CLASSROOM CONTACT PROGRAMME

(Academic Session : 2024 - 2025)

ENTHUSIAST COURSE
PHASE : ALL PHASE
TARGET : PRE MEDICAL 2025
Test Type : MAJOR Test Pattern : NEET (UG)
TEST DATE : 28-01-2025
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. 3 2 4 4 1 2 3 1 4 1 3 3 2 2 1 1 4 2 1 4 4 3 4 2 2 4 4 4 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 3 2 1 1 4 1 1 1 3 3 2 3 4 2 1 2 1 1 4 2 4 2 1 3 2 3 1 3 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. 2 4 1 4 4 2 4 3 4 3 4 2 3 4 4 2 1 2 3 2 1 1 2 4 2 3 2 4 1 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. 1 4 2 3 3 3 3 3 4 3 3 3 4 1 4 1 1 1 2 2 2 2 4 3 4 3 2 1 4 4
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. 1 2 4 3 1 3 3 3 3 3 3 1 4 4 2 3 4 1 1 1 3 1 3 3 3 1 4 3 3 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. 1 2 3 4 3 3 2 3 4 1 3 4 4 2 1 3 3 2 2 1 2 3 2 2 2 3 1 3 3 1
Q. 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200
A. 4 4 4 4 1 1 4 1 4 3 1 1 1 2 2 1 4 4 3 1

HINT – SHEET

SUBJECT : PHYSICS 3. Ans ( 4 )

In Ist case : H = I2Rt ⇒ H ∝ I2t ; t1 = 20 seconds,
SECTION-A t2 = 18 seconds
1. Ans ( 3 )
2
H2 I2 t2
⇒ =( )
H1 I1 t1
(2M)L2 2
I0 = ( ×3 H2 4 18
3
)
⇒ =( ) × ⇒ H2 = 960J
600 3 20
I0 = 2ML2 4. Ans ( 4 )
2. Ans ( 2 )
Angular momentum will remain conserved,
L2
we, know, KE =
2I
∵ L = constant 4R
⇒R=2+ ⇒ 4R + R2 = 8 + 6R
4+R
1
∴ KE ∝ ⇒ R2 − 2R − 8 = 0 ⇒ R2 − 4R + 2R − 8 = 0
I ⇒ R(R − 4) + 2(R − 4) = 0
KE1 I2
= ⇒ K′ = 2I ⇒ K ′ = K ⇒ R = 4, R = – 2 (not possible)
KE2 I1 K I 2
⇒ R = 4Ω .
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5. Ans ( 1 ) 12. Ans ( 3 )
3

T.Ei = T.Ef By LMC →

K.Ei + P.Ei = K.Ef + P.Ef mv = (7m) V' + mv'
1 2 kQ2 kQ2 V' = V/8
2
mv
1 2kQ2
+
r
=
kQ2
O +
kQ2
x 13. Ans ( 2 )
m + =
2 rm r x
2 2
2kQ kQ r
= ⇒x=
r x 2
6. Ans ( 2 )
4.8 × 10−6
t= −19 10
= 3 × 103 sec 6 = 12 × i
1.6 × 10 × 10
1
7. Ans ( 3 ) i=
2
A

yBright − ydark =
nλD
−
(2n − 1)λD 14. Ans ( 2 )
d 2d Conceptual / theory
λD (2n − 2n + 1)
=
d
λD
2 15. Ans ( 1 )
=
2d
8. Ans ( 1 )
∈0 A
C= = 8pF
d
∈ × 6A
C1 = 0 = 12 × 8 = 96pF
d/2

9. Ans ( 4 )
By work-energy theorem Tmax – mg = mamax.
WF = Δ K.E. 800 – 500 = 50a
Work done by gravity 300
a= = 6 m/s2
Mgy0 = Δ K.E. = Kf – Ki 50
Here, Ki = 0
So Kf = K = mgy0
16. Ans ( 1 )
Work = T × Area
10. Ans ( 1 ) = 0.02 × 0.05
= 0.001 J
Unit of force
= (109) (10 cm) (0.1 sec) – 2 17. Ans ( 4 )
= (10 × 10 – 3 × 10 × 10 – 2 × 102) N 1
= 0.1 N h∝
r
11. Ans ( 3 ) h2
h1
r
= 1
r2
h2 = r h1 = 3h1
(r/3)
Given h1 = 3 mm
∴ h2 = 9mm

18. Ans ( 2 )
2
V = 13 π D2 H
( )

∴ % Error in V = 2(% error in D) + % error in H. x 2 y 2

( ) + ( ) = sin2 ω t + cos2 ω t
∵ Least count is 2mm. a a
x2 y2
∴ % error in D = 2mm × 100% = 1% ⇒ +=1
20cm a2 a2
2mm
& % error in H = 20cm × 100% = 1% ⇒ x2 + y2 = a2
So % error in V = 2 × 1% + 1% = 3% Hence, it is a circular path.
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19. Ans ( 1 ) 25. Ans ( 2 )
C
ν= B1 = B3 = 0
√μr ε r
C2 C2 μ0 IQ
μr = = = 4.5 B2 =
V 2 εr C2
×2 4πR
9
μ0 I π μ0 I
20. Ans ( 4 ) =
4πR
× =
2 8R

Iavg =
1
2
× 50 × 2
= 25A
26. Ans ( 4 )
(2 − 0) F = qvB
1 2
21. Ans ( 4 ) k = qV =
2
mv
v ∝ √V ′
∵ P = NhC
tλ
∴ F ∝ v ∝ √V ′
⇒ No of incident photons per unit 't'
N Pλ 5 × 10−3 × 500 × 10−9 F V
= = = 125 × 1014 ∴ =√
t hC 20 × 10−26 F′ 6V
1 ′
∴ F = √6F
No. of photo electrons emitted per second = (No.
of photons emitted)
n
100
27. Ans ( 4 )
⇒ = 125 × 1012 2πm
t T=
then i = ne = 125 × 1012 × 1.6 × 10 – 19 T1
qB
m q
t
–5 ⇒ = 1 × 2
= 2 × 10 A = 20 μ A T2 m2 q 1

22. Ans ( 3 ) ⇒ T1
T2
=
4m
m
×
e
2e
=
2
1
From momentum conservation, ⇒ T1 : T2 = 2 : 1
|patom | = |pphoton |
∵ pphoton =
E
C
& patom = mv 28. Ans ( 4 )
1 2 For a molecule in solid no degree of freedom
∴ Eatom = mv
2 associated with translation and rotational, but
2Eatom
∴ mv = patom =
V
degree of freedom associated to vibration along x,
Also, E
=
2Eatom y and z axis, f = 2 × 3 = 6
∴E =
C
atom 2C
EV
V
29. Ans ( 2 )
23. Ans ( 4 ) Cp – CV = R
Cp – Cv = 2 cal/mole°C
256 256−4(m) 256−(4m) 224
A −−−−−−−→ X
96 m no. of α
−−−−−−−→ Y =B
88
8 – Cv = 2 cal/mol°C ⇒ Cv = 6 cal/mole°C
96−(2m) n no. of β 96−(2m)+n
So that 256 – 4m = 224 ⇒ 4m = 32, m = 8 no. of α Δ U = nCv Δ T
96 – (2m) + n = 88 ⇒ 96 – 16 + n = 88 = 5 × 6 × (20 – 10)
⇒ n = 7 no. of Beta
= 300 cal.
24. Ans ( 2 ) th 30. Ans ( 1 )
For the 4 member lyman : n1 = 1, n2 = 5
for the 3rd member of balmer : n1 = 2, n2 = 5 V
−
V
=4
−R ( 14 1
)
4ℓc 2ℓ0
λ1 25
V V
= − =x
λ2 R ( 11 − 1
) 4(2ℓc ) 2(2ℓ0 )
25
21 1 V V
λ1 100
[ − ] = x
= 2 4ℓc 2ℓ0
λ2 24
1
25
λ1 21 21
⇒ ×4=x
= = 2
λ2 24 × 4 96 ⇒x=2
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31. Ans ( 4 ) 38. Ans ( 1 )
f0 = 100
0.5
= 200 cm

fe = 100
20
= 5 cm

M = f0 = 2005
= 40
fe
33. Ans ( 2 ) The forward biased diode will conduct while the
d g d reverse biased will not
g 1 = g (1 − ) ⇒ = g (1 − )
R 5 R
1 d 4 d
=1− ⇒ =
5 R 5 R ∴ Equivalent resistance = 10 + 15 = 25 Ω
4R
d=
5 39. Ans ( 1 )
34. Ans ( 1 ) Fcos37° = μ (mg – F sin 37°)
By applying conservation of energy μmg 0.5 × 20 × 10
KEi + PEi = KEf + PEf F= o o
= = 90.90 N
cos 37 + μ sin 37 4 3
1 2 GMe m GMe m 5
+ 0.5 × 5
mv − =0−
2 R 2R ∴ F ≈ 91N
40. Ans ( 3 )
1 2 GMe m 1
mv = [− + 1]
2 R 2
1 2 GMe m GMe
mv = ⇒v=√
2 2R R
35. Ans ( 1 )
F=6π η ρ ν
= 6 × 3.14 ×–0.9
5
× 5 × 10 – 3 × 10 × 10 – 2
= 847.8 × 10– 3 N
= 8.48 × 10 N
For first stone (A → G)
SECTION-B
1 2
h – 5 = vA t +
36. Ans ( 4 )
g. t [vA = √2 × 10 × 5]
2
1
20 10 h – 5 = 10t + g t2 ____(1)
Ig = = 0.4A ; new I ′g = = 0.2A 2
50 50 For second stone (B → G)
Ig has to be decreased by = 0.4 – 0.2 = 0.2 A ; 0.2 ampere
current needs to be passed by shunt (half the current) h – 25 = 1 gt2 ____(2)
2
Solve (1) & (2)
t = 2s
2×5
Also time taken for O → A ; t′ = √ = 1s
10
So total time taken by first stone = t + t' = 3s.
so it needs to be shunted by resistance = 50Ω . 41. Ans ( 3 )
37. Ans ( 1 ) ϕ=
1
2
(4 − t2 )
R1 R2 200 20
R= = = Ω
at t = 2, ϕ = 0
R1 + R2 30 3
dR dR dR
= 21 + 22 −dϕ 1
R 2 R R2 ε= = − (0 − 2t) = t
1 dt 2
dR 1 0.5 2 2 2
= + ε2 t 1 t3
400 100 400 H=∫ . dt = ∫ . dt = . ( )
9 R 0 8 8 3 0
4 1 9 1 1 1 1
dR = + = = Ω = × × (8 − 0) = J .
9 18 18 2 8 3 3

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42. Ans ( 2 ) 47. Ans ( 2 )
228
X →Y +Z +Q
118 110
cos2 60° + cos2 60° + cos2 γ =1
Q = ((BE)y + (BE)z)total – (BEx)total ⇒ γ = 45°
= (118 + 110)6.7 – (228) (4.2 MeV) Vertical component of velocity = v cos γ = 50 1
= (228) (6.7 – 4.2) √ 2
= 570 MeV = 25 √ 2

43. Ans ( 3 ) 48. Ans ( 1 )

P = ρ Hg gh F= Gm1 m2
= 13.6 × 103 × 9.8 × 10 – 10 × 10 – 3 r2
= 133.3 × 10 – 10 N/m2 ⇒F∝
1
PV = NKT r2
⇒ F ∝ m1m2
N = PV
KT ⇒ This force provides centripetal force and acts
133.3 × 10−10 × 10−6
N= towards sun
1.38 ×6 10−23 × 300
= 3.2 × 10 ⇒ T2 ∝ a3 (Kepler’s third law)
44. Ans ( 4 ) 49. Ans ( 1 )
∵ PT4 = C Angular width of central maxima
nRT 4 2λ 2 × 500 × 10−9
⇒ T =C = = = 0.1 rad
V
nR 5 d 10 × 10−6
⇒V= T = 5.75 degree
C
∴ dV = 5nR T4 ∴ 2.86 degree on both side of central maxima
dT C which is at θ = 30°
1 dV 5nRT 4 C
∴
V dT
5
=
C
×
nRT 5 50. Ans ( 4 )
⇒ 2 r21 v 9 r 3
T ∵ v T ∝ r ⇒ 2 = T1 = ⇒ 1 =
45. Ans ( 2 ) r2 vT2
3
4 r2 2
V1 r 27
1
K (A2 – x2) = 1 ( 1 kx2 ) ∴ =( 1 ) =
V2 r2 8
2 3 2
2
x
⇒ A2 – x2 = SUBJECT : CHEMISTRY
3
2
4x
⇒ A2 = SECTION-A
3
⇒x=
√3
A = 0.87 A 55. Ans ( 3 )
2
x
∴ × 100% = 87
A
⇒ x = 87
46. Ans ( 1 )
From Above are aromatic in nature.
x – 3t = 0 59. Ans ( 3 )
x = 3t
dx
∴ =3
dt
⇒ wave speed = 3 m/s

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64. Ans ( 4 ) SUBJECT : BOTANY
CO + H2O CO2 + H2) × 2 Keq=K12
⇌

2H2 + O2 2H2O
⇌ Keq=K2 SECTION-A
___________________
2CO + O2 2CO2
⇌ Keq = K12.K2
OR 101. Ans ( 3 )
Target rxn = 2 × eq(1) + eq (2) NCERT Pg. No. # 13
(T.R.) So KT .R. = K12 . K2 102. Ans ( 3 )
67. Ans ( 4 ) NCERT Pg. # 22

h=√
Kh √ Kw
=
103. Ans ( 4 )
C Ka × C NCERT Pg. # 6
h = max., Ka = min., pKa = max.
104. Ans ( 1 )
68. Ans ( 3 ) NCERT Pg. # 60
NCERT-XI, Pg # 222, Part-1
71. Ans ( 4 ) 105. Ans ( 4 )
CuSO4(s) + 5H2O( ℓ ) → CuSO4.5H2O(s) is an NCERT Pg. # 77
example of Δ Hydration H° of CuSO4 106. Ans ( 1 )
73. Ans ( 3 ) NCERT, Pg. # 71
Δ S = nCp ln
T2 107. Ans ( 1 )
T1 NCERT Pg. # 85
5 600
=2×
2
R ln
300 108. Ans ( 1 )
= 5R ln 2 XII NCERT Page No. # 117 (E), 128, 129 (H)
77. Ans ( 1 ) 109. Ans ( 2 )
(1) Glycinato
NCERT-XII, Pg # 98
110. Ans ( 2 )
NCERT Pg. # 228
111. Ans ( 2 )
NCERT-XII, Pg. # 211
(2) Ethylenediamine
112. Ans ( 2 )
NCERT-XII, Pg. # 207
113. Ans ( 4 )
NCERT Pg. # 200
(3) Aqua H2O monodentate.
(4) Oxalato (C2O42 – ) 114. Ans ( 3 )
NCERT, Pg # 220
115. Ans ( 4 )
NCERT, Pg # 224
SECTION-B 116. Ans ( 3 )
96. Ans3 ( 3 ) NCERT Pg. No. # 225
sp hybridisation can only have p π -d π bond as it 117. Ans ( 2 )
can not form p π -p π bond. NCERT Pg#74
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118. Ans ( 1 ) 137. Ans ( 4 )
NCERT Pg. # 98 NCERT Pg. # 95, 98
119. Ans ( 4 ) 138. Ans ( 1 )
NCERT Pg. # 96
NCERT Pg. # 89,90
120. Ans ( 4 ) 139. Ans ( 1 )
NCERT-XI Pg.#11
NCERT, Pg # 216
121. Ans ( 1 ) 140. Ans ( 1 )
NCERT (XI) Pg # 21
NCERT Pg. # 201, 200
123. Ans ( 4 ) 141. Ans ( 3 )
NCERT Pg. # 14
NCERT Pg. # 207 - 210
124. Ans ( 3 ) 142. Ans ( 1 )
NCERT Pg # 23-25
NCERT Pg. No. # 15
125. Ans ( 1 ) 143. Ans ( 3 )
NCERT Pg # 32
NCERT Pg # 33
126. Ans ( 3 ) 144. Ans ( 3 )
NCERT, Pg. # 147
NCERT XI - P. No. - 38,40
127. Ans ( 3 ) 145. Ans ( 3 )
NCERT Pg. # 160
NCERT, Pg. # 146
128. Ans ( 3 ) 146. Ans ( 1 )
NCERT, Pg. # 231
NCERT XII Pg. # 213,214,220,223
129. Ans ( 3 ) 147. Ans ( 4 )
NCERT Pg # 174
NCERT Pg. # 159
130. Ans ( 3 ) 148. Ans ( 3 )
NCERT Pg. No. # 177
131. Ans ( 3 ) NCERT Pg. # 177
149. Ans ( 3 )
NCERT-XI, Pg. # 67
NCERT Pg. # 64
132. Ans ( 1 ) 150. Ans ( 2 )
NCERT Pg. # 64
NCERT XII Pg # 76
133. Ans ( 4 )
NCERT, Pg. # 61, 64, 65 SUBJECT : ZOOLOGY
134. Ans ( 4 ) SECTION-A
NCERT Pg # 73, 74
135. Ans ( 2 ) 151. Ans ( 1 )
Module
NCERT Pg. # 76
152. Ans ( 2 )
SECTION-B NCERT Page No. # 122
136. Ans ( 3 ) 153. Ans ( 3 )
NCERT Pg. No. # 10 NCERT-XII, Pg. # 206
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154. Ans ( 4 ) 176. Ans ( 3 )
NCERT Pg. # 208 NCERT Pg. # 199, 201
156. Ans ( 3 ) 178. Ans ( 3 )
NCERT Pg. # 43 Module-7 Page No. #82
157. Ans ( 2 ) 179. Ans ( 3 )
NCERT, Pg. # 38 XI NCERT Pg. No :- 309, 310, 311
158. Ans ( 3 ) 180. Ans ( 1 )
NCERT, Pg. # 30 NCERT XII Pg # 31
159. Ans ( 4 ) 181. Ans ( 4 )
NCERT, Pg. # (E)-27, (H)-30 NCERT Pg. No. 46
160. Ans ( 1 ) 182. Ans ( 4 )
NCERT Pg#138 NCERT (XII) Pg. # 136
161. Ans ( 3 ) SECTION-B
NCERT (XIth) Pg. # 144
162. Ans ( 4 ) 186. Ans ( 1 )
NCERT Page No. # 116
NCERT Page # 199
187. Ans ( 4 )
163. Ans ( 4 ) NCERT, Pg. # 222
NCERT (XIIth) Pg. # 165
189. Ans ( 4 )
164. Ans ( 2 ) NCERT-XII, Pg. # 179,180,182,183
NCERT XII Pg. # 183, 186, 187.
190. Ans ( 3 )
165. Ans ( 1 ) NCERT XII, Pg. No. 210
191. Ans ( 1 )
th
NCERT (XII ) Pg. # 185
166. Ans ( 3 ) NCERT Pg. # 201
NCERT, Pg. # 47 192. Ans ( 1 )
167. Ans ( 3 ) NCERT, Pg. # 194
NCERT Pg. # 112 193. Ans ( 1 )
168. Ans ( 2 ) NCERT Pg. No. # 46, 47
NCERT Pg. # 82 194. Ans ( 2 )
169. Ans ( 2 ) NCERT, Pg. # 43
NCERT-XI Page No. 233 195. Ans ( 2 )
170. Ans ( 1 ) Module-4 Pg#33
NCERT Pg. # 242,243,245 196. Ans ( 1 )
171. Ans ( 2 ) NCERT-XII, Pg. # 161, 162
NCERT Pg # 242, 243 197. Ans ( 4 )
172. Ans ( 3 ) NCERT Pg. # 173
NCERT XI, Pg. No. # 3 198. Ans ( 4 )
NCERT Pg. # 190
173. Ans ( 2 ) 199. Ans ( 3 )
NCERT Pg # 135 (E), 163 (H) NCERT Pg.No.321
174. Ans ( 2 ) 200. Ans ( 1 )
NCERT-XII, Pg. # 161 NCERT XII Pg # 28,30
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