kerala pet 2014 paper

7/17/2019 Kerala PET 2014 Paper

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1. Five cells each of emf E and internal resistance r send thesame amount of current through an external resistance R

whether the cells are connected in parallel or in series. Then

the ratio R

r is

(a) 2 (b)1

2

(c)1

5(d) 1

(e) 5

2. The power dissipated in the transmission cables carryingcurrent I and voltage V is inversely proportional to(a) V (b) V 2

(c) V (d) I

(e) I

3. A rigid container with thermally insulated walls contains agas and a coil of resistance 50 , carrying a current of l A.The change in internal energy of the gas after 2 minutes will be(a) 6 kJ (b) 10 kJ(c) 3 kJ (d) 12 kJ(e) 1.5 kJ

4. The magnitude of the magnetic field inside a long solenoidis increased by(a) decreasing its radius(b) decreasing the current through it(c) increasing its area of cross-section(d) introducing a medium of higher permeability(e) decreasing the number of turns in it

5. A bar magnet of moment of inertia 9 × l0 –5 kg m2 placed ina vibration magnetometer and oscillating in a uniformmagnetic field l62 × l0 –5T makes 20 oscillations in 15 s.The magnetic moment of the bar magnet is(a) 3 Am2 (b) 2 Am2

(c) 5 Am2 (d) 6 Am2

(e) 4 Am2

6. Identify the correctly matched pair Material Example

(a) Diamagnetic - Gadolinium(b) Soft ferromagnetic - Alnico(c) Hard ferromagnetic - Copper (d) Paramagnetic - Sodium(e) Permanent magnet - Aluminum

7. If the radius of the dees of cyclotron is r , then the kineticenergy of a proton of mass m accelerated by the cyclotronat an oscillating frequency v is(a) 42m2v2r 2 (b) 42mv2r 2

(c) 22mv2r 2 (d) 2mv2r 2

(e) 2m2v2r 2

8. If a magnetic dipole of moment M situated in the directionof a magnetic field B is rotated by 180°, then the amount of work done is(a) MB (b) 2MB

(c)MB

2(d) 0

(e) MB9. The polarity of induced emf is given by

(a) Ampere's circuital law(b) Biot-Savart law(c) Lenz’s law(d) Fleming's right hand rule(e) Fleming's left hand rule

10. In an LCR series circuit, at resonance(a) the current and voltage are in phase(b) the impedance is maximum(c) the current is minimum(d) the quality factor is independent of R (e) the current leads the voltage by /2

11. A conducting ring of radius l m kept in a uniform magneticfield B of 0.01 T , rotates uniformly with an angular velocity100 rad s –1 with its axis of rotation perpendicular to B. The

maximum induced emf in it is(a) 1.5V (b) V(c) 2V (d) 0.5V(e) 4V

12. A step down transformer increases the input current 4 A to24 A at the secondary. If the number of turns in the primarycoil is 330, the number of turns in the secondary coil is(a) 60 (b) 50(c) 65 (d) 45(e) 55

13. In a plane electromagnetic wave, the electric field of amplitude l V m –1 varies with time in free space. The average

energy density of magnetic field is (in J m –2

)(a) 8.86 × 10 –2 (b) 4.43 × l0 –2

(c) 17.72 × 10 –2 (d) 2.21 × 10 –2

(e) 1.11 × 10 –2

14. Which one of the following is the property of amonochromatic, plane electromagnetic wave in free space?(a) Electric and magnetic fields have a phase difference

of /2(b) The energy contribution of both electric and magnetic

fields are equal(c) The direction of propagation is in the direction of

electric filed E

(d) The pressure exerted by the wave is the product of energy density and the speed of the wave

(e) The speed of the wave is B/E

KERALA PET 2014 - SOLVED PAPER

Paper-I : Physics Chemistry



2014-2 2014 SOLVED PAPER

15. The apparent flattening of the sun at sunset and sunrise isdue to(a) refraction (b) diffraction(c) total internal reflection (d) interference(e) polarization

16. The polarising angle for a medium is found to be 60°. The

critical angle of the medium is

(a)1 1

sin2

(b)

1 3sin

2

(c)1 1

sin3

(d)

1 1sin

4

(e)1 2

sin3

17. Identify the mismatch in the following(a) Myopia - Concave lens(b) For rear view - Concave mirror

(c) Hypermetropia - Convex lens(d) Astigmatism - Cylindrical lens(e) Reflecting telescope - Convex mirror

18. In Young's double slit experiment, to increase the fringewidth(a) the wavelength of the source is increased(b) the source is moved towards the slit(c) the source is moved away from the slit(d) the slit separation is increased(e) the screen is moved towards the slit

19. Light of wavelength 5000 A° is incident normally on a slit of width 2.5 × l0 –4 cm. The angular position of second minimum

from the central maximum is

(a) 1 1sin

5

(b) 1 2

sin5

(c)3

(d)6

(e)4

20. An electron of mass me and a proton of mass m p areaccelerated through the same potential. Then the ratio of their de Broglie wavelengths is

(a) 1 (b) e

p

m

m

(c) e

p

m

m (d) p

e

m

m

(e) p

e

m

m

21. The half-life of a radioactive substance is 20 minutes. Thetime taken between: 5% decay and 87.5% decay of thesubstance will be

(a) 20 minutes (b) 30 minutes(c) 40 Minutes (d) 25 minutes(e) 10 minutes

22. The ratio of the surface area of the nuclei 52Te125 to that of

13Al27 is

(a)5

3(b)

125

17

(c)

1

4 (d)

25

9

(e)3

5

23. If the frequency of incident light falling on a photosensitivemetal is doubled, the kinetic energy of the emitted photoelectron is

(a) unchanged

(b) halved

(c) doubled

(d) more than twice its initial value

(e) reduced to 14

th

24. The significant result deduced from the Rutherford'sscattering experiment is that

(a) whole of the positive charge is concentrated at thecentre of atom

(b) there are neutrons inside the nucleus

(c) -particles are helium nuclei

(d) electrons are embedded in the atom

(e) electrons are revolving around the nucleus

25. On an average, the number of neutrons and the energy of a

neutron released per fission of a uranium atom arerespectively

(a) 2.5 and 2 keV (b) 3 and l keV

(c) 2.5 and 2 MeV (d) 2 and 2 keV

(e) l and 2 MeV

26. The inputs A, B and C to be given in order to get an outputY = 1 from the following circuit are

(a) 0, 1, 0

(b) 1, 0, 0

(c) 1, 0, l

A

B

C

Y

(d) 1, 1, 0

(e) 0 , 0, 1

27. The collector resistance and the input resistance of a CEamplifier are respectively 10 k and 2 k . If of the transistor is 49, the voltage gain of the amplifier is

(a) 125 (b) 150

(c) 175 (d) 200

(e) 24528. The light emitting diode (LED) is

(a) a heavily doped p-n junction with no external bias

(b) a heavily doped p-n junction with reverse bias

(c) a heavily doped p-n junction with forward bias

(d) a lightly doped p-n junction with no external bias

(e) a lightly doped p-n junction with reverse bias



2014 SOLVED PAPER 2014-3

29. A point-to-point communication mode is seen in

(a) Satellite cable communication(b) Television transmission

(c) FM radio transmission

(d) AM radio transmission

(e) Fax transmission30. If the heights of transmitting and the receiving antennasare each equal to h, the maximum line-of-sight distance between them is (R is the radius of earth)

(a) 2R h (b) 4R h

(c) 6R h (d) 8R h

(e) R h

31. The ionospheric layer acts as a reflector for the frequencyrange

(a) l kHz to 10 kHz (b) 3 MHz to 30 MHz

(c) 3 kHz to 30 kHz (d) 100 kHz to 1 MHz(e) 3 GHz to 30 GHz

32. ln a simple pendulum experiment, the maximum percentageerror in the measurement of length is 2% and that in theobservation of the time-period is 3%. Then the maximum percentage error in determination of the acceleration due togravity g is(a) 5% (b) 6%

(c) 1% (d) 8%(e) 10%

33. The pitch and the number of circular scale divisions in ascrew gauge with least count 0.02 mm are respectively

(a) l mm and 100 (b) 0.5 mm and 50(c) 1 mm and 50 (d) 0.5 mm and 100

(e) l mm and 20034. A ball is dropped from the top of a tower of height 100 m

and at the same time another ball is projected verticallyupwards from ground with a velocity 25 ms –1. Then thedistance from the top of the tower, at which the two ballsmeet is(a) 68.4 m (b) 48.4 m

(c) 18.4 m (d) 28.4 m

(e) 78.4 m

35. The ratio of distances traversed in successive intervals of time when a body falls freely under gravity from certainheight is

(a) l : 2 : 3 (b) l : 5 : 9

(c) 1 : 3 : 5 (d) 1 : 2 : 3

(e) l : 4 : 936. A particle starting with certain initial velocity and uniform

acceleration covers a distance of 12 m in first 3 seconds anda distance of 30 m in next 3 seconds. The initial velocity of the particle is

(a) 3 ms –1 (b) 2.5 ms –1

(c) 2 ms –1 (d) 1.5 ms –1

(e) 1 ms –1

37. A ball of mass 10 g moving perpendicular to the plane of thewall strikes it and rebounds in the same line with the samevelocity. If the impulse experienced by the wall is 0.54 Ns,the velocity of the ball is

(a) 27 ms –1 (b) 3.7 ms –1

(c) 54 ms –1 (d) 37 ms –1

(e) 5.4 ms –1

38. A particle has the position vector ˆˆ ˆ2r i j k

and the

linear momentum ˆˆ ˆ2 p i j k

. Its angular momentum

about the origin is

(a) ˆˆ ˆ 3i j k (b) ˆˆ ˆ 3i j k

(c) ˆˆ ˆ 3i j k (d) ˆˆ ˆ 5i j k

(e) ˆˆ ˆ 5i j k

39. The vertical component of velocity of a projectile at itsmaximum height (u - velocity of projection, -angle of projection) is

(a) u sin (b) u cost

(c)sin

u

(d) 0

(e)cos

u

40. The coordinates of a particle moving in x–y plane at anyinstant of time t are x = 4t2; y = 3t2. The speed of the particleat that instant is

(a) 10 t (b) 5 t(c) 3 t (d) 2 t

(e) 13t

41. A cyclist bends while taking turn in order to

(a) reduce friction(b) provide required centripetal force

(c) reduce apparent weight(d) reduce speed

(e) sit comfortably

42. Two blocks of masses 2 kg and 4 kg are attached by an

inextensible light string as shown in the figure. If a force of 120 N pulls the blocks vertically upward, the tension in thestring is (take g = 10 ms –2)

(a) 20 N

(b) 15 N

(c) 35 N

F = 120 N

4 kg

2 kg(d) 40 N(e) 30 N

43. The total energy of a solid sphere of mass 300 g which rollswithout slipping with a constant velocity of 5 ms –1 along astraight line is

(a) 5.25 J (b) 3.25 J(c) 0.25 J (d) l.25 J

(e) 0.625J




44. A bullet when fired into a target loses half of its velocityafter penetrating 20 cm. Further distance of penetration before it comes to rest is

(a) 6.66 cm (b) 3.33 cm

(c) 12.5 cm (d) 10 cm(e) 5 cm

45. In elastic collision(a) both momentum and kinetic energy are conserved

(b) neither momentum nor kinetic energy is conserved(c) only momentum is conserved

(d) only kinetic energy is conserved(e) forces involved in the interaction are non-conservative

46. Two discs rotating about their respective axis of rotationwith angular speeds 2 rads –1 and 5 rads –1 are brought intocontact such that their axes of rotation coincide. Now, theangular speed of the system becomes 4 rads –1. If the momentof inertia of the second disc is 1 × 10 –3 kg m2, then the

moment of inertia of the first disc (in kg m2) is(a) 0.25 × 10 –3 (b) 1.5 × 10 –3

(c) 1.25 × 10 –3 (d) 0.75 × 10 –3

(e) 0.5 × 10 –3

47. A wheel is rotating at 1800 rpm about its own axis. Whenthe power is switched off, it comes to rest in 2 minutes.Then the angular retardation in rad s –1 is

(a) 2 (b)

(c)2

(d)

4

(e)6

48. If the angular momentum of a particle of mass m rotatingalong a circular path of radius r with uniform speed is L, thecentripetal force acting on the particle is

(a)2

2

L

mr (b)

2L

mr

(c)L

mr (d)

2L m

r

(e) 2Lmr

49. Pick out the wrong statement from the following(a) The Sl unit of universal gravitational constant is

Nm2kg –2

(b) The gravitational force is a conservative force

(c) The force of attraction due to a hollow spherical shellof uniform density on a point mass inside it is zero

(d) The centripetal acceleration of the satellite is equal toacceleration due to gravitational potential

(e) Gravitational potential energy

=gravitational potential

mass of the body

50. If a body of mass m has to be taken from the surface of earthto a height h = R, then the amount of energy required is(R : radius of earth)

(a) mgR (b)R

3

mg

(c)R

2

mg(d)

R 12

mg

(e)R

9

mg

51. The total energy of an artificial satellite of massm revolvingin a circular orbit around the earth with a speed v is

(a) 21

2mv (b) 21

4mv

(c) 21

4

mv (d) – mv2

(e)21

2 mv

52. Two soap bubbles each with radius r 1 and r 2 coalesce invacuum under isothermal conditions to form a bigger bubbleof radius R. Then R is equal to

(a) 2 21 2r r (b) 2 2

1 2r r

(c) r 1 – r 2 (d)2 2

1 2

2

r r

(e) 2 21 22 r r

53. The ratio of hydraulic stress to the corresponding strain isknown as(a) compressibility (b) bulk modulus

(c) Young's modulus (d) rigidity modulus(e) expansion coefficient

54. A boy can reduce the pressure in his lungs to 750 mm of mercury. Using a straw he can drink water from a glass uptothe maximum depth of (atmospheric pressure = 760 mm of mercury; density of mercury = 13.6 gcm –3)

(a) 13.6 cm (b) 9.8 cm(c) 10 cm (d) 76 cm

(e) 1.36 cm55. A spring stores l J of energy for a compression of l mm. The

additional work to be done to compress it further by l mm is(a) l J (b) 2 J(c) 3 J (d) 4 J

(e) 0.5 J56. If m represents the mass of each molecule of a gas and T, its

absolute temperature then the root mean square velocity of the gaseous molecule is proportional to

(a) m T (b) m1/2

T1/2

(c) m –1/2 T1/2 (d) m –1/2 T1

(e) Tm1/2




57. A Carnot engine operating between temperatures T1 andT2 has efficiency 0.2. When T2 is reduced by 50 K, itsefficiency increases to 0.4. Then T1 and T2 are respectively(a) 200 K, 150 K (b) 250 K, 200 K (c) 300 K, 250 K (d) 300 K, 200 K (e) 300 K, 150 K

58. A molecule of a gas has six degrees of freedom. Then themolar specific heat of the gas at constant volume is

(a)R

2(b) R

(c)3R

2(d) 2 R

(e) 3 R 59. Total number of degrees of freedom of a rigid diatomic

molecule is(a) 3 (b) 6(c) 5 (d) 2(e) 7

60. If the differential equation for a simple harmonic motion is2 2

2

d y

dt + 2 y = 0, the time-period of the motion is

(a) 2s (b)2s

s

(c)2

s

(d) 2s

(e)2

s

61. Identify the wrong statement from the following

(a) If the length of a spring is halved, the time period of

each part becomes1

2 times the original

(b) The effective spring constant K of springs in parallel

is given by1 2

1 1

K K K

(c) The time period of a stiffer spring is less than that of asoft spring

(d) The spring constant is inversely proportional to thespring length

(e) The unit of spring constant is Nm –1

62. The total energy of the particle executing simple harmonicmotion of amplitude A is 100 J. At a distance of 0.707 A fromthe mean position, its kinetic energy is(a) 25 J (b) 50 J(c) 100 J (d) 12.5 J(e) 70 J

63. Two travelling waves, yl = A sin [k ( x + ct ) ] and y2 = A sin[k ( x – ct )] are superposed on a string. The distance betweenadjacent antinodes is

(a) c

(b)

2

ct

(c)2k

(d)

k

(e)

k

64. If a stretched wire is vibrating in the second overtone, thenthe number of nodes and antinodes between the ends of the string are respectively(a) 2 and 2 (b) l and 2(c) 4 and 3 (d) 2 and 3(e) 3 and 4

65. Pick out the correct statement in the following with referenceto stationary wave pattern(a) In a tube closed at one end, all the harmonics are

present(b) In a tube open at one end, only even harmonics are

present(c) The distance between successive nodes is equal to

the wavelength(d) In a stretched string, the first overtone is the same as

the second harmonic(e) Reflection of a wave from a rigid wall changes the

phase by 45°

66. A plane square sheet of charge of side 0.5 m has uniformsurface charge density. An electron at 1 cm from the centreof the sheet experiences a force of l.6 × 10 –2 N directedaway from the sheet. The total charge on the plane squaresheet is(0 = 8.854 × 10 –12 C2m –2 N –1)(a) 16.25 µC (b) –22.15 µC(c) – 44.27 µC (d) 144.27 µC(e) 8.854 µC

67. The energy stored in a capacitor of capacitance C having acharge Q under a potential V is

(a)

21

Q V2 (b)

21

C V2

(c)21 Q

2 V(d)

1QV

2

(e)1

CV2

68. The electrostatic force between two point charges is directly proportional to the(a) sum of the charges(b) distance between the charges(c) permittivity of the medium(d) square of the distance between the charges

(e) product of the charges69. The time period of revolution of a charge q1 and of mass m

moving in a circular path of radius r due to Coulomb forceof attraction with another charge q2 at its centre is

(a)3

0

1 2

16 mr

q q

(b)

2 30

1 2

8 mr

q q

(c)3

0

1 216

mr

q q

(d)

3 30

1 2

16 mr

q q

(e)2 3

0

1 28mr

q q




70. A point charge of 2 C experiences a constant force of 1000 Nwhen moved between two points separated by a distanceof 2 cm in a uniform electric field. The potential difference between the two points is(a) 12 V (b) 8 V(c) 10 V (d) 16 V

(e) 5 V71. In the network shown below, if potential across XY is 4 V,

then the input potential across AB is

A

B

2

4

4

8

2

X

Y

4 4 V

8

(a) 16 V (b) 20 V(c) 8 V (d) 12 V(e) 24 V

72. If the ammeter A shows a zero reading in the circuit shown below, the value of resistance R is

10 2 V

500 R A

(a) 500 (b) 125 (c) 100 (d) 41.5(e) 4

73. In a reaction 2A + B 3C, the concentration of A decreasesfrom 0.5 mol L –1 to 0.3 mol L –1 in 10 minutes. The rate of production of 'C' during this period is(a) 0.01 mol L –1 min –1 (b) 0.04 mol L –1 min –1

(c) 0.05 mol L –1 min –1 (d) 0.03 mol L –1 min –1

(e) 0.02 mol L –1 min –1

74. Ammonium ion (NH4+) reacts with nitrite ion (NO2

– ) in

aqueous solution according to the equation NH4

+ (aq) + NO2 – (aq) N2(g) + 2H2O(l)

The following initial rates of reaction have been measuredfor the given reactant concentrations.Expt. No. [NH4

+], (M) [NO2 – ], (M) Rate (M/hr)

1 0.010 0.020 0.0202 0.015 0.020 0.0303 0.010 0.010 0.005Which of the following is the rate law for this reaction?(a) rate = k [NH4

+ ] [NO2 – ]4

(b) rate = k [NH4+] [NO2

– ]

(c) rate =k [NH4

+

] [NO2

–

]

2

(d) rate = k [NH4+]2 [NO2

– ]

(e) rate = k [NH4+]1/2 [NO2

– ]1/4

75. Gold sol can be prepared by(a) hydrolysis of gold (III) chloride(b) oxidation of gold by aquaregia(c) peptization(d) treating gold (III) chloride with metallic zinc

(e) reduction of gold (III) chloride with formalin solution76. The IUPAC name of the complex [Co(NH3)2(H2O)4]Cl3 is

(a) Diaminetetraaquacobalt (III) trichloride(b) Diaminetetraaquacobalt (II) chloride(c) Diaminetetraaquacobalt (III) chloride(d) Tetraaquadiaminecobalt (III) trichloride

(e) Tetraaquadiaminecobalt (II) chloride77. The products obtained by the ozonolysis of 2-ethylbut-

l-ene are

(a) propanone and ethanol(b) ethanal and 3-pentanone(c) butanal and ethanol(d) methanal and 2-pentanone(e) methanal and 3-pentanone

78. When but-2-yne is treated with Na in liquid ammonia(a) cis-2-butene is obtained(b) trans-2-butene is formed(c) n-butane is the major product(d) it rearranges to but-l-yne(e) there is no reaction

79. The correct decreasing order of reactivity for a given alkyl(R) group in both S Nl and S N2 reaction mechanisms is(a) R-I > R-Br > R-C1 > R-F

(b) R-I > R-Cl > R-Br > R-F(c) R-F > R-Cl > R-Br > R-I(d) R-F > R-I > R-Cl > R-Br (e) R-Br > R-I > R-F > R-Cl

80. The compound of molecular formula C5H10O(A) reacts withTol1en's reagent to give silver mirror but does not undergoaldol condensation. The compound A is(a) 3-pentanone (b) 2,2-dimethylpropanal

(c) 3-hydroxy-2-pentene (d) 3-methylbutanal(e) 3-methyl-2-butanone

81. When n-hexane is heated with anhydrous AlCl3 and HCl

gas the major product obtained is(a) l-chlorohexane (b) 2-chlorohexane(c) 3-chlorohexane (d) hex-3-ene(e) mixture of 2-methylpentane and 3-methylpentane

82. How many monochloro structural isomers are expected infree radical monochlorination of 2-methylbutane?(a) 2 (b) 3(c) 4 (d) 5(e) 6

83. Chloroform reacts with oxygen in the presence of light togive

(a) carbon tetrachloride (b) carbonyl chloride(c) methyl chloride (d) methylene dichloride(e) acetaldehyde




84. Which one of the following is not expected to undergoiodoform reaction?(a) Propan-2-ol (b) 1-Phenylethanol(c) 2-Butanol (d) Ethanol(e) Diphenyl methanol

85. Identify the combination of compounds that undergo

Aldol condensation followed by dehydration to produce but-2-enal(a) methanal and ethanol(b) two moles of ethanal(c) methanal and propanone(d) ethanal and propanone(e) two moles of ethanol

86. The correct increasing order of the acid strength of benzoicacid (I), 4-nitrobenzoic acid (II), 3,4-dinitrobenzoic acid (III)and 4-mcthoxybenzoic acid(IV) is(a) I < II < III < IV (b) II < I < IV < III(c) IV < I < II < III (d) IV < II < I < III

(e) I < IV < II < III87. An organic compound with the molecular formula C8H8O

forms 2,4-DNP derivative, reduces Tollen's reagent andundergoes Cannizzaro reaction. On vigorous oxidation, itgives l, 2-benzenedicarboxylic acid. The organic compoundis(a) 2-ethylbenzaldehyde (b) 2-methylbenzaldehyde(c) acetophenone (d) 3-methylbenzaldehyde(e) phenylacetaldehyde

88. Phenyl isocyanide is prepared from aniline by(a) Rosenmund's reaction(b) Kolbe's reaction

(c) Reimer-Tiemann reaction(d) Wurtz reaction(e) Carbylamine reaction

89. Choose the correct order of decreasing basic strength of the following compounds in aqueous solution(i) C6H5 NH2 (ii) C2H5 NH2(iii) NH3 (iv) (CH3)2 NH(a) (i) > (ii) > (iii) > (iv) (b) (iv) > (ii) > (iii) > (i)(c) (ii) > (i) > (iii) > (iv) (d) (iv) > (iii) > (ii) > (i)(e) (ii) > (iv) > (iii) > (i)

90. Gabriel's phthalimide synthesis can be used to prepare(a) ethanamine

(b) N-methylmethanamine(c) benzeneamine(d) N,N-dimethylmethanamine(e) p-toluidine

91. The sugar moiety present in RNA molecule is(a) -D-2-deoxyribose (b) -D-galactose(c) -D-fructofuranose (d) -D-ribose(e) -D-glucopyranose

92. Novolac, the linear polymer used in paints is(a) copolymer of 1,3-butadiene and styrene(b) obtained by the polymerization of methyl methacrylate(c) initial product obtained in the condensation of phenol

and formaldehyde in the presence of acid catalyst(d) obtained by the polymerization of caprolactum(e) copolymer of melamine and formaldehyde

93. The carbohydrate used as storage molecules in animals is(a) sucrose (b) glycogen

(c) maltose (d) glucose(e) fructose

94. Green chemistry deals with

(a) study of plant physiology

(b) study of extraction of natural products from plants(c) detailed study of reactions involved in the synthesis

of chlorophyll

(d) utilization of existing knowledge base for reducing thechemical hazards along with developmental activities

(e) synthesis of chemical compounds using green light

95. A 250 W electric bulb of 80% efficiency emits a light of 6626Å wavelength. The number of photons emitted per second by the lamp is (h = 6.626 × 10 –34 J s)

(a) 1.42 × 1017 (b) 2.l8 × 1016

(c) 6.66 × 1020 (d) 2.83 × 1016

(e) 4.25 × 1016

96. The shortest wavelength of the line in hydrogen atomicspectrum of Lyman series when R H = 109678 cm –1 is

(a) 1002.7 Å (b) 1215.67 Å

(c) 1127.30 Å (d) 911.7 Å(e) 1234.7 Å

97. The work function of a metal is 5 eV. What is the kineticenergy of the photoelectron ejected from the metal surfaceif the energy of the incident radiation is 6.2 eV?

(1 eV = 1.6 × 10 –19 J)

(a) 6.626 × 10 –19 J (b) 8.01 × 10 –19J

(c) 1.92 × 10 –18 J (d) 8.010 × 10 –18J

(e) 1.92 × 10 –19J98. The lattice energy of NaCl is 788 kJ mol –1. This means that

788 kJ of energy is required

(a) to separate one mole of solid NaCl into one mole of Na(g) and one mole of Cl(g) to infinite distance

(b) to separate one mole of solid NaCl into one mole of Na+(g) and one mole of Cl – (g) to infinite distance

(c) to convert one mole of solid NaCl into one mole of gaseous NaCl

(d) to convert one mole of gaseous NaCl into one mole of solid NaCl

(e) to separate one mole of gaseous NaCl into one moleof Na+(g) and one mole of Cl – (g) to infinite distance

99. Arrange the following species in the correct order of their stability C2, Li2, O2

+, He2+

(a) Li2 < He2+ < O2

+ < C2 (b) C2 < O2+ < Li2 < He2

+

(c) He2+ < Li2 < C2 < O2

+ (d) O2+ < C2 < Li2 < He2

+

(e) C2 < Li2 < He2+ < O2

+

100. Molecular formulae and shapes of some molecules are given below. Choose the incorrect match

Formula Shape

(a) NH3 – Trigonal pyramidal(b) SF4 – Tetrahedral

(c) ClF3 – T-shaped(d) PCl5 – Trigonal bipyramidal

(e) BF3 – Trigonal planar




101. Potassium dichromate belongs to which crystal system?(a) Tetragonal (b) Orthorhombic(c) Triclinic (d) Hexagonal(e) Monoclinic

102. If two moles of an ideal gas at 500 K occupies a volumeof 41 litres, the pressure of the gas is (R = 0.082 L atm K –1

mol –1)(a) 2 atm (b) 3 atm(c) 4 atm (d) 5 atm(e) l atm

103. At 273 K, the density of a certain gaseous oxide at 2atmosphere is same as that of dioxygen at 5 atmosphere.The molecular mass of the oxide (in g mol –1) is(a) 80 (b) 64(c) 32 (d) 160(e) 70

104. The reaction of H2 is given belowH2 + CO + R – CH = CH2 R – CH2 – CH2 – CHO

is specifically called as(a) hydrogenation (b) reduction(c) hydroformylation (d) dehydration(e) formylation

105. Which of the following are isoelectronic species?(i) NH3 (ii) CH3

+

(iii) NH2 – (iv) NH4

+

Choose the correct answer from the codes given below(a) (i), (ii), (iii) (b) (ii), (iii), (iv)(c) (i), (ii), (iv) (d) (i), (iii), (iv)(e) (ii), (iii)

106. The salt of an alkali metal gives violet colour in the flametest. Its aqueous solution gives a white precipitate with barium chloride in hydrochloric acid medium. The salt is(a) K 2SO4 (b) KCl(c) Na2SO4 (d) K 2CO3(e) Li2SO4

107. In which one of the following the central atom is sp3

hybridized?(a) NH4

+ (b) BF3(c) SF6 (d) PCl5(e) XeF4

108. Which one of the following statements is not true in respect

of properties of interhalogen compounds?(a) They are all covalent compounds(b) They are volatile solids or liquids except ClF(c) IF5 has square pyramidal structure(d) They are all paramagnetic in nature(e) BrF3 is used in the preparation of UF6 in the

enrichment of 235U109. Which one of the following is an incorrect statement?

(a) O3 oxidises PbS to PbSO4(b) O3 dioxide nitric oxide to nitrogen dioxide(c) O3 oxidises aqueous KI at pH = 9.2(d) The two oxygen-oxygen bond lengths in O3 are

different(e) O3 is used as an oxidizing agent in the manufacture of KMnO4

110. The correct descending order of oxidizing power of thefollowing is(a) Cr 2O7

2– > MnO4 – > VO2

+

(b) MnO4 – > Cr 2O7

2– > VO2+

(c) VO2+ > MnO4

– > Cr 2O72–

(d) MnO4 > VO2+ > Cr 2O7

2–

(e) Cr 2O72– > VO2

+ > MnO4 –

111. The number of electrons that are involved in the reductionof permanganate to manganese(II) salt, manganate andmanganese dioxide respectively are(a) 5, 1, 3 (b) 5, 3, 1(c) 2, 7, 1 (d) 5, 2, 3(e) 2, 3; 1

112. The calculated magnetic moment of a divalent ion of anatom with atomic number 24 in aqueous solution is(a) 4.90 BM (b) 5.92 BM(c) 3.87 BM (d) 2.84 BM

(e) 1.73 BM113. The entropy of vaporization of a liquid is 58 JK –1 mol –1.

If 100 g of its vapour condenses at its boiling point of l23°C,the value of entropy change for the process is(Molar mass of the liquid = 58 g mol –1)(a) –100 JK –1 (b) 100 JK –1

(c) –123 JK –1 (d) 123 JK –1

(e) 1230 JK –1

114. The values of limiting ionic conductance of H+ and HCOO –

ions are respectively 347 and 53 Scm2mol –1 at 298 K, If themolar conductance of 0.025M methanoic acid at 298 K is 40Scm2mol –1, the dissociation constant of methanoic acid at

298 K is(a) l × l0 –5 (b) 2 × 10 –5

(c) l.5 × l0 –4 (d) 2.5 × l0 –5

(e) 2.5 × l0 –4

115. In a closed cylinder of capacity 24.6 L the following reactionoccurs at 27°C

A2(s) B2(s) + 2C(g). At equilibrium lg of B2(s)(molar mass = 50 g mol –1) is present. The equilibriumconstant K p for the equlibrium in atm2 unit is(R = 0.082 L atm K –1 mol –1)(a) 1.6 × 10 –2 (b) 1.6 × 10 –5

(c) 1.6 × l0 –3

(d) 1.6 × 10 –4

(e) l.6 × 10 –1

116. The pH of a saturated solution of a metal hydroxide of formula X(OH)2 is 12.0 at 298 K. What is the solubility product of the metal hydroxide at 298 K (in mol3 L –3)?(a) 2 × 10 –6 (b) 1 × l0 –7

(c) 5 × 10 –5 (d) 2 × 10 –5

(e) 5 × l0 –7

117. An aqueous solution containing 3g of a solute of molar mass 111.6 g mo1 –1 in a certain mass of water freezes at – 0.125°C. The mass of water in grams present in the solutionis (K f = 1.86 K kg mol –1)

(a) 300 (b) 600(c) 500 (d) 400(e) 250




118. A sample of sea water contains 5 × 10 –3 g of dissolvedoxygen in 1 kilogram of the sample. The concentration of O2 in that sea water sample in ppm is

(a) 5 × 10 –4 (b) 5 × l0 –3

(c) 5 × 10 –2 (d) 5 × 10 –1

(e) 5119. The change in potential of the half-cell Cu2+|Cu, when

aqueous Cu2+ solution is diluted 100 times at 298 K?

2.303 RT0.06

F

(a) increases by 120 mV (b) decreases by 120 mV(c) increases by 60 mV (d) decreases by 60 mV(e) no change

120. Consider the following electrolytic cells(i) M(s) | M2+(aq), 0.1M || X2+(aq), 0.01M | X(s)(ii) M(s) | M2+(aq), 0.1M || X2+(aq), 0.1M | X(s) and

(iii) M(s) | M2+(aq), 0.011M || X2+(aq), 0.1M | X(s)The cell EMF of the above cells are E1, E2 and E3respectively. Which one of the following is true?(a) E1> E2 > E3 (b) E2 >E3 > E1(c) E3> E1> E2 (d) E1 >E3 > E2(e) E3 >E2 > E1

Paper-II : Mathematics

1. If the operation is defined by a b = a2 +b2 for all real

numbers a and b, then (2 3) 4 =

(a) 120 (b) 185(c) 175 (d) 129

(e) 3122. The number of students who take both the subjects

mathematics and chemistry is 30. This represents 10% of the enrolment in mathematics and 12% of the enrolment inchemistry. How many students take at least one of thesetwo subjects?(a) 520 (b) 490

(c) 560 (d) 480(e) 540

3. Let f(x) = |x – 2| , where x is a real number. Which one of thefollowing is true?(a) f is Periodic. (b) f (x + y) = f (x) + f (y)

(c) f is an odd function (d) f is not a 1–l function

(e) f is an even function4. lf A = {l,3,5,7} and B = {l,2,3,4,5,6, 7,8}, then the number of

one–to–one functions from A into B is

(a) 1340 (b) 1860(c) 1430 (d) 1880 (e) 1680

5. The range of the function f (x) = x2 + 2x+ 2 is

(a) (1, ) (b) (2, )

(c) (0, ) (d) [1, ) (e) ( – , )

6. If f (x) = x and g(x) = 2x – 3, then domain of (f o g) (x) is

(a) ( , 3) (b)3

,2

(c)3

,02

(d)3

0,2

(e)3

,2

7. If z =3

2

( 3 ) (3 4)

(8 6 )

i i

i, then |z| is equal to

(a) 8 (b) 2(c) 5 (d) 4(e) 10

8. Let 1 be a complex number. If || = l and z =1

1

,

then Re(z) is equal to

(a) l (b)1

| 1|

(c) Re() (d) 0

(e)

9. If z =2 i3e

, then l + z + 3z2 + 2z3 + 2z4 +3z5 is equal to(a) –3ei/3 (b) 3e i/3

(c) 3e 2i/3 (d) –3e2i/3

(e) 0

10. If z1= 2 2 (l + i) and z2 =l + i 3 , then zl2 z2

3 is equal to(a) 128 i (b) 64i(c) – 64 i (d) –128 i(e) 256

11. If the complex numbers zl, z2 and z3 denote the vertices of an isosceles triangle, right angled at zl, then

(zl –z2)2 + (z1 –z3)2 is equal to(a) 0 (b) (Z2 + Z3 )2

(c) 2 (d) 3(e) (z2 – z3 )2

12. If the roots of x2 – ax + b = 0 are two consecutive oddintegers, then a2 – 4b is(a) 3 (b) 4(c) 5 (d) 6(e) 7

13. lf and are the roots of x2 – ax + b2 = 0, then 2 + 2 isequal to(a) a2 + 2b2 (b) a2 – 2b2

(c) a2 – 2b (d) a2 +2b(e) a2 – b2




14. If and , are the roots of the equation x2 + 3x – 4 = 0, then

1 1

is equal to

(a)3

4

(b)

3

4

(c)4

3

(d)

4

3

(e)3

2

15. The value of x such that 32x – 2(3x+2) + 81 = 0 is(a) 1 (b) 2

(c) 3 (d) 4

(e) 516. If the roots of the equation x2 + 2bx + c = 0 are and ,

then b2 – c =

(a)2( )

4

(b) 2( )

(c) 2( ) (d)2( )

2

(e)2( )

2

17. The equation whose roots are the squares of the roots of the equation 2x2 + 3x + 1 = 0 is

(a) 4x2 + 5x + 1 = 0 (b) 4x2 – x + 1 = 0 (c) 4x2 – 5x – 1=0 (d) 4x2 – 5x + 1 = 0 (e) 4x2 + 5x –1= 0

18. The sum of the series17

n 8

1

(n 2)(n 3) is equal to

(a)1

17(b)

1

18

(c)1

19(d)

1

20

(e)1

21

19. If two positive numbers are in the ratio 3+2 2 : 3 – 2 2 ,then the ratio between their A.M. and G.M. is(a) 6 : 1 (b) 3 : 2

(c) 2 : 1 (d) 3 : l(e) 1 : 6

20. Let x1, x2 + x3……,xn be in an A.P. If xl + x4 + x9 + x11 + x20+ x22 + x27+x30 = 272, then x1 +x2 +x3 +…….+x30 is equal to

(a) 1020 (b) 1200(c) 716 (d) 2720(e) 2072

21. If the second and fifth terms of a G.P. are 24 and 3respectively, then the sum of first six terms is

(a) 181 (b)181

2

(c) 189 (d)189

2(e) 19122. If the sum of first 75 terms of an A.P. is 2625, then the 38th

term of the A.P. is(a) 39 (b) 37(c) 36 (d) 38(e) 35

23. If –5, k, –1 are in A.P., then the value of k is equal to(a) –5 (b) –3(c) –1 (d) 3(e) 5

24. Let Tn denote the number of triangles which can be formed

by using the vertices of a regular polygon of n sides. If Tn+1 – Tn = 36, then n is equal to(a) 2 (b) 5(c) 6 (d) 8(e) 9

25. The middle term in the expansion of1010

10

x

x is

(a) 10C5 (b) 10C6

(c) 10C5 10

1

x(d) 10C5 x10

(e) 10C5

1010

26. The coefficient of x49 in the product (x – 1)(x –2)(x – 3)……… (x –50) is(a) –2250 (b) –1275(c) 1275 (d) 2250(e) –49

27. The sum of the coefficients in the binomial expansion of 6

12x

x

is equal to

(a) 1024 (b) 729(c) 243 (d) 512(e) 64

28. The value of 2P1+3P1+...+ nP1 is equal to

(a)2 2

2

n n(b)

2 2

2

n n

(c)2 1

2

n n(d)

2 1

2

n n

(e)2 2

2

n n

29. How many four digit numbers abcd exist such that a is odd,b is divisible by 3, c is even and d is prime?(a) 380 (b) 360

(c) 400 (d) 520(e) 480




30. If a1, a2, a3,….are in A.P., then the value of

1 2

2 3

3 4

1

1

1

a a

a a

a a

is

equal to

(a) a4 – a1 (b) 1 42a a

(c) 1 (d) 2 3

2

a a

(e) 0

31. If1 1

2 2

3 3

2

2

2

a x y

b x y

c x y

= 02

abc

, then the area of the triangle

whose vertices are 1 1 2 2, , ,

x y x y

a a b b and 3 3,

x y

c c is

(a)1

4 abc (b)

1

8 abc

(c)1

4(d)

1

8

(e)1

12

32. The system of linear equations 3x+y –z = 2, x–z =l and2x + 2y + az = 5 has unique solution when

(a) a 3 (b) a 4(c) a 5 (d) a 2(e) a 1

33. If A =2 2

1 3

k

k is a singular matrix, then the value of

5k–k 2 is equal to(a) 0 (b) 6(c) – 6 (d) – 4(e) 4

34. If a,b,c are non–zero and different from l, then the value of

log 1 log log1 1

log log 1 log

1log log log 1

a a a

a b a

a a c

b c

b c

cc

is

(a) 0 (b) l+loga(a+b+c)(c) loga (ab + bc + ca) (d) 1(e) loga (a+b+c)

35. The number of solutions for the system of equations2x + y = 4, 3x+ 2y = 2 and x + y = – 2 is(a) 1 (b) 2

(c) 3 (d) infinitely many(e) 0

36. The number of solutions of the inequation|x – 2 | + |x + 2| < 4 is(a) 1 (b) 2(c) 4 (d) 0(e) infinite

37. The shaded region shown in the figure is given by the

inequations

(0, 14) (19, 14)

(15, 0)

Y

X

(a) 14x + 5y 70, y 14 and x – y 5

(b) 14x + 5y 70, y 14 and x – y 5

(c) 14x + 5y 70, y 14 and x – y 5

(d) 14x + 5y 70, y 14 and x – y 5(e) 14x + 5y 70, y 14 and x – y 5

38. Let p, q and r be any three logical statements. Which one of the following is true?

(a) ~ [ (~ )] (~ ) p q p q

(b) ~ [( ) (~ ) (~ ) (~ ) (~ ) p q r p q r

(c) ~ [ (~ )] (~ ) p q p q

(d) ~ [ (~ )] (~ ) ~ p q p q

(e) ~ [ (~ )] p q p q

39. The truth values of p, q and r for which ( p q) (~ r ) hastruth value F are respectively(a) F, T, F (b) F, F, F(c) T, T, T (d) T, F, F(e) F, F, T

40. ~ [(~ ) )] p q is logically equivalent to

(a) ~ ( ) p q (b) ~ [ (~ )] p q

(c) (~ ) p q (d) (~ ) p q

(e) (~ ) (~ ) p q

41. Let 0,2

. Which one of the following is true?

(a) sin2 >cos2 (b) sin2 < cos2 (c) sin > cos (d) cos >sin

(e) sin + cos 2

42. The value of sin –1 12 2 1sin

3 3

is equal to

(a)6

(b)

4

(c) 2

(d)

2

3

(e) 0




43. If ab < l and cos –1

2

2

1

1

a

a + cos –1

2

2

1 b

1 b

= 2tan –1 x,

then x is equal to

(a) 1

a

ab (b) 1

a

ab

(c)1

a b

ab(d)

1

a b

ab

(e)1a+b

ab

44. The value of tan(1°) +tan (89°) is equal to

(a)1

sin1 (b)

2

sin2

(c) 2sin1 (d) 1sin2

(e)sin2

2

45. Let sn = cos10

n, n = 1, 2, 3, .... Then the value of

1 2 10

1 2 10

. .....

.... s s s

s s s is equal to

(a)

1

2 (b)

3

2

(c) 2 2 (d) 0

(e)1

2

46. cos –1 7

cos5

=

(a)3

5

(b)

2

5

(c)75

(d)75

(e)2

5

47. The value of sec2 (tan –13) + cosec2 (cot –1 2) is equal to(a) 5 (b) 13(c) 15 (d) 23(e) 25

48. If sin + cosec = 2, then the value of sin 6 + cosec6 isequal to

(a) 0 (b) 1(c) 2 (d) 23

(e) 26

49. If 0 < x < , then

sin8 7 sin 6 18sin 4 12sin 2

sin 7 6sin 5 12sin 3

x x x x

x x x =

(a) 2 sin x (b) sin x(c) sin 2x (d) 2 cos x

(e) cos x50. The points (2,5) and (5,l) are the two opposite vertices of arectangle. If the other two vertices are points on the straightline y = 2x + k, then the value of k is(a) 4 (b) 3(c) – 4 (d) –3(e) 1

51. The circumcentre of the triangle with vertices (8, 6), (8,–2)and (2, –2) is at the point(a) (2, –1) (b) (1, –2)(c) (5, 2) (d) (2, 5)(e) (4, 5)

52. The ratio by which the line 2x + 5y – 7 = 0 divides thestraight line joining the points (–4, 7) and (6,–5) is(a) 1 : 4 (b) 1 : 2(c) 1 : 1 (d) 2 : 3(e) 1 : 3

53. The number of points (a,b), Where a and b are positiveintegers lying on the hyperbola x2 – y2 = 512 is(a) 3 (b) 4(c) 5 (d) 6(e) 7

54. If p is the length of the perpendicular from the origin to the

line whose intercepts with the coordinate axes are1

3

and

1

4 then the value of p is

(a)3

4(b)

1

12(c) 5 (d) 12

(e)1

555. The slope of the straight line joining the centre of the

circle x2 + y2 –8x+ 2 y = 0 and the vertex of the parabolay = x2 – 4x +10 is

(a) 52

(b) 72

(c)3

2

(d)

5

2

(e)7

256. A straight line perpendicular to the line 2x + y = 3 is passing

through (l,1). Its y–intercept is(a) 1 (b) 2

(c) 3 (d)1

2

(e)1

3




57. If p and q are respectively the perpendiculars from the originupon the straight lines whose equations arex sec + ycosec = a and xcos –ysin = a cos 2, then4p2 + q 2 is equal to(a) 5a2 (b) 4a2

(c) 3a2 (d) 2a2

(e) a2

58. The shortest distance between the circles(x–1)2 +(y+2)2 =1 and (x + 2)2 + (y – 2)2 = 4 is(a) 1 (b) 2(c) 3 (d) 4(e) 5

59. The centre of the circle whose radius is 5 and which touchesthe circle x2 + y2 – 2x – 4y – 20 = 0 at (5, 5) is(a) (l0, 5) (b) (5 , 3)(c) (5 , 10) (d) (8 , 9)(e) (9, 8)

60. A circle passes through the points (0,0) and (0,l) and alsotouches the circle x2 + y2 = 16. The radius of the circle is(a) 1 (b) 2(c) 3 (d) 4(e) 5

61. A circle of radius 8 is passing through origin and the

point (4, 0). If the centre lies on the line y = x, then theequation of the circle is(a) (x–2)2 + (y–2)2 = 8 (b) (x+2)2 + (y+2)2 = 8(c) (x–3)2 + (y–3)2 =8 (d) (x+3)2 + (y+3)2 = 8(e) (x–4)2 + (y–4)2 = 8

62. The parametric form of the ellipse 4(x+ l)2 + (y –l)2 = 4 is

(a) x = cos – 1, y = 2 sin – 1(b) x = 2cos – 1, y = 2 sin +1(c) x = cos – 1, y = 2 sin + 1(d) x = cos + 1, y = 2 sin + 1(e) x = cos + 1, y = 2 sin – 1

63. A point P on an ellipse is at a distance of 6 units from a

focus. lf the eccentricity of the ellipse is3

5, then the distance

of P from the corresponding directrix is

(a)8

5

(b)5

8(c) 10 (d) 12(e) 15

64. If the length of the latus rectum and the length of transverse

axis of a hyperbola are 4 3 and 2 3 respectively, then

the equation of the hyperbola is

(a)2 2

13 4

x y

(b)2 2

13 9

x y

(c)2 2

16 9

x y

(d)2 2

16 3

x y

(e)2 2

13 6

x y

65. lf the eccentricity of the hyperbola2 2

2 21

x y

a bis 5

4 and

2x + 3y – 6 = 0 is a focal chord of the hyperbola, then thelength of transverse axis is equal to

(a)

12

5 (b) 6

(c)24

7(d)

24

5

(e)12

7

66. The length of the transverse axis of a hyperbola is 2 cos . The foci of the hyperbola are the same as that of the ellipse9x2 + 16y2 = 144. The equation of the hyperbola is

(a)2 2

2 21

cos 7 cos

x y

(b)2 2

2 21

cos 7 cos

x y

(c)2 2

2 21

1 cos 7 cos

x y

(d)2 2

2 21

1 cos 7 cos

x y

(e)2 2

2 21

cos 5 cos

x y

67. lf ˆˆ ˆ2 2 ,| |

a i j k b = 5 and the angle between a and

b

is6

, then the area of the triangle formed by these two

vectors as two sides is

(a)15

4(b)

15

2

(c) 15 (d)15 3

2

(e) 15 3

68. If .

a b = 0 and

a b makes an angle of 60° with

a , then(a) | | 2 | |

a b (b) 2 | | | |

a b

(c) | | 3 | |

a b (d) | | | |

a b

(e) 3 | | | |

a b

69. If ˆ ˆˆ ˆ ˆ ˆ, , i j j k i k are the position vectors of the vertices

of a triangle ABC taken in order, then A is equal to

(a)2

(b)

5

(c)

6

(d)

4

(e)3




70. Let ˆˆ ˆˆ 2 3 a i j k . If b is a vector such that 2ˆ ˆˆ. | |a b b

and ˆˆ| | 7 a b , then ˆ| |b =

(a) 7 (b) 3(c) 7 (d) 3

(e) 7 3

71. If ˆˆ,a b and c are three non–zero vectors such that each

one of them being perpendicular to the sum of the other

two vectors, then the value of 2ˆˆ ˆ| | a b c is

(a) 2 2 2ˆˆ ˆ| | | | | | a b c (b) ˆˆ ˆ| | | | | | a b c

(c) 2 2 2ˆˆ ˆ2(| | | | | | ) a b c (d)2 2 2ˆˆ ˆ(| | | | | | )

2

a b c

(e) 0

72. Let, , u v and

w be vectors such that 0

u v w . If | |u = 3,

v = 4 and | |w = 5 then . . . u v v w w u =

(a) 0 (b) – 25(c) 25 (d) 50(e) 47

73. If ˆˆ ˆ(3 2 6 ) i j k is a unit vector, then the values of are

(a) 1

7 (b) 7

(c) 43 (d)1

43

(e) 1

7

74. If the direction cosines of a vector of magnitude 3 are

2 2, , , 0

3 3 3

aa , then the vector is

(a) ˆˆ ˆ2 2 i j k (b) ˆˆ ˆ2 2 i j k

(c) ˆˆ ˆ2 2 i j k (d) ˆˆ ˆ2 2 i j k

(e) ˆˆ ˆ2 2 i j k

75. Equation of the plane through the mid–point of the linesegment joining the points P(4,5,–10) and Q(–l,2,l) and perpendicular to PQ is

(a)3 7 9 ˆˆ ˆ. 452 2 2

r i j k

(b) 135ˆˆ ˆ. 22

r i j k

(c) 135ˆˆ ˆ. 5 3 11 02

r i j k

(d) ˆˆ ˆ. 4 5 10 85 r i j k

(e) 135ˆˆ ˆ. 5 3 11

2

r i j k

76. The angle between the straight lines x–1=2 3 5

3 2

y z

and x = 3r + 2; y = – 2r – 1, z = 2, where r is a parameter, is

(a)

4

(b) 1 3cos

182

(c) 1 3sin

182

(d)2

(e) 077. Equation of the line through the point (2,3,l) and parallel to

the line of intersection of the planes x – 2y – z + 5 = 0 and x+ y + 3z = 6 is

(a)2 3 1

5 4 3

x y z

(b) 2 3 1

5 4 3

x y z

(c) 2 3 15 4 3 x y z (d) 2 3 14 3 2 x y z

(e) 2 3 1

4 3 2

x y z

78. A unit vector parallel to the straight line

2 3 2

3 1 4

x y z

is

(a)1 ˆˆ ˆ(3 4 )26

i j k (b)1 ˆˆ ˆ( 3 )26

i j k

(c)1 ˆˆ ˆ(3 4 )26

i j k (d)1 ˆˆ ˆ(3 4 )26

i j k

(e)1 ˆˆ ˆ( 3 4 )26

i j k

79. The angle between a normal to the plane 2x – y + 2z – 1 = 0and the z–axis is

(a) 1 1cos

3

(b) 1 2sin

3

(c)1 2

cos 3

(d)1 1

sin 3

(e) 1 3sin

5

80. Foot of the perpendicular drawn from the origin to the plane2x – 3 y + 4z = 29 is

(a) (5, –1, 4) (b) (7, –1, 3)(c) (5, –2, 3) (d) (2, –3, 4)(e) (1, –3, 4)

81. The distance between the x–axis and the point (3, 12, 5) is(a) 3 (b) 13

(c) 14 (d) 12(e) 5




82. If9

1 i (xi – 5) = 9 and

9

1 i (xi – 5)2 = 45, then the

standard deviation of the 9 items x1, x2, ...., x9 is

(a) 9 (b) 4

(c) 3 (d) 2

(e) 1

83. If two dice are thrown simultaneously, then the probabilitythat the sum of the numbers which come up on the dice to be more than 5 is

(a) 5

36(b)

1

6

(c) 5

18(d)

7

18

(e) 13

18

84. Let A and B be two events such thatP(A B) = P(A) + P(B) – P(A) P(B). If 0 < P(A) < 1 and 0 <P(B) < 1, then P(A B)' =(a) 1– P(A) (b) 1 – P(A')

(c) 1– P (A) P(B) (d) [1– P (A)] P (B')

(e) 1

85. The standard deviation of 9, 16, 23, 30, 37, 44, 51 is

(a) 7 (b) 9

(c) 12 (d) 14

(e) 16

86. The value of 5 58 83

3lim3

x

x

x is equal to

(a)5

8(b)

5

64

(c)5

216(d) 1

27

(e)1

63

87. Let f (x) = (x5 – 1)(x3+1), g(x) = (x2 – 1)(x2 – x+1) and

let h (x) be such that f(x) = g(x)h(x). Then 1lim x h(x) is

(a) 0 (b) 1

(c) 3 (d) 4(e) 5

88.2

50

log(1 3 )lim

( 1)

x x

x

x e

(a)3

5(b)

5

3

(c)3

5

(d)

5

3

(e) 1

89. If f (x) =2

3 1

x

x, then f (f(x)) is

(a) x (b) –x

(c)1

x(d)

1

x

(e) 0

90. Let f (x) =2

3, 2

1, 2

ax x

a x x. Then the values of a for which f

is continuous for all x are(a) 1and –2 (b) 1 and 2(c) –1 and 2 (d) –1 and –2(e) 0 and 3

91. Let R be the set of all real numbers. Let f : R R be a

function such that |f (x)– f (y)|2 |x–y|3 , x y R. Then

f ' (x) =

(a) f (x) (b) 1(c) 0 (d) x2

(e) x

92. Let f (x) = 2

1sin

2

x t dt. Then the value of

0

( ) ( )lim

x

f x f

x is equal to

(a)1

4(b)

1

2

(c) 34

(d) 1

(e) 0

93. If y = f (x2 + 2) and f (3) = 5, thendy

dx at x = 1 is

(a) 5 (b) 25(c) 15 (d) 20(e) 10

94. Let f(x) = x2 + bx + 7. If f (5) = 2f 7

2

, then the value of b

is(a) 4 (b) 3(c) –4 (d) –3(e) 2

95. If x= sin t and y = tan t, thendy

dx=

(a) cos3 t (b) 3

1

cos t

(c) 2

1

cos t (d) sin2 t

(e) 2

1

sin t




96. lf x = a cos3 and y = a sin3 , then l +2

dy

dx is

(a) tan (b) tan2 (c) 1 (d) sec2

(e) sec

97. If y = sin –l 2(2 1 ) x x , –

1 1

2 2 x , then

dy

dx is equal to

(a) 21

x

x(b) 2

1

1 x

(c) 2

2

1 x(d) 2

2

1

x

x

(e) 2

2

1

x

x98. A straight line parallel to the line 2x – y + 5 = 0 is also a

tangent to the curve y2 = 4x + 5. Then the point of contactis(a) (2, l) (b) ( –1, 1)(c) (1, 3) (d) (3, 4)(e) (– 1, 2)

99. The function f (x) = 2x3 – l5x2 + 36x + 6 is strictly decreasingin the interval(a) (2, 3) (b) (– , 2)(c) (3, 4) (d) (– , 3) (4, )

(e) (– , 2) (3, )100. The slope of the tangent to the curve y2exy = 9e –3x2 at( –1, 3) is

(a)15

2

(b)

9

2

(c) 15 (d)15

2

(e)9

2

101. The radius of a cylinder is increasing at the rate of 5 cm/min

so that its volume is constant. When its radius is 5 cm andheight is 3 cm, the rate of decreasing of its height is(a) 6 cm/min (b) 3 cm/min(c) 4 cm/min (d) 5 cm/min(e) 2 cm/min

102. The function f (x) =22 1 1 4

151 30 4 5

x if x

x if x is not suitable

to apply Rolle’s theorem since(a) f (x) is not continuous on [1, 5]

(b) f (1) f (5)

(c) f (x) is continuous only at x = 4(d) f (x) is not differentiable in (4, 5)(e) f (x) is not differentiable at x = 4

103. The slope of the normal to the curve y = x2 –2

1

x at ( –1, 0)

is

(a)1

4(b)

1

4

(c) 4 (d) –4

(e) 0

104. The minimum value of sin x + cos x is

(a) 2 (b) 2

(c)1

2(d)

1

2

(e) 1

105.3

2 4 4

1

( 1) dx

x x

is equal to

(a)

34 4(1 )

x

C x

(b)

14 4(1 )

2

xC

x

(c)

14 4(1 )

x

C x

(d)

14 4

2

(1 )

xC

x

(e)

14 2

(1 ) x C x

106.2

(1 )

sin ( )

x

x

x e

xe dx is equal to

(a) – cot (ex) + C (b) tan (xex) + C

(c) tan (ex) + C (d) cot (xex) + C

(e) – cot (xex) + C

107.2(1 )

x xedx

x is equal to

(a)( 1)

xeC

x(b)

1

xe C

x

(c)1

x xe

C x

(d)1

x xe

C x

(e) 2( 1)

xe

C x

108. xe (sin x + 2 cos x) sin x dx is equal to

(a) ex cos x + C (b) ex sin x + C

(c) ex sin2 x + C (d) ex sin 2x + C

(e) ex (cos x + sin x) + C




109. 1 cos x dx is equal to

(a) 2sin2

xC (b) 2sin

2

xC

(c)1

sin2 2 x

C (d) 2 sin2 2 x C

(e) 2 2 sin2

xC

110.2 1

x

dx x

is equal to

(a) 2 11 sec x x C (b) 2 11 tan x x C

(c) 2 11 sec x x C (d) 2 1 tan x x C

(e) 2 1 sec x x C

111.2

4

5

xdx

xis equal to

(a)

3

2

2

1 51

15

C x

(b)

3

2

2

1 11

15

C x

(c)

32

2

1 51

15

C x

(d)

3

2

2

1 11

15

C x

(e)

32

2

1 11

10

C x

112. The value of1

0 x

dx

e e is equal to

(a)1 1

log2

e

e(b)

1log

2

e

(c)1

log(1 ) e

e

(d)2

log1

e

(e)1 2

log1

e e

113. Area bounded by the curves y = ex, y = e –x and the straightline x = 1 is (in sq. units)

(a)1

ee

(b)1

2 ee

(c)1

2 ee

(d)1

2 ee

(e)1ee

114. The value of the integral1

1 log

3

e

xdx

x is equal to

(a)1

4(b)

1

2

(c)3

4 (d) e

(e)1

e

115. The value of the integral1 3

80 1

xdx

x is equal to

(a)8

(b)

4

(c)16

(d)

6

(e) 12

116. The value of the integral4

2

log

t dt

t is equal to

(a)1

2(log 2)2 (b)

5

2(log 2)2

(c)3

2(log 2)2 (d) (log 2)2

(e)3

2 (log 2)

117. The solution of the differential equation(kx – y2)dy = (x2 – ky) dx is(a) x3 – y3 = 3kxy + C (b) x3 + y3 = 3kxy+C(c) x2 – y2 = 2kxy + C (d) x2 + y2 = 2kxy + C(e) x3 – y2 = 3kxy+C

118. The solution of the differential equation dy

dx = ex + l is

(a) y = ex + C (b) y = x + ex + C(c) y = xex +C (d) y = x(ex + 1) + C(e) y = ex + Cx

119. The order and degree of the differential equation3

2

22

d y dy

dxdx= y are respectively

(a) l, l (b) 1, 2(c) 1, 3 (d) 2, 1(e) 2, 2

120. An integrating factor of the differential equation

sin xdy

dx + 2 y cos x = 1 is

(a) sin2 x (b)2

sin x

(c) log |sin x| (d) 2

1

sin x

(e) 2 sinx




Paper-I : Physics Chemistry

1. (d) Given : Number of cells, n = 5, emf of each cell = E Internal resistance of each cell = r

In series, current through resistance R

I =5

5

nE E

nr R r R

In parallel, current through resistance R

I =5

5

E nE E

r r nR r R R

n

According to question, I = I'

5 5

5 5 5

E E

r R r R

5r + R = r + 5 R

or R = r 1 R

r

2. (b) If V is the voltage across R and I the current through it,then power P delivered via transmission cables

P = VI or, I =P

V The power dissipated in the transmission cables

Pc = I 2 R =2

2

P R

V i.e., Pc

2

1

V

3. (a) Change in internal energy of the gas = Heat produced

due to current flowingGiven, I = 1A, R = 5t = 2 min = 120 sU = I 2 Rt = (1A)2 (50 ) (120 s) = 6 × 103 J = 6 kJ

4. (d) Inside a long solenoid, magnitude of magnetic field isincreased by introducing higher permeability medium.

5. (e) Given, I = 9 × 10 –5 kg m2, B = 162 × 10 –5 T

T =15 3

s20 4

In a vibration magnetometer

Time period, T = 2 I

MB or M =

2

2

4 I

BT

2 5

22

2 5

4 9 104 A m

316 10

4

6. (d)7. (c) Kinetic energy of the proton accelerated by cyclotron,

K =2 2 2

2

q B r

m

As =

2

qB

m

or qB = 2m

K =(2 )

2

m r

m

2mr

8. (b) Work done in rotating the magnetic dipole from position 1= 0° to 2= 180°

W = MB (cos1 – cos2) W = MB (cos – cos 180°) = 2 MB

9. (c) Lenz's law gives the polarity of induced emf.10. (a) As we know,

At resonance, inductive reactance X L = capacitivereactance X C

In an LCR series circuit, the phase difference betweenthe current and voltage

tan C L X X

R

[ Here X C – X L = 0]

i.e., the current and voltage are in phase.11. (b) Given, B = 0.01 T , A = R2 = × (1 m)2 = m2

= 100 rads –1

The maximum induced emf max= BA= 0.01××100 V = V

12. (e) Given I p = 4 A, I s, = 24 A, N p = 330, N s = ?

As p ps

s p p s s

N I I N N

I N I

N s =4

330 5524

13. (d) The average energy density due to magnetic field

u B =22

rms 0

0 0

1 1

2 2 2

B B

0rms

2

B B

=2 20 0 0

00 0

( / )1 1

4 4

B E c E B

c

=2 20 0

0 0 0 0 00

1 1 1

4 4 (1/ )

E E c

c

= 2 12 20 0

1 18.854 10 (1)

4 4 E

= 2.21 × 10 –12 J m –3

14. (b) The energy is equally divided between electric andmagnetic fields.The electric and magnetic field are in the same phase.The direction of propagation is perpendicular to bothelectric and magnetic fields. The pressure exerted bythe wave is equal to the energy density of the wave.The speed of the wave Vwave = E/B.

15. (a) The apparent flattening (oval shape) of the sun atsunset and sunrise is due to refraction.

16. (c) Given,polarising angle i p = 60°We have to find critical angle C

According to Brewster's law, = tani p

= tan 60° = 3

HINTS SOLUTIONS




And, sin C =1 1

3

or C = sin –1 1

3 17. (b) For rear view, Convex mirror is used in vehicle

18. (a) As we know, Fringe width, = D

d

Hence with increase of wavelength of the sourcefringe width increases.

19. (b) Given wavelength = 5000 Å = 5 × 10 –7 mslit width a = 2.5 × 10 –4 cm = 2.5 × 10 –6 m, n (secondminima) = 2

sinn =n

a

Angular position of nth minimum from the centralmaximum

sin 2 =7

6

2 × 5 ×10 m 2

52.5 ×10 m

or 2 = sin –1 2

5

20. (e) As we know, de Broglie wavelength, =2

h

mqV

Since potential (V ) is same for both the particles

1

mq

Thus, pe

p e

m

m

( q p = qe)

21. (c) Given, Half life period, T 1/2 = 20 minutesFor 50% decay 1 half lifeFor 87.5% decay= 50% + 25% + 12.5%decay 3 half lifeHence interval(50% – 87.5% decay)

= 2 half life= 2 × 20 = 40 minutes22. (d) As we know, radius of nucleus, R = R0 A

1/3

where R0 is a constant and A is the mass number

1/3 1/3Te Te

Al Al

125 5

27 3

R A

R A

2 2

Te Te

Al Al

5 25

3 9

S R

S R

23. (d) According to Einstein's photoelectric equation, the

kinetic energy (k) of the emitted photoelectronK = h – 00 is a work function of the metal.

If the frequency of incident radiation () is doubled,thenKinetic energy K = 2hv – 0 = 2(hv – 0) + 0= 2K + 0i.e., K > 2K

24. (a) The significant result deduced from the Rutherford's

scattering is that whole of the positive charge isconcentrated at the centre of atom i.e. nucleus.

25. (c) On an average 2.5 neutrons are released per fission of the uranium atom.And the energy of the neutron released per fission of the uranium atom is 2 MeV.

26. (e) A

B

C

Y

1

1

1

0

1

0

OR GateAND Gate

27. (e) According to question, output resistance, R0 = 10 k

Input resistance, Ri = 2 k and = 49

Voltage gain, AV = × 0 1049 245

2i

R

R

28. (c) The light emitting diode is a heavily doped p-n junctionwhich emits spontaneous radiation under forward bias.

29. (e) A point-to-point communication mode is seen in faxtransmission. This mode of communication takes placeover a link between a single transmitter and a receiver.

30. (d) The maximum line-of-sight distance between thetransmitting and receiving antennas is

d M = 2 2T R Rh Rhwhere hT and h R are the heights of transmitting andreceiving antennas respectively.

d M = 2 2 2 2 8 Rh Rh Rh Rh

( hT = h R = h)

31. (b) The ionospheric layer acts as a reflector for a certainrange of frequencies (3 MHz to 30 MHz). Beyond 30MHz em. waves penetrate the ionosphere and escape.

32. (d) As we know, time period of a simple pendulum

T = 22

2

4 L Lg

g T

The maximum percentage error in g

100 100 2 100g L T

g L T

= 2% + 2(3%) = 8%33. (c) Least count of a screw gauge

=Pitch

Number of circular scale divisions

=1 mm

50 = 0.02 mm

Therefore the pitch and no. of circular scale divisionsare 1mm and 50 respectively.




34. (e) Let the two balls Pand Q meet at height x m from theground after time t s from the start.We have to find distance, BC = (100 – x)

100 m

A

B

x m

(100 – ) x m

C

25 m s –1

P

Q

For ball PS = x m, u = 25 m s –1, a = – g

From S = ut + 21

2at

x = 25t – 21

2 gt .......... (i)

For ball QS = (100 – x) m, u = 0, a = g

100 – x = 0 + 21

2 gt .......... (ii)

Adding eqns. (i) and (ii), we get100 = 25t or t = 4 sFrom eqn. (i),

x = 25 × 4 – 219.8 (4) 21.6 m

2

Hence distance from the top of the tower = (100 – x) m = (100 – 21.6 m) = 78.4 m

35. (c) As we know, distance traversed in nth second

S n = u +1

(2 1)2 a n

Here, u = 0, a =g

S n =1

(2 1)2

g n

Distance traversed in 1st second i.e., n = 1

S 1 =1 1

(2 1 1)2 2

g g

Distance traversed in 2nd second i.e., n = 2

S 2 =1 3

(2 2 1)2 2

g g

Distance traversed in 3rd second i.e., n = 3

S 3 =1 5

(2 3 1)2 2

g g

S 1 : S 2 : S 3 =1 3 5

: : 1:3 : 52 2 2

g g g

36. (e) Let u be the initial velocity that have to find and a bethe uniform acceleration of the particle.For t = 3s, distance travelled S = 12 m andfor t = 3 + 3 = 6 s distance travelled S = 12 + 30 = 42 mFrom, S =ut + 1/2 at 2

12 = u × 3 +21

32 a or 24 = 6u + 9a ...(i)

Similarly, 42 = u × 6 + 216

2 a

or 42 = 6u + 18a ...(ii)On solving, we get u = 1 m s –1

37. (a) As the ball, m = 10 g = 0.01 kg rebounds after striking

the wall Change in momentum = mv – (– mv) = 2 mv

Inpulse = Change in momentum = 2mv

= 1Impulse 0.54 N s2 7 m s

2 2 × 0.01 kgm

38. (b) Given, position vector ˆˆ ˆ2r i j k

and momentum ˆˆ ˆ2 p i j k

Angular momentum L r p

= ˆ ˆˆ ˆ ˆ ˆ(2 ) ( 2 )i j k i j k

= ˆˆ ˆ( 2 1) (2 1) ( 1 4)i j k

= ˆˆ ˆ 3i j k 39. (d) At maximum height ( H ) i.e. at point P the vertical

component of the projectile u sin = 0 whereas itshorizontal component i.e. u cos remains the same.

H

u

xu cos

u sin

u cosP

y

40. (a) According to the question, at any instant t , x = 4t 2, y = 3t 2

v x = 2(4 ) 8dx d

t t dt dt

and v y = 2(3 ) 6dy d

t t dt dt

The speed of the particle at instant t.

v =2 2 2 2(8 ) (6 ) 10 x yv v t t t

41. (b) The cyclist bends while taking turn in order to providenecessary centripetal force.

42. (d) Acceleration of the system

a =4 2 120 40 20

4 2 6

F g g

210 ms

From figure

a2 kg

T

2 gFBD of block

T – 2g = 2a

T = 2 (a + g) = 2 (10 + 10) = 40 N




43. (a) Given, mass of the sphere, M = 300 g = 0.3 kgSpeed of the sphere V = 5 m s –1

In case of rolling motion without slippingTotal energy, K = K trans + K rot

R

v

i.e., K = 2 21 1

2 2 Mv I

= 2 2 21 1 2

2 2 5 Mv MR

22solid sphere

5 I MR

= 2 21 1

2 5 Mv Mv ( v = R)

= 2 27 70.3 5 5.25 J

10 10 Mv

44. (a) Let the bullet be fired with velocity.

For 20 cm penetration of bulletusing v2 – u2 = 2as

22( ) 2 (20)

2 a

–2

23 340 or

4 160a a

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

For latter part of penetration,Let before it comes to rest distance travelled by the bullet be x

Again, using v2 – u2 = 2as we get

0 – 2 22

ax

or x = –2 2

2

1606.66 cm

8 8 3a

(using eq. (i))

Therefore, the distance travelled by the bullet before itcomes to rest = 6.66 cm

45. (a) In elastic collision both momentum and kinetic energyare conserved.

46. (e) According to law of conservation of angular momentum,

1 1 2 2I I =

1 2

I I

Substituting the values of 1 = 2 rad s –1

2 = 5 rad s –1

2 = I 10 –3 kg m2

I1 2 + 1 10 –3, 5 = (I1 + 1 10 –3) 4 2I1 + 5 10 –3 = 4I1 + 4 10 –3

2I1 = 1 10 –3

I1 = –3

–3 21 100.5 10 kg m

2

47. (c) Given, no. of rotation n = 1800 rpm = 1800 rpsTime, t = 2 minutes = 120s

Initial angular speed 0= –12 1800

rad s60

= 60 rad s –1

Final angular speed (as wheel comes to rest) = 0

Angular retardation = 0 –

t

= –260 0

rad s

120 2

48. (a) From the relation, angular momentum, L = mvr

v = L

mr

Centripetal force acting on the particle

F =

2

2 2

3

Lm

mv Lmr

r r mr

49. (e) Gravitational potential energy= gravitation potential × mass of body.Hence statement (e) is wrong.

50. (c) Gravitational potential energy of the body on thesurface of earth.

U s =GMm

R

where M = mass of the earthm = mass of the bodyGravitational potential energy of the body at a heighth (= R) form the surface of earth

U h =2

GMm GMm GMm

R h R R R

Required amount of energy = U h – U s

= 112 2

GMm GMm GMm R R R

=2 2

GMm mgR

R 2

GM g

R

51. (e) Total energy of the satellite = – Kinetic energy of thesatelite

= 21mv

2

52. (a) Volumes of two soap bubbles

3 31 1 2 2

4 4 and

3 3V r V r

where r 1 and r 2 are the radii of soap bubbles.Let s be the surface tension of the soap solution. Theexcess pressure inside the two soap bubbles, then

1 21 2

4 4 and

S S P P

r r

When these two bubbles coalesce under isothermalconditions a bigger bubble of radius R is formed. If V

and P be the volume and excess pressure inside this bigger bubble, then

34 4and

3

S V R P

R

Here Boyle's law holds as the bigger bubble is formedunder isothermal conditionsi.e., P1V 1 + P2V 2 = PV




3 3 31 2

1 2

4 4 4 4 4 4

3 3 3

S S S r r R

r r R

2 2 2 2 21 2 1 2or r r R R r r

53. (b) Bulk modulus is the ratio of hydraulic stress to the

corresponding strain.54. (a) Pressure difference between lungs and atmosphere

= (760 –750) mm of Hg= 10 mm of Hg = 1 cm of HgLet the boy can suck water from depth h. ThenPressure difference = hwater g = 1 cm of Hg

or, h × 1g cm –3 × 980 cm s –2

= 1 cm × 13.6g cm –3 × 980 cm s –2

h = 13.6 cmThe boy can suck water from the depth of 13.6 cm

55. (c) As we know, energy stored in a spring

21

2U kx x = extension (or compression) in the spring.k = spring constant of the springAs per question, for x = 1mm = 1 × 10 –3m

–3 21(1 10 m) 1 J

2U k .......(i)

If spring is further compressed by 1 mm then

3 21(2 10 m)

2U k

.......(ii)

Dividing eqn. (ii) by (i), we get

4 or 4

U

U U U

Work doneW = U – U = 4U – U

= 3U = 3 × 1 J = 3J56. (d) Root mean square (rms) velocity of gas molecules

rms3 Bk T

vm

where m = mass of each molecule of gas, T = absolutetemperature and k B = Boltzmann constant.

–1/2 1/2rms rmsor

T v v m T

m

57. (b) When efficiency of carnot engine, = 0.2Efficiency of a Carnot engine,

2 2

1 11– or, 0.2 1–

T T

T T

2

1or 0.8

T

T ...(i)

When T 2 is reduced by 50 K, its efficiency becomes 0.4

2

1

– 500.4 1–

T

T

2

1

– 50or 0.6

T

T ...(ii)

Dividing eqn. (i) by (ii)

2

2

0.8 4

– 50 0.6 3

T

T

3T 2 = 4T 2 – 200 or T 2 = 200 K

From eqn. (ii), 21 – 50 200 – 50

250K 0.6 0.6

T T

58. (e) Molar specific heat of the gas at constant volume,

63 [ 6 given]

2 2V f

C R R R f

59. (c) Total 5 degrees of freedom : 3 translational and 2rotational for a rigid diatomic molecule.

60. (a) The differential equation of simple harmonic motion is

2 2

2 22 0 or –2

d y d y y y

dt dt ...(i)

Standard equation of simple harmonic motion is

22

2 –

d y y

dt ...(ii)

Comparing eq. (i) and (ii),

2 = 2 or = 2

As we know,2

T

Time period,2 2

22

T s

61. (b) In parallel combination, the effective spring constant(K) of springsK = K 1 + K 2 + ...Hence, option (b) is wrong.

62. (b) In SHM, Total energy, 2 2total

1E

2m A

and, Kinetic energy, 2 2 2K

1E ( – )

2 m A x

where x is the distance from the mean position.At x = 0.707 A

2 2 2 2 2K

1 1E ( – (0.707 ) (0.5 )

2 2

m A A m A

As per question, Etotal = 100 J

2 2K

1E 0.5 m A 0.5 100J 50J

2

63. (e) Given : y1 = A sin [k ( x + ct )] = A sin (kx – kct )and, y2 = A sin [k ( x – ct )] = A sin (kx – kct )According to principle of superposition, the resultantwave on the string y = y1 + y2= A sin (kx + kct ) + A sin (kx – kct )

= A [sin (kx + kct ) + sin (kx – kct )]= 2 A sin kx cos kct

[ sin (C + D) + sin (C – D) = 2sin C cos D]




The amplitude of this wave is Ak = 2 A sin kx

The positions of antinodes (where amplitude ismaximum) are given by|sin kx| = 1

1

2kx n

or 1

where 0,1,2,...2

x n nk

3 5

, , ,....2 2 2

xk k k

Therefore the distance between adjacent antinodes is/k.

64. (d) In the third harmonics or second overtone the vibrationof a stretched wire is as shown in the figure.

A N

A N

ATotal number of nodes N = 2Total number of antinodes A = 3

65. (d) In a tube closed at one end, only odd harmonics are present but in a tube open at both ends all theharmonics are present.The distance betwen successive nodes is equal to half the wavelength i.e. /2.Reflection of a wave from a rigid wall changes the phase by 180º or .

66. (c) Electric field due to the sheet02

E

Here = uniform surface density of charge.Force experienced by the electron

0

0

2or

2

F eF eE

e

–12 –12

–19

2 1.6 10 8.854 10

1.6 10

= 17.708 × 10 –5C m –2 = 177.08 × 10 –6 C m –2

= 177.08 C m –2

Area of the sheet, A = (0.5m)2 = 0.25m2 (given)Total charge on the sheetQ = A = (177.08 C m –2) (0.25 m2) = 44.27C

As the electron experiences the force directed awayfrom the sheet therefore the sheet must be negativelycharged.

–44.27 CQ

67. (d) Energy stored in a capacitor

221 1 1

2 2 2

QU QV CV

C ( Q = CV )

68. (e) According to Coulomb's law the electrostatic force between two point charges

1 21 22

q q 1F i.e. F q q

4 r

where is the permittivity of the medium, q1 and q2 aretwo point charges and r distance between two charges.

69. (d) The force of attraction between two charges q1 and q2 provides the required centripetal force for circular motion.

21 2

20

1. .,

4

q qmvi e

r r

O

q m1,

r

q2

F

or 1 2

0

1

4

q qv

mr

Time period of revolution

1 2

0

2 2

1

4

r r T

v q q

mr

3 30 0

1 2 1 2

4 162

mr mr r

q q q q

70. (c) Given, F = 1000 N, q = 2C and d = 2 cm = 2 × 10 –2 m

–11000500NC

2

F N E

q C

Also or V

E V Ed d

V = (500 NC –1) (2 × 10 –2m) = 10V

71. (a) The equivalent circuit of given network is as shown inthe figure.

C X Y B

4

8

8

4

4

2

A

2

4V

From the network

Resistance between C and X

1 1 1 1 2 1 1 4 1

4 8 8 8 8 2CX R

or RCX = 2Resistance between X and Y

1 1 1 1 1 2 1

4 4 4 4 2 XY R

or R XY = 2Total resistance between A and B

R AB = R AC + RCX + R XY + RYB

= 2 + 2 + 2 + 2 = 8Current through X to Y

XY

XY

P.d.across 4V2

resistance R 2

XY I A

Potential difference across AB

= I × R AB = 2( A) (8) = 16V




72. (b) The ammeter will show a zero reading, if the potential

difference across R = 2V

10

2500 500

E R R

R R

10 R = 1000 + 2 R

8 R = 1000 or R = 125

73. (d) Accor ding to rate law

1d[A] d[B] 1 d[C]

2dt dt 3 dt

d[C] 3 d[A]

dt 2 dt

= –3 (0.3 0.5) 0.02 3

2 10 2

= 0.03 mol L –1 min –1

74. (c) 4 2 2 2 NH ( ) NO ( ) N ( ) 2H O( )aq aq g l

Let the rate law be

Rate = k [NH4+] x [NO2

– ] y

From expt. (1), 0.02 = k (0.01) x (0.02) y ... (i)

From expt. (2), 0.03 = k (0.015) x (0.02) y .. (ii)

Dividing equation (ii) by (i), we get

0.03 (0.015) (0.02)

0.02 (0.01) (0.02)

x y

x y

k

k

(1.5)1 = (1.5) x x = 1

From expt. (3), 0.005 = k (0.01) x (0.01) y ... (iii)

Dividing equation (iii) by (i), we get

0.005 (0.01) (0.01)

0.02 (0.01) (0.02)

x y

x y

k

k

(0.25) = (0.5) y or (0.5)2 = (0.5) y

y = 2

Now, putting the values of x and y in eqn. (i), we get

rate = k [NH4+]1 [NO2

– ]2

75. (e) 2AuCl3 + 3HCHO + 3H2O Reduction

2Au(sol) + 3HCOOH + 6HCl

76. (c)

77. (e)

H C –3 CH – C = CH2 2

C H52

O3 CH3

CH2

CH3CH2

C

O

O

C

O

H

H

HCHO + CH3CH2

CH3CH2

C = O

Zn/ H O2

78. (b) CH3 – C C – CH3

Na

liq.NH3

H

H C3 H

CH3

trans-2-butene

CC

79. (a) The reactivity order of alkyl halides is as follows.

R – I > R-Br > R-Cl > R-F which is due to decrease in

C – X bond length.

Smaller the energy required to break the R – X bond,

better is the leaving group (X) and faster is the rate of

reaction.80. (b) Compound A reacts with Tollen's reagent thus it should

be aldehyde moreover it does not undergo aldol

condensation. The compounds which do not have

-H atom does not undergo aldol condensation. Thus

compound A is

CH3

CH3

C CHO

CH3

(no- -H-atom present)

2, 2-dimethyl propanal

81. (e) CH3CH2CH2CH2 CH2CH3 CH3 CH CH2

CH2 CH +3

anhy. AlCl + HCl3

isomerisation 2-methylpentane

CH3

+ CH3CH3CH CCH

CH3CH3 CH3

CH3

CH2CH3CH3 + CH3

CH

CH3

CH2CH2CH3

2, 3-dimethylbutane 2, 2-dimethylbutane 3-methylpentane




86. (c) Presence of electron withdrawing group (e.g. –NO2group) in benzoic acid increases the acid strength while presence of electron donating group (e.g. –OCH3)decreases the acid strength.

COOH COOH COOH

NO2 NO2

NO2

COOH

OCH3

( )IV (I) (II) (III)

< < <

87. (b) Given organic compound will be 2-methylbenzaldehyde as it gives the following reactions.

CHOCH3

+ 2[Ag(NH3) ] + 3OH2+ –

COO –

CH3

+ 2Ag + 4NH3+ 2H O2

C H–

O

CH3+ H N2 NO2 –H2O

NO2 C H–

N NO2

NO2

CHO COOH

CH3 [O] COOH

2-methyl benzaldehyde

1, 2-benzenedicarboxylic acid

88. (e) 6 5 2 3C H NH CHCl 3KOH

6 5 2(carbylamine

reaction)

C H NC 3KCl 3H O

89. (b) All aliphatic amines are stronger than NH3. Aromaticamines are less basic then aliphatic amines. Amongaliphatic amines 2° amines are more basic than 1° and

3° amines hence the correct order of basic strengthwill be

(CH3)2 NH > C2H5 NH2 > NH3 > C6H5 NH2

82. (c) H C3 CH2 CH CH3

CH32-Methylbutane

H C3 CH2 CH CH2Cl

CH32-Methyl-1-chlorobutane

H C3 CH2 C CH3


Cl

H C3 CH CH CH3

CH3

3-Methyl-2-chlorobutane

Cl

H C3 CH CH2 CH2Cl


h

Cl2

83. (b) On exposure to air and sunlight, chloroform, acolourless heavy liquid, oxidises to carbonyl chloride

(phosgene), a highly poisonous gas used in warfare.

light3 2 3 2

unstable

1CHCl O CCl (OH) COCl HCl

2

84. (e) Diphenyl methanol (C6H5CHOHC6H5) does not haveCH(OH)CH3 group hence it will not form yellow

precipitate with an alkaline solution of iodine (iodoform reaction)

85. (b)|| || | ||

OH3 3 3 2

Aldol

O O OH O

CH CH H C CH CH CHCH C H

23H O

But-2-enal

CH CH CH CHO

It is the - H that adds on the carbonyl oxygen of thesecond carbonyl compound.

90. (a) Gabriel's phthalimide synthesis is used for synthesis of 1° amines,

C C C

C C C NH NK NR + RNH2

O O O

O O O

alc. KOH RX

(– KX)

HOH/H+

(–H O)2 or OH –

COOH

COOH

Phthalimide Pot. phthalimide N-alkyl phthalimide Phthalic acid

1º amine




91. (d) -D-ribose is present in RNA molecule

92. (c)

Phenol

OH

n2

2

H O

Formaldehyde

2nH C O

Novolac

OHH

C

H

H

C

Hn

93. (b) Glycogen is known as animal starch. This remain storedin liver and muscles and acts as a reserve food materialin animals.

94. (e) Green chemistry may be defined as the programme of developing new chemical products and chemical processes or making improvements in the already

existing compounds and processes so as to make lessharmful to human health and environment. This meansthe same as to reduce the use and production of hazardous chemicals.

95. (c) E =34 6

1910

hc 6.626 10 3 103 10 J

6626 10

Energy emitted by bulb =250 80

200J100

n × 3 × 10 –19 = 200 n = 6.66 × 1020 (where n = no. of photons)

96. (a) 2 21 2

1 1 1

H R n n

For Lyman series, n1 = 1 and n2 =

then,2

1 1 1

(1)

H R

or, =1 1

or 109678

H R

= 911.7 × 10 –8 cm = 911.7A

97. (e) K.E. = h – h0 = 6.2 – 5.0 = 1.2 eV1 eV = 1.6 × 10 –19 Jthen 1.2 eV = 1.2 × 1.6 × 10 –19 J

= 1.92 × 10 –19 J98. (b) Lattice energy is defined as the energy required to

separate one mole of anionic compound into its ionsin gaseous state.

lattice NaCl( ) Na ( ) Cl ( )s g g

99. (c) Bond length decreases and hence the stabilityincreases with the bond order C2 (12); Bond order = 2Li2 (6); Bond order = 1O2

+ (15); Bond order = 2.5

He2+

(3); Bond order = 0.5

Hence the correct order of their stability is

2 2 22He Li C O

100. (b) In SF4 the hybridisation is sp3d and the shape of

molecule is trigonal bipyramidal.

S

F

FF

F

101. (c) Potassium dichromate belongs to triclinic system.102. (a) From ideal gas equation.

PV = nRT

P =nRT 2 0.082 500

2atmV 41

103. (a) P n

P w

M

1 2 2

2 1 1

P M (Molecular mass of O ) 2 32

;P M (Molecular mass of oxide) 5 M

5 32

80 g/mol2

104. (c)105. (b) No. of electrons in

NH3 = 7 + 3 = 10CH3

+ = 6 + 3 – 1 = 8 NH2

– = 7 + 2 + 1 = 10 NH4

+ = 7 + 4 – 1 = 10106. (a) K 2SO4 give violet colour in the flame test.

K 2SO4 + BaCl2 — BaSO4 + 2KCl

white ppt107. (a) For neutral molecules:

No. of electron pairs = No. of atoms bonded to it +1

2[Group number of central atom – Valency of the centralatom].For ions : No. of electron pairs = No. of atoms bonded to it +

1

2[Group no. of central atom – valency of the central

atom no. of electrons]

On calcuting no. of electron pairs in given molecules.We find that in the given molecules hybridisation is NH4

+ – sp3 SF6 – sp3d 2

BF3 – sp2 PCl5 – sp3d

XeF4 – sp3d 2

108. (d) Interhalogen compounds are diamagnetic in nature.109. (d) The two bond lengths in the ozone molecule are same

1 . 2 7

8 Å 1

. 2 7 8 Å

O O

O

110. (a) The higher the oxidation number of the central elementmore is the oxidising power.

Oxidation state of Mn in 4MnO is x + 4(–2) = –1 x = + 7




Oxidation state of Cr in 22 7Cr O is

2 x + 7(–2) = –2 x = + 6

Oxidation state of V in 2VO is

x + 2(–2) = + 1 x = + 5Hence, the correct descending order of oxidizing power

is : 24 2 7 2MnO Cr O VO .

111. (a) 5e 24

Permanganate ion Manganese(II) salt

7

MnO Mn

24

Manganate ion

6MnO

2

Manganese dioxide

4MnO

+1e –

+3e –

112. (a) Electronic configuration of divalent ion M2+ ( Z = 24)= [Ar] 3d 4 4s0

Number of unpaired electrons (n) = 4

= ( 2)B.M 4(4 2) 4.89 4.90 B.M. n n

113. (a) Let A be the given liquid.A() A(g), S = 58 JK –1 mol –1

For A(g) A()

S = 110058 JK

58

= – 100 JK –1

114. (a) m(HCOOH) H HCOO = 347 + 53 = 400 S cm2 mol –1

Degree of dissociation, =cm

m

400.1

400

So, dissociation constant,K a = c2 (for weak electrolytes) = 0.025 × (0.1)2 = 2.5 × 10 –4

115. (c) For equation

2 2A (s) B (s) 2C(g)

No.of moles of C =1

250

= 0.04 moles

Partial pressure of C =nRT

V

=0.04 0.082 300

24.6

= 0.04 atmK p = PC

2 = (0.04)2 = 1.6 × 10 –3 atm2

116. (a) 22X(OH) X 2OH ... (i)

K sp = [X2+] [OH – ]2 ... (ii) pH + pOH = 1412 + pOH = 14 pOH = 2

pOH = – log [OH – ]2 = –log [OH – ][OH – ] = 1 × 10 –2

From the chemical equation (i), the [X2+] is half of the [OH – ] ion, therefore,

[X2+] =2

31 105 10

2

Now putting the value of [OH – ] and [X2+] ions ineqn. (ii), we get

K sp = [X2+] [OH – ]2 = (5 × 10 –3 ) (1 × 10 –2)2

= 5 × 10 –3 × 1 × 10 –4 = 5 × 10 –7

117. (d) T f =1000

f K w

W M

... (i)

Given, K f = 1.86 K kg mol –1, w = 3g, W = ? M = 111.6 g mol –1, T f = –0.125°C

T f = 0 C ( 0.125 C) 0.125 C f f T T

On putting the values in eqn. (i), we get

0.125 =1000 1.86 3

111.6W

W =1000 1.86 3

400 g0.125 111.6

118. (e) Dissolved oxygen = 5 × 10 –3 g

= 5 mg/kg of sea water Concentration of O2 (in ppm) = 5 ppm

119. (d) Cu2+ + 2e – Cu

Eel =2

el2.303 RT

E log [Cu ]nF

Eel =2

el0.06

E log [Cu ]2

Eel1 – Eel2

= 0.03 log2

12

2

[Cu ]

[Cu ]

= 0.03 × 2 = 0.06 V Potential of half cell decreases by 60 mV

120. (e) For cell reaction,

2 2M( ) X ( ) M ( ) X( ) s aq aq s

Nernst equation,

Ecell =2

cell 2

0.0591 [X ]E log

[M ]

n

E1 = cell0.0591 0.01

E log 0.1 n

E1 = cellE 0.02955

E2 = cell0.0591 0.1

E log2 0.1

E2 = cellE (as log 1 = 0)

E3 = cell0.0591 0.1

E log2 0.011

= cell0.1

E 0.0295 log

0.011

Therefore, E3 > E2 > E1




Paper-II : Mathematics

1. (b) Consider 2 3 = 4 + 9 = 13

(2 3) 4 = 13 4 = 132 + 42

= 169 + 16= 185

2. (a) Let the number of students who take only Math be xand only Chemistry be y.

M C

x y30

So, from the Venn diagram, we have total number of students who take Math = x + 30and take Chemistry = y + 30.According to question, we have

30 = 10

30

100

x

x = 270 and

30 = 12

30100

y

y = 220x + y + 30 = 270 + 220 + 30 = 520.

3. (d) Let f ( x) = | x – 2|Put x = 1 and x = 3 in f ( x), we get

f (1) = |1 – 2| = 1 f (3) = |3 – 2| = 1

Since f ( x) = f ( y) but x y, therefore f is Not one-one.

4. (b) Let A = {1, 3, 5, 7 }, n(A) = 4 = m (say)and B = {1, 2, 3, 4, 5, 6, 7, 8}, n(B) = 8 = n (say)Total number of one to onefunction = nCm (m)!, n m

= 8C4 (4)! =8!

4!4!4!

= 8 × 7 × 6 × 5

= 16805. (d) Let f ( x) = y = x2 + 2 x + 2 = ( x + 1)2 + 1

y – 1 = ( x + 1)2

x + 1 = 1 y

x = 1 1 y Since, y – 1 0 y 1 range is [1, ).

6. (e) Let f ( x) = , x g( x) = 2 x – 3

Now, ( fog) ( x) = f [g( x)]

= f [2 x – 3] = 2 3 x

Now, for domain of ( fog) ( x), we have 2 3 x 0

x 3

2

x 3

,2

7. (b) Consider

3 2

2

3 3 4

8 6

i i

i

=

3 2

2

3 3 3 3 3 9 16 24

64 36 96

i i i i i

i i

= 3 3 9 3 3 9 24 16

64 96 36

i i i

i

=

2 2

56 192 56 192 28 96

28 96 28 96

i i i

i

=1568 5376 5376 18432

784 9216

i i

=20000

210000

i

i

| z| = 22 = 2.

8. (d) Let = u + iv

z =1 1

1 1

u iv

u iv

=

1 1 1

1 1 1

u iv u iv u iv

u iv u iv u iv

=2 2 2

2 2

1 ( 1) ( 1)

2 1

u iv u iv u v

u u v

=2 2

2 21

2 1u v iv iv

u v u

= 2 20 21

2 2

ivu v

u

=1

iv

u Re( z) = 0

9. (a) Let z2

3 2 2cos sin

3 3

i

e i

Consider 1 + z + 3 z2 + 2 z3 + 2 z4 + 3 z5

= 1 + + 32 + 23 + 24 + 35

= 1 + + 32 + 2 + 2 + 32

( 3 = 1)= 3 + 3 + 62 = 3(1 + + 2) + 32

= 32 = 34

3

i

e

( 1 + + 2 = 0)

=4 4

3 cos sin3 3

i

= 3 cos sin3 3

i

= 33 cos sin 33 3

i

i e




10. (d) Let z1 = 22 2 1 , 1 3i z i

21

z = 4 × 2 (1 + i)2

= 8 (1 + 2i – 1)= 16i

32 z = 31 3i

= 31 3 3 3 3 3 3i i

= 1 3 3 9 3 3 i i

= – 8

2 321 z z = 16i × (– 8) = –128 i

11. (a) ( z3 – z1) = ( z2 – z1) [cos 90 + i sin 90]( z3 – z1) = i ( z2 – z1)( z3 – z1)2 = – ( z2 – z1)2

( z3 – z1)2 + ( z2 – z1)2 = 0.

12. (b) Let , + 2 be two consecutive odd rootsof x2 – ax + b = 0 + + 2 = a and ( + 2) = b

2 + 2 = a

=2

2

a and 2 + 2 =b

Since, 2 + 2 =b

2

2 22

4 2

a a = b

2 4 4

24

a aa b

a2 – 4a + 4 + 4a – 8 = 4b

a2 – 4b = 4.13. (b) Since and are the roots of x2 – ax + b2 = 0, therefore

+ = a, = b2

+ = ( + )2 – 2 = a2 – 2b2

14. (b) Let and be the root of x2 + 3 x – 4 = 0,therefore sum and product of roots are + = – 3 = – 4

1 1 =

3 3.4 4

15. (b) Consider 32 x – 22 3 81 x = 0 32 x – 2.3 x. 32 + 81 = 0 32 x – 18.3 x + 81 = 0

2

3 x – 18.3 x + 81 = 0

(3 x – 9)2 = 0 3 x – 9 = 0 3 x = 32

x = 2.16. (a) Let and be the root of the equation x2 + 2bx + c = 0

+ = – 2b b =2

and = c

Consider b2 – c =2

2

= 2 24

4 4

17. (d) Given equation is 2 x2 + 3 x + 1 = 0Let and be the roots of 2 x2 + 3 x + 1 = 0

+ =3 1

,2 2

2 + 2 = ( + )2 – 2

=23 1

22 2

=9 5

14 4

and 2 2 = 14

Required quadratic equation is

2 5 1

4 4 x x = 0.

4 x2 – 5 x + 1 = 0

18. (d)

17

8

1

2 3n

n n =

17

8

3 2

2 3n

n n

=

17

8

3 2

2 3n

n n

n n

=

17

8

1 1

2 3n

n n

=1 1 1 1

...10 11 11 12

1 1...

19 20

=1 1 2 1 1

10 20 20 20

19. (d) Let the two positive numbers be ‘a’ and ‘b’.Therefore,

a : b = 3 + 2 2 : 3 2 2

A. M. =6

32 2

a b

G.M. = 1 1ab

A.M

G.M=

3

1 A.M : G.M = 3 : 1.

20. (a) Given x1 + x4 + x9 + x11 + x20 + x22 + x27 + x30 = 272 a + a + 3d + a + 8d +a + 10d + a +19d + a + 21d

+ a + 26d + a + 29d = 272 8a + 116d = 272 2a + 29d = 68




Consider, x1 + x2 + ..... + x30 = S30 =30

[2 29 ]2

a d

= 15 × 68= 1020.

21. (d) Let ar = 24, ar 4 = 34

ar ar

= 33 1 124 8 2r r

Since r =1

2, therefore

124

2 a

a = 48S6 =

6

6

1148 11

1892 248 .1 1 212 2

22. (e) Given S75 = 2625

75 2 74 26252

a d

75 [a + 37d ] = 2625

a + 37d =2625

3575

38th term = 3523. (b) Since –5, k , –1 are in A.P

Therefore, k – (– 5) = – 1 – k

k + 5 = – 1 – k 2k = – 6

k = –3.24. (e) Tn = nC3

Tn + 1 – Tn = (n + 1)C3 – nC3

= nC2= 36

!

2! ( 2)!

n

n = 36

( 1)

1.2

n n = 36

n(n – 1) = 72

n2 – n – 72 = 0

(n – 9) (n + 8) = 0

n = 9 ( n – 8)

25. (a) General term = Tr + 1 =10

10 10C

10

r r

r

x

x

Here n = 10, which is an even number.

Now,th10

12

term i.e. 6th term is the middle term.

Hence, middle term = T6

T5 + 1 =10 5 5

105

10C

10

x

x

=

5 510

5

10

C 10

x

x

= 105C .

26. (b) x49 (–1 – 2 – 3 – ....... – 50)= – x49 (1 + 2 + 3 + ......... + 50 )

= – x49 50 51

2

( 1)Sum of natural numbers

2

n n

n

= – 1275 x49.Hence, coeff. of x49 = – 1275

27. (b) Put x = 1 in the given expansion

Then,61

2 x x

= (1 + 2)6 = 36 = 729.

28. (e) Consider 2P1 + 3P1 + ... + nP1= 2 + 3 + ...... + n

=( 1)

12

n n

=2 2

2

n n

29. (c) ‘a’ can take the values 1, 3, 5, 7, 9‘b’ is divisible by 3 b can take 0, 3, 6, 9

c can take 0, 2, 4, 6, 8 ‘d ’ is prime therefore d can take 2, 3, 5, 7 Total numbers of 4 digit numbers= 5C1 × 4C1 × 5C1 × 4C1= 5 × 4 × 5 × 4 = 400.

30. (e) Consider1 2 1 2

2 3 2 1 3 2

3 4 3 2 4 3

1 1

1 0

1 0

a a a a

a a a a a a

a a a a a a

R 2 R 2 – R 1R 3 R 3 – R 2

=

1 2 1

0 0

0

a a

d d

d d

Where ‘d’ is the common difference.31. (d) Let the vertices of a triangle be

3 31 1 2 2, , , , x y x y x y

and a a b b x c

Area of the triangle =

1 1

2 2

3 3

1

11

2

1

x y

a a x y

b b

x y

c c

=1 1

2 2

3 3

1

2

x y a

x y babc

x y c

=

1 1

2 2 3 3

3 3

21 1

2 2C4

2

x y a

x y b C abc

x y c

=1 1 1

.4 2 8

abc

abc




32. (d) Given system of equations are3x + y – z = 2x + 0 y – z = 12x + 2y + az = 5Since system has unique solution thereforedeterminant is non-zero.

3 1 1

1 0 1 0

2 2 a

3(0 + 2) – 1(a + 2) – 1(2 – 0) 0 6 – a – 2 – 2 0 2 – a 0 a 2.

33. (e) Since matrix is singular therefore2 2

01 3

k

k

(2 – k) (3 – k) – 2 = 0 6 – 5k + k 2 – 2 = 0

k 2 – 5k + 4 = 0 – k 2 + 5k – 4 = 0 5k – k 2 = 4.

34. (a) Consider

log 1 log log

1 1log loglog 1

log log 11log

a a a

a ab

a ca

b c

b c

c

c

=

log1 log log

log log1 log

log log log1

b c

b c

c c

=

0 log log

log 0 log

log log 0

b c

b c

c c

= – log b [– log c × log c] + log c [– log b × log c]= 2 log b log c – 2 log b log c = 0Alternatively:

On solving determinant, we get

0 log log

log 0 log

log log 0

b c

b c

c c

which is the determinant of a skew symmetric matrixof odd order.Hence, its value is zero.

35. (a) Given system of equations can be re-written as2x + y – 4 = 0 ......... . (1)3x + 2y – 2 = 0 ........ (2)x + y + 2 = 0 ............ (3)

Consider

2 1 4

3 2 2

1 1 2

= 2 (4 + 2) – 1 (6 + 2) – 4 (3 – 2)

= 12 – 8 – 4 = 0Since, determinant is zero therefore system of equations has unique solution.

36. (d) Given : |x – 2| + | x + 2| < 4

|x – 2| =( 2) if 2

( 2) if 2

x x

x x

| x + 2| =( 2) if 2

( 2) if 2

x x

x xSolution for x > 2 :

( x – 2) + ( x + 2) < 42 x < 4 x < 2 (Contradiction)

Solution for x < –2 –( x – 2) – ( x + 2) < 4 – x + 2 – x – 2 < 4 –2 x < 4 x > – 2 (Contradiction)

Hence, there is no solution of the given inequation.37. (e) From the figure it is clear that there are 3 lines.

Line which passes from (0, 14) and (19, 14) is y = 14 In the shaded region 0 y 14Line which passes from (5, 0) and (0, 14) is14 x + 5y = 70 In the shaded region 14 x + 5 y 70Line which passes from (5, 0) and (19, 14) isx – y – 5 = 0 In the shaded region x – y 5Thus, inequations are 14 x + 5y 70, x – y 5, y 14.

38. (c) Statement given in option (c) is correct. [p (~ q) ] = (~ p) ~ (~ q)= (~ p) q

39. (e) F, F, T

40. (d) Consider ~ [(~ p) q] = ~ (~ p) ~ q = p ~ q.

41. (e) We know a sin + b cos 2 2a b

sin + cos 2 21 1 2

42. (c) 23

1

13

2

1 12 2 1sin sin3 3

= 1 1 1tan 2 2 tan

2 2

= 1 1tan 2 2 cot 2 2 .2

43. (e) Let ab < 1. Given sum = 2 tan –1 a + 2 tan –1 b= 2 [tan –1 a + tan –1 b]

=1 12 tan 2tan

1

a b x

ab

x =1

a b

ab




44. (b) Consider tan(1°) + tan (89°)= tan 1° + tan (90° – 1°)= tan 1° + cot 1°

=sin1 cos1

cos1 sin1

=2 2

sin 1 cos 1sin1 . cos1

=2

2sin1 . cos1

=2

.sin2

45. (d) Let sn = cos10

n

For n = 5

s5 = cos5

cos 0.10 2

Hence, 1 2 10

1 2 10

s s ... s0

s s ... s

46. (a) 1 17 3cos cos cos cos 2

5 5

= 1 3 3cos cos .

5 5

47. (c) Given sum = 1 + tan2 (tan –1 3) + 1 + cot2 (cot –1 2)( sec2 = 1 + tan2 and cosec2 = 1 + cot2 ) = 1 + 32 + 1 + 22 = 15.

48. (c) Let sin + cosec = 2

sin +1

2sin sin2 – 2 sin + 1 = 0 (sin – 1)2 = 0 sin – 1 = 0 sin = 1 sin6 + cosec6 = 16 + 16 = 2.

49. (d) Given expression is equal to

[sin8 sin 6 ] [6sin6 ] 6sin 4 12sin 4 12sin 2

sin 7 6sin5 12sin3

x x x x x x

x x x

=2sin7 cos 6[sin 6 sin 4 ] 12[sin 4 sin 2 ]

sin 7 6sin 5 12sin 3

x x x x x x

x x x

=2sin7 cos 6[2sin5 cos ] 12[2sin3 cos ]

sin 7 6sin5 12sin3

x x x x x x

x x x

=2cos [sin 7 6sin 5 12sin 3 ]

[sin 7 6sin5 12sin3 ]

x x x x

x x x

= 2 cos x

50. (c) Let ABCD be a rectangle with vertices A (2, 5) andC (5, 1).

D

AM

B

C

(2, 5)

(5, 1)

Now, M is the mid point of (Diagonal) line AC.

M is2 5 5 1 7

, ,32 2 2

Since M is on the given straight line y = 2 x + k

therefore M satisfies the equation of straight line.

Put x = 72

and y = 3 in y = 2 x +k

we get 3 =7

2 3 7 42

k k

51. (c) Let PQR be a triangle with vertices P(2, – 2),Q(8, – 2) and R(8, 6).

C

P Q

R

(8, – 2)

(8, 6)

(2, – 2)

Since circumcentre (C) is the mid point of Hypotenusetherefore

C =2 8 2 6

, (5,2)2 2

52. (c) Given equation is 2 x + 5 y – 7 = 0 and given points are(– 4, 7) and (6, – 5).Thus a = 2, b = 5, c = – 7and x1 = –4, y1 = 7

x2 = 6, y2 = – 5The required ratio is

= – 1 1

2 2

2( 4) 5 (7) 7

2(6) 5( 5) 7

ax by c

ax by c

= 1: 1

53. (b) a2 – b2 = 512 (a + b) (a – b) = 29

(a + b, a – b) = (28, 2), (27, 22), (26, 23), (25, 24)Since, a > b, a + b > a – b therefore the other combinations like (24, 25) etc cannot be accepted.(29, 1) also cannot be accepted since a and b are positive integers.

54. (e) Equation of line in intercept form is 1

1 13 4

x y

3 x + 4 y – 1 = 0

p = 2 2

3(0) 4(0) 1 1.

53 4

55. (b) Given eq n of circle is x2 + y2 – 8 x + 2 y = 0

So, centre: (4, – 1)Given eq n of parabola can be rewritten as y = ( x – 2)2 + 6

( x – 2)2 = 4.1

( 6)4

y

Vertex of parabola is (2, 6)

Required slope =6 ( 1) 7

2 4 2




56. (d) Since the straight line is perpendicular to the line2 x + y = 3 therefore its equation is x – 2 y + k = 0

It passes through (1, 1) 1 – 2 + k = 0 k = 1 Required equation is x – 2 y + 1 = 0

y– intercept = 1 1 .2 2

cb

57. (e) p = 2 2

0(sec ) 0(cos ec )

sec cosec

a

= 2 2sec cosec

a

=2 2

2 2

sin cos

sin cos

a

= asin cos = sin22

a

Similarly,

q = 2 2

cos2cos2

cos sin

aa

4 p2 + q2 = 2

24 sin 2 cos2

2

a

a

= a2 (sin2 2 + cos2 2) = a2.58. (b) Let C1 and r 1 be the centre and radius of the circle

(x – 1)2 + (y + 2)2 = 1 respectively and let C2 and r 2 bethe centre and radius of the circle

(x + 2)2 + (y – 2)2 = 4 respectively.C1 = (1, – 2), C2 = (– 2, 2)

r 1 = 1 r 2 = 2

C1 C2 = 1 29 16 5 r r

Two circles do not intersect dmin = C1 C2 – (r 1 + r 2)

= 5 – (1 + 2) = 2.59. (e) Let the required centre be (h, k )

5

(1, 2)

5

( , )h k

1 2, (5, 5)

2 2

h k

(By using mid-point formula) h = 9 and k = 8.

60. (b) Let centre of bigger circle be (0, 0) and centre of smaller circle be (h, k )

(0, 1)

(0, 0)

(h, k)

x y+ = 1622

The two circles touch internally

C1 C2 = |r 1 – r 2|

2 2 2 24h k h k

2 2 2 22 4 2h k h k

r = 2.61. (a) Let the required circle be

x2 + y2 + 2gx + 2 fy + c = 0 .... (1)

Since circle is passing through origin therefore c = 0Radius = 2 2 8g f c g2 + f 2 = 8 .... (2)

Since circle is passing through (4, 0) therefore we have16 + 8g = 0 g = – 2 f 2 = 4 f = – 2 Required circle is x2 + y2 – 4 x – 4 y = 0or ( x – 2)2 + ( y – 2)2 = 8

62. (c) Given equation of ellipse can be re-written as2 2

2

( 1) ( 1)1

1 2

x y

x + 1 = 1 cos and y – 1 = 2 sin

x = cos – 1 and y = 2 sin + 163. (c) Let P be any ponit on ellipse and F be the focus. Then

PF 3

PM 5e

PM =5 5

PF 6 103 3

64. (e) Length of latus rectum of hyperbola is2

224 3 2 3

bb a

a

Length of transverse axis is 2a = 2 3 3 a

b2 = 2 3 3 6

2 2

13 6

x y

65. (d) Focus : (ae, 0) =5

,04

a

2 x + 3 y – 6 = 0 passes through5

,04

a

5

2 3(0) 6 04

a

5 246 2 52 a a




66. (b) Length of the transverse axis of a hyperbola = 2 cos Þ 2a = 2 cos a = cos Given equation of ellipse can be rewritten as

2 2

116 9

x y

Foci = 16 9,0 7,0

Since focus of hyperbola = 7,0

e cos =7

7cos

e

Now, a2 + b2 = 7 7 = b2 + cos2 Now, required equation of hyperbola is

2 2

2 21

cos 7 cos

x y

67. (a) Let a

= ˆˆ ˆ2 2i j k

1 4 4a = 3

Given b

= 5

Area =1 1 1 1

| | | | . | | sin 3 52 2 2 2

a b a b

=15

4

6

68. (e) .

a b a = cos60

a b a

cos 60° =1

2 =

.

a b a

a b a =

2

a

a b a

1

2=

a

a b...(1)

as . 0, a b a b

. a b b = cos30

a b b

cos 30° =

3 .

2

a b b

a b b

a b

60°

30°

b

a

3

2=

2

b

a b b=

b

a b...(2)

Dividing (2) by (1)

32

12

=

b

a

3 =

b

ab = 3

a

69. (e)

A B

C

j + k i + j

i + k

Now, ˆ ˆAC k j

and ˆ ˆAB k i

Let be the angle between AC

and AB.

cos =1 0 0 0 1

22 2

= 60° =3

70. (a) Consider ˆˆ| | 7a b

Squaring on both the sides, we get2 2 2

2 .a b a a b b

7 = 14 – 22 2

b b

21 4 9 and .

a a b b

2

7b

7b 71. (a) Consider

2 2 2 2

2 . . .a b c a b c a b a c b c

Since |a| b + c i. e. a. (b + c) = 0 .......... (1)Similarly b. (c + a) = 0 .......... (2)and c. (a + b) = 0 .......... (3)(1) + (2) + (3) gives2 (a. b + a. c + b. c) = 0

2 2 2 2

a b c a b c

72. (b) We know2 2 2 2

u v w u v w

+ 2 [u. v +v. w + u. w] = 0

9 + 16 + 25 = – 2 [u. v +v. w +u. w]

[u. v +v. w + u. w] = 50 252

73. (a) Since ˆˆ ˆ3 2 6i j k is a unit vector therefore

ˆˆ ˆ3 2 6 1i j k

ˆˆ ˆ3 2 6 1i j k

9 4 36 1

49 1

1

49

= 1

7




74. (b) Given =2 2

, ,3 3 3

am n

We have 2 + m2 + n2 = 1

24 4

19 9 9

a

2 8

1 19 9

aa

Thus =2 1 2

, ,3 3 3

m n

The required vector = ˆˆ ˆ3 i mj nk

vector is2 1 2 ˆˆ ˆ33 3 3

i j k

= ˆˆ ˆ2 2i j k

75. (e) Mid point of PQ is = 3 7 9, ,2 2 2

DR of the normal is = (4 – (– 1), 5 – 2, – 10 – 1)= 5, 3, – 11

Equation of plane is

3 7 95 3 11 0

2 2 2 x y z

5 x + 3 y – 11 z =135

2

r . 135ˆˆ ˆ5 3 112

i j k

76. (d) Given straight lines can be re-written as

1st line is

31 52

31 22

y x z

2nd line is 2 1 2

3 2 0

x y z

Let be the angle between them.

cos =3 3 0

09

1 4 9 4 04

= 90° =2

77. (a) Let DR of the line of intersection of the planes be a, b, c

a – 2b – c = 0 .......... (1)and a + b + 3c = 0 .......... (2)on solving, we get

5 4 3

a b c

a = – 5 k , b = – 4 k , c = 3k

Thus required eq n of line is

2 3 15 4 3

x y z

78. (c) Given equation of straight line is

2 3

3 1

x y

=

2

4

z

DR of the line : 3, – 1, – 4Direction cosines are

2 2 2 2 2 2

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

2 2 2

4

(3) ( 1) ( 4)

=3 1 4

, ,26 26 26

Required unit vector is1 ˆˆ ˆ3 426

i j k

79. (c) DR’s of the normal : 2, – 1, 2

DR’s of z axis : 0, 0, 1

cos =(2)(0) ( 1)(0) (2)(1) 2

34 1 4 0 0 1

= cos –1 2

3 .

80. (d) Equation of foot of perpendicular drawn from originto the given plane is

1 1 1 x x y y z z

a b c

=

1 1 12 2 2

( )ax by cz

a b c

29

12 3 4 29

x y z

x = 2, y = – 3, z = 4

81. (b) Required distance = 2 2 2(3 3) (12 0) (5 0)

= 2 2(12) (5)

= 144 25 13

82. (d) Let

9

1

( 5)

i

i

x = 9

9 9

1 1

5 9

i

i i

x

9

1

(9 5) 9

i

i

x

45 54i i x x Similarly,

2 10 54 25 9 45i

x 2 360

i x

=2

360 54 3242

9 9 81




83. (e) Total outcome = 36Favourable cases are :(1, 5), (1, 6), (2, 4), (2, 5), (2, 6)(3, 3), (3, 4), (3, 5), (3, 6), (4, 2)(4, 3), (4, 4), (4, 5), (4, 6), (5, 1)(5, 2), (5, 3), (5, 4), (5, 5), (5, 6)

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) Number of favourable cases = 26

Probability =26 13

6 6 18

84. (d) Consider P (A B) = P (A B) (By De’morgan law)

= P(A) P(B)= [1 – P(A)]. P(B)

85. (d) Standard deviation = =2 1

12

nd

d = size between each observation = 7n = total number of observation = 7

= 72(7) 1 49 1

712 12

=48

7 7 2 1412

86. (c) Consider5 5

8 83

3lim

3 x

x

x

=5 5

8 83

3 3lim

3 3 x

x x

x x

=

4

75 3 52168 3 87. (e) Given f ( x) = g( x) h( x)

h ( x) =( )

( )

f x

g x

1

lim ( ) x

h x

=1

( )lim

( ) x

f x

g x

1lim

x

5 3

2 2

1 1

1 1

x x

x x x

=5 5

4

1

1lim 5 1 5

1 x

x

x

88. (a) Consider2

50

log(1 3 )lim

( 1) x x

x

x e

=

22

2

50

log(1 3 )3

3lim1

55

x x

x x

x

e x x

x

=2 2

2 20 0

log (1 3 ) 3lim lim

3 5 x x

x x

x x

× 50

1lim

1

5

x x e

x

=3 3

1 15 5

89. (a)2

3 1

x f

x

=

22

3 13 6

13 1

x

x x

x

[By defn of f ( x)]

=2 6 2

3 6 3 1

x x

x x

=7

7

x x

90. (c) Since f ( x) is continuous for all x, thereforeL.H.L = R.H.L

2

2 2lim 3 lim 1

x x

ax a x

2a + 3 = 2a2 – 1 2a2 – 2a – 4 = 0 a2 – a – 2 = 0 a = – 1, 2

91. (c) Consider | f ( x) – f ( y)|2 | x – y|3

| f ( y) – f ( x)|2 | y – x|3

2

2

( ) ( ) f y f x y x

y x

Put y = x + x,2

0 0

( ) ( )lim lim x x

f x x f x x

x

[ f (x)]2 0 [ f (x)] 0

92. (d) Let f ( x) =2

1

sin2

xt

dt

f ( x) =1

1 (1 cos )2

x

t dt

= 1

1sin

2

xt t

= 1

sin 1 sin12

x x

f ( x) = 1

1 cos2

x

f () = 1

1 1 12

93. (e) Let y = f ( x2 + 2)

dy

dx= f ( x2 + 2) 2 x

dy

dxat x = 1 is = f (3).2 = 5 × 2 × 1

= 1094. (c) Let f ( x) = x2 + bx + 7

f ( x) = 2 x + b

f (5) = 2 × 5 + b = 10 + b

f 7

2 = 2 ×

7

2 + b = 7 + b

According to the question10 + b = 2 × (7 +b) = 14 + 2b

b = – 4




95. (b) Let x = sin t and y = tan t

dx

dt = cos t and

2secdy

t dt

dy

dx

=2

3

sec 1

cos cos

dy dt t

dt dx t t

96. (d) Let x =a cos3 , y = a sin3

dx

d 3 a cos2 (– sin ) and

dy

d = 3 a sin2 (cos )

dy

dx=

dy d

d dx

dy

dx =

2

2

3 sin cos

3 cos sin

a

a

= – tan

Consider2

2 21 1 tan sec

dy

dx

97. (c) Let y = sin –1 22 1 x x

put x = sin

Þ y = sin –1 22sin 1 sin

= sin –1 [2 sin cos ] = sin –1 [sin 2]

= 2

y = 2 sin –1 x

2

2

1

dy

dx x

98. (b) Given curve is y2 = 4 x + 5on differentiating, we get

22 4

dy dy y

dx dx y

Given line is 2 x – y + 5 = 0 y = 2 x + 5

slope of line is 2. Therefore,

22 1 y

y

put y = 1 in the equation of curve, we get1 = 4 x + 5 x = – 1Hence, point of contact is (– 1, 1)

99. (a) Let f ( x) = 2 x3 – 15 x2 + 36 x + 6 f ( x) = 6 x2 – 30 x + 36= 6( x2 – 5 x + 6)

= 6( x – 2) (x – 3) < 0 x (2, 3)Hence, f ( x) is strictly decreasing in the interval (2, 3)

100. (c) Given curveis y2 e xy = 9e –3 x2

on differentiating both side, we get

y2 e xy 3.2 9 . 2 xydy dy x y e y e x

dx dx

9.e –3

3 3

3 .6 18

dy dye e

dx dx

3 3 3 39 27 6 18

dy dye e e e

dx dx

45.e –3 = 3 39 6 dy

e edx

15 =dy

dx

101. (a) Volume of cylinder = V = r 2h

2 .2 0

dV dh dr r h r

dt dt dt

Given 5,dr

dt height (h) = 3cm

25. 30.(5) 0dh

dt

150

625

dh

dt

102. (e) Since left hand and right hand derivative are not equalat x = 4 f ( x) is not differentiable at x = 4.Hence, Rolle’s theorem is not suitable on f ( x).

103. (a) Let the curve be y = x2 – 21 x

3

22

dy x

dx x

at 1,0 2 ( 2) 4 dy

dx

slope =

1 1 1.

4 4dydx

104. (b) Required minimum value = 2 21 1 2

105. (c) Let I = 2 4 3/ 41

( 1)dx

x x

= 3/ 42 4

4

1

11

dx

x x x

= 3/ 42 3

4

1.

1. 1

dx

x x x

= 3/ 45

4

11

dx

x x




Put4

11 t

x

5

4dx dt

x

I =5

3/4 51 ( )

4 ( ). x dt

t x

= 3/ 4

1

4 ( )

dt

t

=

1/ 4

14

1

4

t c

=1/ 4

4

11 c

x

= 1/ 441

xc

x

106. (e) Let I = 2

(1 )

sin ( )

x

x

x edx

xe

Put xe x = t

( xe x + e x) dx = dt

[e x (1 + x)] dx = dt

I =2

2cosec

sin

dt t dt

t

= – cot t + C= – cot ( x e x) + C

107. (b) Let I = 2(1 )

x xe

dx x

= 2

1 1

1 (1 ) xe dx

x x

= C1

xe

x

108. (c) Let I = sin 2cos sin xe x x xdx

= 2sin 2sin cos xe x x x dx

= ex sin2x + C

[ ( ) ( )] ( ) C x xe f x f x dx e f x

109. (e) Let I = 1 cos x dx

=21 2cos 1

2

xdx

= 22cos 2 x dx

=1

2

sin 22 C x

= 2 2 sin C2

x

110. (a) Let I =2 1 x

dx x

x = sec dx = sec tan d

I =

2sec 1

.sec tansec d

=tan sec tan

sec

d

= 2sec 1 d = tan – + C

= 2 11 sec C x x

111. (c) Let I =2

4

5 xdx

x

put x = 5tan

2

4

5

xdx

x

=

22

4

5 5tan5.sec

5 tand

=2

22 4

5(1 tan )5.sec

(5) tan

d

=2

24

5sec5sec

25tan d

=2

4

5sec 5sec

25tan

d

= 4

1 cos

5 sind

Put sin = t cos d = dt

= 415

dt

t

= 3

1 1C

15 t

= 3

1 1C

15 sin

= 21cosec . cosec C

15

= 2 211 cot . 1 cot C

15

=2 2

2 21 1 tan 1 tan C

15 tan tan




=

3/ 22

2

1 1 tanC

15 tan

=

3/ 22

2

11 5 C15

5

x

x

= 2

321 51 C

15 x

112. (a) Let I =

1 1

0 0

x

x x x

dx e dx

e e e e e

put e x = t

e x dx = dt

I =

1 1

1 1 1

( )

e edt

t t e e t t e

=1 1

1 1 1 1e e

dt dt e t e t e

= 11

1 1log log ( )

eet t e

e e

= 1

1log log ( )

et t e

e

=1

1 log

e

t e t e

=1 1

log log2 1

e

e e e

=

12

11

1log

ee

=1 1

log2

e

e

113. (c) Given curves are y = e x and y = e –x

Now, e x = e – x x = 0

Area = A = 1

0

x xe e dx

= 1

0 x xe e

= 1 0 0 11 1e e e e e

e

= e +1

2.e

114. (b) Let I =1

1 log

3

e x

dx x

put 1 + log x = t

1

dx dt x

when x = 1, t = 1

when x = e, t = 2

I =

2

1

1

3 t dt

=

22

1

1 1 4 1 1 3

3 2 3 2 2 3 2

t

=1

2

115. (c) Let I =

1 3

80 1

x dx

x

put x4 = t 4 x3 dx = dt

I =

1

20

1

4 1

dt

t

= 1

10

1tan

4 t

= 14 4 16

116. (c) Let I =

4

2

logt dt

t

Put log t = x

1dt

t = dx

I =

log4

log2 x dx

=

2log22

log22

x

= 2 214 log 2 log 2

2

= 213 log 2

2

= 23log2

2




117. (b) Given differential equation is

(kx – y2) dy = ( x2 – ky)dx

(kx) dy – y2 dy = x2 dx – ky dx

kx dy + ky dx = x2 dx + y2 dy

k [ x dy + y dx] = x2 dx + y2 dy

k [d ( x y)] = 3 31

3d x y

kxy =3 3

13

x yK

x3 + y3 = 3k xy + K 1

x3 + y3 = 3k xy + C

118. (b) Given differential equation is 1 xdye

dx

dy = (e x + 1) dx

on integrating both side, we get

y = ex + x + C

119. (e) Given differential equation is2

2

32

d y dy y

dxdx

2

2

32dy d y

ydx dx

On squaring both sides, we get

23 2

2

dy d y y

dx dx

order is 2, degree is 2

120. (a) Given differential equation is sin 2 cos 1 dy

x y xdx

cos 1

2sin sin

dy x y

dx x x

2cot cosecdy

x y xdx

I.F = 2cot x dxe

=2log(sin ) 2sin xe x

kerala pet 2014 paper

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