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Test Type :MAJOR Test Pattern :NEET (UG)
TEST DATE : 17-03-2024
ANSWER KEY
Q. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
A. 3 2 2 1 2 2 2 2 3 2 3 1 3 2 1 4 1 1 1 2 1 3 1 3 3 1 2 3 2 1
Q. 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60
A. 4 2 4 2 4 2 3 1 3 1 4 1 1 3 2 2 2 3 2 1 2 1 3 4 2 3 3 3 1 2
Q. 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90

A. 3 4 3 3 4 3 4 2 1 3 2 2 3 3 4 3 3 4 3 4 3 4 2 4 2 2 3 3 4 3
Q. 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120
A. 3 2 2 3 2 2 3 4 4 3 3 2 1 3 1 1 3 4 3 4 2 1 3 1 3 2 3 4 2 3
Q. 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150

A. 2 4 4 2 2 2 3 3 4 4 3 1 3 4 3 3 4 1 4 2 3 1 1 1 1 2 1 1 2 2
Q. 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180

A. 4 3 1 1 2 4 4 3 1 1 2 4 3 3 3 3 4 2 3 3 3 4 2 3 3 3 4 4 4 3
Q. 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200
A. 2 3 1 1 2 3 3 3 1 2 4 1 4 3 2 3 2 1 4 3

HINT – SHEET

SUBJECT : PHYSICS 3. Ans ( 2 )

Time taken by proton to traverse an arc of 90o is
SECTION - A T 1 2πM
t= = ( )
4 4 qB
1. Ans ( 3 )
→vb = →vbr + →vr
4. Ans ( 1 )
In direction of stream h
λ=
√2mqv

t1 = 2 = 0.4 hr λd = λ∝ given
5
Opposite to direction of stream h h
=
t2 = 2 hr = 2hr √ 2 m e × 100 √ 2 × 4m × 2e × v2
1 ∴ v2 = 25 volt
Total time = 2.4 hr
2. Ans ( 2 ) 5. Ans ( 2 )
For a damped oscillation, graph given in option
T = m(g – a) (2) is correct.
slope of v-t curve gives acceleration
First 2 sec a1 = 2m/s2, T1= m(g – a1) 6. Ans ( 2 )
= 100(10 – 2) = 800 Q = AV = A√2gh ; 70 = 1√2 × 980 × h
Two to six second a2 = 0, T2 = mg = 1000 h = 2.5 cm
Six to seven second a3 = – 4m/s2, T3 = m(g – a3) 7. Ans ( 2 )
= 100(10 + 4) 4T
= 1400 Pex =
R
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9. Ans ( 3 ) 15. Ans ( 1 )
λ 1 ΔV 0.12
r= ⇒r= = γΔT ⇒ = (3α) × 20
√ μ2 − 1 4
2 V 100
√
−1 0.12
= 2 × 10−5 ∘ C −1
( )
3 ⇒α =
3 3 6 100 × 60
r= ⇒r= ⇒ d=2×r= m
√ 16 − 9 √ 7 √ 7
16. Ans ( 4 )
10. Ans ( 2 ) Given ; Δ T = 5°C = 5K and ΔP × 100 = 0.5
P
1 1 Also closed container ⇒ V = constant
P =( + ) × 100 D if f & f are in cm
1 2
f1 f2 ΔP
1 3 1 1
∴ = ΔT
= ( − 1) (0 − ) =
P T
100
f1 2 −10 20 ⇒ T = P × ΔT = × 5 = 1000 K
1 1 1 ΔP 0.5
= (2 − 1) (− − 0) = −
f2 10 10 17. Ans ( 1 )
1 1 Isothermal process : PV = constant
D=( − ) × 100 = −5D
20 10 P → 2P,
11. Ans ( 3 ) V → V1 ⇒ V1 = V (T = constant)
2
W = ΔK Adiabatic process : PV γ = constant
1 K2 2P → P2
W= MV 2 (1 + ) γ 1.4
2 R2 V V1 V /2
1 1 1 3
→ V ⇒ P2 = ( ) ⋅ P1 = ( ) ⋅ 2P
2 V2 V
= × 100 × 1 (1 + ) = × 100 × = 75 J
2 2 2 2 ⇒ P2 = 2 × 2 – 1.4P = 2 × 0.38 P
12. Ans ( 1 ) ⇒ P2 = 0.76 P
τm1 g = τm2 g Therefore, P2 = 0.76P = 0.76
P0 P 1
6 × 16 g = 8 × mg ; m = 6 × 2 m = 12 kg
18. Ans ( 1 )
13. Ans ( 3 ) 10λD λD
( ) = [(2 × 6 − 1) ]
Cut in voltage of diode VC = 0.4 V d medium 2d air
V − VC 10λair 11
I= = λair
R μ 2
(Forward resistance is very small in comparison to R) 20
μ= = 1.8
10 − 0.4 11
I=
12000 19. Ans ( 1 )
I = 0.8 mA π
As the current i leads the voltage by it is an RC
4
14. Ans ( 2 ) XC π 1
circuit, hence tan ϕ = ⇒ tan =
R 4 ωCR
⇒ ω CR=1 as ω = 100 rad/sec
1
⇒ CR = sec – 1.
¯¯¯¯¯¯¯¯¯¯ ¯¯¯ ¯¯¯ 100
¯¯¯ ¯¯¯
A. B = A + B = (A + B) (A + B) From all the given options only option (1) is correct.
¯ + AB
= AA ¯ + BA
¯ + BB
¯ = AB
¯ + BA
¯ = EX – OR
20. Ans ( 2 )
OR
¯¯¯¯¯¯¯¯¯¯¯¯¯¯
A B → G1 → A + B
¯¯¯¯¯¯¯¯¯¯
A B → G2 → A. B
¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯
G1 G2 → G3 A + B + A. B 2Q
¯¯¯ ¯¯¯ ¯¯¯¯¯¯¯¯¯¯ ¯¯¯ ¯¯¯ V =
A. B + A. B ; AB + A B C

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21. Ans ( 1 ) 25. Ans ( 3 )
ω 1 = α t α = 2 θ = 100 rad t'→(t + 5) 2Kλ E(r)
ω 1 = 2t ω 2 = 2t' = 2(t+5) E = ⇒ λ =
r 2k
ω1 + ω2 ′ 18 × 105 × 4 × 10−2
θ= . (t − t) = c/m
2 2 × 9 × 10 9
2t + 2(t + 5)
100 = ×5 = 4 × 10 – 6 c/m
2
t = 7.5 sec = 4 μ c/m
22. Ans ( 3 ) 26. Ans ( 1 )
Electric field from the distance R from the centre
kQ kQ
is = = 20V/m,
r2 4R2
V2 kQ
Here R = ⇒ = 80
P R2
Total power supplied Electric field from the distance R/2 from the
V2 2 V2 kQ
= =( )( )
centre is = x
3
( )R
3 R R3
2
kQ R kQ 80
2
= × 60 = 40W
= × = = = 40V/m
R3 2 2R2 2
3
23. Ans ( 1 ) 27. Ans ( 2 )
d g
ϕ −ϕ g (1 − ) = , d = R/2
e = −( f i ) R h
2
t ( 1+ R
)

e = − ( 4A0 B0 − A0 B0 )
1+
h √
= 2 , h = R ( √2 − 1 )
t R
e = − 3A0 B0 h ≅R(1.4 – 1) = 0.4R
t
|e| = 3A0 B0 28. Ans ( 3 )
t W = Uf – Vi
24. Ans ( 3 ) GMm GMm
=− − (− )
Re + 3Re Re
3 GMm 3
= = (gR) m
4 Re 4
29. Ans ( 2 )
R 80
=
Required is 55 20
Fq0 = 0 R = 220 Ω
F1 = F2 30. Ans ( 1 )
K(1 × 10 ) (q0 )−9 −9
K(9 × 10 ) (q0 ) 1 1
= Shift = t ( 1− ) ⇒ =3 ( 1− )
μ 1.5
r21 r22
= 1 cm upwards
r21 1
=
2
r2 9 31. Ans ( 4 )
r1 1 V12 V22
= × t1 = Heat = × t2
r3 3 R R
r → r1 : r3 → 1 : 3 V12 t
2
V12 t1 = V22 t2 ; = 2 ; [ 220 ] = t2
r1 = 1 r r2 = 3 r V22 t1 110 5
4 4 t2
r1 = 0.25 r r2 = 0.75 r 4= ; t2 = 20 minutes
5
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32. Ans ( 2 ) 37. Ans ( 3 )
e = NB 2 π r dr
dt
N = 1 B = 0.025 r = 2 cm = 2 × 10 – 2 m
dr dr
= 1mm/s ⇒ = 10−3 m/ sec
dt dt –2 –3
e = (1) (0.025) (2 π ) (2×10 ) (10 )
e = 4 π × 25 × 10 – 3 – 2 – 3 B.2πr = μ0 (
I
πr2 ⇒ B =
μ0 Ir
= 100 π × 10 – 8 = π × 10 – 6 volts = e = π μ v
)
πR2 2πR2
2R μ I
34. Ans ( 2 ) (∵ r = ); B= 0
3 3πR
38. Ans ( 1 )
da 3 3
= − tan 37∘ = − ⇒ a=− x
dx 4 4
3 2π √3 4π
2
so, ω = ⇒ = ⇒T= s
4 T 2 √3

39. Ans ( 3 )
Block will rest against vertical wall d(y1 ) π
V1 = = (0.1 × 100 π ) cos ( 100πt + )
Mg ≤ F ℓ dt 3
Mg ≤ μ N {F ℓ = μ sN} d(y2 )
V2 = = (0.1 × 100 π ) [ – sin (100 π t)]
Mg ≤ μ F dt
π
F ≥ mg ; Fmin = mg = 1 × 9.8 = 49N OR V2 = (0.1 × 100 π ) [ cos(100πt +
2
)]

μ μ 0.2
35. Ans ( 4 ) Phase difference of V1 with respect to V2
For hydrogen-like atom, the radius of nth orbit is π π π
n2 is Δ ϕ = − =−
rn = r0 3 2 6
Z
Where r0 = 0.51 × 10 – 10 metre 40. Ans ( 1 )
0.51 × 10−10
Here, rn = metre Work = Energy stored
4
In the ground state, n = 1 1
0.51 × 10−10 12 = × (Force) × ( Δ l)
∴ = × 0.51 × 10−10 2
4 Z
∴ Z=4 F L1 F L2
So, the atom is triply ionised beryllium. Δl1 = = 0.8mm & Δl2 = = 0.2mm
Ycu A Ysteel A

SECTION - B Net Δ l = Δ l1 + Δ l2 = 1mm = 10 – 3 m,

1 1
36. Ans ( 2 ) Elastic potential energy = × 500 × 10 – 3 = J
2 4
41. Ans ( 4 )
0.5
L.C. = mm = 0.01 mm
50
zero error = VSD × LC = 2 × 0.01 mm
SRel = uRel t + 1 aRel t2 ; 96 = 20 t + 1 ( – 2)t2
2 2 = 0.02 mm (+ve)
t2 – 20 t + 96 = 0
20 ± √(20)2 − 4 × 1 × 96 d = MSD+(VSD) × LC ⊝ zero error
20 ± 4
t= ;t=
2×1 2 d = 1.00 mm + 0.20 mm – 0.02 mm
t = 8s, 12s
t = 8 sec {He will be over take the bus} = 1.18 mm = 0.118 cm
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42. Ans ( 1 ) 48. Ans ( 3 )
V 8 Given wheatstone bridge is balanced
R= = = 4Ω
I 2
ΔR ΔV ΔI R1 C 4Ω 8μF
and ( ) × 100 = [ + ] × 100 As = 2 ⇒ =
R V I R2 C1 5Ω 10μF
0.5 8
= × 100 + 0.2 × 100 = 6.25 + 10 V10μF =
8 + 10
× 9V = 4V
8 2
ΔR Q10 μ F = C.V. = 10 μ × 4V = 40 μ C
× 100 = 16.25%
R
R = 4 ± 16.25% 49. Ans ( 2 )
2
R2 u sin 2θ2
43. Ans ( 1 ) =( 2)
R1 u 1 sin 2θ1
2
Increment in kinetic energy = work done 2u sin 60∘
=( ) ×
u sin 30∘
x2 10
R2
1 = 4 √3
⇒ m(v2 − u2 ) = ∫ F . dx = ∫ 3x. dx R
2
x1 2 50. Ans ( 1 )
1 2 3 2 10 3
mv = [x ]2 = [100 − 4]
2 2 2
1 3
× 8 × v2 = × 96 ⇒ v = 6m/ s
2 2

45. Ans ( 2 )
u = √μS rg ⇒ v = √75 × 60 × 9.8 = 21 m/s
46. Ans ( 2 ) R1 = 2R0, R2 = R0
m (8gR) R2 2i
TL = + mg = 9 mg i1 = × i = i , i2 = i – i1 =
R R1 + R2 3 3
TH 1
TH = TL – 6 mg = 3 mg ;
TL
=
3
B 1 = μ0 i1 =μ0 i
,⊗
4R 12R
47. Ans ( 2 ) μ0 i2 μ0 i
B2 = = ,⊙
4R 6R
Let mass of neutron = m, then mass of deutron = 2m μ i
B0 = B2 – B1 = 0 , ⊙
12R

SUBJECT : CHEMISTRY
By momentum conservation
SECTION-A
mu = mv1 + 2mV2 ⇒ u = v1 + 2v2 ..... (1)
52. Ans ( 1 )
v2 − v1 NCERT-XI, Pg # 112, Part-I
e= =1 (Elastic collision) u = v2 – v1 ..... (2)
u
u 63. Ans ( 3 )
by equation (1) & (2) v1 = −
3 NCERT-XI, Pg # 50,51, Part-I
KEi − KEf
Fractional loss of KE of neutron = 64. Ans ( 3 )
KEi
1 2 1 u
2 NCERT-XI, Pg # 48,57,52,44, Part-1
mu − m( )
2 2 3 1 8 65. Ans ( 4 )
= =1− =
1 2 9 9 NCERT-XI, Pg # 204, Part-1
mu
2
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68. Ans ( 2 ) 82. Ans ( 4 )
H2 → 2H+ + 2e – NCERT 12th Page No. # 355, 335
2
[H + ]
83. Ans ( 2 )
Ecell = 0 − 0.0591 log NCERT 12th Page No. # 356
2 PH 2
0.0591 0.0591 84. Ans ( 4 )
=0− log 10−8 = 8 × = 0.236V
2 2 NCERT 12th Page No. # 374, 375
69. Ans ( 1 )
85. Ans ( 2 )
1 [R]1
K= ln NCERT 11th Page No. # 393, 394, 394, (13.17)
t2 − t1 [R]2
1 0.06
K= ln = 1.83 × 10−3 sec−1 SECTION-B
(20 − 10) × 60 0.02
70. Ans ( 3 ) 86. Ans ( 2 )
NCERT-XI, Pg # 105, Part-I

+2+2=+4 91. Ans ( 3 )

NCERT-XI, Pg # 9, 10, Part-I
NCERT 23-24, Page No. # 245
92. Ans ( 2 )
Part-2, unit-7
NCERT-XI, Pg # 195, 197, Part-1
71. Ans ( 2 ) It is necessary that while calculating the value of
NCERT 23-24, XI, Page No. # 240 Part-2 Unit-7 Kp, pressure should be expressed in bar as
72. Ans ( 2 ) standard state is 1 bar.
NCERT 23 - 24 Page No. # 136; Part-1, Unit-5 95. Ans ( 2 )
73. Ans ( 3 ) NCERT-XI 23-24, Page No. # 195, Part-1, Unit-6.
NCERT 23-24, Page No. # 138, Part-1, Unit-5. 96. Ans ( 2 )
74. Ans ( 3 ) NCERT 11th Page No. # 375
NCERT-XI, 23-24, Page No. 202, Part-1, Unit-6. 97. Ans ( 3 )
75. Ans ( 4 ) NCERT 12th Page No. # 373
NCERT 11th Page No. # 336, 338, 340 98. Ans ( 4 )
76. Ans ( 3 ) NCERT 12th Page No. # 402, 394, 393
NCERT 11th Page No. # 341 99. Ans ( 4 )
77. Ans ( 3 ) NCERT 11th Page No. # 374
NCERT 12th Page No. # 384, 341 100. Ans ( 3 )
78. Ans ( 4 ) NCERT 12th Page No. # 304
NCERT 11th Page No. # 392 SUBJECT : BOTANY
79. Ans ( 3 )
SECTION - A
NCERT 12th Page No. # 401, 343
80. Ans ( 4 ) 101. Ans ( 3 )
NCERT 12th Page No. # 343, 363, 340 NCERT-XI Pg.# 18,19,20,21
81. Ans ( 3 ) 102. Ans ( 2 )
NCERT 12th Page No. # 317 NCERT XI Page # 23
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103. Ans ( 1 ) 122. Ans ( 4 )
NCERT-XI Pg.# 32 NCERT-XI, Pg.#220
104. Ans ( 3 ) Photorespiration involves peroxisomes
NCERT-XI Pg.# 33 mitochondria chloroplast
105. Ans ( 1 )
123. Ans ( 4 )
NCERT-XI Pg.# 38,39
NCERT-XI Pg.# 233, Fig.-14.4
106. Ans ( 1 )
124. Ans ( 2 )
NCERT-XI Pg.# 78,79
NCERT XI Pg.# 179 (E & H)
108. Ans ( 4 )
125. Ans ( 2 )
NCERT-XI Pg.# 89
NCERT-XI, Pg.# 159
109. Ans ( 3 )
126. Ans ( 2 )
NCERT XI Pg # 87
NCERT-XI, Pg.#24
110. Ans ( 4 )
127. Ans ( 3 )
NCERT-XII, Pg. # 248
NCERT-XI Pg. # 229
111. Ans ( 2 )
128. Ans ( 3 )
NCERT-XII Pg.# 242
NCERT XI Pg # 240
112. Ans ( 1 )
129. Ans ( 4 )
NCERT-XII Pg.# 260
NCERT-XII, Pg.# 109
113. Ans ( 3 )
130. Ans ( 4 )
NCERT Pg # 284 (E), 309 (H) NCERT-XII, Pg.# 182
114. Ans ( 1 ) 131. Ans ( 3 )
NCERT Pg. # 265 NCERT XII Pg. # 186
115. Ans ( 3 ) 132. Ans ( 1 )
NCERT-XII Pg.# 231 NCERT-XII, Pg.# 183
116. Ans ( 2 ) 133. Ans ( 3 )
NCERT-XII Pg.# 230
NCERT-XII, Pg.# 96, 97
117. Ans ( 3 )
134. Ans ( 4 )
NCERT-XII Pg.# 227
NCERT-XII, Pg.# 102,103
118. Ans ( 4 )
NCERT-XII Pg.# 227 135. Ans ( 3 )
NCERT-XII
119. Ans ( 2 )
NCERT (XII) Page # 26,27 SECTION - B

120. Ans ( 3 ) 136. Ans ( 3 )

NCERT, Page#23(E), 25(H) Module no. 1, Pg. # 115(E), 129(H)
121. Ans ( 2 ) 137. Ans ( 4 )
NCERT (XII) Pg. # 25 Figure NCERT-XI Pg.# 73
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138. Ans ( 1 ) 159. Ans ( 1 )
NCERT (XI) Pg. # 92,93 NCERT - Page No. 290
139. Ans ( 4 ) 164. Ans ( 3 )
NCERT-XI Pg.# 90 NCERT XI, Pg. # 303
140. Ans ( 2 ) 166. Ans ( 3 )
NCERT (XIth) Pg. # 74 NCERT Pg. No. 34
141. Ans ( 3 ) 167. Ans ( 4 )
NCERT-XII Pg.# 243 NCERT XII, Pg # 67
142. Ans ( 1 ) 169. Ans ( 3 )
NCERT XII Pg # 261, 262 NCERT Pg. # 150
143. Ans ( 1 ) 170. Ans ( 3 )
NCERT-XII Pg.# 265 NCERT Pg. No. 147-149
144. Ans ( 1 ) 171. Ans ( 3 )
NCERT(XII) Pg. # 26(E) & 27(H) NCERT XII Pg. # 151
145. Ans ( 1 ) 174. Ans ( 3 )
NCERT – XI, Pg. # 215 NCERT XI, Pg. # 136
146. Ans ( 2 ) 178. Ans ( 4 )
NCERT-XI, Pg.# 248, 249, 250 NCERT (XIth) Pg. # 164 Para : I
147. Ans ( 1 ) 180. Ans ( 3 )
NCERT XI Pg. # 245 NCERT Pg. # 167

148. Ans ( 1 ) 181. Ans ( 2 )

NCERT XI Pg.# 147
NCERT-XII, Pg.# 96
SECTION - B
149. Ans ( 2 )
NCERT-XII, Pg.# 77 186. Ans ( 3 )
NCERT,Pg.# 201
150. Ans ( 2 )
NCERT-XII, Pg.# 87 189. Ans ( 1 )
NCERT Pg # 208
SUBJECT : ZOOLOGY
190. Ans ( 2 )
SECTION - A NCERT-XII-Page No. # 211 (E), 231, 232 (H)
151. Ans ( 4 ) 191. Ans ( 4 )
NCERT Pg #45 NCERT Pg. No. # 54
153. Ans ( 1 ) 193. Ans ( 4 )
Pg. No. 59 NCERT 2022 - 2023 Edition NCERT (XIth) Pg. # 295
157. Ans ( 4 ) 200. Ans ( 3 )
NCERT XI, Pg.# 275,17.6 NCERT – XII, Pg. # 130

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