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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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