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FIITJEE Ltd., ICES House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi-110016, Ph: 26515949, 26569493, Fax: 011-26513942.

PAPER -2003

1. A particle of mass M and charge Q moving with velocity v

describes a circular path of radiusR when subjected to a uniform transverse magnetic field of induction B. The work done bythe field when the particle completes one full circle is

(A)2Mv

2 R

R

(B) zero

(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-axisenters a region where a magnetic field of induction B is along the y –axis, and an electricfield of induction B is along the y-axis, and an electric field of magnitude 104 V/m is along thenegative 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

(C) 1016 Wb/m2 (D) 103 Wb/m2

2. A.

F q(E v B)

…(1)

The solution of this problem can be obtained by resolving the motion along the threecoordinate axes namely

x

x x y z z y

F qa (E v B v B )

m m

y

y y z x x z

F qa (E v B v B )

m m

z

z z x y y z

F qa (E v B v B )

m m

For the given problem,

x yE E 0, y zv v 0 and

x zB B 0

Substituting in equation (2), we get

x za a 0 and

y y x za E v B

If the particle passes through the region undeflected ay is also zero, then

y x zE v B

4

y 3 2

z

z

E 10B 10 Wb / m

v 10

3. A thin rectangular magnet suspended freely has a period of oscillation equal to T. Now it isbroken 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

(A)1

2 2(B) 1/2

(C) 2 (D) 1/4

3. B.

When the magnet is divided into 2 equal parts, the magnetic dipole movement

itin M Sir (physics-iitjee.blogspot.com)


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M = pole strength lengthM

2 and moment of inertia

21I' mass (length)

12

=

2

1 m

12 2 2

II'8

Time periodI'

2M'B

I / 8

2M

B2

TT '

2

T ' 1

T 2

4. A magnetic needle lying parallel to a magnetic field requires W units of work to turn itthrough 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

10,

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 themagnet.

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 becomesparamagnetic.



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7. A spring balance is attached to the ceiling of a lift. A man hangs his bag on the spring andthe spring reads 49 N, when the lift is stationary. If the lift moves downward with anacceleration 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

(A)30 E

100.5

(B)30 E

100 0.5

(C)30(E 0.5i)

100

, where I is the current in the potentiometer wire.

(D) 30 E

100

8. A.

E E 30 30EV .

L 100 100

9. A strip of copper and another germanium are cooled from room temperature to 80 K. Theresistance 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 isnegative, therefore when copper and germanium are cooled, resistance of copper stripdecreases and that of germanium increases.

10. Consider telecommunication through optical fibres. Which of the following statements is nottrue?(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 10 5 A, is connected with thethermocouple. The smallest temperature difference that can be detected by this system is

(A) 16C (B) 12C(C) 8C (D) 20C



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11. A.

E = 25 106 V

IR = 105 40 = 4 104 V4

6

4 1016 C

25 10

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 Cuare 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.

Zn zx

Cu Cx

m Z

m Z

I and t are same for both Cu and Zn electrodes

Cu

0.13 31.5

m 32.5

Cu 0.13 32.5m 0.126 g.32.5

13. Dimensions of 0 0

1

, where symbols have their usual meaning, are

(A) [L1T] (B) [L2T2]

(C) [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)Y XI 32I (B)

Y XI 16I

(C)Y XI 32 I (D)

Y XI 64 I

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,

4

x

1I R t

2 …(i)

Moment of inertia of disc Y,4

yI 32 R t …(ii)

From equation (i) and (ii)

y xI 64 I

15. The time period of a satellite of earth is 5 hours. If the separation between the earth and thesatellite 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.



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Time period of a satellite T =3

e

2 r

R g

r = distance between satellite and the earth.3/ 2T r

3/ 2

1 1

2 2

T r

T r

2 1T 8T 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

Kinetic energy, 21K I

2

Therefore,2

2K 2KL

1 1 2

2 2 1

L K

L K

1

2

L2 2 4

L

2

LL

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

(A)4u

238(B)

4u

234

(C)4u

234(D)

4u

238

18. B.

According to principle of conservation of linear momentum the momentum of the systemremains 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

v234

.

19. Two spherical bodies of mass M and 5M and radii R and 2R respectively are released in freespace with initial separation between their centres equal to 12R. If they attract each other



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due to gravitational force only, then the distance covered by the smaller body just beforecollision 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 thesame 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 heightof 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 horizontalrange.

Maximum horizontal range2u sin 2

Rg

2 3 1(10) 2

2 2R 8.6610

m.

23. An ammeter reads upto 1 ampere. Its internal resistance is 0.81 ohm. To increase the rangeto 10 A the value of the required shunt is

(A) 0.03 (B) 0.3

(C) 0.9 (D) 0.09 23. D.

S =g

g

I G 1 0.810.09

I I 10 1

24. The physical quantities not having same dimensions are(A) torque and work (B) momentum and Planck’s constant



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(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 bythe three sides of a triangle ABC(as shown). The particle will now movewith 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 arerepresented 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, theelectric charge inside the surface will be

(A)2 1 0

( ) (B) 2 1

0

( )

(C) 2 1

0

( )

(D)

2 1 0( )

26. A.

According to Gauss’s theorem, charge in flux =0

charge enclosed by the surface

2 1 0q ( ) .

27. A horizontal force of 10 N is necessary to just hold a block stationaryagainst a wall. The coefficient of friction between the block and thewall 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 frictionin 10 s. then the coefficient of friction is(A) 002 (B) 0.03

(C) 0.04 (D) 0.01

28. C.

Retardationu 6

0.6t 10

m/sec2

Frictional force = mg = ma

a 0.6

0.06.g 10

29. Consider the following two statements.



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(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 dependsupon(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, distancebetween 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 forceP is applied at the free end of the rope, the force exerted by the rope on the block is

(A)Pm

M m(B)

Pm

M m

(C) P (D)Pm

M m

31. D.

Force on block = mass acceleration =PM

M m

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 tothe lower end. The weight stretches the wire by 1 mm. Then the elastic energy stored in thewire 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,

1U stress strain volume

2

1 F A

2 A

=

1F

2 = 31

200 10 0.1 J2



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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 andthen released so that the mass executes SHM of time period T. If the mass is increased bym, 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 ' M m

9

25

M

M m

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 springconstants k1 and k2 respectively. If the maximum velocities, during oscillations, are equal, theratio of amplitudes of A and B is

(A) 1

2

k

k(B) 2

1

k

k

(C) 2

1

k

k(D) 1

2

k

k

37. C.

1 2

2 1

a k

a k .

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.



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Time period of simple pendulum is given by.

T 2g

New length21 121

'100 199

T ' ' 21

T 100

T ' 11

T 10

1

T ' T T10

T 10% of T.

39. The displacement y of wave travelling in the x-direction is given by

y = 104sin 600t 2x3

metres,

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

2 2

= 300 m/sec.

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, thendi

e Ldt

Therefore,e

Ldi/dt

Substituting values, we get8 0.05

L 0.1 H4

41. In an oscillating LC circuit the maximum charge on the capacitor is Q. The charge on thecapacitor 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 =21 Q

E2 C

2 21 1 Q 1 q

2 2 C 2 C

q =Q

2.

42. The core of any transformer is laminated so as to



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(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. A.

43. Let F

be the force acting on a particle having position vector r

and T

be the torque of thisforce about the origin. Then

(A) r T 0

and T 0F

(B) r T 0

and T 0F

(C) r T 0

and T 0F

(D) r T 0

and T 0F

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.

2In20.4 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 f 1 and f 2. if the velocities of thephotoelectrons (of mass m) coming out are respectively v1 and v2, then

(A) 2 2

1 2 1 2

2hv v (f f )

m (B)

1/ 2

1 2 1 2

2hv v (f f )

m

(C) 2 2

1 2 1 2

2hv v (f f )

m

(D)

1/ 2

1 2 1 2

2hv v (f f )

m

46. A.

2 2

1 2 1 2

1m(v v ) h(f f )

2

2 2

1 2 1 2

2hv v (f f )

m .

47. Which of the following cannot be emitted by radioactive substance during their decay?



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(A) protons (B) neutrinos(C) helium nuclei (D) electrons

47. A.

48. A 3 volt battery with negligible internal resistance is connected in acircuit 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

3V 3

3

3

i

48. B.

The current through the circuit,V 3

I 1.5 AR 2

49. A sheet of aluminium foil of negligible thickness is introduced between the plates of acapacitor. 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 acapacitor, the capacitance of capacitor remains unchanged.

50. The displacement of a particle varies according to the relation x = 4(cos t + sin t). theamplitude 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 thecentre of the shell is

(A)0

2Q

4 R(B)

0

2Q

4 R

0

2q

4 R

(C)0 0

2Q q

4 R 4 R

(D)

0

(q Q) 2

4 R

51. C.

The total potential at P =0

1 1(q 2Q)

4 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

(C) 4 1010 joule (D) 32 1032 joule

52. D.

Required work done is2

2

1 Qw

2 C

=18 2

32

4

1 (8 10 )32 10 J

2 10



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53. The co-ordinates of a moving particle at any time t are given by x = t3 and y = t3. Thespeed to the particle at time t is given by

(A) 2 23t (B) 2 2 23t

(C) 2 2 2t (D) 2 2

53. B.

Speed = 2 2 2 2 2 2 2v (3 t ) (3 t ) 3t .

54. During an adiabatic process, the pressure of a gas is found to be proportional to the cube of

its absolute temperature. The ratiop

v

C

Cfor the gas is

(A) 4/3 (B) 2(C) 5/3 (D) 3/2

54. A.

P

V

C 4

C 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 arelationship 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

(C) 16.8 106 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 unstretchedposition. 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 betweentwo 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 n1 T

2L m

2 2

3

1 10 9.8 1 110 10 50Hz

2L 9.8 10 2L 2 1



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59. A tuning fork of known frequency 256 Hz makes 5 beats per second with the vibrating stringof a piano. The beat frequency decreases to 2 beats per second when the tension in thepiano string is slightly increased. The frequency of the piano string before increasing thetension 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 thefollowing 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

1 1 2H H He n given that the repulsive potential energy between the two nuclei is 7.7

1014 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(C) 103 K (D) 109 K

61. D.14

9

23

7.7 10 2T 3.7 10 K.

3 1.38 10

62. Which of the following atoms has the lowest ionization potential?

(A) 14

7N (B) 133

55Cs

(C) 40

18 Ar (D) 16

8O

62. B.Since 133

55Cs has larger size among the four atoms given, thus the electrons present in the

outermost orbit will be away from the nucleus and the electrostatic force experienced byelectrons due to nucleus will be minimum. Therefore the energy required to liberate electron

from outer orbit will be minimum in the case of 133

55Cs.

63. The wavelengths involved in the spectrum of deuterium 2

1( 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 toremove 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.



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65. A.

The energy of the first excited state of Li is2 2

0

2 2 2

Z E 3 13.6E 30.6 eV.

n 2

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

(C) t1/4 (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 10m/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 emitradiation 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

(A)2 3

2 2

q qcos

b a (B)

2 3

2 2

q qsin

b a

(C)2 3

2 2

q qcos

b a (D)

2 3

2 2

q qsin

b a

q3

q1 q2x

ba

y

69. B.

2 3

x 2 2

q qF sin

b a

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.2 2

consumed 2

V (110)P 250 watt.

R (220) /1000

71. 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



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71. C.

The objective of compound microscope is a convex lens. We know that a convex lens formsreal 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

904

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.

Rate of coolingd

dt

Therefore n = 1.

75. The length of a given cylindrical wire is increased by 100%. Due to the consequent decreasein 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

(C) ML1T1 (D) ML2T2

1. C.

Dimensions of (coefficient of viscosity)2

0 0 0 1

MLT

M L M LT

= ML1

T1

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.

2

f i

mkK K x

2

2f iK k x .

3. A ball is released from the top of a tower of height h metres. It takes T seconds to reach theground. 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 timeof 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.

1

2u sinT

g

and

2

2u sin (90 )T

g

and

2u sin 2

R g

1 2

2u sin 2u cos 2RT T .

g g g

6. Which of the following statements is false for a particle moving in a circle with a constantangular 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.



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AIEEE PAPER-04-PH-2


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 m7. 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 canexert a maximum force of 144 N on the gun. How many bullets can he fire per second at themost?(A) one (B) four (C) two (D) three

8. D.

Change in momentum for each bullet fired is

40

1200 48 N1000 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. Two masses m1 = 5 kg and m2 = 4.8 kg tied to a string are hangingover a light frictionless pulley. What is the acceleration of the masseswhen lift free to move?(g = 9.8 m/s2)(A) 0.2 m/s2 (B) 9.8 m/s2

(C) 5 m/s2 (D) 4.8 m/s2

9. A.

21 2

1 2

m ma g 0.2 m / sm m

m1

m2

10. A uniform chain of length 2 m is kept on a table such that a length of 60 cm hangs freelyfrom the edge of the table. The total mass of the chain is 4 kg. What is the work done inpulling 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. Thecoefficient of static friction between the block and the plane is 0.8. If the frictional force onthe block is 10 N, the mass of the block (in kg) is (take g = 10 m/s 2)(A) 2.0 (B) 4.0(C) 1.6 (D) 2.5

11. A.

m = 2 kg



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AIEEE PAPER-04-PH-3


12. A force ˆ ˆ ˆF (5i 3 j 2k)N

is applied over a particle which displaces it from its origin to the

point ˆ ˆr (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

Heref i

ˆ ˆs r r (2i j)

ˆ ˆ ˆ ˆ ˆW (5i 3 j 2k)(2i j) = 10 3 = 7 J.

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

(A) 1

1

mv t

t(B)

2

1

2

1

mv t

t

(C)2

1

1

mv t

t(D)

2

1

1

mv t

t

13. B.

Power 2

1 1 1

2

1 1 1

v v mv tP F v mav m t

t t t

14. A particle is acted upon by a force of constant magnitude which is always perpendicular tothe 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 theparticle 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 alsothere are no external torque acting but also there are no external torque acting on thesphere. 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 theball should be same as the speed of the person.

0

0

vv cos

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 I A and IB such that(A) I A = IB (B) I A > IB(C) I A < IB (D) I A/IB = d A/dB

Where d A and dB are their densities.

17. C.

Moment of inertia of a uniform density solid sphere, A = 22 MR5

And of hollow sphere B = 22MR

3Since M and R are same, I A < IB.

18. A satellite of mass m revolves around the earth of radius R at a height x from its surface. If gis the acceleration due to gravity on the surface of the earth, the orbital speed of the satelliteis

(A) gx (B)gR

R x

(C)

2gR

R x (D)

1/ 22gR

R x

18. D.

For the satellite, the gravitational force provides the necessary centripetal force i.e.2

e 0

2

GM m Mv

(R X) (R X)

and e

2

GMg

R

1/ 2

2

0

gRv

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 by3(R h)

T 2GM

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 energyof object of mass m raised from the surface of the earth to a height equal to the radius R of

the earth is

(A) 2 mgR (B)1

mgR2

(C)1

mgR4

(D) mgR

20. B.



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21. Suppose the gravitational force varies inversely as the nth power of distance. Then the timeperiod planet in circular orbit of radius R around the sun will be proportional to

(A)

n 1

2R

(B)

n 1

2R

(C) Rn (D)

n 2

2R

21. A.( n 1) / 2T R

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.

Work done = 2 21 1kx k

2 2 where is the total extensions.

1 1(k ) F

2 2

23. Spherical balls of radius R are falling in a viscous fluid of viscosity with a velocity v. Theretarding 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 = 6 Rv

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.

The pressure inside the smaller bubble will be morei 0

4TP P

r

Therefore, if the bubbles are connected by a tube, the air will flow from smaller bubble to thebigger.

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

given that the density of the bob is4

10003

kg/m3. What relationship between t and t0 is

true?(A) t = t0 (B) t = t0/2(C) t = 2t0 (D) t = 4t0

25. C.



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AIEEE PAPER-04-PH-6


0

T 1 1

1T '11

3

0

T2

T

or, T = 2T0

26. A particle at the end of a spring executes simple harmonic motion with a period t1, while thecorresponding period for another spring is t2. If the period of oscillation with the two springsin series is t, then

(A) T = t1 + t2 (B) 2 2 2

1 2T t t

(C) 1 1 1

1 2T t t (D) 2 2 2

1 2T t t

26. B.2 2 2

1 2t t T

27. The total energy of particle, executing simple harmonic motion is

(A) x (B) x

2

(C) independent of x (D) x1/2

27. C.

In simple harmonic motion, as a particle is displaced from its mean position, its kineticenergy 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 thewave is(A) 2000 m/s (B) 5 m/s

(C) 20 m/s (D) 5 m/s.

28. B.

1v 5 msk

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

(A)2 2

0

m

(B)

2 2

0

1

m( )

(C) 2 20

1

m( ) (D) 2 20

m

29. B.

For forced oscillations, the displacement is given by

x A sin( t ) with 0

2 2

0

F / m A



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AIEEE PAPER-04-PH-7


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 naturalfrequency. 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

1 2 1 2

m 1 2

n n n n

1 1 1

m

3

2

32. If the temperature of the sun were to increase from T to 2T and its radius from R to 2R, thenthe 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 r 2

P r 2

T4

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, thetemperature inside the vessel at equilibrium will be

(A) T1 + T2 (B) (T1 + T2)/2(C) 1 2 1 1 2 2

1 1 2 2 2 1

T T (P V P V )

P V T P V T

(D) 1 2 1 1 2 2

1 1 1 2 2 2

T T (P V P V )

P V T P T T

34. C.

The number of moles of system remains same According to Boyle’s law,P1V1 + P2V2 = P(V1 + V2)

1 2 1 1 2 2

1 1 2 2 2 1

T T (P V P V )T

P V T P V T



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35. A radiation of energy E falls normally on a perfectly reflecting surface. The momentumtransferred to the surface is(A) E/c (B) 2E/c(C) Ec (D) E/c2

35. B.

surface

2EP P .

c

36. The temperature of two outer surfaces of a composite slab,consisting of two materials having coefficients of thermalconductivity K and 2K and thickness x and 4x, respectively areT2 and T1 (T2 > T1). The rate of heat transfer through the slab,

in a steady state is 2 1 A(T T )K

f,x

with f equal to

(A) 1 (B) ½(C) 2/3 (D) 1/3

x 4x

2KK T1T2

36. D.

2 1

22T TkAq T

x 3

= 2 1

kAT T

3x

37. A light ray is incident perpendicular to one face of a 90 prism and istotally internally reflected at the glass-air interface. If the angle of

reflection is 45, we conclude that the refractive index n

(A)1

n2

(B) n 2

(C)1

n

2

(D) n 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 thecurved surface. Now this lens has been used to form the image of an object. At whatdistance from this lens an object be placed in order to have a real image of the size of theobject?(A) 20 cm (B) 30 cm

(C) 60 cm (D) 80 cm

38. A.

1 m

1 2 1

F f f

and1

1 1 1 1(1.5 1)

f 30 60

andmf 15 cm.

F = 10 cm.Object should be placed at 20 cm from the lens.



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AIEEE PAPER-04-PH-9


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)

(C) tan1(1/n) (D) tan1(n)

39. D.

Brewster’s law: According to this law the ordinary light is completely polarised in the planeof incidence when it gets reflected from transparent medium at a particular angle known asthe angle of polarisation.n = tan ip.

40. The maximum number of possible interference maxima for slit-separation equal to twice thewavelength in Young’s double-slit experiment is(A) infinite (B) five(C) three (D) zero

40. B.

For interference maxima, d sin = nHere d = 2

sin = n/2 and is satisfied by 5 integral values of n (2, 1, 0, 1, 2), as the maximumvalue 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,0

2

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 themrepel each other with a force F when kept apart at some distance. A third sphericalconductor having same radius as that of B but uncharged brought in contact with B, thenbrought 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.

2

0

1 (q / 2)(3q / 4) 3FF' .

4 d 8

43. A charged particle q is shot towards another charged particle Q which is fixed, with a speedv it approaches Q upto a closest distance r and then returns. If q were given a speed 2v, theclosest distances of approach would be(A) r (B) 2r (C) r/2 (D) r/4



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43. D.

By principle of conservation of energy

21 KqQmv

2 r …(i)

Finally, 2

2

1 KqQm(2v)

2 r …(ii)

Equation (i) (ii),

1 r '

4 r

r

r '4

.

44. Four charges equal to Q are placed at the four corners of a square and a charge q is at itscentre. If the system is in equilibrium the value of q is

(A)Q

(1 2 2)4

(B)Q

(1 2 2)4

(C)Q

(1 2 2)2

(D)Q

(1 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

46. C.

The given circuit can be written as

6 VI 4A

1.5

.

6 V

2

3

1.5

6

47. The resistance of the series combination of two resistances is S. When they are joined inparallel 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;

1 2

1 2

R RP

R R

S = nP

1 2

1 2

1 2

nR RR R

R R

2

1 2 1 2(R R ) nR R If R1 = R2, so minimum value of n = 4.



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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, thenthe ratio of the currents passing through the wire will be(A) 3 (B) 1/3(C) 8/9 (D) 2.

48. B.1 2

2 1

I R

I R

[current divider rule since voltage is same in parallel]2

1 2 1

2

2 1 2

I L r

I L r

2

1

2

I 3 2 1

I 4 3 3

.

49. In a metre bridge experiment null point is obtained at 20 cm from one end of the wire whenresistance 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 4Xagainst Y?(A) 50 cm (B) 80 cm(C) 40 cm (D) 70 cm

49. A.

We have from meter bridge experiment,

1 1

2 2

R

R

, where 2 = (100 1) cm

In the first case, X/Y = 20/80

In the second case4X

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 havinghigh 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 andcobalt with small amounts of other substance. The positive temperature coefficient types arebased 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. C.



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Let t be the time taken, then

836 t1000 1 (40 10)

4.2

[using Q = mst]

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

n

dEa 2b 0

d

;

n

a

2b

Now

a

700 Cb (given)

n =700/2 = 350CNow

c0 C.

So,n

0 C

But mathematicallyn

0 C .

53. The electrochemical equivalent of a metal is 3.3 107 kg per coulomb. The mass of themetal 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 magneticinduction at any point inside the tube is(A) infinite (B) zero

(C) 0 2i

4 r

tesla (D)

2i

r tesla

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 fieldat the centre of the coil is B. It is then bent into a circular loop of n turns. The magnetic fieldat the centre of the coil will be(A) nB (B) n2 B(C) 2nB (D) 2n2B

55. B.

0n i

B '2r '

= 2 20

in n B

.



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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 theloop?

(A) 250 T (B) 150 T

(C) 125 T (D) 75 T

56. A.

Using formula2

0

2 2 3 / 2iRB

2(R X )

, we get

2

0

2 2 3 / 2

i(3)54

2[(3) (4) ]

…(i)

At the centre of the coil, X = 0 and B = 0i

2(3)

Using equation (i)3

2

54 5B

(3) 3

B = 250 T.

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 timesand its direction reversed. The distance is also increased to 3d. The new value of the forcebetween them is

(A) 2F (B) F/3

(C) 2F/3 (D) F/3

57. C.

Force between two long conductor carrying current

0 1 2I I

F2 d

According to question

0 1 2( 2I )(I )

F '2 d

From equation (i) and (ii), 3F ' F.2

58. The length of a magnet is large compared to its width and breadth. The time period of itswidth and breadth. The time period of its oscillation in a vibration magnetometer is 2 s. Themagnet is cut along its length into three equal parts and three parts are then placed on eachother 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.

Time period of vibration,T

T 2MB

Where = moment of inertia of magnet, M = magnetic moment2m

I12

and M = pole strength

21 m I

I' 312 3 3 9

and M’ = pole strength (will remain the same) (/3) 3 = M.

T 2T ' s.

99



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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 inopposite phase.

61. A coil having n turns and resistance 4R . This combination is moved in time t seconds froma magnetic field W1 weber to W2 weber. The induced current in the circuit is

(A) 2 1W W

5Rnt

(B) 2 1

(W W )

5Rt

(C) 2 1W W

Rnt

(D) 2 1

n(W W )

Rt

61. B.

n dI

R ' dt

or, 2 1

2 1

W W1I n

R ' t t

(W1 and W2 are not the magnetic field, but the values of flux associated with one turn of coil)

2 1n(W W )1

I(R 4R) t

or, 2 1n(W W )

I

5Rt

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 isperpendicular to the field. If the total resistance of the circuit is R the mean power generatedper period of rotation is

(A)2B r

2R

(B)

2 2(B r )

2R

(C)2(B r )

2R

(D)

2 2(B r )

8R

62. B.

Magnetic flux = BA cos

= B2r

cos t2

2

ind

d 1B r

dt 2

sin t

2 2 2 4 2 2

ind B r sin tP

R 4R

Now, <sin2 t> = ½ (mean value)

2 2(B r )

P .8R



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63. In a LCR circuit capacitance is changed from C to 2C. For the resonant frequency to remainunchanged, 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, thenthe 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

2

ind

1B R 50 V.

2

65. According to Einstein’s photoelectric equation, the plot of the kinetic energy of the emittedphoto electrons from a metal Vs the frequency, of the incident radiation gives straight linewhose 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.

maxKE h W y = mx + C

Slope of the line in the graph is h, thePlanck’s constant.

K.E.

= h

W

66. The work function of a substance is 4.0 eV. Then longest wavelength of light that can causephotoelectron emission from this substance approximately(A) 540 nm (B) 400 nm

(C) 310 nm (D) 220 nm

66. C.

hcW

34 8

longest 19

hc 6.6 10 3 10

W 4.0 1.6 10

longest 310 nm.



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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

m g

qE

15

4

9.9 10 10

3 10

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

(C) 31/2 : 1 (D) 1 : 21/3

68. B.1/ 3

1 2

2 2

R m

R 2m

1/31

2

R1: 2

R .

69. The binding energy per nucleon of deuteron 2

1( H) and helium nucleus 4

2( He) is 1.1 MeV and

7 MeV respectively. If two deuteron nuclei react to form a single helium nucleus, then theenergy 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. Thedistance of the closest approach is of the order of

(A) 1 Å (B) 1010 cm

(C) 1012 cm (D) 1015 cm

70. C.

At closest approach, all the kinetic energy of the -particle will converted into the potentialenergy of the system, K.E. = P.E.

1 2

0

q q15 MeV

4 r

5 106 e = 9 1092

1 2Z Z e

r

9 19

6

9 10 92 2 1.6 10r

5 10

r = 5.3 1014 m = 5.3 1012 cm.



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71. When npn transistor is used as amplifier (A) electrons move from base to collector (B) holes move from emitter to base(C) electrons move from collector 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, fe

i

0e L

h A

1 h R

=6 3

5048.78

1 25 10 1 10

73. A piece of copper and another of germanium are cooled from room temperature to 77 K, theresistance 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 temperaturedecreases 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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FIITJEESOLUTION 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

(1) R2 (2) 1/R2

(3) 1/R (4) R

1. (4)2

1 2 2

2u sin2 2Rt t

gg

θ= =

2. An annular ring with inner and outer radii R1 and R2 is rolling without slipping with auniform 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) 2

1

R

R(2)

2

1

2

R

R

(3) 1 (4) 1

2

R

R

2. (4)2

1 1 1

2

2 2 2

F R R

F R R

ω= =

ω

3. A smooth block is released at rest on a 45° incline and then slides a distance d. Thetime taken to slide is n times as much to slide on rough incline than on a smooth

incline. The coefficient of friction is

(1)k 2

11

nµ = − (2)

k 2

11

nµ = −

(3)s 2

11

nµ = − (4)

s 2

11

nµ = −

3. (1)

2

1

1 gd t

2 2=

( ) 2

k 21 gd 1 t2 2

= − µ

2

22

2

1 k

t 1n

t 1= =

− µ

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 (2) 2cosφ

(3) 2tanφ (4) tanφ

4. (3)

mg s sin φ =s

mgcos .2

µ φ

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)

F.32

21 1 vmv m

2 2 4= −

F(3+x) 21mv

2

=

x = 1 cm

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=ax

2

+bx where a and b are constants.The acceleration is

(1) −2abv2 (2) 2bv3

(3) −2av3 (4) 2av2

7. (3)t = ax2 + bxby 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 ft

2

(3) S = 1/2 ft2 (4) S = 1/4 ft2

8. (none)2

1ft

S2

=

0v 2Sf =

During retardation t1 t2 t

V0

V

S2 = 2S



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During constant velocity15S -3S = 12S = vot

2ftS

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) 21m/ s

2towards north-east (2) 21

m/ s2

towards north.

(3) zero (4) 21m/ s

2towards north-west

9. (4)

f iV V

at

−=

r rr

( )ˆ ˆ5 j 5i 1 ˆ ˆ j i10 2

−= = −

21a ms2

−∴ = towards north west

(S)

(W) (E)

(N) j

i

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)

s = 50 +( )

( )

23 2 10 50

2 2

− × × −

= 293 m.

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

a

α

11. (4)

mg sinα = ma cos α a gtan∴ = α

ma

mg

N α

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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12. (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

cma 0∴ =

sincecm

v 0=

cmx 0∴∆ =

14. The moment of inertia of a uniform semicircular disc of mass M and radius r about aline perpendicular to the plane of the disc through the centre is

(1) 21Mr

4(2) 22

Mr 5

(3) Mr 2 (4) 21Mr

2

14. (4)2R

2I 2M2

=

2MRI

2∴ =

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?

(1) 3 m/s2 (2) 15 m/s2

(3) 5 m/s2 (4) 10 m/s2

15. (4)

2kxa 10m / s

m= =

16. 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. Themaximum momentum of the block after collision is

(1) MK L (2)2KL

2M

(3) zero (4)2ML

K

M

16. (1)2

21 PKL

2 2m= P MK L∴ =



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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

mm

v

Before

collision

After

collision

v/√3

(1) v (2) √3 v(3) 2v/√3 (4) v/√3

17. (3)

1mv mv cos= θ

1

mv0 mv sin

3= − θ

1

2v v

3∴ =

m

V1

θ

v

3

m

v

Before

collision

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 thecapillary 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

(1) 2S2Y (2) S2/2Y

(3) 2Y/S2 (4) S/2Y

19. (2)

1U

2= stress × strain

2S

2Y=

20. Average density of the earth(1) does not depend on g (2) is a complex function of g(3) is directly proportional to g (4) is inversely proportional to g

20. (3)

g =av

G4R

3

πρ

21. A body of mass m is accelerated uniformly from rest to a speed v in a time T. Theinstantaneous power delivered to the body as a function time is given by

(1)2

2

mvt

T⋅ (2)

2

2

2

mvt

T⋅

(3)2

2

1 mvt

2 T⋅ (4)

22

2

1 mvt

2 T⋅

21. (1)P= (ma).v



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= m a2 t

= m2

2

Vt

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

[ ]k

0.5µ =

(1) 800 m (2) 1000 m(3) 100 m (4) 400 m

22. (2)

2

k

1mgs mu

2µ =

2

k

us 1000m

2 g= =

µ

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 inthe figure, is lying on a smooth floor. A force F isapplied at the point P parallel to AB, such thatthe object has only the translational motionwithout rotation. Find the location of P withrespect to C

P

l

2 l Fr

A B

C

(1)2

3l (2)

3

2l

(3)4

3l (4) l

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 thesame as at a depth ‘d’ below the surface of earth. When both ‘d’ and ‘h’ are muchsmaller than the radius of earth, then which one of the following is correct?

(1)

h

d 2= (2)

3h

d 2= (3) d 2h= (4) d = h

25. (3)

( )2 3

GM GM

RR h=

+( )R d−

⇒ 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 betweenthem to take the particle far away from the sphere(you may take G = 6 . 67 × 10-11 Nm2 / kg2)

(1) 1013.34 10 J−× (2) 103.33 10 J−×

(3)

9

6.67 10 J

−

× (4)

10

6.67 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 P

v

C

Cof the

mixture is(1) 1.59 (2) 1.62(3) 1.4 (4) 1.54

27. (2)

1 v1 2 v 2

v

1 2

n c n cc

n n

+=

+=

29R

18

P

47Rc

18= , P

v

c1.62

c=

28. The intensity of gamma radiation from a given source is I. On passing through 36 mm

of lead, it is reduced toI

8. The thickness of lead which will reduce the intensity to

I

2will be

(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 electromagneticradiation 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)

g hcE 0.5eV= =λ

30. A photocell is illuminated by a small bright source placed 1 m away. When the same

source of light is placed1

2m away, the number of electrons emitted by photo

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)

2

1I

r ∝

31 Starting with a sample of pure 66Cu, 7/8 of it decays into Zn in 15 minutes. The

corresponding half-life is(1) 10 minutes (2) 15 minutes

(3) 5 minutes (4)1

72

minutes

31. (3)

1/ 2 1/ 2 1/ 2t t t0 0 0

0

N N NN

2 4 8 → → →

1/ 23t 15=

1/ 2t 5∴ =

32. If radius of 27

13 Al nucleus is estimated to be 3.6 Fermi then the radius 125

52Te nucleus

be nearly(1) 6 fermi (2) 8 fermi(3) 4 fermi (4) 5 fermi

32. (1)1

3R 125R 6

3.6 27

= ⇒ =

fermi

33. The temperature-entropy diagram of a reversible enginecycle is given in the figure. Its efficiency is(1) 1/2 (2) 1/4(3) 1/3 (4) 2/3

2So

2To

T

To

SS

33. (3)

BC

W

Q

∆η =

0 0

0 0

S T

2 1/ 33S T

2

= =

S

T

A C

B

S0 2S0

2T0

T0

34. The figure shows a system of two concentric spheres of radii

r 1 and r 2 and kept at temperatures T1 and T2 respectively. Theradial rate of flow of heat in a substance between the twoconcentric sphere is proportional to

T1

T2

r 1

r 2

(1) 2 1

1 2

r r

r r

−(2) 2

1

r ln

r

(3) 1 2

2 1

r r

r r −(4) ( )2 1

ln r r −



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34. (3)

( )( )

1 2

1 2

2 1

4 r r KdQT T

dt r r

π = − −

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 energiesin 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

P

V

II

I

A B

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)

( )1 cos2 ty

2

− ω=

37. A Young’s double slit experiment uses a monochromatic source. The shape of theinterference fringes formed on a screen is(1) hyperbola (2) circle(3) straight line (4) parabola

37. (3)Straight lineNote: If instead of young’s double slit experiment, young’s double hole experimentwas given shape would have been hyperbola.

38. Two simple harmonic motions are represented by the equation y1 = 0.1

sin 100 t3

π π +

and y2 = 0.1 cosπt. The phase difference of the velocity of particle 1

w.r.t. the velocity of the particle 2 is

(1) −π/6 (2) π/3

(3) −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

(1) 36 7 (2) 36 / 7

(3) 36 5 (4) 4 5



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39. (2)

2

h 36r

71= =

µ −

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 thesedots can be resolved by the eye? [ Take wavelength of light = 500 nm ](1) 5 m (2) 1m(3) 6 m (4) 3m

40. (1)

( )

( )1mm1.22Resolution limit

3mm R

λ= =

∴ R = 5 m

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)

am

air

m

1P

P1

µ− µ =

µ− µ

l

l

Pm=5/8 D

42. The diagram shows the energy levels for an electron

in a certain atom. Which transition shown representsthe emission of a photon with the most energy ?(1) III (2) IV(3) I (4) II

n=4

n=3

n=2

n=1I II III IV

42. (1)

∆E ∝ 2 2

1 2

1 1

n n

−

43. If the kinetic energy of a free electron doubles. Its deBroglie wavelength changes bythe factor

(1)1

2(2) 2

(3)1

2(4) 2

43. (3)

h

2Kmλ =



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44. In a common base amplifier, the phase difference between the input signal voltageand output voltage is

(1)4

π(2) π

(3) 0 (4)2

π

44. (3)

No phase difference between input and output signal.

45. In a full wave rectifier circuit operating from 50 Hz mains frequency, the fundamentalfrequency 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) 7

3Li . Which of the following is the

nucleus of element X ?

(1) 12C6 (2) 10

5B

(3) 9

5B (4) 11

4Be

46. (2)1 4 7

0 2 3X n He Li+ → +

47. A moving coil galvanometer has 150 equal divisions. Its current sensitivity is10 divisions per milliampere and voltage sensitivity is 2 divisions per millivolt. Inorder that each division reads 1 volt, the resistance in ohms needed to be connectedin series with the coil will be(1) 103 (2) 105 (3) 99995 (4) 9995

47. (4)

Ig=15mA Vg = 75mV

g

g g

VVR

I I= −

48. Two voltameters one of copper and another of silver, are joined in parallel. When atotal 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 thecharge which flows through the silver voltameter is

(1) 1

2

q

z1

z+

(2) 2

1

q

z1

z+

(3) q 1

2

z

z(4) q 2

1

z

z

48. (2)q1Z1=q2Z2 q=q1+q2



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∴ 2

2

1

qq

Z1

Z

=+

49. In the circuit, the galvanometer G showszero deflection. If the batteries A and Bhave negligible internal resistance, the

value of the resistor R will be

2V

12V

500Ω

R

A

G

B

(1) 200 Ω (2) 100 Ω

(3) 500 Ω (4) 1000 Ω

49. (2)

12R2

500 R=

+

50. Two sources of equal emf are connected to an external resistance R. The internalresistance of the two sources are R1 and R2 (R2 > R1). If the potential differenceacross 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)

1 2

2EI

R R R=

+ +

2E R I 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 potentialdifference V across the capacitance is

(1) 2mC Ts

∆ (2) mC Ts∆

(3)ms T

C

∆(4)

2ms T

C

∆

51. (4)

Dimensionally only 4th option is correct.

52. One conducting U tube can slide inside another asshown in figure, maintaining electrical contactsbetween the tubes. The magnetic field B isperpendicular 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

A

D x

B

x x x x

x x x x

x x x x

x x

CD

VV

(1) BlV (2) – BlV

(3) zero (4) 2 BlV

52. (4)

d2B v

dt

φ= l



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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)2V t

HR

∆=

2

'

VH' t

R= ∆ Given R’ = R/2

54. Two thin, long parallel wires separated by a distance ‘d’ carry a current of ‘i’ A in thesame 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 intensityof the light which does not get transmitted is

(1)0

1I

2(2)

0

1I

4

(3) zero (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 isproportional to

(1) cos θ (2) cot θ

(3) sin θ (4) tan θ

+

+

+

+

+

+

+

S

B

θ

P

56. (4)

tanθ = ( )o

q

2 mg

σ

ε

Mg

o

q

2

σε

θ

57. Two point charges + 8q and – 2q are located at x = 0 and x = L respectively. Thelocation of a point on the x axis at which the net electric field due to these two pointcharges is zero is(1) 2L (2) L/4



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(3) 8L (4) 4L

57. (1)

( )2 2

k2q k8q0

xx L− + =

−

⇒ x = 2L

58. Two thin wires rings each having a radius R are placed at a distance d apart withtheir axes coinciding. The charges on the two rings are +q and –q. The potentialdifference between the centres of the two rings is

(1) QR/4πε0d2 (2)

2 20

Q 1 1

2 R R d

−

πε +

(3) zero (4)2 2

0

Q 1 1

4 R R d

−

πε +

58. (2)

12 2

kq kqv

R R d

= −

+

22 2

kq kqv

R R d

−= +

+

59. A parallel plate capacitor is made by stacking n equally spaced plates connectedalternatively. If the capacitance between any two adjacent plates is C then theresultant 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 thetuning 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)

|f 1−f 2| =4

Since mass of second tuning fork increases so f 2 decrease and beats increase so

f 1>f 2

⇒ f 2=f 1−4 = 196

61. If a simple harmonic motion is represented by2

2

d x

dt+ αx = 0, its time period is

(1)2πα

(2)2π

α

(3) 2πα (4) 2π α



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61. (2)

ω2=α

ω = √α

2T

π=

α

62. The bob of a simple pendulum is a spherical hollow ball filled with water. A pluggedhole near the bottom of the oscillation bob gets suddenly unplugged. Duringobservation, 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)

v v /5 6f f f

v 5

+= =

% increase in frequency = 20%

64. If I0 is the intensity of the principal maximum in the single slit diffraction pattern, thenwhat will be its intensity when the slit width is doubled?(1) 2I0 (2) 4I0 (3) I0 (4) I0/2

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 eachother. 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)

2 2o 1 2B I I2r

µ= +

7

2

4 10B 5

2 2 10

−

−

π ×= ×

× π ×

B = 5×10−5

66. A coil of inductance 300 mH and resistance 2Ω is connected to a source of voltage2V. 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 (4) 0.3 s

66. (2)R

tL

oI I 1 e

− = −

R0.693 t

L=

.3 0.693t 0.1sec

2

×= =

67. The self inductance of the motor of an electric fan is 10 H. In order to impartmaximum power at 50 Hz, it should be connected to a capacitance of

(1) 4µF (2) 8µF

(3) 1µF (4) 2µF

67. (3)

1f

2 LC=

π

2 21C4 f 10

= × π ×

C = 1µF

68. An energy source will supply a constant current into the load of its internal resistanceis(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)

oE E

I if R r R r r = <<+

69. A circuit has a resistance of 12 Ω and an impedance of 15 Ω. The power factor of thecircuit will be(1) 0.8 (2) 0.4(3) 1.25 (4) 0.125

69. (1)

R 12 4cos 0.8

Z 15 5φ = = = =

70. The phase difference between the alternating current and emf is π/2. Which of the

following cannot be the constituent of the circuit?(1) C alone (2) R.L(3) L. C (4) L alone

70. (2)

0<phase difference for R-L circuit < π/2



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71. A uniform electric field and a uniform magnetic field are acting along the samedirection in a certain region. If an electron is projected along the direction of the fieldswith a certain velocity then(1) its velocity will decrease. (2) its velocity will increase.(3) it will turn towards right of direction of motion. (4) it will turn towards left of direction of motion.

71. (1)

F e E v B eE = − + × = − r r r rr

eEa

m= −

rr

( ) o

eEv t v t

m= −

72. A charged particle of mass m and charge q travels on a circular path of radius r thatis perpendicular to a magnetic field B. The time taken by the particle to complete onerevolution is

(1)2 mq

B

π(2)

22 q B

m

π

(3)2 qB

m

π(4)

2 m

qB

π

72. (4)

mω2r=Bqωr

ω = Bq/m

2 mT

qB

π=

73. In a potentiometer experiment the balancing with a cell is at length 240 cm. On

shunting the cell with a resistance of 2Ω the balancing length becomes 120 cm. The

internal resistance of the cell is(1) 1 Ω (2) 0.5 Ω

(3) 4 Ω (4) 2 Ω

73. (4)

1

2

240r R 1 2 1 2

120

= − = − = Ω

l

l

74. The resistance of hot tungsten filament is about 10 times the cold resistance. Whatwill be the resistance of 100 W and 200 V lamp when not in use?

(1) 40 Ω (2) 20 Ω

(3) 400 Ω (4) 200 Ω

74. (1)2

hot

V 200 200R 400

P 100

×= = = Ω

cold resistance = Rhot/10 = 400/10 = 40 Ω

75. A magnetic needle is kept in a non-uniform magnetic field. It experiences(1) a torque but not a force (2) neither a force nor a torque(3) a force and a torque. (4) a force but not a torque.



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75. (3)

In non uniform magnetic field, dipole experiences both force and torque.



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FIITJEE Solutions to AIEEE−2006- PHYSICS

FIITJEE Ltd. , ICES H ouse , 29 – A, Kalu Sarai, Sarvapr iya Vihar , Ne w Del hi -110016, Ph 2651 5949, 26569493, Fa x: 011-26513942.

41. A particle located at x = 0 at time t = 0, starts moving along the positive x-direction with a velocity

‘v’ that varies as v = xα . The displacement of the particle varies with time as(1) t3 (2) t2 (3) t (4) t1/2

Ans: (2)

Sol. dx dx dtdt x

= α = α ⇒∫ ∫ x α t2

42. A mass of M kg is suspended by a weightless string. The horizontal force that is required todisplace it until the string makes an angle of 45° with the initial vertical direction is

(1) Mg( 2 1)− (2) Mg( 2 1)+

(3) Mg 2 (4)Mg

2

Ans: (1)

Sol. F A sin 45 = Mg (A − A cos 45)

F = Mg (√2 − 1)

43. A bomb of mass 16 kg at rest explodes into two pieces of masses of 4 kg and 12 kg. The velocityof the 12 kg mass is 4 ms−1. The kinetic energy of the other mass is(1) 96 J (2) 144 J(3) 288 J (4) 192 J

Ans: (3)

Sol. m1v1 = m2v2

22 2

1 1KE m v 4 144 288J

2 2= = × × =

44. A particle of mass 100 g is thrown vertically upwards with a speed of 5 m/s. the work done by theforce of gravity during the time the particle goes up is

(1) 0.5 J (2) −0.5 J(3) −1.25 J (4) 1.25 J

Ans: (3)

Sol. − mgh = − mg2v

1.25J2g

= −

.

45. A whistle producing sound waves of frequencies 9500 Hz and above is approaching a stationaryperson with speed v ms−1. The velocity of sound in air is 300 ms−1. If the person can hear frequencies upto a maximum of 10,000 Hz, the maximum value of v upto which he can hear thewhistle is

(1) 30 ms

−1

(2) 15 2 ms

−1

(3) 15 / 2 ms−1 (4) 15 ms−1

Ans: (4)

Sol.( )

app

f 300f v 15m / s

300 v= ⇒ =

−

46. A electric dipole is placed at an angle of 30° to a non-uniform electric field. The dipole willexperience



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(1) a torque only(2) a translational force only in the direction of the field(3) a translational force only in a direction normal to the direction of the field(4) a torque as well as a translational force

Ans: (4)

Sol. A torque as well as a translational force

47. A material ‘B’ has twice the specific resistance of ‘A’. A circular wire made of ‘B’ has twice the

diameter of a wire made of ‘A’. Then for the two wires to have the same resistance, the ratio A A /AB

of their respective lengths must be(1) 2 (2) 1

(3)1

2(4)

1

4

Ans: (1)

Sol. A A1 2

A

RR

ρ=

π

A B B

2 2B

RR

ρ=

π

A

2 2 A B A A A2 2

B A B A A

R 2 R

R 4Rρ ρ= =ρ ρ ⋅

A

A ⇒ B

A

2=A

A

48. The Kirchhoff’s f irst law ( i 0)=∑ and second law ( )iR E=∑ ∑ , where the symbols have their

usual meanings, are respectively based on(1) conservation of charge, conservation of energy(2) conservation of charge, conservation of momentum(3) conservation of energy, conservation of charge(4) conservation of momentum, conservation of charge

Ans: (1)

Sol. Conservation of charge, conservation of energy49. In a region, steady and uniform electric and magnetic fields are present. These two fields are

parallel to each other. A charged particle is released from rest in this region. The path of theparticle will be a(1) circle (2) helix(3) straight line (4) ellipse

Ans: (3)

Sol. Straight line

50. Needles N1, N2 and N3 are made of a ferromagnetic, a paramagnetic and a diamagneticsubstance respectively. A magnet when brought close to them will(1) attract all three of them(2) attract N1 and N2 strongly but repel N3 (3) attract N1 strongly, N2 weakly and repel N3 weakly(4) attract N1 strongly, but repel N2 and N3 weakly

Ans: (3)

Sol. attracts N1 strongly, N2 weakly and Repel N3 weakly

51. Which of the following units denotes the dimensions ML2/Q2, where Q denotes the electriccharge?



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(1) Weber (Wb) (2) Wb/m2 (3) Henry (H) (4) H/m2

Ans: (3)

Sol. Henry (H)

52. A player caught a cricket ball of mass 150 g moving at a rate of 20 m/s. If the catching process iscompleted in 0.1 s, the force of the blow exerted by the ball on the hand of the player is equal to(1) 300 N (2) 150 N(3) 3 N (4) 30 N

Ans: (4)

Sol. ( )mv 0 0.15 20− ⇒ ×

3F 30N

0.1= =

53. A ball of mass 0.2 kg is thrown vertically upwards by applying a force by hand. If the hand moves0.2 m which applying the force and the ball goes upto 2 m height further, find the magnitude of theforce. Consider g = 10 m/s2

(1) 22 N (2) 4 N(3) 16 N (4) 20 N

Ans: (4)

Sol. mgh = FsF = 20 N

54. Consider a two particle system with particles having masses m1 and m2. If the first particle ispushed towards the centre of mass through a distance d, by what distance should the secondparticle be moved, so as to keep the centre of mass at the same position?

(1) d (2) 2

1

m

md

(3) 1

1 2

mm m+

d (4) 1

2

m dm

Ans: (4)

Sol. m1 d + m2 x = 0

1

2

m dx

m=

55. Starting from the origin, a body oscillates simple harmonically with a period of 2 s. After what timewill its kinetic energy be 75% of the total energy?

(1)1

s12

(2)1

s6

(3)1

s4

(4)1

s3

Ans: (2)

Sol. 2 2max

1 3 1mv mv

2 4 2

=

2 2 2 2 23 A cos t A

4ω ω ⇒ ω



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3cos t

2ω =

1t t sec

6 6

πω = ⇒ =

56. The maximum velocity of a particle, executing simple harmonic motion with an amplitude 7 mm, is

4.4 m/s. The period of oscillation is(1) 100 s (2) 0.01 s(3) 10 s (4) 0.1 s

Ans: (2)

Sol. Aω = vmax

max

2 2 AT 0.01sec

v

π π= = =

ω

57. A string is stretched between fixed points separated by 75 cm. It is observed to have resonantfrequencies of 420 Hz and 315 Hz. There are no other resonant frequencies between these two.Then, the lowest resonant frequency for this string is(1) 10.5 Hz (2) 105 Hz

(3) 1.05 Hz (4) 1050 Hz

Ans: (2)

Sol. ( )( )n 1n

v 315, v 4202 2

+= =

A A

Solvingv

1052

=A

58. Assuming the sun to be a spherical body of radius R at a temperature of T K, evaluate the totalradiant power, incident on Earth, at a distance r from the Sun.

(1)2 4

2

R T

r

σ(2)

2 2 40

2

4 r R T

r

π σ

(3)2 2 40

2

r R T

r

π σ(4)

2 2 40

2

r R T

4 r

σ

π

where r 0 is the radius of the Earth and σ is Stefan’s constant.

Ans: (3)

Sol. 2

4 202

r ( T 4 R )

4 r

πσ ⋅ π

π

4 2 20

2

T R r

r

σπ=

59. The refractive index of glass is 1.520 for red light and 1.525 for blue light. Let D1 and D2 be an of minimum deviation for red and blue light respectively in a prism of this glass. Then(1) D1 > D2

(2) D1 < D2 (3) D1 = D2 (4) D1 can be less than or greater than depending upon the angle of prism

Ans: (2)

Sol. D = (µ − 1)AD2 > D1



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60. In a Wheatstone’s bridge, there resistances P, Q and R connected in the three arms and thefourth arm is formed by two resistances S1 and S2 connected in parallel. The condition for bridgeto be balanced will be

(1)1 2

P R

Q S S=

+(2)

1 2

P 2R

Q S S=

+

(3) 1 2

1 2

R(S S )P

Q S S

+= (4) 1 2

1 2

R(S S )P

Q 2S S

+=

Ans: (3)

Sol. 1 2

1 2

R(S S )P

Q S S

+=

61. 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. 5

i 0.5

10

= =

5Ω 10Ω 5Ω

5 Volt

10Ω

10Ω I

+ −

62. In a series resonant LCR circuit, the voltage across R is 100 volts and R = 1 kΩ with C = 2µF. Theresonant 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. 100

i 0.1 A1000

= =

L C 6

0.1V V 250

200 2 10−= = =

× ×V

63 Two insulating plates are both uniformly charged in such a way thatthe potential difference between them is V2 −V1 = 20 V. (i.e. plate 2is at a higher potential). The plates are separated by d = 0.1 m andcan be treated as infinitely large. An electron is released from rest onthe 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(3) 7.02 × 1012 m/s (4) 1.87 × 106 m/s

0.1 m

Y

X

2 1

Ans: (2)

Sol. 21mv eV

2

=

2eVv

m= = 2.65 × 106 m/s

64. The resistance of a bulb filament is 100 Ω at a temperature of 100°C. If its temperature coefficientof 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

Ans: (2)



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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 ‘(1) N.A.B.ω (2) N.A.B.R.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 by210t 50t 250φ = − +

The induced emf at t = 3 s is(1) 190 V (2) −190 V(3) −10 V (4) 10 V

Ans: (3)

Sol. d

e (20 t 50) 10dt

φ= − = − − = − volt

67. A thermocouple is made from two metals, Antimony and Bismuth. If one junction of the couple iskept 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(1) 10−1 s (2) 10−4 s(3) 10−10 s (4) 10−16 s

Ans: (3)

Sol. 10−10 sec.

69. An alpha nucleus of energy21

mv2 bombards a heavy nuclear target of charge Ze. Then thedistance of closest approach for the alpha nucleus will be proportional to

(1)1

Ze(2) v2

(3)1

m(4)

4

1

v

Ans: (3)



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Sol. 1

m

70. The threshold frequency for a metallic surface corresponds to an energy of 6.2 eV, and thestopping 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. 1242eVnm

110011.2

λ = ≈ Å

Ultraviolet region

71. The energy spectrum of β-particles [number N(E) as a function of β-energy E] emitted from aradioactive source is(1)

N(E)

E0

E

(2)

N(E)

E0

E

(3)

N(E)

E0

E

(4)

N(E)

E0

E

Ans: (4)

Sol.

N(E)

E0

E

72. When 3Li7 nuclei are bombarded by protons, and the resultant nuclei are 4Be8, the emittedparticles 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 isformed by(1) Metallic binding (2) Ionic binding(3) Covalent binding (4) Van der Waals binding

Ans: (3)

Sol. Covalent binding



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74. If the ratio of the concentration of electrons that of holes in a semiconductor is7

5and the ratio of

currents is7

,4

then what is the ratio of their drift velocities?

(1)4

7(2)

5

8

(3) 45

(4) 54

Ans: (4)

Sol. e

n

n 7

n 5= e

n

I 7

I 4=

d e

d n

(V )

(V ) ⇒ e n

n e

I n 5

I n 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. b e cI I I= −

β = c

b

I49

I=

76. The potential energy of a 1 kg particle free move along the x-axis is given by4 2x x

V(x) J4 2

= −

The total mechanical energy of the particle 2 J. Then, the maximum speed (in m/s) is

(1) 2 (2) 3 / 2

(3) 2 (4) 1/ 2

Ans: (2)

Sol. k Emax = ET − Umin Umin (±1) = −1/4 J

KEmax = 9/4 J ⇒ U =3

2J

77. A force of ˆFk− acts on O, the origin of the coordinate system. The torque about the point (1, −1)is

(1) ( )ˆ ˆF i j− − (2) ( )ˆ ˆF i j−

(3) ( )ˆ ˆF i j− + (4) ( )ˆ ˆF i j+

Ans: (3)

Sol. ˆ ˆ ˆ( i j) ( Fk)τ = − + × −G

= - F( ˆ î j)+



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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′ =

(1)m

(m 2M)

ω+

(2)(m 2M)

m

ω +

(3)(m 2M)

(m 2M)

ω −

+

(4)m

(m M)

ω

+

Ans: (1)

Sol. Li = Lf mR2ω = (mR2 + 2MR2)ω′

ω′ =m

m 2M

ω +

79. If the terminal speed of a sphere of gold (density = 19.5 kg/m3) 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)

Sol. s s

g g

v ( )

v ( )

ρ − ρ=

ρ − ρA

A

vs = 0.1 m/s

80. The work of 146 kJ is performed in order to compress one kilo mole of gas adiabatically and inthis 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 totalenergy density of the electromagnetic wave is(1) 3.3 × 10−3 J/m3 (2) 4.58 × 10−6 J/m3 (3) 6.37 × 10−9 J/m3 (4) 81.35 × 10−12 J/m3

Ans: (2)

Sol. Uav = 2 60 rmsE 4.58 10−ε = × J/m3

2 6 3rmsE 4.58 10 J/m−= ×

82. 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 leavecontact with the platform for the first time(1) at the highest position of the platform (2) at the mean position of the platform

(3) for an amplitude of 2

g

ω(4) for an amplitude of

2

2

g

ω

Ans: (3)

Sol. Aω2 = g⇒ A = g/ω2



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83. An electric bulb is rated 220 volt − 100 watt. The power consumed by it when operated on 110volt will be(1) 50 watt (2) 75 watt(3) 40 watt (4) 25 watt

Ans: (4)

Sol. 2 2

1 2

1 2

V VResistance

P P= =

⇒ P2 = 25 W

84. The anode voltage of a photocell is kept fixed. The wavelength λ of the light falling on the cathodeis gradually changed. The plate current I of the photocell varies as follows :(1)

O λ

I

(2)

O λ

I

(3)

O λ

I

(4)

O λ

I

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. If the binding energy per nucleon in 73Li and 4

2He nuclei are 5.60 MeV and 7.06 MeV

respectively, then in the reaction7 43 2p Li 2 He+ →

energy of proton must be(1) 39.2 MeV (2) 28.24 MeV(3) 17.28 MeV (4) 1.46 MeV

Ans: (3)

Sol. EP = (8 × 7.06 – 7 × 5.60) MeV = 17.28 MeV

87. If the lattice constant of thissemiconductor is decreased, then which

of the following is correct?

EC

EV

Eg

Condutton band width

Band gapValence badn width

(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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88. In the following, which one of the diodes is reverse biased?(1)

R

+ 5 V

(2)

R

+ 10 V

+ 5 V

(3)

R10 V

− 5 V

(4)

R

−10 V

Ans: (1)

89. The circuit has two oppositely connect ideal diodesin parallel. What is the current following in thecircuit?

4 Ω

3 Ω 2 Ω

D1

12 V

D2

(1) 1.33 A (2) 1.71 A(3) 2.00 A (4) 2.31 A

Ans: (3)

Sol. D1 is reverse biased therefore it will act like an open circuit.12

i 2.00 A6

= =

90. A long solenoid has 200 turns per cm and carries a current i. The magnetic field at its centre is6.28 × 10−2 Weber/m2. Another long solenoid has 100 turns per cm and it carries a current i/3. Thevalue of the magnetic field at its centre is(1) 1.05 × 10−4 Weber/m2 (2) 1.05 × 10−2 Weber/m2 (3) 1.05 × 10−5 Weber/m2 (4) 1.05 × 10−3 Weber/m2

Ans: (2)

Sol. B2 =2

21 2 2

1 1

B n i (6.28 10 )(100 i / 3)1.05 10

n i 200(i)

−−× ×

= = × W/m2

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 mA

2

(3) 23 mA (4) 3 mA2

Ans: (4)

Sol. I = 2m (A/ 2 )2 = 3 mA2

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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Ans: (2)

93. Two rigid boxes containing different ideal gases are placed on a table. Box A contains one moleof 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 untilthe 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(1) f 0

5T T

2= (2) f 0

3T T

7=

(3) f 07

T T3

= (4) f 03

T T2

=

Ans: (4)

Sol. ∆U = 0

⇒ f 0 f 0

3 5 7R(T T ) 1 R(T T ) 0

2 2 3− + × − =

Tf = 0

3T

2

94. Two spherical conductors A and B of radii 1 mm and 2 mm are separated by a distance of 5 cmand are uniformly charged. If the spheres are connected by a conducting wire then in equilibriumcondition, 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)

Sol. A B

B A

E r 2

E r 1= =

95. An inductor (L = 100 mH), a resistor (R = 100 Ω) and abattery (E = 100 V) are initially connected in series asshown in the figure. After a long time the battery is

disconnected after short circuiting the points A and B. Thecurrent in the circuit 1 mm after the circuit isE

A B

R

L

(1) 1 A (2) 1/e A(3) e A (4) 0.1 A

Ans: (2)

Sol. I = I0 Rt/Le− =

1 A

e

* * * * * *



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FIITJEE Solutions to AIEEE - 2007 -PHYSICS

FIITJEE Ltd. , ICES Hous e, 2 9 – A, Kalu Sara i, Sarvapri ya Vihar, N ew Delhi -110 016, Ph 2651 5949, 2 6569 493, Fax: 01 1-26 5139 42.

1

P Phh y yssiiccss CCooddee--OO

41. The displacement of an object attached to a spring and executing simple harmonic motion is

given by x = 2 × 10−2cos 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 × 10−2cos πt

v = −0.02π sin πt

v is maximum at t =1

0.5sec2

=

42. In an a.c. circuit the voltage applied is E = E0 sin ωt. The resulting current in the circuit is

I = I0 sin t2

π ω −

. The power consumption in the circuit is given by

(1) 0 0E IP

2= (2) P = zero

(3) 0 0E IP

2= (4) P 2 E0I0

Sol. (2)

cos φ = 0So 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 theplates becomes equal to the electromotive force of the battery. The ratio of the energy stored inthe capacitor and the work done by the battery will be(1) 1 (2) 2

(3)1

4(4)

1

2

Sol. (4)

1qv

12

qv 2

=

45. An ideal coil of 10H is connected in series with a resistance of 5 Ω and a battery of 5V. 2 secondafter the connection is made the current flowing in amperes in the circuit is

(1) (1 − e) (2) e

(3) 1e− (4)1(1 e )−−



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2

Sol. (4)Rt

L0

i i 1 e

= −

= ( )11 e−−

46. A long straight wire of radius ‘a’ caries a steady current i. The current is uniformly distributed

across its cross section. The ratio of the magnetic field ata

2and 2a is

(1)1

4(2) 4

(3) 1 (4)1

2

Sol. (3)2

0 2

a i aB2

2 4a

ππ = µ

π

01

iB

4 a

µ=

π…(i)

2 0B 2 (2a) iπ = µ

02

iB

4 a

µ=

π…(ii)

1

2

B1

B=

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 isotope17

8 O , Mp and MN are the masses of a proton and a

neutron respectively, the nuclear binding energy of the isotope is

(1) 2

O P(M 8M )C− (2) 2

O P N(M 8M 9M )C− −

(3)2

OM C (4)

2

O N(M 17M )C−

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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3

50. If in a p-n junction diode, a square

input signal of 10V is applied asshown

5V

−5V

RL

Then the output signal across RL will be

(1) 10V (2)

−10V

(3)

−5V

(4) +5V

Sol. (4)

51. Photon of frequency ν has a momentum associated with it. If c is the velocity of light, themomentum is

(1) ν/c (2) hνc

(3) hν/c

2

(4) hν/c

Sol. (4)

h hP

c

ν= =

λ

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

2

0

0 0

dx (V gt ft )dt= + +∫ ∫

x = v0 +1 1

g f 2 3

+

53. For the given uniform square lamina ABCD, whose centre is O,

(1) AC EF2I I=

(2) AD EFI 3I=

(3) AC EFI I=

(4) AC EFI 2I=

A B

CD

F

E

Sol. (3)

AC EFI I= (from ⊥rd

axis theorem)

54. A point mass oscillates along the x-axis according to the law x = x0 cos (ωt − π/4). If theacceleration of the particle is written as

a = A cos(ωt + δ)

(1) A = x0 , δ = −π/4 (2) A = x0ω2, δ = −π/4

(3) A = x0ω2, δ = −π/4 (4) A = x0ω2

, δ = 3π/4



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4

Sol. (4)

v = −x0ω sin (ωt − π/4)

a = −x0ω2cos t

4

π ω + π −

a = A cos(ωt + δ)

A = x0ω2;

3

4

πδ =

55. Charges are placed on the vertices of a square as shown. Let E be theelectric field and V the potential at the centre. If the charges on A and Bare interchanged with those on D and C respectively, then

(1) EG

remains unchanged, V changes

(2) Both EG

and V change

(3) EG

and V remains unchanged

(4) EG

changes, V remains unchanged

A B

CD

q q

−q −q

Sol. (4)

As EG

is a vector quantity

56. The half-life period of a radio-active element X is same as the mean life time of another radio-active 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)

1/ 2

x

ln2t =

λ

mean

y

1τ =

λ;

dNN

dt= −λ

x y

ln2 1

=λ λ ⇒ x yλ = λ (0.6932) ⇒ y xλ > λ

57. A Carnot engine, having an efficiency of η = 1/10 as heat engine, is used as a refrigerator. If thework 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)

12

2

TW Q 1

T

= −

2

1

T1

Tη = −

10 2

10Q 1

9

= −

2

1

T11

10 T= − ⇒ 2

1

T 1 91

T 10 10= − =

2

110 Q

9

=

⇒ 1

2

T 10

T 9=

Q2 = 90 J

58. Carbon, silicon and germanium have four valence electrons each. At room temperature whichone 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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5

Sol. (4)

59. A charged particle with charge q enters a region of constant, uniform and mutually orthogonal

fields EG

and BG

with a velocity vG

perpendicular to both EG

and BG

, and comes out without any

change in magnitude or direction of vG

. Then

(1) 2v E B /B= ×G GG

(2) 2v B E /B= ×G GG

(3) 2v E B /E= ×G G

G (4) 2v B E /E= ×G G

G

Sol. (1)

v B E× = −G GG

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)

x 2

20V

x 4=

−

2 2

dV 20E (2x 0)

dx (x 4)= − = −

−

160 10

144 9= =

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)

2

2 2

1 2

1 1h Rhcz

n n

ν = −

62. A block of mass ‘m’ is connected to another block of mass ‘M’ by a spring (massless) of springconstant ‘k’. The blocks are kept on a smooth horizontal plane. Initially the blocks are at rest andthe spring is unstretched. Then a constant force ‘F’ starts acting on the block of mass ‘M’ to pullit. Find the force on the block of mass ‘m’

(1)mF

M(2)

(M m)F

m

+

(3)mF

(m M)+(4)

MF

(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

1f

P= ⇒ −0.1 m ⇒ −10 cm



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6

64. One end of a thermally insulated rod is kept at atemperature T1 and the other at T2. The rod is

composed of two sections of lengths A1 and A2 and

thermal conductivities k1 and k2 respectively. Thetemperature at the interface of the two sections is

T1T2

k1 k2

A2A1

(1) (k2A2T1+ k1A1T2) / (k1A1 + k2A2) (2) (k2A1 T1 + k1A1T2) / (k2A1 + k1A2)

(3) (k1A2 T1 + k2A1T2) / (k1A2 + k2A1) (4) (k1A1 T1 + k2A2T2) / (k1A1 + k2A2)

Sol. (3)

1 1 2 2

1 2

(T T)k (T T )k− −=

A A

l 1 2 2 2 1

1 2 2 1

Tk T kT

k k

+=

+

A A

A A

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)

1

0

IB 10log I

=

2

0

I'B log

I

=

given 2 1B B 20− =

20 = 10 logI'

I

I' 100I=

66. If Cp and Cv denote the specific heats of nitrogen per unit mass at constant pressure and constantvolume respectively, then

(1) Cp −Cv = R/28 (2) Cp −Cv = R/14

(3) Cp −Cv = R (4) Cp −Cv = 28RSol. (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 lyingat a distance ‘d’ from O, in a direction perpendicular to the plane of the wires AOB and COD, willbe given by

(1)

1 2

0 1 2I I

2 d

µ + π

(2) ( )1 2

2 20

1 2I I

2 d

µ+

π

(3) ( )0

1 2I I

2 d

µ+

π(4) ( )2 20

1 2I I

2 d

µ+

π



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7

Sol. (2)

2 201 2

I(I I )

2 d

µ+

π

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 °Cwill be

(1) 2 ohm (2) 1 ohm

(3) 4 ohm (4) 3 ohm

Sol. (3)

5 1 50

6 1 100

+ α=

+ α

5 = R0(1+ α × 50)RO = 4

70. A parallel plate condenser with a dielectric of dielectric constant K between the plates has acapacity C and is charged to a potential V volts. The dielectric slab is slowly removed frombetween 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 moonrespectively and if Millikan’s oil drop experiment could be performed on the two surfaces, one will

find the ratioelectronic charge on the moon

electronic charge on the earthto be

(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

(1) 1/3 (2) 1/2

(3) 1/6 (4) 1/4

Sol. (1)

In this question distance of centre of mass of new disc is αR notRα

.

3M MR R 0

4 4− α + =

⇒ 1

3α =



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8

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)

2

gsin

I1

MR

θ

+(2)

2

gsin

MR1

I

θ

+

(3)

2

gsin

I1

MR

θ

−(4) 2

gsin

MR1

I

θ

−

Sol. (1)

Mg sin θ − f = Ma

fR = Ia

R

⇒ a =

2

gsin

I1

MR

θ

+

θ

f

Mg

N

74. 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 constantis 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 thehighest 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 6

λ(λ

being the wavelength of the light used) is I. If I0 denotes the maximum intensity,0

I

Iis equal to

(1)1

2(2)

3

2

(3) 1/2 (4) 3/4



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9

Sol. (4)

2

max

Icos

I 2

φ =

78. Two springs, of force constants k1 and k2, are connected to amass 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

(1) f/2 (2) f/4

(3) 4f (4) 2f

m

k2k1

Sol. (4)

1 2k k1f

2 m

+=

π

1 2k k1

f ' 2 2f 2 m

+= =

π

79. When a system is taken from state i to state f along the path iaf, itis 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

i b

f

Sol. (1)

80. A particle of mass m executes simple harmonic motion with amplitude ‘a’ and frequency ‘ν’. Theaverage kinetic energy during its motion from the position of equilibrium to the end is

(1) π2ma

2ν2(2)

1

4π2

ma2ν2

(3) 4π2ma

2ν2(4) 2π2

ma2ν2

Sol. (1)

2 2 2 2 21ma f ma

4ω = π

*************



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FIITJEE Solutions to AIEEE - 2008

FIITJEE Ltd., ICES Ho use, 29 – A, Kalu Sarai , Sa rvapriya Vihar, New Delhi - 110016 , Ph 26515949, 26569493, Fax: 011-26513942.

1

A A IIEEEEEE––22000088,, PP A A PPEERR((CC−55)) 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 deductedfor 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.

M M aatthheemmaattiiccss 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 polealong 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

(1)7 3 1

2 3 1⋅

−m (2) ( )7 3

3 12

⋅ + m

(3)( )7 3

3 12 ⋅ − m (4)

7 3 1

2 3 1⋅ +

Sol: (2) BD = AB = 7 + x

Also AB = x tan 60° = x 3

∴ x 3 7 x= +

x =7

3 1−

AB = ( )7 33 1

2+ .

45° 60°

A

BCD 7 x

2. It is given that the events A and B are such that P (A) =1

4,

A 1P

B 2 =

andB 2

P A 3

=

. Then P (B) is

(1)1

6(2)

1

3

(3)2

3(4)

1

2

Sol: (2) ( )

( )

P A B 1

2P B

∩= ,

( )

( )

P A B 2

3P A

∩=

Hence

( )

( )

P A 3

4P B = . (But P (A) = 1/4)

⇒ ( )1

P B3

= .



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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

(1)3

5(2) 0

(3) 1 (4)

2

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 thelength of the semi−major axis is

(1)8

3(2)

2

3

(3)4

3

(4)5

3

Sol: (1) Major axis is along x-axis.a

ae 4e

− =

1a 2 4

2 − =

a =8

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)

(2, 0)O

X =2

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

1

2

α += − 1 and

02

2

β += − .

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

(1) g (y) =

3y 4

3

+(2) g (y) =

y 3

4 4

+

+

(3) g (y) =y 3

4

+(4) g (y) =

y 3

4

−



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Sol: (4)

Function is increasing

x = ( )y 3

g y4

−= .

8. The conjugate of a complex number is 1i 1−

. Then the complex number is

(1)1

i 1

−−

(2)1

i 1+

(3)1

i 1

−+

(4)1

i 1−

Sol: (3) Put − i in place of i

Hence1

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 ∈ Ias 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. Thena possible value of k is(1) 1 (2) 2(3) − 2 (4) − 4

Sol: (4)

Slope of bisector = k − 1Middle point =

k 1 7,

2 2

+

Equation of bisector is

y − 7

2= (k − 1)

( )k 1x

2

+−

Put x = 0 and y = − 4.⇒ k = ± 4.

11. The solution of the differential equationdy x y

dx x

+= satisfying the condition y (1) = 1 is

(1) y = ln x + x (2) y = x ln x + x2

(3) y = xe(x−1) (4) y = x ln x + x

Sol: (4) y = vx



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dy dv

v xdx dx

= +

v +dv

x 1 vdx

= +

⇒ dv =dx

x

∴ v = log x + c

⇒ y

log x cx

= +

Since, y (1) = 1, we havey = 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 followinggives 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

65

+ + + +=

⇒ a + b = 7 … (1)Given that Variance is 6.8

∴ Variance =( )

2iX A

n

−∑

=( ) ( )2 2a 6 b 6 4 1 16

6.85

− + − + + +=

⇒ a2 + b2 = 25a2 + (7 − a)2 = 25 (Using (1))⇒ a2 − 7a + 12 = 0∴ a = 4, 3 and b = 3, 4.

13. The vector ˆ ˆ â i 2 j k= α + + βG

lies in the plane of the vectors bG

= ˆ î j+ and ˆ ˆc j k= +G

and bisects the

angle between bG

and cG

. Then which one of the following gives possible values of α and β?(1) α = 2, β = 2 (2) α = 1, β = 2(3) α = 2, β = 1 (4) α = 1, β = 1

Sol: (4) ( )ˆ â b c= λ +

G

⇒

ˆ ˆ î 2 j kˆ ˆ î 2 j k 2

+ +α + + β = λ

λ = 2α and λ = 2 and λ = 2β

⇒ α = 1 and β = 1.

14. The non−zero verctors a, bGG

and cG

are related by a 8b=GG

and c 7b= −GG

. Then the angle between aG

and cG

is(1) 0 (2) π/4(3) π/2 (4) π

Sol: (4)

Since a 8b=GG

c 7b= − GG

∴ aG

and bG

are like vectors and bG

and cG

are unlike.⇒ aG

and cG

will be unlike



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Hence, angle between aG

and cG

= π.

15. The line passing through the points (5, 1, a) and (3, b, 1) crosses the yz−plane at the point17 13

0, ,2 2

−

. Then

(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

y 1x 5 z a

2 1 b a 1

−− −= = = λ

− −.

If line crosses yz−plane i.e., x = 0x = 2λ + 5 = 0⇒ λ = −5/2,

Since, y = λ (1 − b) + 1 =17

2

( )5 17

1 b 12 2− − + = b = 4.

Also, z = λ (a − 1) + a =13

2−

( )5 13

a 1 a2 2

− − + = −

⇒ a = 6.

16. If the straight linesy 2x 1 z 3

k 2 3

−− −= = and

y 3x 2 z 1

3 k 2

−− −= = intersect at a point, then the

integer k is equal to

(1) − 5 (2) 5(3) 2 (4) − 2

Sol: (1) y 2x 1 z 3

k 2 3

−− −= = and

y 3x 2 z 1

3 k 2

−− −= =

Since lines intersect in a pointk 2 3

3 k 2

1 1 2−= 0

∴ 2k2 + 5k − 25 = 0

k = − 5, 5/2.Directions: Questions number 17 to 21 are Assertion−Reason type questions. Each of these questionscontains two statements : Statement − 1 (Assertion) and Statement−2 (Reason). Each of these questions alsohas 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

... n1 2 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)



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P (n) =1 1 1

...1 2 n

+ + +

P (2) =1 1

21 2

+ >

Let us assume that P (k) =1 1 1

... k1 2 k

+ + + > is true

∴ P (k + 1) =1 1 1 1

... k 11 2 k k 1

+ + + + > ++

has to be true.

L.H.S. >( )1 k k 1 1

kk 1 k 1

+ ++ =

+ +

Since ( )k k 1 k+ > (∀ k ≥ 0)

∴ ( )k k 1 1 k 1

k 1k 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 trueSince (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


Sol: (4)

Let A =a b

c d

so that A2 =2

2

a bc ab bd 1 0

0 1ac dc bc d

+ + =

+ +

⇒ 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.

19. Statement −1: ( ) ( )n

n n 1r

r 0

r 1 C n 2 2 −

=

+ = +∑ .

Statement −2: ( ) ( ) ( )n

n n 1n r r

r 0

r 1 C x 1 x nx 1 x−

=

+ = + + +∑ .

(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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Sol: (2)

( )n

nr

r 0

r 1 C=

+∑ =n

n nr r

r 0

r C C=

+∑

=n n

n 1 nr 1 r

r 0 r 0

nr C C

r −

−

= =

+∑ ∑ = n 1 nn2 2− +

= 2n−1 (n + 2)Statement −1 is true

( ) n r n r n r r r r r 1 C x r C x C x+ = +∑ ∑ ∑

=n n

n 1 r n r r 1 r

r 0 r 0

n C x C x−−

= =

+∑ ∑ = nx (1 + x)n−1 + (1 + x)n

Substituting x = 1( ) n n 1 n

r r 1 C n 2 2−+ = +∑

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 pStatement –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


Sol: (4) Given statement r = ∼ p ↔ qStatement −1 : r 1 = (p ∧ ∼ q) ∨ (∼ p ∧ q)

Statement −2 : r 2 = ∼ (p ↔ ∼ q) = (p ∧ q) ∨ (∼ q ∧ ∼ p)From the truth table of r, r 1 and r 2,r = r 1.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 thenumber 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


Sol: (1) x1 + x2 + x3 + x4 + x5 = 65 + 6 – 1C5 – 1 = 10C4.

22. Let f(x) =( )

1x 1 sin , if x 1

x 1

0, if x 1

− ≠ −

=

. Then which one of the following is true?

(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

Sol: (1)

f ′(1) =( ) ( )

h 0

f 1 h f 1lim

h→

+ −



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⇒ f ′(1) =

( )

h 0 h 0

11 h 1 sin 0

h 11 h 1lim lim sinh h h→ →

+ − − + − =

⇒ f ′(1) =h 0

1lim sin

h→

∴ f is not differentiable at x = 1.Similarly, f ′(0) =

( ) ( )

h 0

f h f 0lim

h→

−

⇒ f ′(0) =

( ) ( )

h 0

1h 1 sin sin 1

h 1limh→

− − −

⇒ 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 termsis 48. If the terms of the geometric progression are alternately positive and negative, then the firstterm is(1) –4 (2) –12

(3) 12 (4) 4Sol: (2)

Let a, ar, ar 2, …a + ar = 12 …(1)ar 2 + ar 3= 48 …(2)dividing (2) by (1), we have

( )

( )

2ar 1 r

a r 1

++

= 4

⇒ r 2 = 4 if r ≠ – 1∴ r = – 2also, 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?

(1) The cubic has minima atp

3and maxima at –

p

3

(2) The cubic has minima at –p

3and maxima at

p

3

(3) The cubic has minima at bothp

3and –

p

3

(4) The cubic has maxima at bothp

3and –

p

3

Sol: (1) Let f(x) = x3 – px + qNow for maxima/minimaf ′(x) = 0⇒ 3x2 – p = 0

⇒ x2 =p

3

∴ x = ± p

3.

–√(p/3)

√(p/3)

25. 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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Sol: (2) x7 + 14x5 + 16x3 + 30x – 560 = 0Let 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) (2) p → (p ∨ q)(3) p → (p ∧ q) (4) 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. The value of 1 15 2cot cosec tan

3 3− − +

is

(1)6

17

(2)3

17

(3)4

17(4)

5

17

Sol: (1)

Let E = 1 15 2cot cosec tan

3 3− − +

⇒ E = 1 13 2cot tan tan

4 3− − +

⇒ E = 1

3 2

4 3cot tan

3 214 3

−

+

− ⋅

⇒ E = 1 17 6cot tan

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 …(1)

⇒ 2(x – h) + 2(y – 2)dy

dx= 0

⇒ (x – h) = – (y – 2)dy

dx

substituting in (1), we have

( ) ( )2

2 2dyy 2 y 2 25

dx − + − =

(y – 2)2y′2 = 25 – (y – 2)2.

29. Let I =1

0

sinxdx

x∫ and J =1

0

cosxdx

x∫ . Then which one of the following is true?

(1) I > 23 and J > 2 (2) I < 23 and J < 2

(3) I <2

3and J > 2 (4) I >

2

3and J < 2



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Sol: (2)

I =1

0

sinxdx

x∫ <1 1 1

3 / 2

00 0

x 2 2dx xdx x

3 3x= = =∫ ∫

⇒ I <2

3

J =1

0

cosxdx

x∫ <1

1

0

0

1dx 2 x 2

x= =∫

∴ J ≤ 2.

30. The area of the plane region bounded by the curves x + 2y2 = 0 and x + 3y2 = 1 is equal to

(1)5

3(2)

1

3

(3)2

3(4)

4

3

Sol: (4) Solving the equations we get the points of intersection (–2, 1) and (–2, –1)The bounded region is shown as shadedregion.

The required area = 2 ( ) ( )1

2 2

0

1 3y 2y− − −∫

= ( )11 3

2

00

y 2 42 1 y dy 2 y 2

3 3 3

− = − = × =

∫ .

(–2, 1)

(–2, –1)x + 2y = 0

x + 3y = 1

(1, 0) x

y

31. The value of sinxdx

2

sin x4

π −

∫ is

(1) x + log cos x c4

π − +

(2) x – log sin x c4

π − +

(3) x + log sin x c4

π − +

(4) x – log cos x c4

π − +

Sol: (3)

sin x dxsinxdx 4 42 2

sin x sin x4 4

π π − + =π π − −

∫ ∫

= 2 cos cot x sin dx4 4 4

π π π + − ∫

= dx cot x dx4

π + − ∫ ∫

= x + ln sin x c4

π − +

.

32. How many different words can be formed by jumbling the letters in the word MISSISSIPPI in which notwo S are adjacent?(1) 8 . 6C4 . 7C4 (2) 6 . 7 . 8C4 (3) 6 . 8 . 7C4 (4) 7 . 6C4 . 8C4



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Sol: (4)

Other than S, seven letters M, I, I, I, P, P, I can be arranged in7!

2! 4!= 7 . 5 . 3.

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

c 1 a

b a 1

− −−

−= 0 ⇒ 1(1 – a2) + c(–c – ab) – b(ca + b) = 0

⇒ 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 theadjoint of matrix A is integer.

Now detA = ± 1 and A –1 =1

det(A)(adj 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 rootsof 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, thenα + 4β = 6 and 4αβ = aα + 3β = c and 3αβ = 6.We get αβ = 2 ⇒ a = 8So the first equation is x2 – 6x + 8 = 0 ⇒ x = 2, 4

If α = 2 and 4β = 4 then 3β = 3If α = 4 and 4β = 2, then 3β = 3/2 (non-integer)∴ common root is x = 2.

CChheemmiissttr r y y

PART − B

36. The organic chloro compound, which shows complete stereochemical inversion during a SN2 reaction,is(1) (C2H5)2CHCl (2) (CH3)3CCl(3) (CH3)2CHCl (4) CH3Cl

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.



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37. Toluene is nitrated and the resulting product is reduced with tin and hydrochloric acid. The product soobtained is diazotised and then heated with cuprous bromide. The reaction mixture so formedcontains(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

NO2

CH3

NO2

+

Sn/HClSn/HCl

CH3

NH2

CH3

NH2

NaNO2/HClNaNO2/HCl

CH3

N2Cl

CuBr

CH3

Br

CH3

N2

ClCuBr

CH3

Br

CH3

Nitration →

38. 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 3Sol. (4)

E

en

en

oxNO2

Coordination no. = 6 and Oxidation no. = 3



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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

2 4Conc.H SO →

OH

SO3H

3Conc.HNO

→

OH

NO2

41. In the following sequence of reactions, the alkene affords the compound ‘B’3 2O H O

3 3 ZnCH CH CHCH A B.= → →

The compound B is(1) CH3CH2CHO (2) CH3COCH3 (3) CH3CH2COCH3 (4) CH3CHO

Sol. (4)

CH3 CH CH CH3 CH3 CH

O O

CH

O

CH3

(A)

H2O/Zn

C

HCH3

O

(B) 42. Larger number of oxidation states are exhibited by the actinoids than those by the lanthanoids, themain 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) [Co(H2O)6]3+ (4) [Co(NH3)6]

3+



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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 mmHg. If a mixture solution of ‘A’ and ‘B’ boils at 80oC and 1 atm pressure, the amount of ‘A’ in themixture is (1 atm = 760 mm Hg)

(1) 52 mol percent (2) 34 mol percent(3) 48 mol percent (4) 50 mol percent

Sol. (4)o o

T A A B BP P X P X= +

760 = ( )o A B A520X P 1 X+ −

AX 0.5⇒ =

Thus, mole% of A = 50%

45. For a reaction1 A 2B,

2→ rate of disappearance of ‘A’ is related to the rate of appearance of ‘B’ by the

expression

(1) [ ] [ ]d A d B1dt 2 dt

− = (2) [ ] [ ]d A d B1dt 4 dt

− =

(3)[ ] [ ]d A d B

dt dt− = (4)

[ ] [ ]d A d B4

dt dt− =

Sol. (2)

1 A 2B

2→

[ ] [ ]2d A d B

dt 2dt

−= +

[ ] [ ]d A d B1

dt 4 dt

−=

46. The equilibrium constants1PK and

2PK for the reactions X 2YU and Z P Q,+U 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 atthese equilibria is(1) 1 : 36 (2) 1 : 1(3) 1 : 3 (4) 1 : 9

Sol. (1)

( )

X 2Y

1 0

1 x 2x−

ZZZXYZZZ

( )

( )1

2 1

1p

2x Pk

1 x 1 x

= − +

( )

Z P Q

1 0 0

1 x x x

+

−

ZZZXYZZZ

( )2

122

p

Pxk

1 x 1 x

= − +

1 1

2 2

4 P P1 1

P 9 P 36

×= ⇒ =

47. Oxidising power of chlorine in aqueous solution can be determined by the parameters indicatedbelow:



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( ) ( ) ( ) ( )

diss eg hyd

1H H H2

2

1Cl g Cl g Cl g Cl aq .

2

∆ ∆ ∆− − → → →

The energy involved in the conversion of ( )2

1Cl g

2to Cl−(g)

(using the data,

2

1 1 1

diss Cl eg Cl hyd ClH 240 kJmol , H 349 kJmol , H 381kJmol )− − −∆ = ∆ = − ∆ = −

will be(1) +152 kJmol−1 (2) −610 kJmol−1 (3) −850 kJmol−1 (4) +120 kJmol−1

Sol. (2)

For the process ( )2 aq

1Cl g Cl

2− →

diss 2 eg hyd

1H H of Cl Cl Cl

2−∆ = ∆ + ∆ + ∆

240349 381

2= + − −

= − 610 kJ mol−1

48. Which of the following factors is of no significance for roasting sulphide ores to the oxides and notsubjecting 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

HCHO+ →

OH

CH2OH

CH2OH

Polymerize → CH2

CH2

O

n

50. For the following three reactions a, b and c, equilibrium constants are given:

a. ( ) ( ) ( ) ( )2 2 2 1CO g H O g CO g H g ; K+ +U b. ( ) ( ) ( ) ( )4 2 2 2CH g H O g CO g 3H g ; K+ +U

c. ( ) ( ) ( ) ( )4 2 2 2 3CH g 2H O g CO g 4H g ; K+ +U

Which of the following relations is correct?

(1) 1 2 3K K K= (2) K2K3 = K1

(3) K3 = K1K2 (4) 3 23 2 1K .K K=

Sol. (3)Equation (c) = equation (a) + equation (b)Thus K3 = K1.K2

51. The absolute configuration of



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HO2C

OH OH

CO2H

HH is

(1) S, S (2) R, R(3) R, S (4) S, R

Sol. (2)HO2C

OH OH

CO2H

HH

1 2

Both C1 and C2 have R – configuration.

52. The electrophile, E⊕ attacks the benzene ring to generate the intermediate σ-complex. Of thefollowing, which σ-complex is of lowest energy?

(1)

NO2

H E

(2)

H

E

(3)

H

E

NO2

(4)

NO2

H

E

Sol. (2)

NO2 is electron withdrawing which will destabilize σ - complex.

53. α-D-(+)-glucose and β-D-(+)-glucose are(1) conformers (2) epimers(3) anomers (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,

2 2 3

1 3X Y XY , H 30 kJ,

2 2+ → ∆ = − to be at equilibrium, the temperature will be

(1) 1250 K (2) 500 K(3) 750 K (4) 1000 K

Sol. (3)

2 2 3

1 3X Y XY

2 2+ →

1

reaction

3 1

S 50 40 60 40 Jmol2 2

−

∆ = − × + × = − ∆G = ∆H - T∆Sat equilibrium ∆G = 0



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∆H = T∆S30 × 103 = T × 40⇒ T = 750 K

55. Four species are listed belowi. 3HCO− ii. H3O

+

iii. 4HSO− iv. HSO3FWhich 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?(1) 2

2 2C , O , CO, NO− − (2) 22 2NO , C , CN , N+ − −

(3) 2 22 2 2CN , N , O , C− − − (4) 2 2N , O , NO , CO− +

Sol. (2)22 2NO , C , CN and N+ − −

all have fourteen electrons.

57. Which one of the following pairs of species have the same bond order?(1) CN− and NO+ (2) CN− and CN+ (3) 2O− and CN− (4) NO+ 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(1) 8.51 × 105 Jmol−1 (2) 6.56 × 105 Jmol−1 (3) 7.56 × 105 Jmol−1 (4) 9.84 × 105 Jmol−1

Sol. (4)6 6

2 1 2

1.312 10 1.312 10E E E

12

× ×∆ = − = − − −

5 19.84 10 J mol−= ×

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)

Al

Cl Cl

Cl Cl

Al

Cl

Cl

Be

Cl

Cl Cl

Be

Cl

Cl

Be

Cl

60. Given 3 2Cr / Cr Fe /FeE 0.72 V, E 0.42 V.+ +

° °= − = − The potential for the cell

Cr Cr 3+ (0.1 M)Fe2+ (0.01 M)Fe is(1) 0.26 V (2) 0.399 V(3) −0.339 V (4) −0.26 V

Sol. (1)



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3 2

0 0Cr / Cr Fe /Fe

As E 0.72 V and E 0.42 V+ += − = − 2 32Cr 3Fe 3Fe 2Cr + ++ → +

( )

( )

23

0cell cell 32

Cr 0.0591E E log

6 Fe

+

+= −

( )( )

( )

2

3

0.10.05910.42 0.72 log

6 0.01= − + −

( )

( )

2

3

0.10.05910.30 log

6 0.01= −

2

6

0.0591 100.30 log

6 10

−

−= − 40.05910.30 log10

6= −

Ecell = 0.2606 V

61. Amount of oxalic acid present in a solution can be determined by its titration with KMnO4 solution inthe 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 gof 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)0

ssolute

s

P P XP− =

s

s

17.5 P 0.1

P 10

−=

s

s

17.5 P0.01

P

−=

⇒ Ps = 17.325 mm Hg

63. Among the following substituted silanes the one which will give rise to cross linked silicone polymer onhydrolysis is(1) R4Si (2) RSiCl3

(3) R2SiCl2 (4) R3SiCl

Sol. (2)

R Si

Cl

Cl

Cl 2H O → R Si

OH

OH

OH Condensationpolymerization

→ R Si O Si R

O

O

Si

O

Si

Si

O

Si

n

64. In context with the industrial preparation of hydrogen from water gas (CO + H2), which of the followingis the correct statement?



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(1) CO and H2 are fractionally separated using differences in their densities(2) CO is removed by absorption in aqueous Cu2Cl2 solution(3) H2 is removed through occlusion with Pd(4) CO is oxidised to CO2 with steam in the presence of a catalyst followed by absorption of CO2 in

alkali

Sol. (4) 2H O2 2 2CO H CO 2H+ → +

KOH

K2CO3

65. In a compound atoms of element Y from ccp lattice and those of element X occupy 2/3rd of tetrahedralvoids. The formula of the compound will be(1) X4Y3 (2) X2Y3 (3) X2Y (4) X3Y4

Sol. (1)

No. of atoms of Y = 4

No. of atoms of X =2

83

×

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. Thecorrect 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(1) CH3CH2CH2C≡CCH2CH2CH3 (2) CH3CH2C≡CH(3) CH3CH=CHCH3 (4) CH3CH2C≡CCH2CH3

Sol. (2)

3Na/Liq.NH3 2 3 2CH CH C CH CH CH C CNa

Θ⊕

∆− ≡ → ≡

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(1) CH3−CH=CH2 (2) CH3C≡C−CH3

(3) CH3 C

H

C

H

CH3 (4) CH4

Sol. (4)

3 3 4CH MgX CH C C H CH− + − ≡ − →

69. The correct decreasing order of priority for the functional groups of organic compounds in the IUPACsystem of nomenclature is(1) −COOH, −SO3H, −CONH2, −CHO (2) −SO3H, −COOH, −CONH2, −CHO(3) −CHO, −COOH, −SO3H, −CONH2 (4) −CONH2, −CHO, −SO3H, −COOH

Sol. (2)3 2SO H, COOH, CONH , CHO− − − −

70. The pKa of a weak acid, HA, is 4.80. The pKb of a weak base, BOH, is 4.78. The pH of an aqueoussolution of the corresponding salt, BA, will be



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(1) 9.58 (2) 4.79(3) 7.01 (4) 9.22

Sol. (3)It is a salt of weak acid and weak base

w a

b

K KH

K

+ × =

pH = 7.01

P Phh y yssiiccss

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 byrequiring that electron waves reflected from the planes of atoms in a crystal interfere constructively (see infigure).

Incoming

Electrons

Outgoing

Electrons

d

i

Crystal plane

71. Electrons accelerated by potential V are diffracted from a crystal. If d = 1Å and i = 30°, V should beabout (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

2meV

v = 50 volt

i

72. If a strong diffraction peak is observed when electrons are incident at an angle ‘i’ from the normal tothe crystal planes with distance ‘d’ between them (see figure), de Broglie wavelength λdB of electronscan be calculated by the relationship (n is an integer)(1) d sin i = nλdB (2) 2d cos 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 deBroglie wavelength. They are detected on a screen at a distance ‘D’ from the slit (see figure).



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D

d y = 0

Which of the following graph can be expected to represent the number of electrons ‘N’ detected as afunction of the detector position ‘y’(y = 0 corresponds to the middle of the slit)?(1) y

d N

(2) y

d N

(3) y

d N

(4) y

d N

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 timessmaller. Given that the escape velocity from the earth is 11 kms−1, the escape velocity from thesurface of the planet would be(1) 1.1 kms−1 (2) 11 kms−1 (3) 110 kms−1 (4) 0.11 kms−1

Sol. (3)

vesc =2GM 2G 10M

R R 10

×= = 10 × 11 = 110 km/s

75. 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 <ρ1). Assume that the liquid applies a viscous force on the ball that is proportional to thesquare of its speed v, i.e., Fviscous = −kv2(k>0). The terminal speed of the ball is

(1) 1 2Vg( )

k

ρ − ρ(2) 1Vg

k

ρ

(3) 1Vg

k

ρ(4) 1 2Vg( )

k

ρ − ρ

Sol. (1)

ρ1Vg − ρ2Vg =

2

Tkv

⇒ vT =( )1 2Vg

k

ρ − ρ

76. Shown in the figure below is a meter-bridge set up with null deflection in the galvanometer.

G

55Ω R

20 cm

The value of the unknown resistor R is(1) 13.75 Ω (2) 220 Ω (3) 110 Ω (4) 55 Ω



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Sol. (2) 55 R 55 8

R 22020 80 2

×= ⇒ = = Ω

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

(mass/length) varies with x as

nx

k L , where n can be zero or any positive number. If the position xCM of the centre of mass of the rod is plotted against ‘n’, which of the following graphs best approximatesthe dependence of xCM on n?(1)

L

L/2

O n

xCM (2)

L/2

O n

xCM

(3)

L

L/2

O n

xCM (4)

L

L/2

O n

xCM

Sol. (1)

xcm =

λ = =

∫∫ ∫∫ ∫ ∫

n

n

xk .xdxdmx dx.x L

dm dm xk dx

L

( )

( )

( )

+

+

++ = = + +

Ln 2

Ln

n 10

n0

kx

x n 1n 2 Ln 2kx

n 1 L

xcm =L 2L 3L 4L 5L

, , , , , . . .2 3 4 5 6

78. While measuring the speed of sound by performing a resonance column experiment, a student getsthe first resonance condition at a column length of 18 cm during winter. Repeating the sameexperiment 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 RT

4x M

γ

xn =1 RT

4 M

γ

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 (3) MT−1C−1 (4) MT−2C−1

Sol. (3) F = qvBB = 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 anaxis perpendicular to its plane and passing through one of its corners is

(1)

5

6 ma2

(2)

1

12 ma2

(3)7

12ma2 (4)

2

3ma2



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Sol. (4)

I = Icm + m

22 2

2a 2 ma ma 2ma

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(1) 17.6 kg ms−1 (2) 17.565 kg ms−1 (3) 17.56 kg ms−1 (4) 17.57 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 beestimated 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

Approximate KE =1

60 100 3000 J2

× × =

KE range ⇒ 2000 to 5000 joule

83. A parallel plate capacitor with air between the plates has a capacitance of 9 pF. The separationbetween its plates is ‘d’. The space between the plates is now filled with two dielectrics. One of the

dielectrics has dielectric constant k1 = 3 and thicknessd

3while the other one has dielectric constant

k2 = 6 and thickness 2d3

. Capacitance of the capacitor is now

(1) 1.8 pF (2) 45 pF(3) 40.5 pF (4) 20.25 pF

Sol. (3)

C′ = 0 0

1 2

A A

d d d 2d

9 183 6

ε ε=

++= 018A

4d

ε

C′ = 40.5 PF

3 6

C = 9 PF

84. The speed of sound in oxygen (O2) at a certain temperature is 460 ms−1. The speed of sound inhelium (He) at the same temperature will be (assumed both gases to be ideal)(1) 460 ms−1 (2) 500 ms−1 (3) 650 ms−1 (4) 330 ms−1

Sol. No option is correct

v =RT

M

γ

1 1 2

2 2 1

74V M 5

5V M

323

×γ= =

γ

×

2

460 21

V 25 8=

× ⇒ v2 =

460 5 2 2

21

× ×= 1420



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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.andStatement – II

For heavy nuclei, binding energy per nucleon increases with increasing Z while for light nuclei itdecrease 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 itssides is 4π GM.andStatement – IIIf the direction of a field due to a point source is radial and its dependence on the distance ‘r’ for thesource is given as 1/r 2, its flux through a closed surface depends only on the strength of the sourceenclosed 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/r 2

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

Sol. (4) As liquid 1 floats above liquid 2,

ρ1 < ρ2

Liquid 1

Liquid 2

ρ3

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. Noconduction is found between P and Q. By connecting the common (negative) terminal of themultimeter to R and the other (positive) terminal to P or Q, some resistance is seen on themultimeter. 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. (2)



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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’ andleaves from ‘D’. We apply superposition principle to find voltage ‘∆V’ developed between ‘B’ and ‘C’. Thecalculation 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 = ρ j, 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’.

A B C D

aa b

∆v

I I

89. ∆V measured between B and C is

(1)I I

a (a b)

ρ ρ−

π π +(2)

I I

a (a b)

ρ ρ−

+

(3)I I

2 a 2 (a b)

ρ ρ−

π π +(4)

I

2 (a b)

ρπ −

Sol. (3) Choosing A as origin,

E = ρ j = ρ2

I

2 r π

VC − VB = −

( )a b

2a

I 1

dr 2 r

+ρ

π ∫ = ( )

I 1 1

2 aa b

ρ

− π +

VB − VC =( )

I 1 1

2 a a b

ρ − π +

90. For current entering at A, the electric field at a distance ‘r’ from A is

(1)2

I

8 r

ρπ

(2)2

I

r

ρ

(3)2

I

2 r

ρπ

(4)2

I

4 r

ρπ

Sol. (3)

91. A student measures the focal length of convex lens by putting an object pin at a distance ‘u’ from thelens and measuring the distance ‘v’ of the image pin. The graph between ‘u’ and ‘v’ plotted by thestudent should look like

O u (cm)

v (cm)

(1)

O u (cm)

v (cm)

(2)

O u (cm)

v (cm)

(3)

O u (cm)

v (cm)

(4)



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Sol. (3) 1 1 1

constantv 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)vv = 2/3 m/s

Energy loss = ( ) ( ) ( )2

21 1 20.5 2 1.5

2 2 3 × − ×

= 0.67 J

93. 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?

(1)

A B

(2) A B

(3)A B

(4)

A

B

Sol. (3)

Capillary rise h =2Tcos

gr

θρ

. As soap solution has lower T, h will be low.

94. Suppose an electron is attracted towards the origin by a force k/r where ‘k’ is a constant and ‘r’ is thedistance 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 ‘r n’ and the kinetic energy of the electron to be Tn. Then which of

the following is true?(1) Tn ∝ 1/n2, r n ∝ n2 (2) Tn independent of n, r n ∝ n(3) Tn ∝ 1/n, r n ∝ n (4) Tn ∝ 1/n, r n ∝ n2



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Sol. (2) 2k mv

r r =

mv2 = k (independent or r)

nh

mvr 2

= π

⇒ r ∝ n and T = 21mv

2is independent of n.

95. A wave travelling along the x-axis is described by the equation y(x, t) = 0.005 cos (αx −βt). If thewavelength and the time period of the wave are 0.08 m and 2.0 s, respectively, then α and β inappropriate units are

(1) α = 25.00 π, β = π (2) α =0.08 2.0

,π π

(3) α =0.04 1.0

,β =π π

(4) α = 12.50 π, β =2.0

π

Sol. (1) y = 0.005 cos (αx − βt)comparing the equation with the standard form,

y = A cosx t

2T

− π λ

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(3) 4.8 π × 10-5 H (4) 2.4 π × 10-4 H

Sol. (4)

M = 0 1 2N N AµA

= 2.4 π × 10−4 H

97. In the circuit below, A and B represent two inputs and Crepresents the output.The circuit represents(1) NOR gate(2) AND gate(3) NAND gate(4) OR gate

A

B

C

Sol. (4) A B C0 0 00 1 11 0 11 1 1

98. A body is at rest at x = 0. At t = 0, it starts moving in the positive x-direction with a constantacceleration. At the same instant another body passes through x = 0 moving in the positive x-direction 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 correctlydescribes (x1 – x2)as a function of time ‘t’?



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(1)

O t

(x1 −x2) (2)

O t

(x1 −x2)

(3)

O t

(x1 −x2) (4)

O t

(x1 −x2)

Sol. (2)

x1(t) =21

at2

x2(t) = vt

x1 − x2 = 21at

2− vt

99. An experiment is performed to find the refractive index of glass using a travelling microscope. In thisexperiment 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 thefollowing 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?(1)

O r

E(r)

R

(2)

O r

E(r)

R

(3)

O r

E(r)

R

(4)

O r

E(r)

R

Sol. (1)

E(r) =

≥ πε

20

0 if r < R

Qif r R

4 r

101. A 5V battery with internal resistance 2Ω and a 2V battery with internal resistance 1Ω are connected toa 10Ω resistor as shown in the figure. The current in the 10 Ω resistor is

2V

1 Ω 2 Ω

5V

P1

P2

10 Ω

(1) 0.27 A P2 to P1 (2) 0.03 A P1 to P2 (3) 0.03 A P2 to P1 (4) 0.27 A P1 to P2



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Sol. (3)

2 1P P

5 0 2

2 10 1V V1 1 1

2 10 1

+ −− =

+ +

2 1

P PV V

I 0.0310

−= = from P2 → P1

5 V

10 Ω

P1

2 V

1 Ω 2 Ω

P2

i

102. A horizontal overhead power line is at a height of 4m from the ground and carries a current of 100 Afrom east to west. The magnetic field directly below it on the ground is (µ0 = 4π × 10-7 T m A-1)(1) 2.5 × 10-7 T southward (2) 5 × 10-6 T northward(3) 5 × 10-6 T southward (4) 2.5 × 10-7 northward

Sol. (3)

B =7

0 i 4 10 100

2 R 2 4

−µ π ×= ×

π π= 5 × 10−6 T southward

103. Relative permittivity and permeability of a material are εr and µr , respectively. Which of the followingvalues of these quantities are allowed for a diamagnetic material?(1) εr = 0.5, µr = 1.5 (2) εr = 1.5, µr = 0.5(3) εr = 0.5, µr = 0.5 (4) εr = 1.5, µr = 1.5

Sol. (2)

104. Two full turns of the circular scale of a screw gauge cover a distance of 1 mm on its main scale. Thetotal number of divisions on the circular scale is 50. Further, it is found that the screw gauge has azero error of − 0.03 mm while measuring the diameter of a thin wire, a student notes the main scalereading of 3 mm and the number of circular scale divisions in line with the main scale as 35. Thediameter 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 thechambers 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 isremoved without doing any work on the gas, the final equilibrium temperature of the gas in thecontainer will be

(1) 1 2 1 1 2 2

1 1 2 2 2 1

T T (P V P V )

P V T P V T

++

(2) 1 1 1 2 2 2

1 1 2 2

P V T P V T

P V P V

++

(3) 1 1 2 2 2 1

1 1 2 2

P V T P V TP V P V

++(4) 1 2 1 1 2 2

1 1 1 2 2 2

T T (P V P V )P V T P V T

++

Sol. (1) U = U1 + U2

T =( )

( )1 1 2 2 1 2

1 1 2 2 2 1

P V P V T T

P V T P V T

+

+



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A A IIEEEEEE––22000088,, PP A A PPEERR((CC−55)) A A NNSSWWEERRSS

1. (2) 2. (2) 3. (3) 4. (1) 5. (2) 6. (3) 7. (4) 8. (3) 9. (4) 10. (4) 11. (4) 12. (4) 13. (4) 14. (4) 15. (3) 16. (1) 17. (3) 18. (4) 19. (2) 20. (4) 21. (1) 22. (1) 23. (2) 24. (1) 25. (2) 26. (2) 27. (1) 28. (3) 29. (2) 30. (4) 31. (3) 32. (4)33. (4) 34. (3) 35. (4) 36. (4) 37. (1) 38. (4) 39. (3) 40. (2)

41. (4) 42. (2) 43. (1) 44. (4) 45. (2) 46. (1) 47. (2) 48. (1) 49. (4) 50. (3) 51. (2) 52. (2) 53. (3) 54. (3) 55. (3) 56. (2) 57. (1) 58. (4) 59. (3) 60. (1) 61. (3) 62. (4) 63. (2) 64. (4) 65. (1) 66. (3) 67. (2) 68. (4) 69. (2) 70. (3) 71. (2) 72. (4)73. (4) 74. (3) 75. (1) 76. (2)77. (1) 78. (2) 79. (3) 80. (4)81. (1) 82. (4) 83. (3) 84. no option is correct85. (4) 86. (2) 87. (4) 88. (2)89. (3) 90. (3) 91. (3) 92. (3)

93. (3) 94. (2) 95. (1) 96. (4)97. (4) 98. (2) 99. (1) 100. (1)101. (3) 102. (3) 103. (2) 104. (4)105. (1)



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FIITJEE Ltd. , FI ITJE E Hou se, Kalu Sara i, Sarvapr iya Vihar, New D elhi -110 016, Ph 4610 6000 , 26 515949, 2656 9493 , Fa x: 0 11-2651 3942 .

1

A A IIEEEEEE––22000099,, BBOOOOKKLLEETT CCOODDEE(( A 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 correctresponse.

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. 1/4 th 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.

P Phh y yssiiccss

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 anelectrostatic 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 iszero(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 notdepend on the path. Electrostatic force is aconservative force.

2. The above is a plot of binding energy per nucleonEb, against the nuclear mass M; A, B, C, D, E, Fcorrespond 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)1

streaction is fusion and 4

threaction is fission.



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2

3. A p-n junction (D) shown in the figure can actas 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 correctoutput waveform.

(1) (2)

(3) (4)

Sol: (1)Truth Table

A B Y

1 1 1

1 0 0

0 1 0

0 0 0

*5. 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) a2T

2+ 4π2

v2

(2)aT

x

(3) aT + 2πv (4)aT

v

Sol: (2)2 2 2

2

aT xT 4 4T

x x TT

ω π π= = × = = constant.



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6. In an optics experiment, with the position of the object fixed, a student varies the position of a convexlens and for each position, the screen is adjusted to get a clear image of the object. A graph betweenthe object distance u and the image distance v, from the lens, is plotted using the same scale for thetwo axes. A straight line passing through the origin and making an angle of 45

owith the x-axis meets

the experimental curve at P. The coordinates of P will be

(1) (2f, 2f) (2)f f

,2 2

(3) (f, f) (4) (4f, 4f)

Sol: (1)It is possible when object kept at centre of curvature.u = vu = 2f, v = 2f.

*7. A thin uniform rod of length l and mass m is swinging freely about a horizontal axis passing throughits end. Its maximum angular speed is ω. Its centre of mass rises to a maximum height of

(1)

2 21

3 g

ωl(2)

1

6 g

ωl

(3)2 21

2 g

ωl(4)

2 21

6 g

ωl

Sol: (4)T.Ei = T.Ef

21I mgh

2ω =

2 21 1m mgh2 3

× ω =l ⇒ 2 2

1h6 g

ω= l

h

•

8. Let4

QP(r) r

R=

πbe the charge density distribution for a solid sphere of radius R and total charge Q.

for a point ‘p’ inside the sphere at distance r 1 from the centre of the sphere, the magnitude of electricfield is

(1) 0 (2)2

o 1

Q

4 r πε

(3)

21

4o

Qr

4 Rπε(4)

21

4o

Qr

3 Rπε

Sol: (3)1r

2

4 22 0 1

1 4

0 0

Qr4 r dr

R Qr E4 r E

4 R

ππ

π = ⇒ =ε πε

∫. r

P

R



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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)

IR corresponds to least value of 2 2

1 2

1 1n n

−

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?

(1) 3 × 104

J (2) 5 × 104

J

(3) 6 × 104

J (4) 7 × 104

J

Sol: (2) Thermal energy corresponds to internal energyMass = 1 kgdensity = 8 kg/m

3

⇒ Volume = 3mass 1mdensity 8

=

Pressure = 8 × 104

N/m2

∴ Internal Energy =5

2P × V = 5 × 10

4J

11. 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 = R o(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

3 o2.5 10 / C−α = × .

Statement 2: R = Ri (1 + α∆T) is valid only when the change in the temperature ∆T is small and ∆R =

(R - Ro) << Ro.(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: (1)

Directions: Question numbers 12 and 13 are based on the following paragraph. A current loop ABCD is held fixed on the plane of the paper as shownin the figure. The arcs BC (radius = b) and DA (radius = a) of the loopare joined by two straight wires AB and CD. A steady current I isflowing in the loop. Angle made by AB and CD at the origin O is 30

o.

Another straight thin wire with steady current I1 flowing out of the planeof the paper is kept at the origin.



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12. The magnitude of the magnetic field (B) due to loop ABCD at the origin (O) is

(1) zero (2)( )o b a

24ab

µ −

(3) oI b a

4 ab

µ − π

(4) ( ) ( )oI 2 b a a b4 3

µ π − + + π

Sol: (2)Net magnetic field due to loop ABCD at O isB = B AB + BBC + BCD + BDA

= o oI I0 0

4 a 6 4 b 6

µ µπ π+ × + − ×

π π= o oI I

24a 24b

µ µ− = ( )oI b a

24ab

µ−

13. Due to the presence of the current I1 at the origin(1) The forces on AB and DC are zero(2) The forces on AD and BC are zero

(3) The magnitude of the net force on the loop is given by ( ) ( )o 1II2 b a a b

4 3

µ π − + + π

(4) The magnitude of the net force on the loop is given by ( )o 1IIb a

24ab

µ−

Sol: (2)The forces on AD and BC are zero because magnetic field due to a straight wire on AD and BC isparallel to elementary length of the loop.

14. A mixture of light, consisting of wavelength 590 nm and an unknown wavelength, illuminates Young’sdouble slit and gives rise to two overlapping interference patterns on the screen. The centralmaximum of both lights coincide. Further, it is observed that the third bright fringe of known lightcoincides with the 4

thbright fringe of the unknown light. From this data, the wavelength of the

unknown light is(1) 393.4 nm (2) 885.0 nm(3) 442.5 nm (4) 776.8 nm

Sol: (3)

3λ1 = 4λ2

⇒ 2 1

3 3590

4 4λ = λ = × =

1770

4= 442.5 nm

15. Two points P and Q are maintained at the potentials of 10V and -4V respectively. The work done inmoving 100 electrons from P to Q is

(1)1719 10 J−− × (2)

179.60 10 J−×

(3)162.24 10 J−− × (4)

162.24 10 J−×

Sol: (4)

W = QdV = Q(Vq - VP) = -100 × (1.6 × 10-19

) × (– 4 – 10)

= + 100 × 1.6 × 10-19

× 14 = +2.24 10-16

J.

16. The surface of a metal is illuminated with the light of 400 nm. The kinetic energy of the ejectedphotoelectrons was found to be 1.68 eV. The work function of the metal is (hc = 1240 eV nm)(1) 3.09 eV (2) 1.41 eV(3) 151 eV (4) 1.68 eV

Sol: (2)

2o

1mv eV 1.68eV

2= = ⇒

hc 1240evnmh

400nmν = =

λ= 3.1 eV ⇒ 3.1 eV = Wo + 1.6 eV

∴ Wo = 1.42 eV



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*17. A particle has an initial velocity ˆ ˆ3i 4 j+ and an acceleration of ˆ ˆ0.4i 0.3 j+ . Its speed after 10 s is

(1) 10 units (2) 7 2 units

(3) 7 units (4) 8.5 units

Sol: (2)

ˆ û 3i 4 j= +r

; ˆ â 0.4i 0.3 j= +r

u u at= +r r r

= ( )ˆ ˆ ˆ ˆ3i 4 j 0.4i 0.3j 10+ + + = ˆ ˆ ˆ ˆ3i j 4i 3 j+ + + = ˆ ˆ7i 7 j+

Speed is2 27 7 7 2+ = units

*18. A motor cycle starts from rest and accelerates along a straight path at 2 m/s2. At the starting point of

the motor cycle there is a stationary electric sire. How far has the motor cycle gone when the driver hears the frequency of the siren at 94% of its value when the motor cycle was at rest? (speed of sound = 330 ms

-1).

(1) 49 m (2) 98 m(3) 147 m (4) 196 m

Sol: (2)

Motor cycle, u = 0, a = 2 m/s

2

Observer is in motion and source is at rest.

⇒ O

S

v vn n

v v

−′ =

+ ⇒ O330 v94

n n100 330

−= ⇒ 330 – vO =

330 94

100

×

⇒ O

94 33v 330

10

×= − =

33 6m/ s

10

×

s =2 2v u 9 33 33 9 1089

2a 100 100

− × × ×= = ≈ 98 m.

*19. Consider a rubber ball freely falling from a height h = 4.9 m onto a horizontal elastic plate. Assumethat the duration of collision is negligible and the collision with the plate is totally elastic. Then thevelocity as a function of time the height as function of time will be

(1) (2)

(3) (4)

Sol: (3)

21h gt2= , v = - gt and after the collision, v = gt.

(parabolic) (straight line)Collision is perfectly elastic then ball reaches to same height again and again with same velocity.

t3t12t2t1

-v1

+v1

v

t

h

y



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20. A charge Q is placed at each of the opposite corners of a square. A charge q is placed at each of theother two corners. If the net electrical force on Q is zero, then the Q/q equals

(1) 2 2− (2) -1

(3) 1 (4)1

2−

Sol: (1)Three forces F41, F42 and f 43 acting on Q are shownResultant of F41 + F43

= 2 Feach

=2

o

1 Qq2

4 dπε

Resultant on Q becomes zero only when ‘q’ charges are of negative nature.

( )4,2 2

o

1 Q QF

4 2d

×=

πε

⇒ 2 2

dQ Q Q2

d 2d

×=

⇒ Q Q2 q2×× =

∴ Q

q2 2

= − or Q

2 2q

= −

Qq

Q

F42

F43

F41 q

*21. A long metallic bar is carrying heat from one of its ends to the other end under steady-state. The

variation of temperature θ along the length x of the bar from its hot end is best described by which of the following figure.

(1) (2)

(3) (4)

Sol: (2)

We know thatdQ d

KAdt dx

θ=

In steady state flow of heat

dQ 1d . .dx

dt kAθ =

⇒ θH - θ = k x′ ⇒ θ = θH - k x′

Equation θ = θH - k ′ x represents a straight line.



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22. A transparent solid cylindrical rod has a refractive index of

2

3. It is surrounded by air. A light ray is incident at the mid

point of one end of the rod as shown in the figure.

θ

The incident angle θ for which the light ray grazes along the wall of the rod is

(1) 1 1sin2

− (2) 1 3sin

2−

(3)1 2

sin3

−

(4)1 1

sin3

−

Sol: (4)

SinC =3

2….. (1)

Sin r = sin (90 – C) = cosC =1

2

2

1

sinsinr µθ = µ

2 1sin

23θ = ×

1 1sin

3

− θ =

θ

C

*23. Three sound waves of equal amplitudes have frequencies (v – 1), v, (v + 1). They superpose to givebeats. The number of beats produced per second will be(1) 4 (2) 3(3) 2 (4) 1

Sol: (3)

Maximum number of beats = 1 ( 1)ν + − ν − = 2

*24. The height at which the acceleration due to gravity becomesg

9(where g = the acceleration due to

gravity on the surface of the earth) in terms of R, the radius of the earth is

(1) 2R (2)R

2

(3)R

2(4) 2 R

Sol: (1)

( )2

GMg

R h′ =

+, acceleration due to gravity at height h

⇒

( )

2

2 2

g GM R.

9 R R h=

+=

2R

gR h

+

⇒

21 R

9 R h

= +

⇒ R 1

R h 3=

+

⇒ 3R = R + h ⇒ 2R = h



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*25. Two wires are made of the same material and have the same volume. However wire 1 has cross-

sectional area A and wire-2 has cross-sectional area 3A. If the length of wire 1 increases by ∆x onapplying force F, how much force is needed to stretch wire 2 by the same amount?(1) F (2) 4F(3) 6F (4) 9F

Sol: (4)

1 1 2 2 A A=l l ⇒ 1 1 1 12

2

A A A 3A 3×= = =

l l ll ⇒ 1

2

3=l

l

∆x1 =1

1

F

A×

γl ….. (i)

22 2

Fx

3A∆ =

γl …..(ii)

Here ∆x1 = ∆x2

2 12 1

F F

3A A=

γ γl l

F2 = 11 1

2

3F 3F 3× = ×l

l= 9F

*26. In an experiment the angles are required to be measured using an instrument. 29 divisions of themain scale exactly coincide with the 30 divisions of the vernier scale. If the smallest division of themain scale is half-a-degree(=0.5

o), then the least count of the instrument is

(1) one minute (2) half minute(3) one degree (4) half degree

Sol: (1)

Least count =value of main scale division

No of divisions on vernier scale=

1MSD

30=

o o1 1 1

30 2 60× = = 1 minute

27. An inductor of inductance L = 400 mH and resistors of resistances R1 = 2Ω and R2 = 2Ω are connected to a battery of emf 12V as shown inthe figure. The internal resistance of the battery is negligible. Theswitch S is closed at t = 0. The potential drop across L as a functionof time is

(1)5t6e V−

(2)3t12

e Vt

−

(3) ( )t /0.26 1 e V−− (4)5t12e V−

Sol: (4)

1

1

FI

R= =

126A

2=

22 2

dIE L R I

dt= + ×

( )ct / t2 oI I 1 e−= − ⇒ o

2

E 12I 6A

R 2= = =

3

c

L 400 10t 0.2

R 2

−×= = =

( )t /0.22I 6 1 e−= −

Potential drop across L = E – R2I2 = 12 – 2 × 6 (1 – e-bt

) = 12 e-5t



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Directions: Question numbers 28, 29 and 30 are based on the following paragraph.Two moles of helium gas are taken over the cycle ABCDA, as shown in the P – T diagram.

*28. Assuming the gas to be ideal the work done on the gas in taking it from A to B is(1) 200 R (2) 300 R(3) 400 R (4) 500 R

Sol: (3)

W AB = ∆Q - ∆U = nCpdT – nCvdT (atconstant pressure)= n(Cp – Cv)dt

= nRdT = 2 × R × (500 – 300) = 400 R

A B

CD

P

n = 2, γ = 1.67

300 K 500 K T

2 × 105

Pa

1 × 105

Pa

*29. The work done on the gas in taking it from D to A is(1) – 414 R (2) + 414 R(3) – 690 R (4) + 690 R

Sol: (1) At constant temperature (isothermal process)

WDA =

1

2

P

nRTln P

=

5

5

10

2.303 2R 300 log 2 10

× × ×

=1

2.303 600Rlog2

×

= 0.693 × 600 R = - 414 R.

*30. The net work done on the gas in the cycle ABCDA is(1) Zero (2) 276 R(3) 1076 R (4) 1904 R

Sol: (2)

Net work done in a cycle = W AB + WBC + WCB + WBA

= 400 R + 2 × 2.303 × 500 R ln 2 – 400R – 414 R= 1000R x ln 2 – 600R x ln 2 = 400R x ln 2 = 276R



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CC H H E E M M I I SST T R RY Y

PART − B

31. Knowing that the Chemistry of lanthanoids (Ln) is dominated by its +3 oxidation state, which of thefollowing statements in incorrect ?

(1) Because of the large size of the Ln (III) ions the bonding in its compounds is predominantly ionic incharacter.

(2) The ionic sizes of Ln (III) decrease in general with increasing atomic number.(3) Ln (III) compounds are generally colourless.(4) Ln (III) hydroxides are mainly basic in character.

Sol: (3) 3Ln+ compounds are mostly coloured.

32. A liquid was mixed with ethanol and a drop of concentrated 2 4H SO was added. A compound with a

fruity smell was formed. The liquid was :

(1) 3CH OH (2) HCHO

(3) 3 3CH COCH (4) 3CH COOH

Sol: (4) Esterification reaction is involved

( )3CH COOH

l+

)2 5 (

HC H OH

+

→l

3 2 5( ) 2 ( )

CH COOC H H O+l l

*33. Arrange the carbanions, ( )3 33CH C, CCl , ( )3 6 5 22

CH CH, C H CH , in order of their decreasing stability :

(1) ( ) ( )6 5 2 3 3 33 2C H CH CCl CH C CH CH> > > (2) ( ) ( )3 3 6 5 2 32 3

CH CH CCl C H CH CH C> > >

(3) ( ) ( )3 6 5 2 3 32 3CCl C H CH CH CH CH C> > > (4) ( ) ( )3 3 6 5 2 33 2

CH C CH CH C H CH CCl> > >

Sol: (3)

o2 carbanion is more stable than o3 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)

C=C

H

CH3 CH3

H

C=C

H

H3C H

CH3

cis Trans

*35. In which of the following arrangements, the sequence is not strictly according to the property writtenagainst it ?

(1)2 2 2 2CO SiO SnO PbO< < < : increasing oxidising power

(2) HF< HCl < HBr < HI : increasing acid strength

(3) 3 3 3 3NH PH AsH SbH< < < : increasing basic strength

(4) B < C < O < N : increasing first ionization enthalpy.

Sol: (3)

Correct basic strength is 3 3 3 3NH PH AsH BiH> > >



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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 OH

NaOH → 2

o

CO

6atm, 140 C →

COONa

3H O+

→

COOH

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 ( )adsorptionH∆ 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) 3CH COCl (2) 3 2

CH CH Cl

(3) 2 2CH Cl CH Cl (4) 3 2

CH CHCl

Sol: (4)

3 2

aq.KOHCH CHCl → 3

OH/

CH CH\OH

3

2

CH CHOH O

→−

*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 34 2 1(h 6.6 10 kg m s ,− −= ×

mass of electron, 31

me 9.1 10 kg−= × )

(1) 41.52 10 m−× (2) 35.10 10 m−×

(3) 31.92 10 m−× (4) 33.84 10 m−×

Sol: (3)

hx.m v

4∆ ∆ =

π

x∆ =h

4 m vπ ∆

v∆ =0.005

600 0.03

100

× =

⇒ 34

3

31

6.625 10x 1.92 10 m

4 3.14 9.1 10 0.03

−−

−

×∆ = = ×

× × × ×



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40. In a fuel cell methanol is used as fuel and oxygen gas is used as an oxidizer. The reaction is

3 2 2 2

3CH OH( ) O (g) CO (g) 2H O( )

2+ → +l l At 298K standard Gibb’s energies of formation for

3 2CH OH( ), H O( )l l and

2CO (g) are -166.2, -237.2 and -394.4 kJ 1mol− respectively. If standard

enthalpy of combustion of methanol is -726kJ 1mol− , efficiency of the fuel cell will be

(1) 80 % (2) 87%(3) 90% (4) 97%

Sol: (4)

3CH OH( )l + 2 2 2

3O (g) CO (g) 2H O( )

2→ + l

1H 726kJ mol−∆ = −

Also o

f 3G CH OH( )∆ l = -166.2 kJ mol

-1

o

f 2G H O( )∆ l = -237.2 kJ mol

-1

o

f 2G CO ( )∆ l = -394.4 kJ mol-1

Q G∆ = o

f GΣ∆ products o

f G−Σ∆ reactants.

= -394.4 -2 (237.2) + 166.2

= 702.6− kJ mol-1

now Efficiency of fuel cell =G

100H

∆×

∆

=702.6

100726

×

= 97%

41. Two liquids X and Y form an ideal solution. At 300K, vapour pressure of the solution containing 1 molof X and 3 mol of Y is 550 mm Hg. At the same temperature, if 1 mol of Y is further added to thissolution, vapour pressure of the solution increases by 10 mm Hg. Vapour pressure (in mmHg) of Xand 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) o o

T X X Y YP P x P x= +

Xx = mol fraction of X

Yx = mol fraction of Y

∴ 550 = o o

x Y

1 3P P

1 3 1 3

+ + +

=o o

X YP 3P

4 4+

∴ 550 (4) = o o

X YP 3P+ ………….. (1)

Further 1 mol of Y is added and total pressure increases by 10 mm Hg.

∴ 550 + 10 = o o

X Y

1 4P P

1 4 1 4

+ + +

∴ 560 (5) =o o

X YP 4P+ ………….(2)

By solving (1) and (2)

We get, o

XP = 400 mm Hg

o

YP = 600 mm Hg



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42. The half life period of a first order chemical reaction is 6.93 minutes. The time required for thecompletion 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)

1

1/ 2

0.6932 0.6932min

t 6.93

−λ = =Q

Also t =[ ]

[ ]o A2.303

log Aλ

[ ]o A = initial concentration (amount)

[ ] A = final concentration (amount)

∴ t =2.303 6.93 100

log0.6932 1

×

= 46.06 minutes

43. Given : 3

o

Fe / FeE 0.036V,+ = − 2

o

Fe / FeE + = -0.439V. The value of standard electrode potential for the

change, 3 2

(aq)Fe e Fe (aq)+ − ++ → will be :

(1) -0.072 V (2) 0.385 V(3) 0.770 V (4) -0.270

Sol: (3)

Q 3 oFe 3e Fe; E 0.036V+ −+ → = −

∴ O o

1G nFE 3F( 0.036)∆ = − = − −

= +0.108 F

Also 2Fe 2e Fe;+ −+ → oE = -0.439 V

∴ O

2G∆ = -nF oE

= -2 F( -0.439)= 0.878 F

To find oE for 3 2

(aq)Fe e Fe (aq)+ − ++ →

O

G∆ = -nFEo

= -1FE o

Qo o o

1 2G G G= −

∴ oG 0.108F= - 0.878F

∴ -FEo

= +0.108F – 0.878F

∴ OE = 0.878 - 0.108= 0.77v

*44. On the basis of the following thermochemical data : ( o

(aq)fG H 0)+∆ =

2H O( ) H (aq) OH (aq); H 57.32kJ+ −→ + ∆ =l

2 2 2

1H (g) O (g) H O( )

2+ → l ; H 286.20kJ∆ = −

The value of enthalpy of formation of OH− ion at o25 C is :(1) -22.88 kJ (2) -228.88 kJ(3) +228.88 kJ (4) -343.52 kJ

Sol: (2) By adding the two given equations, we have

2(g)H + 2(g) (aq) (aq)

1O H OH

2

+ −→ + ; H∆ =-228.88 Kj

Here o

f H∆ of (aq)

H 0+ =

∴ o

f H∆ of OH− = -228.88 kJ



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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)

r = 2a 2 3614 4

×=

= 127 pm

46. Which of the following has an optical isomer ?

(1) ( )3 3CO NH Cl

+ (2) ( ) ( )

2

3 2CO en NH

+

(3) ( ) ( )3

2 4CO H O en

+ (4) ( ) ( )

3

32 2CO en NH

+

Sol: (4)

It is an octahedral complex of the type ( ) 22M AA X

Where AA is bidentate ligand.

*47. Solid Ba ( )3 2NO is gradually dissolved in a 41.0 10 M−× 2 3Na CO solution. At what concentration of

2Ba + will a precipitate begin to form ?(Ksp for Ba CO3 = 5.1 × 910− ).

(1) 4.1 510 M−× (2) 55.1 10 M−×

(3) 88.1 10 M−× (4) 78.1 10 M−×

Sol: (2)

( )3 3 3 32Ba NO CaCO BaCO 2NaNO+ → +

Here [ ]2 4

3 2 3CO Na CO 10 M− − = =

2 2

sp 3K Ba CO+ − = ⇒ ( )9 2 4 2 55.1 10 Ba 10 Ba 5.1 10− + − + − × = ⇒ = ×

At this value, just precipitation starts.

48. Which one of the following reactions of Xenon compounds is not feasible ?(1)

3XeO 6HF+ → Xe

6 2F 3H O+

(2) 4 2 33Xe F 6H O 2 Xe XeO 12+ → + + HF + 1.5 2O

(3) 2 2 22XeF 2H O 2Xe 4HF O+ → + +

(4) 6 7XeF RbF Rb(XeF ]+ →

Sol: (1)Remaining are feasible

*49. Using MO theory predict which of the following species has the shortest bond length ?

(1) 2

2O + (2) 2O+

(3) 2O− (4) 2

2O −

Sol: (1)

Bond length1

bond order α

Bond order =no..of bonding e no.of antibonding e

2

−

Bond orders of 2

2 2 2O , O , O+ − − and2

2O+ are respectively

2.5, 1.5, 1 and 3.

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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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 5d 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 3 11.0 10 ms−×

(Mass of proton = 1.67 2710 kg−× and 34h 6.63 10 Js−= × ) :

(1) 0.032 nm (2) 0.40 nm(3) 2.5 nm (4) 14.0 nm

Sol: (2)

h

mvλ =

34

27 3

6.63 100.40

1.67 10 10

−

−

×= ≡

× ×nm

52. A binary liquid solution is prepared by mixing n-heptane and ethanol. Which one of the followingstatements 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 for a compound of the molecular formula

( )3CH CH CH CH OH Me− = − − is :

(1) 3 (2) 2(3) 4 (4) 6

Sol: (3) About the double bond, two geometrical isomers are possible and the compound is having one chiralcarbon.

*54. The IUPAC name of neopentane is(1) 2-methylbutane (2) 2, 2-dimethylpropane(3) 2-methylpropane (4) 2,2-dimethylbutane

Sol: (2)

Neopentane is

3

3 3

3

CH|

H C C CH|

CH

− −

*55. The set representing the correct order of ionic radius is :

(1)2 2Li Be Na Mg+ + + +> > > (2)

2 2Na Li Mg Be+ + + +> > >

(3) 2 2Li Na Mg Be+ + + +> > > (4) 2 2Mg Be Li Na+ + + +> > >

Sol: (2) Follow the periodic trends

56. 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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Sol: (3)

Carbohydrates are polyhydroxy carbonyl compounds.

*57. The bond dissociation energy of B – F in 3BF is 646 kJ 1mol− whereas that of C-F in 4CF is 515kJ1mol− . 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 3BF as compared to that between C and F in 4CF(3) significant pπ - pπ interaction between B and F in 3BF whereas there is no possibility of such

interaction between C and F in 4CF .

(4) lower degree of pπ - pπ interaction between B and F in 3BF than that between C and F in 4CF .

Sol: (3) option itself is the reason

58. In Cannizzaro reaction given below

2 Ph CHO ( )

2 2

( ): OH

Ph CH OH PhCO −

−

→ + && the slowest step is :

(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 2CH OH

Sol: (2) Hydride transfer is the slowest step.

59. Which of the following pairs represents linkage isomers ?

(1) ( ) [ ]3 44Cu NH Pt Cl and ( ) [ ]3 44

Pt NH CuCl

(2) ( ) ( )3 2 2Pd P Ph NCS and ( ) ( )3 2 2

Pd P Ph SCN

(3) ( )3 3 45

CO NH NO SO

and ( )3 4 3

5

CO NH SO NO

(4) ( )2 3 24Pt Cl NH Br and ( )2 3 24

Pt Br NH Cl

Sol: (2) NCS

-is ambidentate ligand and it can be linked through N (or) S

60. Buna-N synthetic rubber is a copolymer of :

(1) 2 2

Cl|

H C CH C CH= − = and 2 2H C CH CH CH= − =

(2) 2 2H C CH CH CH= − = and 5 6 2

H C CH CH− =

(3) 2H C CH CN= − and 2 2H C CH CH CH= − =

(4) 2H C CH CN= − and 2 2

3

H C CH C CH|CH

= − =

Sol: (3)



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M M aatthh e emmaattiiccss

PART − C

61. Let a, b, c be such that ( )b a c 0+ ≠ . If ( ) ( ) ( )

n 2 n 1 n

a a 1 a 1 a 1 b 1 c 1

b b 1 b 1 a 1 b 1 c 1 0,c c 1 c 1 1 a 1 b 1 c

+ +

+ − + + −

− + − + − − + =− + − − −

then the

value of ‘n’ is(1) zero (2) any even integer (3) any odd integer (4) any integer

Sol: (3)

( )n

a a 1 a 1 a 1 b 1 c 1

b b 1 b 1 1 a 1 b 1 c 1

c c 1 c 1 a b c

+ − + + −

− + − + − − − +

− + −( )

n

a a 1 a 1 a 1 a 1 a

b b 1 b 1 1 b 1 b 1 b

c c 1 c 1 c 1 c 1 c

+ − + −

= − + − + − + − −

− + − +

( )n 1

a a 1 a 1 a 1 a a 1

b b 1 b 1 1 b 1 b b 1

c c 1 c 1 c 1 c c 1

+

+ − + −

= − + − + − + − −

− + − +

( )n 2

a a 1 a 1 a a 1 a 1

b b 1 b 1 1 b b 1 b 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 isequal to(1) 10.0 (2) 20.0(3) 10.1 (4) 20.2

Sol: (3)

( )

nsum of quantities

Mean xn

= =( )a l

2

n

+[ ]

11 1 100d 1 50d

2= + + = +

[ ]i1 1 2M.D. x x 255 50d 49d 48d .... d 0 d ...... 50dn 101= − ⇒ = + + + + + + + + =∑ d 50 x 51

101 2

255 x 101d 10.1

50 x 51⇒ = =

*63. If the roots of the equation 2bx cx a 0+ + = be imaginary, then for all real values of x, the expression2 2 23b x 6bcx 2c+ + is

(1) greater than 4ab (2) less than 4ab(3) greater than – 4ab (4) less than – 4ab

Sol: (3)2bx cx a 0+ + =

Roots are imaginary ⇒ 2 2 2c 4ab 0 c 4ab c 4ab− < ⇒ < ⇒ − > − 2 2 23b x 6bcx 2c+ +

since 23b 0>

Given expression has minimum value

Minimum value =( )( )

( )

2 2 2 2 2 22

22

4 3b 2c 36b c 12b cc 4ab

12b4 3b

−= − = − > − .



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*64. Let A and B denote the statements

A: cos cos cos 0α + β + γ =

B: sin sin sin 0α + β + γ =

If ( ) ( ) ( )3

cos cos cos2

β − γ + γ − α + α − β = − , then

(1) A is true and B is false (2) A is false and B is true(3) both A and B are true (4) both A and B are false

Sol: (3)

( ) ( ) ( )3

cos cos cos2

β − γ + γ − α + α − β = −

⇒ ( ) ( ) ( )2 cos cos cos 3 0 β − γ + γ − α + α − β + =

⇒ ( ) ( ) ( ) 2 2 2 2 2 22 cos cos cos sin cos sin cos sin cos 0 β − γ + γ − α + α − β + α + α + β + β + γ + γ =

⇒ ( ) ( )2 2

sin sin sin cos cos cos 0α + β + γ + α + β + γ =

*65. The lines ( )2p p 1 x y q 0+ − + = and ( ) ( )2

2 2p 1 x p 1 y 2q 0+ + + + = are perpendicular to a common line

for (1) no value of p (2) exactly one value of p

(3) exactly two values of p (4) more than two values of p

Sol: (2)

Lines must be parallel, therefore slopes are equal ⇒ ( ) ( )2 2p p 1 p 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. If u, v, wr r uur

are non-coplanar vectors and p, q are real numbers, then the equality

3u pv pw pv w qu 2w qv qu 0 − − =

r r uur r uur r uur r r

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)

Sol: (1)

( )2 23p pq 2q u v w 0 − + = r r uur

But u v w 0 ≠ r r uur

2 23p pq 2q− + = 0

2 222 2 2 2q 7q q 7

2p p pq 0 2p p q 02 4 2 4

+ − + + = ⇒ + − + =

⇒ p = 0, q = 0,q

p 2=

This possible only when p = 0, q = 0 exactly one value of (p, q)

68. Let the linex 2 y 1 z 2

3 5 2

− − += =

−lies in the plane x 3y z 0+ − α + β = . Then ( ),α β equals

(1) (6, - 17) (2) ( - 6, 7)(3) (5, - 15) (4) ( - 5, 15)

Sol: (2)

Dr’s of line = ( )3, 5, 2−



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Dr’s of normal to the plane = ( )1, 3, − α

Line is perpendicular to normal ⇒ ( ) ( ) ( )3 1 5 3 2 0− + −α = 3 15 2 0 2 12 6⇒ − − α = ⇒ α = − ⇒ α = −

Also ( )2, 1, 2− lies on the plane

( )2 3 6 2 0 7+ + − + β = ⇒ β =

∴ ( ) ( ), 6, 7α β = −

*69. From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are to be selected andarranged in a row on the shelf so that the dictionary is always in the middle. Then the number of sucharrangements 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

4C ways. 1 dictionary can be selected from 3 dictionaries

in 3

1C 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 3

4 1C x C x 4! = 1080

70. [ ]0

cot x dxπ

∫ , [ ]• denotes the greatest integer function, is equal to

(1)2

π(2) 1

(3) – 1 (4)2

π−

Sol: (4)

Let [ ]0

I cot x dxπ

= ∫ …(1)

( )0

cot x dxπ

= π − ∫ [ ]0

cot x dxπ

= −∫ …(2)

Adding (1) and (2)

2I = [ ]0

cot x dxπ

∫ + [ ]0

cot x dxπ

−∫ = ( )0

1 dxπ

= −∫ [ ] [ ]x x 1 if x Z

0 if x Z

+ − = − ∉

= ∈

Q

[ ]0

xπ

= − = −π

∴ I2

π= −

71. For real x, let ( ) 3f x x 5x 1= + + , then

(1) f is one-one but not onto R (2) f is onto R but not one-one(3) f is one-one and onto R (4) f is neither one-one nor onto R

Sol: (3)Given ( ) 3f x x 5x 1= + +

Now ( ) 2f ' x 3x 5 0, x R= + > ∀ ∈

∴ f(x) is strictly increasing function

∴ It is one-oneClearly, f(x) is a continuous function and also increasing on R,

Ltx → −∞

( )f x = −∞ and Ltx → ∞

( )f x = ∞

∴ f(x) takes every value between −∞ and ∞ .Thus, f(x) is onto function.



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72. In a binomial distribution1

B n, p4

=

, if the probability of at least one success is greater than or

equal to9

10, then n is greater than

(1)4 3

10 10

1

log log−(2)

4 3

10 10

1

log log+

(3)4 3

10 10

9log log−

(4)4 3

10 10

4log log−

Sol: (1)n

n

3

4

9 3 11 q n log 10

10 4 10

− ≥ ⇒ ≤ ⇒ ≥ −

4 3

10 10

1n

log log⇒ ≥

−

*73. If P and Q are the points of intersection of the circles 2 2x y 3x 7y 2p 5 0+ + + + − = and2 2 2x 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

Sol: (1)Given circles 2 2S x y 3x 7y 2p 5 0= + + + + − =

2 2 2S' x y 2x 2y p 0= + + + − =

Equation of required circle is S S' 0+ λ =

As it passes through (1, 1) the value of ( )

( )2

7 2p

6 p

− +λ =

−

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 directioncosines of the vector are

(1) 6, 3, 2− (2)

6 3 2

, ,5 5 5−

(3)6 3 2

, ,7 7 7

− (4)6 3 2

, ,7 7 7

− −

Sol: (3)

Projection of a vector on coordinate axis are 2 1 2 1 2 1x x , y y , z z− − −

2 1 2 1 2 1x x 6, y y 3, z z 2− = − = − − =

( ) ( ) ( )2 2 2

2 1 2 1 2 1x x y y z z 36 9 4 7− + − + − = + + =

The D.C’s of the vector are6 3 2

, ,7 7 7

−

*75. If 4Z 2z− = , then the maximum value of Z is equal to

(1) 3 1+ (2) 5 1+

(3) 2 (4) 2 2+

Sol: (2)

4 4Z Z

Z Z

= − +

4 4Z Z

Z Z⇒ = − +

4 4Z Z

Z Z⇒ ≤ − +

4Z 2

Z⇒ ≤ +



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2Z 2 Z 4 0⇒ − − ≤

( )( ) ( )( )Z 5 1 Z 1 5 0 1 5 Z 5 1− + − − ≤ ⇒ − ≤ ≤ +

*76. Three distinct points A, B and C are given in the 2 – dimensional coordinate plane such that the ratioof the distance of any one of them from the point (1, 0) to the distance from the point ( - 1, 0) is equal

to

1

3 . Then the circumcentre of the triangle ABC is at the point

(1) ( )0, 0 (2)5

, 04

(3)5

, 02

(4)5

, 03

Sol: (2)

( )P 1, 0= ; ( )Q 1, 0−

Let ( ) A x, y=

AP BP CP 1

AQ BQ CQ 3= = = ..(1)

( ) ( )2 2

2 2 2 23AP AQ 9AP AQ 9 x 1 9y x 1 y⇒ = ⇒ = ⇒ − + = + + 2 2 2 2 2 29x 18x 9 9y x 2x 1 y 8x 20x 8y 8 0⇒ − + + = + + + ⇒ − + + =

2 2 5x y x 1 0

2⇒ + − + = …(2)

∴ A lies on the circleSimilarly B, C are also lies on the same circle

∴ Circumcentre of ABC = Centre of Circle (1) =5

, 04

*77. The remainder left out when ( )2n 12n8 62

+− is divided by 9 is

(1) 0 (2) 2(3) 7 (4) 8

Sol: (2)

( )2n 12n8 62

+− = ( ) ( )

n 2n 11 63 63 1

++ − −

( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( )( )( )n 2n 1 2 n 2 2n 12n 1 2n 1n n

1 2 1 21 63 1 63 1 c 63 c 63 .... 63 1 c 63 c 63 .... 1 63+ ++ += + + − = + + + + + − + + + −

( ) ( ) ( ) ( ) ( ) ( )( )( )n 1 2n2n 1 2n 1n n

1 2 1 22 63 c c 63 .... 63 c c 63 .... 63− + += + + + + − + + −

∴ Reminder is 2

*78. The ellipse 2 2x 4y 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

(1) 2 2x 16y 16+ = (2) 2 2x 12y 16+ =

(3) 2 24x 48y 48+ = (4) 2 24x 64y 48+ =

Sol: (2)



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23

( )2 2

2 2 x yx 4y 4 1 a 2, b 1 P 2, 1

4 1+ = ⇒ + = ⇒ = = ⇒ =

Required Ellipse is2 2 2 2

2 2 2 2

x y x y1 1

a b 4 b+ = ⇒ + =

(2, 1) lies on it

2

2 2

4 1 1 1 3 41 1 b

16 4 4 3b b

⇒ + = ⇒ = − = ⇒ =

∴ 2 2 2 2

2 2x y x 3y1 1 x 12y 16

416 16 4

3

+ = ⇒ + = ⇒ + =

1

2 A 2 A’

V’ V

P (2, 1)

(4, 0)

*79. The sum to the infinity of the series2 3 4

2 6 10 141 ......

3 3 3 3+ + + + + is

(1) 2 (2) 3(3) 4 (4) 6

Sol: (2)

Let2 3 4

2 6 10 14S 1 ....

3 3 3 3= + + + + + …(1)

2 3 4

1 1 2 6 10S ....3 3 3 3 3= + + + + …(2)

Dividing (1) & (2)

2 3 4

1 1 4 4 4S 1 1 ....

3 3 3 3 3

− = + + + + +

2 2

2 4 4 1 1S 1 ......

3 3 33 3

= + + + +

2 2

2 4 4 1 4 4 3 4 2 6S

13 3 3 2 3 3 23 31

3

⇒ = + = + = + = −

2 6S S 3

3 3⇒ = ⇒ =

80. The differential equation which represents the family of curves 2c x

1y c e= , where1

c and2

c are

arbitrary constants is

(1) 2y ' y= (2) y" y 'y=

(3) yy " y '= (4) ( )2

yy" y '=

Sol: (4)2c x

1y c e= …(1)

2c x

2 1y ' c c e=

2y ' c y= …(2)

2y" c y '=

From (2)

2

y 'c

y=

So,( )

( )2

2y 'y" yy" y '

y

= ⇒ =

81. One ticket is selected at random from 50 tickets numbered 00, 01, 02, …., 49. Then the probabilitythat the sum of the digits on the selected ticket is 8, given that the product of these digits is zero,equals

(1)1

14(2)

1

7

(3)5

14(4)

1

50

Sol: (1)



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24

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 zeroB = 00, 01, 02, 03, …., 09, 10, 20, 30, 40

A B 8∩ =

Required probability = ( ) ( )( )

1

P A B 150P A /B14P B 14

50

∩= = =

82. Let y be an implicit function of x defined by2x xx 2x cot y 1 0− − = . Then ( )y ' 1 equals

(1) – 1 (2) 1(3) log 2 (4) – log 2

Sol: (1)2x xx 2x cot y 1 0− − = …(1)

Now x = 1,

1 – 2 coty – 1 = 0 ⇒ coty = 0 ⇒ y =2

π

Now differentiating eq. (1) w.r.t. ‘x’

( ) ( ) ( )2x x 2 xdy2x 1 logx 2 x c osec y cot y x 1 logx 0

dx

+ − − + + =

Now at 1,2

π

( ) ( )1,

2

dy2 1 log1 2 1 1 0 0

dx π

+ − − + =

1, 1,2 2

dy dy2 2 0 1

dx dxπ π

⇒ + = ⇒ = −

83. The area of the region bounded by the parabola

( )

2y 2 x 1− = − , the tangent to the parabola at the

point (2, 3) and the x-axis is(1) 3 (2) 6(3) 9 (4) 12

Sol: (3)

Equation of tangent at (2, 3) to

( )2

y 2 x 1− = − is 1S 0=

x 2y 4 0⇒ − + =

Required Area = Area of ∆OCB + Area of

OAPD – Area of ∆PCD

( ) ( ) ( )

3

2

0

1 14 x 2 y 4y 5 dy 1 x 2

2 2= + − + −

∫

33

2

0

y4 2y 5y 1 4 9 18 15 1

3

= + − + − = − − + −

28 19 9 sq. units= − =

0

D (0, 3)

P (2, 3)

2y = x + 4

A (1, 2)

B (-4, 0)

C (0, 2) A (1, 2)

(or)



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25

Area = ( ) ( )3 3

2 2

0 0

2y 4 y 4y 5 dy y 6y 5 dy− − + − = − + −∫ ∫ ( )( )

333

2

00

y 3 273 y dy 9 sq.units

3 3

− = − − = = =

∫

84. Given ( ) 4 3 2P x x ax bx 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)

( ) 4 3 2P x x ax bx cx d= + + + +

( ) 3 2P' x 4x 3ax 2bx c= + + +

Q x = 0 is a solution for ( )P' x 0= , ⇒ c = 0

∴ ( ) 4 3 2P x x ax bx 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 ],

( ) ( )3 2 2P' x 4x 3ax 2bx x 4x 3ax 2b= + + = + +

( )P' xQ has only one root x = 0, 24x 3ax 2b 0+ + = has no real roots.

∴ ( )2

2 3a3a 32b 0 b

32− < ⇒ <

∴ b > 0Thus, we have a > 0 and b > 0

∴ ( ) ( )3 2P' x 4x 3ax 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. The shortest distance between the line y x 1− = and the curve 2x y= is

(1)3 2

8(2)

2 3

8

(3)3 2

5

(4)3

4

Sol: (1)

x y 1 0− + = …(1)2x y=

dy1 2y

dx=

dy 1

dx 2y⇒ = = Slope of given line (1)

1 11 y

2y 2= ⇒ = ⇒

21 1 1

y x2 2 4

= ⇒ = =

⇒ ( )

1 1x, y ,

4 2

=



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26

∴ The shortest distance is

1 11

3 3 24 2

81 1 4 2

− += =

+

Directions: Question number 86 to 90 are Assertion – Reason type questions. Each of these questionscontains 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 haveto select the correct choice

86. Let ( ) ( )2

f x x 1 1, x 1= + − ≥ −

Statement-1 : The set ( ) ( ) 1x : f x f 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 sinx= .

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 sinx=

( ) ( )gof x sin x x=

2

2

sinx ,x 0

sinx ,x 0

− <= ≥

∴ ( ) ( )2

2

2xcosx ,x 0gof ' x

2xcosx ,x 0

− <=

≥

Clearly, ( ) ( ) ( ) ( )L gof ' 0 0 R gof ' 0= =

∴ gof is differentiable at x = 0 and also its derivative is continuous at x = 0

Now, ( ) ( )2 2 2

2 2 2

2cos x 4x sinx ,x 0gof " x

2cosx 4x sinx ,x 0

− + <=

− ≥

( ) ( )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 is2n 1

4

−

Statement-2 : The sum of first n natural numbers is( )n n 1

2

+and the sum of squares of first n natural

numbers is( ) ( )n n 1 2n 1

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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27

(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 1n

+= = +

( ) ( ) ( ) ( )22 2 22 2

i

1 1Variance x x 2 4 ..... 2n n 1

n n

= − = + + + − + ∑

( )( ) ( )

( )2 22 2 2 2

n n 1 2n 11 42 1 2 ..... n n 1 n 1

n n 6

+ + = + + + − + = − +

( ) ( ) ( ) ( )[ ] ( ) ( ) 2n 1 2 2n 1 3 n 1 n 1 4n 2 3n 3 n 1 n 1 n 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 q p q↔ ~q p ~ q↔ ( )~ p ~ q↔

T T T F F T

T F F T T F

F T F F T F

F F T T F T

90. 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)n 1 2 1

adj A A A A− −

= = =

( )n 2 0

adj adj A A A A A A−

= = =



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28

A A IIEEEEEE––22000099,, A A NNSSWWEERR KKEE Y Y

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)

2. (1)

3. (3)

4. (1)

5. (2)

6. (1)

7. (4)

8. (3)

9. (4)

10. (2)11. (1)

12. (2)

13. (2)

14. (3)

15. (4)

16. (2)

17. (2)

18. (2)

19. (3)

20. (1)

21. (2)

22. (4)

23. (3)

24. (1)

25. (4)

26. (1)

27. (4)

28. (3)

29. (1)

30. (2)

31. (3)

32. (4)

33. (3)

34. (3)

35. (3)

36. (3)

37. (4)

38. (4)

39. (3)

40. (4)41. (3)

42. (3)

43. (3)

44. (2)

45. (2)

46. (4)

47. (2)

48. (1)

49. (1)

50. (2)

51. (2)

52. (2)

53. (3)

54. (2)

55. (2)

56. (3)

57. (3)

58. (2)

59. (2)

60. (3)

61. (3)

62. (3)

63. (3)

64. (3)

65. (2)

66. (3)

67. (1)

68. (2)

69. (4)

70. (4)71. (3)

72. (1)

73. (1)

74. (3)

75. (2)

76. (2)

77. (2)

78. (2)

79. (2)

80. (4)

81. (1)

82. (1)

83. (3)

84. (2)

85. (1)

86. (3)

87. (3)

88. (4)

89. (3)

90. (2)

1. (1)

2. (4)

3. (1)

4. (3)

5. (3)

6. (1)

7. (1)

8. (1)

9. (2)

10. (1)11. (2)

12. (4)

13. (3)

14. (3)

15. (2)

16. (2)

17. (2)

18. (1)

19. (1)

20. (2)

21. (2)

22. (2)

23. (2)

24. (1)

25. (1)

26. (1)

27. (3)

28. (2)

29. (2)

30. (2)

31. (2)

32. (1)

33. (2)

34. (3)

35. (2)

36. (2)

37. (1)

38. (4)

39. (1)

40. (3) 41. (4)

42. (1)

43. (2)

44. (2)

45. (3)

46. (2)

47. (1)

48. (4)

49. (2)

50. (1)

51. (2)

52. (2)

53. (4)

54. (1)

55. (2)

56. (1)

57. (1)

58. (4)

59. (4)

60. (3)

61. (2)

62. (2)

63. (1)

64. (3)

65. (1)

66. (1)

67. (2)

68. (4)

69. (1)

70. (4)71. (4)

72. (1)

73. (3)

74. (4)

75. (3)

76. (2)

77. (1)

78. (1)

79. (2)

80. (1)

81. (4)

82. (3)

83. (3)

84. (1)

85. (4)

86. (1)

87. (3)

88. (4)

89. (2)

90. (2)

1. (4)

2. (4)

3. (4)

4. (3)

5. (2)

6. (1)

7. (4)

8. (4)

9. (4)

10. (3)11. (1)

12. (3)

13. (4)

14. (3)

15. (2)

16. (4)

17. (3)

18. (3)

19. (4)

20. (1)

21. (2)

22. (3)

23. (3)

24. (3)

25. (4)

26. (1)

27. (2)

28. (3)

29. (2)

30. (2)

31. (4)

32. (2)

33. (3)

34. (4)

35. (2)

36. (3)

37. (1)

38. (4)

39. (2)

40. (2) 41. (1)

42. (4)

43. (2)

44. (1)

45. (4)

46. (3)

47. (1)

48. (2)

49. (3)

50. (3)

51. (3)

52. (4)

53. (1)

54. (3)

55. (1)

56. (2)

57. (1)

58. (3)

59. (3)

60. (2)

61. (4)

62. (3)

63. (4)

64. (1)

65. (3)

66. (4)

67. (3)

68. (4)

69. (4)

70. (4)71. (3)

72. (4)

73. (1)

74. (4)

75. (4)

76. (3)

77. (4)

78. (4)

79. (3)

80. (2)

81. (2)

82. (3)

83. (3)

84. (4)

85. (1)

86. (1)

87. (2)

88. (3)

89. (4)

90. (1)

1. (1)

2. (1)

3. (1)

4. (1)

5. (4)

6. (2)

7. (1)

8. (1)

9. (1)

10. (4)11. (4)

12. (4)

13. (4)

14. (1)

15. (3)

16. (3)

17. (1)

18. (1)

19. (1)

20. (4)

21. (4)

22. (1)

23. (2)

24. (4)

25. (2)

26. (4)

27. (2)

28. (1)

29. (2)

30. (4)

31. (4)

32. (3)

33. (4)

34. (4)

35. (1)

36. (3)

37. (3)

38. (4)

39. (3)

40. (1) 41. (4)

42. (4)

43. (2)

44. (4)

45. (2)

46. (2)

47. (2)

48. (2)

49. (1)

50. (1)

51. (3)

52. (3)

53. (1)

54. (3)

55. (4)

56. (4)

57. (1)

58. (2)

59. (3)

60. (2)

61. (4)

62. (4)

63. (1)

64. (1)

65. (2)

66. (3)

67. (2)

68. (4)

69. (4)

70. (2)71. (3)

72. (3)

73. (1)

74. (4)

75. (4)

76. (1)

77. (2)

78. (1)

79. (1)

80. (3)

81. (1)

82. (4)

83. (4)

84. (1)

85. (1)

86. (3)

87. (4)

88. (3)

89. (1)

90. (1)



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(Hyderabad Classes) Limited. 5-9-14/B, Saifabad, (Opp. Secretariat) Hyderabad. 500 063. Phone: 040-66777000 – 03 Fax: 040-66777004

A A I I E E E E E E – – 2 2 0 0 110 0

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7/14/2019 Aieee


AIEEE−−−−2010−−−−2


1. The standard enthalpy of formation of NH3 is –46.0 kJ mol

–1. If the enthalpy of formation of H2 from

its atoms is –436 kJ mol –1 and that of N2 is –712 kJ mol –1, the average bond enthalpy of N–H bond inNH3 is(1) –964 kJ mol

–1(2) +352 kJ mol

–1(3) + 1056 kJ mol

–1(4) –1102 kJ mol

–1

1. (2)Sol : Enthalpy of formation of NH3 = –46 kJ/mole

∴ N2 + 3H2 → 2NH3 ∆Hf = – 2 x 46 kJ molBond breaking is endothermic and Bond formation is exothermicAssuming ‘x’ is the bond energy of N–H bond (kJ mol

–1)

∴ 712 + (3 x 436)– 6x = –46 x 2

∴ x = 352 kJ/mol

2. The time for half life period of a certain reaction A → products is 1 hour. When the initialconcentration of the reactant ‘A’, is 2.0 mol L

–1, how much time does it take for its concentration to

come from 0.50 to 0.25 mol L –1 if it is a zero order reaction ?(1) 4 h (2) 0.5 h (3) 0.25 h (4) 1 h

2. (3)

Sol : For a zero order reactionx

kt

= → (1)

Where x = amount decomposedk = zero order rate constant

for a zero order reaction

[ ]0

1

2

Ak

2t= → (2)

Since [A0] = 2M , t1/2 = 1 hr; k = 1

∴ from equation (1)0.25

t 0.25hr1

= =

3. A solution containing 2.675 g of CoCl3. 6 NH3 (molar mass = 267.5 g mol –1

) is passed through acation exchanger. The chloride ions obtained in solution were treated with excess of AgNO 3 to give4.78 g of AgCl (molar mass = 143.5 g mol

–1). 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 3. (1)

Sol : CoCl3. 6NH3 → xCl – AgNO3 → x AgCl ↓

n(AgCl) = x n(CoCl3. 6NH3)

4.78 2.675x

143.5 267.5

= ∴ x = 3

∴ The complex is ( )3 36Co NH Cl

4. Consider the reaction :

Cl2(aq) + H2S(aq) → S(s) + 2H+(aq) + 2Cl

– (aq)

The rate equation for this reaction is rate = k [Cl 2] [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) A only

4. (4)




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Sol: Rate equation is to be derived wrt slow

Step ∴ from mechanism (A)Rate = k[Cl2] [H2S]

5. If 10 –4

dm3

of water is introduced into a 1.0 dm3

flask to 300 K, how many moles of water are in thevapour phase when equilibrium is established ?(Given : Vapour pressure of H2O at 300 K is 3170 Pa ; R = 8.314 J K

–1mol

–1)

(1) 5.56 x 10 –3

mol (2) 1.53 x 10 –2

mol (3) 4.46 x 10 –2

mol (4) 1.27 x 10 –3

mol5. (4)

Sol :PV

nRT

= =

= 128 x 10 –5

moles

=5

1 1

3170 10 atm 1L

0.0821 L atm k mol 300K

−

− −

× ×

× ≈ 1.27 x 10

–3mol

6. One mole of a symmetrical alkene on ozonolysis gives two moles of an aldehyde having a molecularmass of 44 u. The alkene is(1) propene (2) 1–butene (3) 2–butene (4) ethene

6. (3)Sol : 2–butene is symmetrical alkene

CH3 –CH=CH–CH3 O3

3Zn /H O22.CH CHO →

Molar mass of CH3CHO is 44 u.

7. 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 dissolvedin 1 kg of water, is (K f = 1.86 K kg mol

–1)

(1) 0.0372 K (2) 0.0558 K (3) 0.0744 K (4) 0.0186 K7. (2)

Sol : Vant Hoff’s factor (i) for Na2SO4 = 3∴ ∆Tf = (i) kf m

= 3 x 1.80 x0.01

0.0558 K1

=

8. From amongst the following alcohols the one that would react fastest with conc. HCl and anhydrousZnCl2, is(1) 2–Butanol (2) 2–Methylpropan–2–ol (3) 2–Methylpropanol (4) 1–Butanol

8. (2)

Sol : 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

9. In the chemical reactions,

NH2

NaNO2

HCl, 278 KA

HBF4B

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

9. (3)Sol :




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NH2 N2 Cl

NaNO2

HCl, 278 K

HBF4

F

N2 BF3 HCl

(A) (B)benzene diazonium

chloride

fluorobenzene

10. 29.5 mg of an organic compound containing nitrogen was digested according to Kjeldahl’s methodand the evolved ammonia was absorbed in 20 mL of 0.1 M HCl solution. The excess of the acidrequired 15 mL of 0.1 M NaOH solution for complete neutralization. The percentage of nitrogen inthe compound is(1) 59.0 (2) 47.4 (3) 23.7 (4) 29.5

10. (3)Sol : Moles of HCl reacting with

ammonia = (moles of HCl absorbed ) – (moles of NaOH solution required)= (20 x 0.1 x 10

–3) – (15 x 0.1 x 10

–3)

= moles of NH3 evolved.= moles of nitrogen in organic compound

∴ wt. of nitrogen in org. comp = 0.5 x 10 –3

x 14= 7 x 10

–3g

% wt =3

3

7 1023.7%

29.5 10

−

−

×=

×

11. The energy required to break one mole of Cl–Cl bonds in Cl2 is 242 kJ mol –1

. The longestwavelength of light capable of breaking a single Cl – Cl bond is(c = 3 x 10

8ms

–1and NA = 6.02 x 10

23mol

–1)

(1) 594 nm (2) 640 nm (3) 700 nm (4) 494 nm11. (4)

Sol : Energy required for 1 Cl2 molecule =3

A

242 10

N

×Joules.

This energy is contained in photon of wavelength ‘λ’.34 8 3

23

hc 6.626 10 3 10 242 10E

6.022 10

−× × × ×= =

λ λ ×

λ = 49470

A ≈ 494 nm

12. Ionisation energy of He+

is 19.6 x 10 –18

J atom –1

. The energy of the first stationary state (n = 1) of Li2+

is(1) 4.41 x 10

–16J atom

–1(2) –4.41 x 10

–17J atom

–1

(3) –2.2 x 10 –15 J atom –1 (4) 8.82 x 10 –17 J atom –1 12. (2)

Sol : 2

2 2He He

1 1IE 13.6 Z

1+ +

= − ∞

= ( )2

He He13.6Z where Z 2+ + =

Hence 2 18

He13.6 Z 19.6 10−

+× = × J atom –1

.

( ) 221 2 2Li Li

1E 13.6 Z

1+ += − × =

2

22 Li

2He

He

Z13.6 Z

Z

+

+

+

− ×

= –19.6 x 10 –18

x 1794.41 10 J /atom

4

−= − ×

13. Consider the following bromides :




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Me Br

Br

Me Me

Br

Me

(A) (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

13. (1)Sol : 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.

MeMe

Me

Me> >

14. Which one of the following has an optical isomer ?

(1) ( ) ( )2

32

Zn en NH+

(2) ( )3

3

Co en+

(3) ( ) ( )3

24

Co H O en+

(4) ( )2

2

Zn en+

(en = ethylenediamine)14. (2)Sol : Only option (2) is having non–super imposable mirror image & hence one optical isomer.

Zn+

NH3

NH3

en

no optical isomer. It is

Tetrahedral with a plane of symmetry

Co+

Co+

en en

en

enen

en

optical isomer

( 1) ( 2)

2

3 3

H2O

H2O

Co+

H2O

H2O

en

Horizontal plane is plane of symmtry

Zn+

en

en

no optical isomer, it is

tetrahedral with a plane of symmetry

3) 4)

3

2

15. On mixing, heptane and octane form an ideal solution. At 373 K, the vapour pressures of the twoliquid components (heptane and octane) are 105 kPa and 45 kPa respectively. Vapour pressure ofthe solution obtained by mixing 25.0g of heptane and 35 g of octane will be (molar mass of heptane= 100 g mol

–1an dof octane = 114 g mol

–1).

(1) 72.0 kPa (2) 36.1 kPa (3) 96.2 kPa (4) 144.5 kPa15. (1)




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Sol : Mole fraction of Heptane =25 /100 0.25

0.4525 35 0.557

100 114

= =+

HeptaneX 0.45= .

∴ Mole fraction of octane = 0.55 = Xoctane

Total pressure = 0

i iXP

= (105 x 0.45) + (45 x 0.55) kPa = 72.0 KPa

16. The main product of the following reaction is C6H5CH2CH(OH)CH(CH3)2 conc.H SO2 4 → ?

(1) C C

H5C6

H CH(CH3)2

H

(2) C C

C6H5CH2

H CH3

CH3

(3) C C

C6H

5

H H

CH(CH3)2

(4) C CH2

H5

C6

CH2

CH2

H3C

16. (1)Sol :

CH2 CH

OH

CH

CH3

CH3

CH2 CH CH

CH3

CH3

CH CH HC

CH3

CH3

H2SO4conc.

loss of proton

(conjugated system)

Trans isomers is more stable & main product here

C C

H

CH(CH3)2H

(trans isomer)

17. Three reactions involving2 4H PO− are given below :

(i) H3PO4 + H2O → H3O+

+2 4

H PO− (ii)2 4

H PO− + H2O → 2

4HPO − + H3O

+

(iii) 2

2 4 3 4H PO OH H PO O− − −+ → +




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In which of the above does2 4

H PO− act as an acid ?

(1) (ii) only (2) (i) and (ii) (3) (iii) only (4) (i) only17. (1)

Sol : (i) 3 4acid

H PO + H2 → 3H O+ + 2 4conjugate base

H PO−

(ii)2 4acid

H PO− + H2O → 2

4conjugate base

HPO− + 3H O+

(iii)2 4acid

H PO− +acid

OH− → 3 4

conjugate acid

H PO + O –2

Only in reaction (ii)2 4H PO− acids as ‘acid’.

18. In aqueous solution the ionization constants for carbonic acid areK1 = 4.2 x 10

–7and K2 = 4.8 x 10

–11

Select the correct statement for a saturated 0.034 M solution of the carbonic acid.

(1) The concentration of 2

3CO − is 0.034 M.

(2) The concentration of2

3CO−

is greater than that of 3HCO−

.(3) The concentration of H

+and

3HCO− are approximately equal.

(4) The concentration of H+

is double that of 2

3CO − .

18. (3)

Sol : A → H2CO3 H+

+3

HCO− K1 = 4.2 x 10 –7

B → 3

HCO− H+

+ 2

3CO− K2 = 4.8 x 10

–11

As K2 << K1

All majortotal A

H H+ + ≈

and from I equilibrium,3A total

H HCO H+ − + ≈ ≈

2

3CO−

is negligible compared to3

total

HCO or H− +

19. The edge length of a face centered cubic cell of an ionic substance is 508 pm. If the radius of thecation is 110 pm, the radius of the anion is(1) 288 pm (2) 398 pm (3) 618 pm (4) 144 pm

19. (4)Sol : For an ionic substance in FCC arrangement,

( )2 r r+ −+ = edge length

( )2 110 r−+ = 508

r –

= 144 pm

20. The correct order of increasing basicity of the given conjugate bases (R = CH3) is

(1) 2RCOO HC C R NH< = < < (2) 2R HC C RCOO NH< ≡ < <

(3)2

RCOO NH HC C R< < ≡ < (4)2

RCOO HC C NH R< ≡ < <

20. (4)Sol : Correct order of increasing basic strength is

R–COO(–)

< CH≡C(–)

< ( )2

NH−

< R(–)

21. The correct sequence which shows decreasing order of the ionic radii of the elements is(1) Al

3+> Mg

2+> Na

+> F

– > O

2– (2) Na

+> Mg

2+> Al

3+> O

2– > F

–

(3) Na+

> F –

> Mg2+

> O2–

> Al3+

(4) O2–

> F –

> Na+

> Mg2+

> Al3+

21. (4)

Sol : For isoelectronic species higher theZ

eratio , smaller the ionic radius




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z

efor 2 8

O 0.810

− = =

9

F 0.910

−

= =

11Na 1.1

10

+ = =

2 12Mg 1.2

10

+ = =

3 13Al 1.3

10

+ = =

22. Solubility product of silver bromide is 5.0 x 10 –13

. The quantity of potassium bromide (molar masstaken as 120 g of mol

–1) to be added to 1 litre of 0.05 M solution of silver nitrate to start the

precipitation of AgBr is

(1) 1.2 x 10

–10

g (2) 1.2 x 10

–9

g (3) 6.2 x 10

–5

g (4) 5.0 x 10

–8

g22. (2)

Sol : Ag+

+ Br – AgBr

Precipitation starts when ionic product just exceeds solubility product

spK Ag Br+ − =

13sp 11

K 5 10Br 10

0.05Ag

−− −

+

× = = =

i.e., precipitation just starts when 10 –11

moles of KBr is added to 1L of AgNO3 solution.No. of moles of KBr to be added = 10

–11

∴ weight of KBr to be added = 10 –11

x 120= 1.2 x 10

–9g

23. The Gibbs energy for the decomposition of Al2O3 at 500°C is as follows :

2

3Al2O3 →

4

3Al + O2, ∆rG = + 966 kJ mol

–1

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

23. (3)

Sol : ∆G = – nFE G

EnF

−∆=

3966 10E

4 96500

×= −

×

= –2.5 V

∴ The potential difference needed for the reduction = 2.5 V

24. At 25°C, the solubility product of Mg(OH)2 is 1.0 x 10 –11

. At which pH, will Mg2+

ions start precipitatingin the form of Mg(OH)2 from a solution of 0.001 M Mg

2+ions ?

(1) 9 (2) 10 (3) 11 (4) 824. (2)

Sol : 2

2Mg 2OH Mg(OH)+ −+

22

sp

sp 4

2

OH H

K Mg OH

KOH 10

Mg

p 4 and p 10

+ −

− −

+

=

= =

∴ = =




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25. Percentage of free space in cubic close packed structure and in body centred packed structure arerespectively

(1) 30% and 26% (2) 26% and 32% (3) 32% and 48% (4) 48% and 26%25. (2)Sol : packing fraction of cubic close packing and body centred packing are 0.74 and 0.68 respectively.

26. 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

26. (3)Sol :

H2C=HC C2H5

H

CH3

only 3–methyl–1–pentene has a chiral carbon

27. Biuret test is not given by(1) carbohydrates (2) polypeptides (3) urea (4) proteins

27. (1)Sol : It is a test characteristic of amide linkage. Urea also has amide linkage like proteins.

28. The correct order of 0

2M /ME + values with negative sign for the four successive elements Cr, Mn, Fe

and Co is(1) Mn > Cr > Fe > Co (2) Cr > Fe > Mn > Co (3) Fe > Mn > Cr > Co (4) Cr > Mn > Fe > Co

28. (1)

29. The polymer containing strong intermolecular forces e.g. hydrogen bonding, is(1) teflon (2) nylon 6,6 (3) polystyrene (4) natural rubber

29. (2)Sol : nylon 6,6 is a polymer of adipic acid and hexamethylene diamine

C

O

(CH2)4 C

O

NH (CH2)6 NHn

30. For a particular reversible reaction at temperature T, ∆H and ∆S were found to be both +ve. If Te isthe temperature at equilibrium, the reaction would be spontaneous when(1) Te > T (2) T > Te (3) Te is 5 times T (4) T = Te

30. (2)

Sol : G H T S∆ = ∆ − ∆

at equilibrium, ∆G = 0

for a reaction to be spontaneous ∆G should be negative

eT T∴ >

31. A rectangular loop has a sliding connector PQ of length and resistance R Ω and it is moving with aspeed 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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(1)1 2

B v 2B vI I , I

R R= − = =

(2) 1 2

B v 2B v

I I , I3R 3R= = =

(3) 1 2

B vI I I

R= = =

(4) 1 2

B v B vI I , I

6R 3R= = =

31. 2Sol. A moving conductor is equivalent to a battery of emf = v B (motion emf)

Equivalent circuit

1 2I I I= +

applying Kirchoff’s law

1I R IR vB 0+ − = ……………(1)

2I R IR vB 0+ − = ……………(2)adding (1) & (2)

RR

I1 I2

2IR IR 2vB+ =

2vBI

3R=

1 2

vBI I

3R= =

32. Let C be the capacitance of a capacitor discharging through a resistor R. Suppose t1 is the timetaken for the energy stored in the capacitor to reduce to half its initial value and t 2 is the time takenfor the charge to reduce to one-fourth its initial value. Then the ratio t1 /t2 will be

(1) 1 (2) 12

(3) 14

(4) 2

32. 3

Sol.22

t / T 2 2t / T00

q1 q 1U (q e ) e

2 C 2C 2C

− −= = = (where CRτ = )

2t /

iU U e− τ=

12t /

i i

1U Ue

2

− τ=

12t / 1e

2

− τ= 1

Tt ln2

2=

Now t / T

0q q e−=

t /2T

0 01 q q e4

−=

2t T ln4 2Tln2= =

∴ 1

2

t 1

t 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 acompletely inelastic collision.Statement-2 : Principle of conservation of momentum holds true for all kinds of collisions.




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(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.33. 1Sol. m2 m1

v1 v2

If it is a completely inelastic collision then

1 1 2 2 1 2m v m v m v m v+ = +

1 1 2 2

1 2

m v m vv

m m

+=

+

2 2

1 2

1 2

p pK.E

2m 2m= +

as 1 2p and p

both simultaneously cannot be zerotherefore total KE cannot be lost.

34. Statement-1 : When ultraviolet light is incident on a photocell, its stopping potential is V0 and themaximum kinetic energy of the photoelectrons is Kmax. When the ultraviolet light is replaced by X-rays, both V0 and Kmax increase.Statement-2 : Photoelectrons are emitted with speeds ranging from zero to a maximum valuebecause 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 Statement-1.(3) Statement-1 is false, Statement-2 is true.(4) Statement-1 is true, Statement-2 is false.

34. 4Sol. 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 =h ν − ϕ

Stopping potential is to stop the fastest photoelectron

0

hV

e e

ν ϕ= −

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 beforegetting ejected and not due to range of frequencies in the incident light.

35. 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 inequilibrium in a mixture of this oil and water, which of the following pictures represents its equilibriumposition ?(1) (2)




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(3) (4)

35. 2

Sol. oil waterρ < ρ < ρ

Oil is the least dense of them so it should settle at the top with water at the base. Now the ball isdenser than oil but less denser than water. So, it will sink through oil but will not sink in water. So itwill stay at the oil–water interface.

36. A particle is moving with velocity ˆ ˆv K(y i x j)= +

, where K is a constant. The general equation for its

path is(1) y = x

2+ constant (2) y

2= x + constant (3) xy = constant (4) y

2= x

2+ constant

36. 4Sol. ˆ ˆv Ky i Kx j= +

dx dyKy, Kx

dt dt= =

dy dy dt Kx

dx dt dx Ky= × =

y dy = x dxy

2= x

2+ 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(1) (2)

(3) (4)

37. 1

Sol. The magnetic field in between because of each will be in opposite directionBin between = 0 0i iˆ ˆ j ( j)

2 x 2 (2d x)

µ µ− −

π π −

0i 1 1 ˆ( j)2 x 2d x

µ = − π −

at x = d, Bin between = 0

for x < d, Bin between = ˆ( j)

for x > d, Bin between = ˆ( j)−

towards x net magnetic field will add up and direction will be ˆ( j)−

towards x ' net magnetic field will add up and direction will be ˆ( j)




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38. In the circuit shown below, the key K is closed at t = 0. The current through the battery is

(1) 1 2

2 2

1 2

VR R

R R+at t = 0 and

2

V

Rat t = ∞

(2)2

V

Rat t = 0 and 1 2

1 2

V(R R )

R R

+at t = ∞

(3)2

V

Rat t = 0 and 1 2

2 2

1 2

VR R

R R+at t = ∞

(4) 1 2

1 2

V(R R )

R R

+at t = 0 and

2

V

Rat t = ∞

38. 2Sol. At t = 0, inductor behaves like an infinite resistance

So at t = 0,2

Vi

R=

and at t = ∞ , inductor behaves like a conducting wire

1 2

e q 1 2

V(R R )Vi

R R R

+= =

39. The figure shows the position – time (x – t)graph of one-dimensional motion of a bodyof mass 0.4 kg. The magnitude of eachimpulse is(1) 0.4 Ns (2) 0.8 Ns(3) 1.6 Ns (4) 0.2 Ns

39. 2Sol. 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.

Initial velocity =2

1 m/s2

=

Final velocity =2

1 m/s2

− = −

iP 0.4=

N – s

iPj 0.4= −

N – s

f iJ P P= −

= – 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 massM

2each.

Speed of light is c.

40. 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

40. 3Sol. 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)m

cM m

∆

+ ∆(2)

2 mc

M

∆(3)

mc

M

∆(4)

mc

M m

∆

+ ∆

41. 2Sol. Conserving the momentum

1 2

M M0 V V

2 2= −

1 2V V= …………….(1)

2 2 2

1 2

1 M 1 Mmc . V . .V

2 2 2 2∆ = + …………….(2)

2 2

1

Mmc V

2∆ =

22

1

2 mcV

M

∆=

12 mV c

M∆=

42. A radioactive nucleus (initial mass number A and atomic number Z) emits 3 α-particles and 2positrons. The ratio of number of neutrons to that of protons in the final nucleus will be

(1)A Z 8

Z 4

− −

−(2)

A Z 4

Z 8

− −

−(3)

A Z 12

Z 4

− −

−(4)

A Z 4

Z 2

− −

−

42. 2Sol. In positive beta decay a proton is transformed into a neutron and a positron is emitted.

0p n e+ + → +

no. of neutrons initially was A – Zno. 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.

43. 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

(1)2 2

0

q j

4 rπ ε(2) –

2 2

0

q j

4 rπ ε

(3) – 2 2

0

q j

2 rπ ε(4)

2 2

0

q j

2 rπ ε

43. 3

Sol. Linear charge density qr

λ = π

2

K.dqˆ Ê dEsin ( j) sin ( j)r

= θ − = θ −

2

K qr Ê d sin ( j)r r

= θ θ −π

y

θ xθ

dθ

20

K q ˆsin ( j)r

π

= θ −π

2 2

0

q ˆ( j)2 r

= −π ε




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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. 1Sol. Truth table for given combination is

A B X0 0 0

0 1 1

1 0 11 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

45. 2Sol. The efficiency of cycle is

2

1

T1

Tη = −

for adiabatic process

TVγ –1= constant

For diatomic gas7

5γ =

1 1

1 1 2 2T V T Vγ− γ−=

1

21 2

1

VT TV

γ−

=

71

51 2

T T (32)−

= 5 2/ 5

2T (2 )=

= T2 x 4T1 = 4T2.

1 31 0.75

4 4

η = − = =

46. If a source of power 4 kW produces 1020

photons/second, the radiation belong to a part of thespectrum called

(1) X–rays (2) ultraviolet rays (3) microwaves (4) γ –rays46. 1Sol. 4 x 10

3= 10

20x hf

3

20 34

4 10f

10 6.023 10−

×=

× ×

f = 6.03 x 1016

HzThe obtained frequency lies in the band of X–rays.

47. 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

47. 1




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48. In a series LCR circuit R = 200 Ω and the voltage and the frequency of the main supply is 220 V and50 Hz respectively. On taking out the capacitance from the circuit the current lags behind the voltageby 30°. On taking out the inductor from the circuit the current leads the voltage by 30

o. The power

dissipated in the LCR circuit is(1) 305 W (2) 210 W (3) Zero W (4) 242 W48. 4Sol. The given circuit is under resonance as XL = XC

Hence power dissipated in the circuit is2V

PR

= = 242 W

49. Let there be a spherically symmetric charge distribution with charge density varying as

0

5 r(r)

4 R

ρ = ρ −

upto r = R, and ρ(r) = 0 for r > R, where r is the distance from the origin. The

electric field at a distance r(r < R) from the origin is given by

(1) 0

0

4 r5 r

3 3 R

πρ − ε

(2) 0

0

r5 r

4 3 R

ρ − ε

(3) 0

0

4 r5 r

3 4 R

ρ − ε

(4) 0

0

r5 r

3 4 R

ρ − ε

49. 2Sol. Apply shell theorem the total charge upto distance r can be calculated as followed

2dq 4 r .dr.= π ρ

2

0

5 r4 r .dr.

4 R

= π ρ −

32

0

5 r4 r dr dr

4 R

= πρ −

r 32

0

0

5 rdq q 4 r dr dr

4 R

= = πρ −

3 4

0

5 r 1 r44 3 R 4

= πρ −

2

kqE

r=

3 4

02

0

1 1 5 r r.4

4 r 4 3 4R

= πρ − πε

0

0

r 5 rE

4 3 R

ρ = − ε

50. The potential energy function for the force between two atoms in a diatomic molecule is

approximately given by 12 6

a bU(x) x x= − , where a and b are constants and x is the distance between

the atoms. If the dissociation energy of the molecule is D = [U(x = ∞) – Uat equilibrium], D is

(1)2b

2a(2)

2b

12a(3)

2b

4a(4)

2b

6a

50. 3

Sol. 12 6

a bU(x)

x x= −

U(x = ∞) = 0

as,13 7

dU 12a 6bF

dx x x

= − = − +

at equilibrium, F = 0




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∴ 6 2ax

b=

∴

2

at equilibrium 2

a b b

U 2a 4a2abb

−= − =

∴ 2

at equilibrium

bD U(x ) U

4a = = ∞ − =

51. Two identical charged spheres are suspended by strings of equal lengths. The strings make anangle of 30° with each other. When suspended in a liquid of density 0.8 g cm

–3, the angle remains

the same. If density of the material of the sphere is 16 g cm –3

, the dielectric constant of the liquid is(1) 4 (2) 3 (3) 2 (4) 1

51. 3Sol. From F.B.D of sphere, using Lami’s theorem

Ftanmg = θ ………………(i)

when suspended in liquid, as θ remains same,

∴ F'

tan

mg 1d

= θρ

−

………………(ii)

θ T

F

mg

using (i) and (ii)

F F' Fwhere, F'

mg Kmg 1

d

= =ρ

−

∴ F F'

mg mg K 1 d

=

ρ

−

or1

K 2

1d

= =ρ

−

52. 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 arenearly

(1) 1 21 2,

2

α + αα + α (2) 1 2

1 2,2

α + αα + α (3) 1 2

1 2

1 2

,α α

α + αα + α

(4) 1 2 1 2,2 2

α + α α + α

52. 4

Sol. Let R0 be the initial resistance of both conductors∴ At temperature θ their resistance will be,

1 0 1 2 0 2R R (1 ) and R R (1 )= + α θ = + α θ

for, series combination, Rs = R1 + R2

s0 s 0 1 0 2R (1 ) R (1 ) R (1 )+ α θ = + α θ + + α θ

wheres0 0 0 0

R R R 2R= + =

∴ 0 s 0 0 1 2

2R (1 ) 2R R ( )+ α θ = + θ α + α

or 1 2s

2

α + αα =

for parallel combination, 1 2p

1 2

R RR

R R=

+




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0 1 0 2p0 p

0 1 0 2

R (1 )R (1 )R (1 )

R (1 ) R (1 )

+ α θ + α θ+ α θ =

+ α θ + + α θ

where, 0 0 0p0

0 0

R R RR R R 2= =+

∴ 2

0 0 1 2 1 2p

0 1 2

R R (1 )(1 )

2 R (2 )

+ α θ + α θ + α α θ+ α θ =

+ α θ + α θ

as1 2andα α are small quantities

∴ 1 2

α α is negligible

or 1 2 1 2p 1 2

1 2

[1 ( ) ]2 ( ) 2

α + α α + αα = = − α + α θ

+ α + α θ

as 2

1 2( )α + α is negligible

∴ 1 2p

2

α + αα =

53. A point P moves in counter-clockwise direction on a circular pathas shown in the figure. The movement of ‘P’ is such that itsweeps out a length s = t

3+ 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(1) 13 m/s

2(2) 12 m/s

2

(3) 7.2 m/s2

(4) 14 m/s2

53. 4Sol. S = t

3+ 5

∴ speed, 2dsv 3t

dt= =

and rate of change of speeddv

6tdt

= =

∴ tangential acceleration at t = 2s, at = 6 x 2 = 12 m/s2

at t = 2s, v = 3(2)2

= 12 m/s

∴ centripetal acceleration,2

2

c

v 144a m/ s

R 20= =

∴ net acceleration = 2 2

t ia a+

214 m /s≈

54. Two fixed frictionless inclined plane making an angle 30o

and 60o

with the vertical are shown in the figure. Twoblock A and B are placed on the two planes. What is therelative vertical acceleration of A with respect to B ?(1) 4.9 ms

–2in horizontal direction

(2) 9.8 ms –2

in vertical direction(3) zero(4) 4.9 ms

–2in vertical direction

54. 4Sol. mg sin θ = ma

∴ a = g sin θ where a is along the inclined plane

∴ vertical component of acceleration is g sin2 θ

∴ relative vertical acceleration of A with respect to B is




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2 2 2gg[sin 60 sin 30] 4.9 m/ s

2− = = in vertical direction.

55. 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)

(1)2 2v vˆ ˆcos i sin j

R R− θ + θ (2)

2 2v vˆ ˆsin i cos jR R

− θ + θ

(3)2 2v vˆ ˆcos i sin j

R R− θ − θ (4)

2 2v vˆ î jR R

+

55. 3Sol. For a particle in uniform circular motion,

2va

R=

towards centre of circle

∴ 2v ˆ â ( cos i sin j)

R

= − θ − θ

or2 2v vˆ â cos i sin j

R R= − θ − θ

x

y

P (R, θ)

ac

ac

Directions: Questions number 56 – 58 are based on the following paragraph.

An initially parallel cylindrical beam travels in a medium of refractive index0 2(I) Iµ = µ + µ , where µ0

and µ2 are positive constants and I is the intensity of the light beam. The intensity of the beam isdecreasing with increasing radius.

56. 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

56. 2Sol. As intensity is maximum at axis,

∴ µ will be maximum and speed will be minimum on the axis of the beam.

∴ beam will converge.

57. 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

57. 4

Sol. For a parallel cylinderical beam, wavefront will be planar.

58. The speed of light in the medium is(1) minimum on the axis of the beam (2) the same everywhere in the beam(3) directly proportional to the intensity I (4) maximum on the axis of the beam

58. 1




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59. A small particle of mass m is projected at an angle θ with thex-axis with an initial velocity v0 in the x-y plane as shown in the

figure. At a time 0v sint

g

θ< , the angular momentum of the

particle is

(1) 2

0ˆmgv t cos j− θ (2)

0ˆmgv t cos kθ

(3) 2

0

1 ˆmgv t cos k2

− θ (4) 2

0

1 ˆmgv t cos i2

θ

where ˆ ˆ î, j and k are unit vectors along x, y and z–axis respectively.

59. 3

Sol. L m(r v)= ×

2

0 0 0 0

1ˆ ˆ ˆ ˆL m v cos t i (v sin t gt ) j v cos i (v sin gt) j2

= θ + θ − × θ + θ −

0 1 ˆmv cos t gt k2 = θ −

2

0

1 ˆmgv t cos k2

= − θ

60. The equation of a wave on a string of linear mass density 0.04 kg m –1

is given by

t xy 0.02(m)sin 2

0.04(s) 0.50(m)

= π −

. The tension in the string is

(1) 4.0 N (2) 12.5 N (3) 0.5 N (4) 6.25 N60. 4

Sol.2 2

2

2 2

(2 /0.004)T v 0.04 6.25 N

k (2 / 0.50)

ω π= µ = µ = =

π

61. Let cos(α + β) =4

5and let sin(α – β) =

5

13, where 0 ≤ α, β ≤

4

π, then tan 2α =

(1)56

33(2)

19

12(3)

20

7(4)

25

16

61. 1

cos (α + β) =4

5 tan(α + β) =

3

4

sin(α – β) = 513

tan(α – β) = 512

tan 2α = tan(α + β + α – β) =

3 5564 12

3 5 331

4 12

+=

−

62. 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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(3) x ∈ S and x ≤ 0 x is not rational

(4) There is a rational number x ∈ S such that x ≤ 0

62. 2

P: there is a rational number x ∈ S such that x > 0

~P: Every rational number x ∈ S satisfies x ≤ 0

63. Let ˆ ˆ ˆ ˆ â j k and c i j k= − = − −

. Then vector b

satisfying a b c 0 and a b× + = ⋅

= 3 is

(1) ˆ ˆ ˆ2i j 2k− + (2) ˆ ˆ î j 2k− − (3) ˆ ˆ î j 2k+ − (4) ˆ ˆ î j 2k− + −

63. 4

c b a= ×

b c 0⋅ =

( ) ( )1 2 3ˆ ˆ ˆ ˆ ˆ ˆb i b j b k i j k+ + ⋅ − − = 0

b1 – b2 – b3 = 0

and a b⋅ = 3

b2 – b3 = 3

b1 = b2 + b3 = 3 + 2b3

( ) ( )3 3 3ˆ ˆ ˆb 3 2b i 3 b j b k= + + + +

.

64. The equation of the tangent to the curve y = x +2

4

x, that is parallel to the x-axis, is

(1) y = 1 (2) y = 2 (3) y = 3 (4) y = 0

64. 3

Parallel to x-axis dy

dx= 0

3

81

x− = 0

x = 2 y = 3

Equation of tangent is y – 3 = 0(x – 2) y – 3 = 0

65. Solution of the differential equation cos x dy = y(sin x – y) dx, 0 < x <2

πis

(1) y sec x = tan x + c (2) y tan x = sec x + c (3) tan x = (sec x + c)y (4) sec x = (tan x + c)y

65. 4

cos x dy = y(sin x – y) dx

2dyy tanx y sec x

dx= −

2

1 dy 1

tanx sec xdx yy − = −

Let1

ty

=

2

1 dy dt

dx dxy− =

dy

dx− – t tan x = –sec x

dt

dx+ (tan x) t = sec x.

I.F. = e tan x dx= sec x

Solution is t(I.F) = (I.F) sec x dx




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1

ysec x = tan x + c

66. The area bounded by the curves y = cos x and y = sin x between the ordinates x = 0 and x =3

2

πis

(1) 4 2 + 2 (2) 4 2 – 1 (3) 4 2 + 1 (4) 4 2 – 2

66. 4

( ) ( ) ( )

5 3

4 4 2

50

4 4

cosx sinx dx sinx cosx dx cosx sinx 4 2 2

π π π

π π

− + − + − = −

cos x sin x

4

π

5

4

π

3

2

π

2π 0

π

67. If two tangents drawn from a point P to the parabola y2

= 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

67. 2

The locus of perpendicular tangents is directrix

i.e, x = –a; x = –1

68. If the vectors ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ â i j 2k, b 2i 4 j k and c i j k= − + = + + = λ + + µ

are mutually orthogonal, then (λ, µ) =

(1) (2, –3) (2) (–2, 3) (3) (3, –2) (4) (–3, 2)

68. 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

69. Consider the following relations:

R = (x, y) | x, y are real numbers and x = wy for some rational number w;

S =m p

, m, n, p and q are integers such that n, q 0 and qm = pnn q

≠

. 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

69. 2

xRy need not implies yRx

S:m p

sn q

⇔ qm = pn

m ms

n nreflexive




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m ps

n q

p ms

q nsymmetric

m p p rs , sn q q s 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

2x 3, if x 1

− ≤ −

+ > −. If f has a local minimum at x = –1, then a

possible value of k is

(1) 0 (2)1

2− (3) –1 (4) 1

70. 3

f(x) = k – 2x if x ≤ –1

= 2x + 3 if x > –1

2x + 3k – 2x

1

–1

x 1lim→− −

f(x) ≤ –1This is truewhere k = –1

71. 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

71. 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. 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

is1

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

72. 2




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N(S) =20

C4

Statement-1: common difference is 1; total number of cases = 17

common difference is 2; total number of cases = 14





Prob. =20

4

17 14 11 8 5 2 1

85C

+ + + + += .

73. 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

73. 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.

74. Let S1 = ( )

1010

j j 1 j j 1 C= − , S2 =

1010

j j 1 j C= and S3 =

102 10

j j 1 j C= .

Statement-1: S3 = 55 × 29

Statement-2: S1 = 90 × 28

and S2 = 10 × 28.

(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

74. 2

S1 = ( )( ) ( ) ( ) ( ) ( )( )

10 108

j 1 j 2

10! 8! j j 1 90 90 2

j j 1 j 2 ! 10 j ! j 2 ! 8 j 2 != =

− = = ⋅− − − − − −

.

S2 =

( ) ( )( ) ( ) ( )( )

10 109

j 1 j 1

10! 9! j 10 10 2 j j 1 ! 9 j 1 ! j 1 ! 9 j 1 != =

= = ⋅− − − − − −

.

S3 = ( )( )

( )10 10 10

10 10

j j j 1 j 1 j 1

10! j j 1 j j j 1 C j C

j! 10 j != = =

− + = − = − = 90 . 2

8+ 10 . 2

9

= 90 . 28

+ 20 . 28

= 110 . 28

= 55 . 29.

75. Let A be a 2 × 2 matrix with non-zero entries and let A2

= 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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(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

75. 2

Let A = a bc d

, abcd ≠ 0

A2

=a b a b

c d c d

⋅

A2

=2

2

a bc ab bd

ac cd bc d

+ +

+ +

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 = –a

2

– bc = –1.

76. Let f: R → R be a continuous function defined by f(x) =x x

1

e 2e−+.

Statement-1: f(c) =1

3, for some c ∈ R.

Statement-2: 0 < f(x) ≤ 1

2 2, for all x ∈ R

(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

76. 4

f(x) =x

x x 2x

1 e

e 2e e 2−=

+ +

f′(x) =( )

( )

2x x 2x x

22x 2

e 2 e 2e e

e +

+ − ⋅

f′(x) = 0 e2x

+ 2 = 2e2x

e2x

= 2 ex

= 2

maximum f(x) =2 1

4 2 2=

0 < f(x) ≤

1

2 2 ∀ x ∈ R

Since 0 <1 1

3 2 2< for some c ∈ R

f(c) =1

3

77. 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

(1) There is a regular polygon withr 1

R 2= (2) There is a regular polygon with

r 2

R 3=




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(3) There is a regular polygon withr 3

R 2= (4) There is a regular polygon with

r 1

R 2=

77. 2

r =a

cot2 n

π

‘a’ is side of polygon.

R =a

cosec2 n

π

cotr n cosR n

cosecn

ππ

= =π

2cos

n 3

π≠ for any n ∈ N.

78. If α and β are the roots of the equation x2

– x + 1 = 0, then α2009+ β2009

=

(1) –1 (2) 1 (3) 2 (4) –2

78. 2

x2

– x + 1 = 0 x =1 1 4

2

± −

x =1 3 i

2

±

α =1 3

i2 2

+ , β =1 i 3

2 2−

α = cos isin3 3

π π

+ , β = cos isin3 3

π π

−

α2009+ β2009

= 2cos20093

π

=2 2

2cos 668 2cos3 3

π π π + π + = π +

=2 1

2cos 2 13 2

π − = − − =

79. The number of complex numbers z such that |z – 1| = |z + 1| = |z – i| equals

(1) 1 (2) 2 (3) ∞ (4) 0

79. 1

Let z = x + iy

|z – 1| = |z + 1| Re z = 0 x = 0

|z – 1| = |z – i| x = y

|z + 1| = |z – i| y = –x

Only (0, 0) will satisfy all conditions.

Number of complex number z = 1

80. 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° 80. 2




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AIEEE−−−−2010−−−−27


= cos 45° =1

2

m = cos 120° =1

2−

n = cos θ

where θ is the angle which line makes with positive z-axis.

Now 2

+ m2

+ n2

= 1

1 1

2 4+ + cos

2θ = 1

cos2θ =

1

4

cos θ =1

2(θ Being acute)

θ = 3

π.

81. The line L given byx y

5 b+ = 1 passes through the point (13, 32). The line K is parallel to L and has

the equationx y

c 3+ = 1. Then the distance between L and K is

(1) 17 (2)17

15(3)

23

17(4)

23

15

81. 3

Slope of line L =b

5−

Slope of line K =3

c−

Line L is parallel to line k.

b 3

5 c= bc = 15

(13, 32) is a point on L.

13 32 32 8

15 b b 5

+ = = −

b = –20 c =3

4−

Equation of K: y – 4x = 3

Distance between L and K =52 32 3 23

17 17

− +=

82. A person is to count 4500 currency notes. Let an denote the number of notes he counts in the nth

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

82. 1

Till 10th

minute number of counted notes = 1500

3000 =n

2[2 × 148 + (n – 1)(–2)] = n[148 – n + 1]




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AIEEE−−−−2010−−−−28


n2

– 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 withx

f(3x)lim

f(x)→∞= 1. Then

x

f(2x)lim

f(x)→∞=

(1)2

3(2)

3

2(3) 3 (4) 1

83. 4

f(x) is a positive increasing function

0 < f(x) < f(2x) < f(3x)

0 < 1 <f(2x) f(3x)

f(x) f(x)<

x x x

f(2x) f(3x)lim1 lim lim

f(x) f(x)→∞ →∞ →∞≤ ≤

By sandwich theorem.

x

f(2x)lim

f(x)→∞= 1

84. Let p(x) be a function defined on R such that p′(x) = p′(1 – x), for all x ∈ [0, 1], p(0) = 1 and p(1) = 41.

Then1

0

p(x) dx equals

(1) 21 (2) 41 (3) 42 (4) 41

84. 1 p′(x) = p′(1 – x)

p(x) = –p(1 – x) + c

at x = 0

p(0) = –p(1) + c 42 = c

now p(x) = –p(1 – x) + 42

p(x) + p(1 – x) = 42

I =1 1

0 0

p(x) dx p(1 x) dx= −

2 I =1

0

(42) dx I = 21.

85. Let f: (–1, 1) → R be a differentiable function with f(0) = –1 and f′(0) = 1. Let g(x) = [f(2f(x) + 2)]2.

Then g′(0) =

(1) –4 (2) 0 (3) –2 (4) 4

85. 1

g′(x) = 2(f(2f(x) + 2)) ( )( )d

f 2f(x) 2dx

+

= 2f(2f(x) + 2) f′(2f(x) + 2) . (2f′(x))

g′(0) = 2f(2f(0) + 2) . f′(2f(0) + 2) . 2(f′(0) = 4f(0) f′(0)

= 4(–1) (1) = –4




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86. There are two urns. Urn A has 3 distinct red balls and urn B has 9 distinct blue balls. From each urn

two balls are taken out at random and then transferred to the other. The number of ways in which

this can be done is

(1) 36 (2) 66 (3) 108 (4) 3

86. 3

Total number of ways =3C2 ×

9C2

= 3 × 9 8

2

×= 3 × 36 = 108

87. Consider the system of linear equations:

x1 + 2x2 + x3 = 3

2x1 + 3x2 + x3 = 3

3x1 + 5x2 + 2x3 = 1

The system has

(1) exactly 3 solutions (2) a unique solution

(3) no solution (4) infinite number of solutions

87. 3

D =

1 2 1

2 3 1

3 5 2

= 0

D1 =

3 2 1

3 3 1

1 5 2

≠ 0

Given system, does not have any solution.

No solution.

88. An urn contains nine balls of which three are red, four are blue and two are green. Three balls are

drawn at random without replacement from the urn. The probability that the three balls have different

colour is

(1)2

7(2)

1

21(3)

2

23(4)

1

3

88. 1

n(S) =9C3

n(E) =3C1 ×

4C1 ×

2C1

Probability =9

3

3 4 2 24 3! 24 6 26!

9! 9 8 7 7C

× × × ×= × = =

× ×

.

89. For two data sets, each of size 5, the variances are given to be 4 and 5 and the corresponding

means are given to be 2 and 4, respectively. The variance of the combined data set is

(1)11

2(2) 6 (3)

13

2(4)

5

2

89. 1

σx2

= 4

σy2

= 5

x = 2

y = 4




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AIEEE−−−−2010−−−−30


ix

5

= 2 xi = 10; yi = 20

σx

2

= ( ) ( )22 2

i i

1 1

x x y 162 5

− = −

xi2

= 40

yi2

= 105

σz2

= ( ) ( )2

2 2

i i

1 x y 1 145 90 55 11x y 40 105 9

10 2 10 10 10 2

+ − + − = + − = = =

90. The circle x2

+ y2

= 4x + 8y + 5 intersects the line 3x – 4y = m at two distinct points if

(1) –35 < m < 15 (2) 15 < m < 65 (3) 35 < m < 85 (4) –85 < m < –3590. 1

Circle x2

+ y2

– 4x – 8y – 5 = 0

Centre = (2, 4), Radius = 4 16 5+ + = 5If circle is intersecting line 3x – 4y = m

at two distinct points.

length of perpendicular from centre < radius

6 16 m

5

− −< 5

|10 + m| < 25

–25 < m + 10 < 25

–35 < m < 15.

* * *




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AIEEE−−−−2010−−−−31


READ THE FOLLOWING INSTRUCTIONS CAREFULLY:

(!.!.!.!.!,,,,////

2!!!!>>>>!#

$ 41.)/

!

% 3<(#

* 2 # # # # >>>> .%/ (

1 1 1 1 # <#

!

9 7 ! # .)

-!-/#<

; !

<

#@!4"A<%

.',$/

8 3#<!<

477<#7<#7<#7<#

& 5-<

' 1 # !<# <

<5)7<!

<!-

! <<!

( 7: ( 7: ( 7: ( 7:

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5: - 5 #

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5)744

% 1!

* - )# # # #- ) # # # #- )# # # #- ) # # # #

#<#<#<#<)-))-))-))-)




7/14/2019 Aieee


FIITJEE (Hyderabad Classes) Limited. 5-9-14/B, Saifabad, (Opp. Secretariat) Hyderabad. 500 063. Phone: 040-66777000 – 03 Fax: 040-66777004

A A I I E E E E E E – – 2 2 0 0 1111 ( ( S S e e t t – – Q 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.

6. 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.

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, mobilephone, 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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AIEEE−2011−2


PART A : PHYSICS

1. The transverse displacement ( )y x,t of a wave on a string is given by ( ) ( )2 2ax bt 2 ab xty x,t e

− + += .

This represents a

(1) wave moving in – x direction with speed

b

a (2) standing wave of frequency b

(3) standing wave of frequency1

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 mmCircular scale reading : 52 divisionsGiven 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 themass m, if the string does not slip on the pulley, is

(1) g (2)2

g3

(3)g

3(4)

3g

2

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 atrest at a point near the rim of the disc. The insect now moves along a diameter of the disc to reach itsother 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 ( )0 0X X A> . If the maximum separation between

them is ( )0X A+ , the phase difference between their motion is :

(1)3

π(2)

4

π(3)

6

π(4)

2

π

7. Two bodies of masses m and 4 m are placed at a distance r. The gravitational potential at a point on theline joining them where the gravitational field is zero is:

(1)4Gm

r

− (2)6Gm

r

− (3)9Gm

r

− (4) zero

8. Two identical charged spheres suspended from a common point by two massless strings of length l are

initially a distance ( )d d 1<< apart because of their mutual repulsion. The charge begins to leak from

both the spheres at a constant rate. As a result the charges approach each other with a velocity v. Thenas a function of distance x between them,

(1) 1v x−∝ (2) 1/ 2v x∝ (3) v x∝ (4) 1/ 2v x−∝



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AIEEE−2011−3


9. A boat is moving due east in a region where the earth’s magnetic field is 5 1 15.0 10 NA m− − −× due north and

horizontal. The boat carries a vertical aerial 2m long. If the speed of the boat is 1.50 ms−1, the

magnitude of the induced emf in the wire of aerial is :(1) 0.75 mV (2) 0.50 mV (3) 0.15 mV (4) 1 mV

10. An object, moving with a speed of 6.25 m/s, is decelerated at a rate given by :

dv 2.5 vdt

= −

where v is the instantaneous speed. The time taken by the object, to come to rest, would be :(1) 2 s (2) 4 s (3) 8 s (4) 1 s

11. A fully charged capacitor C with initial charge q0 is connected to a coil of self inductance L at t = 0. Thetime at which the energy is stored equally between the electric and the magnetic field is :

(1) LC4

π(2) 2 LCπ (3) LC (4) LCπ

12. Let the x – z plane be the boundary between two transparent media. Medium 1 in z ≥ 0 has a refractive

index of 2 and medium 2 with z < 0 has a refractive index of 3 . A ray of light in medium 1 given by

the vector A 6 3 i 8 3 j 10k→ ∧ ∧ ∧

= + − is incident on the plane of separation. The angle of refraction in

medium 2 is(1) 45

0(2) 60

0(3) 75

0(4) 30

0

13. A current I flows in an infinitely long wire with cross section in the form of a semicircular ring of radius R.The magnitude of the magnetic induction along its axis is

(1) 0

2

I

2 R

µ

π(2) 0

I

2 R

µ

π(3) 0

2

I

4 R

µ

π(4) 0

2

I

R

µ

π

14. A thermally insulated vessel contains an ideal gas of molecular mass M and ratio of specific heats γ. It is

moving with speed υ and is suddenly brought to rest. Assuming no heat is lost to the surroundings, itstemperature increases by :

(1) ( ) 21 M K2 Rγ − υ

γ(2)

2M K2R

γ υ (3) ( ) 21 M K2R

γ − υ (4) ( )( )

21 M K2 1 R

γ − υγ +

15. A mass M, attached to a horizontal spring, executes S.H.M. with amplitude A1. When the mass M passesthrough its mean position then a smaller mass m is placed over it and both of them move together with

amplitude A2. The ratio of 1

2

A

A

is :

(1)M m

M

+(2)

1/ 2M

M m

+

(3)

1/ 2M m

M

+

(4)M

M m+

16. Water is flowing continuously from a tap having an internal diameter 38 10 m−× . The water velocity as it

leaves the tap is 0.4 ms−1. The diameter of the water stream at a distance 12 10 m−× below the lap is

close to :

(1) 37.5 10 m−× (2) 39.6 10 m−× (3) 33.6 10 m−× (4) 35.0 10 m−×



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AIEEE−2011−4


17. 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 : Sky wave signals are used for long distance radio communication. These signals are ingeneral, less stable than ground wave signals.Statement-2 : The state of ionosphere varies from hour to hour, day to day and season to season.(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.

18. Three perfect gases at absolute temperatures T1, T2 and T3 are mixed. The masses of molecules are m1,m2 and m3 and the number of molecules are n1, n2 and n3 respectively. Assuming no loss of energy, thefinal temperature of the mixture is :

(1) 1 1 2 2 3 3

1 2 3

n T n T n T

n n n

+ +

+ +(2)

2 2

1 1 2 2 3 3

1 1 2 2 3 3

n T n T n T

n T n T n T

+ +

+ +(3)

2 2 2 2 2 2

1 1 2 2 3 3

1 1 2 2 3 3

n T n T n T

n T n T n T

+ +

+ +(4)

( )1 2 3T T T

3

+ +

19. A pulley of radius 2 m is rotated about its axis by a force ( )2F 20t 5t= − Newton (where t is measured in

seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation made by the

pulley before its direction of motion if reversed, is :(1) more than 3 but less than 6 (2) more than 6 but less than 9(3) more than 9 (4) less than 3

20. A resistor ‘R’ and 2µF capacitor in series is connected through a switch to 200 V direct supply. Acrossthe capacitor is a neon bulb that lights up at 120 V. Calculate the value of R to make the bulb light up 5 s

after the switch has been closed. ( )10log 2.5 0.4=

(1) 51.7 10× Ω (2) 62.7 10× Ω (3) 73.3 10× Ω (4) 41.3 10× Ω

21. A Carnot engine operating between temperatures T1 and T2 has efficiency1

.6

When T2 is lowered by 62

K, its efficiency increases to1

3

. Then T1 and T2 are, respectively :

(1) 372 K and 330 K (2) 330 K and 268 K (3) 310 K and 248 K (4) 372 K and 310 K

22. If a wire is stretched to make it 0.1% longer, its resistance will :(1) increase by 0.2% (2) decrease by 0.2% (3) decrease by 0.05% (4) increases by 0.05%

23. Direction:

The question has a paragraph followed by two statements, Statement – 1 and statement – 2. Of thegiven four alternatives after the statements, choose the one that describes the statements.

A thin air film is formed by putting the convex surface of a plane – convex lens over a plane glass plate.With monochromatic light, this film gives an interference pattern due to light reflected from the top(convex) surface and the bottom (glass plate) surface of the film.Statement-1 : When light reflects from the air-glass plate interface, the reflected wave suffers a phase

change of π.Statement-2 : The centre of the interference pattern is dark.(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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AIEEE−2011−5


24. A car is fitted with a convex side-view mirror of focal length 20cm. A second car 2.8 m behind the firstcar is overtaking the first car at relative speed of 15 m/s. The speed of the image of the second car asseen in the mirror of the first one is :

(1)1

m/ s15

(2) 10m/ s (3) 15m/ s (4)1

m/ s10

25. Energy required for the electron excitation in Li++

from the first to the third Bohr orbit is :(1) 36.3 eV (2) 108.8 eV (3) 122.4 eV (4) 12.1 eV

26. The electrostatic potential inside a charged spherical ball is given by b2φ = αρ + where r is the distance

from the centre; a, b are constants. Then the charge density inside ball is

(1)06a r − ε (2)

024 a r − π ε (3)06a− ε (4)

024 a r − π ε

27. A water fountain on the ground sprinkles water all around it. If the speed of water coming out of thefountain is v, the total area around the fountain that gets wet is :

(1)4

2

v

gπ (2)

4

2

v

2 g

π(3)

2

2

v

gπ (4)

4v

gπ

28. 100g of water is heated from 300

C to 500

C. Ignoring the slight expansion of the water, the change in itsinternal energy is (specific heat of water is 4148 J/kg/K):(1) 8.4 kJ (2) 84 kJ (3) 2.1 kJ (4) 4.2 kJ

29. The half life of a radioactive substance is 20 minutes. The approximate time interval ( )2 1t t− between

the time t2 when2

3of it has decayed and time t1 and

1

3of it had decayed is :

(1) 14 min (2) 20 min (3) 28 min (4) 7 min

30. 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 : A metallic surface is irradiated by a monochromatic light of frequency v > v0 (the threshold

frequency). The maximum kinetic energy and the stopping potential are Kmax and V0 respectively. If thefrequency 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 asurface 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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AIEEE−2011−6


PART B: MATHEMATICS

31. The lines 1L : y x 0− = and 2L : 2x y 0+ = intersect the line 3L : y 2 0+ = at P and Q respectively. The

bisector of the acute angle between 1L and 2L intersect 3L 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. If 2 4 A sin x cos x= + , then for all real x

(1)13

A 116

≤ ≤ (2) 1 A 2≤ ≤ (3)3 13

A4 16

≤ ≤ (4)3

A 14

≤ ≤

33. The coefficient of 7x in the expansion of ( )6

2 31 x x x− − + is

(1) -132 (2) -144 (3)132 (4) 144

34.( )

x 2

1 cos 2 x 2lim

x 2→

− − −

(1) equals 2 (2) equals 2− (3) equals1

2(4) does not exist

35. Statement – 1 : The number of ways of distributing 10 identical balls in 4 distinct boxes such that nobox is empty is 9

3C

Statement – 2 : The number of ways of choosing any 3 places from 9 different places is 93C .


– 1




36.2

2

d x

dyequals

(1)

1 32

2

d y dy

dxdx

− − −

(2)

22

2

d y dy

dxdx

−

(3)

32

2

d y dy

dxdx

− −

(4)

12

2

d y

dx

−



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AIEEE−2011−7


37. If dy

y 3 0dx

= + > and ( )y 0 2= , then ( )y ln2 is equal to

(1) 5 (2) 13 (3) -2 (4) 7

38. 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 y for some rational number = ∈ × = α α is an equivalence relation on

R .


– 1




39. The value of ( )

1

2

0

8log 1 xdx

1 x

+

+∫ is

(1) log28

π(2) log2

2

π(3) log2 (4) log2π

40. Let ,α β be real and z be a complex number. If 2z z 0+ α + β = has two distinct roots on the line

Rez 1= , then it is necessary that

(1) ( )1, 0β ∈ − (2) 1β = (3) ( )1,β ∈ ∞ (4) ( )0,1β ∈

41. Consider 5 independent Bernoulli’s trials each with probability of success p. If the probability of at least

one failure is greater than or equal to31

32, then p lies in the interval

(1)3 11

,4 12

(2)1

0,2

(3)11

, 112

(4)1 3

,2 4

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 ( )1

f xx x

=−

is

(1) ( )0, ∞ (2) ( ), 0−∞ (3) ( ) , 0−∞ ∞ − (4) ( ),−∞ ∞

44. If the angle between the liney 1 z 3

x2

− −= =

λand the plane x 2y 3z 4+ + = is 1 5

cos14

−

, then λ

equals

(1)3

2(2)

2

5(3)

5

3(4)

2

3



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AIEEE−2011−8


45. If $( )1a 3i k

10= +

r$ and $( )1

b 2i 3j 6k7

= + −r

$ $ , then the value of ( ) ( ) ( )2a b . a b a 2b − × × +

r r r r r ris

(1) -3 (2) 5 (3) 3 (4) -5

46. Equation of the ellipse whose axes are the axes of coordinates and which passes through the point

( )3,1− and has eccentricity 25

is

(1) 2 25x 3y 48 0+ − = (2) 2 23x 5y 15 0+ − = (3) 2 25x 3y 32 0+ − = (4) 2 23x 5y 32 0+ − =

47. Let I be the purchase value of an equipment and ( )V t be the value after it has been used for t years.

The value ( )V t depreciates at a rate given by differential equation( )

( )dV t

k T tdt

= − − , where k 0> is a

constant and T is the total life in years of the equipment. Then the scrap value ( )V T of the equipment is

(1)2kT

I2

− (2)( )

2k T t

I2

−− (3) kTe− (4) 2 I

Tk

−

48. The vector ar

and br

are not perpendicular and cr

and dr

are two vectors satisfying: b c b d× = ×r r r r

and

a.d 0=r r

. Then the vector dr

is equal to

(1)a.c

c ba.b

+

r rr r

r r (2)b.c

b ca.b

+

r rr r

r r (3)a.c

c ba.b

−

r rr r

r r (4)b.c

b ca.b

−

r rr r

r r

49. The two circles 2 2x y ax+ = and ( )2 2 2x y c c 0+ = > touch each other if

(1) a c= (2) a 2c= (3) a 2c= (4) 2 a c=

50. If C and D are two events such that C D⊂ and ( )P D 0≠ , then the correct statement among the

following is

(1) ( ) ( )P C| D P C≥ (2) ( ) ( )P C| D P C< (3) ( )( )

( )

P DP C | D

P C= (4) ( ) ( )P C| D P C=

51. 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

52. Consider the following statements

P : Suman is brilliant

Q : Suman is rich

R : Suman is honestThe negation of the statement “Suman is brilliant and dishonest if and only if Suman is rich” can be

expressed as

(1) ( )( )~ Q P ~ R↔ ∧ (2) ~ Q ~ P R↔ ∧ (3) ( )~ P ~ R Q∧ ↔ (4) ( )~ P Q ~ R∧ ↔

53. The shortest distance between line y x 1− = and curve 2x y= is

(1)3 2

8(2)

8

3 2(3)

4

3(4)

3

4



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54. If the mean deviation about the median of the numbers a, 2a, …, 50a is 50, then a equals

(1) 3 (2) 4 (3) 5 (4) 2

55. Statement – 1 : 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

− −= = .

Statement – 2 : The line:x y 1 z 2

1 2 3

− −= = bisects the line segment joining ( ) A 1, 0, 7 and ( )B 1, 6, 3 .

(1) Statement – 1 is true, Statement–2 is true; Statement–2 is not a correct explanation for Statement – 1




56. Let A and B be two symmetric matrices of order 3.

Statement – 1 : ( ) A BA and ( ) AB A are symmetric matrices.

Statement – 2 : AB is symmetric matrix if matrix multiplication of A and B is commutative.


– 1




57. If ( )1ω ≠ is a cube root of unity, and ( )7

1 A B+ ω = + ω . Then ( ) A, B equals

(1) (1, 1) (2) (1, 0) (3) (-1, 1) (4) (0, 1)

58. The value of p and q for which the function ( )

( )

2

3 / 2

sin p 1 x sinx, x 0

xf x q , x 0

x x x, x 0

x

+ + <

= =

+ − >

is continuous for all x in R, is

(1)5 1

p , q2 2

= = (2)3 1

p , q2 2

= − = (3)1 3

p , q2 2

= = (4)1 3

p , q2 2

= = −

59. The area of the region enclosed by the curves1

y x, x e, yx

= = = and the positive x-axis is

(1) 1 square units (2)3

2square units (3)

5

2square units (4)

1

2square units

60. For 5

x 0,2

π ∈

, define ( )x

0

f x t sin t dt= ∫ . Then f has

(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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PART C: CHEMISTRY

61. Among the following the maximum covalent character is shown by the compound :

(1) 2SnCl (2) 3 AlCl (3) 2MgCl (4) 2FeCl

62. The presence or absence of hydroxyl group on which carbon atom of sugar differentiates RNA and DNA?(1) 2

nd(2) 3

rd(3) 4

th(4) 1

st

63. Trichloroacetaldehyde was subjected to Cannizzaro’s reaction by using NaOH. The mixture of theproducts 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 abovereaction 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) ( )2p H = 1 atm and H+ =1.0 M (2) p ( )2H =2 atm and H+ =1.0 M

(3) ( )2p H = 2 atm and H+ = 2.0 M (4) ( )2p H =1 atm and H+ = 2.0 M

66. The strongest acid amongst the following compounds is :

(1) HCOOH (2) ( )3 2 2CH CH CH Cl CO H

(3)2 2 2ClCH CH CH COOH (4)

3CH COOH

67. The degree of dissociation ( )α of a weak electrolyte,x y

A B is related to van’t Hoff factor (i) by the

expression :

(1)i 1

x y 1

−α =

+ +

(2)x y 1

i 1

+ −α =

−

(3)x y 1

i 1

+ +α =

−

(4)

( )

i 1

x y 1

−α =

+ −

68. à’ and `b’ are van der Waals’ constants for gases. Chlorine is more easily liquefied than ethane because

(1) a and b for 2Cl a< and b for 2 6C H

(2) a for 2Cl a< for

2 6C H but b for 2Cl b> for

2 6C H

(3) a for 2Cl a> for 2 6C H but b for 2Cl < b for 2 6C H

(4) a and b for 2Cl > a and b for

2 6C H

69. A vessel at 1000 K contains2CO with a pressure of 0.5 atm. Some of the

2CO 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) 4BH− (2) ( )4

B OH−

(3) 2BO− (4) 3

6BF −

71. Which of the following facts about the complex ( )3 36Cr NH Cl 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 2 3d sp hybridization and is octahedral in shape.



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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 o6 C− will be :

[ f K 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) 2 2 3 2MgO K O Al O Na O< < < (2) 2 2 2 3Na O K O MgO Al O< < <

(3) 2 2 2 3K O Na O Al O MgO< < < (4) 2 3 2 2 Al O MgO Na O K O< < <

74. The rate of a chemical reaction doubles for every o10 C rise of temperature. If the temperature is raised

by o50 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 [ ]2

4NiCl

−is

(1) 5.46 BM (2) 2.83 BM (3) 1.41 BM (4) 1.82 BM

76. The hybridization of orbitals of N atom in 3 2NO ,NO

− +

and 4NH

+

are respectively :

(1) 2 3sp , sp, sp (2) 3 2sp, sp , sp (3) 2 3sp , sp , sp (4) 2 3sp, sp , sp

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, 3CH OH , 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) 2 4N O has two resonance structures

(4) The stability of hydrides increases from 3NH to 3BiH in group 15 of the periodic table

80. The outer electron configuration of Gd (Atomic No : 64 is :

(1) 8 0 24f 5d 6s (2) 4 4 24f 5d 6s (3) 7 1 24f 5d 6s (4) 3 5 24f 4d 6s

81. Which of the following statements regarding sulphur is incorrect ?

(1) The vapour at o200 C consists mostly of 8S rings

(2) At o600 C the gas mainly consists of 2S molecules

(3) The oxidation state of sulphur is never less than +4 in its compounds

(4) 2S molecule is paramagnetic.

82. The structure of 7IF is :

(1) trigonal bipyramid (2) octahedral (3) pentagonal bipyramid (4) square pyramid



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83. Ozonolysis of an organic compound gives formaldehyde as one of the products. This confirms thepresence 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 at680 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 3Cl (4) Aqueous NaOH

87. Phenol is heated with a solution of mixture of KBr and KBrO3. The major product obtained in the abovereaction 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 centrepositions. If one atom of B is missing from one of the face centred points, the formula of the compound is:

(1) 2 AB (2) 2 3 A B (3) 2 5 A B (4) 2 A B

89. The entropy change involved in the isothermal reversible expansion of 2 moles of an ideal gas from a

volume of 10 3dm to a volume of 100 3dm at o27 C is :

(1) 35.8J mol 1 1K− − (2) 32.3J 1 1mol K− − (3) 42.3J 1 1mol K− − (4) 38.3J 1 1mol K− −

90. Identify the compound that exhibits tautomerism.

(1) Lactic acid (2) 2-Pentanone (3) Phenol (4) 2- Butene



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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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SOLUTIONS

PART A

PHYSICS1. 1

Sol. ( ) ( )

2

x,ty e a x b t−= +

bV

a=

Wave moving in − ve x –direction.

2. 1

Sol. Diameter of wire1

52 0.52mm 0.052cm100

= × = =

3. 2Sol. Mg –T = Ma ….. (1)

21T R I MR

2

× = α = α

1T Ma

2= ( )a R= α ….. (2)

From (1) and (2)2g

a3

=

α

T

a

Mg

R

4. 3

Sol. ( )2 2

2 1W T A T 8 r r 0.4 mJ= × ∆ = × π − = π

5. 3

Sol. 0τ =

Angular momentum is conserve

1 11 1 2 2 2

2

II I

I

ωω = ω ⇒ ω =

I2 first decreases and then increases

∴ω first increases and then decreases.

6. 4

Sol. 1 0φ =

22

πφ =

X0

X0+A

A

7. 3

Sol. Position of the null point from mass m,r r

x34m

1m

= =

+

3 12 GmV Gm 9

r 2r r

= − + = −



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8. 4Sol. At any instant of separation between charges is x.

equilibrium condition =2

2

Q xK

2x= ω

l

2 3Q Cx⇒ =

2dQ dx2Q C3x

dt dt⇒ =

3 / 21/ 2

2

dx xx

dt x

−⇒ ∝ ∝

9. 3

Sol. HE B V 0.15mV= =l

10. 1

Sol. dv

2.5 vdt

= −

Integrating the above equation.2 v 2.5t C⇒ = − +

at t 0,v 6.25 C 5= = ⇒ =

at5

v 0 t 2s2.5

= ⇒ = =

11. 1

Sol. Charge oscillates simple harmonic motion q = q0 sin ωt,21 q

U2 C

=

0qq t

42

π= ⇒ ω =

T 2t LC LC8 8 4

π π⇒ = = =

12.Sol. Normal to the plane is z –axis

z

1 1

A 10 1cos , 60

A 20 2θ = = = θ =

0

1 1 2 2 2 2

3sin sin 2 3 sin 45

2µ θ = µ θ ⇒ × = θ ⇒ θ =

13. 4

Sol. 0didB cos i sin j

2 R

→ ∧ ∧µ = − θ − θ π

Tdi Rd

R= θ

π

Id= θ

π

y

θxθ

dθ

0

2

IdB cos i sin j

2 R

→ ∧ ∧µ = − θ − θ π



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0

2

IB j

R

→ ∧µ= −

π

14. 3

Sol. W U= ∆

2

v

1

mv nC dT2 =

m RdT

M 1=

γ −

( ) 2M 1 vdT K

2R

γ −=

15. 3Sol. Energy of simple harmonic oscillator is constant.

( )2 2 2 2

1 2

1 1M A m M A

2 2⇒ ω = + ω

2

1

2

2

A M m

M A

+=

1

2

A M m

A M

+∴ =

16. 3Sol. Equation of continuity

( ) ( )a v top a v bottom⇒ × = ×

( )22 2 2

bv 0.4 2 9.8 0.2 v u 2gh is used − = × × − =

vb = 2m/s (nearly)3 28 10 0.4 d 4− π × × = π ×

3

d 3.6 10 m−

≈ ×

17. 1Sol. Since ionospheric properties change with time, these signals are in general less stable than ground wave

signals.

18. 1

Sol. Data ( )1 2 2 3 3 1 2 3n,k,t n kT n kT n n n kT⇒ + + = + +

1 1 2 2 3 3

1 2 3

n T n T n TT

n n n

+ +∴ =

+ +

19. 1

Sol.r × F = I × α

( )2 22 20t 5t 10 4t t− = α ⇒ α = −

2d4t t

dt

ω= −

( )2 2d 4t t dtω = −

32 t

2t3

ω = − (on integration)

0 t 6sω = ⇒ =



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32d t

2tdt 3

θω = = −

32 t

d 2t dt3

θ = −

3 42t t

3 12⇒ θ = − (on integration)

( )in 6s 36radθ =

2 n 36⇒ π =

36n 6

2= =<

π

20. 2

Sol. ( )t /Rc

cV E 1 e−= −

t/Rc 120 31 e

200 5

−− = =

6

6

5R 2.7 101.84 10−⇒ = = × Ω×

21. 4

Sol. 1 21

1

T T 1

T 6

−η = =

( )1 2

2

1

T T 62 1

T 3

− −η = =

1 2

1 1

T T 62 1

T T 3

−⇒ + =

1

1 62 1

6 T 3

+ =

1

62 1

T 6=

1T 62 6 372K∴ = × =

1 2

1

T T 1

T 6

−=

2

1

T 11

T 6− =

2T 5

372 6=

2T 310K⇒ =

22. 1

Sol. 2R ∝ l (for a given volume)

R 2% %

R

∆ ∆⇒ =

l

l

Thus when wire is stretched by 0.1% resistance increases by 0.2%



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23. 1

Sol. As light enters from air to glass it suffers a phase change on π and therefore at centre there will bedestructive interference.

24. 1

Sol. 1 1 1

v u f

+ =

2 2

1 dv 1 du0

dt dtv u− − =

2

2

dv v du

dt dtu

= −

f = 20 cm

1 1 1

u 280 20+ =

−

280v cm

15⇒ =

2

I

280

v 1515 280

= − × ×

1m/ s

15=

25. 2

Sol. 2

n 2

ZE 13.6

n= −

Li

9E 13.6 122.4eV

1

++ = − × = −

Li

9E 13.6 13.6eV

9

++ + = − × = −

( )E 13.6 122.4∆ = − − −

= 108.8 eV

26. 3

Sol. Potential inside (φ) = ar 2

+ b

r

vE 2ar

r

δ∴ = − = −

δ

Electric field inside uniformly charged solid volume varies with ‘r’. So charge density is constant

( ) 2 3

net 2ar 4 r 8 ar φ = − π = − π

3

3

0

4r

38 ar σ × π

− π =ε

06a∴ σ = − ε

27. 1

Sol. Max. range2 2u v

i.e.,g g

= (radius of circle)

Area occupied

22 4

2

v v

g g

π= π =



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28. 1

Sol. Q U W∆ = ∆ + ∆ (ignoring expansion)

U ms T 0.1 4.184 20 8.368kJ∆ = ∆ = × × =

29. 2

Sol.1

2

t 20= minutes

2t

0N N e λ−= 1t ln3λ =

2t

0 0 1

2 1N N e t ln3

3

λ−= =λ

2t

0 0

2N N e

3

λ−=

2

1 3t ln

2=

λ

2 1

1 3t t ln ln3

2

− = − λ

1 1 0.693

ln 2

= = λ λ

= 20 min

30. 3

Sol. max 0KE h h= υ − υ

0h h e vυ − υ = × ∆

00

hhV

e e

υυ= −

‘υ’ is doubled

KEmax = 2hυ − hυ0

( ) 00

h2hV V

e e

υυ′′ = ∆ = −

max

max

KE

KEmay not be equal to 2

0

0

V

V

′⇒ may not equal to 2

KE max = hυ - hυ0

0hhV

e e

υυ= −



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PART B: MATHEMATICS

31. 2

Sol:

1L 0=OP

3L 0=

Q

2L 0=R

y+2=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: 2 4 7 cos4x 3 A sin x cos x A 1

8 4

+= + = ⇒ ≤ ≤

33. 2

Sol: ( ) ( ) ( )66 62 21 x x 1 x 1 x 1 x − − − = − −

6 6 6 2 6 3 6 4 6 5 6 6 6 6 2 6 4 6 60 1 2 3 4 5 6 0 1 2 3C C x C x C x C x C x C x C C x C x C x .... = − + − + − + × − + − +

Coefficient of 7 6 6 6 6 6 61 3 3 2 5 1x C C C C C C 120 300 36 144= − + = − + = −

34. 4

Sol: ( )2

x 2

2sin x 2lim

x 2→

−

−

( )x 2

2 sin x 2lim

x 2→

−

−

R.H.L. 2= , L.H.L. 2= −

Limit does not exist.

35. 4

Sol: (n 1) (10 1) 9(r 1) (4 1) 3C C C− −

− −= =

Statement 1 is correct

Statement 2 is also correct

From 9 we can select 3 in 93C ways. It is correct explanation.



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36. 3

Sol: 2

d dx d 1 1 d dy

dydy dy dy dy dxdydx dx

= = −

2 32

2

dy 1 d dy d y dy

dydx dx dx dxdx

dx

− − = − = −

37. 4

Sol: dy dy

y 3 dxdx y 3

= + ⇒ =+

( )ln y 3 x c+ = +

x 0 y 2= ⇒ =

ln5 0 c⇒ = +

c ln5= ( )ln y 3 x ln5+ = +

x ln5y 3 e ++ = ln 2 ln 5y 3 e +⇒ + =

y 3 10 y 7+ = ⇒ =

38. 2 Sol: 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.

Alsox

1x

= is a rational number B⇒ is reflexive

x

y= α is rational

y

x⇒ need not be rational

i.e.,0

1is rational

1

0⇒ is not rational

Hence B is not symmetric

⇒ B is not an equivalence relation.



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39. 4

Sol: ( )

1

2

0

log 1 xI 8 dx

1 x

+=

+∫

( )

( )

42

20

log 1 tan

8 sec d let x tan1 tan

π

+ θ

= θ θ = θ+ θ∫

4

0

8 log 1 tan d4

π

π = + − θ θ

∫4

0

1 tan8 log 1 d

1 tan

π

− θ = + θ + θ ∫ ( )

4 4

0 0

8 log2 d 8 log 1 tan d

π π

= θ− + θ θ∫ ∫

8log2 I4

π= −

2I 2 log2= π

I log2= π

40. 3 Sol: 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 ( )21 p 1,= β = + ∈ ∞

p 0≠ since roots are distinct.

41. 2

Sol: n 5=

Success = p

Failure = q

P (at least one failure)31

32≥

1 – P (no failure)31

32≥

( )31

1 P x 532

− = ≥

5 55

311 C p

32− ≥

5 1p

32− ≥ −

51

p 32≤

1p

2≤

1p 0,

2

∈



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AIEEE−2011−23


42. 3

Sol:

1 2 3 4 5 6 ……

200 200 200 240 280 ……. ……..

Sum = 11040

120 + 80 + 160 + 40 + 200 + 240 + … = 11040

( )n

2a n 1 d 80 40 110402

+ − + + =

( )n

240 n 1 40 109202

+ − =

n 6 n 1 546+ − =

( )n n 5 546+ =

n 21=

43. 2

Sol: 1

x x 0 x x xx x

⇒ − > ⇒ > ⇒−

is negative

( )x , 0∈ −∞

44. 4

Sol: 5

cos14

θ =

3sin

14θ =

2

1 4 3sin

1 4 1 4 9

+ + λθ =

+ + λ + +

2

3 5 3 2

314 5 14

+ λ= ⇒ λ =+ λ

45. 4

Sol: ( ) ( ) ( ) ( ) ( ) ( ) 2a b . a b a 2b 2a b . a. a 2b b b. a 2b a − × × + = − + − +

( ) ( ) ( )2 2 2

5 a b 5 a.b 5= − + = −

46. 4

Sol: ( )

22 2 2 2 22 3 3a

b a 1 e a 1 a5 5 5

= − = − = =

2 2

2 2

x y1

a b+ =

2 2

9 51

a 3a⇒ + =

2 32a

3=

2 32b

5=

∴ Required equation of ellipse 2 23x 5y 32 0+ − =



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AIEEE−2011−24


47. 1

Sol: ( ) ( )dV

k T t dV k T t dtdt

= − − ⇒ = − −

Integrate( )

( )

( )2 2

k T t k T tV c V c

2 2

− − −= + ⇒ = +

−

at t 0 V I= ⇒ =

( )2 2 2kT kT kT

I c c I c V T I2 2 2

= + ⇒ = − ⇒ = = −

48. 3

Sol: b c b d× = ×

( ) ( )a b c a b d⇒ × × = × ×

( ) ( ) ( ) ( )a.c b a.b c a.d b a.b d⇒ − = −

( ) ( ) ( )a.c b a.b c a.b d⇒ − = −

a.cd c b

a.b

∴ = −

49. 1

Sol: ( )1 2

ac , 0 ; c 0, 0

2

= =

1 2ar ; r c2

= =

1 2 1 2

a ac c r r c c a

2 2= − ⇒ = − ⇒ =

50. 1

Sol: ( ) ( )( )

( )( )

P C DCC D C P C D P C P P C

D P D

∩ ∩ = ⇒ ∩ = ⇒ = ≥

51. 1

Sol: 2

4 k 2

k 4 1 0 k 6k 8 0 k 4, 2

2 2 1

= ⇒ − + = ⇒ =

52. 1

Sol: ( ) ( ) ~ P ~ R Q ~ Q P ~ R∧ ↔ = ↔ ∧



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AIEEE−2011−25


53. 1

Sol: ( )2P y , y=

Perpendicular distance from P to x y 1 0− + = is

2y y 1

2

− +

2

y y 1 0 y R− + > ∀ ∈

∴ Coefficient 2y 0>

∴ Min value21 4ac b 3

4a2 4 2

−= =

54. 2

Sol: i

1x A

n−∑

A = Median =25a 26a

25.5a2

+=

Mean deviation ( ) ( ) 1 2a 25.5a 2a 25.5a 24.5a 23.5a ... 0.5a50 50= − + − = + +

2

312.5a 5050

= = (Given)

625a 2500 a 4⇒ = ⇒ =

55. 1

Sol:

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. 1

Sol: T T A A, B B= =

( )( ) ( ) ( ) ( ) ( )T T T T T A BA BA A A B A AB A A BA= = = =

( )( ) ( ) ( ) ( ) ( )T TT T T AB A A AB A B A A BA AB A= = = =

∴ Statement – 1 is correct

Statement – 2

( )T T T AB B A 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. 1

Sol: 21+ ω = −ω

( ) ( ) ( ) ( )77 2 14 21 1 A B A, B 1, 1+ ω = −ω = −ω = −ω = + ω = + ω ⇒ =

58. 2

( ) 2

3 / 2x 0 x 0

sin p 1 sinx x x xlim q lim

x x→ →

+ + + −= =

( ) ( )x 0

1lim p 1 cos p 1 x cos x q

2→+ + + = =

1 3 1p 1 1 p ; q

2 2 2⇒ + + = ⇒ = − =

59. 2

Sol:

O 1 e

1 e

0 1

1 1 3 Area xdx dx 1

x 2 2= + = + =∫ ∫

60. 3

Sol: ( )f ' x x sinx=

Given5

x 0,2

π ∈

( )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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AIEEE−2011−27


PART C: CHEMISTRY

61. (2)Sol : Greater charge and small size of cation cause more polarization and more covalent is that compound

62. (1)

Sol : In RNA, the sugar is Dβ − − Ribose, where as in DNA the Sugar is β -D-2-deoxy Ribose

63. (4)

Sol : ( )

3 3 3 2

OH2CCl CHO CCl COONa CCl CH OH

−

→ +

Cannizaro reaction is a disproportionation reactionOne aldehyde molecule is oxidized to salt of the carboxylic Acid, other one is reduced to

Alcohol. So the compound is3 2CCl CH OH

IUPAC Name is 2, 2, 2, - Trichloro ethanol

64. (3)

Sol : 2 5 3 3 2 5

( )C H O N a CH C Cl CH C O C H

|| ||O O

− ⊕+ − − → − − − Ethyl ethanoate

65. (2)

Sol : 22H 2e H (g)+ −+ →

oE E 0.059= − log 2H

2

P

H+

(here E is –ve when2

2

HP H+ > )

=10

0.0591 2log

2 1

−

=.0591

.30102

−× = negative value

66. (2)Sol : Electron releasing groups (Alkyl groups) de stabilizes conjugate base.

The +I effect of 3 7C H is less than - I effect of Cl

Ka of HCOOH is 517.9 10−×

aK of 3 2

CH CH CH COOH|Cl

− is 5139 10−×

67. (4)

Sol : i = 1- ( )n 1 n 1α + α = + α −

i 1

n 1

−= α

−

y x

x y A B xA yB+ −→ +

n = x+y

Soi 1

x y 1

−α =

+ −



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AIEEE−2011−28


68. (3)

Sol : ease of liquefactiona

b∝

for ethane a = 5.49, b=0.0638

for 2Cl a = 6.49, b = 0.0562

69. (4)

Sol : 2CO (g) C+ 2CO(g)

Initial moles p OEquilibriumm moles p-x 2x

Total pressure at equilibrium = 0.8 atm ; Total no.of moles = p + x.

Therefore p n∝ ;0.5 p

x 0.30.8 p x

= ⇒ =+

2

2

COp

CO

P 0.6 0.6K 1.8

P 0.2

×= = = atm

70. (4)

Sol : As Boron has only four orbitals in the valence shell ( i.e. 2s, 2p x, 2py & 2pz) it can show a maximumvalency of four only.

Therefore [ ]3

6BF

−is not possible

71. (2)

Sol : ( )3 36Cr NH Cl involves 2 3d sp hybridization and it is an inner orbital complex.

72. (4)

Sol : 2f f f

1 2

w 1000T K m K

w m

×∆ = × = ×

×

1w & 2w = wt of solvent & solute respecting

2m = mw of solute

f T∆ = ( )o 0 2w 10000 6 6 1.86

4000 62

×− − = = ×

×

Therefore 2w 800g=

73. (4)Sol : Across a period metallic strength decreases & down the group it increases

74. (2)Sol : Temperature coefficient µ =2;

T

210

1

kk

∆

µ = ;

505 210

1

k2 2 32

k= = =

Therefore 32 1 2k k=



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AIEEE−2011−29


75. (2)

Sol : In [ ]2

4NiCl

−, n = 2

µ = ( )n n 2+ BM

= ( )2 2 2 2.82BM+ =

76. (1)Sol :

77. (3)Sol : The general o.s of lanthanides is +3, only few elements exhibit +4 o.s.

78. (2)

Sol : Molefraction of solute ( )2X in aqueous solution =m

1000m

18+

=5.2

1000

5.2 18+

= 0.09

79. (4)

Sol : Stability of hydrides decreases down the group from 3NH to 3BiH as M-H bond energy decreases.

80. (3)

81. (3)Sol : `S’ can exhibit a minimum oxidation state of -2

(Ex. 2H S)

82. (3)

Sol : In7

IF , I undergoes 3 3sp d hybridisation

83. (1)Sol : Vinyl group

2CH CH

= −

on ozonolosys give formaldehyde

84. (2)

Sol : absorbed 1 2

1 1 1= +

λ λ λ

⇒ 2

1 1 1355 680= + λ

⇒ 2 742.8λ = ≅ 743 nm

85. (2 , 4)Sol : Formaldehyde and Acetaldehyde can be oxidized by tollen’s reagent to give silver mirror.



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AIEEE−2011−30


86. (3)

Sol : Phenol gives violet coloured comlex compound with neutral 3FeCl , benzoic acid gives pale dull yellow

ppt. with neutral3FeCl

87. (3)

Sol : In acidic medium, KBr + KBrO3 in turn produces2

Br . Phenol reacts with2

Br (aq) to give 2, 4, 6-

trinitrophenol

88. (3)

Sol : Effective no.of A atoms =1

8 18

× =

Effective no.of B atoms =1

52

× ( One is missing) =5

2

Therefore formula is 1 5 2 5

2

A B A B=

89. (4)Sol : For an ideal gas, for isothermal reversible process,

S∆ = 2.303 nR log2

1

v

v

= 2.303100

2 8.314 log10

× × ×

= 38.3 J 1 1mol .k− −

90. 2, (2, 3)Sol : both 2-pentanone, phenol can exhibit tautomerism



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Ltd., FIITJEE House, 29 – A, Kalu Sarai, Sarvapriya Vihar, New Delhi – 110016, Ph: 46106000, 26569493, Fax: 26513942website: www.fiitjee.com

A A I I E E E E E E – – 2 2 0 0 112 2 ( ( S S e e t t – – C 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 TestBooklet, 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 andChemistry 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 ineach question will be treated as wrong response and marks for wrong response will bededucted 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 examinationhall/room.

10. Rough work is to be done on the space provided for this purpose in the Test Booklet only. Thisspace is given at the bottom of each page and in 3 pages (Pages 21 – 23) at the end of thebooklet.

11. On completion of the test, the candidate must hand over the Answer Sheet to the Invigilator onduty 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 AnswerSheet is the same as that on this booklet. In case of discrepancy, the candidate shouldimmediately report the matter to the Invigilator for replacement of both the Test Booklet and theAnswer Sheet.

13. Do not fold or make any stray marks on the Answer Sheet.



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AIEEE−−−−2012−−−−2


1. The equation esin x

– e –sin x

– 4 = 0 has(1) infinite number of real roots (2) no real roots(3) exactly one real root (4) exactly four real roots

1. 2

Sol. esin x

– e –sin x

= 4 esin x

= t

t – 1

t= 4

t2

– 4t – 1 = 0 t =4 16 4

2

± +

t =4 2 5

2

± t = 2 5±

esin x

= 2 ± 5 –1 ≤ sin x ≤ 11

e≤ e

sin x ≤ e

esin x

= 2 + 5 not possible

esin x

= 2 – 5 not possible ∴ hence no solution

2. Let â and b be two unit vectors. If the vectors ˆ ˆˆ ˆc a 2b and d 5a 4b= + = −

are perpendicular to each other,

then the angle between â and b is

(1)6

π(2)

2

π(3)

3

π(4)

4

π

2. 3

Sol. c d 0⋅ =

22

5 a 6a b 8 b 0+ ⋅ − =

1

6a b 3 a b2

⋅ = ⋅ =

( )a b3

π⋅ =

3. 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 theradius of the balloon decreases 49 minutes after the leakage began is

(1)9

7(2)

7

9(3)

2

9(4)

9

2

3. 3

Sol. v = 24r

3π

After 49 minutes volume = 4500π – 49 (72π) = 972π

4

3πr

3= 972π r

3= 729 r = 9

v =4

3πr

3 2dv 4 dr

3rdt 3 dt

= π 72π = 4π 2 drr

dt

dr 72 2

dt 4 9 9 9= =

⋅ ⋅

4. Statement 1: The sum of the series 1 + (1 + 2 + 4) + (4 + 6 + 9) + (9 + 12 + 16) + ...... + (361 + 380 +400) is 8000.

Statement 2: ( )( )n

33 3

k 1

k k 1 n=

− − = 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



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AIEEE−−−−2012−−−−3


(4) Statement 1 is true, statement 2 is false4. 2Sol. Statement 1 has 20 terms whose sum is 8000

And statement 2 is true and supporting statement 1. k

thbracket is (k – 1)

2+ k(k – 1) + k

2= 3k

2– 3k + 1.

5. 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

5. 1

Sol. ~(~p ∨ q) = p ∧ ~q

6. If the integral5tanx

tanx 2− dx = x + a ln |sin x – 2 cos x| + k, then a is equal to

(1) –1 (2) –2 (3) 1 (4) 26. 4

Sol.5 tanx 5sinx

dx dxtanx 2 sinx 2cosx

=− −

( ) ( )2 cos x 2sinx sinx 2cos xdx

sin x 2cos x

+ + −

−

=cos x 2 sin x

2 dx dx ksin x 2 cos x

+ + +

− = 2 log |sin x – 2 cos x| + x + k ∴ a = 2

7. Statement 1: An equation of a common tangent to the parabola y2

=16 3x and the ellipse 2x2

+ y2

= 4 is

y = 2x + 2 3 .

Statement 2: If the line y = mx +4 3

m, (m ≠ 0) is a common tangent to the parabola

y2

= 16 3x and the ellipse 2x2

+ y2

= 4, then m satisfies m4

+ 2m2

= 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

Sol. y2

= 16 3x 2 2x y

12 4

+ =

y = mx +4 3

mis tangent to parabola

which is tangent to ellipse c2

= a2m

2+ b

2

2

48

m= 2m

2+ 4 m

4+ 2m

2= 24 m

2= 4

8. Let A =

1 0 0

2 1 0

3 2 1

. If u1 and u2 are column matrices such that Au1 =

1

0

0

and Au2 =

0

1

0

, then u1 + u2 is

equal to

(1)

1

1

0

−

(2)

1

1

1

− −

(3)

1

1

0

−

−

(4)

1

1

1

− −

8. 4



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AIEEE−−−−2012−−−−4


Sol. A =

1 0 0

2 1 0

3 2 1

Let u1 =

a

b

c

; u2 =

d

e

f

Au1 = 1

1 1

0 u 2

0 1

= −

Au2 =

0

1

0

u2 =

0

1

2

−

u1 + u2 =

1

1

1

− −

9. If n is a positive integer, then ( ) ( )2n 2n

3 1 3 1+ − − is

(1) an irrational number (2) an odd positive integer(3) an even positive integer (4) a rational number other than positive integers

9. 1

Sol. ( ) ( ) ( ) ( ) ( ) ( )n n

2n 2n 2 2 n n3 1 3 1 3 1 3 1 4 2 3 4 2 3

+ − − = + − − = + − −

= ( ) ( )n n

n2 2 3 2 3

+ − −

= n n n n n 1 n n 2 n n n n 1 n n 20 1 2 0 1 22 C 2 C 2 3 C 2 3 C 2 C 2 3 C 2 3− − − − + + + ⋅ ⋅ ⋅ ⋅ ⋅ − − + − ⋅ ⋅ ⋅ ⋅ ⋅

= n 1 n n 1 n n 3 n 11 32 C 2 3 C 2 3 3 2 3+ − − + + + ⋅ ⋅ ⋅ ⋅ =

(some integer)

Which is irrational

10. If 100 times the 100th

term of an AP with non zero common difference equals the 50 times its 50th

term,then the 150

thterm of this AP is

(1) –150 (2) 150 times its 50th

term(3) 150 (4) zero

10. 4

Sol. 100(T100) = 50(T50) 2[a + 99d] = a + 49d a + 149d = 0 T150 = 0

11. 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

(1)

5

6

π

(2) 6

π

(3) 4

π

(4)

3

4

π

11. 2Sol. 3 sin P + 4 cos Q = 6 ...... (1)

4 sin Q + 3 cos P = 1 ...... (2)

From (1) and (2) ∠P is obtuse.(3 sin P + 4 cos Q)

2+ (4 sin Q + 3 cos P)

2= 37

9 + 16 + 24 (sin P cos Q + cos P sin Q) = 37

24 sin (P + Q) = 12

sin (P + Q) =1

2 P + Q =

5

6

π R =

6

π

12. 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



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AIEEE−−−−2012−−−−5


(3) x – 2y + 2z – 1 = 0 (4) x – 2y + 2z + 5 = 012. 1Sol. Equation of plane parallel to x – 2y + 2z – 5 = 0 is x – 2y + 2z + k = 0 ...... (1)

perpendicular distance from O(0, 0, 0) to (1) is 1k

1 4 4+ += 1 |k| = 3 k = ±3 ∴ x – 2y + 2z – 3 = 0

13. 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

(1)29

5(2) 5 (3) 6 (4)

11

5

13. 3

Sol. Point p =6 2 12 2

,5 5

+ +

p =8 14

,5 5

8 14p ,

5 5

lies on 2x + y = k 16 14

k5 5

+ = k =30

5= 6

14. Let x1, x2, ......, xn be n observations, and let x be their arithematic mean and σ2be their variance.

Statement 1: Variance of 2x1, 2x2, ......, 2xn is 4 σ2.

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

14. 4

Sol. σ2

=

22i ix x

n n

−

Variance of 2x1, 2x2, ....., 2xn =( )

2 2i i

2x 2x

n n

−

=

22i ix x

4n n

− = 4σ

2

Statement 1 is true.

A.M. of 2x1, 2x2, ......, 2xn = 1 2 n 1 2 n2x 2x 2x x x x2 2x

n n

+ + ⋅ ⋅ ⋅ ⋅ + + + ⋅ ⋅ ⋅ ⋅ + = =

Statement 2 is false.

15. The population p(t) at time t of a certain mouse species satisfies the differential equation

dp(t)

dt = 0.5 p(t)

– 450. If p(0) = 850, then the time at which the population becomes zero is

(1) 2 ln 18 (2) ln 9 (3)1

ln182

(4) ln 18

15. 1

Sol.d(p(t)) 1

dt 2= p(t) – 450

d(p(t)) p(t) 900

dt 2

−=

d(p(t))2 dt

p(t) 900=

−



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AIEEE−−−−2012−−−−6


2 ln |p(t) – 900| = t + c

t = 0 2 ln 50 = 0 + c c = 2 ln 50

∴ 2 ln |p(t) – 900| = t + 2 ln 50

P(t) = 0 2 ln 900 = t + 2 ln 50

t = 2 (ln 900 – ln 50) =900

2ln50

= 2 ln 18.

16. Let a, b ∈ R be such that the function f given by f(x) = ln |x| + bx2

+ ax, x ≠ 0 has extreme values at x = –1and x = 2.Statement 1: f has local maximum at x = –1 and at x = 2.

Statement 2: a =1

2and b =

1

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 false16. 2

Sol. f′(x) =1

x+ 2b x + a

f has extremevalues and differentiable

f′(–1) = 0 a – 2b = 1

f′(2) = 0 a + 4b =1

2− a =

1

2, b =

1

4−

f′′(–1), f′′(2) are negative. f has local maxima at –1, 2

17. The area bounded between the parabolas x2

=y

4and x

2= 9y, and the straight line y = 2 is

(1) 20 2 (2) 10 23

(3) 20 23

(4) 10 2

17. 3Sol. Required area

A =

2 2

0 0

y 5 y2 3 y dy 2 dy

2 2

− =

=

23 / 2

3 / 2

0

y 10 20 25 2 0

3 / 2 3 3

= − =

O

y = 2

x2

=9

4

x = 9y

18. Assuming the balls to be identical except for difference in colours, the number of ways in which one ormore balls can be selected from 10 white, 9 green and 7 black balls is(1) 880 (2) 629 (3) 630 (4) 879

18. 4Sol. 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.

19. If f: R → R is a function defined by f(x) = [x]2x 1

cos2

−

π, where [x] denotes the greatest integer

function, then f is(1) continuous for every real x (2) discontinuous only at x = 0(3) discontinuous only at non-zero integral values of x (4) continuous only at x = 0

19. 1



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Sol. f(x) =2x 1 1

x cos x cos x2 2

− π = − π

= [x] sin π x is continuous for every real x.

20. If the linesx 1 y 1 z 1 x 3 y k z

and2 3 4 1 2 1

− + − − −= = = = intersect, then k is equal to

(1) –1 (2)2

9(3)

9

2(4) 0

20. 3

Sol. Any point onx 1 y 1 z 1

2 3 4

− + −= = = t is (2t + 1, 3t – 1, 4t + 1)

And any point onx 3 y k z

1 2 1

− −= = = s is (s + 3, 2s + k, s)

Given lines are intersecting t =3

2

− and s = –5 ∴ k =9

2

21. Three numbers are chosen at random without replacement from 1, 2, 3, ...... 8. The probability that theirminimum is 3, given that their maximum is 6, is

(1)3

8(2)

1

5(3)

1

4(4)

2

5

21. 2Sol. Let A be the event that maximum is 6.

B be event that minimum is 3

P(A) =5

28

3

C

C(the numbers < 6 are 5)

P(B) =

52

83

C

C (the numbers > 3 are 5)

P(A ∩ B) =2

1

83

C

C

Required probability is2

1

52

CB P(A B) 2 1P

A P(A) 10 5C

∩ = = = =

.

22. If z ≠ 1 and2z

z 1−is real, then the point represented by the complex number z lies

(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 axis22. 1

Sol. Let z = x + iy ( x ≠ 1 as z ≠ 1)

z2

= (x2

– y2) + i(2xy)

2z

z 1−is real its imaginary part = 0

2xy (x – 1) – y(x2

– y2) = 0

y(x2

+ y2

– 2x) = 0

y = 0; x2

+ y2

– 2x = 0

∴ z lies either on real axis or on a circle through origin.



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23. Let P and Q be 3 × 3 matrices with P ≠ Q. If P3

= Q3

and P2Q = Q

2P, then determinant of

(P2

+ Q2) is equal to

(1) –2 (2) 1 (3) 0 (4) –1

23. 3Sol. P3

= Q3

P3

– P2Q = Q

3– Q

2P

P2(P – Q) = Q

2(Q – P)

P2(P – Q) + Q

2(P – Q) = O

(P2

+ Q2)(P – Q) = O |P

2+ Q

2| = 0

24. If g(x) =x

0cos4t dt , then g(x + π) equals

(1)g(x)

g( )π(2) g(x) + g(π) (3) g(x) – g(π) (4) g(x) . g(π)

24. 2 or 4

Sol. g(x) =

x

0

cos4t dt

g′(x) = cos 4x g(x) =sin4x

k4

+ g(x) =sin4x

4[ g(0) = 0]

g(x + π) = g(x) + g(π) = g(x) – g(π) ( g(π) = 0)

25. The length of the diameter of the circle which touches the x-axis at the point (1, 0) and passes throughthe point (2, 3) is

(1)10

3(2)

3

5(3)

6

5(4)

5

3

25. 1

Sol. Let (h, k) be centre.(h – 1)2

+ (k – 0)2

= k2 h = 1

(h – 2)2

+ (k – 3)2

= k2 k =

5

3

∴ diameter is 2k =10

3

kk

(h, k)(2, 3)

(1, 0)

26. 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(1) 5

2(2) 3

5(3) 2

5(4) 5

3

26. 2

Sol. 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.

Hence for five elements of X, the number of required chances is 3 × 3 × 3 × 3 × 3 = 35

27. An ellipse is drawn by taking a diameter of the circle (x – 1)2

+ y2

= 1 as its semiminor axis and adiameter of the circle x

2+ (y – 2)

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



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(1) 4x2

+ y2

= 4 (2) x2

+ 4y2

= 8 (3) 4x2

+ y2

= 8 (4) x2

+ 4y2

= 1627. 4Sol. Semi minor axis b = 2

Semi major axis a = 4

Equation of ellipse =2 2

2 2

x y

a b+ = 1

2 2x y

16 4+ = 1

x2

+ 4y2

= 16.

28. Consider the function f(x) = |x – 2| + |x – 5|, x ∈ R.

Statement 1: f′(4) = 0Statement 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.

29. A line is drawn through the point (1, 2) to meet the coordinate axes at P and Q such that it forms atriangle OPQ, where O is the origin. If the area of the triangle OPQ is least, then the slope of the line PQis

(1)1

4

− (2) –4 (3) –2 (4)1

2

−

29. 3Sol. Equation of line passing through (1, 2) with slope m is y – 2 = m(x – 1)

Area of ∆OPQ =2(m 2)

2 m

−

∆ =2m 4 4m

2m

+ − ∆ =

m 22

2 m+ −

∆ is least ifm 2

2 m= m

2= 4 m = ±2 m = –2

30. 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

(1)( )

( )

3 p qr 3q p

p p

⋅= −

⋅

(2)p q

r q pp p

⋅= − +

⋅

(3)p q

r q pp p

⋅= −

⋅

(4)( )

( )

3 p qr 3q p

p p

⋅= − +

⋅

30. 2



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Sol. AE

= vector component of q o n p

AE

=( )

( )

p qp

p q

⋅

⋅

∴ From ∆ABE; AB BE AE+ =

( )

( )

p q pq r

p q

⋅+ =

⋅

( )

( )

p qr q p

p p

⋅= − +

⋅

A B

CD

E

r

q

p

31. 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 tosurrounding temperature, it presses the semicircular parts together. If the

coefficient of linear expansion of the metal is α, and its Youngs' modulus is Y, theforce that one part of the wheel applies on the other part is :

(1) 2 SY Tπ α ∆ (2) SY Tα ∆ (3) SY Tπ α ∆ (4) 2SY Tα ∆

31. 4

Sol. If temperature increases by ∆T,

Increase in length L, L L T∆ = α ∆

∴ L

TL

∆= α ∆

Let tension developed in the ring is T.

∴ T L

Y Y TS L

∆= = α ∆

∴ T S Y T= α ∆

From FBD of one part of the wheel,F = 2T

F

T T

Where, F is the force that one part of the wheel applies on the other part.

∴ F 2S Y T= α ∆

32. The figure shows an experimental plot for discharging of a

capacitor in an R-C circuit. The time constant τ of this circuitlies 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. 4Sol. For discharging of an RC circuit,

t /

0V V e− τ=

So, when 0V

V2

=

t / 00

VV e

2

− τ=

1 t tln

2 ln2= − τ =

τ



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From graph when 0V

V2

= , t = 100 s ∴ 100

144.3 secln2

τ = =

33. In a uniformly charged sphere of total charge Q and radius R, the electric field E is plotted as a function ofdistance from the centre. The graph which would correspond to the above will be

rR

E

rR

E

rR

E

rR

E

(1) (2) (3) (4)33. 3

Sol. inside3

0

1 QE r

4 R

=

π ε

outside3

0

1 QE r

4 r

π ε

∴

rR

E

34. 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×

34. 3Sol. Direction of polarization is parallel to magnetic field,

∴ X| |B

and direction of wave propagation is parallel to E B×

∴ K ||

E B×

35. 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 thependulum suffers a retardation (due to viscous drag) proportional to its velocity, with 'b' as the constant ofproportionality, the average life time of the pendulum is (assuming damping is small) in seconds:

(1)0.693

b(2) b (3)

1

b(4)

2

b

35. 4Sol. As retardation = bv

∴ retarding force = mbv

∴ net restoring torque when angular displacement is θ is given by

= – mg sinθ + mbv

∴ Iα = – mg sinθ + mbv where, I = m

2

∴ 2

2

d g bvsin

dt

θ= α = − θ +

θ mbv

v

m

for small damping, the solution of the above differential equation will be



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bt

20e sin(wt )

−∴ θ = θ + φ

∴ angular amplitude will be =

bt

2

.e

−

θ According to question, in τ time (average life–time),

angular amplitude drops to1

evalue of its original value (θ)

∴ 6

0 20e

e

τ−θ

= θ

61

2

τ=

∴ 2

bτ =

36. 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

36. 4Sol. Number of spectral lines from a state n to ground state is

n(n 1)6

2

−= = .

37. A coil is suspended in a uniform magnetic field, with the plane of the coil parallel to the magnetic lines offorce. When a current is passed through the coil it starts oscillating; it is very difficult to stop. But if analuminium 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.

37. 4Sol. 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.

38. The mass of a spaceship is 1000 kg. It is to be launched from the earth's surface out into free space. Thevalue of 'g' and 'R' (radius of earth) are 10 m/s

2and 6400km respectively. The required energy for this

work will be ;(1) 6.4 x 10

11Joules (2) 6.4 x 10

8Joules (3) 6.4 x 10

9Joules (4) 6.4 x 10

10Joules

38. 4Sol. To launch the spaceship out into free space, from energy conservation,

GMmE 0

R

−+ =

2

GMm GME mR mgR

R R

= = =

= 6.4 x 1010

J

39. Helium gas goes through a cycle ABCDA (consisting of two isochoric andtwo 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

V0 2V0

DA

B C

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Sol. Work done in complete cycle = Area under P–V graph= P0V0

from A to B, heat given to the gas

= v 0 0 03 3 3nC T n R T V P P V2 2 2

∆ = ∆ = ∆ =

from B to C, heat given to the system

p

5nC T n R T

2

= ∆ = ∆

0 0 0

5(2P ) V 5P V

2= ∆ =

from C to D and D to A, heat is rejected.

efficiency, η =work done by gas

100heat given to the gas

×

0 0

0 0 0 0

P V15.4%

3P V 5P V2

η = =

+

40. In Young's double slit experiment, one of the slit is wider than other, so that the amplitude of the light fromone 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

(1) mI (4 5cos )9

+ φ (2) 2mI

1 2cos3 2

φ +

(3) 2m

I1 4cos

5 2

φ +

(4) 2m

I1 8cos

9 2

φ +

40. 4Sol. Let A1 = A0, A2 = 2A0

If amplitude of resultant wave is A then2 2 2

1 2 1 2A A A 2A A cos= + + φ

For maximum intensity,2 2 2

max 1 2 1 2A A A 2A A= + +

∴ 2 22

1 2 1 2

2 2 2

max 1 2 1 2

A A 2A A cosA

A A A 2A A

+ + φ=

+ +

2 2

0 0 0 0

2 2

0 0 0 0

A 4A 2(A )(2A )cos

A 4A 2(A )(2A )

+ + φ=

+ +

2

m

I 5 4cos 1 8cos ( / 2)

I 9 9

+ φ + φ= =

41. 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) (3) (4)41. 1Sol. According to Newtons law of cooling.

0

d( )

dt

θ∝ − θ − θ



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0

dk( )

dt

θ= − θ − θ

0

dk dt

θ= −θ − θ

ln(θ – θ0) = –kt + c

Hence the plot of ln(θ – θ0) vs t should be a straight line with negative slope.

42. A particle of mass m is at rest at the origin at time t = 0. It is subjected to a force F (t) = F 0e –bt

in the x direction. Its speed v(t) is depicted by which of the following curves?

(1) (2)

(3) (4)

42. 3

Sol. bt

0F F e−=

bt0FF

a em m

−= =

bt0Fdve

dt m

−=

tbt

0

Fdv e dt

m

−=

t

bt

0

F 1v e

m b

−− =

btFv e

mb

− =

v = 0 at t = 0

andF

v as tmb

→ → ∞

So, velocity increases continuously and attains a maximum value of

F

v as tmb= → ∞ .

43. 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

43. 3Sol. Resistances of both the bulbs are

2 2

1

1

V 220R

P 25= =

2 2

2

2

V 220R

P 100= =

Hence R1

> R2



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When connected in series, the voltages divide in them in the ratio of their resistances. The voltage of 440V devides in such a way that voltage across 25 w bulb will be more than 220 V.Hence 25 w bulb will fuse.

44. Resistance of a given wire is obtained by measuring the current flowing in it and the voltage differenceapplied across it. If the percentage errors in the measurement of the current and the voltage differenceare 3% each, then error in the value of resistance of the wire is(1) 6% (2) zero (3) 1% (4) 3%

44. 1

Sol.V

Ri

=

R V i

R V i

∆ ∆ ∆= +

V100 3

V

∆× =

V 0.03V

∆ =

Similarly,i

0.03i

∆=

HenceR

0.06R

∆=

So percentage error isR

100 6%R

∆× =

45. A boy can throw a stone up to a maximum height of 10 m. The maximum horizontal distance that the boycan throw the same stone up to will be

(1) 20 2 m (2) 10 m (3) 10 2 m (4) 20 m

45. 4

Sol. maximum vertical height =2u

10m2g

=

Horizontal range of a projectile =2u sin2

g

θ

Range is maximum when θ = 450

Maximum horizontal range =2u

g

Hence maximum horizontal distance = 20 m.

46. This question has statement 1 and statement 2. Of the four choices given after the statements, choose

the one that best describes the two statementsStatement 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.

46. 3Sol. 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 anddiffraction.



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47. A thin liquid film formed between a U-shaped wire and a light slider supports aweight 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) 0.0125 Nm

–1

(2) 0.1 Nm

–1

(3) 0.05 Nm –1

(4) 0.025 Nm –1

w

FILM

47. 4Sol. The force of surface tension acting on the

slider balances the force due to the weight.

F = 2T = w

2T(0.3) = 1.5 x 10 –2

T = 2.5 x 10 –2

N/m

F = 2Tl

w

48. 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 aresult of this rotation a magnetic field of induction B is obtained at the centre of the disc. If we keep boththe amount of charge placed on the disc and its angular velocity to be constant and vary the radius of thedisc then the variation of the magnetic induction at the centre of the disc will be represented by the figure

B

R

B

R

B

R

B

R(1) (2) (3) (4)

48. 1Sol. 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 = (2 r dr)σ π current ‘i’ associated with rotating charge dq is

(dq)wi w r dr

2= = σ

π

Magnetic field dB at center due to element

0 0i dr

dB2r 2

µ µ σ ω= =

R

0net

0

B dB dr2

µ σ ω= = 0

R

2

µ σ ω=

dr

r

20

net

QB Q R

2 R

µ ω = = σ π π

So if Q and w are unchanged then

net

1B

R∝

Hence variation of Bnet with R should be a rectangular hyperbola as represented in (1).

49. Truth table for system of four NAND gates as shown in figure is

A

B

Y



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A B Y

0

0

1

1

0

1

0

1

0

1

1

0

A B Y

0

0

1

1

0

1

0

1

0

0

1

1

A B Y

0

0

1

1

0

1

0

1

1

1

0

0

A B Y

0

0

1

1

0

1

0

1

1

0

0

1

(1) (2) (3) (4)49. 1Sol.

A B y y1 y2 y

0 0 1 1 1 00 1 1 1 0 11 0 1 0 1 1

1 1 0 1 1 0

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 ofheight 500 m. The maximum distance upto which it can detect object located on the surface of the earth(Radius of earth = 6.4 x 10

6m) is

(1) 80 km (2) 16 km (3) 40 km (4) 64 km50. 1Sol. Maximum distance on earth where object can be

detected is d, then2 2 2(h R) d R+ = +

2 2d h 2Rh= +

since h << R, d2

= 2hR

6d 2(500)(6.4 10 ) 80= × = km

R

hd

R

51. Assume that a neutron breaks into a proton and an electron. The energy released during this process is(Mass of neutron = 1.6725 x 10

–27kg; mass of proton = 1.6725 x 10

–27kg; mass of electron = 9 x 10

–31

kg)(1) 0.73 MeV (2) 7.10 MeV (3) 6.30 MeV (4) 5.4 MeV

51. 1Sol.

p e nm (m m ) m∆ = + −

= 9 x 10 –31

kg.Energy released = (9 x 10

–31kg)c

2joules

31 8 2

13

9 10 (3 10 )MeV

1.6 10

−

−

× × ×=

×

= 0.73 MeV.

52. A Carnot engine, whose efficiency is 40%, takes in heat from a source maintained at a temperature of500 K It is desired to have an engine of efficiency 60%. Then, the intake temperature for the sameexhaust (sink) temperature must be(1) efficiency of Carnot engine cannot be made larger than 50%(2) 1200 K (3) 750 K (4) 600 K

52. 3

Sol. S500 T40

100 500

−= , TS = 300 K

600 T 300

100 T

−= T = 750 K

53. 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.



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If two springs S1 and S2 of force constants k1 and k2, respectively, are stretched by the same force, it isfound that more work is done on spring S1 than on spring S2.Statement 1 : If stretched by the same amount, work done on S1, will be more than that on S2

Statement 2 : k1 < k2 (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.

53. 1Sol. F = K1S1 = K2 S2

W1 = FS1, W2 = FS2 K1S1

2> K2S2

2

S1 > S2 K1 < K2

W ∝ KW1 < W2

54. Two cars of masses m1 and m2 are moving in circles of radii r1 and r2, respectively. Their speeds aresuch 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

54. 3Sol. a r∝

55. A cylindrical tube, open at both ends, has a fundamental frequency, f, in air. The tube is dipped verticallyin water so that half of it is in water. The fundamental frequency of the air-column is now

(1) f (2)f

2(3)

3f

4(4) 2f

55. 1

Sol. f0 =v

2

fC =v

2

56. 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 1cmthick, of refractive index 1.50 is interposed between lens and film with its plane faces parallel to film. Atwhat 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

56. 4Sol. Case I: u = –240cm, v = 12, by Lens formula

1 7

f 80=

Case II: v = 12 – 1 353 3

= (normal shift = 2 113 3

− = )

f =7

80

u = 5.6

57. A diatomic molecule is made of two masses m1 and m2 which are separated by a distance r. If wecalculate its rotational energy by applying Bohr's rule of angular momentum quantization, its energy willbe given by (n is an integer)

(1)2 2 2

1 2

2 2 2

1 2

(m m ) n h

2m m r

+(2)

2 2

2

1 2

n h

2(m m )r+(3)

2 2

2

1 2

2n h

(m m )r+(4)

2 2

1 2

2

1 2

(m m )n h

2m m r

+

57. 4



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Sol. r1 = 2

1 2

m r

m m+; r2 = 1

1 2

m r

m m+

(I1 + I2)ω =nh

n2 =π

K.E =1

2(I1 + I2) ω

2=

2 21 2

21 2

n (m m )

2m m r

+

58. A spectrometer gives the following reading when used to measure the angle of a prism.Main scale reading: 58.5 degreeVernier scale reading : 09 divisionsGiven that 1 division on main scale corresponds to 0.5 degree. Total divisions on the vernier scale is 30and match with 29 divisions of the main scale. The angle of the prism from the above data(1) 58.59

o(2) 58.77

o(3) 58.65

o(4) 59

o

58. 3

Sol. L.C =1

60

Total Reading = 585 +9

60= 58.65

59. This question has statement 1 and statement 2. Of the four choices given after the statements, choosethe 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 thisuniform charge distribution there is a finite value of electric potential at the centre of the sphere, at thesurface 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

changes by0

qp

3ε

Statement 2 : The electric field at a distance r(r < R) from the centre of the sphere is0

r

3

ρ

ε

(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 the correct explanation for statement 1

59. 3

Sol. 3

0

1 4E dA r

3

⋅ = ρ × π

ε

E =0

r

3

ρ

ε

Statement 2 is correct

∆PE = (Vsur – Vcent)q = 2

0

qR

6− ρ

ε

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)p dr r rα

= = (2)p dr r rα

= < (3)d pr r rα

> > (4)d pr r rα

= >

60. 2

Sol. r =2mK

Bq



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r ∝ m

q

rα = rp < rd

61. Which among the following will be named as dibromidobis(ethylene diamine)chromium(III) bromide ?

(1) ( ) 33Cr en Br

(2) ( ) 22Cr en Br Br

(3) ( ) 4Cr en Br−

(4) ( ) 2Cr en Br Br

61. 2

Sol. ( ) 22Cr en Br Br

– dibromido bis (ethylene diamine)chromium(III) bromide

62. Which method of purification is represented by the following equation :

( ) ( ) ( ) ( ) ( )523K 1700K2 4 2Ti s 2I g TiI g Ti s 2I g+ → → +

(1) zone refining (2) cupellation (3) Poling (4) Van Arkel62. 4Sol. Van Arkel method

( ) ( ) ( )523K2 4Ti s 2I g TiI g+ →

( ) ( ) ( )1700 K4 2TiI g Ti s 2I g → +

63. Lithium forms body centred cubic structure. The length of the side of its unit cell is 351 pm. Atomic radiusof the lithium will be :(1) 75 pm (2) 300 pm (3) 240 pm (4) 152 pm

63. 4

Sol. For BCC, =3a 4r

3 351r 152pm4×= =

64. The molecule having smallest bond angle is :(1) NCl3 (2) AsCl3 (3) SbCl3 (4) PCl3

64. 3Sol. As the size of central atom increases lone pair bond pair repulsions increases so, bond angle

decreases

65. Which of the following compounds can be detected by Molisch’s test ?(1) Nitro compounds (2) Sugars (3) Amines (4) Primary alcohols

65. 2

Sol. 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 junctionof two liquids.

66. The incorrect expression among the following is :

(1)system

total

GT

S

∆= −

∆(2) In isothermal process f

reversiblei

Vw nRTln

V= −

(3)0 0H T S

lnKRT

∆ − ∆= (4)

0G /RTK e−∆=

66. 3

Sol. ∆G° = –RTln K and ∆ = ∆ − ∆0 0 0G H T S



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67. The density of a solution prepared by dissolving 120 g of urea (mol. Mass = 60 u ) in 1000g of water is1.15 g/mL. The molarity of this solution is :(1) 0.50 M (2) 1.78 M (3) 1.02 M (4) 2.05 M

67. 4Sol. Total weight of solution = 1000 + 120 = 1120 g

Molarity =120 1000

2.05M60 1120 / 1.15

× =

68. The species which can best serve as an initiator for the cationic polymerization is :(1) LiAlH4 (2) HNO3 (3) AlCl3 (4) BuLi

68. 3Sol. lewis acids can initiate the cationic polymerization.

69. Which of the following on thermal decomposition yields a basic as well as an acidic oxide ?(1) NaNO3 (2) KClO3 (3) CaCO3 (4) NH4NO3

69. 3

Sol. CaCO3 → 2Basic AcidicCaO CO+

70. The standard reduction potentials for Zn2+ / Zn, Ni

2+ / Ni, and Fe

2+ / Fe are –0.76, –0.23 and –0.44 V

respectively. The reaction X + Y2+

→ X2+

+ 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

70. 4

Sol. Zn + Fe+2

→ Zn+2

+ Fe

Fe + Ni+2

→ Fe2+

+ Ni

Zn + Ni2+

→ Zn+2

+ NiAll these are spontaneous

71. According to Freundlich adsorption isotherm, which of the following is correct ?

(1) 0x Pm

∝ (2) 1x pm

∝ (3) 1/ nx pm

∝

(4) All the above are correct for different ranges of pressure71. 4

Sol. 0xP

m∝ is true at extremely high pressures

1xp

m∝ ; 1/ nx

pm

∝ are true at low and moderate pressures

72. The equilibrium constant (KC) for the reaction N2(g) + O2(g) → 2NO(g) at temperature T is 4 x 10 –4

. The

value of KC for the reaction, NO(g) → ½ N2(g) + ½ O2(g) at the same temperature is :(1) 0.02 (2) 2.5 x 10

2(3) 4 x 10

–4(4) 50.0

72. 4Sol. 4

2 2 CN O 2NO K 4 10−+ = ×

12 2 C

C

1 1 1NO N O K

2 2 K+ =

1C

4

1K 50

4 10−= =

×

73. The compressibility factor for a real gas at high pressure is :(1) 1 + RT/pb (2) 1 (3) 1 + pb/RT (4) 1–pb/RT

73. 3



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Sol. At high pressure Z =Pb

1RT

+

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

74. 3

Sol.

CH2

NH2

COOHGlycine

75. Aspirin is known as :(1) Acetyl salicylic acid (2) Phenyl salicylate(3) Acetyl salicylate (4) Methyl salicylic acid

75. 1

Sol.

O

COOH

C

O

CH3

Aspirin

Acetyl salicylic acid

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.76. 2

Sol.

N

OO

H

O

Intramolecular H–bonding decreases water solubility.

77. How many chiral compounds are possible on monochlorination of 2–methyl butane ?(1) 8 (2) 2 (3) 4 (4) 6

77. 2Sol. ( )3 2 3 3H C CH CH CH CH− − − on monochlorination gives

( ) ( )( )

2 2 3 3I

Achiral

H C Cl CH CH CH CH− − − ( ) ( )( )

3 3 3II

Chiral

H C CH Cl CH CH CH− − −

CH3 CH2 C

CH3

Cl

CH3CH2 CH

CH2

CH3CH3

Cl

(IV)

chiral

(III)

Achiral



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78. 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

78. 3Sol. Highly pure hydrogen is obtained by the electrolysis of water.

79. The electrons identified by quantum numbers n and l :(a) n = 4, l = 1 (b) n = 4, l = 0 (c) n = 3, l = 2 (d) n = 3 , l = 1Can be placed in order of increasing energy as :(1) (c) < (d) < (b) < (a) (2) (d) < (b) < (c) < (a) (3) (b) < (d) < (a) < (c) (4) (a) < (c) < (b) < (d)

79. 2Sol. (a) (n + l) = 4 + 1 = 5 (b) (n + l) = 4 + 0 = 4 (c) (n + 1) = 3 + 2 = 5 (d) (n + 1) = 3 + 1 = 4

80. For a first order reaction, (A) → products, the concentration of A changes from 0.1 M to 0.025 M in 40minutes. The rate of reaction when the concentration of A is 0.01 M is :(1) 1.73 x 10

–5M/ min (2) 3.47 x 10

–4M/min

(3) 3.47 x 10 –5

M/min (4) 1.73 x 10 –4

M/min80. 2

Sol. =2.303 0.1

k log40 0.025

=0.693

k20

For a F.O.R., rate=k[A]; rate = 2 40.69310 3.47 10 M/min.

20

− −× = ×

81. 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.

81. 4

Sol. FeO → More basic, more ionic, less volatile

82. 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 x 10

–1(2) 1 x 10

–3(3) 1 x 10

—5(4) 1 x 10

–7

82. 3

Sol. 3 1a aH K .C 10 K .10+ − − = =

Ka = 10 –5

83. Which branched chain isomer of the hydrocarbon with molecular mass 72u gives only one isomer ofmono substituted alky halide ?(1) Tertiary butyl chloride (2) Neopentane(3) Isohexane (4) Neohexane

83. 2

Sol.

CH3 C

CH3

CH3

CH3

CH3 C

CH3

CH3

CH2 Cl

Neopentane

Mol. wt = 72uonly one compound

mono chlorination



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84. 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) 27g84. 2

Sol. f fT K .m∆ =

wt 10002.8 1.86

62 1000= × ×

Wt = 93g

85. What is DDT among the following :(1) Greenhouse gas (2) A fertilizer(3) Biodegradable pollutant (4) Non–biodegradable pollutant

85. 4Sol. DDT – non–biodegradable pollutant.

86. The increasing order of the ionic radii of the given isoelectronic species is :(1) Cl

– , Ca

2+, K

+, S

2– (2) S

2– , Cl

– , Ca

2+, K

+(3) Ca

2+, K

+, Cl

– , S

2– (4) K

+, S

2– , Ca

2+, Cl

–

86. 3Sol. For isoelectronic species, as the z/e decreases, ionic radius increases

87. 2–Hexyne gives trans–2–Hexene on treatment with :(1) Pt/H2 (2) Li/NH3 (3) Pd/BaSO4 (4) LiAlH4

87. 2

Sol.

CH3 CH2 CH2 C C CH3C C

H7C3

H CH3

H

2-Hexyne

Li/NH3

Birch reduction

Trans-2-Hexene

88. Iodoform can be prepared from all except :(1) Ethyl methyl ketone (2) Isopropyl alcohol(3) 3–Methyl – 2– butanone (4) Isobutyl alcohol

88. 4Sol. 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

CH3 C

O

C2H5

CH3 CH

CH3

OH

CH3 C

O

CH

CH3

CH3

Isopropyl alchol

ethyl methyl ketone

3-methyl 2-butanone

can give Iodoform Test



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CH3 CH

CH3

CH2 OH

Isobutyl alcohol

can't give

89. In which of the following pairs the two species are not isostructural ?

(1) 23 3CO and NO− − (2) 4 4PCl and SiCl+ (3) PF5 and BrF5 (4) 3

6 6AlF and SF−

89. 3

Sol. (1) 2 23 3CO & NO Sp− − → hybridized, Trigonal planar

(2) 34 4PCl & SiCl Sp+ → hybridized, Tetrahedral

(3) PF5 → Sp3d hybridized, Trigonal bipyramidal

BrF5 → Sp3d

2hybridized, square pyramidal

(4) 3 3 26 6AlF &SF Sp d− → hybridized, octahedral

90. In the given transformation, which of the following is the most appropriate reagent ?CH

HO

CHCOCH3

CH

HO

CHCH2CH3

Reagent

(1)( )

2 2NH NH , O H−

(2) Zn Hg /HCl− (3) 3Na,Liq.NH (4) NaBH4

90. 1Sol. ZnHg/Hcl can’t be used due to the presence of acid sensitive group i.e. OH

CH

HO

CH C

O

CH3 CH

Cl

CH CH2 CH3

Zn-Hg/HCl

and Na/Liq. NH3 and NaBH4 convert – CO – into – CH(OH)–



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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 specifiedspace) 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 wouldbe 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. Allcalculations/writing work are to be done in the space provided for this purpose, in the TestBooklet itself, marked 'Space for Rough Work'. This space is given at the bottom of each pageand 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 onduty in the Room/Hall. However, the candidates are allowed to take away this Test Bookletwith 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 Sheetto the Invigilator on duty and sign the Attendance Sheet again. Cases where a candidate hasnot signed the Attendance Sheet a second time will be deemed not to have handed over theAnswer Sheet and dealt with as an unfair means case. The candidates are also required toput 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. isprohibited.

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 andRegulations 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 theexamination hall/room.



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1. 2 2. 3 3. 3 4. 2

5. 1 6. 4 7. 2 8. 4

9. 1 10. 4 11. 1 12. 1

13. 3 14. 4 15. 1 16. 3

17. 3 18. 4 19. 1 20. 3

21. 2 22. 1 23. 3 24. 2 or 4

25. 1 26. 2 27. 4 28. 2


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