PCB-5 (1st April, 2017) ANSWER KEY
CHEMISTRY
Vcorrection  nb
1. C
Sol. A =
; B = CH3CH2OH;
O
+ CH3CH2OH
OH
OC2H5
O
2. B
3. D
Sol. (c) Exception although it has  - hydrogen it
undergo cannizaro reaction.
(d) Ph – CH2 – CHO (due to presence of  hydrogen it will not under go cannizaro
reaction)
4. C
5. C
6. C
Sol. BF3  sp2
NO 2  sp2
NH 2  sp3
H2O  sp3
7. D
8. C
Sol. It is cyclic condensation to give mesitylene
(Trimethyl Benzene)
9. C
Sol. Ozone & PAN are eye irritant; NO irritate throat
& nose (NCERT) classically smog having SO2
is reducing smog.
10. B
2r1
Sol Time period in first orbit (T1) =
;
v1
2r2
Time period in second orbit (T2) =
v2
T1 2r1 v 2
r v


 1 2
T2
v1 2r2 r2 v1
Velocity of electron in first orbit
2.188  10 8
cm s 1
=
1
Velocity of electron in second orbit
2.188  10 8
cm s -1
=
2
Radius of first orbit = 0.528 × 10-8 cm
Radius of second orbit = 0.528 × 10–8 × 4 cm.
T1
0.528  10 8
2.188  10 8
1



T2 0.528  10  8  4 2  2.188  10 8 8
Alternatively use T  n3
11. A
Sol.
Pm = dRT
PA m A d A RT

PBmB
dBRT
PA 1
 2
PB 3
PA : PB = 6 : 1
12. B
an 2
Sol. Pcorrection 
Unit of a = atm litre2 mol–2
2
V
Unit of b = litre mol–1
 a  atm litre 2 mol-2
Unit of   
b
litre mol-1
= atm litre mol–1
13. C
92
x
92
(2.3)2 =
x
92
x=
x  17
5.29
14. A
Sol. O2 contains an unpaired electron (M.O.T.)
15. A
16. D
17. B
18. A
19. B
Sol. TI3 have T+ & I 3 ion
20. A
21. C
Sol. Hydrolysis of substituted chlorosilanes yields
corresponding
silanols
which
undergo
polymerisation. Out of the given chlorosilanes,
only (CH3)2SiCl2 will give linear polymer on
hydrolysis followed by polymerisation.
Sol. 2.3 =
H OH
Cl
H3C
-2HCl
+
Si
Cl
H3C
H3C
OH
Si
H OH
H3C
Dimethyl dichlorosilane
CH3
OH
Dimethyl silanol
CH3
|
|
HO  S i  O H  HO  S i  OH 
|
|
CH3
CH3
|
CH3
CH3
|
HO  S i  O  S i  OH 
|
|
CH3
CH3
22. D
Sol. i = 1 -  +
i=1–

2
0.54 = 1 -

2

2

 0.46
2
 = 0.92
23. C
24. A
Coordinated by: Dr. Sangeeta Khanna (Chemistry), Harpal Singh (Biology), KP Singh (Physics)
1
+2
Sol. When excess of an electrolyte is added, the
colloidal particles are precipitated. This is
because colloidal particles take up ions
carrying charge opposite to that present on
themselves. This causes neutralization leading
to their coagulation
25. B
Sol. Hsolubility = HL.E. + HH.E.
Gsolubility = Hsolubility - TSsolubility
when Hsolubility
becomes more –ve then
Gsolubility becomes more –ve,
Further Gsolubility = -2.303 RT log Ksp hence if
Ksolubility increases, solubility increases.
Down the group for oxides of 1st group, the
change in lattice energy is more than the
change in hydration energy. Hence, Hsolubility
becomes more –ve down the group.
Hence, Gsolubility become more -ve down the
group hence solubility increases.
Choice (b) is correct.
Choice (a) is false as G becomes more –ve
down the group and not +ve down the group.
Choice (c) is incorrect because Ssolubility can’t
be zero, it is +ve.
Choice (d) is ruled out from explanation above.
26. D
Sol.
NBa(OH)2 = 2 × M [ it is diacidic base]
N1V1 = N2V2
2 × M1 × 25 = 35 × 0.1
3. 5
M1 =
= 0.07 M
50
27. B
28. D
Sol. The characteristics of a concentration cell is
that each half cell is made of same metal and
metal salt solution and only the concentration
of metal salt solution in each half cell is
different.
e. g Zn(s) | Zn 2  || Zn 2  | Zn(s)
C1M
C2M
Here E 0cell  0; E cell  E 0cell 
as Ecell = -
2.303 RT
log Q
nF
2.303 RT
log Q
nF
29. B
Sol. In this process nutrient enriched water bodies
support a dense plant population, which kill
animal life by depriving it of oxygen.
30. C
Sol. It is group 12 – 16 compound
31. A
Sol. Starch is a polymer of -D-glucose
32. B
A
2.303
Sol. K 
log 0
t
A
A
2.3  10 3 2.303

log 0
2.303
1000
A
0.29
1 = log
A
0.29
 10
A
A = 0.029 M
33. B
Sol. 4 mole e– means 4F
4F will deposit
1 mole O2
4F will deposit
2 mole H2
3 mole gas
Volume = 3 × 22.4 lit
= 67.2 lit.
34. A
Sol. [Ag(CN)2]– is linear (sp) with no unpaired
electron, hence magnetic moment = 0.
[Cu(CN)4]3– is tetrahedral (sp3) with no unpaired
electron, hence magnetic moment = 0
[Cu(CN)6]4– is octahedral (sp3d2) with one
unpaired electron, hence magnetic moment =
1(1  2) =1.73BM
[Cu(NH3)4]2+ is square planar (dsp2) with one
unpaired electron, hence magnetic =
1(1  2)
=1.73BM
[Fe(CN)6]4– is octahedral (d2sp3) with no
unpaired electron, hence magnetic moment =
0.
35. C
Sol. CO is a strong ligand & carbonyls have
backdoantion or synergic effect.
36. D

B
C 
D
Sol. A
0.8 M

0.8 M
At equilibrium,
0.8 – 0.6
0.8 – 0.6
0.6 M
0.2 M
0.2 M
0.6  0.6 36
K

 9.0
0.2  0.2
4
37. C
Mg 2   2OH 
Sol. Mg(OH)2
K sp  [Mg 2 ]

0.6 M
[OH ] 2
1  10 -14  [Mg 2  ] [10 -2 ]2
[pH  12  [H ]  10 12 mol L-1
[OH  ]  10 2 mol L-1]
[Mg 2  ] 
10 14
10
4
 10 10 mol L-1
Coordinated by: Dr. Sangeeta Khanna (Chemistry), Harpal Singh (Biology), KP Singh (Physics)
2
+2
Solubility = [Mg 2  ]  10 10 mol L-1
38. A
Sol. If intensity of incident light is doubled, there will
not be any change in the energy (Z) of
photoelectrons ejected but the number of
photoelectrons ejected (Y) will increase.
39. D
1
Sol. B.O. in H2 = (2  0)  1
2
1
B.O. in H 2  (1 – 0) = 1/2
2
1
B.O. in H2  (2 – 1) = ½
2
Although H2 and H 2 have the same bond
order, H 2 has longer bond length than H2
because of the presence of one electron in the
antibonding orbital which repels the two Hatoms from coming close. Further, because of
larger B.O., H2 molecule has the shortest bond
length. Thus, the actual order is:
H2  H2  H2
40. A
41. C
RB  dA

R A  dB




4
4
 1
RB  24  
2
24

 1. 5 
16
47. Potential difference between the point where
electron is released and the plate is
80   20 
 50 V
2
Using conservation of energy,
1
mv 2  V
2
= 1.6 x 1019 x 10
= 50 eV
48. 1 = 310
2 = 316

Hence, beat frequency = f2  f1
K a 1.85  10 5
= 1.85 × 109

14
Kw
10
This reaction is reverse of salt hydrolysis.
CH3COOH + NaOH
CH3COONa + H2O
42. C
43. C
44. D
45. A
r  0.65
Sol.
(MgO) Octahedral

 0.464
1 .4
r

Sol. K =
r
r


0.65
 0.35
1.84

MgS (Tetrahedral)

48. b
54. c
60. a
66. d
72. b
78. a
84. b
90. a
49. b
55. c
61. a
67. d
73. d
79. a
85. d
50. Applying lens formula to objective, we have
1
1
1


v 0  200 50
A
2
200
cm
3
For eyepiece,

ue  
1
1
1


 25 ue 5
25
cm
6
So, the length of tube,
L = v0 + ue = 70.83 cm
3
4 R 
4 3
R  G.   
3 2
3


2
2
(2 R)
R

2 R  
2

4
1 1 4
 GR   . 
3
4 8 9
G.
For same mass & material
1
51. c
57. a
63. c
69. b
75. a
81. b
87. d
v0 
51. E = Esphere + Ecavity
L
46. R  
A
m = dAL
 m
R
A dA
R
50. c
56. c
62. b
68. b
74. c
80. a
86. c
3
units

remains constant.
PHYSICS
47. d
53. d
59. d
65. c
71. c
77. b
83. c
89. b
316 310

2
2
49. Use pV = nRT, T is absolute temperature, n

46. a
52. c
58. d
64. c
70. c
76. a
82. c
88. d
310
unit
2
316
f2 
unit
2
 f1 
or
1
d4
Coordinated by: Dr. Sangeeta Khanna (Chemistry), Harpal Singh (Biology), KP Singh (Physics)
3
+2
2 
1
 GR  

 3 27 
7

GR
27
Resistance of each part,
R1 = R2 =
Time constant,
L
L L
R  R2
  net  1 2  1
Rnet L1  L2
R1 R2
dp
 1.5 t  3
52. I 
dt
5
 dp   1.2 t  3 dt
58. In the absence of battery charge & field remain
constant.
0
 1. 2
t2
3t
2
R
=3
2
5
d
p
59. 1   1
p 2  d2
0
= 0.6 52 + 3 (5)  0
6

 25  15
10
= 30 C




60. The resultant power of the combination,
53. To observe diffraction wavelength should e
comparable to slit size
P = P1 + P2 = 3 D – 1 D = 2 D
As power is positive in sign, so it behaves as, as
converging or convex lens. Now, focal length of
the combined lens is given by
54. Force = q (v x B)
f
q = charge of the electron. V = 2 x 105 m/s and

B  ( î  4 ĵ  3 k̂) T
î
ĵ
F  q 2  10
1
5
61. If v is the velocity of the particle, then
1
E  mv 2
2
k̂
0
0
4
3
Or v 
|F | q| î(0)  ĵ (6  105)  k̂ (8  105) |
q
 36  64 .105= q.10
1
1
 m  50 cm
P (ln D) 2
6
F = 1.6 x 1019 x 106
= 1.6 x 1013 N
55. x = 6 + 3 sin 2 t + 4 cos 2 t
= 6 + 5 sin (2 t + 530)
Here equilibrium position is at x = 6 m and
maximum displacement from equilibrium
position is 5 m 50 amplitude of oscillation is 5 m.
2E
m
As the particle describes a circular path of
radius r in a perpendicular magnetic field.
mv 2
 vqB
r
i.e,
r
mv m

q.B qB

2E
m
2 Em
q2 B2
62. For a particular value of V, higher the temp.
higher will be the PV product.
63. The momentum imparted to one bullet is,
56. Initial phase difference due to phase constant

0 = A  B =
2
2
  2
Due to path difference,  

5
So, total path difference =
2
57. Inductance of each part,
L
L1 = L2 =
= 0.9 x 104 H
2
mv = 10 x 103 x 800
= 8 kgm/s
In is a total of
120
 2 bullets are fired, so
60
momentum transferred to bullets in 1 s is,
p0 = 8 x 2 kgm/s2
=16 N
i.e., we can say 16 N is the average force
exerted by gun on bullets, which is same as
average recoil force experienced by gun.
Coordinated by: Dr. Sangeeta Khanna (Chemistry), Harpal Singh (Biology), KP Singh (Physics)
4
+2

|e|
64. | e |
,  = f  i and i 
t
R
65. U 
f
nRT
2
(f = degree of freedom)


4
i.e., Current leads the voltage
Or
Voltage lags the current
U1 = U2
f1 n1 T1 = f2 n2 T2

n1 f2 T2 32 6



51 5
n2
f1 T1
1 2C 2

V
2 3 
1 3C 2
Eb  
V
2 2 
67. Ea 
_ _ _ (1)
_ _ _ (2)
Ea 4

Eb 9
68. Now, phase difference =
2
 path difference

2
 path difference

600  
180 0

25
 path difference
3
Or Path difference 
3  60  
 0.1
2   5  180
69. Use work-energy theorem.
2

1

3
1
| a | 1
Or | a | 1  3

| a | 1

| a | 3  1 unit
73. Distance between adjacent nodes (or
 1 2

antinodes)     
2 2 k  k
74. Use  = BA and B = 0 r MI
75. For circle
 I
B1  0 1 .L1
4  R2
For arc
 I
B2  0 2 .L2
4  R2
For B1 = B2
I1 = I2.L2
I1 L2 30


I2 L1 40
76. Field along axis of coil B 

 0 iR 2
2 R2  x2
3 / 2
 i
At the centre of coil, B'  0
3 / 2

2
2
B'  0 i 2 R  x


B
2R
 0 iR 2
1

T
T1

T2

2R
3
kT
2
h
h
 
p
2 Km
70. KE 

3


| a | 2 cos 60 0
dv
 ks1/ 3
ds
Or
v2  s2/3
or v 1/3
Now P = F.v
Or
P  s1/3.s1/3
Or
P  s0 i.e., power is independent of s.
Or

2 sin60 0
tan 45 0 
i.e., acceleration a  S
600 

| a |  | b |cos 
1/3

| b |sin 
72. tan  
66. F  s1/3
or

4

Thus
927  273
2
27  273
B' 
 2 = 2 1
X  XL 2 L  L

1
71. tan   C
R
L

R 2  x 2 3/ 2
R3

B  R2  x2

R3
3 / 2
54  32  4 2
33
3 / 2
Coordinated by: Dr. Sangeeta Khanna (Chemistry), Harpal Singh (Biology), KP Singh (Physics)
5
+2

54  125
27
1 
1
1
 1.5  1  

   10  20
= 250 T
When plane surface is silvered,
r 
1
4   10 7  150

105
6
86.  
T
where m = mass per unit length of
m
string.
v
500

0.2
500  10
2
 2500  50 m/ s
79.
80. v  180  7 x ,
v 2  180  7 x
From the dimensions, 180 has square of
velocity,
 7 x = 2 ax
a =  3.5 m/s2
81. Since dQ = dU + dW

87.  
dW = dQ  dU
= 110  40 = 70 J
82. 2 x   A 2  x 2
2 .1   4  1
h
2 km


(3)
2


9
h
2 qVm
i + r = 900
as r = i  2 i = 900
90 0
 450
2
89. Latent heat of vaporisation
Q = mL = 1 x 540 = 540 cal
V = V2 — V1 = 1681  1 = 1680 cm3
= 1680 x 106 m3
and W = PV = 105 x 1680 x 106
 168 J 
168
cal  40 cal
4.2
Change in internal energy.
U = Q  W
= 540  40 = 500 cal



1
1 
1 

  t2 1 
  t3  1 
90. t1 1 




1 
2 
3 



1 
1 
1 



 4 1 
  6 1 
  8 1 

1
.
5
1
.
4
1
.3 






 3
Z2
88. Reflected ray is normal to the incident ray,
i
mp v p
mp

1


.2 
 p m v  4 mp
2
1
For lithium,  ' 
78. Velocity of transverse wave along string = v
v
F 20

 10 cm
2
2
f
77. B = 0 r H
B
r 
0 H
4  0.5 6  0.4 8  0.3


1.5
1.4
1.3
= 1.33 + 1.71 + 1.85
= 4.89 cm
2

3
2 2
T


3
2 kze2
x
2 m
84. Use T 
Bq
83. Use K 
85.
 1
1
1
   1 

F
 R1 R 2



Coordinated by: Dr. Sangeeta Khanna (Chemistry), Harpal Singh (Biology), KP Singh (Physics)
6
PCB-5 (1st April, 2017) ANSWER KEY
BIOLOGY
91
a
92
d
93
d
94
b
95
d
96
a
97
d
98
d
99
b
100
a
101
b
102
b
103
a
104
b
105
d
106
a
107
b
108
d
109
d
110
c
111
a
112
c
113
a
114
b
115
a
116
b
117
b
118
c
119
a
120
b
121
b
122
a
123
d
124
c
125
c
126
d
127
b
128
d
129
d
130
a
131
d
132
d
133
d
134
d
135
b
136
b
137
a
138
b
139
d
140
a
141
c
142
b
143
d
144
a
145
a
146
a
147
c
148
b
149
c
150
b
151
d
152
a
153
c
154
d
155
d
156
a
157
a
158
a
159
a
160
a
161
c
162
d
163
c
164
d
165
d
166
b
167
d
168
c
169
d
170
a
171
c
172
b
173
a
174
d
175
b
176
d
177
d
178
d
179
a
180
c
Coordinated by: Dr. Sangeeta Khanna (Chemistry), Harpal Singh (Biology), KP Singh (Physics)
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