Answer Key
Physics
1. (1) 10. (3) 19. (4) 28. (3) 37. (3)
2. (4) 11. (2) 20. (4) 29. (3) 38. (4)
3. (1) 12. (4) 21. (2) 30. (2) 39. (2)
4. (1) 13. (4) 22. (2) 31. (3) 40. (3)
5. (2) 14. (2) 23. (4) 32. (3) 41. (4)
6. (3) 15. (1) 24. (4) 33. (3) 42. (1)
7. (1) 16. (3) 25. (4) 34. (2) 43. (4)
8. (1) 17. (2) 26. (4) 35. (2) 44. (3)
9. (4) 18. (4) 27. (2) 36. (3) 45. (3)
Chemistry
46. (4) 55. (1) 64. (3) 73. (2) 82. (4)
47. (1) 56. (4) 65. (3) 74. (1) 83. (4)
48. (2) 57. (3) 66. (4) 75. (2) 84. (3)
49. (3) 58. (3) 67. (2) 76. (1) 85. (4)
50. (1) 59. (2) 68. (2) 77. (3) 86. (2)
51. (2) 60. (2) 69. (3) 78. (2) 87. (3)
52. (3) 61. (3) 70. (2) 79. (1) 88. (3)
53. (3) 62. (3) 71. (3) 80. (1) 89. (1)
54. (1) 63. (3) 72. (1) 81. (2) 90. (3)
Biology
91. (3) 109. (4) 127. (4) 145. (4) 163. (1)
92. (2) 110. (1) 128. (4 146. (3) 164. (2)
93. (1) 111. (2) 129. (2) 147. (3) 165. (4)
94. (2) 112. (1) 130. (2) 148. (3) 166. (1)
95. (3) 113. (3) 131. (2) 149. (2) 167. (4)
96. (4) 114. (3) 132. (1) 150. (2) 168. (1)
97. (3) 115. (3) 133. (3) 151. (3) 169. (2)
98. (1) 116. (2) 134. (2) 152. (2) 170. (1)
99. (2) 117. (4) 135. (4) 153. (1) 171. (4)
100. (1) 118. (2) 136. (3) 154. (1) 172. (3)
101. (3) 119. (2) 137. (4) 155. (3) 173. (3)
102. (1) 120. (1) 138. (1) 156. (4) 174. (3)
103. (2) 121. (1) 139. (3) 157. (3) 175. (4)
104. (3) 122. (3) 140. (3) 158. (1) 176. (4)
105. (1) 123. (3) 141. (2) 159. (3) 177. (3)
106. (1) 124. (1) 142. (3) 160. (4) 178 .(1)
107. (3) 125. (3) 143. (3) 161. (1) 179. (4)
108. (1) 126. (2) 144. (2) 162. (2) 180. (4)
, HINTS & SOLUTIONS
SECTION-I (PHYSICS)
1. (1) The force vectors on the circular loop are shown. As
2 K1K 2 the magnetic field near the infinite wire is large the
K eq =
K1 + K 2 force vectors are of greater magnitude. The force
vector components in Y directions will cancel. There
2 × (3 × 6) K 2 is a resultant force in –x direction. The torque on the
= = 4K
(3 + 6) K loop given by M × B is zero.
2. (4) 9. (4)
In vacuum, all have same velocity equal to c. Speed of block at bottom of incline = 2gH
3. (1)
By work energy theorem Wall = ∆KE
=
pi =
3p 3 mu
1 H
µmgx = × m × 2 gH ⇒ x =
2 µ
10. (3)
p f = 2mv 9 × 109 × 2 × 2 × 10−12
=Ui = 1.2 J
3 × 10−2
∴ pi = p f
9 × 109 × 2 × 2 × 10−12
3 Uf = −2
=36 × 10−2 =0.36 J
3mu = 2mv ⇒ v = u 10 × 10
2
11. (2)
4. (1)
=
F m( g + a ) =m(10 + 5) =5 × 15 N = 75 N
Both Statements I and II are correct.
1
5. (2) S= × 5 × 2 × 2 = 10 m
Minimum number of vector of unequal magnitude to
2
produce zero resultant are 3 such that all of the would W = 75 × 10 = 750 J
be in a single plane. 12. (4)
6. (3) Along the electric field at any time t
Light ray moving from rarer to denser medium bends
towards the normal.
7. (1)
Kinetic energy of system is conserved in elastic
collision only.
F = qE
8. (1)
qE
a=
m
For A
vy vy
tan θ A = =
vx a At
1
, For B 1
R1 R0 (27) 3 3
vy vy = =
tan θ B = = R2 1 2
vx aB t R0 (8) 3
As vy is same for both A and B 19. (4)
And θ B > θ A ⇒ q A > qB U 1 1 2 1 x2
= = xy y=z
V 2 2 2 z
∴ q A > qB
20. (4)
A has negative charge and B has positive charge.
13. (4) V2 V2
P= ⇒ R=
R P
=
WT (Td ) m + (−Td=
)M 0
14. (2) (100) 2 (100) 2
R1 = = 200 Ω, R2 = = 400 Ω
50 25
If v0 is threshold frequency, then work function is
equal to hv0 and photosensitivity of a metal is high if
its work function is small.
15. (1)
∆U = nCv ∆T
∆Q= nC p ∆T
∆W = nR∆T
V 100 1
∆U : ∆Q : ∆W =
Cv : C p : R =
i ⇒=
i = A
Req 600 6
5R 7 R
= : :R Total power = P1 + P2 = i 2 R1 + i 2 R2
2 2
=5:7:2 2
1 600 50
16. (3) PT = × (200 + 400)
= = W
6 6×6 3
Path difference between interfering wave for
21. (2)
(2n − 1)λ
minimum intensity is .
2
17. (2)
Ldi
|ε| =
dt
L(4 − 2)
8=
0.02
L = 0.08 H
18. (4) V nR
Slope= =
1 T P
R = R0 ( A) 3
1
Slope ∝
P
θ3 > θ2 > θ1
2
Physics
1. (1) 10. (3) 19. (4) 28. (3) 37. (3)
2. (4) 11. (2) 20. (4) 29. (3) 38. (4)
3. (1) 12. (4) 21. (2) 30. (2) 39. (2)
4. (1) 13. (4) 22. (2) 31. (3) 40. (3)
5. (2) 14. (2) 23. (4) 32. (3) 41. (4)
6. (3) 15. (1) 24. (4) 33. (3) 42. (1)
7. (1) 16. (3) 25. (4) 34. (2) 43. (4)
8. (1) 17. (2) 26. (4) 35. (2) 44. (3)
9. (4) 18. (4) 27. (2) 36. (3) 45. (3)
Chemistry
46. (4) 55. (1) 64. (3) 73. (2) 82. (4)
47. (1) 56. (4) 65. (3) 74. (1) 83. (4)
48. (2) 57. (3) 66. (4) 75. (2) 84. (3)
49. (3) 58. (3) 67. (2) 76. (1) 85. (4)
50. (1) 59. (2) 68. (2) 77. (3) 86. (2)
51. (2) 60. (2) 69. (3) 78. (2) 87. (3)
52. (3) 61. (3) 70. (2) 79. (1) 88. (3)
53. (3) 62. (3) 71. (3) 80. (1) 89. (1)
54. (1) 63. (3) 72. (1) 81. (2) 90. (3)
Biology
91. (3) 109. (4) 127. (4) 145. (4) 163. (1)
92. (2) 110. (1) 128. (4 146. (3) 164. (2)
93. (1) 111. (2) 129. (2) 147. (3) 165. (4)
94. (2) 112. (1) 130. (2) 148. (3) 166. (1)
95. (3) 113. (3) 131. (2) 149. (2) 167. (4)
96. (4) 114. (3) 132. (1) 150. (2) 168. (1)
97. (3) 115. (3) 133. (3) 151. (3) 169. (2)
98. (1) 116. (2) 134. (2) 152. (2) 170. (1)
99. (2) 117. (4) 135. (4) 153. (1) 171. (4)
100. (1) 118. (2) 136. (3) 154. (1) 172. (3)
101. (3) 119. (2) 137. (4) 155. (3) 173. (3)
102. (1) 120. (1) 138. (1) 156. (4) 174. (3)
103. (2) 121. (1) 139. (3) 157. (3) 175. (4)
104. (3) 122. (3) 140. (3) 158. (1) 176. (4)
105. (1) 123. (3) 141. (2) 159. (3) 177. (3)
106. (1) 124. (1) 142. (3) 160. (4) 178 .(1)
107. (3) 125. (3) 143. (3) 161. (1) 179. (4)
108. (1) 126. (2) 144. (2) 162. (2) 180. (4)
, HINTS & SOLUTIONS
SECTION-I (PHYSICS)
1. (1) The force vectors on the circular loop are shown. As
2 K1K 2 the magnetic field near the infinite wire is large the
K eq =
K1 + K 2 force vectors are of greater magnitude. The force
vector components in Y directions will cancel. There
2 × (3 × 6) K 2 is a resultant force in –x direction. The torque on the
= = 4K
(3 + 6) K loop given by M × B is zero.
2. (4) 9. (4)
In vacuum, all have same velocity equal to c. Speed of block at bottom of incline = 2gH
3. (1)
By work energy theorem Wall = ∆KE
=
pi =
3p 3 mu
1 H
µmgx = × m × 2 gH ⇒ x =
2 µ
10. (3)
p f = 2mv 9 × 109 × 2 × 2 × 10−12
=Ui = 1.2 J
3 × 10−2
∴ pi = p f
9 × 109 × 2 × 2 × 10−12
3 Uf = −2
=36 × 10−2 =0.36 J
3mu = 2mv ⇒ v = u 10 × 10
2
11. (2)
4. (1)
=
F m( g + a ) =m(10 + 5) =5 × 15 N = 75 N
Both Statements I and II are correct.
1
5. (2) S= × 5 × 2 × 2 = 10 m
Minimum number of vector of unequal magnitude to
2
produce zero resultant are 3 such that all of the would W = 75 × 10 = 750 J
be in a single plane. 12. (4)
6. (3) Along the electric field at any time t
Light ray moving from rarer to denser medium bends
towards the normal.
7. (1)
Kinetic energy of system is conserved in elastic
collision only.
F = qE
8. (1)
qE
a=
m
For A
vy vy
tan θ A = =
vx a At
1
, For B 1
R1 R0 (27) 3 3
vy vy = =
tan θ B = = R2 1 2
vx aB t R0 (8) 3
As vy is same for both A and B 19. (4)
And θ B > θ A ⇒ q A > qB U 1 1 2 1 x2
= = xy y=z
V 2 2 2 z
∴ q A > qB
20. (4)
A has negative charge and B has positive charge.
13. (4) V2 V2
P= ⇒ R=
R P
=
WT (Td ) m + (−Td=
)M 0
14. (2) (100) 2 (100) 2
R1 = = 200 Ω, R2 = = 400 Ω
50 25
If v0 is threshold frequency, then work function is
equal to hv0 and photosensitivity of a metal is high if
its work function is small.
15. (1)
∆U = nCv ∆T
∆Q= nC p ∆T
∆W = nR∆T
V 100 1
∆U : ∆Q : ∆W =
Cv : C p : R =
i ⇒=
i = A
Req 600 6
5R 7 R
= : :R Total power = P1 + P2 = i 2 R1 + i 2 R2
2 2
=5:7:2 2
1 600 50
16. (3) PT = × (200 + 400)
= = W
6 6×6 3
Path difference between interfering wave for
21. (2)
(2n − 1)λ
minimum intensity is .
2
17. (2)
Ldi
|ε| =
dt
L(4 − 2)
8=
0.02
L = 0.08 H
18. (4) V nR
Slope= =
1 T P
R = R0 ( A) 3
1
Slope ∝
P
θ3 > θ2 > θ1
2
