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Test - 3 (Code C) (Answers & Hints) All India Aakash Test Series for Medical-2019
1/12
1. (3)
2. (3)
3. (2)
4. (2)
5. (4)
6. (2)
7. (2)
8. (4)
9. (1)
10. (2)
11. (2)
12. (3)
13. (2)
14. (2)
15. (4)
16. (1)
17. (4)
18. (3)
19. (2)
20. (1)
21. (4)
22. (3)
23. (1)
24. (1)
25. (3)
26. (1)
27. (4)
28. (3)
29. (1)
30. (4)
31. (1)
32. (1)
33. (3)
34. (3)
35. (2)
36. (2)
Test Date : 03/12/2017
ANSWERS
TEST - 3 (Code C)
All India Aakash Test Series for Medical - 2019
37. (2)
38. (2)
39. (2)
40. (3)
41. (1)
42. (1)
43. (2)
44. (2)
45. (4)
46. (3)
47. (1)
48. (4)
49. (1)
50. (4)
51. (1)
52. (1)
53. (1)
54. (3)
55. (1)
56. (4)
57. (4)
58. (2)
59. (4)
60. (1)
61. (1)
62. (3)
63. (Deleted)
64. (3)
65. (4)
66. (3)
67. (4)
68. (1)
69. (2)
70. (3)
71. (3)
72. (4)
73. (4)
74. (1)
75. (1)
76. (2)
77. (4)
78. (4)
79. (2)
80. (2)
81. (3)
82. (3)
83. (4)
84. (2)
85. (3)
86. (1)
87. (2)
88. (2)
89. (3)
90. (2)
91. (3)
92. (4)
93. (4)
94. (2)
95. (2)
96. (3)
97. (4)
98. (4)
99. (1)
100. (2)
101. (4)
102. (1)
103. (4)
104. (4)
105. (4)
106. (4)
107. (1)
108. (3)
109. (3)
110. (2)
111. (3)
112. (1)
113. (4)
114. (4)
115. (3)
116. (4)
117. (1)
118. (4)
119. (2)
120. (2)
121. (2)
122. (2)
123. (4)
124. (1)
125. (1)
126. (3)
127. (3)
128. (4)
129. (2)
130. (4)
131. (3)
132. (1)
133. (4)
134. (3)
135. (3)
136. (2)
137. (1)
138. (3)
139. (3)
140. (3)
141. (4)
142. (2)
143. (4)
144. (4)
145. (3)
146. (3)
147. (2)
148. (1)
149. (2)
150. (3)
151. (2)
152. (2)
153. (3)
154. (2)
155. (3)
156. (3)
157. (3)
158. (2)
159. (3)
160. (2)
161. (4)
162. (3)
163. (1)
164. (4)
165. (2)
166. (3)
167. (3)
168. (4)
169. (2)
170. (4)
171. (2)
172. (4)
173. (4)
174. (2)
175. (1)
176. (4)
177. (3)
178. (3)
179. (3)
180. (4)
All India Aakash Test Series for Medical-2019 Test - 3 (Code C) (Answers & Hints)
2/12
ANSWERS & HINTS
1. Answer (3)
As downward acceleration of elevator a > g, block
loses the contact with the elevator and its
acceleration is g downward. So speed of block at
t = 1 s,
v = u + gt
= 10 × 1 = 10 m/s
2. Answer (3)
When trolley is sliding with acceleration gsin, string
becomes perpendicular to the ceiling.
mgsin
mgcosmg
T
T = mgcos
3. Answer (2)
In case (I), acceleration of block in horizontal
direction
aI = (gsin30°) cos30° ...(1)
In case (II), acceleration of block in horizontal
direction is acceleration of wedge.
aII = gtan30° ...(2)
Ratio 2I
II
3cos 30
4
⎛ ⎞ ⎜ ⎟⎝ ⎠
a
a
4. Answer (2)
Acceleration 21 2
305 m/s
6
F F
aM
Mass of 10 cm part of uniform rod.
610 1.2 kg
50 m
Now,
10 cm
TF2
a
[ PHYSICS]
T – F2 = ma
T = F2 + ma
= 10 + 1.2 × 5 = 16 N
5. Answer (4)
Just before t = 1 s
Velocity v1 = +2 m/s
Just after t = 1 s
Velocity, v2 = –4 m/s
Change in velocity, v = v2 – v
1 = –6 m/s
Impulse = change in linear momentum.
= 2 × (–6) = –12 Ns
6. Answer (2)
Maximum common acceleration of two blocks, when
force is applied on the upper block.
21
max
2
0.5 20 1010 m/s
10
m g
a
m
Fmax
= (m1 + m
2) a
max = 30 × 10 = 300 N
As F < Fmax
, both blocks move together with
common acceleration.
21505 m/s
30 a
Fs = m
2a = 10 × 5 = 50 N
7. Answer (2)
Limiting force of friction.
For m1 ; (f
1)max
= 1m1g = 25 N
m2 ; (f
2)max
= 2m2g = 40 N
Maximum frictional force fmax
= 65 N
60 N
m1
T
( ) = 25 Nf1 max
As F = 60 N, both blocks will be at rest. Frictional
force on block of mass m1 be (f
1)max
. Tension in the
string.
T = 60 – 25 = 35 N
Test - 3 (Code C) (Answers & Hints) All India Aakash Test Series for Medical-2019
3/12
8. Answer (4)
2
2 2( )2
x
mT L x
L
T(x = 0) 21
2 m L
2
2 2
( 0) 4
( ) 1( )
T x L
T x x L x
4L2 – 4x
2 = L2 4x2 = 3L
2
3
2 L
x
9. Answer (1)
mgcosmg
FcosN
F
Fsin
mgsin
N
Block begins to slide when
Fcos = mgsin + N
Fcos = mgsin + (mgcos + Fsin)
F(cos – sin) = mg(sin + cos)
(sin cos )
(cos sin )
mgF
10. Answer (2)
2
12 sin 2 sin45 v g S g S ...(1)
2
22 (sin45 cos45 )
kv g S ...(2)
2
1
2
2
1 116
1511
16
k
v
v
1
2
4 :1v
v
11. Answer (2)
As 1 >
2, both blocks will start sliding together,
when
(m1 + m
2)gsin =
1m1gcos +
2m2gcos
Or (m1 + m
2)gsin = gcos(
1m1 +
2m2)
1 1 2 2
1 2
tan
m m
m m
12. Answer (3)
60°
u
v
From conservation of linear momentum
m(v cos60° + u) + Mu = 0
13. Answer (2)
mgcosmg
N
v
As block is at rest w.r.t. lift
N = mgcos ...(1)
Displacement of block w.r.t. ground
S = vt vertically upward ...(2)
Work done by normal contact force
W = N × S × cos
= (mgcos)(vt)cos
= mgvt cos2
14. Answer (2)
10 kg
kxm
100 N
N
Normal contact force is minimum when spring force
is maximum.
4 × 10 × xmax
= 21
2m
kx [K = 0]
kxm
= 80 N
Nmin
= 100 – 80 = 20 N
All India Aakash Test Series for Medical-2019 Test - 3 (Code C) (Answers & Hints)
4/12
15. Answer (4)
m1
m2
a1 a
2
2T
T T2T
(2T + T) a1 = Ta
2
16. Answer (1)
17. Answer (4)
e p2
ep
ep
p
+ve
Momentum of ball after first collision = +ep
Change in momentum of ball after first collision.
(p)1 = ep – (–p) = p(1 + e)
Similarly after second collision
(p)2 = e2p – (–ep) = ep(1 + e)
So total momentum imparted
1 2( ) ( )p p p
2(1 ) (1 ) (1 ) terms p e ep e e p e
(1 )
(1 )
p e
e
18. Answer (3)
v
90 -
u
Line of impact
ucos
Let particle strikes with speed u and moves
horizontally with speed v after the collision.
Momentum is conserved along the plane.
So vcos = usin ...(1)
and vcos(90 – ) = eucos
or, vsin = eucos ...(2)
Dividing equation (2) by equation (1)
2 2 1tan cot tan tan 30
3e e ⇒
19. Answer (2)
After collision velocity of heavy block remains the
same.
8 m/sv1
Now,
Relative velocity of separation = Relative velocity of
approach
8 – v1 = (10 – 8)
v1 = +6 m/s (in positive x direction)
20. Answer (1)
At the highest point, speed
v = 100 cos60° = 50 m/s horizontally
Conservation of linear momentum.
2ˆ ˆ2 50 1 100 1i j v
�
2
ˆ ˆ100 100v i j �
2100 2 m/sv
�
So kinetic energy of this part.
2 411 100 2 10 J
2
21. Answer (4)
A
ˆui
B2
4u g i ��
Velocity at point A.
ˆui ...(1)
Speed at the highest point.
2 24 �
Bv u g
24 �
Bv u g
Velocity 2 ˆ4 �
�B
v u g i ...(2)
Change in velocity
2 ˆ4 4 � � �
�B A
v v v u g i i
24
��v u u g
Test - 3 (Code C) (Answers & Hints) All India Aakash Test Series for Medical-2019
5/12
22. Answer (3)
F kxs =
mg
From work-energy theorem.
21
2mgx kx
2mgx
k ...(1)
Spring force at this instant
Fs = kx = 2mg
s2
upward
F mg mg mga g
m m
23. Answer (1)
Instantaneous velocity.
23 8 dx
v t tdt
...(1)
At t = 0, v1 = 3 m/s
t = 9 s, v2 = 12 m/s
W = K
2 2
2 1
1
2 m v v
2 21 1[(12) (3) ]
2 2
135J
4
24. Answer (1)
s
dvm mg
dt
25. Answer (3)
mg
hO
v
4u g �
Let string becomes slack when it makes angle
with upward vertical. Then
2
cosmv
mg �
2cosv g � ...(1)
Also,
2 2 22 2 (1 cos )v u gh u g �
or, cos 4 2 2 cosg g g g � � � �
or,2
3 cos 2 cos3
g g ⇒ � �
Tangential acceleration
5sin
3g g
26. Answer (1)
Fv at t
m
⎛ ⎞ ⎜ ⎟⎝ ⎠ ...(1)
2F
P Fv tm
P t
27. Answer (4)
m
m v/2
v
Conservation of mechanical energy
2
21 1
2 2 2
vmv m
⎛ ⎞ ⎜ ⎟⎝ ⎠
2mg mg �
�
2
5 3 12
8 2 5
mv mg gv ⇒ � �
28. Answer (3)
vx = u
x = 3 m/s
vy = eu
y =
34 3 m/s
4
Speed 2 2
3 2 m/sx y
v v v
All India Aakash Test Series for Medical-2019 Test - 3 (Code C) (Answers & Hints)
6/12
29. Answer (1)
3x
uF
x
23 m/s
x
x
Fa
m
4y
uF
y
ay = –4 m/s2
vx = a
xt = –3t; v
y = a
yt = –4t
2 25
x yv v v t
At t = 2 s, v = 5 × 2 = 10 m/s
30. Answer (4)
B
A
ar
vB
g
O
vA
2
, B
r t
va a g
R
2 2( ) ( ) r t
a a a
2 22
B Av v gR
31. Answer (1)
2
cos60 mvN mg
R
2 2(4)cos60 1 5
2
vN m g
R
⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
= 13 newton
fk =
kN = 6.5 N
N
mgmgsin60°
60°
fk
mgcos60°
Tangential acceleration sin60
kmg f
m
25 3 6.5 2.1 m/s
32. Answer (1)
O
Y
X
50,
3
⎛ ⎞⎜ ⎟⎝ ⎠
5, 0
4
⎛ ⎞⎜ ⎟⎝ ⎠
4 5
3 3Y X
5 5ˆ ˆ0 0
4 3r i j
⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
�
= 5 5ˆ ˆ
4 3i j
0r F �
�
33. Answer (3)
Area under F - t graph
= change in linear momentum
1(2 3) 10
2
mv – mu = 25
v = 25 m/s
Kinetic energy 21 625
J2 2
k mv
34. Answer (3)
Speed of ball just before the collision.
2(10) 2 10 2.2
= 12 m/s
Speed of ball just after the collision.
2 10 5 10 m/s
Coefficient of restitution
10 5
12 6e
35. Answer (2)
For equilibrium
F = 0
2x2 – x = 0 2(2x – 1) = 0
Test - 3 (Code C) (Answers & Hints) All India Aakash Test Series for Medical-2019
7/12
Either x = 0 or1
2x
4 1dF
xdx
At x = 0, 1 0dF
dx
So equilibrium is stable at x = 0.
36. Answer (2)
Kinetic energy is maximum when potential energy is
minimum.
U = 2x2 – 2 ...(1)
4dU
xdx
...(2)
For minimum, , 0
dUU
dx
x = 0
2
24
d U
dx U is minimum at x = 0
Umin
= –2 J
Now,
Kmax
+ Umin
= Emech
Kmax
– 2 = 34 2
max
136
2mv
vmax
= 6 m/s
37. Answer (2)
20 W �
�
P F v
38. Answer (2)
W = Pt = (m)gh
8000 60( ) 3200 kg
10 15m
39. Answer (2)
(REF LEVEL)
L/10
GL/5
Mass of hanging part = 5
M
5 10
⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠i
M LU g
50
MgL
Uf = 0
50 MgL
W U
40. Answer (3)
10dp
F tdt
At t = 2 s F = 20 N
41. Answer (1)
( 0)
(1)
p n muF mnu
t
⎛ ⎞ ⎜ ⎟⎝ ⎠
42. Answer (1)
fs
Maximum acceleration upto which block does not
slide on the floor
amax
= g = 0.5 × 10 = 5 m/s2
Given, a = 4 m/s2
So, fs = 2 × 4 = 8 N
Displacement of block in t = 2 s
214 (2) 8 m
2 S
wf = 8 × 8 = + 64 J
43. Answer (2)
Angle of repose R = 45°
So block does not slide
fs = mgsin37° =
310 6 N
5
44. Answer (2)
2 2
mg gmg ma a ⇒
45. Answer (4)
At the highest point �
�
F v
So 0 �
�
P F v
All India Aakash Test Series for Medical-2019 Test - 3 (Code C) (Answers & Hints)
8/12
46. Answer (3)
2 2 2 2SO Cl SO Cl
Hence this is not applicable to Dalton's law of partial
pressure.
47. Answer (1)
Charge on ionLattice energy
Size of ion
48. Answer (4)
Due to resonance in 2 2
CH CH CH , all C atoms
undergo sp2 hybridization.
49. Answer (1)
dRT dRTP M
M P ⇒
3d 2.64 gm / dm
3 1 1 3R 0.0821dm atm mole k ( 1 dm 1 litre)
∵
775P atm
760
�T 310 273 583 K
2.64 0.0821 583M 124
775 / 760
No. of P atoms in a molecule =124 124
At. Wt of P 31
= 4
50. Answer (4)
In 3 3
N(SiH )�� , N-atom donate its lone pair to vacant
d-orbital of Si to form back bonding.
51. Answer- (1)
52. Answer (1)
Due to presence of lone pair on Br in BrF5, bond
angle is not equal to 90°.
53. Answer (1)
%ionic character = 2
A B A B16(X X ) 3.5(X X )
(Hannay – Smith Equation)
=2
16(2.5 2.1) 3.5(2.5 2.1)
= 6.96%.
[ CHEMISTRY]
54. Answer (3)
Fact
55. Answer (1)
O
O O
O
H
HCH3
CH3( 0) ( 0)
56. Answer (4)
Fact
57. Answer (4)
58. Answer (2)
59. Answer (4)
PV PV 1 1Z n 0.0247
nRT ZRT 1.8 0.0821 273
⇒
No. of H2 molecules
23 220.0247 6.022 10 1.487 10
60. Answer (1)
Fact.
61. Answer (1)
2 4 2N O 2NO���⇀
↽���
No. of moles initially 1 0
At equilibrium 1 – 0.4 0.8
0.6 92 0.8 46Molar mass ofmixture
1.4
∵
= 65.71
W PV RT (760 torr 1 atm)
M ∵
W PM 65.71 1d 2.67g / litre
V RT 0.0821 300
62. Answer (3)
Due to presence of 3 lone pairs on Xe in XeF2,
shape is linear.
63. Deleted
Test - 3 (Code C) (Answers & Hints) All India Aakash Test Series for Medical-2019
9/12
64. Answer (3)
Lewis dot structure of 4
NH
is H N H
H+
H
65. Answer (4)
In BF3, B is sp
2 hybridized
In PCl5, P is sp
3d hybridized
In 4
BF , B
is sp3 hybridized
In 6
PCl , P
is sp3d2 hybridized
66. Answer (3)
Cl Cl
Cl Cl
( = 0)
67. Answer (4)
68. Answer (1)
CH3
CN
has high dipole moment than
CH3
C N due to
presence of Electron with drawing (CN–) at para
position.
69. Answer (2)
Greater the value of 'a' i.e greater the intermolecular
forces of attraction for a gas, then the gas more
easily will be liquified.
70. Answer (3)
AlCl3 has more covalent character but more soluble
in H2O. It is due to high hydration enthalpy.
71. Answer (3)
O N O O N O
Bond order = 1 2
1.52
72. Answer (4)
3 3 3 3 3
* * 2 1
2 21 1 2 2 2 x y
z
p ps s s s p
73. Answer (4)
Fact
74. Answer (1)
Fact
75. Answer (1)
Due to formation of intramolecular H-bonding
76. Answer (2)
Fact.
77. Answer (4)
Fact.
78. Answer (4)
a s
Tv
M
4
2
CH
SO
v300 64 3
v 16 400 1
79. Answer (2)
Xe can form compounds with most electronegative
elements XeF2, XeF
4, XeF
6
80. Answer (2)
Fact
81. Answer (3)
Bond order Bond strength
82. Answer (3)
PVZ 1; 1
nRT ,
m V 22.4 L.∵
83. Answer (4)
84. Answer (2)
Density of a gas = PM
RT; density P; density
1
T
85. Answer (3)
In cyclooctatetraene all are p – p bonds.
86. Answer (1)
4CH
T
P
P =
4CH
1
116X
1 1 9
2 16
All India Aakash Test Series for Medical-2019 Test - 3 (Code C) (Answers & Hints)
10/12
[ BIOLOGY]
91. Answer (3)
92. Answer (4)
Natural system of classification is based on
phytochemistry also.
93. Answer (4)
Iodine is obtained from brown algae (Phaeophyceae)
94. Answer (2)
Non-flagellated male gametes are found in Spirogyra
and flagellated male gametes are found in Fucus.
95. Answer (2)
Sex organs are non-Jacketed
96. Answer (3)
Generally in algae haplontic life cycle is found but
Fucus shows diplontic life cycle.
97. Answer (4)
Both Ectocarpus and Gelidium have chl-a and
cellulosic cell wall
98. Answer (4)
Majority of the red algae are marine and generally
found in warmer areas.
99. Answer (1)
Kelps are large sized brown algae
100. Answer (2)
101. Answer (4)
102. Answer (1)
Gemmae, haploid asexual bud which helps in
asexual reproduction of Marchantia.
103. Answer (4)
Archegonia are found in Cycas
Female cone is not found in Cycas
104. Answer (4)
Gametophytes in gymnosperms are dependent on
sporophytes.
105. Answer (4)
106. Answer (4)
In gymnosperm, veins and veinlets are absent in
leaves.
107. Answer (1)
108. Answer (3)
Pinus and Polytrichum both reproduce by oogamous
type sexual reproduction.
109. Answer (3)
All gymnosperms are heterosporous and produce
haploid microspores and megaspores.
110. Answer (2)
In bryophytes, gametophyte is dominant.
111. Answer (3)
Rhizoids of mosses are multicellular and branched
Prothallus is free-living inconspicuous and
photosynthetic gametophyte
112. Answer (1)
113. Answer (4)
In mosses, spores are formed after meiosis
Synergids and antipodals degenerate after
fertilisation in most of the flowering plant
Gemmae are asexual buds
114. Answer (4)
Zygote does not undergo reduction division
immediately
Sporophyte is dependent on gametophyte
87. Answer (2)
Polarising power of cation charge on cation
88. Answer (2)
dRTP
M
d M
P RT
d 1
P T ⇒
1 2
1 2
d P
P d
1 373
273 1
2 1
2 1
d d
P P
273
373
2
2
d
P
273 273x
373 373x
89. Answer (3)
In any resonance structure atom should obey octet
rule and should have symmetrical charge separation.
90. Answer (2)
In SF4, 'S' atom undergoes sp
3d hybridisation.
Test - 3 (Code C) (Answers & Hints) All India Aakash Test Series for Medical-2019
11/12
115. Answer (3)
In bryophyta zygote does not undergo reduction
division immediately
eg. Polytrichum and Sphagnum
116. Answer (4)
Double fertilisation is found in only angiosperms only
but Cedrus, Pinus are Gymnosperm
117. Answer (1)
'a' and 'c' features are correct.
118. Answer (4)
Only 'b' statement is correct.
'a', 'c' and 'd' statements are incorrect.
Gymnospermic root are generally tap root.
Egg apparatus is group of three cell (two synergid
and one female gamete)
Generally algae show haplontic life cycle
119. Answer (2)
Ploidy of PEN is triploid in nature (3n), e.g. maize.
120. Answer (2)
121. Answer (2)
Synergids(n), Antipodal(n), Pollen grain(n)
Embryo (2n) Integument (2n)
122. Answer (2)
123. Answer (4)
124. Answer (1)
Club moss - Lycopodium (pteridophytes)
Cord moss - Funaria (Bryophytes)
Hair cap moss - Polytrichum (Bryophytes)
Peat moss - Sphagnum (Bryophytes)
125. Answer (1)
126. Answer (3)
In numerical taxonomy, all characters are given equal
importance
127. Answer (3)
128. Answer (4)
The sporophyte in mosses is more elaborate than
that in liverwort.
Spores are formed after meiosis.
129. Answer (2)
Vascular tissue is present in vascular cryptogams
and some are heterosporous.
130. Answer (4)
Naked ovule is the important feature of gymnosperm
131. Answer (3)
‘‘Meiosis in the zygote results in the formation of
haploid spores in algae’’
132. Answer (1)
Diplontic life cycle found in all flowering and
gymnosperm and these both are seed bearing
plants.
133. Answer (4)
During double fertilisation one male gamete fuses
with diploid secondary nucleus.
Synergids and antipodals degenerate after
fertilisation.
134. Answer (3)
Thalloid body organisation is found in Marchantia and
algae.
135. Answer (3)
Brown algae possess the photosynthetic pigments
chl-a and chl-c and fucoxanthin.
136. Answer (2)
Dorsal blood vessel is both collecting & distributing
blood vessel with valves.
137. Answer (1)
Lumbricus has a closed circulatory system and
nucleated leucocytes.
138. Answer (3)
Respiratory exchange occurs through moist body
surface.
139. Answer (3)
Blood glands are present in 4th, 5th & 6th segment.
Typhlosole starts from 27th segment of earthworm.
140. Answer (3)
Lateral hearts are present in segment 7 & 9.
141. Answer (4)
142. Answer (2)
143. Answer (4)
144. Answer (4)
Setae & dorsal pores are absent in last segment of
earthworm while integumentary nephridiopores are
present.
145. Answer (3)
Spermathecae are present in segment 6, 7, 8 & 9.
All India Aakash Test Series for Medical-2019 Test - 3 (Code C) (Answers & Hints)
12/12
146. Answer (3)
Fertilization occurs in cocoon of earthworm,
therefore, it is external. Calciferous glands neutralize
humic acid. Chloragogen cells are analogous to
human liver.
147. Answer (2)
148. Answer (1)
Integumentary nephridia are exonephric and play no
role in osmoregulation.
149. Answer (2)
150. Answer (3)
Blood glands are present in segment 4th, 5th & 6th.
151. Answer (2)
Antennae are cephalic appendages.
152. Answer (2)
Fused, segmentally arranged ganglia are present
ventrally in cockroach.
153. Answer (3)
Cockroaches are active at night hence, nocturnal.
154. Answer (2)
Midgut/mesenteron is lined by peritrophic membrane
while foregut is lined by chitinous cuticle.
155. Answer (3)
Gradual metamorphosis i.e. paurometabolous
condition is observed in Periplaneta.
156. Answer (3)
Titillator is associated with left phallomere spiracles
are present on lateral sides of the body.
157. Answer (3)
Maxilla & mandible are paired appendages in
cockroach.
158. Answer (2)
159. Answer (3)
Ejaculatory duct, pharynx, conglobate gland &
phallic duct are four unpaired structures in
cockroach.
160. Answer (2)
161. Answer (4)
In cockroach, brain is represented by supra-
oesophageal ganglia; mosaic vision has more
sensitivity & sperms are glued together to form a
spermatophore.
162. Answer (3)
163. Answer (1)
Abdomen is long & narrow in males and caudal
styles are present in 9th segment of a male only.
164. Answer (4)
Alveolar sacs are absent in cockroach.
165. Answer (2)
Each ovary contains 8 ovarioles.
166. Answer (3)
Corpora allata is source of juvenile hormone. Its
absence leads to moulting.
167. Answer (3)
Mushroom gland is an unpaired structure in male
cockroach only.
168. Answer (4)
Anal styles are present in male cockroach only.
169. Answer (2)
Uricose gland are excretory structures found in
males of some species. Conglobate gland are
reproductive gland found in male cockroach only.
170. Answer (4)
Anal styles are unjointed structures.
171. Answer (2)
In frog RBCs are nucleated.
172. Answer (4)
173. Answer (4)
In frog teeth are not meant for chewing, digestion of
food begins in stomach & bile juices help in
emulsification.
174. Answer (2)
175. Answer (1)
Corpus callosum, ribs & salivary glands are absent
in frog.
176. Answer (4)
177. Answer (3)
Tadpole is a herbivore while frog is a carnivore.
Tadpole respire by gills while frog shifts to
pulmonary, cutaneous etc. form of respiration.
178. Answer (3)
179. Answer (3)
180. Answer (4)
� � �
Test - 3 (Code D) (Answers & Hints) All India Aakash Test Series for Medical-2019
1/12
1. (4)
2. (2)
3. (2)
4. (1)
5. (1)
6. (3)
7. (2)
8. (2)
9. (2)
10. (2)
11. (2)
12. (3)
13. (3)
14. (1)
15. (1)
16. (4)
17. (1)
18. (3)
19. (4)
20. (1)
21. (3)
22. (1)
23. (1)
24. (3)
25. (4)
26. (1)
27. (2)
28. (3)
29. (4)
30. (1)
31. (4)
32. (2)
33. (2)
34. (3)
35. (2)
36. (2)
Test Date : 03/12/2017
ANSWERS
TEST - 3 (Code D)
All India Aakash Test Series for Medical - 2019
37. (1)
38. (4)
39. (2)
40. (2)
41. (4)
42. (2)
43. (2)
44. (3)
45. (3)
46. (2)
47. (3)
48. (2)
49. (2)
50. (1)
51. (3)
52. (2)
53. (4)
54. (3)
55. (3)
56. (2)
57. (2)
58. (4)
59. (4)
60. (2)
61. (1)
62. (1)
63. (4)
64. (4)
65. (3)
66. (3)
67. (2)
68. (1)
69. (4)
70. (3)
71. (4)
72. (3)
73. (Deleted)
74. (3)
75. (1)
76. (1)
77. (4)
78. (2)
79. (4)
80. (4)
81. (1)
82. (3)
83. (1)
84. (1)
85. (1)
86. (4)
87. (1)
88. (4)
89. (1)
90. (3)
91. (3)
92. (3)
93. (4)
94. (1)
95. (3)
96. (4)
97. (2)
98. (4)
99. (3)
100. (3)
101. (1)
102. (1)
103. (4)
104. (2)
105. (2)
106. (2)
107. (2)
108. (4)
109. (1)
110. (4)
111. (3)
112. (4)
113. (4)
114. (1)
115. (3)
116. (2)
117. (3)
118. (3)
119. (1)
120. (4)
121. (4)
122. (4)
123. (4)
124. (1)
125. (4)
126. (2)
127. (1)
128. (4)
129. (4)
130. (3)
131. (2)
132. (2)
133. (4)
134. (4)
135. (3)
136. (4)
137. (3)
138. (3)
139. (3)
140. (4)
141. (1)
142. (2)
143. (4)
144. (4)
145. (2)
146. (4)
147. (2)
148. (4)
149. (3)
150. (3)
151. (2)
152. (4)
153. (1)
154. (3)
155. (4)
156. (2)
157. (3)
158. (2)
159. (3)
160. (3)
161. (3)
162. (2)
163. (3)
164. (2)
165. (2)
166. (3)
167. (2)
168. (1)
169. (2)
170. (3)
171. (3)
172. (4)
173. (4)
174. (2)
175. (4)
176. (3)
177. (3)
178. (3)
179. (1)
180. (2)
All India Aakash Test Series for Medical-2019 Test - 3 (Code D) (Answers & Hints)
2/12
ANSWERS & HINTS
1. Answer (4)
At the highest point �
�
F v
So 0 �
�
P F v
2. Answer (2)
2 2
mg gmg ma a ⇒
3. Answer (2)
Angle of repose R = 45°
So block does not slide
fs = mgsin37° =
310 6 N
5
4. Answer (1)
fs
Maximum acceleration upto which block does not
slide on the floor
amax
= g = 0.5 × 10 = 5 m/s2
Given, a = 4 m/s2
So, fs = 2 × 4 = 8 N
Displacement of block in t = 2 s
214 (2) 8 m
2 S
wf = 8 × 8 = + 64 J
5. Answer (1)
( 0)
(1)
p n muF mnu
t
⎛ ⎞ ⎜ ⎟⎝ ⎠
6. Answer (3)
10dp
F tdt
At t = 2 s F = 20 N
7. Answer (2)
(REF LEVEL)
L/10
GL/5
[ PHYSICS]
Mass of hanging part = 5
M
5 10
⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠i
M LU g
50
MgL
Uf = 0
50 MgL
W U
8. Answer (2)
W = Pt = (m)gh
8000 60( ) 3200 kg
10 15m
9. Answer (2)
20 W �
�
P F v
10. Answer (2)
Kinetic energy is maximum when potential energy is
minimum.
U = 2x2 – 2 ...(1)
4dU
xdx
...(2)
For minimum, , 0
dUU
dx
x = 0
2
24
d U
dx U is minimum at x = 0
Umin
= –2 J
Now,
Kmax
+ Umin
= Emech
Kmax
– 2 = 34 2
max
136
2mv
vmax
= 6 m/s
Test - 3 (Code D) (Answers & Hints) All India Aakash Test Series for Medical-2019
3/12
11. Answer (2)
For equilibrium
F = 0
2x2 – x = 0 2(2x – 1) = 0
Either x = 0 or1
2x
4 1dF
xdx
At x = 0, 1 0dF
dx
So equilibrium is stable at x = 0.
12. Answer (3)
Speed of ball just before the collision.
2(10) 2 10 2.2
= 12 m/s
Speed of ball just after the collision.
2 10 5 10 m/s
Coefficient of restitution
10 5
12 6e
13. Answer (3)
Area under F - t graph
= change in linear momentum
1(2 3) 10
2
mv – mu = 25
v = 25 m/s
Kinetic energy 21 625
J2 2
k mv
14. Answer (1)
O
Y
X
50,
3
⎛ ⎞⎜ ⎟⎝ ⎠
5, 0
4
⎛ ⎞⎜ ⎟⎝ ⎠
4 5
3 3Y X
5 5ˆ ˆ0 0
4 3r i j
⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
�
= 5 5ˆ ˆ
4 3i j
0r F �
�
15. Answer (1)
2
cos60 mvN mg
R
2 2(4)cos60 1 5
2
vN m g
R
⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
= 13 newton
fk =
kN = 6.5 N
N
mgmgsin60°
60°
fk
mgcos60°
Tangential acceleration sin60
kmg f
m
25 3 6.5 2.1 m/s
16. Answer (4)
B
A
ar
vB
g
O
vA
2
, B
r t
va a g
R
2 2( ) ( ) r t
a a a
2 22
B Av v gR
17. Answer (1)
3x
uF
x
23 m/s
x
x
Fa
m
4y
uF
y
All India Aakash Test Series for Medical-2019 Test - 3 (Code D) (Answers & Hints)
4/12
ay = –4 m/s2
vx = a
xt = –3t; v
y = a
yt = –4t
2 25
x yv v v t
At t = 2 s, v = 5 × 2 = 10 m/s
18. Answer (3)
vx = u
x = 3 m/s
vy = eu
y =
34 3 m/s
4
Speed 2 2
3 2 m/sx y
v v v
19. Answer (4)
m
m v/2
v
Conservation of mechanical energy
2
21 1
2 2 2
vmv m
⎛ ⎞ ⎜ ⎟⎝ ⎠
2mg mg �
�
2
5 3 12
8 2 5
mv mg gv ⇒ � �
20. Answer (1)
Fv at t
m
⎛ ⎞ ⎜ ⎟⎝ ⎠ ...(1)
2F
P Fv tm
P t
21. Answer (3)
mg
hO
v
4u g �
Let string becomes slack when it makes angle
with upward vertical. Then
2
cosmv
mg �
2cosv g � ...(1)
Also,
2 2 22 2 (1 cos )v u gh u g �
or, cos 4 2 2 cosg g g g � � � �
or,2
3 cos 2 cos3
g g ⇒ � �
Tangential acceleration
5sin
3g g
22. Answer (1)
s
dvm mg
dt
23. Answer (1)
Instantaneous velocity.
23 8 dx
v t tdt
...(1)
At t = 0, v1 = 3 m/s
t = 9 s, v2 = 12 m/s
W = K
2 2
2 1
1
2 m v v
2 21 1[(12) (3) ]
2 2
135J
4
24. Answer (3)
F kxs =
mg
Test - 3 (Code D) (Answers & Hints) All India Aakash Test Series for Medical-2019
5/12
From work-energy theorem.
21
2mgx kx
2mgx
k ...(1)
Spring force at this instant
Fs = kx = 2mg
s2
upward
F mg mg mga g
m m
25. Answer (4)
A
ˆui
B2
4u g i ��
Velocity at point A.
ˆui ...(1)
Speed at the highest point.
2 24 �
Bv u g
24 �
Bv u g
Velocity 2 ˆ4 �
�B
v u g i ...(2)
Change in velocity
2 ˆ4 4 � � �
�B A
v v v u g i i
24
��v u u g
26. Answer (1)
At the highest point, speed
v = 100 cos60° = 50 m/s horizontally
Conservation of linear momentum.
2ˆ ˆ2 50 1 100 1i j v
�
2
ˆ ˆ100 100v i j �
2100 2 m/sv
�
So kinetic energy of this part.
2 411 100 2 10 J
2
27. Answer (2)
After collision velocity of heavy block remains the
same.
8 m/sv1
Now,
Relative velocity of separation = Relative velocity of
approach
8 – v1 = (10 – 8)
v1 = +6 m/s (in positive x direction)
28. Answer (3)
v
90 -
u
Line of impact
ucos
Let particle strikes with speed u and moves
horizontally with speed v after the collision.
Momentum is conserved along the plane.
So vcos = usin ...(1)
and vcos(90 – ) = eucos
or, vsin = eucos ...(2)
Dividing equation (2) by equation (1)
2 2 1tan cot tan tan 30
3e e ⇒
29. Answer (4)
e p2
ep
ep
p
+ve
Momentum of ball after first collision = +ep
Change in momentum of ball after first collision.
(p)1 = ep – (–p) = p(1 + e)
Similarly after second collision
(p)2 = e2p – (–ep) = ep(1 + e)
So total momentum imparted
1 2( ) ( )p p p
2(1 ) (1 ) (1 ) terms p e ep e e p e
(1 )
(1 )
p e
e
30. Answer (1)
All India Aakash Test Series for Medical-2019 Test - 3 (Code D) (Answers & Hints)
6/12
31. Answer (4)
m1
m2
a1 a
2
2T
T T2T
(2T + T) a1 = Ta
2
32. Answer (2)
10 kg
kxm
100 N
N
Normal contact force is minimum when spring force
is maximum.
4 × 10 × xmax
= 21
2m
kx [K = 0]
kxm
= 80 N
Nmin
= 100 – 80 = 20 N
33. Answer (2)
mgcosmg
N
v
As block is at rest w.r.t. lift
N = mgcos ...(1)
Displacement of block w.r.t. ground
S = vt vertically upward ...(2)
Work done by normal contact force
W = N × S × cos
= (mgcos)(vt)cos
= mgvt cos2
34. Answer (3)
60°
u
v
From conservation of linear momentum
m(v cos60° + u) + Mu = 0
35. Answer (2)
As 1 >
2, both blocks will start sliding together,
when
(m1 + m
2)gsin =
1m1gcos +
2m2gcos
Or (m1 + m
2)gsin = gcos(
1m1 +
2m2)
1 1 2 2
1 2
tan
m m
m m
36. Answer (2)
2
12 sin 2 sin45 v g S g S ...(1)
2
22 (sin45 cos45 )
kv g S ...(2)
2
1
2
2
1 116
1511
16
k
v
v
1
2
4 :1v
v
37. Answer (1)
mgcosmg
FcosN
F
Fsin
mgsin
N
Block begins to slide when
Fcos = mgsin + N
Fcos = mgsin + (mgcos + Fsin)
F(cos – sin) = mg(sin + cos)
(sin cos )
(cos sin )
mgF
38. Answer (4)
2
2 2( )2
x
mT L x
L
Test - 3 (Code D) (Answers & Hints) All India Aakash Test Series for Medical-2019
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T(x = 0) 21
2 m L
2
2 2
( 0) 4
( ) 1( )
T x L
T x x L x
4L2 – 4x2 = L2 4x2 = 3L2
3
2 L
x
39. Answer (2)
Limiting force of friction.
For m1 ; (f
1)max
= 1m1g = 25 N
m2 ; (f
2)max
= 2m2g = 40 N
Maximum frictional force fmax
= 65 N
60 N
m1
T
( ) = 25 Nf1 max
As F = 60 N, both blocks will be at rest. Frictional
force on block of mass m1 be (f
1)max
. Tension in the
string.
T = 60 – 25 = 35 N
40. Answer (2)
Maximum common acceleration of two blocks, when
force is applied on the upper block.
21
max
2
0.5 20 1010 m/s
10
m g
a
m
Fmax
= (m1 + m
2) a
max = 30 × 10 = 300 N
As F < Fmax
, both blocks move together with
common acceleration.
21505 m/s
30 a
Fs = m
2a = 10 × 5 = 50 N
41. Answer (4)
Just before t = 1 s
Velocity v1 = +2 m/s
Just after t = 1 s
Velocity, v2 = –4 m/s
Change in velocity, v = v2 – v
1 = –6 m/s
Impulse = change in linear momentum.
= 2 × (–6) = –12 Ns
42. Answer (2)
Acceleration 21 2
305 m/s
6
F F
aM
Mass of 10 cm part of uniform rod.
610 1.2 kg
50 m
Now,
10 cm
TF2
a
T – F2 = ma
T = F2 + ma
= 10 + 1.2 × 5 = 16 N
43. Answer (2)
In case (I), acceleration of block in horizontal
direction
aI = (gsin30°) cos30° ...(1)
In case (II), acceleration of block in horizontal
direction is acceleration of wedge.
aII = gtan30° ...(2)
Ratio 2I
II
3cos 30
4
⎛ ⎞ ⎜ ⎟⎝ ⎠
a
a
44. Answer (3)
When trolley is sliding with acceleration gsin, string
becomes perpendicular to the ceiling.
mgsin
mgcosmg
T
T = mgcos
45. Answer (3)
As downward acceleration of elevator a > g, block
loses the contact with the elevator and its
acceleration is g downward. So speed of block at
t = 1 s,
v = u + gt
= 10 × 1 = 10 m/s
All India Aakash Test Series for Medical-2019 Test - 3 (Code D) (Answers & Hints)
8/12
46. Answer (2)
In SF4, 'S' atom undergoes sp3d hybridisation.
47. Answer (3)
In any resonance structure atom should obey octet
rule and should have symmetrical charge separation.
48. Answer (2)
dRTP
M
d M
P RT
d 1
P T ⇒
1 2
1 2
d P
P d
1 373
273 1
2 1
2 1
d d
P P
273
373
2
2
d
P
273 273x
373 373x
49. Answer (2)
Polarising power of cation charge on cation
50. Answer (1)
4CH
T
P
P =
4CH
1
116X
1 1 9
2 16
51. Answer (3)
In cyclooctatetraene all are p – p bonds.
52. Answer (2)
Density of a gas = PM
RT; density P; density
1
T
53. Answer (4)
54. Answer (3)
PVZ 1; 1
nRT ,
m V 22.4 L.∵
55. Answer (3)
Bond order Bond strength
56. Answer (2)
Fact
[ CHEMISTRY]
57. Answer (2)
Xe can form compounds with most electronegative
elements XeF2, XeF
4, XeF
6
58. Answer (4)
a s
Tv
M
4
2
CH
SO
v300 64 3
v 16 400 1
59. Answer (4)
Fact.
60. Answer (2)
Fact.
61. Answer (1)
Due to formation of intramolecular H-bonding
62. Answer (1)
Fact
63. Answer (4)
Fact
64. Answer (4)
3 3 3 3 3
* * 2 1
2 21 1 2 2 2 x y
z
p ps s s s p
65. Answer (3)
O N O O N O
Bond order = 1 2
1.52
66. Answer (3)
AlCl3 has more covalent character but more soluble
in H2O. It is due to high hydration enthalpy.
67. Answer (2)
Greater the value of 'a' i.e greater the intermolecular
forces of attraction for a gas, then the gas more
easily will be liquified.
68. Answer (1)
CH3
CN
has high dipole moment than
CH3
C N due to
presence of Electron with drawing (CN–) at para
position.
Test - 3 (Code D) (Answers & Hints) All India Aakash Test Series for Medical-2019
9/12
69. Answer (4)
70. Answer (3)
Cl Cl
Cl Cl
( = 0)
71. Answer (4)
In BF3, B is sp2 hybridized
In PCl5, P is sp3d hybridized
In 4
BF , B
is sp3 hybridized
In 6
PCl , P
is sp3d2 hybridized
72. Answer (3)
Lewis dot structure of 4
NH
is H N H
H+
H
73. Deleted
74. Answer (3)
Due to presence of 3 lone pairs on Xe in XeF2,
shape is linear.
75. Answer (1)
2 4 2N O 2NO���⇀
↽���
No. of moles initially 1 0
At equilibrium 1 – 0.4 0.8
0.6 92 0.8 46Molar mass ofmixture
1.4
∵
= 65.71
W PV RT (760 torr 1 atm)
M ∵
W PM 65.71 1d 2.67g / litre
V RT 0.0821 300
76. Answer (1)
Fact.
77. Answer (4)
PV PV 1 1Z n 0.0247
nRT ZRT 1.8 0.0821 273
⇒
No. of H2 molecules
23 220.0247 6.022 10 1.487 10
78. Answer (2)
79. Answer (4)
80. Answer (4)
Fact
81. Answer (1)
O
O O
O
H
HCH3
CH3( 0) ( 0)
82. Answer (3)
Fact
83. Answer (1)
%ionic character = 2
A B A B16(X X ) 3.5(X X )
(Hannay – Smith Equation)
=2
16(2.5 2.1) 3.5(2.5 2.1)
= 6.96%.
84. Answer (1)
Due to presence of lone pair on Br in BrF5, bond
angle is not equal to 90°.
85. Answer- (1)
86. Answer (4)
In 3 3
N(SiH )�� , N-atom donate its lone pair to vacant
d-orbital of Si to form back bonding.
87. Answer (1)
dRT dRTP M
M P ⇒
3d 2.64 gm / dm
3 1 1 3R 0.0821dm atm mole k ( 1 dm 1 litre)
∵
All India Aakash Test Series for Medical-2019 Test - 3 (Code D) (Answers & Hints)
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[ BIOLOGY]
91. Answer (3)
Brown algae possess the photosynthetic pigments
chl-a and chl-c and fucoxanthin.
92. Answer (3)
Thalloid body organisation is found in Marchantia and
algae.
93. Answer (4)
During double fertilisation one male gamete fuses
with diploid secondary nucleus.
Synergids and antipodals degenerate after
fertilisation.
94. Answer (1)
Diplontic life cycle found in all flowering and
gymnosperm and these both are seed bearing
plants.
95. Answer (3)
‘‘Meiosis in the zygote results in the formation of
haploid spores in algae’’
96. Answer (4)
Naked ovule is the important feature of gymnosperm
97. Answer (2)
Vascular tissue is present in vascular cryptogams
and some are heterosporous.
98. Answer (4)
The sporophyte in mosses is more elaborate than
that in liverwort.
Spores are formed after meiosis.
99. Answer (3)
100. Answer (3)
In numerical taxonomy, all characters are given equal
importance
101. Answer (1)
102. Answer (1)
Club moss - Lycopodium (pteridophytes)
Cord moss - Funaria (Bryophytes)
Hair cap moss - Polytrichum (Bryophytes)
Peat moss - Sphagnum (Bryophytes)
103. Answer (4)
104. Answer (2)
105. Answer (2)
Synergids(n), Antipodal(n), Pollen grain(n)
Embryo (2n) Integument (2n)
106. Answer (2)
107. Answer (2)
Ploidy of PEN is triploid in nature (3n), e.g. maize.
108. Answer (4)
Only 'b' statement is correct.
'a', 'c' and 'd' statements are incorrect.
Gymnospermic root are generally tap root.
Egg apparatus is group of three cell (two synergid
and one female gamete)
Generally algae show haplontic life cycle
109. Answer (1)
'a' and 'c' features are correct.
110. Answer (4)
Double fertilisation is found in only angiosperms only
but Cedrus, Pinus are Gymnosperm
775P atm
760
�T 310 273 583 K
2.64 0.0821 583M 124
775 / 760
No. of P atoms in a molecule =124 124
At. Wt of P 31
= 4
88. Answer (4)
Due to resonance in 2 2
CH CH CH , all C atoms
undergo sp2 hybridization.
89. Answer (1)
Charge on ionLattice energy
Size of ion
90. Answer (3)
2 2 2 2SO Cl SO Cl
Hence this is not applicable to Dalton's law of partial
pressure.
Test - 3 (Code D) (Answers & Hints) All India Aakash Test Series for Medical-2019
11/12
111. Answer (3)
In bryophyta zygote does not undergo reduction
division immediately
eg. Polytrichum and Sphagnum
112. Answer (4)
Zygote does not undergo reduction division
immediately
Sporophyte is dependent on gametophyte
113. Answer (4)
In mosses, spores are formed after meiosis
Synergids and antipodals degenerate after
fertilisation in most of the flowering plant
Gemmae are asexual buds
114. Answer (1)
115. Answer (3)
Rhizoids of mosses are multicellular and branched
Prothallus is free-living inconspicuous and
photosynthetic gametophyte
116. Answer (2)
In bryophytes, gametophyte is dominant.
117. Answer (3)
All gymnosperms are heterosporous and produce
haploid microspores and megaspores.
118. Answer (3)
Pinus and Polytrichum both reproduce by oogamous
type sexual reproduction.
119. Answer (1)
120. Answer (4)
In gymnosperm, veins and veinlets are absent in
leaves.
121. Answer (4)
122. Answer (4)
Gametophytes in gymnosperms are dependent on
sporophytes.
123. Answer (4)
Archegonia are found in Cycas
Female cone is not found in Cycas
124. Answer (1)
Gemmae, haploid asexual bud which helps in
asexual reproduction of Marchantia.
125. Answer (4)
126. Answer (2)
127. Answer (1)
Kelps are large sized brown algae
128. Answer (4)
Majority of the red algae are marine and generally
found in warmer areas.
129. Answer (4)
Both Ectocarpus and Gelidium have chl-a and
cellulosic cell wall
130. Answer (3)
Generally in algae haplontic life cycle is found but
Fucus shows diplontic life cycle.
131. Answer (2)
Sex organs are non-Jacketed
132. Answer (2)
Non-flagellated male gametes are found in Spirogyra
and flagellated male gametes are found in Fucus.
133. Answer (4)
Iodine is obtained from brown algae (Phaeophyceae)
134. Answer (4)
Natural system of classification is based on
phytochemistry also.
135. Answer (3)
136. Answer (4)
137. Answer (3)
138. Answer (3)
139. Answer (3)
Tadpole is a herbivore while frog is a carnivore.
Tadpole respire by gills while frog shifts to
pulmonary, cutaneous etc. form of respiration.
140. Answer (4)
141. Answer (1)
Corpus callosum, ribs & salivary glands are absent
in frog.
142. Answer (2)
143. Answer (4)
In frog teeth are not meant for chewing, digestion of
food begins in stomach & bile juices help in
emulsification.
144. Answer (4)
145. Answer (2)
In frog RBCs are nucleated.
146. Answer (4)
Anal styles are unjointed structures.
All India Aakash Test Series for Medical-2019 Test - 3 (Code D) (Answers & Hints)
12/12
� � �
147. Answer (2)
Uricose gland are excretory structures found in
males of some species. Conglobate gland are
reproductive gland found in male cockroach only.
148. Answer (4)
Anal styles are present in male cockroach only.
149. Answer (3)
Mushroom gland is an unpaired structure in male
cockroach only.
150. Answer (3)
Corpora allata is source of juvenile hormone. Its
absence leads to moulting.
151. Answer (2)
Each ovary contains 8 ovarioles.
152. Answer (4)
Alveolar sacs are absent in cockroach.
153. Answer (1)
Abdomen is long & narrow in males and caudal
styles are present in 9th segment of a male only.
154. Answer (3)
155. Answer (4)
In cockroach, brain is represented by supra-
oesophageal ganglia; mosaic vision has more
sensitivity & sperms are glued together to form a
spermatophore.
156. Answer (2)
157. Answer (3)
Ejaculatory duct, pharynx, conglobate gland &
phallic duct are four unpaired structures in
cockroach.
158. Answer (2)
159. Answer (3)
Maxilla & mandible are paired appendages in
cockroach.
160. Answer (3)
Titillator is associated with left phallomere spiracles
are present on lateral sides of the body.
161. Answer (3)
Gradual metamorphosis i.e. paurometabolous
condition is observed in Periplaneta.
162. Answer (2)
Midgut/mesenteron is lined by peritrophic membrane
while foregut is lined by chitinous cuticle.
163. Answer (3)
Cockroaches are active at night hence, nocturnal.
164. Answer (2)
Fused, segmentally arranged ganglia are present
ventrally in cockroach.
165. Answer (2)
Antennae are cephalic appendages.
166. Answer (3)
Blood glands are present in segment 4th, 5th & 6th.
167. Answer (2)
168. Answer (1)
Integumentary nephridia are exonephric and play no
role in osmoregulation.
169. Answer (2)
170. Answer (3)
Fertilization occurs in cocoon of earthworm,
therefore, it is external. Calciferous glands neutralize
humic acid. Chloragogen cells are analogous to
human liver.
171. Answer (3)
Spermathecae are present in segment 6, 7, 8 & 9.
172. Answer (4)
Setae & dorsal pores are absent in last segment of
earthworm while integumentary nephridiopores are
present.
173. Answer (4)
174. Answer (2)
175. Answer (4)
176. Answer (3)
Lateral hearts are present in segment 7 & 9.
177. Answer (3)
Blood glands are present in 4th, 5th & 6th segment.
Typhlosole starts from 27th segment of earthworm.
178. Answer (3)
Respiratory exchange occurs through moist body
surface.
179. Answer (1)
Lumbricus has a closed circulatory system and
nucleated leucocytes.
180. Answer (2)
Dorsal blood vessel is both collecting & distributing
blood vessel with valves.