt, where A is amplitude.

We know sinθ increases fast first and then increases slowly. ( For example sin 0 = 0, sin 300= 1/2 and sin 900 =1). So the displacement increases from 0 to A/2 quicker than that from A/2 to A. So time necessary to cover the first half of the amplitude is less than that for the second half. ( If T is the period of S.H.M, we can easily prove that T1 = T/12 and T2 = T/6.) Ans.A

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