CLASSROOM CONTACT PROGRAMME
(Academic Session : 2023 - 2024)
PRE-MEDICAL_LEADER_PHASE-1 & 2_IT-7
TARGET : 2024-25
Test Type :MINOR
TEST DATE : 25-10-2023
Test Pattern :NEET (UG)
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
3
4
4
1
2
3
1
1
3
3
4
2
2
3
4
1
1
4
3
4
2
1
2
2
3
1
4
1
2
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.
1
1
1
4
3
3
2
4
1
4
1
1
4
1
2
1
2
3
1
2
3
2
2
3
1
1
1
3
4
1
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
2
4
3
2
2
4
2
1
3
1
1
2
3
3
2
1
4
2
2
2
3
3
3
2
3
4
3
2
1
4
Q. 91
A.
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
3
2
4
4
2
1
1
3
4
4
4
4
3
1
4
4
4
2
1
3
4
2
3
3
4
2
2
4
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.
1
4
2
2
3
2
3
2
2
3
2
4
4
2
3
1
1
1
2
4
4
4
4
4
4
2
3
4
4
3
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
2
1
1
3
1
2
3
4
3
3
3
2
4
3
4
4
3
4
1
2
3
4
3
2
4
3
3
4
2
Q. 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200
A.
1
2
2
3
1
2
1
2
3
4
3
2
4
3
3
4
3
2
2
2
HINT – SHEET
SUBJECT : PHYSICS
2.
dQ
dQ
dQ
KA
)
=(
)
[
=
(TH − TL )]
L
dt rod (1)
dt rod (2)
dt
3KA
KA
=
(100 − ϕ) =
(ϕ − 0)
2
1
(
SECTION-A
1.
Ans ( 2 )
Let final temperature is q then m1 s1 ( θ 1 – θ 0)
+m2s2( θ 2 – θ 0) + m3s3( θ 3 – θ 0) = (m1s1 + m2s2
+ m3s3) × ( θ – θ 0)
θ=
⇒ ϕ = 60°C
4.
m1 s 1 θ 1 + m2 s 2 θ 2 + m3 s 3 θ 3
m1 s 1 + m2 s 2 + m3 s 3
Now,
m1=Vd1, m2 = Vd2 and m3 = Vd3
∴ θ=
d1 s 1 θ1 + d2 s 2 θ2 + d3 s 3 θ3
d1 s1 + d2 s2 + d3 s3
6001CMD30303123003J
Ans ( 3 )
Ans ( 4 )
∘
x − MP
C
=
BP − MP
100
x − 39
39
⇒
=
239 − 39
100
⇒ x = 117 °W
5.
Ans ( 1 )
Heat transfer will take place only if there is
temperature difference.
HS-1/8
Target:2024-25/25-10-2023
6.
Ans ( 2 )
Let R be the rate of heat supply
For body I
Q = ms Δ T
2R = m × c1 × 60 ..... (1)
For body II
4R = m × c2 × 40 ..... (2)
2R
mC1 60
=
4R
mC2 40
Divide expression (1) by (2)
1
C
3
= 1 ×
2
C2
2
C1
1
=
C2
3
mL1
4R
=
mL2
3R
L1
4
=
L2
3
Ans ( 3 )
good absorbers are good emitters.
8.
Ans ( 1 )
=
T = 80°C
9.
11.
=
V γr Δθ
V γa Δθ
=
γr
γa
=
7γs
6γs
=
7
6
15.
Ans ( 3 )
At 4 °C, density is maximum hence volume is
minimum. Therefore it overflows on both
cooling and heating.
16.
Ans ( 4 )
If A and B are mixed
m×sA×(15 – 10) = m×sB×(25 – 15) ⇒ sA = 2sB
If B and C are mixed
= 574.25
Ans ( 3 )
ΔL = L a ΔT
m×sB×(30 – 25)=m×sC×(40 – 30) ⇒ sC =
If A and C are mixed
m×sA×( θ – 10) = m × sC× (40 – θ )
⇒
θ = 16°C
ΔL
= Lα
ΔT
(Slope)A < (Slope)B ; LA α A < LB α B
⇒
∵ LA = LB therefore ∴ ( α A < α B)
Re al exp ansion
Apparent exp ansion
Ans ( 2 )
For mercury Δ Vm = V0 γ Hg Δ T
(1 lit. = 103 cc) = (103 cc) (1.82 × 10 – 4) (100)
= 18.2 cc
For Flask Δ Vg = V0(3 α g) Δ T
= (103cc) (3 × 0.1 × 10 – 4) (100) = 3cc
Amount which spill out = 18.2 – 3 = 15.2 cc
L
Ans ( 3 )
γ = αx + αy + αZ
γ = α1 + α2 + α2
= α1 + 2 α2
slope
γr = 7γs
γa = γr – γs = 7γs – γs = 6γs
14.
T − 32
T − 273
=
180
100
2297
4
Ans ( 2 )
∴
(T − 0)KA
5T – 160 = 9T – 2457
or 4T = 2457 – 160 = 2297
10.
13.
Ans ( 1 )
T=
ΔT
1
= ∝ Δθ
2
T
1
= × 1.2 × 10−5 × 10
2
ΔT
=
= 6.0 × 10−5
T
Hence,
7.
L
Ans ( 4 )
Time lost in 1 week
= 6.0 × 10 – 5 × T
= 6.0 × 10 – 5 × 7 × 24 × 3600
= 36.285
Now for latest heat
For body I
Q = mL
4R = mL1 ..... (3)
For body II
3R = mL2 ..... (4)
Divide expression (3) by expression (4)
4KA(100 − T )
12.
17.
1
sB
2
Ans ( 1 )
1 1
(
Mv2 ) = J(m.c. Δ θ )
2 2
1
× 1 × (50)2 = 4.2[200 × 0.105 × Δ θ ]
4
W = JQ ⇒
⇒
⇒
HS-2/8
Δ θ = 7.1°C
6001CMD30303123003J
PRE-MEDICAL_LEADER_PHASE-1 & 2_IT-7_25-10-2023
18.
19.
Ans ( 1 )
graph(a) looks exactly same as actual graph for
this scenario as flat vertical lines are
representing latent heat of fusion &
vapourization respectively.
Ans ( 1 )
(RC)y > (RC)x
So ey > ex
Ay > Ax
30.
Ans ( 2 )
t+r+a=1
Ans ( 4 )
(110 + 10) × 1 × (t – 10) = 220 (70 – t)
340 t = 1200 + 15400 = 16600
t=
20.
27.
1660
≅49∘ C
34
Ans ( 3 )
Equivalent electrical circuit, will be as shown
in figure.
Temperature difference between A and D is
180ºC, which is equally
a=1–
(
1 1
13
+ )=
9 6
18
31.
Ans ( 1 )
Black bulb will radiate more energy to the air
column causing more expansion of air column in
limb X. so, alcohol level in limb X will fall.
34.
Ans ( 4 )
σ × 4πR2 × T 4
4 × D2
2
4
∝RT
S=
S
2
S1
2R
2T
=(
) × (
S1
R
T
distributed in all the rods. Therefore,
temperature difference between A and B will be
60ºC, or temperature of B should be 140ºC.
21.
Ans ( 4 )
dQ
; when K = ∞ , dθ = 0
dt
dx
i.e., θ is independent of x, i.e., constant or
uniform.
24.
25.
26.
= −KA
Ans ( 2 )
Convection is a mode of heat transfer by actual
motion of matter. If is possible only in fluids.
convection can be natural or forced.
Ans ( 3 )
It is hotter over the top of a fire than at the
same distance on the sides due the convection
currents rising upwards because of the
difference in densities.
6001CMD30303123003J
)
= 64
SECTION-B
36.
Ans ( 3 )
Spectral energy distribution curve is a continous
curve for ideal black body (IBB)
37.
Ans ( 2 )
Amount of radiation emitted and absorbed by it
will be equal.
38.
Ans ( 4 )
E ∝ T4
dθ
dx
Ans ( 2 )
t1 : t2 : : 1 : 3
t2 = 24 × 3 = 72 h
4
4
300
E1
=(
) = 1 : 256
E2
1200
39.
Ans ( 1 )
dT
eAσ 4
=
[T − T04 ]
ms
dt
dT
⇒−
∝A
dt
−
Now, as for a given mass, area of the sphere is
minimum, hence it will have the lowest rate of
cooling.
HS-3/8
Target:2024-25/25-10-2023
40.
Ans ( 4 )
41.
Ans ( 1 )
T' = 3 T ⇒ E' =
81
E
2
16
E′
81
(
− 1) × 100 = (
− 1) × 100
E
16
= 81 − 16 × 100 = 406.25 ≈ 400
16
θ1 − θ2
=K
t
[
47.
θ1 + θ2
− θ0 ]
2
FL
FL
4F L
=
=
2
AΔL
(πD /4) ΔL
πD2 ΔL
Y πD2 ΔL
F (tension in the wire) =
4L
11
−3 2
−3
(0.91 × 10 ) × 3.142 × (2 × 10
) × (2.376 × 10
)
=
4 × 1.8
= 3.8 × 102 N
As
61ºC to 59ºC ⇒
61 − 59
= K
10
K= 1
150
61 + 59
− 30]
2
[
51ºC to 49ºC ⇒
51 − 49
1
=
t
150
[
51 + 49
− 30]
2
t = 15 min
43.
Ans ( 2 )
As the wire is not free to contract, the thermal
stress is developed at the wire.
The change in temperature, Δ T = 27°C – ( –
39°C) = 66°C
Let Δ L be the change in length of the wire.
Δ L = α L Δ T = (2 × 10 – 5) × 1.8 × 66 = 2.376 ×
10 – 3 m
48.
Ans ( 4 )
Y =
Ans ( 3 )
The given arrangement of rods can be redrawn
as follows
λ1 T 1 = λ2 T 2
–6
0.5 × 10 × T1 = 10 – 4 × T2
T1
= 200
T2
45.
Ans ( 2 )
R = R0 (1 + α T + β T3)
3
3
= 1 + α (10) + β(10)
2
1
∴ 10 (α + 100β) =
2
α + 100β = 0.05
46.
Ans ( 1 )
ΔV
0.12
= γΔT ⇒
= (3α) × 20
100
V
0.12
⇒α =
= 2 × 10−5 ∘ C −1
100 × 60
It is given that H1 = H2
⇒
KA(θ1 − θ2 )
⇒ K3 =
49.
=
K3 A(θ1 − θ2 )
2l
K
K1 K2
=
2
K1 + K2
l
Ans ( 1 )
TB − 90
T −0
T − 20
+ B
+ B
=0
R
R
R
2
2
SUBJECT : CHEMISTRY
SECTION-A
HS-4/8
54.
Ans ( 3 )
Reactivity ∝ stability of carbocation.
55.
Ans ( 1 )
6001CMD30303123003J
PRE-MEDICAL_LEADER_PHASE-1 & 2_IT-7_25-10-2023
56.
Ans ( 1 )
57.
Ans ( 1 )
H2, Pd/C, quinoline
58.
Ans ( 3 )
3°H
is
more
photohalogenation.
72.
Ans ( 3 )
75.
Ans ( 4 )
59.
Ans ( 4 )
cis-2-butene
µ=0
dipole moment (µ)
High EN
EN = sp>sp2>sp3
Stat. (1) ×
60.
Ans ( 1 )
Stat (2) →
reactive
61.
Ans ( 3 )
Rate of SN1 ∝ stability of carbocation
63.
Ans ( 2 )
NCERT-XII Pg#292
64.
Ans ( 3 )
66.
Ans ( 3 )
68.
Ans ( 1 )
Named reactions
6001CMD30303123003J
towards
79.
Ans ( 1 )
85.
Ans ( 1 )
Name reaction
trans-2-butene
µ=0
Low EN →
SECTION-B
88.
Ans ( 2 )
89.
Ans ( 2 )
HS-5/8
Target:2024-25/25-10-2023
91.
Ans ( 3 )
SUBJECT : BOTANY
SECTION-A
101. Ans ( 4 )
NCERT-XII Pg # 27
92.
Ans ( 1 )
Rate of E2 ∝ Stability of product.
102. Ans ( 4 )
NCERT XII, Page # 21
104. Ans ( 3 )
NCERT (XII) Pg. # 23
106. Ans ( 4 )
NCERT Pg. # 35(E) & 37(H)
107. Ans ( 4 )
NCERT Pg. # 35(E) & 37(H)
93.
Ans ( 3 )
108. Ans ( 4 )
NCERT-XII, Page#34
110. Ans ( 1 )
NCERT (XIIth) Pg. # 23
112. Ans ( 4 )
NCERT XII Pg. # 29,31
94.
Ans ( 2 )
116. Ans ( 4 )
NCERT Pg. # 34,35,38,39
117. Ans ( 2 )
NCERT (XII) Pg # 36,37(E), 40(H)
118. Ans ( 2 )
NCERT XII Pg.# 25, 36, para 2.2.2, 2.4.3
98.
Ans ( 1 )
Answer (1)
CH4 can’t be prepared by Wurtz reaction.
119. Ans ( 4 )
NCERT Page # 22-24
121. Ans ( 1 )
NCERT (XII) Pg # 26
126. Ans ( 2 )
NCERT (XII) Pg. # 31
130. Ans ( 3 )
NCERT XII Pg.# 27
131. Ans ( 2 )
NCERT (XIIth), Pg. # 25,26
HS-6/8
6001CMD30303123003J
PRE-MEDICAL_LEADER_PHASE-1 & 2_IT-7_25-10-2023
133. Ans ( 4 )
NCERT-XII, Pg # 34 & 35
134. Ans ( 2 )
NCERT-XII, PG-29
SECTION-B
136. Ans ( 1 )
NCERT Pg .# 28
137. Ans ( 1 )
NCERT-XI Pg. # 36
138. Ans ( 1 )
NCERT-XII Pg#24
139. Ans ( 2 )
NCERT-XII, Pg. # 31
140. Ans ( 4 )
NCERT-XII, Pg. # 37
141. Ans ( 4 )
NCERT Pg # 23
142. Ans ( 4 )
NCERT P.No. 29 Ist para
143. Ans ( 4 )
NCERT Pg. # 11
144. Ans ( 4 )
NCERT-XII, Pg.# 38
146. Ans ( 2 )
NCERT XII Pg # 36
SUBJECT : ZOOLOGY
SECTION-A
151. Ans ( 4 )
NCERT Pg.No. 43
152. Ans ( 2 )
NCERT Pg.No. 49
153. Ans ( 1 )
NCERT Pg.No. 49
154. Ans ( 1 )
NCERT Pg.No. 54
155. Ans ( 3 )
NCERT Pg.No. 43
156. Ans ( 1 )
NCERT Pg.No. 43
157. Ans ( 2 )
NCERT Pg. No. 51
158. Ans ( 3 )
NCERT Pg.No. 52
159. Ans ( 4 )
NCERT Pg.No. 54
160. Ans ( 3 )
NCERT Pg.No. 54
161. Ans ( 3 )
NCERT Pg.No. 49
162. Ans ( 3 )
NCERT Pg.No. 54
163. Ans ( 2 )
NCERT Pg.No. 48
164. Ans ( 4 )
NCERT Pg. No. 49
165. Ans ( 3 )
NCERT Pg.No. 51
166. Ans ( 4 )
NCERT Pg.No. 51
6001CMD30303123003J
HS-7/8
Target:2024-25/25-10-2023
167. Ans ( 4 )
NCERT Pg. No. 52
168. Ans ( 3 )
Allen Module
169. Ans ( 4 )
NCERT Pg.No. 54
170. Ans ( 1 )
NCERT Pg.No. 48
171. Ans ( 2 )
NCERT Pg.No. 48
172. Ans ( 3 )
NCERT Pg.No. 49
174. Ans ( 3 )
NCERT Pg.No. 51
175. Ans ( 2 )
NCERT Pg.No. 54
176. Ans ( 4 )
NCERT Pg.No. 43
177. Ans ( 3 )
Allen Module
178. Ans ( 3 )
NCERT Pg.No. 43
179. Ans ( 4 )
NCERT Pg. No.43
180. Ans ( 2 )
NCERT Pg.No. 49
181. Ans ( 1 )
NCERT Pg.No. 49
182. Ans ( 2 )
NCERT Pg.No. 50
183. Ans ( 2 )
Allen Module
SECTION-B
186. Ans ( 2 )
Allen Module
187. Ans ( 1 )
NCERT Pg.No. 49
188. Ans ( 2 )
NCERT Pg. No. 50
189. Ans ( 3 )
NCERT Pg.No. 54
190. Ans ( 4 )
NCERT Pg.No. 54
191. Ans ( 3 )
Allen Module
192. Ans ( 2 )
Allen Module
193. Ans ( 4 )
NCERT Pg.No. 46
194. Ans ( 3 )
NCERT Pg.No. 46, 47
195. Ans ( 3 )
NCERT Pg. No. 49
196. Ans ( 4 )
NCERT Pg. No. 44
197. Ans ( 3 )
NCERT Pg.No. 43, 49
198. Ans ( 2 )
Allen Module
199. Ans ( 2 )
NCERT Pg.No. 44
200. Ans ( 2 )
NCERT Pg.No. 49
185. Ans ( 1 )
Allen Module
HS-8/8
6001CMD30303123003J