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ANSWERS TO PAPER I

PHYSICS

1. As angle between voltage and current is 90

Power = 0

(d)

2. As they collide elastically so there velocity hence kinetic energy gets exchanged. SoB will

rise to 4h andA will rise to h

(a)

3. Electric field = 1000 N, V101011000. 2drEdV

(a)

4. As in + particle emission proton is converted into neutron

neutron to proton ratio increases

(a)

5. Applying Kirchoffs voltage law in a loop

04122 II

I162

81I

833I A

(b)

AI

4

4

4

4

I

I

3I

2V

6. (a)

7. For a pure inductor circuit potential , tVtVVV sinsin 00 and current

2sin0 tii thus angle =

2

(c)

8. Frequency of 1st source [ 1 = 100 cm]

1

1

vf

Frequency of second source [ 2 = 90 cm]2

2

vf

Beat frequency =21

21

vvff

= 4410090

10010396100

100

1

90

1396 Hz.

(d)

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9.M

RTvrms

3. MandR are constant for the same gas.

Tvrms Tv

v 300

2, T= 1200 K.

(d)

10. We know that,2

2

0

2

4

1

nn r

Ze

r

mv(i) and according to Bohrs model

2

nhmvrn

putting v = in (i) we get2

2

0

322

22

4

1

4 nn r

Ze

rm

hmn

2

2

0

2

mZe

hnrn

2nrn

where h= Planks constant, m = mass of electron

Z= atomic weight, n = principal quantum number

(d)

11. Refractive index,2

BA (Cauchys law)

So, refractive index depends upon the wavelength of the light.(b)

12. We know that,E= V/d

V/m101100.5

5.0 67

E

(a)

13. As the capacitors are identical, they will finally have charge Q/2 each.

Initial energy of the system =Ei = C

Q

2

2

Final energy of the system =Ef= 2C

Q

C

Q

42

)2/( 22

Heat produced = loss in energy =EiEf=C

Q

4

2

(b)

14. E=xd

dV8x at (1, 0, 2) iE 8

(a)

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15. Equation of trajectory,22

2

cos2

1tan

u

gxxy

1tan (i)

gucos (ii)

gg

g

g

uT

22sin2

(b)

16. Pressure at depth h = hgdP

Pressure inside the bubble will ber

TdhP

2

(b)

17. (c)

18. As 2/3RT ,2/3

1 KRT ,2/32/3

2 )01.1(KRT , %5.11001

12

T

TT

(b)

19. Rmmgmg 2'

Rmmgmg 2

5

3

5

22 mgRm

R

g

5

2

(a)

mg m2R

20. When the endA is stopped, reaction force is acted upon

endA hence angular momentum about endA is

conserved

3

4 2mamva

a

v

4

3

(d)

A B

R

v

2a

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21.

21

21

mm

rmr and

21

12

mm

rmr

2

22

2

11 rmrmI

2

21

212

2

21

2

12

2

21

2

21

)()(r

mm

mmr

mm

mm

mm

mmI

(a)

r11 m2

r2

22. (c)

23.g

v

g

v

2

sin2cossin2 222

2tan

Range =g

v

g

v

g

v

5

4cossin22sin 222

(a)

v

24. As9

325

9

5FC

9

5

dF

dC

(b)

25. As internal energy can be increased by doing work also.(c)

26. (c)

27. As temperature and surface area is same

(a)

28. As density is maximum at 4C and volume in both case increases

(a)

29. 100010332

10332',' 0 ff

vv

vvf

s

(c)

vs vo

30. The particle may move alongx-axis andy-axis

(a)

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31. Intensity2

40

r, t

22 )50(

403

)25(

40 12)2(3 2t s

(c)

32. As eqPPP 21 , 220

100

80

100P ,

80

3100

20

1

80

11002P 3.75 D

(d)

33. (c)

34. LetFis the upward force then MFMg

In second case

)()( mMgmMF , MFMg

)2()( mMgmMMg , mMmg 2

g

Mm

2

(b)

35. As force is maximum at extreme position Acceleration is maximum at that point

(a)

36. 7 1 = d1 = 7 d

D

2

1

and 7 2 = d2 = 7 d

D

2

2

2

1

2

1

d

d

(a)

37. Work done is area underPVdiagram =PV

(a)

38. (b)

39. iVD 4 , jVM 4 , jiVDM 34 , 5|| DMV

(c)

40. As net external force on the system is zero hence centre of mass does not move

(b)

41. Current through loop = ne

Magnetic field at centre =r

ne

2

0

(b)

e

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42. As2

1

2

1

n

n

V

V, 2401

1

2

2 Vn

nV V

(a)

43. (d)

44. Taking potential of pointA to be zero

Potential ofE=1V

Potential ofC=1 + 2 = 1V

Potential ofD = 31 = 4V

(a) C

A B

D3V

1V

1V

2V

2

2

22

1

E

45. Ashc

eV ,eV

hc

So if voltage is doubled cut off wavelength is halved.

(b)

46. LetN0 is initial no of nucleus ofX

2/0)2( tt

NX , 2/0 )2(

11 tt NY , 2/02/0 )2(7)2(

11 tt

NN

2/)2(

81

t, 8)2( 2/t , t= 6

(a)

47.8

531

34

106.61010

1063.6

mv

hm

(b)

48. As K.E. of an electron does not depend on intensity

(b)

49. Drawing the phasor diagram, potential drop across

Resistor = 100 V

I= 2 amp

(a)

400

400

100

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50. IfI1 is the current in coil of radiusR1 then magnetic field at the centre of small coil =

1

110

2R

NI

Flux through circular coil 22

2

1

110

2NR

R

NI

1

2

2210

1 2R

RNN

IM

(b)

51. Potential difference across the ring of radiusx and width dx

2

2

0

RBxBdx

R

(d)

x

D

dx

52. As2

1

TM , Mis pole strength and Tis time period

221

211 4 TMTM ,

2211 44 TMM , 12T

(c)

53. (c)

54. As cyclotron frequencyM

qBf

20

(c)

55.2

2qI

2

2

2

2 22 RqRq

M

q q

w

Angular momentum 222 RmmRRL

m

q

Rm

Rq

L

M

22 2

2

(a)

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56. Magnetic field can change only direction(b)

57. If iT is inversion temperature then 20270270iT , 520iT C

(b)

58. As potential difference across each resistance is same

sR

II

34

33

343663 , sR33

3663

111sR

(b)

33I/34

Rs

I/34I Soj = 3663

59. Assuming potential at Cto be zero

Potential atA = 3V

Potential atB = 2V

VVV BA 1

(b)

A

B

D C

2 amp2

23

3

60. Let internal resistance be ri

Current in loopirr

EI

iIrEV

i

i

rr

ErEV

r

riI E

rV

VEri

(c)

CHEMISTRY

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61. (d)

62. (a) :O=O: O:

.. .. ..

.. O=O O:

..

.. ....

+ ..

:O O O:

..

.. ..

+ ....

63. (b)

64. (c) 3NO + 4H+ + 3e 2H2O + NO

(+5) (+2)

65. (a) In ionic reactions only ions are exchanged and oxidation states remain same.

CaCO3(s) CaO(s) + CO2(g)

(+2) (+4) ( 2) (+2) ( 2) (+4)

66. (d)

67. (b)32

2

n

n

2

2

n

O

4

1

x

125

x = 0.5, wH2 = 0.5 2 = 1 gm

68. (a)

69. (b) Greater the s character, more acidic is the hybridised carbon atom.

70. (c) Alkali metals form oxides, peroxides and superoxides. Reactivity with waterincreases due to rapid decrease in ionisation energy on moving down the group.

71. (c)

72. (c)

73. (b)

74. (a)

75. (d) Conjugated diene are generally highly stable; hence in the case of such types of diene

heat of hydrogenation is minimum.

76. (b) Terminal alkynes react with amm. AgNO3 or OH)NH(Ag 23 to give white

precipitate.

77. (c) CH CH + H2O2Hg/H

CH=CH2

OH

CH CH3

O

78. (b)

79. (b)

80. (a)

81. (b)

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82. (c) In AgNO3 + NaCl AgCl + NaNO3, only ions are exchanged hence this

reaction is an ionic reaction.

83. (a)

84. (a)

85. (a)

86. (b)

87. (b)

88. (a) K2Cr2O7 K2CrO4

2 + 2x + 4 = 0 2 +x 8 = 0

x = +6 x = +6

89. (d) Electronegativity difference for O H bond is maximum among given compounds.

90. (d) Decreasing order of stability of free radicals obtained in halogenation of alkanes is

(3 > 2 > 1).

91. (a)

92. (c)

93. (b)

94. (c)

95. (c) 4

3

2

2

OCFe + 4MnO H

2

43

OCFe + Mn2+

n = 3 n = 5

5 mole of 4MnO = 3 mole of FeC2O4

Mole of 4MnO = 15

3=

5

3.

96. (c) Fe+2 Fe+3 + e2

2S 2S+4 + 10e

FeS2 2S+4 + Fe+3 + 11e

Equivalent mass of FeS2 =

11

massmolar.

97. (d)

98. (b)

99. (b) For a bcc structure, Z = 2

M =Z

Na A3

=2

1002.6)10250(0.82310

= 37.6 g mol 1.

100. (a) Kp =

5

32

PCl

PClCl

P

PP

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Since, at equilibrium moles of each component is same, so partial pressure be same

i.e.2

ClP =3

ClP =5

ClP = 36

2= 1

Kp= 1.

101. (b)

102. (c) t =t

o

a

alog

K

303.2=

5.2

20log

K

303.2= 8log

0693.0

303.2= 30 min.

103. (c) o

o

P

PP= Xsolute or o

o

P

PP1 = 1 Xsolute = Xsolvent = oP

P.

105. (a) H = U + ng RT

H U = ng RT =1000

298314.83 = 7.43 kJ mol 1.

106. (d) Ecell =ocellE

1.0

1log

2

059.0; ocellE = 0.54V.

107. (b) Mn7+ + 5e Mn2+

Thus, 5 moles of electron = 5 faraday.

108. (a) Using the relation rate (conc. of reactant)nwhere n is order of the reaction.

In the question, since rate is directly proportional to the conc. of reactant, hence

the reaction is of first order.

109. (b) Total geometrical isomers = 2n

= 22

= 4.

110. (c)

111. (d)

112. (b) H+ will attack the double bond and would give

OHCH3CH2 CH+

which would be resonance stabilized.

113. (d)

114. (c)

115. (b)

116. (c)

118. (c)

119. (b)

120. (c)

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