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Test - 4 (Code-C) (Answers) All India Aakash Test Series for Medical-2020
1/16
1. (4)
2. (2)
3. (4)
4. (3)
5. (3)
6. (2)
7. (2)
8. (4)
9. (4)
10. (3)
11. (4)
12. (2)
13. (3)
14. (1)
15. (2)
16. (2)
17. (2)
18. (2)
19. (3)
20. (4)
21. (2)
22. (2)
23. (Delete)
24. (3)
25. (3)
26. (4)
27. (2)
28. (4)
29. (4)
30. (2)
31. (4)
32. (4)
33. (3)
34. (1)
35. (4)
36. (3)
Test Date : 30/12/2018
ANSWERS
TEST - 4 (Code-C)
All India Aakash Test Series for Medical-2020
37. (2)
38. (3)
39. (3)
40. (2)
41. (2)
42. (4)
43. (4)
44. (2)
45. (3)
46. (2)
47. (4)
48. (2)
49. (3)
50. (4)
51. (2)
52. (4)
53. (4)
54. (2)
55. (2)
56. (4)
57. (4)
58. (3)
59. (4)
60. (3)
61. (2)
62. (4)
63. (3)
64. (3)
65. (3)
66. (2)
67. (2)
68. (3)
69. (2)
70. (3)
71. (4)
72. (3)
73. (3)
74. (2)
75. (2)
76. (1)
77. (3)
78. (2)
79. (4)
80. (4)
81. (3)
82. (3)
83. (3)
84. (1)
85. (3)
86. (1)
87. (1)
88. (4)
89. (4)
90. (1)
91. (3)
92. (3)
93. (4)
94. (4)
95. (1)
96. (4)
97. (3)
98. (3)
99. (4)
100. (3)
101. (3)
102. (3)
103. (4)
104. (3)
105. (3)
106. (4)
107. (2)
108. (4)
109. (2)
110. (4)
111. (3)
112. (2)
113. (3)
114. (3)
115. (2)
116. (2)
117. (3)
118. (4)
119. (3)
120. (3)
121. (3)
122. (2)
123. (3)
124. (1)
125. (2)
126. (4)
127. (3)
128. (3)
129. (2)
130. (3)
131. (3)
132. (2)
133. (3)
134. (3)
135. (4)
136. (2)
137. (4)
138. (4)
139. (3)
140. (3)
141. (4)
142. (2)
143. (2)
144. (2)
145. (4)
146. (2)
147. (3)
148. (3)
149. (4)
150. (4)
151. (2)
152. (3)
153. (1)
154. (3)
155. (3)
156. (4)
157. (2)
158. (1)
159. (3)
160. (2)
161. (2)
162. (4)
163. (3)
164. (1)
165. (1)
166. (2)
167. (3)
168. (3)
169. (4)
170. (3)
171. (1)
172. (4)
173. (4)
174. (2)
175. (1)
176. (2)
177. (4)
178. (2)
179. (3)
180. (1)
All India Aakash Test Series for Medical-2020 Test - 4 (Code-C) (Answers & Hints)
2/16
ANSWERS & HINTS
1. Answer (4)
Hint : = mgx
Sol. :
60°
u
x
y
xomg
At t = 2 s
1cos 20 2 20 m
2 x u t
= mgx = 2 × 10 × 20 = 400 Nm
2. Answer (2)
Hint : I = mr 2
Sol. : Moment of inertia depends on axis of rotation,
mass of the body and distribution of mass about
axis of rotation.
3. Answer (4)
Hint : F + f = macm
Sol. :
f
R
F
Ov
F + f = macm
...(i)
cm
= Icm
(F – f)R = mR2
or F – f = mR = macm
...(ii) (∵ acm
= R)
From (i) & (ii)
f = 0
4. Answer (3)
Hint : L r p �
� �
Sol. : L�
about O is constant as r⊥ and v both are
constant.
[ PHYSICS]
5. Answer (3)
Hint : r F �
��
Sol. :
P(1, 2)
O(0, 0)
r
ˆ4F k
ˆ ˆ(– – 2 ) mr i j�
ˆ4 �
F k
�
��
r F
ˆ ˆ4( 2 ) �
i j Nm
6. Answer (2)
Hint : If ext
= 0 then L = constant
If ext
F
����
= 0 the cm
0a �
Sol. : For a rigid body rotating about a fixed axis L
is always parallel to .
Torque produced by central force is always zero.
7. Answer (2)
Hint : vp = 2v
0 cos (/2).
Sol. :
O60° v
0
r v = 0
120°
Here = 120°
vnet
= 2v0cos60°
vnet
= v0
8. Answer (4)
Hint : Formula for moment of inertia of any
symmetrical part of a body = Formula for moment of
inertia of body.
Sol. : 2
Semidisc Disc2
MRI I
9. Answer (4)
Hint : = ISol. : = II = mk 2
= mk 2
Test - 4 (Code-C) (Answers & Hints) All India Aakash Test Series for Medical-2020
3/16
10. Answer (3)
Hint : Mechanical energy is conserved.
Sol. : In pure rolling on inclined plane potential
energy is converted into translational and rotational
kinetic energy. Net work done by friction and normal
force is zero.
11. Answer (4)
Hint : �
� �
L r p
Sol. : �
L is always prependicular to �
r and �
p .
12. Answer (2)
Hint : In pure rolling on fixed surface net tangential
acceleration of point of contact is zero.
Sol. :
Ov
R2
Rv
For pure rolling with constant angular velocity.
v = R
Radial acceleration of point is R2.
13. Answer (3)
Hint : 2
21
kt
R
⎛ ⎞ ⎜ ⎟
⎝ ⎠
Sol. : 2
2
k
Ris same for disc and solid cylinder.
14. Answer (1)
Hint : Conservation of angular momentum.
Sol. : Li = Lf
Li = mr2
Lf = (mr2 + 2mr2)
2 23mr mr
3
15. Answer (2)
Hint : Conservation of angular momentum about the
point of contact.
Sol. :
Pure rolling
0
v
Applying conservation of angular momentum about
the point of contact,
Li = L
f
mR20 = mR2 + mvR
2 0
02
2
RmR mvR v
⇒
16. Answer (2)
Hint and Sol. : Kepler’s law of area is based on
conservation of angular momentum.
17. Answer (2)
Hint : and � �
� � ��
L r p r F
Sol. : Net torque about C is non zero and net
torque about O is zero.
∴ L is conserve about O but variable about C.
18. Answer (2)
Hint : L r p �
� �
Sol. :
v
y
x
v cos
r H =
L = rp
r = H
L = H mv cos
vL mv
g
2 2
sincos
2
mv
g
3 2
sin cos
2
19. Answer (3)
Hint : .P ��
Sol. : ˆ ˆ ˆ( – 3 2 ) Nmi j k �
ˆ ˆ ˆ(4 2 5 ) rad/si j k �
ˆ ˆ ˆ ˆ ˆ ˆ( – 3 2 ) (4 2 5 )P i j k i j k P = 4 – 6 + 10
P = 8 W
20. Answer (4)
Hint : Use parallel axis theorem.
Sol. : L/2
rL
O
All India Aakash Test Series for Medical-2020 Test - 4 (Code-C) (Answers & Hints)
4/16
2 2
2
3 12
ML MLI Mr
2
2 5
4
Lr
2
5
3 ML
I
21. Answer (2)
Hint : Fraction of kinetic energy which is rotational
2
2 2.
k
k R
Sol. :
2
2
2
21
k
Rf
k
R
for solid sphere 2
2
2
5
k
R .
2 / 5% 100 28%
21
5
22. Answer (2)
Hint and Sol. : I = mk2.
23. Answer (Delete)
24. Answer (3)
Hint : net r t
a a a � � �
Sol. :
ar
v
at
anet
25. Answer (3)
Hint : , d d
dt dt
Sol. : = 3t2 – 4t + 3
6 – 4d
tdt
= 6 rad/s2
at = R, a
r = R2
At t = 1 s = 2 rad/s
at = 6 m/s2 a
r = 4 m/s2
2 2 2
net6 4 52 m/sa
26. Answer (4)
Hint : Use Newton’s law of gravitation.
Sol.: To make M in equilibrium resultant of F1 and
F2 should be equal in magnitude and opposite in
direction to T.
∴ From the figure,
1
2
tan30F
F 30°
30°
F1
F2
m1
T
m2
∴
1
2
1
3
m
m
27. Answer (2)
Hint : Mechanical energy is conserved.
Sol. :
1 = 0
U1 + K
1 = U
2 + K
2
210 cos
2 2 2
L LMg mg I
3(1– cos )
g
L
3
2 g
L
28. Answer (4)
Hint and Sol. : I = mr2.
29. Answer (4)
Hint and Sol. : MI is same for any symmetrical part
of body.
I1 = I
2 = I
3
30. Answer (2)
Hint : rot total
1for a disc.
3K KE
Sol. : Total KE = mg h = mgl sin.
So, rot
1sin
3K mgl .
Test - 4 (Code-C) (Answers & Hints) All India Aakash Test Series for Medical-2020
5/16
31. Answer (4)
Hint : Both particles rotate about their centre of mass.
Sol. : Both particles are always opposite to each
other along the line joining them so 1 = 2.
32. Answer (4)
Hint : Conservation of mechanical energy.
Sol. :
H
v = 0
u
U1 + K1 = U2 + K2
20 0
0
– –10
2
GMm GMmm u
R R H
Now, 2 2 0
0 0
1 1
2 4 2E
GMmm u m v
R
H = R
33. Answer (3)
Hint : 2
E
GMv
R
Sol. : 32 4
3
E
Gv R
R
28
3
Ev GR
E
v R
1
2 2
E E E
p p p
v R
v R
34. Answer (1)
Hint : Angular momentum is conserved.
Sol. : r is minimum at point A and maximum at
point C.
So, KA > K
B > K
C
35. Answer (4)
Hint and Sol.: Total energy of bound system is
negative.
36. Answer (3)
Hint : g = g – R2 cos 2Sol. : = 90° for pole g
pole = g(always) if = 0,
ge = g i.e., increases.
37. Answer (2)
Hint : orbital
1V
r .
Sol. : Orbital speed is independent of mass of
satellite.
38. Answer (3)
Hint : Conservation of mechanical energy.
Sol. : 2 2
0
1 1– –
2 2 ( )
GMm GMm
mv mvR R h
2 2
0
1 1–
2 21
mghmv mv
h
R
2
0
2–
1
⎛ ⎞⎜ ⎟⎝ ⎠
ghv v
h
R
39. Answer (3)
Hint : 2 2 2
interstellar–1
E Ev v v v n
Sol. : v = nvE
n = 3
infinity 9 – 1 8 2 2 E E E
v v v v
40. Answer (2)
Hint : –f i
E E E
Sol. : –
2
GMmE
r
1 1 1 1– –
2 2 8 14i f E E
E GMm GMmr r R R
⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦
7 – 4 3
56 56
⎡ ⎤ ⎢ ⎥
⎣ ⎦E E
GMmGMm
R R
41. Answer (2)
Hint : L r p �
� �
Sol. : L = m0vr
r = R + h
0
GMv
r
0 0( ) ( )
GML m R h m GM R h
R h
All India Aakash Test Series for Medical-2020 Test - 4 (Code-C) (Answers & Hints)
6/16
42. Answer (4)
Hint : 2
1
⎛ ⎞⎜ ⎟⎝ ⎠
h
gg
h
R
Sol. : gh = g /2
2
1 1
21
h
R
⎛ ⎞⎜ ⎟⎝ ⎠
2 – 1h R
43. Answer (4)
Hint : 1n
Fr
Sol. : 2
k mvF
r r
v r°
44. Answer (2)
Hint : vara = v
prp
Sol. : vara = v
prp
2 2
a a p pr r
2
pa
p a
r
r
⎛ ⎞ ⎜ ⎟ ⎝ ⎠
ra = a + ae r
p = a – ae
2
1
1–
p
a
e
e
⎛ ⎞ ⎜ ⎟ ⎝ ⎠
45. Answer (3)
Hint : Mechanical energy and angular momentum
are conserved.
Sol. : Wgravity
= KE
0– GMm
PEr
.
46. Answer (2)
Hint: In a compound algebraic sum of oxidation
numbers of all atoms is zero.
Sol. : NH3 x + 3(+1) = 0 x = – 3
HNO3 (+1) + x + 3(–2) = 0 x = + 5
N2 x = 0
N2H
4 2(x) + 4(+1) = 0 x = – 2
N3H 3x + 1 = 0 x = –1/3
–3 –1/3 0 5
2 33 3NH < N H < N < HNO
47. Answer (4)
Hint: Balance the reaction by ion electron method.
Sol. : 2+ 2– + 3+ 3
2 7 26Fe + Cr O + 14H 6Fe 2Cr 7H O
Sum of coefficient of reactants = 6 + 1 + 14 = 21
48. Answer (2)
Hint: Ozone is better oxidising agent than H2O
2.
Sol.: O3 O
2 + [O]
H2O
2 + [O] H
2O + O
2
H2O
2 is oxidised here.
49. Answer (3)
Hint: When O.N. of an element in a compound is
intermediate, then compound can act as oxidant and
reductant both.
[ CHEMISTRY]
Sol. : HNO2 Oxidation number of N = +3
H2O
2 Oxidation number of O = –1
H2S Oxidation number of S = –2
SO2
Oxidation number of S = +4
50. Answer (4)
Hint: 2– 2– + 3+
2 7 2 4 2 2Cr O +3C O + 14H 2Cr +7H O + 6CO
Sol. :
1 mole Cr2O
72– requires = 3 mole C
2O
42–
So, 1 mole C2O
42–
requires 2–
2 7
1mole Cr O
3
51. Answer (2)
Hint: Sum of the oxidation numbers of all atoms in
a compound is zero.
Sol. : K4[Fe(CN)
6] 4(+1) + x + 6(–1) = 0
x = +2
Fe (CO)5
x + 5(0) = 0 x = 0
[Fe(H2O)
6]Cl
3 x + 6 x (0) + 3 x (–1) = 0
x = +3
52. Answer (4)
Hint: Electrode with higher o
redE value will act as
cathode.
Test - 4 (Code-C) (Answers & Hints) All India Aakash Test Series for Medical-2020
7/16
53. Answer (4)
Hint: o o ocell cathode anodeE E E
Sol. : o o ocell cathode anodeE E E = 2 2
o o
Cu /Cu Zn /ZnE E
= +0.34 – (–0.76) = 1.10 V
54. Answer (2)
Hint: Sum of O.N. of all atoms in a compound is
zero.
Sol. : A2 (BC
4)3
[2(+3) + 3 [(+6) + 4(–2)] = 0
55. Answer (2)
Hint: Ksp
= s2
Sol. : –3
0.07175 10s
143.5
= 5 × 10–7 mol L–1
–
ss
AgCl Ag Cl �
Ksp
= s × s = s²
= (5 × 10–7)2 = 2.5 × 10–13 mol2 L–2
56. Answer (4)
Hint: For sparingly soluble salt, Ksp
<< 1.
Sol. : 1
2spAB s= K
1
3sp2
KAB s=
4
⎛ ⎞ ⎜ ⎟
⎝ ⎠
1
4sp3
KAB s=
27
⎛ ⎞ ⎜ ⎟
⎝ ⎠
1
5sp2 3
KA B s=
108
⎛ ⎞ ⎜ ⎟
⎝ ⎠
57. Answer (4)
Hint: With increase in OH concentration, pH will
increase.
Sol. :
(i)
1 12 100 1 100
10 10N
200
= 0.05 [H+]
So, pH < 7
(ii)
1 12 100 1 100
10 5N200
= 0 [Neutral]
So, pH = 7
(iii)
1 12 100 1 100
5 5N
200
+200.1(H )
200
So, pH 7
(iv)
1 11 100 2 100
2 10N
200
–
300.15(OH )
200
So, pH > 7
58. Answer (3)
Hint: When equations are added, then their
equilibrium constants are multiplied
Sol. : A � 2B K1 ...(i)
B � C+D K2 ...(ii)
2C + 2D � E K3 ...(iii)
2 (ii) + (i) + (iii)
A � E, Keq
= K1 K
22 K
3,
On reversing, E � A
eq'K =
2
1 2 3
1
K K K
59. Answer (4)
Hint: wh
a b
KK =
K K
Sol. : Salt of (WA + WB) will have higher KH value.
60. Answer (3)
Hint: Acidic buffer contains weak acid & its salt with
strong base.
Sol. : HClO4 is a strong acid.
61. Answer (2)
Hint: Larger is the Keq
, greater is the extent of
reaction.
Sol. : If Keq
<< 1, then (Reactant) >> (Product)
62. Answer (4)
Hint:
2w
h
a
K chK =
K 1 h
,
Sol. : On increasing the concentration of salt, ‘h’
decreases. On adding base, common ion effect is
applied therefore, h decreases and on increasing
temperature Kw and K
a changes.
63. Answer (3)
Hint: a b
1 1pH 7 pK pK
2 2
Sol. : pH of salt of weak acid and weak base is
concentration independent.
All India Aakash Test Series for Medical-2020 Test - 4 (Code-C) (Answers & Hints)
8/16
64. Answer (3)
Hint: In presence of common ion solubility decreases
while in complex formation solubility increases.
Sol. : More the concentration of common ion, lesser
will be the solubility.
3 4 1 2 3 3
(Complex formation) No common ion
0.1 M NH (s ) Water (s ) 0.1 M HCl (s ) 0.2 M AgNO (s )
65. Answer (3)
Hint : G° = –2.303RTlogKeq
Sol.: G° = –2.303 × R × 298 × log10
= –2.303 × 298 × R
66. Answer (2)
Hint:
½ ½2 2
c
(H ) (I )K =
(HI)
Sol. : 2 2
1 IHI H I
2 2
1 0 0
12 2
���⇀↽���
½ ½
2 2
C
(H ) (I )K =
(HI)=
½ ½
2 2
(1– )
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
2(1 )
0.5 1
2(1 0.5) 2
67. Answer (2)
Hint: Ksp
= [Ba2+][OH–]2
Sol. : 2
2Ba(OH) Ba 2OH
s 2s
���⇀↽���
Ksp
= (s) (2s)2 5.0 × 10–7 = 4s3
3 –75s = 10
4
= 1.25 × 10–7
s = 5 × 10–3
[OH–] = 2s = 2 × 5 × 10–3 = 10–2
pH = 14 – pOH = 12
68. Answer (3)
Hint: Thermoneutral reactions (H = 0) are
temperature independent.
Sol. : On increasing pressure the equilibrium is
shifted towards the lesser number of gaseous moles.
69. Answer (2)
Hint: Eq. mass = molar mass
n-factor
Sol. : 3
2
–6eN 2N
2
14 2 14Eq. mass
6 3(N )
70. Answer (3)
Hint: [H+]total
= [H+]acid
+ [H+]H2O
Sol. : pH = 5 [H+] = 10–5 M
N1 V
1 = N
2 V
2
10–5 × 1 = N2 × 1000
N2 = 10–8
Total H+ = H+ ion from acid + H+ ion from H2O
[H+] = 10–8 + 10–7 = 1.1 × 10–7
pH = –log(1.1 × 10–7) = 6.95
71. Answer (4)
Hint: [H+][OH–] = 10–14
Sol. :
[OH–] = 1
–14+ –1410
[H ] = 101
pH = 14
72. Answer (3)
Hint: Lewis acid is a substance which can accept
a pair of electrons.
Sol. : Al in AlCl3 and Sn in SnCl
4 have vacant d-
orbitals
In CO2, oxygen atoms are more
electronegative than the carbon atom so
carbon atom become electron deficient.
73. Answer (3)
Hint: Keq
depends only on temperature.
Sol. : On changing concentration Keq
remains same.
74. Answer (2)
Hint: For an exothermic reaction, on increasing
temperature Keq
decreases.
Sol. : Increasing temperature in exothermic reaction,
rate of forward & backward both reaction increases,
but rate of backward reaction increases more.
Test - 4 (Code-C) (Answers & Hints) All India Aakash Test Series for Medical-2020
9/16
75. Answer (2)
Hint: Active mass of solid substance remains
constant.
Sol. : 3 2
p NH H SK (P )(P )
On adding 3
3 NHNH ,P increases and
2H S
P decreases.
76. Answer (1)
Hint: Conjugate base is obtained by removing a
proton of the species.
Sol. : 2
4SO
is conjugate base of 4
HSO .
77. Answer (3)
Hint: PB = P
C and P
E = P
F.
Sol. : A(s) B(g) C(g)���⇀↽���
PB = P
C P
B + P
C = 80 atm
PB = P
C = 40 atm
1P
K 40 40 = 1600 atm2
D(s) E(g) F(g)�
PE = P
F P
E + P
F
= 40
atm
PE = P
F = 20 atm
2P
K 20 20 400 atm2
1
2
P
P
K 1600= = 4
K 400
78. Answer (2)
Hint: Electron deficient species are Lewis acid.
Sol. : B2H
6 is a Lewis acid.
79. Answer (4)
Hint: w
1pH = pK
2
Sol. : 14
w
1 1pH = pK log(2 10 ) 6.85
2 2
80. Answer (4)
Hint. : 4
Acidic Salt Salt Salt
(SA WB) (SB WA)
pH 7 pH 7 pH 7 pH 7
HCl NH Cl NaCl NaCN
81. Answer (3)
Hint: Applicable only for conjugate acid & base pair.
Sol. : HCOOH – Acid
HCOO– – Conjugate base
82. Answer (3)
Hint: In presence of common ion solubility of salt
decreases.
Sol. : + –
+ –
Ag CN Ag + CN
s s
KCN K + CN
0.02 M 0.02 M
�
Ksp
= [Ag+] [CN–]
1.0×10–16 = (s) (s+0.02) s(0.02)�
–161.0 10
s =0.02
= 5 × 10–15 M
83. Answer (3)
Hint: For precipitation, Ksp
< Ionic product (IP)
Sol. :
(i)
–4 –6 –10–102 10 10 10
I.P 0.5 102 2 2
⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
(ii)
–5 –5–1010 2 10
I.P 0.5 102 2
⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
(iii)
–4 –4–1010 10
I.P 25 102 2
⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
84. Answer (1)
Hint: For an acidic buffer solution a weak acid its
salt with strong base is taken.
Sol. : (H3PO
4 + NaH
2PO
4), (NaH
2PO
4 + Na
2HPO
4)
and (Na2HPO
4 + Na
3PO
4)
85. Answer (3)
Hint: Salt of weak acid & strong base undergoes
anionic hydrolysis.
Sol. : HCOOH, CH3COOH and HCN are weak acid.
86. Answer (1)
Hint: pH of acidic buffer is given by
a
saltpH pK log
acid
⎛ ⎞ ⎜ ⎟⎝ ⎠
Sol. :
Initial, meq,
final, meq, = 100
==
All India Aakash Test Series for Medical-2020 Test - 4 (Code-C) (Answers & Hints)
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[ BIOLOGY]
91. Answer (3)
Sol. : DNA sequence, chemical nature of proteins,
crystals & aromatic compounds are used in
chemotaxonomy by scientists to resolve confusions
in classification.
92. Answer (3)
Hint : Natural system of classification is based on
morphology, anatomy, embryology & phytochemistry.
Sol. : Phylogenetic system of classification includes
evolutionary relationships between organisms, as a
basis of classification.
93. Answer (4)
Hint : Oogamous type of sexual reproduction
involves fusion between non-motile female gamete
and motile male gamete in some green and brown
algae.
Sol. : Ulothrix and Spirogyra have isogamous type of
reproduction whereas Volvox and Polysiphonia have
oogamous reproduction but Polysiphonia has non-
motile male gametes.
94. Answer (4)
Hint : Porphyra is a red alga.
Sol. : The stored food of red algae is floridean starch.
95. Answer (1)
Hint : Peptidoglycan is constituent of bacterial cell
wall.
Sol. : In chlorophyceae outer layer of cell wall is
made up of pectose, while inner layer is constituted
by cellulose.
96. Answer (4)
Hint : Pear shaped zoospores & pyriform motile
gametes are found in brown algae.
Sol. : Brown algae are found mostly in marine
habitats.
a
SaltpH pK log
Acid
⎛ ⎞ ⎜ ⎟⎝ ⎠
–3
3
100 10
800= 4.74 log
200 10
800
⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠
= 4.74 – 0.30 = 4.44
87. Answer (1)
Hint: KP = K
C (RT)ng
Sol. : KP = K
C (RT)ng
– nC
P
K(RT) g
K
(i)1C
P
K(RT)
K
(ii)C
P
K1
K
(iii)–1C
P
K= (RT)
K
(iv)–2C
P
K= (RT)
K
88. Answer (4)
Hint: Adding inert gas at constant volume has no
effect on equilibrium.
89. Answer (4)
Hint: eq
H llogK logA
2.303R T
⎛ ⎞ ⎜ ⎟⎝ ⎠
Sol. : For exothermic reaction.
Hslope ( )ve
2.303R
log Keq
1/T
90. Answer (1)
Hint: 1 1 2 2
1 2
N V + N VN =
V + V
Sol. : pH = 2 pH = 3
N1 = 10–2 N
2 = 10–3
–2 –310 × V +10 × V
N =V + V
0.011=
2= 55 × 10–4
4pH = – log(55 10 ) 2.26
Test - 4 (Code-C) (Answers & Hints) All India Aakash Test Series for Medical-2020
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97. Answer (3)
Hint : Funaria is a moss.
Sol. : In mosses, leafy gametophytes have
multicellular and branched rhizoids.
98. Answer (3)
Hint : In liverworts, sporophytes are diploid
structures which develop from embryo and are
dependent on gametophytes.
Sol. : All bryophytes, including Marchantia are
homosporous.
99. Answer (4)
Sol. : A - Dictyota (Brown alga)
B - Porphyra (Red alga)
C - Selaginella (Pteridophyte)
D - Salvinia (Aquatic fern)
100. Answer (3)
Sol. : Bryophytes have gametophytic main plant body.
Pteridophytes, gymnosperms and angiosperms have
sporophytic main plant body.
101. Answer (3)
Hint : In bryophytes, the sporophyte is parasitic over
gametophyte.
Sol. : Members of pteridophytes may be
homosporous (one kind of spores) or heterosporous
(two kinds of spores).
102. Answer (3)
Hint : Adiantum is a fern which belongs to the class
Pteropsida.
Sol. : Psilopsida – Psilotum
Lycopsida – Selaginella
Sphenopsida – Equisetum
103. Answer (4)
Hint : Angiosperms are called flowering plants.
Sol. : Occurrence of triploid endosperm, double
fertilisation and presence of fruits are exclusive
features of angiosperms.
104. Answer (3)
Hint : Bryophytes have haplo-diplontic life cycle.
Sol. : Fucus, Cycas and mango all have diplontic life
cycle. Funaria being a bryophyte has haplo-diplontic
life cycle.
105. Answer (3)
Hint : It is group of antiparasitic drugs that expel
parasitic worms, i.e., helminths hence called
anthelmintics or vermifuges.
Sol. : Dryopteris is used to obtain anthelmintic drug
to treat helmenthiasis.
106. Answer (4)
Hint : Red wood tree (Sequoia) – tallest gymnosperm
Sol. :
Polysiphonia shows haplo-diplontic life cycle
pattern whereas Volvox has haplontic life cycle
pattern.
Statements (a) & (d) are correct.
107. Answer (2)
Hint : Stems are usually branched in orders
Coniferales and Gnetales of gymnosperms.
Sol. : Cedrus – branched stem.
Cycas – unbranched stem.
108. Answer (4)
Hint : Pteridophytes are vascular plants with
dominant diploid sporophyte.
Sol. : In pteridophytes, the gametophyte is short-
lived, haploid and independent.
109. Answer (2)
Hint : Gymnosperms have archegonia in their ovule.
Sol. : Pinus has 2-8 archegonia inside its ovule,
which is absent in the ovule of papaya as it is an
angiosperm.
110. Answer (4)
Hint :
Cycas
Eucalyptus
(gymnosperm) = nEndosperm in
(angiosperm) = 3n
⎡⎢⎣
Sol. : Wolfia is an angiosperm, within ovules of
which are present highly reduced female
gametophytes i.e., embryo sacs.
111. Answer (3)
Hint : All bryophytes are called amphibians of plant
kingdom.
Sol. : Mosses like Funaria, have sporophyte which
is partially dependent on gametophyte.
112. Answer (2)
Sol. : Primary protonema is a filamentous structure
develops from spore germination in Funaria.
113. Answer (3)
Sol. : Antherozoids of bryophytes are biflagellated.
114. Answer (3)
Hint : In red algae, the major pigments are chl a and
d, carotenoids and xanthophylls.
Sol. : Porphyra is a red alga, while Fucus, Dictyota
& Laminaria are brown algae.
All India Aakash Test Series for Medical-2020 Test - 4 (Code-C) (Answers & Hints)
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115. Answer (2)
Hint : Carrageen is sulphated polysaccharide.
Sol. : Sulphated polysaccharides or hydrocolloids
are found in the cell wall of rhodophyceae.
116. Answer (2)
Sol. : Brown alga – Sargassum,
Green alga – Chara,
Red alga - Gracilaria
117. Answer (3)
Hint : Pteridophytes are the first vascular
embryophytes.
Sol. : Dryopteris – Pteridophyte
Funaria – Bryophyte
Cycas – Gymnosperm
Mangifera - Angiosperm
118. Answer (4)
Sol. : Gametophytes of mosses have multicellular
and branched rhizoids.
119. Answer (3)
Hint : Homosporous pteridophytes have monoecious
gametophyte.
Sol. : Heterospory in pteridophytes is precursor to
the seed habit.
120. Answer (3)
Hint : Coralloid roots of Cycas are irregular & do not
possess root hairs.
Sol. : Coralloid roots of Cycas are symbiotically
associated to Anabaena (cyanobacteria).
121. Answer (3)
Hint : Gymnosperms have integumented ovules.
Sol. : In ovules of gymnosperms, nucellus is
protected by single integument.
122. Answer (2)
Hint : Endosperm represents female gametophyte in
gymnospermic seed.
Sol. :
Plumule, radicle, suspensor & cotyledons
represent future sporophyte.
Testa, tegmen & perisperm represent parental
sporophyte.
123. Answer (3)
Hint : Ephedra, Gnetum & Welwitschia are placed in
the most advanced order of gymnosperms.
Sol. : Ephedra is placed in order Gnetales.
124. Answer (1)
Sol. : Tonoplast, membrane of sap vacuole is a
selectively permeable membrane.
125. Answer (2)
Hint : Water/solvent always moves from lower DPD
to higher DPD of solution.
Sol. : OP = 13TP = 5DPD = 8
AOP = 6TP = 2DPD = 4
B
Low DPDHigh DPD
126. Answer (4)
Hint : Most of the absorbed water is lost through
stomata of the leaves.
Sol. : Less than 1 percent of water, reaching the
leaves is used in photosynthesis & plant growth.
127. Answer (3)
Hint : Elements most readily mobilised are
phosphorus, nitrogen & potassium.
Sol. :
Mobile elements are frequently remobilised from
older parts to younger parts.
A C4 plant loses only half as much water as a
C3 plant for the same amount of CO
2 fixed.
128. Answer (3)
Hint : Water in the adjacent xylem moves into the
phloem by osmosis.
Sol. : Sucrose transport in the living phloem sieve
tube cells occurs through osmotic pressure gradient.
Loading of sucrose into sieve tube cells is an active
process.
129. Answer (2)
Hint : Water potential is chemical potential of water
molecules which regulates water movement as water
always moves from high water potenitial to low water
potential.
Sol. : Water potential of a cell = solute potential +
pressure potential.
Water potential of pure water is zero at atmospheric
pressure but by adding solutes it decreases and
become negative.
130. Answer (3)
Hint : In both facilitated diffusion and active transport,
movement of molecules occur with the help of
membrane proteins.
Sol. : Both facilitated diffusion and active transport,
show saturation of the transport and response to
protein inhibitors.
Test - 4 (Code-C) (Answers & Hints) All India Aakash Test Series for Medical-2020
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Facilitated diffusion is a downhill process as it does
not require ATP while active transport is an uphill
process.
131. Answer (3)
Hint : Osmotic pressure of an electrolyte is higher
than a non-electrolyte.
Sol. : A cell which does not show any change in
0.7 M sugar solution because it is isotonic, would
show decreased volume of cell due to 0.7 M NaCl
solution by exosmosis. As 0.7 M NaCl solution due
to dissociation has double osmotic pressure than
0.7 M sugar solution, hence become hypertonic in
comparison to cytoplasm of cell.
132. Answer (2)
Hint : In dry atmosphere when the relative humidity
is low, a plant has higher rate of transpiration.
Sol. : In humid atmosphere i.e., at high relative
humidity, the rate of transpiration decreases.
133. Answer (3)
Hint : Opening of stomata starts with incomplete
oxidation of carbohydrate.
Sol. :
Step I – In light, starch in guard cells is
incompletely oxidised into phosphoenol pyruvate
(PEP).
Step II – H+ from guard cells are transported to the
epidermal cells.
Step III – K+ ions are absorbed into guard cell.
Step IV – Increased K+ & malate ion form
potassium malate & store it in vacuoles.
134. Answer (3)
Hint : The chief sinks for the mineral elements are
the growing regions of the plants.
Sol. : Mature leaves are the chief source of
mobilised minerals while young leaves, developing
flowers, fruits, seeds, apical & lateral meristems are
the chief sinks for the mineral elements.
135. Answer (4)
Hint : Guttation is the process which occurs in early
morning and night while transpiration occurs in day
time.
Sol. : Due to transpiration a negative pressure is
generated in xylem vessels while guttation occurs
due to positive root pressure.
136. Answer (2)
Hint : The structure is responsible for propulsion/
whip like movement in tail of male gamete.
Sol. : Male gamete called sperm in humans is
flagellated. Flagella is the propulsion equipment that
pushes the sperm towards the ovum. Amoeboid
movement shown by leucocytes involves
pseudopodia formation.
137. Answer (4)
Hint : Filaments of this protein run close to the
F-actin throughout its length.
Sol. : Troponin - Tropomyosin complex masks the
myosin binding site on actin. Troponin binds to
calcium which results in unmasking of myosin
binding site on actin.
138. Answer (4)
Hint : Leaky channels are always open.
Sol. : Efflux of K+ is essential for repolarisation,
which occurs through opening of K+ voltage gated
channels.
139. Answer (3)
Hint : All muscle fibres contain actin and myosin.
Sol. : Smooth muscle fibres appear non-striated as
actin and myosin filaments are not regularly arrayed
along the length of the cell.
140. Answer (3)
Hint : Identify an enzyme.
Sol. : Acetylcholinesterase breaks acetylcholine in
synaptic cleft into acetate and choline.
141. Answer (4)
Hint : Actin filaments slide towards the M-line when
a sarcomere contracts.
Sol. : Thin and thick filaments in a myofibre are
composed of actin and myosin respectively. Thin
filaments slide over thick filaments during muscle
contraction.
142. Answer (2)
Hint : Myosin head has more affinity for ATP than for
actin.
Sol. : Cross bridge between actin and myosin will
not break until a new ATP attaches to myosin head.
143. Answer (2)
Hint : Dark meat is rich in myoglobin.
Sol. : White meat has fast contracting, glycolytic/
anaerobic white muscle fibres. They have more
developed sarcoplasmic reticulum than red muscle
fibres. Red muscle fibres rich in myoglobin have more
mitochondria than white muscle fibres.
All India Aakash Test Series for Medical-2020 Test - 4 (Code-C) (Answers & Hints)
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144. Answer (2)
Hint : Extension is observed in contractile fibres.
Sol. : Nerve fibres do not exhibit contractility and
elasticity.
145. Answer (4)
Hint : A common collagenous connective tissue
layer holds together a number of muscle bundles.
Sol. : Each muscle bundle is called fascicle. Many
fascicles are together wrapped around by fascia.
Sarcolemma is plasma membrane of a muscle fibre.
Endomysium is connective tissue covering of
individual muscle fibre.
146 Answer (2)
Hint : Deficiency of estrogen adds to the increased
chances of fracture in this case.
Sol. : Bones become weak due to demineralisation
caused by enhanced activity of bone dissolving cells
called osteoclasts.
Rickets is observed in children in case of calcium
deficiency. Accumulation of uric acid crystals in
joints is seen in gouty arthritis while rheumatoid
arthritis is an autoimmune disorder.
147. Answer (3)
Hint : In humans, the 11th and 12th pair of ribs are
not attached ventrally to sternum.
Sol. : Some mammals like sloth and manatee have
9 and 6 cervical vertebrae respectively. Cranium in
humans is composed of eight bones. There are
usually 14 phalanges in a limb of a human.
148. Answer (3)
Hint : Muscles associated with heart are striped.
Sol. : All types of muscles have actin and myosin.
Both skeletal and cardiac muscle fibres are striated/
striped. Longer refractory period belongs to cardiac
muscle fibres.
149. Answer (4)
Hint : These are most common type of synovial
joints in humans.
Sol. : In humans, carpo-carpal and tarso-tarsal joints
are gliding joints; atlas-axis joint is a pivot joint; saddle
joint is present between carpal and metacarpal of
thumb; cartilaginous joints are found in between
vertebrae.
150. Answer (4)
Hint : Sternum is placed on same side as human
heart.
Sol. : Sternum is placed ventrally in the human body.
Myasthenia gravis affects skeletal muscles of the
human body.
151. Answer (2)
Hint : Select the triangular flat bone that is also
called shoulder bone.
Sol. : Scapula and clavicle of both side together form
pectoral girdle. Pelvic girdle involves innominate/coxal
bones.
152. Answer (3)
Hint : 8 carpals are arranged equally in two rows in
one wrist.
Sol.: Cranium has 8 bones in humans fixed together
by sutures. Carpals interact with each other through
gliding joints. Carpals are 16 while tarsals are 14 in
number. 2 coxal bones form pelvic girdle while 4
bones form pectoral girdle. 26 vertebrae occur in
vertebral column of an adult and 30 bones are
present in an upper limb of man.
153. Answer (1)
Hint : It is a supporting bone of lower limb.
Sol. : Fibula does not interact with femur at knee
joint. Fibula interacts with tibia at two points.
154. Answer (3)
Hint : Hammer, anvil and stirrup shaped bones are
part of middle ear in man.
Sol. : Incus is anvil shaped bone of middle ear that
interacts with stapes and malleus. Human skull has
two occipital condyles through which it interacts with
atlas vertebra.
155. Answer (3)
Hint : Coordination is the process through which two
or more organs interact and complement the
functions of one another.
Sol. : The neural system provides an organised
network of point to point connections for a quick
coordination. The endocrine system provides
chemical coordination through hormones.
156. Answer (4)
Hint : The neural organisation is very simple in lower
invertebrates.
Sol. : Poriferans, simplest invertebrates lack neural
system. In coelenterates such as Hydra, it appeared
for the first time as a network of neurons.
157. Answer (2)
Hint : Neurons respond to threshold stimulus
generated externally or internally in body.
Sol. : Neurons do not produce stimulus. They detect,
receive and transmit the stimulus.
Test - 4 (Code-C) (Answers & Hints) All India Aakash Test Series for Medical-2020
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158. Answer (1)
Hint : These neuroglial cells form myelin sheath in
PNS.
Sol. : Neurilemma is formed by Schwann cells by
wrapping only one time around unmyelinated
neurons. Oligodendrocytes form myelin sheath in
CNS. Microglial cells are macrophages of neural
system.
159. Answer (3)
Hint : These neurons contain one afferent and one
efferent process.
Sol. : Bipolar neurons contain one dendrite (afferent)
and one axon (one efferent) process. Unipolar
neurons are found in embryonic stage, while
multipolar neurons are found in CNS of adult human.
Dorsal root ganglion of spinal cord has
pseudounipolar neurons.
160. Answer (2)
Hint : Unconditioned reflex is inborn and not an
acquired reaction.
Sol. : Watering of mouth on seeing food is
conditioned reflex as prior exposure has already
occurred and memory of taste has been formed.
161. Answer (2)
Hint : Extensor muscle ‘Quadriceps’ is involved.
Sol. : Bending of knee involves the flexor muscle
Hamstring. This is a stretch reflex where the lower
leg moves outwards and upwards. Interneurons are
absent. Patella is the knee bone/cap involved.
162. Answer (4)
Hint : It appears like sea horse in a cross section
of forebrain.
Sol. : Hippocampus is a part of a complex structure
called limbic system of forebrain. Hippocampus
converts short term memory to long term memory in
man. Amygdala is also a part of limbic system
involved in emotional reactions such as anger and
rage.
163. Answer (3)
Hint : Canal of midbrain. This duct is also known as
‘‘Aqueduct of Sylvius’’.
Sol. : Iter/cerebral aqueduct canal is a part of the
midbrain through which CSF flows.
164. Answer (1)
Hint : These ions are essential for impulse generation
and transmission.
Sol. : Voltage gated Na+ channels opens upon
reaching threshold stimulus. Opening of calcium ion
channels promotes exocytosis of neurotransmitter
rich synaptic vesicles in synaptic cleft.
165. Answer (1)
Hint : Communication channels are present in
synapses where size of synaptic cleft is smaller
than 6 nm.
Sol. : Acetylcholine is a neurotransmitter in a
chemical synapse. In case of electrical synapse,
impulse jumps from pre-synaptic membrane to
postsynaptic membrane rapidly through gap
junctions.
166. Answer (2)
Hint : This centre is also responsible for
cardiovascular reflex and gastric secretions.
Sol. : Medulla is responsible for setting respiratory
rhythm. Thirst, hunger, satiety and body temperature
are regulated by hypothalamus.
167. Answer (3)
Hint : This tract of nerve fibres is a part of forebrain.
Sol. : Cerebral hemispheres are connected by
corpus callosum. Round swellings in midbrain form
corpora quadrigemina. Cerebellum is also divided into
two hemispheres.
168. Answer (3)
Hint : Gray matter of brain lies closer to pia mater
Sol. : Orientation of white and gray matter is
reversed in spinal cord in comparision to brain.
Hence, white matter is closest to pia mater followed
by arachnoid and outermost dura mater.
169. Answer (4)
Hint : Total change in potential difference across a
neuronal surface upon receiving threshold stimulus.
Sol. : Resting potential of neurons is around -70 mV.
If the overshoot after depolarisation is +40 mV, spike
potential = (70 + 40) = 110 mV.
170. Answer (3)
Hint : ECF contains 30 times less K+ than
axoplasm while ECF contains 10 times more Na+
than ICF.
Sol. : Potential difference around -70 mV exists
across axolemma due to differential distribution of
ions. Neural membrane is impermeable to movement
of negatively charged proteins. The exterior surface
of axolemma is positively charged.
171. Answer (1)
Hint : These granules are associated with non
dividing cells of neural tissue.
Sol. : Nissl’s granules are absent in axon of neurons
and glial cells.
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� � �
172. Answer (4)
Hint : Presence of nodes of Ranvier promote
saltatory and consequently faster conduction of
impulse.
Sol. : Myelin sheath is an effective insulator around
axons. Impulse jumps from one node to another at
node of Ranvier. Threshold stimulus triggers the
impulse but is not directly responsible for its
conduction in a nerve fibre. Increase in temperature
increases speed of impulse conduction.
173. Answer (4)
Hint : This part acts as relay center.
Sol. : Amygdala in forebrain is responsible for anger
and rage.
174. Answer (2)
Hint : These are the apertures in wall of 4th ventricle.
Sol. : Each lateral ventricle is connected to third
ventricle by an interventricular foramen (Foramen of
Monro). The third ventricle is connected by iter to the
4th ventricle.
175. Answer (1)
Hint : This lobe is recognised as seat of intelligence
and creativity.
Sol. : Damage to motor speech area i.e, Broca's
area can result in aphasia.
176. Answer (2)
Hint : The neurotransmitter involved is a catecholamine
and derivative of tyrosine.
Sol. : Schizophrenia is a personality disorder
resulting from excessive secretion of dopamine.
177. Answer (4)
Hint : Function of autonomic nervous system is to
control and coordinate the activities of visceral
organs.
Sol. : Parasympathetic branch of nervous system
promotes secretion of intestinal juice.
178. Answer (2)
Hint : In conditions of fear, flight and fight
sympathetic system is activated.
Sol. : In emergencies, stimulation of sympathetic
branch of nervous system promotes relaxation of
breathing pathways. Parasympathetic system
hinders breathing and narrows respiratory pathways
by promoting bronchoconstriction. Therefore,
inhibition of bronchoconstrictions occurs as effect of
sympathetic branch.
179. Answer (3)
Hint : This part is associated with maintaining body
equilibrium.
Sol. : Purkinje cells are a feature of cerebellum part
of hindbrain.
180. Answer (1)
Hint : Transport of 3Na+ and 2K+ occurs in different
directions at the same time by the pump.
Sol. : One ATP is spent by Na+/K+ ATPase.
Test - 4 (Code-D) (Answers) All India Aakash Test Series for Medical-2020
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1. (3)
2. (2)
3. (4)
4. (4)
5. (2)
6. (2)
7. (3)
8. (3)
9. (2)
10. (3)
11. (4)
12. (1)
13. (3)
14. (4)
15. (4)
16. (2)
17. (4)
18. (4)
19. (2)
20. (4)
21. (3)
22. (3)
23. (Delete)
24. (2)
25. (2)
26. (4)
27. (3)
28. (2)
29. (2)
30. (2)
31. (2)
32. (1)
33. (3)
34. (2)
35. (4)
36. (3)
Test Date : 30/12/2018
ANSWERS
TEST - 4 (Code-D)
All India Aakash Test Series for Medical-2020
37. (4)
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124. (3)
125. (3)
126. (3)
127. (4)
128. (3)
129. (3)
130. (4)
131. (1)
132. (4)
133. (4)
134. (3)
135. (3)
136. (1)
137. (3)
138. (2)
139. (4)
140. (2)
141. (1)
142. (2)
143. (4)
144. (4)
145. (1)
146. (3)
147. (4)
148. (3)
149. (3)
150. (2)
151. (1)
152. (1)
153. (3)
154. (4)
155. (2)
156. (2)
157. (3)
158. (1)
159. (2)
160. (4)
161. (3)
162. (3)
163. (1)
164. (3)
165. (2)
166. (4)
167. (4)
168. (3)
169. (3)
170. (2)
171. (4)
172. (2)
173. (2)
174. (2)
175. (4)
176. (3)
177. (3)
178. (4)
179. (4)
180. (2)
All India Aakash Test Series for Medical-2020 Test - 4 (Code-D) (Answers & Hints)
2/16
ANSWERS & HINTS
1. Answer (3)
Hint : Mechanical energy and angular momentum
are conserved.
Sol. : Wgravity
= KE
0– GMm
PEr
.
2. Answer (2)
Hint : vara = v
prp
Sol. : vara = v
prp
2 2
a a p pr r
2
pa
p a
r
r
⎛ ⎞ ⎜ ⎟ ⎝ ⎠
ra = a + ae r
p = a – ae
2
1
1–
p
a
e
e
⎛ ⎞ ⎜ ⎟ ⎝ ⎠
3. Answer (4)
Hint : 1n
Fr
Sol. : 2
k mvF
r r
v r°
4. Answer (4)
Hint : 2
1
⎛ ⎞⎜ ⎟⎝ ⎠
h
gg
h
R
Sol. : gh = g /2
2
1 1
21
h
R
⎛ ⎞⎜ ⎟⎝ ⎠
2 – 1h R
5. Answer (2)
Hint : L r p �
� �
Sol. : L = m0vr
r = R + h
[ PHYSICS]
0
GMv
r
0 0( ) ( )
GML m R h m GM R h
R h
6. Answer (2)
Hint : –f i
E E E
Sol. : –
2
GMmE
r
1 1 1 1– –
2 2 8 14i f E E
E GMm GMmr r R R
⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦
7 – 4 3
56 56
⎡ ⎤ ⎢ ⎥
⎣ ⎦E E
GMmGMm
R R
7. Answer (3)
Hint : 2 2 2
interstellar–1
E Ev v v v n
Sol. : v = nvE
n = 3
infinity 9 –1 8 2 2 E E E
v v v v
8. Answer (3)
Hint : Conservation of mechanical energy.
Sol. : 2 2
0
1 1– –
2 2 ( )
GMm GMm
mv mvR R h
2 2
0
1 1–
2 21
mghmv mv
h
R
2
0
2–
1
⎛ ⎞⎜ ⎟⎝ ⎠
ghv v
h
R
9. Answer (2)
Hint : orbital
1V
r .
Sol. : Orbital speed is independent of mass of
satellite.
10. Answer (3)
Hint : g = g – R2 cos 2Sol. : = 90° for pole g
pole = g(always) if = 0,
ge = g i.e., increases.
Test - 4 (Code-D) (Answers & Hints) All India Aakash Test Series for Medical-2020
3/16
11. Answer (4)
Hint and Sol.: Total energy of bound system is
negative.
12. Answer (1)
Hint : Angular momentum is conserved.
Sol. : r is minimum at point A and maximum at
point C.
So, KA > K
B > K
C
13. Answer (3)
Hint : 2
E
GMv
R
Sol. : 32 4
3
E
Gv R
R
28
3
Ev GR
E
v R
1
2 2
E E E
p p p
v R
v R
14. Answer (4)
Hint : Conservation of mechanical energy.
Sol. :
H
v = 0
u
U1 + K
1 = U
2 + K
2
20 0
0
– –10
2
GMm GMmm u
R R H
Now, 2 2 0
0 0
1 1
2 4 2E
GMmm u m v
R
H = R
15. Answer (4)
Hint : Both particles rotate about their centre of mass.
Sol. : Both particles are always opposite to each
other along the line joining them so 1 =
2.
16. Answer (2)
Hint : rot total
1for a disc.
3K KE
Sol. : Total KE = mg h = mgl sin.
So, rot
1sin
3K mgl .
17. Answer (4)
Hint and Sol. : MI is same for any symmetrical part
of body.
I1 = I2 = I318. Answer (4)
Hint and Sol. : I = mr2.
19. Answer (2)
Hint : Mechanical energy is conserved.
Sol. :
1 = 0
U1 + K
1 = U
2 + K
2
210 cos
2 2 2
L LMg mg I
3(1– cos )
g
L
3
2 g
L
20. Answer (4)
Hint : Use Newton’s law of gravitation.
Sol.: To make M in equilibrium resultant of F1 and
F2 should be equal in magnitude and opposite in
direction to T.
∴ From the figure,
1
2
tan30F
F 30°
30°
F1
F2
m1
T
m2
∴
1
2
1
3
m
m
21. Answer (3)
Hint : , d d
dt dt
Sol. : = 3t 2 – 4t + 3
6 – 4d
tdt
= 6 rad/s2
at = R, a
r = R2
At t = 1 s = 2 rad/s
at = 6 m/s2 a
r = 4 m/s2
2 2 2
net6 4 52 m/sa
All India Aakash Test Series for Medical-2020 Test - 4 (Code-D) (Answers & Hints)
4/16
22. Answer (3)
Hint : net r t
a a a � � �
Sol. :
ar
v
at
anet
23. Answer (Delete)
24. Answer (2)
Hint and Sol. : I = mk2.
25. Answer (2)
Hint : Fraction of kinetic energy which is rotational
2
2 2.
k
k R
Sol. :
2
2
2
21
k
Rf
k
R
for solid sphere 2
2
2
5
k
R .
2 / 5% 100 28%
21
5
26. Answer (4)
Hint : Use parallel axis theorem.
Sol. : L/2
rL
O
2 2
2
3 12
ML MLI Mr
2
2 5
4
Lr
2
5
3 ML
I
27. Answer (3)
Hint : .P ��
Sol. : ˆ ˆ ˆ( – 3 2 ) Nmi j k �
ˆ ˆ ˆ(4 2 5 ) rad/si j k �
ˆ ˆ ˆ ˆ ˆ ˆ( – 3 2 ) (4 2 5 )P i j k i j k P = 4 – 6 + 10
P = 8 W
28. Answer (2)
Hint : L r p �
� �
Sol. :
v
y
x
v cos
r H =
L = rp
r = H
L = H mv cos
vL mv
g
2 2
sincos
2
mv
g
3 2
sin cos
2
29. Answer (2)
Hint : and � �
� � ��
L r p r F
Sol. : Net torque about C is non zero and net
torque about O is zero.
∴ L is conserve about O but variable about C.
30. Answer (2)
Hint and Sol. : Kepler’s law of area is based on
conservation of angular momentum.
31. Answer (2)
Hint : Conservation of angular momentum about the
point of contact.
Sol. :
Pure rolling
0
v
Applying conservation of angular momentum about
the point of contact,
Li = L
f
mR20 = mR2 + mvR
2 0
02
2
RmR mvR v
⇒
Test - 4 (Code-D) (Answers & Hints) All India Aakash Test Series for Medical-2020
5/16
32. Answer (1)
Hint : Conservation of angular momentum.
Sol. : Li = L
f
Li = mr2
Lf = (mr2 + 2mr2)
2 23mr mr
3
33. Answer (3)
Hint : 2
21
kt
R
⎛ ⎞ ⎜ ⎟
⎝ ⎠
Sol. : 2
2
k
Ris same for disc and solid cylinder.
34. Answer (2)
Hint : In pure rolling on fixed surface net tangential
acceleration of point of contact is zero.
Sol. :
Ov
R2
Rv
For pure rolling with constant angular velocity.
v = R
Radial acceleration of point is R2.
35. Answer (4)
Hint : �
� �
L r p
Sol. : �
L is always prependicular to �r and
�
p .
36. Answer (3)
Hint : Mechanical energy is conserved.
Sol. : In pure rolling on inclined plane potential
energy is converted into translational and rotational
kinetic energy. Net work done by friction and normal
force is zero.
37. Answer (4)
Hint : = I
Sol. : = I
I = mk 2
= mk 2
38. Answer (4)
Hint : Formula for moment of inertia of any
symmetrical part of a body = Formula for moment of
inertia of body.
Sol. : 2
Semidisc Disc2
MRI I
39. Answer (2)
Hint : vp = 2v
0 cos (/2).
Sol. :
O60° v
0
r v = 0
120°
Here = 120°
vnet
= 2v0cos60°
vnet
= v0
40. Answer (2)
Hint : If ext
= 0 then L = constant
If ext
F����
= 0 the cm
0a �
Sol. : For a rigid body rotating about a fixed axis L
is always parallel to .
Torque produced by central force is always zero.
41. Answer (3)
Hint : r F �
��
Sol. :
P(1, 2)
O(0, 0)
r
ˆ4F k
ˆ ˆ(– – 2 ) mr i j�
ˆ4 �
F k
�
��
r F
ˆ ˆ4( 2 ) �
i j Nm
42. Answer (3)
Hint : L r p �
� �
Sol. : L�
about O is constant as r⊥ and v both are
constant.
All India Aakash Test Series for Medical-2020 Test - 4 (Code-D) (Answers & Hints)
6/16
46. Answer (1)
Hint: 1 1 2 2
1 2
N V + N VN =
V + V
Sol. : pH = 2 pH = 3
N1 = 10–2 N
2 = 10–3
–2 –310 × V +10 × V
N =V + V
0.011=
2= 55 × 10–4
4pH = – log(55 10 ) 2.26 47. Answer (4)
Hint: eq
H llogK logA
2.303R T
⎛ ⎞ ⎜ ⎟⎝ ⎠
Sol. : For exothermic reaction.
Hslope ( )ve
2.303R
log Keq
1/T
48. Answer (4)
Hint: Adding inert gas at constant volume has no
effect on equilibrium.
[ CHEMISTRY]
49. Answer (1)
Hint: KP = K
C (RT)ng
Sol. : KP = K
C (RT)ng
– nC
P
K(RT) g
K
(i)1C
P
K(RT)
K
(ii)C
P
K1
K
(iii)–1C
P
K= (RT)
K
(iv)–2C
P
K= (RT)
K
50. Answer (1)
Hint: pH of acidic buffer is given by
a
saltpH pK log
acid
⎛ ⎞ ⎜ ⎟⎝ ⎠
Sol. :
Initial, meq,
final, meq, = 100
==
43. Answer (4)
Hint : F + f = macm
Sol. :
f
R
F
Ov
F + f = macm
...(i)
cm
= Icm
(F – f)R = mR2
or F – f = mR = macm
...(ii) (∵ acm
= R)
From (i) & (ii)
f = 0
44. Answer (2)
Hint : I = mr2
Sol. : Moment of inertia depends on axis of rotation,
mass of the body and distribution of mass about
axis of rotation.
45. Answer (4)
Hint : = mgx
Sol. :
60°
u
x
y
xomg
At t = 2 s
1cos 20 2 20 m
2 x u t
= mgx = 2 × 10 × 20 = 400 Nm
Test - 4 (Code-D) (Answers & Hints) All India Aakash Test Series for Medical-2020
7/16
a
SaltpH pK log
Acid
⎛ ⎞ ⎜ ⎟⎝ ⎠
–3
3
100 10
800= 4.74 log
200 10
800
⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠
= 4.74 – 0.30 = 4.44
51. Answer (3)
Hint: Salt of weak acid & strong base undergoes
anionic hydrolysis.
Sol. : HCOOH, CH3COOH and HCN are weak acid.
52. Answer (1)
Hint: For an acidic buffer solution a weak acid its
salt with strong base is taken.
Sol. : (H3PO
4 + NaH
2PO
4), (NaH
2PO
4 + Na
2HPO
4)
and (Na2HPO
4 + Na
3PO
4)
53. Answer (3)
Hint: For precipitation, Ksp
< Ionic product (IP)
Sol. :
(i)–4 –6 –10
–102 10 10 10I.P 0.5 10
2 2 2
⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
(ii)–5 –5
–1010 2 10I.P 0.5 10
2 2
⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
(iii)–4 –4
–1010 10I.P 25 10
2 2
⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
54. Answer (3)
Hint: In presence of common ion solubility of salt
decreases.
Sol. : + –
+ –
Ag CN Ag + CN
s s
KCN K + CN
0.02 M 0.02 M
�
Ksp
= [Ag+] [CN–]
1.0×10–16 = (s) (s+0.02) s(0.02)�
–161.0 10
s =0.02
= 5 × 10–15 M
55. Answer (3)
Hint: Applicable only for conjugate acid & base pair.
Sol. : HCOOH – Acid
HCOO– – Conjugate base
56. Answer (4)
Hint. : 4
Acidic Salt Salt Salt
(SA WB) (SB WA)
pH 7 pH 7 pH 7 pH 7
HCl NH Cl NaCl NaCN
57. Answer (4)
Hint: w
1pH = pK
2
Sol. : 14
w
1 1pH = pK log(2 10 ) 6.85
2 2
58. Answer (2)
Hint: Electron deficient species are Lewis acid.
Sol. : B2H
6 is a Lewis acid.
59. Answer (3)
Hint: PB = P
C and P
E = P
F.
Sol. : A(s) B(g) C(g)���⇀↽���
PB = P
C P
B + P
C = 80 atm
PB = P
C = 40 atm
1P
K 40 40 = 1600 atm2
D(s) E(g) F(g)�
PE = P
F P
E + P
F
= 40
atm
PE = P
F = 20 atm
2P
K 20 20 400 atm2
1
2
P
P
K 1600= = 4
K 400
60. Answer (1)
Hint: Conjugate base is obtained by removing a
proton of the species.
Sol. : 2
4SO
is conjugate base of 4
HSO .
61. Answer (2)
Hint: Active mass of solid substance remains
constant.
Sol. : 3 2
p NH H SK (P )(P )
On adding 3
3 NHNH , P increases and
2H S
P decreases.
62. Answer (2)
Hint: For an exothermic reaction, on increasing
temperature Keq
decreases.
Sol. : Increasing temperature in exothermic reaction,
rate of forward & backward both reaction increases,
but rate of backward reaction increases more.
All India Aakash Test Series for Medical-2020 Test - 4 (Code-D) (Answers & Hints)
8/16
63. Answer (3)
Hint: Keq
depends only on temperature.
Sol. : On changing concentration Keq
remains same.
64. Answer (3)
Hint: Lewis acid is a substance which can accept
a pair of electrons.
Sol. : Al in AlCl3 and Sn in SnCl
4 have vacant d-
orbitals
In CO2, oxygen atoms are more
electronegative than the carbon atom so
carbon atom become electron deficient.
65. Answer (4)
Hint: [H+][OH–] = 10–14
Sol. :
[OH–] = 1
–14+ –1410
[H ] = 101
pH = 14
66. Answer (3)
Hint: [H+]total
= [H+]acid
+ [H+]H2O
Sol. : pH = 5 [H+] = 10–5 M
N1 V
1 = N
2 V
2
10–5 × 1 = N2 × 1000
N2 = 10–8
Total H+ = H+ ion from acid + H+ ion from H2O
[H+] = 10–8 + 10–7 = 1.1 × 10–7
pH = –log(1.1 × 10–7) = 6.95
67. Answer (2)
Hint: Eq. mass = molar mass
n-factor
Sol. : 3
2
–6eN 2N
2
14 2 14Eq. mass
6 3(N )
68. Answer (3)
Hint: Thermoneutral reactions (H = 0) are
temperature independent.
Sol. : On increasing pressure the equilibrium is
shifted towards the lesser number of gaseous moles.
69. Answer (2)
Hint: Ksp
= [Ba2+][OH–]2
Sol. : 22Ba(OH) Ba 2OH
s 2s
���⇀↽���
Ksp
= (s) (2s)2 5.0 × 10–7 = 4s3
3 –75s = 10
4
= 1.25 × 10–7
s = 5 × 10–3
[OH–] = 2s = 2 × 5 × 10–3 = 10–2
pH = 14 – pOH = 12
70. Answer (2)
Hint:
½ ½2 2
c
(H ) (I )K =
(HI)
Sol. : 2 2
1 IHI H I
2 2
1 0 0
12 2
���⇀↽���
½ ½
2 2
C
(H ) (I )K =
(HI)=
½ ½
2 2
(1– )
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
2(1 )
0.5 1
2(1 0.5) 2
71. Answer (3)
Hint : G° = –2.303RTlogKeq
Sol.: G° = –2.303 × R × 298 × log10
= –2.303 × 298 × R
72. Answer (3)
Hint: In presence of common ion solubility decreases
while in complex formation solubility increases.
Sol. : More the concentration of common ion, lesser
will be the solubility.
3 4 1 2 3 3
(Complex formation) No common ion
0.1 M NH (s ) Water (s ) 0.1 M HCl (s ) 0.2 M AgNO (s )
73. Answer (3)
Hint: a b
1 1pH 7 pK pK
2 2
Sol. : pH of salt of weak acid and weak base is
concentration independent.
Test - 4 (Code-D) (Answers & Hints) All India Aakash Test Series for Medical-2020
9/16
74. Answer (4)
Hint:
2w
h
a
K chK =
K 1 h
,
Sol. : On increasing the concentration of salt, ‘h’
decreases. On adding base, common ion effect is
applied therefore, h decreases and on increasing
temperature Kw and K
a changes.
75. Answer (2)
Hint: Larger is the Keq
, greater is the extent of
reaction.
Sol. : If Keq
<< 1, then (Reactant) >> (Product)
76. Answer (3)
Hint: Acidic buffer contains weak acid & its salt with
strong base.
Sol. : HClO4 is a strong acid.
77. Answer (4)
Hint: wh
a b
KK =
K K
Sol. : Salt of (WA + WB) will have higher KH value.
78. Answer (3)
Hint: When equations are added, then their
equilibrium constants are multiplied
Sol. : A � 2B K1 ...(i)
B � C+D K2 ...(ii)
2C + 2D � E K3 ...(iii)
2 (ii) + (i) + (iii)
A � E, Keq
= K1 K
22 K
3,
On reversing, E � A
eq'K =
2
1 2 3
1
K K K
79. Answer (4)
Hint: With increase in OH concentration, pH will
increase.
Sol. :
(i)
1 12 100 1 100
10 10N
200
= 0.05 [H+]
So, pH < 7
(ii)
1 12 100 1 100
10 5N200
= 0 [Neutral]
So, pH = 7
(iii)
1 12 100 1 100
5 5N
200
+200.1(H )
200
So, pH 7
(iv)
1 11 100 2 100
2 10N
200
–
300.15(OH )
200
So, pH > 7
80. Answer (4)
Hint: For sparingly soluble salt, Ksp
<< 1.
Sol. : 1
2spAB s= K
1
3sp2
KAB s=
4
⎛ ⎞ ⎜ ⎟
⎝ ⎠
1
4sp3
KAB s=
27
⎛ ⎞ ⎜ ⎟
⎝ ⎠
1
5sp2 3
KA B s=
108
⎛ ⎞ ⎜ ⎟
⎝ ⎠
81. Answer (2)
Hint: Ksp
= s2
Sol. : –3
0.07175 10s
143.5
= 5 × 10–7 mol L–1
–
ss
AgCl Ag Cl �
Ksp
= s × s = s²
= (5 × 10–7)2 = 2.5 × 10–13 mol2 L–2
82. Answer (2)
Hint: Sum of O.N. of all atoms in a compound is
zero.
Sol. : A2 (BC
4)3
[2(+3) + 3 [(+6) + 4(–2)] = 0
83. Answer (4)
Hint: o o ocell cathode anodeE E E
Sol. : o o ocell cathode anodeE E E = 2 2
o o
Cu /Cu Zn /ZnE E
= +0.34 – (–0.76) = 1.10 V
84. Answer (4)
Hint: Electrode with higher o
redE value will act as
cathode.
All India Aakash Test Series for Medical-2020 Test - 4 (Code-D) (Answers & Hints)
10/16
[ BIOLOGY]
91. Answer (4)
Hint : Guttation is the process which occurs in early
morning and night while transpiration occurs in day
time.
Sol. : Due to transpiration a negative pressure is
generated in xylem vessels while guttation occurs
due to positive root pressure.
92. Answer (3)
Hint : The chief sinks for the mineral elements are
the growing regions of the plants.
Sol. : Mature leaves are the chief source of
mobilised minerals while young leaves, developing
flowers, fruits, seeds, apical & lateral meristems are
the chief sinks for the mineral elements.
93. Answer (3)
Hint : Opening of stomata starts with incomplete
oxidation of carbohydrate.
Sol. :
Step I – In light, starch in guard cells is
incompletely oxidised into phosphoenol pyruvate
(PEP).
Step II – H+ from guard cells are transported to the
epidermal cells.
Step III – K+ ions are absorbed into guard cell.
Step IV – Increased K+ & malate ion form
potassium malate & store it in vacuoles.
94. Answer (2)
Hint : In dry atmosphere when the relative humidity
is low, a plant has higher rate of transpiration.
Sol. : In humid atmosphere i.e., at high relative
humidity, the rate of transpiration decreases.
95. Answer (3)
Hint : Osmotic pressure of an electrolyte is higher
than a non-electrolyte.
Sol. : A cell which does not show any change in
0.7 M sugar solution because it is isotonic, would
show decreased volume of cell due to 0.7 M NaCl
solution by exosmosis. As 0.7 M NaCl solution due
to dissociation has double osmotic pressure than
0.7 M sugar solution, hence become hypertonic in
comparison to cytoplasm of cell.
85. Answer (2)
Hint: Sum of the oxidation numbers of all atoms in
a compound is zero.
Sol. : K4[Fe(CN)
6] 4(+1) + x + 6(–1) = 0
x = +2
Fe (CO)5
x + 5(0) = 0 x = 0
[Fe(H2O)
6]Cl
3 x + 6 x (0) + 3 x (–1) = 0
x = +3
86. Answer (4)
Hint: 2– 2– + 3+
2 7 2 4 2 2Cr O +3C O + 14H 2Cr +7H O + 6CO
Sol. :
1 mole Cr2O
72– requires = 3 mole C
2O
42–
So, 1 mole C2O
42–
requires 2–
2 7
1mole Cr O
3
87. Answer (3)
Hint: When O.N. of an element in a compound is
intermediate, then compound can act as oxidant and
reductant both.
Sol. : HNO2 Oxidation number of N = +3
H2O
2 Oxidation number of O = –1
H2S Oxidation number of S = –2
SO2
Oxidation number of S = +4
88. Answer (2)
Hint: Ozone is better oxidising agent than H2O
2.
Sol.: O3 O
2 + [O]
H2O
2 + [O] H
2O + O
2
H2O
2 is oxidised here.
89. Answer (4)
Hint: Balance the reaction by ion electron method.
Sol. : 2+ 2– + 3+ 3
2 7 26Fe + Cr O + 14H 6Fe 2Cr 7H O
Sum of coefficient of reactants = 6 + 1 + 14 = 21
90. Answer (2)
Hint: In a compound algebraic sum of oxidation
numbers of all atoms is zero.
Sol. : NH3 x + 3(+1) = 0 x = – 3
HNO3 (+1) + x + 3(–2) = 0 x = + 5
N2 x = 0
N2H
4 2(x) + 4(+1) = 0 x = – 2
N3H 3x + 1 = 0 x = –1/3
–3 –1/3 0 5
2 33 3NH < N H < N < HNO
Test - 4 (Code-D) (Answers & Hints) All India Aakash Test Series for Medical-2020
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96. Answer (3)
Hint : In both facilitated diffusion and active transport,
movement of molecules occur with the help of
membrane proteins.
Sol. : Both facilitated diffusion and active transport,
show saturation of the transport and response to
protein inhibitors.
Facilitated diffusion is a downhill process as it does
not require ATP while active transport is an uphill
process.
97. Answer (2)
Hint : Water potential is chemical potential of water
molecules which regulates water movement as water
always moves from high water potenitial to low water
potential.
Sol. : Water potential of a cell = solute potential +
pressure potential.
Water potential of pure water is zero at atmospheric
pressure but by adding solutes it decreases and
become negative.
98. Answer (3)
Hint : Water in the adjacent xylem moves into the
phloem by osmosis.
Sol. : Sucrose transport in the living phloem sieve
tube cells occurs through osmotic pressure gradient.
Loading of sucrose into sieve tube cells is an active
process.
99. Answer (3)
Hint : Elements most readily mobilised are
phosphorus, nitrogen & potassium.
Sol. :
Mobile elements are frequently remobilised from
older parts to younger parts.
A C4 plant loses only half as much water as a
C3 plant for the same amount of CO
2 fixed.
100. Answer (4)
Hint : Most of the absorbed water is lost through
stomata of the leaves.
Sol. : Less than 1 percent of water, reaching the
leaves is used in photosynthesis & plant growth.
101. Answer (2)
Hint : Water/solvent always moves from lower DPD
to higher DPD of solution.
Sol. : OP = 13TP = 5DPD = 8
AOP = 6TP = 2DPD = 4
B
Low DPDHigh DPD
102. Answer (1)
Sol. : Tonoplast, membrane of sap vacuole is a
selectively permeable membrane.
103. Answer (3)
Hint : Ephedra, Gnetum & Welwitschia are placed in
the most advanced order of gymnosperms.
Sol. : Ephedra is placed in order Gnetales.
104. Answer (2)
Hint : Endosperm represents female gametophyte in
gymnospermic seed.
Sol. :
Plumule, radicle, suspensor & cotyledons
represent future sporophyte.
Testa, tegmen & perisperm represent parental
sporophyte.
105. Answer (3)
Hint : Gymnosperms have integumented ovules.
Sol. : In ovules of gymnosperms, nucellus is
protected by single integument.
106. Answer (3)
Hint : Coralloid roots of Cycas are irregular & do not
possess root hairs.
Sol. : Coralloid roots of Cycas are symbiotically
associated to Anabaena (cyanobacteria).
107. Answer (3)
Hint : Homosporous pteridophytes have monoecious
gametophyte.
Sol. : Heterospory in pteridophytes is precursor to
the seed habit.
108. Answer (4)
Sol. : Gametophytes of mosses have multicellular
and branched rhizoids.
109. Answer (3)
Hint : Pteridophytes are the first vascular
embryophytes.
Sol. : Dryopteris – Pteridophyte
Funaria – Bryophyte
Cycas – Gymnosperm
Mangifera - Angiosperm
110. Answer (2)
Sol. : Brown alga – Sargassum,
Green alga – Chara,
Red alga - Gracilaria
111. Answer (2)
Hint : Carrageen is sulphated polysaccharide.
Sol. : Sulphated polysaccharides or hydrocolloids
are found in the cell wall of rhodophyceae.
All India Aakash Test Series for Medical-2020 Test - 4 (Code-D) (Answers & Hints)
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112. Answer (3)
Hint : In red algae, the major pigments are chl a and
d, carotenoids and xanthophylls.
Sol. : Porphyra is a red alga, while Fucus, Dictyota
& Laminaria are brown algae.
113. Answer (3)
Sol. : Antherozoids of bryophytes are biflagellated.
114. Answer (2)
Sol. : Primary protonema is a filamentous structure
develops from spore germination in Funaria.
115. Answer (3)
Hint : All bryophytes are called amphibians of plant
kingdom.
Sol. : Mosses like Funaria, have sporophyte which
is partially dependent on gametophyte.
116. Answer (4)
Hint :
Cycas
Eucalyptus
(gymnosperm) = nEndosperm in
(angiosperm) = 3n
⎡⎢⎣
Sol. : Wolfia is an angiosperm, within ovules of
which are present highly reduced female
gametophytes i.e., embryo sacs.
117. Answer (2)
Hint : Gymnosperms have archegonia in their ovule.
Sol. : Pinus has 2-8 archegonia inside its ovule,
which is absent in the ovule of papaya as it is an
angiosperm.
118. Answer (4)
Hint : Pteridophytes are vascular plants with
dominant diploid sporophyte.
Sol. : In pteridophytes, the gametophyte is short-
lived, haploid and independent.
119. Answer (2)
Hint : Stems are usually branched in orders
Coniferales and Gnetales of gymnosperms.
Sol. : Cedrus – branched stem.
Cycas – unbranched stem.
120. Answer (4)
Hint : Red wood tree (Sequoia) – tallest gymnosperm
Sol. :
Polysiphonia shows haplo-diplontic life cycle
pattern whereas Volvox has haplontic life cycle
pattern.
Statements (a) & (d) are correct.
121. Answer (3)
Hint : It is group of antiparasitic drugs that expel
parasitic worms, i.e., helminths hence called
anthelmintics or vermifuges.
Sol. : Dryopteris is used to obtain anthelmintic drug
to treat helmenthiasis.
122. Answer (3)
Hint : Bryophytes have haplo-diplontic life cycle.
Sol. : Fucus, Cycas and mango all have diplontic life
cycle. Funaria being a bryophyte has haplo-diplontic
life cycle.
123. Answer (4)
Hint : Angiosperms are called flowering plants.
Sol. : Occurrence of triploid endosperm, double
fertilisation and presence of fruits are exclusive
features of angiosperms.
124. Answer (3)
Hint : Adiantum is a fern which belongs to the class
Pteropsida.
Sol. : Psilopsida – Psilotum
Lycopsida – Selaginella
Sphenopsida – Equisetum
125. Answer (3)
Hint : In bryophytes, the sporophyte is parasitic over
gametophyte.
Sol. : Members of pteridophytes may be
homosporous (one kind of spores) or heterosporous
(two kinds of spores).
126. Answer (3)
Sol. : Bryophytes have gametophytic main plant body.
Pteridophytes, gymnosperms and angiosperms have
sporophytic main plant body.
127. Answer (4)
Sol. : A - Dictyota (Brown alga)
B - Porphyra (Red alga)
C - Selaginella (Pteridophyte)
D - Salvinia (Aquatic fern)
128. Answer (3)
Hint : In liverworts, sporophytes are diploid
structures which develop from embryo and are
dependent on gametophytes.
Sol. : All bryophytes, including Marchantia are
homosporous.
129. Answer (3)
Hint : Funaria is a moss.
Sol. : In mosses, leafy gametophytes have
multicellular and branched rhizoids.
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130. Answer (4)
Hint : Pear shaped zoospores & pyriform motile
gametes are found in brown algae.
Sol. : Brown algae are found mostly in marine
habitats.
131. Answer (1)
Hint : Peptidoglycan is constituent of bacterial cell
wall.
Sol. : In chlorophyceae outer layer of cell wall is
made up of pectose, while inner layer is constituted
by cellulose.
132. Answer (4)
Hint : Porphyra is a red alga.
Sol. : The stored food of red algae is floridean starch.
133. Answer (4)
Hint : Oogamous type of sexual reproduction
involves fusion between non-motile female gamete
and motile male gamete in some green and brown
algae.
Sol. : Ulothrix and Spirogyra have isogamous type of
reproduction whereas Volvox and Polysiphonia have
oogamous reproduction but Polysiphonia has non-
motile male gametes.
134. Answer (3)
Hint : Natural system of classification is based on
morphology, anatomy, embryology & phytochemistry.
Sol. : Phylogenetic system of classification includes
evolutionary relationships between organisms, as a
basis of classification.
135. Answer (3)
Sol. : DNA sequence, chemical nature of proteins,
crystals & aromatic compounds are used in
chemotaxonomy by scientists to resolve confusions
in classification.
136. Answer (1)
Hint : Transport of 3Na+ and 2K+ occurs in different
directions at the same time by the pump.
Sol. : One ATP is spent by Na+/K+ ATPase.
137. Answer (3)
Hint : This part is associated with maintaining body
equilibrium.
Sol. : Purkinje cells are a feature of cerebellum part
of hindbrain.
138. Answer (2)
Hint : In conditions of fear, flight and fight
sympathetic system is activated.
Sol. : In emergencies, stimulation of sympathetic
branch of nervous system promotes relaxation of
breathing pathways. Parasympathetic system
hinders breathing and narrows respiratory pathways
by promoting bronchoconstriction. Therefore,
inhibition of bronchoconstrictions occurs as effect of
sympathetic branch.
139. Answer (4)
Hint : Function of autonomic nervous system is to
control and coordinate the activities of visceral
organs.
Sol. : Parasympathetic branch of nervous system
promotes secretion of intestinal juice.
140. Answer (2)
Hint : The neurotransmitter involved is a catecholamine
and derivative of tyrosine.
Sol. : Schizophrenia is a personality disorder
resulting from excessive secretion of dopamine.
141. Answer (1)
Hint : This lobe is recognised as seat of intelligence
and creativity.
Sol. : Damage to motor speech area i.e, Broca's
area can result in aphasia.
142. Answer (2)
Hint : These are the apertures in wall of 4th ventricle.
Sol. : Each lateral ventricle is connected to third
ventricle by an interventricular foramen (Foramen of
Monro). The third ventricle is connected by iter to the
4th ventricle.
143. Answer (4)
Hint : This part acts as relay center.
Sol. : Amygdala in forebrain is responsible for anger
and rage.
144. Answer (4)
Hint : Presence of nodes of Ranvier promote
saltatory and consequently faster conduction of
impulse.
Sol. : Myelin sheath is an effective insulator around
axons. Impulse jumps from one node to another at
node of Ranvier. Threshold stimulus triggers the
impulse but is not directly responsible for its
conduction in a nerve fibre. Increase in temperature
increases speed of impulse conduction.
145. Answer (1)
Hint : These granules are associated with non
dividing cells of neural tissue.
Sol. : Nissl’s granules are absent in axon of neurons
and glial cells.
146. Answer (3)
Hint : ECF contains 30 times less K+ than
axoplasm while ECF contains 10 times more Na+
than ICF.
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Sol. : Potential difference around -70 mV exists
across axolemma due to differential distribution of
ions. Neural membrane is impermeable to movement
of negatively charged proteins. The exterior surface
of axolemma is positively charged.
147. Answer (4)
Hint : Total change in potential difference across a
neuronal surface upon receiving threshold stimulus.
Sol. : Resting potential of neurons is around -70 mV.
If the overshoot after depolarisation is +40 mV, spike
potential = (70 + 40) = 110 mV.
148. Answer (3)
Hint : Gray matter of brain lies closer to pia mater
Sol. : Orientation of white and gray matter is
reversed in spinal cord in comparision to brain.
Hence, white matter is closest to pia mater followed
by arachnoid and outermost dura mater.
149. Answer (3)
Hint : This tract of nerve fibres is a part of forebrain.
Sol. : Cerebral hemispheres are connected by
corpus callosum. Round swellings in midbrain form
corpora quadrigemina. Cerebellum is also divided into
two hemispheres.
150. Answer (2)
Hint : This centre is also responsible for
cardiovascular reflex and gastric secretions.
Sol. : Medulla is responsible for setting respiratory
rhythm. Thirst, hunger, satiety and body temperature
are regulated by hypothalamus.
151. Answer (1)
Hint : Communication channels are present in
synapses where size of synaptic cleft is smaller
than 6 nm.
Sol. : Acetylcholine is a neurotransmitter in a
chemical synapse. In case of electrical synapse,
impulse jumps from pre-synaptic membrane to
postsynaptic membrane rapidly through gap
junctions.
152. Answer (1)
Hint : These ions are essential for impulse generation
and transmission.
Sol. : Voltage gated Na+ channels opens upon
reaching threshold stimulus. Opening of calcium ion
channels promotes exocytosis of neurotransmitter
rich synaptic vesicles in synaptic cleft.
153. Answer (3)
Hint : Canal of midbrain. This duct is also known as
‘‘Aqueduct of Sylvius’’.
Sol. : Iter/cerebral aqueduct canal is a part of the
midbrain through which CSF flows.
154. Answer (4)
Hint : It appears like sea horse in a cross section
of forebrain.
Sol. : Hippocampus is a part of a complex structure
called limbic system of forebrain. Hippocampus
converts short term memory to long term memory in
man. Amygdala is also a part of limbic system
involved in emotional reactions such as anger and
rage.
155. Answer (2)
Hint : Extensor muscle ‘Quadriceps’ is involved.
Sol. : Bending of knee involves the flexor muscle
Hamstring. This is a stretch reflex where the lower
leg moves outwards and upwards. Interneurons are
absent. Patella is the knee bone/cap involved.
156. Answer (2)
Hint : Unconditioned reflex is inborn and not an
acquired reaction.
Sol. : Watering of mouth on seeing food is
conditioned reflex as prior exposure has already
occurred and memory of taste has been formed.
157. Answer (3)
Hint : These neurons contain one afferent and one
efferent process.
Sol. : Bipolar neurons contain one dendrite (afferent)
and one axon (one efferent) process. Unipolar
neurons are found in embryonic stage, while
multipolar neurons are found in CNS of adult human.
Dorsal root ganglion of spinal cord has
pseudounipolar neurons.
158. Answer (1)
Hint : These neuroglial cells form myelin sheath in
PNS.
Sol. : Neurilemma is formed by Schwann cells by
wrapping only one time around unmyelinated
neurons. Oligodendrocytes form myelin sheath in
CNS. Microglial cells are macrophages of neural
system.
159. Answer (2)
Hint : Neurons respond to threshold stimulus
generated externally or internally in body.
Sol. : Neurons do not produce stimulus. They detect,
receive and transmit the stimulus.
160. Answer (4)
Hint : The neural organisation is very simple in lower
invertebrates.
Sol. : Poriferans, simplest invertebrates lack neural
system. In coelenterates such as Hydra, it appeared
for the first time as a network of neurons.
Test - 4 (Code-D) (Answers & Hints) All India Aakash Test Series for Medical-2020
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161. Answer (3)
Hint : Coordination is the process through which two
or more organs interact and complement the
functions of one another.
Sol. : The neural system provides an organised
network of point to point connections for a quick
coordination. The endocrine system provides
chemical coordination through hormones.
162. Answer (3)
Hint : Hammer, anvil and stirrup shaped bones are
part of middle ear in man.
Sol. : Incus is anvil shaped bone of middle ear that
interacts with stapes and malleus. Human skull has
two occipital condyles through which it interacts with
atlas vertebra.
163. Answer (1)
Hint : It is a supporting bone of lower limb.
Sol. : Fibula does not interact with femur at knee
joint. Fibula interacts with tibia at two points.
164. Answer (3)
Hint : 8 carpals are arranged equally in two rows in
one wrist.
Sol.: Cranium has 8 bones in humans fixed together
by sutures. Carpals interact with each other through
gliding joints. Carpals are 16 while tarsals are 14 in
number. 2 coxal bones form pelvic girdle while 4
bones form pectoral girdle. 26 vertebrae occur in
vertebral column of an adult and 30 bones are
present in an upper limb of man.
165. Answer (2)
Hint : Select the triangular flat bone that is also
called shoulder bone.
Sol. : Scapula and clavicle of both side together form
pectoral girdle. Pelvic girdle involves innominate/coxal
bones.
166. Answer (4)
Hint : Sternum is placed on same side as human
heart.
Sol. : Sternum is placed ventrally in the human body.
Myasthenia gravis affects skeletal muscles of the
human body.
167. Answer (4)
Hint : These are most common type of synovial
joints in humans.
Sol. : In humans, carpo-carpal and tarso-tarsal joints
are gliding joints; atlas-axis joint is a pivot joint; saddle
joint is present between carpal and metacarpal of
thumb; cartilaginous joints are found in between
vertebrae.
168. Answer (3)
Hint : Muscles associated with heart are striped.
Sol. : All types of muscles have actin and myosin.
Both skeletal and cardiac muscle fibres are striated/
striped. Longer refractory period belongs to cardiac
muscle fibres.
169. Answer (3)
Hint : In humans, the 11th and 12th pair of ribs are
not attached ventrally to sternum.
Sol. : Some mammals like sloth and manatee have
9 and 6 cervical vertebrae respectively. Cranium in
humans is composed of eight bones. There are
usually 14 phalanges in a limb of a human.
170 Answer (2)
Hint : Deficiency of estrogen adds to the increased
chances of fracture in this case.
Sol. : Bones become weak due to demineralisation
caused by enhanced activity of bone dissolving cells
called osteoclasts.
Rickets is observed in children in case of calcium
deficiency. Accumulation of uric acid crystals in
joints is seen in gouty arthritis while rheumatoid
arthritis is an autoimmune disorder.
171. Answer (4)
Hint : A common collagenous connective tissue
layer holds together a number of muscle bundles.
Sol. : Each muscle bundle is called fascicle. Many
fascicles are together wrapped around by fascia.
Sarcolemma is plasma membrane of a muscle fibre.
Endomysium is connective tissue covering of
individual muscle fibre.
172. Answer (2)
Hint : Extension is observed in contractile fibres.
Sol. : Nerve fibres do not exhibit contractility and
elasticity.
173. Answer (2)
Hint : Dark meat is rich in myoglobin.
Sol. : White meat has fast contracting, glycolytic/
anaerobic white muscle fibres. They have more
developed sarcoplasmic reticulum than red muscle
fibres. Red muscle fibres rich in myoglobin have more
mitochondria than white muscle fibres.
174. Answer (2)
Hint : Myosin head has more affinity for ATP than for
actin.
Sol. : Cross bridge between actin and myosin will
not break until a new ATP attaches to myosin head.
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175. Answer (4)
Hint : Actin filaments slide towards the M-line when
a sarcomere contracts.
Sol. : Thin and thick filaments in a myofibre are
composed of actin and myosin respectively. Thin
filaments slide over thick filaments during muscle
contraction.
176. Answer (3)
Hint : Identify an enzyme.
Sol. : Acetylcholinesterase breaks acetylcholine in
synaptic cleft into acetate and choline.
177. Answer (3)
Hint : All muscle fibres contain actin and myosin.
Sol. : Smooth muscle fibres appear non-striated as
actin and myosin filaments are not regularly arrayed
along the length of the cell.
178. Answer (4)
Hint : Leaky channels are always open.
Sol. : Efflux of K+ is essential for repolarisation,
which occurs through opening of K+ voltage gated
channels.
179. Answer (4)
Hint : Filaments of this protein run close to the
F-actin throughout its length.
Sol. : Troponin - Tropomyosin complex masks the
myosin binding site on actin. Troponin binds to
calcium which results in unmasking of myosin
binding site on actin.
180. Answer (2)
Hint : The structure is responsible for propulsion/
whip like movement in tail of male gamete.
Sol. : Male gamete called sperm in humans is
flagellated. Flagella is the propulsion equipment that
pushes the sperm towards the ovum. Amoeboid
movement shown by leucocytes involves
pseudopodia formation.