CLASSROOM CONTACT PROGRAMME

(Academic Session : 2024 - 2025)

ENTHUSIAST COURSE
PHASE : ALL
TARGET : PRE MEDICAL 2025
Test Type : MAJOR Test Pattern : NEET (UG)
TEST DATE : 05-02-2025
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. 2 4 3 1 3 4 1 2 3 2 2 3 1 3 3 2 2 2 3 4 2 2 2 1 2 1 2 1 4 3
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. 3 1 2 3 2 2 2 3 1 2 1 4 4 4 1 2 4 2 2 1 4 4 2 1 4 1 4 1 1 4
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. 4 4 4 2 2 3 4 4 4 3 3 4 1 3 1 1 1 3 1 1 3 1 2 2 4 4 3 3 3 4
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 1 3 4 2 1 1 3 2 1 3 2 3 2 4 4 3 2 2 2 1 1 3 4 2 3 1 3 3 1
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 3 1 4 1 2 4 3 1 2 1 2 2 3 2 4 1 2 4 4 3 2 4 4 3 4 2 1 1 1
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. 1 2 2 4 1 3 2 2 3 4 3 4 4 2 1 1 2 4 3 4 3 2 1 1 4 3 3 1 1 1

HINT – SHEET

1. Ans ( 2 ) 3. Ans ( 3 )

If voltmeter has resistance : Rv 2

then, measured P.D. across 'R' = IReq For each resistor : Imax R = Pmax
RRV Pmax √ 20
where Req = ( ) < R
Imax = √ = = 2A
R + RV R 5
3 Each resistor can have maximum current of 2A, before
Hence Vmeas <R⇒ <R it get damage. So maximum current in branch AB = 2A.
I 0.6
⇒ R > 5Ω . These each resistor of branch BC will carry 1A as
2. Ans ( 4 ) shown.
Pnet = PAB+ PBC = (2)2(5) + (2)(1)2(5) = 20+10 = 30W.
4. Ans ( 1 )
ke2 Gme ⋅ mp
Fe = Fg =
r2 r2
Simplified circuit Fe ke2
= ≃ 2.4 × 1039
10 − 6 Fg Gme ⋅ mp
i= = 1A
3+1
Applying KVL between A and B 5. Ans ( 3 )
VB – 6 – (1×1) = VA As the field is along x-direction, equipotential surface
VB – VA = 7 V. must be parallel to y-z plane and perpendicular to x-axis.
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7. Ans ( 1 )
b

from (1) & (2)

Distance of a nth bright fringe from central fringe sin θ =
m2
=
6
nλD m1 10
yn = ⇒ θ = 37°
d
∵ λ blue < λ yellow from (3)
∴ yn(blue) < yn(yellow) N = 10 × 10 × cos 37°
8. Ans ( 2 ) N = 80 N
(force exerted by inclined on block m1)
This is a wheat stone bridge which is balanced
hence circuit can be reduced to following 12. Ans ( 3 )
8×8 Speed of train A : VA = 1 km/hr
Ceq. = = 4μF
8+8 Speed of train B : VB = 2 km/hr
speed of person w.r.t. train A ;
⇒ VPA = 0.2
⇒ VP – VA = 0.2 ⇒ VP = 1.2 km/hr
Rel. speed of person wrt train B;
VPB = VP + VB = 1.2 + 2 = 3.2 km/hr
9. Ans ( 3 ) 13. Ans ( 1 )
P1 m1 gh/t1 m1 /t1 300/5 600 12
= = = = = = 2.4 Total distance
P2 m2 gh/t2 m2 /t2 50/2 250 5 Average speed =
total time
10. Ans ( 2 ) (
0+30
2
) t + (30)(2t)
Minimum value of velocity of pendulum to desires = = 25 km/hr
3t
a circle is :
v2 = √5gl = √100 = 10m/s
14. Ans ( 3 )
Ei = 4 π R2T … (i)
By linear momentum conservation,
4 3 4
m ν 1 = – mv1 + mv2 πR = n πr3
3 3
50 × 10 – 3 v = – 50 × 10 – 3 × 100 + 1 × 10 r = R(n) – 1/3
v = 100 m/s Ef = n4 π r2T
11. Ans ( 2 ) = n1/3 4 π R2T … (ii)
Δ E = W = Ef – Ei
= 4 π R2T [n1/3 – 1]
15. Ans ( 3 )
FT
= nϕ
A
F 1000
T = m2g ___(1) ϕ= =
Aη (0.04)(2 × 109 )
T = mg sinθ 10 −4
ϕ= = 0.125 × 10−4
N = mg cosθ ___(2) 4×2
ϕ = 1.25 × 10 – 5
16. Ans ( 2 )
T = m ω 2r
1
450 = × ω2 × (1)
2
∴ ω = 30 rad/sec

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17. Ans ( 2 ) 22. Ans ( 2 )
E0 mv
= C in vaccum r=
B0 qB
E0 = B0C v → same
B → same
= 3 × 10 – 6 × 3 × 108 m
r∝
q
= 900 V/m r1 (m/q)1
=
18. Ans ( 2 ) r2
r1
(m/q)2
4m q
All in parallel, = ×
r2 m 2q
r2= r/2
After redraw,
23. Ans ( 2 )
P1V1 γ = P2V2 γ
γ
V1
P2 = P1 ( )
V2
5/3
2 1 16
LP Q = = H P2 = 1 × ( )

4 2 2

19. Ans ( 3 ) Y = 5/3

P2 = 32 atm
V 2 = VR2 + VL2
24. Ans ( 1 )
∴ VL2 = V 2 − VR2 ∵ ℓ Ai = ℓ Bi and ℓ Af = ℓ Bf
∴ Δ ℓA = Δ ℓB
2 2
VL = √(220) − (132) ℓ ∝ A Δ T1 = ℓ α B Δ T2
= √88 × 352 = 176V
α A(180° – 30°) = α B (T – 30°)
4 α × 150° = 3 α (T – 30°)
20. Ans ( 4 ) T = 230°C
In presence of magnetic field 25. Ans ( 2 )
→ = q(→v × B)
F → 5
U= RT ∴ PV = RT
2
⇒ F→ ⊥B
→ U = 5 PV V= m
2 ρ
→
⇒ →a ⊥B U = × P = × 8 × 104 × 1
5 m 5
2 ρ 2 4
→ =0 4
= 5 × 10 J
∴ →a ⊥B

3 – 12 α = 0
26. Ans ( 1 )
U = n1f1RT + n2f2RT
α = 3/12 = 1/4 = 2 × 5 RT + 4 × 3
RT
2 2
21. Ans ( 2 ) = 5 RT + 6 RT
F μ0 i, i2 = 11 RT
=
ℓ
F
2πd
2 × 10−7 × 5 × 5
27. Ans ( 2 )
⇒ = The second pendulum is a simple pendulum whose
ℓ 0.5
–6 time period is constant and 2 sec, which does not
= 10 N/m and repulsive depend on mass of the bob
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28. Ans ( 1 ) 36. Ans ( 2 )
I1 (4)2 2
(2π) 4
2
2π
2
dA πR2
I ∝ w2A2 So, = × =( ) ( ) =
I2 (8) 2
(6π) 2 8 6π dt T
I1 1 1 2 4π 2 3
= × = 1 : 36 T = R
I2 4 9 GM
29. Ans ( 4 ) T=
2π
√GM
R3/2 ⇒
dA
dT
=
πR2
2π
R3/2
K = 6 π × 0.01 dA1
√
GM

2π dt R1 R
⇒ = 6π × 0.01 =√ = n ⇒ 1 = n2
λ dA2 R2 R2
1
∴ λ = (m)
dt
2 0.06
100
37. Ans ( 2 )
⇒ dC.T = (m) = 16.67m Wext + Wg = 4K = 0
6
30. Ans ( 3 ) Wext – m4V = 0
4
When light ray travels parallel to the base, the light Wext = 2 × =4J
suffers minimum deviation 2
δ min = 40° 38. Ans ( 3 )
from graph,
i = e = 45°
31. Ans ( 3 )
1 1 1
+ =−
−20 v 10 Diode 2 is in reverse bias
v = – 20 cm So current will not flow in branch of 2nd diode, So
32. Ans ( 1 ) we can assume it to be broken wire.
[ML2T – 2] → Energy and Torque Diode 1 is in forward bias
[ML2T – 3] → Power So it will behave like conducting wire. So new
[M0L0T – 1] → Frequency circuit will be
[MLT – 2] → Force
33. Ans ( 2 )
For (A) : A and AB may have same dimension.
3
15 × 10 15 × 10
For (B) : As A and B have different dimension so Req =
15 + 10
Correct answer (C)
=
25
= 6Ω

A
exp ( −
B
) is meaningless.
39. Ans ( 1 )
For (C) : AB2 is meaningful. For emission of photo electron λ ≤ λ 0
For (D) : AB – 4 is meaningful. λ 0 → threshold wavelength
34. Ans ( 3 ) 40. Ans ( 2 )
T = Iα 1
5g × 0.5 – 2g × 0.5 F∝ [Electrostatic force]
r2
= (5 × 0.52 + 2 × 0.52) α r ∝ n2
α = 8.4 ra/s2 F∝
1

35. Ans ( 2 ) n4
41. Ans ( 1 )
Forces will be such that door will be in rotational
Only Lyman series is absorbed as higher energy
as well as translational equilibrium
levels are unstable.
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42. Ans ( 4 ) 54. Ans ( 1 )
N MSD = (N + 1) VSD ⇒ 1VSD = (
N
) MSD H2 S, BF4− , P Cl5 can exist

55. Ans ( 4 )
N +1
[Given = 1MSD = a units]
1
LC = 1MSD – 1VSD = 1 − N ( ) a=( ) a unit
(N + 1) N +1

43. Ans ( 4 ) P-H bond is non-polar i.e. is PH3 dipole is only due
MSR = 0 mm, CSR = 52 div,
to L.P.
LC = 1 = 0.01 mm
100
Diameter = (0 + 52 × 0.01)mm = 0.052 cm
56. Ans ( 1 )
44. Ans ( 4 )
r21 vT1 9 r 3
∵ vT ∝ r2 ⇒ = = ⇒ 1 =
r22 vT2 4 r 2 2
3
V1 r 27
∴ =( 1 ) =
V2 r2 8
46. Ans ( 2 )
for isoelectronic species
Cation < Neutral < anion
47. Ans ( 4 ) 57. Ans ( 4 )
Trans – [CoCl2(en)2]+
Concept
48. Ans ( 2 )
NCERT, Pg # 90
Electron gain enthalpy order is Cl > F > Br POS present optically inactive
49. Ans ( 2 ) 58. Ans ( 1 )
CO2 has linear shape. KMnO4
50. Ans ( 1 ) K+ MnO−4
[NCERT-XI, Part-1, Chapter – Chemical Bonding
Pg.no. = 128 – 29]
As pre M.O.T. (iso e – species)
51. Ans ( 4 ) Mn+7 → d° configuration
(1) NH4 +Cl – (2) K+ MnO4 – diamagnetic

(3) Na+(O – H) – (4)

59. Ans ( 1 )
KMnO4 is strong O.A. so HCl can't be taken in
titration.
52. Ans ( 4 ) 60. Ans ( 4 )
All sp3 hybridised species show p π -d π type of
Zn has fulfilled configuration
bonding.
53. Ans ( 2 ) 61. Ans ( 4 )
BCl3, PCl5, CO32 – , SO3

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64. Ans ( 2 ) 88. Ans ( 3 )
inB
RLVP =
nA + inB
i
0.4 =
3+i
i=2
⇒ α = 100%

91. Ans ( 3 )
66. Ans ( 3 ) NCERT, Pg. # 22
Sucrose is a disaccharidc of α – D – Glucose and
92. Ans ( 1 )
β – D – Fructose. NCERT XII Pg. # 15, 16
68. Ans ( 4 ) 93. Ans ( 3 )
CH3 MgBr NCERT, Pg. # 7
CH3COOH + PCl5 → −−−−−−→
dry ether 94. Ans ( 4 )
NCERT-XII, Pg. # (E)-63
(1) CH3 MgBr 95. Ans ( 2 )
−−−−−−−−−→ NCERT Pg. # 62
(2) H2 O

96. Ans ( 1 )
NCERT Pg. No. # 64
69. Ans ( 4 ) 97. Ans ( 1 )
NCERT Pg. # 72
Reagent :
98. Ans ( 3 )
Semicarbazide NCERT Pg. # 73
72. Ans ( 4 ) 99. Ans ( 2 )
NCERT 11th P.No. 374 NCERT Pg. # 83
73. Ans ( 1 ) 100. Ans ( 1 )
Rate of SN2 reaction α 1 NCERT Pg. # 92

74. Ans ( 3 )
steric hindrance
101. Ans ( 3 )
NCERT Pg. # 90
Fact.
102. Ans ( 2 )
75. Ans ( 1 ) NCERT XII Pg. # 95
Because they have same molecular formula but
103. Ans ( 3 )
different F-G. NCERT XII, Pg. # 100, 101
83. Ans ( 2 ) 104. Ans ( 2 )
More reactive metal displaces less reactive metal NCERT Pg. # 106
from their salt solution. 105. Ans ( 4 )
NCERT Pg. # 104

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106. Ans ( 4 ) 124. Ans ( 4 )
NCERT Pg. # 193 XI NCERT Page. No :- 207, 13.2
107. Ans ( 3 ) 125. Ans ( 1 )
NCERT, Pg. # 202 NCERT XI Pg. # 228, 231, 233
108. Ans ( 2 ) 126. Ans ( 2 )
NCERT Pg. # 198 NCERT (XI) Pg # 80, 81
109. Ans ( 2 ) 127. Ans ( 4 )
NCERT Pg. # 223 NCERT (XI) Pg # 78, 79, 80, 81
110. Ans ( 2 ) 128. Ans ( 3 )
NCERT Pg.# 213 NCERT-XI, Pg. # 76
111. Ans ( 1 ) 129. Ans ( 1 )
Module-1 Both A & R are correct and R is correct
112. Ans ( 1 ) explaination of assertion.
NCERT-XI, Pg. # 19, 129 130. Ans ( 2 )
113. Ans ( 3 ) NCERT Page No. 87, 90
NCERT (XI) Pg # 09 131. Ans ( 1 )
114. Ans ( 4 ) NCERT Page No. 86
NCERT XI Pg.# 9 132. Ans ( 2 )
115. Ans ( 2 ) XIth NCERT Page No. – 87, 90
NCERT Pg. No. # 19,21,26,39 133. Ans ( 2 )
116. Ans ( 3 ) NCERT, Pg. # 193
NCERT Pg. No. # 223 134. Ans ( 3 )
117. Ans ( 1 ) NCERT, Pg # 225
Module Pg. #234 135. Ans ( 2 )
118. Ans ( 3 ) NCERT Pg#61,69,73,(E);68,76,77,82(H)
NCERT XI, Page No. - 227,230 + Module 138. Ans ( 2 )
119. Ans ( 3 ) NCERT, Pg. # 44, 45
NCERT XI Page No. 233 139. Ans ( 4 )
120. Ans ( 1 ) NCERT, Pg. # 42
NCERT XI Pg. # 223 140. Ans ( 4 )
121. Ans ( 2 ) NCERT XII Pg # 124
NCERT XI Page No. 218, 220 141. Ans ( 3 )
122. Ans ( 3 ) NCERT Pg. # 200
NCERT Pg. No. # 243 142. Ans ( 2 )
123. Ans ( 1 ) NCERT Pg # 194
NCERT Pg. No. # 239 143. Ans ( 4 )
NCERT Pg#199
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144. Ans ( 4 ) 164. Ans ( 2 )
NCERT-XI Page No. 234 NCERT (XII) Pg. # 195
145. Ans ( 3 ) 165. Ans ( 1 )
NCERT, Pg. # 234 NCERT-XII Pg. # 202
146. Ans ( 4 ) 166. Ans ( 1 )
NCERT Pg # 48 NCERT Pg # 212
147. Ans ( 2 ) 168. Ans ( 4 )
NCERT Pg. # 188 (Fig. 14.3) NCERT XI Pg # 134
148. Ans ( 1 ) 169. Ans ( 3 )
NCERT Pg. # 183 NCERT-XI, Pg. # 94
149. Ans ( 1 ) 170. Ans ( 4 )
NCERT Pg. # 185, 186, 187 NCERT-XI, Pg # 125
150. Ans ( 1 ) 171. Ans ( 3 )
NCERT Page No. # 138 NCERT Pg#236
152. Ans ( 2 ) 172. Ans ( 2 )
NCERT Pg. # 18 NCERT Pg. No. # 186, 187
153. Ans ( 2 ) 173. Ans ( 1 )
NCERT, Pg. # 332 NCERT-XI Pg. # 278, 279
154. Ans ( 4 ) 174. Ans ( 1 )
NCERT Pg # 242 NCERT Pg. # 241
155. Ans ( 1 ) 175. Ans ( 4 )
NCERT Pg. # 211,212,213 NCERT Pg. No. # 223
156. Ans ( 3 ) 176. Ans ( 3 )
NCERT, Pg. # 208 Module-120
157. Ans ( 2 ) 177. Ans ( 3 )
NCERT, Pg. # 35 Module-5 Page No. # 103
158. Ans ( 2 ) 178. Ans ( 1 )
NCERT Pg. # (E)-36, (H)-39 NCERT Pg. # 83
159. Ans ( 3 ) 180. Ans ( 1 )
Module-4 Pg#100 Module-8 Page No. #61
160. Ans ( 4 )
NCERT Pg. No. # 222-223
161. Ans ( 3 )
NCERT Pg. # 30
163. Ans ( 4 )
NCERT XI, Pg.# 107

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