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Test - 7 (Code-C) (Answers & Hints) All India Aakash Test Series for Medical-2018
1/9
1. (1)
2. (3)
3. (4)
4. (3)
5. (2)
6. (1)
7. (2)
8. (1)
9. (1)
10. (3)
11. (2)
12. (4)
13. (4)
14. (3)
15. (2)
16. (4)
17. (4)
18. (2)
19. (1)
20. (2)
21. (3)
22. (1)
23. (1)
24. (3)
25. (3)
26. (1)
27. (1)
28. (3)
29. (3)
30. (2)
31. (3)
32. (1)
33. (2)
34. (3)
35. (1)
36. (3)
37. (1)
38. (4)
39. (2)
40. (1)
41. (3)
42. (1)
43. (3)
44. (3)
45. (1)
46. (3)
47. (3)
48. (1)
49. (3)
50. (1)
51. (2)
52. (4)
53. (4)
54. (3)
55. (1)
56. (2)
57. (2)
58. (3)
59. (3)
60. (2)
61. (1)
62. (1)
63. (Deleted)
64. (4)
65. (1)
66. (3)
67. (3)
68. (3)
69. (4)
70. (1)
71. (3)
72. (3)
73. (1)
74. (1)
75. (4)
76. (4)
77. (4)
78. (3)
79. (3)
80. (2)
81. (1)
82. (4)
83. (3)
84. (2)
85. (2)
86. (4)
87. (3)
88. (4)
89. (3)
90. (2)
91. (2)
92. (3)
93. (2)
94. (1)
95. (3)
96. (2)
97. (2)
98. (2)
99. (1)
100. (4)
101. (2)
102. (2)
103. (3)
104. (3)
105. (1)
106. (2)
107. (1)
108. (1)
109. (3)
110. (4)
111. (4)
112. (2)
113. (3)
114. (3)
115. (3)
116. (3)
117. (4)
118. (4)
119. (4)
120. (2)
121. (1)
122. (2)
123. (2)
124. (2)
125. (2)
126. (4)
127. (2)
128. (3)
129. (4)
130. (3)
131. (3)
132. (1)
133. (2)
134. (1)
135. (3)
136. (1)
137. (1)
138. (3)
139. (2)
140. (1)
141. (2)
142. (1)
143. (4)
144. (3)
145. (2)
146. (2)
147. (4)
148. (3)
149. (1)
150. (3)
151. (3)
152. (1)
153. (1)
154. (3)
155. (1)
156. (2)
157. (4)
158. (3)
159. (1)
160. (3)
161. (2)
162. (3)
163. (4)
164. (2)
165. (1)
166. (2)
167. (4)
168. (3)
169. (2)
170. (4)
171. (1)
172. (3)
173. (2)
174. (1)
175. (1)
176. (2)
177. (4)
178. (3)
179. (3)
180. (3)
ANSWERS
TEST - 7 (Code-C)
All India Aakash Test Series for Medical-2018
Test Date : 19-02-2017
All India Aakash Test Series for Medical-2018 Test - 7 (Code-C) (Answers & Hints)
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Hints to Selected Questions
[ PHYSICS]
1. Answer (1)
Efficiency of Carnot engine and coefficient of
performance of refrigerator are related as
1
...(i)
Here, percentage = 10%
or = 0.1 ...(ii)
Putting value of in (i),
1 0.1 0.99
0.1 0.1
= 9
2. Answer (3)
1 2
v
700 Hz 700 Hz
0
1
⎧ ⎫ ⎨ ⎬⎩ ⎭
v v
n n
v
0
2
⎧ ⎫ ⎨ ⎬⎩ ⎭
v v
n n
v
Given 2 1
8 n n
0 0{ } 8 nv v v v
v
02
8nv
v
02 700
8350
v
v0 = 2 m/s
3. Answer (4)
By law of mixture,
Heat given = Heat taken
18(1)(50 – ) = (10)(0.5)(0 + 20) + 10(80)
900 – 18 = 100 + 800
18 = 0
= 0°C
Temperature of mixture will be 0°C and whole ice
will melt.
4. Answer (3)
2FP F PA P VT
A ⇒
5. Answer (2)
rel boat water v v v
ˆ ˆ ˆ ˆ3 5 2 3i j i j
ˆ ˆ ˆ ˆ3 5 2 3i j i j ˆ ˆ5 8i j
6. Answer (1)
F T MA F T
AM
∵T ma
F ma
M
7. Answer (2)
8. Answer (1)
3 22 2 4 8
3 3
e
e e e
e e
GM Gv R G R
R R
ve R
e
9. Answer (1)
230
dUF r
dr
|F | = 30r2 ...(i)
2
mvF
r
2 230 m
r v
r
mv2 = 30r
3 = 30000
2 4115000 1.5 10 J
2 mv
10. Answer (3)
For simple pendulum in liquid,
11
2
⎛ ⎞⎜ ⎟⎝ ⎠
gd
nl
For spring pendulum in liquid, frequency remain
unchanged.
Test - 7 (Code-C) (Answers & Hints) All India Aakash Test Series for Medical-2018
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11. Answer (2)
Given, l1 = 1.44 m
l2 = 1.00 m
When two pendulums are in out of phase, set number
of vibration completed by them are n1 and n
2
respectively. We can say if n1 = n, then n
2 = n +
1
2.
Now, by question set they are out of phase after time
t. So,
t = n1T1 = n
2T2
1 2
2 1
n T
n T
2
1
1 10
1 1.44 12
2
n
n
l
l
12n = 10n + 5
2n = 5
n = 2.5
Now, t = n1T1
= 1(2.5)(2 )g
l
= 1.44
(2.5)(2)( )10
6 s
12. Answer (4)
Here, H1 = H
2 = H
3
1 1 2 2(120 ) 1.5 ( ) 3 ( 0)
KA T KA T T KA T
l l l
(120 – ) = 1.5( – ) = 3T T T T1 1 2 2
120 – = 3T T1 2
1.5( – ) = 3T T1 2
T2
T T1 2 + 3 = 120 ...(i) T T
1 2 = 3 ...(ii)
Solving (i) and (ii),
T1 = 60°C
T2 = 20°C
13. Answer (4)
E T4
4
1 1
2 2
⎛ ⎞ ⎜ ⎟⎝ ⎠
E T
E T
Put T1 = T
T2 = 1.5T
1
2
1
5
E
E
E2 5E
1
E2 = 500% of E
1
14. Answer (3)
2
12 m
Il
I l2
2I
I
l
l
⎛ ⎞ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟
⎝ ⎠
l
l
l
l 2 2I I I l
l
I = 2I
15. Answer (2)
2
' 1 9
hg g h R
R
⎛ ⎞ ⇒ ⎜ ⎟⎝ ⎠
16. Answer (4)
Final sphere
Final particle
90°
30° = 1
60° = 1
1 2+ = 90°
Initial particle
17. Answer (4)
18. Answer (2)
Temporary loss =
2
21 1(2 )
2 2 2
⎛ ⎞ ⎜ ⎟⎝ ⎠
vmv m
= 2 21 1 1
2 2 2
⎛ ⎞ ⎜ ⎟⎝ ⎠
mv mv
= 2
EE
= 2
E
19. Answer (1)
m1
m2
l1
l2
l
COM
All India Aakash Test Series for Medical-2018 Test - 7 (Code-C) (Answers & Hints)
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I = m1l1
2 + m2l2
2 ...(i)
Here,2
1
1 2
m
m m
ll
1
2
1 2
m
m m
ll
Put value of l1 and l
2 in (i) and get answer.
20. Answer (2)
O
45°
Hmax
u cos 45°
L = mvr
= m(u cos)Hmax
=
2 2sin 45
22u u
mg
L =
3
4 2
mu
g
21. Answer (3)
f iE ME ME ME
22. Answer (1)
23. Answer (1)
A
Bmg
L/2
L cos
2
I 2
cos2 3
L MLmg
3cos
2
g
L
24. Answer (3)
nB – n
A = n
1...(A)
nC – n
A = n
2...(B)
From (A) and (B),
nC – n
B = n
2 – n
1
25. Answer (3)
2 kg block do not tendency to move. So there is no
force between them.
26. Answer (1)
27. Answer (1)
2 21 1
2 2 mv mgx Kx
28. Answer (3)
2
2 2 23 33
2 2 2 MR
MR MR MK
2 273
2MR MK
7
6K R
29. Answer (3)
Work = Change in rotational kinetic energy
W = 2 2
2 1
1( )
2 I
=
2
2 2 2
2 1
1(4 )( )
2 2
⎛ ⎞ ⎜ ⎟
⎝ ⎠
MRn n
= 1500 J
30. Answer (2)
3
5A
B...(i)
2 235 2 cos 60 A B AB ...(ii)
Solving (i) and (ii), we get
A = 15, B = 25
31. Answer (3)
32. Answer (1)
a
1x
x2 – 1
For block to remain stationary over wedge,
mg sin = ma cos
2
1tan
1
⎛ ⎞ ⎜ ⎟⎝ ⎠
a g g
x
21
ga
x
Test - 7 (Code-C) (Answers & Hints) All India Aakash Test Series for Medical-2018
5/9
33. Answer (2)
For fourth harmonic, the string will vibrate in four
loops.
L/4
L/8Pressed
Plucked
34. Answer (3)
Given, 2
2
v
l = n
or n = v
l...(i)
For closed pipe, frequency of second overtone is
5 5
4 4 v n
n
l.
35. Answer (1)
2 23 38 2 8 m/sv n v
36. Answer (3)
U = (N1/3 – 1)U
37. Answer (1)
P h g
V V
V V
100 V
V= 100
h g
=
3
8
100 10 9.8 100
9.8 10
= 0.1
38. Answer (4)
39. Answer (2)
40. Answer (1)
L = 10
0
10 logI
I
= 10 log10
105
= 50 log10
10
= 50 dB
41. Answer (3)
22
1 2max 1 2
min 1 2 1 2
9
1
⎛ ⎞⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠
I II a a
I a a I I
42. Answer (1)
43. Answer (3)
44. Answer (3)
L = MvR + Ic
45. Answer (1)
The elongation produced by m1 at equilibrium is the
amplitude for m2
1m g Ka
1m g
aK
⇒
[ CHEMISTRY]
46. Answer (3)
N1V1 = N
2V2
2
1 1200 V
2 10
V2 = 1000
Water added = 1000 – 200 = 800 cc
47. Answer (3)
N = M × nfactor
48. Answer (1)
Atomic mass × specific heat = 6.4
Atomic mass = 6.4 6.4
Specific heat 0.16 = 40
49. Answer (3)
3 2 2 21mol2 mol100 g
CaCO 2HCl CaCl H O CO
Number of moles of HCl taken = 1 × 1
1 mol HCl will produce 1
2 mol CO
2.
50. Answer (1)
8
6
c 3 100.75 m
v 400 10
All India Aakash Test Series for Medical-2018 Test - 7 (Code-C) (Answers & Hints)
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51. Answer (2)
2 1
4 3
(E E ) ( 3.4) ( 13.6) 10.2eV
(E E ) ( 0.85) ( 1.5) 0.66
52. Answer (4)
2
2
ZE
n
2n
rZ
Zv
n
53. Answer (4)
B
O O – H
H
O
H
54. Answer (3)
BrF5 (sp
3d2)
Br
F
F
F
F
F
Due to lp–bp repulsion none of the bond angle is at
90°.
55. Answer (1)
Fact.
56. Answer (2)
2
1Cl Cl (aq)
2
H = 1
2 × 240 + (–349) + (–381)
= –610 kJ mol–1
57. Answer (2)
Fact.
58. Answer (3)
2O
W 16n
M 32
W = 2
10
1
V2.303 nRT log
V
= 10
16 502.303 8.314 300 log joule
32 5
= –2.87 × 103 J
= –2.87 kJ
59. Answer (3)
Fact.
60. Answer (2)
rH° =
i f i fi i
a H (Product) b H (Reactant)
= 6 × (–94) + 3 × (–68.3) – 11.7
= –564 – 204.9 – 11.7
rH° = –780.6 kcal
61. Answer (1)
Fact.
62. Answer (1)
2CO
P
CO
pK
p
21.4 0.800
1.4 p 0.80 p
FeO CO(g) Fe CO (g)
���⇀↽���
0.80 p0.625
1.4 p
0.875 – 0.625p = 0.80 + p
0.875 – 0.80 = p + 0.625p
0.075 = 1.625p
p = 0.046
pCO
= 1.4 – 0.046 = 1.354
2CO
p 0.80 0.046 0.846
63. Deleted
64. Answer (4)
H+ = 14
9.62 10
= 3.10 × 10–7
pH = –log[3.10 × 10–7]
= 7 – 0.491
= 6.509
65. Answer (1)
Fact.
66. Answer (3)
67. Answer (3)
Fact.
68. Answer (3)
5 3
6 6 3 6 5 2C H HNO C H NO
69. Answer (4)
2 3
2 7Cr O 6e 2Cr
Test - 7 (Code-C) (Answers & Hints) All India Aakash Test Series for Medical-2018
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70. Answer (1)
Gram equivalent KMnO4
= Gram equivalent FeSO4
Gram equivalent KMnO4
= N × V (L)
= 1 × 200
1000 = 0.2
Gram equivalent FeSO4 =
Weight
Equivalent weight
Weight0.2 30.4 g
152
71. Answer (3)
Fact.
72. Answer (3)
Na, Al, Si, P, S, Cl are paramagnetic.
73. Answer (1)
NaHCO3 + NaOH Na
2CO
3 + H
2O
74. Answer (1)
Fact.
75. Answer (4)
Fact.
76. Answer (4)
Both acidic oxide and amphoteric oxide react with
NaOH.
77. Answer (4)
N
H H
H
H+
78. Answer (3)
Zeolites are 3-D silicates.
79. Answer (3)
CH
H
CH3
CH3
C
, ,
CH – C – C – CH3 3
H CH3
H CH3
(Alkanes does not exhibitgeometrical isomerism)
,
80. Answer (2)
Homolytic fission produces free radical
3 3 3CH CH 2CH
81. Answer (1)
H w.r.t. carbonyl group should be present for
tautomerism.
82. Answer (4)
Fact.
83. Answer (3)
12 0.396Percentage of C 100 21.95%
44 0.492
2 0.2028Percentage of H 100 4.5%
18 0.492
Percentage of O = (100 – 21.95 – 4.5) = 73.47%
84. Answer (2)
Percentage of N = 1.4 m.eq. of acid used
Weight of compound
85. Answer (2)
SO3 is a neutral electrophile due to presence of
vacant d-orbital.
86. Answer (4)
Fact.
87. Answer (3)
Br
Cl
2Na/ether
Wurtz reaction
88. Answer (4)
CH = CH2
CH CH
OH
Hg /H SO2 2 4
+
CH CHO3
(X)
89. Answer (3)
CH –CH=CH2 2
CH–CH –CH2 2
CH –CH–CH2 3
HBr
Br–
Br
CH–CH –CH2 3
H shifting–
Benzyl carbocation
90. Answer (2)
Br
Br2
H O2
Br
BrOH
Br
H O2
All India Aakash Test Series for Medical-2018 Test - 7 (Code-C) (Answers & Hints)
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[ BIOLOGY
]
91. Answer (2)
Thermoacidophiles
92. Answer (3)
Protista
93. Answer (2)
PS-II is present.
94. Answer (1)
Zygote
95. Answer (3)
Dinoflagellates
96. Answer (2)
97. Answer (2)
98. Answer (2)
99. Answer (1)
Heterospory – Pteridophytes
Seed habit – Gymnosperms
Fruit habit – Angiosperms
100. Answer (4)
Red algae
101. Answer (2)
102. Answer (2)
Fabaceae
103. Answer (3)
Analogous to tuber of sweet potato and possess
axillary buds.
104. Answer (3)
105. Answer (1)
All the tissue inside endodermis.
106. Answer (2)
Completely secondary in origin.
107. Answer (1)
108. Answer (1)
Oxidative phosphorylation
109. Answer (3)
Ovum (n), meiocyte (2n)
110. Answer (4)
Metaphase is followed by anaphase.
111. Answer (4)
112. Answer (2)
Consists of living part.
113. Answer (3)
ABA closes stomata.
114. Answer (3)
115. Answer (3)
116. Answer (3)
Legumes
117. Answer (4)
118. Answer (4)
C3
119. Answer (4)
C4 – 5, 2
120. Answer (2)
Glycolysis – 1 NADH
Link – 1 NADH
TCA cycle – 3 NADH and 1 FADH2
121. Answer (1)
122. Answer (2)
Fats are poor in oxygen.
123. Answer (2)
124. Answer (2)
Secondary meristem
125. Answer (2)
126. Answer (4)
Precursor to seed habit – Selaginella.
127. Answer (2)
128. Answer (3)
Tuber aestivum
129. Answer (4)
130. Answer (3)
Herbarium is quick referral system.
131. Answer (3)
Polymoniales
132. Answer (1)
Test - 7 (Code-C) (Answers & Hints) All India Aakash Test Series for Medical-2018
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133. Answer (2)
134. Answer (1)
135. Answer (3)
136. Answer (1)
Cnidaria and ctenophora are diploblastic.
137. Answer (1)
Echinoderms and hemichordates are also
deuterostomes.
138. Answer (3)
Sea-anemones are carnivores.
139. Answer (2)
140. Answer (1)
141. Answer (2)
142. Answer (1)
Sperms are formed in testes and stored in seminal
vesicles.
143. Answer (4)
144. Answer (3)
Pancreatic acini secrete pancreatic enzymes.
145. Answer (2)
146. Answer (2)
Carbohydrates are absorbed as monosaccharides
and amino acids are absorbed actively.
147. Answer (4)
Enterokinase activates trypsinogen.
148. Answer (3)
149. Answer (1)
150. Answer (3)
During inspiration, the intrapleural pressure becomes
negative.
151. Answer (3)
152. Answer (1)
153. Answer (1)
154. Answer (3)
Cardiac output to muscles increases during
exercise.
The resistance to flow in the pulmonary system is
much lower than in systemic circulation, hence
pressure in pulmonary artery is lower than in aorta.
155. Answer (1)
156. Answer (2)
157. Answer (4)
Descending limb of the loop of Henle is impermeable
to salts.
158. Answer (3)
These are in the cortex.
159. Answer (1)
160. Answer (3)
Alcohol suppresses ADH secretion.
161. Answer (2)
162. Answer (3)
163. Answer (4)
164. Answer (2)
165. Answer (1)
166. Answer (2)
167. Answer (4)
168. Answer (3)
Vitamin A itself is retinol and its aldehyde is retinal.
169. Answer (2)
170. Answer (4)
171. Answer (1)
172. Answer (3)
Hormones are required in traces.
173. Answer (2)
174. Answer (1)
175. Answer (1)
Pineal gland
176. Answer (2)
177. Answer (4)
178. Answer (3)
179. Answer (3)
180. Answer (3)
� � �
Test - 7 (Code-D) (Answers & Hints) All India Aakash Test Series for Medical-2018
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1. (3)
2. (1)
3. (1)
4. (3)
5. (1)
6. (3)
7. (4)
8. (2)
9. (3)
10. (1)
11. (3)
12. (1)
13. (4)
14. (3)
15. (1)
16. (4)
17. (1)
18. (1)
19. (3)
20. (3)
21. (1)
22. (1)
23. (3)
24. (3)
25. (1)
26. (4)
27. (3)
28. (4)
29. (2)
30. (2)
31. (4)
32. (1)
33. (2)
34. (2)
35. (4)
36. (1)
37. (3)
38. (3)
39. (4)
40. (3)
41. (4)
42. (1)
43. (2)
44. (1)
45. (3)
46. (4)
47. (1)
48. (2)
49. (1)
50. (2)
51. (4)
52. (4)
53. (1)
54. (4)
55. (3)
56. (4)
57. (1)
58. (1)
59. (2)
60. (4)
61. (2)
62. (3)
63. (3)
64. (1)
65. (1)
66. (3)
67. (2)
68. (1)
69. (1)
70. (1)
71. (3)
72. (2)
73. (Deleted)
74. (3)
75. (3)
76. (4)
77. (1)
78. (1)
79. (4)
80. (4)
81. (3)
82. (1)
83. (2)
84. (2)
85. (4)
86. (3)
87. (1)
88. (3)
89. (1)
90. (1)
91. (1)
92. (3)
93. (4)
94. (3)
95. (1)
96. (1)
97. (2)
98. (1)
99. (4)
100. (2)
101. (4)
102. (4)
103. (4)
104. (4)
105. (3)
106. (4)
107. (2)
108. (2)
109. (4)
110. (1)
111. (1)
112. (1)
113. (1)
114. (4)
115. (4)
116. (2)
117. (1)
118. (3)
119. (3)
120. (4)
121. (3)
122. (1)
123. (1)
124. (4)
125. (4)
126. (2)
127. (3)
128. (2)
129. (4)
130. (4)
131. (1)
132. (3)
133. (4)
134. (1)
135. (4)
136. (1)
137. (1)
138. (1)
139. (4)
140. (4)
141. (3)
142. (3)
143. (4)
144. (1)
145. (3)
146. (2)
147. (4)
148. (1)
149. (2)
150. (4)
151. (3)
152. (4)
153. (4)
154. (1)
155. (4)
156. (1)
157. (3)
158. (1)
159. (2)
160. (4)
161. (3)
162. (1)
163. (1)
164. (3)
165. (1)
166. (3)
167. (3)
168. (1)
169. (4)
170. (4)
171. (2)
172. (1)
173. (4)
174. (3)
175. (4)
176. (3)
177. (4)
178. (1)
179. (3)
180. (3)
ANSWERS
TEST - 7 (Code-D)
All India Aakash Test Series for Medical-2018
Test Date : 19-02-2017
All India Aakash Test Series for Medical-2018 Test - 7 (Code-D) (Answers & Hints)
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Hints to Selected Questions
[ PHYSICS]
1. Answer (3)
The elongation produced by m1 at equilibrium is the
amplitude for m2
1m g Ka
1m g
aK
⇒
2. Answer (1)
L = MvR + Ic
3. Answer (1)
4. Answer (3)
5. Answer (1)
22
1 2max 1 2
min 1 2 1 2
9
1
⎛ ⎞⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠
I II a a
I a a I I
6. Answer (3)
L = 10
0
10 logI
I
= 10 log10
105
= 50 log10
10
= 50 dB
7. Answer (4)
8. Answer (2)
9. Answer (3)
P h g
V V
V V
100 V
V= 100
h g
=
3
8
100 10 9.8 100
9.8 10
= 0.1
10. Answer (1)
U = (N1/3 – 1)U
11. Answer (3)
2 23 38 2 8 m/sv n v
12. Answer (1)
Given, 2
2
v
l = n
or n = v
l...(i)
For closed pipe, frequency of second overtone is
5 5
4 4 v n
n
l.
13. Answer (4)
For fourth harmonic, the string will vibrate in four
loops.
L/4
L/8Pressed
Plucked
14. Answer (3)
a
1x
x2 – 1
For block to remain stationary over wedge,
mg sin = ma cos
2
1tan
1
⎛ ⎞ ⎜ ⎟⎝ ⎠
a g g
x
21
ga
x
15. Answer (1)
16. Answer (4)
3
5A
B...(i)
2 235 2 cos 60 A B AB ...(ii)
Solving (i) and (ii), we get
A = 15, B = 25
17. Answer (1)
Work = Change in rotational kinetic energy
W = 2 2
2 1
1( )
2 I
=
2
2 2 2
2 1
1(4 )( )
2 2
⎛ ⎞ ⎜ ⎟
⎝ ⎠
MRn n
= 1500 J
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18. Answer (1)
2
2 2 23 33
2 2 2 MR
MR MR MK
2 273
2MR MK
7
6K R
19. Answer (3)
2 21 1
2 2 mv mgx Kx
20. Answer (3)
21. Answer (1)
2 kg block do not tendency to move. So there is no
force between them.
22. Answer (1)
nB – n
A = n
1...(A)
nC – n
A = n
2...(B)
From (A) and (B),
nC – n
B = n
2 – n
1
23. Answer (3)
A
Bmg
L/2
L cos
2
I 2
cos2 3
L MLmg
3cos
2
g
L
24. Answer (3)
25. Answer (1)
f iE ME ME ME
26. Answer (4)
O
45°
Hmax
u cos 45°
L = mvr
= m(u cos)Hmax
=
2 2sin 45
22u u
mg
L =
3
4 2
mu
g
27. Answer (3)
m1
m2
l1
l2
l
COM
I = m1l1
2 + m2l2
2 ...(i)
Here,2
1
1 2
m
m m
ll
1
2
1 2
m
m m
ll
Put value of l1 and l
2 in (i) and get answer.
28. Answer (4)
Temporary loss =
2
21 1(2 )
2 2 2
⎛ ⎞ ⎜ ⎟⎝ ⎠
vmv m
= 2 21 1 1
2 2 2
⎛ ⎞ ⎜ ⎟⎝ ⎠
mv mv
= 2
EE
= 2
E
29. Answer (2)
30. Answer (2)
Final sphere
Final particle
90°
30° = 1
60° = 1
1 2+ = 90°
Initial particle
31. Answer (4)
2
' 1 9
hg g h R
R
⎛ ⎞ ⇒ ⎜ ⎟⎝ ⎠
All India Aakash Test Series for Medical-2018 Test - 7 (Code-D) (Answers & Hints)
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32. Answer (1)
2
12 m
Il
I l2
2I
I
l
l
⎛ ⎞ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟
⎝ ⎠
l
l
l
l 2 2I I I l
l
I = 2I
33. Answer (2)
E T4
4
1 1
2 2
⎛ ⎞ ⎜ ⎟⎝ ⎠
E T
E T
Put T1 = T
T2 = 1.5T
1
2
1
5
E
E
E2 5E
1
E2 = 500% of E
1
34. Answer (2)
Here, H1 = H
2 = H
3
1 1 2 2(120 ) 1.5 ( ) 3 ( 0)
KA T KA T T KA T
l l l
(120 – ) = 1.5( – ) = 3T T T T1 1 2 2
120 – = 3T T1 2
1.5( – ) = 3T T1 2
T2
T T1 2 + 3 = 120 ...(i) T T
1 2 = 3 ...(ii)
Solving (i) and (ii),
T1 = 60°C
T2 = 20°C
35. Answer (4)
Given, l1 = 1.44 m
l2 = 1.00 m
When two pendulums are in out of phase, set number
of vibration completed by them are n1 and n
2
respectively. We can say if n1 = n, then n
2 = n +
1
2.
Now, by question set they are out of phase after time
t. So,
t = n1T1 = n
2T2
1 2
2 1
n T
n T
2
1
1 10
1 1.44 12
2
n
n
l
l
12n = 10n + 5
2n = 5
n = 2.5
Now, t = n1T1
= 1(2.5)(2 )g
l
= 1.44
(2.5)(2)( )10
6 s
36. Answer (1)
For simple pendulum in liquid,
11
2
⎛ ⎞⎜ ⎟⎝ ⎠
gd
nl
For spring pendulum in liquid, frequency remain
unchanged.
37. Answer (3)
230
dUF r
dr
|F | = 30r2 ...(i)
2
mvF
r
2 230 m
r v
r
mv2 = 30r
3 = 30000
2 4115000 1.5 10 J
2 mv
38. Answer (3)
3 22 2 4 8
3 3
e
e e e
e e
GM Gv R G R
R R
ve R
e
39. Answer (4)
40. Answer (3)
F T MA F T
AM
∵T ma
F ma
M
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[ CHEMISTRY]
46. Answer (4)
Br
Br2
H O2
Br
BrOH
Br
H O2
47. Answer (1)
CH –CH=CH2 2
CH–CH –CH2 2
CH –CH–CH2 3
HBr
Br–
Br
CH–CH –CH2 3
H shifting–
Benzyl carbocation
48. Answer (2)
CH = CH2
CH CH
OH
Hg /H SO2 2 4
+
CH CHO3
(X)
49. Answer (1)
Br
Cl
2Na/ether
Wurtz reaction
50. Answer (2)
Fact.
41. Answer (4)
rel boat water v v v
ˆ ˆ ˆ ˆ3 5 2 3i j i j
ˆ ˆ ˆ ˆ3 5 2 3i j i j ˆ ˆ5 8i j
42. Answer (1)
2FP F PA P VT
A ⇒
43. Answer (2)
By law of mixture,
Heat given = Heat taken
18(1)(50 – ) = (10)(0.5)(0 + 20) + 10(80)
900 – 18 = 100 + 800
18 = 0
= 0°C
Temperature of mixture will be 0°C and whole ice
will melt.
44. Answer (1)
1 2
v
700 Hz 700 Hz
0
1
⎧ ⎫ ⎨ ⎬⎩ ⎭
v v
n n
v
0
2
⎧ ⎫ ⎨ ⎬⎩ ⎭
v v
n n
v
Given 2 1
8 n n
0 0{ } 8 nv v v v
v
02
8nv
v
02 700
8350
v
v0 = 2 m/s
45. Answer (3)
Efficiency of Carnot engine and coefficient of
performance of refrigerator are related as
1
...(i)
Here, percentage = 10%
or = 0.1 ...(ii)
Putting value of in (i),
1 0.1 0.99
0.1 0.1
= 9
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51. Answer (4)
SO3 is a neutral electrophile due to presence of
vacant d-orbital.
52. Answer (4)
Percentage of N = 1.4 m.eq. of acid used
Weight of compound
53. Answer (1)
12 0.396Percentage of C 100 21.95%
44 0.492
2 0.2028Percentage of H 100 4.5%
18 0.492
Percentage of O = (100 – 21.95 – 4.5) = 73.47%
54. Answer (4)
Fact.
55. Answer (3)
H w.r.t. carbonyl group should be present for
tautomerism.
56. Answer (4)
Homolytic fission produces free radical
3 3 3CH CH 2CH
57. Answer (1)
CH
H
CH3
CH3
C
, ,
CH – C – C – CH3 3
H CH3
H CH3
(Alkanes does not exhibitgeometrical isomerism)
,
58. Answer (1)
Zeolites are 3-D silicates.
59. Answer (2)
N
H H
H
H+
60. Answer (4)
Both acidic oxide and amphoteric oxide react with
NaOH.
61. Answer (2)
Fact.
62. Answer (3)
Fact.
63. Answer (3)
NaHCO3 + NaOH Na
2CO
3 + H
2O
64. Answer (1)
Na, Al, Si, P, S, Cl are paramagnetic.
65. Answer (1)
Fact.
66. Answer (3)
Gram equivalent KMnO4
= Gram equivalent FeSO4
Gram equivalent KMnO4
= N × V (L)
= 1 × 200
1000 = 0.2
Gram equivalent FeSO4 =
Weight
Equivalent weight
Weight0.2 30.4 g
152
67. Answer (2)
2 3
2 7Cr O 6e 2Cr
68. Answer (1)
5 3
6 6 3 6 5 2C H HNO C H NO
69. Answer (1)
Fact.
70. Answer (1)
71. Answer (3)
Fact.
72. Answer (2)
H+ = 14
9.62 10
= 3.10 × 10–7
pH = –log[3.10 × 10–7]
= 7 – 0.491
= 6.509
73. Deleted
74. Answer (3)
2CO
P
CO
pK
p
21.4 0.800
1.4 p 0.80 p
FeO CO(g) Fe CO (g)
���⇀↽���
0.80 p0.625
1.4 p
0.875 – 0.625p = 0.80 + p
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0.875 – 0.80 = p + 0.625p
0.075 = 1.625p
p = 0.046
pCO
= 1.4 – 0.046 = 1.354
2CO
p 0.80 0.046 0.846
75. Answer (3)
Fact.
76. Answer (4)
rH° =
i f i fi i
a H (Product) b H (Reactant)
= 6 × (–94) + 3 × (–68.3) – 11.7
= –564 – 204.9 – 11.7
rH° = –780.6 kcal
77. Answer (1)
Fact.
78. Answer (1)
2O
W 16n
M 32
W = 2
10
1
V2.303 nRT log
V
= 10
16 502.303 8.314 300 log joule
32 5
= –2.87 × 103 J
= –2.87 kJ
79. Answer (4)
Fact.
80. Answer (4)
2
1Cl Cl (aq)
2
H = 1
2 × 240 + (–349) + (–381)
= –610 kJ mol–1
81. Answer (3)
Fact.
82. Answer (1)
BrF5 (sp
3d2)
Br
F
F
F
F
F
Due to lp–bp repulsion none of the bond angle is at
90°.
83. Answer (2)
B
O O – H
H
O
H
84. Answer (2)
2
2
ZE
n
2n
rZ
Zv
n
85. Answer (4)
2 1
4 3
(E E ) ( 3.4) ( 13.6) 10.2eV
(E E ) ( 0.85) ( 1.5) 0.66
86. Answer (3)
8
6
c 3 100.75 m
v 400 10
87. Answer (1)
3 2 2 21mol2 mol100 g
CaCO 2HCl CaCl H O CO
Number of moles of HCl taken = 1 × 1
1 mol HCl will produce 1
2 mol CO
2.
88. Answer (3)
Atomic mass × specific heat = 6.4
Atomic mass = 6.4 6.4
Specific heat 0.16 = 40
89. Answer (1)
N = M × nfactor
90. Answer (1)
N1V1 = N
2V2
2
1 1200 V
2 10
V2 = 1000
Water added = 1000 – 200 = 800 cc
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[ BIOLOGY
]
91. Answer (1)
92. Answer (3)
93. Answer (4)
94. Answer (3)
95. Answer (1)
Polymoniales
96. Answer (1)
Herbarium is quick referral system.
97. Answer (2)
98. Answer (1)
Tuber aestivum
99. Answer (4)
100. Answer (2)
Precursor to seed habit – Selaginella.
101. Answer (4)
102. Answer (4)
Secondary meristem
103. Answer (4)
104. Answer (4)
Fats are poor in oxygen.
105. Answer (3)
106. Answer (4)
Glycolysis – 1 NADH
Link – 1 NADH
TCA cycle – 3 NADH and 1 FADH2
107. Answer (2)
C4 – 5, 2
108. Answer (2)
C3
109. Answer (4)
110. Answer (1)
Legumes
111. Answer (1)
112. Answer (1)
113. Answer (1)
ABA closes stomata.
114. Answer (4)
Consists of living part.
115. Answer (4)
116. Answer (2)
Metaphase is followed by anaphase.
117. Answer (1)
Ovum (n), meiocyte (2n)
118. Answer (3)
Oxidative phosphorylation
119. Answer (3)
120. Answer (4)
Completely secondary in origin.
121. Answer (3)
All the tissue inside endodermis.
122. Answer (1)
123. Answer (1)
Analogous to tuber of sweet potato and possess
axillary buds.
124. Answer (4)
Fabaceae
125. Answer (4)
126. Answer (2)
Red algae
127. Answer (3)
Heterospory – Pteridophytes
Seed habit – Gymnosperms
Fruit habit – Angiosperms
128. Answer (2)
129. Answer (4)
130. Answer (4)
131. Answer (1)
Dinoflagellates
132. Answer (3)
Zygote
133. Answer (4)
PS-II is present.
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134. Answer (1)
Protista
135. Answer (4)
Thermoacidophiles
136. Answer (1)
137. Answer (1)
138. Answer (1)
139. Answer (4)
140. Answer (4)
141. Answer (3)
Pineal gland
142. Answer (3)
143. Answer (4)
144. Answer (1)
Hormones are required in traces.
145. Answer (3)
146. Answer (2)
147. Answer (4)
148. Answer (1)
Vitamin A itself is retinol and its aldehyde is retinal.
149. Answer (2)
150. Answer (4)
151. Answer (3)
152. Answer (4)
153. Answer (4)
154. Answer (1)
155. Answer (4)
156. Answer (1)
Alcohol suppresses ADH secretion.
157. Answer (3)
158. Answer (1)
These are in the cortex.
159. Answer (2)
Descending limb of the loop of Henle is impermeable
to salts.
160. Answer (4)
161. Answer (3)
162. Answer (1)
Cardiac output to muscles increases during
exercise.
The resistance to flow in the pulmonary system is
much lower than in systemic circulation, hence
pressure in pulmonary artery is lower than in aorta.
163. Answer (1)
164. Answer (3)
165. Answer (1)
166. Answer (3)
During inspiration, the intrapleural pressure becomes
negative.
167. Answer (3)
168. Answer (1)
169. Answer (4)
Enterokinase activates trypsinogen.
170. Answer (4)
Carbohydrates are absorbed as monosaccharides
and amino acids are absorbed actively.
171. Answer (2)
172. Answer (1)
Pancreatic acini secrete pancreatic enzymes.
173. Answer (4)
174. Answer (3)
Sperms are formed in testes and stored in seminal
vesicles.
175. Answer (4)
176. Answer (3)
177. Answer (4)
178. Answer (1)
Sea-anemones are carnivores.
179. Answer (3)
Echinoderms and hemichordates are also
deuterostomes.
180. Answer (3)
Cnidaria and ctenophora are diploblastic.
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