Physics Spectrum

(b) [M- 1L5T 2 ] (d) [ML5T]

2. A man throws the balls with the same speed vertically upwards one after the other at an interval of 2 s. What should be the speed of the throw, so that more than two balls are in the sky at any time? (take g = 9.8 m/s) (a) Any speed less than 19.6 m/s (b) Only with speed 19.6 m/s (c) More than 19.6 m/s (d) Atleast 9.8 m/s

given by y = 5 sin (4.0t - 0.02x ) where, y and x are expressed in cm and time in seconds. Calculate the amplitude, frequency and velocity of the wave. (a) 8 cm, 0.8673 cycle s - 1, 200 cms - 1 (b) 5 cm, 0.673 cycle s - 1, 200 cms - 1 (c) 5.8 cm, 0.673 cycle s - 1, 250 cms - 1 (d) None of the above

6. A ball of 0.5 kg moving with a speed of 12 m/s strikes a hard wall at an angle of 30° with the wall. It is reflected with the same speed and at the same angle. If the ball is in contact with the wall for 0.25s, the average force acting on the wall is (a) 48 N (c) 12 N

30°

(a) [ML5T - 2 ] (c) [ML-5T - 1 ]

5. Equation of a transverse wave travelling in a rope is

30°

aö q ÷ = b , where, V V2 ø p = pressure, V = volume and q = absolute temperature. If a and b are constants, then dimensions of a will be æ è

1. An equation is given as ç p +

(b) 24 N (d) 96 N

3. A stone is attached to one end of a string and rotated

7. A bomb of mass 30 kg at rest explodes into two

in a vertical circle. If string breaks at the position of maximum tension, then it will break at

pieces of masses 18 kg and 12 kg. The velocity of 18 kg mass is 6 ms -1. The kinetic energy of the other mass is

A

(a) 256 J (c) 524 J C

D

(b) B (d) D

4. A body of 1 kg explodes into three fragments. The ratio of their masses is 1 :1 : 3. The fragments of same mass move perpendicular to each other with the speed 30 m/s, while the heavier part remains in the intial direction. The speed of heavier part is 10 m/s 2 (c) 20 2 m/s (a)

8. A wheel of bicycle is rolling without slipping on a level road. The velocity of the centre of mass is v cm , then true statement is

B (a) A (c) C

(b) 486 J (d) 324 J

(b) 10 2 m/s (d) 30 2 m/s

A

cm

Vcm

B (a) the velocity of point A is 2 vcm and velocity of point B is zero (b) the velocity of point A is zero and velocity of point B is 2 vcm (c) the velocity of point A is 2 vcm and velocity of point B is - vcm (d) the velocities of both A and B are Vcm

67

9. Imagine a new planet having the same density as that of Earth but it is 3 times bigger than the Earth in size. If the acceleration due to gravity on the surface of Earth is g and that on the surface of the new planet is g’, then, g 9 (d) g ¢ = 27 g

(a) g ¢ = 3g

(b) g ¢ =

(c) g ¢ = 9g

10. The ratio of the radii of gyration of a circular disc about a tangential axis in the plane of the disc and of a circular ring of the same radius about a tangential axis in the plane of the ring is (a) 2 : 3 (c) 5 : 6

(b) 2 : 1 (d) 1 : 2

11. The escape velocity of a body on the surface of the Earth is 11.2 km/s. If the Earth’s mass increases to twice its present value and the radius of the Earth becomes half, then the escape velocity would become

(d) 5.6 km/s

incompressible liquid of mass M and closed at both the ends. The tube is then rotated in a horizontal plane about one of its end with a uniform angular velocity w. The force exerted by the liquid at the other end is ML2 w 2 ML2 w 2 (d) 2 (b)

(c) MLw 2

16 l 81 4 (d) l 3 (b)

respectively are given small linear displacement in one direction at the same time. They will again be in the same phase when the pendulum of shorter length has completed oscillations. (b) 1 (d) 3

1 6 ature of sink is reduced by 62°C, then its efficiency will be doubled. Temperature of the source is

15. An engine has an efficiency of . When the temper-

(a) 124°C

68

(b) 37°C

(c) 62°C

(b) circular clockwise

(c) elliptical anti-clockwise

(d) elliptical clockwise

17. A mass of 2.0 kg is put on a flat pan attached

m

to a vertical spring fixed on the ground as shown in the figure. The masses of the spring and the pan are negligible. When pressed slightly and released the mass executes a simple harmonic motion. The spring constant m is 200 N/m. What should be the minimum amplitude of the motion, so that the mass gets detached from the pan? (take g = 10 m /s2 ) (a) 8.0 cm (b) 10.0 cm (c) Any value less than 12.0 cm(d) 4.0 cm

(a) 120 W (c) 304 W

(b) 240 W (d) 320 W

19. A point performs simple harmonic oscillation of period T and the equation of motion is given by x = a sin (wt + p /6). After the elapse of what fraction of the time period the velocity of the point will be equal to half of its maximum velocity? T 8

(b)

T 6

(c)

T 3

(d)

T 12

constant 2) between the plates has capacitance C. If oil is removed, the capacitance of capacitor becomes (a) 2 C C (c) 2

(b) 2C C (d) 2

21. A hospital uses an ultrasonic scanner to locate

14. Two simple pendulums of lengths 0.5 m and 2.0 m,

(a) 5 (c) 2

(a) circular anti-clockwise

20. A parallel plate capacitor with oil (dielectric

corresponding to maximum intensity of radiation emitted by a source at temperature 2000 K is l, then what is the wavelength corresponding to maximum intensity of radiation at temperature 3000 K? 2 l 3 81 (c) l 16

motions given by, pö æ x = a sin (wt + d) and y = a sin ç wt + d + ÷ act on a è 2ø particle simultaneously. Then, the motion of particle will be

(a)

13. The wavelength

(a)

harmonic

power is 60 W and temperature of surrounding is 227° C. If the temperature of the black body is changed to 1227° C, then its radiating power will be

12. A tube of length L is filled completely with an

MLw 2 2

simple

18. For a black body at temperature 727° C, its radiating

(a) 44.8 km/s (b) 22.4 km/s (c) 11.2 km/s (remain unchanged)

(a)

16. Two

(d) 99°C

tumour 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. The wavelength of sound in tissue is close to (a) 4 ´ 10 -4 m (c) 4 ´ 10

-3

m

(b) 8 ´ 10 -4 m (d) 8 ´ 10 -3 m

22. A charged wire is bent in the form of a semi-circular arc of radius a. If charge per unit length is l coloumb/m, then the electric field at the centre O (a)

l

2 pa 2 e 0 l (c) 2 pe0 a

(b)

l 4p 2 e 0 a

(d) zero

AIPMT RIDE 1 23. Two trains move towards each other with the same speed. The speed of sound is 340 m/s. If the height of the tone of the whistle of one of them heard on the 9 other changes times, then the speed of each train 8 should be (a) 20 m/s

(b) 2 m/s

(c) 200 m/s

(d) 2000 m/s

24. Three concentric spherical shells have radii a, b and c (a < b < c) and have surface charge densities s, - s and s respectively. If V A , VB and VC denote the potentials of the three shells, then for c = a + b, we have (a) VC = VA ¹ VB (c) VC ¹ VB ¹ VA

-1

towards a hill sounds a horn of frequency 600 Hz. If the velocity of sound in air is 330 ms -1, then the frequency of reflected sound as heard by driver is (a) 550 Hz (c) 720 Hz

(a) in an elliptical orbit (c) along a parabolic path

(b) 555.5 Hz (d) 500 Hz

31. Two bar magnets having same geometry with magnetic moments M and 2M, are firstly placed in such a way that their similar poles are on the same side, then its period of oscillation is T1. Now, the polarity of one of the magnets is reversed the time period of oscillations becomes T2 . Then, (a) T1 < T2 (c) T1 = T2

(b) T1 > T2 (d) T2 = ¥

placed at right angles to each other. The wire AOB carries an electric current I1 and COD carries a current I2 . The magnetic field on a point lying at a distance d from O, in a direction perpendicular to the plane of the wire AOB and COD, will be given 1/ 2

m 0 æ I1 + I2 ö ç ÷ 2p è d ø m (c) 0 ( I1 + I2 ) 2 pd

cross-sectional area (L, A) [2L, ( A / 2)], [(L / 2), 2 A]. In which case is the resistance minimum?

(b)

33. For a given incident ray as shown in figure, the condition of total internal reflection of the ray will be satisfied if the refractive index of block will be

(a) It is the same in all three cases (b) Wire of cross-sectional area 2 A (c) Wire of cross-sectional area A 1 (d) Wire of cross-sectional area A 2

45° incident ray

27. Two batteries, one of emf 18 V and internal

r

resistance 2 W and the other emf of 12 V and internal resistance 1 W are connected as shown in the figure. The voltmeter V will record a recording of V

3+1 2 3 (c) 2

18 V 1W

2 +1 2 7 (d) 6 (b)

34. A car is fitted with a convex side view mirror of focal

12 V (c) 14 V

(d) 18 V

28. Fuse wire is a wire of (a) low resistance and low melting point (b) low resistance and high melting point (c) high resistance and high melting point (d) high resistance and low melting point

29. A 10 eV electron is circulating in a plane at right angles to a uniform field of magnetic induction 10- 4 Wb/m2 . The orbital radius of the electron is (a) 12 cm (c) 11 cm

q

(a)

2W

(b) 30 V

m0 2 ( I1 + I22 )1/ 2 2 pd m (d) 0 ( I12 + I22 ) 2 pd

(a)

26. There are three copper wires of length and

(a) 15 V

(b) in a circular orbit (d) along a straight line

32. Two identical conducting wires AOB and COD are

(b) VC = VB ¹ VA (d) VC = VB = VA

25. The driver of a car travelling with speed 30 ms

same magnetic field is maintained, then the electron move

(b) 16 cm (d) 18 cm

30. A beam of electrons passes undeflected through mutually perpendicular electric and magnetic fields. If the electric field is switched off and the

length 20 cm. A second car 2.8 m behind the first car is overtaking the first car at a relative speed of 15 m/s. The speed of the image of the second car as seen in the mirror of the first one is (a)

1 m/s 15

(c) 15 m/s

(b) 10 m/s (d)

1 m/s 10

35. A body is located on a wall. Its image of equal size is to be obtained on a parallel wall with the help of a convex lens. The lens is placed at a distance d ahead of second wall, then the required focal length will be (a) only d/4 (b) only d/2 (c) more than d/4 but less than d/2 (d) less then d/4

69

is 2 cm. Focal length of lens when immersed in a liquid of refractive index of 1.25 will be

capacitor C in series produce oscillations of frequency F. If L is doubled and C is changed to 4C, the frequency will be

(a) 10 cm (c) 5 cm

(a) f / 4 (c) f / 2 2

36. Focal length of a convex lens of refractive index 1.5 (b) 2.5 cm (d) 7.5 cm

(b) 8f (d) f / 2

37. The photoelectric work function for a metal

44. The logic circuit shown below has the input

surface is 4.125 eV. The cut-off wavelength for this surface is

waveforms A and B. Pick out the correct output waveforms.

(a) 4125 Å

(b) 3000 Å

(c) 6000 Å

(d) 2062.5 Å

A y

38. The work function for metals A, B and C are 1.92 eV,

B

2.0 eV and 5 eV respectively. According to Einstein’s equation, the metals which will emit photoelectrons for a radiation of wavelength 4100 Å is/are

Input A

(a) None (b) A only (c) A and B only (d) All the three metals

Input B

39. The total energy of electron in the ground state of

(a)

hydrogen atom is - 13.6 eV. The kinetic energy of an electron in the first excited state is (a) 3.4 eV (c) 13.6 eV

(b)

(b) 6.8 eV (d) 1.7 eV

40. The binding energy of deutron is 2.2 MeV and that of

(c)

4 2 He

is 28 MeV. If two deuterons are fused to form 4 one 2 He, then the energy released is (a) 25.8 MeV (c) 19.2 MeV

(b) 23.6 MeV (d) 30.2 MeV

41. The radius of germanium (Ge) nuclide is measured

(d)

45. For a given circuit of ideal p-n junction diode, which of the following is correct?

to be twice the radius of 94 Be. The number of nucleon in Ge are (a) 73 (c) 75

Diode

R

(b) 74 (d) 72

42. Zener diode is used for (a) producing oscillations in an oscillator (b) amplification (c) stabilisation (d) rectification

V (a) In forward biasing, the voltage across R is V (b) In reverse biasing, the voltage across R is V (c) In forward biasing, the voltage across R is 2 V (d) In reverse biasing, the voltage across R is 2 V

43. A transistor-oscillator using a resonant circuit with an inductor L (of negligible resistance) and a

Answers 1. 11. 21. 31. 41.

(a) (b) (a) (a) (d)

2. 12. 22. 32. 42.

(c) (a) (c) (b) (c)

3. 13. 23. 33. 43.

(b) (a) (a) (c) (c)

4. 14. 24. 34. 44.

(b) (c) (a) (a) (a)

5. 15. 25. 35. 45.

(b) (d) (c) (b) (a)

6. 16. 26. 36.

(b) (b) (b) (c)

7. 17. 27. 37.

(b) (b) (c) (b)

8. 18. 28. 38.

(a) (d) (d) (c)

9. 19. 29. 39.

(a) (d) (c) (a)

Detailed solutions of these questions are available on http://www.arihantbooks.com/Physics%20Spectrum.pdf

70

10. 20. 30. 40.

(c) (d) (b) (b)

Dimensions of p = dimensions of p= \

2

V2

a V2

- 1 - 2

a = pV = [ML

a

T

][L3 ]2 = [ML5T - 2 ]

2. (c) Time taken by ball to reach maximum height, v = u - gt at maximum height, final speed is zero i.e. v = 0 So, u = gT u T= g In 2 s, u = 2 ´ 9.8 = 19.6 m/s If man throws the ball with velocity of 19.6 m/s, then after 2 s it will reach the maximum height. When he throws 2nd ball, 1st is at top. When he throws third ball, 1st will come to ground and 2nd will be at the top. Therefore only 2 balls are in air. If he wants to keep more than 2 balls in air he should throw the ball with a speed greater than 19.6 m/s.

3. (b) When string makes an angle q with the vertical in a vertical circle, then

or

T - mg cos q =

mv 2 l

T = mg cos q +

mv 2 l

qmg cos q B mg mg sin q Tension is maximum when cos q = + 1 i.e, q= 0 Thus, q is zero at lowest point B. At this point tension is maximum. So, string will break at point B.

4. (b) Let v be the velocity and q the direction of the third peice as shown.

m B

C

5. (b) Given, y = 5 sin ( 4.0 t - 0.02 x ) comparing this with the 2p ù é standard equation of wave motion, y = A sin ê2 pft x where l úû ë A, f and l are amplitude, frequency and wavelength respectively. Thus, amplitude, A = 5 cm, 2 pf = 4 4 frequency, f = = 0.673 cycle s -1 2p 2p Again, = 0.02 l 2p or wavelength, l = = 100p cm 0.02 Velocity of the transverse wave, v = f ´ l 4 2p = ´ = 200 cms - 1 2 p 0.02

60° O

As the magnitudes of OA and OB are equal, 60° the components of OA and OB along the wall are equal and in the same direction, while those perpendicular to the wall are A equal and opposite. Thus, the change in momentum is due to only the change in direction of the perpendicular components. Hence, DP = OB sin 30 ° - ( - OA sin 30 ° ) = mv sin 30 ° - ( - mv sin 30 ° ) = 2 mv sin 30 ° Its time rate will appear in the form of average force acting on the wall. \ F ´ t = 2 mv sin 30 ° 2 mv sin 30 ° F= t Given,

q

B

30°

q

O

v = 10 2 m/s So, speed of heavier part of a fragment is 10 2 m/s.

The vector OA represents the momentum of the object before the collision and the vector OB that after the collision. The vector AB represents the change in momentum of the object DP

D

O

cos q = sin q q = 45° Thus, Ð AOC = Ð BOC = 180 ° - 45° = 135° Putting the value of q in Eq. (i), we get 3 mv 30 m = 3 mv cos 45° = 2

6. (b)

A

C

or

30°

1. (a) According to principle of homogeneity of dimensions,

A m

3m

Equating the momentum of the system along OA and OB to zero, we get ... (i) m ´ 30 - 3 m ´ v cos q = 0 ... (ii) m ´ 30 - 3 m ´ v sin q = 0 From Eqs (i) and (ii), we get 3 mv cos q = 3 mv sin q

m = 0.5 kg, v = 12 m/s, t = 0.25 s

q = 30 ° Hence, Net average froce on well 2 ´ 0.5 ´ 12 sin 30 ° F= = 24 N 0.25

7. (b) Applying conservation of linear momentum m1 u1 = m 2 u 2 Here, m1 = 18 kg, m 2 = 12 kg, u 1 = 6 ms -1 u 2 =? \

18 ´ 6 = 12 u 2 18 ´ 6 u2 = = 9 ms - 1 12 Thus, kinetic energy of 12 kg mass 1 1 K 2 = m 2 u 22 = ´ 12 ´ 9 2 = 6 ´ 81 = 486 J 2 2 Þ

71

8. (a) Velocity of point A is

9. (a) The acceleration due to gravity on the new planet can be found using the relation GM

...(i)

R2

4 ´ R 3r, where r is density of the planet 3 Thus, Eq. (i) becomes 4 G ´ pR 3r 4 3 \ g= = G ´ ´ pRr 3 R2 Þ g µR g ¢ R¢ \ = g R g ¢ 3R Þ = =3 g R But M =

l mT = constant or æT ö l 2 = l1 ç 1 ÷ è T2 ø

axis in their planes are respectively 5 Id = Md R 2 4 3 Ir = M r R 2 2

or \

Kd = Kr

Id M ´ r Ir Md 5 6

11. (b) Escape velocity on the Earth’s surface is given by 2 GMe Re

M ¢e = 2 Me and R ¢e =

Re 2

ves = 11.2 km/s \

v ¢es 2 Me Re = ´ = 4 =2 ves Me Re / 2

\

v ¢es = 2 ves = 2 ´ 11.2 = 22.4 km/s

12. (a) Let the length of a small element of tube be dx. Mass of this element dm =

72

14. (c) For the pendulum to be again in the same phase, there should be difference of one complete oscillation. If smaller pendulum complete n oscillations the larger pendulum will complete ( n - 1) oscillations.

or

n l1 = ( n - 1) l 2

or

n = n -1

or

n =2 n =2n -2 n -1

M dx L

\

l2 2 .0 = l1 0.5

n =2

15. (d) Efficiency of engine is given by h =1-

where, G is gravitational constant, Me and Re are the mass and radius of Earth respectively. ¢ ves Me¢ R = ´ e \ ves Me R ¢e but,

Here, T1 = 2000 K , T2 = 3000 K , l1 = l Then, wavelength corresponding to maximum intensity of 2000 2 radiation, l 2 = l ´ = l. 3000 3

Id : Ir = 5 : 6

¢ = ves

l1T1 = l 2T2

So, time period of n oscillations of first = time period of ( n - 1) oscillation of second i.e. nT1 = ( n - 1) T2 l l n2 p 1 = ( n - 1) 2 p 2 g g

I M

Id ( 5 / 4 ) Md R 2 M r = ´ = Ir ( 3 / 2 ) M r R 2 Md

MLw2 2

13. (a) Wien’s displacement law is given by

10. (c) Moment of inertia of a disc and circular ring about a tangential

\

where, M is mass of filled liquid and L is length of tube. Force on this element dF = dM ´ nw2 F M 2 2 ò0 dF = L w ò0 xdx M 2 é L2 ù or F= w ê ú L êë 2 úû F=

Þ g ¢ = 3g So, acceleration due to gravity g ¢ on the new planet is 3g.

I = MK 2 Þ K =

F+dF r

Thus, the velocity of point A is 2 vcm and velocity of point B is zero.

But,

w

F

= 2 vcm Velocity of point B is, v B = vcm - Rw = vcm - vcm = 0

g=

L dx

(Q vcm = Rw)

v A = vcm + Rw = vcm + vcm

\

T2 T1

T2 1 5 =1- h =1- = T1 6 6

...(i)

In other case, T2 - 62 2 2 =1- h =1- = T1 6 3

...(ii)

Using Eq. (i), T2 - 62 =

2 2 6 T1 = ´ T2 3 3 3

1 T2 = 62 5 T2 = 310 K = 310 - 273 °C = 37 °C 6 6 Here, T1 = T2 = ´ 310 = 372 K = 372 - 273 = 99 °C 5 5 Thus, initial temperature of the source is T1 = 99 °C or

aw pö æ = awcos ç wt + ÷; è 2 6ø

16. (b) Two simple harmonic motions can be written as, ... (i)

x = a sin (wt + d) pö æ y = a sin ç wt + d + ÷ è 2ø

and

... (ii)

x 2 + y 2 = a 2 [sin 2 ( wt + d) + cos 2 ( wt + d)] x2 + y2 = a 2

(Qsin 2 q + cos 2 q = 1)

aY

This is the equation of a circle At wt + d = 0 , x = 0, y = a p At wt + d = , x = a, y = 0 2 At wt + d = p, x = 0, y = - a 3p At wt + d = , x = - a, y = 0 2 At wt + d = 2 p, x = 0, y = a

–a

O

wt+dx a

–a

spring F = ka Restoring force is balanced by weight Mg of block. For mass to execute simple harmonic motion of amplitude a, Mg \ Ka = Mg or a = K where, M = 2 kg, K = 200 N/m g = 10 m/s2 2 ´ 10 10 10 a= = m= ´ 100 cm 200 100 100 a = 10 cm

...(i)

...(ii)

Here, P1 = 60 W,

T1 = 727 °C = 1000 K T0 = 227 °C = 500 K, T2 = 1227 °C = 1500 K Substituting in Eq. (ii) Radiated power by a block body 4

(1000 ) - ( 500 )

´ 60 =

\

l=

1.7 ´ 10 3 4.2 ´ 10 6

= 4 ´ 10 -

4

m

22. (c) Considering symmetric elements each of length dl at A and B. We note that electric field perpendicular to PO are cancelled and those along PO are added. The electric field due to an element of length dl (= adq) along PO, 1 dq dE = cos q (Qdl = adq) 4pe 0 a 2 =

1 ldl 1 l( adq) cos q = cos q 4pe 0 a 2 4pe 0 a 2

E =ò

p/ 2 - p/ 2

= 2. ( 500 )4

æ 3 4 - 1ö ´ç ÷ ´ 60 ( 500 ) è 2 4 - 1ø 4

19. (d) When a object performs SHM of period T and equation of

dx pö æ = a wcos ç wt + ÷ è dt 6ø

the distance travelled by the wave during the time, any one particle of the medium completes one vibration about the mean v position. Wavelength of a wave is l = n

Net electric field at O

80 = ´ 60 = 320 W 15 pö æ motion is given by x = a sin ç wt + ÷ è 6ø

21. (a) Wavelength of a wave is the length of one wave. It is equal to

Wavelength of sound in tissue is

of radiations emitted by the body, not only depends upon the temperature of the body but also on the temperature of the surrounding. The power radiated by the body is given by, ... (i) P = s (T 4 - T04 )

(1500 )4 - ( 500 )4

(K = 2 )

n = frequency of wave (sound) Here, v = 1.7 ´ 10 3 m/s, n = 4.2 ´ 10 6 Hz

18. (d) Boltzmann corrected Stefan’s law and stated that the amount

where, T0 is the absolute temperature of the surrounding. P2 æ T24 - T04 ö =ç \ ÷ P1 è T14 - T04 ø

...(ii)

where v = velocity of wave (sound)

Hence, minimum amplitude of the motion should be 10 cm, that the mass gets detached from the pan.

v=

Ke 0 A d where, e 0 = electric permittivity of free space K = dielectric constant A = area of each plate of capacitor d = distance between two plates When dielectric (oil) is removed, so capacitance, e A C0 = 0 d Comparing Eqs. (i) and (ii) C = KC 0 C C Net capacitance becomes C 0 = = K 2 C=

17. (b) Let the minimum amplitude of SHM is a restoring force on

4

20. (d) The capacitance of a parallel plate capacitor with dielectric (oil) between its plates is

Thus, it is obvious that motion of particle is traversed in clockwise direction.

P2 =

p p p = Þ wt = 6 3 6 2p p T t= , t= T 6 12 Time period of the velocity of the point will be equal to half it maximum velocity is T/12 wt +

or y = a cos (wt + d) Squaring and adding Eqs. (i) and (ii), we get or

1 pö æ = cos ç wt + ÷ è 2 6ø

dE = 2 ò

p/ 2 0

1 l a cos qdq 4pe 0 a2

1 l 1 l l 2 [sin q]p/ . .1= 0 = 2. 4pe 0 a 4pe 0 a 2 pe 0 a

23. (a) According to Doppler’s effect, whenever there is a relative motion between a source of sound and listner, the apparent frequency of sound heard by the listner is different from the actual frequency of sound emitted by the source. Apparent frequency of sound wave heard by the listner is v - vl v¢ = ´v v - vs where, V is actual frequency of sound emitted by the source, Vs is the velocity of source and Vl is velocity of listner. According to

73

9 v and source and observer are moving in 8 opposite direction with same speed (say V), then apparent æ v + vl ö frequency v ¢ = v ´ ç ÷ è v + vs ø problem v ¢ =

\

9 340 + v v=v´ 8 340 - v

\

17 v = 340 or v =

340 = 20m/s 17

r ( L / 2 ) rL = 2A 4A

(ii) Length = L, area = A Putting in Eq. (i)

s s (a - b + c ) = (2 a ) e0 e0

rL A

(iii) Length = 2 L, area = A / 2 2L 4rL Putting in Eq. (i), R = r = A/2 A As it is understood from above, resistance is minimum only in option (b).

1 s 4pa 2 1 s 4pb2 1 s 4pc 2 + . 4pe 0 a 4pe 0 b 4pe 0 c =

R=

R=

24. (a) Here, potential at point A VA =

Putting in Eq. (i)

(Qc = a + b)

27. (c) It is clear that the two cells oppose each other hence, the effective emf inclosed in circuit is 18 - 12 = 6 V and net resistance is 1 + 2 = 3 W. The current in circuit will be in direction of arrow shown in figure.

c

V

b a

18 V 2W 12 V

Potential at B, 2

VB =

2

1 s 4pa 1 s 4pb 1 s 4pc . + . 4pe 0 a 4pe 0 b 4pe 0 c

2

1W effective emf 6 I= = =2 A total resistance 3

Potential at c = and

VC =

s e0

æ a2 ö s - b +c÷ = (2 a ) ç c è ø e0

(Qc = a + b)

1 s4pa 2 1 s 4pb2 1 s 4pc 2 . + 4pe 0 c 4pe 0 c 4pe 0 c ö s s æ a 2 b2 +c÷ = (2 a ) ç e0 è c c ø e0

The potential difference across V will be same as the terminal voltage of either cell. Since current is drawn from the cell of 18 V, hence, V1 = E1 - i r1 = 18 - (2 ´ 2 ) = 18 - 4 = 14 V Similarly, current enters in the cell of 12 V, hence V2 = E 2 + i r2 = 12 + 2 ´ 1 = 14 V

(Qc = a + b)

28. (d) The electric fuse is a device which is used to limit the current in

25. (c) Whenever there is a relative motion between a source of

an electric circuit. Thus, the use of fuse safeguards the circuit and the appliances connected in the circuit from being damaged. It is always connected with the live (or phase) wire. The fuse wire is a short piece of wire made of a material of high resistance and low melting point so, that it may easily melt due to over heating when excessive current passes through it.

=

Hence, VA = VC ¹ VB sound and the observer (listner), the frequency of sound heard by the observer is different from the actual frequency of sound emitted by source.

29. (c) If charged particle is moving perpendicular to the direction of

hill

Case I Case II

n · ¾ ¾ ¾ ¾-¾ ® ... | n ¢ 1 30 ms

S

0

n ¢¢ · ¾ ¾ ¾¾ ® ¬¾¾ n ¢ -1 S

30 ms

V n V - 30 V + 30 For Case II, n ¢¢ = n¢ V From Eqs. (ii) and (i), we get, V + 30 360 n ¢¢ = n= ´ 600 = 720Hz V - 30 300 For Case I,

n¢ =

... (i) ... (ii)

26. (b) The relation between length and area is R=

rL A

...(i)

r being specific resistance is the proportionality constant and depends on nature of material. L (i) length = , area = 2 A 2

74

B, it experiences a maximum force which acts perpendicular to the direction of B as well as V. Hence, this force will provide the required centripetal force and the charged particle will describe a circular path in the magnetic field of radius r, mv 2 = qB r Now, KE of electron = 10 eV 1 mV 2 = 10 eV 2 1 \ ´ ( 9.1 ´ 10 - 31 ) v 2 = 10 ´ 1.6 ´ 10 - 19 2 2 ´ 10 ´ 1.6 ´ 10 - 19 v2 = 9.1 ´ 10 - 31 v 2 = 3.52 ´ 1012 v = 1.88 ´ 10 6 m Now, radius of circular path, mv 9.1 ´ 10 - 31 ´ 1.88 ´ 10 6 r= = = 11 cm qB 1.6 ´ 10 - 19 ´ 10 - 4

30. (b) If both electric and magnetic fields are present and perpendicular to each other and the particle is moving perpendicular to both of them with Fe = Fm. In this situation E ¹ 0 and B ¹ 0. But if electric field becomes zero, then only force due to magnetic field exists. Under this force the charge moves along a circle.

\

cos r = 1 - sin 2 r = 1 -

Thus, Eq. (i) becomes, 1

m>

1-

31. (a) The time period of bar magnet I MH

T = 2p

\

I

T1 = 2 p

\

= 2p

M1H

I 3 MH

I I = 2p M 2H MH

T2 = 2 p

m=

...(ii)

From Eqs. (i) and (ii),

O in a direction perpendicular to the plane ABCD due to currents through AOB and COD are perpendicular to each other is

A

P

q I2

D

34. (a) According to mirror formula, we get 1 1 1 1 du 1 dv + = Þ =0 u v f u 2 dt v 2 dt du - v 2 æ du ö = ç ÷ dt u 2 è dt ø v f = u u -f

But, 2

\

2ù

éæm 2 I ö æm 2 I ö B = B12 + B22 = ê ç 0 1 ÷ + ç 0 2 ÷ ú è ø è 4p d ø ú 4 p d êë û

1/ 2

=

m 0 2 2 1/ 2 ( I1 + I2 ) 2pd

33. (c) For total internal reflection to take place, angle of incidence > critical angle i.e. q>C sin q > sin C 1 sin C = m 1 m

1 cos r

...(i)

...(i)

æ1 1 1 ö = ( l m g - 1) ç ÷ fl è R1 R2 ø

...(i)

öæ1 1 æm g 1 ö =ç - 1÷ ç ÷ fl è m l ø è R1 R2 ø

...(ii)

or

(m - 1) fl = g fa æ m g ö - 1÷ ç è ml ø

fl (1.5 - 1) 1/2 5 = = = fa æ 1.5 ö 1/ 5 2 - 1÷ ç è 1.25 ø Focal length of a conven lens when immersed in a liquid, 5 5 \ fl = fa = ´ 2 = 5 cm 2 2

From Snell’s law, sin 45° =m sin r sin r =

...(i)

Focal length of convex lens when immersed in liquid of refractive index m l

From Eqs. (i) and (ii) 1 (m - 1) fa = g 1 æm g ö - 1÷ ç fl è m l ø

From figure, q = 90 ° - r

m>

Given, For equal size image

1 1 1 = f v u v =d

æ1 1 1 ö = (m g - 1) ç ÷ fa è R1 R2 ø

2

i.e.

2

dv 0.2 1 æ f ö du æ ö -1 =-ç =ç ÷ ÷ ´ 15 = ms è u - t ø dt è - 2.8 - 0.2 ø dt 15

36. (c) Focal length of convex lens

I1 B

sin ( 90 - r ) >

2m 2

v = u =d By sign convention, u = - d 1 1 1 d or f = \ = + f d d 2

d

So,

1 =1 2

3 2

32. (b) The magnetic field induction at a point P, at a distance d from

O

or m 2 -

35. (b) The lens formula can be written as

T1 < T2

C

1

1 2m 2

Refractive index of a block.

...(i)

When opposite poles of magnets are placed on same side, then net magnetic moment. M2 = 2 M - M = M \

1

m2 = 1-

where, M = magnetic moment of magnet I = moment of inertia H = horizontal component of magnetic field When the same poles of magnets are placed on same side, then net magnetic moment M1 = M + 2 M = 3 M

1 2m 2

1 2m

37. (b) The maximum wavelength above which no photoelectron can emit from metal surface is called cutt-off wavelength and is given by

75

R = R0 A1/ 3

hc Cutt -off wave length hc or Cut-off wavelength = Work function hc \ l0 = W0 Work function =

h = 6.6 ´ 10 -

Given,

34

R1 æ A1 ö =ç ÷ R2 è A2 ø ...(i)

Substituting the given values in Eq. (ii), Å = 3 ´ 10

- 7

m

38. (c) Work function for wavelength of 4100 Å is - 34

´ 3 ´ 10 hc 6.62 ´ 10 = l 4100 ´ 10 - 10

= 4.8 ´ 10 - 19 J =

4.8 ´ 10 - 19 1.6 ´ 10

42. (c) Zener diode is a silicon crystal diode having an reverse current characteristic which is particularly suitable for voltage regulatory purposes. Due to this characteristic, it is used as voltage stabiliser in many applications in electronics appliances.

Cut off vave length to be used l 0 = 3000 Å

W0 =

A2 = 9 ´ (2 )3 = 72

Hence,

Thus in germanium (Ge) nucleus number of nucleons is 72.

´ 3 ´ 10 8

4.125 ´ 1.6 ´ 10 - 19

- 19

43. (c) In a series L-C circuit, frequency of LC oscillations is given by

8

eV = 3 eV

f= or

fµ

WA = 1.92 eV WB = 2 .0 eV WC = 5eV Since, WA < W WB < W Hence, A and B will emit photoelectrons

39. (a) The energy of hydrogen atom when the electron revolves in 2

f 2 L ´ 4C = = 8 f2 LC

\

f2 = f / 2 2

44. (a) Truth Table

40. (b) The reaction can be written as 1H

2

4

+ 1H ¾® 2He + energy

The energy released in the reaction is difference of binding energies of daughter and parent nuclei. Hence, energy released, = binding energy of 2He 4 - 2 ´ binding energy of 1H2 = 28 - 2´2.2 = 23.6 MeV

41. (d) Let radius of 94 Be nucleus be r. Then radius of germanium (Ge) nucleus will be 2 r. Radius of a nucleus is given by

76

A

B A¢ B ¢ y = A ¢ + B ¢

1 0 0 1

1 0 1 0

= eV

n In the ground state, n = 1 - 13.6 E= = - 13.6 eV 12 - 13.6 for n = 2, E = = - 3.4 eV 22 So, kinetic energy of electron in the first excited state (i.e. for n = 2) is KE = - E = - ( - 3.4 ) = 3.4 eV 2

LC L 2C 2 L1C1

Given, L1 = L, C1 = C, L 2 = 2 L, C 2 = 4C, f1 = f

nth orbit is - 13.6

1 2p LC 1

f1 = f2

Now,

E=

(Q A1 = 9 )

3

W0 = 4.125 eV = 4.125 ´ 1.6 ´ 10 - 19 J

l0 =

1/ 3

9 æ 1ö ç ÷ = è2 ø A2

or

C = 3 ´ 10 m/s

34

r æ 9 ö =ç ÷ 2 r è A2 ø

Þ

J-s

8

6.6 ´ 10 -

1/ 3

0 1 1 0

0 1 0 1

1 0 0 0

A

A

y B

B

Output waveform 1

1 0

0

0

0

0

45. (a) In forward biasing the diode conducts. For ideal junction diode, the forward resistance is zero. Therefore, entire applied voltage occurs across resistance R i.e. there occurs no voltage drop. While in reverse biasing, the diode does not conduct. So, it has infinite resistance, thus, voltage across R is zero in reverse biasing.

Paper 1 with the horizontal into a uniform gravitational field. The slope of the trajectory of the particle varies not according to which of the following curves?

(d) the centripetal acceleration is given by a = T sina where, T = tension in the string

5. Consider the situation given below.

t

(d)

Slope

t

(b)

x

Slope

(c)

Slope

(a)

Slope

y

x

2. The half lives of two radioactive samples A and B are 15 min and 30 min, respectively. If A and B have same number of particles, then ratio of residual particles in the samples A and B, (a) after 60 min is 1/8 (c) after 80 min is 4.( 4 )1/ 3

(b) after 60 min is 1/4 (d) after 80 min is 88

3. Two boats A and B are moving in a river. Boat A moves normal to the river currents as observed by an observer moving with river currents. Boat B moves normal to the river as observed by the observer on the ground. Let u and v be the velocities of boat with respect to water and boat with respect to the ground respectively. Then, to stationary observer on the ground (a) boat B reaches exactly opposite point on the bank ub (b) boat A reaches a point æç ö÷ distance downstream from the è v ø starting point (c) boat A travels a longer distance (d) to a ground observer, boat A moves faster than boat B.

4. A particle of mass m is suspended by the mean of a string of length l and describes a circle of radius, where a is the angle the string makes with the vertical axis. For this situation, (a) tension developed in the string is mg / cos a. l g sin2 a cos a l cos a (c) the time period is 2 p g (b) speed of mass m is

42

x q

d Fixe ge wed

A pulley is smooth, massless and string is inextensible light mass. The frictional coefficient between block x and y is m while friction between block x and contact surface of wedge is zero. For this situation, choose the correct alternatives. (a) The acceleration of the system if (m y - m x ) m³ tan q, where m y > m x 2 my (b) The friction between x and y is zero when m X = mY (c) Block will move up if m x > m y (d) Tension in the string is mg (sin q - mcos q ) if m x = m y = m.

6. A block of mass m is kept on a smooth surface and connected to the spring of stiffness K which in turn connected to a fixed wall as shown in the figure below. m0

K v

m

A bullet of mass m 0 is moving with the horizontal velocity v, strikes the block gets embedded to it and their common velocity is V. The spring is compressed by an amount x then, (a) V = [ K ( m + m0 )]1/ 2 × æ K ö (c) v = ç ÷ è m + m0 ø

1/ 2

×x

x m0

(b) V = [2 K (m + m0 )]1/ 2 × æ 3K ö (d) v = ç ÷ è m + m0 ø

x m0

1/ 2

×x

7. The diagram shows a cyclic process ABCA for a sample of 5 mole ideal gas. The temperatures of the gas at A and B are 250K and 450 K C respectively. An amount of heat 1000 J is withdrawn from the gas in the process. Take the gas constant, A B R = 8.3 J K -1 mol -1 for this cyclic Volume(V) process. Pressure(p)

One or More than One Option Correct Type 1. A heavy particle is projected with a velocity at an angle

JEE ADVANCED RIDE 1 (a) Work done during process AB is 8300 J (b) Work done during the process BC is - 9300 J (c) Work done during process CA is zero (d) Work done during process is + 9300 J

8. Some amount of an ideal gas is enclosed in a vertical cylindrical container support a piston of mass M. The piston is of area of cross-section A and is free to move. At equilibrium of the piston the volume of the gas is V0 , while the pressure is p0 . Now, piston is slightly displaced from its mean position x and then released. Assuming the system is completely isolated. (a) The restoring force of the piston is

A 2g p0 x in magnitude V0

A 3g p0 (b) The restoring force of the piston is x in magnitude V0 (c) The frequency of oscillations is (d) The time period would be

A 2g p0 MV0

is inserted into a long solenoid having 1000 turn per metre and carrying a current of 4 A. If the magnetic field induction within the core of solenoid is set to be 1.7 T, then neglecting the end effects, the order of magnitude of pole strength while expressed in unit (Am-1 ) is

14. A moving coil galvanometer having a coil of resistance 15 W needs 30 mA current for showing full deflection on the scale. In order to pass a maximum current of 3A through the galvanometer. If R is the resistance R should be added as a shunt then what is the order of magnitude of R ?

15. Two parallel wires A and B are separated at a distance of 6 cm. The current in the wires A and B are 5A and 2A respectively. The distance in cm from the wire B where, the magnetic field induction is zero, would be?

16. In a capacitive circuit, the electric field between plates

A 3g p0 3 M V0

9. The temperature of a body falls from 50°C to 40°C in 10 min when its surrounding temperature is constant at 20°C. Then, (a) the time taken by the body to cool upto temperature 30°C is 10 ln (0.50) given by t = ln(0.66) (b) the time taken by a body is given by t =

13. Long cylindrical iron core of area of cross-section 4 cm2

10 ln (0.20) ln (0.33)

(c) the constant used to determine the above meintioned time ln (0.5) is given by K = 10 ln (0.66) (d) the constant is given by K = 10

10. A proton enters a uniform magnetic field in direction 5

region of 0.2T with a velocity 2 ´ 10 m/s. The velocity makes an angle of 45° with the magnetic field induction. Then, (a) radius of path followed by proton is 7.34 ´ 10 -2 m (b) radius of path followed by proton is 7.34 ´ 10 -5 m (c) radius of path followed by proton is 3.26 ´ 10 -7 s (d) the time period to complete revolution of the proton on the circular track is 3.26 ms

Integer Type 11. An inductance 20 mH, a capacitor 100 mF, a resistor 50 W and an AC source of 12V, 50Hz are forms an series AC circuit. The average power dissipated in the circuit expressed as the nearest integer is

12. An inductor having inductance L and a resistance R = 5 W are connected in series with a battery of emf 10V. The maximum rate at which the energy is stored in the magnetic field would be

of a parallel plate capacitor is drops from 2 ´ 10 -6 F to its one third in 4.4 ´ 10 -6 s while the plates are connected through a thin wire. After disconnecting the supply, the resistance of the connecting wire would be.....(in ohm).

17. A block of mass 20 kg is kept on a horizontal surface. The coefficient of friction exerting between block and surface is 0.6. A horizontal force is applied on the block which varies with the graph shown below. F(N) 200 0

5

t(s)

10

If the speed of the block after 10 s is 8v then find the speed of a block, v is (take, g =10ms -2 )

18. Two soap bubbles x and y are in a closed chamber with

air pressure 8N / m2 . The radii of the bubbles are 2cm and 4 cm respectively. The surface tension of soap water is 0.004 N/m. If N A and N B are the number of moles of air in the bubbles x and y, respectively, then Nx will be...... (neglecting the effect of gravity) Ny

19. A uniform string having mass 0.1 kg and length 2.45 m hangs from a ceiling. If the speed of transverse wave produced in string at a point x m(x < 2.45 m) distant from its lower end is v. Then calculate the time taken by transverse wave travel the full length of the string?

20. The band energy gap in germanium is DE = 0.68 eV. The numbers of hole-electron pair for the material is -

DE

directly proportional to e 2 kt . If the percentage increment in number of charge carriers in pure germanium as the temperature increased from 300K to 320 K is given by 63.5 x, then the value of x would be

APRIL 2015 Physics SPECTRUM

43

JEE ADVANCED RIDE 1

Paper 2 1. A boy releases an arrow with the help of bow horizontally from the top of a 20 m height of a tower. What should be the minimum initial horizontal velocity of arrow, so as to hit the ground at a distance of 30 m from the bottom of the tower (take g = 10 m/s2 ) (a) 15 m/s

(b) 20 m/s

(c) 30 m/s

time t = 0, three particles A, B and C are the corners P, Q and R respectively. Now particles are moving along the sides of the triangle in anti-clockwise with a constant speed v. The time after which the particles meet one another is given by 3a 2v

(b)

2a 3v

(c)

3a 2v

2a v

(d)

increased from 20.0 cm to 20.1 cm is [assume that the rate does not appreciably change during the given change in radius] (c) 16 p cm

(b) 16 . ´ 10 4 p cm3

3

(c)

FL2 - H 2

(b)

FH - WL F L2 - H 2

(d)

WL - FH

WL - FH F L2 - H 2 WL - FH FL2 - H 2F

9. There is a mass m which is oscillating inside the spherical shell along its diameter whose radius is R. The kinetic energy of oscillating mass at any time is K, the force applied by the mass on the shell at the same instant is (a)

3. The change in volume of an sphere while the radius is

(a) 1600 p cm3

(a)

(d) 7.5 m/s

2. There is an equilateral triangle PQR of side length a. At

(a)

the rope and ground is H. If the thickness of the log is assumed to be negligible, the frictional coefficient between log and ground is

7R 9K

(b)

3K R

(c)

2K R

(d)

2R 3K

10. Figure shows three blocks of masses m1, m2 and m3 are connected with one another. All the surfaces are frictionless. The string and pulleys are light in mass. The acceleration of mass m1 is

(d) 160 p cm3

m1

A

4. A bob of mass m is attached to light inextensible string of length L is kept in the horizontal position. What will be the angle made by spring with the vertical axis so, as to the weight of the bob balances the tension produced in the string? m2

æ 2 ö - 1 æ 1ö - 1 æ 1ö - 1 æ2 ö (a) cos - 1 ç ÷ (b) cos ç ÷ (c) cos ç ÷ (d) cos ç ÷ è 6ø è 3ø è 3ø è 3ø

5. A skate board negotiating the circular surface of 3.5m as shown in the figure. At the angle a = 30° the speed of centre of mass is 4m/s. If the mass of the person along O with the skate board is 60 kg while, the X centre of mass is 0.60m from the a V surface. The normal reaction exerting between the surface and the skate board wheel is (take, g = 10 m/s 2 ) (a) 512 N (d) 794 N

(b) 691 N

Y

(c) 864 N

6. An inclined plane having angle q with the horizontal. A body starts from rest to slide down the incline. If the height of incline is h then the time taken by the body to reach the bottom of the incline is æ2 h ö 1 (a) sin q ç ÷ (b) sin q è g ø

2h g

2h g cos q

(c)

2 h cos q g

(d)

7. As starting from rest a body slides down to an inclined plane having angle of inclination 30° is thrice the time taken by it to slide down the same distance in absence of friction. In this situation, the coefficient of friction between the body and the inclined surface would be (a)

7 9

(b)

5 6

(c)

8 9

(d)

3 8

8. A log of weight W connected to a rope of length L which is pulled by a force in such a way that the velocity of the log is constant. The distance between the free end of

44

APRIL 2015 Physics SPECTRUM

B

(a)

m1g

æ 1 1 1 ö + + ç ÷ m2 m3 ø è m1 g (c) m1 æ 1 1 ö 1+ + ç ÷ 4 è m2 m3 ø

(b)

m3

4 m1g

æ m2 + m3 ö ç ÷ è m2m3 ø m1g (d) æ 1 1 ö 4ç + ÷ m3 ø è m2

Passage 1 A heavy particle whirling by means of a incompressible light string of length L. The particle is given to speed v0 at its minimum height. The string becomes slack at some angle, and the particle proceeds on a parabola as shown in the figure.

Parabolic path v P(Point to Slack) Circular path v0

11. Determine the angle a if particle passes through point of suspension. (a) tan- 1 ( 3 )

(b) tan- 1 ( 2 )

(c) tan- 1 ( 5 )

æ 3ö (d) tan- 1 ç ÷ è 2 ø

12. Determine the value of v0 if particle passes through point of suspension. 1

1

(a) ( g l + 2 3 )2

(b) [g l (2 +

3 )]2

(d) [ 2 g l +

3 ]2

1

(c) [ 3 g l +

2 ]2

1

18. Match the following column regarding physical

Passage 2 A wooden plank of length 2 m and of uniform area of cross section is hinged at one end to the bottom F of a tank as shown in figure. R The tank is filled with water upto a height I a P of 1m. The specific gravity of the plank is 0.6. Excluding the case for which q is C zero. 133 . dg l cos q

(b)

133 . d lg sin q

(c)

166 . d lg cos q

(d)

14 . d l g cos q

14. Find the angle q that the plank makes with vertical in the equilibrium position. (a) cos -1( 0.64)

A (a) (1, 3) (b)(1, 3) (c) 3 (d)(1, 3)

B 2 3 1 (1, 4)

C 1 1 3 1

(d) tan ( 070 . )

as shown below in pV-diagram. Pressures(p)

A parallel plate capacitor is placed horizontally in such a way, that its linear plate is dipped into a liquid having dielectric constant K and density d. The area of plates is A. Now, the plates are connected to a battery that supplies a charge +Q to the upper plate of capacitor.

A

pA=32p0

B

pB=p0

C

15. The force on upper plate of the capacitor will be? ( K 2 - 1) Q 2 2 A e 0K

2

(b)

( K - 1) Q 2 2 KAe0

(c)

2 Ae 0 2

( K - 1) Q

2

(d)

2 K 2 Ae 0 ( K - 1) Q 2

Relate to the above diagram match the column Column I

capacitor, is 2 e0 A 2d 2g 2

( K - 1) Q

2

(b)

2 Ae0d 2g ( K - 1)Q

2

(c)

( K 2 - 1)Q 2 2

2

2 A K e0dg

(d)

2 AK 2e0dg

radiating light of wavelength 5890 Å uniformly in all directions then match the following columns (all the magnitudes are in their SI units) Column I

Column II 7

A. Rate of emission of photon from the lamp (i.e., 1. 4.9 ´ 10 photon per second) B. The distance from lamp where average flux is 1 2. 5.9 ´ 1014 photon/cm 2-s C. Flux of photon at a distance 2 m from the lamp 3. 2 ´ 10 4 D. Average density (photon/cm 2) of photons at a distance 2 m from the lamp.

A (a) 4 (c) 3

B 1 1

C D 2 3 4 2

A (b) 4 (d) 1

B 1 2

Column II 1. 2. 3. 4.

D® B D® C A® C A® D

A (a) 4 (c) 3

B 3 1

160 p0V0 ln 2 36 p0V0 24 p0V0 31 p0V0

C D 2 1 2 4

A (b) 4 (d) 1

B 3 3

C D 1 2 2 4

20. Match the following columns with reference to expressions for the physical quantities. Column I

Column II

A. Half- life of radioactive sample

1. N 0(1 / 2 )n

B. Atoms remaining after n half-lives 2. A0 e - lt 3. 0.693 log N 0 t1/ 2 N

C. Decay constant, K

4. 3 ´ 10 20

D. Radioactivity

C D 3 2 3 4

A (a) 4 (c) 1

B 1 4

4. 0.693 / l

C D 3 2 3 2

A (b) 4 (d) 3

B 2 4

C D 3 1 2 1

Answers

Paper 1 1. (b,c,d) 11. (3)

A. B. C. D.

( K - 1)Q 2

17. Match the Column Type If a 100 W sodium lamp is

D

V0 VC VD Volume(V)

16. The rise in liquid level h, is the space between plates of (a)

4. b-decay

D (1, 4) (1, 4) (1, 4) (1, 3)

B ® A ® D ® B and B ® A ® C ® B

Passage 3

(a)

2. Characteristics of X-ray 3. Hydrogen spectrum

taken along in a cyclic process.

-1

(c) cos ( 075 . )

Column II 1. Photoelectric effect

19. An amount of 1 mole of a monoatomic ideal gas is

(b) sin-1( 0.85)

-1

Column I A. Transition between two atomic energy levels B. Moseley’s law C. Change of photon energy into kinetic energy of electron. D. Electron emission.

Q mg

13. Determine the buoyant force applied by the liquid. (a)

Situations

2. (b,c) 12. (5)

3. (a,b,c,d) 13. (2)

4. (a,b,c) 14. (1)

5. (a,b) 15. (4)

4. (c) 14. (a)

5. (d) 15. (a)

6. (a,c) 16. (2)

7. (a,b,c) 17. (3)

8. (a,c) 18. (6)

9. (a,d) 19. (1)

10. (a,d) 20. (2)

8. (c) 18. (a)

9. (b) 19. (a)

10. (c) 20. (a)

Paper 2 1. (a) 11. (b)

2. (b) 12. (b)

3. (d) 13. (c)

6. (b) 16. (c)

7. (c) 17. (a)

Detailed solutions of these questions are available on http://www.arihantbooks.com/Physics%20Spectrum.pdf

45

Answer with Explanations Paper 1

3. (a,b,c,d) The speed of boat A with respect to ground vA = v 2 + u 2

1. (b,c,d) If the particle is projected with velocity u at an angle q, then equation of its trajectory will be

The speed of boat B with respect to ground v B = v 2 - u 2

gx 2

y = x tan q -

2 u 2 cos 2 q dy We know slope is given by, m = dx gx Therefore, slope, m = tan q u 2 cos 2 q It implies that the graph between slope and x will be straight line having negative slope and a non-zero positive intercept on y-axis. But x is directly proportional to the time t, therefore the shape of graph between slope and time will be same as that of the graph between slope and x. Hence, only option (a) is correct i.e. options (b), (c) and (d) are incorrect.

Þ vA > vB To the ground observer, since the boat moves normal to the river currents therefore, it reaches the point exactly opposite point to the starting one. The velocity of boat A with respect to ground is more that of boat B. Therefore A travels longer distance with respect to ground frame. The downward distance travelled by A æ b ö ub x = ut = u ç ÷ = èvø v

4. (a, b, c) Consider the diagram

2. (b,c) As we know that,

a

æ 1ö half-life of radioactive sample, N = No ç ÷ è2 ø

n

T cos a T 2 mwr

for the time 60 min, and for sample A æ 1ö N A = No ç ÷ è2 ø

For equilibrium,

where,

total time 60 nA = = =4 half life 15

Þ

N æ 1ö N A = No ç ÷ = o è2 ø 16

mg cos a Now the centripetal force acting along horizontal direction, we get

T cos a = mg Þ T =

4

mv 2 T r sin a = T sin a Þ v = r m

For sample B, æ 1ö N B = No ç ÷ è2 ø where,

nB =

nB

T=

N æ 1ö N B = No ç ÷ = o è2 ø 4

nA =

T y g my

nA

80 16 æ 1ö = Þ N A = No ç ÷ è2 ø 15 3

16 / 3

for sample B, æ 1ö N B = No ç ÷ è2 ø

nB

æ 1ö = No ç ÷ è2 ø

The ratio of residual sample i.e.

= 4.( 4 )1/ 3

46

NA = NB

2 pr = v

2 p l sin a lg sin 2 a / cos a

= 2p

l cos a g

5. (a, b) Consider the free body diagram of blocks x and y

Now the ratio of residual praticles in the samples A and B NA N 1 = o = N B 16 4 No 4 Similarly for the time 80 min and for sample A,

where,

(\ r = l sin a)

For the time period,

total time 60 = =2 half life 30

æ 1ö N A = No ç ÷ è2 ø

l g sin 2 a cos a

=

2

Þ

O T sina mg

nA

8/ 9

[as, n B =

æ 1ö N0 ç ÷ è2 ø

16 / 3

æ 1ö N0 ç ÷ è2 ø

8/ 3

80 ] 30

q Sin

mg

T

f

gc my

o

x

sq

g mx

f q n Si

gc mx

os

q

mg

where, f is the frictional force exerting between x and y is required to keep the blocks stationary and T is tension in the string. Then, for the equilibrium of the blocks …(i) m y g sin q = T + f and …(ii) m x g sin q = T - f Solving Eqs (i) and (ii), we get, ( m - m x ) g sin q F= y 2 while f £ fmax , then ( m y - m x ) g sin q £ mm y g cos q 2 ( m - m x ) tan q or, m³ y 2 my

The blocks will remain stationary for this condition otherwise not. When, m x = m y , f = 0 and T = m x g sin q and the system is in equilibrium. If m y < m x , the blocks will move only when m £ ( m x - m y )tan q / 2 m x .

Now frequency, n =

The time period, T =

6. (a, c) Applying law of conservation of momentum for bullet-mass system, we get m 0v ( m + m0 )

Now from energy conservation principle, we know é ù K 1 1 ( m + m 0 ) V 2 = Kx 2 Þ V = ê úx 2 2 ë ( m0 + m ) û

…(i)

…(ii)

æ K ö m 0v ç ÷ x= ( m0 + m ) è m + m0 ø So, the velocity of spring which is compressed by an distance x, x we get v = [ K( m 0 + m )]1/ 2 m0

7. (a, b, c) As we know that the change in internal energy during a

Pressure(p)

cyclic process is zero. This implies, that the heat supplied to the gas is equal to work done by the gas. …(i) Þ WAB + WBC + WCA = - 1000 J During process AB, the work done, WAB = pA ( VB - VA ) C = nR (TB - TA ) = 5 ´ 8.3 ´ (450 - 250) A B = 5 ´ 8.3 ´ 200 Volume(V) = 8300 J Now, the work done during process CA is zero, because the volume is constant Therefore, 8300 + WBC = - 1000 J Þ WBC = - 9300 J

Þ

mean or equilibrium position. So, increase in volume of the gas DV = Ax As the process is adiabatic pV g = constant Þ on differentiating it, we get. Dp DV ln p + g ln V = constant Þ +g =0 p V gp Dp = DV V Now, we can write gp Dp = 0 DV V0 The resultant force acting on the piston Ag p0 F = AD p = DV V0 A2g p0 x = -Kx V0 2

A g p0 K= V0

Thus, it is clear that the motion of the piston is simple harmonic with angular frequency w=

K = M

A2g p0 MV0

dq = - K( q - q0 ) dt dq = - K dt q - q0

…(i)

where, K is a constant on integrating, 40°c dq ò50°c q - q0 = K (10 min) Þ

é 40 °C - 20 °C ù ln ê ú = 10 K ë 50 °C - 20 °C û

ln 2 / 3 = 10 K ln(2 / 3 ) K= …(ii) 10 Let t be the time to cool the body upto temperature 30°C from 50°C. 30°C dq Then [from Eq (i)] ò40°C q - q = Kt 30 °C - 20 °C Þ ln = Kt 40 °C - 20 °C æ 1ö ln ç ÷ = Kt Þ è2 ø Þ

\

æ 1 ö ln (2 / 3) ln ç ÷ = ´t è2 ø 10

Þ

10ln (1 / 2) =t ln(2 / 3)

8. (a, c) Let the piston is displaced through a distance x above its

=-

MV0 A2g p0

9. (a, d) According to Newton‘s law of cooling

From Eqs (i) and (ii) we get.

where,

1 n

T = 2p

m 0v = ( m + m 0 ) V Þ V =

A2g p0 MV0

1 w = 2p 2p

[from Eq (ii)]

10. (a, d) When proton enters the magnetic field induction region with some inclination with it. The velocity has two components i.e. the component of velocity parallel to induction is v|| = 2 ´ 10 5 cos 45° = 2 ´ 10 5 m/s and the component of velocity perpendicular and the magnetic induction, v ^ = 2 ´ 10 5 cos 45° = 2 ´ 10 5 As, the magnetic, force f = q ( v ´ B ) is perpendicular to the magnetic induction So, v|| is remains constant Now, for radius r of the path followed by proton m v ^2 qv ^ B = r mv ^ (1.67 ´ 10 -27 ) ´ 2 ´ 10 5 = Þ r= qB 1.6 ´ 10 -19 ´ 0.2 =

2 .35 ´ 10 -22 3.2 ´ 10 -19

= 0.734 ´ 10 -3 = 7.34 ´ 10 -2m

Now, time taken by proton to complete one revolution 2pr T= v^ =

2 ´ 3.14 ´ 7.34 ´ 10 -2 2 ´ 10 5

=

46.09 ´ 10 -7 1.41

= 32 .69 ´ 10 -7 = 3.27 ´ 10 -6s» 3.27ms

47

11. (3) The time period of source is given by

13. (2) The magnetic intensity inside the solenoid is H = ni = 1000 ´ 4 = 4000 Am -1

1 1 T= = = 20 ms n 50 The capacitive reactance 1 1 100 Xc = = = W wC 2 p ´ 50 ´ 100 ´ 10 -6 p

Again, magnetic field of solenoid B = m 0( H + I ) Þ l = =

The inductive reactance, X L = wL = 2 p ´ 50 ´ 20 ´ 10 -3 = 2p W

= 13460 ´ 10 2 Am -1 = 1.34 ´ 10 6 Am -1

100 X = | Xc - X L| = - 2p p

Now, pole strength devoloped at the ends of the solenoid, m = IA = 1.34 ´ 10 6 ´ 4 ´ 10 -4 = 5.36 ´ 10 2 Am -1

14. (1) For a moving coil galvanometer, the current through the coil ...(i)

Now, the average power dissipated in the circuit Pav = Vrms i rms cos f where, cos f = power factor = R / Z V R RV 2 Pav = Vrms × rms × = rms Þ Z Z Z2 =

( 50 W)(12 V )2

= 7200 Z 2 WV 2

Z2 Substituting the value of Z 2 from Eq (i) into Eq (ii) we get Pav =

7200 W - V 2

315W 2 ~ –3W

Þ

…(ii)

= 2 .286 W

1 2 Li 2 1 [ I = I0(1 - e -t / t )] U = LI02 (1 - e -t / t )2 2 The rate at which energy is stored is dU æ 1ö P= = Li 02(1 - e -t / t ) ´ ( - e -t / t ) ´ ç - ÷ è tø dt LI02 -t / t (e - e -2 t / t ) t dP The rate will be maximum when =0 dt Ll 02 æ 1 -t / t 2 -2 t / t ö + e Þ ç- e ÷=0 ø t è t t Þ

e -t / t =

1 2

1 in Eq (i) we get, 2 L l2 æ 1 1 ö = 0 ç - ÷ t è2 4 ø

Putting the value of e -t / t = Pmax

=

Le 2 2

4R ( L / R)

=

e2 4R

Now, given that, e = 10V and R = 5 W (10 )2 100 Pmax = = \ 4 ´ 5 20 = 5W

48

Rs I R s + Rg Rs ´ 3A R s + 15W

30 mA Rs = Þ (10 -2 )( R s + 15) = R s 3A R s + 15

Rs 15 R 15 + = Rs Þ Rs - s = 100 100 100 100 15 Þ 99 R s = 15 Þ R s = » 0.15 W 99 The value of shunt resistance R s = 1.5 ´ 10 -1 W Þ

15. (4) Let there is a point P at a distance of x cm from wire B, where

U=

P=

Ig =

30 mA =

12. (5) The energy stored in the magnetic field at time t is

Þ

- 4000 = 0.135 ´ 10 7

= 13500 ´ 10 2 - 40 ´ 10 2

The net reactance

= 25.5 W Therefore, the impedance (squared) Z 2 = ( 50 )2 + (25.5)2 = 3150 W 2

17 . 4p ´ 10 -7

B -H m0

(i)

the net magnetic field induction is zero, clearly distance of this point from wire A will be ( 6 + d ) cm The magnetic field induction at point P due to wire A having current 5A is m 0 i1 5 m0 B1 = = 2 p (d + x ) 2 p ( 6 + x ) Also, magnetic field induction at point P due to wire B having current 2A is m i 2m 0 B2 = 0 2 = 2 pd 12 p As B1 and B2 are in mutually opposite direction, So, net megnetic induction at point P is Bnet = |B1 - B2| = 0 5m 0 2m 0 Þ B1 = B2 Þ = ( 6 + x )2 p 12 p The distance from the wire B where the magnetic field induction is zero, x = 4 cm.

16. (2) The electric field between the plates of capacitor is given by E= Þ

Q Q = e -t / Rc A Î0 A Î0

E = E 0e -t / Rc Þ

E = e -t / Rc E0

As, finally the electric field drops to 4.4 ´ 10 -6s, thus Eq (i) becomes E0 3 = e -t / RC E0

…(i)

E0 after time interval 3

E0 ù é ê\ E = 3 ú ë û

Þ

Þ

1 - 4.4 ´ 10 -6 e RC 3 4.4 ´ 10 -6 = ln 3 = 11 . Rc

=

R=

py Vy = n y RT

i.e.

n x py v y 12 4 = = ´ =6 n y px v x 16 2

3

19. (1) As shown in figure

4.4 ´ 10 -6 11 . ´ 2 ´ 10

and

The mass of x length of string, m =

-6

Thus, the resistance of a connecting wire is 2 W Þ R =2W

17. (3) Force of limiting friction flim = 0.6N = 0.6 mg = 0.6 ´ 20 ´ 10 = 120 N. From the graph given in the question F = 40t = 200 for t £ 5, 5 £ t £ 10 The limiting friction acts at t ³ 35 Using impulse equation, t

T = m

Þ

5

ò3

10

( 40 t - 120 )dt + ò (200 - 120 )dt = 20 V

v=

dx gx

Þt =

1

L

ò x g 0

-

1 2dx

Time taken by transverse wave on the string

4T R For bubble x, excess pressure is given by 4T pX - po = RX 4T pX = p0 + Rx =8+ Þ

M L

xg

ò dt = ò

p - po =

Þ

Lower end

Now if at point x, the wave travels a distance dx in time dt then, dx dx v= Þ gx = dt dt On integrating we get,

18. (6) As we know that, for the soap bubbles the excess pressure is given by

x

where, m = linear mass density of the string =

5

40 (25 - 9 ) - 120(2 ) + 80 ´ (10 - 5) = 20 V 2 Þ 320 - 240 + 400 = 20 V1 Þ V1 = 24 = 8v m/s The speed of a block Þ v = 3 m/s Þ

pulse

L=2.45m

æMö ç ÷ xg èLø æMö ç ÷ èLø

o

where, Pi and Pf are initial and final momentum.

T

So, weight of x length of string M W = mg = xg L This weight produces the tension in string Þ W =T M = xg L

Now, speed of the wave v =

ò Fdt = Pf - Pi

M x L

-2

and T2 are N1 and N 2, respectively.

= 16

px = 16 N/m 2

For bubble y the excess pressure in given by 4T py - po = R 4T Þ py = po + R 4 ´ 0.04 =8+ 4 ´ 10 -2 = 12 N / m 2 Also, the volume of the bubbles are 4 Vx = p Rx3 3 and similarly 4 v y = p Ry3 3 Now, ideal gas equation we know px Vx = n x RT

L 2 .45 Þt = 2 Þ t = 1s g 9.8

20. (2) Suppose, the number of charge carriers at temperatures T1

Then,

4 ´ 0.04 2 ´ 10

t =2

-

DE 2KT1

-

DE 2KT2

N1 = N 0e N 2 = N 0e

where, N 0 = constant For the calculation of, percentage increment during, rise in temperature from T1 to T2. æN ö N - N1 f= 2 ´ 100 = ç 2 - 1÷ ´ 100 N1 è N1 ø Þ

Now,

é DE æç 1 - 1 ö÷ ù 2k è T1 T2 ø …(i) f = 100 êe - 1ú ê ú êë úû DE é 1 1ù ê - ú 2 K ë T1 T2 û 0.68 eV 1 ù é 1 = ´ê ú = 0.82 2 ´ 8.62 ´ 10 -15 eVK -1 ë 300K 320K û

Thus equation becomes f = 100[e 0. 82 - 1] ~ - 127 Now, according to question, f = 63.5 x = 124 Þ x = 2

49

Paper 2

Þ

1. (a) The initial vertical velocity of the arrow is zero. The time taken by the arrow to reach the ground 2h g

Þ

where, h = height of tower

5. (d) The figure for the situation, can be drawn as below.

2 ´ 20 =2 s 10

C

2. (b) The velocity of particle A is v along PQ. Also the velocity of v 2 Now, the rate of decrement of separation between the particles A v 3v and B is, v + = 2 2 This rate of decrease of separation is constant. So, the time taken in reducing separation between the particles A and B from a to 0 is given by a 2a t= = 3v 3v 2 particle B is along QR. Its component along QP is v cos 60 ° =

3. (d) Suppose, the radius of the sphere be, R, then the volume of sphere is 4 p R3 3 Differentiating it with respect to R, we get dV 4 d ( R )3 4 = p = p . 3 R 2 = 4pR 2 dR 3 dR 3 when radius is R = 20cm then, dV = 4p (20cm )2 = 1600 p cm 2 dR Now, the change in volume as the radius is changed from 20.0cm to 20.1 cm is dV DV = DR dR = 1600 p cm 2 ´ 0.1 cm = 160 p cm 3 V=

4. (c) According to free body diagram as shown in figure (Initial position horizontal) m C L po fina A s l itio L-H q n T B H Vertical

mg cosq mg

In above figure q is angle made by string from the vertical. The centripetal force at position B mv 2 (from figure) = T - mg cos q L where, v = velocity of the bob at position B.

50

a O a mg mg cosq

The centripetal force, mv 2 mv 2 = N - mg cos a Þ N = + mg cos a r r Given that, r = 3.5 m a = 30 ° m = 60 kg and v = 4 m/s Therefore, OC = 3.5m - 0.6m 60 ´ 4 2 + 60 ´ 10 ´ cos 30 ° = 2.9 m Þ N = 3.5 60 ´ 16 3 + 600 ´ 3.5 2 960 = + 300 3 = 274.2 + 519.6 3.5 The normal reaction exerting between the surface and the skate board is = 793.8 ~ - 794 N =

6. (b) The situation can be figured as

q

Now, the equation for horizontal distance covered by the arrow. Xhorizontal = Vhorizontal ´ t X Þ Vhorizontal = horizontal t 30m = = 15m/s 2s

N

sin

t=

q

h

mg cos q mg

g

Þ

æ 1ö 3 cos q = 1 Þ q = cos - 1 ç ÷ è 3ø

m

t=

Þ

mv 2 L æ mö mg = mg cos q ç ÷ 2 g ( L - H ) è l ø

T = mg cos q +

q

The force downward to the incline F = mg sin q Þ ma = mg sin q Now, from equation of motion, v = u + at, we get, v = 0 + at \ a =t / v But, (from, v 2 = u 2 + 2 as v = 2 gh 2 gh . sin q Time taken by the body toreach the bottom of the incline is So,

t=

=

1 2h sin q g

7. (c) In case of frictionless surface, the acceleration along the incline, g 2 \ distance covered by the body along the incline 1 g l = ´ ´t2 2 2 a = g sin 30 =

gt 2 …(i) 4 Now when friction is present, the acceleration along the incline g mg g a¢ = = (1 - m ) 2 2 2 \ distance covered by the body along the incline 1 g l= (1 - m )t 2…(ii) 2 2 On dividing Eq (i) by Eq (ii), we get gt ¢ 2 t2 4 1= = (1 - m ) 2 gt ¢ t ¢2 (1 - m ) 4 According to the question, t ¢ = 3 t t2 1 1= Þ Þ 1-m = 2 9 ( 3 t ) (1 - m ) Þ

The coefficient of friction between the body and the inclined 1 8 surface, m = 1 - = 9 9

R f

F sina F L a H F cosa

W

Now, and or

L2 - H 2 H and sin a = L L F L2 - H 2 f = F cos a = L F sin a + R = W R = W - F sin a æHö R =W - F ç ÷ èLø

... (i)

... (ii)

Further ... (iii) f = mR é L2 - H 2 ù é F L2 - H 2 1 æ H öù ú = m êW - F ç ÷ ú Þ m = Fê ´ è L øû L L é æ H öù êë úû ë êW - F çè L ÷ø ú ë û The coefficient of friction lies between log and ground is m=

F L2 - H 2 WL - FH

K 2K 2 K K 3K = ÞN = + = R R R R R Hence, N is force applied by mass m on the spherical shell. N-

10. (c) Suppose, the acceleration of block of mass m1 be a 0. The acceleration of mass m 2 with respect to ground is a 0 - a (downward) and the acceleration of mass m 3 with respect to ground is a 0 + a (downward) Now, consider the figure with proper direction of acceleration.

As pulley is of negligible mass Þ 2T ¢ = T T Þ T¢ = 2 For horizontal motion of m1 T = m1 a 0 For vertical motion of m 2 T m2 g - = m2 ( a0 - a ) 2 For vertical motion of m 3 T m 3g - = m 3 ( a 0 + a ) 2 From Eqs (ii), (iii) and (iv) we get, a m 2g - m1 0 2 = g - m1a 0 a0 - a = m2 2 m2 a0 m 3g - m1 2 = g - m1a 0 and a0 + a = m3 2 m3

Þ O

R q

B

Þ

P v

... (ii) ... (iii)

... (iv)

... (v)

... (vi)

a0 = g -

m1a 0 æ 1 1 ö + ç ÷ 4 è m2 m3 ø

é m æ 1 1 öù a 0 ê1 + 1 ç + ÷ú = g 4 è m1 m 3 ø û ë

So, acceleration of mass, m

h

mg N

Assume that the velocity of particle at P be v. Now, from conservation of total mechanical energy 1 k = mv 2 = mgh 2 mv 2 Also, from the figure N - mg sin q = R From Eqs (i) and (ii), we get

... (i)

On adding Eqs (v) and (vi), we get ma æ 1 1 ö 2 a0 = 2 g - 1 0 ç + ÷ 2 è m2 m3 ø Þ

9. (b) The situation can be figured as below.

A

hö æ çQsin q = ÷ è Rø

a0-a

8. (c) As, figure shown below

As, cos a =

æ hö 2k N - mg ç ÷ = è Rø R

l=

a0 =

g m1 æ 1 1 ö 1+ + ç ÷ 4 è m2 m3 ø

11. (b) The particle goes in a parabola under the action of gravity. As ... (i) ... (ii)

it passes through point of suspension O, the equations for horizontal vertical motion gives ... (i) l sin q = ( v cos q) t where, v = speed of particle at point p and 1 - l cos q = ( v sin q) t - gt 2 2

51

æ l sin q ö 1 æ l sin q ö or - l cos q = v sin q ç ÷- gç ÷ è v cos q ø 2 è v cos q ø Þ Þ This gives

2

1 sin 2 q g. l 2 v 2 cos q 1 g l sin 2 q - cos 2 q = 1 - cos 2 q 2 g l cos 2 q

[from Eq (i)]

- cos 2 q = sin 2 q -

[from Eq (i)]

tan q = 2

Circular path v0

Angle suspended by a particle when passes through point of suspension, q = tan - 1( 2 )

...(ii)

12. (b) There is only force acting on particle at point is its weight mg the radial component of force is mg cos q. As the particle goes on mv 2 circle upto point P, thus, mg cos q = l ... (iii) v 2 = g l cos q Þ From conservation of energy, we get 1 1 mv 02 = mv 2 + mg l (1 + cos q) 2 2 v 2 = v 02 - 2 g l (1 + cos q) Þ

Þ

cos q =

1.66 4

Angle made by plank with vertical in equilibrium position Þ cos q = 0.64 Þ q = cos -1( 0.64 )

Consider the volume of liquid in space in between the plates of capacitor. The electric field at surface of the liquid due to charge +Q to the Q upper plate of capacitor E1 = in downward direction. 2 Ae 0

....(iv)

æ The force on charge - Q ç1 è

1

[Q tan q = 2 = cos q =

Q is K

Q 2 Ae 0 K

\ Net electric field, E = E1 + E 2 =

v 02 = g l (2 + 3 cos q)

v 0 = [ g l (2 + 3 )]2

Thus, net charge on lower plate is 1ö -Q æ - Q + Q ç1 - ÷ = è Kø K

E2 =

Velocity of a particle passes through point of suspenstion is Velocity of particle passes through point of suspension is

( K + 1)Q 2 Ae 0 K

1ö ÷ at the surface of liquid is kø

1 ö ( K + 1)Q ( K 2 - 1)Q 2 æ F = Q ç1 - ÷ = è K ø 2 Ae 0 K 2 Ae 0 K 2 1 3

]

16. (c) As the upper plate of capacitor is given to charge + Q and an

The length of plank, 2 l = 2m the centre of gravity of the plank is Q. So, we have OQ = l = 1m CR l Now, CP = = 2 2 cos q Let the linear mass density of the plank is d then its weight is mg = 2 ldg The mass of part CR of the plank æ l ö =ç ÷d è cos q ø

1ö æ amount of charge Q ç1 - ÷ is appeared at the surface closer to è Kø the plate which implies an electric field will exists in between these two. Due to appearance of this electric field a force is exerted on liquid surface and thus there is a rise in liquid level. For the equilibrium condition, the weight of the raised volume, the space between the plates is balanced by electric force. Also, weight of raised volume of the liquid is W = hAdg ( K 2 - 1)Q 2 Now, hAdg = 2 Ae 0 K 2

Thus water displaced by the plank 1 l 1.66 ld = d= 0.6 cos q cos q

Rise in liquid level between the plates of capacitor ( K 2 - 1)Q 2 h= 2 A2K 2e 0dg

13. (c) The height of water level is l = 1m

This implies that the buoyant force, F =

1.66 ldg cos q

14. (a) For the equilibrium, the torque of mg about C should balance torque of F about C Þ mg(CQ )sin q = F(Cp)sin q

52

4 cos 2 q = 1.66

Field at the same surface of liquid due to charge -

From Eqs (iii) and (iv), we get v 02 - 2 g l (1 + cos q) = g l cos q

Þ

Þ

plate, an amount of negative will be induced to the surface of liquid in between the plates of capacitor. 1ö æ The amount of induced charge will be - Q ç1 - ÷ and charge è Kø 1ö æ + Q ç1 - ÷ will appear to surface in contact with lower plate. è Kø

P(Point to Slack)

Þ

æ 1.66d l ö æ l ö (2 ld ) l = ç ÷ç ÷ è cos q ø è 2 cos q ø

15. (a) As the battery supplies an amount of charge +Q to the upper

Parabolic path v

or

Þ

-28

17. (a) (A) As energy emit out form lamp E = hv = hc = 1990 ´ 10 -10 l

Now, suppose number of proton is n Þ nE = 100 J/s 100 \ n= = 3 ´ 10 +20 photon/s 3376 ´ 10 -22

5890 ´ 10

(B) As, if flux in photon is 1 photon / cm 2 -s n Þ 1= 4p r 2 where, r = distance from the lamp 3 ´ 10 20 = 4.9 ´ 10 4 Km 4 ´ 3.14

r=

Þ

= 4.9 ´ 10 7 m (C) When r = 2 m then, n ´ 4p (200 cm )2 = 3 ´ 10 20 Flux of photons emit out from the lamp n = 5.9 ´ 1014 photon / cm 2 Þ (D) l Average density at r = 2 m is given by 3 ´ 10 20 r= 4 ´ 3.14 ´ (200 )2 ´ ( 3 ´ 1010 ) » 2 ´ 10 4 photon / cm 2

18. (a) (A) Transition between two atomic energy levels ® on photo-electric effect and hydrogen spectrum. (B) Moseley’s Law ® characteristics of x-rays. (C) Change of photon energy into kinetic energy of electron ® photoelectric effect. (D) Electron emission ® Photoelectric effect and b-decay

19. (a) As shown in figure Pressures(p) pA=32p0

pB=p0

AD is an isothermal process So, ( 32 p0 ) V0 = p0V0 V0 = 32 V0 AC is an adiabatic process So,

5 5 æ 32 P0 V0 3 = P0VC 3 ças pV g = constant and g = è

5ö ÷ 3ø

Þ VC = 8 V0 Work done in process DB = 31 p0V0 Work done in process. DC = 24 p0V0 Work done in process AC p V - pA VA = C C = 36 p0V0 2 f Work done in process AD æV ö = RT loge ç D ÷ = 160 p0V0 loge 2 è VF ø = 160 p0V0 ln 2

20. (a) (A) Half-life of radioactive substance, T1/ 2 = 0.693 l

æ 1ö (B) Atoms remaining after n half lives, N = N 0 ç ÷ è2 ø (C) Decay constant, k =

A

(as pV = constant)

n

0.693 N log 0 t1/ 2 N

(D) Radioactivity of a substance, A = A 0 e - lt B C

D

V0 VC VD Volume(V)

53

TEST RIDER The Simulator Test Series Towards JEE Main and Advanced

JEE ADVANCED RIDE 2 More than One Option Correct Type

1. A pendulum is constructed from two identical uniform thin rods A and B each of length l and mass m connected at right angle forming T shape. It suspended free end and swings in vertical plane.

l

13 2 ml 12 17 l (c) Time period of small angular oscillation of T is 2 π 18g (b) Moment of inertia of T about axis of rotation is

18g (d) Angular frequency of T for small oscillation is 17 l

2. A photon moves vertically up in a region with gravitational field g downwards. The frequency of photon at an instant is ν0 . After it has moved up by height h (a) its speed decreases (b) its energy decreases  1 − gh  − gh/c 2 (c) its frequency is ν 0   (d) its frequency is νe  c2 

3. A point isotropic source of sound power 1 mW emits sound of frequency 170 Hz in all directions. Velocity of sound is 340 m/s, then (a) intensity at any point at distance r depends on 1/r 2 250 (b) at, r = m, the loudness of source is 60 dB π (c) amplitude of sound wave depends on 1/ r (d) amplitude of oscillation of a point at 4 m is 2A and at 55 m it is −8 A at t = 0 55

4. A block of mass m moving

m

u

m

(a) horizontal momentum of system is conserved (b) final speed of block is zero and wedge is u towards right (c) at maximum height from ground block will be moving with speed u /2 (d) maximum height attained by block is u 2 / 4g

74

The charged on the capacitor is varying with time as Q = Q0 (1 − e − βt ). Obtain the value(s) of Q0 and β in the given circuit. CVR 2 R1 + R 2 CVR1 (b) Q 0 = R1 + R 2 R + R2 (c) β = 1 CR1R 2

S V

C

R2

(a) Q 0 =

l

(a) Moment of inertia of T about axis of 17 2 rotation is ml 12

with speed u on smooth horizontal surface towards stationary wedge of same mass initially as given all surfaces are smooth, then

R1

5. At t = 0, switch S is closed.

Paper 1

(d) β =

R1 C ( R1 + R 2 )2

6. A wire having uniform mass density λ, is bent in the form of a ring of radius R as shown in figure is in the plane of ring but not at centre. Two elements of the ring a1 and a2 substend very small equal angles at point A. They are at distances r1 and r2 from the ring, then

r1 A C

R r2

(a) ratio of mass of elements a1 and a2 is r1 / r2 (b) the elements a1 and a2 produce same potential at A (c) the element a1 produces greater magnitude of gravitational field at A, than element a2 (d) the element a1 produces greater potential than element a2

7. A small solid cylinder of

m, R

mass m and radius R h slides down a smooth M curve from height h. It gets onto plank of mass Smooth Negligible friction M, which is resting on a smooth surface. If µ is coefficient of friction between cylinder and plank, then (a) cylinder is performing impure rolling initially on plank and finally performs pure rolling motion (b) friction on cylinder is initially backward and finally it will be zero (c) velocity of COM of plank + cylinder at moment of pure rolling v is 0 , where v 0 = 2 gh 2 (d) minimum length required to attain pure rolling on plank is 3v 20 / 16 µg , where v 0 = 2 gh

8. A given mass of ideal gas is taken at constant pressure from state A (pressure p, volume V) to state B (pressure p, volume 4V). Subsequently, the gas is taken at constant volume from state B to state C (pressure p / 4 , volume 4V). Select the correct statements from the following. (a) A quantity of heat (say Q1) is given to the system is going A to B (b) A quantity of heat (say Q 2) is taken out of the system is going from B to C (c) Q1 > Q 2 (d) Q1 = Q 2

9. Two radioactive nuclei A and B are initially in the ratio 1 : 4. Also initial activities of the nuclei are in the ratio 1 : 8. Given that, half-life of A is 2 yr, choose the correct alternatives. Half-life of nuclei B is 1 yr At t = 4 yr, activities of A and B are equal At t = 6 yr, ratio of numbers of nuclei of A to that of B is 2 : 1 Fractions of nuclei decayed in one mean life for A and B are f1 and f2 respectively (f1 > f2)

bottom of a container of area A. The liquid is filled upto height h from base. As liquid comes out A  then,  a =   3 (a) level of liquid in container falls at rate of

h

gh 2

g m/s 2 4 g (c) acceleration of top surface of liquid is 3 (d) None of the above (b) acceleration of top surface of liquid is

11. Water is flowing in

k

x

D

k 2m

1 kg A

µ=0.3

B

µ=0

15. A steel bar of length L is held between rigid supports and heated non-uniformly in such a manner that the temperature increases ∆T at distance x from one end T x3 is given by ∆T = 0 2 as shown in the figure. The L stress in the bar (assume that modulus of elasticity for steel is E and thermal coefficient of expansion is E α T0 . Find the value of x. α) is given as σ = x x-axis of the solid spherical segment of mass M (see in figure) 53MR2 is . The value of x is 40x

R/2

R/2 O

x

p moles of a monoatomic ideal gas is 2×103 A B taken through a cyclic process shown in P - T 103 C diagram. The process CA is T(k) represented as 300K 600K PT = constant. If efficiency 3x of cycle is 1 − , then find the value of x. 12 ln 2 + 15

17. Two

37.5 cm

35 cm varying cross-section 200 cm pipe the areas of cross-sections 1, 2 1 2 and 3 are 1cm2 , 2 cm2 3 and A cm2 respectively. Water levels are shown in different vertical tubes. The speed of water at cross-section 3 1 is m/s. The value of x is x

12. Consider a horizontal

y=x2

attached to the left and right walls as shown in figure. A 1 kg block is initially held against, the left hand spring compressing the spring by 0.l m. The block is released to move. The floor is frictionless except for the section AB. Coefficient of friction for AB is 0.3. Find the distance (in cm) from point B where, block finally comes to rest.

16. The moment of inertia about

Integer Based Type

C

14. Massless springs, each with k = 1350 Nm−1 are

µ=0

10. A small hole of area a is at the

y

given in figure. One of the B refracting surface is given by y = x2 . A ray of light travelling parallel to x-axis is incident normally on face AB and h refracted. Find the minimum A distance of incidence ray from surface AD. Refractive index for prism is 3.

y

surface moving vertically upward with velocity 2 m/s. A small 2m/s ball of mass 2 kg is O moving with velocity 2$i − $j (m/s). If coefficient of restitution is 1/2 and coefficient of friction is 1/3. Find horizontal component of velocity of ball after collision.

p(N/m2)

(a) (b) (c) (d)

13. The cross-section of a prism as

18. A photoelectric plate is initially exposed to a spectrum of hydrogen gas excited to second energy level. Later when the same photoelectric plate is exposed to a spectrum of some unknown hydrogen like gas excited to second energy level, it is found that the de-Broglie wavelength of the photoelectrons now ejected has decreased 61 . times. For this new gas difference of energies of first Lyman line and Balmer series, limit is found to be two times the ionisation potential of the hydrogen atom. Detect the atom and find the value of Z.

75

19. The linear charge density of the circumference of a

U

ring of radius R is λ = λ 0 sin θ, where θ is defined as shown in figure. Find the electric dipole moment 1 (in C-m) of the ring, if λ 0 = C/m and R = 1 m π y

O θ

a

b

d

c

Vo

(a) 225 log 2 (c) 400 ln 2

x

2Vo

V

(b) 357 log 2 (d) 300 ln 2

5. An adiabatic vessel containing 3 moles of a 20. Find the power factor for the circuit as shown in figure. 20 Ω L=5/πH

XC=20 Ω

C=20/πµF XL=30 Ω

(a) 6 × 1012 rad/s (c) 4 × 1013 rad/s

V=220 V, f=50 Hz

Paper 2 temperature of the body T (not very different from temperature, T0 of surrounding) and time is given by (a) T

(d) T

(c) T

(b) T T0 t

O

O

t

t

O

t

welded with two beads of masses m1 and m2 , such that the ratio of the lengths of spring between the beads is 5 : 2. If the stiffness of the original spring is k , then the frequency of oscillations of the beads in a smooth horizontal rigid tube is (assume m1 = m and m2 = 2m) (a)

2. A floating ice slab of thickness is remained on temperature 0°C. If the atmospheric temperature is T °C, then time taken for thickness of the layer to increases from x1 to x2 is proportional to (take ρ as the density of ice, K as the thermal conductivity, L as the latent heat of fusion and A on the x-sectional area of slab). (a) (ρ2L ) (c)

ρ K 2T

ρL 2 ( x2 − x12 ) 2 KT ρL2 (d) ( x22 − x12 ) 2 KT (b)

2

3. Metal disc of radius 5 × 10 mm is set to rotate about central axis normal to its plane. Upon raising its temperature gradually, the radius is increased upto 5.075 × 102 mm. The percentage change in the rotational kinetic energy will must be (a) − 3% (c) − 2%

(b) − 9% (d) 4%

4. Consider the diagram of internal energy U versus V for 2 moles of an ideal gas under cyclic process. The temperature of the gas at b and c are 500 K and 300 K respectively. The total heat absorbed by the gas during the process is

76

(b) 3 × 10 9 rad/s (d) 9 × 10 8 rad/s

6. A circular spring having natural length l0 is cut and

1. The cooling curve of a body in reference to the

O

diatomic gas having moment of inertia of each molecule is I = 2.76 × 10−46 kg-m2 , while the root mean angular velocity is ω0 = 5 × 1010 rad/s. An another adiabatic vessel contains 5 moles of a monoatomic gas at temperature 470 K. The root mean square angular velocity of diatomic, molecules when the two vessels are connected by a thin tube of negligible volume is (assuming all the gases are ideal and Boltzmann’s constant K = 1.38 × 10−23 joule per molecule)

k m 17

(b) 7

k 15 m

(c)

1 k 105 m

(d)

17 15

k m

7. Consider a liquid filled in a uniform U-tube upto height has shown in figure. If the liquid is displaced through a small distance x, then the frequency of oscillation of the liquid will be (a)

2g h

(b)

g h

(c)

1 2

g /h

(d)

h

2g 3h

8. A monoatomic light beam of wavelength λ is incident on an isolated metallic sphere of radius a. The threshold wavelength λ 0 which is larger than λ . The number of photoelectrons emitted before the emission of photoelectron will stop is 4 πε0h2c  1 1 −   3ae  2 λ λ 0  2 4 πε0h ac  1 1 (c)  −  7e 2  λ λ0  (a)

4 πε0a 2h2c  1 1   −  3e 2  λ 2λ0  4 πε0ahc  1 1 (d)  −  e2  λ λ0  (b)

9. A wire having length l and mass m can freely slide on a pair of parallel smooth horizontal rails placed in a vertical magnetic field B. The rails are connected by a capacitor having capacitance C. The electric resistance of the rails and the wire is

zero. Suppose, a force F is applied to the wire place between the rails, then the acceleration of the wire will be (a)

F 3F (b) m + CBl 2 m + C 2 Bl

(c)

F F (d) m + C 2B 2l 2 m + CB2l 2

Passage 3 A hollow sphere of mass m is released from the top of an inclined plane of inclination θ as shown below

10. For the magnet of oscillation magnetometer, the frequency of oscillation is 40 oscillation per minute. A short bar magnet is placed to the North of the magnetometer, at a separation of 40 cm from oscillating magnet with its North pole pointing towards North. The frequency of oscillation is found to be 60 oscillation per minute. If the horizontal component of the Earth’s magnetic field is 12µ T, then the magnetic moment of this short bar magnet, will be (a) 32 . Am

2

(b) 18 . Am

2

(c) 4.8 Am

2

(d) 2.2 Am

2

Passage 1

11. Find the magnetic field at a distance x from the axis. µ 0 i0x 2 πa

(b)

µ 0 i0 2 πa

(c)

µ 0 i0x 2 πa

(d)

2

µ 0 i 0 x2 2 πa

12. The magnetic field at a distance x, where, b < x < c is (a) (c)

15. What should be the minimum coefficient of friction between the sphere and the inclined plane to just prevent the sliding of sphere? 4 (a) cosθ 5

µ 0i (c 2 − x 2 ) 2

(b)

2

2 π x(c − b )

µ 0 i 0(c 2 − x 2 ) 2

2

2 π(c − b )

µ i  c + x (d) 0 0   2 π c − b

µ 0 i 0(c − x ) 2 πx(c 2 − b 2 )

2 (b) sinθ 3

(c)

2 tanθ 5

2 (d) cosθ 3

16. The angular acceleration of sphere, while friction between sphere and plane becomes half of its previous value is (a)

There is a coaxial cable which consists of an inner wire of radius a surrounded by an outer shell of inner and outer radii b and c, respectively. The inner wire carries an electric current i0 and outer shell carries an equal current in opposite direction.

(a)

θ

3 g sinθ 10 R

(b)

3 gR sinθ 5

(c)

3R g sinθ 4

(d)

4 g cosθ 9 R

17. Match the following columns. Column I

Column II

A. Emission of radiation

1. Nucleon property

B. Automatic open and closing of doors

2. Photoelectric effect

C. Radioactivity

3. Surface property

D. Breader reactor

4. Generation of fuel itself

Codes A (a) 3 (c) 3

B 2 2

C 1 4

D 4 1

(b) (d)

A 1 2

B 2 1

C 3 3

D 4 4

18. Match the set of parameters given in Column I with the graph given in Column II.

Passage 2

Column I

A table with smooth horizontal surface is placed in a cabin which moves in a circle of a large radius R as shown below. A smooth pulley of small radius is fastened to the m table. Two masses m and 2m 2 m placed on the table are connected through a string going over the pulley. Initially the masses are held by a person with the string along the outward radius and then the system is released from rest with respect to cabin.

13. With respect to the cabin, the modulus of initial

Column II

A. Potential energy of a simple pendulum (y-axis a function of displacement ( x-axis)

y

1.

x

B. Displacement ( y-axis) as a function 2. of time ( x-axis) for one dimensional motion at zero or constant acceleration when the body is moving along the positive x-direction

y

C. Range of a projectile ( y-axis) as a 3. function of its velocity ( x-axis) when projected at a fixed angle

y

D. The square of the time period ( y-axis) of a simple pendulum as a function of its length ( x-axis).

y

x

x

acceleration of the masses as seen from the cabin is (a)

ω 2R 3

(b)

ω 2R 2

(c)

ω 2R 3

(d)

ω 2R 5

4.

14. The tension developed in the string is given by mω 2R 3 4 (c) mω 2R 5 (a)

(b)

4 mω 2R 3

(d) 4ω 2Rm

x

Codes A (a) p (c) q, r

B q, s q

C s p

D q s

A (b) q, s (d) r

B p, r s

C q p

D s q

77

19. The materials against its Poisson’s ratio is given in the columns which has to be matched. Column I

20. The quantities are given along with its dimension in two column which has to be matched.

Column II

Column I

Column II

1.

0.27

A.

Coefficient viscosity

B. Iron

2.

0.16

B.

Gravitational constant 2.

[ML−1T −1]

C. Steel

3.

0.19

C.

Thermal conductivity

3.

[M0 L0T 0 K 0]

D. Tungsten

4.

0.20

D.

A unit vector in a particular direction

4.

[MLT −3 K −1]

(b) (d)

A 2 2

Codes A (a) 2 (c) 3

B 1 1

C 4 2

D 3 4

(b) (d)

A 1 2

B 2 1

C 3 3

D 4 4

Codes A (a) 2 (c) 1

B 1 2

C 4 3

D 3 4

1.

[M−1L3T −2]

A. Aluminium

B 1 1

C 3 4

D 4 3

Answers Paper 1 1. (a,c) 11. (2)

2. (b,d) 12. (0)

3. (a,b,d) 13. (2)

2. (b) 12. (a)

3. (a) 13. (a)

4. (a,b,c,d) 14. (3)

5. (a,c) 15. (3)

6. (a,b,c) 16. (5)

7. (a,b,c,d) 17. (7)

4. (c) 14. (b)

5. (a) 15. (c)

6. (b) 16. (a)

7. (b) 17. (a)

8. (a,b,c) 18. (2)

9. (a,c) 19. (1)

10. (a,b) 20. (0)

8. (d) 18. (a)

9. (d) 19. (d)

10. (c) 20. (d)

Paper 2 1. (c) 11. (c)

Detailed solutions of these questions are available on http://www.arihantbooks.com/Physics%20Spectrum.pdf

78

Answer with Explanations 3. (a,b,d) According to wave motion, intensity and loudness of a

Paper 1 1. (a,c) According to the SHM, the basic idea is same as of simple pendulum its just there are two forces, one weight of vertical rod and other weight of horizontal rod due to which torque will be produced. ml 2 Moment of inertia for AB rod about A is and rod CD about A 3 is A =

ml 2 + ml 2 12

=

13 ml 2 12

17 ml 2 Total moment of inertia of T is 12 Restoring torque about A l τ A = mg sin θ + mgl sin θ 2 3 mgl = sin θ 2

B

C

As, sin θ ≈ θ i.e.

D

wave i.e. L = 10 log10

where, I0 is the intensity of minimum audible sound which is 10 −12 Wm −2 . For point source, sound will spread in the spherical region. So, intensity at any point A is P W I = 02 2 4πr m 1 So, I∝ 2 A P0 r r loudness of sound in decible will be  I L(dB ) = 10 log10    I0  (where, I0 = reference intensity)  I 60 = 10 log10    I0  I = I0 × 10 6

⇒

I = 10 −12 × 10 6 = 10 −6 W/m 2 P

θ

4 πr 2

 3 mgl  τA =  θ  2  mg

=

2

17 ml × 2 17 l =2π 12 × 3 mgl 18 g

force acting on a body equal to change in kinetic energy of the same body i.e. ∆KE = Wnet Loss in the energy of photon = hν − h( ν − dν ) = h(dν ) = work done by gravity hν Let say, m = mass of photon = 2 c −GMm   u =  x hdν = − du 0   du GMm   =+ 2   dx x  hdν =

 hν  − Gm  2  c 

dx = x2 GMhν hdν = − 2 2 dx c r ν dν − Gm R + h dx = ∫ν 0 ν c 2 ∫R x 2

x2

dx

250 π

∴ Due to opposite phase that would be written as −

8A . 55

4. (a,b,c,d) As conservation of linear momentum and energy, maximum height and velocity attained by block moving on smooth horizontal surface i= n

i.e.

i= n

∑ Pi = ∑ mi vi

(i)

c

2

=

250 m π 1 As, we know intensity of sound, I α A2 ∝ 2 , r 1 So A∝ r A 1 r 2 2 A 55 8A So, = , = , A2 = A2 r1 A2 4 55

u

2 2

v = ν 0 e − gh /c

4 × 3.14 × 10 −6

i =1

− GMh

Frequency of a photon,

1 × 10 −3

i =1

where, P is the momentum of N-particle system. Taking block and wedge as a system, we can analyse that there is no horizontal force acting on system. So, horizontal momentum is conserved.

 ν Gm  1 1  −GMh h  = + 2  −  = 2 2  ν0 c  R + h R Rc v = ν 0e R

P 4 π × 10 −6

Radius of an isotropic point source, r =

2. (b,d) According to the work-energy theorem, work done by all the

−GMm

(I0 = 10 −12 W/m 2 )

= I = 10 −6

r2 =

mg

l Time period of pendulum, T = 2 π K T =2π

I I0

GM   Q g = 2  R  

(ii)

v2 v1 Hmax v2

So, conservation of linear momentum, mu = mv1 − mv 2 and conservation of kinetic energy, then

v1 (iii) …(i)

79

mu 2 1 = mv12 + mv 22 2 2 Solving Eqs. (i) and (ii), we get v1 = u , v 2 = 0

…(ii)

u 2 So, conservation of mechanical energy upto maximum height, we get

∴ From beginning to maximum height mu = m(2 v ) ⇒ v =

2

1 1 u  u2 mu 2 = m   × 2 + mgh, h = 4g 2 2 2 

5. (a,c) We have to use the steady state concept, i.e. in steady state capacitor will work as open circuit. We will calculate the potential difference across R2 and accordingly charge will be calculated The charge will be = C (potential difference across R2 ) Q 0 is the steady state charge stored in the capacitor Q 0 = C (potential difference across capacitor in steady state) = C(steady state current through R2 ) ( R2 )  V  R1 = C  R2  R1 + R2  Q0 = As, where,

β= Rnet = β=

β=

CVR2 R1 + R2

V

R2

C

1 CRnet R1 R2 R1 + R2

R1

1  R1 R2  C   R1 + R2 

R2

R1 + R2 CR1 R2

6. (a,b,c) The mass of the portion of the ring, determine field and potential due to each part and then find the net potential and net field at point O due to elementary masses dm a1 , dm a2 while comparing, we get Elementary masses, dm a1 = λ ( r1dθ ) and dm a2 = λ ( r2 d θ ) dm a1 r = 1 dm a2 r2 Gravitational field at A dE1 = − and

dE 2 =

Gdm1 r12

r1 a2

−Gdm 2 r22

dθ r2

 dm   r  dE1 dm r2 = − 21 × 2 2 =  1  ×  2  dE 2  dm 2   r1  r1 dm

2

2

r r  r = 1 ×  2 = 2 r2  r1  r1 dE1 As, r2 > r1 , So > 1, dE1 > dE 2 dE 2 ∴ Gravitational potential −Gdm1 Gdm 2 , dv 2 = − dV1 = r1 r2 dV1 dm1 r2 r1 r2 = × = × = 1 ⇒ dV1 = dV2 dV2 dm 2 r1 r2 r1 i.e. element a, produces equal potential to element a 2

80

A

dθ

a1

7. (a,b,c,d) First conservation of mechanical energy is used i.e. loss in potential energy mgh is converted in to KE 1 mgh = mv 2 2 ⇒ v = 2 gh When the cylinder starts moving on the plank rolling as well as rotational motion will be retarted. Initially block is having translation as well as rotation and after sometime it will start pure rolling on the plank. Form conservation of energy we can say that the velocity of cylinder just before getting onto plank is 1 mgh = mv 02 2 where, v 0 = 2 gh FBD for motion on rough plank. So, cylinder is having retardation of linear motion. v = u 0 − µgt and it is having angular acceleration. µmgR α= mR 2 / g µ mg 2µg = R 2 µgt So, ω=0+ R Plank will be having accelerated motion v ′ = 0 + µgt At pure rolling v ′ = v − ωR µgt = v 0 − µgt − 2 µgt v t= 0 4µg 2v v Velocities of cylinder and wedge v = 0 = v ′ = 0 4 4 mv + mv ′ v + v ′ v 0 vCOM = = = 2m 2 2 For minimum length of plank let us consider cylinder (w.r.t.) plank, µ mg for linear motion distance moved by COM in time of pure rolling 1 S = vot − (2 µg ) t 2 µ mg 2  v  1  v  = v 0  0  − × 2 µg ×  0   4µg  2  4µg  =

va µ mg

vo

2

v 02 v2 3 v 02 − 0 = 4 mg 16µg 16µg

8. (a,b,c) As first law of thermodynamics and its processes i.e. isochoric and isobaric Q = U + ∆W where, Q is heat energy, U is potential p energy and ∆W is the work done by the A p B system. In AB process, W > 0 and ∆U > 0 because due to volume increment temperature also increases so, ∆Q for p/4 C AB will also be positive. So heat is given in process AB, in BC work done V 4V V is zero and temperature decreases due to ∆U and ∆Q will be negative that means heat is rejected by gas.

Q1 = W1 + ∆U1

=

 ( 4 V ) pV  = p( 3 V ) + nC V  p −   nR nR  p   ( 4 V ) p( 4 V )  Q 2 = ∆U 2 = nC V  4 −  nR   nR   So,

=−

log 2 λ

( N 0 )A 1 = ( N 0 )B 4 R0A λ A N 0 A 1 = = R0B λ B N 0 B 8 λA 1 = λB 2

λB =2λA (t1 / 2 )A = 2 × (t1 / 2 )B = 2 × 1 = 2 yr 1 1 At 4 yr, A will be and B will be (4 half-lives) 4 16 1 N 0A ( Nt )A 1 = 4 = 4 × =1 1 ( Nt )B 4 N 0B 16 ( Rt )A NtA ⋅ λ B 1 1 = =1× = ( Rt )B NtB ⋅ λ B 2 2 1 1 and B will be 8 64 1 N 0A ( Nt )A = 8 =2 ( Nt )B 1 / 64 N 0B

At 6 yr, A will be

of continuity at two levels v1 considered. The height of the considered level will be taken as zero, hence the term containing height i.e. ρgh = 0 Apply Bernoulli’s equation between P and Q points 1 1 p0 + ρgh + ρv12 = p0 + 0 + ρv 22 2 2 ρ ρgh = ( v 22 − v12 ) 2 v 22 − v12 = 2 gh From equation of continuity, we get A Av1 = av 2 = v 2 3 v 2 = 3 v1 9 v12 − v12 = 2 gh 8 v12 = 2 gh v1 =

gh gh = 4 2

Acceleration of top layer of liquid, dv 1 1 dh a= 1 = g × × dt 2 2 h dt

P h Q

g h

×

gh g = − ms −2 2 8

1 2 ρv + ρgh = constant 2 1 where, p is pressure energy, ρv 2 is kinetic energy per unit 2 volume and ρgh is the potential energy per unit volume Apply Bernoulli’s equation at pipe 1, 2, and 3 1 1 ρ p3 + ρv 32 = p2 + ρv 22 = p1 + v12 2 2 2 1 ρ p0 + ρgh3 + ρv 32 = p0 + ρgh2 + v 22 2 2 ρ = po + ρgh1 + v12 2 v 32 v 22 v2 …(i) gh3 + = gh2 + = gh, + 1 2 2 2 Equation of continuity at point 3 and 2 are v 3 A3 = v 2 A2 …(ii) v3 × A = v2 × 2 …(iii) v 2 × 2 = v1 × 1 From Eq. (i) v2 v2 gh2 + 2 = gh1 + 1 2 2 35 v 22 20 v12 10 × + = 10 × + 100 2 100 2 2 2 15 v12 − v 22 4 v 2 − v 2 3 v 22 = = = 10 2 2 2 v 2 = 1 m/s, v1 = 2 m/s From Eq.(i), v2 v2 gh3 + 3 = gh2 + 2 2 2 37.5 v 32 35 1 10 × + = 10 × + ⇒ 100 2 100 2 2 v 3 1 35 37.5 1 2 .5 1 25 = + − = − = − 2 2 10 10 2 10 2 100 v 32 1 = 2 4 1 1 ⇒ v 32 = , v 3 = 2 2 i.e.

10. (a,b) Apply Bernoulli’s equation

1 4

or streamline flow.

9. (a,c) According to radioactive law, we know

⇒

g × ( − v1 ) h

11. (2) It is based on Bernoulli’s equation for flowing liquid for steady

Q1 > Q 2

R = λN 0 and t1 / 2 =

1 4

PE=0

v2

…(i)

p+

The value of x is

x =2

12. (0) According to impulse and conservation of energy in elastic collision. Ball with respect to surface upward impulse on ball ∫ Ndt = Pf − Pi

∫ Ndt = 2 vy − ( − 2 × 4 ) ∫ Ndt = 2 vy + 8

Vy 2m/s2 4m/s

…(i)

The coefficient of restitution e along normal direction v e= y 4 v g = 4 e = 2 m/s

Vr

2

N

…(ii)

81

So, ∫ Ndt = 4 + 8 = 12 Horizontal impulse − ∫ µNdt = Pf − Pi −µ ∫ Ndt = 2 v x − 2 × 2 1 − × 12 = 2 v x − 4 3 i.e. vx = 0

13. (2) Consider the rays as

1 shown in figure, for ray i1 1 with angle of 2 incidence i1 , (i1 < θC ) i2 where, (θC is critical H1 angle). The ray will get 3 i3 refracted, but H1 is not H2 the minimum height. H3 For ray 3 with angle of incidence i 3( i 3 < θC ) the ray will get reflected (due to phenomenon of TIR), now for light ray 2 with angle of incidence i 2 ( i 2 = θC ) the ray will just refracted (limiting ray) and corresponding H 2 will be minimum height so, that light will just refracted. For TIR at that point of incidence on curved surface. We can early say that, as h increases i decreases and for TIP at the point of incidence minimum value of i should be θC and corresponding to that h will be minimum for refraction.

According to Snell’s law, we get 3 sin i = 1sin 90 ° 1 sin i = 3 So, slope of tangent will be 90° − i dy = tan( 90 ° − i ) dx 2 2 = cot i = 1 At, y=h dy =2 x dx 2 x =2 2

y=x2 h

i (90° – i) A

(90° – i) D

x = 2m y = h = ( 2 )2 = 2 m The minimum distance of incident ray from surface AD is 2 m.

14. (3) It is based on conservation of mechanical energy for the two springs and the non-conservative force i.e. friction will dissipate same part of mechanical energy. Velocity of block at A is according to conservation of energy mv 2 Kx 2 = 2 2 1 × v A2 1350 × ( 0.1)2 = 2 2 1350 1350 2 m/s vA = ⇒ vA = 100 100

Now, with this KE, it will compress right spring, we get mv B2 kx f2 = 2 2 1 150 1350 × = × x f2 2 100 2 150 1 1 x f2 = × = 1300 100 900 1 100 xf = = = 3.33 cm ~ −3 30 30

15. (3) It is based on the thermal conductivity of the material and bulk modulus of elasticity. We can directly relate change in length of steel bar with stress applied in bar and easily. Consider an element of the bar of the length dx shown in figure.

x

Change in length of element dx due to temperature difference is dU = dx × α × ∆T T x2 dU = dx × α × 0 2 L Integrating both sides, we get Lα T x2 ∫ dU = ∫0 L02 dx αT L3 U = 02 3L Bar is rigidly field stress in given by σ = E × strain U = ×E L α T0 L3 E α T0 σ= 2 ×E = 3 3L ⋅ L E α T0 Stress in bar, σ = 3 As per given in question, we get E α T0 E α T0 = x 3 So, x= 3

16. (5) Consider small segment that will be disc at distance x as shown dm = ρ ⋅ dV For the moment of inertia of the disc about then given axis 1 dI = dm ⋅ y 2 2 where, y is the radius of a solid spherical body The solid sphere is generated by revolving circle about diameter. Mass of spherical segment x2+y2=R2

A

During motion of a block from A to B, v B2 = v A2 + 2 as a = − µ g = − 0.3 × 10 = − 3 m/s 1350 1350 = −2 × 3 ×2 = − 12 100 100 1350 − 1200 150 = = m/s 2 100 100

82

dx

O

R

y

x

C

x

dx

M = density × volume of segment

ρ=

M = V

M 2 ∫ π y dx

R/ 2

M

ρ=

3

=

M

=

R

where, λ is wavelength and E is energy of an electron. E1 = W0 + KE1

R

2 2 ∫ π ( R − x )dx

R/ 2

E1=(2 → 1 in H-atom)

24 M 3

5πR 5πR 24 Consider a small disc of radius y and thickness dx as shown is figure. Mass of small disc dm = ρ ⋅ π y 2dx Moment of inertia of disc about O, x-axis 1 1 ρπ 4 dI = dm ⋅ y 2dl = ρπ y 2dx × y 2 = y dx 2 2 2 Total moment of inertia ρπ R 4 I = ∫ dl = y dx 2 ∫R / 2 ρπ R I= ( R 2 − x 2 )2 dx 2 ∫R / 2 ρπ R 4 = ( R − 2 R 2 x 2 + x 4 )dx 2 ∫R2 I= =

⇒

24 M 5πR 3 24 M 3

π  4 −2 R 2 x 3 x 5  × R × +  2  3 5  R / 2

×

π 53 R 5 × 2 480

5πR 53 53 MR 2 I= MR 2 = 200 40 x x=5 Total work done Total heat supplied WAB = p ( ∆V ) = nR∆T (isochoric process) 25 =2 × × 300 = 5000 J 3 T  p WBC = nRT ln  f  = nRT ln  i  (isothermal process) T  pf   i  2 × 10 5  = 2 × R × 600 ln   = 1200 R ln(2 ) 5  1 × 10  WCA =

nR∆T 2 R × 300 =− = − 1200R 1− y 1 − y2

W = 600 R + 1200 R ln(2 ) − 1200 R W = 600 R [2 ln(2 ) − 1] Heat supplied to the monoatomic gas from A to B, 5R Q AB = nC p ∆T = 2 × × 300 = 1500R 2 Similarly, Q BC = WBC = 1200 R ln(2 ) 600 R [2 ln (2 ) − 1] Efficiency of a cycle, η = 1500 R + 1200 R ln(2 ) 12 ln(2 ) − 6 21 = =1− 15 + 12 ln(2 ) 15 + 12 ln(2 ) 3 ×7 =1 − 15 + 12 ln(2 ) ∴

e–

W0

KE1

E2=(2 → 1 in unknown gas)

e–

W0

KE2

…(ii)

Also,

E 2 = W0 + KE 2 1 λe ∝ KE

⇒

KE 2  λ 1  =   = 6.1 KE1  λ 2 

Also,

1 1 E1 = 13.61 × 12  2 − 2  1 2 

2

R

×

17. (7) Net work done for the process, Efficiency =

…(i)

1   λ1  Q λ 2 = 6.1  

= 10.2 eV Now, ionisation energy of H-atoms = − E1 = 13.6 eV Energy of first line in Lyman series 1 1 = 13.6 Z 2  2 − 2  eV =10.2 Z 2 eV 1 2 

…(iii)

and energy of series limit of Balmer series  1 1  2 = 13.6 × Z 2  2 −  eV = 3.4 Z eV ( ∞ )2  2 Given,

13.6 × 2 = 10.2 Z 2 − 3.4 Z 2

⇒

Z = 2 ⇒ He + ion

19. (1) As the charge density vary with sine function, then half part of ring will be positive and half part will be negative. We can consider positive element of find the dipole moment of this small part. Then, p=∫

π /2

0

=∫

2 dpsin φ [direction − j]

π /2

0

2 [ λ 0 sin θ Rd θ ⋅ 2 R ]sin θ

= 2 λ 0R 2 ∫

π /2

0

(1 − cos 2 θ )dθ

 π = 2 λ 0 R 2   = πR 2 λ = 1 2 

20. (0) The concept of complex number in which we calculate impedance of circuit by treating as pure real number and capacitance, inductance as pure imaginary number

20 Ω L=5/πH

XC=20 Ω

C=20/πµF XL=30 Ω

x =7

18. (2) It is based on photoelectric effect and ionisation energy of an atom. So, using de-Broglie relation h i.e. λ = 2 mE

V=220 V, f=50 Hz

83

1 = 500 Ω 20 2 π × 50 × × 10 −6 π 5 For L1 X L = ωL = 2 π × 50 × = 500 Ω π For 1st part of circuit, Z1 = R || Z1′ Z1′ = j ( X L − X C ) = j ( 500 − 500 ) = 0 20 + j × 0 Z1 = =0 20 + j × 0 For

C1 XC =

1 = ωC

As, we know the that moment of inertia, I for the disc is where, R is radius of disc. ⇒

L2 L2 = ( R −2 )  MR 2  M 2    2 

Thus, % change in rotational kinetic energy, dK 2 dR K r = r × 100 = − × 100 = − 3% Kr R

4. (c) For a cyclic process ∆U = 0

Paper 2

⇒

1. (c) As final temperature of a body T = T0 + ∆T

   ∆T =  1 +      T0   = T04

4

 − 1   

T 4 − T04 = 4T03 ∆T

...(i)

T0 t

O

by dx in time dt. Then amount of heat flowing through the slab in time dt is given by. KA [ 0 − (T )] dt KAT dt ...(i) Q= = x x Now, if dm is the mass of water frozen into ice then, Q = dm × L but dm = Aρ dx ...(ii) ⇒ Q = AρL dx Equating Eqs. (i) and (ii), we get KATdt ρL = AρL dx ⇒ dt = xdx x KT By integration, we get, ρL

x2 1

xdx ⇒ t =

ρL KT

nRT V

⇒

p=

Therefore,

W =∫

⇒

W = nRT [ln V ]vv f = nRT ln

x2

 x2     2  x1

Time taken to increase the thickness of the layer ρL t= [ x 22 − x12 ] 2 KT

3. (a) As, percentage change in radius of a metal disc, dR ( 507.5 − 500.0 ) × 100 = × 100 = 1.5% R 500.0 L2 1 1 Rotational kinetic energy K r = Iω 2 = × ( Iω 2 ) = 2I 2 2I

vf

vi

nRT dV V

[from Eq. (i)]

i

For process ab, Wab = nRTb ln

For process cd , wcd

T

2. (b) Let the thickness of ice at 0°C is x. If the thickness is increased

t

...(i)

pdV

We have pV = nRT

  ∆T − 1 1 + 4 T0  

∫0 dt = KT ∫x

vf

vi

(using binomial theorem)

By Stefan’s law dT eAσ 4 = [T − T04 ] dt mc Form Eq (i), we get dT eAσ 3 = 4T0 ∆T dt mc dT dT So, ∝ ∆T or ∝ (T − T0 ) dt dt Thus, curve will be given as showin in figure

∆T = 0 ⇒ i.e. T = constant.

Now, the work done, W = ∫

According to Newton’s law of cooling, we get T 4 − T04 = [(T0 + ∆T ) 4 − To4 )] ⇒

84

Kr =

On differentiating, we get dK r 2 dR =− Kr R

For 2nd part, Z 2 = j ( X L − X C ) = j × 10 Ω Equivalent is Z = Z1 + Z 2 = 0 + 10 j R 0 So, power factor = = =0 Z 10 j

⇒

MR 2 2

Vf Vi

2 V0 V0

= 1000 R ln 2 V 1 = nRTc ln 0 = 2 R × 300 ln 2 V0 2

(QTb = 500 K )

= − 600 Rln 2 As the process bc and da is isochoric so, Wbc = 0 and Wda = 0 Work done in complete cycle W = Wab + Wbc + Wcd + Wda = 1000 R ln 2 + 0 − 600 R ln 2 + 0 = 400 Rln 2 This is the value of heat involved in the given cyclic process.

5. (a) Equipartition of energy tells that energy associated with each 1 KT. As the diatomic gas molecule has two 2 rotational degree of freedom so, final temperature of the system. f n T +f n T Tf = 1 1 1 2 2 2 f1 n1 + f2 n 2 degree of freedom is

For diatomic gas, F1 = 5, n = 3 and T1 = 250 K For monoatomic gas, f2 = 3, n 2 = 5 and T2 = 470 K 5 × 3 × 250 + 3 × 5 × 470 ∴ Tf = = 360 K 5× 3 + 3 × 5 Suppose, the final angular velocity of diatomic gas molecule is ωrms, f then according to law of equipartition of energy 1 2 Iω rms, f = KT 2 2 KT Angular velocity of diatomic molecule, ω rms. f = I = ⇒

2 × 1.38 × 10 −23 × 360 2 .76 × 10 −46 12

ω rms f = 6 × 10 rad/s

6. (b) The circular spring is given below.

The two springs can be assumed to be parallel between two masses m1 and m 2 . K1 The equivalent spring constant m1 m2 Keq = K1 + K 2 K2 m1 m 2 while, µ = m1 + m 2 1 If the spring cut, the force constant of spring K ∝ . l ⇒ K 2 l 2 = K1 l1 = Kl 5l 2l Substituting l1 = and l 2 = , we get 7 7 5l K1 = Kl 7 7K 2l 7K and K 2. = Kl ⇒ K 2 = ⇒ K1 = 5 7 2 7K 7K Therefore, Keq = K1 + K 2 = + 5 2 14 K + 35 K 49 = = K 10 10 The reduced mass, m m m × 2 m 2 m2 2 = = m µ1 = 1 2 = m1 + m 2 m + 2 m 3 m 3 Now, frequency of oscillation. 49 k keq 49 2 k k 10 ω= = = × =7 3 µ 10 3 m 15 m m 2

7. (b) Let the density of liquid be ρ. The total mechanical energy of

9. (d) Let the speed of the wire at any time t be v. The emf in induced between the ends of wire. E = vB l As rails and wire are of zero resistance then, charge on capacitor. q = C E = CvBl dq dv The current through the circuit i = = CBl = CBla dt dt The current is the cause to force on the wire. Thus, the magnetic force F ′ = i lB = CB 2 l 2 a So, the net force F − F ′ = ma F − CB 2 l 2 a = ma ⇒ Acceleration of the wire,

2

h+ x h − x  +  A ( h + x )ρ g + A ( h − x ) × ρ g. 2 2   2

⇒

E=

1  dx  (2 A h ρ )   + 2 Aρg ( h 2 + x 2 )  dt  2

On differentiating, we get This implies ω 2 =

d 2x dt 2

= acceleration = −

2 Aρg x 2 Aρh

2 Aρg 2 Aρh

Thus, the frequency of oscillation of the liquid, 2 Apg g or ω= = 2 Ahρ h

8. (d) If the potential of sphere is raised to V, the electron should have a minimum energy φ + eV to be able to come out. Thus, emission of photoelectron will stop when hc hc = φ + eV = + eV λ λ0

a=

F m + CB 2 l 2

10. (c) Let the magnetic field produced by short bar magnet be B. As short bar magnet is oscillating, hence the resultant horizontal field is BH + B. Now, suppose, M and M′ denote the magnetic moments of oscillating magnet and other magnet, respectively. 1 MBH Then, ν= 2π I and ⇒

ν′ = ν ′2 v2

=

1 2π

M( BH + B ) I

BH + B BH

2

⇒

B  60   =1+  40 BH

⇒

B = 1.25 BH

liquid column is 1  dx  { A ( h + x ) ρ + A ( h − x ) ρ}    dt  2

1 1   −  λ λ0 

The change on the sphere needed to take its potential to V is φ = ( 4 πε 0 a ) V The number of electron emitted is 1 φ 4 πε 0 aV 4 πε 0 ahc  1 = n= =  −  e e  λ λ 0 e2

m2

E=

hc e

V=

⇒

m1

∴

B = 125 . × 12 µT = 15 × 10 −6 T

As the magnetic field can be written as µ 2 m′ B= 0 4π d 3 2π m′ = Bd 3 = 0.5 × 10 7 × 15 × 10 −6 × ( 40 × 10 −2 )3 µ0 = 0.5 × 15 × 40 × 40 × 40 × 10 −5 1 = 64 × 15 × × 10 −2 = 480 × 10 −2 = 4.8 Am 2 2

11. (c) Let m assume a circle of radius x with the centre at the axis of the cable. The magnetic field at each point of the circle will have the same magnitude and will be tangential to it. The circulation of field B along the circle is ∫ B ⋅ d l = B 2 πx The current enclosed within the circle is i0 i . πx 2 = 02 x 2 πa 2 a

85

Ampere’s law tells that

Thus, B2 πx =

µ 0 i0 x 2

Therefore, Eq (i) becomes 3 mg sin θ − f = mg sin θ 5 2 ⇒ f = mg sin θ 5

∫B ⋅ d l = µ 0 i

a2 B=

⇒

µ 0i0 x

2 mg sin θ 5 2 Minimum coefficient of friction, µ = tan θ 5

2 πa 2 The direction is along the tangent to the circle.

12. (a) The area of x-section of the outer shell is πc 2 − πb2 . The area of x-section of outer shell within part c is πx 2 − πb2 . Thus, current i ( x 2 − b2 ) through this part is 0 2 . The net current enclosed by the c − b2 circle is i ( x 2 − b2 ) i 0(c 2 − x 2 ) i0 − 0 2 = c − b2 c 2 − b2 From Ampere’s law, B 2 πx = B=

⇒

16. (a) The friction becomes its half when the coefficient of friction becomes half of its original value, then 1 2 1  2 µ =  tan θ = tan θ = tan θ  10 2 5 5 Then, net torque = Iα (given) µmg cos θR =

µ 0 i(c 2 − x 2 )

2

µN = µmg cos θ =

⇒

2 mR 2 α 3

1 2 × tan θ × ( mg cos θ ) × R = mR 2 α 10 3 Angular acceleration of a sphere 3  g sin θ  α= ⇒   10  R 

c 2 − b2

⇒

2

µ 0 i 0(c − x ) 2 πx (c 2 − b2 )

13. (a) Consider the diagram below. Let the bigger mass of 2 m accelerates towards right with acceleration be a. Then, free body diagram can be given as

17. (a) (A) Emission of radiation → Surface property.

M1 m

(B) Automatic open and closing of doors → Photoelectric effect

M2 2m

(C) Radio activity → Nucleon property

a

(D) Breader reactor → Generatiuon of fuel itself

T

m

mω2R

T

2m

2mω2R

18. (a) (i) Potential energy of a simple pendulum y

a

For mass along M 2 2 mω 2 R − T = 2 ma

...(i)

For mass along M1 , T − mω 2 R = ma

...(ii) x

Adding Eqs (i) and (ii), we get, mω 2 R = 3 ma ⇒ a =

(ii)

2

ω R 3

Range of a projectile when projected at a fixed angle y

14. (b) Now, substituting the value of a in Eq (i), we get ω 2R ⇒ 3 4 Tension developed in the string T = mω 2 R 3 2 mω 2 R − T = 2 m.

15. (c) Consider the diagram.

x

(iii) a

The hollow sphere is released from mg sin θ the top of incline having angle of θ inclination θ. mg mg cos θ θ To prevent the sliding, the sphere will make only pure rolling in this condition. ...(i) mg sin θ − f = ma where, f is friction force Now, torque produced by the friction about the centre of sphere (let radius of sphere is R) 2 2  a ...(ii) f × R = mR 2 ×   ⇒ f = ma  R 3 3 where, a = acceleration Now, from Eqs. (i) and (ii), we get 2 3 mg sin θ − ma = ma ⇒ a = g sin θ 3 5

86

As, square of time period of a simple pendulum as a function of its length y

x

19. (d) (A) Aluminium → 0.16 (B) Iron → 0.27 (C) Steel → 0.19 (D) Tungsten → 0.2 4

20. (d) (A) Coefficient of viscosity → [ML−1 T −1 ] (B) Gravitational consatant → [M -1L3 T −2 ] (C) Thermal conductivity → [MLT −3K −1 ] (D) A unit vector in a particular direction → [M ° L ° T ° K ° ]

TEST RIDER The Simulator Test Series Towards JEE Main and Advanced (a) 5 V

Instructions p p p

p

This test consists of 30 questions. Each question is allotted 4 marks for correct response. Candidates will be awarded marks as stated above for correct response of each question. 1/4 marks will be deducted for indicating incorrect response of each question. No deduction from the total score will be made if no response is indicated for an item in the answer sheet. There is only one correct response for each question. Filling up more than one response in any question will be treated as wrong response and marks for wrong response will be deducted according as per instructions.

1. The potential energy U of a particle varies with distance x from a fixed origin as U =

A x , where A x2 + B

and B are dimensional constants. The dimensional formula for AB is

2. A car has to go from station A to station B (the car will start from rest from station A and comes to halt at station B ), which are at a distance of 500 m in minimum possible time. Find this minimum possible time. The car has a limiting acceleration of 7 m/s2 and it can have a maximum speed of 35 m/s. (b) 20 s

(c) 10 s

(d) 19.28 s

3. A block of mass 1 kg is attached to one end of a spring of force constant k = 20 N/m. The other end of the spring is attached to a fixed rigid support. This spring block system is made to oscillate on a rough horizontal surface The initial ( µ = 0.04). displacement of the block from the equilibrium position a = 30 cm. How many times the block passes from the mean position before coming to rest? (g = 10 m/s2 ) (a) 11

(b) 7

(c) 6

(d) 15

4. What is the potential difference between points A

6V

5V 6Ω

3Ω 1Ω

4Ω B

22

4V

C

A 2Ω

D

(d) 11.4 V

Q1 and Q2 (< Q1 ) respectively. If they are now brought close together to form a parallel plate capacitor C, the potential difference between them is (a)

(Q1 + Q 2 ) 2C

(b)

(Q1 + Q 2 ) C

(c)

(Q1 − Q 2 ) C

(d)

6. A particle of charge −q and mass m P

enters a uniform magnetic field B (perpendicular to paper inwards) at P with a velocity v0 at an angle α and leaves the field at Q with velocity v at angle β as shown in figure. Then, 2 mV0 sinα (c) PQ = Bq

v

(Q1 − Q 2 ) 2C B

α v0

β Q

(b) v = v 0

(d) particle remains in the filed for time t =

2 m( π − α ) Bq

7. In the circuit shown in figure, L and switch S is closed at C time t = 0. The current through capacitor and inductor will be equal after a time t equal to R=

(a) CR

(b) CR ln (2)

(c)

L R ln (2 )

L

R

C

R

V

S

(d) LR

8. Radiation falls on a target within a solenoid of 2000 turns per metre, carrying a current of 2.5 A. Electrons emitted move in a circle of radius 10 mm. What is the wavelength of radiation? The work function of the target is 0.5 eV, e = 16 . × 10−19 C, −34 −31 h = 6.63 × 10 J - s, m = 91 . × 10 kg. (a) 1.76 nm

and B of circuit shown in figure?

(c) 10.4 V

5. Two identical metal plates are given positive charge

(a) α = β

(a) [M1L7/ 2T −2 ] (b) [M1L11/ 2T −2 ] (c) [M1L5/ 2T −2 ] (d) [M1La / 2T −2 ]

(a) 15.65 s

(b) 9 V

(b) 3.52 nm

(c) 4.28 nm

(d) 7.04 nm

9. A radioactive element X converts into another stable element Y. Half-life of X is 2 hr. Initially, only X is present. After time t, the ratio of atoms of X and Y is found to be 1 : 4. Then, t in hours is (a) 2 (c) 6

(b) 4 (d) between 4 and 6

10. What is the base resistence RB in the circuit as RC

(a) be at rest  h1 − h2  (b) be moving with an acceleration of g    h1 + h2 + h 

shown in figure, if hFE = 90? 2 kΩ

9V

g 2( h1 + h2 + 2 )

(c) be moving with a velocity of ( h1 − h2 )

RB 4V

(d) exert a net force to the right on the tube

16. A solid whose volume does not change with

3V

(a) 29 kΩ

(b) 82 kΩ

(c) 108 kΩ

(d) 55 kΩ y

11. A disc of mass M and radius R is rolling with angular speed ω on a horizontal plane as shown in the figure. The magnitude of angular momentum of the disc about the origin O is (a)

1 MR 2ω 2

(b) MR 2ω

(c)

temperature floats in a liquid. For two different temperatures t1 and t2 , the fraction f1 and f2 of volume of solid remain submerged. What is the coefficient of volume expansion of liquid? (a) ω

3 MR 2ω 2

x

(d) 2 MR 2ω

12. A satellite is revolving around the Earth with orbital speed v0 . If it stops suddenly, the speed with which it will strike the surface of Earth would be (v e = escape velocity of a particle on Earth’s surface). (a)

ve2 v0

(b) v 0

(c) ve2 − v 02

(d) ve2 − 2 v 02

B are two points at which its velocity is zero. It passes through a certain point P ( AP < BP) at successive intervals of 0.5 s and 1.5 s with a speed of 3 m/s. Then, (a) the maximum speed of particle is 3 2 m/s (b) the maximum speed of particle is 2 m/s 2 −1 AP (c) the ratio is BP 2 +1 AP 1 (d) the ratio is BP 2

(c)

f1 + f2 f2 t 1 + f1t 2

(d)

f1 + f2 f1t 1 + f2 t 2

taken through a cyclic process shown in the V-T diagram. Which of the following statement is/are true?

A

2V0

B

V V0

C

D

(a) The magnitude of work done by the gas is RT0 loge 2 (b) Work done by gas is V0T0 (c) Net work done by the gas is zero (d) Work done by the gas is 2 RT0 loge 2

T0

T

2T0

a plane surface of glass. The angle of incidence is 60° and refractive index of glass is µ = 3 / 2. The diameter of the refracted beam is (a) 4.00 cm (c) 3.26 cm

(b) 3.0 cm (d) 2.52 cm

19. Two identical circular loops of metal wire are lying on a table without touching each other. Loop A carries a current which increases with time in response, the loop B

14. Water from a tap emerges vertically downwards with an initial speed of 1.0 m/s. The cross-sectional area of a tap is 10−4 m2 . Assume that the pressure is constant throughout the stream of water and that the flow is steady, the cross-sectional area of stream 0.15 m below the tap is (b) 10 . × 10 −4 m2 (d) 2.0 × 10 −5 m2

15. The U-tube has a uniform cross-section as shown in the figure. A liquid is filled in the two arms upto heights h1 and h2 and then the liquid is allowed to move. Neglect viscosity and surface tension. When the level equalise in the two arms, the liquid will

f1 − f2 f1t 1 − f2 t 2

18. A circular beam of light of diameter d = 2 cm falls on

13. A particle is excluding SHM on a straight line A and

(a) 5.0 × 10 −4 m2 (c) 5.0 × 10 −5 m2

(b)

17. One mole of an ideal gas is

m O

f1 − f2 f2 t 1 − f1t 2

(a) remains stationary (b) is attracted by the loop A (c) is repelled by the loop A (d) rotates about its CM, with CM fixed

20. Two long parallel wires are at a distance 2d apart. They carry steady equal currents flowing out of the plane of the paper as shown in the figure. The variation of the magnetic field B along the line XX′ is given by

(a)

X

X′

(b) X

X′

d

h1 X

h2 (c)

X

X′

d

d d

X′

(d)

h d

d

23

21. A tuning fork of 512 Hz is used to produce resonance in a resonance tube experiment. The level of water at first resonance is 30.7 cm and at second resonance is 63.2 cm. The error in calculating velocity of sound is (a) 204.8 cm/s (c) 51.2 cm/s

(b) 1024 cm/s (d) 161.3 cm/s

the graph of log I versus time is as shown by the dotted line in log I the adjoining diagram, where I is the current. When the value of resistance is doubled, which of the solid curves best represents the variation of log I versus time?

S R Q P t

(b) Q (d) S

23. In a transistor used as an amplifier β = 60, RL = 6 kΩ and internal resistance of the transistor is 600 Ω . The ratio of power amplification to the voltage amplification would be (a) 120 (c) 600

(b) 60 (d) 300

24. The index of refraction of heavy flint glass is 1.68 at 434 nm and 1.65 at 671 nm. Calculate the difference in the angle of deviation of blue (434 nm) and red (671 nm) light incident at 65° on one side of a heavy flint glass prism with apex angle 60°. (a) 3.9° (c) 3.1°

(a) 8.57 cm, 17.5 cm and 1.4, 0.4 (b) 9.64 cm, 12.5 cm and 2.4, 3.1 (c) 7.50 cm, 10.4 cm and 1.5, 0.6 (d) 8.65 cm, 18.6 cm and 1.0, 0.8

27. From the figure A, missile of

22. In an RC circuit while charging,

(a) P (c) R

the lens and still be focused on the film and what are the magnifications of the images of an object at the nearest and farthest distances?

(b) 2.9° (d) 2.5°

25. When an atom of U235 undergoes fission reaction in a reactor, about 200 MeV of energy is liberated, suppose that a reactor using Uranium−235 has an output of 700 MW and is 20% efficient. How many uranium atoms does it consume in one day? What mass of uranium does it consume each day? (a) 9.4 × 10 23d −1 and 3.5 kg (b) 9.3 × 10 24 d −1 and 37 . kg (c) 9.5 × 10 24 d −1 and 37 . kg (d) 9.2 × 10 24 d −1 and 3.8 kg

400 m/s

mass m moving with velocity v (v = 200 m/s), explodes in 60° mid air, breaking in two parts θ M 3M of masses and . If the 4 4 v2 smaller piece flies off at an angle of 60° with respect to the original direction of motion with an initial speed of 400 m/s, then what is the initial velocity of the other piece? (a) 225 m/s (c) 220 m/s

(b) 231 m/s (d) 227 m/s

28. In hydrogen spectrum, the wavelength of H2 line is 656 nm, whereas in the spectrum of a distant galaxy, the wavelength of H2 line is 706 nm. Estimated speed of galaxy with respect to Earth is (a) 2 × 10 8 m/s (c) 2 × 10 6 m/s

(b) 2 × 10 7 m/s (d) 2 × 10 5 m/s

29. A block and a pan of equal masses are connected by a string going over a smooth light pulley as shown in the figure. Initially the system is at rest. A particle of mass m falls on the pan and strikes the pan with a speed v. Find the speed with which the system moved just after the collision. v 3 v (c) 9

v 5 v (d) 7

(a)

(b)

30. The half-life of

m m

m

215

At is 100 µs. If a sample initially contains 6 mg of the element, what is its activity initially and after 200 µs ?

26. A camera is fitted with a bellows in order to be able to vary the lens to film distance from 7 to 12 cm. With a lens of focal length 50 mm, what are the nearest and farthest distances an object can be from

(a) 1.4 × 1017 Bq (b) 1.7 × 1018 Bq (c) 1.16 × 10 23 Bq (d) 1.9 × 10 24 Bq

and 2.8 × 1015 Bq and 2.5 × 1017 Bq and 2.9 × 10 22 Bq and 3.4 × 10 20 Bq

Answers 1. (b) 11. (c) 21. (c)

2. (d) 12. (d) 22. (b)

3. (b) 13. (c) 23. (b)

4. (c) 14. (c) 24. (b)

5. (d) 15. (c) 25. (c)

6. (d) 16. (a) 26. (a)

7. (b) 17. (a) 27. (b)

8. (b) 18. (c) 28. (b)

9. (d) 19. (c) 29. (a)

Detailed solutions of these questions are available on http://www.arihantbooks.com/Physics%20Spectrum.pdf

24

10. (b) 20. (b) 30. (c)

Answer with Explanations 1. (b) As, potential energy U of a particle varies with distance x from a fixed origin is given by =

Dimensions of A = ∴

x

=

(2 n + 1) =

x2 + B

ka 20 × 0. 3 = = 15 µmg 0.04 × 1 × 10

2 n + 1 = 15

here, B must have the dimension of x 2 i.e. [ L2 ] 4 x2

or

A x

[ML2 T −2 ][L2 ] L1/ 2

7/ 2 −2

= [ML

n =7

∴ T ]

AB = [ML7/ 2 T −2 ] [L2 ] = [M1L11/ 2 T −2 ]

4. (c) Let I1 and I2 be the current drawn from cells of emf 6V and 4 V in the circuit. Then,

I1 =

6 =1A 2 + 3 +1

I2 =

4 = 0.4 A 6+4

2. (d) Let t is the time in which the car is acquiring maximum speed 35 =7 ⇒ t = 5s t this time the car will cover a distance of 1 1 s = at 2 = × 7 × 52 = 87.5 m. If the car starts deaccelerating at 2 2 t = 5 s, then it will travel a total distance of 2s = 175m, before coming to rest.

VA − VB = 1 × 3 = 3V, VB − VC = 5 V

In

Since, this distance is not the 500 m distance required, the car will travel for some distance with constant maximum speed. Let this time be t1 . Then the velocity-time graph can be drawn as shown in figure.

VC − VD = 0.4 × 6 = 2 .4 V = 3 + 5 + 2 .4 = 10.4 V

5. (d) Electric field within the plates E = EQ 1 + EQ 2

v(m/s) 35

+Q1

7 t

VA − VD = ( VA − VB ) + ( VB − VC ) + ( VC − VD )

∴

+Q2

E2

E1

7 t1

t

or

t(s)

E = E1 − E 2 Q1 Q2 = − 2 Aε 0 2 Aε 0

The distance travelled in time t 0 = 500 − 175 = 325 m 325 time taken, t1 = = 9.28 s 35

E=

Hence, total time of travel = t + t1 + t = 5 + 9.28 + 5 = 19.28 s

3. (b) Let the initial amplitude a decreases to a1 , to the other side i.e.

∴ Potential difference between the plates. Q − Q 2  Q1 − Q 2 Q1 − Q 2 VA − VB = E ⋅ d =  1 = d = 2C  2 Aε 0   Aε  2  0  d 

after the first sweep, decreases in elastic potential energy = work done against friction 1 2 1 2 ka − ka1 = µmg( a + a1 ) 2 2 1 or k( a + a1 )( a − a1 ) = µmg( a + a1 ) 2 2µmg or …(i) a − a1 = k 2 µmg Similarly …(ii) a1 − a 2 = k M 2µmg …(iii) a n −1 − a n = k

or a n =

Substituting in the above equation,  µmg  (2 n + 1)   =a  k 

6. (d) Velocity vector is perpendicular to magnetic field. Therefore path of the particle is a circle of radius r = v0

α P v

90

R β Q

mv 0 Bq

°

C S 90°

v = v 0 as the speed of the particle does not change in magnetic field.

On adding Eqs. (i), (ii) and (iii) we get, 2µmg a − an = k the block stops when µmg = ka n

Q1 − Q 2 2 Aε 0

Centre of the circle is C µmg k

CP = CQ ∴

∠CPQ = ∠CQP

or

90 − α = 90 − β

∴( α = β )

For the PQ = 2 PR = 2 r cos( 90 − α ) = 2 r sin α 2 mv 0 sin α ∴ PQ = Bq

25

∠PCR = ∠RCQ = α

10. (b) here, hFE = forward current ratio i.e. β = 90 °, VCE = 4 V

∠PCQ = 2 α

∴

arc PSQ = (2 π − 2 α )r = t PSQ =

∴

If IC is the collector current, then

2 mv 0( π − α ) Bq

9 − 4 = IC RC or IC = 2 .5 mA I 2 .5mA = 2 .78 × 10 −5 A IB = C = β 90

PSQ 2 m( π − α ) = V0 Bq

Since, the transistor operates in active region therefore,

7. (b) As in L-C-R circuit, resistance R is given by L R= C L L or CR = R2 = C R

∴

∴

RB = 82 kΩ

11. (c) From the theorem

Hence, time constants of both the circuits are equal τC = τ L = τ V i L = (1 − e −t / τ ) R V iC = (1 − e −t / τ ) R

Now,

VBE = 0.7 V 3 − 0.7 2 .3 = 82 × 10 3 Ω RB = = Ib 2 .78 × 10 −5

y

…(i)

v=Rω

…(ii)

Equating Eqs. (i) and (ii), we get 1 e −t / τ = , 2 or

ω

O

(1 − e −t / τ ) = e −t / τ

…(i)

L 0 = L com + M(r × v )

or 2 e −t / τ = 1 t = ln (2) τ

t = τ ln (2) = CR ln (2)

x

Angular momentum about O = Angular momentum about center of mass + Angular momentum of center of mass about origin 1 3 = L 0 = Iω + MRv = MR 2 ω + MR( Rω ) = MR 2 ω 2 2

8. (b) Magnetic field inside the solenoid is

y

B = µ 0 nI = ( 4 π × 10 −7 ) × 2000 × 2 .5 = 2 π × 10 −3 T Bevmax = =

2 mvmax r

or vmax =

Ber m

ω O

(2 π × 10 −3 ) × (1.6 × 10 −19 ) × 0.01 9.1 × 10

−31

x

In this case both the terms in Eq. (i) i.e. L com and M(r × v ) have the same direction X. That’s why we have to use L 0 = Iω + mrv.

= 1.112 × 10 7 m/s hc 1 2 = mvmax + W0 λ 2 1 = × ( 9.1 × 10 −31 ) × (1.112 × 10 7 )2 + 0.5 2 × 1.6 × 10 −19

12. (d) Let r be the radius of the satellite from centre of earth. Then, orbital velocity is given v 02 =

GM r

…(i)

If, R = radius of earth = 5.64 × 10 −17

λ=

v

hc 5.64 × 10 −17

=

( 6.63 × 10 −34 ) × ( 3 × 10 8 ) 5.64 × 10 −17

= 3.52 × 10 −9 m

Applying conservation of mechanical energy between points A and B, kinetic energy = change in potential energy

9. (d) At t = 0, suppose number of atoms of X is N 0 u=0

At t = 4 hours (i.e. two half lives) N Number of atoms of X left undecayed N x = 0 4 3N0 Number of atoms of Y formed, N y = 4 Nx 1 ∴ = Ny 3 At, t = 6 hours (i.e. three half lives), N 7N0 N x = 0 and N y = 8 8 Nx 1 = Ny 7 As the given ratio (1/4) lies between (1/3) and (1/7), therefore, time t lies between 4 and 6 hours.

26

v B

R

A

r

1 GMm  GMm  mv 2 = − − −   2 r R 

∴

v2 =

2 Gm 2 Gm = ve2 − 2 v 02 = ve2 − 2 v 02 − R r

13. (c) Let the displacement equation of particle is x=asinωt a t=0 O P

B a

→

x

A a−x

 h − h2  U-tube =  1  A where, A = cross-sectional area of tube,  2 

Time period of particle would be T = (t PA + t AP ) + (t PB

2π + t BP ) = ( 0.5s ) + (1.5s ) = 2 s = ω ω = π s −1

∴

x = a sin πt

…(i)

v = ( aπ )cos πt 1 = t then, tOAP = t + 2

…(ii)

∴ and Let,

tOP

 1 π  then, 3 = aπ cos πt = aπ cos π t +  = aπ cos  + πt = aπ sin πt  2 2  πt =

∴ or

aπ =

π 4

2

 h − h2  loss in potential energy of this liquid =  1  Aρg  2  The mass of the entire liquid in U-tube = ( h1 + h2 + h ) ρA If this liquid moves with velocity v, then 1 KE = ( h1 + h2 + h )ρAv 2 2 using law of conservation of energy.

3 = 3 2 m/s = vmax cos π / 4

2

1  h − h2  ( h1 + h2 + h ) ρAv 2 =  1  Aρg  2  2

π a = 4 2 a a− AP a − x 2 = 2 −1 = = a BP a + x 2 +1 a+ 2 x = a sin πt = a sin

∴

ρ = density of liquid. The decrease in height of this liquid  h − h2  = 1   2 

or

if ρ is coefficient of volume expansion of liquid, then d0 Density at temperature t1 is, d 1 = 1 + ρt1 …(i)

v1

A1

Density at temperature, t 2 is d 2 =

or

f1 + f1ρt 2 = f2 + f2ρt1

A2

f1 − f2 = ρ( f2 t1 − f1t 2 )

v2

The coefficient of volumetric expansion of liquid is (f − f ) ρ= 1 2 f2 t1 − f1 t 2

Applying continuity equation between 1 and point A1 v1 = A2 v 2 A v 2 =  1  v1  A2 

…(ii)

Substituting value of v 2 from Eq. (ii) in Eq (i) A12 A22

∴

A2 =

17. (a) In the processes AB and CD, volume is constant therefore, no work is done WDA = RT0 loge 2

v12 = v12 + 2 gh A22 =

or

A12 v12 v12

 V  WBC = R(2 T0 )loge  0  = − 2 RT0 loge 2  2 V0  Total work done by the gas

+ 2 gh

W = WDA + WBC

A1 v1

= RT0 loge 2 − 2 RT0 loge 2

v12 + 2 gh

= − RT0 loge 2 |W| = RT0 loge 2

Substituting the given values, A2 =

d0 1 + ρt 2

According to Archimede’s principle d 1 f2 d 0 1 + ρt 0 f1 Vd 1 = m = f2 Vd 2 or = = d 2 f1 (1 + ρt1 ) d 0

h

we get

g 2 ( h1 + h2 + 2 )

16. (a) Let V be the volume of solid, d be its density and m be its mass,

14. (c) Applying Bernoullis theorem between point 1 and 2 v 22 = v12 + 2 gh

v = ( h1 − h2 )

(10 −4 m2 )(1.0 m / s )

18. (c) Let d ′ be the diameter of refracted beam

(1.0 m / s2 ) + 2 × 10 × 0.15

As, cross-sectional area of stream of water 0.015 m below the tap is A2 = 5.0 × 10 −5 m2 .

60° d 60° p r

15. (c) When there is equal level of liquid in two arms of U-tube, then h1 + h2 2 ( h1 + h2 ) h1 − h2 We may consider that a length h1 − of the = 2 2 liquid has been transferred from left arm to right arm of U-tube. The mass of the liquid transferred from left arm to right arm of

r

Q r

height of liquid in each arms of U-tube =

Then, d = PQ cos 60 °

i.e.

d ′ = PQ cos r d′ cos r = = 2 cos r, d ′ = 2 d cos r d cos 60 °

27

sin r =

cos r = 1 − sin 2 µ = d ′ = (2 )(2 )

∴

Therefore,

3 sin i 1 = 2 = 3 µ 3 2

v = ν × λ = 512 ( 65.2 ± 0.1) = ( 33980 ± 51.2 ) cm/s Therefore, error in velocity = 51.2 cm/s

2 3

22. (b) While charging in a RC circuit, I = E e −t / RC R

2 3

Taking log of both sides log I = log

2 The diameter of refracted beam is = 4 cm ≈ 3.26 cm 3

19. (c) For understanding, let us assume that the two loops are lying in the plane of paper as shown. The current in loop 1 will produced, • magnetic field in loop 2. Therefore,

1

Perpendicular to paper outwards

2

This is the equation of a straight line, with 1 slope = − RC When R is doubled, slope becomes less negative i.e. more also at t = 0, current will be less, graph Q represents these facts. 3

600

RC

β 2 RL Now, power amplification, AP = RC

F

F

E t − R RC

23. (b) The voltage amplification AV = βRL = 60 × 6 × 10 = 600

Perpendicular to paper inwards F

λ = ( 65.0 ± 0.1) cm

As, velocity of sound is

( 60 )2 × 6000 = 36000 600 A 36000 The required ratio, P = = 60 AV 600 AP =

⇒

increase in current loop 1 will produced an induced current in loop 2 which produces ⊗ magnetic field passing through it i.e. induced current in loop 2 will also be clockwise as shown above. The loop will now repel each other as the current at the nearest and farthest points of the two loops.

20. (b) If the current flows out of the paper, the magnetic field at points to the right of the wire will be upwards and to the left will be downwards as shown in figure.

24. (b) The minimum deviation formula Snell’s law applied to the first surface n sin θ1 1.00 sin 65° sin θ 2 = 1 = = 0.549 n2 1.65 or

θ 2 = 33.3 °

from geometry B B

i°

A

Now, let us come to the problem, magnetic field at C = 0 Magnetic field in region BX′ will be upwards (+ve) because all points lying in this region are to the right of both the wires.

θ1

C

B

X′

Similarly, magnetic field in region AX will be downwards (–ve) magnetic field in region AC will be upwards (+ve), because points are closer to A compared to B. Similarly magnetic field in region, BC will be downwards (–ve) graph B satisfies all the conditions.

21. (c) For first resoance,

θD

180 –A

or

φ1 = 47.8 °

Figure shows a ray traced through the prism. The angle of deviation is θ D = θ1 + φ1 − A = 65 + 47.8 ° − 60 ° = 52 .8 ° The difference in θ D for blue and red light is

30.7 =

λ +n 4

…(i)

63.2 =

3λ +n 4

…(ii)

∆θ D = 557 . ° − 52 .8 ° = 2 .9 °

25. (c) Each fission yield 200 MeV = (200 × 10 6 ) × (1.6 × 10 −19 ) J of

for second resonance,

On solving Eqs. (i) and (ii), we get λ = 65 cm The error in measuring the length using metric scale would be 0.1 cm which is least count of metric scale.

28

φ1

Snells law applied to the second surfaces thus gives n sin φ 2 1.65 × sin 26.7 ° sin φ1 = 2 = = 0.741 n1 1.00

B

A

φ2

φ 2 = A − θ 2 = 60 − 33.3 ° = 26.7 °

i

X

θ2

energy Only 20% of this is utilised efficiently and so, Energy generated per fission = (200 × 10 6 ) × 1.6 × 10 −19 × 0.20 = 6.4 × 10 −12 J Because the reactors output is 700 × 106 J/s, the number of fission required per second is

fission/s =

7 × 10 8 J / s 6.4 × 10

−12

= 1.1 × 10 20 s −1

and fission/day = 86400 S/d × 11 . × 10 20 s −1 = 9.5 × 10 24 d −1 There are 6.02 × 10

26

= and

48 × 10 4 9 400 = 231 m/s 3

v2 =

atoms in 235 kg of Uranium-235. Therefore

the mass of uranium 235 consumed in one day is

28. Since, the wavelength ( λ ) is increasing, we can say that the galaxy is receding. Doppler effect can be given by C+v λ′ = λ C−v

 9.5 × 10 24  Mass =   × 235 = 3.7 kg  6.02 × 10 26 

26. (a) Let S and S′ refer to object and image distances. The nearest distance S occer when S′ is largest, So 1 1 1 1 1 1 = − = − = S f S ′ 5.0cm 12 cm 60 cm or

or

706 = 656

or

2

C + v = 1.16 C − 1.16 v

∴

v=

magnification of the images of an object at the nearest S′ 12 cm m= = = 1.4 S 8.57 cm At the farthest distance S′ 7 cm = = 0.4 S 17.5cm

27. (b) Over a very small time interval surrounding the moment of explosion, the effect of gravity, (an external force) can be neglected. Then, all forces are internal and momentum is conserved. The vector diagram for momentum conservation is shown in figure. 3M M v 2 = Mv − v1 4 4 4 1 or v 2 = v − v1 3 3 M v1 4

29. (a) Let N = magnitude of the contact force between the particle and the pan. T = tension in the string Consider the impulse imparted to the particle, the force is N in upward direction and the impulse is ∫ N dt . This should be equal to the change in its momentum. Thus, …(i) ∫ N dt = mv − mV Similarly considering the impusle imparted to the pan, …(ii)

∫ Tdt = mV

…(iii)

On adding Eqs. (ii) and (iii), we get

∫ N dt = 2 mV Comparing with Eq. (i) mv − mV = 2 mV or V =

v 3

30. (c) The number of radioactive atoms initially present is

60° Mv

3M v2 4

Then,

∫ ( N − T )dt = mV and that to the block,

N0 = θ

0.16C 0.16 × 3 × 10 8 = 2 .16 2 .16

v = 0.22 × 10 8 m/s ≈ 2 × 10 7 m/s

S = 17.5 cm

m=

C+v C−v

C + v  706  =  = 1.16 C − v  656

or

S = 8.57 cm

The farthest distance S occurs when S′ is smallest, so 1 1 1 1 1 2 = − = − = S f S ′ 5.0cm 7 cm 35cm

…(i)

16 8 1 v 22 = v 2 ⋅ v 2 = ( v ⋅ v ) + ( v ⋅ v1 ) + ( v1 ⋅ v1 ) 9 9 9 16 8 1 2 = (200 ) − (200 )( 400 )cos 60 ° + ( 400 )2 9 9 9

( 6 × 10 −3g ) ( 6.03 × 10 23 atoms / mol ) 215g / mol

= 1.68 × 1019 atoms and the decay constant of 215 At is ln 2 0.693 λ= = = 6930 s −1 T1 / 2 100 × 10 −6 Hence, the initial activity is A0 = λN 0 = 6930 × 1.68 × 1019 = 1.6 × 10 23 Bq At

t = T1 / 2  1 A = A0e − λt = 1.16 × 10 23 ×   2 

2

= 2.9 × 10 22 Bq

29

TEST RIDER The Simulator Test Series Towards JEE Main and Advanced Instructions p p p

p

This test consists of 30 questions. Each question is allotted 4 marks for correct response. Candidates will be awarded marks as stated above for correct response of each question. 1/4 marks will be deducted for indicating incorrect response of each question. No deduction from the total score will be made if no response is indicated for an item in the answer sheet. There is only one correct response for each question. Filling up more than one response in any question will be treated as wrong response and marks for wrong response will be deducted according as per instructions.

1. The heat produced in a wire carrying an electric current depends on the current, the resistance and the time. Assuming that the dependance is of the product of powers type, guess an equation between these quantities using dimensional analysis. The dimensional formula of resistance is [ML2 I−2 T −3 ] and heat is a form of energy. (a) KRt (c) KI 2Rt

(b) KIRt (d) KIR 2t

2. A sphere of mass m and radius r as shown in figure slips on a rough horizontal plane. At some instant, it has translational velocity v0 and ω=v0 /2r r rotational velocity about the v0 centre Find the f v0 /2r . A translational velocity after the sphere starts pure rolling. 5 v0 6 4 (c) v 0 3 (a)

6 v0 7 2 (d) v 0 5 (b)

3. The density inside a solid sphere of radius a is given

a , where ρ0 is the density at the surface and r r denotes the distance from the centre. Find the gravitational field due to this sphere at a distance 2a from its centre. by ρ = ρ0

1 πGρ0a 2 2 (c) πGρ0a 3 (a)

3 πGρ0a 4 5 (d) πGρ0a 4 (b)

4. A shunt-wound generator delivers 48 A at a brush potential of 120 V. The field coils have a resistance of 60 Ω and the armature has a resistance of 0.14 Ω.

If the stray power loss is 500W, what is the efficiency of the generator? (a) 85% (c) 87%

(b) 84% (d) 89%

5. A source emitting sound of frequency 180 Hz is placed in front of a wall at a distance of 2 m from it. A detector is also placed in front of the wall at the same distance from it. Find the minimum distance between the source and the detector for which the detector detects a maximum of sound. Speed of sound in air = 360 m/s. (a) 4 m (c) 7 m

(b) 6 m (d) 3 m

6. An object is kept in front of a concave mirror at a distance n times the focal length from focus. The magnification of image will be 1 n (c) n

1 n (d) − n

(a)

(b) −

7. A particle moves in the X-Y plane with a constant

acceleration of 1.5 m/s2 in the direction making an angle of 37° with the X-axis. At t = 0, the particle is at the origin and its velocity is 8.0 m/s along the X-axis. Find the velocity and the Y position of the particle at t = 4.0 s. (a) 15 m/s and (42 m, 6.2 m) (b) 16 m/s and (45 m, 7.5 m) (c) 18 m/s and (48 m, 8.5 m) (d) 13.3 m/s and (41.6 m, 7.2 m)

a=1.5m/s2

37° v=8.0 m/s

8. A 10 µF capacitor is charged

X

through a resistance of 0.1 MΩ from a battery of 1.5 V. Calculate the time required for the capacitor to get charged upto 0.75 V for situations shown in Fig. (a) and (b). R

E

K

C

K (a)

(a) 0.693 s, 0 s (c) 0 s, 1 s

E

C

R

(b)

(b) 1 s, 0 s (d) 0.533 s, 1 s

25

9. Two long straight wires with the same cross-section

14. What amount of heat will be generated in a coil of

are arranged in air parallel to one another. The distance between the axis of the wire is η times larger than the radius of wires crosssection. The capacitance of the wires per unit length would be [ consider, η >> 1]

resistance R due to a charge q passing through it, if the current in the coil decreases down to zero uniformly during a time interval ∆t?

2 πε0 1nη πε (c) 0 1nη (a)

(b)

πε0 21nη

area A and length L is hanging from a fixed support. Young’s modulus of the material of the rod is Y. Neglect the lateral contraction. Find the elongation of the rod. WL 2 AY

(b)

WL 4 AY

(c)

WL 3 AY

(d)

WL 5 AY

11. White light is incident normally on a glass plate of

thickness 0.50 × 10−6 and refracting index of 1.50. Which wavelength in the visible region (400 nm-700 nm) are strongly reflected by the plate? (a) 429 nm and 600 nm (c) 475 nm and 640 nm

2 q 2R 3 ∆t 3 q 2R (d) 2 ∆t (b)

15. A uniform chain of length l and mass m over changes

(d) Data is insufficient

10. A uniform heavy rod of weight W, cross-sectional

(a)

4 q 2R 3 ∆t 3 q 2R (c) 4 ∆t (a)

(b) 575 nm and 650 nm (d) 520 nm and 690 nm

12. The variation of u and v for a concave lens is shown

a smooth table with its two third part lying on the table. Find the kinetic energy of the chain as it completely slips off the table. (a)

3 mg l 8

(b)

y

26

(b)

101 99

(c)

99 100

2 mg l 7

(a) 173 . × 10 4 NC −1 and 34.7° (c) 428 . × 10 9 NC −1 and 45°

(b) 2.38 × 10 8 NC −1 and 40.9° (d) 4.9 × 1010 NC −1 and 34.7°

17. A particle is projected under gravity with velocity v = 2ag from a point at a height h above the level plane making an angle θ. The maximum range R on the ground is v=√2ag O

θ

P f

R

x

2

(a) ( a 2 + 1)h (b) a 2h −f

(d)

(d)

h

3

cross-sectional area 20 × 10−6 m2 as shown in figure. The gas occupies a height L, at the bottom of the tube and is separated from air at atmospheric pressure by a L2 Gas Air mercury column of mass Gas Air 0.002 kg. If the tube is quickly L turned isothermially, upside 1 down so that mercury column encloses the gas from below. The gas now occupies height L1 in the tube. The L ratio 2 is (take, atmospheric pressure = 105 N-m−2 ) L1 102 101

4 mg l 9

the vertices A and B of a right angled triangle ABC, respectively. B is the right angle, AC = 2 × 10−3 m and BC = 10−2 m. Find the magnitude and direction of the resultant electric intensity at C.

1

13. An ideal gas is trapped inside a test tube of

(a)

(c)

16. Point charge 4 × 10−6 C and 2 × 10−6 C are placed at

in figure. Find correct statement about (1), (2) and (3) parts of graph. (a) 1. Real image of virtual object 2. Virtual image of real object 3. Real image of virtual object (b) 1. Real image of real object 2. Virtual image of real object 3. Virtual image of real object (c) 1. Virtual image of real object 2. Real image of real object 3. Virtual image of virtual object (d) None of the above

5 mg l 7

100 99

(c) ah

(d) 2 a ( a + h)

18. As shown in figure, a metal rod makes contact with a partial circuit and completes the circuit. The circuit area is perpendicular to a B=0.15T magnetic field with B = 015 . T. The resistance 50 cm of the total circuit is 3 Ω. v=2m/s How large a force is needed to move the rod −x as indicated with a constant speed of 2 m/s? (a) 375 . × 10 −3 N (c) 2.98 × 10 −2 N

(b) 475 . × 10 −4 N (d) 4.99 × 10 −3 N

19. A man whose weight is 667 N and whose density

980 kg / m3 can just float in water h is head above the surface with the help of life jacket which is wholly immersed. Assuming that the volume of his head is 1/15 of his whole volume and that the specific gravity of the life jacket is 0.25, find the volume of the life jacket. (a) 3 L

(b) 5 L

(c) 4 L

(d) 7 L

1.000 kg of water just above the freezing point. A similar can contains 1.000 kg of water just below the boiling point. The two cans are brought into thermal contact. Find the change in entropy of the system. (a) 96 J/K (c) 100 J/K

(b) 101 J/K (d) 105 J/K

21. Find the centre of mass of a metal circular sheet of

2 cm

diameter 6 cm, which has 2 cm square cut out of it. C The two sides of the square lie along diameters of the circle. OM is bisector of ∠AOD in figure. A

(a) At 5 cm from O towards B (b) At the centre of the circle (c) At 0.233 cm from O towards M (d) At 3 cm from O towards A

O

45°

M 2cm

B

D

22. Two pendulums of lengths 1 m and 1.21 m respectively starts swinging together with same amplitude. The number of vibrations that will be executed by the longer pendulum before the two will swing together again are (a) 9

(b) 10

(c) 11

required to raise the temperature of a 15 g block of Cu from 5 to 30 K? (a) 0.665 cal (c) 0.987 cal

(b) 0.771 cal (d) 0.957 cal

27. A train sounds its whistle as it approaches and leaves a rail road crossing. An observer at the crossing measures a frequency of 219 Hz as the train approaches and a frequency of 184 Hz as the train leaves. The speed of sound is 340 m/s. Find the speed of the train and the frequency of its whistle. (a) 30 m/s and 100 Hz (c) 35 m/s and 250 Hz

(b) 29.5 m/s and 200 Hz (d) 25 m/s and 60 Hz

28. In figure, a wire perpendicular to long straight wire is moving parallel to the latter with a speed v = 10 m/s in the direction of the current flowing in the latter. The current is 10A, what is the magnitude of the potential difference between the ends of the moving wire? i 10 A r

10cm

20. A copper can of negligible heat capacity, contains

(d) 12

v=10m/s

(a) 60°

(b) 15°

(c) 45°

(d) 30°

24. From the Zener diode circuit shown in figure, the current through the Zener diode is + 220v 50v −

(a) 34 mA (c) 36.5 mA

20kΩ

5kΩ R1

Output Voltage

(b) 31.5 mA (d) 2.5 mA

25. Find the magnitude of the Hall voltage induced across a silver wire of square cross-section 1 mm on a side, when it carries a current of 1.5 A and a transverse magnetic field B of strength 0.1 T is applied. (a) 15 . × 10 −7 V (c) 14 . × 10 9 V

(b) 17 . × 10 −8 V (d) 16 . × 10 −8 V

26. At low temperatures, the specific heat capacity of

metals can be expressed as C = k1T + k3 T 3 , where T is in kelvin. For Cu, k3 = 2.48 × 10−7 cal/g-K4 , k1 = 2.75 × 10−6 cal/g-K2 . How much heat energy is

2. (b) 12. (a) 22. (b)

3. (a) 13. (b) 23. (d)

4. (b) 14. (a) 24. (b)

5. (d) 15. (c) 25. (d)

V dr

×B B

(a) 47.2 µV

(b) 46.1 µV

(c) 45.3 µV

(d) 43.4 µV

29. A particle free to move along the x-axis has potential

energy given by u(x) = k[1 − exp(− x2 )] for − ∞ ≤ x ≤ + ∞, where k is a positive constant of appropriate dimensions. Then, (a) at points away from the origin, the particle is in unstable equilibrium (b) for any finite non-zero value of x, there is a force directed away from the origin k (c) if its total mechanical energy is , it has its minimum kinetic 2 energy at the origin (d) for small displacements from x = 0, the motion is simple harmonic

30. One plate of a capacitor is connected to a spring as shown in figure. Area of both the plates is A. In steady state separation between the plates is 6.8 d (spring was unstretched and the distance between the plates was d when the capacitor was unchanged). The force constant of the E spring is approximately 4ε0 AE 2 d3 6ε E 2 (c) 0 3 Ad

2 ε0 AE d2 ε AE 3 (d) 0 3 2d

(a)

(b)

Answers 1. (c) 11. (a) 21. (c)

A

r2

23. A sky wave with a frequency 55 MHz is incident on the D-region of Earth’s atmosphere at 30°. The angle of refraction is (electron/density for D-region is 400 electron/cm3)

r1

6. (b) 16. (b) 26. (b)

7. (d) 17. (d) 27. (b)

8. (a) 18. (a) 28. (b)

9. (c) 19. (c) 29. (d)

10. (a) 20. (c) 30. (a)

Detailed solutions of these questions are available on http://www.arihantbooks.com/Physics%20Spectrum.pdf

27

Answer with Explanations 1. (c) Let the heat produced be H, the current through the wire be I, the resistance be R and the time be t. Since heat is a form of energy, its dimensional formula is [ML2 T −2 ]. Let us assume that the required equation is H = KI a R bt c , where, K is a dimensional constant. Writing dimensions of both sides, [ML2 T −2 ] =[ I a ][ML2I −2 T −3 ]b [Tc ] = [M bL2b T − 3b + c I a − 2 b ] Equating the exponents, we get b =1 2b =2 − 3b + c = − 2 a −2b = 0 Solving these, we get, a = 2, b = 1 and c = 1 Thus, the required equation is H = KI 2 Rt

2. (b) Velocity of the centre = v0 and the angular velocity about the v0 . Thus v 0 > ω 0 r. The sphere slips forward and thus the 2r friction by the plane on the sphere will act backward. As the friction is kinetic, its value is µN = µMg and the sphere will be f deacelerated by α cm = m Hence, f ...(i) v(t ) = v 0 − t m This friction will also have torque τ = fr about the centre. This torque is clockwise and in the direction of ω 0 . Hence, the angular acceleration about the centre will be r 5t α =f =  2  2 gmr mr    5 centre =

and the clockwise angular velocity at time t will be 5f v 5t ω(t ) = ω 0 + t= 0 + t 2 mr 2 r 2 mr Pure rolling starts when v(t ) = rω(t ) v 5t i.e. v(t ) = 0 + t ...(ii) 2 2m Eliminating t from Eqs. (i) and (ii), we get 5 5 v v(t ) + v(t ) = v 0 + 0 2 2 2 Translational velocity after the sphere starts pure rolling, 2 6 or v(t ) = × 3 v 0 = v 0 7 7

3. (a) The field is required at a point outside the sphere. Dividing the sphere in concentric shells, each shell can be replaced by a point particle at its centre having mass equal to the mass of the shell. Thus, the whole sphere can be replaced by a point particle at its centre having mass equal to the mass of the given sphere. If the mass of the sphere is M, the gravitational field at the given point is GM GM ...(i) E= = (2 a )2 4 a 2 The mass M may be calculated as follows. Consider a concentric shell of radius r and thickness dr

28

Its volume is dV = ( 4 πr 2 )d r  a and its mass is dM = ρdv = ρ 0  ( 4 πr 2dr )  r = 4 πρ 0 ardr The mass of the whole sphere is a

M = ∫ 4πρ 0 ardr 0

= 2 πρ 0 a 3 Thus, by Eq. (i) by the gravitational field is 2 πGρ 0 a 3 1 E= = πGρ 0 a 2 4a 2

4. (b) The terminal voltage is Vt = 120 V The delivered power output and the power losses in the field coils, PFC = IF2 Rf and the armature, Pa = Ia2 Ra The efficiency of power output/power input where the denominator includes the stray losses. Thus, Poutput = Vt Ia = 120 × 48 = 5760 W 2

 120 PFC =   × 60 = 240 W  60  Pstray = 500 W Parmature = ( 48 )2 × 0.14 = 322 .56 W Efficiency of the generator =

5760 × 100 5760 + 240 + 322 .56 + 520

= 0.84 = 84%

5. (d) The situation is shown in figure. Suppose the detector is placed at a distance of 4 m from the source. The direct wave received from the sources travels a distance of S x metre. The wave reaching the detector after reflection from the wall has a travelled h 1

 x2  2 distance of =2 (2)2 +  m. 4  

D

The path difference between the two waves is 1     2 x2  2  ∆ = 2 (2 ) +  − x  metre 4       Constructive interference will take when ∆ = λ , 2 λ , ... The minimum distance x for a maximum corresponds to ∆=λ c 360 m / s The wavelength is λ = = =2m ν 180 s −1 According to equation (i), we get 1

 x2  2 2 (2 )2 +  − x = 2 4  

...(i)

1

or

 x2  2 x 4 +  = 1 + 4  2 

or

4+

x2 x2 =1+ + x or x = 3 4 4

6. (b) For concave mirror, 1 1 1 + = v u f and use sign convention, also the magnification in case of −v concave mirror is given by m = u Distance of image 1 1 1 + = 1 v u f 2 Distance from focus is nf, so distance from 3 1 1 1 pole nf + f1 from the formula, + = v u f 1 1 1 + =− v − ( nf + f ) f 1 1 1 n =− + =− v f ( n + 1)f ( n + 1)f ( n + 1)f v=− n Magnification of image  ( n + 1)f  − n  v 1 m=− =− =− u − ( n + 1 ) f n     Using this relation

7. (d) A particle accelerate along x-component and y-component, a x = (1.5 m/s 2 )(cos37 ° ) 4 = 1.2 m/s 2 5 a y = (1.5 m/s 2 )(sin 37 ° ) = (1.5 m/s 2 ) × and

3 = 0.90 m/s 2 5 The initial velocity has components 4 x = 8.0 m/s and vy = 0 At t = 0, x = 0 and y = 0 The x-component of the velocity at time t = 4.0 s is given by v x = u x + a xt = 8.0 m/s + (12 . m/s 2 )( 4.0 s ) = 1.5 ×

= 8.0 + 4.8 = 12 .8 m/s The y-component of velocity at t = 4.0 s is given by v y = u y + a yt = 0 + ( 0.90 m/s 2 )( 4.0 s ) = 3.6 m/s 2 The velocity of the particle at t = 4.0 s is v = v x2 + v y2 = (12 .8 m/s)2 + ( 3.6 m/s)2 = 13.3 m/s The velocity makes an angle θ with x-axis where, v 3.6 m/s 9 tan θ = y = = v x 12 .8 m/s 32 The x-coordinate at t = 4.0 s is 1 x = u xt + a xt 2 2

1 × 1.2 × (4.0)2 2 = 32 m + 9.6 m = 41.6 m The y-coordinate at t = 4.0 s is 1 y = u yt + a yt 2 2 1 = × 0.90 × ( 4.0 )2 2 = 7.2 m Thus, the particle is at (41.6 m, 7.2 m) at t = 4.0 s = 8.0 × 4.0 +

8. (a) In case of charging of a capacitor through a resistance q = q 0[1 − e i.e.

e

−

t RC

−

t RC

]

 q  = 1 −   q0 

...(i)

But as for a capacitor, q = CV q V 0.75 1 = = = q 0 V0 1.5 2 q So, substituting the value of   in Eq. (i) q0  e t / RC = 2, i.e. t = CR loge 2 So for circuit (a) C = 10 µF and R = 0.1 MΩ So, t = 10 −5 × 10 5 loge 2 = 0.693 s However in circuit (b) as capacitor is connected directly to the battery initially it acts like short circuit and hence, it will change instantaneously, i.e. t = 0 s.

9. (c) It is based on the combination of two long charged wires will act as a capacitor. Let us give equal and opposite charges to two wires so that they would have linear charge densities as + λ and − λ +λ

−λ

P

dx x Electric field at point p

Electric field at point P λ λ E= + 2 πε 0 x 2 πε 0( ηa − x ) ηa − x

∫ dV = − ∫ Edx = − ∫ Edx a

where, a is radius of wire. θ πε 0 C= = |v| 1nη

10. (a) Consider a small length dx of the rod at a distance x from the fixed end. The part below this small element has length (L − x). The tension T of the rod at the element equals to the weight of the rod below it. W T = (L − x) L x Elongation in the element is given by dx Elongation = Original length × Stress/Y Tdx ( L − x )Wdx = = AY LAY

29

L

Mg  Mg     p0 +  ( AL1 ) =  p0 −  ( AL 2 )   A A

( L − x ) W dx LAY 0

The total elongation = ∫

Mg 10 5 + 0.002 × 10 101 20 × 10 −6 A = = Mg 0 .002 × 10 99 5 p0 − 10 − A 20 × 10 −6

L

 WL W  x2 W  2 L2  Lx −  =  L − − 0 + 0 = LAY  2  0 LAY  2  2 AY

p0 +

L2 = L1

11. (a) The light of wavelength λ is strongly reflected if 1  2 µd =  n +  λ  2

...(i)

14. (a) Suppose initial current is i 0 then ∆t

t  t    i(t ) = io 1 −  ⇒ q = i 0 ∫ 1 −  dt   ∆t  ∆t  0

where, n is a non-negative integer. 2 µd = 2 × 1.50 × 0.5 × 10 −6 m = 1.5 × 10 −6 m

...(ii)

Putting λ = 400 nm in Eq. (i) and using Eq. (ii) 1  1.5 × 10 −6 m =  n +  ( 400 × 10 −9 m ) or n = 3.25  2 Putting λ = 700 nm in Eq. (ii), we have 1  1.5 × 10 −6 m =  n +  (700 × 10 −9m ) or n = 1.66  2 Thus, within 400 nm to 700 nm, the integer n can have the values 2 and 3. Putting these values of n in (i), the wavelengths becomes 4µd λ= = 600 nm and 429 nm 2n +1

12.

Thus, light of wavelengths 429 nm and 600 nm are strongly reflected. 1 1 1 (a) It is based on the mirror formula i.e. + = , then the sign v u f convention can be decided accordingly. Then, take the cases such as 1. u > 0, v > 0, 2. u < 0, v < 0 3. u > 0 and v < 0 For part (1) u > 0, v > 0, so object is virtual and real image. For part (2) u < 0, v < 0, then it is real object and virtual image. For part (3) u > 0, v < 0, so it is virtual object and real image.

13. (b) If pressure near any area A of mercury column is A, then force on the column will be pA. The force on the column will be perpendicular to the surface. PO A

i0 =

PL A

∆t

mg P0 A

t 

2

Rdt

2

∆t

=

4q 2 R t2 2t   2q   1 + −  dt × R = 3 ∆t 2 ∆ t ∆ t   0

∫  ∆t 

15. (c) Let us take the zero of potential energy at the table. Consider a part dx of the chain at a depth x below the surface of the table. m The mass of this part is dx = dx and hence its potential energy is l m  −  dx gx.  l  The potential energy of the

l of the chain 3

x

that overhangs is l/ 3

U1 =

∫− 0

m gxdx l

dx l

 m  x2  3 1 = −  g    = − mg l l 2 18     0 This is also the potential energy of the full chain in the initial position because the part lying on the table has zero potential energy. The potential energy of the chain when it completely ships off the table is l

V2 = ∫ −

m 1 gx dx = − mg l l 2

 1   1  4 The loss in potential energy =  − mg l −  − mg l = mgl  18   2  9

16. (b) Figure shows the right angle triangle ABC, such that

p2 A mg

p1 A

 2q 

∫  ∆t 1 − ∆t   0

0

Force on the column are shown in above diagram. For the equilibrium, forces must be balanced. In first stage equilibrium of mercury column p0 A

2q ∆t

Amount of heat generated in a coil, H =

P2 A mg

mg p0 A

Similarly in second stage, p0 A + Mg = p0 A Mg So, p1 = p0 + A Mg p2 = p0 − A For isothermal changes. p1 V1 = p2 V2

30

So,

AC = 2 × 10 −2 m and BC = 10 −2 m. The charges of q A = 4 × 10 −6 C and q B = 2 × 10 −6 C are placed at the vertices A and B respectively. Let E A and E B be electric intensity at point C due to charges q A and q B , respectively. Then, 1 qA EA = × = 9 × 10 9 4 πε 0 ( AC )2 ×

4 × 10 −6

E θ α EB EA C −2

10 2×

m

A −2 qA=4×10 C

(2 × 10 −2 )2 7

−1

= 9 × 10 NC (along AC) 1 qB 2 × 10 −6 EB = × = 9 × 10 9 × 4 πε 0 ( BC )2 (10 −2 )2 = 18 × 10 7 NC −1 (along BC)

θ

qB=2×10-2m

=

B

If θ is angle between the direction of E A and E B , then E = E A2 + E B2 + 2 E A E B cos θ In right angled ∆ABC, ∠ACB = θ BC 10 −2 cos θ = = = 0.5 ∴ AC 2 × 10 −2 or

 667   14     gV ρ   15 Volume of the life jacket, Vj = water 0.75 g  14 0.667 −   ( 9.8 )( 0.07 )  15 = ( 0.75)( 9.8 )

θ = 60 °

Hence, E = ( 9 × 10 7 )2 + (18 × 10 7 )2 + 2 × 9 × 10 7 × 18 × 10 7 × 0.5

= 3.637 × 10 −3 m 3 ≈ 4 × 10 −3 m 3

= 9 × 10 7 1 + 4 + 2 = 2 .38 × 10 8 NC −1 Suppose that the resultant electric intensity E makes an angle α with line AC. Then, E B sin θ 18 × 10 7 sin 60 ° tan α = = E A + E B cos θ 9 × 10 7 + 18 × 10 7 cos 60 ° = or

18 × 10 7 × 0.866 9 × 10 7 + 18 × 10 7 × 0.5

= 0.866

α = 40.9 °

17. (d) The coordinate of point P are ( R1 − h ). These coordinates should satisfy the equation of projectile, i.e. gR 2 − h = R tan θ − (1 + tan 2 θ ) 2 (2 ag ) or

2

2

=4L

20. (c) For the contents of either can, dH = CmdT , where for water, Cm = 4184 J/K Since, the heat capacities of the two masses are equal, the final temperature will be average of the initial temperature. T + T2i T1f = T2f = 1i 2 273 K + 373 K = = 323 K 2 Then, change in entropy of the system, ∆S = ∆S1 + ∆S 2 T1 f T1 i

R tan θ − 4 aR tan θ + ( R − 4 ah ) = 0

4 a ≥ ( R − 4 ah ) R 2 ≤ 4a ( a + h)

or

R ≤ 2 a ( a + h)

OP = r1 = 12 + 12 = 2 at 45° with OB Mass of remaining portion, M 2 = π ( 3 )2 − 4 = ( 9 π − 4 ) sq cm

18. (a) The induced emf in the rod causes a current of flow counter clockwise in the circuit because of this current in the rod, it experiences a force to the left due to the magnetic field. In order to the pull the rod to the right with constant speed, this force must be balanced by the puller. The emf induced in the loop is ∆φ B∆A B( L∆x ) |ε| = N =1 = = BLv ∆t ∆t ∆t Mechanical power supplied to circuit = Rate at which electric work is done on charge |ε|2 ( ∆q )( ε ) Eq. (i) Fv = = I|ε| = ∆t R Substitute the value of |ε | in Eq. (i), we get B 2 L2 v F= R ( 0.15 T )2 × ( 0.50m )2 × (2 m / s ) = 3Ω F = 3.75 × 10 −3 N Force is needed to move a rod is 3.75 × 10 −3 N ρg

= + 100 J/K Mass of square, m1 = 4 σ. Its centre of mass is at P where

∴ The maximum range is Rmax = 2 a ( a + h )

19. (c) The man’s volume is V = W =

2

21. (c) Let σ be the mass per unit area (1 sq cm) of disc

2

or

dT1 2 dT + Cm 2 T1 T∫ i T2

  323 K   323 K   = (4184 J/K) ln   + ln    273 K     273 K 

2

or

T f

= ∫ Cm

For θ to be real ( 4 aR )2 ≥ 4 R 2( R 2 − 4 ah ) 2

[1 L = 10 −3 m 3 ]

667 = 0.07 m 3 980 × 9.8

Equating the buoyant force to the weight of the man plus the weight of the life jacket  14  ρ water g  V + Vj  = 667 + ( 0.25 ρ water )g Vj  15 

∴ M 2 = ( 9 π − 4 )σ, r2 = ? As centre of mass of circular plate is at O, r=0 ∴ M1 r1 = − M 2 r2 Mr − 4σ × 2 r2 = − 1 1 = = − 0.233 cm M2 ( 9 π − 4 )σ i.e. At 0.233 cm from point 0 towards M.

22. (b) Let T1 and T2 be the time periods of the pendulum with lengths 1.0 m and 1.21 m, respectively. T2 l 121 . = 2 = = 11 . T1 l1 1

...(i)

Let V1 and V2 be the vibrations made by two pendulum to swing together. …(ii) VT 1 1 = V2T2 For the two pendulums to swing together, required condition is V1 − V2 = 1 or V1 = V2 + 1 ∴ ( V2 + 1)T1 = V2T2 ( V2 + 1) T2 or = = 11 . V2 T1 or or

1+

1 = 11 . V2

1 = 11 . − 1 = 0.1 V2 1 1 = V2 10

or

V2 = 10

31

As the train recedes from him. Dividing Eq. (i) by Eq. (ii), we find that ν a |v| + |vT| ...(iii) = νT |v| − |VT|

23. (d) For D-region, ν = 55 × 10 6 Hz i = 30 ° N = 400 × 10 6 electron/cm 3 As

µ = 1− = 1−

Also, or

81.45N ν2

81.45 × 400 × 10 6 ( 55 × 10 6 )2

Solving this equation for the speed of the train, ν − νr |vT| = |v| a = 29.5 m/s νa + νr

≈1

sin i or sin i = sin r sin r i = r = 30 °

Using Eqs. (iv) and (i), we get  |v | 2 ν a ν r ν 0 = ν a 1 − T  =  |v|  ν a + ν r

µ=

24. (b) R = 5 × 10 3 Ω, Vi = 220 V, zener voltage, V2 = 50 V

= 200 Hz

V 50 Load current, IL = 2 = = 2 .5 × 10 −3 A RL 20 × 10 3

28. (b) According to right-hand rule, the magnetic field due to the

Current through R, I = I − I2 = 34 × 10 −3 − 2 .5 × 10 −3 = 31.5 × 10 −3 A = 31.5 mA

25. (d) Let the wire run in the X-direction and the magnetic field in the Z-direction. In the Hall effect, the Lorentz force, which is due to B2 and the induced electric field E y vanishes ...(i) eE y = eVx Bz Allowing the conduction electrons to move down the wire at drift speed v n . I I ...(ii) vn = = Ane l 2 ne and the Hall voltage is given by V Ey = H l Substituting (ii) and (iii) in Eq. (i), we get IB VH = 2 nel Transverse magnetic field BH 1.5 A × 0.1 T = 5.85 × 10 28 m −3 × 1.6 × 10 −19 C × 1 × 10 −3 m = 1.6 × 10

−8

current carrying wire is directed down into the plane of the figure all along the wire segment AB. µ i The magnitude of the field is B = 0 2 πr Since v is perpendicular to B, |v × B | = Bv Lorentz force to be zero (as required for the equilibrium of a mobile charge carries in the wire segment) There must be an electric field Eopp = − v × B Eopp is directed from A towards B along the wire, so that the electric potential is higher at A than at B, VA > VB , specifically r2 r1

...(iii)

r1

r2

µ iv dr µ iv  r  = 0 ∫ = 0 ln  2  2π r r 2 π  r1  1 Using

i = 10 A, v = 10 m/s,

r2 10 cm = = 10 r1 100 cm

VA − VB = 2 × 10 −7 × 10 × 10 ln 10 = 46.1 µV 2

29. (d) As, potential energy of a particle U( x ) = k(1 − e − x ) It is an exponentially increasing graph of potential energy |U| with x 2 . Therefore, U versus x graph will be as shown. U From the graph, it is clear that origin potential K energy U is minimum (therefore, kinetic energy O will be maximum) and force acting on the n particle is also zero because dU F=− = − (slope of U- x graph) = 0 dx Therefore, origin is the stable equilibrium position. Hence, particle will oscillate simple harmonically about n = 0 for small displacement.

26. (b) The heat energy required is given by Tf

∆H = m ∫ c (T ) dT Ti

Tf

= m ∫ (k1T + k 3T 3 ) dT Ti

k k  = m  1 (Tf2 − Ti 2 ) + 3 (Tf4 − Ti 4 ) 4 2  Inserting the given numerical values, we find that −6  2 .48 × 10 −7    2 .75 × 10   2 2 ∆H = 150    ( 30 4 − 54 ) ( 30 − 5 ) +  2 4      

30. (a) In equilibrium, electrostatic attraction between the plates = spring force ∴

= 0.0180 + 0.7527 = 0.771 cal

27. (b) In case of only the source is moving with respect to the medium. Let |vT| denote the speed of the train and ν 0 denote the emitted frequency, the observer measure a frequency |v| ...(i) νa = ν0 = 219 Hz |v| − |vT|

32

r2

VA − VB = ∫|Eopp|( r )dr = ∫ Bvdr

V

As the train approaches and a frequency |v| νT = ν 0 = 184 Hz |v| + |vT|

...(iv)

...(ii)

q2 = kn 2 ε0 A (CE )2 = k(d − 0.8 d ) 2 ε0 A 2

∴

 ε0 A  2   E  0.8 d  = 0 . 2 dk 2 ε0 A

Force constant of the spring, k =

ε 0 AE 2 0.256d

3

≅

4 ∈0 AE 2 d3

TEST RIDER The Simulator Test Series Towards JEE Main and Advanced 4. A stationary observer receives sonic oscillation from

Instructions p p p

p

This test consists of 30 questions. Each question is allotted 4 marks for correct response. Candidates will be awarded marks as stated above for correct response of each question. 1/4 marks will be deducted for indicating incorrect response of each question. No deduction from the total score will be made if no response is indicated for an item in the answer sheet. There is only one correct response for each question. Filling up more than one response in any question will be treated as wrong response and marks for wrong response will be deducted according as per instructions.

1. In the figure shown, a parallel capacitor having square plates of edge a and plate separation d. The gap between the plates is filled with a dielectric constant k which varies parallel to an edge a, where k and α are constants and x is the distance from the left end. k Calculate the capacitance. k = k0 + α x (a)

ε0 a 2  aα   k0 +  d  2 

(b)

3ε 0a  aα   k0 +  4d  4 

(d)

ε0 a  aα   k0 +  6d  2 

directions are placed at x = ± a parallel to y-axis with z = 0. Magnetic field at origin O is B1 and at B P (2a, 0, 0) B2 . Then, the ratio 1 is B2 (b) −

1 2

(c) −

1 3

(d) 2

(a)

A B

(b)

A 3 1 −  B 2 

(c)

1 3 A −  B 2 

(d)

3A 2B

5. A ball of mass m is thrown upwards with a velocity v. If air exerts an average resisting force F, then the velocity with which the ball returns to the thrower is (a) v

mg mg + F

(b) v

F mg + F

(c) v

mg − F mg + F

(d) v

mg + F mg

6. Two bodies of masses m1 and m2 are placed at a distance r apart. Then at the position, where gravitational field due to them is zero, the gravitational potential is (b) − G m1 / r (d) − G( m1 +

m2 )2 / r

a spring attached to the bottom of the vessel. In equilibrium spring is compressed. The vessel now moves downwards with an acceleration The a ( < g). spring length (a) will become zero (b) may increase, decrease or remain constant (c) will decrease (d) will increase

3. A ray of light enters into a glass slab from air as shown in the figure. Refractive index of glass is given by µ = A − Bt, where A and B are constants, thickness t is from top of slab. Maximum distance travelled by light in medium is

(b) 2 m/s (d) 1.5 m/s

7. A block is submerged in a vessel filled with water by

2. Two parallel wires carrying equal currents in opposite

(a) −3

(a) 1 m/s (c) 0.5 m/s

(a) zero (c) −G m2 / r

2

2

(c)

ε0 2 a 2  aα   3k 0 +   d 2 

two tuning forks, one which approaches and other recedes with same speed. As this take place the observer hears the beat frequency 2 Hz. Find the speed of each tuning fork, if their oscillation is 680Hz and velocity of sound in air is 340m/s.

60° Air Slab

8. A uniform disk of mass M and radius R is supported vertically by a pivot at its centre. As shown in figure, a small dense object (also of mass m) is attached to the rim and raised to the Mass m highest point above the centre. The (unstable) Mass R system is then released. M What is the angular speed of

71

the system, when the attached object passes directly beneath the pivot? 6g (a) 5R 4g (c) 3R

8g (b) 3R 10g (d) 7R

9. A conducting circular loop of radius a and resistance R is kept on a horizontal plane. A vertical time varying magnetic field B = 2t is switched on at time t = 0. Then, (a) power generated in the coil at any time t is constant (b) flow of charge per unit time from any section of the coil is constant (c) total charge passed through any section between time  4 πa 2  t = 0 to t = 2 is    R  (d) All of the above

10. A metal ring of mass m and radius R is placed on a smooth horizontal table and is set rotating about its own axis in such a way that each part of the ring moves with a speed v. Find the tension in the ring. mv 2 2 πR m 2v 2 (d) T = 2 πR

mv 2 πR mv (c) T = πR (a) T =

(b) T =

between C and D, so that the resistance of the entire circuit between A and B does not change with the number of elementary sets used, is R A

R

R

B

(a) R (c) R( 3 − 1)

R R

R

R

R

R C

R R

R

D

(b) 3R (d) R( 3 + 1)

12. Two equal point charges 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, is approximately proportional to (a) x (c) x 3

(b) x 2 (d) 1 / x

13. An infinitely long conductor PQR is bent to form a right angle as shown in figure. A current I flows through PQR. The magnetic field due to this current at the point M is H1 , now another infinitely long straight conductor QS is connected at Q, so that the I current is in QR as well as in QS, the current in PQ 2 remaining unchanged.

72

1 (a) 2 2 (c) 3

90° P

S

Q

90°

(b) 1

R

(d) 2

+∞

14. A thin plano-convex lens of focal length f is split into two halves, one is placed at O x 1 x2 certain distance to other. Magnification of image by 1.8m one half is 2 and object and image are at a distance of 1.8 m, then (a) x1 = 0.6 m (c) f = 0.4 m

I

L2

(b) x2 = 12 . m (d) None of these

15. A 4 g bullet is fired horizontally with a speed of 300 m/s into a 0.8 kg block of wood at rest on a table. If the coefficient of friction between the block and the table is 0.3, how far will the block slide? What fraction of the bullets, energy is dissipated in the collision itself? (a) 0.815 m and 98.6% (c) 0.675 m and 85.7%

11. In the figure, the value of resistors to be connected

M

The magnetic field at M is now H H2 . The ratio 1 is given by H2 –∞

(b) 0.379 m and 99.5% (d) 0.425 and 96.4%

16. In the adjacent diagram, CP represents a wavefront and AO and BP are the corresponding two rays. Find the condition on Q for constructive interference at P between the ray BP and reflected ray OP. 3λ (a) cosθ = 2d (c) sec θ − cos θ =

Q

O

R

θ θ d

C A

P

λ (b) cosθ = 4d λ d

B

(d) sec θ − cos θ =

4λ d

17. The radius of a typical atomic nucleus is about 5 × 10−15 m. Assuming the position uncertainity of a proton in the nucleus to be 5 × 10−15 m. What will be the smallest uncertainity in the proton’s momentum? What is its energy in eV? (a) 4 × 10 −19 kg m/s, 415 keV (b) 1.05 × 10 -20kg m/s, 412 keV (c) 2 × 10 -14 kg m/s, 310 keV (d) 9 × 10 -14 kg m/s, 225 keV

18. In the given circuit R1 = 20Ω, R2 = 40Ω, R3 = 60Ω, R4 = 180Ω, C = 5 µF and E = 6 V. The switch has been closed for a long time. What is C′ R1 D R2 E′ the charge on the capacitor? (a) Zero (b) 2.5 µC (c) 5µC (d) 10µC

C

R3 B

R4 F

G S A

E=6V

H

JEE MAIN RIDE 3 3Ω

x

19. A 4 µF capacitor and a resistance of 2.5mΩ are in series with 12V battery. Find the time after which the potential difference across the capacitor is 3 times the potential difference across the resistor. (given ln 2 = 0.693) (a) 13.86s

(b) 0.693s

(c) 7s

20. The magnetic field in a region  x is given by B = B0   k. A  a

(a) B0v 0 d

(b)

B0v 0d 2 2a

(c)

y d d x

(d)

B0v 0 d 2 a

21. When modulation percentage is 75, an AM transmitter produces 10 kW. What would be percentage power saving, if the carrier and one of the sidebands were suppressed before transmission took place? (a) 89.1%

(b) 9.1%

24Ω

(c) 7.81%

60Ω

8Ω

48V

y 1Ω

temperature difference θ between its two junctions θ2 in accordance with the relation E = 70 θ − , where 20 E is in micro volt, θ is in degree celcius and one junction is at zero degree celcius. If E may be determined to be ±100µV, the possible error in °C, when measuring a temperature of 200° C is (a) ±14 .

B0v 0d 3 a2

30Ω

20Ω

25. The emf E of a certain thermocouple depends on the

(d) 18s

square loop of side d is placed with its edge along the x and y-axes. The loop is moved with a constant velocity v = v0 . The emf induced in the loop is

(a) 160 V (b) 128 V (c) 80 V (d) 62 V

(d) 100%

(b) ±17 .

(c) ±2.0

(d) ±5.0

26. In the ideal double slit experiment, where a glass plate (refractive index 1.5) of thickness t is introduced in the path of one of the interfering beams (wavelength) the intensity at the position where the central maximum occurred previously remains unchanged. The minimum thickness of the glass plate is (a) 2 λ

(b)

2λ 3

(c)

λ 3

(d) λ

22. A 10 kg instrument packet is fixed vertically to a

27. A mixture of violet light of wavelength 3800 Å and

height of 637 km above the Earth’s surface. Which corresponds to a distance from centre of the Earth of 1.1 times the Earth radii. Find the weight (W) and potential energy (PE) of the instrument packet at its maximum altitude. (take the Earth’s surface as the zero level of PE)

blue light of wavelength 4000 Å is incident normally on an air film of 0.00029 mm thickness. The region of the refracted light is

(a) 90 N and 65 mJ (c) 81 N and 57 mJ

(b) 95 N and 70 mJ (d) 75 N and 60 mJ

23. The dispersive power of crown and flint glasses are 0.03 and 0.05 respectively. The refractive indices for yellow light of these glasses are 1.57 and 1.64 respectively. It is desired from a chromatic combination of prisms of crown and flint glasses which can produce a deviation of 1° in the yellow ray. Find the refracting angles of the two prism needed. (a) 5.6° and 7.8° (c) 6.1° and 3.4°

(a) infrared

(b) visible

(d) electromagnetic

28. A 120V, 60W lamp is run from a 240V, 50Hz mains supply using a capacitor connected in series with the lamp and supply. What is theoretical value of the capacitor required to operate the lamp at its normal rating? (a) 3.8 µF

(b) 6.6 µF

(c) 7.7 µF

(d) 13.3 µF

29. A 100 eV electron is fired directly towards a large metal plate having surface charge density − 2 × 10−6 cm−2 . The distance from where the electron be project so that it just falls to strike the plate is (a) 0.22 mm

(b) 4.8° and 2.4° (d) 3.9° and 5.3°

(c) ultraviolet

(b) 0.44 mm

(c) 0.66 mm

(d) 0.88 mm

30. For a transistor amplifier in common-emitter

24. The potenial difference across 8Ω resistance is 48 V as shown in figure. The value of potential difference across X and Y points will be

configurations for load impedance of 1 kΩ (hfe = 50µs and hoe = 25µs), then current gain is (a) −5.2

(b) −157 .

(c) −24.8

(d) −4878 .

Answers 1. (a) 11. (c) 21. (a)

2. (a) 12. (b) 22. (c)

3. (c) 13. (c) 23. (b)

4. (c) 14. (c) 24. (a)

5. (c) 15. (b) 25. (c)

6. (d) 16. (b) 26. (a)

7. (d) 17. (b) 27. (c)

8. (b) 18. (b) 28. (c)

9. (d) 19. (a) 29. (b)

10. (b) 20. (d) 30. (d)

Detailed solutions of these questions are available on http://www.arihantbooks.com/Physics%20Spectrum.pdf

APRIL 2015

73

Answer with Explanations 1. (a) It is based on the parallel plate capacitor with plate area A and separation d between the plates. Then dielectric slab of dielectric constant k is inserted in the space between the plates i.e. Q k ε0 A c= = = K C0 V d

4. (c) It is based on Doppler effect. When relative separation

a

x

dx

ε A where, C = 0 is the capacitance without the dielectric. d Consider a small strip of width dx at a separation x from the left as in figure. The strip from a small capacitor of plate area xdx its capacitance is ( k + αx ) ε 0 adx dC = 0 d The given capacitor may be divided into such strips with x varying form o to a. All these strips are connected in parallel. The capacitance of the given capacitor is a( k + ax ) ε adx 0 C=∫ 0 0 d ε a2  aα  C = 0 k0 +  d  2 

2. (a) Case I If the currents are in the direction shown in figure. Then, µ i µ i B1 = − 2  0  k$ = − 0 k$ = magnetic  2 πa  πa

y

field at o µ i µ i B 2 = 0 k$ − 0 k$ = magnetic field 2 πa 2 π 3a at P µ i = 0 k$ 3 πa B1 ∴ =−3 B2 Case II In this case B1 =

P O

a

a

a

y i

a

∴

a

x

a

B1 =−3 B2

3. (c) It is based on TIR and refraction of light. There will be refraction from different parts. When the angle of incidence in denser medium will more than critical angle for the consecutive layers, TIR will occur and light will be reflected back. Path of light followed in slab should be a curve and at maximum depth it should be horizontal. By Snell’s law at maximum depth, 1 × sin 60 ° = µ sin 90 °

74

where,

v 0 = observed speed v s = source speed

 v ± v0  According to Doppler’s effect, f ′ =   f0  v ± vs For given arrangement,  v  Frequency due to (1) f1 =   f0  v − vs  v  Frequency due to (2) f2 =   f0  v + vs Q

1

2

f1 − f2 = beat frequency = 2Hz vf0 vf0 − =2 v − vs v + vs

 2v  vf0  2 s 2  = 2 , i.e, [ v s