Aieee
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Aieee...
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Nitin M Sir (physics-iitjee.blogspot.com)
PAPER -2003 1.
A particle of mass M and charge Q moving with velocity v describes a circular path of radius R when subjected to a uniform transverse magnetic field of induction B. The work done by the field when the particle completes one full circle is Mv 2 (A) (B) zero 2R R (C) BQ2R (D) BQv2R
1.
B. Since the particle completes one full circle, therefore displacement of particle = 0 Work done = force displacement = 0
2.
A particle of charge 16 1018 coulomb moving with velocity 10 ms1 along the x-axis enters a region where a magnetic field of induction B is along the y –axis, and an electric field of induction B is along the y-axis, and an electric field of magnitude 104 V/m is along the negative z-axis. If the charged particle continues moving along the x-axis, the magnitude of B is (A) 103 Wb/m2 (B) 105 Wb/m2 16 2 (C) 10 Wb/m (D) 103 Wb/m2
2.
A. …(1) F q(E v B) The solution of this problem can be obtained by resolving the motion along the three coordinate axes namely F q a x x (E x v yB z v zB y ) m m Fy q ay (E y v zB x v xB z ) m m F q a z z (E z v xB y v yB z ) m m For the given problem, E x E y 0, v y v z 0 and B x B z 0 Substituting in equation (2), we get a x a z 0 and a y E y v xB z If the particle passes through the region undeflected ay is also zero, then E y v xB z Bz
Ey vz
10 4 103 Wb / m2 10
3.
A thin rectangular magnet suspended freely has a period of oscillation equal to T. Now it is broken into two equal halves (each having half of the original length) and one piece is made to oscillate freely in the same field. If its period of oscillation is T’, the ratio T/T is 1 (A) (B) 1/2 2 2 (C) 2 (D) 1/4
3.
B. When the magnet is divided into 2 equal parts, the magnetic dipole movement
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Nitin M Sir (physics-iitjee.blogspot.com)
Nitin M Sir (physics-iitjee.blogspot.com) AIEEE-PAPER-03-PH-2
M = pole strength length
M and moment of inertia 2
1 mass (length)2 12 2 1 m = 12 2 2 I I' 8 I'
Time period 2
I' I/ 8 2 M M'B B 2
T 2 T' 1 T 2 T'
4.
A magnetic needle lying parallel to a magnetic field requires W units of work to turn it through 60. The torque needed to maintain the needle in this position will be (A) 3 W (B) W (C) (3/2) W (D) 2W
4.
A. W = MB(cos 2 cos 1) Initially magnetic needle is parallel to a magnet field, therefore 1 0, 2 60 W MB(cos 60 cos 0) MB e MB sin 60 ZW 3 / 2 3W
5.
The magnetic lines of force inside a bar magnet (A) are from north-pole to south-pole of the magnet (B) do not exist (C) depend upon the area of cross-section of the bar magnet (D) are from south-pole to north-pole of the magnet.
5.
D. The magnetic lines of force inside a bar magnet are from south pole to north pole of the magnet.
6.
Curie temperature is the temperature above which (A) a ferromagnetic material becomes paramagnetic (B) a paramagnetic material becomes diamagnetic (C) a ferromagnetic material becomes diamagnetic (D) a paramagnetic material becomes ferromagnetic.
6.
A. Curie temperature is the temperature above which a ferromagnetic material becomes paramagnetic.
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Nitin M Sir (physics-iitjee.blogspot.com)
Nitin M Sir (physics-iitjee.blogspot.com) AIEEE PAPER-03-PH-3
7.
A spring balance is attached to the ceiling of a lift. A man hangs his bag on the spring and the spring reads 49 N, when the lift is stationary. If the lift moves downward with an acceleration of 5 m/s2, the reading of the spring balance will be (A) 24 N (B) 74 N (C) 15 N (D) 49 N
7.
A. Reading of spring balance = m(g a) = 5 4.8 = 24 N
8.
The length of a wire of a potentiometer is 100 cm, and the e.m.f. of its stand and cell is E volt. It is employed to measure the e.m.f. of a battery whose internal resistance is 0.5 . If the balance point is obtained at =30 cm from the positive end, the e.m.f. of the battery is 30 E (A) 100.5 30 E (B) 100 0.5 30(E 0.5i) (C) , where I is the current in the potentiometer wire. 100 30 E (D) 100
8.
A. V
E E 30 30E . L 100 100
9.
A strip of copper and another germanium are cooled from room temperature to 80 K. The resistance of (A) each of these decreases (B) copper strip increases and that of germanium decreases (C) copper strip decreases and that of germanium increases (D) each of these increases.
9.
C. The temperature coefficient of resistance of copper is positive and that of germanium is negative, therefore when copper and germanium are cooled, resistance of copper strip decreases and that of germanium increases.
10.
Consider telecommunication through optical fibres. Which of the following statements is not true? (A) Optical fibres can be of graded refractive index. (B) Optical fibres are subject to electromagnetic interference from outside. (C) Optical fibres have extremely low transmission loss. (D) Optical fibres may have homogeneous core with a suitable cladding
10.
B. Optical fibres are subject to electromagnetic interference from outside.
11.
The thermo e.m.f. of a thermo-couple is 25 V/C at room temperature. A galvanometer of 40 ohm resistance, capable of detecting current as low as 105 A, is connected with the thermocouple. The smallest temperature difference that can be detected by this system is (A) 16C (B) 12C (C) 8C (D) 20C
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Nitin M Sir (physics-iitjee.blogspot.com)
Nitin M Sir (physics-iitjee.blogspot.com) AIEEE-PAPER-03-PH-4
11.
A. E = 25 106 V IR = 105 40 = 4 104 V 4 10 4 16C 25 10 6
12.
The negative Zn pole of a Daniell cell, sending a constant current through a circuit, decreases in mass by 0.13 g in 30 minutes. If the electrochemical equivalent of Zn and Cu are 32.5 and 31.5 respectively, the increase in the mass of the positive Cu pole in this time is (A) 0.180 g (B) 0.141 g (C) 0.126 g (D) 0.242 g
12.
C. mZn Z zx mCu ZCx I and t are same for both Cu and Zn electrodes 0.13 31.5 mCu 32.5 0.13 32.5 mCu 0.126 g. 32.5
13.
Dimensions of (A) [L1T] (C) [L2T2]
1 , where symbols have their usual meaning, are 0 0 (B) [L2T2] (D) [LT1]
13.
C.
14.
A circular disc X of radius R is made from an iron pole of thickness t, and another disc Y of radius 4R is made from an iron plate of thickness t/4. then the relation between the moment of inertia IX and IY is (A) IY 32IX (B) IY 16IX (C) IY 32 IX (D) IY 64 IX
14.
D. If t is the thickness and R is the radius of the disc, then mass = R2t = density of the material of the disc. Moment of inertia of disc X, 1 …(i) Ix R 4 t 2 Moment of inertia of disc Y, …(ii) Iy 32 R 4 t From equation (i) and (ii) Iy 64 Ix
15.
The time period of a satellite of earth is 5 hours. If the separation between the earth and the satellite is increased to 4 times the previous value, the new time period will become (A) 10 hours (B) 80 hours (C) 40 hours (D) 20 hours
15.
C. FIITJEE Ltd., ICES House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi-110016, Ph: 26515949, 26569493, Fax: 011-26513942.
Nitin M Sir (physics-iitjee.blogspot.com)
Nitin M Sir (physics-iitjee.blogspot.com) AIEEE PAPER-03-PH-5
2 r 3 Re g r = distance between satellite and the earth. T r3/2 3/2 T r 1 1 T2 r2
Time period of a satellite T =
T2 8T1 8 5 40 hours
16.
A particle performing uniform circular motion has angular momentum L. If its angular frequency is doubled and its kinetic energy halved, then the new angular momentum is (A) L/4 (B) 2L (C) 4L (D) L/2
16.
A. Angular momentum of a particle performing uniform circular motion L = I 1 Kinetic energy, K I2 2 2K 2K Therefore, L 2 L1 K12 L 2 K 2 1 L1 22 4 L2 L L2 . 4
17.
Which of the following radiations has the least wavelength? (A) -rays (B) -rays (C) -rays (D) X-rays
17.
D.
18.
When U238 nucleus originally at rest, decays by emitting an alpha particle having a speed u, the recoil speed of the residual nucleus is 4u 4u (A) (B) 238 234 4u 4u (C) (D) 234 238
18.
B. According to principle of conservation of linear momentum the momentum of the system remains the same before and after the decay. Atomic mass of uranium = 238 and after emitting an alpha particle. = 238 4 = 234 238 0 = 4u + 234 v 4u v . 234
19.
Two spherical bodies of mass M and 5M and radii R and 2R respectively are released in free space with initial separation between their centres equal to 12R. If they attract each other FIITJEE Ltd., ICES House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi-110016, Ph: 26515949, 26569493, Fax: 011-26513942.
Nitin M Sir (physics-iitjee.blogspot.com)
Nitin M Sir (physics-iitjee.blogspot.com) AIEEE-PAPER-03-PH-6
due to gravitational force only, then the distance covered by the smaller body just before collision is (A) 2.5R (B) 4.5R (C) 7.5R (D) 1.5R 19.
C. The two spheres collide when the smaller sphere covered the distance of 7.5 R.
20.
The difference in the variation of resistance with temperature in a metal and a semiconductor arises essentially due to the difference in the (A) crystal structure (B) variation of the number of charge carries with temperature (C) type of bonding (D) variation for scattering mechanism with temperature.
20.
B. Variation of the number charge carriers with temperature.
21.
A car moving with a speed of 50 km/hr, can be stopped by brakes after at least 6 m. If the same car is moving at a speed of 100 km/hr, the minimum stopping distance is (A) 12 m (B) 18 m (C) 24 m (D) 6 m
21.
C.
22.
A boy playing on the roof of a 10 m high building throws a ball with a speed of 10 m/s at an angle of 30 with the horizontal. How far from the throwing point will the ball be at the height of 10 m from the ground? [g = 10 m/s2, sin 30 = ½, cos 30 = 3/2] (A) 5.20 m (B) 4.33 m (C) 2.60 m (d) 8.66 m.
22.
D. The ball will be at the height of 10 m from the ground when it cover its maximum horizontal range. u2 sin 2 Maximum horizontal range R g R
(10)2 2 10
3 1 2 2 8.66 m.
23.
An ammeter reads upto 1 ampere. Its internal resistance is 0.81 ohm. To increase the range to 10 A the value of the required shunt is (A) 0.03 (B) 0.3 (C) 0.9 (D) 0.09
23.
D. S=
24.
IgG I Ig
1 0.81 0.09 10 1
The physical quantities not having same dimensions are (A) torque and work (B) momentum and Planck’s constant FIITJEE Ltd., ICES House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi-110016, Ph: 26515949, 26569493, Fax: 011-26513942.
Nitin M Sir (physics-iitjee.blogspot.com)
Nitin M Sir (physics-iitjee.blogspot.com) AIEEE PAPER-03-PH-7
(C) stress and Young’s modulus
(D) speed and (00)1/2
24.
B. Dimensions of momentum = kg m/sec = [MLT2] Dimensions of Planck’s constant = joule sec = [ML2T1] Dimensions of momentum dimensions of Planck’s constant.
25.
Three forces start acting simultaneously on a particle moving with velocity v . These forces are represented in magnitude and direction by the three sides of a triangle ABC(as shown). The particle will now move with velocity (A) less than v (B) greater than v (C) v in the direction of the largest force BC (D) v , remaining unchanged.
A
B
C
25.
D. According to triangle law of vector addition if three vectors addition if three vectors are represented by three sides of a triangle taken in same order, then their resultant is zero. Therefore resultant of the forces acting on the particle is zero, so the particles velocity remains unchanged.
26.
If the electric flux entering and leaving an enclosed surface respectively is 1 and 2, the electric charge inside the surface will be ( 1 ) (A) (2 1 ) 0 (B) 2 0 ( 1 ) (C) 2 (D) (2 1 )0 0
26.
A. According to Gauss’s theorem, charge in flux =
ch arg e enclosed by the surface 0
q (2 1 ) 0 . 27.
A horizontal force of 10 N is necessary to just hold a block stationary against a wall. The coefficient of friction between the block and the wall is 0.2. The weight of the block is (A) 20 N (B) 50 N (C) 100 N (D) 2 N
10 N
27.
D. Weight of the block = R = 0.2 10 = 2N.
28.
A marble block of mass 2 kg lying on ice when given a velocity of 6 m/s is stopped by friction in 10 s. then the coefficient of friction is (A) 002 (B) 0.03 (C) 0.04 (D) 0.01
28.
C. u 6 0.6 m/sec2 t 10 Frictional force = mg = ma a 0.6 0.06. g 10
Retardation
29.
Consider the following two statements. FIITJEE Ltd., ICES House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi-110016, Ph: 26515949, 26569493, Fax: 011-26513942.
Nitin M Sir (physics-iitjee.blogspot.com)
Nitin M Sir (physics-iitjee.blogspot.com) AIEEE-PAPER-03-PH-8
(1) Linear momentum of a system of particles is zero. (2) Kinetic energy of system of particles is zero. (A) A does not imply B and B does not imply A. (B) A implies B but B does not imply A (C) A does not imply B but b implies A’ (D) A implies B and B implies A. 29.
C.
30.
Two coils are placed close to each other. The mutual inductance of the pair of coils depends upon (A) the rates at which current are changing in the two coils (B) relative position and orientation of the two coils (C) the materials of the wires of the coils (D) the currents in the two coils
30.
C. The mutual inductance of the pair of coils depends on geometry of two coils, distance between two coils, distance between two coils, relative placement of two coils etc.
31.
A block of mass M is pulled along a horizontal friction surface by a rope of mass m. If a force P is applied at the free end of the rope, the force exerted by the rope on the block is Pm Pm (A) (B) Mm Mm Pm (C) P (D) Mm
31.
D. Force on block = mass acceleration =
PM Mm
32.
A light spring balance hangs from the hook of the other light spring balance and a block of mass M kg hangs from the former one. Then the true statement about scale reading is (A) both the scales read M kg each (B) the scale of the lower one reads M kg and of upper one zero (C) the reading of the two scales can be anything but sum of the reading will be M kg (D) both the scales read M/2 kg.
32.
A. Both the scales read M kg each.
33.
A wire suspended vertically from one of its ends stretched by attaching weight of 200 N to the lower end. The weight stretches the wire by 1 mm. Then the elastic energy stored in the wire is (A) 0.2 J (B) 10 J (C) 20 J (D) 0.1 J
33.
D. The elastic potential energy stored in the wire, 1 U stress strain volume 2 1 F 1 1 A = F = 200 10 3 0.1 J 2 A 2 2
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Nitin M Sir (physics-iitjee.blogspot.com)
Nitin M Sir (physics-iitjee.blogspot.com) AIEEE PAPER-03-PH-9
34.
The escape velocity for a body projected vertically upwards from the surface of earth is 11 km/s. If the body is projected at an angle of 45 with the vertical, the escape velocity will be (A) 112 km/s (B) 22 km/s (C) 11 km/s (D) 11/2 m/s
34.
C. The escape velocity of a body is independent of the angle of projection.
35.
A mass M is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes SHM of time period T. If the mass is increased by m, the time period becomes 5T/3. then the ratio of m/M is (A) 3/5 (B) 25/9 (C) 16/9 (D) 5/3
35.
C. T M T' Mm 9 M 25 Mm 9M + 9m = 25 M m 16 M 9
36.
“Heat cannot by itself flow from a body at lower temperature to a body at higher temperature” is a statement of consequence of (A) second law of thermodynamics (B) conservation of momentum (C) conservation of mass (D) first law of thermodynamics.
36.
A. Second law of thermodynamics.
37.
Two particles A and B of equal masses are suspended from two massless springs of spring constants k1 and k2 respectively. If the maximum velocities, during oscillations, are equal, the ratio of amplitudes of A and B is k1 k (A) (B) 2 k2 k1 (C)
k2 k1
(D)
k1 k2
37.
C. a1 k2 . a2 k1
38.
The length of a simple pendulum executing simple harmonic motion is increased by 21%. The percentage increase in the time period of the pendulum of increased length is (A) 11% (B) 21% (C) 42% (D) 10%
38.
D. FIITJEE Ltd., ICES House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi-110016, Ph: 26515949, 26569493, Fax: 011-26513942.
Nitin M Sir (physics-iitjee.blogspot.com)
Nitin M Sir (physics-iitjee.blogspot.com) AIEEE-PAPER-03-PH-10
Time period of simple pendulum is given by. T 2 g 21 121 New length ' 100 199 T' ' 21 T 100 T ' 11 T 10 1 T' T T 10 T 10% of T. 39.
The displacement y of wave travelling in the x-direction is given by y = 104sin 600t 2x metres, 3 where x is expressed in metres and t in seconds. The speed of the wave-motion, in ms1 is (A) 300 (B) 600 (C) 1200 (D) 200
39.
A. Velocity of wave = n=
600 2 = 300 m/sec. 2 2
40.
When the current changes from +2 A to 2 A in 0.05 second, an e.m.f. of 8 V is induced in a coil. The coefficient of self-induction of the coil is (A) 0.2 H (B) 0.4 H (C) 0.8 H (D) 0.1 H
40.
D. If e is the induced e.m.f. in the coil, then e L Therefore, L
di dt
e di / dt
Substituting values, we get L
8 0.05 0.1 H 4
41.
In an oscillating LC circuit the maximum charge on the capacitor is Q. The charge on the capacitor when the energy is stored equally between the electric and magnetic field is (A) Q/2 (B) Q/3 (C) Q/2 (D) Q
41.
C. energy stored in capacitor = E
1 Q2 2 C
1 1 Q2 1 q2 2 2 C 2C Q q= . 2
42.
The core of any transformer is laminated so as to FIITJEE Ltd., ICES House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi-110016, Ph: 26515949, 26569493, Fax: 011-26513942.
Nitin M Sir (physics-iitjee.blogspot.com)
Nitin M Sir (physics-iitjee.blogspot.com) AIEEE PAPER-03-PH-11
(A) reduce the energy loss due to eddy currents (B) make it light weight (C) make it robust and strong (D) increase the secondary voltage.
42. 43.
A. Let F be the force acting on a particle having position vector r and T be the torque of this force about the origin. Then (A) r T 0 and F T 0 (B) r T 0 and F T 0 (C) r T 0 and F T 0 (D) r T 0 and F T 0
43.
D. Torque = Force Position vector T F r r T r (F r ) 0 F T F (F r ) 0
44.
A radioactive sample at any instant has its disintegration rate 5000 disintegrations per minute. After 5 minutes, the rate is 1250 disintegrations per minute. Then, the decay constant (per minute) is (A) 0.4 In 2 (B) 0.2 In 2 (C) 0.1 In 2 (D) 0.8 In 2.
44.
A.
2In2 0.4 In 2. 5
45.
A nucleus with Z = 92 emits the following in a sequence; , , , , , , , , , , , +, +, . The Z of the resulting nucleus is (A) 76 (B) 78 (C) 82 (D) 74
45.
B. The Z of resultant nucleus = 92 16 + 4 2 = 78
46.
Two identical photo cathodes receive light of frequencies f1 and f2. if the velocities of the photoelectrons (of mass m) coming out are respectively v1 and v2, then 1/ 2 2h 2h (A) v12 v 22 (B) v1 v 2 (f1 f2 ) (f1 f2 ) m m (C) v12 v 22
2h (f1 f2 ) m
2h (D) v1 v 2 (f1 f2 ) m
1/ 2
46.
A. 1 m(v12 v 22 ) h(f1 f2 ) 2 2h v12 v 22 (f1 f2 ) . m
47.
Which of the following cannot be emitted by radioactive substance during their decay? FIITJEE Ltd., ICES House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi-110016, Ph: 26515949, 26569493, Fax: 011-26513942.
Nitin M Sir (physics-iitjee.blogspot.com)
Nitin M Sir (physics-iitjee.blogspot.com) AIEEE-PAPER-03-PH-12
(A) protons (C) helium nuclei
(B) neutrinos (D) electrons
47.
A.
48.
A 3 volt battery with negligible internal resistance is connected in a circuit as shown in the figure. The current I, in the circuit will be (A) 1 A (B) 1.5 A (C) 2 A (D) 1/3 A
48.
i
3V
3
3 3
B. The current through the circuit, I
V 3 1.5 A R 2
49.
A sheet of aluminium foil of negligible thickness is introduced between the plates of a capacitor. The capacitance of the capacitor (A) decreases (B) remains unchanged (C) becomes infinite (D) increases.
49.
B. When a sheet of aluminium foil of negligible thickness is introduced between the plates of a capacitor, the capacitance of capacitor remains unchanged.
50.
The displacement of a particle varies according to the relation x = 4(cos t + sin t). the amplitude of the particle is (A) 4 (B) 4 (C) 42 (D) 8
50.
C. The amplitude of given wave equation = 4 2 .
51.
A thin spherical conduction shell of radius R has a charge q. another charge Q is placed at the centre of the shell. The electrostatic potential at a point P at a distance R/2 from the centre of the shell is 2Q 2Q 2q (A) (B) 40R 40R 40R 2Q q (q Q) 2 (C) (D) 40R 40R 40 R
51.
C. The total potential at P =
1 1 (q 2Q) 40 R
52.
The work done in placing a charge of 8 1018 coulomb on a condenser of capacity 100 micro-farad is (A) 16 1032 joule (B) 3.2 1026 joule 10 (C) 4 10 joule (D) 32 1032 joule
52.
D. Required work done is w =
1 Q2 2 C2
1 (8 10 18 )2 32 10 32 J 2 10 4
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Nitin M Sir (physics-iitjee.blogspot.com)
Nitin M Sir (physics-iitjee.blogspot.com) AIEEE PAPER-03-PH-13
53.
The co-ordinates of a moving particle at any time t are given by x = t3 and y = t3. The speed to the particle at time t is given by (A) 3t 2 2 (B) 3t 2 2 2 (C) t 2 2 2
(D)
2 2
53.
B. Speed = v (3t 2 )2 (3t 2 )2 3t 2 2 2 .
54.
During an adiabatic process, the pressure of a gas is found to be proportional to the cube of C its absolute temperature. The ratio p for the gas is Cv (A) 4/3 (B) 2 (C) 5/3 (D) 3/2
54.
A. CP 4 . CV 3
55.
Which of the following parameters does not characterize the thermodynamic state of matter? (A) temperature (B) pressure (C) work (D) volume
55.
C. The work done does not characterize a thermodynamic state of matter. It gives only a relationship between two different thermodynamic state.
56.
A carnot engine takes 3 106 cal of heat from a reservoir at 627C, and gives it to a sink at 27C. The work done by the engine is (A) 4.2 106 J (B) 8.4 106 J 6 (C) 16.8 10 J (D) zero.
56.
B. Work done by the engine while taking heat Q = 3 106 cal is W = 2 106 4.2 = 8.4 106 J.
57.
A spring of spring constant 5 103 N/m is stretched initially by 5 cm from the unstretched position. Then the work required to stretch is further by another 5 cm is (A) 12.50 N-m (B) 18.75 N-m (C) 25.00 N-m (D) 6.25 N-m
57.
B. Required work done = 25 6.25 = 18.75 N–m.
58.
A metal wire of linear mass density of 9.8 gm is stretched with a tension of 10 kg-wt between two rigid supports 1 metre apart. The wire passes at its middle point between the poles of a per magnet and it vibrates in resonance when carrying an alternating current of frequency n. The frequency n of the alternating source is (A) 50 Hz (B) 100 Hz (C) 200 Hz (D) 25 Hz
58.
A. Frequency of oscillation n
1 T 2L m
1 10 9.8 1 1 102 102 50Hz 2L 9.8 10 3 2L 2 1
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Nitin M Sir (physics-iitjee.blogspot.com)
Nitin M Sir (physics-iitjee.blogspot.com) AIEEE-PAPER-03-PH-14
59.
A tuning fork of known frequency 256 Hz makes 5 beats per second with the vibrating string of a piano. The beat frequency decreases to 2 beats per second when the tension in the piano string is slightly increased. The frequency of the piano string before increasing the tension was (A) (256 + 2) Hz (B) (256 2) Hz (C) (256 5) Hz (D) (256 + 5) Hz
59.
C.
60.
A body executes simple harmonic motion. The potential energy (P.E.), the kinetic energy (K.E.) and total energy (T.E.) are measured as function of displacement x. Which of the following statement is true? (A) K.E. is maximum when x = 0 (B) T.E. is zero when x = 0 (C) K.E. is maximum when x is maximum (D) P.E. is maximum when x = 0.
60.
A. Since at x = 0, the potential energy is minimum, the kinetic energy is maximum.
61.
In the nuclear fusion reaction, 2 3 4 given that the repulsive potential energy between the two nuclei is 7.7 1 H 1 H 2 He n 14 10 J, the temperature at which the gases must be heated to initiate the reaction is nearly [Boltzmann’s constant k = 1.38 1023 J/K] (A) 107K (B) 105 K 3 (C) 10 K (D) 109 K
61.
D. T
62.
7.7 10 14 2 3.7 10 9 K. 3 1.38 10 23
Which of the following atoms has the lowest ionization potential? (A) 14 (B) 133 7 N 55 Cs (C)
40 18
Ar
(D)
16 8
O
62.
B. Since 133 has larger size among the four atoms given, thus the electrons present in the 55 Cs outermost orbit will be away from the nucleus and the electrostatic force experienced by electrons due to nucleus will be minimum. Therefore the energy required to liberate electron from outer orbit will be minimum in the case of 133 55 Cs.
63.
The wavelengths involved in the spectrum of deuterium (12 D) are slightly different from that of hydrogen spectrum, because (A) size of the two nuclei are different (B) nuclear forces are different in the two cases (C) masses of the two nuclei are different (D) attraction between the electron and the nucleus is different in the two cases.
63.
C.
64.
In the middle of the depletion layer of a reverse-biased p-n junction, the (A) electric field is zero (B) potential is maximum (C) electric field is maximum (D) potential is zero
64.
A.
65.
If the binding energy of the electron in a hydrogen atom is 13.6 eV, the energy required to remove the electron from the first excited state of Li++ is (A) 30.6 eV (B) 13.6 eV (C) 3.4 eV (D) 122.4 eV. FIITJEE Ltd., ICES House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi-110016, Ph: 26515949, 26569493, Fax: 011-26513942.
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65.
A. The energy of the first excited state of Li is Z 2E 32 13.6 E2 2 0 30.6 eV. n 22
66.
A body is moved along a straight line by a machine delivering a constant power. The distance moved by the body in time t is proportional to (A) t3/4 (B) t3/2 1/4 (C) t (D) t1/2
66.
B. Distance goes as t3/2
67.
A rocket with a lift-off mass 3.5 104 kg is blasted upwards with an initial acceleration of 10 m/s2. Then the initial thrust of the blast is (A) 3.5 105 N (B) 7.0 105 N (C) 14.0 105 N (D) 1.75 105 N
67.
A.
68.
To demonstrate the phenomenon of interference we require two soruces which emit radiation of (A) nearly the same frequency (B) the same frequency (C) different wavelength (D) the same frequency and having a definite phase relationship.
68.
A. Initial thrust of the blast = m a = 3.5 104 10 = 3.5 105 N
69.
Three charges q1, +q2 and q3 are placed as shown in the figure. The x-component of the force on q1 is proportional to q2 q3 q2 q3 (A) 2 2 cos (B) 2 2 sin b a b a q2 q3 q2 q3 (C) 2 2 cos (D) 2 2 sin b a b a
69.
y
q3 a
q1
b q2
x
B. Fx
q2 b
2
q3 a2
sin
70.
A 220 volt, 1000 watt bulb is connected across a 110 volt mains supply. The power consumed will be (A) 750 watt (B) 500 watt (C) 250 watt (D) 1000 watt
70.
C. Pconsumed
71.
V2 (110)2 250 watt. R (220)2 /1000
The image formed by an objective of a compound microscope is (A) virtual and diminished (B) real and diminished (C) real and enlarged (D) virtual and enlarged FIITJEE Ltd., ICES House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi-110016, Ph: 26515949, 26569493, Fax: 011-26513942.
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71.
C. The objective of compound microscope is a convex lens. We know that a convex lens forms real and enlarged image when an object is placed between its focus and lens.
72.
The earth radiates in the infra-red region of the spectrum. The spectrum is correctly given by (A) Rayleigh Jeans law (B) Planck’s law of radiation (C) Stefan’s law of radiation (D) Wien’s law
72.
D.
73.
To get three images of a single object, one should have two plane mirrors at an angle of (A) 60 (B) 90 (C) 120 (D) 30
73.
B. The number of images formed of two plane mirrors are placed at an angle is n =
360 1
Here n = 3 360 3 1 360 90 4 74.
According to Newton’s law of cooling, the rate of cooling of a body is proportional to ()n, where is the difference of the temperature of the body and the surroundings, and n is equal to (A) two (B) three (C) four (D) one
74.
D. According to Newton’s law of cooling. d Rate of cooling dt Therefore n = 1.
75.
The length of a given cylindrical wire is increased by 100%. Due to the consequent decrease in diameter the change in the resistance of the wire will be (A) 200% (B) 100% (C) 50% (D) 300%
75.
D. %change =
3R 100% 300% . R
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PAPER -2004 1.
Which one of the following represents the correct dimensions of the coefficient of viscosity? (A) ML1T2 (B) MLT1 1 1 (C) ML T (D) ML2T2
1.
C. Dimensions of (coefficient of viscosity)
MLT 2 = ML1T1 M L M0LT 1 0 0
2.
A particle moves in a straight line with retardation proportional to its displacement. Its loss of kinetic energy for any displacement x is proportional to (A) x2 (B) ex (C) x (D) logex
2.
A.
mk 2 x 2 K f ki x2 . K f Ki
3.
A ball is released from the top of a tower of height h metres. It takes T seconds to reach the ground. What is the position of the ball in T/3 seconds? (A) h/9 metres from the ground (B) 7h/9 metres from the ground (C) 8h/9 metres from the ground (D) 17h/18 metres from the ground.
3.
C.
4.
If A B B A, then the angle between A and B is (A) (B) /3 (C) /2 (D) /4
4.
A.
5.
A projectile can have the same range R for two angles of projection. If T1 and T2 be the time of flights in the two cases, then the product of the two time of flights is directly proportional to (A) 1/R2 (B) 1/R (C) R (D) R2
5.
C. Range is same for complimentary angles.
T1
2u sin 2u sin (90 ) and T2 g g
u2 sin 2 g 2u sin 2u cos 2R T1T2 . g g g and R
6.
Which of the following statements is false for a particle moving in a circle with a constant angular speed? (A) The velocity vector is tangent to the circle. (B) The acceleration vector is tangent to the circle. (C) The acceleration vector points to the centre of the circle. (D) The velocity and acceleration vectors are perpendicular to each other. FIITJEE Ltd., ICES House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi-110016, Ph: 26515949, 26569493, Fax: 011-26513942.
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6.
B. The acceleration vector is along the radius of circle.
7.
An automobile travelling with speed of 60 km/h, can brake to stop within a distance of 20 cm. If the car is going twice as fast, i.e 120 km/h, the stopping distance will be (A) 20 m (B) 40 m (C) 60 m (D) 80 m
7.
D. If the initial speed is doubled, the stopping distance becomes four times, i.e. 80 m.
8.
A machine gun fires a bullet of mass 40 g with a velocity 1200 ms1. The man holding it can exert a maximum force of 144 N on the gun. How many bullets can he fire per second at the most? (A) one (B) four (C) two (D) three
8.
D. Change in momentum for each bullet fired is
40 1200 48 N 1000
If a bullet fired exerts a force of 48 N on man’s hand so man can exert maximum force of 144 N, number of bullets that can be fired = 144/48 = 3 bullets.
9.
9.
Two masses m1 = 5 kg and m2 = 4.8 kg tied to a string are hanging over a light frictionless pulley. What is the acceleration of the masses when lift free to move? (g = 9.8 m/s2) (A) 0.2 m/s2 (B) 9.8 m/s2 2 (C) 5 m/s (D) 4.8 m/s2 A.
m1 m2
m m2 2 a 1 g 0.2 m / s m1 m2
10.
A uniform chain of length 2 m is kept on a table such that a length of 60 cm hangs freely from the edge of the table. The total mass of the chain is 4 kg. What is the work done in pulling the entire chain on the table? (A) 7.2 J (B) 3.6 J (C) 120 J (D) 1200 J
10.
B. Work done = mgh = 1.2 0.3 10 = 3.6 J.
11.
A block rests on a rough inclined plane making an angle of 30 with the horizontal. The coefficient of static friction between the block and the plane is 0.8. If the frictional force on the block is 10 N, the mass of the block (in kg) is (take g = 10 m/s2) (A) 2.0 (B) 4.0 (C) 1.6 (D) 2.5
11.
A. m = 2 kg
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12.
ˆ A force F (5iˆ 3ˆj 2k)N is applied over a particle which displaces it from its origin to the ˆ ˆ point r (2i j) m. The work done on the particle in joules is (A) 7 (B) +7 (C) +10 (D) +13
12.
B. Work done, W = F s Here s rf ri (2iˆ ˆj) ˆ ˆ ˆj) = 10 3 = 7 J. W (5iˆ 3ˆj 2k)(2i
13.
A body of mass m, accelerates uniformly from rest to v1 in time t1. The instantaneous power delivered to the body as a function of time t is mv 2 t mv1t (A) (B) 2 1 t1 t1 (C)
13.
mv1t 2 t1
(D)
mv12 t t1
B. v v mv 2 t Power P F v mav m 1 1 t 2 1 t1 t1 t1
14.
A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle, the motion of the particle takes place in a plane. It follows that (A) its velocity is constant (B) its acceleration is constant (C) its kinetic energy is constant (D) it moves in a straight line.
14.
C. When a force of constant magnitude which is always perpendicular to the velocity of the particle acts on a particle, the work done and hence change in kinetic energy is zero.
15.
A solid sphere is rotating in free space. If the radius of the sphere is increased keeping mass same which one of the following will not be affected? (A) moment of inertia (B) angular momentum (C) angular velocity (D) rotational kinetic energy.
15.
B. Let it be assume that in “free space” not only the acceleration due to gravity it acting but also there are no external torque acting but also there are no external torque acting on the sphere. If due to internal changes in the system, the radius has increased, then the law of the conservation of angular momentum holds good.
16.
A ball is thrown from a point with a speed 0 at an angle of projection . From the same point and at the same instant person starts running with a constant speed 0/2 to catch the ball. Will the person be able to catch the ball? If yes, what should be the angle of projection? (A) yes, 60 (B) yes, 30 (C) no (D) yes, 45
16.
A. For the person to be able to catch the ball, the horizontal component of the velocity of the ball should be same as the speed of the person. v v 0 cos 0 2 = 60.
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17.
One solid sphere A and another hollow sphere B are of same mass and same outer radii. Their moment of inertia about their diameters are respectively IA and IB such that (A) IA = IB (B) IA > IB (C) IA < IB (D) IA/IB = dA/dB Where dA and dB are their densities.
17.
C. 2 MR 2 5
Moment of inertia of a uniform density solid sphere, A = 2 MR 2 3 Since M and R are same, IA < IB.
And of hollow sphere B =
18.
A satellite of mass m revolves around the earth of radius R at a height x from its surface. If g is the acceleration due to gravity on the surface of the earth, the orbital speed of the satellite is gR (A) gx (B) Rx (C)
18.
1/ 2
gR 2 (D) R x
gR 2 Rx
D. For the satellite, the gravitational force provides the necessary centripetal force i.e. GMe GMem Mv 02 and g 2 (R X) (R X) R2 1/ 2
gR 2 v0 R X
19.
The time period of an earth satellite in circular orbit is independent of (A) the mass of the satellite (B) radius of its orbit (C) both the mass and radius of the orbit (D) neither the mass of the satellite nor the radius of its orbit.
19.
A. The time period of satellite is given by (R h)3 T 2 GM where, R + h = radius of orbit satellite, M = mass of earth.
20.
If g is the acceleration due to gravity on the earth’s surface, the gain in the potential energy of object of mass m raised from the surface of the earth to a height equal to the radius R of the earth is 1 (A) 2 mgR (B) mgR 2 1 (C) mgR (D) mgR 4
20.
B.
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21.
Suppose the gravitational force varies inversely as the nth power of distance. Then the time period planet in circular orbit of radius R around the sun will be proportional to n 1 2
n 1 2
(A) R
(B) R
(C) Rn
(D) R
n2 2
21.
A. T R(n 1) / 2
22.
A wire fixed at the upper end stretches by length by applying a force F. The work done in stretching is (A) F/2 (B) F (C) 2F (D) F/2
22.
D. 1 2 1 2 kx k where is the total extensions. 2 2 1 1 (k ) F 2 2
Work done =
23.
Spherical balls of radius R are falling in a viscous fluid of viscosity with a velocity v. The retarding viscous force acting on the spherical ball is (A) directly proportional to R but inversely proportional to v. (B) directly proportional to both radius R and velocity v. (C) inversely proportional to both radius R and velocity v. (D) inversely proportional to R but directly proportional to velocity v.
23.
B. Retarding viscous force = 6Rv
24.
If two soap bubbles of different radii are connected by a tube, (A) air flows from the bigger bubble to the smaller bubble till the sizes are interchanged. (B) air flows from bigger bubble to the smaller bubble till the sizes are interchanged (C) air flows from the smaller bubble to the bigger. (D) there is no flow of air.
24.
C. 4T The pressure inside the smaller bubble will be more Pi P0 r Therefore, if the bubbles are connected by a tube, the air will flow from smaller bubble to the bigger.
25.
The bob of a simple pendulum executes simple harmonic motion in water with a period t, while the period of oscillation of the bob is t0 in air. Neglecting frictional force of water and 4 given that the density of the bob is 1000 kg/m3. What relationship between t and t0 is 3 true? (A) t = t0 (B) t = t0/2 (C) t = 2t0 (D) t = 4t0
25.
C.
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T 1 1 1 T0 ' 1 1 3 T 2 T0 or, T = 2T0
26.
A particle at the end of a spring executes simple harmonic motion with a period t1, while the corresponding period for another spring is t2. If the period of oscillation with the two springs in series is t, then (A) T = t1 + t2 (B) T 2 t12 t 22 (C) T 1 t11 t 21
(D) T 2 t12 t 22
26.
B. t12 t 22 T 2
27.
The total energy of particle, executing simple harmonic motion is (A) x (B) x2 (C) independent of x (D) x1/2
27.
C. In simple harmonic motion, as a particle is displaced from its mean position, its kinetic energy is converted to potential energy and vice versa and total energy remains constant. The total energy of simple harmonic motion is independent of x.
28.
The displacement y of a particle in a medium can be expressed as y = 106sin(110t + 20 x + /4) m, where t is in seconds and x in meter. The speed of the wave is (A) 2000 m/s (B) 5 m/s (C) 20 m/s (D) 5 m/s.
28.
B. v
5 ms1 k
29.
A particle of mass m is attached to a spring (of spring constant k) and has a natural angular frequency 0. An external force F(t) proportional to cost (0) is applied to the oscillator. The time displacement of the oscillator will be proportional to m 1 (A) 2 (B) 2 2 0 m(0 2 ) 1 m (C) (D) 2 m(02 2 ) 0 2
29.
B. For forced oscillations, the displacement is given by F /m x A sin(t ) with A 20 0 2
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30.
In forced oscillation of a particle the amplitude is maximum for a frequency 1 of the force, while the energy is maximum for a frequency 2 of the force, then (A) 1 = 2 (B) 1 > 2 (C) 1 < 2 when damping is small and 1 > 2 when damping is large (D) 1 < 2
30.
A. Both amplitude and energy get maximised when the frequency is equal to the natural frequency. This is the condition of resonance. 1 = 2
31.
One mole of ideal monoatomic gas ( = 5/30) is mixed with one mole of diatomic gas ( = 7/5). What is for the mixture? denotes the ratio of specific heat at constant pressure, to that at constant volume. (A) 3/2 (B) 23/15 (C) 35/23 (D) 4/3
31.
A. Q = Q1 + Q2 n1 n2 n n 1 2 m 1 1 1 2 1 3 m 2
32.
If the temperature of the sun were to increase from T to 2T and its radius from R to 2R, then the ratio of the radiant energy received on earth to what it was previously will be (A) 4 (B) 16 (C) 32 (D) 64.
32.
D. According to Stefan’s law, P AT4 and A r2 P r2T4
33.
Which of the following statements is correct for any thermodynamic system? (A) The internal energy changes in all processes. (B) Internal energy and entropy are state functions. (C) The change in entropy can never be zero. (D) The work done in an adiabatic process is always zero.
33.
B.
34.
Two thermally insulated vessels 1 and 2 are filled with air at temperatures (T1, T2), volume (V1, V2) and pressure (P1, P2) respectively. If the valve joining two vessels is opened, the temperature inside the vessel at equilibrium will be (A) T1 + T2 (B) (T1 + T2)/2 T1T2 (P1V1 P2 V2 ) T T (P V P2 V2 ) (C) (D) 1 2 1 1 P1V1T2 P2 V2 T1 P1V1T1 P2 T2 T2
34.
C. The number of moles of system remains same According to Boyle’s law, P1V1 + P2V2 = P(V1 + V2) T T (P V P2 V2 ) T 1 2 1 1 P1V1T2 P2 V2 T1 FIITJEE Ltd., ICES House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi-110016, Ph: 26515949, 26569493, Fax: 011-26513942.
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35.
A radiation of energy E falls normally on a perfectly reflecting surface. The momentum transferred to the surface is (A) E/c (B) 2E/c (C) Ec (D) E/c2
35.
B. Psurface P
36.
36.
The temperature of two outer surfaces of a composite slab, consisting of two materials having coefficients of thermal conductivity K and 2K and thickness x and 4x, respectively are T2 and T1 (T2 > T1). The rate of heat transfer through the slab, A(T2 T1 )K in a steady state is f, with f equal to x (A) 1 (B) ½ (C) 2/3 (D) 1/3
x
T2
4x
K
2K
T1
D. q
= 37.
2E . c
2T T1 kA T2 2 x 3
kA T2 T1 3x
A light ray is incident perpendicular to one face of a 90 prism and is totally internally reflected at the glass-air interface. If the angle of reflection is 45, we conclude that the refractive index n 1 (A) n (B) n 2 2 1 (C) n (D) n 2 2
45 45
37.
B. Angle of incidence i > C for total internal reflection. Here i = 45 inside the medium. 45 > sin1(1/n) n > 2.
38.
A plane convex lens of refractive index 1.5 and radius of curvature 30 cm is silvered at the curved surface. Now this lens has been used to form the image of an object. At what distance from this lens an object be placed in order to have a real image of the size of the object? (A) 20 cm (B) 30 cm (C) 60 cm (D) 80 cm
38.
A. 1 2 1 F f1 fm 1 1 1 1 (1.5 1) 60 f1 30 and fm 15 cm. F = 10 cm. Object should be placed at 20 cm from the lens.
and
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39.
The angle of incidence at which reflected light totally polarized for reflection from air to glass (refractive index n), is (A) sin1(n) (B) sin1(1/n) 1 (C) tan (1/n) (D) tan1(n)
39.
D. Brewster’s law: According to this law the ordinary light is completely polarised in the plane of incidence when it gets reflected from transparent medium at a particular angle known as the angle of polarisation. n = tan ip.
40.
The maximum number of possible interference maxima for slit-separation equal to twice the wavelength in Young’s double-slit experiment is (A) infinite (B) five (C) three (D) zero
40.
B. For interference maxima, d sin = n Here d = 2 sin = n/2 and is satisfied by 5 integral values of n (2, 1, 0, 1, 2), as the maximum value of sin can only be 1.
41.
An electromagnetic wave of frequency = 3.0 MHz passes from vacuum into a dielectric medium with permittivity = 4.0. Then (A) wavelength is doubled and the frequency remains unchanged (B) wavelength is doubled and frequency becomes half (C) wavelength is halved and frequency remains unchanged (D) wavelength and frequency both remain unchanged.
41.
C. Refractive index,
2 0
Speed and wavelength of wave will becomes half, the frequency remaining unchanged (frequency of a wave depends on the source as due to refraction, it is assumed that the energy is conserved. h remains the same) 42.
Two spherical conductor B and C having equal radii and carrying equal charges in them repel each other with a force F when kept apart at some distance. A third spherical conductor having same radius as that of B but uncharged brought in contact with B, then brought in contact with C and finally removed away from both. The new force of repulsion, between B and C is (A) F/4 (B) 3F/4 (C) F/8 (D) 3F/8.
42.
D. F'
43.
1 (q / 2)(3q / 4) 3F . 40 d2 8
A charged particle q is shot towards another charged particle Q which is fixed, with a speed v it approaches Q upto a closest distance r and then returns. If q were given a speed 2v, the closest distances of approach would be (A) r (B) 2r (C) r/2 (D) r/4 FIITJEE Ltd., ICES House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi-110016, Ph: 26515949, 26569493, Fax: 011-26513942.
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43.
D. By principle of conservation of energy 1 KqQ mv 2 2 r 1 KqQ Finally, m(2v)2 2 2 r Equation (i) (ii), 1 r' 4 r r r' . 4
…(i) …(ii)
44.
Four charges equal to Q are placed at the four corners of a square and a charge q is at its centre. If the system is in equilibrium the value of q is Q Q (A) (1 2 2) (B) (1 2 2) 4 4 Q Q (C) (1 2 2) (D) (1 2 2) 2 2
44.
B. q=
Q (1 2 2) 4
45.
Alternating current can not be measured by D.C. ammeter because (A) A.C. cannot pass through D.C. (B) A.C. changes direction (C) average value of current for complete cycle is zero (D) D.C. ammeter will get damaged.
45.
C.
46.
The total current supplied to the circuit by the battery is (A) 1 A (B) 2 A (C) 4 A (D) 6 A
2 6V
6 1.5 3
46.
C. The given circuit can be written as 6V I 4A . 1.5
47.
The resistance of the series combination of two resistances is S. When they are joined in parallel through total resistance is P. If S = nP, then the minimum possible value of n is (A) 4 (B) 3 (C) 2 (D) 1
47.
A. Let resistances be R1 and R2 So, S = R1 + R2; R1R 2 P R1 R 2 S = nP nR1R 2 R1 R 2 R1 R 2
(R1 R2 )2 nR1R2 If R1 = R2, so minimum value of n = 4. FIITJEE Ltd., ICES House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi-110016, Ph: 26515949, 26569493, Fax: 011-26513942.
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48.
An electric current is passed through a circuit containing two wires of the same material, connected in parallel. If the length and radii of the wires are in the ratio of 4/3 and 2/3, then the ratio of the currents passing through the wire will be (A) 3 (B) 1/3 (C) 8/9 (D) 2.
48.
B. I1 R 2 I2 R1 [current divider rule since voltage is same in parallel] I1 L 2 r12 I2 L1 r22 2
I1 3 2 1 . I2 4 3 3
49.
In a metre bridge experiment null point is obtained at 20 cm from one end of the wire when resistance X is balanced against another resistance Y. If X < Y, then where will be the new position of the null point from the same end, if one decides to balance a resistance of 4X against Y? (A) 50 cm (B) 80 cm (C) 40 cm (D) 70 cm
49.
A. We have from meter bridge experiment, R1 1 , where 2 = (100 1) cm R2 2 In the first case, X/Y = 20/80 4X In the second case Y 100 = 50 cm.
50.
The thermistors are usually made of (A) metals with low temperature coefficient of resistivity (B) metals with high temperature coefficient of resistivity (C) metal oxides with high temperature coefficient of resistivity ‘ (D) semiconducting materials having low temperature coefficient of resistivity.
50.
C. These are devices whose resistance varies quite markedly with temperature mean having high temperature coefficient of resistivity. [Their name are derived from thermal resistors]. Depending on their composition they can have either negative temperature coefficient or positive temperature coefficient or positive temperature coefficient or positive temperature coefficient characteristics. The negative temperature coefficient types consists of a mixture of oxides of iorn, nickel and cobalt with small amounts of other substance. The positive temperature coefficient types are based on barium titanate.
51.
Time taken by a 836 W heater to heat one litre of water from 10C to 40C is (A) 50 s (B) 100 s (C) 150 s (D) 200 s
51.
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Let t be the time taken, then 836 t 1000 1 (40 10) [using Q = mst] 4.2 t = 150 sec. 52.
The thermo emf of a thermocouple varies with the temperature of the hot junction as E = a + b2 in volts where the ratio a/b is 700C. If the cold junction is kept at 0C, then the neutral temperature is (A) 700C (B) 350C (C) 1400C (D) no neutral temperature is possible for this thermocouple.
52.
D. E = a + b2 At neutral temperature dE/d = 0 dE a a 2bn 0 ; n d 2b a Now 700C (given) b n =700/2 = 350C Now c 0C. So, n 0C But mathematically n 0C .
53.
The electrochemical equivalent of a metal is 3.3 107 kg per coulomb. The mass of the metal liberated at the cathode when a 3 A current is passed for 2 seconds will be (A) 19.8 107 kg (B) 9.9 107 kg (C) 6.6 107 kg (D) 1.1 107 kg
53.
A. m = Zit, m = 3.3 107 3 2 = 19.8 107 kg.
54.
A current I ampere flows along an infinitely long straight thin walled tube, then the magnetic induction at any point inside the tube is (A) infinite (B) zero 0 2i 2i (C) tesla (D) tesla 4 r r
54.
B. Considering Ampere’s loop (shown by dotted line), no current is enclosed by this loop. Therefore, the magnetic field will be zero inside the tube.
55.
A long wire carries a steady current. It is bent into a circle of one turn and the magnetic field at the centre of the coil is B. It is then bent into a circular loop of n turns. The magnetic field at the centre of the coil will be (A) nB (B) n2 B (C) 2nB (D) 2n2B
55.
B. B'
n 0 i i = n2 0 n2B . 2r '
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56.
The magnetic field due to a current carrying circular loop of radius 3 cm at a point on the axis at a distance of 4 cm from the centre is 54 T. What will be its value at the centre of the loop? (A) 250 T (B) 150 T (C) 125 T (D) 75 T
56.
A. Using formula B 54
0iR 2 , we get 2(R X2 )3 / 2 2
0i(3)2 2[(3)2 (4)2 ]3 / 2
At the centre of the coil, X = 0 and B =
…(i) 0i 2(3)
Using equation (i) 54 53 B = 250 T. B (3)2 3 57.
Two long conductors, separated by a distance d carry current I1 and I2 in the same direction. They exert a force F on each other. Now the current in one of them increased to two times and its direction reversed. The distance is also increased to 3d. The new value of the force between them is (A) 2F (B) F/3 (C) 2F/3 (D) F/3
57.
C. Force between two long conductor carrying current II F 0 12 2 d According to question ( 2I1 )(I2 ) F' 0 2 d 3 From equation (i) and (ii), F ' F. 2
58.
The length of a magnet is large compared to its width and breadth. The time period of its width and breadth. The time period of its oscillation in a vibration magnetometer is 2 s. The magnet is cut along its length into three equal parts and three parts are then placed on each other with their like poles together. The time period of this combination will be (A) 2 s (B) 2/3 s (C) 23 s (D) 2/3 s .
58.
B. T MB Where = moment of inertia of magnet, M = magnetic moment m 2 and M = pole strength I 12 2 1 m I I' 3 12 3 3 9 and M’ = pole strength (will remain the same) (/3) 3 = M. T 2 T' s. 9 9
Time period of vibration, T 2
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59.
The materials suitable for making electromagnets should have (A) high retentivity and high coercivity (B) low retentivity and low coercivity (C) high retentivity and low coercivity (D) low retentivity and high coercivity
59.
B.
60.
In an LCR series a.c. circuit, the voltage across each of the components, L, C and R is 50 V. The voltage across the LC combination will be (A) 50 V (B) 502 V (C) 100 V (D) 0 V(zero)
60.
D. In series LCR circuit, the voltage across the inductor (L) and the capacitor (C) are in opposite phase.
61.
A coil having n turns and resistance 4R . This combination is moved in time t seconds from a magnetic field W 1 weber to W 2 weber. The induced current in the circuit is W W1 (W W1 ) (A) 2 (B) 2 5Rnt 5Rt W2 W1 n(W2 W1 ) (C) (D) Rnt Rt
61.
B. n d R ' dt 1 W W1 or, I n 2 R ' t 2 t1 (W 1 and W 2 are not the magnetic field, but the values of flux associated with one turn of coil) 1 n(W2 W1 ) I (R 4R) t n(W2 W1 ) or, I 5Rt I
62.
In a uniform magnetic field of induction B a wire in the form of semicircle of radius r rotates about the diameter of the circle with angular frequency . The axis of rotation is perpendicular to the field. If the total resistance of the circuit is R the mean power generated per period of rotation is Br 2 (Br 2 )2 (A) (B) 2R 2R (Br)2 (Br2 )2 (C) (D) 2R 8R
62.
B. Magnetic flux = BA cos = B
r 2 cos t 2
d 1 Br 2 sin t dt 2 2ind B2 2r 4 2 sin2 t P R 4R Now, = ½ (mean value) (Br 2 )2 P . 8R
ind
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63.
In a LCR circuit capacitance is changed from C to 2C. For the resonant frequency to remain unchanged, the inductance should be changed from L to (A) 4L (B) 2L (C) L/2 (D) L/4
63.
C. res
1
LC if res is to remain same, the product LC should also not change. LC = LC LC = L2C L = L/2
64.
A metal conductor of length 1 m rotates vertically about one of its ends at angular velocity 5 radians per second. If the horizontal component of earth’s magnetic field is 0.3 104 T, then the e.m.f. developed between the two ends of the conductor is (A) depends on the nature of the metal used (B) depends on the intensity of the radiation (C) depends both on the intensity of the radiation and the metal used (D) is the same for all metals and independent of the intensity of the radiation.
64.
B. emf. developed is given by 1 ind BR 2 50 V. 2
65.
According to Einstein’s photoelectric equation, the plot of the kinetic energy of the emitted photo electrons from a metal Vs the frequency, of the incident radiation gives straight line whose slope (A) depends on the nature of the metal used (B) depends on the intensity of the radiation (C) depends both on the intensity of the radiation and the metal used (D) is the same for all metals and independent of the intensity of the radiation.
65.
D. KEmax h W {y = mx + C} Slope of the line in the graph is h, the Planck’s constant.
K.E.
=h W
66.
The work function of a substance is 4.0 eV. Then longest wavelength of light that can cause photoelectron emission from this substance approximately (A) 540 nm (B) 400 nm (C) 310 nm (D) 220 nm
66.
C. hc W hc 6.6 10 34 3 108 W 4.0 1.6 10 19 longest 310 nm. longest
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67.
A charged oil drop is suspended in a uniform field of 3 104 V/m so that it neither falls nor rises. The charge on the drop will be (take the mass of the charge = 9.9 1015 kg and g = 10 m/s2) (A) 3.3 1018 C (B) 3.2 1018 C (C) 1.6 1018 C (D) 4.8 1018 C.
67.
A. Since ball is hanging in equilibrium, force by gravity is balanced by electric force. qE = mg mg q E 9.9 10 15 10 3 10 4 q = 3.3 1018 C
68.
A nucleus disintegrates into two nuclear parts which have their velocities in the ratio 2 : 1. The ratio of their nuclear sizes will be (A) 21/3 : 1 (B) 1 : 3 1/2 1/2 (C) 3 : 1 (D) 1 : 21/3
68.
B. 1/ 3
R1 m2 R 2 2m2 R 1 1: 21/ 3 . R2
69.
The binding energy per nucleon of deuteron (12 H) and helium nucleus ( 24 He) is 1.1 MeV and 7 MeV respectively. If two deuteron nuclei react to form a single helium nucleus, then the energy released is (A) 13.9 MeV (B) 26.9 MeV (C) 23.6 MeV (D) 19.2 MeV
69.
C. Energy released = total binding energy of product total binding energy of reactants 28 (2 2.2) = 28 4.4 = 236 MeV.
70.
An -particle of energy 5 MeV is scattered through 180 by a fixed uranium nucleus. The distance of the closest approach is of the order of (A) 1 Å (B) 1010 cm 12 (C) 10 cm (D) 1015 cm
70.
C. At closest approach, all the kinetic energy of the -particle will converted into the potential energy of the system, K.E. = P.E. 1 q1q2 5 MeV 40 r Z1 Z 2 e2 r 9 9 10 92 2 1.6 10 19 r 5 106 r = 5.3 1014 m = 5.3 1012 cm.
5 106 e = 9 109
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71.
When npn transistor is used as amplifier (A) electrons move from base to collector (C) electrons move from collector to base
(B) holes move from emitter to base (D) holes move from base to emitter.
71.
A. When npn transistor is used, majority charge carrier electrons of n type emitter move from emitter to base and then base to collector.
72.
For a transistor amplifier in common emitter configuration having load impedance of 1 k (hfe = 50 and hoe = 25) the current gain is (A) 5.2 (B) 15.7 (C) 24.8 (D) 48.78
72.
D. In CE configuration, A i =
hfe 1 h0eRL
50 48.78 1 25 10 6 1 103
73.
A piece of copper and another of germanium are cooled from room temperature to 77 K, the resistance of (A) each of them increases (B) each of them decreases (C) copper decreases and germanium increases (D) copper increases and germanium decreases.
73.
D. Copper is metallic conductor and germanium is semiconductor therefore as temperature decreases resistance of good conductor decreases while for semiconductor it increases.
74.
The manifestation of band structure in solids is due to (A) Heisenberg’s uncertainty principle (B) Pauli’s exclusion principle (C) Bohr’s correspondence principle (D) Boltzmann’s law
74.
B.
75.
When p-n junction diode is forward biased (A) the depletion region is reduced and barrier height is increased (B) the depletion region is widened and barrier height is reduced. (C) both the depletion region and barrier height reduced (D) both the depletion region and barrier height increased.
75.
C.
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FIITJEE SOLUTION TO AIEEE-2005 PHYSICS 1.
A projectile can have the same range R for two angles of projection. If t1 and t2 be the times of flights in the two cases, then the product of the two time of flights is proportional to (2) 1/R2 (1) R2 (3) 1/R (4) R
1.
(4)
t1 t 2 = 2.
2u2 sin 2θ 2R = g g2
An annular ring with inner and outer radii R1 and R2 is rolling without slipping with a uniform angular speed. The ratio of the forces experienced by the two particles situated on the inner and outer parts of the ring, F1/F2 is (1)
R1 R2
R2 R1
2
(2)
(3) 1
(4)
R1 R2
2.
(4) F1 R1ω2 R1 = = F2 R 2 ω2 R 2
3.
A smooth block is released at rest on a 45° incline and then slides a distance d. The time taken to slide is n times as much to slide on rough incline than on a smooth incline. The coefficient of friction is
3.
(1) µk = 1 −
1 n2
(2) µk = 1 −
1 n2
(3) µ s = 1 −
1 n2
(4) µ s = 1 −
1 n2
(1)
1 g 2 t1 2 2 1 g d= (1 − µk ) t 22 2 2 t 22 1 = n2 = 2 t1 1 − µk d=
4.
The upper half of an inclined plane with inclination φ is perfectly smooth while the lower half is rough. A body starting from rest at the top will again come to rest at the bottom if the coefficient of friction for the lower half is given by
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(1) 2sinφ (3) 2tanφ 4.
(2) 2cosφ (4) tanφ
(3)
mg s sin φ = µmgcos φ.
S/2
s 2
S/2
φ
5.
A bullet fired into a fixed target loses half of its velocity after penetrating 3 cm. How much further it will penetrate before coming to rest assuming that it faces constant resistance to motion? (1) 3.0 cm (2) 2.0 cm (3) 1.5 cm (4) 1.0 cm
5.
(4)
1 1 v2 mv 2 − m 2 2 4 1 F(3+x) = mv 2 2 x = 1 cm
F.3 =
6.
Out of the following pair, which one does NOT have identical dimensions is (1) angular momentum and Planck’s constant (2) impulse and momentum (3) moment of inertia and moment of a force (4) work and torque
6.
(3) Using dimension
7.
The relation between time t and distance x is t=ax2+bx where a and b are constants. The acceleration is (2) 2bv3 (1) −2abv2 (4) 2av2 (3) −2av3
7.
(3) t = ax2 + bx by differentiating acceleration = - 2av3
8.
A car starting from rest accelerates at the rate f through a distance S, then continues at constant speed for time t and then decelerates at the rate f/2 to come to rest. If the total distance traversed is 15 S, then (1) S=ft (2) S = 1/6 ft2 2 (4) S = 1/4 ft2 (3) S = 1/2 ft
8.
(none) ft 2 S= 1 2 v 0 = 2Sf During retardation S2 = 2S
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V V0
t1
t2
t
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During constant velocity 15S -3S = 12S = vot ft 2 ⇒S= 72 9.
A particle is moving eastwards with a velocity of 5 m/s in 10 seconds the velocity changes to 5 m/s northwards. The average acceleration in this time is (1)
1 2
m / s2 towards north-east
(3) zero 9.
(4) r r r Vf − Vi a= t 5ˆj − 5iˆ 1 ˆ ˆ = = j−i 10 2 1 ∴a = ms−2 towards north west 2
(
)
1 m / s2 towards north. 2 1 (4) m / s2 towards north-west 2
(2)
(N)
ˆj
ˆi (E)
(W) (S)
10.
A parachutist after bailing out falls 50 m without friction. When parachute opens, it decelerates at 2 m/s2. He reaches the ground with a speed of 3 m/s. At what height, did he bail out? (1) 91 m (2) 182 m (3) 293 m (4) 111 m
10.
(3)
11.
A block is kept on a frictionless inclined surface with angle of inclination α. The incline is given an acceleration a to keep the block stationary. Then a is equal to (1) g/tanα (2) g cosecα (3) g (4) g tanα
11.
32 − ( 2 × 10 × 50 ) s = 50 + 2 ( −2 ) = 293 m.
(4) mg sinα = ma cos α ∴ a = g tan α
a α
ma
N
α
mg
12.
A spherical ball of mass 20 kg is stationary at the top of a hill of height 100 m. It rolls down a smooth surface to the ground, then climbs up another hill of height 30 m and finally rolls down to a horizontal base at a height of 20 m above the ground. The velocity attained by the ball is (1) 40 m/s (2) 20 m/s (3) 10 m/s (4) 10 30 m/s
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(1) mgh = ½ mv2 v = 2gh = 2 × 10 × 80 = 40 m/s
13.
A body A of mass M while falling vertically downwards under gravity breaks into two parts; a body B of mass 1/3 M and a body C of mass 2/3 M. The centre of mass of bodies B and C taken together shifts compared to that of body A towards (1) depends on height of breaking (2) does not shift (3) body C (4) body B
13.
(2) No horizontal external force is acting ∴ acm = 0 since v cm = 0 ∴ ∆x cm = 0
14.
The moment of inertia of a uniform semicircular disc of mass M and radius r about a line perpendicular to the plane of the disc through the centre is (1)
2 2 Mr 5 1 (4) Mr 2 2
1 2 Mr 4
(2)
(3) Mr2 14.
(4)
R2 2 MR 2 ∴I = 2 2I = 2M
15.
A particle of mass 0.3 kg is subjected to a force F=−kx with k=15 N/m. What will be its initial acceleration if it is released from a point 20 cm away from the origin? (2) 15 m/s2 (1) 3 m/s2 2 (4) 10 m/s2 (3) 5 m/s
15.
(4)
a= 16.
kx = 10m / s2 m
The block of mass M moving on the frictionless horizontal surface collides with a spring of spring constant K and compresses it by length L. The maximum momentum of the block after collision is (1)
KL2 2M ML2 (4) K
(2)
MK L
(3) zero 16.
(1) 1 2 P2 KL = 2 2m
∴ P = MK L
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M
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17.
17.
A mass ‘m’ moves with a velocity v and collides inelastically with another identical mass. After collision the 1st mass moves with velocity v/√3 in a direction perpendicular to the initial direction of motion. Find the speed of the 2nd mass after collision (1) v (2) √3 v (3) 2v/√3 (4) v/√3 (3) mv = mv1 cos θ mv 0= − mv1 sin θ 3 2 ∴ v1 = v 3
v/√3 m
m v
Before collision
After collision
v 3
m
m
v
Before collision
θ
V1
After collision
18.
A 20 cm long capillary tube is dipped in water. The water rises up to 8 cm. If the entire arrangement is put in a freely falling elevator the length of water column in the capillary tube will be (1) 8 cm (2) 10 cm (3) 4 cm (4) 20 cm
18.
(4) Water will rise to the full length of capillary tube
19.
If S is stress and Y is Young’s modulus of material of a wire, the energy stored in the wire per unit volume is (2) S2/2Y (1) 2S2Y 2 (4) S/2Y (3) 2Y/S
19.
(2) U=
20.
20.
1 S2 stress × strain = 2 2Y
Average density of the earth (1) does not depend on g (3) is directly proportional to g (3)
g=
(2) is a complex function of g (4) is inversely proportional to g
G4π ρav R 3
21.
A body of mass m is accelerated uniformly from rest to a speed v in a time T. The instantaneous power delivered to the body as a function time is given by mv 2 mv 2 2 ⋅t ⋅ t (2) (1) T2 T2 1 mv 2 1 mv 2 2 ⋅t (3) ⋅ t (4) 2 T2 2 T2
21.
(1) P= (ma).v
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= m a2 t V2 =m 2 t T 22.
Consider a car moving on a straight road with a speed of 100 m/s. The distance at which car can be stopped is [µk = 0.5] (1) 800 m (3) 100 m
22.
(2) 1000 m (4) 400 m
(2)
1 mu2 2
µk mgs = s=
u2 = 1000m 2µk g
23.
Which of the following is incorrect regarding the first law of thermodynamics? (1) It is not applicable to any cyclic process (2) It is a restatement of the principle of conservation of energy (3) It introduces the concept of the internal energy (4) It introduces the concept of the entropy
23.
(none) More than one statements are incorrect
24.
A ‘T’ shaped object with dimensions shown in the figure, is lying on a smooth floor. A force F is applied at the point P parallel to AB, such that the object has only the translational motion without rotation. Find the location of P with respect to C 2 3 (1) l (2) l 3 2 4 (4) l (3) l 3
l A
r F
B P 2l
C
24.
(3) P will be the centre of mass of system
25.
The change in the value of g at a height ‘h’ above the surface of the earth is the same as at a depth ‘d’ below the surface of earth. When both ‘d’ and ‘h’ are much smaller than the radius of earth, then which one of the following is correct? h 3h (1) d = (2) d = 2 2 (4) d = h (3) d = 2h
25.
(3) GM
(R + h )
2
=
GM (R − d) R3
⇒ d = 2h
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26.
A particle of mass 10 g is kept on the surface of a uniform sphere of mass 100 kg and radius 10 cm. Find the work to be done against the gravitational force between them to take the particle far away from the sphere (you may take G = 6 . 67 × 10-11 Nm2 / kg2) (2) 3.33 × 10 −10 J (1) 13.34 × 10 −10 J (3) 6.67 × 10 −9 J (4) 6.67 × 10−10 J
26.
(4) w = GMm / R = 6.67 × 10-10 J
27.
A gaseous mixture consists of 16 g of helium and 16 g of oxygen. The ratio mixture is (1) 1.59 (3) 1.4
27.
CP of the Cv
(2) 1.62 (4) 1.54
(2)
n1c v1 + n2 c v 2 29R = n1 + n2 18 cP 47R = 1.62 cP = , cv 18 cv =
28.
The intensity of gamma radiation from a given source is I. On passing through 36 mm I of lead, it is reduced to . The thickness of lead which will reduce the intensity to 8 I will be 2 (1) 6 mm (2) 9 mm (3) 18 mm (4) 12 mm
28.
(4) Use I = I0 e-µx
29.
The electrical conductivity of a semiconductor increases when electromagnetic radiation of wavelength shorter than 2480 nm is incident on it. The band gap in (eV) for the semiconductor is (1) 1.1 eV (2) 2.5 eV (3) 0.5 eV (4) 0.7 eV
29.
(3) Eg =
30.
hc = 0.5 eV λ
A photocell is illuminated by a small bright source placed 1 m away. When the same 1 source of light is placed m away, the number of electrons emitted by photo 2 cathode would (1) decrease by a factor of 4 (2) increase by a factor of 4 (3) decrease by a factor of 2 (4) increase by a factor of 2
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30.
(2)
I∝ 31
1 r2
Starting with a sample of pure corresponding half-life is (1) 10 minutes
66
(3) 5 minutes 31.
Cu, 7/8 of it decays into Zn in 15 minutes. The (2) 15 minutes (4) 7
1 minutes 2
(3)
N0 t1/ 2 N0 t1/ 2 N0 → → 2 4 8 = 15 ∴ t1/ 2 = 5
t1/ 2 N0 →
3t1/ 2 32.
32.
If radius of be nearly (1) 6 fermi (3) 4 fermi
27 13
Al nucleus is estimated to be 3.6 Fermi then the radius
125 52
Te nucleus
(2) 8 fermi (4) 5 fermi
(1) 1
R 125 3 = ⇒ R = 6 fermi 3.6 27 33.
The temperature-entropy diagram of a reversible engine cycle is given in the figure. Its efficiency is (1) 1/2 (2) 1/4 (3) 1/3 (4) 2/3
T 2To To
S
33.
(3) η=
34.
T
∆W QBC
S0 T0 = 2 = 1/ 3 3S0 T0 2
B
2T0
T0 A S0
The figure shows a system of two concentric spheres of radii r1 and r2 and kept at temperatures T1 and T2 respectively. The radial rate of flow of heat in a substance between the two concentric sphere is proportional to r r1
(1)
r2 − r1 r1r2
(2) ln 2
(3)
r1r2 r2 − r1
(4) ln ( r2 − r1 )
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2So
C 2S0 S
r1 r2
T1 T2
S
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34.
(3) 4πr1r2K dQ dt = ( T1 − T2 ) r − r ( 2 1)
35.
A system goes from A to B via two processes I and II as shown in the figure. If ∆U1 and ∆U2 are the changes in internal energies in the processes I and II respectively, the (1) ∆U1 = ∆U2 (2) relation between ∆U1 and ∆U2 can not be determined (3) ∆U2 > ∆U1 (4) ∆U2 < ∆U1
35.
(1) Internal energy is state function
36.
The function sin2(ωt) represents (1) a periodic, but not simple harmonic motion with a period 2π/ω (2) a periodic, but not simple harmonic motion with a period π/ω (3) a simple harmonic motion with a period 2π/ω (4) a simple harmonic motion with a period π/ω.
36.
(4)
y=
II P
A
B I
V
(1 − cos 2ωt ) 2
37.
A Young’s double slit experiment uses a monochromatic source. The shape of the interference fringes formed on a screen is (1) hyperbola (2) circle (3) straight line (4) parabola
37.
(3) Straight line Note: If instead of young’s double slit experiment, young’s double hole experiment was given shape would have been hyperbola.
38.
Two simple harmonic motions are represented by the equation y1 = 0.1
π
sin 100πt + and y2 = 0.1 cosπt. The phase difference of the velocity of particle 1 3
w.r.t. the velocity of the particle 2 is (1) −π/6 (3) −π/3
(2) π/3 (4) π/6
38.
(1) Phase difference (φ) = 99πt + π/3 −π/2 at t = 0 φ = −π/6.
39.
A fish looking up through the water sees the outside world contained in a circular horizon. If the refractive index of water is 4/3 and the fish is 12 cm below the surface, the radius of this circle in cm is (2) 36 / 7 (1) 36 7 (4) 4 5 (3) 36 5
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39.
(2) r=
h µ −1 2
=
36 7
40.
Two point white dots are 1 mm apart on a black paper. They are viewed by eye of pupil diameter 3 mm. Approximately, what is the maximum distance at which these dots can be resolved by the eye? [ Take wavelength of light = 500 nm ] (1) 5 m (2) 1m (3) 6 m (4) 3m
40.
(1)
(1mm ) 1.22λ = Re solution limit = R ( 3mm )
∴R=5m 41.
A thin glass (refractive index 1.5) lens has optical power of – 5D in air. Its optical power in a liquid medium with refractive index 1.6 will be (1) 1 D (2) -1D (3) 25 D (4) – 25 D
41.
(none) µl − 1 µ Pm = a Pair µ l − 1 µ m
Pm=5/8 D 42.
The diagram shows the energy levels for an electron in a certain atom. Which transition shown represents the emission of a photon with the most energy ? (1) III (2) IV (3) I (4) II
n=4 n=3 n=2
I
42.
II
III
IV
n=1
(1) 1 1 − 2 2 n1 n2
∆E ∝ 43.
If the kinetic energy of a free electron doubles. Its deBroglie wavelength changes by the factor (1) (3)
43.
1 2 1
(2) 2
2
(3) λ=
h 2Km
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(4)
2
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44.
In a common base amplifier, the phase difference between the input signal voltage and output voltage is (1)
π 4
(2) π
(3) 0
(4)
π 2
44.
(3) No phase difference between input and output signal.
45.
In a full wave rectifier circuit operating from 50 Hz mains frequency, the fundamental frequency in the ripple would be (1) 50 Hz (2) 25 Hz (3) 100 Hz (4) 70.7 Hz
45.
(3) frequency = 2 (frequency of input signal).
46.
A nuclear transformation is denoted by X(n, α) 73 Li . Which of the following is the nucleus of element X ? (1) 12C6 (2) 105 B (4) 114Be (3) 95 B
46.
(2) X +10 n →24 He + 73 Li
47.
A moving coil galvanometer has 150 equal divisions. Its current sensitivity is 10 divisions per milliampere and voltage sensitivity is 2 divisions per millivolt. In order that each division reads 1 volt, the resistance in ohms needed to be connected in series with the coil will be (2) 105 (1) 103 (3) 99995 (4) 9995
47.
(4) Ig=15mA
Vg = 75mV
V Vg R= − Ig Ig
48.
Two voltameters one of copper and another of silver, are joined in parallel. When a total charge q flows through the voltameters, equal amount of metals are deposited. If the electrochemical equivalents of copper and silver are z1 and z2 respectively the charge which flows through the silver voltameter is (1)
q z 1+ 1 z2
(3) q 48.
z1 z2
(2) q1Z1=q2Z2 q=q1+q2
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(2)
q z 1+ 2 z1
(4) q
z2 z1
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∴ q2 = 49.
q Z 1+ 2 Z1
(1) 200 Ω (3) 500 Ω 49.
500Ω
In the circuit, the galvanometer G shows zero deflection. If the batteries A and B have negligible internal resistance, the value of the resistor R will be
G 2V
R B
A
12V
(2) 100 Ω (4) 1000 Ω
(2) 12R =2 500 + R
50.
Two sources of equal emf are connected to an external resistance R. The internal resistance of the two sources are R1 and R2 (R2 > R1). If the potential difference across the source having internal resistance R2 is zero, then (1) R = R2 × (R1 + R2)/R2 – R1) (2) R = R2 – R1 (3) R = R1R2 / (R1 + R2) (4) R = R1R2 / (R2 – R1)
50.
(2) I=
2E R1 + R2 + R
E − R2I = 0
⇒ R = R2−R1 51.
A fully charged capacitor has a capacitance ‘C’ it is discharged through a small coil of resistance wire embedded in a thermally insulated block of specific heat capacity ‘s’ and mass ‘m’. If the temperature of the block is raised by ‘∆T’. The potential difference V across the capacitance is (1) (3)
2mC∆T s ms∆T C
(2) (4)
mC∆T s 2ms∆T C
51.
(4) Dimensionally only 4th option is correct.
52.
One conducting U tube can slide inside another as shown in figure, maintaining electrical contacts between the tubes. The magnetic field B is perpendicular to the plane of the figure. if each tube moves towards the other at a constant speed V, then the emf induced in the circuit in terms of B, l and V where l is the width of each tube will be (1) BlV (2) – BlV (3) zero (4) 2 BlV
52.
(4) dφ = 2Blv dt
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x x x x
D
x A x V x
x
x
x V x C
x
D
x
x
B x
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53.
A heater coil is cut into two equal parts and only one part is now used in the heater. The heat generated will now be (1) doubled (2) four times (3) one fourth (4) halved
53.
(1) V 2 ∆t R V2 H ' = ' ∆t R H=
Given R’ = R/2
54.
Two thin, long parallel wires separated by a distance ‘d’ carry a current of ‘i’ A in the same direction. They will (1) attract each other with a force of µ0i2/(2πd) (2) repel each other with a force of µ0i2 / (2πd) (3) attract each other with a force of µ0i2(2πd2) (4) repel each other with a force of µ0i2/(2πd2)
54.
(1) Using the definition of force per unit length due to two long parallel wires carrying currents.
55.
When an unpolarized light of intensity I0 is incident on a polarizing sheet, the intensity of the light which does not get transmitted is (1)
1 I0 2
(2)
(3) zero
1 I0 4
(4) I0
55.
(1) When unpolarised light of intensity Io is incident on a polarizing sheet, only Io/2 is transmitted.
56.
A charged ball B hangs from a silk thread S which makes an angle θ with a large charged conducting sheet P, as show in the figure. The surface charge density σ of the sheet is proportional to (1) cos θ (2) cot θ (3) sin θ (4) tan θ
56.
+ + P + + + + +
θ
S
B
(4)
tanθ =
qσ 2 ε ( o ) mg
θ
qσ 2εo Mg
57.
Two point charges + 8q and – 2q are located at x = 0 and x = L respectively. The location of a point on the x axis at which the net electric field due to these two point charges is zero is (1) 2L (2) L/4
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(3) 8L 57.
(4) 4L
(1) −
k2q
( x − L)
2
+
k8q =0 x2
⇒ x = 2L 58.
Two thin wires rings each having a radius R are placed at a distance d apart with their axes coinciding. The charges on the two rings are +q and –q. The potential difference between the centres of the two rings is (1) QR/4πε0d2 (3) zero
58.
Q 1 1 − 2 2πε0 R R + d2 Q 1 1 (4) − 2 4πε0 R R + d2
(2)
(2) kq kq − R R2 + d2 −kq kq v2 = + 2 R R + d2
v1 =
59.
A parallel plate capacitor is made by stacking n equally spaced plates connected alternatively. If the capacitance between any two adjacent plates is C then the resultant capacitance is (1) (n − 1)C (2) (n + 1)C (3) C (4) nC
59.
(1) Ceq=(n−1) C (Q all capacitors are in parallel)
60.
When two tuning forks (fork 1 and fork 2) are sounded simultaneously, 4 beats per second are heard. Now, some tape is attached on the prong of the fork 2. When the tuning forks are sounded again, 6 beats per seconds are heard. If the frequency of fork 1 is 200 Hz, then what was the original frequency of fork 2? (1) 200 Hz (2) 202 Hz (3) 196 Hz (4) 204 Hz
60.
(3) |f1−f2| =4 Since mass of second tuning fork increases so f2 decrease and beats increase so f1>f2 ⇒ f2=f1−4 = 196
61.
If a simple harmonic motion is represented by (1)
2π α
(3) 2πα
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d2 x + αx = 0, its time period is dt 2 2π
(2)
α
(4) 2π α
Nitin M Sir (physics-iitjee.blogspot.com) -1561.
(2) ω2=α ω = √α T=
2π
α
62.
The bob of a simple pendulum is a spherical hollow ball filled with water. A plugged hole near the bottom of the oscillation bob gets suddenly unplugged. During observation, till water is coming out, the time period of oscillation would (1) first increase and then decrease to the original value. (2) first decreased then increase to the original value. (3) remain unchanged. (4) increase towards a saturation value.
62.
(1) First CM goes down and then comes to its initial position.
63.
An observer moves towards a stationary source of sound, with a velocity one fifth of the velocity of sound. What is the percentage increase in the apparent frequency? (1) zero. (2) 0.5% (3) 5% (4) 20%
63.
(4) f=
v + v /5 6f f= v 5
% increase in frequency = 20% 64.
If I0 is the intensity of the principal maximum in the single slit diffraction pattern, then what will be its intensity when the slit width is doubled? (2) 4I0 (1) 2I0 (4) I0/2 (3) I0
64.
(3) Maximum intensity is independent of slit width.
65.
Two concentric coils each of radius equal to 2π cm are placed at right angles to each other. 3 Ampere and 4 ampere are the currents flowing in each coil respectively. The magnetic induction in Weber/m2 at the centre of the coils will be (µ0 = 4π × 10−7 Wb/A-m) (1) 12 × 10−5 (2) 10−5 (3) 5 × 10−5 (4) 7 × 10−5
65.
(3) µo 2 2 I1 + I2 2r 4π × 10 −7 ×5 B= 2 × 2π × 10 −2 B=
B = 5×10−5 66.
A coil of inductance 300 mH and resistance 2Ω is connected to a source of voltage 2V. The current reaches half of its steady state value in (1) 0.05 s (2) 0.1 s
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(3) 0.15 s 66.
(4) 0.3 s
(2) R − t I = Io 1 − e L R 0.693 = t L .3 × 0.693 = 0.1sec t= 2
67.
The self inductance of the motor of an electric fan is 10 H. In order to impart maximum power at 50 Hz, it should be connected to a capacitance of (1) 4µF (2) 8µF (3) 1µF (4) 2µF
67.
(3) f=
1
2π LC 1 C= 4 × π2 f 2 × 10
C = 1µF 68.
An energy source will supply a constant current into the load of its internal resistance is (1) equal to the resistance of the load. (2) very large as compared to the load resistance. (3) zero. (4) non-zero but less than the resistance of the load.
68.
(2) I=
Eo E if R D1
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FIITJEE Solutions to AIEEE−2006- PHYSICS 60.
In a Wheatstone’s bridge, there resistances P, Q and R connected in the three arms and the fourth arm is formed by two resistances S1 and S2 connected in parallel. The condition for bridge to be balanced will be P R P 2R (2) (1) = = Q S1 + S2 Q S1 + S2 (3)
P R(S1 + S2 ) = Q S1S2
Ans:
(3)
Sol.
P R(S1 + S2 ) = Q S1S2
61.
(4)
P R(S1 + S2 ) = Q 2S1S2
The current I drawn from the 5 volt source will be (1) 0.17 A (2) 0.33 A (3) 0.5 A (4) 0.67 A
Ans:
(3)
Sol.
i=
10Ω 5Ω
10Ω
5Ω
10Ω
I
5 = 0.5 10
+
− 5 Volt
62.
In a series resonant LCR circuit, the voltage across R is 100 volts and R = 1 kΩ with C = 2µF. The resonant frequency ω is 200 rad/s. At resonance the voltage across L is (1) 4 × 10−3 V (2) 2.5 × 10−2 V (3) 40 V (4) 250 V
Ans:
(4)
Sol.
i=
100 = 0.1 A 1000
VL = VC = 63
0.1 200 × 2 × 10−6
= 250 V
Two insulating plates are both uniformly charged in such a way that the potential difference between them is V2 −V1 = 20 V. (i.e. plate 2 is at a higher potential). The plates are separated by d = 0.1 m and can be treated as infinitely large. An electron is released from rest on the inner surface of plate 1. What is its speed when it hits plate 2? (e = 1.6 × 10−19 C, me = 9.11 × 10−31 kg) (1) 32 × 10−19 m/s (2) 2.65 × 106 m/s 12 (3) 7.02 × 10 m/s (4) 1.87 × 106 m/s
Ans: Sol.
64.
Ans:
Y 0.1 m X 1
2
(2)
1 mv 2 = eV 2 2eV = 2.65 × 106 m/s v= m The resistance of a bulb filament is 100 Ω at a temperature of 100°C. If its temperature coefficient of resistance be 0.005 per °C, its resistance will become 200 Ω at a temperature of (1) 200°C (2) 300°C (3) 400°C (4) 500°C (2)
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FIITJEE Solutions to AIEEE−2006- PHYSICS Sol.
200 = 100[1 +(0.005 × ∆t)] T− 100 = 200 T = 300° C
65.
In an AC generator, a coil with N turns, all of the same area A and total resistance R, rotates with frequency ω in a magnetic field B. The maximum value of emf generated in the coil is ‘ (2) N.A.B.R.ω (1) N.A.B.ω (3) N.A.B (4) N.A.B.R
Ans:
(1)
Sol.
NBAω
66.
The flux linked with a coil at any instant ‘t’ is given by φ = 10t 2 − 50t + 250 The induced emf at t = 3 s is (1) 190 V (2) −190 V (3) −10 V (4) 10 V
Ans:
(3)
Sol.
e=−
67.
A thermocouple is made from two metals, Antimony and Bismuth. If one junction of the couple is kept hot and the other is kept cold then, an electric current will (1) flow from Antimony to Bismuth at the cold junction (2) flow from Antimony to Bismuth at the hot junction (3) flow from Bismuth to Antimony at the cold junction (4) not flow through the thermocouple
Ans:
(1)
Sol.
Flow from Antimony to Bismuth at cold junction
68.
The time by a photoelectron to come out after the photon strikes is approximately (2) 10−4 s (1) 10−1 s −10 (4) 10−16 s (3) 10 s
Ans:
(3)
Sol.
10−10 sec.
69.
An alpha nucleus of energy
Ans:
dφ = −(20 t − 50) = −10 volt dt
1 mv 2 bombards a heavy nuclear target of charge Ze. Then the 2 distance of closest approach for the alpha nucleus will be proportional to 1 (1) (2) v2 Ze 1 1 (3) (4) 4 m v (3)
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FIITJEE Solutions to AIEEE−2006- PHYSICS Sol.
1 m
70.
The threshold frequency for a metallic surface corresponds to an energy of 6.2 eV, and the stopping potential for a radiation incident on this surface 5 V. The incident radiation lies in (1) X-ray region (2) ultra-violet region (3) infra-red region (4) visible region
Ans:
(2)
Sol.
λ=
1242eVnm ≈ 1100 Å 11.2
Ultraviolet region The energy spectrum of β-particles [number N(E) as a function of β-energy E] emitted from a radioactive source is (1) (2)
71.
N(E)
N(E) E
E0
E0
(3)
(4)
N(E)
N(E)
E0
Ans:
E
E
E0
E
(4)
Sol. N(E)
E0
E
72.
When 3Li7 nuclei are bombarded by protons, and the resultant nuclei are 4Be8, the emitted particles will be (1) neutrons (2) alpha particles (3) beta particles (4) gamma photons
Ans:
(4)
Sol.
Gamma-photon
73.
A solid which is transparent to visible light and whose conductivity increases with temperature is formed by (1) Metallic binding (2) Ionic binding (3) Covalent binding (4) Van der Waals binding
Ans:
(3)
Sol.
Covalent binding
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FIITJEE Solutions to AIEEE−2006- PHYSICS 74.
If the ratio of the concentration of electrons that of holes in a semiconductor is currents is 4 7 4 (3) 5
(1)
7 and the ratio of 5
7 , then what is the ratio of their drift velocities? 4 5 (2) 8 5 (4) 4
Ans:
(4)
Sol.
ne 7 Ie 7 = = nn 5 In 4
(Vd )e I n 5 ⇒ e× n = (Vd )n In ne 4 75.
In a common base mode of a transistor, t collector current is 5.488 mA for an emit current of 5.60 mA. The value of the base current amplification factor (β) will be (1) 48 (2) 49 (3) 50 (4) 51
Ans:
(2)
Sol.
Ib = Ie − Ic
β=
Ic = 49 Ib
76.
The potential energy of a 1 kg particle free move along the x-axis is given by x 4 x2 V(x) = − J 4 2 The total mechanical energy of the particle 2 J. Then, the maximum speed (in m/s) is (1) 2 (2) 3 / 2 (4) 1/ 2 (3) 2
Ans:
(2)
Sol.
k Emax = ET − Umin Umin (±1) = −1/4 J KEmax = 9/4 J ⇒ U =
3 2
J
A force of −Fkˆ acts on O, the origin of the coordinate system. The torque about the point (1, −1) is (1) −F ˆi − ˆj (2) F ˆi − ˆj
77.
( ) (3) −F ( ˆi + ˆj )
Ans: Sol.
( ) (4) F ( ˆi + ˆj )
(3) G ˆ τ = ( −ˆi + ˆj) × ( −Fk) = - F( ˆi + ˆj)
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FIITJEE Solutions to AIEEE−2006- PHYSICS 78.
A thin circular ring of mass m and radius R is rotating about its axis with a constant angular velocity ω. Two objects each of mass M are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with an angular velocity ω′ = ωm ω(m + 2M) (2) (1) (m + 2M) m ω(m − 2M) ωm (3) (4) (m + 2M) (m + M)
Ans:
(1)
Sol.
Li = Lf mR2ω = (mR2 + 2MR2)ω′ mω ω′ = m + 2M
79.
If the terminal speed of a sphere of gold (density = 19.5 kg/m ) is 0.2 m/s in a viscous liquid (density = 1.5 kg/m3) of the same size in the same liquid. (1) 0.2 m/s (2) 0.4 m/s (3) 0.133 m/s (4) 0.1 m/s
Ans:
(4) v s (ρs − ρ A ) = v g (ρg − ρ A )
3
Sol.
vs = 0.1 m/s 80.
The work of 146 kJ is performed in order to compress one kilo mole of gas adiabatically and in this process the temperature of the gas increases by 7° C. The gas is (R = 8.3 J mol−1 K−1) (1) monoatomic (2) diatomic (3) triatomic (4) a mixture of monoatomic and diatomic
Ans:
(2)
Sol.
146 = Cv∆T ⇒ Cv = 21 J/mol K
81.
The rms value of the electric field of the light coming from the Sun is 720 N/C. The average total energy density of the electromagnetic wave is (2) 4.58 × 10−6 J/m3 (1) 3.3 × 10−3 J/m3 3 −9 (4) 81.35 × 10−12 J/m3 (3) 6.37 × 10 J/m
Ans:
(2)
Sol.
2 = 4.58 × 10−6 J/m3 Uav = ε0Erms 2 Erms = 4.58 × 10 −6 J / m3
82.
Ans: Sol.
A coin is placed on a horizontal platform which undergoes vertical simple harmonic motion of angular frequency ω. The amplitude of oscillation is gradually increased. The coin will leave contact with the platform for the first time (1) at the highest position of the platform (2) at the mean position of the platform g g2 (3) for an amplitude of 2 (4) for an amplitude of 2 ω ω (3) Aω2 = g ⇒ A = g/ω2
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FIITJEE Solutions to AIEEE−2006- PHYSICS 83.
An electric bulb is rated 220 volt − 100 watt. The power consumed by it when operated on 110 volt will be (1) 50 watt (2) 75 watt (3) 40 watt (4) 25 watt
Ans:
(4)
Sol.
V12 V22 = = Re sis tance P1 P2 ⇒ P2 = 25 W The anode voltage of a photocell is kept fixed. The wavelength λ of the light falling on the cathode is gradually changed. The plate current I of the photocell varies as follows : (1) (2)
84.
I
I
O
λ
λ
O
(3)
(4) I
I
O
λ
λ
O
Ans:
(3)
85.
The ‘rad’ is the correct unit used to report the measurement of (1) the rate of decay of radioactive source (2) the ability of a beam of gamma ray photons to produce ions in a target (3) the energy delivered by radiation to a target. (4) the biological effect of radiation
Ans:
(4)
86.
Ans:
If the binding energy per nucleon in respectively, then in the reaction p + 73 Li → 2 24He energy of proton must be (1) 39.2 MeV (3) 17.28 MeV (3)
Sol.
EP = (8 × 7.06 – 7 × 5.60) MeV = 17.28 MeV
87.
If the lattice constant of this semiconductor is decreased, then which of the following is correct?
7 3 Li
and
4 2 He
nuclei are 5.60 MeV and 7.06 MeV
(2) 28.24 MeV (4) 1.46 MeV
EC
Condutton band width Band gap Valence badn width
Eg EV
(1) All Ec, Eg, Ev decrease (2) All Ec, Eg, Ev increase (3) Ec, and Ev increase but Eg decreases (4) Ec, and Ev, decrease Eg increases
Ans:
(4)
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FIITJEE Solutions to AIEEE−2006- PHYSICS 88.
In the following, which one of the diodes is reverse biased? (1) (2) +5V
+ 10 V
R R +5V
(3)
(4) − 10 V
R
R −10 V
−5V
Ans:
(1)
89.
4Ω
The circuit has two oppositely connect ideal diodes in parallel. What is the current following in the circuit? 12 V
(1) 1.33 A (3) 2.00 A
D1
D2
3Ω
2Ω
(2) 1.71 A (4) 2.31 A
Ans:
(3)
Sol.
D1 is reverse biased therefore it will act like an open circuit. 12 i= = 2.00 A 6
90.
A long solenoid has 200 turns per cm and carries a current i. The magnetic field at its centre is 6.28 × 10−2 Weber/m2. Another long solenoid has 100 turns per cm and it carries a current i/3. The value of the magnetic field at its centre is (1) 1.05 × 10−4 Weber/m2 (2) 1.05 × 10−2 Weber/m2 2 −5 (4) 1.05 × 10−3 Weber/m2 (3) 1.05 × 10 Weber/m
Ans:
(2)
Sol.
B2 =
91.
Four point masses, each of value m, are placed at the corners of a square ABCD of side A. The moment of inertia through A and parallel to BD is (1) mA2 (2) 2 mA2
B1n2i2 (6.28 × 10 −2 )(100 × i / 3) 2 = = 1.05 × 10 −2 W/m n1i1 200(i)
(3) 3 mA 2
(4) 3 mA2
Ans:
(4)
Sol.
2 2 I = 2m (A/ 2 ) = 3 mA
92.
A wire elongates by A mm when a load W is hanged from it. If the wire goes over a pulley and two weights W each are hung at the two ends, the elongation of the wire will be (in mm) (1) A/2 (2) A (3) 2A
(4) zero
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FIITJEE Solutions to AIEEE−2006- PHYSICS Ans:
(2)
93.
Two rigid boxes containing different ideal gases are placed on a table. Box A contains one mole of nitrogen at temperature T0, while Box B contains one mole of helium at temperature (7/3) T0. The boxes are then put into thermal contact with each other and heat flows between them until the gases reach a common final temperature. (Ignore the heat capacity of boxes). Then, the final temperature of the gases, Tf, in terms of T0 is 5 3 (1) Tf = T0 (2) Tf = T0 2 7 7 3 (4) Tf = T0 (3) Tf = T0 3 2 (4)
Ans: Sol.
∆U = 0 3 5 7 ⇒ R(Tf − T0 ) + 1× R(Tf − T0 ) = 0 2 2 3 3 Tf = T0 2
94.
Two spherical conductors A and B of radii 1 mm and 2 mm are separated by a distance of 5 cm and are uniformly charged. If the spheres are connected by a conducting wire then in equilibrium condition, the ratio of the magnitude of the electric fields at the surface of spheres A and B is (1) 1 : 4 (2) 4 : 1 (3) 1 : 2 (4) 2 : 1
Ans:
(4) E A rB 2 = = EB rA 1
Sol. 95.
An inductor (L = 100 mH), a resistor (R = 100 Ω) and a battery (E = 100 V) are initially connected in series as shown in the figure. After a long time the battery is disconnected after short circuiting the points A and B. The current in the circuit 1 mm after the circuit is (1) 1 A (3) e A
Ans:
(2)
Sol.
I = I0 e−Rt / L =
L
B
A
R
E
(2) 1/e A (4) 0.1 A
1 A e
******
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FIITJEE Solutions to AIEEE - 2007 -PHYSICS
Physics
1
Code-O
41.
The displacement of an object attached to a spring and executing simple harmonic motion is given by x = 2 × 10−2 cos πt metres. The time at which the maximum speed first occurs is (1) 0.5 s (2) 0.75 s (3) 0.125 s (4) 0.25 s
Sol.
(1) x = 2 × 10−2 cos πt v = −0.02π sin πt v is maximum at t =
1 = 0.5 sec 2
42.
In an a.c. circuit the voltage applied is E = E0 sin ωt. The resulting current in the circuit is π I = I0 sin ωt − . The power consumption in the circuit is given by 2 E0I0 (1) P = (2) P = zero 2 EI (4) P 2 E0I0 (3) P = 0 0 2
Sol.
(2) cos φ = 0 So power = 0
43.
An electric charge 10−3 µC is placed at the origin (0, 0) of X–Y co-ordinate system. Two points A and B are situated at ( 2, 2 ) and (2, 0) respectively. The potential difference between the points A and B will be (1) 9 volt (2) zero (3) 2 volt (4) 4.5 volt
Sol.
(2) Both points are at same distance from the charge
44.
A battery is used to charge a parallel plate capacitor till the potential difference between the plates becomes equal to the electromotive force of the battery. The ratio of the energy stored in the capacitor and the work done by the battery will be (1) 1 (2) 2 1 1 (3) (4) 4 2
Sol.
(4) 1 qv 1 2 = qv 2
45.
An ideal coil of 10H is connected in series with a resistance of 5 Ω and a battery of 5V. 2 second after the connection is made the current flowing in amperes in the circuit is (1) (1 − e) (2) e (3) e −1 (4) (1 − e−1 )
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FIITJEE Solutions to AIEEE - 2007 -PHYSICS Sol.
2
(4) Rt i = i0 1 − e L
= (1 − e−1 ) 46.
A long straight wire of radius ‘a’ caries a steady current i. The current is uniformly distributed a across its cross section. The ratio of the magnetic field at and 2a is 2 1 (2) 4 (1) 4 1 (3) 1 (4) 2
Sol.
(3) a i πa2 = µ0 2 2 πa 4 µi …(i) B1 = 0 4πa B2 2π(2a) = µ0i B2π
B2 =
µ0i 4 πa
…(ii)
B1 =1 B2
47.
A current I flows along the length of an infinitely long, straight, thin walled pipe. Then (1) the magnetic field is zero only on the axis of the pipe (2) the magnetic field is different at different points inside the pipe (3) the magnetic field at any point inside the pipe is zero (4) the magnetic field at all points inside the pipe is the same, but not zero
Sol.
(3) Use Ampere’s law
48.
If MO is the mass of an oxygen isotope 8 O17 , Mp and MN are the masses of a proton and a neutron respectively, the nuclear binding energy of the isotope is (2) (MO − 8MP − 9MN )C2 (1) (MO − 8MP )C2 (3) MO C2
(4) (MO − 17MN )C2
Sol.
(2) Binding energy = (MO − 8MP − 9 MN)C2
49.
In gamma ray emission from a nucleus (1) both the neutron number and the proton number change (2) there is no change in the proton number and the neutron number. (3) only the neutron number changes (4) only the proton number changes
Sol.
(2)
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FIITJEE Solutions to AIEEE - 2007 -PHYSICS 50.
If in a p-n junction diode, a square input signal of 10V is applied as shown
5V RL −5V
Then the output signal across RL will be (1)
10V
(2) −10V
(3)
(4)
+5V
−5V
Sol.
(4)
51.
Photon of frequency ν has a momentum associated with it. If c is the velocity of light, the momentum is (1) ν/c (2) hνc (3) hν/c2 (4) hν/c
Sol.
(4) P=
h hν = λ c
52.
The velocity of a particle is v = v0 + gt + ft2. If its position is x = 0 at t = 0, then its displacement after unit time (t = 1) is (1) v0 + 2g + 3f (2) v0 + g/2 + f/3 (3) v0 + g + f (4) v0 + g/2 + f
Sol.
(2) x
1
∫ dx = ∫ (V
0
0
+ gt + ft 2 )dt
0
1 1 x = v0 + g + f 2 3 53.
For the given uniform square lamina ABCD, whose centre is O, (1) 2IAC = IEF (2) IAD = 3IEF (3) IAC = IEF (4) IAC = 2IEF
E A
D
B
F
C
Sol.
(3) IAC = IEF (from ⊥rd axis theorem)
54.
A point mass oscillates along the x-axis according to the law x = x0 cos (ωt − π/4). If the acceleration of the particle is written as a = A cos(ωt + δ) (2) A = x0ω2, δ = −π/4 (1) A = x0 , δ = −π/4 2 (4) A = x0ω2, δ = 3π/4 (3) A = x0ω , δ = −π/4
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FIITJEE Solutions to AIEEE - 2007 -PHYSICS Sol.
(4) v = −x0ω sin (ωt − π/4) π 2 a = −x0ω cos ωt + π − 4 a = A cos(ωt + δ) 3π A = x0ω2; δ = 4
55.
Charges are placed on the vertices of a square as shown. Let E be the electric field and V the potential at the centre. If the charges on A and B are interchanged with those on D and C respectively, then G (1) E remains unchanged, V changes G (2) Both E and V change G (3) E and V remains unchanged G (4) E changes, V remains unchanged
q
−q
A
B
D
C
q
−q
Sol.
(4) G As E is a vector quantity
56.
The half-life period of a radio-active element X is same as the mean life time of another radioactive element Y. Initially they have the same number of atoms. Then (1) X will decay faster than Y (2) Y will decay faster than X (3) X and Y have same decay rate initially (4) X and Y decay at same rate always.
Sol.
(2) t1/ 2 =
ln2 λx
τmean =
1 dN ; = −λN λ y dt
ln2 1 = ⇒ λ x = λ y (0.6932) ⇒ λ y > λ x λx λy
57.
A Carnot engine, having an efficiency of η = 1/10 as heat engine, is used as a refrigerator. If the work done on the system is 10 J, the amount of energy absorbed from the reservoir at lower temperature is (1) 99 J (2) 90 J (3) 1 J (4) 100 J
Sol.
(2) T W = Q2 1 − 1 T2 10 − 1 10 = Q2 9
1 10 = Q2 9 Q2 = 90 J 58.
η = 1−
T2 T1
T T 1 1 9 = 1− 2 ⇒ 2 = 1− = 10 T1 T1 10 10
⇒
T1 10 = T2 9
Carbon, silicon and germanium have four valence electrons each. At room temperature which one of the following statements is most appropriate? (1) The number of free conduction electrons is significant in C but small in Si and Ge. (2) The number of free conduction electrons is negligible small in all the three. (3) The number of free electrons for conduction is significant in all the three. (4) The number of free electrons for conduction is significant only in Si and Ge but small in C.
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FIITJEE Solutions to AIEEE - 2007 -PHYSICS
5
Sol.
(4)
59.
A charged particle with charge q enters a region of constant, uniform and mutually orthogonal G G G G G fields E and B with a velocity v perpendicular to both E and B , and comes out without any G change in magnitude or direction of v . Then G G G G G G (2) v = B × E / B2 (1) v = E × B / B2 G G G G G G (4) v = B × E / E2 (3) v = E × B / E2
Sol.
(1) G G G v × B = −E
60.
The potential at a point x (measured in µm) due to some charges situated on the x-axis is given by V(x) = 20/(x2 − 4) Volts. The electric field E at x = 4 µm is given by (1) 5/3 Volt/µm and in the −ve x direction (2) 5/3 Volt/µm and in the +ve x direction. (3) 10/9 Volt /µm and in the −ve x direction (4) 10/9 Volt/µm and in the +ve x direction.
Sol.
(4) 20 x2 − 4 dV 20 160 10 E=− = 2 (2x − 0) = = 2 144 9 dx (x − 4) Vx =
61.
Which of the following transitions in hydrogen atoms emit photons of highest frequency? (1) n = 2 to n = 6 (2) n = 6 to n = 2 (3) n = 2 to n = 1 (4) n = 1 to n = 2
Sol.
(3) 1 1 hν = Rhcz 2 2 − 2 n n 2 1
62.
A block of mass ‘m’ is connected to another block of mass ‘M’ by a spring (massless) of spring constant ‘k’. The blocks are kept on a smooth horizontal plane. Initially the blocks are at rest and the spring is unstretched. Then a constant force ‘F’ starts acting on the block of mass ‘M’ to pull it. Find the force on the block of mass ‘m’ mF (M + m)F (1) (2) M m mF MF (3) (4) (m + M) (m + M)
Sol.
(3) Kx = ma =
mF m+M
63.
Two lenses of power -15 D and + 5D are in contact with each other. The focal length of the combination is (1) −20 cm (2) − 10 cm (3) + 20 cm (4) + 10 cm
Sol.
(2) P = P1 + P2 = −10 1 f= ⇒ −0.1 m ⇒ −10 cm P
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FIITJEE Solutions to AIEEE - 2007 -PHYSICS 64.
One end of a thermally insulated rod is kept at a A1 T1 temperature T1 and the other at T2. The rod is composed of two sections of lengths A1 and A2 and k1 thermal conductivities k1 and k2 respectively. The temperature at the interface of the two sections is (1) (k2A2T1+ k1A1T2) / (k1A1 + k2A2) (2) (k2A1 T1 + k1A1T2) / (k2A1 + k1A2) (3) (k1A2 T1 + k2A1T2) / (k1A2 + k2A1)
Sol.
A2
T2
k2
(4) (k1A1 T1 + k2A2T2) / (k1A1 + k2A2)
(3) (T1 − T)k1 (T − T2 )k 2 = A1 A2 T=
Tk l 1A 2 + T2k 2 A 1 k1A 2 + k 2 A 1
65.
A sound absorber attenuates the sound level by 20 dB. The intensity decreases by a factor of (1) 1000 (2) 10000 (3) 10 (4) 100
Sol.
(4) I B1 = 10log I0 I' B2 = log I0 given B2 − B1 = 20
I' 20 = 10 log I I' = 100I 66.
If Cp and Cv denote the specific heats of nitrogen per unit mass at constant pressure and constant volume respectively, then (2) Cp −Cv = R/14 (1) Cp −Cv = R/28 (4) Cp −Cv = 28R (3) Cp −Cv = R
Sol.
(1) Mayer Formula
67.
A charged particle moves through a magnetic field perpendicular to its direction. Then (1) the momentum changes but the kinetic energy is constant (2) both momentum and kinetic energy of the particle are not constant (3) both, momentum and kinetic energy of the particle are constant (4) kinetic energy changes but the momentum is constant
Sol.
(1)
68.
Two identical conducting wires AOB and COD are 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 wires AOB and COD, will be given by 12
(1)
µ0 I1 + I2 2π d
(2)
µ0 2 2 1 2 (I1 + I2 ) 2πd
(3)
µ0 (I1 + I2 ) 2πd
(4)
µ0 2 2 (I1 + I2 ) 2πd
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7
Sol.
(2) µ 0I (I12 + I22 ) 2πd
69.
The resistance of a wire is 5 ohm at 50°C and 6 ohm at 100°C. The resistance of the wire at 0 °C will be (1) 2 ohm
(2) 1 ohm
(3) 4 ohm
(4) 3 ohm
Sol.
(3) 5 1 + 50α = 6 1 + 100α 5 = R0(1+ α × 50) RO = 4
70.
A parallel plate condenser with a dielectric of dielectric constant K between the plates has a capacity C and is charged to a potential V volts. The dielectric slab is slowly removed from between the plates and then reinserted. The net work done by the system in this process is (1) ½ (K−1)CV2
(2) CV2(K − 1)/K
(3) (K−1)CV2
(4) zero
Sol.
(4)
71.
If gE and gm are the accelerations due to gravity on the surfaces of the earth and the moon respectively and if Millikan’s oil drop experiment could be performed on the two surfaces, one will electronic charge on the moon find the ratio to be electronic charge on the earth (1) 1
(2) 0
(3) gE/gm
(4) gm/gE
Sol.
(1)
72.
A circular disc of radius R is removed from a bigger circular disc of radius 2R such that the circumferences of the discs coincide. The centre of mass of the new disc is α/R from the centre of the bigger disc. The value of α is
Sol.
(1) 1/3
(2) 1/2
(3) 1/6
(4) 1/4
(1) In this question distance of centre of mass of new disc is αR not
α . R
3M M αR + R = 0 4 4 1 ⇒ α= 3 −
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FIITJEE Solutions to AIEEE - 2007 -PHYSICS 73.
A round uniform body of radius R, mass M and moment of inertia ‘I’, rolls down (without slipping) an inclined plane making an angle θ with the horizontal. Then its acceleration is (1)
(3)
Sol.
74.
gsin θ I 1+ MR2 g sin θ I 1− MR2
(1) Mg sin θ − f = Ma a fR = I R gsin θ ⇒a= I 1 + 2 MR
gsin θ MR2 1+ I
(2)
g sin θ MR2 1− I
(4)
N
f Mg θ
Angular momentum of the particle rotating with a central force is constant due to (1) Constant Force
(2) Constant linear momentum.
(3) Zero Torque
(4) Constant Torque
Sol.
(3)
75.
A 2 kg block slides on a horizontal floor with a speed of 4 m/s. It strikes a uncompressed spring, and compresses it till the block is motionless. The kinetic friction force is 15 N and spring constant is 10,000. N/m. The spring compresses by (1) 5.5 cm
(2) 2.5 cm
(3) 11.0 cm
(4) 8.5 cm
Sol.
(1)
76.
A particle is projected at 60° to the horizontal with a kinetic energy K. The kinetic energy at the highest point is (1) K
(2) Zero
(3) K/2
(4) K/4
Sol.
(4)
77.
In a Young’s double slit experiment the intensity at a point where the path difference is being the wavelength of the light used) is I. If I0 denotes the maximum intensity,
(1)
1 2
(3) 1/2
(2)
I I0
λ (λ 6
is equal to
3 2
(4) 3/4
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FIITJEE Solutions to AIEEE - 2007 -PHYSICS Sol.
(4) I φ = cos2 Imax 2
78.
Two springs, of force constants k1 and k2, are connected to a mass m as shown. The frequency of oscillation of the mass is f. If both k1 and k2 are made four times their original values, the frequency of oscillation becomes
Sol.
79.
(1) f/2
(2) f/4
(3) 4f
(4) 2f
k1
k2 m
(4) f=
1 k1 + k 2 2π m
f' =
k + k2 1 2 1 = 2f 2π m
When a system is taken from state i to state f along the path iaf, it is found that Q = 50 cal and W = 20 cal. Along the path’ ibf Q = 36 cal. W along the path ibf is (1) 6 cal
(2) 16 cal.
(3) 66 cal.
(4) 14 cal.
a
f
i
b
Sol.
(1)
80.
A particle of mass m executes simple harmonic motion with amplitude ‘a’ and frequency ‘ν’. The average kinetic energy during its motion from the position of equilibrium to the end is
Sol.
1 2 2 2 π ma ν 4
(1) π2ma2ν2
(2)
(3) 4π2ma2ν2
(4) 2π2ma2ν2
(1) 1 ma2 ω2 = π2 f 2ma2 4 *************
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FIITJEE Solutions to AIEEE - 2008
AIEEE–2008, PAPER(C−5) Note: (i) The test is of 3 hours duration. (ii) The test consists of 105 questions of 3 marks each. The maximum marks are 315. (iii) There are three parts in the question paper. The distribution of marks subjectwise in each part is as under for each correct response. Part A − Mathematics (105 marks) − 35 Questions Part B − Chemistry (105 marks) − 35 Questions Part C − Physics (105 marks) − 35 Questions (iv) Candidates will be awarded three marks each for indicated correct response of each question. One mark will be deducted for indicated 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.
Mathematics PART − A 1.
AB is a vertical pole with B at the ground level and A at the top. A man finds that the angle of elevation of the point A from a certain point C on the ground is 60°. He moves away from the pole along the line BC to a point D such that CD = 7 m. From D the angle of elevation of the point A is 45°. Then the height of the pole is 7 3 1 7 3 m (2) ⋅ ( 3 + 1) m (1) ⋅ 2 2 3 −1 (3)
Sol:
7 3 ⋅ ( 3 − 1) m 2
A
D
45° 7
It is given that the events A and B are such that P (A) = 1 6 2 (3) 3
(1)
Sol:
7 3 1 ⋅ 2 3 +1
(2) BD = AB = 7 + x Also AB = x tan 60° = x 3 ∴ x 3 =7+x 7 x= 3 −1
7 3 ( 3 + 1) . AB = 2 2.
(4)
60° C
x
B
1 B 2 A 1 , P = and P = . Then P (B) is B 2 A 3 4
1 3 1 (4) 2
(2)
(2) P ( A ∩ B) 1 P ( A ∩ B) 2 = , = 2 3 P (B ) P (A) P (A) 3 Hence (But P (A) = 1/4) = . P (B ) 4 1 ⇒ P (B ) = . 3 FIITJEE Ltd., ICES House, 29 – A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 26515949, 26569493, Fax: 011-26513942.
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FIITJEE Solutions to AIEEE - 2008
2
3.
A die is thrown. Let A be the event that the number obtained is greater than 3. Let B be the event that the number obtained is less than 5. Then P (A ∪ B) is 3 (1) (2) 0 5 2 (3) 1 (4) 5
Sol:
(3) A = {4, 5, 6} , B = {1, 2, 3, 4} . Obviously P (A ∪ B) = 1.
4.
A focus of an ellipse is at the origin. The directrix is the line x = 4 and the eccentricity is 1/2. Then the length of the semi−major axis is 8 2 (1) (2) 3 3 4 5 (3) (4) 3 3
Sol:
(1) Major axis is along x-axis. a − ae = 4 e 1 a2 − = 4 2 8 a= . 3
5.
A parabola has the origin as its focus and the line x = 2 as the directrix. Then the vertex of the parabola is at (1) (0, 2) (2) (1, 0) (3) (0, 1) (4) (2, 0)
Sol:
(2) Vertex is (1, 0)
X =2
O
(2, 0)
6.
The point diametrically opposite to the point P (1, 0) on the circle x2 + y2 + 2x + 4y − 3 = 0 is (1) (3, − 4) (2) (− 3, 4) (3) (− 3, − 4) (4) (3, 4)
Sol:
(3) Centre (− 1, − 2) Let (α, β) is the required point α +1 β+0 = −2 . = − 1 and 2 2
7.
Let f : N → Y be a function defined as f (x) = 4x + 3, where Y = {y ∈ N : y = 4x + 3 for some x ∈ N}. Show that f is invertible and its inverse is 3y + 4 y+3 (1) g (y) = (2) g (y) = 4 + 4 3 y+3 y−3 (4) g (y) = (3) g (y) = 4 4 FIITJEE Ltd., ICES House, 29 – A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 26515949, 26569493, Fax: 011-26513942.
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FIITJEE Solutions to AIEEE - 2008 Sol:
(4) Function is increasing y−3 = g(y) . x= 4
8.
The conjugate of a complex number is −1 i −1 −1 (3) i+1
(1)
3
1 . Then the complex number is i −1 1 (2) i+1 1 (4) i −1
Sol:
(3) Put − i in place of i −1 . Hence i+1
9.
Let R be the real line. Consider the following subsets of the plane R × R. S = {(x, y) : y = x + 1 and 0 < x < 2}, T = {(x, y) : x − y is an integer}. Which one of the following is true? (1) neither S nor T is an equivalence relation on R (2) both S and T are equivalence relations on R (3) S is an equivalence relation on R but T is not (4) T is an equivalence relation on R but S is not
Sol:
(4) T = {(x, y) : x−y ∈ I} as 0 ∈ I T is a reflexive relation. If x − y ∈ I ⇒ y − x ∈ I ∴ T is symmetrical also If x − y = I1 and y − z = I2 Then x − z = (x − y) + (y − z) = I1 + I2 ∈ I ∴ T is also transitive. Hence T is an equivalence relation. Clearly x ≠ x + 1 ⇒ (x, x) ∉ S ∴ S is not reflexive.
10.
The perpendicular bisector of the line segment joining P (1, 4) and Q (k, 3) has y−intercept − 4. Then a possible value of k is (1) 1 (2) 2 (3) − 2 (4) − 4
Sol:
(4) Slope of bisector = k − 1 k +1 7 Middle point = , 2 2 Equation of bisector is ( k + 1) 7 y− = (k − 1) x − 2 2 Put x = 0 and y = − 4. ⇒ k = ± 4.
11.
The solution of the differential equation (1) y = ln x + x (3) y = xe(x−1)
Sol:
dy x + y = satisfying the condition y (1) = 1 is dx x (2) y = x ln x + x2 (4) y = x ln x + x
(4) y = vx FIITJEE Ltd., ICES House, 29 – A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 26515949, 26569493, Fax: 011-26513942.
Nitin M Sir (physics-iitjee.blogspot.com)
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FIITJEE Solutions to AIEEE - 2008
4
dy dv =v+x dx dx dv v+ x = 1+ v dx dx ⇒ dv = x ∴ v = log x + c y ⇒ = log x + c x Since, y (1) = 1, we have y = x log x + x
12.
The mean of the numbers a, b, 8, 5, 10 is 6 and the variance is 6.80. Then which one of the following gives possible values of a and b? (1) a = 0, b = 7 (2) a = 5, b = 2 (3) a = 1, b = 6 (4) a = 3, b = 4
Sol:
(4) Mean of a, b, 8, 5, 10 is 6 a + b + 8 + 5 + 10 ⇒ =6 5 ⇒a+b=7 Given that Variance is 6.8
∴ Variance = 2
∑(X − A)
… (1)
2
i
n 2
( a − 6 ) + ( b − 6 ) + 4 + 1 + 16 = 6.8 5 ⇒ a2 + b2 = 25 a2 + (7 − a)2 = 25 (Using (1)) ⇒ a2 − 7a + 12 = 0 ∴ a = 4, 3 and b = 3, 4. =
13.
Sol:
G G G The vector a = α ˆi + 2ˆj + β kˆ lies in the plane of the vectors b = ˆi + ˆj and c = ˆj + kˆ and bisects the G G angle between b and c . Then which one of the following gives possible values of α and β? (1) α = 2, β = 2 (2) α = 1, β = 2 (4) α = 1, β = 1 (3) α = 2, β = 1
(4) G a = λ ( bˆ + cˆ ) ˆi + 2ˆj + kˆ ⇒ α ˆi + 2ˆj + β kˆ = λ 2
λ = 2α and λ = 2 and λ = ⇒ α = 1 and β = 1.
14.
Sol:
2β
G G G G G G G G The non−zero verctors a, b and c are related by a = 8b and c = −7b . Then the angle between a G and c is (1) 0 (2) π/4 (3) π/2 (4) π
(4) G G Since a = 8b G G c = −7b G G G G ∴ a and b are like vectors and b and c are unlike. G G ⇒ a and c will be unlike FIITJEE Ltd., ICES House, 29 – A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 26515949, 26569493, Fax: 011-26513942.
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FIITJEE Solutions to AIEEE - 2008
5
G G Hence, angle between a and c = π.
15.
The line passing through the points (5, 1, a) and (3, b, 1) crosses the yz−plane at the point 17 −13 , 0, . Then 2 2 (1) a = 2, b = 8 (2) a = 4, b = 6 (3) a = 6, b = 4 (4) a = 8, b = 2
Sol:
(3) Equation of line passing through (5, 1, a) and (3, b, 1) is x − 5 y −1 z − a = = =λ. 2 1− b a − 1 If line crosses yz−plane i.e., x = 0 x = 2λ + 5 = 0 ⇒ λ = −5/2, 17 Since, y = λ (1 − b) + 1 = 2 5 17 − (1 − b ) + 1 = 2 2 b = 4. 13 Also, z = λ (a − 1) + a = − 2 5 13 − ( a − 1) + a = − 2 2 ⇒ a = 6.
16.
If the straight lines
x −1 y − 2 z − 3 x − 2 y − 3 z −1 = = = = and intersect at a point, then the k 2 3 3 k 2
integer k is equal to (1) − 5 (3) 2 Sol:
(2) 5 (4) − 2
(1) x −1 y − 2 z − 3 x − 2 y − 3 z −1 and = = = = k 2 3 3 k 2 Since lines intersect in a point k 2 3 3 k 2 =0 1 1 −2 ∴ 2k2 + 5k − 25 = 0 k = − 5, 5/2.
Directions: Questions number 17 to 21 are Assertion−Reason type questions. Each of these questions contains two statements : Statement − 1 (Assertion) and Statement−2 (Reason). Each of these questions also has four alternative choices, only one of which is the correct answer. You have to select the correct choice.
17.
Statement − 1: For every natural number n ≥ 2,
1 1
+
1 2
+ ... +
1 n
> n.
Statement −2: For every natural number n ≥ 2, n ( n + 1) < n + 1 . (1) Statement −1 is false, Statement −2 is true (2) Statement −1 is true, Statement −2 is true, Statement −2 is a correct explanation for Statement −1 (3) Statement −1 is true, Statement −2 is true; Statement −2 is not a correct explanation for Statement −1. (4) Statement − 1 is true, Statement − 2 is false. Sol:
(3) FIITJEE Ltd., ICES House, 29 – A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 26515949, 26569493, Fax: 011-26513942.
Nitin M Sir (physics-iitjee.blogspot.com)
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FIITJEE Solutions to AIEEE - 2008 1
P (n) =
1 1
P (2) =
1
+ +
1
+ ... +
2 1
1 n
> 2
2
1
Let us assume that P (k) = ∴ P (k + 1) =
L.H.S. > Since ∴
1 1
k+
1
+
2
1 k +1
+ ... +
k +1
1 1
k
+ ... +
2 +
1 k +1
1 k
> k is true
> k + 1 has to be true.
k +1
(∀ k ≥ 0)
k +1
>
1
+
k ( k + 1) + 1
=
k ( k + 1) > k
k ( k + 1) + 1
6
k +1
= k +1
Let P (n) = n ( n + 1) < n + 1 Statement −1 is correct. P (2) = 2 × 3 < 3 If P (k) = k ( k + 1) < (k + 1) is true Now P (k + 1) = ( k + 1) ( k + 2 ) < k + 2 has to be true Since (k + 1) < k + 2 ∴ ( k + 1) ( k + 2 ) < ( k + 2 ) Hence Statement −2 is not a correct explanation of Statement −1. 18.
Let A be a 2 × 2 matrix with real entries. Let I be the 2 × 2 identity matrix. Denote by tr (A), the sum of diagonal entries of A. Assume that A2 = I. Statement −1: If A ≠ I and A ≠ − I, then det A = − 1. Statement −2: If A ≠ I and A ≠ − I, then tr (A) ≠ 0. (1) Statement −1 is false, Statement −2 is true (2) Statement −1 is true, Statement −2 is true, Statement −2 is a correct explanation for Statement −1 (3) Statement −1 is true, Statement −2 is true; Statement −2 is not a correct explanation for Statement −1. (4) Statement − 1 is true, Statement − 2 is false.
Sol:
(4) a b 2 Let A = so that A = c d
a2 + bc ab + bd 1 0 = 2 ac + dc bc + d 0 1
⇒ a2 + bc = 1 = bc + d2 and (a + d)c = 0 = (a + d)b. Since A ≠ I, A ≠ 1, a = – d and hence detA =
1 − bc
b
c
− 1 − bc
= – 1 + bc – bc = – 1
Statement 1 is true. But tr. A = 0 and hence statement 2 is false. n
19.
Statement −1:
∑ (r + 1)
n
Cr = ( n + 2 ) 2n−1 .
n
Cr xr = (1 + x ) + nx (1 + x )
r =0 n
Statement −2:
∑ (r + 1)
n
n −1
.
r =0
(1) Statement −1 is false, Statement −2 is true (2) Statement −1 is true, Statement −2 is true, Statement −2 is a correct explanation for Statement −1 (3) Statement −1 is true, Statement −2 is true; Statement −2 is not a correct explanation for Statement −1. (4) Statement − 1 is true, Statement − 2 is false.
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FIITJEE Solutions to AIEEE - 2008 Sol:
7
(2) n
∑ (r + 1)
n
r =0
n
=
∑
n
Cr =
∑r
n
Cr + n Cr
r =0
n
r
r =0
n n−1 n Cr −1 + Cr = n 2n−1 + 2n r r =0
∑
= 2n−1 (n + 2) Statement −1 is true ( r + 1) n Cr xr = r n Cr xr +
∑
∑
n
∑
= n
n −1
Cr −1 xr +
r =0
n
∑
n
∑
n
Cr xr
Cr xr = nx (1 + x)n−1 + (1 + x)n
r =0
Substituting x = 1 ( r + 1) n Cr = n 2n −1 + 2n
∑
Hence Statement −2 is also true and is a correct explanation of Statement −1. 20.
Let p be the statement “x is an irrational number”, q be the statement “y is a transcendental number”, and r be the statement “x is a rational number iff y is a transcendental number”. Statement –1: r is equivalent to either q or p Statement –2: r is equivalent to ∼ (p ↔ ∼ q). (1) Statement −1 is false, Statement −2 is true (2) Statement −1 is true, Statement −2 is true, Statement −2 is a correct explanation for Statement −1 (3) Statement −1 is true, Statement −2 is true; Statement −2 is not a correct explanation for Statement −1. (4) Statement − 1 is true, Statement − 2 is false.
Sol:
(4) Given statement r = ∼ p ↔ q Statement −1 : r1 = (p ∧ ∼ q) ∨ (∼ p ∧ q) Statement −2 : r2 = ∼ (p ↔ ∼ q) = (p ∧ q) ∨ (∼ q ∧ ∼ p) From the truth table of r, r1 and r2, r = r1. Hence Statement − 1 is true and Statement −2 is false.
21.
In a shop there are five types of ice-creams available. A child buys six ice-creams. Statement -1: The number of different ways the child can buy the six ice-creams is 10C5. Statement -2: The number of different ways the child can buy the six ice-creams is equal to the number of different ways of arranging 6 A’s and 4 B’s in a row. (1) Statement −1 is false, Statement −2 is true (2) Statement −1 is true, Statement −2 is true, Statement −2 is a correct explanation for Statement −1 (3) Statement −1 is true, Statement −2 is true; Statement −2 is not a correct explanation for Statement −1. (4) Statement − 1 is true, Statement − 2 is false.
Sol:
(1) x1 + x2 + x3 + x4 + x5 = 6 5+6–1 C5 – 1 = 10C4.
22.
Sol:
1 if x ≠ 1 ( x − 1) sin , . Then which one of the following is true? Let f(x) = x − 1 0, if x = 1 (1) f is neither differentiable at x = 0 nor at x = 1 (2) f is differentiable at x = 0 and at x = 1 (3) f is differentiable at x = 0 but not at x = 1 (4) f is differentiable at x = 1 but not at x = 0
(1)
f (1 + h ) − f (1) h→0 h
f′(1) = lim
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FIITJEE Solutions to AIEEE - 2008 (1 + h − 1) sin 1 − 0 h 1 1 + h − 1 = lim sin ⇒ f′(1) = lim h →0 h →0 h h h 1 ⇒ f′(1) = lim sin h→0 h ∴ f is not differentiable at x = 1. f (h) − f ( 0 ) Similarly, f′(0) = lim h →0 h ( h − 1) sin 1 − sin (1) h − 1 ⇒ f′(0) = lim h→0 h ⇒ f is also not differentiable at x = 0. 23.
The first two terms of a geometric progression add up to 12. The sum of the third and the fourth terms is 48. If the terms of the geometric progression are alternately positive and negative, then the first term is (1) –4 (2) –12 (3) 12 (4) 4
Sol:
(2) Let a, ar, ar2, … a + ar = 12 ar2 + ar3= 48 dividing (2) by (1), we have ar 2 (1 + r ) =4 a ( r + 1) ⇒ r2 = 4 if r ≠ – 1 ∴r=–2 also, a = – 12 (using (1)).
24.
Suppose the cube x3 – px + q has three distinct real roots where p > 0 and q > 0. Then which one of the following holds? p p and maxima at – (1) The cubic has minima at 3 3 (2) The cubic has minima at –
Sol:
p and maxima at 3
(3) The cubic has minima at both
p and – 3
p 3
(4) The cubic has maxima at both
p and – 3
p 3
(1) Let f(x) = x3 – px + q Now for maxima/minima f′(x) = 0 ⇒ 3x2 – p = 0 p ⇒ x2 = 3
∴x=± 25.
…(1) …(2)
p 3
√(p/3) –√(p/3)
p . 3
How many real solutions does the equation x7 + 14x5 + 16x3 + 30x – 560 = 0 have? (1) 7 (2) 1 (3) 3 (4) 5
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FIITJEE Solutions to AIEEE - 2008 Sol:
(2) x7 + 14x5 + 16x3 + 30x – 560 = 0 Let f(x) = x7 + 14x5 + 16x3 + 30x ⇒ f′(x) = 7x6 + 70x4 + 48x2 + 30 > 0 ∀ x. ∴ f (x) is an increasing function ∀ x.
26.
The statement p → (q → p) is equivalent to (1) p → (p → q) (3) p → (p ∧ q)
Sol:
(2) p → (q → p) = ~ p ∨ (q → p) = ~ p ∨ (~ q ∨ p) since p ∨ ~ p is always true = ~ p ∨ p ∨ q = p → (p ∨ q).
27.
Sol:
5 2 The value of cot cos ec −1 + tan−1 is 3 3 6 (1) 17 4 (3) 17
(2) p → (p ∨ q) (4) p → (p ↔ q)
3 17 5 (4) 17
(2)
(1) 5 2 Let E = cot cos ec −1 + tan−1 3 3 3 2 ⇒ E = cot tan−1 + tan−1 4 3
3 2 + −1 ⇒ E = cot tan 4 3 1 − 3 ⋅ 2 4 3 17 6 . ⇒ E = cot tan−1 = 6 17
28.
The differential equation of the family of circles with fixed radius 5 units and centre on the line y = 2 is (1) (x – 2)y′2 = 25 – (y – 2)2 (2) (y – 2)y′2 = 25 – (y – 2)2 (3) (y – 2)2y′2 = 25 – (y – 2)2 (4) (x – 2)2y′2 = 25 – (y – 2)2
Sol:
(3) (x – h)2 + (y – 2)2 = 25 dy =0 ⇒ 2(x – h) + 2(y – 2) dx dy ⇒ (x – h) = – (y – 2) dx substituting in (1), we have
…(1)
2
( y − 2 )2
dy 2 + ( y − 2 ) = 25 dx (y – 2)2y′2 = 25 – (y – 2)2. 1
29.
Let I =
∫ 0
sin x x
1
dx and J =
∫ 0
2 and J > 2 3 2 (3) I < and J > 2 3
(1) I >
cos x x
dx . Then which one of the following is true? 2 and J < 2 3 2 (4) I > and J < 2 3
(2) I <
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9
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FIITJEE Solutions to AIEEE - 2008 Sol:
(2) 1
∫
I=
sin x x
0
1
dx <
1
x
∫
x
0
dx =
1
∫
xdx =
0
2 3/2 2 x = 3 3 0
2 ⇒I< 3 1
∫
J=
cos x x
0
1
dx <
∫ 0
1
1
x
dx = 2 x 0 = 2
∴ J ≤ 2. 30.
The area of the plane region bounded by the curves x + 2y2 = 0 and x + 3y2 = 1 is equal to 5 1 (2) (1) 3 3 2 4 (4) (3) 3 3
Sol:
(4) Solving the equations we get the points of intersection (–2, 1) and (–2, –1) The bounded region is shown as shaded region.
y (–2, 1)
1
The required area = 2
∫ (1 − 3y ) − ( −2y ) 2
2
0
(1, 0)
1
1
y3 2 4 = 2 (1 − y 2 ) dy = 2 y − = 2 × = . 3 0 3 3
∫ 0
x + 2y2 = 0
x
(–2, –1)
x + 3y2 = 1
31.
The value of
2
sin xdx
∫ sin x − π
Sol:
is
4
π (1) x + log cos x − + c 4
π (2) x – log sin x − + c 4
π (3) x + log sin x − + c 4
π (4) x – log cos x − + c 4
(3) π π sin x − + dx sin xdx 4 4 2 = 2 π π sin x − sin x − 4 4
∫
∫
π π π 2 cos + cot x − sin dx 4 4 4 π = dx + cot x − dx 4 π = x + ln sin x − + c . 4
∫
=
∫
32.
∫
How many different words can be formed by jumbling the letters in the word MISSISSIPPI in which no two S are adjacent? (2) 6 . 7 . 8C4 (1) 8 . 6C4 . 7C4 7 (4) 7 . 6C4 . 8C4 (3) 6 . 8 . C4 FIITJEE Ltd., ICES House, 29 – A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 26515949, 26569493, Fax: 011-26513942.
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FIITJEE Solutions to AIEEE - 2008 Sol:
11
(4)
Other than S, seven letters M, I, I, I, P, P, I can be arranged in
7! = 7 . 5 . 3. 2! 4!
Now four S can be placed in 8 spaces in 8C4 ways. Desired number of ways = 7 . 5 . 3 . 8C4 = 7 . 6C4 . 8C4. 33.
Let a, b, c be any real numbers. Suppose that there are real numbers x, y, z not all zero such that x = cy + bz, y = az + cx and z = bx + ay. Then a2 + b2 + c2 + 2abc is equal to (1) 2 (2) – 1 (3) 0 (4) 1
Sol:
(4) The system of equations x – cy – bz = 0, cx – y + az = 0 and bx + ay – z = 0 have non-trivial solution if 1 −c −b 2 c −1 a = 0 ⇒ 1(1 – a ) + c(–c – ab) – b(ca + b) = 0 b a −1
⇒ a2 + b2 + c2 + 2abc = 1. 34.
Let A be a square matrix all of whose entries are integers. Then which one of the following is true? (1) If detA = ± 1, then A–1 exists but all its entries are not necessarily integers (2) If detA ≠ ± 1, then A–1 exists and all its entries are non-integers (3) If detA = ± 1, then A–1 exists and all its entries are integers (4) If detA = ± 1, then A–1 need not exist
Sol:
(3) Each entry of A is integer, so the cofactor of every entry is an integer and hence each entry in the adjoint of matrix A is integer. 1 Now detA = ± 1 and A–1 = (adj A) det(A) ⇒ all entries in A–1 are integers.
35.
The quadratic equations x2 – 6x + a = 0 and x2 – cx + 6 = 0 have one root in common. The other roots of the first and second equations are integers in the ratio 4 : 3. Then the common root is (1) 1 (2) 4 (3) 3 (4) 2
Sol:
(4) Let α and 4β be roots of x2 – 6x + a = 0 and α, 3β be the roots of x2 – cx + 6 = 0, then α + 4β = 6 and 4αβ = a α + 3β = c and 3αβ = 6. We get αβ = 2 ⇒ a = 8 So the first equation is x2 – 6x + 8 = 0 ⇒ x = 2, 4 If α = 2 and 4β = 4 then 3β = 3 If α = 4 and 4β = 2, then 3β = 3/2 (non-integer) ∴ common root is x = 2.
Chemistry PART − B 36.
The organic chloro compound, which shows complete stereochemical inversion during a SN2 reaction, is (2) (CH3)3CCl (1) (C2H5)2CHCl (4) CH3Cl (3) (CH3)2CHCl
Sol.
(4) For SN2 reaction, the C atom is least hindered towards the attack of nucleophile in the case of (CH3Cl). Hence, (4) is the correct answer. FIITJEE Ltd., ICES House, 29 – A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 26515949, 26569493, Fax: 011-26513942.
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FIITJEE Solutions to AIEEE - 2008
12
37.
Toluene is nitrated and the resulting product is reduced with tin and hydrochloric acid. The product so obtained is diazotised and then heated with cuprous bromide. The reaction mixture so formed contains (1) mixture of o− and p−bromotoluenes (2) mixture of o− and p−dibromobenzenes (3) mixture of o− and p−bromoanilines (4) mixture of o− and m−bromotoluenes
Sol.
(1) CH3
CH3
CH3 NO2 +
→ Nitration
NO2 Sn/HCl Sn/HCl CH3 NH2
CH3
NH2 NaNO2/HCl
CH3 N2 Cl
CuBr
NaNO2/HCl CH3
N2 Cl CuBr
CH3 Br
CH3
Br
38.
Sol.
The coordination number and the oxidation state of the element ‘E’ in the complex [E(en)2(C2O4)]NO2 (where (en) is ethylene diamine) are, respectively, (1) 6 and 2 (2) 4 and 2 (3) 4 and 3 (4) 6 and 3 (4) en
E
ox
NO2
en
Coordination no. = 6 and Oxidation no. = 3 FIITJEE Ltd., ICES House, 29 – A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 26515949, 26569493, Fax: 011-26513942.
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FIITJEE Solutions to AIEEE - 2008
13
39.
Identify the wrong statements in the following: (1) Chlorofluorocarbons are responsible for ozone layer depletion (2) Greenhouse effect is responsible for global warming (3) Ozone layer does not permit infrared radiation from the sun to reach the earth (4) Acid rains is mostly because of oxides of nitrogen and sulphur
Sol.
(3) Ozone layer does not allow ultraviolet radiation from sun to reach earth.
40.
Phenol, when it first reacts with concentrated sulphuric acid and then with concentrated nitric acid, gives (1) 2,4,6-trinitrobenzene (2) o-nitrophenol (3) p-nitrophenol (4) nitrobenzene
Sol.
(2)
OH
OH
OH NO2
Conc.H2 SO4 →
Conc.HNO3 →
SO3H 41.
In the following sequence of reactions, the alkene affords the compound ‘B’ O3 H2 O CH3 CH = CHCH3 → A → B. Zn The compound B is (1) CH3CH2CHO (3) CH3CH2COCH3
Sol.
(2) CH3COCH3 (4) CH3CHO
(4)
H3C
O CH
CH
CH3
H3C
CH
CH
O
O
CH3 (A)
H2O/Zn O H3C
42.
C
H (B) Larger number of oxidation states are exhibited by the actinoids than those by the lanthanoids, the main reason being (1) 4f orbitals more diffused than the 5f orbitals (2) lesser energy difference between 5f and 6d than between 4f and 5d orbitals (3) more energy difference between 5f and 6d than between 4f and 5d orbitals (4) more reactive nature of the actinoids than the lanthanoids
Sol.
(2) Being lesser energy difference between 5f and 6d than 4f and 5d orbitals.
43.
In which of the following octahedral complexes of Co (at. no. 27), will the magnitude of ∆ o be the highest? (1) [Co(CN)6]3− (2) [Co(C2O4)3]3− 3+ (4) [Co(NH3)6]3+ (3) [Co(H2O)6]
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FIITJEE Solutions to AIEEE - 2008
14
Sol.
(1) CNΘ is stronger ligand hence ∆ o is highest.
44.
At 80oC, the vapour pressure of pure liquid ‘A’ is 520 mm Hg and that of pure liquid ‘B’ is 1000 mm Hg. If a mixture solution of ‘A’ and ‘B’ boils at 80oC and 1 atm pressure, the amount of ‘A’ in the mixture is (1 atm = 760 mm Hg) (1) 52 mol percent (2) 34 mol percent (3) 48 mol percent (4) 50 mol percent
Sol.
(4) PT = PAo X A + PBo XB
760 = 520XA + PBo (1 − XA ) ⇒ X A = 0.5 Thus, mole% of A = 50%
45.
1 A → 2B, rate of disappearance of ‘A’ is related to the rate of appearance of ‘B’ by the 2
For a reaction expression d[ A ] (1) − = dt d[ A ] (3) − = dt
1 d [B ] 2 dt d [B] dt
1 d [B ] dt 4 dt d[ A ] d [B ] (4) − =4 dt dt
(2) −
d[ A ]
=
Sol.
(2) 1 A → 2B 2 −2d [ A ] d [B ] =+ dt 2dt −d [ A ] 1 d [B ] = dt 4 dt
46.
The equilibrium constants KP1 and K P2 for the reactions X U 2Y and Z U P + Q, respectively are in the ratio of 1 : 9. If the degree of dissociation of X and Z be equal then the ratio of total pressure at these equilibria is (1) 1 : 36 (2) 1 : 1 (3) 1 : 3 (4) 1 : 9
Sol.
(1) ZZZ X X YZZ Z 2Y 1
0
2x (1 − x ) 2 ( 2x ) P1 1 kp = (1 − x ) 1 + x 1
ZZZ X Z YZZ Z P+Q 1
0
(1 − x )
0
x x 1
k p2 =
x 2 P2 (1 − x ) 1 + x
4 × P1 1 P 1 = ⇒ 1 = P2 9 P2 36
47.
Oxidising power of chlorine in aqueous solution can be determined by the parameters indicated below: FIITJEE Ltd., ICES House, 29 – A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 26515949, 26569493, Fax: 011-26513942.
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FIITJEE Solutions to AIEEE - 2008
15
1 ∆dissH 1 ∆egH ∆hydH 2 Cl2 ( g) Cl− ( aq) . → Cl ( g) → Cl− ( g) → 2 1 The energy involved in the conversion of Cl2 ( g) to Cl−(g) 2 (using the data, ∆ dissHCl2 = 240 kJmol−1, ∆ egHCl = −349 kJmol−1, ∆ hydHCl = −381 kJmol−1 )
will be (1) +152 kJmol−1 (3) −850 kJmol−1 Sol.
(2) −610 kJmol−1 (4) +120 kJmol−1
(2)
For the process
1 − → Claq Cl2 ( g) 2
1 ∆Hdiss of Cl2 + ∆ egCl + ∆ hyd Cl− 2 240 =+ − 349 − 381 2 = − 610 kJ mol−1 ∆H =
48.
Which of the following factors is of no significance for roasting sulphide ores to the oxides and not subjecting the sulphide ores to carbon reduction directly? (1) Metal sulphides are thermodynamically more stable than CS2 (2) CO2 is thermodynamically more stable than CS2 (3) Metal sulphides are less stable than the corresponding oxides (4) CO2 is more volatile than CS2
Sol.
(1)
49.
Bakelite is obtained from phenol by reacting with (1) (CH2OH)2 (2) CH3CHO (3) CH3COCH3 (4) HCHO
Sol.
(4) OH
OH
CH2 CH2OH
+ HCHO →
Polymerize →
CH2
O n
CH2OH
50.
For the following three reactions a, b and c, equilibrium constants are given: K1 a. CO ( g) + H2O ( g) U CO2 ( g) + H2 ( g) ; b.
CH4 ( g) + H2O ( g) U CO ( g) + 3H2 ( g) ;
c.
CH4 ( g) + 2H2O ( g) U CO2 ( g) + 4H2 ( g) ; K 3
K2
Which of the following relations is correct? (1) K1 K 2 = K 3
(2) K2K3 = K1
(3) K3 = K1K2
(4) K 3 .K 32 = K12
Sol.
(3) Equation (c) = equation (a) + equation (b) Thus K3 = K1.K2
51.
The absolute configuration of FIITJEE Ltd., ICES House, 29 – A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 26515949, 26569493, Fax: 011-26513942.
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FIITJEE Solutions to AIEEE - 2008 HO2C
CO2H
HO H
OH
H
is (1) S, S (3) R, S Sol.
16
(2) R, R (4) S, R
(2) HO2C
CO2H 1
2
OH HO H H Both C1 and C2 have R – configuration. 52.
The electrophile, E ⊕ attacks the benzene ring to generate the intermediate σ-complex. Of the following, which σ-complex is of lowest energy? NO2 H
(1)
E
(2)
H
E NO2
NO2 H
(3)
E
(4)
H E
Sol.
(2) NO2 is electron withdrawing which will destabilize σ - complex.
53.
α-D-(+)-glucose and β-D-(+)-glucose are (1) conformers (3) anomers
(2) epimers (4) enantiomers
Sol.
(3) α - D (+) glucose and β - D (+) glucose are anomers.
54.
Standard entropy of X2, Y2 and XY3 are 60, 40 and 50 JK−1mol−1, respectively. For the reaction, 1 3 X2 + Y2 → XY3 , ∆H = −30 kJ, to be at equilibrium, the temperature will be 2 2 (1) 1250 K (2) 500 K (3) 750 K (4) 1000 K
Sol.
(3) 1 3 X2 + Y2 → XY3 2 2 1 3 ∆Sreaction = 50 − × 40 + × 60 = −40 Jmol−1 2 2 ∆G = ∆H - T∆S at equilibrium ∆G = 0 FIITJEE Ltd., ICES House, 29 – A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 26515949, 26569493, Fax: 011-26513942.
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17
∆H = T∆S 30 × 103 = T × 40 ⇒ T = 750 K
55.
Four species are listed below i. HCO 3−
ii.
H3O+
iv. HSO3F iii. HSO −4 Which one of the following is the correct sequence of their acid strength? (1) iv < ii < iii < I (2) ii < iii < i < iv (3) i < iii < ii < iv (4) iii < i < iv < ii Sol.
(3) (iv) > (ii) > (iii) > (i)
56.
Which one of the following constitutes a group of the isoelectronic species? (2) NO + , C22 − , CN− , N2 (1) C22 − , O 2− , CO, NO (3) CN− , N2 , O22 − , C22 −
(4) N2 , O2− , NO + , CO
Sol.
(2) NO + , C22 − , CN− and N2 all have fourteen electrons.
57.
Which one of the following pairs of species have the same bond order? (2) CN− and CN+ (1) CN− and NO+ − − (4) NO+ and CN+ (3) O2 and CN
Sol.
(1) Both are isoelectronic and have same bond order.
58.
The ionization enthalpy of hydrogen atom is 1.312 × 106 Jmol−1. The energy required to excite the electron in the atom from n = 1 to n = 2 is (2) 6.56 × 105 Jmol−1 (1) 8.51 × 105 Jmol−1 5 −1 (3) 7.56 × 10 Jmol (4) 9.84 × 105 Jmol−1
Sol.
(4) 1.312 × 106 1.312 × 106 − − 1 22 5 −1 = 9.84 × 10 J mol
∆E = E2 − E1 = −
59.
Which one of the following is the correct statement? (1) Boric acid is a protonic acid (2) Beryllium exhibits coordination number of six (3) Chlorides of both beryllium and aluminium have bridged chloride structures in solid phase (4) B2H6.2NH3 is known as ‘inorganic benzene’
Sol.
(3)
Cl Al Cl 60.
Sol.
Cl
Cl
Be
Al Cl
Cl
Cl
Cl
Cl
Be Cl
Cl Be
Cl
° Given E°Cr 3+ / Cr = −0.72 V, EFe = −0.42 V. The potential for the cell 2+ / Fe
CrCr3+ (0.1 M)Fe2+ (0.01 M)Fe is (1) 0.26 V (3) −0.339 V (1)
(2) 0.399 V (4) −0.26 V
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18
0 As E0Cr / Cr3+ = −0.72 V and EFe = −0.42 V 2+ / Fe
2Cr + 3Fe2 + → 3Fe + 2Cr 3 + Ecell = E
0 cell
( (
Cr 3 + 0.0591 − log 6 Fe2 +
= ( −0.42 + 0.72 ) −
) )
2 3
( 0.1) ( 0.1) 0.0591 0.0591 = 0.30 − log log 3 3 6 6 ( 0.01) ( 0.01) 2
2
0.0591 10−2 0.0591 log −6 = 0.30 − log10 4 6 6 10 Ecell = 0.2606 V = 0.30 −
61.
Amount of oxalic acid present in a solution can be determined by its titration with KMnO4 solution in the presence of H2SO4. The titration gives unsatisfactory result when carried out in the presence of HCl, because HCl (1) gets oxidised by oxalic acid to chlorine (2) furnishes H+ ions in addition to those from oxalic acid (3) reduces permanganate to Mn2+ (4) oxidises oxalic acid to carbon dioxide and water
Sol.
(3) HCl being stronger reducing agent reduces MnO4− to Mn2+ and result of the titration becomes unsatisfactory.
62.
The vapour pressure of water at 20oC is 17.5 mm Hg. If 18 g of glucose (C6H12O6) is added to 178.2 g of water at 20oC, the vapour pressure of the resulting solution will be (1) 17.675 mm Hg (2) 15.750 mm Hg (3) 16.500 mm Hg (4) 17.325 mm Hg
Sol.
(4) P0 − Ps = Xsolute Ps 17.5 − Ps 0.1 = Ps 10 17.5 − Ps = 0.01 Ps
⇒ Ps = 17.325 mm Hg 63.
Among the following substituted silanes the one which will give rise to cross linked silicone polymer on hydrolysis is (2) RSiCl3 (1) R4Si (4) R3SiCl (3) R2SiCl2
Sol.
(2)
Cl R
OH H2 O Cl →R
Si
Si
Cl
64.
Condensation → R OH polymerization
OH
Si
Si
O
O
Si
O
Si
O
O
Si
Si
R
n
In context with the industrial preparation of hydrogen from water gas (CO + H2), which of the following is the correct statement? FIITJEE Ltd., ICES House, 29 – A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 26515949, 26569493, Fax: 011-26513942.
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FIITJEE Solutions to AIEEE - 2008 (1) (2) (3) (4) Sol.
19
CO and H2 are fractionally separated using differences in their densities CO is removed by absorption in aqueous Cu2Cl2 solution H2 is removed through occlusion with Pd CO is oxidised to CO2 with steam in the presence of a catalyst followed by absorption of CO2 in alkali
(4) H2 O CO + H2 → CO2 + 2H2 KOH
K2CO3 65.
In a compound atoms of element Y from ccp lattice and those of element X occupy 2/3rd of tetrahedral voids. The formula of the compound will be (2) X2Y3 (1) X4Y3 (3) X2Y (4) X3Y4
Sol.
(1) No. of atoms of Y = 4 2 No. of atoms of X = × 8 3 Formula of compound will be X4Y3
66.
Gold numbers of protective colloids A, B, C and D are 0.50, 0.01, 0.10 and 0.005, respectively. The correct order of their protective powers is (1) D < A < C < B (2) C < B < D < A (3) A < C < B < D (4) B < D < A < C
Sol.
(3) Higher the gold number lesser will be the protective power of colloid.
67.
The hydrocarbon which can react with sodium in liquid ammonia is (2) CH3CH2C≡CH (1) CH3CH2CH2C≡CCH2CH2CH3 (4) CH3CH2C≡CCH2CH3 (3) CH3CH=CHCH3
Sol.
(2) Θ
Na / Liq.NH3 CH3 CH2 − C ≡ CH → CH3 CH2 C ≡ CNa⊕ ∆ It is a terminal alkyne, having acidic hydrogen. Note: Solve it as a case of terminal alkynes, otherwise all alkynes react with Na in liq. NH3.
68.
The treatment of CH3MgX with CH3C≡C−H produces (2) CH3C≡C−CH3 (1) CH3−CH=CH2 H H (3) CH3
C
C
CH3
(4) CH4
Sol.
(4) CH3 − MgX + CH3 − C ≡ C − H → CH4
69.
The correct decreasing order of priority for the functional groups of organic compounds in the IUPAC system of nomenclature is (2) −SO3H, −COOH, −CONH2, −CHO (1) −COOH, −SO3H, −CONH2, −CHO (4) −CONH2, −CHO, −SO3H, −COOH (3) −CHO, −COOH, −SO3H, −CONH2
Sol.
(2) −SO3H, − COOH, − CONH2 , − CHO The pKa of a weak acid, HA, is 4.80. The pKb of a weak base, BOH, is 4.78. The pH of an aqueous solution of the corresponding salt, BA, will be
70.
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FIITJEE Solutions to AIEEE - 2008 (1) 9.58 (3) 7.01 Sol.
20
(2) 4.79 (4) 9.22
(3) It is a salt of weak acid and weak base K w × Ka H+ = Kb
pH = 7.01
Physics PART − C Directions: Questions No. 71, 72 and 73 are based on the following paragraph.
Wave property of electrons implies that they will show diffraction effects. Davisson and Germer demonstrated this by diffracting electrons from crystals. The law governing the diffraction from a crystal is obtained by requiring that electron waves reflected from the planes of atoms in a crystal interfere constructively (see in figure). Incoming Electrons
Outgoing Electrons
i
d
Crystal plane
71.
Electrons accelerated by potential V are diffracted from a crystal. If d = 1Å and i = 30°, V should be about (h = 6.6 × 10−34 Js, me = 9.1 × 10−31 kg, e = 1.6 × 10−19 C) (1) 2000 V (2) 50 V (3) 500 V (4) 1000 V
Sol.
(2) 2d cos i = nλ
2d cos i =
h
i
2meV
v = 50 volt 72.
If a strong diffraction peak is observed when electrons are incident at an angle ‘i’ from the normal to the crystal planes with distance ‘d’ between them (see figure), de Broglie wavelength λdB of electrons can be calculated by the relationship (n is an integer) (2) 2d cos i = nλdB (1) d sin i = nλdB (3) 2d sin i = nλdB (4) d cos i = nλdB
Sol.
(4) 2d cos i = nλdB
73.
In an experiment, electrons are made to pass through a narrow slit of width ‘d’ comparable to their de Broglie wavelength. They are detected on a screen at a distance ‘D’ from the slit (see figure).
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FIITJEE Solutions to AIEEE - 2008
y=0
d D
Which of the following graph can be expected to represent the number of electrons ‘N’ detected as a function of the detector position ‘y’(y = 0 corresponds to the middle of the slit)? y y (2) (1) d
N y
(3)
y
(4) d
N
d
N
N
d
Sol.
(4) Diffraction pattern will be wider than the slit.
74.
A planet in a distant solar system is 10 times more massive than the earth and its radius is 10 times smaller. Given that the escape velocity from the earth is 11 kms−1, the escape velocity from the surface of the planet would be (2) 11 kms−1 (1) 1.1 kms−1 (4) 0.11 kms−1 (3) 110 kms−1
Sol.
(3)
vesc = 75.
Vgρ1 k
(4)
Vg(ρ1 − ρ2 ) k
(1) ρ1Vg − ρ2Vg = kv 2T
⇒
76.
2G × 10M = 10 × 11 = 110 km/s R 10
A spherical solid ball of volume V is made of a material of density ρ1. It is falling through a liquid of density ρ2(ρ2 0). The terminal speed of the ball is Vg(ρ1 − ρ2 ) Vgρ1 (2) (1) k k (3)
Sol.
2GM = R
vT =
Vg ( ρ1 − ρ2 ) k
Shown in the figure below is a meter-bridge set up with null deflection in the galvanometer. 55 Ω
R
G 20 cm
The value of the unknown resistor R is (1) 13.75 Ω (3) 110 Ω
(2) 220 Ω (4) 55 Ω
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FIITJEE Solutions to AIEEE - 2008 Sol.
(2) 55 R 55 × 8 = ⇒R= = 220 Ω 20 80 2
77.
A thin rod of length ‘L’ is lying along the x-axis with its ends at x = 0 and x = L. Its linear density n
x (mass/length) varies with x as k , where n can be zero or any positive number. If the position xCM L of the centre of mass of the rod is plotted against ‘n’, which of the following graphs best approximates the dependence of xCM on n? xCM xCM (1) (2) L L/2
(3)
L/2
O xCM
(4)
O xCM
L
L
L/2
L/2
O
Sol.
n
O
n
n
n
(1)
kxn + 2 x k .xdx ( n + 2 ) Ln ∫ dmx = ∫ λdx.x = ∫ L = xcm = n n +1 ∫ dm ∫ dm ∫ k x dx kx n L ( n + 1) L L 2L 3L 4L 5L xcm = , , , , ,... 2 3 4 5 6 n
L
L ) ( = x n +1 n + 2 0 0
78.
While measuring the speed of sound by performing a resonance column experiment, a student gets the first resonance condition at a column length of 18 cm during winter. Repeating the same experiment during summer, she measures the column length to be x cm for the second resonance. Then (1) 18 > x (2) x >54 (3) 54 > x > 36 (4) 36 > x > 18
Sol.
(2)
n=
1 4x
γRT M
xn =
1 γRT 4 M
x∝
T
79.
The dimension of magnetic field in M, L, T and C (Coulomb) is given as (1) MLT−1C−1 (2) MT2C−2 −1 −1 (3) MT C (4) MT−2C−1
Sol.
(3) F = qvB B = F/qv = MC−1T−1
80.
Consider a uniform square plate of side ‘a’ and mass ‘m’. The moment of inertia of this plate about an axis perpendicular to its plane and passing through one of its corners is 1 5 (2) ma2 (1) ma2 6 12 7 2 ma2 (4) ma2 (3) 3 12 FIITJEE Ltd., ICES House, 29 – A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 26515949, 26569493, Fax: 011-26513942.
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FIITJEE Solutions to AIEEE - 2008 Sol.
(4) 2
a 2 ma2 ma2 2 + = ma2 I = Icm + m = 2 6 2 3
81.
A body of mass m = 3.513 kg is moving along the x-axis with a speed of 5.00 ms−1. The magnitude of its momentum is recorded as (2) 17.565 kg ms−1 (1) 17.6 kg ms−1 (4) 17.57 kg ms−1 (3) 17.56 kg ms−1
Sol.
(1) P = mv = 3.513 × 5.00 ≈ 17.6
82.
An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range (1) 200 J − 500 J (2) 2 × 105 J − 3 × 105 J (3) 20,000 J − 50,000 J (4) 2,000 J − 5,000 J
Sol.
(4) Approximate mass = 60 kg Approximate velocity = 10 m/s 1 Approximate KE = × 60 × 100 = 3000 J 2 KE range ⇒ 2000 to 5000 joule
83.
A parallel plate capacitor with air between the plates has a capacitance of 9 pF. The separation between its plates is ‘d’. The space between the plates is now filled with two dielectrics. One of the d while the other one has dielectric constant dielectrics has dielectric constant k1 = 3 and thickness 3 2d . Capacitance of the capacitor is now k2 = 6 and thickness 3 (1) 1.8 pF (2) 45 pF (3) 40.5 pF (4) 20.25 pF
Sol.
(3)
C = 9 PF
Aε 0 Aε 0 18Aε0 = C′ = = d1 d2 d 2d 4d + + 9 18 3 6 C′ = 40.5 PF
3
6
84.
The speed of sound in oxygen (O2) at a certain temperature is 460 ms−1. The speed of sound in helium (He) at the same temperature will be (assumed both gases to be ideal) (2) 500 ms−1 (1) 460 ms−1 −1 (4) 330 ms−1 (3) 650 ms
Sol.
No option is correct γRT v= M V1 = V2
460 = V2
γ1M2 = γ 2 M1
7 ×4 5 5 × 32 3
21 ⇒ 25 × 8
v2 =
460 × 5 × 2 2 21
= 1420
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FIITJEE Solutions to AIEEE - 2008 85.
This question contains Statement -1 and Statement-2. Of the four choices given after the statements, choose the one that best describes the two statements. Statement – I: Energy is released when heavy nuclei undergo fission or light nuclei undergo fusion. and Statement – II For heavy nuclei, binding energy per nucleon increases with increasing Z while for light nuclei it decrease with increasing Z. (1) Statement – 1is false, Statement – 2 is true. (2) Statement – 1is true, Statement – 2 is true; Statement -2 is correct explanation for Statement-1. (3) Statement – 1is true, Statement – 2 is true; Statement -2 is not a correct explanation for Statement-1. (4) Statement – 1 is true, Statement – 2 is False.
Sol.
(4)
86.
This question contains Statement -1 and Statement-2. Of the four choices given after the statements, choose the one that best describes the two statements. Statement – I: For a mass M kept at the centre of a cube of side ‘a’, the flux of gravitational field passing through its sides is 4π GM. and Statement – II If the direction of a field due to a point source is radial and its dependence on the distance ‘r’ for the source is given as 1/r2, its flux through a closed surface depends only on the strength of the source enclosed by the surface and not on the size or shape of the surface (1) Statement – 1is false, Statement – 2 is true. (2) Statement – 1is true, Statement – 2 is true; Statement -2 is correct explanation for Statement-1. (3) Statement – 1is true, Statement – 2 is true; Statement -2 is not a correct explanation for Statement-1. (4) Statement – 1 is true, Statement – 2 is False.
Sol.
(2) g = GM/r2
87.
A jar filled with two non mixing liquids 1 and 2 having densities ρ1 and ρ2 respectively. A solid ball, made of a material of density ρ3, is dropped in the jar. It comes to equilibrium in the position shown in the figure. Which of the following is true for ρ1, ρ2 and ρ3? (1) ρ3 < ρ1 < ρ2 (2) ρ1 < ρ3 < ρ2 (3) ρ1 < ρ2 < ρ3 (4) ρ1 < ρ3 < ρ2
Liquid 1
ρ1
ρ3 Liquid 2
ρ2
Sol.
(4) As liquid 1 floats above liquid 2, ρ1 < ρ2 The ball is unable to sink into liquid 2, ρ3 < ρ2 The ball is unable to rise over liquid 1, ρ1 < ρ3 Thus, ρ1 < ρ3 < ρ2
88.
A working transistor with its three legs marked P, Q and R is tested using a multimeter. No conduction is found between P and Q. By connecting the common (negative) terminal of the multimeter to R and the other (positive) terminal to P or Q, some resistance is seen on the multimeter. Which of the following is true for the transistor? (1) It is an npn transistor with R as base (2) It is a pnp transistor with R as collector (3) It is a pnp transistor with R as emitter (4) It is an npn transistor with R as collector
Sol.
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FIITJEE Solutions to AIEEE - 2008 Directions: Question No. 89 and 90 are based on the following paragraph.
Consider a block of conducting material of resistivity ‘ρ’ shown in the figure. Current ‘I’ enters at ‘A’ and leaves from ‘D’. We apply superposition principle to find voltage ‘∆V’ developed between ‘B’ and ‘C’. The calculation is done in the following steps: (i) Take current ‘I’ entering from ‘A’ and assume it to spread over a hemispherical surface in the block. (ii) Calculate field E(r) at distance ‘r’ from A by using Ohm’s law E = ρj, where j is the current per unit area at ‘r’. (iii) From the ‘r’ dependence of E(r), obtain the potential V(r) at r. (iv) Repeat (i), (ii) and (iii) for current ‘I’ leaving ‘D’ and superpose results for ‘A’ and ‘D’. I
a A
89.
Sol.
∆V measured between B and C is ρI ρI − (1) πa π(a + b) ρI ρI − (3) 2πa 2π(a + b)
I
∆v b B
a C
D
ρI ρI − a (a + b) ρI (4) 2π(a − b)
(2)
(3) Choosing A as origin, I E = ρj = ρ 2πr 2
ρI VC − VB = − 2π
( a +b)
∫ a
ρI 1 1 1 dr = − 2 ( ) 2 a π r a+b
ρI 1 1 − VB − VC = 2π a ( a + b ) 90.
For current entering at A, the electric field at a distance ‘r’ from A is ρI ρI (2) 2 (1) 2 8πr r ρI ρI (4) (3) 2 2πr 4πr 2
Sol.
(3)
91.
A student measures the focal length of convex lens by putting an object pin at a distance ‘u’ from the lens and measuring the distance ‘v’ of the image pin. The graph between ‘u’ and ‘v’ plotted by the student should look like v (cm)
v (cm)
(2)
(1)
O
O
u (cm)
v (cm)
u (cm)
v (cm)
(3)
(4)
O
u (cm)
O
u (cm)
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FIITJEE Solutions to AIEEE - 2008
26
Sol.
(3) 1 1 1 − = = constant v u f
92.
A block of mass 0.50 kg is moving with a speed of 2.00 m/s on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is (1) 0.16 J (2) 1.00 J (3) 0.67 J (4) 0.34 J
Sol.
(3) m1u1 + m2u2 = (m1 + m2)v v = 2/3 m/s 2
Energy loss = 93.
1 ( 0.5 ) × ( 2 )2 − 1 (1.5 ) × 2 = 0.67 J 2 2 3
A capillary tube (A) is dropped in water. Another identical tube (B) is dipped in a soap water solution. Which of the following shows the relative nature of the liquid columns in the two tubes? A
B
(2)
A
B
(3)
A
B
(1)
(4)
B
A
Sol.
(3)
Capillary rise h = 94.
2T cos θ . As soap solution has lower T, h will be low. ρgr
Suppose an electron is attracted towards the origin by a force k/r where ‘k’ is a constant and ‘r’ is the distance of the electron from the origin. By applying Bohr model to this system, the radius of the nth orbital of the electron is found to be ‘rn’ and the kinetic energy of the electron to be Tn. Then which of the following is true? (1) Tn ∝ 1/n2, rn ∝ n2 (2) Tn independent of n, rn ∝ n (3) Tn ∝ 1/n, rn ∝ n (4) Tn ∝ 1/n, rn ∝ n2
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FIITJEE Solutions to AIEEE - 2008 Sol.
(2) k mv 2 = r r (independent or r) mv2 = k 1 h n = mvr ⇒ r ∝ n and T = mv 2 is independent of n. 2 2 π
95.
A wave travelling along the x-axis is described by the equation y(x, t) = 0.005 cos (αx −βt). If the wavelength and the time period of the wave are 0.08 m and 2.0 s, respectively, then α and β in appropriate units are 0.08 2.0 (1) α = 25.00 π, β = π (2) α = , π π 0.04 1.0 π (3) α = (4) α = 12.50 π, β = ,β = π π 2.0
Sol.
(1) y = 0.005 cos (αx − βt) comparing the equation with the standard form, x t y = A cos − 2π λ T 2π/λ = α and 2π/T = β α = 2π/0.08 = 25.00 π β=π
96.
Two coaxial solenoids are made by winding thin insulated wire over a pipe of cross sectional area A = 10 cm2 and length = 20 cm. If one of the solenoids has 300 turns and the other 400 turns, their mutual inductance is (µ0 = 4π × 10-7 Tm A-1) (1) 2.4 π × 10-5 H (2) 4.8 π × 10-4 H -5 (3) 4.8 π × 10 H (4) 2.4 π × 10-4 H
Sol.
(4)
M= 97.
Sol.
98.
µ0 N1N2 A = 2.4 π × 10−4 H A
In the circuit below, A and B represent two inputs and C represents the output. The circuit represents (1) NOR gate (2) AND gate (3) NAND gate (4) OR gate (4) A 0 0 1 1
B 0 1 0 1
A C B
C 0 1 1 1
A body is at rest at x = 0. At t = 0, it starts moving in the positive x-direction with a constant acceleration. At the same instant another body passes through x = 0 moving in the positive xdirection with a constant speed. The position of the first body is given by x1(t) after time ‘t’ and that of the second body by x2(t) after the same time interval. Which of the following graphs correctly describes (x1 – x2)as a function of time ‘t’?
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FIITJEE Solutions to AIEEE - 2008 (1)
(x1 −x2)
(3)
O
t
(x1 −x2)
t
(x1 −x2)
(4)
O
Sol.
(x1 −x2)
(2)
O
28
O
t
t
(2) 1 2 at 2 x2(t) = vt 1 x1 − x2 = at 2 − vt 2
x1(t) =
99.
An experiment is performed to find the refractive index of glass using a travelling microscope. In this experiment distance are measured by (1) a vernier scale provided on the microscope (2) a standard laboratory scale (3) a meter scale provided on the microscope (4) a screw gauage provided on the microscope
Sol.
(1)
100.
A thin spherical shell of radius R has charge Q spread uniformly over its surface. Which of the following graphs most closely represents the electric field E(r) produced by the shell in the range 0 ≤ r< ∞ , where r is the distance from the centre of the shell? E(r) E(r) (1) (2)
(3)
Sol.
O R E(r)
r
O
r
R
(4)
O R E(r)
r
O
r
R
(1)
0 if r < R E(r) = Q 4πε r 2 if r ≥ R 0 101.
A 5V battery with internal resistance 2Ω and a 2V battery with internal resistance 1Ω are connected to a 10Ω resistor as shown in the figure. The current in the 10 Ω resistor is P2 5V 10 Ω
2Ω
2V 1Ω
P1
(1) 0.27 A P2 to P1 (3) 0.03 A P2 to P1
(2) 0.03 A P1 to P2 (4) 0.27 A P1 to P2
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FIITJEE Solutions to AIEEE - 2008 Sol.
P2
(3)
5 0 2 + − VP2 − VP1 = 2 10 1 1 1 1 + + 2 10 1 VP − VP1 I= 2 = 0.03 from P2 → P1 10
5V
10 Ω 2Ω
2V 1Ω
i P1
102.
A horizontal overhead power line is at a height of 4m from the ground and carries a current of 100 A from east to west. The magnetic field directly below it on the ground is (µ0 = 4π × 10-7 T m A-1) (2) 5 × 10-6 T northward (1) 2.5 × 10-7 T southward -6 (4) 2.5 × 10-7 northward (3) 5 × 10 T southward
Sol.
(3)
B=
µ0 i 4π × 10−7 100 = × = 5 × 10−6 T southward 2π R 2π 4
103.
Relative permittivity and permeability of a material are εr and µr, respectively. Which of the following values of these quantities are allowed for a diamagnetic material? (1) εr = 0.5, µr = 1.5 (2) εr = 1.5, µr = 0.5 (4) εr = 1.5, µr = 1.5 (3) εr = 0.5, µr = 0.5
Sol.
(2)
104.
Two full turns of the circular scale of a screw gauge cover a distance of 1 mm on its main scale. The total number of divisions on the circular scale is 50. Further, it is found that the screw gauge has a zero error of − 0.03 mm while measuring the diameter of a thin wire, a student notes the main scale reading of 3 mm and the number of circular scale divisions in line with the main scale as 35. The diameter of the wire is (1) 3.32 mm (2) 3.73 mm (3) 3.67 mm (4) 3.38 mm
Sol.
(4) Diameter = M.S.R. + C.S.R × L.C. + Z.E. = 3 + 35 × (0.5/50) + 0.03 = 3.38 mm
105.
An insulated container of gas has two chambers separated by an insulating partition. One of the chambers has volume V1 and contains ideal gas at pressure P1 and temperature T1. The other chamber has volume V2 and contains ideal gas at pressure P2 and temperature T2. If the partition is removed without doing any work on the gas, the final equilibrium temperature of the gas in the container will be T T (P V + P2 V2 ) P V T + P2 V2 T2 (2) 1 1 1 (1) 1 2 1 1 P1V1T2 + P2 V2 T1 P1V1 + P2 V2 (3)
Sol.
P1V1T2 + P2 V2 T1 P1V1 + P2 V2
(4)
T1T2 (P1V1 + P2 V2 ) P1V1T1 + P2 V2 T2
(1) U = U1 + U2 (P1V1 + P2 V2 ) T1T2 T = (P1V1T2 + P2 V2 T1 )
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FIITJEE Solutions to AIEEE - 2008
AIEEE–2008, PAPER(C−5) ANSW ER S 1. 5. 9. 13. 17. 21. 25. 29. 33. 37. 41. 45. 49. 53. 57. 61. 65. 69. 73. 77. 81. 85. 89. 93. 97. 101. 105.
(2) (2) (4) (4) (3) (1) (2) (2) (4) (1) (4) (2) (4) (3) (1) (3) (1) (2) (4) (1) (1) (4) (3) (3) (4) (3) (1)
2. 6. 10. 14. 18. 22. 26. 30. 34. 38. 42. 46. 50. 54. 58. 62. 66. 70. 74. 78. 82. 86. 90. 94. 98. 102.
(2) (3) (4) (4) (4) (1) (2) (4) (3) (4) (2) (1) (3) (3) (4) (4) (3) (3) (3) (2) (4) (2) (3) (2) (2) (3)
3. 7. 11. 15. 19. 23. 27. 31. 35. 39. 43. 47. 51. 55. 59. 63. 67. 71. 75. 79. 83. 87. 91. 95. 99. 103.
(3) (4) (4) (3) (2) (2) (1) (3) (4) (3) (1) (2) (2) (3) (3) (2) (2) (2) (1) (3) (3) (4) (3) (1) (1) (2)
4. 8. 12. 16. 20. 24. 28. 32. 36. 40. 44. 48. 52. 56. 60. 64. 68. 72. 76. 80. 84. 88. 92. 96. 100. 104.
(1) (3) (4) (1) (4) (1) (3) (4) (4) (2) (4) (1) (2) (2) (1) (4) (4) (4) (2) (4) no option is correct (2) (3) (4) (1) (4)
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FIITJEE Solutions to AIEEE - 2009
1
AIEEE–2009, BOOKLET CODE(A) Note: (i) The test is of 3 hours duration. (ii) The test consists of 90 questions. The maximum marks are 432. (iii) There are three parts in the question paper. The distribution of marks subjectwise in each part is as under for each correct response. Part A − Physics (144 marks)
−
Question No. 1 to 2 and 9 to 30 consists FOUR (4) marks each and Question No. 3 to 8 consists EIGHT (8) marks each for each correct response.
Part B − Chemistry (144 marks)
−
Question No. 31 to 39 and 46 to 60 consists FOUR (4) marks each and Question No. 40 to 45 consists EIGHT (8) marks each for each correct response.
Part C − Mathematics(144 marks)
−
Question No. 61 to 82 and 89 to 90 consists FOUR (4) marks each and Question No. 83 to 88 consists EIGHT (8) marks each for each correct response.
(iv) Candidates will be awarded marks as stated above for correct response of each question. 1/4th 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. (v) * marked questions are from syllabus of class XI CBSE.
Physics PART − A 1.
This question contains Statement-1 and Statement-2. Of the four choices given after the statements, choose the one that best describes the two statements. Statement – 1: For a charged particle moving from point P to point Q, the net work done by an electrostatic field on the particle is independent of the path connecting point P to point Q. Statement-2: The net work done by a conservative force on an object moving along a closed loop is zero (1) Statement-1 is true, Statement-2 is false (2) Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1. (3) Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation of Statement-1. (4) Statement-1 is false, Statement-2 is true
Sol:
(2) Work done by conservative force does not depend on the path. Electrostatic force is a conservative force.
2.
The above is a plot of binding energy per nucleon Eb, against the nuclear mass M; A, B, C, D, E, F correspond to different nuclei. Consider four reactions: (i) A + B → C + ε (ii) C → A + B + ε (iii) D + E → F + ε and (iv) F → D + E + ε where ε is the energy released? In which reactions is ε positive? (1) (i) and (iv) (2) (i) and (iii) (3) (ii) and (iv) (4) (ii) and (iii)
Sol:
(1) 1st reaction is fusion and 4th reaction is fission.
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FIITJEE Solutions to AIEEE - 2009 3.
2
A p-n junction (D) shown in the figure can act as a rectifier. An alternating current source (V) is connected in the circuit.
(1)
(2)
(3)
(4)
Sol:
(3) Given figure is half wave rectifier
4.
The logic circuit shown below has the input waveforms ‘A’ and ‘B’ as shown. Pick out the correct output waveform.
Sol:
(1)
(2)
(3)
(4)
(1) A 1 1 0 0
*5.
Truth Table B Y 1 1 0 0 1 0 0 0
If x, v and a denote the displacement, the velocity and the acceleration of a particle executing simple harmonic motion of time period T, then, which of the following does not change with time? (1) a2T2 + 4π2v2 (3) aT + 2πv
Sol:
aT x aT (4) v (2)
(2)
aT ω2 xT 4π2 4 π2 = = 2 ×T = = constant. x x T T
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FIITJEE Solutions to AIEEE - 2009 6.
In an optics experiment, with the position of the object fixed, a student varies the position of a convex lens and for each position, the screen is adjusted to get a clear image of the object. A graph between the object distance u and the image distance v, from the lens, is plotted using the same scale for the two axes. A straight line passing through the origin and making an angle of 45o with the x-axis meets the experimental curve at P. The coordinates of P will be
f f 2 2
(1) (2f, 2f)
(2) ,
(3) (f, f)
(4) (4f, 4f)
Sol:
(1) It is possible when object kept at centre of curvature. u=v u = 2f, v = 2f.
*7.
A thin uniform rod of length l and mass m is swinging freely about a horizontal axis passing through its end. Its maximum angular speed is ω. Its centre of mass rises to a maximum height of
Sol:
(1)
1 l 2 ω2 3 g
(2)
1 lω 6 g
(3)
1 l 2 ω2 2 g
(4)
1 l 2 ω2 6 g
(4) T.Ei = T.Ef •
1 2 Iω = mgh 2 1 1 2 2 1 l 2 ω2 × ml ω = mgh ⇒ h = 2 3 6 g Let P(r) =
8.
h
Q r be the charge density distribution for a solid sphere of radius R and total charge Q. πR 4
for a point ‘p’ inside the sphere at distance r1 from the centre of the sphere, the magnitude of electric field is (1) 0 (3) Sol:
(2)
Qr12
(4)
4πεoR 4
Q 4πεo r12 Qr12 3πεoR 4
(3) r1
E4πr12 =
Q
∫ πR
4
r4πr 2 dr
0
ε0
P
Qr12 ⇒E= . 4πε0R 4
r
R
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FIITJEE Solutions to AIEEE - 2009
4
9.
The transition from the state n = 4 to n = 3 in a hydrogen like atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition from (1) 2 → 1 (2) 3 → 2 (3) 4 → 2 (4) 5→ 4
Sol:
(4)
1
IR corresponds to least value of
2 n1
−
1 n22
i.e. from Paschen, Bracket and Pfund series. Thus the transition corresponds to 5 → 3. *10.
One kg of a diatomic gas is at a pressure of 8 × 104 N/m2. The density of the gas is 4 kg/m-3. What is the energy of the gas due to its thermal motion? (2) 5 × 104 J (1) 3 × 104 J 4 (3) 6 × 10 J (4) 7 × 104 J
Sol:
(2) Thermal energy corresponds to internal energy Mass = 1 kg density = 8 kg/m3 ⇒ Volume =
mass 1 = m3 density 8
Pressure = 8 × 104 N/m2 ∴ Internal Energy = 11.
5 P × V = 5 × 104 J 2
This question contains Statement-1 and Statement-2. Of the four choices given after the statements, choose the one that best describes the two statements. Statement-1: The temperature dependence of resistance is usually given as R = Ro(1 + α∆t). The resistance of a wire changes from 100 Ω to 150 Ω when its temperature is increased from 27oC to 227oC. This implies that α = 2.5 × 10 −3 / o C . Statement 2: R = Ri (1 + α∆T) is valid only when the change in the temperature ∆T is small and ∆R = (R - Ro) CCl3 > ( CH3 )3 C > ( CH3 )2 CH
(2) ( CH3 )2 CH > CCl3 > C6 H5 CH2 > ( CH3 )3 C
(3) CCl3 > C6 H5 CH2 > ( CH3 )2 CH > ( CH3 )3 C
(4)
( CH3 )3 C > ( CH3 )2 CH > C6H5 CH2
Sol:
(3) 2o carbanion is more stable than 3o and Cl is –I effect group.
*34.
The alkene that exhibits geometrical isomerism is : (1) propene (2) 2-methyl propene (3) 2-butene (4) 2- methyl -2- butene
Sol:
(3) H
H
CH3
H C=C
C=C CH3
CH3
> CCl3
cis
H3C
H Trans
*35.
In which of the following arrangements, the sequence is not strictly according to the property written against it ? (1) CO2 < SiO2 < SnO2 < PbO2 : increasing oxidising power (2) HF< HCl < HBr < HI : increasing acid strength (3) NH3 < PH3 < AsH3 < SbH3 : increasing basic strength (4) B < C < O < N : increasing first ionization enthalpy.
Sol:
(3) Correct basic strength is NH3 > PH3 > AsH3 > BiH3
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FIITJEE Solutions to AIEEE - 2009 36.
The major product obtained on interaction of phenol with sodium hydroxide and carbon dioxide is : (1) benzoic acid (2) salicylaldehyde (3) salicylic acid (4) phthalic acid
Sol:
(3) Kolbe – Schmidt reaction is OH ONa
OH COONa
CO2 → 6atm, 140o C
NaOH →
OH COOH
H3 O+ → Salicylic Acid
37.
Which of the following statements is incorrect regarding physissorptions ? (1) It occurs because of vander Waal’s forces. (2) More easily liquefiable gases are adsorbed readily. (3) Under high pressure it results into multi molecular layer on adsorbent surface. (4) Enthalpy of adsorption ( ∆Hadsorption ) is low and positive.
Sol:
(4) Enthalpy of adsorption regarding physissorption is not positive and it is negative.
38.
Which of the following on heating with aqueous KOH, produces acetaldehyde ? (1) CH3 COCl (2) CH3 CH2 Cl (4) CH3 CHCl2 (3) CH2 Cl CH2 Cl
Sol:
(4) OH / aq.KOH → CH3 CH CH3 CHCl2 \ OH
*39.
In an atom, an electron is moving with a speed of 600m/s with an accuracy of 0.005%. Certainity with which the position of the electron can be located is (h = 6.6 × 10 −34 kg m2 s−1 , mass of electron,
em = 9.1 × 10 −31 kg )
(1) 1.52 × 10 −4 m (3) 1.92 × 10 −3 m Sol:
→ CH3 CHO −H2 O
(2) 5.10 × 10 −3 m (4) 3.84 × 10 −3 m
(3) ∆x.m ∆v = ∆x =
h 4π
h 4π m∆v
0.005 = 0.03 100 6.625 × 10 −34 = 1.92 × 10 −3 m ⇒ ∆x = 4 × 3.14 × 9.1× 10 −31 × 0.03 ∆v = 600 ×
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FIITJEE Solutions to AIEEE - 2009 40.
13
In a fuel cell methanol is used as fuel and oxygen gas is used as an oxidizer. The reaction is 3 CH3 OH(l ) + O2 (g) → CO2 (g) + 2H2 O(l ) At 298K standard Gibb’s energies of formation for 2 CH3 OH( l ), H2 O( l ) and CO2 (g) are -166.2, -237.2 and -394.4 kJ mol−1 respectively. If standard enthalpy of combustion of methanol is -726kJ mol−1 , efficiency of the fuel cell will be (1) 80 % (2) 87% (3) 90% (4) 97%
Sol:
(4) 3 O2 (g) → CO2 (g) + 2H2 O(l ) ∆H = −726kJ mol−1 2 Also ∆Gof CH3 OH(l ) = -166.2 kJ mol-1 CH3 OH(l ) +
∆Gof H2 O(l ) = -237.2 kJ mol-1
∆Gof CO2 (l) = -394.4 kJ mol-1 Q ∆G = Σ∆Gof products −Σ∆Gof reactants. = -394.4 -2 (237.2) + 166.2 = −702.6 kJ mol-1 ∆G × 100 now Efficiency of fuel cell = ∆H 702.6 = × 100 726 = 97%
41.
Two liquids X and Y form an ideal solution. At 300K, vapour pressure of the solution containing 1 mol of X and 3 mol of Y is 550 mm Hg. At the same temperature, if 1 mol of Y is further added to this solution, vapour pressure of the solution increases by 10 mm Hg. Vapour pressure (in mmHg) of X and Y in their pure states will be, respectively : (1) 200 and 300 (2) 300 and 400 (3) 400 and 600 (4) 500 and 600
Sol:
(3) PT = PXo x X + PYo x Y x X = mol fraction of X x Y = mol fraction of Y 1 o 3 ∴ 550 = Pxo + PY 1 + 3 1+ 3 o o P 3P = X + Y 4 4 ∴ 550 (4) = PXo + 3PYo ………….. (1) Further 1 mol of Y is added and total pressure increases by 10 mm Hg. 1 4 ∴ 550 + 10 = PXo + PYo 1+ 4 1+ 4 o o ∴ 560 (5) = PX + 4PY ………….(2) By solving (1) and (2)
We get, PXo = 400 mm Hg PYo = 600 mm Hg
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FIITJEE Solutions to AIEEE - 2009
14
42.
The half life period of a first order chemical reaction is 6.93 minutes. The time required for the completion of 99% of the chemical reaction will be (log 2=0.301) : (1) 230.3 minutes (2) 23.03 minutes (3) 46.06 minutes (4) 460.6 minutes
Sol:
(3) Qλ =
0.6932 0.6932 min−1 = t1/ 2 6.93
Also t =
[ Ao ] 2.303 log λ [A]
[ A o ] = initial concentration (amount) [ A ] = final concentration (amount) 2.303 × 6.93 100 log 0.6932 1 = 46.06 minutes
∴ t=
43.
o Given : EFe = −0.036V, 3+ / Fe
o = -0.439V. The value of standard electrode potential for the EFe 2+ / Fe
3+ + e − → Fe2 + (aq) will be : change, Fe(aq)
(1) -0.072 V (3) 0.770 V
Sol:
(2) 0.385 V (4) -0.270
(3) Q Fe3 + + 3e− → Fe; Eo = −0.036V ∴ ∆G1O = −nFEo = −3F( −0.036) = +0.108 F Also Fe2+ + 2e − → Fe; Eo = -0.439 V ∴ ∆GO2 = -nF Eo = -2 F( -0.439) = 0.878 F 3+ To find Eo for Fe(aq) + e− → Fe2+ (aq)
∆GO = -nFE o = -1FE o o o Q G = G1 − Go2 ∴ Go = 0.108F - 0.878F ∴ -FEo = +0.108F – 0.878F ∴ EO = 0.878 - 0.108 = 0.77v
*44.
+ = 0) On the basis of the following thermochemical data : ( ∆fGo H(aq)
H2 O(l ) → H+ (aq) + OH− (aq); ∆H = 57.32kJ 1 O2 (g) → H2 O(l ) ; ∆H = −286.20kJ 2 The value of enthalpy of formation of OH− ion at 25o C is : (1) -22.88 kJ (2) -228.88 kJ (3) +228.88 kJ (4) -343.52 kJ H2 (g) +
Sol:
(2) By adding the two given equations, we have 1 + − + OH(aq) ; ∆H =-228.88 Kj H2(g) + O2(g) → H(aq) 2 + =0 Here ∆Hof of H(aq) ∴ ∆Hof of OH− = -228.88 kJ
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FIITJEE Solutions to AIEEE - 2009 45.
Copper crystallizes in fcc with a unit cell length of 361 pm. What is the radius of copper atom ? (1) 108 pm (2) 127 pm (3) 157 pm (4) 181 pm
Sol:
(2) For FCC, 2a = 4r (the atoms touches each other along the face- diagonal) 2a = 4 = 127 pm
r=
46.
15
2 × 361 4
Which of the following has an optical isomer ? (1) CO (NH3 )3 Cl
+
(3) CO (H2 O )4 ( en )
(2) CO ( en )(NH3 )2 3+
2+
(4) CO ( en )2 (NH3 )2
3+
Sol:
(4) It is an octahedral complex of the type M ( AA )2 X2 Where AA is bidentate ligand.
*47.
Solid Ba (NO3 )2 is gradually dissolved in a 1.0 × 10 −4 M Na2 CO3 solution. At what concentration of Ba2 + will a precipitate begin to form ?(Ksp for Ba CO3 = 5.1 × 10 −9 ). (1) 4.1 ×10 −5 M (2) 5.1 × 10 −5 M (4) 8.1 × 10 −7 M (3) 8.1 × 10 −8 M
Sol:
(2) Ba (NO3 )2 + CaCO3 → BaCO3 + 2NaNO3 Here CO3−2 = [Na2 CO3 ] = 10 −4 M K sp = Ba +2 CO3−2 ⇒ 5.1× 10 −9 = Ba2+ 10 −4 ⇒ Ba +2 = 5.1× 10 −5 At this value, just precipitation starts.
(
48.
)
Which one of the following reactions of Xenon compounds is not feasible ? (1) XeO3 + 6HF → Xe F6 + 3H2 O (2) 3Xe F4 + 6H2 O → 2 Xe + XeO3 + 12 HF + 1.5 O2 (3) 2XeF2 + 2H2 O → 2Xe + 4HF + O2 (4) XeF6 + RbF → Rb(XeF7 ]
Sol:
(1) Remaining are feasible
*49.
Using MO theory predict which of the following species has the shortest bond length ? (1) O22 + (2) O2+ (3) O2−
Sol:
(4) O22 −
(1) Bond length α
1 bond order
no..of bonding e − no.of antibonding e 2 Bond orders of O2+ , O2− , O2−2 and O2+2 are respectively 2.5, 1.5, 1 and 3.
Bond order =
50.
In context with the transition elements, which of the following statements is incorrect ? (1) In addition to the normal oxidation states, the zero oxidation state is also shown by these elements
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FIITJEE Solutions to AIEEE - 2009 in complexes. (2) In the highest oxidation states, the transition metal show basic character and form cationic complexes. (3) In the highest oxidation states of the first five transition elements (Sc to Mn), all the 4s and 3d electrons are used for bonding. (4) Once the d5 configuration is exceeded, the tendency to involve all the 3d electrons in bonding decreases.
Sol:
(2) In higher Oxidation states transition elements show acidic nature
*51.
Calculate the wavelength (in nanometer) associated with a proton moving at 1.0 × 103 ms −1 (Mass of proton = 1.67 ×10 −27 kg and h = 6.63 × 10 −34 Js ) : (1) 0.032 nm (2) 0.40 nm (3) 2.5 nm (4) 14.0 nm
Sol:
(2) λ=
h 6.63 × 10 −34 = ≡ 0.40 nm mv 1.67 × 10−27 × 103
52.
A binary liquid solution is prepared by mixing n-heptane and ethanol. Which one of the following statements is correct regarding the behaviour of the solution ? (1) The solution formed is an ideal solution (2) The solution is non-ideal, showing +ve deviation from Raoult’s law. (3) The solution is non-ideal, showing –ve deviation from Raoult’s law. (4) n-heptane shows +ve deviation while ethanol shows –ve deviation from Raoult’s law.
Sol:
(2) The interactions between n –heptane and ethanol are weaker than that in pure components.
*53.
The number of stereoisomers possible CH3 − CH = CH − CH ( OH) − Me is :
for
(1) 3 (3) 4
(2) 2 (4) 6
a
compound
of
the
molecular
formula
Sol:
(3) About the double bond, two geometrical isomers are possible and the compound is having one chiral carbon.
*54.
The IUPAC name of neopentane is (1) 2-methylbutane (3) 2-methylpropane
Sol:
(2) 2, 2-dimethylpropane (4) 2,2-dimethylbutane
(2) CH3 | Neopentane is H3 C − C − CH3 | CH3
*55.
The set representing the correct order of ionic radius is : (2) Na+ > Li+ > Mg2+ > Be2+ (1) Li+ > Be2 + > Na+ > Mg2+ (3) Li+ > Na + > Mg2+ > Be2+
Sol: 56.
(4) Mg2+ > Be2+ > Li+ > Na +
(2) Follow the periodic trends The two functional groups present in a typical carbohydrate are : (1) -OH and -COOH (2) -CHO and -COOH (3) > C = O and - OH (4) - OH and -CHO
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FIITJEE Solutions to AIEEE - 2009
17
Sol:
(3) Carbohydrates are polyhydroxy carbonyl compounds.
*57.
The bond dissociation energy of B – F in BF3 is 646 kJ mol−1 whereas that of C-F in CF4 is 515kJ mol−1 . The correct reason for higher B-F bond dissociation energy as compared to that of C- F is : (1) smaller size of B-atom as compared to that of C- atom (2) stronger σ bond between B and F in BF3 as compared to that between C and F in CF4 (3) significant pπ - pπ interaction between B and F in BF3 whereas there is no possibility of such interaction between C and F in CF4 .
(4) lower degree of pπ - pπ interaction between B and F in BF3 than that between C and F in CF4 .
Sol:
(3) option itself is the reason
58.
In Cannizzaro reaction given below (−) : OH && ( − ) the slowest step is : 2 Ph CHO → Ph CH2 OH + PhCO 2 (−)
(1) the attack of : OH at the carboxyl group (2) the transfer of hydride to the carbonyl group (3) the abstraction of proton from the carboxylic group (4) the deprotonation of Ph CH2 OH
Sol:
(2) Hydride transfer is the slowest step.
59.
Which of the following pairs represents linkage isomers ? (1) Cu (NH3 )4 [Pt Cl4 ] and Pt (NH3 )4 [CuCl4 ] (2) Pd (P Ph3 )2 (NCS )2 and Pd (P Ph3 )2 ( SCN)2 (3) CO (NH3 )5 NO3 SO4 and CO (NH3 )5 SO4 NO3 (4) Pt Cl2 (NH3 )4 Br2 and Pt Br2 (NH3 )4 Cl2
Sol:
(2) NCS- is ambidentate ligand and it can be linked through N (or) S
60.
Buna-N synthetic rubber is a copolymer of : Cl | (1) H2 C = CH − C = CH2 and H2C = CH − CH = CH2
(2) H2 C = CH − CH = CH2 and H5 C6 − CH = CH2 (3) H2 C = CH − CN and H2 C = CH − CH = CH2 (4) H2 C = CH − CN and H2 C = CH − C = CH2 | CH3
Sol:
(3)
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FIITJEE Solutions to AIEEE - 2009
Mathematics PART − C a
61.
a +1
Let a, b, c be such that b ( a + c ) ≠ 0 . If −b b + 1 b − 1 + a − 1 c c − 1 c + 1 ( −1)n + 2 a value of ‘n’ is (1) zero (3) any odd integer
Sol:
a +1 a −1
b +1
c −1
b −1
c + 1 = 0, then the
( −1)
n +1
b
( −1)
n
c
(2) any even integer (4) any integer
(3) a a +1 a −1 a +1 b +1 c −1 a a +1 a −1 a +1 a −1 a n n −b b + 1 b − 1 + ( −1) a − 1 b − 1 c + 1 = −b b + 1 b − 1 + ( −1) b + 1 b − 1 −b c c −1 c +1 a −b c c c −1 c +1 c −1 c +1 c a a +1 a −1 a +1 a a −1 a a +1 a −1 a a +1 a −1 n +1 n+ 2 = −b b + 1 b − 1 + ( −1) b + 1 −b b − 1 = −b b + 1 b − 1 + ( −1) −b b + 1 b − 1 c c −1 c +1 c −1 c c +1 c c −1 c +1 c c −1 c +1
This is equal to zero only if n + 2 is odd i.e. n is odd integer. 62.
If the mean deviation of number 1, 1 + d, 1 + 2d, ….. , 1 + 100d from their mean is 255, then the d is equal to (1) 10.0 (2) 20.0 (3) 10.1 (4) 20.2
Sol:
(3) n a + l) 1 sum of quantities 2 ( = [1 + 1 + 100d] = 1 + 50d = Mean ( x ) = 2 n n
1 1 2 d 50 x 51 ∑ xi − x ⇒ 255 = 101[50d + 49d + 48d + .... + d + 0 + d + ...... + 50d] = 101 2 n 255 x 101 ⇒d= = 10.1 50 x 51
M.D. =
*63.
If the roots of the equation bx 2 + cx + a = 0 be imaginary, then for all real values of x, the expression 3b2 x 2 + 6bcx + 2c 2 is (1) greater than 4ab (2) less than 4ab (3) greater than – 4ab (4) less than – 4ab
Sol:
(3) bx 2 + cx + a = 0 Roots are imaginary ⇒ c 2 − 4ab < 0 ⇒ c 2 < 4ab ⇒ −c 2 > −4ab 3b2 x 2 + 6bcx + 2c 2 since 3b2 > 0 Given expression has minimum value 4 3b2 2c 2 − 36b2 c 2 12b2c 2 =− = −c 2 > −4ab . Minimum value = 2 12b2 4 3b
(
)(
(
)
)
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FIITJEE Solutions to AIEEE - 2009 *64.
Let A and B denote the statements A: cos α + cos β + cos γ = 0 B: sin α + sin β + sin γ = 0 If cos ( β − γ ) + cos ( γ − α ) + cos ( α − β ) = −
3 , then 2
(1) A is true and B is false (3) both A and B are true
Sol:
19
(2) A is false and B is true (4) both A and B are false
(3) 3 2 ⇒ 2 cos ( β − γ ) + cos ( γ − α ) + cos ( α − β ) + 3 = 0 cos ( β − γ ) + cos ( γ − α ) + cos ( α − β ) = −
⇒ 2 cos ( β − γ ) + cos ( γ − α ) + cos ( α − β ) + sin2 α + cos2 α + sin2 β + cos2 β + sin2 γ + cos2 γ = 0 ⇒ ( sin α + sin β + sin γ ) + ( cos α + cos β + cos γ ) = 0 2
*65.
(
2
)
(
)
(
2
)
The lines p p2 + 1 x − y + q = 0 and p2 + 1 x + p2 + 1 y + 2q = 0 are perpendicular to a common line for (1) no value of p (3) exactly two values of p
(2) exactly one value of p (4) more than two values of p
Sol:
(2) Lines must be parallel, therefore slopes are equal ⇒ p p2 + 1 = − p2 + 1 ⇒ p = - 1
66.
If A, B and C are three sets such that A ∩ B = A ∩ C and A ∪ B = A ∪ C , then (1) A = B (2) A = C (3) B = C (4) A ∩ B = φ
Sol:
(3)
67.
Sol:
(
)
(
)
r r uur If u, v, w are non-coplanar vectors and p, q are real numbers, then the equality r r uur r uur r uur r r 3u pv pw − pv w qu − 2w qv qu = 0 holds for (1) exactly one value of (p, q) (2) exactly two values of (p, q) (3) more than two but not all values of (p , q) (4) all values of (p, q)
(1)
r r − pq + 2q2 u v r r uur But u v w ≠ 0 3p2 − pq + 2q2 = 0
( 3p
2
)
2
uur w = 0
2
7q2 q 7 q = 0 ⇒ 2p2 + p − + q2 = 0 2p2 + p2 − pq + + 4 2 4 2 q ⇒ p = 0, q = 0, p = 2 This possible only when p = 0, q = 0 exactly one value of (p, q) x − 2 y −1 z + 2 = = lies in the plane x + 3y − αz + β = 0 . Then ( α, β ) equals 3 −5 2 (1) (6, - 17) (2) ( - 6, 7) (3) (5, - 15) (4) ( - 5, 15)
68.
Let the line
Sol:
(2) Dr’s of line = ( 3, − 5, 2 )
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FIITJEE Solutions to AIEEE - 2009
20
Dr’s of normal to the plane = (1, 3, − α ) Line is perpendicular to normal ⇒ 3 (1) − 5 ( 3 ) + 2 ( −α ) = 0 ⇒ 3 − 15 − 2α = 0 ⇒ 2α = −12 ⇒ α = −6 Also ( 2, 1, − 2 ) lies on the plane 2 + 3 + 6 ( −2 ) + β = 0 ⇒ β = 7
∴ ( α, β ) = ( −6, 7 ) *69.
From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are to be selected and arranged in a row on the shelf so that the dictionary is always in the middle. Then the number of such arrangements is (1) less than 500 (2) at least 500 but less than 750 (3) at least 750 but less than 1000 (4) at least 1000
Sol:
(4) 4 novels can be selected from 6 novels in 6 C4 ways. 1 dictionary can be selected from 3 dictionaries in 3 C1 ways. As the dictionary selected is fixed in the middle, the remaining 4 novels can be arranged in 4! ways. ∴ The required number of ways of arrangement = 6 C4 x 3 C1 x 4! = 1080 π
70.
∫ [cot x ] dx , [•] denotes the greatest integer function, is equal to 0
(1)
π 2
(2) 1 (4) −
(3) – 1
Sol:
π 2
(4) π
Let I = ∫ [cot x ] dx
…(1)
0
π
= ∫ cot ( π − x ) dx = 0
π
∫ [ − cot x ] dx
…(2)
0
Adding (1) and (2) π
2I =
∫ [cot x ] dx 0
π
+
∫ [ − cot x ] dx 0
π
= =
∫ ( −1) dx 0
Q [ x ] + [ − x ] = −1 if x ∉ Z = 0 if x ∈ Z
= [ − x ]0 = −π π
∴I = − 71.
π 2
For real x, let f ( x ) = x 3 + 5x + 1 , then (1) f is one-one but not onto R (3) f is one-one and onto R
Sol:
(2) f is onto R but not one-one (4) f is neither one-one nor onto R
(3) Given f ( x ) = x 3 + 5x + 1 Now f ' ( x ) = 3x 2 + 5 > 0, ∀x ∈ R ∴ f(x) is strictly increasing function ∴ It is one-one Clearly, f(x) is a continuous function and also increasing on R, f ( x ) = −∞ and Lt f ( x ) = ∞ Lt x → −∞ x→∞ ∴ f(x) takes every value between −∞ and ∞ . Thus, f(x) is onto function.
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FIITJEE Solutions to AIEEE - 2009 72.
1 In a binomial distribution B n, p = , if the probability of at least one success is greater than or 4 9 equal to , then n is greater than 10 1 1 (1) (2) 4 3 4 log10 − log10 log10 + log103
(3)
Sol:
log10
4
9 − log103
(4)
log10
4
4 − log103
(1) n
1 − qn ≥
*73.
9 1 1 3 ⇒ ≤ ⇒ n ≥ − log 3 10 ⇒ n ≥ 4 10 10 log − log103 4 10 4
If P and Q are the points of intersection of the circles
x 2 + y 2 + 3x + 7y + 2p − 5 = 0
and
x + y + 2x + 2y − p = 0 , then there is a circle passing through P, Q and (1, 1) for (1) all values of p (2) all except one value of p (3) all except two values of p (4) exactly one value of p 2
Sol:
2
2
(1) Given circles S = x 2 + y 2 + 3x + 7y + 2p − 5 = 0 S' = x 2 + y 2 + 2x + 2y − p2 = 0
Equation of required circle is S + λS' = 0 As it passes through (1, 1) the value of λ =
− ( 7 + 2p )
(6 − p ) 2
If 7 + 2p = 0, it becomes the second circle ∴it is true for all values of p 74.
The projections of a vector on the three coordinate axis are 6, - 3, 2 respectively. The direction cosines of the vector are 6 3 2 (1) 6, − 3, 2 (2) , − , 5 5 5 6 3 2 6 3 2 (3) , − , (4) − , − , 7 7 7 7 7 7
Sol:
(3) Projection of a vector on coordinate axis are x 2 − x1, y 2 − y1, z2 − z1 x 2 − x1 = 6, y 2 − y1 = −3, z 2 − z1 = 2
( x 2 − x1 )
2
+ ( y 2 − y1 ) + ( z 2 − z1 ) = 36 + 9 + 4 = 7 2
The D.C’s of the vector are
*75.
If Z − (1)
2
6 3 2 ,− , 7 7 7
4 = 2 , then the maximum value of Z is equal to z
3 +1
(3) 2
Sol:
(2)
5 +1
(4) 2 + 2
(2) 4 4 4 4 Z = Z − + ⇒ Z = Z− + Z Z Z Z 4 4 4 ⇒ Z ≤ Z− + ⇒ Z ≤ 2+ Z Z Z
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FIITJEE Solutions to AIEEE - 2009 2
⇒ Z −2 Z −4 ≤0
( Z −(
) ) ( Z − (1 − 5 )) ≤ 0 ⇒ 1 −
5 +1
5 ≤ Z ≤ 5 +1
*76.
Three distinct points A, B and C are given in the 2 – dimensional coordinate plane such that the ratio of the distance of any one of them from the point (1, 0) to the distance from the point ( - 1, 0) is equal 1 to . Then the circumcentre of the triangle ABC is at the point 3 5 (2) , 0 (1) ( 0, 0 ) 4 5 5 (3) , 0 (4) , 0 2 3
Sol:
(2) P = (1, 0 ) ; Q ( −1, 0 )
Let A = ( x, y ) AP BP CP 1 = = = AQ BQ CQ 3
..(1)
⇒ 3AP = AQ ⇒ 9AP2 = AQ2 ⇒ 9 ( x − 1) + 9y 2 = ( x + 1) + y 2 2
2
⇒ 9x 2 − 18x + 9 + 9y 2 = x 2 + 2x + 1 + y 2 ⇒ 8x 2 − 20x + 8y 2 + 8 = 0 5 …(2) x +1= 0 2 ∴ A lies on the circle Similarly B, C are also lies on the same circle ⇒ x2 + y2 −
5 ∴ Circumcentre of ABC = Centre of Circle (1) = , 0 4
*77.
The remainder left out when 82n − ( 62 )
2n +1
is divided by 9 is
(1) 0 (3) 7 Sol:
(2) 2 (4) 8
(2) 82n − ( 62 )
2n +1
= (1 + 63 ) − ( 63 − 1)
= (1 + 63 ) + (1 − 63 ) n
n
2n +1
(
2n +1
(
= 1 + nc1 63 + nc 2 ( 63 ) + .... + ( 63 ) 2
= 2 + 63 n c1 + nc 2 ( 63 ) + .... + ( 63 )
n −1
−(
2n +1)
c1 + (
n
2n +1)
) + (1−
( 2n +1)
c 2 ( 63 ) + .... − ( 63 )
c1 63 + (
( 2n )
2n +1)
c 2 ( 63 ) + .... + ( −1)( 63 ) 2
( 2n +1)
)
∴ Reminder is 2 *78.
The ellipse x 2 + 4y 2 = 4 is inscribed in a rectangle aligned with the coordinate axes, which in turn in inscribed in another ellipse that passes through the point (4, 0). Then the equation of the ellipse is (2) x 2 + 12y 2 = 16 (1) x 2 + 16y 2 = 16 (3) 4x 2 + 48y 2 = 48
Sol:
(4) 4x 2 + 64y 2 = 48
(2)
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)
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FIITJEE Solutions to AIEEE - 2009 x2 y2 + = 1 ⇒ a = 2, b = 1 ⇒ P = ( 2, 1) 4 1 x2 y2 x2 y2 Required Ellipse is 2 + 2 = 1 ⇒ 2 + 2 = 1 a b 4 b (2, 1) lies on it V’ 4 1 1 1 3 4 ⇒ + = 1 ⇒ 2 = 1 − = ⇒ b2 = 16 b2 4 4 3 b 2 2 2 2 x y x 3y ∴ + = 1⇒ + = 1 ⇒ x 2 + 12y 2 = 16 16 4 16 4 3 2 6 10 14 The sum to the infinity of the series 1 + + 2 + 3 + 4 + ...... is 3 3 3 3 (1) 2 (2) 3 (3) 4 (4) 6 x 2 + 4y 2 = 4 ⇒
*79.
Sol:
P (2, 1) 1 A’
2
A
2
V (4, 0)
(2)
2 6 10 14 + + + + .... 3 32 33 34 1 1 2 6 10 S = + 2 + 3 + 4 + .... 3 3 3 3 3 Dividing (1) & (2) 1 1 4 4 4 S 1 − = 1 + + 2 + 3 + 4 + .... 3 3 3 3 3
Let S = 1 +
…(1) …(2)
2 4 4 1 1 2 4 4 1 4 4 3 4 2 6 2 6 S = + 2 1 + + 2 + ...... ⇒ S = + 2 = + 2 = + = ⇒ S= ⇒S=3 1 3 3 3 3 3 3 3 3 3 3 2 3 3 2 3 3 1 − 3
80.
The differential equation which represents the family of curves y = c1ec 2 x , where c1 and c 2 are arbitrary constants is (2) y " = y ' y (1) y ' = y 2 (4) yy " = ( y ' )
(3) yy " = y ' Sol:
(4) y = c1ec 2 x
…(1)
y ' = c 2 c1e
c2 x
y ' = c2 y y " = c2 y ' From (2) y' c2 = y So, y " =
2
…(2)
( y ') y
2
⇒ yy " = ( y ' )
2
81.
One ticket is selected at random from 50 tickets numbered 00, 01, 02, …., 49. Then the probability that the sum of the digits on the selected ticket is 8, given that the product of these digits is zero, equals 1 1 (2) (1) 14 7 1 5 (3) (4) 50 14
Sol:
(1)
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FIITJEE Solutions to AIEEE - 2009 S = { 00, 01, 02, …., 49 } Let A be the even that sum of the digits on the selected ticket is 8 then A = { 08, 17, 26, 35, 44 } Let B be the event that the product of the digits is zero B = { 00, 01, 02, 03, …., 09, 10, 20, 30, 40 } A ∩ B = {8} 1 1 = 50 = Required probability = P ( A / B ) = 14 14 P (B ) 50 2x Let y be an implicit function of x defined by x − 2x x cot y − 1 = 0 . Then y ' (1) equals P ( A ∩ B)
82.
(1) – 1 (3) log 2 Sol:
(2) 1 (4) – log 2
(1) x 2x − 2x x cot y − 1 = 0 Now x = 1,
…(1)
1 – 2 coty – 1 = 0 ⇒ coty = 0 ⇒ y =
π 2
Now differentiating eq. (1) w.r.t. ‘x’ dy 2x 2x (1 + log x ) − 2 x x ( − c osec 2 y ) + cot y x x (1 + log x ) = 0 dx π Now at 1, 2 dy 2 (1 + log1) − 2 1( −1) + 0 = 0 dx 1, π 2 dy dy ⇒ 2 + 2 =0⇒ = −1 dx 1, π dx 1, π
83.
2
2
The area of the region bounded by the parabola ( y − 2 ) = x − 1 , the tangent to the parabola at the 2
point (2, 3) and the x-axis is (1) 3 (3) 9 Sol:
(2) 6 (4) 12
(3)
Equation
( y − 2)
2
of
tangent
at
(2,
3)
to 2y = x + 4
= x − 1 is S1 = 0
⇒ x − 2y + 4 = 0 Required Area = Area of ∆OCB + Area of OAPD – Area of ∆PCD 3 1 1 = ( 4 x 2 ) + ∫ y 2 − 4y + 5 dy − (1 x 2 ) 2 2 0
(
)
3
y3 = 4 + − 2y 2 + 5y − 1 = 4 − 9 − 18 + 15 − 1 3 0 = 28 − 19 = 9 sq. units
D (0, 3) P (2, 3) C (0, 2)
B (-4, 0)
A (1, 2) A (1, 2) 0
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FIITJEE Solutions to AIEEE - 2009 3
Area =
∫ ( 2y − 4 − y
2
0
84.
)
3
(
3
)
+ 4y − 5 dy = ∫ − y + 6y − 5 dy = − ∫ ( 3 − y ) 2
0
0
3
2
( y − 3 )3 27 = dy = = 9 sq.units 3 3 0
Given P ( x ) = x 4 + ax 3 + bx 2 + cx + d such that x = 0 is the only real root of P' ( x ) = 0 . If P ( −1) < P (1) , then in the interval [ −1, 1] (1) P ( −1) is the minimum and P (1) is the maximum of P (2) P ( −1) is not minimum but P (1) is the maximum of P (3) P ( −1) is the minimum and P (1) is not the maximum of P (4) neither P ( −1) is the minimum nor P (1) is the maximum of P
Sol:
(2) P ( x ) = x 4 + ax 3 + bx 2 + cx + d P ' ( x ) = 4x 3 + 3ax 2 + 2bx + c
Q x = 0 is a solution for P' ( x ) = 0 , ⇒ c = 0
∴ P ( x ) = x 4 + ax 3 + bx 2 + d
…(1)
Also, we have P ( −1) < P (1)
⇒ 1− a + b + d < 1+ a + b + d ⇒ a > 0 Q P ' ( x ) = 0 , only when x = 0 and P(x) is differentiable in ( - 1, 1), we should have the maximum and minimum at the points x = - 1, 0 and 1 only Also, we have P ( −1) < P (1) ∴ Max. of P(x) = Max. { P(0), P(1) } & Min. of P(x) = Min. { P(-1), P(0) } In the interval [ 0 , 1 ], P' ( x ) = 4x 3 + 3ax 2 + 2bx = x 4x 2 + 3ax + 2b
(
)
Q P ' ( x ) has only one root x = 0, 4x + 3ax + 2b = 0 has no real roots. 2
∴ ( 3a ) − 32b < 0 ⇒ 2
3a2 0 Thus, we have a > 0 and b > 0 ∴ P' ( x ) = 4x 3 + 3ax 2 + 2bx > 0, ∀x ∈ ( 0, 1) Hence P(x) is increasing in [ 0, 1 ] ∴ Max. of P(x) = P(1) Similarly, P(x) is decreasing in [-1 , 0] Therefore Min. P(x) does not occur at x = - 1 85.
Sol:
The shortest distance between the line y − x = 1 and the curve x = y 2 is (1)
3 2 8
(2)
2 3 8
(3)
3 2 5
(4)
3 4
(1) x − y +1= 0 x=y
…(1)
2
1 = 2y
dy dy 1 = Slope of given line (1) ⇒ = dx dx 2y 2
1 1 1 1 1 1 1 = 1 ⇒ y = ⇒ y = ⇒ x = = ⇒ ( x, y ) = , 2 2 4 2y 2 4 2
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Nitin M Sir (physics-iitjee.blogspot.com)
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FIITJEE Solutions to AIEEE - 2009 ∴ The shortest distance is
1 1 − +1 4 2 1+ 1
=
3 4 2
=
26
3 2 8
Directions: Question number 86 to 90 are Assertion – Reason type questions. Each of these questions contains two statements Statement-1 (Assertion) and Statement-2 (Reason). Each of these questions also have four alternative choices, only one of which is the correct answer. You have to select the correct choice
86.
Let f ( x ) = ( x + 1) − 1, x ≥ −1 2
{
}
Statement-1 : The set x : f ( x ) = f −1 ( x ) = {0, − 1} Statement-2 : f is a bijection. (1) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1 (2) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1 (3) Statement-1 is true, Statement-2 is false (4) Statement-1 is false, Statement-2 is true Sol:
(3) There is no information about co-domain therefore f(x) is not necessarily onto.
87.
Let f ( x ) = x x and g ( x ) = sin x . Statement-1 : gof is differentiable at x = 0 and its derivative is continuous at that point. Statement-2 : gof is twice differentiable at x = 0. (1) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1 (2) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1 (3) Statement-1 is true, Statement-2 is false (4) Statement-1 is false, Statement-2 is true
Sol:
(3) f ( x ) = x x and g ( x ) = sin x − sin x 2 ,x < 0 gof ( x ) = sin ( x x ) = 2 ,x ≥ 0 sin x −2x cos x 2 ,x < 0 ∴ ( gof ) ' ( x ) = 2 ,x ≥ 0 2x cos x Clearly, L ( gof ) ' ( 0 ) = 0 = R ( gof ) ' ( 0 )
∴ gof is differentiable at x = 0 and also its derivative is continuous at x = 0 −2cos x 2 + 4x 2 sin x 2 ,x < 0 Now, ( gof ) " ( x ) = 2 2 2 ,x ≥ 0 2cos x − 4x sin x ∴ L ( gof ) " ( 0 ) = −2 and R ( gof ) " ( 0 ) = 2 ∴ L ( gof ) " ( 0 ) ≠ R ( gof ) " ( 0 )
∴ gof(x) is not twice differentiable at x = 0. *88.
Statement-1 : The variance of first n even natural numbers is Statement-2 : The sum of first n natural numbers is numbers is
n ( n + 1)( 2n + 1)
n ( n + 1) 2
n2 − 1 4
and the sum of squares of first n natural
6 (1) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1 (2) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1 (3) Statement-1 is true, Statement-2 is false
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FIITJEE Solutions to AIEEE - 2009 (4) Statement-1 is false, Statement-2 is true Sol:
(4) Statement-2 is true Statement-1: Sum of n even natural numbers = n (n + 1) n ( n + 1) Mean x = = n +1 n 2 1 2 2 2 1 Variance = ∑ ( x i ) − x = 22 + 42 + ..... + ( 2n ) − ( n + 1) n n
()
()
1 2 2 4 n ( n + 1)( 2n + 1) 2 2 2 1 + 22 + ..... + n2 − ( n + 1) = − ( n + 1) n n 6 (n + 1) 2 ( 2n + 1) − 3 (n + 1) (n + 1) [ 4n + 2 − 3n − 3] (n + 1)(n − 1) n2 − 1 = = = = 3 3 3 3 ∴ Statement 1 is false. =
89.
Statement-1 : ~ ( p ↔~ q ) is equivalent to p ↔ q . Statement-2 : ~ ( p ↔~ q) is a tautology. (1) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1 (2) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1 (3) Statement-1 is true, Statement-2 is false (4) Statement-1 is false, Statement-2 is true
Sol:
(3)
p T T F F
90.
p↔q
q T F T F
T F F T
~q F T F T
p ↔~ q
~ ( p ↔~ q)
F T T F
T F F T
Let A be a 2 x 2 matrix Statement-1 : adj ( adj A ) = A Statement-2 : adj A = A (1) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1 (2) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1 (3) Statement-1 is true, Statement-2 is false (4) Statement-1 is false, Statement-2 is true
Sol:
(2)
adj A = A
n −1
= A
adj ( adj A ) = A
n−2
2 −1
= A 0
A= A A=A
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FIITJEE Solutions to AIEEE - 2009
AIEEE–2009, ANSWER KEY Test Booklet Code-A
Test Booklet Code-B
Test Booklet Code-C
Test Booklet Code-D
PHY
CHE
MAT
CHE
MAT
PHY
MAT
PHY
CHE
CHE
PHY
MAT
1. (2)
31. (3)
61. (3)
1. (1)
31. (2)
61. (2)
1. (4)
31. (4)
61. (4)
1. (1)
31. (4)
61. (4)
2. (1)
32. (4)
62. (3)
2. (4)
32. (1)
62. (2)
2. (4)
32. (2)
62. (3)
2. (1)
32. (3)
62. (4)
3. (3)
33. (3)
63. (3)
3. (1)
33. (2)
63. (1)
3. (4)
33. (3)
63. (4)
3. (1)
33. (4)
63. (1)
4. (1)
34. (3)
64. (3)
4. (3)
34. (3)
64. (3)
4. (3)
34. (4)
64. (1)
4. (1)
34. (4)
64. (1)
5. (2)
35. (3)
65. (2)
5. (3)
35. (2)
65. (1)
5. (2)
35. (2)
65. (3)
5. (4)
35. (1)
65. (2)
6. (1)
36. (3)
66. (3)
6. (1)
36. (2)
66. (1)
6. (1)
36. (3)
66. (4)
6. (2)
36. (3)
66. (3)
7. (4)
37. (4)
67. (1)
7. (1)
37. (1)
67. (2)
7. (4)
37. (1)
67. (3)
7. (1)
37. (3)
67. (2)
8. (3)
38. (4)
68. (2)
8. (1)
38. (4)
68. (4)
8. (4)
38. (4)
68. (4)
8. (1)
38. (4)
68. (4)
9. (4)
39. (3)
69. (4)
9. (2)
39. (1)
69. (1)
9. (4)
39. (2)
69. (4)
9. (1)
39. (3)
69. (4)
10. (2)
40. (4)
70. (4)
10. (1)
40. (3)
70. (4)
10. (3)
40. (2)
70. (4)
10. (4)
40. (1)
70. (2)
11. (1)
41. (3)
71. (3)
11. (2)
41. (4)
71. (4)
11. (1)
41. (1)
71. (3)
11. (4)
41. (4)
71. (3)
12. (2)
42. (3)
72. (1)
12. (4)
42. (1)
72. (1)
12. (3)
42. (4)
72. (4)
12. (4)
42. (4)
72. (3)
13. (2)
43. (3)
73. (1)
13. (3)
43. (2)
73. (3)
13. (4)
43. (2)
73. (1)
13. (4)
43. (2)
73. (1)
14. (3)
44. (2)
74. (3)
14. (3)
44. (2)
74. (4)
14. (3)
44. (1)
74. (4)
14. (1)
44. (4)
74. (4)
15. (4)
45. (2)
75. (2)
15. (2)
45. (3)
75. (3)
15. (2)
45. (4)
75. (4)
15. (3)
45. (2)
75. (4)
16. (2)
46. (4)
76. (2)
16. (2)
46. (2)
76. (2)
16. (4)
46. (3)
76. (3)
16. (3)
46. (2)
76. (1)
17. (2)
47. (2)
77. (2)
17. (2)
47. (1)
77. (1)
17. (3)
47. (1)
77. (4)
17. (1)
47. (2)
77. (2)
18. (2)
48. (1)
78. (2)
18. (1)
48. (4)
78. (1)
18. (3)
48. (2)
78. (4)
18. (1)
48. (2)
78. (1)
19. (3)
49. (1)
79. (2)
19. (1)
49. (2)
79. (2)
19. (4)
49. (3)
79. (3)
19. (1)
49. (1)
79. (1)
20. (1)
50. (2)
80. (4)
20. (2)
50. (1)
80. (1)
20. (1)
50. (3)
80. (2)
20. (4)
50. (1)
80. (3)
21. (2)
51. (2)
81. (1)
21. (2)
51. (2)
81. (4)
21. (2)
51. (3)
81. (2)
21. (4)
51. (3)
81. (1)
22. (4)
52. (2)
82. (1)
22. (2)
52. (2)
82. (3)
22. (3)
52. (4)
82. (3)
22. (1)
52. (3)
82. (4)
23. (3)
53. (3)
83. (3)
23. (2)
53. (4)
83. (3)
23. (3)
53. (1)
83. (3)
23. (2)
53. (1)
83. (4)
24. (1)
54. (2)
84. (2)
24. (1)
54. (1)
84. (1)
24. (3)
54. (3)
84. (4)
24. (4)
54. (3)
84. (1)
25. (4)
55. (2)
85. (1)
25. (1)
55. (2)
85. (4)
25. (4)
55. (1)
85. (1)
25. (2)
55. (4)
85. (1)
26. (1)
56. (3)
86. (3)
26. (1)
56. (1)
86. (1)
26. (1)
56. (2)
86. (1)
26. (4)
56. (4)
86. (3)
27. (4)
57. (3)
87. (3)
27. (3)
57. (1)
87. (3)
27. (2)
57. (1)
87. (2)
27. (2)
57. (1)
87. (4)
28. (3)
58. (2)
88. (4)
28. (2)
58. (4)
88. (4)
28. (3)
58. (3)
88. (3)
28. (1)
58. (2)
88. (3)
29. (1)
59. (2)
89. (3)
29. (2)
59. (4)
89. (2)
29. (2)
59. (3)
89. (4)
29. (2)
59. (3)
89. (1)
30. (2)
60. (3)
90. (2)
30. (2)
60. (3)
90. (2)
30. (2)
60. (2)
90. (1)
30. (4)
60. (2)
90. (1)
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AIEEE–2010 IMPORTANT INSTRUCTIONS
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AIEEE− −2010− −2
1.
1. Sol :
2.
2. Sol :
–1
The standard enthalpy of formation of NH3 is –46.0 kJ mol . If the enthalpy of formation of H2 from –1 –1 its atoms is –436 kJ mol and that of N2 is –712 kJ mol , the average bond enthalpy of N–H bond in NH3 is (1) –964 kJ mol–1 (2) +352 kJ mol–1 (3) + 1056 kJ mol–1 (4) –1102 kJ mol–1 (2) Enthalpy of formation of NH3 = –46 kJ/mole ∴ N2 + 3H2 → 2NH3 ∆Hf = – 2 x 46 kJ mol Bond breaking is endothermic and Bond formation is exothermic –1 Assuming ‘x’ is the bond energy of N–H bond (kJ mol ) ∴ 712 + (3 x 436)– 6x = –46 x 2 ∴ x = 352 kJ/mol The time for half life period of a certain reaction A → products is 1 hour. When the initial –1 concentration of the reactant ‘A’, is 2.0 mol L , how much time does it take for its concentration to –1 come from 0.50 to 0.25 mol L if it is a zero order reaction ? (1) 4 h (2) 0.5 h (3) 0.25 h (4) 1 h (3) x → (1) For a zero order reaction k = t Where x = amount decomposed k = zero order rate constant for a zero order reaction [ A ]0 → (2) k= 2t1 2
Since [A0] = 2M , t1/2 = 1 hr; k = 1 ∴ from equation (1) 0.25 t= = 0.25hr 1 3.
3. Sol :
A solution containing 2.675 g of CoCl3. 6 NH3 (molar mass = 267.5 g mol–1) is passed through a cation exchanger. The chloride ions obtained in solution were treated with excess of AgNO3 to give –1 4.78 g of AgCl (molar mass = 143.5 g mol ). The formula of the complex is (At. Mass of Ag = 108 u) (1) [Co(NH3)6]Cl3 (2) [CoCl2(NH3)4]Cl (3) [CoCl3(NH3)3] (4) [CoCl(NH3)5]Cl2 (1) – AgNO3 CoCl3. 6NH3 → xCl → x AgCl ↓ n(AgCl) = x n(CoCl3. 6NH3) 4.78 2.675 ∴x=3 =x 143.5 267.5 ∴ The complex is Co (NH3 )6 Cl3
4.
4.
Consider the reaction : + – Cl2(aq) + H2S(aq) → S(s) + 2H (aq) + 2Cl (aq) The rate equation for this reaction is rate = k [Cl2] [H2S] Which of these mechanisms is/are consistent with this rate equation ? + – + – (A) Cl2 + H2 → H + Cl + Cl + HS (slow) + – + – Cl + HS → H + Cl + S (fast) + – (B) H2S ⇔ H + HS (fast equilibrium) – – + Cl2 + HS → 2Cl + H + S (slow) (1) B only (2) Both A and B (3) Neither A nor B (4)
(4) A only
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AIEEE− −2010− −3
Sol:
Rate equation is to be derived wrt slow Step ∴ from mechanism (A) Rate = k[Cl2] [H2S]
5.
If 10 dm of water is introduced into a 1.0 dm flask to 300 K, how many moles of water are in the vapour phase when equilibrium is established ? (Given : Vapour pressure of H2O at 300 K is 3170 Pa ; R = 8.314 J K–1 mol–1) –3 –2 –2 –3 (2) 1.53 x 10 mol (3) 4.46 x 10 mol (4) 1.27 x 10 mol (1) 5.56 x 10 mol (4) PV n= = RT –5 = 128 x 10 moles 3170 × 10 −5 atm × 1 L –3 = ≈ 1.27 x 10 mol 0.0821 L atm k −1mol−1 × 300K
5. Sol :
6. 6. Sol :
–4
3
3
One mole of a symmetrical alkene on ozonolysis gives two moles of an aldehyde having a molecular mass of 44 u. The alkene is (1) propene (2) 1–butene (3) 2–butene (4) ethene (3) 2–butene is symmetrical alkene O3 CH3–CH=CH–CH3 → 2.CH3 CHO Zn / H2 O Molar mass of CH3CHO is 44 u.
7.
7. Sol :
8. 8. Sol : 9.
If sodium sulphate is considered to be completely dissociated into cations and anions in aqueous solution, the change in freezing point of water (∆Tf), when 0.01 mol of sodium sulphate is dissolved in 1 kg of water, is (Kf = 1.86 K kg mol–1) (1) 0.0372 K (2) 0.0558 K (3) 0.0744 K (4) 0.0186 K (2) Vant Hoff’s factor (i) for Na2SO4 = 3 ∴ ∆Tf = (i) kf m 0.01 = 3 x 1.80 x = 0.0558 K 1 From amongst the following alcohols the one that would react fastest with conc. HCl and anhydrous ZnCl2, is (1) 2–Butanol (2) 2–Methylpropan–2–ol (3) 2–Methylpropanol (4) 1–Butanol (2) 3° alcohols react fastest with ZnCl2/conc.HCl due to formation of 3° carbocation and ∴ 2–methyl propan–2–ol is the only 3° alcohol In the chemical reactions, NH2
NaNO2 HCl, 278 K
9. Sol :
A
HBF4
B
the compounds ‘A’ and ‘B’ respectively are (1) nitrobenzene and fluorobenzene (2) phenol and benzene (3) benzene diazonium chloride and fluorobenzene (4) nitrobenzene and chlorobenzene (3)
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AIEEE− −2010− −4
N2 Cl
NH2 NaNO2
F HBF4
N2
HCl, 278 K (A) benzene diazonium
BF3
HCl
(B) fluorobenzene
chloride
10.
10. Sol :
11.
11. Sol :
29.5 mg of an organic compound containing nitrogen was digested according to Kjeldahl’s method and the evolved ammonia was absorbed in 20 mL of 0.1 M HCl solution. The excess of the acid required 15 mL of 0.1 M NaOH solution for complete neutralization. The percentage of nitrogen in the compound is (1) 59.0 (2) 47.4 (3) 23.7 (4) 29.5 (3) Moles of HCl reacting with ammonia = (moles of HCl absorbed ) – (moles of NaOH solution required) –3 –3 = (20 x 0.1 x 10 ) – (15 x 0.1 x 10 ) = moles of NH3 evolved. = moles of nitrogen in organic compound ∴ wt. of nitrogen in org. comp = 0.5 x 10–3 x 14 –3 = 7 x 10 g 7 × 10−3 % wt = = 23.7% 29.5 × 10 −3 –1
The energy required to break one mole of Cl–Cl bonds in Cl2 is 242 kJ mol . The longest wavelength of light capable of breaking a single Cl – Cl bond is 8 –1 23 –1 (c = 3 x 10 ms and NA = 6.02 x 10 mol ) (1) 594 nm (2) 640 nm (3) 700 nm (4) 494 nm (4) 242 × 103 Energy required for 1 Cl2 molecule = Joules. NA This energy is contained in photon of wavelength ‘λ’. hc 6.626 × 10−34 × 3 × 108 242 × 103 =E = λ λ 6.022 × 1023 0
λ = 4947 A ≈ 494 nm 12.
12. Sol :
+
–18
–1
2+
Ionisation energy of He is 19.6 x 10 J atom . The energy of the first stationary state (n = 1) of Li is –16 –1 –17 –1 (1) 4.41 x 10 J atom (2) –4.41 x 10 J atom –15 –1 –17 –1 (3) –2.2 x 10 J atom (4) 8.82 x 10 J atom (2) 1 1 IE + = 13.6 Z 2 + 2 − 2 = 13.6Z2 + where Z + = 2 He He He He 1 ∞
(
–1
Hence 13.6 × Z2
He+
(E1 )Li+2
= −13.6 Z
)
= 19.6 × 10 −18 J atom .
2 Li+2
Z 2 +2 1 2 × 2 = −13.6 Z + × Li2 He 1 Z +
–18
= –19.6 x 10
He
13.
x
9 = −4.41× 10−17 J / atom 4
Consider the following bromides :
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Me
Me
Br
Me
Me Br
Br (A)
13. Sol :
(B) (C) The correct order of SN1 reactivity is (1) B > C > A (2) B > A > C (3) C > B > A (4) A > B > C (1) SN1 proceeds via carbocation intermediate, the most stable one forming the product faster. Hence reactivity order for A, B, C depends on stability of carbocation created. Me Me > > Me Me
14.
Which one of the following has an optical isomer ? (1) Zn ( en )(NH3 )2
14. Sol :
2+
(2) Co ( en )3
3+
(3) Co (H2 O )4 ( en )
3+
(4) Zn ( en )2
2+
(en = ethylenediamine) (2) Only option (2) is having non–super imposable mirror image & hence one optical isomer. en en ( 2) ( 1) en Zn
+2
en
NH3
Co
+3
Co
+3
en
NH3 no optical isomer. It is Tetrahedral with a plane of symmetry en
en
optical isomer
H2O 3)
4)
en
H 2O Co
+3
Zn en
H2O
en
H2O
no optical isomer, it is
Horizontal plane is plane of symmtry
15.
15.
+2
tetrahedral with a plane of symmetry
On mixing, heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components (heptane and octane) are 105 kPa and 45 kPa respectively. Vapour pressure of the solution obtained by mixing 25.0g of heptane and 35 g of octane will be (molar mass of heptane –1 –1 = 100 g mol an dof octane = 114 g mol ). (1) 72.0 kPa (2) 36.1 kPa (3) 96.2 kPa (4) 144.5 kPa (1)
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Sol :
25 /100 0.25 = = 0.45 25 35 0.557 + 100 114 XHep tan e = 0.45 .
Mole fraction of Heptane =
∴ Mole fraction of octane = 0.55 = Xoctane Total pressure = XiPi0 = (105 x 0.45) + (45 x 0.55) kPa = 72.0 KPa 16.
conc. H2 SO4 The main product of the following reaction is C6H5CH2CH(OH)CH(CH3)2 → ? H CH3 H5C 6 C6H5CH2
C
(1)
C
H
CH(CH3)2
C6H 5 C
16. Sol :
C
(4)
H
C
H CH3 H5C6CH2CH2
CH(CH3)2
C
(3)
C
(2)
CH2
H3C
H
(1)
CH2 CH CH CH3 OH CH3 conc.
H2SO4
CH2 CH CH CH3 CH3 loss of proton CH3 CH
CH
HC
(conjugated system) CH3
Trans isomers is more stable & main product here H CH(CH3) 2
C
C
(trans isomer) H
17.
Three reactions involving H2PO −4 are given below : +
(i) H3PO4 + H2O → H3O + H2PO −4
(ii) H2PO −4 + H2O → HPO24− + H3O
+
(iii) H2PO −4 + OH− → H3PO4 + O2 −
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17. Sol :
In which of the above does H2PO −4 act as an acid ? (1) (ii) only (2) (i) and (ii) (3) (iii) only (1) (i) H3PO 4 + H2 → H3 O + + H2PO −4 conjugate base
acid
(ii) H2PO acid
(4) (i) only
− 4
+ H2O → HPO
−2 4 conjugate base
(iii) H2PO−4 + OH− → H3PO 4 acid
acid
+ H3 O+ +O
–2
conjugate acid
Only in reaction (ii) H2PO −4 acids as ‘acid’. 18.
In aqueous solution the ionization constants for carbonic acid are –7 –11 K1 = 4.2 x 10 and K2 = 4.8 x 10 Select the correct statement for a saturated 0.034 M solution of the carbonic acid. (1) The concentration of CO32 − is 0.034 M. (2) The concentration of CO32− is greater than that of HCO3− . +
(3) The concentration of H and HCO3− are approximately equal. +
18. Sol :
(4) The concentration of H is double that of CO32 − . (3) + –7 A→ H2CO3 H + HCO3− K1 = 4.2 x 10 B→
+
HCO3−
As K2 Mg > Na > F > O (2) Na > Mg > Al > O > F + – 2+ 2– 3+ 2– – + 2+ 3+ (3) Na > F > Mg > O > Al (4) O > F > Na > Mg > Al (4) Z For isoelectronic species higher the ratio , smaller the ionic radius e (Hyderabad Classes) Limited. 5-9-14/B, Saifabad, (Opp. Secretariat) Hyderabad. 500 063. Phone: 040-66777000 – 03 Fax: 040-66777004
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z for e
22.
22. Sol :
8 = 0.8 10 9 F− = = 0.9 10 11 Na + = = 1.1 10 12 Mg2 + = = 1.2 10 13 Al3 + = = 1.3 10 O2 − =
–13
Solubility product of silver bromide is 5.0 x 10 . The quantity of potassium bromide (molar mass –1 taken as 120 g of mol ) to be added to 1 litre of 0.05 M solution of silver nitrate to start the precipitation of AgBr is –10 –9 –5 –8 (2) 1.2 x 10 g (3) 6.2 x 10 g (4) 5.0 x 10 g (1) 1.2 x 10 g (2) + – Ag + Br AgBr Precipitation starts when ionic product just exceeds solubility product K sp = Ag+ Br −
Br − =
K sp Ag
+
=
5 × 10−13 = 10 −11 0.05 –11
i.e., precipitation just starts when 10 No. of moles of KBr to be added ∴ weight of KBr to be added 23.
23. Sol :
24. 24. Sol :
moles of KBr is added to 1L of AgNO3 solution. = 10–11 –11 = 10 x 120 –9 = 1.2 x 10 g
The Gibbs energy for the decomposition of Al2O3 at 500°C is as follows : 2 4 –1 Al2O3 → Al + O2, ∆rG = + 966 kJ mol 3 3 The potential difference needed for electrolytic reduction of Al2O3 at 500°C is at least (1) 4.5 V (2) 3.0 V (3) 2.5 V (4) 5.0 V (3) −∆G ∆G = – nFE E= nF 966 × 103 E=− 4 × 96500 = –2.5 V ∴ The potential difference needed for the reduction = 2.5 V –11
2+
At 25°C, the solubility product of Mg(OH)2 is 1.0 x 10 . At which pH, will Mg ions start precipitating 2+ in the form of Mg(OH)2 from a solution of 0.001 M Mg ions ? (1) 9 (2) 10 (3) 11 (4) 8 (2) Mg2 + + 2OH− Mg(OH)2
K sp = Mg2 + OH− = ∴ pOH = 4
OH− K sp Mg2 +
2
= 10 −4
and pH = 10
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25. 25. Sol : 26. 26. Sol :
Percentage of free space in cubic close packed structure and in body centred packed structure are respectively (1) 30% and 26% (2) 26% and 32% (3) 32% and 48% (4) 48% and 26% (2) packing fraction of cubic close packing and body centred packing are 0.74 and 0.68 respectively. Out of the following, the alkene that exhibits optical isomerism is (1) 3–methyl–2–pentene (2) 4–methyl–1–pentene (3) 3–methyl–1–pentene (4) 2–methyl–2–pentene (3) H H2C=HC
C2 H5
only 3–methyl–1–pentene has a chiral carbon
CH3
27. 27. Sol :
Biuret test is not given by (1) carbohydrates (2) polypeptides (3) urea (4) proteins (1) It is a test characteristic of amide linkage. Urea also has amide linkage like proteins.
28.
The correct order of E0 2 +
28.
and Co is (1) Mn > Cr > Fe > Co (2) Cr > Fe > Mn > Co (3) Fe > Mn > Cr > Co (4) Cr > Mn > Fe > Co (1)
29. 29. Sol :
M
30. Sol :
31.
values with negative sign for the four successive elements Cr, Mn, Fe
The polymer containing strong intermolecular forces e.g. hydrogen bonding, is (1) teflon (2) nylon 6,6 (3) polystyrene (4) natural rubber (2) nylon 6,6 is a polymer of adipic acid and hexamethylene diamine O O
C (CH2)4
30.
/M
C
NH (CH2)6 NH
n
For a particular reversible reaction at temperature T, ∆H and ∆S were found to be both +ve. If Te is the temperature at equilibrium, the reaction would be spontaneous when (1) Te > T (2) T > Te (3) Te is 5 times T (4) T = Te (2) ∆G = ∆H − T∆S at equilibrium, ∆G = 0 for a reaction to be spontaneous ∆G should be negative ∴ T > Te
A rectangular loop has a sliding connector PQ of length and resistance R Ω and it is moving with a speed v as shown. The set-up is placed in a uniform magnetic field going into the plane of the paper. The three currents I1, I2 and I are
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B v 2B v , I= R R B v 2B v (2) I1 = I2 = , I= 3R 3R
(1) I1 = −I2 =
B v R B v B v (4) I1 = I2 = , I= 6R 3R 2 A moving conductor is equivalent to a battery of emf = v B Equivalent circuit I = I1 + I2 applying Kirchoff’s law ……………(1) I1R + IR − v B = 0 (3) I1 = I2 = I =
31. Sol.
I2R + IR − v B = 0 adding (1) & (2)
(motion emf)
R
R
……………(2)
I1
I2
2IR + IR = 2v B
I=
2vB 3R
I1 = I2 = 32.
32.
Sol.
vB 3R
Let C be the capacitance of a capacitor discharging through a resistor R. Suppose t1 is the time taken for the energy stored in the capacitor to reduce to half its initial value and t2 is the time taken for the charge to reduce to one-fourth its initial value. Then the ratio t1/t2 will be 1 1 (1) 1 (2) (3) (4) 2 2 4 3 q2 1 q2 1 U= (q0 e− t / T )2 = 0 e−2t / T (where τ = CR ) = 2 C 2C 2C U = Ui e −2t / τ
1 Ui = Ui e−2t 2 1 = e −2t / τ 2 q = q0 e − t / T
1
1
Now
/τ
t1 =
T ln 2 2
1 q0 = q0 e − t / 2T 4 t 2 = T ln 4 = 2T ln 2 ∴
t1 1 = t2 4
Directions: Questions number 33 – 34 contain Statement-1 and Statement-2. Of the four choices given after the statements, choose the one that best describes the two statements. 33.
Statement-1 : Two particles moving in the same direction do not lose all their energy in a completely inelastic collision. Statement-2 : Principle of conservation of momentum holds true for all kinds of collisions.
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33. Sol.
(1) Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1. (2) Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation of Statement1 (3) Statement-1 is false, Statement-2 is true. (4) Statement-1 is true, Statement-2 is false. 1
m1
m2 v1
v2
If it is a completely inelastic collision then m1v1 + m2 v 2 = m1v + m2 v
v= K.E =
m1v1 + m2 v 2 m1 + m2
p12 p2 + 2 2m1 2m2
as p1 and p2 both simultaneously cannot be zero therefore total KE cannot be lost. 34.
34. Sol.
35.
Statement-1 : When ultraviolet light is incident on a photocell, its stopping potential is V0 and the maximum kinetic energy of the photoelectrons is Kmax. When the ultraviolet light is replaced by Xrays, both V0 and Kmax increase. Statement-2 : Photoelectrons are emitted with speeds ranging from zero to a maximum value because of the range of frequencies present in the incident light. (1) Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1. (2) Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation of Statement1. (3) Statement-1 is false, Statement-2 is true. (4) Statement-1 is true, Statement-2 is false. 4 Since the frequency of ultraviolet light is less than the frequency of X–rays, the energy of each incident photon will be more for X–rays K.E photoelectron = hν − ϕ Stopping potential is to stop the fastest photoelectron hν ϕ − V0 = e e so, K.Emax and V0 both increases. But K.E ranges from zero to K.Emax because of loss of energy due to subsequent collisions before getting ejected and not due to range of frequencies in the incident light. A ball is made of a material of density ρ where ρoil < ρ < ρwater with ρoil and ρwater representing the densities of oil and water, respectively. The oil and water are immiscible. If the above ball is in equilibrium in a mixture of this oil and water, which of the following pictures represents its equilibrium position ? (1) (2)
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(3)
(4)
35. Sol.
2 ρoil < ρ < ρwater Oil is the least dense of them so it should settle at the top with water at the base. Now the ball is denser than oil but less denser than water. So, it will sink through oil but will not sink in water. So it will stay at the oil–water interface.
36.
A particle is moving with velocity v = K(y ˆi + x ˆj) , where K is a constant. The general equation for its path is 2 2 2 2 (1) y = x + constant (2) y = x + constant (3) xy = constant (4) y = x + constant 4 v = Ky ˆi + Kx ˆj
36. Sol.
dx dy = Ky, = Kx dt dt dy dy dt Kx = × = dx dt dx Ky y dy = x dx 2 2 y = x + c. 37.
Two long parallel wires are at a distance 2d apart. They carry steady equal current flowing out of the plane of the paper as shown. The variation of the magnetic field along the line XX 'is given by (2) (1)
(3)
37. Sol.
(4)
1 The magnetic field in between because of each will be in opposite direction µ i µ0i Bin between = 0 ˆj − ( −ˆj) 2πx 2π(2d − x) µi 1 1 = 0 − ( ˆj) 2π x 2d − x at x = d, Bin between = 0 for x < d, Bin between = ( ˆj) for x > d, Bin between = ( − ˆj)
towards x net magnetic field will add up and direction will be ( −ˆj) towards x 'net magnetic field will add up and direction will be ( ˆj)
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38.
In the circuit shown below, the key K is closed at t = 0. The current through the battery is V R1 R2 V (1) at t = 0 and at t = ∞ 2 2 R2 R1 + R2 (2)
V (R1 + R2 ) V at t = 0 and at t = ∞ R2 R1 R 2
(3)
V at t = 0 and R2
(4)
V (R1 + R2 ) V at t = 0 and at t = ∞ R1 R 2 R2
V R1 R2 R12 + R22
at t = ∞
38. Sol.
2 At t = 0, inductor behaves like an infinite resistance V So at t = 0, i = R2 and at t = ∞ , inductor behaves like a conducting wire V(R1 + R 2 ) V i= = Re q R1 R 2
39.
The figure shows the position – time (x – t) graph of one-dimensional motion of a body of mass 0.4 kg. The magnitude of each impulse is (1) 0.4 Ns (2) 0.8 Ns (3) 1.6 Ns (4) 0.2 Ns
39. Sol.
2 From the graph, it is a straight line so, uniform motion. Because of impulse direction of velocity changes as can be seen from the slope of the graph. 2 Initial velocity = = 1 m / s 2 2 Final velocity = − = −1 m / s 2 Pi = 0.4 N – s
Pji = −0.4 N – s J = Pf − Pi = – 0.4 – 0.4 = – 0.8 N – s ( J = impulse) J = 0.8 N–s Directions : Questions number 40 – 41 are based on the following paragraph. A nucleus of mass M + ∆m is at rest and decays into two daughter nuclei of equal mass Speed of light is c. 40. 40. Sol.
M each. 2
The binding energy per nucleon for the parent nucleus is E1 and that for the daughter nuclei is E2. Then (1) E2 = 2E1 (2) E1 > E2 (3) E2 > E1 (4) E1 = 2E2 3 After decay, the daughter nuclei will be more stable hence binding energy per nucleon will be more than that of their parent nucleus.
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41.
The speed of daughter nuclei is (1) c
41. Sol.
∆m M + ∆m
(2) c
2 Conserving the momentum M M 0 = V1 − V2 2 2 V1 = V2
42. Sol.
43.
(3) c
∆m M
(4) c
∆m M + ∆m
…………….(1)
1 M 1 M ∆mc 2 = . V12 + . .V22 2 2 2 2 M 2 2 ∆mc = V1 2 2∆mc 2 = V12 M 2∆m V1 = c M 42.
2∆m M
…………….(2)
A radioactive nucleus (initial mass number A and atomic number Z) emits 3 α-particles and 2 positrons. The ratio of number of neutrons to that of protons in the final nucleus will be A −Z−8 A − Z − 12 A−Z−4 A−Z−4 (2) (3) (4) (1) Z−4 Z−4 Z−2 Z−8 2 In positive beta decay a proton is transformed into a neutron and a positron is emitted. p + → n0 + e + no. of neutrons initially was A – Z no. of neutrons after decay (A – Z) – 3 x 2 (due to alpha particles) + 2 x 1 (due to positive beta decay) The no. of proton will reduce by 8. [as 3 x 2 (due to alpha particles) + 2(due to positive beta decay)] Hence atomic number reduces by 8. A thin semi-circular ring of radius r has a positive charge q distributed uniformly over it. The net field E at the centre O is q q ˆj ˆj (1) (2) – 4 π2 ε0 r 2 4 π2 ε 0 r 2 (3) –
q ˆj 2 π2 ε0 r 2
(4)
43.
3
Sol.
Linear charge density λ =
q ˆj 2 π2 ε 0 r 2
y
q πr
K.dq E = dE sin θ( −ˆj) = sin θ( −ˆj) r2 K qr E= 2 d θ sin θ( −ˆj) r πr =
K qπ sin θ( − ˆj) r2 π 0
=
q ( −ˆj) 2π ε 0 r 2
dθ θ
θ
x
2
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44.
The combination of gates shown below yields (1) OR gate (2) NOT gate (3) XOR gate (4) NAND gate
44. Sol.
1 Truth table for given combination is A B X 0 0 0 0 1 1 1 0 1 1 1 1 This comes out to be truth table of OR gate
45.
A diatomic ideal gas is used in a Car engine as the working substance. If during the adiabatic expansion part of the cycle, volume of the gas increases from V to 32V the efficiency of the engine is (1) 0.5 (2) 0.75 (3) 0.99 (4) 0.25 2 The efficiency of cycle is T η = 1− 2 T1 for adiabatic process TVγ–1 = constant 7 For diatomic gas γ = 5 γ −1 γ −1 T1V1 = T2 V2
45. Sol.
T1 = T2
γ −1
V2 V1 7
T1 = T2 (32)5
−1
= T2 (25 )2 / 5 = T2 x 4 T1 = 4T2. 1 3 η = 1− = = 0.75 4 4
46. 46. Sol.
47. 47.
20
If a source of power 4 kW produces 10 photons/second, the radiation belong to a part of the spectrum called (1) X–rays (2) ultraviolet rays (3) microwaves (4) γ–rays 1 3 20 4 x 10 = 10 x hf 4 × 103 f = 20 10 × 6.023 × 10−34 16 f = 6.03 x 10 Hz The obtained frequency lies in the band of X–rays. The respective number of significant figures for the numbers 23.023, 0.0003 and 2.1 x 10–3 are (1) 5, 1, 2 (2) 5, 1, 5 (3) 5, 5, 2 (4) 4, 4, 2 1
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48.
48. Sol.
49.
49. Sol.
In a series LCR circuit R = 200 Ω and the voltage and the frequency of the main supply is 220 V and 50 Hz respectively. On taking out the capacitance from the circuit the current lags behind the voltage o by 30°. On taking out the inductor from the circuit the current leads the voltage by 30 . The power dissipated in the LCR circuit is (1) 305 W (2) 210 W (3) Zero W (4) 242 W 4 The given circuit is under resonance as XL = XC Hence power dissipated in the circuit is V2 = 242 W P= R Let there be a spherically symmetric charge distribution with charge density varying as 5 r upto r = R, and ρ(r) = 0 for r > R, where r is the distance from the origin. The ρ(r) = ρ0 − 4 R electric field at a distance r(r < R) from the origin is given by 4πρ0 r 5 r ρ r 5 r 4 ρ0 r 5 r ρ r 5 r (1) (2) 0 (3) (4) 0 − − − − 3 ε0 3 R 4 ε0 3 R 3 ε0 4 R 3 ε0 4 R 2 Apply shell theorem the total charge upto distance r can be calculated as followed dq = 4 πr 2 .dr.ρ
= 4πr 2 .dr.ρ0
5 2 r3 r dr − dr 4 R
= 4πρ0 dq = q = 4πρ0
r
0
= 4πρ0
5 r − 4 R
5 2 r3 r dr − dr 4 R 5 r3 1 r4 − 43 R 4
kq r2 1 1 5 r3 r4 = .4πρ0 − 2 4πε0 r 4 3 4R
E=
E= 50.
50. Sol.
ρ0 r 5 r − 4ε 0 3 R
The potential energy function for the force between two atoms in a diatomic molecule is a b approximately given by U(x) = 12 − 6 , where a and b are constants and x is the distance between x x the atoms. If the dissociation energy of the molecule is D = [U(x = ∞) – Uat equilibrium], D is b2 b2 b2 b2 (1) (2) (3) (4) 2a 12a 4a 6a 3 a b U(x) = 12 − 6 x x U(x = ∞) = 0 dU 12a 6b as, F=− = − 13 + 7 dx x x at equilibrium, F = 0
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∴
51.
51. Sol.
b −b2 = 2a 4a b
Uat equilibrium =
∴
D = U(x = ∞) − Uat equilibrium =
2
2a b
−
b2 4a
Two identical charged spheres are suspended by strings of equal lengths. The strings make an –3 angle of 30° with each other. When suspended in a liquid of density 0.8 g cm , the angle remains –3 the same. If density of the material of the sphere is 16 g cm , the dielectric constant of the liquid is (1) 4 (2) 3 (3) 2 (4) 1 3 From F.B.D of sphere, using Lami’s theorem F = tan θ ………………(i) T θ mg F when suspended in liquid, as θ remains same, F' ………………(ii) ∴ = tan θ ρ mg 1 − mg d using (i) and (ii) F F' F = where, F'= ρ mg K mg 1 − d
or
52. Sol.
2a b a
∴
∴
52.
x6 =
F = mg
K=
F' mg K 1 −
1 1−
ρ d
ρ d
=2
Two conductors have the same resistance at 0oC but their temperature coefficients of resistance are α1 and α2. The respective temperature coefficients of their series and parallel combinations are nearly α + α2 α + α2 α1 α 2 α + α 2 α1 + α 2 (1) 1 , α1 + α 2 (2) α1 + α 2 , 1 (3) α1 + α 2 , (4) 1 , 2 2 α1 + α 2 2 2 4 Let R0 be the initial resistance of both conductors ∴ At temperature θ their resistance will be, R1 = R0 (1 + α1θ) and R2 = R0 (1 + α 2 θ) for, series combination, Rs = R1 + R2 R s 0 (1 + α s θ) = R0 (1 + α1θ) + R0 (1 + α 2 θ) where R s 0 = R 0 + R0 = 2R0 ∴
2R0 (1 + α s θ) = 2R0 + R0 θ(α1 + α 2 )
or
αs =
α1 + α 2 2
for parallel combination,
Rp =
R1 R 2 R1 + R 2
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Rp0 (1 + αp θ) = where, Rp0 =
R0 (1 + α1θ)R0 (1 + α 2 θ) R0 (1 + α1θ) + R0 (1 + α 2 θ)
R0 R0 R = 0 R0 + R 0 2
∴
R0 R2 (1 + α1θ + α 2 θ + α1α 2 θ) (1 + αp θ) = 0 2 R 0 (2 + α1θ + α 2 θ)
as ∴
α1 and α 2 are small quantities α1 α 2 is negligible
or
αp =
as
( α1 + α 2 )2 is negligible
∴
αp =
α1 + α 2 α + α2 = 1 [1 − (α1 + α 2 )θ] 2 + (α1 + α 2 )θ 2
α1 + α 2 2
53.
A point P moves in counter-clockwise direction on a circular path as shown in the figure. The movement of ‘P’ is such that it 3 sweeps out a length s = t + 5, where s is in metres and t is in seconds. The radius of the path is 20 m. The acceleration of ‘P’ when t = 2 s is nearly 2 2 (2) 12 m/s (1) 13 m/s 2 2 (3) 7.2 m/s (4) 14 m/s
53. Sol.
4 S = t3 + 5 ∴
speed, v =
and ∴ at ∴ ∴
ds = 3t 2 dt
dv = 6t dt 2 tangential acceleration at t = 2s, at = 6 x 2 = 12 m/s 2 t = 2s, v = 3(2) = 12 m/s v 2 144 centripetal acceleration, ac = = m / s2 R 20 rate of change of speed =
net acceleration =
a2t + ai2
≈ 14 m / s2 o
54.
Two fixed frictionless inclined plane making an angle 30 and 60o with the vertical are shown in the figure. Two block A and B are placed on the two planes. What is the relative vertical acceleration of A with respect to B ? –2 (1) 4.9 ms in horizontal direction –2 (2) 9.8 ms in vertical direction (3) zero –2 (4) 4.9 ms in vertical direction
54. Sol.
4 mg sin θ = ma ∴ a = g sin θ where a is along the inclined plane 2 ∴ vertical component of acceleration is g sin θ ∴ relative vertical acceleration of A with respect to B is (Hyderabad Classes) Limited. 5-9-14/B, Saifabad, (Opp. Secretariat) Hyderabad. 500 063. Phone: 040-66777000 – 03 Fax: 040-66777004
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g[sin2 60 − sin2 30] = 55.
55. Sol.
g = 4.9 m / s2 in vertical direction. 2
For a particle in uniform circular motion the acceleration a at a point P(R, θ) on the circle of radius R is (here θ is measured from the x–axis) v2 v2 v2 v2 (2) − sin θ ˆi + cos θ ˆj (1) − cos θ ˆi + sin θ ˆj R R R R 2 2 2 2 v v v ˆ v ˆ (3) − cos θ ˆi − sin θ ˆj (4) i+ j R R R R 3 For a particle in uniform circular motion, y v2 towards centre of circle a= R P (R, θ) ac v2 ∴ a= ( − cos θ ˆi − sin θ ˆj) R x ac v2 v2 ˆ ˆ or a = − cos θ i − sin θ j R R
Directions: Questions number 56 – 58 are based on the following paragraph. An initially parallel cylindrical beam travels in a medium of refractive index µ(I) = µ0 + µ 2I , where µ0
and µ2 are positive constants and I is the intensity of the light beam. The intensity of the beam is decreasing with increasing radius. 56.
56. Sol.
57.
57. Sol. 58. 58.
As the beam enters the medium, it will (1) diverge (2) converge (3) diverge near the axis and converge near the periphery (4) travel as a cylindrical beam 2 As intensity is maximum at axis, ∴ µ will be maximum and speed will be minimum on the axis of the beam. ∴ beam will converge. The initial shape of the wave front of the beam is (1) convex (2) concave (3) convex near the axis and concave near the periphery (4) planar 4 For a parallel cylinderical beam, wavefront will be planar. The speed of light in the medium is (1) minimum on the axis of the beam (3) directly proportional to the intensity I 1
(2) the same everywhere in the beam (4) maximum on the axis of the beam
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59.
59. Sol.
A small particle of mass m is projected at an angle θ with the x-axis with an initial velocity v0 in the x-y plane as shown in the v sin θ figure. At a time t < 0 , the angular momentum of the g particle is (1) −mgv 0 t 2 cos θ ˆj (2) mgv 0 t cos θ kˆ
1 1 (3) − mgv 0 t 2 cos θ kˆ (4) mgv 0 t 2 cos θ ˆi 2 2 where ˆi, ˆj and kˆ are unit vectors along x, y and z–axis respectively. 3 L = m(r × v)
1 L = m v 0 cos θt ˆi + (v 0 sin θt − gt 2 )jˆ × v 0 cos θ ˆi + (v 0 sin θ − gt)jˆ 2 1 = mv 0 cos θt − gt kˆ 2 1 = − mgv 0 t 2 cos θkˆ 2 60.
–1
The equation of a wave on a string of linear mass density 0.04 kg m
t x y = 0.02(m)sin 2π − 0.04(s) 0.50(m) 60.
(1) 4.0 N 4
Sol.
T = µv 2 = µ
61.
Let cos(α + β) = (1)
61.
is given by
. The tension in the string is
(2) 12.5 N
(3) 0.5 N
(4) 6.25 N
ω2 (2π / 0.004)2 = = 6.25 N 0.04 k2 (2π / 0.50)2
56 33
π 4 5 and let sin(α – β) = , where 0 ≤ α, β ≤ , then tan 2α = 5 13 4 19 20 25 (2) (3) (4) 12 7 16
1
3 4 5 tan(α – β) = 12 3 5 + 56 tan 2α = tan(α + β + α – β) = 4 12 = 3 5 33 1− 4 12 4 5 5 sin(α – β) = 13 cos (α + β) =
62.
tan(α + β) =
Let S be a non-empty subset of R. Consider the following statement: P: There is a rational number x ∈ S such that x > 0. Which of the following statements is the negation of the statement P ? (1) There is no rational number x ∈ S such that x ≤ 0 (2) Every rational number x ∈ S satisfies x ≤ 0
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62.
(3) x ∈ S and x ≤ 0 x is not rational (4) There is a rational number x ∈ S such that x ≤ 0 2 P: there is a rational number x ∈ S such that x > 0 ~P: Every rational number x ∈ S satisfies x ≤ 0
63.
Let a = ˆj − kˆ and c = ˆi − ˆj − kˆ . Then vector b satisfying a × b + c = 0 and a ⋅ b = 3 is (1) 2iˆ − ˆj + 2kˆ (2) ˆi − ˆj − 2kˆ (3) ˆi + ˆj − 2kˆ (4) −ˆi + ˆj − 2kˆ
63.
4 c = b×a b⋅c = 0 b1ˆi + b2 ˆj + b3kˆ ⋅ ˆi − ˆj − kˆ = 0
(
)(
)
b1 – b2 – b3 = 0 and a ⋅ b = 3 b2 – b3 = 3 b1 = b2 + b3 = 3 + 2b3 b = ( 3 + 2b3 ) ˆi + ( 3 + b3 ) ˆj + b3kˆ . 64.
The equation of the tangent to the curve y = x +
64.
(1) y = 1 3
(2) y = 2
Parallel to x-axis
4 , that is parallel to the x-axis, is x2 (3) y = 3 (4) y = 0
dy =0 dx
1−
x=2 y=3 Equation of tangent is y – 3 = 0(x – 2) 65. 65.
8 =0 x3
y–3=0
π is 2 (3) tan x = (sec x + c)y (4) sec x = (tan x + c)y
Solution of the differential equation cos x dy = y(sin x – y) dx, 0 < x < (1) y sec x = tan x + c (2) y tan x = sec x + c 4 cos x dy = y(sin x – y) dx dy = y tan x − y 2 sec x dx 1 dy 1 − tan x = − sec x y 2 dx y Let −
1 =t y
1 dy dt = y 2 dx dx
dy dt – t tan x = –sec x + (tan x) t = sec x. dx dx tan x dx I.F. = e = sec x Solution is t(I.F) = (I.F) sec x dx −
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1 sec x = tan x + c y 66.
The area bounded by the curves y = cos x and y = sin x between the ordinates x = 0 and x =
66.
(1) 4 2 + 2 4 π 4
(2) 4 2 – 1
(3) 4 2 + 1
5π 4
3π 2
π 4
5π 4
( cos x − sin x ) dx + ( sin x − cos x ) dx + ( cos x − sin x ) = 4
0
cos x
(4) 4 2 – 2
2 −2
sin x
π 0
3π is 2
5π
3π
4
2
π
2π
4
2
67.
If two tangents drawn from a point P to the parabola y = 4x are at right angles, then the locus of P is (1) 2x + 1 = 0 (2) x = –1 (3) 2x – 1 = 0 (4) x = 1 2 The locus of perpendicular tangents is directrix i.e, x = –a; x = –1
68.
ˆ b = 2iˆ + 4ˆj + kˆ and c = λ ˆi + ˆj + µkˆ are mutually orthogonal, then (λ, µ) = If the vectors a = ˆi − ˆj + 2k,
67.
68.
69.
(1) (2, –3) (2) (–2, 3) 4 a ⋅ b = 0, b ⋅ c = 0, c ⋅a = 0 2λ + 4 + µ = 0 λ – 1 + 2µ = 0 Solving we get: λ = –3, µ = 2
(4) (–3, 2)
Consider the following relations: R = {(x, y) | x, y are real numbers and x = wy for some rational number w}; S=
69.
(3) (3, –2)
m p , n q
m, n, p and q are integers such that n, q ≠ 0 and qm = pn . Then
(1) neither R nor S is an equivalence relation (2) S is an equivalence relation but R is not an equivalence relation (3) R and S both are equivalence relations (4) R is an equivalence relation but S is not an equivalence relation 2 xRy need not implies yRx m p S: s ⇔ qm = pn n q m m s reflexive n n
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m p s n q m p p r s , s n q q s
p m s symmetric q n qm = pn, ps = rq
ms = rn transitive.
S is an equivalence relation. 70.
Let f: R → R be defined by f(x) =
k − 2x, if x ≤ −1 . If f has a local minimum at x = –1, then a 2x + 3, if x > −1
possible value of k is (2) −
(1) 0 70.
3 f(x) = k – 2x = 2x + 3
1 2
(3) –1
(4) 1
if x ≤ –1 if x > –1 2x + 3
k – 2x 1 –1
lim f(x) ≤ –1
x →−1−
71. 71.
This is true where k = –1
The number of 3 × 3 non-singular matrices, with four entries as 1 and all other entries as 0, is (1) 5 (2) 6 (3) at least 7 (4) less than 4 3 First row with exactly one zero; total number of cases = 6 First row 2 zeros we get more cases Total we get more than 7.
Directions: Questions Number 72 to 76 are Assertion – Reason type questions. Each of these questions contains two statements. Statement-1: (Assertion) and Statement-2: (Reason) Each of these questions also has four alternative choices, only one of which is the correct answer. You have to select the correct choice.
72.
72.
Four numbers are chosen at random (without replacement) from the set {1, 2, 3, ....., 20}. Statement-1: The probability that the chosen numbers when arranged in some order will form an AP 1 is . 85 Statement-2: If the four chosen numbers from an AP, then the set of all possible values of common difference is {±1, ±2, ±3, ±4, ±5}. (1) Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation for Statement-1 (2) Statement-1 is true, Statement-2 is false (3) Statement-1 is false, Statement-2 is true (4) Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation for Statement-1 2
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20
N(S) = C4 Statement-1: common difference is 1; total number of cases = 17 common difference is 2; total number of cases = 14 common difference is 3; total number of cases = 11 common difference is 4; total number of cases = 8 common difference is 5; total number of cases = 5 common difference is 6; total number of cases = 2 17 + 14 + 11 + 8 + 5 + 2 1 Prob. = . = 20 85 C4 73.
73.
74.
Statement-1: The point A(3, 1, 6) is the mirror image of the point B(1, 3, 4) in the plane x – y + z = 5. Statement-2: The plane x – y + z = 5 bisects the line segment joining A(3, 1, 6) and B(1, 3, 4). (1) Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation for Statement-1 (2) Statement-1 is true, Statement-2 is false (3) Statement-1 is false, Statement-2 is true (4) Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation for Statement-1 1 A(3, 1, 6); B = (1, 3, 4) Mid-point of AB = (2, 2, 5) lies on the plane. and d.r’s of AB = (2, –2, 2) d.r’s Of normal to plane = (1, –1, 1). AB is perpendicular bisector ∴ A is image of B Statement-2 is correct but it is not correct explanation. 10
Let S1 =
j =1
j ( j − 1) 10 C j , S2 =
10
j
10
j =1
C j and S3 =
10
j2
j =1
10
Cj .
9
74.
Statement-1: S3 = 55 × 2 8 8 Statement-2: S1 = 90 × 2 and S2 = 10 × 2 . (1) Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation for Statement-1 (2) Statement-1 is true, Statement-2 is false (3) Statement-1 is false, Statement-2 is true (4) Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation for Statement-1 2 10 10 10! 8! S1 = j ( j − 1) = 90 = 90 ⋅ 28 . j j − 1 j − 2 ! 10 − j ! j − 2 ! 8 − j − 2 ! ( )( ) ( ) ) ( ( )) j =1 j= 2 (
S2 =
10 j =1
S3 =
10 j =1
j
10 10! 9! = 10 = 10 ⋅ 29 . j ( j − 1)! ( 9 − ( j − 1) )! j − 1 ! 9 − j − 1 ! ) ( ( )) j =1 (
j ( j − 1) + j
10! = j! (10 − j )!
10 j =1
8
j ( j − 1) 10 C j = 8
10 j =1
8
j 10 C j = 90 . 2 + 10 . 2 8
9
9
= 90 . 2 + 20 . 2 = 110 . 2 = 55 . 2 . 75.
2
Let A be a 2 × 2 matrix with non-zero entries and let A = I, where I is 2 × 2 identity matrix. Define Tr(A) = sum of diagonal elements of A and |A| = determinant of matrix A. Statement-1: Tr(A) = 0 Statement-2: |A| = 1 (1) Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation for Statement-1 (2) Statement-1 is true, Statement-2 is false
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75.
(3) Statement-1 is false, Statement-2 is true (4) Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation for Statement-1 2 a b Let A = , abcd ≠ 0 c d a b a b ⋅ c d c d
2
A = 2
A =
a2 + bc ab + bd ac + cd bc + d2
a2 + bc = 1, bc + d2 = 1 ab + bd = ac + cd = 0 c ≠ 0 and b ≠ 0 a+d=0 Trace A = a + d = 0 |A| = ad – bc = –a2 – bc = –1. 76.
Let f: R → R be a continuous function defined by f(x) =
1 . e + 2e − x x
1 , for some c ∈ R. 3 1 Statement-2: 0 < f(x) ≤ , for all x ∈ R 2 2 (1) Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation for Statement-1 (2) Statement-1 is true, Statement-2 is false (3) Statement-1 is false, Statement-2 is true (4) Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation for Statement-1 4 1 ex f(x) = x = e + 2e− x e2x + 2 Statement-1: f(c) =
76.
f′(x) =
(e
2x
)
+ 2 e x − 2e2x ⋅ e x
(e )
2x + 2 2
2x
f′(x) = 0 2x e =2
e + 2 = 2e x e = 2
maximum f(x) = 0 < f(x) ≤
2 1 = 4 2 2
1
2 2 1 1 Since 0 < < 3 2 2 1 f(c) = 3
77.
2x
∀x∈R for some c ∈ R
For a regular polygon, let r and R be the radii of the inscribed and the circumscribed circles. A false statement among the following is r 1 r 2 (1) There is a regular polygon with = (2) There is a regular polygon with = R R 3 2
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(3) There is a regular polygon with 77.
r 3 = R 2
(4) There is a regular polygon with
r 1 = R 2
2
a π cot 2 n ‘a’ is side of polygon. a π R = cosec 2 n π cot r n = cos π = π R n cosec n π 2 for any n ∈ N. cos ≠ n 3 r=
78. 78.
2
2009
If α and β are the roots of the equation x – x + 1 = 0, then α (1) –1 (2) 1 (3) 2 2 2
x –x+1=0
1± 3 i 2 1 3 α = +i , 2 2 π π α = cos + isin , 3 3
x=
+β
2009
= (4) –2
1± 1− 4 2
x=
1 i 3 − 2 2 π π β = cos − isin 3 3 π 2009 2009 α +β = 2cos2009 3 2π 2π = 2cos 668π + π + = 2cos π + 3 3 2π 1 = −2 − =1 = −2cos 3 2 79. 79.
80.
80.
β=
The number of complex numbers z such that |z – 1| = |z + 1| = |z – i| equals (1) 1 (2) 2 (3) ∞ (4) 0 1 Let z = x + iy |z – 1| = |z + 1| Re z = 0 x=0 x=y |z – 1| = |z – i| |z + 1| = |z – i| y = –x Only (0, 0) will satisfy all conditions. Number of complex number z = 1 A line AB in three-dimensional space makes angles 45° and 120° with the positive x-axis and the positive y-axis respectively. If AB makes an acute angle θ with the positive z-axis, then θ equals (1) 45° (2) 60° (3) 75° (4) 30° 2
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= cos 45° =
1 2
m = cos 120° = −
1 2
n = cos θ where θ is the angle which line makes with positive z-axis. 2 2 2 Now + m + n = 1
1 1 2 + + cos θ = 1 2 4 1 2 cos θ = 4 1 cos θ = 2 π θ= . 3 81.
The line L given by the equation (1)
81.
17
(θ Being acute)
x y + = 1 passes through the point (13, 32). The line K is parallel to L and has 5 b
x y + = 1. Then the distance between L and K is c 3 17 23 (2) (3) 15 17
(4)
23 15
3
b 5 3 Slope of line K = − c Line L is parallel to line k. b 3 bc = 15 = 5 c (13, 32) is a point on L. 13 32 32 8 + =1 =− 5 b b 5 3 b = –20 c= − 4 Equation of K: y – 4x = 3 52 − 32 + 3 23 Distance between L and K = = 17 17
Slope of line L = −
82.
82.
th
A person is to count 4500 currency notes. Let an denote the number of notes he counts in the n minute. If a1 = a2 = ...... = a10 = 150 and a10, a11, ...... are in A.P. with common difference –2, then the time taken by him to count all notes is (1) 34 minutes (2) 125 minutes (3) 135 minutes (4) 24 minutes 1 th Till 10 minute number of counted notes = 1500 n 3000 = [2 × 148 + (n – 1)(–2)] = n[148 – n + 1] 2 (Hyderabad Classes) Limited. 5-9-14/B, Saifabad, (Opp. Secretariat) Hyderabad. 500 063. Phone: 040-66777000 – 03 Fax: 040-66777004
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2
n – 149n + 3000 = 0 n = 125, 24 n = 125 is not possible. Total time = 24 + 10 = 34 minutes. 83.
Let f: R → R be a positive increasing function with lim
x →∞
(1) 83.
2 3
(2)
3 2
f(3x) f(2x) = 1. Then lim = x →∞ f(x) f(x)
(3) 3
(4) 1
4 f(x) is a positive increasing function 0 < f(x) < f(2x) < f(3x) f(2x) f(3x) < 0
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AIEEE–2011 (Set – Q) IMPORTANT INSTRUCTIONS 1.
Immediately fill in the particulars on this page of the Test Booklet with Blue/Black Ball Point Pen. Use of Pencil is strictly prohibited.
2.
The Answer Sheet is kept inside the Test Booklet. When you are directed to open the Test Booklet, take out the Answer Sheet and fill in the particulars carefully.
3.
The test is of 3 hours duration.
4.
The Test Booklet consists of 90 questions. The maximum marks are 360.
5.
There are three parts in the question paper A, B, C consisting of Physics, Mathematics, Chemistry having 30 questions in each part of equal weight age. Each question is allotted 4(four) marks for each correct response. Candidates will be awarded marks as stated above in instruction No. 5 for correct response of each question ¼ (one fourth) 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.
6.
7.
There is only one correct response for each question. Filling up more than one response in each question will be treated as wrong response and marks for wrong response will be deducted accordingly as per instruction 6 above.
8.
Use Blue/Black Ball Point Pen only for writing particulars/marking responses on Side-1 and Side-2 of the Answer Sheet. Use of pencil is strictly prohibited.
9.
No candidate is allowed to carry any textual material, printed or written, bits of papers, pager, mobile phone, any electronic device, etc., except the Admit Card inside the examination hall/room.
10.
Rough work is to be done on the space provided for this purpose in the Test Booklet only. This space is given at the bottom of each page and in 3 pages (Pages 21 – 23) at the end of the booklet.
11.
On completion of the test, the candidate must hand over the Answer Sheet to the Invigilator on duty in the Room/Hall. However, the candidates are allowed to take away this Test Booklet with them.
12.
The CODE for this Booklet is Q. Make sure that the CODE printed on Side-2 of the Answer Sheet is the same as that on this booklet. In case of discrepancy, the candidate should immediately report the matter to the Invigilator for replacement of both the Test Booklet and the Answer Sheet.
13.
Do not fold or make any stray marks on the Answer Sheet.
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Nitin M Sir (physics-iitjee.blogspot.com)
Nitin M Sir (physics-iitjee.blogspot.com) AIEEE−2011−2
PART A : PHYSICS 1.
The transverse displacement y ( x,t ) of a wave on a string is given by y ( x,t ) = e
(
− ax2 + bt 2 + 2 ab xt
).
This represents a (1) wave moving in – x direction with speed (3) standing wave of frequency
1 b
b (2) standing wave of frequency a
b
(4) wave moving in + x direction with
a b
2.
A screw gauge gives the following reading when used to measure the diameter of a wire. Main scale reading : 0 mm Circular scale reading : 52 divisions Given that 1 mm on main scale corresponds to 100 divisions of the circular scale. The diameter of wire from the above date is : (1) 0.052 cm (2) 0.026 cm (3) 0.005 cm (4) 0.52 cm
3.
A mass m hangs with the help of a string wrapped around a pulley on a frictionless bearing. The pulley has mass m and radius R. Assuming pulley to be a perfect uniform circular disc, the acceleration of the mass m, if the string does not slip on the pulley, is 2 g 3 (1) g (2) (3) (4) g g 2 3 3
4.
Work done in increasing the size of a soap bubble from a radius of 3 cm to 5 cm is nearly (Surface tension of soap solution = 0.03 Nm−1): (1) 0.2π mJ (2) 2π mJ (3) 0.4 π mJ (4) 4π mJ
5.
A thin horizontal circular disc is rotating about a vertical axis passing through its centre. An insect is at rest at a point near the rim of the disc. The insect now moves along a diameter of the disc to reach its other end. During the journey of the insect, the angular speed of the disc: (1) continuously decreases (2) continuously increases (3) first increases and then decreases (4) remains unchanged
6.
Two particles are executing simple harmonic motion of the same amplitude A and frequency ω along the x-axis. Their mean position is separated by distance X0 ( X0 > A ) . If the maximum separation between them is ( X0 + A ) , the phase difference between their motion is : (1)
π 3
(2)
π 4
(3)
π 6
(4)
π 2
7.
Two bodies of masses m and 4 m are placed at a distance r. The gravitational potential at a point on the line joining them where the gravitational field is zero is: 4Gm 6Gm 9Gm (2) − (3) − (4) zero (1) − r r r
8.
Two identical charged spheres suspended from a common point by two massless strings of length l are initially a distance d ( d v0 (the threshold frequency). The maximum kinetic energy and the stopping potential are Kmax and V0 respectively. If the frequency incident on the surface doubled, both the Kmax and V0 are also doubled. Statement-2 : The maximum kinetic energy and the stopping potential of photoelectrons emitted from a surface are linearly dependent on the frequency of incident light. (1) Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1. (2) Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation of Statement-1. (3) Statement-1 is false, Statement-2 is true. (4) Statement-1 is true, Statement-2 is false.
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Nitin M Sir (physics-iitjee.blogspot.com)
Nitin M Sir (physics-iitjee.blogspot.com) AIEEE−2011−6
PART B: MATHEMATICS 31.
The lines L1 : y − x = 0 and L 2 : 2x + y = 0 intersect the line L3 : y + 2 = 0 at P and Q respectively. The bisector of the acute angle between L1 and L 2 intersect L3 at R . Statement – 1 :
The ratio PR : RQ equals 2 2 : 5 .
Statement – 2 : In any triangle, bisector of an angle divides the triangle into two similar triangles. (1) Statement – 1 is true, Statement – 2 is true; Statement – 2 is not a correct explanation for Statement –1 (2) Statement – 1 is true, Statement– 2 is false. (3) Statement – 1 is false, Statement– 2 is true. (4) Statement – 1 is true, Statement – 2 is true; Statement – 2 is a correct explanation for Statement – 1 32.
33.
If A = sin2 x + cos4 x , then for all real x 13 ≤ A ≤1 (2) 1 ≤ A ≤ 2 (1) 16
35.
3 13 ≤A≤ 4 16
(
The coefficient of x 7 in the expansion of 1 − x − x 2 + x 3 (1) -132
34.
(3)
(2) -144
{
)
6
(4)
3 ≤ A ≤1 4
is
(3)132
(4) 144
}
1 − cos 2 ( x − 2 ) lim x →2 x−2
(1) equals
(2) equals − 2
2
(3) equals
1 2
(4) does not exist
Statement – 1 :
The number of ways of distributing 10 identical balls in 4 distinct boxes such that no box is empty is 9 C3
Statement – 2 :
The number of ways of choosing any 3 places from 9 different places is 9 C3 .
(1) Statement – 1 is true, Statement – 2 is true; Statement – 2 is not a correct explanation for Statement –1 (2) Statement – 1 is true, Statement– 2 is false. (3) Statement – 1 is false, Statement– 2 is true. (4) Statement – 1 is true, Statement – 2 is true; Statement – 2 is a correct explanation for Statement – 1 36.
d2 x dy 2
equals −1
d2 y dy −3 (1) − 2 dx dx
d2 y dy −2 (2) 2 dx dx
d2 y dy −3 (3) − 2 dx dx
d2 y (4) 2 dx
−1
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Nitin M Sir (physics-iitjee.blogspot.com) AIEEE−2011−7
37.
38.
dy = y + 3 > 0 and y ( 0 ) = 2 , then y ( ln2 ) is equal to dx (1) 5 (2) 13 (3) -2
If
(4) 7
Let R be the set of real numbers Statement – 1 : A = ( x,y ) ∈ R × R : y − x is an int eger is an equivalence relation on R . Statement – 2 :
{ } B = {( x,y ) ∈ R × R : x = αy for some rational number α} is an equivalence relation on
R. (1) Statement – 1 is true, Statement – 2 is true; Statement – 2 is not a correct explanation for Statement –1 (2) Statement – 1 is true, Statement– 2 is false. (3) Statement – 1 is false, Statement– 2 is true. (4) Statement – 1 is true, Statement – 2 is true; Statement – 2 is a correct explanation for Statement – 1 1
39.
The value of
∫ 0
(1)
40.
8log (1 + x ) 1 + x2
π log2 8
dx is
(2)
π log2 2
(3) log2
(4) π log2
Let α, β be real and z be a complex number. If z2 + αz + β = 0 has two distinct roots on the line Re z = 1 , then it is necessary that (1) β ∈ ( −1, 0 )
(3) β ∈ (1, ∞ )
(2) β = 1
(4) β ∈ ( 0, 1)
41.
Consider 5 independent Bernoulli’s trials each with probability of success p. If the probability of at least 31 one failure is greater than or equal to , then p lies in the interval 32 3 11 1 11 1 3 (1) , (2) 0, (3) , 1 (4) , 4 12 2 2 4 12
42.
A man saves Rs. 200 in each of the first three months of his service. In each of the subsequent months his saving increases by Rs. 40 more than the saving of immediately previous month. His total saving from the start of service will be Rs. 11040 after (1) 19 months (2) 20 months (3) 21 months (4) 18 months
43.
The domain of the function f ( x ) = (1) ( 0, ∞ )
44.
(2) ( −∞, 0 )
If the angle between the line x = equals 3 (1) 2
(2)
2 5
1 x −x
is (3) ( −∞, ∞ ) − {0}
(4) ( −∞, ∞ )
5 y −1 z − 3 = and the plane x + 2y + 3z = 4 is cos−1 , then λ 14 λ 2
(3)
5 3
(4)
2 3
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Nitin M Sir (physics-iitjee.blogspot.com)
Nitin M Sir (physics-iitjee.blogspot.com) AIEEE−2011−8
45.
(3i$ + k$ ) 10
r If a =
1
(
(1) -3 46.
(2) 5
(
) (
(3) 3
) (
)
(4) -5
Equation of the ellipse whose axes are the axes of coordinates and which passes through the point 2 is ( −3, 1) and has eccentricity 5 (1) 5x 2 + 3y 2 − 48 = 0
47.
)
r 1 r r r r r r and b = 2i$ + 3j$ − 6k$ , then the value of 2a − b . a × b × a + 2b is 7
(2) 3x 2 + 5y 2 − 15 = 0
(3) 5x 2 + 3y 2 − 32 = 0
(4) 3x 2 + 5y 2 − 32 = 0
Let I be the purchase value of an equipment and V ( t ) be the value after it has been used for t years. dV ( t )
= −k ( T − t ) , where k > 0 is a dt constant and T is the total life in years of the equipment. Then the scrap value V ( T ) of the equipment is
The value V ( t ) depreciates at a rate given by differential equation
(1) I − 48.
49.
kT 2 2
(2) I −
2
2
(3) e −kT
(2) a = 2c
(3) a = 2c
(2) P ( C | D ) < P ( C )
(3) P ( C | D ) =
P (D )
P (C)
(4) P ( C | D ) = P ( C )
The number of values of k for which the linear equations 4x + ky + 2z = 0 ; kx + 4y + z = 0 ; 2x + 2y + z = 0 possess a non-zero solution is (1) 2 (2) 1 (3) zero (4) 3 Consider the following statements P : Suman is brilliant Q : Suman is rich R : Suman is honest The negation of the statement “Suman is brilliant and dishonest if and only if Suman is rich” can be expressed as (2) ~ Q ↔~ P ∧ R (3) ~ (P∧ ~ R ) ↔ Q (4) ~ P ∧ ( Q ↔~ R ) (1) ~ Q ↔ (P∧ ~ R )
(
53.
(4) 2 a = c
If C and D are two events such that C ⊂ D and P (D ) ≠ 0 , then the correct statement among the (1) P ( C | D ) ≥ P ( C )
52.
I k
The two circles x 2 + y 2 = ax and x 2 + y 2 = c 2 ( c > 0 ) touch each other if
following is
51.
(4) T 2 −
r r r r r r r r The vector a and b are not perpendicular and c and d are two vectors satisfying: b × c = b × d and rr r a.d = 0 . Then the vector d is equal to rr rr rr rr r a.c r r a.c r r b.c r r b.c r (2) b + r r c (3) c − r r b (4) b − r r c (1) c + r r b a.b a.b a.b a.b
(1) a = c 50.
k (T − t)
)
The shortest distance between line y − x = 1 and curve x = y 2 is (1)
3 2 8
(2)
8 3 2
(3)
4 3
(4)
3 4
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Nitin M Sir (physics-iitjee.blogspot.com)
Nitin M Sir (physics-iitjee.blogspot.com) AIEEE−2011−9
54.
If the mean deviation about the median of the numbers a, 2a, …, 50a is 50, then a equals (1) 3
55.
Statement – 1 :
(2) 4
(3) 5
(4) 2
The point A (1, 0, 7 ) is the mirror image of the point B (1, 6, 3 ) in the line
x y −1 z − 2 = = . 1 2 3 x y −1 z − 2 bisects the line segment joining A (1, 0, 7 ) and B (1, 6, 3 ) . Statement – 2 : The line: = = 1 2 3 (1) Statement – 1 is true, Statement–2 is true; Statement–2 is not a correct explanation for Statement – 1 (2) Statement – 1 is true, Statement– 2 is false. (3) Statement – 1 is false, Statement– 2 is true. (4) Statement – 1 is true, Statement – 2 is true; Statement – 2 is a correct explanation for Statement – 1 56.
Let A and B be two symmetric matrices of order 3. A (BA ) and ( AB ) A are symmetric matrices. Statement – 1 : AB is symmetric matrix if matrix multiplication of A and B is commutative. Statement – 2 : (1) Statement – 1 is true, Statement – 2 is true; Statement – 2 is not a correct explanation for Statement –1 (2) Statement – 1 is true, Statement– 2 is false. (3) Statement – 1 is false, Statement– 2 is true. (4) Statement – 1 is true, Statement – 2 is true; Statement – 2 is a correct explanation for Statement – 1
57.
If ω ( ≠ 1) is a cube root of unity, and (1 + ω) = A + Bω . Then ( A, B ) equals 7
(1) (1, 1)
58.
59.
60.
(2) (1, 0)
(3) (-1, 1)
(4) (0, 1)
sin ( p + 1) x + sin x , x0 x3 / 2 is continuous for all x in R, is 5 1 3 1 1 3 1 3 (1) p = , q = (2) p = − , q = (3) p = , q = (4) p = , q = − 2 2 2 2 2 2 2 2 1 and the positive x-axis is The area of the region enclosed by the curves y = x, x = e, y = x 3 5 1 square units (3) square units (4) square units (1) 1 square units (2) 2 2 2
5π , define f ( x ) = For x ∈ 0, 2
x
∫
t sin t dt . Then f has
0
(1) local minimum at π and 2π (2) local minimum at π and local maximum at 2π (3) local maximum at π and local minimum at 2π (4) local maximum at π and 2π
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Nitin M Sir (physics-iitjee.blogspot.com)
Nitin M Sir (physics-iitjee.blogspot.com) AIEEE−2011−10
PART C: CHEMISTRY 61.
Among the following the maximum covalent character is shown by the compound : (1) SnCl2 (2) AlCl3 (3) MgCl2 (4) FeCl2
62.
The presence or absence of hydroxyl group on which carbon atom of sugar differentiates RNA and DNA ? (2) 3rd (3) 4th (4) 1st (1) 2nd
63.
Trichloroacetaldehyde was subjected to Cannizzaro’s reaction by using NaOH. The mixture of the products contains sodium trichloroacetate and another compound. The other compound is : (1) Trichloromethanol (2) 2, 2, 2-Trichloropropanol (3) Chloroform (4) 2, 2, 2-Trichloroethanol
64.
Sodium ethoxide has reacted with ethanoyl chloride. The compound that is produced in the above reaction is : (1) 2-Butanone (2) Ethyl chloride (3) Ethyl ethanoate (4) Diethyl ether
65.
The reduction potential of hydrogen half cell will be negative if : (1) p (H2 ) = 1 atm and H+ =1.0 M (2) p (H2 ) =2 atm and H+ =1.0 M (4) p (H2 ) =1 atm and H+ = 2.0 M (3) p (H2 ) = 2 atm and H+ = 2.0 M
66.
The strongest acid amongst the following compounds is : (1) HCOOH (2) CH3CH2CH ( Cl ) CO2H (4) CH3COOH
(3) ClCH2CH2CH2COOH 67.
The degree of dissociation ( α ) of a weak electrolyte, A xB y is related to van’t Hoff factor (i) by the expression : i −1 (1) α = x + y +1
(2) α =
x + y −1 i −1
(3) α =
x + y +1 i −1
(4) α =
i −1
( x + y − 1)
68.
`a’ and `b’ are van der Waals’ constants for gases. Chlorine is more easily liquefied than ethane because (1) a and b for Cl2 < a and b for C2H6 (2) a for Cl2 < a for C2H6 but b for Cl2 > b for C2H6 (3) a for Cl2 > a for C2H6 but b for Cl2 < b for C2H6 (4) a and b for Cl2 > a and b for C2H6
69.
A vessel at 1000 K contains CO2 with a pressure of 0.5 atm. Some of the CO2 is converted into CO on the addition of graphite. If the total pressure at equilibrium is 0.8 atm, the value of K is (1) 3 atm (2) 0.3 atm (3) 0.18 atm (4) 1.8 atm
70.
Boron cannot form which one of the following anions ? (1) BH−4
71.
(2) B ( OH)4 −
(3) BO2−
(4) BF63−
Which of the following facts about the complex Cr (NH3 )6 Cl3 s wrong ? (1) The complex is paramagnetic (2) The complex is an outer orbital complex (3) The complex gives white precipitate with silver nitrate solution (4) The complex involves d2 sp3 hybridization and is octahedral in shape.
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Nitin M Sir (physics-iitjee.blogspot.com)
Nitin M Sir (physics-iitjee.blogspot.com) AIEEE−2011−11
72.
Ethylene glycol is used as an antifreeze in a cold climate. Mass of ethylene glycol which should be added to 4 kg of water to prevent it from freezing at −6o C will be : [ K f for water = 1.86 K kg mol −1 , and molar mass of ethylene glycol = 62g mol −1 ) (1) 204.30g (2) 400.00 g (3) 304.60 g (4) 804.32g
73.
Which one of the following order represents the correct sequence of the increasing basic nature of the given oxides ? (1) MgO < K 2O < Al2O3 < Na2O (2) Na2O < K 2O < MgO < Al2O3 (4) Al2O3 < MgO < Na2O < K 2O (3) K 2O < Na2O < Al2O3 < MgO
74.
The rate of a chemical reaction doubles for every 10o C rise of temperature. If the temperature is raised by 50o C , the rate of the reaction increases by about : (1) 24 times (2) 32 times (3) 64 times (4) 10 times
75.
The magnetic moment (spin only) of [NiCl4 ]
2−
(1) 5.46 BM 76.
(2) 2.83 BM
is (3) 1.41 BM
(4) 1.82 BM
The hybridization of orbitals of N atom in NO3− ,NO2+ and NH+4 are respectively : (1) sp2 , sp, sp3
(2) sp, sp3 , sp2
(3) sp2 , sp3 , sp
(4) sp, sp2 , sp3
77.
In context of the lanthanoids, which of the following statements is not correct ? (1) All the members exhibit +3 oxidation state (2) Because of similar properties the separation of lanthanoids is not easy. (3) Availability of 4f electrons results in the formation of compounds in +4 state for all the members of the series. (4) There is a gradual decrease in the radii of the members with increasing atomic number in the series.
78.
A 5.2 molal aqueous solution of methyl alcohol, CH3OH , is supplied. What is the mole fraction of methyl alcohol in the solution ? (1) 0.190 (2) 0.086 (3) 0.050 (4) 0.100
79.
Which of the following statement is wrong ? (1) Nitrogen cannot form dπ - pπ bond. (2) Single N- N bond is weaker than the single P – P bond, (3) N2O4 has two resonance structures (4) The stability of hydrides increases from NH3 to BiH3 in group 15 of the periodic table
80.
The outer electron configuration of Gd (Atomic No : 64 is : (1) 4f 8 5d0 6s2 (2) 4f 4 5d4 6s2 (3) 4f 7 5d1 6s2
81.
(4) 4f 3 4d5 6s2
Which of the following statements regarding sulphur is incorrect ? (1) The vapour at 200o C consists mostly of S8 rings (2) At 600o C the gas mainly consists of S2 molecules (3) The oxidation state of sulphur is never less than +4 in its compounds (4) S2 molecule is paramagnetic.
82.
The structure of IF7 is : (1) trigonal bipyramid (2) octahedral
(3) pentagonal bipyramid
(4) square pyramid
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Nitin M Sir (physics-iitjee.blogspot.com) AIEEE−2011−12
83.
Ozonolysis of an organic compound gives formaldehyde as one of the products. This confirms the presence of : (1) a vinyl group (2) an isopropyl group (3) an acetylenic triple bond (4) two ethylenic double bonds
84.
A gas absorbs a photon of 355 nm and emits at two wavelengths. If one of the emissions is at 680 nm, the other is at : (1) 325 nm (2) 743 nm (3) 518 nm (4) 1035 nm
85.
Silver Mirror test is given by which one of the following compounds ? (1) Acetone (2) Formaldehyde (3) Benzophenone
(4) Acetaldehyde
86.
Which of the following reagents may be used to distinguish between phenol and benzoic acid ? (1) Tollen’s reagent (2) Molisch reagent (3) Neutral Fe Cl3 (4) Aqueous NaOH
87.
Phenol is heated with a solution of mixture of KBr and KBrO3. The major product obtained in the above reaction is (1) 3-Bromophenol (2) 4-Bromophenol (3) 2, 4, 6- Tribromophenol (4) 2-Bromophenol
88.
In a face centred cubic lattice, atom A occupies the corner positions and atom B occupies the face centre positions. If one atom of B is missing from one of the face centred points, the formula of the compound is : (1) AB2 (2) A 2B3 (3) A 2B5 (4) A 2B
89.
The entropy change involved in the isothermal reversible expansion of 2 moles of an ideal gas from a volume of 10 dm3 to a volume of 100 dm3 at 27o C is : (1) 35.8J mol −1K −1 (2) 32.3J mol−1K −1 (3) 42.3J mol−1K −1 (4) 38.3J mol−1K −1
90.
Identify the compound that exhibits tautomerism. (1) Lactic acid (2) 2-Pentanone (3) Phenol
(4) 2- Butene
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Nitin M Sir (physics-iitjee.blogspot.com) AIEEE−2011−13
READ THE FOLLOWING INSTRUCTIONS CAREFULLY 1.
The candidates should fill in the required particulars on the Test Booklet and Answer Sheet (Side–1) with Blue/Black Ball Point Pen.
2.
For writing/marking particulars on Side-2 of the Answer Sheet, use Blue/Black Ball Point Pen only.
3.
The candidates should not write their Roll Numbers anywhere else (except in the specified space) on the Test Booklet/Answer Sheet.
4.
Out of the four options given for each question, only one option is the correct answer.
5.
For each incorrect response, one-fourth (1/4) of the total marks allotted to the question would be deducted from the total score. No deduction from the total score, however, will be made if no response is indicated for an item in the Answer Sheet.
6.
Handle the Test Booklet and Answer Sheet with care, as under no circumstances (except for discrepancy in Test Booklet Code and Answer Sheet Code), will another set be provided.
7.
The candidates are not allowed to do any rough work or writing work on the Answer Sheet. All calculations/writing work are to be done in the space provided for this purpose in the Test Booklet itself, marked ‘Space for Rough Work’. This space is given at the bottom of each page and in 4 pages (Pages 20 – 23) at the end of the booklet.
8.
On completion of the test, the candidates must hand over the Answer Sheet to the Invigilator on duty in the Room/Hall. However, the candidates are allowed to take away this Test Booklet with them.
9.
Each candidate must show on demand his/her Admit Card to the Invigilator.
10.
No candidate, without special permission of the Superintendent or Invigilator, should leave his/her seat.
11.
The candidates should not leave the Examination Hall without handing over their Answer Sheet to the Invigilator on duty and sign the Attendance Sheet again. Cases where a candidate has not signed the Attendance Sheet a second time will be deemed not to have handed over the Answer Sheet and dealt with as an unfair means case. The candidates are also required to put their left hand THUMB impression in the space provided in the Attendance Sheet.
12.
Use of Electronic/Manual Calculator and any Electronic Item like mobile phone, pager etc. is prohibited.
13.
The candidates are governed by all Rules and Regulations of the Board with regard to their conduct in the Examination Hall. All cases of unfair means will be dealt with as per Rules and Regulations of the Board.
14.
No part of the Test Booklet and Answer Sheet shall be detached under any circumstances.
15.
Candidates are not allowed to carry any textual material, printed or written, bits of papers, pager, mobile phone, electronic device or any other material except the Admit Card inside the examination hall/room.
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Nitin M Sir (physics-iitjee.blogspot.com) AIEEE−2011−14
SOLUTIONS PART A PHYSICS 1. Sol.
1
(
)
b a Wave moving in − ve x –direction. y ( x,t ) = e −
ax+ bt
2.
1
Sol.
Diameter of wire =
3. Sol.
2 Mg –T = Ma T × R = Iα = 1 T = Ma 2
2
V=
1 × 52 = 0.52mm = 0.052cm 100 α
….. (1)
R
1 MR 2 α 2
T
( a = αR )
From (1) and (2) a =
….. (2) 2g 3
a Mg
4. Sol.
3 W = T × ∆A = T × 8π r22 − r12 = 0.4π mJ
5. Sol.
3 τ=0 Angular momentum is conserve Iω I1ω1 = I2 ω2 ⇒ ω2 = 1 1 I2 I2 first decreases and then increases ∴ω first increases and then decreases.
6. Sol.
4 φ1 = 0
(
φ2 =
)
X0+A
π 2 X0
7.
3
Sol.
Position of the null point from mass m, x =
r 4m 1+ m
=
A
r 3
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8. Sol.
4 At any instant of separation between charges is x. Q2 x equilibrium condition = K 2 = ω 2l x ⇒ Q2 = Cx 3 dQ dx ⇒ 2Q = C3x 2 dt dt 3/2 dx x ⇒ ∝ 2 ∝ x −1/ 2 dt x
9. Sol.
3 E = BH lV = 0.15mV
10.
1 dv = −2.5 v dt Integrating the above equation. ⇒ 2 v = −2.5t + C at t = 0,v = 6.25 ⇒ C = 5
Sol.
at v = 0 ⇒ t =
5 = 2s 2.5
11.
1
Sol.
Charge oscillates simple harmonic motion q = q0 sin ωt, U = q=
q0
⇒ ωt =
1 q2 2C
π 4
2 T 2π π LC = LC ⇒t= = 8 8 4
12. Sol.
Normal to the plane is z –axis A 10 1 cos θ1 = z = = , θ1 = 60 A 20 2 µ1 sin θ1 = µ 2 sin θ2 ⇒ 2 ×
13. Sol.
3 = 3 sin θ2 ⇒ θ2 = 450 2
4 ∧ ∧ µ0 di cos i sin j − θ − θ 2πR T di = Rdθ πR I = dθ π
y
→
dB =
→
dB =
dθ θ
θ
x
∧ ∧ µ0I cos i sin j − θ − θ 2 2π R
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µ0I ∧ j π 2R
→
B=−
14. Sol.
3 W = ∆U 1 mv 2 = nCv dT 2 m R dT = M γ −1 M ( γ − 1) v 2
dT =
15. Sol.
K
3 Energy of simple harmonic oscillator is constant. 1 1 ⇒ Mω2 A12 = ( m + M) ω2 A 22 2 2 A12 M + m = M A 22 ∴
16. Sol.
2R
A1 M+m = A2 M
3 Equation of continuity ⇒ ( a × v ) top = ( a × v ) bottom v b2 − ( 0.4 ) = 2 × 9.8 × 0.2 v 2 − u2 = 2gh is used 2
vb = 2m/s (nearly) π 8 × 10 −3 × 0.4 = πd2 × 4 d ≈ 3.6 × 10 −3 m
17. Sol.
1 Since ionospheric properties change with time, these signals are in general less stable than ground wave signals.
18. Sol.
1 Data ⇒ n,k,t1 + n2kT2 + n3kT3 = ( n1 + n2 + n3 ) kT ∴T =
19. Sol.
n1T1 + n2 T2 + n3 T3 n1 + n2 + n3
1 r×F=I×α 2 20t − 5t 2 = 10α ⇒ α = 4t − t 2
(
)
dω = 4t − t 2 dt
(
)
dω = 4t 2 − t 2 dt 3
t (on integration) 3 ω = 0 ⇒ t = 6s
ω = 2t 2 −
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ω=
dθ t3 = 2t 2 − dt 3
t3 dθ = 2t 2 − dt 3 2t 3 t 4 − (on integration) 3 12 θ ( in 6s ) = 36rad
⇒θ=
⇒ 2πn = 36 36 n= =< 6 2π
20. Sol.
2 Vc = E 1 − e − t / Rc
(
)
120 3 = 200 5 5 ⇒R= = 2.7 × 106 Ω −6 1.84 × 10
1 − e − t / Rc =
21.
4
Sol.
η1 = η2 = ⇒
T1 − T2 1 = T1 6
T1 − ( T2 − 62 ) T1
=
1 3
T1 − T2 62 1 + = T1 T1 3
1 62 1 + = 6 T1 3 62 1 = T1 6
∴ T1 = 62 × 6 = 372K T1 − T2 1 = T1 6 1−
T2 1 = T1 6
T2 5 = 372 6 ⇒ T2 = 310K 22. Sol.
1 R ∝ l 2 (for a given volume) ∆R 2 ∆l ⇒ %= % l R Thus when wire is stretched by 0.1% resistance increases by 0.2%
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23. Sol.
1 As light enters from air to glass it suffers a phase change on π and therefore at centre there will be destructive interference.
24.
1 1 1 1 + = v u f 1 dv 1 du − 2 − =0 v dt u2 dt
Sol.
dv v 2 du =− 2 dt u dt f = 20 cm 1 1 1 + = u −280 20 280 cm ⇒v= 15 2
280 vI = − × 15 15 × 280 1 m/s = 15
25.
2
Sol.
En = −13.6
Z2 n2
9 = −122.4eV 1 9 ELi ++ + = −13.6 × = −13.6eV 9 ∆E = −13.6 − ( −122.4 ) ELi + + = −13.6 ×
= 108.8 eV 26. Sol.
3 Potential inside (φ) = ar2 + b δv ∴ Er = − = −2ar δr Electric field inside uniformly charged solid volume varies with ‘r’. So charge density is constant φnet = ( −2ar ) 4πr 2 = −8πar 3 −8πar = 3
σ×
4 3 πr 3 ε0
∴ σ = −6aε0
27.
1
Sol.
Max. range =
u2 v2 i.e., (radius of circle) g g 2
v2 πv 4 Area occupied = π = 2 g g
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28. Sol.
1 ∆Q = ∆U + ∆W (ignoring expansion) ∆U = ms∆T = 0.1× 4.184 × 20 = 8.368kJ
29. Sol.
2 t 1 = 20 minutes 2
N = N0 e − λt2
λt1 = ln3
2 1 N0 = N0 e − λt2 t1 = ln3 3 λ 2 N0 = N0 e − λt2 3 1 3 t 2 = ln λ 2 1 3 t 2 − t1 = ln − ln3 λ 2 1 1 0.693 = ln = λ 2 λ = 20 min
30. Sol.
3 KEmax = hυ − hυ0 hυ − hυ0 = e × ∆v hυ hυ0 − e e ‘υ’ is doubled KEmax = 2hυ − hυ0 2hυ hυ0 V0′ = ( ∆V )′ = − e e KEmax may not be equal to 2 KEmax V0 =
⇒
V0′ may not equal to 2 V0
KE max = hυ - hυ0 hυ hυ0 V= − e e
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PART B: MATHEMATICS 31.
2
Sol: Q R
L2 = 0
• y+2=0 P
O
L1 = 0 L3 = 0
P ( −2, − 2 ) ; Q = (1, − 2 )
Equation of angular bisector OR is
(
) (
5 +2 2 x =
)
5− 2 y
∴ PR : RQ = 2 2 : 5
32.
4
Sol:
A = sin2 x + cos4 x =
33.
2
Sol:
1 − x − x 2 (1 − x ) = (1 − x )6 1 − x 2
7 + cos 4x 3 ⇒ ≤ A ≤1 8 4
6
(
)
6
= 6 C0 − 6 C1x + 6 C2 x 2 − 6 C3 x 3 + 6 C4 x 4 − 6 C5 x 5 + 6 C6 x 6 × 6 C0 − 6 C1x 2 + 6 C2 x 4 − 6 C3 x 6 + .... 7 6 6 6 6 6 6 Coefficient of x = C1 C3 − C3 C2 + C5 C1 = 120 − 300 + 36 = −144
34.
4
Sol:
lim
x →2
lim
2sin2 ( x − 2 ) x−2 2 sin ( x − 2 )
x−2 R.H.L. = 2 , L.H.L. = − 2 Limit does not exist. x →2
35. Sol:
4 (n −1)
C(r −1) =
(10 −1)
C(4 −1) = 9 C3
Statement 1 is correct Statement 2 is also correct From 9 we can select 3 in 9 C3 ways. It is correct explanation.
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36.
3
Sol:
d dx d 1 1 d dy =− = 2 dy dy dy dy dy dy dx dx dx
dy = − dx
37. Sol:
−2
d2 y dy −3 1 d dy = − 2 dy dx dx dx dx dx
4 dy dy = y+3⇒ = dx dx y+3 ln ( y + 3 ) = x + c x=0⇒y=2 ⇒ ln5 = 0 + c c = ln5 ln ( y + 3 ) = x + ln5 y + 3 = e x +ln 5 ⇒ y + 3 = eln 2 +ln 5
y + 3 = 10 ⇒ y = 7
38. Sol:
2 x − y is an integer x − x = 0 is an integer ⇒ A is Reflexive x − y is an integer ⇒ y − x is an integer ⇒ A is symmetric x − y, y − z are integers
As sum of two integers is an integer. ⇒ ( x − y ) + ( y − z ) = x − z is an integer ⇒ A is transitive. Hence statement – 1 is true. x Also = 1 is a rational number ⇒ B is reflexive x x y = α is rational ⇒ need not be rational y x 1 0 is rational ⇒ is not rational 0 1 Hence B is not symmetric ⇒ B is not an equivalence relation.
i.e.,
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39.
4
Sol:
I=8
1
∫ 0
π 4
=8
∫ 0 π 4
log (1 + x ) 1 + x2
dx
log (1 + tan θ ) 1 + tan2 θ
sec 2 θ dθ ( let x = tan θ ) π 4
π 4
π 4
1 − tan θ π = 8 log 1 + tan − θ dθ = 8 log 1 + dθ = 8 log2 dθ − 8 log (1 + tan θ ) dθ 4 1 + tan θ 0 0 0 0
∫
∫
∫
∫
π −I 4 2I = 2π log2 I = π log2 = 8log2
40. Sol:
3 Suppose roots are 1 + pi, 1 + qi Sum of roots 1 + pi + 1 + qi = −α which is real ⇒ roots of 1 + pi, 1 − pi
Product of roots = β = 1 + p2 ∈ (1, ∞ ) p ≠ 0 since roots are distinct.
41. Sol:
2 n=5 Success = p Failure = q
P (at least one failure) ≥ 1 – P (no failure) ≥
31 32
31 32
31 32 31 1 − 5 C5p5 ≥ 32 1 −p5 ≥ − 32 1 p5 ≤ 32 1 p≤ 2 1 p ∈ 0, 2 1− P ( x = 5) ≥
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42. Sol:
3
1
2
3
4
200 200 200 240 Sum = 11040 120 + 80 + 160 + 40 + 200 + 240 + … = 11040 n 2a + ( n − 1) d + 80 + 40 = 11040 2 n 240 + ( n − 1) 40 = 10920 2 n 6 + n − 1 = 546
5
6
……
280
…….
……..
n ( n + 5 ) = 546
n = 21
43.
2
1
Sol:
⇒ x − x > 0 ⇒ x > x ⇒ x is negative
x −x
x ∈ ( −∞, 0 )
44.
4
Sol:
cos θ =
5 14 3
sin θ =
14 1 + 4 + 3λ
sin θ =
1 + 4 + λ2 1 + 4 + 9 3 5 + 3λ 2 = ⇒λ= 2 3 14 5 + λ 14
45. Sol:
4
( 2a − b).{(a × b) × (a + 2b )} = ( 2a − b ).{a.(a + 2b) b − b.(a + 2b ) a} = −5 ( a ) ( b ) + 5 ( a.b ) = −5 2
2
2
46.
4
Sol:
2 3 3a2 b2 = a 2 1 − e 2 = a 2 1 − = a 2 = 5 5 5
(
x2 a
2
+
y2 2
b 32 a2 = 3 32 b2 = 5
)
=1⇒
9 a
2
+
5 3a2
=1
∴ Required equation of ellipse 3x 2 + 5y 2 − 32 = 0 FIITJEE (Hyderabad Classes) Limited. 5-9-14/B, Saifabad, (Opp. Secretariat) Hyderabad. 500 063. Phone: 040-66777000 – 03 Fax: 040-66777004
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47. Sol:
1 dV = −k ( T − t ) ⇒ dV = −k ( T − t ) dt dt Integrate V=
−k ( T − t )
2
( −2 )
+c ⇒ V =
k (T − t)
2
2
+c
at t = 0 ⇒ V = I I=
kT 2 kT 2 kT 2 + c ⇒ c = I− ⇒ c = V (T) = I − 2 2 2
48.
3
Sol:
b×c = b×d
( ) ( ) ⇒ ( a.c ) b − ( a.b ) c = ( a.d) b − ( a.b ) d ⇒ ( a.c ) b − ( a.b ) c = − ( a.b ) d
⇒ a× b×c = a× b×d
a.c ∴d = c − b a.b
49. Sol:
50. Sol:
51. Sol:
52. Sol:
1 a c1 = , 0 ; c 2 = ( 0, 0 ) 2 a r1 = ; r2 = c 2 a a c1c 2 = r1 − r2 ⇒ = c − ⇒ c = a 2 2
1
C P (C ∩ D) C ∩ D = C ⇒ P (C ∩ D) = P (C) ⇒ P = ≥ P (C) P (D ) D
1 4 k 2 k 4 1 = 0 ⇒ k 2 − 6k + 8 = 0 ⇒ k = 4, 2 2 2 1 1 ~ ( P∧ ~ R ) ↔ Q =~ Q ↔ (P∧ ~ R )
{
} {
}
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53.
1
Sol:
P = y2 , y
(
)
Perpendicular distance from P to x − y + 1 = 0 is
y2 − y + 1 2
y 2 − y + 1 > 0 ∀y ∈ R ∴ Coefficient y 2 > 0 ∴ Min value =
54. Sol:
2 1 n
1 4ac − b2 4a 2
3 = 4 2
∑ x −A i
25a + 26a = 25.5a 2 1 2 a − 25.5a + 2a − 25.5a = Mean deviation = ( 24.5a + 23.5a ) + ... ( 0.5a ) 50 50 2 (Given) = {312.5a} = 50 50 ⇒ 625a = 2500 ⇒ a = 4
A = Median =
{
55. Sol:
}
{
}
1 B (1,6,3)
1, 2, 3
•A (1,0,7)
Statement – 1 : AB is perpendicular to given line and mid point of AB lies on line Statement – 2 is true but it is not correct explanation as it is bisector only. If it is perpendicular bisector then only statement – 2 is correct explanation. 56. Sol:
1 A T = A, BT = B
( A (BA ) ) = (BA )T A T = ( A TBT ) A = ( AB ) A = A (BA ) T ( ( AB ) A ) = A T ( AB )T = A (BT A T ) = A (BA ) = ( AB ) A T
∴ Statement – 1 is correct Statement – 2
( AB )T
= BT A T = BA = AB
(Q AB is commutative)
Statement – 2 is also correct but it is not correct explanation of Statement – 1
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57. Sol:
1 1+ ω = −ω2
(1 + ω)7 = ( −ω2 ) 58.
7
= −ω14 = −ω2 = 1 + ω = A + Bω ⇒ ( A, B ) = (1, 1)
2 lim
sin ( p + 1) + sin x
= q = lim
x + x2 − x
x3 / 2 1 lim ( p + 1) cos ( p + 1) x + cos x = q = x →0 2 1 3 1 ⇒ p + 1+ 1 = ⇒ p = − ; q = 2 2 2 x
x →0
59. Sol:
x →0
2
1
∫
Area = x dx + 0
60. Sol:
e
1
O
e
1
1
3
∫ x dx = 2 + 1 = 2 1
3 f ' ( x ) = x sin x 5π Given x ∈ 0, 2 f ' ( x ) changes sign from +ve to –ve at π
f ' ( x ) changes sign from -ve to +ve at 2π f has local max at π , local min at 2π
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PART C: CHEMISTRY 61. Sol :
(2) Greater charge and small size of cation cause more polarization and more covalent is that compound
62. Sol :
(1) In RNA, the sugar is β − D − Ribose, where as in DNA the Sugar is β -D-2-deoxy Ribose
63.
(4)
Sol :
64. Sol :
65. Sol :
OH( − ) 2CCl3 CHO → CCl3 COONa + CCl3 CH2OH Cannizaro reaction is a disproportionation reaction One aldehyde molecule is oxidized to salt of the carboxylic Acid, other one is reduced to Alcohol. So the compound is CCl3 CH2OH IUPAC Name is 2, 2, 2, - Trichloro ethanol (3) ( −) ⊕ C2H5 O N a + CH3 − C − Cl || O
→ CH3 − C − O − C2H5 Ethyl ethanoate || O
(2) 2H+ + 2e− → H2 (g) P H2 E = Eo − 0.059 log H+ 2
(here E is –ve when P > H+ 2 ) H2 −0.0591 −.0591 2 = log10 = × .3010 = negative value 2 2 1
66. Sol :
(2) Electron releasing groups (Alkyl groups) de stabilizes conjugate base. The +I effect of C3H7 is less than - I effect of Cl Ka of HCOOH is 17.9 × 10−5 K a of CH3 CH2 CH− COOH is 139 × 10 −5 | Cl
67. Sol :
(4) i = 1- α + nα = 1 + α ( n − 1) i −1 =α n −1 A xB y → xA + y + yB− x
n = x+y So α =
i −1 x + y −1
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68. Sol :
69.
(3) a b for ethane a = 5.49, b=0.0638 for Cl2 a = 6.49, b = 0.0562
ease of liquefaction ∝
(4) CO2 (g) + C
Sol :
2CO(g)
Initial moles p O Equilibriumm moles p-x 2x Total pressure at equilibrium = 0.8 atm ; Total no.of moles = p + x. 0.5 p Therefore p ∝ n ; = ⇒ x = 0.3 0.8 p + x Kp =
70. Sol :
2 PCO 0.6 × 0.6 = = 1.8 atm PCO2 0.2
(4) As Boron has only four orbitals in the valence shell ( i.e. 2s, 2px, 2py & 2pz) it can show a maximum valency of four only. Therefore [BF6 ]
3−
is not possible
71. Sol :
(2) Cr (NH3 )6 Cl3 involves d2 sp3 hybridization and it is an inner orbital complex.
72.
(4)
Sol :
∆Tf = K f × m = K f ×
w 2 × 1000 w 1 × m2
w1 & w 2 = wt of solvent & solute respecting m2 = mw of solute
(
)
∆Tf = 0o − −60 = 6 = 1.86 ×
Therefore w 2 = 800g
w 2 × 1000 4000 × 62
73. Sol :
(4) Across a period metallic strength decreases & down the group it increases
74. Sol :
(2) Temperature coefficient µ =2; ∆T
µ 10 =
k2 ; k1
50
210 = 25 = 32 =
k2 k1
Therefore 32 k1 = k 2
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75. Sol :
(2)
In [NiCl4 ] , n = 2 2−
n ( n + 2 ) BM
µ =
2 ( 2 + 2 ) = 2.82BM
= 76. Sol :
(1)
77. Sol :
(3) The general o.s of lanthanides is +3, only few elements exhibit +4 o.s.
78.
(2)
Sol :
Molefraction of solute ( X2 ) in aqueous solution =
79. Sol :
(4) Stability of hydrides decreases down the group from NH3 to BiH3 as M-H bond energy decreases.
80.
(3)
81. Sol :
(3) `S’ can exhibit a minimum oxidation state of -2 (Ex. H2 S )
82. Sol :
(3) In IF7 , I undergoes sp3 d3 hybridisation
83. Sol :
(1) Vinyl group CH2 = CH −
m 1000 m+ 18 5.2 = 0.09 = 1000 5.2 + 18
on ozonolosys give formaldehyde 84. Sol :
(2) 1 λ absorbed ⇒
=
1 1 + λ1 λ 2
1 1 1 = + 355 680 λ 2
⇒ λ 2 = 742.8 ≅ 743 nm
85. Sol :
(2 , 4) Formaldehyde and Acetaldehyde can be oxidized by tollen’s reagent to give silver mirror.
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86. Sol :
(3) Phenol gives violet coloured comlex compound with neutral FeCl3 , benzoic acid gives pale dull yellow ppt. with neutral FeCl3
87. Sol :
(3) In acidic medium, KBr + KBrO3 in turn produces Br2 . Phenol reacts with Br2 (aq) to give 2, 4, 6trinitrophenol
88.
(3)
Sol :
1 ×8 =1 8 1 5 Effective no.of B atoms = × 5 ( One is missing) = 2 2 Therefore formula is A1B 5 = A 2B5
Effective no.of A atoms =
2
89. Sol :
(4) For an ideal gas, for isothermal reversible process, v ∆S = 2.303 nR log 2 v1 100 = 2.303 ×2 × 8.314 × log 10 = 38.3 J mol−1.k −1
90. Sol :
2, (2, 3) both 2-pentanone, phenol can exhibit tautomerism
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AIEEE–2012 (Set – C) IMPORTANT INSTRUCTIONS 1.
Immediately fill in the particulars on this page of the Test Booklet with Blue/Black Ball Point Pen. Use of pencil is strictly prohibited.
2.
The Answer Sheet is kept inside this Test Booklet. When you are directed to open the Test Booklet, take out the Answer Sheet and fill in the particulars carefully.
3.
The test is of 3 hours duration.
4.
The Test Booklet consists of 90 questions. The maximum marks are 360.
5.
There are three parts in the question paper A, B, C consisting of Mathematics, Physics and Chemistry having 30 questions in each part of equal weightage. Each question is allotted 4 (four) marks for each correct response.
6.
Candidates will be awarded marks as stated above in instruction No.5 for correct response of each question. 1/4 (one fourth) 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.
7.
There is only one correct response for each question. Filling up more than one response in each question will be treated as wrong response and marks for wrong response will be deducted accordingly as per instruction 6 above.
8.
Use Blue/Black Ball Point Pen only for writing particulars/marking responses on Side-1 and Side-2 of the Answer Sheet. Use of pencil is strictly prohibited.
9.
No candidate is allowed to carry any textual material, printed or written, bits of papers, pager, mobile phone, any electronic device, etc., except the Admit Card inside the examination hall/room.
10.
Rough work is to be done on the space provided for this purpose in the Test Booklet only. This space is given at the bottom of each page and in 3 pages (Pages 21 – 23) at the end of the booklet.
11.
On completion of the test, the candidate must hand over the Answer Sheet to the Invigilator on duty in the Room/Hall. However, the candidates are allowed to take away this Test Booklet with them.
12.
The CODE for this Booklet is C. Make sure that the CODE printed on Side-2 of the Answer Sheet is the same as that on this booklet. In case of discrepancy, the candidate should immediately report the matter to the Invigilator for replacement of both the Test Booklet and the Answer Sheet.
13.
Do not fold or make any stray marks on the Answer Sheet.
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Nitin M Sir (physics-iitjee.blogspot.com)
Nitin M Sir (physics-iitjee.blogspot.com)
AIEEE− −2012− −2
1. 1. Sol.
sin x
–sin x
The equation e –e – 4 = 0 has (1) infinite number of real roots (3) exactly one real root 2 sin x –sin x sin x e –e =4 e =t 1 t– =4 t 2
t – 4t – 1 = 0 t=
4±2 5 2
=2±
sin x
= 2 + 5 not possible
sin x
= 2 – 5 not possible
e e
4 ± 16 + 4 2
t = 2± 5
sin x
e
2.
t=
(2) no real roots (4) exactly four real roots
1 sin x ≤e ≤e e
–1 ≤ sin x ≤ 1
5
∴ hence no solution
Let aˆ and bˆ be two unit vectors. If the vectors c = aˆ + 2bˆ and d = 5aˆ − 4bˆ are perpendicular to each other, then the angle between aˆ and bˆ is (1)
π 6
2.
3
Sol.
c⋅d = 0
3. Sol.
4.
(3)
π 3
(4)
π 4
2
2
5 a + 6a ⋅ b − 8 b = 0
6a ⋅ b = 3 3.
π 2
(2)
a⋅b =
(a ⋅ b ) = 3π
1 2
A spherical balloon is filled with 4500π cubic meters of helium gas. If a leak in the balloon causes the gas to escape at the rate of 72π cubic meters per minute, then the rate (in meters per minute) at which the radius of the balloon decreases 49 minutes after the leakage began is 9 9 7 2 (1) (2) (3) (4) 9 9 7 2 3 4 v = πr 2 3 After 49 minutes volume = 4500π – 49 (72π) = 972π 4 3 πr = 972π r3 = 729 r=9 3 4 dv 4 dr dr dr 72 2 v = πr3 72π = 4π r 2 = π3r 2 = = 3 dt 3 dt dt dt 4 ⋅ 9 ⋅ 9 9
Statement 1: The sum of the series 1 + (1 + 2 + 4) + (4 + 6 + 9) + (9 + 12 + 16) + ...... + (361 + 380 + 400) is 8000. n
Statement 2: k =1
(k
3
− (k − 1)
3
)=n
3
for any natural number n.
(1) Statement 1 is false, statement 2 is true (2) Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1 (3) Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1 Ltd., FIITJEE House, 29 – A, Kalu Sarai, Sarvapriya Vihar, New Delhi – 110016, Ph: 46106000, 26569493, Fax: 26513942 website: www.fiitjee.com
Nitin M Sir (physics-iitjee.blogspot.com)
Nitin M Sir (physics-iitjee.blogspot.com)
AIEEE− −2012− −3
4. Sol.
5.
(4) Statement 1 is true, statement 2 is false 2 Statement 1 has 20 terms whose sum is 8000 And statement 2 is true and supporting statement 1. th 2 2 2 k bracket is (k – 1) + k(k – 1) + k = 3k – 3k + 1.
5. Sol.
The negation of the statement “If I become a teacher, then I will open a school” is (1) I will become a teacher and I will not open a school (2) Either I will not become a teacher or I will not open a school (3) Neither I will become a teacher nor I will open a school (4) I will not become a teacher or I will open a school 1 ~(~p ∨ q) = p ∧ ~q
6.
If the integral
6.
(1) –1 4
5 tan x dx = tan x − 2
Sol.
= 2 7.
5 tan x dx = x + a ln |sin x – 2 cos x| + k, then a is equal to tan x − 2 (2) –2 (3) 1 (4) 2 5sin x dx sin x − 2cos x
cos x + 2 sin x dx + dx + k sin x − 2cos x
2 ( cos x + 2sin x ) + ( sin x − 2cos x ) sin x − 2cos x
dx
= 2 log |sin x – 2 cos x| + x + k ∴ a = 2 2
2
2
Statement 1: An equation of a common tangent to the parabola y = 16 3x and the ellipse 2x + y = 4 is y = 2x + 2 3 .
4 3 , (m ≠ 0) is a common tangent to the parabola m 2 2 2 4 2 y = 16 3x and the ellipse 2x + y = 4, then m satisfies m + 2m = 24. (1) Statement 1 is false, statement 2 is true (2) Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1 (3) Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1 (4) Statement 1 is true, statement 2 is false 7. 2 x 2 y2 Sol. y2 = 16 3x + =1 2 4 4 3 y = mx + is tangent to parabola m which is tangent to ellipse c2 = a2m2 + b2 48 2 4 2 2 = 2m + 4 m + 2m = 24 m =4 2 m Statement 2: If the line y = mx +
1 0 0 8.
1
3 2 1
0
equal to −1 (1)
1
−1 (2)
1 −1
0 8.
0
Let A = 2 1 0 . If u1 and u2 are column matrices such that Au1 = 0 and Au2 = 1 , then u1 + u2 is
−1 (3) −1
0
0 1 (4) −1
−1
4 Ltd., FIITJEE House, 29 – A, Kalu Sarai, Sarvapriya Vihar, New Delhi – 110016, Ph: 46106000, 26569493, Fax: 26513942 website: www.fiitjee.com
Nitin M Sir (physics-iitjee.blogspot.com)
Nitin M Sir (physics-iitjee.blogspot.com)
AIEEE− −2012− −4
1 0 0 Sol.
A= 2 1 0
3 2 1
a d Let u1 = b ; u2 = e c f 1
1 u1 = −2 1
Au1 = 0
0 0 Au2 = 1 0
u2 =
If n is a positive integer, then
9.
(1) an irrational number (3) an even positive integer 1
Sol.
(
) ( 2n
−
)
3 −1
2n
(
=
(
) −(
3 +1
2n
)
3 −1
2n
is
(2) an odd positive integer (4) a rational number other than positive integers
)
3 +1
2
n
−
(
)
3 −1
2
n
(
= 4+2 3
) − (4 − 2 3 ) n
n
(2 + 3 ) − (2 − 3 ) n
= 2n = 2n
1 u1 + u2 = −1 −1
−2
9.
3 +1
0 1
{
= 2n +1
n
n
C0 2n + n C1 2n−1 3 + n C2 2n − 23 + ⋅ ⋅ ⋅ ⋅ ⋅ −
n
n
C0 2n − n C1 2n −1 3 + n C2 2n− 2 3 − ⋅ ⋅ ⋅ ⋅ ⋅
}
C1 2n−1 3 + n C3 2n −3 3 3 + ⋅ ⋅ ⋅ ⋅ = 2n+1 3 (some integer)
Which is irrational 10.
10. Sol. 11.
11. Sol.
12.
th
th
If 100 times the 100 term of an AP with non zero common difference equals the 50 times its 50 term, th then the 150 term of this AP is th (1) –150 (2) 150 times its 50 term (3) 150 (4) zero 4 100(T100) = 50(T50) 2[a + 99d] = a + 49d a + 149d = 0 T150 = 0 In a ∆PQR, if 3 sin P + 4 cos Q = 6 and 4 sin Q + 3 cos P = 1, then the angle R is equal to π 3π 5π π (1) (2) (3) (4) 4 4 6 6 2 3 sin P + 4 cos Q = 6 ...... (1) 4 sin Q + 3 cos P = 1 ...... (2) From (1) and (2) ∠P is obtuse. 2 2 (3 sin P + 4 cos Q) + (4 sin Q + 3 cos P) = 37 9 + 16 + 24 (sin P cos Q + cos P sin Q) = 37 24 sin (P + Q) = 12 1 5π π sin (P + Q) = P+Q= R= 2 6 6 An equation of a plane parallel to the plane x – 2y + 2z – 5 = 0 and at a unit distance from the origin is (1) x – 2y + 2z – 3 = 0 (2) x – 2y + 2z + 1 = 0 Ltd., FIITJEE House, 29 – A, Kalu Sarai, Sarvapriya Vihar, New Delhi – 110016, Ph: 46106000, 26569493, Fax: 26513942 website: www.fiitjee.com
Nitin M Sir (physics-iitjee.blogspot.com)
Nitin M Sir (physics-iitjee.blogspot.com)
AIEEE− −2012− −5
12. Sol.
13.
13. Sol.
(3) x – 2y + 2z – 1 = 0 (4) x – 2y + 2z + 5 = 0 1 Equation of plane parallel to x – 2y + 2z – 5 = 0 is x – 2y + 2z + k = 0 perpendicular distance from O(0, 0, 0) to (1) is 1 k =1 |k| = 3 k = ±3 1+ 4 + 4
p
14. Sol.
∴ x – 2y + 2z – 3 = 0
If the line 2x + y = k passes through the point which divides the line segment joining the points (1, 1) and (2, 4) in the ratio 3 : 2, then k equals 29 11 (1) (2) 5 (3) 6 (4) 5 5 3 6 + 2 12 + 2 Point p = , 5 5 p=
14.
...... (1)
8 14 , 5 5
8 14 , 5 5
16 14 + =k 5 5
lies on 2x + y = k
k=
30 =6 5
Let x1, x2, ......, xn be n observations, and let x be their arithematic mean and σ2 be their variance. 2 Statement 1: Variance of 2x1, 2x2, ......, 2xn is 4 σ . Statement 2: Arithmetic mean of 2x1, 2x2, ......, 2xn is 4x . (1) Statement 1 is false, statement 2 is true (2) Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1 (3) Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1 (4) Statement 1 is true, statement 2 is false 4 2
σ =
xi2 − n
xi n
2
Variance of 2x1, 2x2, ....., 2xn =
( 2xi ) n
2
−
2xi n
2
= 4
x i2 − n
xi n
2
2
= 4σ
Statement 1 is true. A.M. of 2x1, 2x2, ......, 2xn =
2x1 + 2x 2 + ⋅ ⋅ ⋅ ⋅ +2xn x + x 2 + ⋅ ⋅ ⋅ ⋅ + xn =2 1 n n
= 2x
Statement 2 is false. 15.
15. Sol.
The population p(t) at time t of a certain mouse species satisfies the differential equation – 450. If p(0) = 850, then the time at which the population becomes zero is 1 (1) 2 ln 18 (2) ln 9 (3) ln18 (4) ln 18 2 1 d(p(t)) 1 = p(t) – 450 dt 2 d(p(t)) p(t) − 900 = dt 2 d(p(t)) 2 = dt p(t) − 900
dp(t) = 0.5 p(t) dt
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Nitin M Sir (physics-iitjee.blogspot.com)
AIEEE− −2012− −6
2 ln |p(t) – 900| = t + c t=0 2 ln 50 = 0 + c ∴ 2 ln |p(t) – 900| = t + 2 ln 50 P(t) = 0 2 ln 900 = t + 2 ln 50 900 t = 2 (ln 900 – ln 50) = 2ln = 2 ln 18. 50
c = 2 ln 50
2
Let a, b ∈ R be such that the function f given by f(x) = ln |x| + bx + ax, x ≠ 0 has extreme values at x = –1 and x = 2. Statement 1: f has local maximum at x = –1 and at x = 2. 1 −1 Statement 2: a = and b = 2 4 (1) Statement 1 is false, statement 2 is true (2) Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1 (3) Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1 (4) Statement 1 is true, statement 2 is false 16. 2 1 Sol. f′(x) = + 2b x + a x f has extremevalues and differentiable f′(–1) = 0 a – 2b = 1 1 1 1 f′(2) = 0 a + 4b = − a= ,b= − 2 2 4 f′′(–1), f′′(2) are negative. f has local maxima at –1, 2
16.
17.
2
The area bounded between the parabolas x = (1) 20 2
17. Sol.
(2)
3 Required area
2
2
3 y−
A= 2 0
= 5
18. 18. Sol.
y3 / 2 3/2
2
= 0
10 2 3
y 2 and x = 9y, and the straight line y = 2 is 4 20 2 (3) (4) 10 2 3
y 2
2
dy = 2 0
x = 9y
5 y 2
dy
2
x =
9 4 y=2
10 3 / 2 20 2 2 −0 = 3 3
O
Assuming the balls to be identical except for difference in colours, the number of ways in which one or more balls can be selected from 10 white, 9 green and 7 black balls is (1) 880 (2) 629 (3) 630 (4) 879 4 Number of ways of selecting one or more balls from 10 white, 9 green, and 7 black balls = (10 + 1)(9 + 1)(7 + 1) – 1 = 11 × 10 × 8 – 1 = 879.
2x − 1 π, where [x] denotes the greatest integer 2
19.
If f: R → R is a function defined by f(x) = [x] cos
19.
function, then f is (1) continuous for every real x (3) discontinuous only at non-zero integral values of x 1
(2) discontinuous only at x = 0 (4) continuous only at x = 0
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Nitin M Sir (physics-iitjee.blogspot.com)
AIEEE− −2012− −7
2x − 1 1 π = x cos x − π 2 2 = [x] sin π x is continuous for every real x.
Sol.
f(x) = x cos
20.
If the lines (1) –1
x −1 y +1 z −1 x −3 y −k z = = and = = intersect, then k is equal to 2 3 4 1 2 1 9 2 (3) (4) 0 (2) 9 2
20.
3
Sol.
Any point on
21.
21. Sol.
x −1 y +1 z −1 = = = t is (2t + 1, 3t – 1, 4t + 1) 2 3 4 x −3 y −k z And any point on = = = s is (s + 3, 2s + k, s) 1 2 1 3 Given lines are intersecting t = − and s = –5 2
5 8
C2 C3
P(A ∩ B) =
(the numbers > 3 are 5) 2 8
C1 C3
Required probability is P
22. Sol.
9 2
Three numbers are chosen at random without replacement from {1, 2, 3, ...... 8}. The probability that their minimum is 3, given that their maximum is 6, is 1 3 1 2 (1) (2) (3) (4) 8 5 5 4 2 Let A be the event that maximum is 6. B be event that minimum is 3 5 C P(A) = 8 2 (the numbers < 6 are 5) C3 P(B) =
22.
∴k=
B P(A ∩ B) 2 C1 2 1 = = 5 = = . A P(A) C2 10 5
z2 is real, then the point represented by the complex number z lies z −1 (1) either on the real axis or on a circle passing through the origin (2) on a circle with centre at the origin (3) either on the real axis or on a circle not passing through the origin (4) on the imaginary axis 1 Let z = x + iy ( x ≠ 1 as z ≠ 1) 2 2 2 z = (x – y ) + i(2xy) If z ≠ 1 and
z2 is real z −1
its imaginary part = 0 2
2
2xy (x – 1) – y(x – y ) = 0 2 2 y(x + y – 2x) = 0 2 y = 0; x + y2 – 2x = 0 ∴ z lies either on real axis or on a circle through origin.
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Nitin M Sir (physics-iitjee.blogspot.com)
Nitin M Sir (physics-iitjee.blogspot.com)
AIEEE− −2012− −8
23. 23. Sol.
24.
3
3
2
2
Let P and Q be 3 × 3 matrices with P ≠ Q. If P = Q and P Q = Q P, then determinant of 2 2 (P + Q ) is equal to (1) –2 (2) 1 (3) 0 (4) –1 3 3 3 P =Q 3 2 3 2 P –P Q=Q –Q P 2 2 P (P – Q) = Q (Q – P) 2 2 P (P – Q) + Q (P – Q) = O 2 2 2 2 (P + Q )(P – Q) = O |P + Q | = 0 If g(x) =
x 0
cos 4t dt , then g(x + π) equals
24.
g(x) g( π) 2 or 4
Sol.
g(x) = cos 4t dt
(2) g(x) + g(π)
(1)
(3) g(x) – g(π)
(4) g(x) . g(π)
x
0
sin 4x +k 4 g(x + π) = g(x) + g(π) = g(x) – g(π) ( g(π) = 0) g′(x) = cos 4x
25.
25. Sol.
26. 26. Sol.
27.
g(x) =
g(x) =
sin 4x [ g(0) = 0] 4
The length of the diameter of the circle which touches the x-axis at the point (1, 0) and passes through the point (2, 3) is 10 3 6 5 (1) (2) (3) (4) 3 5 5 3 1 Let (h, k) be centre. 2 2 2 (h – 1) + (k – 0) = k h=1 (h, k) 5 2 2 2 (h – 2) + (k – 3) = k k= (2, 3) k 3 k 10 ∴ diameter is 2k = 3 (1, 0)
Let X = {1, 2, 3, 4, 5}. The number of different ordered pairs (Y, Z) that can be formed such that Y ⊆ X, Z ⊆ X and Y ∩ Z is empty, is 2 5 5 3 (1) 5 (2) 3 (3) 2 (4) 5 2 Y ⊆ X, Z ⊆ X Let a ∈ X, then we have following chances that (1) a ∈ Y, a ∈ Z (2) a ∉ Y, a ∈ Z (3) a ∈ Y, a ∉ Z (4) a ∉ Y, a ∉ Z We require Y ∩ Z = φ Hence (2), (3), (4) are chances for ‘a’ to satisfy Y ∩ Z = φ. ∴ Y ∩ Z = φ has 3 chances for a. 5 Hence for five elements of X, the number of required chances is 3 × 3 × 3 × 3 × 3 = 3 2
2
An ellipse is drawn by taking a diameter of the circle (x – 1) + y = 1 as its semiminor axis and a 2 2 diameter of the circle x + (y – 2) = 4 as its semi-major axis. If the centre of the ellipse is the origin and its axes are the coordinate axes, then the equation of the ellipse is Ltd., FIITJEE House, 29 – A, Kalu Sarai, Sarvapriya Vihar, New Delhi – 110016, Ph: 46106000, 26569493, Fax: 26513942 website: www.fiitjee.com
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Nitin M Sir (physics-iitjee.blogspot.com)
AIEEE− −2012− −9
2
27. Sol.
2
(1) 4x + y = 4 4 Semi minor axis b = 2 Semi major axis a = 4 Equation of ellipse = 2
2
x2 a2
2
2
(2) x + 4y = 8
+
y2 b2
=1
2
2
2
(3) 4x + y = 8
2
(4) x + 4y = 16
x2 y2 =1 + 16 4
x + 4y = 16. Consider the function f(x) = |x – 2| + |x – 5|, x ∈ R. Statement 1: f′(4) = 0 Statement 2: f is continuous in [2, 5], differentiable in (2, 5) and f(2) = f(5). (1) Statement 1 is false, statement 2 is true (2) Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1 (3) Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1 (4) Statement 1 is true, statement 2 is false 28. 2 Sol. f(x) = 7 – 2x; x < 2 = 3; 2≤x≤5 = 2x – 7; x > 5 f(x) is constant function in [2, 5] f is continuous in [2, 5] and differentiable in (2, 5) and f(2) = f(5) by Rolle’s theorem f′(4) = 0 ∴ Statement 2 and statement 1 both are true and statement 2 is correct explanation for statement 1.
28.
29.
29. Sol.
A line is drawn through the point (1, 2) to meet the coordinate axes at P and Q such that it forms a triangle OPQ, where O is the origin. If the area of the triangle OPQ is least, then the slope of the line PQ is 1 1 (1) − (2) –4 (3) –2 (4) − 4 2 3 Equation of line passing through (1, 2) with slope m is y – 2 = m(x – 1) (m − 2)2 Area of ∆OPQ = 2m
m2 + 4 − 4m 2m m 2 ∆ is least if = 2 m ∆=
30.
∆=
m 2 + −2 2 m 2
m =4
m = ±2
m = –2
Let ABCD be a parallelogram such that AB = q,AD = p and ∠BAD be an acute angle. If r is the vector that coincides with the altitude directed from the vertex B to the side AD, then r is given by 3 ( p ⋅ q) p⋅q (1) r = 3q − p (2) r = −q + p p ⋅p p ⋅ p ( ) (3) r = q −
30.
p⋅q p p ⋅p
(4) r = −3q +
3 ( p ⋅ q)
(p ⋅ p )
p
2
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Nitin M Sir (physics-iitjee.blogspot.com)
AIEEE− −2012− −10
Sol.
(p ⋅ q) p (p ⋅ q) (p ⋅ q) p q+ r = (p ⋅ q)
AE =
31.
31. Sol.
D
AE = vector component of q on p ∴ From ∆ABE; AB + BE = AE
r = −q +
(p ⋅ q) p (p ⋅ p )
p
C
E
A
r
q
B
A wooden wheel of radius R is made of two semicircular parts (see figure); The two parts are held together by a ring made of a metal strip of cross sectional area S and length L. L is slightly less than 2πR. To fit the ring on the wheel, it is heated so that its temperature rises by ∆T and it just steps over the wheel. As it cools down to surrounding temperature, it presses the semicircular parts together. If the coefficient of linear expansion of the metal is α, and its Youngs'modulus is Y, the force that one part of the wheel applies on the other part is : (1) 2π SY α ∆T (2) SY α ∆T (3) π SY α ∆T (4) 2SY α ∆T 4 If temperature increases by ∆T, Increase in length L, ∆L = Lα ∆T
∆L = α ∆T F L Let tension developed in the ring is T. T ∆L ∴ =Y = Y α ∆T S L ∴ T = S Y α ∆T T T From FBD of one part of the wheel, F = 2T Where, F is the force that one part of the wheel applies on the other part. ∴ F = 2S Y α ∆T ∴
32.
The figure shows an experimental plot for discharging of a capacitor in an R-C circuit. The time constant τ of this circuit lies between: (1) 150 sec and 200 sec (2) 0 and 50 sec (3) 50 sec and 100 sec (4) 100 sec and 150 sec
32. Sol.
4 For discharging of an RC circuit, V = V0 e − t / τ So, when
V=
V0 2
V0 = V0 e − t / τ 2 1 t t ln = − τ= 2 τ ln 2
Ltd., FIITJEE House, 29 – A, Kalu Sarai, Sarvapriya Vihar, New Delhi – 110016, Ph: 46106000, 26569493, Fax: 26513942 website: www.fiitjee.com
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Nitin M Sir (physics-iitjee.blogspot.com)
AIEEE− −2012− −11
V0 , t = 100 s 2
From graph when V = 33.
∴ τ=
100 = 144.3 sec ln2
In a uniformly charged sphere of total charge Q and radius R, the electric field E is plotted as a function of distance from the centre. The graph which would correspond to the above will be E E E E
33.
3
Sol.
Einside = Eoutside
r
R (1)
R (2)
r
R (3)
r
R (4)
r
1 Q r 4 π ε 0 R3 1 Q r 4π ε0 r 3
E
∴
R
34.
34. Sol.
r
An electromagnetic wave in vacuum has the electric and magnetic fields E and B , which are always perpendicular to each other. The direction of polarization is given by X and that of wave propagation by k . Then : (1) X || B and k || B × E (2) X || E and k || E × B (3) X || B and k || E × B (4) X || E and k || B × E 3 Direction of polarization is parallel to magnetic field, ∴ X || B
and direction of wave propagation is parallel to E × B ∴ K || E × B 35.
35. Sol.
If a simple pendulum has significant amplitude (up to a factor of 1/e of original) only in the period between t = Os to t = τs, then τ may be called the average life of the pendulum. When the spherical bob of the pendulum suffers a retardation (due to viscous drag) proportional to its velocity, with ' b'as the constant of proportionality, the average life time of the pendulum is (assuming damping is small) in seconds: 0.693 1 2 (1) (2) b (3) (4) b b b 4 As retardation = bv ∴ retarding force = mbv θ ∴ net restoring torque when angular displacement is θ is given by mbv = – mg sinθ + mbv ∴ Iα = – mg sinθ + mbv 2 where, I = m v d2 θ g bv mg ∴ = α = − sin θ + dt 2 for small damping, the solution of the above differential equation will be
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AIEEE− −2012− −12
∴
θ = θ0 e
−
bt 2
sin(wt + φ) −bt
∴ angular amplitude will be = θ.e 2 According to question, in τ time (average life–time), 1 value of its original value (θ) angular amplitude drops to e 6τ − θ0 ∴ = θ0 e 2 e 6τ =1 2 2 ∴ τ= b
36. 36. Sol.
37.
37. Sol. 38.
38. Sol.
39.
Hydrogen atom is excited from ground state to another state with principal quantum number equal to 4. Then the number of spectral lines in the emission spectra will be (1) 2 (2) 3 (3) 5 (4) 6 4 Number of spectral lines from a state n to ground state is n(n − 1) = = 6. 2 A coil is suspended in a uniform magnetic field, with the plane of the coil parallel to the magnetic lines of force. When a current is passed through the coil it starts oscillating; it is very difficult to stop. But if an aluminium plate is placed near to the coil, it stops. This is due to : (1) development of air current when the plate is placed. (2) induction of electrical charge on the plate (3) shielding of magnetic lines of force as aluminium is a paramagnetic material. (4) electromagnetic induction in the aluminium plate giving rise to electromagnetic damping. 4 Oscillating coil produces time variable magnetic field. It cause eddy current in the aluminium plate which causes anti–torque on the coil, due to which is stops. The mass of a spaceship is 1000 kg. It is to be launched from the earth' s surface out into free space. The 2 value of ' g'and ' R'(radius of earth) are 10 m/s and 6400km respectively. The required energy for this work will be ; 11 8 9 10 (1) 6.4 x 10 Joules (2) 6.4 x 10 Joules (3) 6.4 x 10 Joules (4) 6.4 x 10 Joules 4 To launch the spaceship out into free space, from energy conservation, −GMm +E = 0 R GMm GM E= = mR = mgR R R2 10 = 6.4 x 10 J Helium gas goes through a cycle ABCDA (consisting of two isochoric and two isobaric lines) as shown in figure. Efficiency of this cycle is nearly: (Assume the gas to be close to ideal gas) (1) 15.4% (2) 9.1% (3) 10.5% (4) 12.5%
2P0 P0
B
A V0
39.
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C
D 2V0
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AIEEE− −2012− −13
Sol.
Work done in complete cycle = Area under P–V graph = P0V0 from A to B, heat given to the gas 3 3 3 = nCv ∆T = n R∆T = V0 ∆P = P0 V0 2 2 2 from B to C, heat given to the system 5 = nCp ∆T = n R ∆T 2 5 = (2P0 )∆V = 5P0 V0 2 from C to D and D to A, heat is rejected. work done by gas efficiency, η = × 100 heat given to the gas η=
40.
40. Sol.
In Young' s double slit experiment, one of the slit is wider than other, so that the amplitude of the light from one slit is double of that from other slit. If Im be the maximum intensity, the resultant intensity I when they interfere at phase difference φ is given by I I I I φ φ φ (2) m 1 + 2cos2 (3) m 1 + 4cos2 (4) m 1 + 8cos2 (1) m (4 + 5 cos φ) 3 2 5 2 9 2 9 4 Let A1 = A0, A2 = 2A0 If amplitude of resultant wave is A then A 2 = A 12 + A 22 + 2A 1A 2 cos φ For maximum intensity, 2 A max = A 12 + A 22 + 2A 1A 2 A12 + A 22 + 2A1A 2 cos φ A2 = 2 A max A12 + A 22 + 2A1A 2
∴
= I Im 41.
41. Sol.
P0 V0 = 15.4% 3 P0 V0 + 5P0 V0 2
=
A 20 + 4A 20 + 2(A 0 )(2A 0 )cos φ A 02 + 4A 02 + 2(A 0 )(2A 0 )
5 + 4 cos φ 1 + 8 cos2 ( φ / 2) = 9 9
A liquid in a beaker has temperature θ(t) at time t and θ0 is temperature of surroundings, then according to Newton' s law of cooling the correct graph between loge (θ – θ0) and t is
(1) (2) 1 According to Newtons law of cooling. dθ ∝ −(θ − θ0 ) dt
(3)
(4)
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AIEEE− −2012− −14
dθ = −k(θ − θ0 ) dt dθ = −k dt θ − θ0 ln(θ – θ0) = –kt + c Hence the plot of ln(θ – θ0) vs t should be a straight line with negative slope. 42.
42. Sol.
–bt
A particle of mass m is at rest at the origin at time t = 0. It is subjected to a force F (t) = F0e direction. Its speed v(t) is depicted by which of the following curves?
(1)
(2)
(3)
(4)
in the x
3 F = F0 e −bt
F F0 −bt = e m m dv F0 −bt = e dt m t F −bt dv = e dt m 0
a=
t F −1 e−bt 0 m b F v= e −bt mb v = 0 at t = 0 F v→ as t → ∞ mb
v=
and
So, velocity increases continuously and attains a maximum value of v = 43. 43. Sol.
F as t → ∞ . mb
Two electric bulbs marked 25W – 220V and 100W – 220V are connected in series to a 440Vsupply. Which of the bulbs will fuse? (1) both (2) 100 W (3) 25 W (4) neither 3 Resistances of both the bulbs are V 2 2202 R1 = = P1 25 V 2 2202 = P2 100 Hence R1 > R2 R2 =
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AIEEE− −2012− −15
When connected in series, the voltages divide in them in the ratio of their resistances. The voltage of 440 V devides in such a way that voltage across 25 w bulb will be more than 220 V. Hence 25 w bulb will fuse. 44.
44. Sol.
45.
45. Sol.
Resistance of a given wire is obtained by measuring the current flowing in it and the voltage difference applied across it. If the percentage errors in the measurement of the current and the voltage difference are 3% each, then error in the value of resistance of the wire is (1) 6% (2) zero (3) 1% (4) 3% 1 V R= i ∆R ∆V ∆i = + R V i ∆V × 100 = 3 V ∆V = 0.03 V ∆i Similarly, = 0.03 i ∆R Hence = 0.06 R ∆R So percentage error is × 100 = 6% R A boy can throw a stone up to a maximum height of 10 m. The maximum horizontal distance that the boy can throw the same stone up to will be (1) 20 2 m (2) 10 m (3) 10 2 m (4) 20 m 4 u2 maximum vertical height = = 10m 2g Horizontal range of a projectile =
u2 sin2θ g
0
Range is maximum when θ = 45 u2 Maximum horizontal range = g Hence maximum horizontal distance = 20 m. 46.
46. Sol.
This question has statement 1 and statement 2. Of the four choices given after the statements, choose the one that best describes the two statements Statement 1 : Davisson – germer experiment established the wave nature of electrons. Statement 2 : If electrons have wave nature, they can interfere and show diffraction. (1) Statement 1 is false, Statement 2 is true (2) Statement 1 is true, Statement 2 is false (3) Statement 1 is true, Statement 2 is the correct explanation for statement 1 (4) Statement 1 is true, Statement 2 is true, Statement 2 is not the correct explanation for statement 1. 3 Davisson – Germer experiment showed that electron beams can undergo diffraction when passed through atomic crystals. This shows the wave nature of electrons as waves can exhibit interference and diffraction.
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AIEEE− −2012− −16
47.
47. Sol.
48.
A thin liquid film formed between a U-shaped wire and a light slider supports a weight of 1.5 x10–2N (see figure). The length of the slider is 30 cm and its weight negligible. The surface tension of the liquid film is –1 –1 (2) 0.1 Nm (1) 0.0125 Nm –1 –1 (3) 0.05 Nm (4) 0.025 Nm
w
4 The force of surface tension acting on the slider balances the force due to the weight. F = 2T = w –2 2T(0.3) = 1.5 x 10 –2 T = 2.5 x 10 N/m
F = 2Tl
w
A charge Q is uniformly distributed over the surface of non conducting disc of radius R. The disc rotates about an axis perpendicular to its plane and passing through its centre with an angular velocity ω. As a result of this rotation a magnetic field of induction B is obtained at the centre of the disc. If we keep both the amount of charge placed on the disc and its angular velocity to be constant and vary the radius of the disc then the variation of the magnetic induction at the centre of the disc will be represented by the figure
B
B R
48. Sol.
FILM
(1)
B R
B R
(2)
(3)
R
1 Consider ring like element of disc of radius r and thickness dr. If σ is charge per unit area, then charge on the element dq = σ(2πr dr) current ‘i’ associated with rotating charge dq is (dq)w i= = σ w r dr 2π r Magnetic field dB at center due to element µ i µ σ ω dr dr dB = 0 = 0 2r 2 µ σ ωR µ0 σ ω R Bnet = dB = dr = 0 2 0 2
(4)
µ 0 Qω Q = σ π R2 2πR So if Q and w are unchanged then 1 Bnet ∝ R Hence variation of Bnet with R should be a rectangular hyperbola as represented in (1). Bnet =
49.
Truth table for system of four NAND gates as shown in figure is
A Y B
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AIEEE− −2012− −17
A 0 0 1 1
B 0 1 0 1 (1)
Y 0 1 1 0
A 0 0 1 1
B 0 1 0 1 (2)
Y 0 0 1 1
A 0 0 1 1
B 0 1 0 1 (3)
Y 1 1 0 0
A
B
Y
0 0
0 1
1 0
1
0
0
1
1
1
(4)
49. Sol.
1
50.
A radar has a power of 1 Kw and is operating at a frequency of 10 GHz. It is located on a mountain top of height 500 m. The maximum distance upto which it can detect object located on the surface of the earth 6 (Radius of earth = 6.4 x 10 m) is (1) 80 km (2) 16 km (3) 40 km (4) 64 km 1 Maximum distance on earth where object can be detected is d, then d h (h + R)2 = d2 + R 2
50. Sol.
A 0 0 1 1
since
B 0 1 0 1
d2 = h2 + 2Rh h K2S2 S1 > S2 K1 < K2 W∝K W1 < W 2 Two cars of masses m1 and m2 are moving in circles of radii r1 and r2, respectively. Their speeds are such that they make complete circles in the same time t. The ratio of their centripetal acceleration is (1) m1r1 : m2r2 (2) m1 : m2 (3) r1 : r2 (4) 1 : 1 3 a∝r A cylindrical tube, open at both ends, has a fundamental frequency, f, in air. The tube is dipped vertically in water so that half of it is in water. The fundamental frequency of the air-column is now f 3f (1) f (2) (3) (4) 2f 2 4 1 v f0 = 2 v fC = 2 An object 2.4 m in front of a lens forms a sharp image on a film 12 cm behind the lens. A glass plate 1cm thick, of refractive index 1.50 is interposed between lens and film with its plane faces parallel to film. At what distance (from lens) should object be shifted to be in sharp focus on film? (1) 7.2 m (2) 2.4 m (3) 3.2 m (4) 5.6 m 4 Case I: u = –240cm, v = 12, by Lens formula 1 7 = f 80 1 35 2 1 Case II: v = 12 – = (normal shift = 1 − = ) 3 3 3 3 7 f= 80 u = 5.6 A diatomic molecule is made of two masses m1 and m2 which are separated by a distance r. If we calculate its rotational energy by applying Bohr' s rule of angular momentum quantization, its energy will be given by (n is an integer) (m1 + m2 )2 n2h2 (m1 + m2 )n2h2 n 2 h2 2n2 h2 (1) (2) (3) (4) 2m12 m22 r 2 2(m1 + m2 )r 2 (m1 + m2 )r 2 2m1 m2 r 2 4 Ltd., FIITJEE House, 29 – A, Kalu Sarai, Sarvapriya Vihar, New Delhi – 110016, Ph: 46106000, 26569493, Fax: 26513942 website: www.fiitjee.com
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AIEEE− −2012− −19
Sol.
r1 =
m2r m1r ; r2 = m1 + m2 m1 + m2
(I1 + I2)ω = K.E = 58.
58. Sol.
59.
nh =n 2π
n2 2 (m1 + m2 ) 1 2 (I1 + I2) ω = 2 2m1m2r 2
A spectrometer gives the following reading when used to measure the angle of a prism. Main scale reading: 58.5 degree Vernier scale reading : 09 divisions Given that 1 division on main scale corresponds to 0.5 degree. Total divisions on the vernier scale is 30 and match with 29 divisions of the main scale. The angle of the prism from the above data (1) 58.59o (2) 58.77o (3) 58.65o (4) 59o 3 1 L.C = 60 9 Total Reading = 585 + = 58.65 60 This question has statement 1 and statement 2. Of the four choices given after the statements, choose the one that best describes the two statements. An insulating solid sphere of radius R has a uniformly positive charge density ρ. As a result of this uniform charge distribution there is a finite value of electric potential at the centre of the sphere, at the surface of the sphere and also at a point out side the sphere. The electric potential at infinity is zero. Statement 1 : When a charge q is taken from the centre to the surface of the sphere, its potential energy qp changes by 3ε0 ρr 3ε 0 1 is true, Statement 2 is true, Statement 2 is not the correct explanation for statement 1. 1 is true, Statement 2 is false 1 is false, Statement 2 is true 1 is true, Statement 2 is the correct explanation for statement 1
Statement 2 : The electric field at a distance r(r < R) from the centre of the sphere is
59. Sol.
(1) Statement (2) Statement (3) Statement (4) Statement 3 1 4 E ⋅ dA = ρ × πr 3 ε0 3 ρr 3ε 0 Statement 2 is correct E=
∆PE = (Vsur – Vcent)q = −
q ρR2 6 ε0
Statement 1 is incorrect 60.
Proton, Deuteron and alpha particle of the same kinetic energy are moving in circular trajectories in a constant magnetic field. The radii of proton, deuteron and alpha particle are respectively rp, rd and rα. Which one of the following relations is correct? (1) rα = rp = rd (2) rα = rp < rd (3) rα > rd > rp (4) rα = rd > rp
60.
2
Sol.
r=
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AIEEE− −2012− −20
m q rα = r p < r d
r∝
61.
Which among the following will be named as dibromidobis(ethylene diamine)chromium(III) bromide ? (1) Cr ( en )3 Br3
61. Sol.
2
(2) Cr ( en )2 Br2 Br
(3) Cr ( en ) Br4
−
(4) Cr ( en ) Br2 Br
Cr ( en )2 Br2 Br – dibromido bis (ethylene diamine)chromium(III) bromide
62.
Which method of purification is represented by the following equation :
62. Sol.
(1) zone refining 4 Van Arkel method
523K 1700K Ti ( s ) + 2I2 ( g) → TiI4 ( g) → Ti ( s ) + 2I2 ( g )
(2) cupellation
(3) Poling
(4) Van Arkel
523K Ti ( s ) + 2I2 ( g ) → TiI4 ( g ) 1700 K TiI4 ( g ) → Ti ( s ) + 2I2 ( g )
63. 63. Sol.
Lithium forms body centred cubic structure. The length of the side of its unit cell is 351 pm. Atomic radius of the lithium will be : (1) 75 pm (2) 300 pm (3) 240 pm (4) 152 pm 4 For BCC, 3a = 4r
r= 64. 64. Sol. 65. 65. Sol.
66.
3 × 351 = 152pm 4
The molecule having smallest bond angle is : (1) NCl3 (2) AsCl3 (3) SbCl3 (4) PCl3 3 As the size of central atom increases lone pair bond pair repulsions increases so, bond angle decreases Which of the following compounds can be detected by Molisch’s test ? (1) Nitro compounds (2) Sugars (3) Amines (4) Primary alcohols 2 Molisch’s Test : when a drop or two of alcoholic solution of α–naphthol is added to sugar solution and then conc. H2SO4 is added along the sides of test tube, formation of violet ring takes place at the junction of two liquids. The incorrect expression among the following is : ∆Gsystem V (1) (2) In isothermal process w reversible = −nRT ln f = −T Vi ∆Stotal (3) lnK =
66. Sol.
∆H0 − T∆S0 RT
0
(4) K = e−∆G
/ RT
3 ∆G° = –RTln K and ∆G0 = ∆H0 − T∆S0
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AIEEE− −2012− −21
67. 67. Sol.
68. 68. Sol. 69. 69. Sol. 70. 70. Sol.
71.
71. Sol.
The density of a solution prepared by dissolving 120 g of urea (mol. Mass = 60 u ) in 1000g of water is 1.15 g/mL. The molarity of this solution is : (1) 0.50 M (2) 1.78 M (3) 1.02 M (4) 2.05 M 4 Total weight of solution = 1000 + 120 = 1120 g 120 1000 Molarity = × = 2.05M 60 1120 /1.15 The species which can best serve as an initiator for the cationic polymerization is : (1) LiAlH4 (2) HNO3 (3) AlCl3 (4) BuLi 3 lewis acids can initiate the cationic polymerization. Which of the following on thermal decomposition yields a basic as well as an acidic oxide ? (1) NaNO3 (2) KClO3 (3) CaCO3 (4) NH4NO3 3 CaCO3 → CaO + CO 2 Basic
Acidic
2+
2+
2+
The standard reduction potentials for Zn / Zn, Ni / Ni, and Fe / Fe are –0.76, –0.23 and –0.44 V 2+ 2+ respectively. The reaction X + Y → X + Y will be spontaneous when : (1) X = Ni, Y = Fe (2) X = Ni, Y = Zn (3) X = Fe, Y = Zn (4) X = Zn, Y = Ni 4 +2 +2 Zn + Fe → Zn + Fe +2 2+ Fe + Ni → Fe + Ni 2+ +2 Zn + Ni → Zn + Ni All these are spontaneous According to Freundlich adsorption isotherm, which of the following is correct ? x x x (1) ∝ P0 (2) ∝ p1 (3) ∝ p1/ n m m m (4) All the above are correct for different ranges of pressure 4 x ∝ P0 is true at extremely high pressures m x x ∝ p1 ; ∝ p1/ n are true at low and moderate pressures m m –4
72.
The equilibrium constant (KC) for the reaction N2(g) + O2(g) → 2NO(g) at temperature T is 4 x 10 . The value of KC for the reaction, NO(g) → ½ N2(g) + ½ O2(g) at the same temperature is : 2 –4 (3) 4 x 10 (4) 50.0 (1) 0.02 (2) 2.5 x 10 4
Sol.
N2 + O2
72.
NO K1C = 73. 73.
2NO
K C = 4 × 10 −4
1 1 N2 + O2 2 2 1 4 × 10 −4
K1C =
1 KC
= 50
The compressibility factor for a real gas at high pressure is : (1) 1 + RT/pb (2) 1 (3) 1 + pb/RT 3
(4) 1–pb/RT
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AIEEE− −2012− −22
Pb RT
Sol.
At high pressure Z = 1 +
74.
Which one of the following statements is correct ? (1) All amino acids except lysine are optically active (2) All amino acids are optically active (3) All amino acids except glycine are optically active (4) All amino acids except glutamic acid are optically active 3 NH2 CH2 Glycine COOH
74.
Sol. 75. 75.
Aspirin is known as : (1) Acetyl salicylic acid (3) Acetyl salicylate 1 COOH
(2) Phenyl salicylate (4) Methyl salicylic acid
O O
C
Aspirin Acetyl salicylic acid
Sol. 76.
76.
Ortho–Nitrophenol is less soluble in water than p– and m– Nitrophenols because : (1) o–Nitrophenol is more volatile in steam than those of m – and p–isomers (2) o–Nitrophenol shows Intramolecular H–bonding (3) o–Nitrophenol shows Intermolecular H–bonding (4) Melting point of o–Nitrophenol is lower than those of m–and p–isomers. 2 H O O
O
N Sol. 77. 77. Sol.
CH3
Intramolecular H–bonding decreases water solubility. How many chiral compounds are possible on monochlorination of 2–methyl butane ? (1) 8 (2) 2 (3) 4 (4) 6 2 H3C − CH2 − CH ( CH3 ) − CH3 on monochlorination gives H2C ( Cl) − CH2 − CH ( CH3 ) − CH3 (I)
H3 C − CH ( Cl ) − CH ( CH3 ) − CH3 (II) Chiral
Achiral
CH2 Cl
CH3 H3C
CH2
C
CH3
H3C
CH2 CH CH3
Cl (III) Achiral
(IV) chiral
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AIEEE− −2012− −23
78.
78. Sol. 79.
79. Sol. 80.
80. Sol.
81.
81. Sol. 82. 82. Sol.
Very pure hydrogen (99.9%) can be made by which of the following processes ? (1) Reaction of methane with steam (2) Mixing natural hydrocarbons of high molecular weight (3) Electrolysis of water (4) Reaction of salt like hydrides with water 3 Highly pure hydrogen is obtained by the electrolysis of water. The electrons identified by quantum numbers n and l : (a) n = 4, l = 1 (b) n = 4, l = 0 (c) n = 3, l = 2 Can be placed in order of increasing energy as : (1) (c) < (d) < (b) < (a) (2) (d) < (b) < (c) < (a) (3) (b) < (d) < (a) < (c) 2 (a) (n + l) = 4 + 1 = 5 (b) (n + l) = 4 + 0 = 4 (c) (n + 1) = 3 + 2 = 5
83.
(d) (n + 1) = 3 + 1 = 4
Iron exhibits + 2 and +3 oxidation states. Which of the following statements about iron is incorrect ? (1) Ferrous oxide is more basic in nature than the ferric oxide. (2) Ferrous compounds are relatively more ionic than the corresponding ferric compounds (3) Ferrous compounds are less volatile than the corresponding ferric compounds. (4) Ferrous compounds are more easily hydrolysed than the corresponding ferric compounds. 4 FeO → More basic, more ionic, less volatile The pH of a 0.1 molar solution of the acid HQ is 3. The value of the ionization constant, Ka of this acid is : –1 –3 —5 –7 (2) 1 x 10 (3) 1 x 10 (4) 1 x 10 (1) 3 x 10 3 H+ = K a .C
10 −3 = K a .10 −1
–5
Which branched chain isomer of the hydrocarbon with molecular mass 72u gives only one isomer of mono substituted alky halide ? (1) Tertiary butyl chloride (2) Neopentane (3) Isohexane (4) Neohexane 2 CH2 Cl CH3 mono chlorination
H3C
C
CH3
CH3 Sol.
(4) (a) < (c) < (b) < (d)
For a first order reaction, (A) → products, the concentration of A changes from 0.1 M to 0.025 M in 40 minutes. The rate of reaction when the concentration of A is 0.01 M is : –5 –4 (1) 1.73 x 10 M/ min (2) 3.47 x 10 M/min –5 –4 (3) 3.47 x 10 M/min (4) 1.73 x 10 M/min 2 2.303 0.1 k= log 40 0.025 0.693 k= 20 0.693 For a F.O.R., rate=k[A]; rate = × 10−2 = 3.47 × 10 −4 M / min. 20
Ka = 10 83.
(d) n = 3 , l = 1
Neopentane Mol. wt = 72u
H3C
C
CH3
CH3 only one compound
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AIEEE− −2012− −24
84. 84. Sol.
Kf for water is 1.86K kg mol–1. If your automobile radiator holds 1.0 kg of water, how many grams of ethylene glycol (C2H6O2) must you add to get the freezing point of the solution lowered to –2.8°C ? (1) 72g (2) 93g (3) 39g (4) 27g 2 ∆Tf = K f .m 2.8 = 1.86 ×
wt 1000 × 62 1000
Wt = 93g 85. 85. Sol. 86. 86. Sol. 87. 87.
What is DDT among the following : (1) Greenhouse gas (3) Biodegradable pollutant 4 DDT – non–biodegradable pollutant.
(2) A fertilizer (4) Non–biodegradable pollutant
The increasing order of the ionic radii of the given isoelectronic species is : – 2+ + 2– 2– – 2+ + 2+ + – 2– + 2– 2+ – (1) Cl , Ca , K , S (2) S , Cl , Ca , K (3) Ca , K , Cl , S (4) K , S , Ca , Cl 3 For isoelectronic species, as the z/e decreases, ionic radius increases 2–Hexyne gives trans–2–Hexene on treatment with : (1) Pt/H2 (2) Li/NH3 (3) Pd/BaSO4 2 H7C3 Li/NH 3 C H3C CH2 CH2 C C CH3
2-Hexyne
Birch reduction
88. Sol.
H C CH3
Trans-2-Hexene
Sol. 88.
H
(4) LiAlH4
Iodoform can be prepared from all except : (1) Ethyl methyl ketone (2) Isopropyl alcohol (3) 3–Methyl – 2– butanone (4) Isobutyl alcohol 4 Iodoform is given by 1) methyl ketones R-CO-CH3 2) alcohols of the type R-CH(OH)CH3 where R can be hydrogen also
O H3C
C
C2H 5
ethyl methyl ketone CH3 H3C CH OH Isopropyl alchol H3C
O
CH3
C
CH CH3
can give Iodoform Test
3-methyl 2-butanone
Ltd., FIITJEE House, 29 – A, Kalu Sarai, Sarvapriya Vihar, New Delhi – 110016, Ph: 46106000, 26569493, Fax: 26513942 website: www.fiitjee.com
Nitin M Sir (physics-iitjee.blogspot.com)
Nitin M Sir (physics-iitjee.blogspot.com)
AIEEE− −2012− −25
CH3 H3C
CH CH2 OH
can' t give
Isobutyl alcohol 89. 89. Sol.
In which of the following pairs the two species are not isostructural ? (2) PCl+4 and SiCl4 (3) PF5 and BrF5 (1) CO32− and NO3− 3 (1) CO32− & NO3− → Sp2 hybridized, Trigonal planar
(4) AlF63− and SF6
(2) PCl+4 & SiCl4 → Sp3 hybridized, Tetrahedral 3
(3) PF5 → Sp d hybridized, Trigonal bipyramidal 3 2 BrF5 → Sp d hybridized, square pyramidal (4) AlF63 − & SF6 → Sp3 d2 hybridized, octahedral 90.
In the given transformation, which of the following is the most appropriate reagent ? CH CHCOCH3
Reagent HO CH CHCH2CH3
HO
90. Sol.
(−)
(1) NH2NH2 , O H (2) Zn − Hg / HCl (3) Na,Liq.NH3 (4) NaBH4 1 ZnHg/Hcl can’t be used due to the presence of acid sensitive group i.e. OH O
CH CH C
CH3
CH CH CH2 CH3 Zn-Hg/HCl
HO Cl and Na/Liq. NH3 and NaBH4 convert – CO – into – CH(OH)–
Ltd., FIITJEE House, 29 – A, Kalu Sarai, Sarvapriya Vihar, New Delhi – 110016, Ph: 46106000, 26569493, Fax: 26513942 website: www.fiitjee.com
Nitin M Sir (physics-iitjee.blogspot.com)
Nitin M Sir (physics-iitjee.blogspot.com)
AIEEE− −2012− −26
READ THE FOLLOWING INSTRUCTIONS CAREFULLY
1.
The candidates should fill in the required particulars on the Test Booklet and Answer Sheet (Side-1) with Blue/Black Ball Point Pen.
2.
For writing/marking particulars on Side-2 of the Answer Sheet, use Blue/Black Ball Point Pen only.
3.
The candidates should not write their Roll Numbers anywhere else (except in the specified space) on the Test Booklet/Answer Sheet.
4.
Out of the four options given for each question, only one option is the correct answer.
5.
For each incorrect response, one-fourth (1/4) of the total marks allotted to the question would be deducted from the total score. No deduction from the total score, however, will be made if no response is indicated for an item in the Answer Sheet
6.
Handle the Test Booklet and Answer Sheet with care, as under no circumstance (except for discrepancy in Test Booklet Code and Answer Sheet Code), will another set be provided.
7.
The candidates are not allowed to do any rough work or writing work on the Answer Sheet. All calculations/writing work are to be done in the space provided for this purpose, in the Test Booklet itself, marked ' Space for Rough Work' . This space is given at the bottom of each page and in 3 pages (Pages 21 - 23) at the end of the booklet.
8.
On completion of the test, the candidates must hand over the Answer Sheet to the Invigilator on duty in the Room/Hall. However, the candidates are allowed to take away this Test Booklet with them.
9.
Each candidate must show on demand his/her Admit Card to the Invigilator.
10.
No candidate, without special permission of the Superintendent or Invigilator, should leave his/her seat.
11.
The candidates should not leave the Examination Hall without handing over their Answer Sheet to the Invigilator on duty and sign the Attendance Sheet again. Cases where a candidate has not signed the Attendance Sheet a second time will be deemed not to have handed over the Answer Sheet and dealt with as an unfair means case. The candidates are also required to put their left hand THUMB impression in the space provided in the Attendance Sheet.
12.
Use of Electronic/Manual Calculator and any Electronic Item like mobile phone, pager etc. is prohibited.
13.
The candidates are governed by all Rules and Regulations of the Board with regard to their conduct in the Examination Hall. All cases of unfair means will be dealt with as per Rules and Regulations of the Board.
14.
No part of the Test Booklet and Answer Sheet shall be detached under any circumstances.
15.
Candidates are not-allowed to carry any textual material, printed or written, bits of papers, pager, mobile phone, electronic device or any other material except the Admit Card inside the examination hall/room.
Ltd., FIITJEE House, 29 – A, Kalu Sarai, Sarvapriya Vihar, New Delhi – 110016, Ph: 46106000, 26569493, Fax: 26513942 website: www.fiitjee.com
Nitin M Sir (physics-iitjee.blogspot.com)
Nitin M Sir (physics-iitjee.blogspot.com)
AIEEE− −2012− −27
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Ltd., FIITJEE House, 29 – A, Kalu Sarai, Sarvapriya Vihar, New Delhi – 110016, Ph: 46106000, 26569493, Fax: 26513942 website: www.fiitjee.com
Nitin M Sir (physics-iitjee.blogspot.com)
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