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PAMITY INSTITUTE FOR COMPETITIVE EXAMINATIONS



PART-A: CHEMISTRY

1. (2) 2. (3) 3. (1) 4. (4) 5. (3)6. (2) 7. (1) 8. (3) 9. (4) 10. (1)11. (1) 12. (3) 13. (4) 14. (1) 15. (3)16. (4) 17. (3) 18. (3) 19. (3) 20. (1)21. (4) 22. (4) 23. (3) 24. (4) 25. (4)26. (1) 27. (2) 28. (4) 29. (1) 30. (1)

PART-B: PHYSICS

31. (2) 32. (3) 33. (4) 34. (3) 35. (2)

36. (3) 37. (1) 38. (1) 39. (2) 40. (4)

41. (2) 42. (2) 43. (2) 44. (1) 45. (4)

46. (2) 47. (4) 48. (4) 49. (2) 50. (1)

51. (3) 52. (4) 53. (2) 54. (4) 55. (4)

56. (3) 57. (2) 58. (4) 59. (2) 60. (4)

PART-C: MATHEMATICS

61. (4) 62. (1) 63. (4) 64. (2) 65. (3)

66. (4) 67. (3) 68. (3) 69. (3) 70. (1)

71. (1) 72. (2) 73. (4) 74. (2) 75. (2)

76. (1) 77. (1) 78. (1) 79. (3) 80. (1)

81. (2) 82. (3) 83. (3) 84. (2) 85. (2)

86. (2) 87. (4) 88. (2) 89. (1) 90. (2)

AMITY INSTITUTE FOR COMPETITIVE EXAMINATIONS



PART-A: CHEMISTRY

1. The presence or absence of hydroxy group on which carbon atom of sugar differentiates RNA and

DNA?

(1) 1st (2) 2nd (3) 3rd (4) 4th

Sol: Ans [2]

OHOH 2C

H H

OH

H

H

H

OH OHOH 2C

H H

OH

H

OH

H

OH

Deoxy Ribose Sugar Ribose Sugar

1

2

1

2

2. Among the following the maximum covalent character is shown by the compound:

(1) FeCl2 (2) SnCl2 (3) AlCl3 (4) MgCl2

Sol: Ans [3] According to Fajan’s rule more is the polarising power of cation more will be covalent

character. Al3+ has more polarising power than. Fe2+, Sn2+, Mg2+.

3. Which of the following statement is wrong?

(1) The stability of hydrides increases from NH3 to BiH3 in group 15 of the periodic table.

(2) Nitrogen cannot form d – p bond.

(3) Single N – N bond is weaker than the single P – P bond.

(4) N2O4 has two resonance structures.

Sol: Ans [1] Stability of Hydrides of 15th group element decreases down the group due to decrease in

bond dissociation enthalpy.

4. Phenol is heated with a solution of mixture of KBr and KBrO3. The major product obtained in the

above reaction is:

(1) 2-Bromophenol (2) 3-Bromophenol

(3) 4-Bromophenol (4) 2,4,6-Tribromophenol

Sol: Ans [4] KBr and KBrO3 react and produce Br2 in water

OH

22 H OBr

OH

Br Br

Br2,4,6-tribromophenol

Test Code-P AIEEE - 2011


Chemistry, Physics & Mathematics (Test Code: P) AIEEE-2011


5. A 5.2 molal aqueous solution of methyl alcohol, CH3OH, is supplied. What is the mole fraction of

methyl alcohol in the solution?

(1) 0.100 (2) 0.190 (3) 0.086 (4) 0.050

Sol: Ans [3] xsolute = 5.2

55.55 5.2= 0.086

6. The hybridisation of orbitals of N atom in 3NO , 2NO and 4NH are respectively:

(1) sp, sp2, sp3 (2) sp2, sp, sp3 (3) sp, sp3, sp2 (4) sp2, sp3, sp

Sol: Ans [2] O NO

OO N O N

H

H HH

sp2sp sp3

7. 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 –6°C will be:

(Kf for water = 1.86 K kg mol–1, and molar mass of ethylene glycol = 62 g mol–1)

(1) 804.32g (2) 204.30 g (3) 400.00g (4) 304.60g

Sol: Ans [1]W 1000T K

W Wsolute

f fsolute solvent

W 10006 1.8662 4000solute

Wsolute = 804.32g

8. The reduction potential of hydrogen half-cell will be negative if:

(1) p(H2) = 1 atm and [H+] = 2.0 M (2) p(H2) = 1 atm and [H+] = 1.0 M

(3) p(H2) = 2 atm and [H+] = 1.0 M (4) p(H2) = 2 atm and [H+] = 2.0 M

Sol: Ans [3] (aq) 2(g)2H 2e H

2 2

0 22H /H H /H

0.591 [p(H )]E E log2 [H ]

For negative reduction potential 22

[pH ]log[H ] must be positive

2 2H / H

0.591 [2]E 0 log2 [1]

= 0.0591log 2

2

= 0.0591 0.3010

2 = –0.0089V


AIEEE-2011 Chemistry, Physics & Mathematics (Test Code: P)


9. Which of the following reagents may be used to distinguish between phenol and benzoic acid?

(1) Aqueous NaOH (2) Tollen’s reagent (3) Molisch reagent (4) Neutral FeCl3

Sol: Ans [4] Phenol on reaction with FeCl3 give violet colour of ferric phenoxide

6 5 3 6 5 3Ferric phenoxide(Violet Colour )

3C H OH FeCl (C H O) 3HCl

10. Trichloroacetaldehyde was subjected to Cannizzaro’s reaction by using NaOH. The mixture of the

products contains sodium trichloroacetate and another compound. The other compound is:

(1) 2, 2, 2-Trichloroethanol (2) Trichloromethanol

(3) 2, 2, 2-Trichloropropanol (4) Chloroform

Sol: Ans [1] Cl

Cl

Cl H

OCannizzar reaction

OH Cl

Cl

Cl O-

O Cl

Cl

Cl OH

2,2,2-trichloroethanol11. Which one of the following orders presents the correct sequence of the increasing basic nature of the

given oxides?

(1) Al2O3 MgO Na2O K2O (2) MgO K2O Al2O3 Na2O

(3) Na2O K2O MgO Al2O3 (4) K2O Na2O Al2O3 MgO

Sol: Ans [1] Basic strength of oxides increases with increase in metallic chracter.

12. A gas absorbs a photon of 355 nm and emits at two wavelengths. If one of the emissions is at

680 nm, the other is at:

(1) 1035 nm (2) 325 nm (3) 743 nm (4) 518 nm

Sol: Ans [3] totalE = E1 + E2

hC

= 1 2

hC hC

1

= 1 2

1 1

1355

= 2

1 1680

2 = 743

13. Which of the following statements regarding sulphur is incorrect?

(1) S2 molecule is paramagnetic

(2) The vapour at 200°C consists mostly of S8 rings

(3) At 600°C the gas mainly consists of S2 molecules

(4) The oxidation state of sulphur is never less than +4 in its compounds.




Sol: Ans [4] Sulphur shows oxidation states from –2 to +6

14. The entropy change involved in the isothermal reversible expansion of 2 moles of an ideal gas from a

volume of 10 dm3 to a volume of 100 dm3 at 27°C is:

(1) 38.3 J mol–1 K–1 (2) 35.8 J mol–1 K–1 (3) 32.3 J mol–1 K–1 (4) 42.3 mol–1 K–1

Sol: Ans [1] s = 2.303 nR log 2

1

vv

= 2.303 × 2 × 8.314 log 10010

= 38.3 J mol–1 K–1

15. Which of the following facts about the complex [Cr(NH3)6]Cl3 is wrong?

(1) The complex involves d2-sp3 hybridisation and is octahedral in shape.

(2) The complex is paramagnetic

(3) The complex is an outer orbital complex

(4) The complex gives white precipitate with silver nitrate solution.

Sol: Ans [3] Cr(NH)6]Cl3

Cr3+ – 3d34s04p0

xx xx xx xx xx xx

3d 4s 4p

NH3 NH3 NH3 NH3 NH3 NH3

33 6[Cr(NH ) ]

d2sp3

Inner orbital complex

16. The structure of IF7 is:

(1) square pyramid (2) trigonal bipyramid

(3) octahedral (4) pentagonal bipyramid

Sol: Ans [4]

F F

F

F

F

T

F

F

Pentagonal bipyramidal




17. The rate of a chemical reaction doubles for every 10°C rise of temperature. If the temperature is

raised by 50°C, the rate of the reaction increases by about:

(1) 10 times (2) 24 times (3) 32 times (4) 64 times

Sol: Ans [3] Temperature increase by 10°C, rate is double, its temperature increase 5 times, rate increase

ri(2)5 (raise to power 5 time) or 32 times.

18. The strongest acid amongst the following compound is:

(1) CH3COOH (2) HCOOH

(3) CH3CH2CH(Cl)CO2H (4) ClCH2CH2CH2COOH

Sol: Ans [3] —Cl is electron withdrawing group, hence increase acidity of acid.

19. Identify the compound that exhibits tautomerism.

(1) 2-Butene (2) Lactic acid (3) 2-Pentanone (4) Phenol

Sol: Ans [3]

O

CH3 CH3

OH

CH3 CH3

Tautomerisation

Keto form(excess)

enol(Trace)

(Less stable)

20. A vessel at 1000 K contains CO2 with a pressure of 0.5 atm. Some of the CO2 is converted into CO

on the addition of graphite. If the total pressure at equilibrium is 0.8 atm, the value of K is:

(1) 1.8 atm (2) 3 atm (3) 0.3 atm (4) 0.18 atm

Sol: Ans [1] 2(g) (g)CO C 2CO

0.5 atm 0–x +2x_____________________________________

At equilibrium 0.5 – x 2xTotal pressure = 0.5 – x + 2x = 0.5 – x0.5 + x = 0.8x = 0.3

2COp = 0.5 – 0.3 = 0.2

pCO = 2 × 0.3 = 0.6

2[0.6]K[0.2]

= 1.8 atm




21. In context of the lanthanoids, which of the following statements is not correct?

(1) There is a gradual decrease in the radii of the members with increasing atomic number in the series.

(2) All the members exhibit +3 oxidation state.

(3) Because of similar properties the separation of lanthanoids is not easy.

(4) Availability of 4f electrons results in the formation of compounds in +4 state for all the members of

the series.

Sol: Ans [4] Factual

22. ‘a’ and ‘b’ are van der Waals’ constants for gases. Chlorine is more easily liquefied than ethane because

(1) a and b for Cl2 a and b for C2H6

(2) a and b for Cl2 a and b for C2H6

(3) a for Cl2 a for C2H6 but b for Cl2 b for C2H6

(4) a for Cl2 a for C2H6 but b for Cl2 b for C2H6

Sol: Ans [4] Value of ‘a’ related to Interaction in between molecules. But for ‘b’ value depend size of

molecule.

23. The magnetic moment (spin only) of [NiCl4]2– is:

(1) 1.82 BM (2) 5.46 BM (3) 2.82 BM (4) 1.41 BM

Sol: Ans [3] [NiCl4]2–, Cl– are weak ligand.

d8 –

( 2) B.Mn n

8 = 2.82 B.M.

24. In a face centred cubic lattice, atom A occupies the corner positions and atom B occupies the face

centre position. If one atom of B is missing from one of the face centred points, the formula of the

compound is:

(1) A2B (2) AB2 (3) A2B3 (4) A2B5

Sol: Ans [4] Total atom at corner (A) = 1 8 18

Total atom at face centre (B) = 1 6 32

If one atom remove from one face centred point

Total atom (B) = 1 552 2

Formula A2B5




25. The outer electron configuration of Gd (Atomic No: 64) is:

(1) 4f3 5d5 6s2 (2) 4f8 5d0 6s2 (3) 4f4 5d4 6s2 (4) 4f7 5d1 6s2


26. Boron cannot form which one of the following anions?

(1) 36BF (2) 4BH (3) 4B(OH) (4) 2BO

Sol: Ans [1] Boron cannot make hexavalent due to absence of vacant d-orbital.

27. Ozonolysis of an organic compound gives formaldehyde as one of the products. This confirms the

presence of:

(1) two ethylenic double bonds (2) a vinyl group

(3) an isopropyl group (4) an acetylenic triple bond


H

H R

Hozonolysis HCHO + O

H

R

28. Sodium ethoxide has reacted with ethanoyl chloride. The compound that is produced in the above

reaction is:

(1) Diethyl ether (2) 2-Butanone (3) Ethyl chloride (4) Ethyl ethanoate

Sol: Ans [4]Cl

O

CH3

++ Na OCH3

CH3O

O

CH3ethyl acetate

+ NaCl

or ethyl ethanoate

29. The degree of dissociation () of a weak electrolyte, AxBy is related to van’t Hoff factor (i) by the

expression.

(1)1

( 1)i

x y

(2)

11

ix y

(3)1

1x y

i

(4)1

1x y

i

Sol: Ans [1]

xA y xyB xA yB

At equilibrium (1 – ) x y

(1 )1

x yi

11

ix y




30. Silver Mirror test is given by which one of the following compounds?

(1) Acetaldehyde (2) Acetone (3) Formaldehyde (4) Benzophenone

Sol: Ans [1] All aldehyde give silver mirror test by tollen’s reagnet. Except formaldehyde

OH

H

-OHCannizzaro CH3 OH H

O-

O

PART-B: PHYSICS

31. 100 g of water is heated from 30°C to 50°C. Ignoring the slight expansion of the water, the change in

its internal energy is (specific heat of water is 4184 J/kg/K)

(1) 4.2 kJ (2) 8.4 kJ (3) 84 kJ (4) 2.1 kJ

Sol: Ans [2] Change in internal energy = (ms T)

0.1 × 4184 × 20 = 8368 J 8.4 kJ

32. The half life of a radioacitve substance is 20 minutes. The approximate time interval (t2 – t1) between

the time t2 when 23

of it has decayed and time t1 when 13

of it had decayed is

(1) 7 min (2) 14 min (3) 20 min (4) 28 min

Sol: Ans [3] 100

23

tN N e .....(i)

2003

tN N e .....(ii)

Dividing (ii) by (i)

2 1( )12

t te

loge2 = (t2 – t1)

(t2 – t1) = 20 min.

33. A mass M, attached to a horizontal spring, executes S.H.M with amplitude A1. When the mass M

passes through its mean position then a smaller mass m is placed over it and both to them move

together with amplitude A2. The ratio of 1

2A

is

(1)M

M m(2)

M mM

(3)1/2M

M m

(4)1/2M m

M




Sol: Ans [4] Let v1 and v2 are velocities at mean position before and after putting mass, m

Mv1 = (M + m). v2

1 2. ( )K kM A M m AM M m

2 1MA A

M m

or 1

2

A M mA M

34. Energy required for the electron excitation in Li++ from the first to the third Bohr orbit is

(1) 12.1 eV (2) 36.3 eV (3) 108.8 eV (4) 122.4 eV

Sol: Ans [3] E = 13.6 × (3)21 11 9

= 13.6 × 9 × 89

= 108.8 eV

35. The transverse displacement y(x, t) of a wave on a string is given by 2 2( 2( , ) ) ax bt ab xty x t e . This

represents a

(1) wave moving in +x direction with speed ab

(2) wave moving in –x direction with speed ba

(3) standing wave of frequeny b (4) standing wave of frequency 1b

Sol: Ans [2] y(x, t) can be written as y(x, t) = 2ax bte

Direction of propogation of wave is negative x-axis

/v b a

36. A resistor ‘R’ and 2 F capacitor in series is connected through a switch to 200 V direct supply.

Across the capacitor is a neon bulb that lights up at 120 V. Calculate the value of R to make the bulb

light up 5s after the switch has been closed. (log102.5 = 0.4)

(1) 1.3 × 104 (2) 1.7 × 105 (3) 2.7 × 106 (4) 3.3 × 107

Sol: Ans [3] Charge on capacitor when P.D. across it 120 V

q = 2 × 120 = 240 m

Using 0 1t

RCq q e

65

2 10240 400 1 Re




62 10 0.4t

Re

or 6 log 2.5 2.303 0.42 10 e

tR

6

52 10 2.303 0.4

R

R = 2.7 × 106

37. 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) 2

o IR

(2) 22

oIR

(3) 2oIR (4) 4

oIR

Sol: Ans [1] Let curent is going inward

Due to symmetry By = 0

.sinxdB dB

. .sin2o dl

R

( ) IR

dB

x

y

. .sin .2

o R dR

0

2sin2 2

o oxB d

or 2B

oy

IR

38. A Carnot engine operating between temperatures T1 and T2 has efficiency 16

. When T2 is lowered by

62 K, its efficiency increases to 13

. Then T1 and T2 are, respectively

(1) 372 K and 310 K (2) 372 K and 330 K (3) 330 K and 268 K (4) 310 K and 248 K

Sol: Ans [1] 2 2

1 1

1 516 6

T TT T

... (i)

2

1

621 13

TT

or 2

1

62 23

TT

...(ii)

Dividing, 2

2

5/ 6 562 2/ 3 4

TT

4T2 = 5T2 – 310T2 = 310 k

From (i) 16 310 3725

T k




39. An object, moving with a speed of 6.25 m/s, is decelerated at a rate given by: 2.5 dv vdt

where v is the instantanous speed. The time taken by the object, to come to rest, would be

(1) 1 s (2) 2 s (3) 4 s (4) 8 s

Sol: Ans [2]0

0

2.5t

u

dv dtv

01/ 2

2.51/ 2 u

v t

–2 v1/2 = –2.5 t

2 6.25 2 sec.

2.5t

40. The electrostatic potential inside a charged spherical ball is given by = a r2 + b where r is the

distance from the centre, a, b are constants. Then the charge density inside the ball is

(1) 24 oa r (2) 6 oa r (3) 24 oa (4) 6 oa

Sol: Ans [4] 2dE ardr

flux = 2

0

( 2 .4 ) ( )iqar r r R

qi = –8 0 ar3

Charge density = 3

00

2

8 643

ar ar

41. A car is fitted with a convex side view mirror of focal length 20 cm. A second car 2.8 m behind the

first car is overtaking the first car at a relative speed of 15 m/s. The speed of the image of the second

car as seen in the mirror of the first one is

(1)1 m/s

10(2)

1 m/s15

(3) 10 m/s (4) 15 m/s

Sol: Ans [2]1 1 1v u f

2dv dumdt dt

...(i)

Also, 20 1

280 20 15fm

u f

21 115 m / sec.

15 15dvdt




42. If a wire is stretched to make it 0.1% longer, its resistance will

(1) increase by 0.05% (2) increase by 0.2%

(3) decrease by 0.2% (4) decrease by 0.05%

Sol: Ans [2] lRA

= 2

. lV

(V = volume of wire)

100 2 100 0.2%

R lR l

(increases)

43. 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, the final temperature of the mixture is

(1) 1 2 3( )3

T T T(2)

1 1 2 2 3 3

1 2 3

n T n T n Tn n n

(3)2 2 2

1 1 2 2 3 3

1 1 2 2 3 3

n T n T n Tn T n T n T (4)

2 2 2 2 2 21 1 2 2 3 3

1 1 2 2 3 3

n T n T n Tn T n T n T

Sol: Ans [2] Let final temperature of mixture is T, 1 1 1 2 3 3( ) ( ) ( ) 0 v vn C T T n C T T n T T

1 1 2 2 3 3

1 2 3( )

n T n T n TTn n n

44. Two identical charged spheres suspended from a common point by two massless strings of length l are

initially a distance d(d << l) 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.

Then as a function of distance x between them,

(1) 1/ 2v x (2) 1v x (3) 1/ 2v x (4) v x

Sol: Ans [1] sin eT F

cos T mg

tan eFmg

Fe

T

x/2x/2

mg

for small , 2

22 .

x kql mg x

23 22

lkqx cqmg




differentiating w.r.t. time

23 2 .dx dqx cqdt dt

1/ 2323 2 .

x dqx v cc dt

1/ 2v x

45. 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) 4 mJ (2) 0.2 mJ (3) 2 mJ (4) 0.4 mJ

Sol: Ans [4] W S A2 2

2 10.03 2 4 [ ] r r

40.24 [25 9] 10

30.384 10 J

= 0.4 mJ

46. A fully charged capacitor C with initial charge q0 is connected to a coil of self inductane L at t = 0.

The time at which the energy is stored equally between the electric and the magnetic field is

(1) LC (2)4 LC (3) 2 LC (4) LC

Sol: Ans [2]2

21 12 2

q LiC

or q LC i q0 cos(wt) = sin( )oLC l wt

also 0 0q LC i

tan( ) 1wt

wt = /4

4

t LC

47. Two bodies of masses m and 4m are placed at a distance r. The gravitational potential at a point on

the line joining them where the gravitational field is zero is

(1) zero (2)4

Gmr

(3)6

Gmr

(4)9

Gmr

Sol: Ans [4] Let gravitational field will be zero at distance x from mass m

2 24

( )

Gm G mx r x




1 2

x r x m 4m

x

C2x = r – x or x = r/3

V at C 4

/ 3 2 / 3

Gm G mr r

9

Gmr

48. A thin horizontal circular disc is rotating about a vertical axis passing through its centre. An insect is at

rest at a point near the rim of the disc. The insect now moves along a diameter of the disc to reach its

other end. During the journey of the insect, the angular speed of the disc

(1) remains unchanged (2) continuously decreases

(3) continuously increases (4) first increases and then decreases

Sol: Ans [4] I constantI first decreases then increases first increses then decreas0es.

49. Let the x – z (correct is x – y plane) 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 ˆˆ ˆ6 3 8 3 10 A i j k is incident on the plane of separation.

The angle of refraction in medium 2 is

(1) 30° (2) 45° (3) 60° (4) 75°

Sol: Ans [2] Let angle of incident = i

cos i = 10 1

236 3 64 3 100

3sin

2i

32 3 sin2

r

1sin2

r or r =45°

50. Two particles are executing simple harmonic motion of the same amplitude A and frequency along

the x-axis. Their mean position is separated by distance X0 (X0 > A). If the maximum separation

between them is (X0 + A), the phase difference between their motion is

(1)2

(2)3

(3)4

(4)6




Sol: Ans [1] Possibility:

x0

M1 M2

Phse difference = /2

51. Direction: The question has a paragraph followed by two statements, Statement-1 and Statement-2.

Of the given 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 phasechange of .

Statement-2: The centre of the interference pattern is dark.

(1) Statement-1 is true, Statement-2 is false

(2) Statement-1 is true, Statement-2 is true and 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: Ans [3] Statement-1- True

Reflection at denser surface surffers phase change of .

Statement-2 - True

52. A thermally insulated vessel contains an ideal gas of molecular mass M and ratio of specific heats . It

is moving with speed v and is suddenly brought to rest. Assuming no heat is lost to the surroundings,

its temperature increases by

(1)2( 1)

2( 1)

Mv KR (2)

2( 1)2

Mv KR (3)

2

2Mv K

R (4)2( 1)

2 Mv K

R

Sol: Ans [4] Kinetic energy of 20

12 CM V change into heat where VC is velocity of centre of mass of gas.

2

01 .2 1

oC

M RM V TM

21 1

2T Mv

R




53. A screw gauge gives the following reading when used to measure the diameter of a wire.

Main scale reading : 0 mm

Circular scale reading: 52 divisions

Given that 1 mm on main scale corresponds to 100 divisions of the circular scale.

The diameter of wire from the above data is

(1) 0.52 cm (2) 0.052 cm (3) 0.026 cm (4) 0.005 cm

Sol: Ans [2] Diameter = Main scale reacding + L.C × (circular scale reading) – zero error.

10 52 0.52100

mm = 0.052 cm

54. A boat is moving due east in a region where the earth’s magnetic field is 5.0 × 10–5 NA–1 m–1 duenorth and horizontal. The boat carries a vertical aerial 2 m long. If the speed of the boat is 1.50 ms–1, themagnitude of the induced emf in the wire of aerial is

(1) 1 mV (2) 0.75 mV (3) 0.50 mV (4) 0.15 mV

Sol: Ans [4] emf lvB

52 1.5 5 10

N

S

W E

B vB=1.5 m/sec

515 10

= 0.15 mV

55. Direction: The question has a Statement-1 and Statement-2. Of the four choices given after thestatements, choose the one that best describes the 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.


(2) Statement-1 is true, Statement-2 is true and Statement-2 is the correct explanation ofStatement-1

(3) Statement-1 is true, Statement-2 is true, Statement-2 is not the correct explanation ofStatement-1


Sol: Ans [4]

56. A mass m hangs with the help of a string wrapped around a pulley on a frictionless bearing. The pulley

has mass m and radius R. Assuming pulley to be a perfect uniform circular disc, the acceleration of the

mass m, if the string does not slip on the pulley is

(1)32

g (2) g (3)23

g (4)3g




Sol: Ans [3] mg – T = ma ... (i)

T.R 2

2

mR... (ii) T

a

mg

a R ...(iii)

2 .3

a g

57. A water fountain on the ground sprinkles water all around it. If the speed of water coming out of the

fountain is v. the total area around the fountain that gets wet is

(1)2

vg (2)

4

2vg (3)

4

22 v

g (4)2

2vg

Sol: Ans [2] Maximum area = (max. range)2

= 22

vg

4

2

v

g

58. Direction: This question has a Statement-1 and Statement-2. Of the four choices given after thestatements, choose the one that best describes the two statements.

Statement-1: A metallic surface is irradiated by a monochromatic light of frequency v > v0 (thethreshold frequency). The maximum kinetic energy and the stopping potential are Kmax

and V0 respectively. If the frequency incident on the surface is doubled, both the Kmax

and V0 are also doubled.

Statement-2: The maximum kinetic energy and the stopping potential of photoelectrons emitted froma surface are linearly dependent on the frequency of incident light.


(2) Statement-1 is true, Statement-2 is true, Statement-2 is the correct explanation ofStatement-1

(3) Statement-1 is true, Statement-2 is true, Statement-2 is not the correct explanation ofStatement-1


Sol: Ans [4] Statement-1- False

max K hv

o

hvVe e




'max 2 2[ ] K hv hv

= max2 K

' 2 2 o

hv hvVe e e e e

2 oV

e

Statement-2 - True

59. A pulley of radius 2m is rotated about its axis by a force F = (20 t – 5t2) newton (where t is measured

in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is

10 kg m2, the number of rotations made by the pulley before its direction of motion if reversed. is

(1) less than 3 (2) more than 3 but less than 6

(3) more than 6 but less than 9 (4) more than 9

Sol: Ans [2] Using I

2(20 5 ). . dt t R Idt

2

0 0. (20 5 )

tI d R t t dt

2 35. (10 )3

I R t t ...(i)

for = 0 2 3510 03

t t

2 5(10 ) 03

t t

t = 6 sec

Using (i)6

2 3

0 0

5. (10 )3

I d R t t dt

63 4

0

10 53 12

I R t t

3 42 10 56 6 36 .

10 3 12

rad

no. of rotations 362




60. Water is flowing continuously from a tap having an internal diameter 8 × 10–3 m. The water velocity

as it leaves the tap is 0.4 ms–1. The diameter of the water stream at a distance 2 × 10–1 m below the

tap is close to

(1) 5.0 × 10–3 m (2) 7.5 × 10–3 m (3) 9.6 × 10–3 m (4) 3.6 × 10–3 m

Sol: Ans [4] Using 212

P v gh const.

2 1 21 1(0.4) 2 10 02 2

g v

4.16v

Using 1 1 2 2A v A v

or 2 21 1 2 2D v D v

31

2 12

0.4 8 102.04

vD Dv

338 10 3.6 10

2.45

m

PART-C: MATHEMATICS

61. Let , be real and z be a complex number. If z2 + z + b = 0 has two distinct roots on the line

Re z = 1, then it is necessary that

(1) (0, 1) (2) (–1, 0) (3) || = 1 (4) (1, )

Sol: Ans [4] Let roots be (1 + ki) and (1 + li)

The sum = 2 + (k + l) i = (real)

k = – l

Product = = 1 + k2 (1, ).

62. The value of 1

20

8log(1 )1

x dxx

is

(1) log 2 (2) log 28

(3) log 22

(4) log 2

Sol: Ans [1] Put x = tan then

I = /4

0

8 log(1 tan )d

....(i)

I = /4

0

8 log 1 tan4

d .....(ii)

Adding we get 2I = /4

0

8 log2 d

I = log2




63.2

2

d xdy equals

(1)12

2

d ydx

(2)1 32

2

d y dydx dx

(3)22

2

d y dydx dx

(4)

32

2

d y dydx dx

Sol: Ans [4]2

2

1/

d x d dx ddy dy dy dy dy dx

= 21 . d dy

dy dxdydx

= – 3 2

2.dx d ydy dx

64. 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 t

dt 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 equipmentis

(1) 2 ITk

(2)2

2kTI (3)

2( )2

k T tI (4) e–kT

Sol: Ans [2]dvdt = – k (T – t)

V = – k (2

2tTt

+ c

I = 0 + c c = I

V (T) = – k 2

2

2TT

+ I = I – 2

2kT

65. The coefficient of x7 in the expansion of (1 – x – x2 + x3)6 is

(1) 144 (2) –132 (3) –144 (4) 132

Sol: Ans [3] (1 – x)6 (1 – x2)6

Tr, s = 6Cr (– x)r 6Cs (–x2)s

r + 2s = 7

coefficient = 6C1 6C3 – 6C36C2 + 6C5

6C1

= 6 × 20 – 20 × 15 + 36 = – 144.




66. For 50, ,2

x

define 0

( ) sinx

f x t t dt . Then f has

(1) local maximum at p and 2 (2) local minimum at and 2

(3) local minimum at and local maximum at 2 (4) local maximum at and local minimum at 2

Sol: Ans [4] f (́x) = x sin x = 0 x = , 2

at x = , local maximum and at x = 2 local minimum.

67. The area of the region enclosed by the curves y = x, x = e, y = 1/x and the positive x-axis is

(1) 1/2 square units (2) 1 square units (3) 3/2 square units (4) 5/2 square units

Sol: Ans [3] Area = 1 + 1

1e

dxx

= 1 312 2

. 1 x

y

x=e

y=x

68. The lines L1 : y – x = 0 and L2 : 2x + y = 0 intersect the line L3 : y + 2 = 0 at P and Q respectively.

The bisector of the acute angle between L1 and L2 intersects L3 at R.

Statement-1: The ratio PR : RQ equals 2 2 : 5

Statement-2: In any triangle, bisector of an angle divides the triangle into two similar triangles.

(1) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation forStatement-1.

(2) Statement-1 is true, Statement-2 is true; Statement-2 is not correct explanation of Statement-1.

(3) Statement-1 is true, Statement-2 is false.

(4) Statement-1 is false, Statement-2 is true

Sol: Ans [3] Solving lines P is (–2, –2) & Q is (1, –2) OP = 2 2 and OQ = 5

so statement 1 is true and 2 is false.

69. The values of p and q for which the function 2

3/ 2

sin( 1) sin , 0

( ) , 0

, 0

p x x xx

f x q x

x x x xx

is continuous for all x in R, are

(1)1 3,2 2

p q (2)5 1,2 2

p q (3)3 1,2 2

p q (4)1 3,2 2

p q




Sol: Ans [3]0

sin( 1) sinlimx

p xx

= p + 2 = q =

12

p = 32

, q = 12

.

70. If the angle between the line 1 3

2y zx

and the plane x + 2y + 3z = 4 is 1 5cos ,

14

then

equals :

(1) 2/3 (2) 3/2 (3) 2/5 (4) 5/3

Sol: Ans [1] sin = 2

1 4 3 3145 14

= 23

.

71. The domain of the function 1( ) is :

| |f x

x x

(1) (–, ) (2) (0, ) (3) (–, 0) (4) (–, ) – {0}

Sol: Ans [3] For domain |x| > x x < 0

72. The shortest distance between line y – x = 1 and curve x = y2 is :

(1)3

4(2)

3 28

(3)8

3 2 (4)43

Sol: Ans [2] The equation of parallel tangent is y = x + 14

shortest distance =

114

1 1

= 3 2

8

73. 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) 18 months (2) 19 months (3) 20 months (4) 21 months

Sol: Ans [4]2n

[200 × 2 + (n –1)40] = 11040– 400 = 10,640

n = 19

so total time taken = 19 + 2 = 21 months.




74. Consider the following statements

P : Suman is brilliant

Q : Suman is rich

R : Suman is honest

The negation of the statement “Suman is brilliant and dishonest if and only if Suman is rich” can beexpressed as :

(1) ~ P ^ (Q ~ R) (2) ~ (Q (P ^ ~ R)) (3) ~ Q ~ P ^ R (4) ~ (P ^ ~ R) Q

Sol: Ans [2] The statement is Q (P ~ R)

so its negation is ~ (Q (P ~ R))

75. If (1) is a cube root of unity, and (1 + )7 = A + B. Then (A, B) equals :

(1) (0, 1) (2) (1, 1) (3) (1, 0) (4) (–1, 1)

Sol: Ans [2] (1 + )7 = A + B

(–2)7 = A + B

–2 = A + B = 1 +

A = 1 and B = 1

76. If 1 ˆˆ310

a i k

and 1 ˆˆ ˆ2 3 6 ,7

b i j k

then the value of 2 2a b a b a b is :

(1) –5 (2) –3 (3) 5 (4) 3

Sol: Ans [1] (2a – b) . [(a × b) × (a + 2b)]

= [2a – b a × b a + 2b]

= [a × b a + 2b 2a – b]

= – 5 2| |a b = – 5

77. If 3 0dy ydx

and y(0) = 2, then y(ln 2) is equal to

(1) 7 (2) 5 (3) 13 (4) –2

Sol: Ans [1] 3dy

y = dx ln (y + 3) = x + c

3

5xy e

y (ln 2) = 7.

78. Equation of the ellipse whose axes are the axes of coordinates and which passes through the point

(–3, 1) and has eccentricity 2 / 5 is :

(1) 3x2 + 5y2 – 32 = 0 (2) 5x2 + 3y2 – 48 = 0 (3) 3x2 + 5y2 – 15 = 0 (4) 5x2 + 3y2 – 32 = 0




Sol: Ans [1] 2 2

9 1 1a b

and 2

2

2 15

ba

a2 = 323

and b2 = 325

79. If the mean deviation about the median of the numbers a, 2a, ....., 50a is 50, then |a| equals :

(1) 2 (2) 3 (3) 4 (4) 5

Sol: Ans [3]51 51 512 .... 50

2 2 2a a aa a a = 50 × 50

| |2a

[49 + 47 + 45 + ... + 5 + 3 + 1 + 3 + ... + 49] = 2500

| |2a

[2 × 25 × 25] = 2500

|a| = 4.

80. 2

1 cos{2( 2)}lim2x

xx

(1) does not exist (2) equals 2 (3) equals – 2 (4) equals 12

Sol: Ans [1]2

2 2

2sin ( 2) | 2 || sin( 2) |lim lim2 2x x

x xx x

= 2 2

| sin( 2) |lim2x

xx

which does not exist.

81. Statement-1: The number of ways of distributing 10 identical balls in 4 distinct boxes such that no

box is empty is 9C3.

Statement-2: The number of ways of choosing any 3 places from 9 different places is 9C3.

(1) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for

Statement-1.




Sol: Ans [2] The number of ways of distributing n identical balls is r distinct boxes such that no box

remains empty is n –1Cr –1.




82. Let R be the set of real numbers

Statement-1: A = {(x, y) R × R: y – x is an integer) is an equivalence reaction on R.

Statement-2: B = {(x, y) R × R : x = y for some rational number } is an equivalence relation onR.





Sol: Ans [3] Statement-1 is true, Statement-2 is false.

83. 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 to 3132

, then p lies in the interval:

(1)1 3,2 4

(2)3 11,4 12

(3)10,2

(4)11 ,112

Sol: Ans [3] 1 – (p)5 3132

1

32 p5

10,2

84. The two circles x2 + y2 = ax and x2 + y2 = c2 (c> 0) touch each other if

(1) 2 |a| = c (2) |a| = c (3) a = 2c (4) |a| = 2c

Sol: Ans [2] |r1 – r2| = C1C2

85. Let A and B be two symmetric materices of order 3.

Statement-1: A(BA) and (AB)A are symmetric materices.

Statement-2: AB is symmetric matrix if matrix multiplication of A with B is commutative.





Sol: Ans [2] [A(BA)]T = (BA)TAT = (ATBT)A

= (AB)A

= A(BA) since A and B are symmetric

Statement 1 is true

(AB)T = BTAT = BA = AB statement 2 is true




86. 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(C|D) < P(C) (4) P(C|D)=P(D)P(C)

Sol: Ans [2]CPD

= (C D) P(C)P

P(D) P(D)

C D.

P(D) 0 0 < P(D) 1

1 1

P(D) <

P(C) P(C)P(D)

P(C) PCD

87. The vectors a and b are not prependicular and c and d

are two vectors satisfying: b c b d

and . 0.a d Then the vector d

is equal to:

(1)..

b cb ca b

(2)..

a cc ba b

(3)..

b cb ca b

(4)..

a cc ba b

Sol: Ans [4] ( ) ( )a b c a b d

( . ) ( . ) ( . ) ( . )a c b a b c a d b a b d [ . 0]a d

( . )( . )a cd c ba b

88. Statement-1: The point A(1, 0, 7) is the mirror image of the point B (1, 6, 3) in the line:

1 21 2 3x y z

Statement-2: The line : 1 2

1 2 3x y z bisects the line segment joining A(1, 0, 7) and B(1, 6, 3).








Sol: Ans [2] Line 1 2

1 2 3x y z is to the line segment AB. Also the line passes through the mid

point of AB.

89. If A = sin2x + cos4x, then for all real x:

(1)34

A 1 (2)1316

A 1 (3) 1 A 2 (4)34

A 1316

Sol: Ans [1] A = sin2x + cos4x = cos4x – cos2x + 1

= t2 – t + 1 for 0 t 1

34

t2 – t +1 1

90. 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) 3 (2) 2 (3) 1 (4) zero

Sol: Ans [2] For non-zero solution = 0

4 24 1 0

2 2 1

kk

4(4 – 2) – k(k –2) + 2 (2k – 8) = 0

+8 – k2 + 2 k + 4k – 16 = 0

– k2 + 6k – 8 = 0 k2 – 6 k + 8 = 0 k = 2, or 4.

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