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Test - 2 (Code E) (Answers) All India Aakash Test Series for Medical-2019
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1. (2)
2. (3)
3. (1)
4. (1)
5. (1)
6. (2)
7. (3)
8. (2)
9. (1)
10. (4)
11. (1)
12. (3)
13. (3)
14. (1)
15. (4)
16. (1)
17. (4)
18. (1)
19. (4)
20. (3)
21. (3)
22. (1)
23. (1)
24. (2)
25. (2)
26. (1)
27. (4)
28. (2)
29. (1)
30. (2)
31. (1)
32. (2)
33. (3)
34. (2)
35. (1)
36. (3)
Test Date : 25/11/2018
ANSWERS
TEST - 2 (Code E)
All India Aakash Test Series for Medical - 2019
37. (2)
38. (1)
39. (2)
40. (3)
41. (4)
42. (1)
43. (4)
44. (1)
45. (2)
46. (3)
47. (2)
48. (4)
49. (3)
50. (3)
51. (4)
52. (4)
53. (2)
54. (4)
55. (3)
56. (4)
57. (1)
58. (1)
59. (3)
60. (4)
61. (3)
62. (2)
63. (4)
64. (2)
65. (3)
66. (2)
67. (1)
68. (4)
69. (4)
70. (1)
71. (2)
72. (1)
73. (1)
74. (4)
75. (2)
76. (3)
77. (2)
78. (4)
79. (1)
80. (3)
81. (2)
82. (4)
83. (4)
84. (1)
85. (4)
86. (1)
87. (2)
88. (3)
89. (2)
90. (4)
91. (4)
92. (4)
93. (2)
94. (3)
95. (4)
96. (4)
97. (3)
98. (2)
99. (4)
100. (2)
101. (4)
102. (1)
103. (4)
104. (4)
105. (3)
106. (2)
107. (4)
108. (1)
109. (2)
110. (3)
111. (1)
112. (2)
113. (4)
114. (4)
115. (3)
116. (3)
117. (4)
118. (3)
119. (3)
120. (1)
121. (2)
122. (2)
123. (1)
124. (4)
125. (1)
126. (4)
127. (2)
128. (3)
129. (3)
130. (4)
131. (4)
132. (4)
133. (1)
134. (4)
135. (3)
136. (3)
137. (1)
138. (3)
139. (1)
140. (4)
141. (3)
142. (3)
143. (3)
144. (4)
145. (3)
146. (1)
147. (4)
148. (2)
149. (2)
150. (2)
151. (1)
152. (4)
153. (3)
154. (1)
155. (4)
156. (4)
157. (3)
158. (4)
159. (2)
160. (3)
161. (2)
162. (3)
163. (1)
164. (3)
165. (2)
166. (4)
167. (2)
168. (2)
169. (3)
170. (2)
171. (3)
172. (4)
173. (4)
174. (2)
175. (3)
176. (2)
177. (2)
178. (3)
179. (1)
180. (3)
All India Aakash Test Series for Medical-2019 Test - 2 (Code E) (Hints and Solutions)
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PHYSICS
1. Answer (2)
Hint: Apply F = ma
Sol.:
30°
10 kg
5 kg
T1
T2
T2
5 kg
T1
aa
a
100 – T1 = 10a
T1 – T
2 – 50 sin30° = 5a
T1 – 50 sin30° = 5a
20 a = 50
a = 5
2 ms–2
T1 = 100 –
50
2
T1 =
150
2N
T2 =
55 25
2
T2 =
75 N
2
1
2
150= = 2
2
75
2
T
T
2. Answer (3)
Hint: Translational equilibrium
F = 0
Sol.:
k
20 kg
T
2
T
2
T
4
T
4
T
HINTS AND SOLUTIONS
T = 200 N
4
T= k x
1000x = 200
4
x = 1
m20
= 5 cmx
3. Answer (1)
Hint: Use concept of pseudo force
Sol.:
A B
F.B.D. of B w.r.t. cart
NAB
m aB B
NAB
= 5 × 2
NAB
= 10 N
F.B.D. of A w.r.t. cart
NAB
NWA
m aA
NWA
= mAa = N
AB = 10 × 2 + 10 = 30 N
30= = 3 :110
WA
AB
N
N
4. Answer (1)
Hint: Use concept of constraint equation.
Sol.:
9 kg
30°
10 kg
60°
T
T
a1
a2
Test - 2 (Code E) (Hints and Solutions) All India Aakash Test Series for Medical-2019
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T cos60° = 10 a2
T = 20 a2
...(i)
9g – T = 9a1
90 – T = 9a1
...(ii)
a2cos60° = a
1
a2 = 2a
1
T = 40a1
...(iii)
From equation (ii) & (iii),
49a1 = 90
or, a1 � 1.8 m/s2
5. Answer (1)
Hint: Properties of the friction force.
Sol.: fs Static friction
0 fs f
s max
0 Ns sf
flim
= s
N
6. Answer (2)
Hint: Concept of static and kinetic friction.
Sol.:
F = 25 N
QP
1maxsf
2maxsf
1 1max 1
= = 0.4 5 10s sf m g
1max
= 20Nsf
2 2max 2
= = 0.5 10 10 = 50Ns sf m g
1 2
max max appAs
s sf f F
So, system will be at rest
F.B.D of P
25 N
N
1maxs
fP
20 + N = 25
N = 5 N
F.B.D. of Q
Nfs Q
N = fs
= 5 Nsf
7. Answer (3)
Hint: Use F = ma
Sol.:
mB P A
10 N
x = 5 m x = 2 m x = 0
T T
Total mass of rope
M = 5 × 2 = 10 kg
2101 m/s
10 a
Mass of length 3 m
m = 3 × 2 = 6 kg
T = ma = 6 N
8. Answer (2)
Hint: Fnet
= 0, fk = N
Sol.:
37°
A
B
f
f
Net force on A will be zero
Wsin 37° = fk
3= cos37
5
BW m g
3 4= 3
5 5W W
1=4
All India Aakash Test Series for Medical-2019 Test - 2 (Code E) (Hints and Solutions)
4/22
9. Answer (1)
Hint: Fmin
= 2
1
mg
; when = tan–1()
Sol.: Fmin
= 2
1
mg
=
2
4 40010 10
3 3=
54
1 33
⎛ ⎞ ⎜ ⎟⎝ ⎠
Fmin
= 80 N
10. Answer (4)
Hint: For equilibrium Fnet
= 0
Sol.:
200 N
T1
T2
T4
T2
100 N
P
T1cos
T1sin
T1cos
T1sin T
2
T3
10 kg
100 N
T3
20 kg
200 N
T2
(F.B.D. ofblock 20 kg)
(F.B.D. ofblock 10 kg)
(F.B.D. ofpoint )P
P
T1cos = 100 N
T1sin = 200 N
tan = 2
–1
= tan 2
2 2 2
1= (100) (200)T
1=100 5 NT
11. Answer (1)
Hint: fmax
= s(m
1 + m
2)g
Sol.:
100 N10 kg
5 kg fk
fk
Maximum force for common acceleration
fmax
= 0.4 × 15 × 10 = 60 N
Force applied is greater than 60 N. So friction will be
kinetic.
fk =
kN = 0.4 × 50 = 20 N
2100 – 208 m/s
10a
12. Answer (3)
Hint: Concept of constraint equation.
Sol.:
P
T
aP
T/2
T/2
Q
T
Let movable pulley displaces by 'x' and block P by 'y '
l1 + l
2 = l
1 + x + l
2 + x – y
y = 2x
aP = 2a
Q
13. Answer (3)
Hint: Use dW = F dr����
Sol.: dW = ˆ ˆ ˆ ˆ( ) ( ) x y
F i F i dx i dyi
=x y
dW F dx F dy∫ ∫ ∫
W =
3 3
1 0
2xdx y dy∫ ∫
W =
32
32
0
1
9 1= – 9
2 2 2
xy
⎡ ⎤ ⎡ ⎤⎡ ⎤ ⎢ ⎥ ⎢ ⎥⎣ ⎦
⎣ ⎦⎣ ⎦= 4 + 9 = 13 J
14. Answer (1)
Hint: Wnet
= Wmg
+ Wf + W
N
Test - 2 (Code E) (Hints and Solutions) All India Aakash Test Series for Medical-2019
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Sol.:
37°
3 m
4 m
5 m10 k
g
Wnet
= Wmg
+ Wf + W
N
Wmg
= mg × 3 cos0° = 10 × 10 × 3 = 300 J
WN
= N × 5 cos 0°
WN
= 0
Wf
= N (5cos180°) = mgcos37°(–1) × 5
= –1 4
10 10 54 5 = –100 J
Wnet
= 300 – 100 net
= 200 JW
15. Answer (4)
Hint: W = 0 = 0F dr����
Sol.: ˆ ˆ ˆ= ( 3 – 4 ) N�
F i i bk
ˆ ˆ ˆ= (2 – 0) (2 – (–1) (3 – 1) �
r i j k
ˆ ˆ ˆ= (2 3 2 ) m �
r i i k
= 0F r�
�
ˆ ˆ ˆ ˆ ˆ ˆ(3 – 4 ) (2 3 2 )i j bk i j k = 0
6 – 12 + 2b = 0
= 3b
16. Answer (1)
Hint: Apply work–energy theorem
Wnet
= K
Sol.:
x m( )2
–5
10
6 8 10
F (N)
0
Wnet
= 1
10 2 10 4 – 2 52
Wnet
= 40 J
Wnet
= Kf – K
i
2 21 12 – 2 (2) = 40
2 2v
v2 – 4 = 40
v2 = 44
= 2 11m/sv
17. Answer (4)
Hint: Work = F ds����
= F(ds) cos
Sol.: W = F dscosWork done by static friction force may either be
positive, negative or zero. It depends on angle
between direction of force and displacement.
18. Answer (1)
Hint: Use ˆ ˆ= – –
U UF i j
x y
�
Sol.: U = –6x – 8y
= –6
U
x
= –8
U
y
ˆ ˆ(6 8 ) N �
F i j
2 2| | = 6 8 = 10 N�
F
2ˆ ˆ= = (3 4 ) m/s⇒ �
� �
F ma a i j
a = 10
2
2= 5 m/sa
x
y
(6, 4) m
All India Aakash Test Series for Medical-2019 Test - 2 (Code E) (Hints and Solutions)
6/22
Particle will move in the direction of acceleration
therefore, it will never cross y-axis and x-axis.
19. Answer (4)
Hint: =P F v� �
�
Sol.: F = constant
0 0
∫ ∫v x
vdv dx
2 v x
½v x
P F v �
�
½P x
20. Answer (3)
Hint: Use concept of coefficient of restitution.
Sol.: e = (Relative velocity of separation)
(Relative velocity of approach)
e = 2
1
v
v
v1
v2
e < 1
0 1e
21. Answer (3)
Hint: Conservation of energy and conservation of
linear momentum.
Sol.:
2mv1
4
m
0
2
v
4
m
v0
By conservation of linear momentum of system
0
0 1
ˆ ˆ ˆ= 24 4 2
vm mv i mv i i
0
116
vv
By conservation energy of the block.
2
1
12 = 2
2m v mgh
h =
2
0
2
1
2 (16)
v
g
2
0=512
vh
g
22. Answer (1)
Hint: Conservation of energy.
Sol.:
v0
R
h
(90° – )
B
RefA
v
2 2
0
1 1= ( )
2 2mv mg R h mv
At point B
mgcos(90° – ) – N =
2mv
R
mgsin =
2mv
R[∵ N = 0]
and h = R sin
2 mgR = 1
sin sin2
mgR mgR mgR
3 2sin = 1 sin =
2 3 ⇒
H = h + R
H = R + R sin
H = R + 2
3
R
5=
3
RH
Test - 2 (Code E) (Hints and Solutions) All India Aakash Test Series for Medical-2019
7/22
23. Answer (1)
Hint: Conservation of mechanical energy
Sol.: At maximum extension of spring, speed of
block will be zero.
x
K
m
MEi = M.E
f
0 = 21
2kx mg
1
2kx2 = mgx
2=
mgx
k
24. Answer (2)
Hint: =W F d� �
Sol.:
T
mg
=2
ga
T – mg = ma
T = 3
2
mg
W = Td cos0°
3=
2
mgdW
25. Answer (2)
Hint: Condition for stable and unstable equilibrium.
Sol.: For equilibrium,
0dU
dx
3 23
– 2 03 2
d x xx
dx
⎛ ⎞ ⎜ ⎟
⎝ ⎠
(x2 – 3x + 2) = 0
x = 1, 2
Now,
2
22 – 3
d Ux
dx
For stable equilibrium,
2
20
d U
dx
For unstable equilibrium,
2
20
d U
dx
at x = 1
2
2–1 0
d U
dx
So unstable equilibrium,
at x = 2
2
21 0
d U
dx
So stable equilibrium.
26. Answer (1)
Hint: W = F S��
Sol.:
B
A
32 N
Let both block move together
32 = 8 a
a = 4 m/s2
fmax
= mg
fmax
= 1
3 10 =15 N2
Pseudo force on block A = ma = 4 × 3 = 12 N
There will be no slipping between the block and
12 N friction force will act on block A in forward
direction.
S = ut + 21
2at
All India Aakash Test Series for Medical-2019 Test - 2 (Code E) (Hints and Solutions)
8/22
S = 21
4 (2) = 8 m2
W = fs
× S
W = 12 × 8
= 96 JW
27. Answer (4)
Hint: Properties of center of mass.
Sol.: Center of mass relative to the body will not
depend on reference of frame. And it may lie inside
or outside the body.
28. Answer (2)
Hint: I = Mk2 k Radius of gyration.
Sol.:
2
2 =6
MaI I
I =
2
12
Ma
2
12
Ma = Mk2
=2 3
ak
29. Answer (1)
Hint: A1r1 = A
2r2
Sol.:
r1
r2
C1C C
2
2
2 2,
4 2
R RA r
A1 = R2 – A
2 = R2 –
2 23
4 4
R R
2 2
1
3
4 4 2
R R Rr
1
6
Rr
30. Answer (2)
Hint: =
dl
dt
�
�
= r F �
��
Sol.:
y
x
z
(2, 0, 0)
y
0
S = 21
2ut at
21– 10(2) = – 20 m2
y
ˆ= – 20 N�
F j
ˆ ˆ= (2 – 20 ) m�
r i j
ˆ ˆ ˆ
= 2 –20 0
0 –20 0
i j k
�
ˆ ˆ ˆ= (0 – 0) – (0 – 0) (–40 – 0)i j k �
ˆ= – 40 N-m� k
31. Answer (1)
Hint: Apply conservation of angular momentum about
the hinged point and conservation of M.E.
Sol.: =i f
L L
2
2
03
⎛ ⎞⎜ ⎟⎝ ⎠
�� �
mmv = m +
03
=4
v�
Test - 2 (Code E) (Hints and Solutions) All India Aakash Test Series for Medical-2019
9/22
Conservation of M E
Ref
21 3= 2
2 2 2
mg mgI mg � �
�
22
0
2
14
92= 3
3 16
mv
mg�
��
2
0= 8v g�
v0 = 2 2g�
32. Answer (2)
Hint: Apply parallel axis theorem.
Sol.:
R
5R
I = 2 2
2(2 )5
3 2
⎡ ⎤ ⎢ ⎥⎣ ⎦
m R mRm R
I =
2 24 11
3 2
⎡ ⎤ ⎢ ⎥⎣ ⎦
mR mR
I =
2 2
28 33 41=
6 6
mR mRmR
33. Answer (3)
Hint: C.M. = 0S��
Sol.:
C.M.
C.M.
x
Initialposition
Final position
0
0
ˆ=
CPS li�
ˆ= –
PGS xi�
SCG
= ˆ= ( – )CP PG
S S l x i� �
ˆ ˆ( – ) – 2=
3
�
CM
l x im x miS
m
ml – 3xm = 0
x = 3
l
2– =
3
ll x
34. Answer (2)
Hint: net 1 2 3
= � � � �
Sol.:
45°
F
F 32 =F F
F
F F2 =
F F1 =
O
net
ˆ ˆ ˆ= (– ) (– ) ( ) 0 �
FR k FR k FR k
net
ˆ= –FRk�
35. Answer (1)
Hint: Use = andF ma = I �
���
.
Sol.:
A
B
T2
T2
T1
T1
All India Aakash Test Series for Medical-2019 Test - 2 (Code E) (Hints and Solutions)
10/22
100 – T1 = 10a ...(i)
T2 = 5a ...(ii)
T1R – T
2R = I
T1R – T
2R =
2
2mR a
R
T1 – T
2 =
2
ma...(iii)
T1 – T
2 = a
From equation (i), (ii) & (iii)
16 a = 100
a = 225
m/s4
= a
R
=
25
4
1
4
2= 25 rad/s
36. Answer (3)
Hint: C.M. C.M.
= �
� � �
L M r v I
Sol.:
x
zL
yM R,
v0
0
0
2 0 0
0
–72ˆ ˆ ˆ= (– ) (– ) =5 5
� v Mv RL Mv R k MR k k
R
37. Answer (2)
Hint: Use conservation of angular momentum and
v2 = u2 + 2as
Sol.:
v0 v
= 0
Conservation of angular momentum about the point
of contact.
mv0R (– ˆk ) =
2
ˆ(– )2
mRmvR k
⎛ ⎞ ⎜ ⎟
⎝ ⎠
2
02
mR vmv R mvR
R
02
=3
v
v
a = g
v2 = u2 + 2as
2
20
0
4 1= – 2
9 2
vv gs
2
05
=9
vs
g
38. Answer (1)
Hint: 2 21 1
=2 2
K E I mv
Sol.: KE1 = KE
2
2 2 2 2
1 1 2
1 1 1=
2 2 2mR mv mv
2 2 2
1 1 2
1 1 1=
2 2 2mv mv mv
2
1
2
1=2
v
v
⎛ ⎞⎜ ⎟⎝ ⎠
1
2
1=
2
v
v
39. Answer (2)
Sol.:
v
Velocity of center of mass decreases. To maintain
pure rolling angular velocity will also decrease. Here
direction of is clockwise.
Therefore, torque due to static friction will be
anticlockwise therefore, it will act upwards.
Test - 2 (Code E) (Hints and Solutions) All India Aakash Test Series for Medical-2019
11/22
40. Answer (3)
Hint: C.M.
= ����
� �
�
Pv v OP
Sol.:
P
v1
vC.M.
60°30°
60°
vC.M.
O
0
C.M. 1| | = | |� � �
Pv v v
2 2
C.M. 1 1 C.M.| | = 2 cos60 �
Pv v v v v
2 2
0 1 0 1
�
Pv v v v v
As v1 =
02
R
v1 =
0
2
v
2 2
2 0 0 0
0
7| | = =
4 2 2
�
P
v v v
v v
41. Answer (4)
Hint: couple
mg
Sol.: fs = F, so they constitute a couple
F3
4
a
mgf = Fs
a/2
P
N
Torque about point P.
3 2
4 2 3
a a mgF mg F ⇒
maxs sf f
mg F
2
3
mgmg
2
3
42. Answer (1)
Hint: =L r P� �
�
Sol.: =L r P� �
�
andr r
L r L P � � �
�
= 0 = 0L r L P � � �
�
43. Answer (4)
Hint: Conservation of angular momentum about Q.
Sol.: =i f
L L� �
Taking anticlockwise direction positive
2
– =2 12
mmv I � �
v
Q
P
3 22
– =3 2 2 3
m m m
l
� � � �
– =3 12 3
4 –=
3 12
3=
4
44. Answer (1)
Hint: 2
0
1=
2t t
Sol.: 2
1
1= (1)2
1=2
21= (3)2
9=
2
2 1
9= – = – = 4
2 2
2
1
= 8
All India Aakash Test Series for Medical-2019 Test - 2 (Code E) (Hints and Solutions)
12/22
45. Answer (2)
Hint: 2
2
2=
1
ghv
k
R
⎛ ⎞⎜ ⎟
⎝ ⎠
Sol.: ring
2=
2
ghv ...(i)
Hollow, Sph.
2 2= =
2 51
3 3
gh ghv
...(ii)
ring
H,sp
2
52= =
62
5
6
gh
v
v gh
CHEMISTRY
46. Answer (3)
Hint: KE T
Sol.: Average molar kinetic energy is 3RT
2
⎛ ⎞⎜ ⎟⎝ ⎠
47. Answer (2)
Hint: Density of ideal PM
gas(d) =RT
Sol.:
d = PM 1.2 atm × 32 (g/mol.)
=RT L-atm
0.0821 600KK-mol.
⎛ ⎞⎜ ⎟⎝ ⎠
= 0.779 � 0.78 g/L
48. Answer (4)
Hint:H = U + ngRT
H > U (if ng > 0)
Sol.:ng of (i) Reaction = –5
ng of (ii) Reaction = 0
ng of (iiii) Reaction = 0
ng of (iv) Reaction = +1
Hence, for reaction (iv) H > U
49. Answer (3)
Hint: G = H – TS
For non-spontaneous reaction, G > 0
Sol.: If H > 0 and S < 0
G > 0 (For all temperature)
50. Answer (3)
Hint: In an isolated system, mass as well as energy
do not exchange with surrounding.
Sol.: Universe is an example of isolated system.
51. Answer (4)
Hint: Active mass of solid is unity.
Sol.: Calcium oxide is solid so its active mass is
unity.
52. Answer (4)
Hint: Keq
=
Coefficient
Coefficient
[Product]
[Reactant]
Sol.: 2 5 2 2 eq
1N O (g) 2NO (g) O (g), K = 0.5
2�
2 2 2 54NO (g) O (g) 2N O (g), � K
eq=
2
eq
1
K
⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠
= 1
0.5 × 0.5 = 4
53. Answer (2)
Hint: Acidic buffer is a mixture of weak acid and its
salt with strong base.
Sol.: (CH3COOH + CH
3COONa) Acidic buffer
54. Answer (4)
Hint: pH = – log[H+]
Sol.:[OH–] obtained from 0.1 N NaOH = 10–1 M
pOH = 1
pH = 13
55. Answer (3)
Hint: pH = a
17 (pK – pK )
2b
Sol.:pH of 0.1 M CH3COONH
4
pH = 1
7 (pKa – pK )2
b
pH = 7 (∵ pKa = pK
b)
56. Answer (4)
Hint: A3B
2(s) + H
2O � 3A2+(aq) + 2B3–(aq)
K = [A2+]3 [B3–]2
Test - 2 (Code E) (Hints and Solutions) All India Aakash Test Series for Medical-2019
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Sol.: K = (3s)3(2s)2 = 27s3 × 4s2 = 108 s5
s =
1/5
K
108
⎛ ⎞⎜ ⎟⎝ ⎠
57. Answer (1)
Hint: Unit of Kp : g
n
(atm) .
Sol.: For reaction PCl5(g) � PCl
3(g) + Cl
2(g),
ng = 1
So, unit of Kp
is atm
58. Answer (1)
Hint: Conjugate acid is formed by addition of one H+
ion to the given species.
Sol.: Conjugate acid of 2– –
4 2 4HPO is H PO
59. Answer (3)
Hint: Interaction energy of dipole-dipole force in
stationary polar molecule 3
1
r
60. Answer (4)
Rate of Volume effused=
effusion Time taken
⎛ ⎞⎜ ⎟⎝ ⎠
1
Molar mass
Sol.:
2
gas
H
r
r
= 2 2 2
2
2
gas
H H Hgas
H gas gas gas
H
V
M t Mt= = =
V M t M
t
=
gas
1 2=
8 M
Mgas
= 64 × 2 = 128 u
61. Answer (3)
Hint: Standard boiling point of water is 99.6°C
62. Answer (2)
Hint: Greater the atomicity of gas, greater is the
entropy.
Sol.: Correct order of entropy is
He < O2 < CO
2 < PCl
3
63. Answer (4)
Hint: SReaction
= (S)Product
– (S)Reactants
(:coefficients of reactants and products)
Sol.:Sreaction
= 6 × 100 – (2 × 30 + 5 × 60)
= 600 – (60 + 300) = 600 – 360 = 240
GReaction
= HReaction
– TSReaction
For spontaneous reaction
GReaction
< 0
HReaction
< TSReaction
300 × 103 < T × 240
T >
3300 10
240
T > 1250 K
64. Answer (2)
Hint: KC = [CO
2]
Sol.: CaCO3(s) � CaO(s) + CO
2(g)
KC = [CO
2]
0.5 2
mol.= [CO ]
⎛ ⎞⎜ ⎟⎝ ⎠�
moles of CO2 = 0.5 × 5 = 2.5
65. Answer (3)
Hint: Salt having minimum value of Ksp
will give ppt
first.
Sol.: Since, ppt appearance order is
AgI AgBr AgCl
Hence, Ksp
(AgI) < Ksp
(AgBr) < Ksp
(AgCI)
So, z < y < x
66. Answer (2)
Hint: Ksp
of Al(OH)3 = [Al3+] [OH–]3
Sol.: Al(OH)3(s) � Al3+(aq) + 3OH–(aq)
S 3S
∵ pH = 9, Hence, pOH = 5 [OH–] = 10–5 M
3S = 10–5 S =
–5
10
3 M
Ksp
(Al(OH)3) = S(3S)3 = 27S4 =
4–5
1027
3
⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠
=
–20 –20
27 10 10=
27 3 3
⎛ ⎞⎜ ⎟ ⎝ ⎠
67. Answer (1)
Hint: [H+]Resulting
(Normality) = 1 1 2 2
1 2
N V +N V
V V
Sol.: [H+] = 0.1 2 V + 0.1×1× V
V + V
= 0.2 + 0.1 0.3
= = 0.152 2
All India Aakash Test Series for Medical-2019 Test - 2 (Code E) (Hints and Solutions)
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pH = – log[H+] = – log(15 × 10–2)
= 2 – log(15) = 0.824 0.82�
68. Answer (4)
Hint: Salt of weak acid and strong base will undergo
anionic hydrolysis.
Sol.: CH3COONa + H
2O CH
3COO– (aq) + Na+(aq)
CH COO3 + H
2O � CH
3COOH + OH (aq)
69. Answer (4)
Hint: Due to common ion effect, main source of Cl–
in solution is strong electrolyte NaCl.
Sol.: Ksp
(AgCl) = [Ag+] [Cl–]
1.6 × 10–10 = [Ag+] × 10–2
[Ag+] = 1.6 × 10–8
70. Answer (1)
Hint: For reaction 2A(g) � B(g) + C(g),
Equilibrium constant KC =
2
[B][C]
[A]
Sol.: 2A(g) � B(g) + C(g)
0.2 0 0
0.2 – 2x x x
KC =
2
2= 0.09
(0.2 – 2 )
x
x
= 0.3(0.2 – 2 )
x
x
x = 0.06 – 0.6x
1.6x = 0.06
x = 0.06
1.6
Concentration of B is 0.06
1.6 10
⎛ ⎞⎜ ⎟⎝ ⎠
= 3.75 × 10–3 M.
71. Answer (2)
Hint: For reaction
A(g) � B(g), KC =
[B] 0.02= = 2
[A] 0.01
Sol.:
A(g) � B(g)
At new eq. 0.02 – x 0.02 + x
KC =
0.02 + x= 2
0.02 – x
0.02 + x = 0.02 × 2 – 2x
3x = 0.02
x = 0.02
= 0.006673
[B] = 0.02 + 0.0067 = 0.0267
72. Answer (1)
Hint: Pure water is neutral and in pure water,
[H+] = [OH–] = w
K
Sol.: For acidic solution, [H+] > w
K
For basic solution, [OH–] > w
K
73. Answer (1)
Hint: Kp = K
c(RT)ng
Sol.: If ng = 0
Kp = K
c
For reaction N2(g) + O
2(g) � 2NO(g), n
g = 0
74. Answer (4)
Hint: pH of salt of weak base and strong acid is
b
17 – (pK log C)
2
⎧ ⎫⎨ ⎬⎩ ⎭
.
Sol.:
4 4 2NH OH HCl NH Cl H O
Initial mol, 0.1 V 0.1 V
Final mol, 0 0 0.1 V
⎧ ⎫⎨ ⎬⎩ ⎭
Concentration of NH4Cl =
–2
0.1= 5 ×10
2
pH = –2
17 – 4.75 log(5 ×10 )
2
= 17 – 4.75 – 2 log5
2
= 5.2755 � 5.28
75. Answer (2)
Hint: Partial pressure of a gas
= Mole fraction × Total pressure
Sol.: Moles of 2
xO
32
(Assume, X is mass of each gas mixed together)
Test - 2 (Code E) (Hints and Solutions) All India Aakash Test Series for Medical-2019
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Moles of x
He4
Total moles x x 9
x32 4 32
pHe
x
84P P
9 9x
32
Fraction of total pressure exerted by He = 8
9.
76. Answer (3)
Hint:
P
L G(L + G)
V
Sol.: At temp. TC and pressure P
2, gas phase
exists.
77. Answer (2)
Hint: Number of mole = Volume at STP(ml)
22400
Sol.: For A vapours, mole = 112 1
mol22400 200
Molar mass of A = 0.48 × 200 = 96 g
2
A(g)
H
r 2 1 1= = =
r 96 48 4 3
78. Answer (4)
Hint: For H2 gas, value of a is negligible.
Sol.: m2
m
aP + (V – b) = RT [ a is negligible]
V
⎛ ⎞⎜ ⎟⎝ ⎠
∵
P(Vm
– b) = RT
PbZ =1+
RT
79. Answer (1)
Hint: Partial pressure = (Total pressure)
× Mole fraction of gas
Sol.: 2
O
20n = = 0.625
32
CO
2.8n = = 0.1
28
Mole fraction oxygen gas = 0.625
0.1 0.625
= 0.625 625
=0.725 725
= 25
29
⎛ ⎞⎜ ⎟⎝ ⎠
pO2
= 25
29
⎛ ⎞⎜ ⎟⎝ ⎠
× 2 = 50
atm29
Partial volume of oxygen gas = 25
1029
250
29 L
80. Answer (3)
Hint: Fusion
2 2
Fusion
n HH O(s) H O(l), S =
T
⎛ ⎞⎜ ⎟
⎝ ⎠�
Sol.: S =
3
–1 –16 10= 21.98 Jmol K
273
At equilibrium, G = 0
81. Answer (2)
Hint: H2O(l) � H
2O(v) (1 atm, 100°C)
U = q + W
Sol.: q = nHVapourization
q = 1 × 40.66 × 103 J = 40.66 × 103 J
w = – ngRT
= –1 × 8.314 × 373 = –3101.12 J
U = q + w = (40.66 × 103 – 3101.12) J
= 37558.878 J = 37.56 × 103 J
82. Answer (4)
Hint: U = q + w and w = –2.303 nRT log2
1
V
V
⎛ ⎞⎜ ⎟⎝ ⎠
Sol.: w = –2.303 × 2 × 8.314 × 300 log20
5
⎛ ⎞⎜ ⎟⎝ ⎠
= –2.303 × 2 × 8.314 × 300 × log4
= –2.303 × 2 × 8.314 × 300 × 2 × 0.30
= – 6892.9 J � –6.9 kJ
U = q + w
q = –w (for isothermal process)
q = +6.9 kJ
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83. Answer (4)
Hint: S = 2 1
p
1 2
T PnC ln + nRln
T P
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
Sol.: S = 1
2
PnR ln
P
⎛ ⎞⎜ ⎟⎝ ⎠
(At constant temperature)
S = 1
2
PnR ln
P
⎛ ⎞⎜ ⎟⎝ ⎠
= 1 × R × 2
ln0.5
⎛ ⎞⎜ ⎟⎝ ⎠
= R ln 4 = 2R ln2
84. Answer (1)
Hint: Overall enthalpy of a process, is the sum of
the enthalpies of all steps involved.
Sol.:
(i) C(s) + O2(g) CO
2(g) + 94 kcal/mol
(ii) H2(g) +
1
2O
2(g) H
2O(l) + 68 kcal/mol
(iii) CH4(g) + 2O
2(g) CO
2(g) + 2H
2O(l) + 213 Kcal/mol.
On applying, (i) + 2 × (ii) – (iii)
Reaction obtained is
C(s) + 2H2(g) CH
4(g)
H = (–94) + 2(–68) –(–213) = –17 kcal.
85. Answer (4)
Hint: Vapour pressure of a liquid depends on
temperature.
86. Answer (1)
Hint:
Enthalpy of
= 3 ×
Enthalpy of hydrogenation –
Resonance hydrogenation of benzene of cyclohexane
energy
Sol.:
+ H2 ; H = –x kJ mol–1
+ 3H2
Enthalpy of hydrogenation of benzene
= 3H – Resonance energy.
= 3(–x) – (–y)
= (y – 3x) kJ mol–1
87. Answer (2)
Hint: pH of acidic buffer pH = pKa + log
Salt
Acid
⎡ ⎤⎢ ⎥⎣ ⎦
Sol.:
CH3COOH + NaOH CH
3COONa + H
2O
(mmol)initial
10 6
(mmol)final
4 0 6
Resulting solution is an acidic buffer
pH = 4.74 + 6
log4
⎛ ⎞⎜ ⎟⎝ ⎠
= 4.74 + 0.18 = 4.92
88. Answer (3)
Hint: Salt having minimum value of 'solubility (mole/l)'
will be least soluble.
Sol.: For binary electrolyte, lower the value of Ksp
,
lower will be its solubility.
89. Answer (2)
Hint: Lewis acids are electron pair acceptors.
Sol.: B2H
6 is an electron deficient species.
90. Answer (4)
Hint: G = G0 + 2.303RT log Qp
Sol.: QP =
2 5
2 2
2 2N O
2 5 2
N O
P 2= =1
P P 2 1
G = G0 + 2.303 RT log10
(1) = G0 = –43 kJ/mol
BIOLOGY
91. Answer (4)
Hint: The body of fungi is not differentiated into root,
stem & leaves.
Sol.: The body of fungi is thalloid & haploid (n).
Each cell contains one set of chromosomes.
92. Answer (4)
Hint: On basidium exogenously produced spores are
basidiospores.
Sol.: Basidiospores are produced by members of
basidiomycetes. These are haploid and sexual
spores.
93. Answer (2)
Hint: Claviceps purpurea causes ergot of rye.
Sol.: Claviceps belongs to the class Ascomycetes.
It is a sac fungus which possesses septate
mycelium and chitinous cell wall.
It reproduces by conidia which are non-motile
asexual spores.
Test - 2 (Code E) (Hints and Solutions) All India Aakash Test Series for Medical-2019
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94. Answer (3)
Hint: Deuteromycetes are called imperfect fungi.
Sol.:Morels
AscomycetesAspergillus
⎤⎥⎦
Bracket fungi ] Basidiomycetes
Deuteromycetes
Alternaria
Colletotrichum
Trichoderma
⎤⎥⎥⎥⎦
95. Answer (4)
Hint: Rhizopus is a conjugation fungus, also known
as bread mould.
Sol.: It reproduces by non-motile asexual spores and
non-motile zygospores.
It is a saprophytic fungus with chitinous cell wall.
96. Answer (4)
Sol.: Bladderwort is a partially heterotrophic
insectivorous plant.
97. Answer (3)
Hint: Phycobiont is algal partner in lichen
association. They are either green algae or blue-
green algae.
Sol.: Algal partner synthesizes food for fungal
partner. Lichens do not grow in air polluted areas.
Animals store their food in form of glycogen or fat.
98. Answer (2)
Hint: A virus is larger than sub viral agents/particles.
Sol.: Viroids are infectious RNA particles. They lack
protein. They cause diseases in plants only.
Viruses have proteinaceous capsid.
99. Answer (4)
Hint: Prions lack genetic material.
Sol.: Prions are proteinaceous infectious particles. It
causes mad cow disease in cattles & scrapie
disease in sheep.
100. Answer (2)
Hint: Viruses are obligate intracellular parasites.
Sol.: Viruses can multiply within the host cell only.
They cannot respire, divide and grow externally.
101. Answer (4)
Hint: All viruses have capsid.
Sol.: Bacterial viruses are known as bacteriophages,
they have capsid (protein coat) & genetic material,
usually dsDNA.
102. Answer (1)
Hint: Mycorrhiza is a mutually beneficial association.
Sol.: Mycorrhiza is a symbiotic association between
roots of higher plants & fungi.
103. Answer (4)
Hint: Dinoflagellates are flagellated, mostly marine
photosynthetic protist.
Sol.: Dinoflagellates have two flagella, one is
transverse and another is longitudinal
104. Answer (4)
Hint: Paramoecium has two nuclei.
Sol.: Paramoecium is a ciliated protozoan
105. Answer (3)
Hint: Cell wall of diatoms is indestructible due to
presence of silica.
Sol.: Due to presence of silica, the fossilized
diatoms get deposited at the sea bed to form
diatomaceous earth.
Diatoms are microscopic, unicellular protists which
float passively in water current.
106. Answer (2)
Hint: Euglenoids lack cell wall.
Sol.: Slime moulds are heterotrophic (Saprophytic).
Protozoans are predators & parasites. Chrysophytes
include diatoms and desmids which are mostly
photosynthetic.
107. Answer (4)
Hint: Protists are unicellular eukaryotes.
Sol.: Eukaryotes have well defined nucleus and
other membrane bound cell organelles.
108. Answer (1)
Hint: Heterocyst is related to N2 fixation which takes
place in anaerobic condition.
Sol.: In heterocyst cell wall is impermeable to
oxygen but permeable to nitrogen. It lacks PS-II but
PS-I remains active and produces ATP required for
N2-fixation.
109. Answer (2)
Hint: Archaebacteria contain pseudomurein in their
cell wall.
Sol.: The cell wall of eubacteria is made up of
mainly peptidoglycan. They are prokaryotes hence
they lack membrane bounded cell organelles.
Archaebacteria include halophiles, methanogens and
thermoacidophiles.
Archaebacteria have introns in their genetic material.
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110. Answer (3)
Hint: Mycoplasma are smallest living organisms.
Sol.: They lack cell wall and can survive without
oxygen.
Methanogens are responsible for production of
biogas.
111. Answer (1)
Hint: Chemoautotrophic bacteria lack photosynthetic
pigments.
Sol.: ChemoautotrophicNitrosomonas
Nitrobacter
⎤⎥⎦
PhotosyntheticNostoc
Chlorobium
⎤⎥⎦
Lactobacillus] Heterotrophic.
112. Answer (2)
Hint: Prokaryotes lack nuclear membrane &
membrane bound cell organelles.
Sol.: Streptococcus is a prokaryote. Remaining
organisms are eukaryotes.
113. Answer (4)
Hint:
Cyanobacteria evolve O2 during photosynthesis.
Sol.: Cyanobacteria have enzyme nitrogenase which
fixes atmospheric nitrogen. Some cyanobacteria
form symbiotic associations with ferns. Such as
Anabaena with Azolla.
Nostoc is a filamentous BGA.
114. Answer (4)
Sol.: Bordeaux mixture is first fungicide discovered
by RMA Millardet. It is commonly known as holy
water of plant pathology.
115. Answer (3)
Hint: Tap root is the primary root.
Sol.: Tap roots (Primary root) arise from radicle
whereas fibrous roots arise from the base of the
stem.
116. Answer (3)
Hint: Carrot & Turnip are modifications of tap root.
Sol.: Ginger is a modified underground stem. Edible
part of sweet potato is a modified adventitious root.
117. Answer (4)
Hint: Pneumatophores help a plant to get oxygen
from the atmosphere.
Sol.: They grow vertically upward hence negatively
geotropic & can be seen in Rhizophora which grow
in marshy/swampy areas.
118. Answer (3)
Hint: Stem bears nodes, internodes and buds.
Sol.: Potato is a tuber which is a modified
underground stem because it bears buds.
119. Answer (3)
Hint: Phylloclade is a modified stem.
Sol.: In Opuntia, phylloclade is flattened and
photosynthetic.
120. Answer (1)
Hint: Albuminous seed contains endosperm.
Sol.: Coconut has endospermic seed where seed
coat is thick & non-membranous.
Groundnut, pea, and soyabean have ex-albuminous
(non-endospermic) seeds.
121. Answer (2)
Hint: Axile placentation has placenta in the axial
position and ovules are attached to it.
Sol.: Tomato & lemon have axile placentation. Pea
has marginal, mustard has parietal & sunflower and
marigold have basal placentation.
122. Answer (2)
Hint: In some seeds nucellus persists in the form of
a layer.
Sol.: Perisperm is persistent nucellus. Seeds having
perisperm are called perispermic seeds.
123. Answer (1)
Hint: Given floral formula belongs to family
Solanaceae.
Sol.: Solanaceae members have persistent calyx,
superior ovary with many ovules. Seeds are
endospermic. Flowers are bisexual and
actinomorphic.
124. Answer (4)
Hint: In monadelphous condition, all stamens are
united in a single bundle but anthers remain free. It
is found in family Malvaceae.
Sol.: In pea flower the stamens are united in two
bundles so it is diadelphous.
125. Answer (1)
Hint: In opposite phyllotaxy, a pair of leaves arise at
each node opposite to each other.
Sol.: Guava, Calotropis have opposite phyllotaxy.
Test - 2 (Code E) (Hints and Solutions) All India Aakash Test Series for Medical-2019
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Mustard, China roseAlternate phyllotaxy
Sunflower
⎤⎥⎦
Alstonia, Nerium ] Whorled phyllotaxy.
126. Answer (4)
Hint: Banana is a triploid plant hence cannot form
gametes.
Sol.: Banana fruit develops without fertilisation of
gametes, hence it is a parthenocarpic fruit. Banana
has suckers.
127. Answer (2)
Hint: In racemose inflorescence the terminal flower
is absent.
Sol.: It (main axis) has unlimited growth and flowers
ever borne in acropetal manner.
128. Answer (3)
Hint: Fused carpels— syncarpous
Sol.: Staminode is sterile anther. Twisted aestivation
is seen in petals of cotton. Unisexual flowers are
seen in maize.
129. Answer (3)
Hint: Papilionaceous corolla shows vexillary
aestivation.
Sol.: Mango is a drupe. Vexillum is largest posterior
petal in pea also called standard petal.
130. Answer (4)
Hint: In members of family Poaceae, the seed coat
is fused with fruit wall (Caryopsis type of fruit).
Sol.: Gram is a dicot. Wheat and maize are
monocots and belong to family Poaceae.
131. Answer (4)
Hint: This plant belongs to the family Fabaceae.
Sol.: It is Indigofera.
132. Answer (4)
Hint: Thorns can be seen in Citrus or Bougainvillea.
Sol.: Spines are modification of leaves.
133. Answer (1)
Sol.: Root hairs arise from epidermal cells of
maturation zone.
134. Answer (4)
Hint: Aleurone layer is a part of endosperm.
Sol.: It is triploid, provides nourishment and seen in
maize grains.
135. Answer (3)
Sol.: Strawberry is an aggregate fruit.
Mulberry & jackfruits are composite fruits or multiple
fruits.
Pomegranate is a simple fleshy or succulent fruit.
136. Answer (3)
Hint: The primary organ involved in this process is
kidney.
Sol.: Removal of metabolic waste products from the
body is known as excretion. Main role of sweating
is to facilitate thermoregulation. Defaecation involves
removal of undigested and unabsorbed food.
137. Answer (1)
Hint: Part of nephron exhibiting brush border
appearance.
Sol.: ADH facilitates facultative/conditional
reabsorption of water from DCT & collecting tubule.
PCT is responsible for obligate reabsorption and is
internally lined by simple cuboidal epithelium
containing microvilli giving brush border appearance
which increase surface area for reabsorption.
138. Answer (3)
Hint: These cells are granular, phagocytic and most
abundant type of WBCs.
Sol.: Differential leukocyte count i.e., DLC reveals
percentage of type of WBCs in blood.
Neutrophils – 60–65%
Lymphocytes – 20–25%
Monocytes – 6–8%
Eosinophils – 2–3%
139. Answer (1)
Hint: Excretory organ in annelids.
Sol.: Respiration in earthworm (annelids) occurs
through moist cuticle.
140. Answer (4)
Hint: Squamous epithelium is found at this surface.
Sol.: Gaseous exchange takes place in the alveoli
in lungs.
141. Answer (3)
Hint: Fluid filled cavity is present around lungs.
Sol.: Pleura is divided into two layers. Outer pleural
membrane is in close contact with thoracic lining
whereas inner pleural membrane is in contact with
lung surface.
All India Aakash Test Series for Medical-2019 Test - 2 (Code E) (Hints and Solutions)
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142. Answer (3)
Hint: TV + IRV + ERV = TLC – RV
Sol.: TLC – RV = VC,
EC = ERV + TV = 1100 + 500 = 1600 ml
143. Answer (3)
Hint: Lungs are situated in cavity above diaphragm.
Sol.: Trachea divides into two primary bronchi at the
level of 5th thoracic vertebra and enters lungs.
144. Answer (4)
Hint: The nitrogenous waste product which requires
minimum loss of water in its elimination.
Sol.: Land snails excrete uric acid whereas most
terrestrial organisms excrete either urea (humans,
amphibians) or uric acid (insects, birds, reptiles).
Aminotelic organisms excrete amino acid (Unio and
echinoderms)
145. Answer (3)
Hint: These animals live on both land and water.
Sol.: Crocodiles are reptiles with completely divided
four chambered heart. Amphibians have three
chambered heart.
146. Answer (1)
Hint: Dialysis helps to counter uremia.
Sol.: Dialysing fluid is isotonic to blood plasma.
147. Answer (4)
Hint: This membrane divides body cavity into
thoracic and abdominal cavity.
Sol.: Pneumotaxic centre is called "switch-off" point
of inspiration. It stimulates neurons of expiratory
centre located in medulla. Diaphragm is stimulated
by respiratory rhythm centre in medulla.
148. Answer (2)
Hint: Enzyme which is released in body injured site
in presence of Ca+2.
Sol.: Prothrombinase/thrombokinase is also
responsible for conversion of inactive plasma protein
prothrombin into an active enzyme thrombin
responsible for conversion of fibrinogen into fibrin.
149. Answer (2)
Hint: This hormone secreted by JG cells stimulates
RAAS.
Sol.: Renin secreted by JG cells in response to low
B.P. stimulates RAAS to maintain GFR. Angiotensin-
II is a vasoconstrictor. Rennin digests casein in
infant stomach.
150. Answer (2)
Hint: These cells give rise to enucleated platelets.
Sol.: Megakaryocytes are specialised large cells in
red bone marrow which divide to form cellular
fragments lacking nucleus known as blood platelets/
thrombocytes.
151. Answer (1)
Hint: WBCs which participate in allergic reaction
during worm infestation.
Sol.: Eosinophils are WBCs that stain with acidic
dye eosin and have bilobed nucleus. They participate
in a number of allergic reactions and fight against
worm infestation.
152. Answer (4)
Hint: Identify the blood group of universal donors.
Sol.: Universal donors (O–) and individuals with
blood group B– and O+ can donate blood to the
patient.
153. Answer (3)
Hint: This heart sound is produced during ventricular
systole.
Sol.: First heart sound 'lub' is produced during
ventricular systole. It has low frequency of about
25–45 Hz and is of comparatively longer duration
(0.15 sec) in comparison to second heart sound
which has higher frequency than 50 Hz but duration
is only 0.12 sec.
154. Answer (1)
Hint: Lymphatic fluid is poured into venous blood.
Sol.: Lymph from right thoracic duct drains into right
subclavian vein whereas left thoracic duct drains
lymph into left subclavian vein
155. Answer (4)
Hint: Identify a layer absent in alveolar capillaries.
Sol.: Capillaries lack tunica media formed by
smooth muscle fibres. So, diffusion membrane does
not contain muscular tissue.
156. Answer (4)
Hint: Affinity of Hb for O2 is inversely proportional to
P50
value.
Sol.: Shift of curve B towards right indicates
increase in P50
value and thus decreased affinity of
haemoglobin for oxygen, leading to greater
dissociation of oxygen from hemoglobin.
Test - 2 (Code E) (Hints and Solutions) All India Aakash Test Series for Medical-2019
21/22
157. Answer (3)
Hint: Maximum percentage of CO2 is transported in
plasma in bicarbonate form.
Sol.: 77% CO2 is transported through blood plasma.
Dissolvedin plasma – 7%
In the form ofbicarbontates
in plasma – 70%
Plasma(77%)
158. Answer (4)
Hint: ADH acts on this region to prevent diuresis.
Sol.: The collecting duct is responsible for secretion
of urea not for absorption. Various nephrons open
into collecting ducts. Collecting duct secretes urea
into interstitium to increase osmolarity for
reabsorption of water.
159. Answer (2)
Hint: Stretch receptors are activated upon filling of
urinary bladder beyond threshold.
Sol.: Hemodialysis is needed when kidneys
malfunction. Urine formation is independent of stretch
reflex.
160. Answer (3)
Hint: Volume of ventricle decreases during their
contraction.
Sol.: During ventricular systole, closure of cuspid/AV
valves occurs. Blood enters ventricles during atrial
systole.
161. Answer (2)
Hint: Volume of blood pumped by each ventricle in
0.8 seconds.
Sol.: Cardiac output = heart rate x stroke volume.
Hypertension is observed if blood pressure exceeds
140/90 or more repeatedly. Opening of pulmonary
artery is guarded by semilunar valves.
162. Answer (3)
Hint: During this phase, blood is not pumped from
ventricles to aorta.
Sol.: During isovolumetric contraction, both
auriculoventricular valves and semilunar valves remain
closed. Pressure within ventricles increases but
amount of blood remains unchanged in isovolumetric
contraction.
163. Answer (1)
Hint: 2/3rd filling of ventricles occurs when all
chambers of heart are relaxing.
Sol.: When auricles are in diastole auriculoventricular
valves remain open but semilunar valves present at
opening of aorta are closed.
164. Answer (3)
Hint: Duration of each auricular cycle is equal to
duration of the cardiac cycle.
Sol.: During each cardiac cycle, an auricular cycle
as well as a ventricular cycle is completed. Each
cycle has duration of 0.8 sec. So time of auricular
cycle is 0.8 sec not 0.6 second.
165. Answer (2)
Hint: Identify the machine use to measure an ECG.
Sol.: Electrical activities of heart is measured by an
instrument known as electrocardiograph and graph
obtained is known as electrocardiogram.
Stethoscope is used to hear heart sounds.
166. Answer (4)
Hint: Well developed region of JG nephrons
extending into medulla in JG nephrons.
Sol.: Loop of Henle of both cortical and juxtamedullary
nephron lie in medullary pyramid.
167. Answer (2)
Hint: This hormone is released by atrial walls.
Sol.: Vasa recta acts as countercurrent exchanger
because they exchange water for ions. ANF
opposes RAAS.
168. Answer (2)
Hint: These are region between medullary pyramids
in kidney.
Sol.: Invagination of cortex into medulla forms
columns of Bertini which divide medullary region of
kidney into renal pyramidals.
169. Answer (3)
Hint: Ascending limb of loop of Henle is permeable
to salt.
Sol.: Descending limb is called concentrating
segment of loop of Henle as it is permeable to water
and impermeable to salts.
170. Answer (2)
Hint: High threshold substance like glucose is
completely absorbed in kidney in a normal healthy
person.
All India Aakash Test Series for Medical-2019 Test - 2 (Code E) (Hints and Solutions)
22/22
Sol.: Diabetes insipidus occurs due to deficiency of
ADH (vasopressin) from hypothalamus. All types of
nephrons have peritubular capillary networks. GFR in
healthy individual is 125 ml/minute or 180 L/day.
171. Answer (3)
Hint: Maximum amount of air which can be inspired
or expired forcefully.
Sol.: Vital capacity is measured for lung function
test by an instrument known as spirometer. With
exception of RV, TLC, FRC. All other lung capacities
and volumes can be measured by spirometer.
172. Answer (4)
Hint: Innermost lining of wall of blood vessels is
constituted by squamous cells.
Sol.: Arteries have narrower lumen than veins and
mostly carry oxygenated blood.
173. Answer (4)
Hint: Sympathetic stimulation alters heart rate.
Sol.: Sympathetic neural signals increase cardiac
output and stroke volume but decrease duration of
cardiac cycle. Cardiac muscles are not under the
control of our will.
174. Answer (2)
Hint: Pressure difference on both sides of filtration
membrane.
Sol.: NFP = (GHP) – (BCOP + CHP)
= 60 – (30 + 20)
= 10 mmHg
175. Answer (3)
Hint: Blood vessel exiting from major osmoregulatory
organ of human body.
Sol.: Largest amount of urea is present in hepatic
vein.
176. Answer (2)
Hint: This hormone is released in response to
increase in blood pressure than normal.
Sol.: During dehydration (profuse sweating)
osmolarity of body fluid increases, increased
secretion of ADH from hypothalamus occurs. GFR
decreases during profuse sweating.
177. Answer (2)
Hint: Ischemic conditions reflect oxygen deficiency
in cardiac muscle fibres.
Sol.: When heart muscle is suddenly damaged by
inadequate blood supply, the condition is called heart
attack.
178. Answer (3)
Hint: This wave indicates ventricular repolarisation.
Sol.: T-wave represents the return of ventricles from
excited to normal state. QRS complex represents
depolarisation of ventricles while P-wave represents
atrial depolarisation.
179. Answer (1)
Hint: pCO2 in alveoli is 40 mmHg.
Sol.:
Atmospheric air
Alveoli Deoxy-generated
blood
Oxy-genated
blood
Tissue
pO2 159 104 40 95 40
pCO2 0.3 40 45 40 45
180. Answer (3)
Hint: Disorder associated with accumulation of dust
particles.
Sol.: Pneumoconiosis is occupational disease of
lungs due to inhalation of dust, characterised by
inflammation, coughing and fibrosis.
� � �
Test - 2 (Code F) (Answers) All India Aakash Test Series for Medical-2019
1/22
Test Date : 25/11/2018
ANSWERS
TEST - 2 (Code F)
All India Aakash Test Series for Medical - 2019
1. (2)
2. (1)
3. (4)
4. (1)
5. (4)
6. (3)
7. (2)
8. (1)
9. (2)
10. (3)
11. (1)
12. (2)
13. (3)
14. (2)
15. (1)
16. (2)
17. (1)
18. (2)
19. (4)
20. (1)
21. (2)
22. (2)
23. (1)
24. (1)
25. (3)
26. (3)
27. (4)
28. (1)
29. (4)
30. (1)
31. (4)
32. (1)
33. (3)
34. (3)
35. (1)
36. (4)
37. (1)
38. (2)
39. (3)
40. (2)
41. (1)
42. (1)
43. (1)
44. (3)
45. (2)
46. (4)
47. (2)
48. (3)
49. (2)
50. (1)
51. (4)
52. (1)
53. (4)
54. (4)
55. (2)
56. (3)
57. (1)
58. (4)
59. (2)
60. (3)
61. (2)
62. (4)
63. (1)
64. (1)
65. (2)
66. (1)
67. (4)
68. (4)
69. (1)
70. (2)
71. (3)
72. (2)
73. (4)
74. (2)
75. (3)
76. (4)
77. (3)
78. (1)
79. (1)
80. (4)
81. (3)
82. (4)
83. (2)
84. (4)
85. (4)
86. (3)
87. (3)
88. (4)
89. (2)
90. (3)
91. (3)
92. (4)
93. (1)
94 (4)
95. (4)
96. (4)
97. (3)
98. (3)
99. (2)
100. (4)
101. (1)
102. (4)
103. (1)
104. (2)
105. (2)
106. (1)
107. (3)
108. (3)
109. (4)
110. (3)
111. (3)
112. (4)
113. (4)
114. (2)
115. (1)
116. (3)
117. (2)
118. (1)
119. (4)
120. (2)
121. (3)
122. (4)
123. (4)
124. (1)
125. (4)
126. (2)
127. (4)
128. (2)
129. (3)
130. (4)
131. (4)
132. (3)
133. (2)
134. (4)
135. (4)
136. (3)
137. (1)
138. (3)
139. (2)
140. (2)
141. (3)
142. (2)
143. (4)
144. (4)
145. (3)
146. (2)
147. (3)
148. (2)
149. (2)
150. (4)
151. (2)
152. (3)
153. (1)
154. (3)
155. (2)
156. (3)
157. (2)
158. (4)
159. (3)
160. (4)
161. (4)
162. (1)
163. (3)
164. (4)
165. (1)
166. (2)
167. (2)
168. (2)
169. (4)
170. (1)
171. (3)
172. (4)
173. (3)
174. (3)
175. (3)
176. (4)
177. (1)
178. (3)
179. (1)
180. (3)
All India Aakash Test Series for Medical-2019 Test - 2 (Code F) (Hints and Solutions)
2/22
PHYSICS
1. Answer (2)
Hint: 2
2
2=
1
ghv
k
R
⎛ ⎞⎜ ⎟
⎝ ⎠
Sol.: ring
2=
2
ghv ...(i)
Hollow, Sph.
2 2= =
2 51
3 3
gh ghv
...(ii)
ring
H,sp
2
52= =
62
5
6
gh
v
v gh
2. Answer (1)
Hint: 2
0
1=
2t t
Sol.: 2
1
1= (1)2
1=2
21= (3)2
9=
2
2 1
9= – = – = 4
2 2
2
1
= 8
3. Answer (4)
Hint: Conservation of angular momentum about Q.
Sol.: =i f
L L� �
HINTS AND SOLUTIONS
v
Q
P
Taking anticlockwise direction positive
2
– =2 12
mmv I � �
3 22
– =3 2 2 3
m m m
l
� � � �
– =3 12 3
4 –=
3 12
3=
4
4. Answer (1)
Hint: =L r P� �
�
Sol.: =L r P� �
�
andr r
L r L P � � �
�
= 0 = 0L r L P � � �
�
5. Answer (4)
Hint: couple
mg
Sol.: fs = F, so they constitute a couple
F3
4
a
mgf = Fs
a/2
P
N
Torque about point P.
3 2
4 2 3
a a mgF mg F ⇒
Test - 2 (Code F) (Hints and Solutions) All India Aakash Test Series for Medical-2019
3/22
maxs sf f
mg F
2
3
mgmg
2
3
6. Answer (3)
Hint: C.M.
= ����
� � �
Pv v OP
Sol.:
P
v1
vC.M.
60°30°
60°
vC.M.
O
0
C.M. 1| | = | |� � �
Pv v v
2 2
C.M. 1 1 C.M.| | = 2 cos60 �
Pv v v v v
2 2
0 1 0 1
�
Pv v v v v
As v1 =
02
R
v1 =
0
2
v
2 2
2 0 0 0
0
7| | = =
4 2 2
�
P
v v v
v v
7. Answer (2)
Sol.:
v
Velocity of center of mass decreases. To maintain
pure rolling angular velocity will also decrease. Here
direction of is clockwise.
Therefore, torque due to static friction will be
anticlockwise therefore, it will act upwards.
8. Answer (1)
Hint: 2 21 1
=2 2
K E I mv
Sol.: KE1 = KE
2
2 2 2 2
1 1 2
1 1 1=
2 2 2mR mv mv
2 2 2
1 1 2
1 1 1=
2 2 2mv mv mv
2
1
2
1=2
v
v
⎛ ⎞⎜ ⎟⎝ ⎠
1
2
1=
2
v
v
9. Answer (2)
Hint: Use conservation of angular momentum and
v2 = u2 + 2as
Sol.:
v0 v
= 0
Conservation of angular momentum about the point
of contact.
mv0R (– ˆk ) =
2
ˆ(– )2
mRmvR k
⎛ ⎞ ⎜ ⎟
⎝ ⎠
2
02
mR vmv R mvR
R
02
=3
v
v
a = g
v2 = u2 + 2as
2
20
0
4 1= – 2
9 2
vv gs
2
05
=9
vs
g
10. Answer (3)
Hint: C.M. C.M.
= �
� � �
L M r v I
All India Aakash Test Series for Medical-2019 Test - 2 (Code F) (Hints and Solutions)
4/22
Sol.:
x
zL
yM R,
v0
0
0
2 0 0
0
–72ˆ ˆ ˆ= (– ) (– ) =5 5
� v Mv RL Mv R k MR k k
R
11. Answer (1)
Hint: Use = andF ma = I �
���
.
Sol.:
A
B
T2
T2
T1
T1
100 – T1 = 10a ...(i)
T2 = 5a ...(ii)
T1R – T
2R = I
T1R – T
2R =
2
2mR a
R
T1 – T
2 =
2
ma...(iii)
T1 – T
2 = a
From equation (i), (ii) & (iii)
16 a = 100
a = 225
m/s4
= a
R
=
25
4
1
4
2= 25 rad/s
12. Answer (2)
Hint: net 1 2 3
= � � � �
Sol.:
45°
F
F 32 =F F
F
F F2 =
F F1 =
O
net
ˆ ˆ ˆ= (– ) (– ) ( ) 0 �
FR k FR k FR k
net
ˆ= –FRk�
13. Answer (3)
Hint: C.M. = 0S��
Sol.:
C.M.
C.M.
x
Initialposition
Final position
0
0
ˆ=
CPS li�
ˆ= –
PGS xi�
SCG
= ˆ= ( – )CP PG
S S l x i� �
ˆ ˆ( – ) – 2=
3
�
CM
l x im x miS
m
ml – 3xm = 0
x = 3
l
2– =
3
ll x
Test - 2 (Code F) (Hints and Solutions) All India Aakash Test Series for Medical-2019
5/22
14. Answer (2)
Hint: Apply parallel axis theorem.
Sol.:
R
5R
I = 2 2
2(2 )5
3 2
⎡ ⎤ ⎢ ⎥⎣ ⎦
m R mRm R
I =
2 24 11
3 2
⎡ ⎤ ⎢ ⎥⎣ ⎦
mR mR
I =
2 2
28 33 41=
6 6
mR mRmR
15. Answer (1)
Hint: Apply conservation of angular momentum about
the hinged point and conservation of M.E.
Sol.: =i f
L L
2
2
03
⎛ ⎞⎜ ⎟⎝ ⎠
�� �
mmv = m +
03
=4
v�
Conservation of M E
Ref
21 3= 2
2 2 2
mg mgI mg � �
�
22
0
2
14
92= 3
3 16
mv
mg�
��
2
0= 8v g�
v0 = 2 2g�
16. Answer (2)
Hint: =
dl
dt
�
�
= r F �
��
Sol.:
y
x
z
(2, 0, 0)
y
0
S = 21
2ut at
21– 10(2) = – 20 m2
y
ˆ= – 20 N�
F j
ˆ ˆ= (2 – 20 ) m�
r i j
ˆ ˆ ˆ
= 2 –20 0
0 –20 0
i j k
�
ˆ ˆ ˆ= (0 – 0) – (0 – 0) (–40 – 0)i j k �
ˆ= – 40 N-m� k
17. Answer (1)
Hint: A1r1 = A
2r2
Sol.:
r1
r2
C1C C
2
All India Aakash Test Series for Medical-2019 Test - 2 (Code F) (Hints and Solutions)
6/22
2
2 2,
4 2
R RA r
A1 = R2 – A
2 = R2 –
2 23
4 4
R R
2 2
1
3
4 4 2
R R Rr
1
6
Rr
18. Answer (2)
Hint: I = Mk2 k Radius of gyration.
Sol.:
2
2 =6
MaI I
I =
2
12
Ma
2
12
Ma = Mk2
=2 3
ak
19. Answer (4)
Hint: Properties of center of mass.
Sol.: Center of mass relative to the body will not
depend on reference of frame. And it may lie inside
or outside the body.
20. Answer (1)
Hint: W = F S��
Sol.:
B
A
32 N
Let both block move together
32 = 8 a
a = 4 m/s2
fmax
= mg
fmax
= 1
3 10 =15 N2
Pseudo force on block A = ma = 4 × 3 = 12 N
There will be no slipping between the block and
12 N friction force will act on block A in forward
direction.
S = ut + 21
2at
S = 21
4 (2) = 8 m2
W = fs
× S
W = 12 × 8
= 96 JW
21. Answer (2)
Hint: Condition for stable and unstable equilibrium.
Sol.: For equilibrium,
0dU
dx
3 23
– 2 03 2
d x xx
dx
⎛ ⎞ ⎜ ⎟
⎝ ⎠
(x2 – 3x + 2) = 0
x = 1, 2
Now,
2
22 – 3
d Ux
dx
For stable equilibrium,
2
20
d U
dx
For unstable equilibrium,
2
20
d U
dx
at x = 1
2
2–1 0
d U
dx
So unstable equilibrium,
at x = 2
2
21 0
d U
dx
So stable equilibrium.
Test - 2 (Code F) (Hints and Solutions) All India Aakash Test Series for Medical-2019
7/22
22. Answer (2)
Hint: =W F d� �
Sol.: T
mg
=2
ga
T – mg = ma
T = 3
2
mg
W = Td cos0°
3=
2
mgdW
23. Answer (1)
Hint: Conservation of mechanical energy
Sol.: At maximum extension of spring, speed of
block will be zero.
x
K
m
MEi = M.E
f
0 = 21
2kx mg
1
2kx2 = mgx
2=
mgx
k
24. Answer (1)
Hint: Conservation of energy.
Sol.:
v0
R
h
(90° – )
B
RefA
v
2 2
0
1 1= ( )
2 2mv mg R h mv
At point B
mgcos(90° – ) – N =
2mv
R
mgsin =
2mv
R[∵ N = 0]
and h = R sin
2 mgR = 1
sin sin2
mgR mgR mgR
3 2sin = 1 sin =
2 3 ⇒
H = h + R
H = R + R sin
H = R + 2
3
R
5=
3
RH
25. Answer (3)
Hint: Conservation of energy and conservation of
linear momentum.
Sol.:
2mv1
4
m
0
2
v
4
m
v0
By conservation of linear momentum of system
0
0 1
ˆ ˆ ˆ= 24 4 2
vm mv i mv i i
0
116
vv
By conservation energy of the block.
2
1
12 = 2
2m v mgh
All India Aakash Test Series for Medical-2019 Test - 2 (Code F) (Hints and Solutions)
8/22
h =
2
0
2
1
2 (16)
v
g
2
0=512
vh
g
26. Answer (3)
Hint: Use concept of coefficient of restitution.
Sol.: e = (Relative velocity of separation)
(Relative velocity of approach)
e = 2
1
v
v
v1
v2
e < 1
0 1e
27. Answer (4)
Hint: =P F v� �
�
Sol.: F = constant
0 0
∫ ∫v x
vdv dx
2 v x
½v x
P F v �
�
½P x
28. Answer (1)
Hint: Use ˆ ˆ= – –
U UF i j
x y
�
Sol.: U = –6x – 8y
= –6
U
x
= –8
U
y
ˆ ˆ(6 8 ) N �
F i j
2 2| | = 6 8 = 10 N�
F
2ˆ ˆ= = (3 4 ) m/s⇒ �
� �
F ma a i j
a = 10
2
2= 5 m/sa
x
y
(6, 4) m
Particle will move in the direction of acceleration
therefore, it will never cross y-axis and x-axis.
29. Answer (4)
Hint: Work = F ds����
= F(ds) cos
Sol.: W = F dscosWork done by static friction force may either be
positive, negative or zero. It depends on angle
between direction of force and displacement.
30. Answer (1)
Hint: Apply work–energy theorem
Wnet
= K
Sol.:
x m( )2
–5
10
6 8 10
F (N)
0
Wnet
= 1
10 2 10 4 – 2 52
Wnet
= 40 J
Wnet
= Kf – K
i
2 21 12 – 2 (2) = 40
2 2v
v2 – 4 = 40
v2 = 44
= 2 11m/sv
Test - 2 (Code F) (Hints and Solutions) All India Aakash Test Series for Medical-2019
9/22
31. Answer (4)
Hint: W = 0 = 0F dr����
Sol.: ˆ ˆ ˆ= ( 3 – 4 ) N�
F i i bk
ˆ ˆ ˆ= (2 – 0) (2 – (–1) (3 – 1) �
r i j k
ˆ ˆ ˆ= (2 3 2 ) m �
r i i k
= 0F r�
�
ˆ ˆ ˆ ˆ ˆ ˆ(3 – 4 ) (2 3 2 )i j bk i j k = 0
6 – 12 + 2b = 0
= 3b
32. Answer (1)
Hint: Wnet
= Wmg
+ Wf + W
N
Sol.:
37°
3 m
4 m
5 m10 k
g
Wnet
= Wmg
+ Wf + W
N
Wmg
= mg × 3 cos0° = 10 × 10 × 3 = 300 J
WN
= N × 5 cos 0°
WN
= 0
Wf
= N (5cos180°) = mgcos37°(–1) × 5
= –1 4
10 10 54 5 = –100 J
Wnet
= 300 – 100 net
= 200 JW
33. Answer (3)
Hint: Use dW = F dr����
Sol.: dW = ˆ ˆ ˆ ˆ( ) ( ) x y
F i F i dx i dyi
=x y
dW F dx F dy∫ ∫ ∫
W =
3 3
1 0
2xdx y dy∫ ∫
W =
32
32
0
1
9 1= – 9
2 2 2
xy
⎡ ⎤ ⎡ ⎤⎡ ⎤ ⎢ ⎥ ⎢ ⎥⎣ ⎦
⎣ ⎦⎣ ⎦= 4 + 9 = 13 J
34. Answer (3)
Hint: Concept of constraint equation.
Sol.:
P
T
aP
T/2
T/2
Q
T
Let movable pulley displaces by 'x' and block P by 'y'
l1 + l
2 = l
1 + x + l
2 + x – y
y = 2x
aP = 2a
Q
35. Answer (1)
Hint: fmax
= s(m
1 + m
2)g
Sol.:
100 N10 kg
5 kg fk
fk
Maximum force for common acceleration
fmax
= 0.4 × 15 × 10 = 60 N
Force applied is greater than 60 N. So friction will be
kinetic.
fk =
kN = 0.4 × 50 = 20 N
2100 – 208 m/s
10a
36. Answer (4)
Hint: For equilibrium Fnet
= 0
Sol.:
200 N
T1
T2
T4
T2
100 N
P
T1cos
T1sin
T1cos
T1sin T
2
T3
10 kg
100 N
T3
20 kg
200 N
T2
(F.B.D. ofblock 20 kg)
(F.B.D. ofblock 10 kg)
(F.B.D. ofpoint )P
P
All India Aakash Test Series for Medical-2019 Test - 2 (Code F) (Hints and Solutions)
10/22
T1cos = 100 N
T1sin = 200 N
tan = 2
–1
= tan 2
2 2 2
1= (100) (200)T
1=100 5 NT
37. Answer (1)
Hint: Fmin
= 2
1
mg
; when = tan–1()
Sol.: Fmin
= 2
1
mg
=
2
4 40010 10
3 3=
54
1 33
⎛ ⎞ ⎜ ⎟⎝ ⎠
Fmin
= 80 N
38. Answer (2)
Hint: Fnet
= 0, fk = N
Sol.:
37°
A
B
f
f
Net force on A will be zero
Wsin 37° = fk
3= cos37
5
BW m g
3 4= 3
5 5W W
1=4
39. Answer (3)
Hint: Use F = ma
Sol.:
mB P A
10 N
x = 5 m x = 2 m x = 0
T T
Total mass of rope
M = 5 × 2 = 10 kg
2101m/s
10 a
Mass of length 3 m
m = 3 × 2 = 6 kg
T = ma = 6 N
40. Answer (2)
Hint: Concept of static and kinetic friction.
Sol.:
F = 25 N
QP
1maxsf
2maxsf
1 1max 1
= = 0.4 5 10s sf m g
1max
= 20Nsf
2 2max 2
= = 0.5 10 10 = 50Ns sf m g
1 2
max max appAs
s sf f F
So, system will be at rest
F.B.D of P
25 N
N
1maxs
fP
20 + N = 25
N = 5 N
F.B.D. of Q
Nfs Q
N = fs
= 5 Nsf
41. Answer (1)
Hint: Properties of the friction force.
Sol.: fs Static friction
0 fs f
s max
0 Ns sf
flim
= s
N
Test - 2 (Code F) (Hints and Solutions) All India Aakash Test Series for Medical-2019
11/22
42. Answer (1)
Hint: Use concept of constraint equation.
Sol.:
9 kg
30°
10 kg
60°
T
T
a1
a2
T cos60° = 10 a2
T = 20 a2
...(i)
9g – T = 9a1
90 – T = 9a1
...(ii)
a2cos60° = a
1
a2 = 2a
1
T = 40a1
...(iii)
From equation (ii) & (iii),
49a1 = 90
or, a1 � 1.8 m/s2
43. Answer (1)
Hint: Use concept of pseudo force
Sol.:
A B
F.B.D. of B w.r.t. cart
NAB
m aB B
NAB
= 5 × 2
NAB
= 10 N
F.B.D. of A w.r.t. cart
NAB
NWA
m aA
NWA
= mAa = N
AB = 10 × 2 + 10 = 30 N
30= = 3 :110
WA
AB
N
N
44. Answer (3)
Hint: Translational equilibrium
F = 0
Sol.:
k
20 kg
T
2
T
2
T
4
T
4
T
T = 200 N
4
T= k x
1000x = 200
4
x = 1
m20
= 5 cmx
45. Answer (2)
Hint: Apply F = ma
Sol.:
30°
10 kg
5 kg
T1
T2
T2
5 kg
T1
aa
a
100 – T1 = 10a
T1 – T
2 – 50 sin30° = 5a
T1 – 50 sin30° = 5a
20 a = 50
a = 5
2 ms–2
T1 = 100 –
50
2
T1 =
150
2N
T2 =
55 25
2
T2 =
75 N
2
1
2
150= = 2
2
75
2
T
T
All India Aakash Test Series for Medical-2019 Test - 2 (Code F) (Hints and Solutions)
12/22
CHEMISTRY
46. Answer (4)
Hint: G = G0 + 2.303RT log Qp
Sol.: QP =
2 5
2 2
2 2N O
2 5 2
N O
P 2= =1
P P 2 1
G = G0 + 2.303 RT log10
(1) = G0 = –43 kJ/mol
47. Answer (2)
Hint: Lewis acids are electron pair acceptors.
Sol.: B2H
6 is an electron deficient species.
48. Answer (3)
Hint: Salt having minimum value of 'solubility (mole/l)'
will be least soluble.
Sol.: For binary electrolyte, lower the value of Ksp
,
lower will be its solubility.
49. Answer (2)
Hint: pH of acidic buffer pH = pKa + log
Salt
Acid
⎡ ⎤⎢ ⎥⎣ ⎦
Sol.:
CH3COOH + NaOH CH
3COONa + H
2O
(mmol)initial
10 6
(mmol)final
4 0 6
Resulting solution is an acidic buffer
pH = 4.74 + 6
log4
⎛ ⎞⎜ ⎟⎝ ⎠
= 4.74 + 0.18 = 4.92
50. Answer (1)
Hint:
Enthalpy of
= 3 ×
Enthalpy of hydrogenation –
Resonance hydrogenation of benzene of cyclohexane
energy
Sol.:
+ H2 ; H = –x kJ mol–1
+ 3H2
Enthalpy of hydrogenation of benzene
= 3H – Resonance energy.
= 3(–x) – (–y)
= (y – 3x) kJ mol–1
51. Answer (4)
Hint: Vapour pressure of a liquid depends on
temperature.
52. Answer (1)
Hint: Overall enthalpy of a process, is the sum of
the enthalpies of all steps involved.
Sol.:
(i) C(s) + O2(g) CO
2(g) + 94 kcal/mol
(ii) H2(g) +
1
2O
2(g) H
2O(l) + 68 kcal/mol
(iii) CH4(g) + 2O
2(g) CO
2(g) + 2H
2O(l) + 213 Kcal/mol.
On applying, (i) + 2 × (ii) – (iii)
Reaction obtained is
C(s) + 2H2(g) CH
4(g)
H = (–94) + 2(–68) –(–213) = –17 kcal.
53. Answer (4)
Hint: S = 2 1
p
1 2
T PnC ln + nRln
T P
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
Sol.: S = 1
2
PnR ln
P
⎛ ⎞⎜ ⎟⎝ ⎠
(At constant temperature)
S = 1
2
PnR ln
P
⎛ ⎞⎜ ⎟⎝ ⎠
= 1 × R × 2
ln0.5
⎛ ⎞⎜ ⎟⎝ ⎠
= R ln 4 = 2R ln2
54. Answer (4)
Hint: U = q + w and w = –2.303 nRT log2
1
V
V
⎛ ⎞⎜ ⎟⎝ ⎠
Sol.: w = –2.303 × 2 × 8.314 × 300 log20
5
⎛ ⎞⎜ ⎟⎝ ⎠
= –2.303 × 2 × 8.314 × 300 × log4
= –2.303 × 2 × 8.314 × 300 × 2 × 0.30
= – 6892.9 J � –6.9 kJ
U = q + w
q = –w (for isothermal process)
q = +6.9 kJ
55. Answer (2)
Hint: H2O(l) � H
2O(v) (1 atm, 100°C)
U = q + W
Sol.: q = nHVapourization
q = 1 × 40.66 × 103 J = 40.66 × 103 J
Test - 2 (Code F) (Hints and Solutions) All India Aakash Test Series for Medical-2019
13/22
w = – ngRT
= –1 × 8.314 × 373 = –3101.12 J
U = q + w = (40.66 × 103 – 3101.12) J
= 37558.878 J = 37.56 × 103 J
56. Answer (3)
Hint: Fusion
2 2
Fusion
n HH O(s) H O(l), S =
T
⎛ ⎞⎜ ⎟
⎝ ⎠�
Sol.: S =
3
–1 –16 10= 21.98 Jmol K
273
At equilibrium, G = 0
57. Answer (1)
Hint:
Partial pressure = (Total pressure) × Mole fraction of gas
Sol.:
2O
20n = = 0.625
32
CO
2.8n = = 0.1
28
Mole fraction oxygen gas = 0.625
0.1 0.625
= 0.625 625
=0.725 725
= 25
29
⎛ ⎞⎜ ⎟⎝ ⎠
pO2
= 25
29
⎛ ⎞⎜ ⎟⎝ ⎠
× 2 = 50
atm29
Partial volume of oxygen gas = 25
1029
250
29 L
58. Answer (4)
Hint: For H2 gas, value of a is negligible.
Sol.:
m2
m
aP + (V – b) = RT [ a is negligible]
V
⎛ ⎞⎜ ⎟⎝ ⎠
∵
P(Vm
– b) = RT
PbZ =1+
RT
59. Answer (2)
Hint: Number of mole = Volume at STP(ml)
22400
Sol.: For A vapours, mole = 112 1
mol22400 200
Molar mass of A = 0.48 × 200 = 96 g
2
A(g)
H
r 2 1 1= = =
r 96 48 4 3
60. Answer (3)
Hint:
P
L G(L + G)
V
Sol.: At temperature TC and pressure P
2, gas phase
exists.
61. Answer (2)
Hint: Partial pressure of a gas
= Mole fraction × Total pressure
Sol.: Moles of 2
xO
32 (Assume, X is mass of
each gas mixed together)
Moles of x
He4
Total moles x x 9
x32 4 32
pHe
x
84P P
9 9x
32
Fraction of total pressure exerted by He = 8
9.
62. Answer (4)
Hint: pH of salt of weak base and strong acid is
b
17 – (pK log C)
2
⎧ ⎫⎨ ⎬⎩ ⎭
.
All India Aakash Test Series for Medical-2019 Test - 2 (Code F) (Hints and Solutions)
14/22
Sol.:
4 4 2NH OH HCl NH Cl H O
Initial mol, 0.1 V 0.1 V
Final mol, 0 0 0.1 V
⎧ ⎫⎨ ⎬⎩ ⎭
Concentration of NH4Cl =
–2
0.1= 5 ×10
2
pH = –2
17 – 4.75 log(5 ×10 )
2
= 17 – 4.75 – 2 log5
2
= 5.2755 � 5.28
63. Answer (1)
Hint: Kp = K
c(RT)ng.
Sol.: If ng = 0
Kp = K
c
For reaction N2(g) + O
2(g) � 2NO(g), n
g = 0
64. Answer (1)
Hint: Pure water is neutral and in pure water,
[H+] = [OH–] = w
K
Sol.: For acidic solution, [H+] > w
K
For basic solution, [OH–] > w
K
65. Answer (2)
Hint: For reaction
A(g) � B(g), KC =
[B] 0.02= = 2
[A] 0.01
Sol.:
A(g) � B(g)
At new eq. 0.02 – x 0.02 + x
KC =
0.02 + x= 2
0.02 – x
0.02 + x = 0.02 × 2 – 2x
3x = 0.02
x = 0.02
= 0.006673
[B] = 0.02 + 0.0067 = 0.0267
66. Answer (1)
Hint: For reaction 2A(g) � B(g) + C(g),
Equilibrium constant KC =
2
[B][C]
[A]
Sol.: 2A(g) � B(g) + C(g)
0.2 0 0
0.2 – 2x x x
KC =
2
2= 0.09
(0.2 – 2 )
x
x
= 0.3(0.2 – 2 )
x
x
x = 0.06 – 0.6x
1.6x = 0.06
x = 0.06
1.6
Concentration of B is 0.06
1.6 10
⎛ ⎞⎜ ⎟⎝ ⎠
= 3.75 × 10–3 M.
67. Answer (4)
Hint: Due to common ion effect, main source of Cl–
in solution is strong electrolyte NaCl.
Sol.: Ksp
(AgCl) = [Ag+] [Cl–]
1.6 × 10–10 = [Ag+] × 10–2
[Ag+] = 1.6 × 10–8
68. Answer (4)
Hint: Salt of weak acid and strong base will undergo
anionic hydrolysis.
Sol.: CH3COONa + H
2O CH
3COO– (aq) + Na+(aq)
CH COO3
+ H2O � CH
3COOH + OH (aq)
69. Answer (1)
Hint: [H+]Resulting
(Normality) = 1 1 2 2
1 2
N V +N V
V V
Sol.: [H+] = 0.1 2 V + 0.1×1× V
V + V
= 0.2 + 0.1 0.3
= = 0.152 2
pH = – log[H+] = – log(15 × 10–2)
= 2 – log(15) = 0.824 0.82�
Test - 2 (Code F) (Hints and Solutions) All India Aakash Test Series for Medical-2019
15/22
70. Answer (2)
Hint: Ksp
of Al(OH)3 = [Al3+] [OH–]3
Sol.: Al(OH)3(s) � Al3+(aq) + 3OH–(aq)
S 3S
∵ pH = 9, Hence, pOH = 5 [OH–] = 10–5 M
3S = 10–5 S =
–5
10
3 M
Ksp
(Al(OH)3)= S(3S)3 = 27S4 =
4–5
1027
3
⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠
=
–20 –20
27 10 10=
27 3 3
⎛ ⎞⎜ ⎟ ⎝ ⎠
71. Answer (3)
Hint: Salt having minimum value of Ksp
will give ppt
first.
Sol.: Since, ppt appearance order is
AgI AgBr AgCl
Hence, Ksp
(AgI) < Ksp
(AgBr) < Ksp
(AgCI)
So, z < y < x
72. Answer (2)
Hint: KC = [CO
2]
Sol.: CaCO3(s) � CaO(s) + CO
2(g)
KC = [CO
2]
0.5 2
mol.= [CO ]
⎛ ⎞⎜ ⎟⎝ ⎠�
moles of CO2 = 0.5 × 5 = 2.5
73. Answer (4)
Hint: SReaction
= (S)Product
– (S)Reactants
(:coefficients of reactants and products)
Sol.:Sreaction
= 6 × 100 – (2 × 30 + 5 × 60)
= 600 – (60 + 300) = 600 – 360 = 240
GReaction
= HReaction
– TSReaction
for spontaneous reaction
GReaction
< 0
HReaction
< TSReaction
300 × 103 < T × 240
T >
3300 10
240
T > 1250 K
74. Answer (2)
Hint: Greater the atomicity of gas, greater is the
entropy.
Sol.: Correct order of entropy is
He < O2 < CO
2 < PCl
3
75. Answer (3)
Hint: Standard boiling point of water is 99.6°C
76. Answer (4)
Rate of Volume effused=
effusion Time taken
⎛ ⎞⎜ ⎟⎝ ⎠
1
Molar mass
Sol.:
2
gas
H
r
r
= 2 2 2
2
2
gas
H H Hgas
H gas gas gas
H
V
M t Mt= = =
V M t M
t
=
gas
1 2=
8 M
Mgas
= 64 × 2 = 128 u
77. Answer (3)
Hint: Interaction energy of dipole-dipole force in
stationary polar molecule 3
1
r
78. Answer (1)
Hint: Conjugate acid is formed by addition of one H+
ion to the given species.
Sol.: Conjugate acid of 2– –
4 2 4HPO is H PO
79. Answer (1)
Hint: Unit of Kp : g
n
(atm) .
Sol.: For reaction PCl5(g) � PCl
3(g) + Cl
2(g),
ng = 1
So, unit of Kp
is atm
80. Answer (4)
Hint: A3B
2(s) + H
2O � 3A2+(aq) + 2B3–(aq)
K = [A2+]3 [B3–]2
Sol.: K = (3s)3(2s)2 = 27s3 × 4s2 = 108 s5
s =
1/5
K
108
⎛ ⎞⎜ ⎟⎝ ⎠
81. Answer (3)
Hint: pH = a
17 (pK – pK )
2b
Sol.:pH of 0.1 M CH3COONH
4
pH = 1
7 (pKa – pK )2
b
pH = 7 (∵ pKa = pK
b)
All India Aakash Test Series for Medical-2019 Test - 2 (Code F) (Hints and Solutions)
16/22
82. Answer (4)
Hint: pH = – log[H+]
Sol.: [OH–] obtained from 0.1 N NaOH = 10–1 M
pOH = 1
pH = 13
83. Answer (2)
Hint: Acidic buffer is a mixture of weak acid and its
salt with strong base.
Sol.: (CH3COOH + CH
3COONa) Acidic buffer
84. Answer (4)
Hint: Keq
=
Coefficient
Coefficient
[Product]
[Reactant]
Sol.: 2 5 2 2 eq
1N O (g) 2NO (g) O (g), K = 0.5
2�
2 2 2 54NO (g) O (g) 2N O (g), � K
eq=
2
eq
1
K
⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠
= 1
0.5 × 0.5 = 4
85. Answer (4)
Hint: Active mass of solid is unity.
Sol.: Calcium oxide is solid so its active mass is
unity.
86. Answer (3)
Hint: In an isolated system, mass as well as energy
do not exchange with surrounding.
Sol.: Universe is an example of isolated system.
87. Answer (3)
Hint: G = H – TS
For non-spontaneous reaction, G > 0
Sol.: If H > 0 and S < 0
G > 0 (For all temperature)
88. Answer (4)
Hint:H = U + ngRT
H > U (if ng > 0)
Sol.:ng of (i) Reaction = –5
ng of (ii) Reaction = 0
ng of (iiii) Reaction = 0
ng of (iv) Reaction = +1
Hence, for reaction (iv) H > U
89. Answer (2)
Hint: Density of ideal PM
gas(d) =RT
Sol.: d = PM 1.2 atm × 32 (g/mol.)
=RT L-atm
0.0821 600KK-mol.
⎛ ⎞⎜ ⎟⎝ ⎠
= 0.779 � 0.78 g/L
90. Answer (3)
Hint: KE T
Sol.: Average molar kinetic energy is 3RT
2
⎛ ⎞⎜ ⎟⎝ ⎠
BIOLOGY
91. Answer (3)
Sol.: Strawberry is an aggregate fruit.
Mulberry & jackfruits are composite fruits or multiple
fruits.
Pomegranate is a simple fleshy or succulent fruit.
92. Answer (4)
Hint: Aleurone layer is a part of endosperm.
Sol.: It is triploid, provides nourishment and seen in
maize grains.
93. Answer (1)
Sol.: Root hairs arise from epidermal cells of
maturation zone.
94. Answer (4)
Hint: Thorns can be seen in Citrus or Bougainvillea.
Sol.: Spines are modification of leaves.
95. Answer (4)
Hint: This plant belongs to the family Fabaceae.
Sol.: It is Indigofera.
96. Answer (4)
Hint: In members of family Poaceae, the seed coat
is fused with fruit wall (Caryopsis type of fruit).
Sol.: Gram is a dicot. Wheat and maize are
monocots and belong to family Poaceae.
97. Answer (3)
Hint: Papilionaceous corolla shows vexillary
aestivation.
Sol.: Mango is a drupe. Vexillum is largest posterior
petal in pea also called standard petal.
Test - 2 (Code F) (Hints and Solutions) All India Aakash Test Series for Medical-2019
17/22
98. Answer (3)
Hint: Fused carpels— syncarpous
Sol.: Staminode is sterile anther. Twisted aestivation
is seen in petals of cotton. Unisexual flowers are
seen in maize.
99. Answer (2)
Hint: In racemose inflorescence the terminal flower
is absent.
Sol.: It (main axis) has unlimited growth and flowers
ever borne in acropetal manner.
100. Answer (4)
Hint: Banana is a triploid plant hence cannot form
gametes.
Sol.: Banana fruit develops without fertilisation of
gametes, hence it is a parthenocarpic fruit. Banana
has suckers.
101. Answer (1)
Hint: In opposite phyllotaxy, a pair of leaves arise at
each node opposite to each other.
Sol.: Guava, Calotropis have opposite phyllotaxy.
Mustard, China roseAlternate phyllotaxy
Sunflower
⎤⎥⎦
Alstonia, Nerium ] Whorled phyllotaxy.
102. Answer (4)
Hint: In monadelphous condition, all stamens are
united in a single bundle but anthers remain free. It
is found in family Malvaceae.
Sol.: In pea flower the stamens are united in two
bundles so it is diadelphous.
103. Answer (1)
Hint: Given floral formula belongs to family
Solanaceae.
Sol.: Solanaceae members have persistent calyx,
superior ovary with many ovules. Seeds are
endospermic. Flowers are bisexual and
actinomorphic.
104. Answer (2)
Hint: In some seeds nucellus persists in the form of
a layer.
Sol.: Perisperm is persistent nucellus. Seeds having
perisperm are called perispermic seeds.
105. Answer (2)
Hint: Axile placentation has placenta in the axial
position and ovules are attached to it.
Sol.: Tomato & lemon have axile placentation. Pea
has marginal, mustard has parietal & sunflower and
marigold have basal placentation.
106. Answer (1)
Hint: Albuminous seed contains endosperm.
Sol.: Coconut has endospermic seed where seed
coat is thick & non-membranous.
Groundnut, pea, and soyabean have ex-albuminous
(non-endospermic) seeds.
107. Answer (3)
Hint: Phylloclade is a modified stem.
Sol.: In Opuntia, phylloclade is flattened and
photosynthetic.
108. Answer (3)
Hint: Stem bears nodes, internodes and buds.
Sol.: Potato is a tuber which is a modified
underground stem because it bears buds.
109. Answer (4)
Hint: Pneumatophores help a plant to get oxygen
from the atmosphere.
Sol.: They grow vertically upward hence negatively
geotropic & can be seen in Rhizophora which grow
in marshy/swampy areas.
110. Answer (3)
Hint: Carrot & Turnip are modifications of tap root.
Sol.: Ginger is a modified underground stem. Edible
part of sweet potato is a modified adventitious root.
111. Answer (3)
Hint: Tap root is the primary root.
Sol.: Tap roots (Primary root) arise from radicle
whereas fibrous roots arise from the base of the
stem.
112. Answer (4)
Sol.: Bordeaux mixture is first fungicide discovered
by RMA Millardet. It is commonly known as holy
water of plant pathology.
113. Answer (4)
Hint: Cyanobacteria evolve O2 during photosynthesis.
Sol.: Cyanobacteria have enzyme nitrogenase which
fixes atmospheric nitrogen. Some cyanobacteria
form symbiotic associations with ferns. Such as
Anabaena with Azolla.
Nostoc is a filamentous BGA.
All India Aakash Test Series for Medical-2019 Test - 2 (Code F) (Hints and Solutions)
18/22
114. Answer (2)
Hint: Prokaryotes lack nuclear membrane &
membrane bound cell organelles.
Sol.: Streptococcus is a prokaryote. Remaining
organisms are eukaryotes.
115. Answer (1)
Hint: Chemoautotrophic bacteria lack photosynthetic
pigments.
Sol.: ChemoautotrophicNitrosomonas
Nitrobacter
⎤⎥⎦
PhotosyntheticNostoc
Chlorobium
⎤⎥⎦
Lactobacillus] Heterotrophic.
116. Answer (3)
Hint: Mycoplasma are smallest living organisms.
Sol.: They lack cell wall and can survive without
oxygen.
Methanogens are responsible for production of
biogas.
117. Answer (2)
Hint: Archaebacteria contain pseudomurein in their
cell wall.
Sol.: The cell wall of eubacteria is made up of
mainly peptidoglycan. They are prokaryotes hence
they lack membrane bounded cell organelles.
Archaebacteria include halophiles, methanogens and
thermoacidophiles.
Archaebacteria have introns in their genetic material.
118. Answer (1)
Hint: Heterocyst is related to N2 fixation which takes
place in anaerobic condition.
Sol.: In heterocyst cell wall is impermeable to
oxygen but permeable to nitrogen. It lacks PS-II but
PS-I remains active and produces ATP required for
N2-fixation.
119. Answer (4)
Hint: Protists are unicellular eukaryotes.
Sol.: Eukaryotes have well defined nucleus and
other membrane bound cell organelles.
120. Answer (2)
Hint: Euglenoids lack cell wall.
Sol.: Slime moulds are heterotrophic (Saprophytic).
Protozoans are predators & parasites. Chrysophytes
include diatoms and desmids which are mostly
photosynthetic.
121. Answer (3)
Hint: Cell wall of diatoms is indestructible due to
presence of silica.
Sol.: Due to presence of silica, the fossilized
diatoms get deposited at the sea bed to form
diatomaceous earth.
Diatoms are microscopic, unicellular protists which
float passively in water current.
122. Answer (4)
Hint: Paramoecium has two nuclei.
Sol.: Paramoecium is a ciliated protozoan
123. Answer (4)
Hint: Dinoflagellates are flagellated, mostly marine
photosynthetic protist.
Sol.: Dinoflagellates have two flagella, one is
transverse and another is longitudinal.
124. Answer (1)
Hint: Mycorrhiza is a mutually beneficial association.
Sol.: Mycorrhiza is a symbiotic association between
roots of higher plants & fungi.
125. Answer (4)
Hint: All viruses have capsid.
Sol.: Bacterial viruses are known as bacteriophages,
they have capsid (protein coat) & genetic material,
usually dsDNA.
126. Answer (2)
Hint: Viruses are obligate intracellular parasites.
Sol.: Viruses can multiply within the host cell only.
They cannot respire, divide and grow externally.
127. Answer (4)
Hint: Prions lack genetic material.
Sol.: Prions are proteinaceous infectious particles. It
causes mad cow disease in cattles & scrapie
disease in sheep.
128. Answer (2)
Hint: A virus is larger than sub viral agents/particles.
Sol.: Viroids are infectious RNA particles. They lack
protein. They cause diseases in plants only.
Viruses have proteinaceous capsid.
129. Answer (3)
Hint: Phycobiont is algal partner in lichen
association. They are either green algae or blue-
green algae.
Sol.: Algal partner synthesizes food for fungal
partner. Lichens do not grow in air polluted areas.
Animals store their food in form of glycogen or fat.
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130. Answer (4)
Sol.: Bladderwort is a partially heterotrophic
insectivorous plant.
131. Answer (4)
Hint: Rhizopus is a conjugation fungus, also known
as bread mould.
Sol.: It reproduces by non-motile asexual spores and
non-motile zygospores.
It is a saprophytic fungus with chitinous cell wall.
132. Answer (3)
Hint: Deuteromycetes are called imperfect fungi.
Sol.:Morels
AscomycetesAspergillus
⎤⎥⎦
Bracket fungi] Basidiomycetes
Deuteromycetes
Alternaria
Colletotrichum
Trichoderma
⎤⎥⎥⎥⎦
133. Answer (2)
Hint: Claviceps purpurea causes ergot of rye.
Sol.: Claviceps belongs to the class Ascomycetes.
It is a sac fungus which possesses septate
mycelium and chitinous cell wall.
It reproduces by conidia which are non-motile
asexual spores.
134. Answer (4)
Hint: On basidium exogenously produced spores are
basidiospores.
Sol.: Basidiospores are produced by members of
basidiomycetes. These are haploid and sexual
spores.
135. Answer (4)
Hint: The body of fungi is not differentiated into root,
stem & leaves.
Sol.: The body of fungi is thalloid & haploid (n).
Each cell contains one set of chromosomes.
136. Answer (3)
Hint: Disorder associated with accumulation of dust
particles.
Sol.: Pneumoconiosis is occupational disease of
lungs due to inhalation of dust, characterised by
inflammation, coughing and fibrosis.
137. Answer (1)
Hint: pCO2 in alveoli is 40 mmHg.
Sol.:
Atmospheric air
Alveoli Deoxy-generated
blood
Oxy-genated
blood
Tissue
pO2 159 104 40 95 40
pCO2 0.3 40 45 40 45
138. Answer (3)
Hint: This wave indicates ventricular repolarisation.
Sol.: T-wave represents the return of ventricles from
excited to normal state. QRS complex represents
depolarisation of ventricles while P-wave represents
atrial depolarisation.
139. Answer (2)
Hint: Ischemic conditions reflect oxygen deficiency
in cardiac muscle fibres.
Sol.: When heart muscle is suddenly damaged by
inadequate blood supply, the condition is called heart
attack.
140. Answer (2)
Hint: This hormone is released in response to
increase in blood pressure than normal.
Sol.: During dehydration (profuse sweating)
osmolarity of body fluid increases, increased
secretion of ADH from hypothalamus occurs. GFR
decreases during profuse sweating.
141. Answer (3)
Hint: Blood vessel exiting from major osmoregulatory
organ of human body.
Sol.: Largest amount of urea is present in hepatic
vein.
142. Answer (2)
Hint: Pressure difference on both sides of filtration
membrane.
Sol.: NFP = (GHP) – (BCOP + CHP)
= 60 – (30 + 20)
= 10 mmHg
143. Answer (4)
Hint: Sympathetic stimulation alters heart rate.
Sol.: Sympathetic neural signals increase cardiac
output and stroke volume but decrease duration of
cardiac cycle. Cardiac muscles are not under the
control of our will.
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144. Answer (4)
Hint: Innermost lining of wall of blood vessels is
constituted by squamous cells.
Sol.: Arteries have narrower lumen than veins and
mostly carry oxygenated blood.
145. Answer (3)
Hint: Maximum amount of air which can be inspired
or expired forcefully.
Sol.: Vital capacity is measured for lung function
test by an instrument known as spirometer. With
exception of RV, TLC, FRC. All other lung capacities
and volumes can be measured by spirometer.
146. Answer (2)
Hint: High threshold substance like glucose is
completely absorbed in kidney in a normal healthy
person.
Sol.: Diabetes insipidus occurs due to deficiency of
ADH (vasopressin) from hypothalamus. All types of
nephrons have peritubular capillary networks. GFR in
healthy individual is 125 ml/minute or 180 L/day.
147. Answer (3)
Hint: Ascending limb of loop of Henle is permeable
to salt.
Sol.: Descending limb is called concentrating
segment of loop of Henle as it is permeable to water
and impermeable to salts.
148. Answer (2)
Hint: These are region between medullary pyramids
in kidney.
Sol.: Invagination of cortex into medulla forms
columns of Bertini which divide medullary region of
kidney into renal pyramidals.
149. Answer (2)
Hint: This hormone is released by atrial walls.
Sol.: Vasa recta acts as countercurrent exchanger
because they exchange water for ions. ANF
opposes RAAS.
150. Answer (4)
Hint: Well developed region of JG nephrons
extending into medulla in JG nephrons.
Sol.: Loop of Henle of both cortical and
juxtamedullary nephron lie in medullary pyramid.
151. Answer (2)
Hint: Identify the machine use to measure an ECG.
Sol.: Electrical activities of heart is measured by an
instrument known as electrocardiograph and graph
obtained is known as electrocardiogram.
Stethoscope is used to hear heart sounds.
152. Answer (3)
Hint: Duration of each auricular cycle is equal to
duration of the cardiac cycle.
Sol.: During each cardiac cycle, an auricular cycle
as well as a ventricular cycle is completed. Each
cycle has duration of 0.8 sec. So time of auricular
cycle is 0.8 sec not 0.6 second.
153. Answer (1)
Hint: 2/3rd filling of ventricles occurs when all
chambers of heart are relaxing.
Sol.: When auricles are in diastole auriculoventricular
valves remain open but semilunar valves present at
opening of aorta are closed.
154. Answer (3)
Hint: During this phase, blood is not pumped from
ventricles to aorta.
Sol.: During isovolumetric contraction, both
auriculoventricular valves and semilunar valves remain
closed. Pressure within ventricles increases but
amount of blood remains unchanged in isovolumetric
contraction.
155. Answer (2)
Hint: Volume of blood pumped by each ventricle in
0.8 seconds.
Sol.: Cardiac output = heart rate x stroke volume.
Hypertension is observed if blood pressure exceeds
140/90 or more repeatedly. Opening of pulmonary
artery is guarded by semilunar valves.
156. Answer (3)
Hint: Volume of ventricle decreases during their
contraction.
Sol.: During ventricular systole, closure of cuspid/AV
valves occurs. Blood enters ventricles during atrial
systole.
157. Answer (2)
Hint: Stretch receptors are activated upon filling of
urinary bladder beyond threshold.
Sol.: Hemodialysis is needed when kidneys
malfunction. Urine formation is independent of stretch
reflex.
158. Answer (4)
Hint: ADH acts on this region to prevent diuresis.
Sol.: The collecting duct is responsible for secretion
of urea not for absorption. Various nephrons open
into collecting ducts. Collecting duct secretes urea
into interstitium to increase osmolarity for
reabsorption of water.
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159. Answer (3)
Hint: Maximum percentage of CO2 is transported in
plasma in bicarbonate form.
Sol.: 77% CO2 is transported through blood plasma.
Dissolvedin plasma – 7%
In the form ofbicarbontates
in plasma – 70%
Plasma(77%)
160. Answer (4)
Hint: Affinity of Hb for O2 is inversely proportional to
P50
value.
Sol.: Shift of curve B towards right indicates
increase in P50
value and thus decreased affinity of
haemoglobin for oxygen, leading to greater
dissociation of oxygen from hemoglobin.
161. Answer (4)
Hint: Identify a layer absent in alveolar capillaries.
Sol.: Capillaries lack tunica media formed by
smooth muscle fibres. So, diffusion membrane does
not contain muscular tissue.
162. Answer (1)
Hint: Lymphatic fluid is poured into venous blood.
Sol.: Lymph from right thoracic duct drains into right
subclavian vein whereas left thoracic duct drains
lymph into left subclavian vein.
163. Answer (3)
Hint: This heart sound is produced during ventricular
systole.
Sol.: First heart sound 'lub' is produced during
ventricular systole. It has low frequency of about
25–45 Hz and is of comparatively longer duration
(0.15 sec) in comparison to second heart sound
which has higher frequency than 50 Hz but duration
is only 0.12 sec.
164. Answer (4)
Hint: Identify the blood group of universal donors.
Sol.: Universal donors (O–) and individuals with
blood group B– and O+ can donate blood to the
patient.
165. Answer (1)
Hint: WBCs which participate in allergic reaction
during worm infestation.
Sol.: Eosinophils are WBCs that stain with acidic
dye eosin and have bilobed nucleus. They participate
in a number of allergic reactions and fight against
worm infestation.
166. Answer (2)
Hint: These cells give rise to enucleated platelets.
Sol.: Megakaryocytes are specialised large cells in
red bone marrow which divide to form cellular
fragments lacking nucleus known as blood platelets/
thrombocytes.
167. Answer (2)
Hint: This hormone secreted by JG cells stimulates
RAAS.
Sol.: Renin secreted by JG cells in response to low
B.P. stimulates RAAS to maintain GFR. Angiotensin-
II is a vasoconstrictor. Rennin digests casein in
infant stomach.
168. Answer (2)
Hint: Enzyme which is released in body injured site
in presence of Ca+2.
Sol.: Prothrombinase/thrombokinase is also
responsible for conversion of inactive plasma protein
prothrombin into an active enzyme thrombin
responsible for conversion of fibrinogen into fibrin.
169. Answer (4)
Hint: This membrane divides body cavity into
thoracic and abdominal cavity.
Sol.: Pneumotaxic centre is called "switch-off" point
of inspiration. It stimulates neurons of expiratory
centre located in medulla. Diaphragm is stimulated
by respiratory rhythm centre in medulla.
170. Answer (1)
Hint: Dialysis helps to counter uremia.
Sol.: Dialysing fluid is isotonic to blood plasma.
171. Answer (3)
Hint: These animals live on both land and water.
Sol.: Crocodiles are reptiles with completely divided
four chambered heart. Amphibians have three
chambered heart.
172. Answer (4)
Hint: The nitrogenous waste product which requires
minimum loss of water in its elimination.
Sol.: Land snails excrete uric acid whereas most
terrestrial organisms excrete either urea (humans,
amphibians) or uric acid (insects, birds, reptiles).
Aminotelic organisms excrete amino acid (Unio and
echinoderms)
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173. Answer (3)
Hint: Lungs are situated in cavity above diaphragm.
Sol.: Trachea divides into two primary bronchi at the
level of 5th thoracic vertebra and enters lungs.
174. Answer (3)
Hint: TV + IRV + ERV = TLC – RV
Sol.: TLC – RV = VC,
EC = ERV + TV = 1100 + 500 = 1600 ml
175. Answer (3)
Hint: Fluid filled cavity is present around lungs.
Sol.: Pleura is divided into two layers. Outer pleural
membrane is in close contact with thoracic lining
whereas inner pleural membrane is in contact with
lung surface.
176. Answer (4)
Hint: Squamous epithelium is found at this surface.
Sol.: Gaseous exchange takes place in the alveoli
in lungs.
177. Answer (1)
Hint: Excretory organ in annelids.
Sol.: Respiration in earthworm (annelids) occurs
through moist cuticle.
178. Answer (3)
Hint: These cells are granular, phagocytic and most
abundant type of WBCs.
Sol.: Differential leukocyte count i.e., DLC reveals
percentage of type of WBCs in blood.
Neutrophils – 60–65%
Lymphocytes – 20–25%
Monocytes – 6–8%
Eosinophils – 2–3%
179. Answer (1)
Hint: Part of nephron exhibiting brush border
appearance.
Sol.: ADH facilitates facultative/conditional
reabsorption of water from DCT & collecting tubule.
PCT is responsible for obligate reabsorption and is
internally lined by simple cuboidal epithelium
containing microvilli giving brush border appearance
which increase surface area for reabsorption.
180. Answer (3)
Hint: The primary organ involved in this process is
kidney.
Sol.: Removal of metabolic waste products from the
body is known as excretion. Main role of sweating
is to facilitate thermoregulation. Defaecation involves
removal of undigested and unabsorbed food.
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