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Test - 3 (Code E) (Answers) All India Aakash Test Series for Medical-2019
1/15
1. (1)
2. (3)
3. (2)
4. (2)
5. (1)
6. (3)
7. (2)
8. (2)
9. (1)
10. (3)
11. (2)
12. (2)
13. (2)
14. (2)
15. (3)
16. (2)
17. (3)
18. (3)
19. (4)
20. (3)
21. (3)
22. (1)
23. (4)
24. (3)
25. (3)
26. (4)
27. (2)
28. (4)
29. (1)
30. (1)
31. (4)
32. (2)
33. (1)
34. (3)
35. (2)
36. (3)
Test Date : 20/01/2019
ANSWERS
TEST - 4 (Code E)
All India Aakash Test Series for Medical - 2019
37. (4)
38. (2)
39. (1)
40. (2)
41. (3)
42. (1)
43.Deleted
44. (3)
45. (4)
46. (2)
47. (3)
48. (2)
49. (3)
50. (4)
51. (3)
52. (3)
53. (3)
54. (1)
55. (4)
56. (4)
57. (1)
58. (3)
59. (3)
60. (4)
61. (3)
62. (1)
63. (2)
64. (4)
65. (3)
66. (3)
67. (2)
68. (1)
69. (4)
70. (3)
71. (2)
72. (4)
73. (2)
74. (2)
75. (1)
76. (4)
77. (1)
78. (4)
79. (2)
80. (2)
81. (2)
82. (2)
83. (4)
84. (2)
85. (2)
86. (4)
87. (4)
88. (3)
89. (3)
90. (2)
91. (1)
92. (3)
93. (4)
94. (2)
95. (1)
96. (2)
97. (4)
98. (4)
99. (1)
100. (3)
101. (2)
102. (2)
103. (4)
104. (1)
105. (2)
106. (3)
107. (4)
108. (3)
109. (4)
110. (2)
111. (4)
112. (3)
113. (2)
114. (4)
115. (4)
116. (1)
117. (2)
118. (3)
119. (3)
120. (3)
121. (1)
122. (2)
123. (3)
124. (1)
125. (4)
126. (2)
127. (3)
128. (3)
129. (2)
130. (2)
131. (4)
132. (3)
133. (2)
134. (3)
135. (4)
136. (1)
137. (1)
138. (4)
139. (3)
140. (3)
141. (1)
142. (1)
143. (2)
144. (3)
145. (2)
146. (3)
147. (2)
148. (2)
149. (3)
150. (4)
151. (3)
152. (2)
153. (1)
154. (3)
155. (2)
156. (3)
157. (2)
158. (4)
159. (1)
160. (3)
161. (1)
162. (2)
163. (2)
164. (4)
165. (1)
166. (2)
167. (2)
168. (1)
169. (3)
170. (4)
171. (2)
172. (4)
173. (2)
174. (3)
175. (1)
176. (3)
177. (4)
178. (4)
179. (1)
180. (3)
All India Aakash Test Series for Medical-2019 Test - 4 (Code E) (Hints and Solutions)
2/15
PHYSICS
1. Answer (1)
Hint: Wave transport energy and momentum.
Sol.: Transverse wave do not propagate inside a
fluid.
2. Answer (3)
Hint: RMS
3RTV
M
Sol.: RMS
3
Mass of gas
PVV
At constant volume RMS
V P
3. Answer (2)
Hint: avg2
fKTE for per molecule
Sol.: rms
3RTV
M
4. Answer (2)
Hint: PV = nRT
Sol.: PV = nRT
Mass( Molecular weight)PV RT M
M
MassPVR
T M
massPV R
T M
1Slope
M
5. Answer (1)
Hint: Sound RMS
3v v
Sol.: RMS
3RTv
M
HINTS AND SOLUTIONS
Sound
RTv
M
RMS av mp > > v v v
6. Answer (3)
Hint: PV = nRT
Sol.: PV = nRT
V nR
T P
Slopen
P
tan60A
A
n
P
tan30B
B
n
P
1A
B
P
P
7. Answer (2)
Hint: (mix)
mix
mix
P
V
C
C
Sol.: 1 21 2
(mix)
1 2
P P
P
nC n CC
n n
1 21 2
(mix)
1 2
V V
V
nC n CC
n n
(mix)
5 7
2 23
2P
R R
C R
CV (mix)
= 2R
2
3
V
P
C
C
8. Answer (2)
Hint: 0
1V
T V
Test - 4 (Code E) (Hints and Solutions) All India Aakash Test Series for Medical-2019
3/15
Sol.: PV = nRT
P V = nRT
V nR
T P
0
nR nR
PV nRT
1
T
n = –1
9. Answer (1)
Hint: PV = nRT
Sol.:nR
P TV
As, SlopenR
V is constant.
So, V = constant, W = 0 and = constant
10. Answer (3)
Hint: a = –2x
Sol.: 2 2 2
2 2–
–
aT x TT
x x
– 2 2constantT
11. Answer (2)
Hint: 2 2 21 1and KE ( – )
2 2U kx k A x
Sol.: As body moves from mean position, speed
decreases. So, KE decreases and PE increases.
12. Answer (2)
Hint: 0
2H
Tg
Sol.: At equilibrium condition
FB = weight
AHg = Ah0g
H = h0
[A is circular area of cylinder]
2h
Tg
13. Answer (2)
Hint: 2 2 2– –a x v A x
Sol.: a = v
2 2 2–x A x
2x2 = A2 – x2
At 21 2 second
2
Ax T
14. Answer (2)
Hint: 2R
Tg
Sol.: Time period of particle is independent of length
of chord.
15. Answer (3)
Hint: If y = f(ax ± bt)
then b
va
Sol.: 2
9
3 ( – 10 )y
x t
v = 10 m/s
16. Answer (2)
Hint: a = –2x
Sol.: a = –2x
2–B A B
A
2 B
T A
2A
TB
17. Answer (3)
Hint: 2 2
Resultant 1 2 1 22 cosA A A A A
Sol.: x = x1 + x
2
3 sin( )x A t
max3 ;A A
2 2 2
net2 cos60A A A A
∵
amax
= 2Amax
23 A
All India Aakash Test Series for Medical-2019 Test - 4 (Code E) (Hints and Solutions)
4/15
18. Answer (3)
Hint: Conservation of ME and linear momentum.
Sol.: 2 21 1( )
2 2M m v Kx
and mv0 = (M + m)v
x = Amplitude of oscillation
On solving
0mv M m
xM m K
19. Answer (4)
Hint: vparticle
= –(Slope) vwave
Sol.: Acceleration of particle always along their mean
position.
vA is upward and v
C is downward.
20. Answer (3)
Hint & Sol.: Two waves of same frequency & moves
in opposite direction.
21. Answer (3)
Hint: v = f
Sol.: v = f1
1 and v = f
2
2
1
1
vf
and
2
2
vf
1 2
1 2
1 1– –f f v
f1 – f
2 = n
1 2
2 1–
n
v
22. Answer (1)
Hint: 0
0
S
v vf f
v v
Sol.:
fref.
f0
v
Reflected frequency
ref. 0–
c vf f
c v
23. Answer (4)
Hint: In closed organ pipe only odd harmonics are
present.
Sol.: 50 × 1 = 50, 50 × 3 = 150, 50 × 5 = 250.
Odd harmonics are present. So, pipe is closed at
one end.
340 34m
50 5
v
f
34 171.7 m
4 20 10l
24. Answer (3)
Hint: 1 2
(2 – 1),
2 4o c
nv nf f v
l l
Sol.: 1 2
1 5;
2 4o c
vf f v
l l
fo = f
c
1 2
1 5
2 4l l
2 1
5
2l l
l1 = 10 cm; l
2 = 25 cm.
25. Answer (3)
Hint and Sol.: Doppler’s effect depends on relative
motion.
26. Answer (4)
Hint: 0
S.L 10logI
I
Sol.: 1
1
0
S.L 10logI
I
2
2
0
S.L 10logI
I
S.L = S.L2 – S.L
1 = 30
2 11000I I
27. Answer (2)
Hint and Sol.: All medium particles between two
successive nodes oscillate in same phase and
particles on one side of a node oscillate in opposite
phase with those on the other side of same node.
Test - 4 (Code E) (Hints and Solutions) All India Aakash Test Series for Medical-2019
5/15
28. Answer (4)
Hint: 2L
Tg
Sol.: L = 2.45 m
g = 9.8 m/s2
2 1sL
Tg
29. Answer (1)
Hint: 2
nf v
l
Sol.: 3
3
2f v
l v = 48 m/s
f3 = 48 Hz l = 1.5 m
1 mv
f
30. Answer (1)
Hint: RT
vM
Sol.: 2 RTv
M
v2 T (T in Kelvin)
31. Answer (4)
Hint: = kx
Sol.: Path difference 3 5
–4 8 8
x
2 5
8
5
4
32. Answer (2)
Hint and Sol.: particle
=
y
vt
33. Answer (1)
Hint: max
waveand
Pv A v
k
Sol.: sin( )y A t kx
= 50; k = 5
10 m/svk
maxp
v = A = 10 × 50 = 500 m/s
max
50p
v
v
34. Answer (3)
Hint: P
v
Sol.: air
air
Pv
2
2
H
H
Pv
2air
4 4 332 1328 m/sH
v v
35. Answer (2)
Hint and Sol.: Phase difference between incident
and reflected wave is when reflected from rigid
boundary.
36. Answer (3)
Hint: Doppler’s effect depends on relative motion.
37. Answer (4)
Hint: fBeat
= |f1 – f
2|
Sol.:
fv
s
Frequency of reflected sound.
–
r
c vf f
c v
Beat
2–
–r
vff f f
c v
All India Aakash Test Series for Medical-2019 Test - 4 (Code E) (Hints and Solutions)
6/15
38. Answer (2)
Hint: fbeat
= |f1 – f
2|
Sol.: fB – f
A = 2
1 1
30 25A Bf f
25
30
A
B
f
f
On solving
fA = 10 Hz; f
B = 12 Hz
39. Answer (1)
Hint: 2 2
net 1 2 1 22 cosA A A A A .
Sol.: 2A
2A
60°
60°
A
Anet
= 2A – A = A
40. Answer (2)
Hint: x = A sin(t + )
Sol.:
45°
45°
t t =
A(1)
t = 0
(2)
(2)
x
mean
(t t) =
t = 0
x = Acos45° 2
A
41. Answer (3)
Hint: 2
AV
1KE
3kA
Sol.: ( )AV
0
1KE KE
A
xdx
A
On solving
2 2
AV
1(KE)
3m A
42. Answer (1)
Hint and Sol.: 2m
Tk
43. Deleted
44. Answer (3)
Hint: Damped and forced oscillation.
Sol.: In damped oscillation amplitude and energy
both decreases exponentially.
At resonance amplitude becomes maximum.
45. Answer (4)
Hint: U = nCvT
Sol.: Since5
2v
RC for diatomic
3
2v
RC for monoatomic.
Slopev
C
CHEMISTRY
46. Answer (2)
Hint: 32
%wt. 100 94.12%34
47. Answer (3)
Hint:
Zn + 2NaOH Na2
ZnO2 + H
2
2Al + 2NaOH + 6H2O 2Na[Al(OH)
4] + 3H
2
48. Answer (2)
Hint: At low temperature para form dominant.
Sol.: At absolute zero only para form of hydrogen
exists.
Test - 4 (Code E) (Hints and Solutions) All India Aakash Test Series for Medical-2019
7/15
49. Answer (3)
Hint:
Reducing :
atomic >
nascent > dihydrogen.
ability hydrogen hydrogen
50. Answer (4)
Hint: Solubility order
CaCO3 < NaHCO
3 < KHCO
3
51. Answer (3)
Hint: Li is hardest alkali metal.
Sol. : Lithium being smallest in size shows
dominating covalent character.
52. Answer (3)
Hint: – O – O – linkage is peroxide linkage.
Sol.: BaO2 and Na
2O
2 are peroxides.
53. Answer (3)
Hint: Higher the lattice energy, higher the thermal
stability.
Sol.: Order of thermal stability : LiF > NaF > KF.
54. Answer (1)
Hint: Phosphorus give disproportionation reaction
with Ca(OH)2.
Sol.:
2P4 + 3Ca(OH)
2 + 6H
2O
3Ca(H
2PO
2)2
+ 2PH3
55. Answer (4)
Hint: Beryllium shows covalent nature in its
compounds.
Sol.: Be shows diagonal relationship with Al.
56. Answer (4)
Hint: NaH is ionic hydride.
Sol.: Li, Be and Mg form covalent hydrides.
57. Answer (1)
Hint: H4P
2O
5 contains only two P – OH bond.
HO – P – O – P – OH
OII
IH
OII
IH
58. Answer (3)
Hint: CNG consist of lower alkane as major
constituent.
59. Answer (3)
Hint: Higher the (charge/radius) value, higher is
hydration.
60. Answer (4)
Hint: Ammoniated electron gives blue colouration.
61. Answer (3)
Hint: Correct order of bond angle is
H2O > H
2S > H
2Te
62. Answer (1)
Hint: Solubility order
BeSO4 > MgSO
4 > BaSO
4
63. Answer (2)
Hint: CaO is a basic oxide
64. Answer (4)
Hint: 4 2 4 2
(gypsum) (Dead burnt plaster)
CaSO 2H O CaSO 2H O
65. Answer (3)
Hint: In carbon and oxygen d-orbitals are absent.
66. Answer (3)
Hint: 1H1,
1D2,
1T3 same electronic configuration
but different atomic masses.
Sol.: BE of D2 > BE of H
2
67. Answer (2)
Hint: Br in BrF5 is sp3d2 hybridised.
Br
F
F
F
F
F
68. Answer (1)
Hint: Higher the proton affinity higher is the
basicity.
Sol.: NH3 is most basic.
69. Answer (4)
Hint: Dielectric constant of H2O and D
2O
respectively are 78.39 and 78.06 C2 N–1 m–2.
70. Answer (3)
Hint: Solvay process
Sol.: NH3 + H
2O + CO
2 NH
4HCO
3
NH4HCO
3 + NaCl
4 3NH Cl + NaHCO
All India Aakash Test Series for Medical-2019 Test - 4 (Code E) (Hints and Solutions)
8/15
71. Answer (2)
Hint: Be(OH)2 is amphoteric in nature.
Sol.: Be(OH)2 + 2NaOH Na
2[Be(OH)
4].
72. Answer (4)
Hint: CaC2 + 2H
2O Ca(OH)
2 + C
2H
2
73. Answer (2)
Hint: NO2 is a brown coloured gas.
Sol.: 2NaNO3 2NaNO
2 + O
2
74. Answer (2)
Hint: Castner-Kellner cell is used to prepare NaOH.
Sol.: At cathode: HgNa e Na Hg
2Na – Hg + 2H2O 2NaOH + H
2 + 2Hg
At anode: 2
1Cl Cl e
2
75. Answer (1)
Hint: Be (OH)2 is amphoteric in nature.
Sol.: Basicity of second group hydroxides increases
down the group.
76. Answer (4)
Hint: Na2CO
3 + H
2O + CO
2 2NaHCO
3
77. Answer (1)
Hint: H, T, D are isotopes of hydrogen.
Sol.: H2, D
2, T
2, H–D, D–T, H–T
78. Answer (4)
Hint: Clark’s method for the removal of temporary
hardness.
79. Answer (2)
Hint: Formula of brown coloured complex is
[Fe(H2O)
5(NO)]2+
80. Answer (2)
Hint: 6.8 ml O2 at STP is obtained by 1 ml H
2O
2
solution by decomposition.
Sol.:
224 ml O2 at STP is obtained by
224ml
6.8H
2O
2.
= 32.94 ml H2O
2
81. Answer (2)
Hint: Cu + 8HNO3(dilute) 3Cu(NO
3)2 + 2NO + 4H
2O.
82. Answer (2)
Hint: CaCN2 is calcium cyanamide.
Sol.: CaCN2 + 3H
2O CaCO
3 + 2NH
3(g)
83. Answer (4)
Hint: NO2
is paramagnetic oxide of nitrogen.
N
O O
84. Answer (2)
Hint: H2O is neutral oxide.
Sol.: Oxides of alkali metals are basic.
85. Answer (2)
Hint: A suspension of magnesium hydroxide in water
is called milk of magnesia.
86. Answer (4)
Hint: 4Zn + 10HNO3(dilute) 4Zn(NO
3)2 + 5H
2O + N
2O.
87. Answer (4)
Hint: Spontaneous combustion of phosphine is used
in Holme’s signal.
88. Answer (3)
Hint: Cobalt is used as a catalyst in preparation of
methanol.
89. Answer (3)
Hint: H2O
2 oxidises PbS into PbSO
4.
90. Answer (2)
Hint: Zn reacts both with acid and base.
Sol.: Zn + 2HCl ZnCl2 + H
2
Zn + 2NaOH Na2ZnO
2 + H
2
BIOLOGY
91. Answer (1)
Hint : More the H+ ions, more acidic the condition
is and less will be the pH.
Sol. : During stomatal opening, due to the activity
of hydrogen-potassium ion-exchange pump, H+ from
guard cells are transported to neighbouring
subsidiary cells while K+ ions are transported into
the guard cells. This increases the pH of guard cells.
92. Answer (3)
Hint : Transpiration pull and root pressure causes
upliftment of water by pulling and pushing
respectively.
Test - 4 (Code E) (Hints and Solutions) All India Aakash Test Series for Medical-2019
9/15
Sol. : Transpiration develops a negative water
potential in the xylem which creates a ‘pull’ for
translocation of water while root pressure is a
positive hydrostatic pressure responsible for pushing
of water.
93. Answer (4)
Hint : In girdling experiment, a ring of bark upto the
depth of the phloem layer is carefully removed.
Sol. : In a girdled plant, root cells die first than the
shoot cells, due to stoppage of translocation of
sugars and other materials to the roots.
94. Answer (2)
Sol. : In guard cells, cellulose microfibrils are
radially oriented.
95. Answer (1)
Hint : Water molecules move from an area of high
water potential (w) to an area of low water potential
(w).
Sol. : w of cell A =
s +
p = – 2 atm
w of cell B = – DPD = – 10 atm
w of cell C = OP – TP = – 3 atm
w of cell D = – 6 atm
So the correct direction of movement of water is
A B
C D
↗
96. Answer (2)
Hint : In this process, loss of water in the form of
liquid droplets occurs through special openings
called hydathodes.
Sol. : Herbaceous plants lose water in the form of
droplets due to high root pressure. This process is
called guttation.
97. Answer (4)
Hint : Movement of sucrose from mesophyll cells to
sieve tube elements via companion cells, requires
energy in the form of ATP.
Sol. : Loading of sucrose in sieve tube cells via
companion cells is an active process.
98. Answer (4)
Hint : Facilitated diffusion is a passive transport.
Sol. : In facilitated diffusion, transport of ions/
molecules occurs with the help of specific membrane
proteins called transporters. This process is highly
specific because transporter proteins are highly
selective. The rate of facilitated transport may
saturate.
99. Answer (1)
Hint : Intercellular movement of water occurs
through cytoplasmic connections between
neighbouring cells.
Sol. : These structures are called plasmodesmata.
100. Answer (3)
Hint : Water potential (w) is free energy of water
molecules which decreases by adding solutes.
Sol. : Water potential of pure water is zero at
atmospheric pressure. By increasing external
pressure on pure water, its water potential increases.
101. Answer (2)
Hint : Mg is the co-factor of RuBisCO enzyme.
Sol. : Zn – Synthesis of auxin
K – Maintains turgidity of cells
Ni – Component of urease
102. Answer (2)
Hint : For immobile minerals, deficiency symptoms
tend to appear first in younger tissues.
Sol. : Among the given minerals, calcium is an
immobile element.
103. Answer (4)
Sol. : Na is a beneficial element.
104. Answer (1)
Hint : Ammonification involves decomposition of
organic nitrogen into ammonia.
Sol. : Ammonifying bacteria convert organic nitrogen
of dead plants and animals into ammonia.
105. Answer (2)
Hint : In green plants, Mn and Cl are involved in
splitting of water to release oxygen during
photosynthesis.
Sol. : Due to Mn toxicity translocation of Ca to
shoot apex is inhibited. Mn deficiency causes grey
spots in oats. Little leaf symptom is caused by the
deficiency of Zn. Transport of carbohydrate in phloem
is facilitated by boron.
106. Answer (3)
Hint : Leghaemoglobin is an O2 scavenger pigment
present in the cytosol of root nodule cells.
All India Aakash Test Series for Medical-2019 Test - 4 (Code E) (Hints and Solutions)
10/15
Sol. : After the formation of root nodule, bacteroids
present in root nodule cells produce a reddish-pink
pigment called leghaemoglobin which protects
nitrogenase from oxygen. Nodule formation is
induced by bacterial and plant signals.
107. Answer (4)
Hint : Nitrifying bacteria have chemoautotrophic
mode of nutrition.
Sol. : Nitrosomonas is a nitrifying bacterium which
converts NH3 into NO
2–.
108. Answer (3)
Hint : During nitrogen fixation atmospheric nitrogen
is converted into ammonia.
Sol. : Nitrogenase enzyme is a Mo – Fe protein. It
requires ATPs, strong reducing agents like FADH2,
NADPH and anaerobic condition for fixation of
nitrogen into ammonia.
109. Answer (4)
Hint : In C3 plants, CO
2 acceptor is a 5-carbon
containing molecule.
Sol. : In C3 plants, RuBP is the primary CO
2
acceptor molecule.
110. Answer (2)
Hint : Non-cyclic photophosphorylation is associated
with formation of assimilatory power and splitting of
water.
Sol. : Non-cyclic phosphorylation occurs only in
grana thylakoids. In this process both photosystems
work in a series, where PS II operates first, followed
by PS I. Products of non-cyclic photophosphorylation
are ATP, NADPH and O2.
111. Answer (4)
Hint : Dictyosomes are unconnected units of Golgi
body found in plant cells.
Sol. : Golgi bodies are not associated with
photorespiration. Organelles involved in photo
respiration are chloroplast, peroxisome and
mitochondria.
112. Answer (3)
Hint : C4 plants lack the wasteful process
responsible for loss of fixed CO2.
Sol. : In C3 plants, when there is high
concentration of O2 in stroma, oxygenase activity of
RuBisCO leads to photorespiratory loss. On the
other hand C4 plants have mechanism to increase
the CO2 concentration in comparison to O
2
concentration at the enzyme site, because of which
C4 plants do not have photorespiration. Thus
productivity of C4 plant is better than C
3 plants.
113. Answer (2)
Hint : Electron transport of non-cyclic
photophosphorylation starts with photosystem II.
Sol. : A = photosystem II
B = Photosystem I
C = Stroma
D = Lumen
114. Answer (4)
Hint : C4 plants have Kranz anatomy.
Sol. : In C4 plant, first carboxylation occurs in
mesophyll cells by PEPcase enzyme which
eventually leads to malic acid formation. Malic acid
gets decarboxylated in bundle sheath cells where
second carboxylation takes place by RuBisCO
enzyme.
115. Answer (4)
Hint : Sorghum shows spatial difference in double
carboxylation.
Sol. : Sorghum is a C4 plant.
116. Answer (1)
Hint : Photorespiration is a process which involves
loss of fixed carbon as CO2 in plants.
Sol. : Carboxylation, reduction and regeneration are
three steps of Calvin cycle.
117. Answer (2)
Hint : CO2 is the major limiting factor, influencing
the rate of photosynthesis.
Sol. : C3 plants show saturation at 450 ppm of
CO2, while C
4 plants show saturation at 360 ppm of
CO2 concentration at high light intensities. C
3 plants
show CO2 fertilization effect as in the CO
2 enriched
atmosphere they show higher yield.
118. Answer (3)
Hint : First action spectrum of photosynthesis was
prepared by T.W. Engelmann.
Sol. : Engelmann used a green alga Cladophora to
describe action spectrum of photosynthesis.
119. Answer (3)
Hint : Wheat is a C3 plant while maize is a C
4
plant.
Test - 4 (Code E) (Hints and Solutions) All India Aakash Test Series for Medical-2019
11/15
Sol. : For formation of 1 molecule of glucose
C3 plants require = 18 ATPs
C4 plants require = 30 ATPs
So for each glucose molecule, C3 plants require 12
ATPs less than C4 plants.
120. Answer (3)
Hint : The process through which electrons pass
from one carrier to another is called electron
transport system (ETS).
Sol. : In mitochondria, ETS is linked to ATP
synthesis (oxidative phosphorylation). Electron
carriers of ETS are present in inner mitochondrial
membrane.
121. Answer (1)
Hint : Fermentation is a kind of anaerobic
respiration, carried out primarily by fungi and
anaerobic bacteria.
Sol. : In fermentation, pyruvic acid is partially
oxidized in the presence of external electron donor
(reducing agent), NADH.
In fermentation, no new ATP or NADH molecules are
produced apart from those produced during
glycolysis.
122. Answer (2)
Hint : Succinyl CoA is an intermediate of Krebs
cycle, used in the biosynthesis of chlorophyll.
Sol. : Terminal electron acceptor of mitochondrial
ETS is O2.
Cofactor of pyruvic acid decarboxylase enzyme = Mg+2
Cofactor of lactic acid dehydrogenase enzyme = Zn+2.
123. Answer (3)
Hint : In human kidney cells, NADH molecules
produced in glycolysis enter the mitochondria via
malate-aspartate shuttle.
Sol. : In aerobic respiration, ATP produced by
oxidative phosphorylation is coupled with ETS.
By ETS and oxidative phosphorylation
1 NADH produces 3 ATPs
and 1 FADH2 produces 2 ATPs
Total NADH molecules produced in aerobic
respiration = 10 (30 ATPs)
Total FADH2 molecules produced in aerobic
respiration = 2 (4 ATPs) so total ATP produced by
ETS and oxidative phosphorylation is 34.
124. Answer (1)
Hint : Respiratory quotient of any respiratory
substrate is the ratio of volume of CO2 released and
volume of O2 consumed during respiration.
Sol. : RQ of carbohydrate = 1
RQ of Proteins (Albumin) = 0.9
RQ of Fats (Palmitic acid) = 0.7
RQ of Organic acids >1 (for oxalic acid it is 4)
So descending order of RQ values for given
substrates is oxalic acid > glucose > albumin >
palmitic acid.
125. Answer (4)
Hint : Respiration mediated breakdown of different
substrates such as carbohydrate, fats and proteins
show interconnection as all produce energy through
Krebs cycle eventually.
Sol. : Acetyl CoA is a metabolite which is a
common product during respiratory breakdown of fats,
proteins and carbohydrates. It is also a raw material
for synthesis of carotenoids, terpenes and gibberellins.
126. Answer (2)
Sol. : Induction of -amylase in barley endosperms
is a bioassay of gibberellins.
127. Answer (3)
Hint : Initially zygotic division shows geometric
growth pattern.
Sol. : In geometric growth, every cell divides with
all the daughter cells growing and dividing again.
128. Answer (3)
Hint : Some synthetic auxins are used as
weedicides.
Sol. : 2, 4-D and 2, 4, 5-T are synthetic auxins
which act as weedicides.
129. Answer (2)
Hint : Abscisic acid is called stress hormone.
Sol. : Abscisic acid induces dormancy in buds and
seeds whereas gibberellin prevents it.
ABA acts as antagonist of gibberellins.
130. Answer (2)
Sol. : Garner and Allard first reported the
photoperiodic response of flowering in tobacco.
131. Answer (4)
Hint : Apical dominance is a phenomenon in which
apical buds do not allow growth of lateral buds.
All India Aakash Test Series for Medical-2019 Test - 4 (Code E) (Hints and Solutions)
12/15
Sol. : Auxin promotes apical dominance while
cytokinin counteracts it.
132. Answer (3)
Hint : Phytohormone that counteracts apical
dominance is best suitable for tea plantation.
Sol. : Promotion of parthenocarpy – Auxin.
Antitranspiratory action – Abscisic acid
Root development in various cuttings – Auxin.
Cytokinin counteracts with auxin and help in making
bushes of tea plants.
133. Answer (2)
Hint : These cells are present just next to the cells
of meristematic zone.
Sol. : Increased vacuolation is a feature of the cells
which are in phase of elongation.
134. Answer (3)
Hint : Cytokinins are modified purines.
Sol. : Apart from cytokinin, auxin, gibberellin and
abscisic acid are acidic in nature.
135. Answer (4)
Hint : Thigmonasty is a contact or touch stimulated
variation of movement in plants like Drosera.
Sol. : Phytochromes are chromoproteins or
photoreceptor pigments which are used to detect
light by plants.
136. Answer (1)
Hint: This animal is also called bookworm.
Sol.: Loligo and Laccifer are economically beneficial
mollusc and insect respectively. Limulus is a living
fossil.
137. Answer (1)
Hint: Identify an arthropod.
Sol.: Pheretima (an annelid), Pavo (a bird) and
Petromyzon (cyclostome) have closed circulatory
system. Capillaries are a feature of closed circulatory
system.
138. Answer (4)
Hint: Metamorphosis involves conversion of larva to
adult form.
Sol.: Direct development occurs in Hirudinaria.
Butterfly, Ascidia and Culex show metamorphosis
and their larval forms are caterpillar, tadpole &
wriggler respectively.
139. Answer (3)
Hint: This parameter is shared by larval
echinoderms with platyhelminthes.
Sol.: Adult echinoderms are radially symmetrical
while their larvae are bilaterally symmetrical. Anus
develops from blastopore in deuterostome body plan.
140. Answer (3)
Hint: These organisms occur in exclusively marine
conditions.
Sol.: Saccoglossus a hemichordate, has proboscis
gland as its excretory organ. Gills are meant for
respiration in hemichordates. In molluscs, gills serve
both the function of respiration and excretion.
141. Answer (1)
Hint: This organism forms gemmules to reproduce.
Sol.: Sponges lack distinct germ layers, has ostia
for entry of water and has collar cells in body.
142. Answer (1)
Hint: Metagenesis.
Sol.: Physalia exists in both polyp and medusa
form. Hydra, Meandrina and Adamsia exist as
polyps.
143. Answer (2)
Hint: Identify organisms with cnidocytes.
Sol.: Adamsia is sea anemone i.e., a cnidarian
whose tentacles are rich in stinging cells.
Euspongia has flagellated choanocytes.
Pleurobrachia does not show metagenesis and
Planaria is a free living worm, hence it lacks hooks.
144. Answer (3)
Hint: Parapodia are lateral appendages that help in
swimming and respiration.
Sol.: Malpighian tubules are found in insects
(arthropods) while parapodia are found in annelids
such as Nereis.
145. Answer (2)
Hint: Identify a roundworm which gives birth to young
worms.
Sol.: Ancylostoma and Ascaris are oviparous round
worms. Fasciola is a flatworm.
146. Answer (3)
Hint: Identify worms that are pseudocoelomate.
Test - 4 (Code E) (Hints and Solutions) All India Aakash Test Series for Medical-2019
13/15
Sol.: Ringworm is a fungus; earthworm is an
annelid; bookworm is an arthropod; Liver fluke &
tapeworm are flatworms. Pinworm, filarial worm,
hookworm and roundworm belong to aschelminthes.
147. Answer (2)
Hint: Comb plates are locomotory structures in
members of this phylum.
Sol.: Annelids, arthropods and molluscs occupy
either aquatic or terrestrial habitats.
148. Answer (2)
Hint: This structure is composed of cellulose in
plants.
Sol.: Animal cells lack a cell wall. All animals are
heterotrophs and show division of labour. Poriferans
lack nervous system.
149. Answer (3)
Hint: Identify first triploblastic animals.
Sol.: Regeneration is a prominent feature of Porifera
Coelenterata, Platyhelminthes and Echinoderms.
150. Answer (4)
Hint: Select the organism that occurs exclusively in
marine water.
Sol.: Asterias has water vascular system and show
external fertilisation. Poriferans have water canal
system.
151. Answer (3)
Hint: Identify a true fish.
Sol.: Flying fish is a bony fish. Hagfish is a
cyclostome. Cuttlefish and devil fish belong to
phylum Mollusca.
152. Answer (2)
Hint: Identify a mollusc with scientific name
Dentalium.
Sol.: Tusk shell has a calcareous shell
(exoskeleton) while chitinous exoskeleton is a feature
of arthropods.
153. Answer (1)
Hint: Metamerism refers to presence of segments
and probable repeat of organs.
Sol.: Wuchereria is a filarial worm, where excretory
pore eliminates nitrogenous waste. Wastes present
in alimentary canal are eliminated through anus.
Pseudocoelom, absence of segmentation and
presence of bilateral symmetry are features of
Aschelminthes.
154. Answer (3)
Hint: This animal lives in loosely organised
community.
Sol.: Locusta is an economically harmful insect
(arthropod).
155. Answer (2)
Hint: Organisms that show bioluminescence.
Sol.: Flatworms like Schistosoma are dioecious but
digestion is only extracellular. In triploblastic animals
such as aschelminthes, extracellular digestion is
observed.
156. Answer (3)
Hint: Arthropods have open circulatory system.
Sol.: Pila is a mollusc with open circulatory system
and lacks segmentation. Anopheles has malpighian
tubules for excretion. Nereis is segmented and has
nephridia.
Presence of solid ventral nerve cord is a character of
non chordates.
157. Answer (2)
Hint: This organism is commonly called lac insect.
Sol.: Laccifer is a useful insect. Its secretion called
lac and shellac are used in bangle industry.
Anopheles, Culex and Aedes act as vectors to
spread diseases such as malaria, filariasis and
dengue respectively.
158. Answer (4)
Hint: Triploblastic organisms show extracellular
digestion.
Sol.: Adult echinoderms are radially symmetrical and
most poriferans are asymmetrical. Notochord is
absent in both Echinoderms and Poriferans. True
coelom is a feature of Echinoderms.
159. Answer (1)
Hint: Select an arthropod.
Sol.: Sea hare is Aplysia with unsegmented body.
Sea cucumber is dioecious. Tapeworm shows
internal fertilisation.
160. Answer (3)
Hint: Another name for this organism is Euspongia.
Sol.: Antedon and Sea urchin exhibit radial
symmetry in adult form. Salpa shows bilateral
symmetry.
All India Aakash Test Series for Medical-2019 Test - 4 (Code E) (Hints and Solutions)
14/15
161. Answer (1)
Hint: This structure divides body cavity into thoracic
and abdominal cavity.
Sol.: Mammals show direct development. Heterodont
dentition is absent in aquatic mammals such as
whale, sea cow, etc. Sloth and sea cow have 9 and
6 cervical vertebrae respectively.
162. Answer (2)
Hint: Select an avian adaptation.
Sol.: Feathers reduce weight and are found in birds.
163. Answer (2)
Hint: Pneumatic bones have air cavities to reduce
weight of flying birds.
Sol.: Neophron i.e. vulture is a flying bird that has
both pneumatic bones and preen/oil gland. Air sacs
in birds are avascular and meant for storage of air
but not exchange of gases.
164. Answer (4)
Hint: Snakes detect vibrations through jaw bones.
Sol.: Vipera has epidermal scales; 3 chambered
heart and is limbless in adult form. It has dry
cornified aglandular skin.
165. Answer (1)
Hint: Such an organism produces internal body heat
through metabolic activity to maintain body
temperature.
Sol.: Mammals are homeotherms and endotherms.
They exhibit internal fertilisation (Felis) as do birds.
Chelone inhabits water. Replites show shedding of
skin (ecdysis).
166. Answer (2)
Hint: Heart and blood vessels are present in open
and closed circulatory system.
Sol.: Heart is dorsal in position in non-chordates
usually while it is ventral in chordates. Nerve cord is
ventral in non-chordates but dorsal in position in
chordates. Post anal tail is a feature of chordates.
Gill slits are lateral in position in chordates.
167. Answer (2)
Hint: Identify a circular mouth scaleless fish like
organism.
Sol.: Myxine is hagfish, a cyclostome. Torpedo,
Pristis and Trygon belong to super class Pisces.
168. Answer (1)
Hint: Select a cartilaginous fish.
Sol.: Air/swim bladder helps to maintain buoyancy in
bony fish. Dog fish has to swim continuously to
avoid sinking.
169. Answer (3)
Hint: Dry, aglandular skin is a feature of reptiles.
Sol.: Ichthyophis is a limbless amphibian
Bangarus is a limbless reptile
Balaenoptera is a limbless mammal
They all have bony vertebral column. Skin is glandular
in mammals.
170. Answer (4)
Hint: Identify an animal where development of
embryo can occur on land.
Sol.: Pisces and amphibians are anamniotes. Their
development is linked to water directly. Reptiles
evolved prior to birds. Mammals, birds and reptiles
are all amniotes.
171. Answer (2)
Hint: Gill cover is absent in cartilaginous fish.
Sol.: Chondrichthyes are cartilaginous fish which
lack operculum generally; their males have pelvic
claspers and show internal fertilisation. Bony fishes
possess operculum, have cycloid/ctenoid scales and
show external fertilisation.
172. Answer (4)
Hint: It is another term for arthrodial membrane.
Sol.: Arthrodial membrane is flexible membrane that
connects tergite with pleurites and sternites.
173. Answer (2)
Hint: First pair of wings is mesothoracic.
Sol.: Tegmina/Elytra or wing cover is the alternative
name for first pair of wings in cockroach.
174. Answer (3)
Hint: Identify mouth parts that are paired in
cockroach.
Sol.: Hypopharynx acts as tongue. Maxillae are
paired structures that help bring food to mandibles.
175. Answer (1)
Hint: Glands whose ducts open into brood or genital
pouch.
Sol.: Seminal vesicles nourish the sperms. Phallic/
conglobate secrete substances that help to form a
protective cover of spermatophore.
Test - 4 (Code E) (Hints and Solutions) All India Aakash Test Series for Medical-2019
15/15
176. Answer (3)
Hint: Sexual dimorphism in cockroach is based on
presence of unjointed filamentous structures attached
to 9th sternum.
Sol.: Anal/caudal styles are unjointed structures
attached to 9th sternum in male cockroach. Anal
cerci are jointed structures present in both male and
female cockroach.
177. Answer (4)
Hint: External segmentation is found in arthropods.
Sol.: Schizocoelom and presence of external
metamerism/segmentation is a feature of Periplaneta.
178. Answer (4)
Hint: Part of foregut lined by cuticle is not secretory
in nature.
Sol.: Crop part of foregut in cockroach does not
secrete enzymes. Haemoglobin is absent in blood of
cockroach. Hepatic caecae are present at the
junction of foregut and midgut.
179. Answer (1)
Hint: These paired structures help in respiration in
cockroach.
Sol.: 10 pairs of spiracles are present in body of
Periplaneta.
180. Answer (3)
Hint: Appositional image.
Sol.: Vision in cockroach has low resolution but high
sensitivity.
�����
Test - 4 (Code F) (Answers) All India Aakash Test Series for Medical-2019
1/15
1. (4)
2. (3)
3. Deleted
4. (1)
5. (3)
6. (2)
7. (1)
8. (2)
9. (4)
10. (3)
11. (2)
12. (3)
13. (1)
14. (2)
15. (4)
16. (1)
17. (1)
18. (4)
19. (2)
20. (4)
21. (3)
22. (3)
23. (4)
24. (1)
25. (3)
26. (3)
27. (4)
28. (3)
29. (3)
30. (2)
31. (3)
32. (2)
33. (2)
34. (2)
35. (2)
36. (3)
Test Date : 20/01/2019
ANSWERS
TEST - 4 (Code F)
All India Aakash Test Series for Medical - 2019
37. (1)
38. (2)
39. (2)
40. (3)
41. (1)
42. (2)
43. (2)
44. (3)
45. (1)
46. (2)
47. (3)
48. (3)
49. (4)
50. (4)
51. (2)
52. (2)
53. (4)
54. (2)
55. (2)
56. (2)
57. (2)
58. (4)
59. (1)
60. (4)
61. (1)
62. (2)
63. (2)
64. (4)
65. (2)
66. (3)
67. (4)
68. (1)
69. (2)
70. (3)
71. (3)
72. (4)
73. (2)
74. (1)
75. (3)
76. (4)
77. (3)
78. (3)
79. (1)
80. (4)
81. (4)
82. (1)
83. (3)
84. (3)
85. (3)
86. (4)
87. (3)
88. (2)
89. (3)
90. (2)
91. (4)
92. (3)
93. (2)
94. (3)
95. (4)
96. (2)
97. (2)
98. (3)
99. (3)
100. (2)
101. (4)
102. (1)
103. (3)
104. (2)
105. (1)
106. (3)
107. (3)
108. (3)
109. (2)
110. (1)
111. (4)
112. (4)
113. (2)
114. (3)
115. (4)
116. (2)
117. (4)
118. (3)
119. (4)
120. (3)
121. (2)
122. (1)
123. (4)
124. (2)
125. (2)
126. (3)
127. (1)
128. (4)
129. (4)
130. (2)
131. (1)
132. (2)
133. (4)
134. (3)
135. (1)
136. (3)
137. (1)
138. (4)
139. (4)
140. (3)
141. (1)
142. (3)
143. (2)
144. (4)
145. (2)
146. (4)
147. (3)
148. (1)
149. (2)
150. (2)
151. (1)
152. (4)
153. (2)
154. (2)
155. (1)
156. (3)
157. (1)
158. (4)
159. (2)
160. (3)
161. (2)
162. (3)
163. (1)
164. (2)
165. (3)
166. (4)
167. (3)
168. (2)
169. (2)
170. (3)
171. (2)
172. (3)
173. (2)
174. (1)
175. (1)
176. (3)
177. (3)
178. (4)
179. (1)
180. (1)
All India Aakash Test Series for Medical-2019 Test - 4 (Code F) (Hints and Solutions)
2/15
PHYSICS
1. Answer (4)
Hint: U = nCvT
Sol.: Since5
2v
RC for diatomic
3
2v
RC for monoatomic.
Slopev
C
2. Answer (3)
Hint: Damped and forced oscillation.
Sol.: In damped oscillation amplitude and energy
both decreases exponentially.
At resonance amplitude becomes maximum.
3. Deleted
4. Answer (1)
Hint and Sol.: 2m
Tk
5. Answer (3)
Hint: 2
AV
1KE
3kA
Sol.: ( )AV
0
1KE KE
A
xdx
A
on solving, 2 2
AV
1(KE)
3m A
6. Answer (2)
Hint: x = A sin(t + )Sol.:
45°
45°
t t =
A(1)
t = 0
(2)
(2)
x
mean
(t t) =
t = 0
x = Acos45° 2
A
HINTS AND SOLUTIONS
7. Answer (1)
Hint: 2 2
net 1 2 1 22 cosA A A A A .
Sol.: 2A
2A
60°
60°
A
Anet
= 2A – A = A
8. Answer (2)
Hint: fbeat
= |f1 – f
2|
Sol.: fB – f
A = 2
1 1
30 25A Bf f
25
30
A
B
f
f
On solving
fA = 10 Hz; f
B = 12 Hz
9. Answer (4)
Hint: fBeat
= |f1 – f
2|
Sol.:
fv
s
Frequency of reflected sound.
–
r
c vf f
c v
Beat
2–
–r
vff f f
c v
10. Answer (3)
Hint: Doppler’s effect depends on relative motion.
11. Answer (2)
Hint and Sol.: Phase difference between incident
and reflected wave is when reflected from rigid
boundary.
Test - 4 (Code F) (Hints and Solutions) All India Aakash Test Series for Medical-2019
3/15
12. Answer (3)
Hint: P
v
Sol.:air
air
Pv
2
2
H
H
Pv
2air
4 4 332 1328 m/sH
v v
13. Answer (1)
Hint: max
waveand
Pv A v
k
Sol.: sin( )y A t kx
= 50; k = 5
10 m/svk
maxp
v = A = 10 × 50 = 500 m/s
max
50p
v
v
14. Answer (2)
Hint and Sol.: particle
=
y
vt
15. Answer (4)
Hint: = kx
Sol.: Path difference 3 5
–4 8 8
x
2 5
8
5
4
16. Answer (1)
Hint: RT
vM
Sol.: 2 RTv
M
v2 T (T in Kelvin)
17. Answer (1)
Hint: 2
nf v
l
Sol.: 3
3
2f v
l v = 48 m/s
f3 = 48 Hz l = 1.5 m
1 mv
f
18. Answer (4)
Hint: 2L
Tg
Sol.: L = 2.45 m
g = 9.8 m/s2
2 1sL
Tg
19. Answer (2)
Hint and Sol.: All medium particles between two
successive nodes oscillate in same phase and
particles on one side of a node oscillate in opposite
phase with those on the other side of same node.
20. Answer (4)
Hint: 0
S.L 10logI
I
Hint: 0
S.L 10logI
I
Sol.: 1
1
0
S.L 10logI
I
2
2
0
S.L 10logI
I
S.L = S.L2 – S.L
1 = 30
2 11000I I
21. Answer (3)
Hint and Sol.: Doppler’s effect depends on relative
motion.
22. Answer (3)
Hint: 1 2
(2 – 1),
2 4o c
nv nf f v
l l
All India Aakash Test Series for Medical-2019 Test - 4 (Code F) (Hints and Solutions)
4/15
Sol.: 1 2
1 5;
2 4o c
vf f v
l l
fo = f
c
1 2
1 5
2 4l l
2 1
5
2l l
l1 = 10 cm; l
2 = 25 cm.
23. Answer (4)
Hint: In closed organ pipe only odd harmonics are
present.
Sol.: 50 × 1 = 50, 50 × 3 = 150, 50 × 5 = 250.
Odd harmonics are present. So, pipe is closed at
one end.
340 34m
50 5
v
f
34 171.7 m
4 20 10l
24. Answer (1)
Hint: 0
0
S
v vf f
v v
Sol.:
fref.
f0
v
Reflected frequency
ref. 0–
c vf f
c v
25. Answer (3)
Hint: v = f
Sol.: v = f1
1 and v = f
2
2
1
1
vf
and
2
2
vf
1 2
1 2
1 1– –f f v
f1 – f
2 = n
1 2
2 1–
n
v
26. Answer (3)
Hint & Sol.: Two waves of same frequency & moves
in opposite direction.
27. Answer (4)
Hint: vparticle
= –(Slope) vwave
Sol.: Acceleration of particle always along their mean
position.
vA is upward and v
C is downward.
28. Answer (3)
Hint: Conservation of ME and linear momentum.
Sol.: 2 21 1( )
2 2M m v Kx
and mv0 = (M + m)v
x = Amplitude of oscillation
On solving
0mv M m
xM m K
29. Answer (3)
Hint: 2 2
Resultant 1 2 1 22 cosA A A A A
Sol.: x = x1 + x
2
3 sin( )x A t
max3 ;A A
2 2 2
net2 cos60A A A A
∵
amax
= 2Amax
23 A
30. Answer (2)
Hint: a = –2x
Sol.: a = –2x
2–B A B
A
2 B
T A
2A
TB
31. Answer (3)
Hint: If y = f(ax ± bt)
then b
va
Test - 4 (Code F) (Hints and Solutions) All India Aakash Test Series for Medical-2019
5/15
Sol.: 2
9
3 ( – 10 )y
x t
v = 10 m/s
32. Answer (2)
Hint: 2R
Tg
Sol.: Time period of particle is independent of length
of chord.
33. Answer (2)
Hint: 2 2 2– –a x v A x
Sol.: a = v
2 2 2–x A x
2x2 = A2 – x2
At 21 2 second
2
Ax T
34. Answer (2)
Hint: 0
2H
Tg
Sol.: At equilibrium condition
FB = weight
AHg = Ah0g
H = h0
[A is circular area of cylinder]
2h
Tg
35. Answer (2)
Hint: 2 2 21 1and KE ( – )
2 2U kx k A x
Sol.: As body moves from mean position, speed
decreases. So, KE decreases and PE increases.
36. Answer (3)
Hint: a = –2x
Sol.: 2 2 2
2 2–
–
aT x TT
x x
– 2 2constantT
37. Answer (1)
Hint: PV = nRT
Sol.:nR
P TV
As, SlopenR
V is constant.
So, V = constant, W = 0 and = constant
38. Answer (2)
Hint: 0
1V
T V
Sol.: PV = nRT
P V = nRT
V nR
T P
0
nR nR
PV nRT
1
T
n = –1
39. Answer (2)
Hint: (mix)
mix
mix
P
V
C
C
Sol.: 1 21 2
(mix)
1 2
P P
P
nC n CC
n n
1 21 2
(mix)
1 2
V V
V
nC n CC
n n
(mix)
5 7
2 23
2P
R R
C R
CV (mix)
= 2R
2
3
V
P
C
C
All India Aakash Test Series for Medical-2019 Test - 4 (Code F) (Hints and Solutions)
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40. Answer (3)
Hint: PV = nRT
Sol.: PV = nRT
V nR
T P
Slopen
P
tan60A
A
n
P
tan30B
B
n
P
1A
B
P
P
41. Answer (1)
Hint: Sound RMS
3v v
Sol.: RMS
3RTv
M
Sound
RTv
M
RMS av mp > > v v v
42. Answer (2)
Hint: PV = nRT
Sol.: PV = nRT
Mass( Molecular weight)PV RT M
M
MassPVR
T M
massPV R
T M
1Slope
M
43. Answer (2)
Hint: avg2
fKTE for per molecule
Sol.: rms
3RTV
M
44. Answer (3)
Hint: RMS
3RTV
M
Sol.: RMS
3
Mass of gas
PVV
At constant volume RMS
V P
45. Answer (1)
Hint: Wave transport energy and momentum.
Sol.: Transverse wave do not propagate inside a
fluid.
CHEMISTRY
46. Answer (2)
Hint: Zn reacts both with acid and base.
Sol.: Zn + 2HCl ZnCl2 + H
2
Zn + 2NaOH Na2ZnO
2 + H
2
47. Answer (3)
Hint: H2O
2 oxidises PbS into PbSO
4.
48. Answer (3)
Hint: Cobalt is used as a catalyst in preparation of
methanol.
49. Answer (4)
Hint: Spontaneous combustion of phosphine is used
in Holme’s signal.
50. Answer (4)
Hint: 4Zn + 10HNO3(dilute) 4Zn(NO
3)2 + 5H
2O + N
2O.
51. Answer (2)
Hint: A suspension of magnesium hydroxide in water
is called milk of magnesia.
52. Answer (2)
Hint: H2O is neutral oxide.
Sol.: Oxides of alkali metals are basic.
Test - 4 (Code F) (Hints and Solutions) All India Aakash Test Series for Medical-2019
7/15
53. Answer (4)
Hint: NO2
is paramagnetic oxide of nitrogen.
N
O O
54. Answer (2)
Hint: CaCN2 is calcium cyanamide.
Sol.: CaCN2 + 3H
2O CaCO
3 + 2NH
3(g)
55. Answer (2)
Hint: Cu + 8HNO3(dilute) 3Cu(NO
3)2 + 2NO + 4H
2O.
56. Answer (2)
Hint: 6.8 ml O2 at STP is obtained by 1 ml H
2O
2
solution by decomposition.
Sol.:
224 ml O2 at STP is obtained by
224ml
6.8H
2O
2.
= 32.94 ml H2O
2
57. Answer (2)
Hint: Formula of brown coloured complex is
[Fe(H2O)
5(NO)]2+
58. Answer (4)
Hint: Clark’s method for the removal of temporary
hardness.
59. Answer (1)
Hint: H, T, D are isotopes of hydrogen.
Sol.: H2, D
2, T
2, H–D, D–T, H–T
60. Answer (4)
Hint: Na2CO
3 + H
2O + CO
2 2NaHCO
3
61. Answer (1)
Hint: Be (OH)2 is amphoteric in nature.
Sol.: Basicity of second group hydroxides increases
down the group.
62. Answer (2)
Hint: Castner-Kellner cell is used to prepare NaOH.
Sol.: At cathode: HgNa e Na Hg
2Na – Hg + 2H2O 2NaOH + H
2 + 2Hg
At anode: 2
1Cl Cl e
2
63. Answer (2)
Hint: NO2 is a brown coloured gas.
Sol.: 2NaNO3 2NaNO
2 + O
2
64. Answer (4)
Hint: CaC2 + 2H
2O Ca(OH)
2 + C
2H
2
65. Answer (2)
Hint: Be(OH)2 is amphoteric in nature.
Sol.: Be(OH)2 + 2NaOH Na
2[Be(OH)
4].
66. Answer (3)
Hint: Solvay process
Sol.: NH3 + H
2O + CO
2 NH
4HCO
3
NH4HCO
3 + NaCl
4 3NH Cl + NaHCO
67. Answer (4)
Hint: Dielectric constant of H2O and D
2O
respectively are 78.39 and 78.06 C2 N–1 m–2.
68. Answer (1)
Hint: Higher the proton affinity higher is the
basicity.
Sol.: NH3 is most basic.
69. Answer (2)
Hint: Br in BrF5 is sp3d 2 hybridised.
Br
F
F
F
F
F
70. Answer (3)
Hint: 1H1,
1D2,
1T3 same electronic configuration
but different atomic masses.
Sol.: BE of D2 > BE of H
2
71. Answer (3)
Hint: In carbon and oxygen d-orbitals are absent.
72. Answer (4)
Hint: 4 2 4 2
(gypsum) (Dead burnt plaster)
CaSO 2H O CaSO 2H O
73. Answer (2)
Hint: CaO is a basic oxide
All India Aakash Test Series for Medical-2019 Test - 4 (Code F) (Hints and Solutions)
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74. Answer (1)
Hint: Solubility order
BeSO4 > MgSO
4 > BaSO
4
75. Answer (3)
Hint: Correct order of bond angle is
H2O > H
2S > H
2Te
76. Answer (4)
Hint: Ammoniated electron gives blue colouration.
77. Answer (3)
Hint: Higher the (charge/radius) value, higher is
hydration.
78. Answer (3)
Hint: CNG consist of lower alkane as major
constituent.
79. Answer (1)
Hint: H4P
2O
5 contains only two P – OH bond.
HO – P – O – P – OH
OII
IH
OII
IH
80. Answer (4)
Hint: NaH is ionic hydride.
Sol.: Li, Be and Mg form covalent hydrides.
81. Answer (4)
Hint: Beryllium shows covalent nature in its
compounds.
Sol.: Be shows diagonal relationship with Al.
82. Answer (1)
Hint: Phosphorus give disproportionation reaction
with Ca(OH)2.
Sol.:
2P4 + 3Ca(OH)
2 + 6H
2O
3Ca(H
2PO
2)2
+ 2PH3
83. Answer (3)
Hint: Higher the lattice energy, higher the thermal
stability.
Sol.: Order of thermal stability : LiF > NaF > KF.
84. Answer (3)
Hint: – O – O – linkage is peroxide linkage.
Sol.: BaO2 and Na
2O
2 are peroxides.
85. Answer (3)
Hint: Li is hardest alkali metal.
Sol. : Lithium being smallest in size shows
dominating covalent character.
86. Answer (4)
Hint: Solubility order
CaCO3 < NaHCO
3 < KHCO
3
87. Answer (3)
Hint:
Reducing :
atomic >
nascent > dihydrogen.
ability hydrogen hydrogen
88. Answer (2)
Hint: At low temperature para form dominant.
Sol.: At absolute zero only para form of hydrogen
exists.
89. Answer (3)
Hint:
Zn + 2NaOH Na2
ZnO2 + H
2
2Al + 2NaOH + 6H2O 2Na[Al(OH)
4] + 3H
2
90. Answer (2)
Hint: 32
%wt. 100 94.12%34
BIOLOGY
91. Answer (4)
Hint: Thigmonasty is a contact or touch stimulated
variation of movement in plants like Drosera.
Sol.: Phytochromes are chromoproteins or
photoreceptor pigments which are used to detect
light by plants.
92. Answer (3)
Hint: Cytokinins are modified purines.
Sol.: Apart from cytokinin, auxin, gibberellin and
abscisic acid are acidic in nature.
93. Answer (2)
Hint: These cells are present just next to the cells
of meristematic zone.
Sol.: Increased vacuolation is a feature of the cells
which are in phase of elongation.
Test - 4 (Code F) (Hints and Solutions) All India Aakash Test Series for Medical-2019
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94. Answer (3)
Hint: Phytohormone that counteracts apical
dominance is best suitable for tea plantation.
Sol.: Promotion of parthenocarpy – Auxin.
Antitranspiratory action – Abscisic acid
Root development in various cuttings – Auxin.
Cytokinin counteracts with auxin and help in making
bushes of tea plants.
95. Answer (4)
Hint: Apical dominance is a phenomenon in which
apical buds do not allow growth of lateral buds.
Sol.: Auxin promotes apical dominance while
cytokinin counteracts it.
96. Answer (2)
Sol.: Garner and Allard first reported the
photoperiodic response of flowering in tobacco.
97. Answer (2)
Hint: Abscisic acid is called stress hormone.
Sol.: Abscisic acid induces dormancy in buds and
seeds whereas gibberellin prevents it.
ABA acts as antagonist of gibberellins.
98. Answer (3)
Hint: Some synthetic auxins are used as
weedicides.
Sol.: 2, 4-D and 2, 4, 5-T are synthetic auxins
which act as weedicides.
99. Answer (3)
Hint: Initially zygotic division shows geometric growth
pattern.
Sol.: In geometric growth, every cell divides with all
the daughter cells growing and dividing again.
100. Answer (2)
Sol.: Induction of -amylase in barley endosperms
is a bioassay of gibberellins.
101. Answer (4)
Hint: Respiration mediated breakdown of different
substrates such as carbohydrate, fats and proteins
show interconnection as all produce energy through
Krebs cycle eventually.
Sol.: Acetyl CoA is a metabolite which is a common
product during respiratory breakdown of fats, proteins
and carbohydrates. It is also a raw material for
synthesis of carotenoids, terpenes and gibberellins.
102. Answer (1)
Hint: Respiratory quotient of any respiratory
substrate is the ratio of volume of CO2 released and
volume of O2 consumed during respiration.
Sol.: RQ of carbohydrate = 1
RQ of Proteins (Albumin) = 0.9
RQ of Fats (Palmitic acid) = 0.7
RQ of Organic acids >1 (for oxalic acid it is 4)
So descending order of RQ values for given
substrates is oxalic acid > glucose > albumin >
palmitic acid.
103. Answer (3)
Hint: In human kidney cells, NADH molecules
produced in glycolysis enter the mitochondria via
malate-aspartate shuttle.
Sol.: In aerobic respiration, ATP produced by
oxidative phosphorylation is coupled with ETS.
By ETS and oxidative phosphorylation
1 NADH produces 3 ATPs
and 1 FADH2 produces 2 ATPs
Total NADH molecules produced in aerobic
respiration = 10 (30 ATPs)
Total FADH2 molecules produced in aerobic
respiration = 2 (4 ATPs) so total ATP produced by
ETS and oxidative phosphorylation is 34.
104. Answer (2)
Hint: Succinyl CoA is an intermediate of Krebs
cycle, used in the biosynthesis of chlorophyll.
Sol.: Terminal electron acceptor of mitochondrial
ETS is O2.
Cofactor of pyruvic acid decarboxylase enzyme = Mg+2
Cofactor of lactic acid dehydrogenase enzyme = Zn+2.
105. Answer (1)
Hint: Fermentation is a kind of anaerobic respiration,
carried out primarily by fungi and anaerobic bacteria.
Sol.: In fermentation, pyruvic acid is partially oxidized
in the presence of external electron donor (reducing
agent), NADH.
In fermentation, no new ATP or NADH molecules are
produced apart from those produced during
glycolysis.
All India Aakash Test Series for Medical-2019 Test - 4 (Code F) (Hints and Solutions)
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106. Answer (3)
Hint: The process through which electrons pass
from one carrier to another is called electron
transport system (ETS).
Sol.: In mitochondria, ETS is linked to ATP
synthesis (oxidative phosphorylation). Electron
carriers of ETS are present in inner mitochondrial
membrane.
107. Answer (3)
Hint: Wheat is a C3 plant while maize is a C
4 plant.
Sol.: For formation of 1 molecule of glucose
C3 plants require = 18 ATPs
C4 plants require = 30 ATPs
So for each glucose molecule, C3 plants require 12
ATPs less than C4 plants.
108. Answer (3)
Hint: First action spectrum of photosynthesis was
prepared by T.W. Engelmann.
Sol.: Engelmann used a green alga Cladophora to
describe action spectrum of photosynthesis.
109. Answer (2)
Hint: CO2 is the major limiting factor, influencing the
rate of photosynthesis.
Sol.: C3 plants show saturation at 450 ppm of CO
2,
while C4 plants show saturation at 360 ppm of CO
2
concentration at high light intensities. C3 plants
show CO2 fertilization effect as in the CO
2 enriched
atmosphere they show higher yield.
110. Answer (1)
Hint: Photorespiration is a process which involves
loss of fixed carbon as CO2 in plants.
Sol.: Carboxylation, reduction and regeneration are
three steps of Calvin cycle.
111. Answer (4)
Hint: Sorghum shows spatial difference in double
carboxylation.
Sol.: Sorghum is a C4 plant.
112. Answer (4)
Hint: C4 plants have Kranz anatomy.
Sol.: In C4 plant, first carboxylation occurs in
mesophyll cells by PEPcase enzyme which
eventually leads to malic acid formation. Malic acid
gets decarboxylated in bundle sheath cells where
second carboxylation takes place by RuBisCO
enzyme.
113. Answer (2)
Hint: Electron transport of non-cyclic
photophosphorylation starts with photosystem II.
Sol.: A = photosystem II
B = Photosystem I
C = Stroma
D = Lumen
114. Answer (3)
Hint: C4 plants lack the wasteful process
responsible for loss of fixed CO2.
Sol.: In C3 plants, when there is high concentration
of O2 in stroma, oxygenase activity of RuBisCO
leads to photorespiratory loss. On the other hand C4
plants have mechanism to increase the CO2
concentration in comparison to O2 concentration at
the enzyme site, because of which C4 plants do not
have photorespiration. Thus productivity of C4 plant is
better than C3 plants.
115. Answer (4)
Hint: Dictyosomes are unconnected units of Golgi
body found in plant cells.
Sol.: Golgi bodies are not associated with
photorespiration. Organelles involved in photo
respiration are chloroplast, peroxisome and
mitochondria.
116. Answer (2)
Hint: Non-cyclic photophosphorylation is associ-ated
with formation of assimilatory power and splitting of
water.
Sol.: Non-cyclic phosphorylation occurs only in
grana thylakoids. In this process both photosystems
work in a series, where PS II operates first, followed
by PS I. Products of non-cyclic photophosphorylation
are ATP, NADPH and O2.
117. Answer (4)
Hint: In C3 plants, CO
2 acceptor is a 5-carbon
containing molecule.
Sol.: In C3 plants, RuBP is the primary CO
2 acceptor
molecule.
118. Answer (3)
Hint: During nitrogen fixation atmospheric nitrogen is
converted into ammonia.
Sol.: Nitrogenase enzyme is a Mo – Fe protein. It
requires ATPs, strong reducing agents like FADH2,
NADPH and anaerobic condition for fixation of
nitrogen into ammonia.
Test - 4 (Code F) (Hints and Solutions) All India Aakash Test Series for Medical-2019
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119. Answer (4)
Hint: Nitrifying bacteria have chemoautotrophic mode
of nutrition.
Sol.: Nitrosomonas is a nitrifying bacterium which
converts NH3 into NO
2–.
120. Answer (3)
Hint: Leghaemoglobin is an O2 scavenger pigment
present in the cytosol of root nodule cells.
Sol.: After the formation of root nodule, bacteroids
present in root nodule cells produce a reddish-pink
pigment called leghaemoglobin which protects
nitrogenase from oxygen. Nodule formation is
induced by bacterial and plant signals.
121. Answer (2)
Hint: In green plants, Mn and Cl are involved in
splitting of water to release oxygen during
photosynthesis.
Sol.: Due to Mn toxicity translocation of Ca to shoot
apex is inhibited. Mn deficiency causes grey spots
in oats. Little leaf symptom is caused by the
deficiency of Zn. Transport of carbohydrate in phloem
is facilitated by boron.
122. Answer (1)
Hint: Ammonification involves decomposition of
organic nitrogen into ammonia.
Sol.: Ammonifying bacteria convert organic nitrogen
of dead plants and animals into ammonia.
123. Answer (4)
Sol.: Na is a beneficial element.
124. Answer (2)
Hint: For immobile minerals, deficiency symptoms
tend to appear first in younger tissues.
Sol.: Among the given minerals, calcium is an
immobile element.
125. Answer (2)
Hint: Mg is the co-factor of RuBisCO enzyme.
Sol.: Zn – Synthesis of auxin
K – Maintains turgidity of cells
Ni – Component of urease
126. Answer (3)
Hint: Water potential (w) is free energy of water
molecules which decreases by adding solutes.
Sol.: Water potential of pure water is zero at
atmospheric pressure. By increasing external
pressure on pure water, its water potential increases.
127. Answer (1)
Hint: Intercellular movement of water occurs through
cytoplasmic connections between neighbouring cells.
Sol.: These structures are called plasmodesmata.
128. Answer (4)
Hint: Facilitated diffusion is a passive transport.
Sol.: In facilitated diffusion, transport of ions/
molecules occurs with the help of specific membrane
proteins called transporters. This process is highly
specific because transporter proteins are highly
selective. The rate of facilitated transport may
saturate.
129. Answer (4)
Hint: Movement of sucrose from mesophyll cells to
sieve tube elements via companion cells, requires
energy in the form of ATP.
Sol.: Loading of sucrose in sieve tube cells via
companion cells is an active process.
130. Answer (2)
Hint: In this process, loss of water in the form of
liquid droplets occurs through special openings
called hydathodes.
Sol.: Herbaceous plants lose water in the form of
droplets due to high root pressure. This process is
called guttation.
131. Answer (1)
Hint: Water molecules move from an area of high
water potential (w) to an area of low water potential
(w).
Sol.: w of cell A =
s +
p = – 2 atm
w of cell B = – DPD = – 10 atm
w of cell C = OP – TP = – 3 atm
w of cell D = – 6 atm
So the correct direction of movement of water is
A B
C D
↗
132. Answer (2)
Sol.: In guard cells, cellulose microfibrils are radially
oriented.
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133. Answer (4)
Hint: In girdling experiment, a ring of bark upto the
depth of the phloem layer is carefully removed.
Sol.: In a girdled plant, root cells die first than the
shoot cells, due to stoppage of translocation of
sugars and other materials to the roots.
134. Answer (3)
Hint: Transpiration pull and root pressure causes
upliftment of water by pulling and pushing
respectively.
Sol.: Transpiration develops a negative water
potential in the xylem which creates a ‘pull’ for
translocation of water while root pressure is a
positive hydrostatic pressure responsible for pushing
of water.
135. Answer (1)
Hint: More the H+ ions, more acidic the condition is
and less will be the pH.
Sol.: During stomatal opening, due to the activity of
hydrogen-potassium ion-exchange pump, H+ from
guard cells are transported to neighbouring
subsidiary cells while K+ ions are transported into
the guard cells. This increases the pH of guard cells.
136. Answer (3)
Hint: Appositional image.
Sol.: Vision in cockroach has low resolution but high
sensitivity.
137. Answer (1)
Hint: These paired structures help in respiration in
cockroach.
Sol.: 10 pairs of spiracles are present in body of
Periplaneta.
138. Answer (4)
Hint: Part of foregut lined by cuticle is not secretory
in nature.
Sol.: Crop part of foregut in cockroach does not
secrete enzymes. Haemoglobin is absent in blood of
cockroach. Hepatic caecae are present at the
junction of foregut and midgut.
139. Answer (4)
Hint: External segmentation is found in arthropods.
Sol.: Schizocoelom and presence of external
metamerism/segmentation is a feature of Periplaneta.
140. Answer (3)
Hint: Sexual dimorphism in cockroach is based on
presence of unjointed filamentous structures attached
to 9th sternum.
Sol.: Anal/caudal styles are unjointed structures
attached to 9th sternum in male cockroach. Anal
cerci are jointed structures present in both male and
female cockroach.
141. Answer (1)
Hint: Glands whose ducts open into brood or genital
pouch.
Sol.: Seminal vesicles nourish the sperms. Phallic/
conglobate secrete substances that help to form a
protective cover of spermatophore.
142. Answer (3)
Hint: Identify mouth parts that are paired in
cockroach.
Sol.: Hypopharynx acts as tongue. Maxillae are
paired structures that help bring food to mandibles.
143. Answer (2)
Hint: First pair of wings is mesothoracic.
Sol.: Tegmina/Elytra or wing cover is the alternative
name for first pair of wings in cockroach.
144. Answer (4)
Hint: It is another term for arthrodial membrane.
Sol.: Arthrodial membrane is flexible membrane that
connects tergite with pleurites and sternites.
145. Answer (2)
Hint: Gill cover is absent in cartilaginous fish.
Sol.: Chondrichthyes are cartilaginous fish which
lack operculum generally; their males have pelvic
claspers and show internal fertilisation. Bony fishes
possess operculum, have cycloid/ctenoid scales and
show external fertilisation.
146. Answer (4)
Hint: Identify an animal where development of
embryo can occur on land.
Sol.: Pisces and amphibians are anamniotes. Their
development is linked to water directly. Reptiles
evolved prior to birds. Mammals, birds and reptiles
are all amniotes.
147. Answer (3)
Hint: Dry, aglandular skin is a feature of reptiles.
Sol.: Ichthyophis is a limbless amphibian.
Bangarus is a limbless reptile
Balaenoptera is a limbless mammal
Test - 4 (Code F) (Hints and Solutions) All India Aakash Test Series for Medical-2019
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They all have bony vertebral column. Skin is glandular
in mammals.
148. Answer (1)
Hint: Select a cartilaginous fish.
Sol.: Air/swim bladder helps to maintain buoyancy in
bony fish. Dog fish has to swim continuously to
avoid sinking.
149. Answer (2)
Hint: Identify a circular mouth scaleless fish like
organism.
Sol.: Myxine is hagfish, a cyclostome. Torpedo,
Pristis and Trygon belong to super class Pisces.
150. Answer (2)
Hint: Heart and blood vessels are present in open
and closed circulatory system.
Sol.: Heart is dorsal in position in non-chordates
usually while it is ventral in chordates. Nerve cord is
ventral in non-chordates but dorsal in position in
chordates. Post anal tail is a feature of chordates.
Gill slits are lateral in position in chordates.
151. Answer (1)
Hint: Such an organism produces internal body heat
through metabolic activity to maintain body
temperature.
Sol.: Mammals are homeotherms and endotherms.
They exhibit internal fertilisation (Felis) as do birds.
Chelone inhabits water. Replites show shedding of
skin (ecdysis).
152. Answer (4)
Hint: Snakes detect vibrations through jaw bones.
Sol.: Vipera has epidermal scales; 3 chambered
heart and is limbless in adult form. It has dry
cornified aglandular skin.
153. Answer (2)
Hint: Pneumatic bones have air cavities to reduce
weight of flying birds.
Sol.: Neophron i.e. vulture is a flying bird that has
both pneumatic bones and preen/oil gland. Air sacs
in birds are avascular and meant for storage of air
but not exchange of gases.
154. Answer (2)
Hint: Select an avian adaptation.
Sol.: Feathers reduce weight and are found in birds.
155. Answer (1)
Hint: This structure divides body cavity into thoracic
and abdominal cavity.
Sol.: Mammals show direct development. Heterodont
dentition is absent in aquatic mammals such as
whale, sea cow, etc. Sloth and sea cow have 9 and
6 cervical vertebrae respectively.
156. Answer (3)
Hint: Another name for this organism is Euspongia.
Sol.: Antedon and Sea urchin exhibit radial
symmetry in adult form. Salpa shows bilateral
symmetry.
157. Answer (1)
Hint: Select an arthropod.
Sol.: Sea hare is Aplysia with unsegmented body.
Sea cucumber is dioecious. Tapeworm shows
internal fertilisation.
158. Answer (4)
Hint: Triploblastic organisms show extracellular
digestion.
Sol.: Adult echinoderms are radially symmetrical and
most poriferans are asymmetrical. Notochord is
absent in both Echinoderms and Poriferans. True
coelom is a feature of Echinoderms.
159. Answer (2)
Hint: This organism is commonly called lac insect.
Sol.: Laccifer is a useful insect. Its secretion called
lac and shellac are used in bangle industry.
Anopheles, Culex and Aedes act as vectors to
spread diseases such as malaria, filariasis and
dengue respectively.
160. Answer (3)
Hint: Arthropods have open circulatory system.
Sol.: Pila is a mollusc with open circulatory system
and lacks segmentation. Anopheles has malpighian
tubules for excretion. Nereis is segmented and has
nephridia.
Presence of solid ventral nerve cord is a character of
non chordates.
161. Answer (2)
Hint: Organisms that show bioluminescence.
Sol.: Flatworms like Schistosoma are dioecious but
digestion is only extracellular. In triploblastic animals
such as aschelminthes, extracellular digestion is
observed.
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162. Answer (3)
Hint: This animal lives in loosely organised
community.
Sol.: Locusta is an economically harmful insect
(arthropod).
163. Answer (1)
Hint: Metamerism refers to presence of segments
and probable repeat of organs.
Sol.: Wuchereria is a filarial worm, where excretory
pore eliminates nitrogenous waste. Wastes present
in alimentary canal are eliminated through anus.
Pseudocoelom, absence of segmentation and
presence of bilateral symmetry are features of
Aschelminthes.
164. Answer (2)
Hint: Identify a mollusc with scientific name
Dentalium.
Sol.: Tusk shell has a calcareous shell
(exoskeleton) while chitinous exoskeleton is a feature
of arthropods.
165. Answer (3)
Hint: Identify a true fish.
Sol.: Flying fish is a bony fish. Hagfish is a
cyclostome. Cuttlefish and devil fish belong to
phylum Mollusca.
166. Answer (4)
Hint: Select the organism that occurs exclusively in
marine water.
Sol.: Asterias has water vascular system and show
external fertilisation. Poriferans have water canal
system.
167. Answer (3)
Hint: Identify first triploblastic animals.
Sol.: Regeneration is a prominent feature of Porifera
Coelenterata, Platyhelminthes and Echinoderms.
168. Answer (2)
Hint: This structure is composed of cellulose in
plants.
Sol.: Animal cells lack a cell wall. All animals are
heterotrophs and show division of labour. Poriferans
lack nervous system.
169. Answer (2)
Hint: Comb plates are locomotory structures in
members of this phylum.
Sol.: Annelids, arthropods and molluscs occupy
either aquatic or terrestrial habitats.
170. Answer (3)
Hint: Identify worms that are pseudocoelomate.
Sol.: Ringworm is a fungus; earthworm is an
annelid; bookworm is an arthropod; Liver fluke &
tapeworm are flatworms. Pinworm, filarial worm,
hookworm and roundworm belong to aschelminthes.
171. Answer (2)
Hint: Identify a roundworm which gives birth to young
worms.
Sol.: Ancylostoma and Ascaris are oviparous round
worms. Fasciola is a flatworm.
172. Answer (3)
Hint: Parapodia are lateral appendages that help in
swimming and respiration.
Sol.: Malpighian tubules are found in insects
(arthropods) while parapodia are found in annelids
such as Nereis.
173. Answer (2)
Hint: Identify organisms with cnidocytes.
Sol.: Adamsia is sea anemone i.e., a cnidarian
whose tentacles are rich in stinging cells.
Euspongia has flagellated choanocytes.
Pleurobrachia does not show metagenesis and
Planaria is a free living worm, hence it lacks hooks.
174. Answer (1)
Hint: Metagenesis.
Sol.: Physalia exists in both polyp and medusa
form. Hydra, Meandrina and Adamsia exist as
polyps.
175. Answer (1)
Hint: This organism forms gemmules to reproduce.
Sol.: Sponges lack distinct germ layers, has ostia
for entry of water and has collar cells in body.
176. Answer (3)
Hint: These organisms occur in exclusively marine
conditions.
Sol.: Saccoglossus a hemichordate, has proboscis
gland as its excretory organ. Gills are meant for
respiration in hemichordates. In molluscs, gills serve
both the function of respiration and excretion.
Test - 4 (Code F) (Hints and Solutions) All India Aakash Test Series for Medical-2019
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177. Answer (3)
Hint: This parameter is shared by larval
echinoderms with platyhelminthes.
Sol.: Adult echinoderms are radially symmetrical
while their larvae are bilaterally symmetrical. Anus
develops from blastopore in deuterostome body plan.
178. Answer (4)
Hint: Metamorphosis involves conversion of larva to
adult form.
Sol.: Direct development occurs in Hirudinaria.
Butterfly, Ascidia and Culex show metamorphosis
and their larval forms are caterpillar, tadpole &
wriggler respectively.
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179. Answer (1)
Hint: Identify an arthropod.
Sol.: Pheretima (an annelid), Pavo (a bird) and
Petromyzon (cyclostome) have closed circulatory
system. Capillaries are a feature of closed circulatory
system.
180. Answer (1)
Hint: This animal is also called bookworm.
Sol.: Loligo and Laccifer are economically beneficial
mollusc and insect respectively. Limulus is a living
fossil.