TM
HINT & SOLUTIONS
CLASS-XII _ NEET - 2020
ONLINE TEST SERIES _ UNIT TEST - 02
[ DATE : 04-08-2019 ]
BIOLOGY
1. C 2. A 3. C 4. C 5. C 6. C 7. B
8. B 9. A 10. B 11. D 12. C 13. B 14. D
15. C 16. D 17. A 18. B 19. B 20. B 21. B
22. B 23. C 24. C 25. B 26. B 27. A 28. D
29. C 30. A 31. B 32. D 33. C 34. A 35. B
36. C 37. B 38. D 39. A 40. C 41. C 42. C
43. D 44. D 45. D 46. D 47. A 48. A 49. C
50. C 51. D 52. C 53. C 54. A 55. B 56. C
57. D 58. A 59. C 60. C 61. D 62. C 63. B
64. A 65. C 66. B 67. B 68. D 69. A 70. A
71. D 72. B 73. A 74. C 75. C 76. C 77. C
78. D 79. A 80. C 81. C 82. B 83. B 84. C
85. B 86. C 87. C 88. D 89. D 90. C
PHYSICS
91. A 92. D 93. B 94. C 95. A 96. C 97. B
98. C 99. A 100. D 101. A 102. C 103. A 104. C
105. A 106. B 107. B 108. B 109. B 110. A 111. B
112. C 113. B 114. A 115. A 116. A 117. B 118. A
119. A 120. C 121. B 122. A 123. D 124. C 125. A
126. B 127. A 128. B 129. A 130. A 131. C 132. B
133. D 134. B 135. D
CHEMISTRY
136. D 137. D 138. B 139. D 140. A 141. B 142. C
143. A 144. A 145. B 146. D 147. C 148. B 149. D
150. A 151. C 152. D 153. B 154. A 155. D 156. D
157. A 158. C 159. D 160. A 161. A 162. D 163. C
164. B 165. A 166. D 167. C 168. C 169. A 170. C
171. D 172. D 173. B 174. D 175. D 176. A 177. C
178. C 179. A 180. C
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PHYSICS
91. A 99. A
Diffraction of sound is more evident in daily
2
= x life than diffraction of light. Because the wave
length of sound waves is greater than that
f = 7.69 2
of light
92. D
Both produce interference 100. D
For easy occurrence of Fresnel's diffraction
93. B obstacle should be of the order of wave
y1 = A1 sin t length, narrow opening should be of the order
y2 = A2 cos (t + ) of wave length, source and screen should be
at finite distance from the obstacle
y2 = A2 sin (t + + )
2
101. A
2 Diffraction of sound is more evident in daily
= x
life than light waves, because wavelength
( s) of sound waves >> wavelength (1) of
= 2 x light waves
2
102. C
x = Distance from central ring to first maxima
2 2
3 D D D
x= – =
2 a a 2a
94. C
I KA2 l = 6000 Å, D = 1m, a = 10–4 m
A= A12 A 22 2A1A 2 cos 6 10 7 1
x= = 3 × 10–3 = 3 mm
2 10 4
A= I1 I2 2 I1 I2 cos
103. A
95. A
conditions of interference 10
= (2 × 6 – 1)
µ 2
96. C
the phenomenon of di ffracti on can be 10 11 20
= µ = = 1.8
observed, when the obstacle is of the same µ 2 11
order as the wavelength of light used
104. C
Fringe width . Therefore, and hence
97. B decreases 1.5 times when immersed in liquid.
Phenomenon of diffraction occurs for all kinds The distance between central maxima and
of waves 10th maxima is 3 cm in vacuum. When immersed
98. C in liquid it will reduce to 2 cm. Position of
central maxima will not change while 10th
Diffraction of light is observed only, when
maxima will be obtained at y = 4cm.
the obstacle size is of the same order that
of wavelength of light
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105. A 112. C
Statement-I is true, Statement-II is true and 113. 8
Statement-II is correct expl anation of
Statement-I. t 1.7 1 D t 1.4 1 D D
Shift S 5
d d d
106. B
By theory t 8 103 mm 8 m
n11 = (2n2 – 1) 2/2
31 = (2 × 5 – 1) 2/2 114. A
9 2
31 = 115. A
2
21 = 32 116. A
10liquidD 11 airD 1
107. B =
d 2d
I
IR = 4I0 cos2 (4I 0 ) 20 air 20 liquid
2 2 = =
11 liquid 11 air
1 1.8 = liquid
cos2 =
2 2 2
117. B
dx
= d sin =
b
=
D
d.x
2 D
0.1 × =
180 f
/2 d.x
2 D 600 10 9 180
d=
0. 1
D
x= = 1.25 × 10–4 m = 0.125 mm
4d 6 18 10 6
d=
108. B
12.27 D d = 34.39 × 10–5
Y
V d d = 3.44 × 10–4
V
118. A
109. B
119. A
12.27 D
Y
V d Imax Imin
f= I
V max Imin
110. A
D 2
= I1
d 1
I2
1
111. B
When light travels through a medium of Imax
I1 I2 2
I1
1
2
1
refractive index n, the wavelength changes f=
Imin
=
I1 I2
2
I2
Imax I1 I2
2
2
1 I1
o Imin 1 1
to . Therefore the path difference I I I2
1 2 1
n 2
I1
1
2 2 I2
x n x
o
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2
1
130. A
1
1 ( 1)2 ( 1)2
I1 = 4I0 cos2
f= = 2
1
2
( 1)2 ( 1)2
1 x=0 I1 = 4I0
1
2x x
Dx= = × =
y y 2
4 2
= = x I1 2
2 2 1 I2 = 4I0 cos2 = 2I0 Þ =
4 I2 1
120. C
131. C
A1 = 2A2 Þ I1 = 4I2 = 4I0
121. B
2 2
d
no. of maxima = 2 + 1
Imax = I1 I2 = I1 4I2 = 9I2
= 9I0
2 I = I1 + I2 I1 I2 cos
no. of maxima = 2 + 1 = 5
= I2 + 4I2 + 2 I2 4I2 cosf
122. A Im
Hyperbolic = 5I2 + 4I2 cos = (5 + 4 cos)
9
123. D Im Im 2
= [1 + 4 (cos)] = 1 8 cos
2 6 9 9 2
= × =
3
132. B
I = I0 cos2 = I0 cos2
2 6
B
I 3
= R
I0 4
124. C S1 d S2 O
31D 42D
= 31 = 41 Path differnce on the circle of radius R around
d d
O on the wall will be same hence concentric
3 3 circle.
2 = 1 = × 590 = 442.5 nm
4 4
133. D
125. A
134. B
Initiancity of parallel beam is cylindrical
therefore the wave front will be planar.
135. D
126. B
= (2n - 1)
2
127. A
D
128. B x5 =
d
Both statements I and II are correct but
statement II does not explain statement I. 9
129. A D=
2
Constant + source
I1 = 4I0 cos2 = 4I0
2
For incomepent source
I2 = I0 + I0 = 2I0
I1
I2
= 2.
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CHEMISTRY
136. D 161. A
2.5 M aqueous solution means 2.5 moles of
137. D the solute is dissolved in 1000 g of water.
nsolute = 2.5 mol
138. B 1000
nwater = 55.55mol
18
139. D nsolute
Xsolute = n
solute nwater
140. A
2.5
0.043
141. B 2.5 55.55
142. C 162. D
2.4 M-KCl and 1M-K3[Fe(CN)6]
143. A
163. C
P n solute
144. A =
Ps n solvent
145. B 10 – 9 (1 / m)
=
9 (20 / 200)
146. D
m = 90
147. C
164. B
The vapour pressure and freezing point are
148. B the highest for urea
149. D 165. A
Molality of solution should be same
150. A M1=M2;
4 1000 10 1000
151. C or MB=160
60 96 MB 90
152. D 166. D
0.1 M Ca(NO3)2 and 0.1 M Na2SO4
153. B
167. C
154. A sugar= x
155. D
w w
156. D mw v sugar mw v X
157. A 5 1
342 100 mw 100
158. C mw=68.4
159. D 168. C
Mole fraction = Osmosis takes place if osmotic pressure of
both sides of solution is different.
moles of alcohol 2 2
0.25
total moles 26 8 169. A
+ —
KBr K + Br
1 – –
160. A
1–
P – Ps n 6 10 1–
x2 2 0.02 i 1
P n1 60 90 1
i=1+0.8=1.8
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174. D
80
0.8 Hsoln = H1+ H2 – H3
100
Tf=i. Kf.m = 1.8×1.86×0.5 175. D
= 1.674 K
F.P. of solution = 273–1.674=271.326 K i
m.wt
170. C
given PS = 550 mm III < IV < I < II
p 0A = 400 p 0B = 600
176. A
1 x
Now 550 = 400 × + 600 A C A 7 58.5
1 x 1 x 0.61
550 + 550x = 400 + 600x B C B 95 7
50x = 150
150 177. C
x= =3 p°–Ps loss in mass of water chamber
50
ps loss in mass of solution chamber
171. D P PS n w solute M
Largest no. of solute particle’s highest will
PS N m W
be B.P.
0.05 10 18
172. D
2.5 msolute 90
p 0A = 100 nA = 1
msolute = 50 × 2 =100
p 0B = 60 nB = 3
178. C
PS = p 0A × XA + p 0B × XB
Tf =Kf.m = 1.86 × 0.01 = 0.0186K .
1 3
PS = 100 × + 60 × 179. A
1 3 1 3
= 25 + 45 = 70 = CRT
given PS = 75 w
Hence calculated PS is less than given PS = M r × RT
So D, all options are correct w
0.6 1000
173. B 2.463 = M 100 × 0.08 × 300
w
HBr H+ + Br– Mw = 60
i = 1 + (2 – 1) × 0.9 = 1.9 180. C
More is the pol arity more wi ll be the
8.1 / 81
Tf = iKf m = 1.9 × 1.86 = 3.53º attraction.
100 / 1000
Tf = 0ºC – Tf = – 3.5ºC
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