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Test - 6 (Code-C) (Answers) All India Aakash Test Series for Medical-2020
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1. (1)
2. (4)
3. (4)
4. (1)
5. (2)
6. (3)
7. (1)
8. (1)
9. (4)
10. (3)
11. (1)
12. (3)
13. (3)
14. (3)
15. (3)
16. (1)
17. (1)
18. (4)
19. (4)
20. (2)
21. (2)
22. (1)
23. (3)
24. (4)
25. (1)
26. (4)
27. (4)
28. (3)
29. (1)
30. (4)
31. (2)
32. (1)
33. (4)
34. (1)
35. (1)
36. (1)
Test Date : 17/02/2019
ANSWERS
TEST - 6 (Code-C)
All India Aakash Test Series for Medical-2020
37. (2)
38. (2)
39. (2)
40. (3)
41. (3)
42. (4)
43. (1)
44. (1)
45. (2)
46. (2)
47. (3)
48. (1)
49. (4)
50. (2)
51. (3)
52. (3)
53. (1)
54. (4)
55. (1)
56. (1)
57. (2)
58. (2)
59. (4)
60. (3)
61. (3)
62. (4)
63. (4)
64. (1)
65. (2)
66. (1)
67. (2)
68. (1)
69. (4)
70. (3)
71. (4)
72. (2)
73. (1)
74. (4)
75. (2)
76. (3)
77. (3)
78. (1)
79. (3)
80. (4)
81. (4)
82. (3)
83. (2)
84. (1)
85. (3)
86. (4)
87. (3)
88. (4)
89. (4)
90. (1)
91. (4)
92. (4)
93. (3)
94. (1)
95. (2)
96. (4)
97. (2)
98. (4)
99. (3)
100. (1)
101. (2)
102. (4)
103. (3)
104. (1)
105. (4)
106. (4)
107. (1)
108. (2)
109. (3)
110. (3)
111. (1)
112. (4)
113. (3)
114. (2)
115. (3)
116. (3)
117. (2)
118. (3)
119. (3)
120. (4)
121. (3)
122. (1)
123. (2)
124. (3)
125. (3)
126. (4)
127. (3)
128. (3)
129. (4)
130. (2)
131. (4)
132. (3)
133. (3)
134. (3)
135. (2)
136. (3)
137. (1)
138. (4)
139. (3)
140. (4)
141. (4)
142. (4)
143. (2)
144. (3)
145. (3)
146. (2)
147. (4)
148. (3)
149. (3)
150. (3)
151. (2)
152. (3)
153. (1)
154. (3)
155. (2)
156. (4)
157. (4)
158. (3)
159. (4)
160. (1)
161. (4)
162. (4)
163. (3)
164. (3)
165. (3)
166. (1)
167. (3)
168. (3)
169. (3)
170. (2)
171. (2)
172. (1)
173. (4)
174. (3)
175. (3)
176. (1)
177. (4)
178. (1)
179. (4)
180. (3)
All India Aakash Test Series for Medical-2020 Test - 6 (Code-C) (Answers & Hints)
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ANSWERS & HINTS
1. Answer (1)
Hint: The least time interval after which particle
repeats its motion is called time period.
Sol.: Least time for given combination is 2
2. Answer (4)
Hint: If function increases or decreases
monotonically with time, function is non-periodic.
Sol.: (1) y = cos3t = 13cos – cos3
4t t
it is periodic but not SHM
(2) 2 sin SHM4
⎛ ⎞ ⎜ ⎟⎝ ⎠
y t
(3) y = cos2t, it is SHM
(4) y = aet is non periodic
3. Answer (4)
Hint: y = Asin(t + )
Sol.: The angle at time 2
( )4
t tT
.
The projection on y axis
2 2( ) sin sin
4 120 4
⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
y t R t R tT
sin60 4
y R t ⎡ ⎤ ⎢ ⎥
⎣ ⎦
4. Answer (1)
Sol.: Time taken from x = A to is .2 6
A Tx
x = Acost
2cos 3
2
AA
T
1 2cos 3
2 T
23
3 T
T = 18 s
[ PHYSICS]
5. Answer (2)
Hint: vmax
= ASol.: x = Asint
(10 cm)sin2
x t
–1 –1
max(10 cm) rad s 5 cm s
2v
6. Answer (3)
Hint: 2m
Tk
Sol.: 3
2 22 2
3
m mT
kk
⎛ ⎞⎜ ⎟⎝ ⎠
7. Answer (1)
Hint: dx
vdt
Sol.: 10 sin5
v t
x vdt ∫
–10 cos5
5
t C
x
–50cos5
x t C
at t = 0, v = 0, x = –50 so, C = 0
50cos5
x t⎛ ⎞ ⎜ ⎟
⎝ ⎠
8. Answer (1)
Hint: 2
20
md x dxb kxdtdt
Sol.:2
20
md x dxb kxdtdt
...(i)
2
22 0d x dx
xdt dt
This is equation of damped oscillation.
Test - 6 (Code-C) (Answers & Hints) All India Aakash Test Series for Medical-2020
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9. Answer (4)
Hint: T
v xg
Sol.: v xg
1
22
dv ga v xg g
dx x (constant)
So a x°
10. Answer (3)
Hint: (vp)max
= A
Sol.:1
2
p wv v
0
1(2 )
2y f f
= 4y0
11. Answer (1)
Hint and Sol.: Open end of a closed organ pipe is
displacement antinode and pressure node.
12. Answer (3)
Hint: 0
1
2
Tf
l
Sol.:10
20
If 100 N100 1
169 N20 169 1.3
⎡ ⎤ ⎢ ⎥ ⎣ ⎦
Tf
Tf
0.3f0 = 20
0
200Hz
3f
13. Answer (3)
Hint:p
yv
t
Sol.: y = 20 sin(100 t – 2x)
20 100 cos(100 – 2 )p
yv t x
t
3 12 10 cos 100 0 – 2
6p
v⎡ ⎤ ⎢ ⎥⎣ ⎦
3 312 10 10 cm/s
2p
v⎛ ⎞ ⎜ ⎟⎝ ⎠
14. Answer (3)
Hint: = kx
Sol.:2
x
0.52
k
[0.5 – 0.25]2 8
15. Answer (3)
Hint: Fundamental frequency, 0
4
vf
l
Sol.:0
330110 Hz
4 0.75f
only odd harmonic are produced in closed pipe.
So frequencies 330 Hz, 550 Hz, 770 Hz are
possible. 660 Hz is not possible.
16. Answer (1)
Hint: B
v
Sol.: 2
B
v
or, 2 2 –6 –2 2
8 3
1 1
8 10 (2.5 10 )
2 10 kg/m
B
v kv
17. Answer (1)
Hint: P RT
vM
Sol.: If temperature is constant v will remain
constant.
Since pressure and density will change in the same
ratio.
18. Answer (4)
Hint: Pitch is related to frequency of sound.
Sol.: Quality is the sensation of human ear due to
wave form of sound.
19. Answer (4)
Hint: T
v
Sol.: Stress
v
11 –5
3
1.6 10 1.6 10 20
8 10
Yv
4
–11.6 1.6 2 1080 m s
8
All India Aakash Test Series for Medical-2020 Test - 6 (Code-C) (Answers & Hints)
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20. Answer (2)
Hint: Resultant amplitude formula.
Sol.: When amplitude of component waves are
equal, amplitude of resultant, 2 cos2
R A⎛ ⎞ ⎜ ⎟
⎝ ⎠
Amplitude, 3
2 cos 2 36 2
R A A A
21. Answer (2)
Hint: Amplitude = 2Asinkx
Sol.: Amplitude = 4sinx ...(i)
22m
⇒
...(ii)
at2m,
3 3x
2Amplitude 4sin 2 3 m
3
22. Answer (1)
Hint: Beat frequency = |f1 – f
2|.
Sol.: 1 = 1020 f
1 = 510 Hz.
2 = 1004 f
2 = 502 Hz
fb = |f
1 – f
2| = 8 Hz
23. Answer (3)
Hint: Wavelength depends on medium.
Sol.: (1) Frequency does not change by changing
the medium of propagation.
(2) If wave is reflected from denser medium,
the phase change is .
24. Answer (4)
Hint: Y
v
Sol.: 9
3 –1
3
8 102 10 m s
2 10
Yv
L
5
4
O
In second overtone
5
4L
41.6 m
5
L
3
32 101.25 10 1250 Hz
1.6
vf
25. Answer (1)
Hint: Only odd harmonic in closed organ pipe.
Sol.: f1 : f
2 : f
3 = 5 : 7 : 9
So pipe is closed
So, fundamental frequency 0
28557 Hz
5f
0
0
380 20m
57 3
v
f
26. Answer (4)
Hint: fB = f
1 ~ f
2
Sol.: f = 4 5t = 1st 2nd
fB
4 5
27. Answer (4)
Hint: m m
a a
v
v
Sol.:1650 5
330 1
m
a
v
v
5
1
m
a
5m a
28. Answer (3)
Hint: End correction, e = 0.6r
Sol.:1 2
3
4( ) 4( )
v vf
l e l e
3(l1 + e) = l
2 + e
2 1– 3
2
l le
72.6 – 720.3 0.6
2e r
3 1cm
6 2r
29. Answer (1)
Hint: 0
Source–
vf f
v v
⎛ ⎞ ⎜ ⎟
⎝ ⎠
Sol.:350
5500 5000350 – v
⎛ ⎞ ⎜ ⎟
⎝ ⎠
11 350
10 350 – v
350 – 11v = 0
350m/s
11v
Test - 6 (Code-C) (Answers & Hints) All India Aakash Test Series for Medical-2020
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30. Answer (4)
Hint: Stress
v .
Sol.: f0 = 100 Hz, v = f
0
0 = 200 m/s
Now, 2Stress v .
Stress = (200)2 × 8 × 103 = 3.2 × 108 N/m2
31. Answer (2)
Hint & Sol.: Sound travel fastest in solids.
32. Answer (1)
Hint: 1 2 3
1 2 3
1 1 1: : : :n n n
l l l
Sol.:1 2 3: : 1: 2 :1n n n
1 2 3: : 2 :1: 2l l l
l1 + l
2 + l
3 = 150 cm 5K = 150 K = 30.
So, l1 = 60 cm
l2 = 30 cm
l3 = 60 cm
so, x1 = 60 cm
x2 = 90 cm
33. Answer (4)
Hint: y = 2A sinkx cost
Sol.: y = 2A sinkx cost
2sin cos(90 )30
xy t
⎛ ⎞ ⎜ ⎟⎝ ⎠
2
30
= 60 cm
30 cm2
Number of loops = 10
Number of nodes = 10 + 1 = 11
Number of antinodes = 10
Amplitude of x = 5 cm
2sin 5 1cm30
34. Answer (1)
Hint: v
f
Sol.:330 1
m 50 cm660 2
3 512.5, 37.5, 62.5
4 4 4
(Not possible)
Only 1st and 2nd resonance can be produced.
So minimum height of water
hmin
= 50 – 37.5 = 12.5 cm
35. Answer (1)
Hint: 1 1
2 2
f v
f v
Sol.: 1 1 2
2 2 1
, If Constantf v m
Tf v m
1
2
600 2 1
32 4
f
f f
f = 2400 Hz
36. Answer (1)
Hint: Use Newton’s formula P
v
Sol.: According to Laplace’s correction
v v Percentage error
v v 1.4 1100 100%
v 1.4
15%�
37. Answer (2)
Hint: nl = constant
Sol.: 2
1
20 4
25 5
A
B
n l
n l n
B > n
A.
So, nB – n
A = 5 on solving,
nA = 20
nB = 25
38. Answer (2)
Hint: f1 – f
n = (n – 1) f
B
Sol.:1– 29 5 145
nf f
1
1– 1452
ff
f1 = 290 Hz 1
145 Hz2
f⇒ .
39. Answer (2)
Hint: 0
0
s
v vf f
v v
⎛ ⎞ ⎜ ⎟⎝ ⎠.
Sol.: When engine approaches man then frequency
is higher and constant, when engine passes man
then frequency heard lowered but remains constant.
All India Aakash Test Series for Medical-2020 Test - 6 (Code-C) (Answers & Hints)
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46. Answer (2)
Hint : 1 2 3 4 5 6
2 2 3CH CH– CH – C C–CH
Hex–1–en–4–yne
47. Answer (3)
Hint : More the s-character in hybrid orbitals, more
will be the electronegativity.
Sol. : C2H
2 : sp hybridised carbon
50% s-character.
48. Answer (1)
Hint : Cyclic, planar and conjugated compounds
containing (4n + 2)e– are aromatic in nature.
[ CHEMISTRY]
Sol. : HH is non planar because of steric
repulsion of inner hydrogen.
49. Answer (4)
Hint : Alkyl group with –H atoms attach to
benzene, activates the ring for SE reactions.
Sol. : does not have –H atom.
50. Answer (2)
Hint :
2Volumeof N (atSTP)28
% of N 10022400 wt.of compound
40. Answer (3)
Hint: 0
0
s
v vf f
v v
⎡ ⎤ ⎢ ⎥⎣ ⎦
Sol.:
Motorist Band
5 ms–1
Wall
10 ms–1
Frequency of reflected sound
1
330 3301000 1000
330 –10 320f
⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
As motorist is moving toward wall
0
2 1
330 3351000
320 330
v vf f
v
⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎣ ⎦
2
3351000
320f
⎡ ⎤ ⎢ ⎥⎣ ⎦
41. Answer (3)
Hint: 1
v
Sol.: Density of moist air is less than that of dry air.
42. Answer (4)
Hint: Pressure variation is maximum at displacement
nodes.
Sol.: For stationary waves, all particles oscillates in
same phase with in a loop. Stationary wave is
formed when two waves are travelling in opposite
direction.
43. Answer (1)
Hint: y = f (x ± vt)
Sol.: y = f (x + vt) when wave is moving towards
negative x-axis
y = f (x – vt) when wave is moving towards
positive x-axis
44. Answer (1)
Hint: F = –kx, k = m2.
Sol.: 2 2
2
4 2 8
k
m
2 2
2
2 2T
or, 4 2 sT
45. Answer (2)
Hint: 2m
Tk
Sol.: 2 kg
2300N/m
T
2s
150T
Test - 6 (Code-C) (Answers & Hints) All India Aakash Test Series for Medical-2020
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Sol. : 1 1 2 2
1 2
P V P V
T T
(Experiment condition) (STP condition)
2760 V(725 – 25) 60
300 273
∵ Volume of N2 (at STP) = 50.29 ml
28 50.29% of N 100 22.45%
22400 0.28
51. Answer (3)
Hint : Octet of N in 4
NH
is complete and no vacant
orbital is present on N atom.
52. Answer (3)
Hint : For tautomerism, acidic H-atom should be
present.
Sol. :
O
does not contain any acidic H-atom.
53. Answer (1)
Hint. : CH3
CH2
– –C
CH3
CH3
––
–CH2– is neohexyl group.
54. Answer (4)
Hint : HNO3 converts NaCN and Na
2S into HCN and
H2S respectively.
55. Answer (1)
Hint : More the delocalization of +ve charge, more
will be the stability of carbocation.
Sol. : CH2
+ has maximum number of
resonating structures, so most stable.
56. Answer (1)
Hint : Order of -I effect.
– CN > – COOH > – OH
57. Answer (2)
Hint : Higher the availability of lone pair on nitrogen
atom higher is its basic nature and hence higher the
pKa
58. Answer (2)
Hint : Glycerol decomposes near its boiling point.
Sol. : Compounds which decompose near its boiling
point are purified by vacuum distillation method.
59. Answer (4)
Hint : Pent–2–ene and 1-chloro-2-bromopropene
show geometrical isomerism.
60. Answer (3)
Hint : Kjeldahl’s method is not applicable to
compounds containing
(i) Nitrogen in ring.
(ii) (–NO2) and azo (– N = N –) groups.
Sol. : Nitrogen present in CN , can be
quantitatively estimated by Kjeldahl’s method.
61. Answer (3)
Hint : Photochemical smog is oxidising smog.
62. Answer (4)
Hint : N2O is a greenhouse gas.
63. Answer (4)
Hint : For highly polluted water, BOD value is more
than 17 ppm.
64. Answer (1)
Hint : H2SO
4 + H
2SO
4 SO
3 + HSO
4
–
+ H
3O+
Sol. : SO3 is an electrophile.
65. Answer (2)
Hint : Carbonium ion intermediate is formed.
Sol. :
CH = CH2
H+
H
+
H – +
CI–
shift
CI
A
66. Answer (1)
Hint : (i) O3
(ii) Zn+H O2
C=C C =O+O =C
Sol. :
2–oxo propanal
CH3 – C – CH = O + O = CH2
O+O = CH2
CH3 – C – CH = CH2
CH2
2-methylbuta-1, 3-diene
2 3 4
(C H5 8)
Methanal
Methanal
1
67. Answer (2)
Hint : When C – CI bond acquires double bond
character then reactivity of C – Cl bond decreases.
Sol. : CH2 = CH
– CI
..CH
2 –CH CI
(vinyl chloride)
– +
All India Aakash Test Series for Medical-2020 Test - 6 (Code-C) (Answers & Hints)
8/16
68. Answer (1)
Hint : 2, 3–dimethylbutane has two type of
H-atoms.
Sol. :
CI
2
h Cl+
Cl
Chiral carbon
69. Answer (4)
Hint : 1-alkynes are acidic in nature.
Sol. :
Ch C CH3 –
NaNH2
CH3– C C Na
CH3 – CH – CI
2
CH3– C CH
2 3–C – CH
Pent–2–yne
– +
70. Answer (3)
Hint : In presence of sunlight addition reaction and
in presence of FeCl3 electrophilic substitution
reaction take place.
Sol. :
+ 3CI2
Sunlight
H CI
CI H
H
CI
H
CI
CI
H
CI
H
( )C H Cl6 6 6
( ) (X)Benzene hexachloride
+ CI2
FeCI3 + HCI
CI
( ) (Y)Chlorobenzene
71. Answer (4)
Hint : More the H-bonding and lesser the torsional
strain, more will be the stability of conformers.
Sol. : Stability order:
(1) Cyclohexane : Chair > Boat > Half chair
(2) Butane (around C2 and C
3) : Anti > Gauche >
Partially eclipsed
(3) Propane : Staggered > Skew > Eclipsed
(4) Ethane -1, 2-diol: Gauche > Anti > Eclipsed
72. Answer (2)
Hint : R of R-MgX on reaction with acidic-H give
alkane.
Sol. :
CH3
CH – CH MgCI2
CH3
– +
CH –3 O – H
+–
CH3
CH3
– CH – CH
3 + Mg
CI
OCH3
73. Answer (1)
Hint : During hydration of alkene in acidic medium
carbocation is formed as intermediate.
Sol. :
CH3
CH – CH = CH2
H2 O H
CH3
C
H
CH
+CH
3
H
shift
CH3
C
CH2 CH
3
+
H2 O
–H+
CH3
C
CH2
OH
CH3
+
–
(3° alcohol)
CH3
CH
CH CH
2
(i) B2 H THF
6,
CH3
CH CH2 CH
2
=
(ii) H2O ,NaOH
2
(1° alcohol)
OH
74. Answer (4)
Hint : By Kolbe’s electrolytic method alkane, alkene
and alkyne all can be prepared.
Sol. :
2 CH3 – COONa
+ H
2 O CH
3 – CH
3 + 2CO
2
(Sodium acetate)ethane
+ 2NaOH + H 2
CH2 – COONa + 2H
2 O CH
2 + 2CO
2 + 2NaOH + H
2
CH
ethene2
CH –2
COONa
(Sodium succinate)
CH – COONa + 2H
2 O CH + 2CO
2
+ 2NaOH + H2
CHethyne
CH –
COONa
(Sodium fumarate)
75. Answer (2)
Hint : H atom attach to sp-hybridised carbon is
acidic in nature.
Sol. : CH C 3 – C– H acidic H atom.
Propyne
76. Answer (3)
Hint : Symmetrical alkene does not show
Markovnikov’s addition of HBr.
Test - 6 (Code-C) (Answers & Hints) All India Aakash Test Series for Medical-2020
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Sol. : is a symmetrical alkene.
77. Answer (3)
Hint : Propyne contains acidic hydrogen.
Sol. : 4
3
NH OH3 3AgNO
whiteppt.
CH C CH CH C CAg
4
3
NH OH3 2 AgNO
CH C CH Noreaction
78. Answer (1)
Hint : Less stable carbocation can rearrange to
more stable carbocation.
Sol. :
OH
H+
OH2
+
–H O2
CH – – CH3 2
C
CH3
CH3
+ CH3
–
shiftCH – – CH – CH
3 3 C
CH3
+
H
CH3 – C = CH – CH
3
CH3
– H+
79. Answer (3)
Hint : Less stable alkenes are more reactive towards
catalytic hydrogenation.
Sol. :
CH = CH2
has only one alpha–H
atom. So, hyperconjugation effect is small and
alkene is least stable.
80. Answer (4)
Hint : More the surface area of Isomeric alkanes,
more will be the boiling point.
Sol. : As the branching in isomeric alkane increases
their surface area decreases so boiling point
decreases
BP : > >
81. Answer (4)
Hint :
3 C H2 2
Red Hot
Fe tube
Ethyne 873 K
COONa + NaOHCaO
+ Na CO2 3
Sodium
benzoate
OH + Zn
Phenol
+ ZnO
82. Answer (3)
Hint :
CH – C C – CH3 3
Lindlar’s catalyst
Na in liq. NH
3
C = C
CH3
HH
CH3
(Cis – but – 2 – ene)
C = CH
H
CH3
(trans – but – 2 – ene)
CH3
(A)
(B)
Sol. : For A, 0
For B, = 0
83. Answer (2)
Hint : Nitration takes place in that benzene ring
which has more electron density.
Sol. :
C – O
O
Conc. HNO3
Conc. H SO2 4
C – O
O
NO2
(Major)
(–M) (+M)
84. Answer (1)
Hint : More the s-character in carbanion, more will
be stability.
Sol. : In ethyne carbanion (CH C–), carbon is sp
hybridised which has more s-character (50%), in
other words can hold negative charge more better, so
more stable.
85. Answer (3)
Hint : The degree of unsaturation (DOU) for C4H
6 is
two.
Sol. : ⇒C H (DOU = 3)4 4
All India Aakash Test Series for Medical-2020 Test - 6 (Code-C) (Answers & Hints)
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[ BIOLOGY]
91. Answer (4)
Hint: ATP is synthesized during respiration.
Sol.: ATP acts as the energy currency of the cell.
The energy released by oxidation of food in
respiration is used to synthesise ATP and not used
directly.
92. Answer (4)
Hint: Plants have lesser demands for gaseous
exchange as compared to animals.
Sol.: Roots, stems and leaves all respire at lower
rates than animals. In plants, cells are closely
packed and located quite close to the surface of the
plant. Hence it is easy for gases to diffuse inside.
93. Answer (3)
Hint: Partial oxidation of glucose without utilization
of oxygen is called glycolysis.
Sol.: Glycolysis takes place in all organisms in the
cytoplasm. Hence all organisms whether anaerobes
or aerobes retain enzymes for glycolysis.
94. Answer (1)
Sol.: The term glycolysis has originated from two
Greek words that are ‘glycon’ which means sugar
and ‘lysis’ which means splitting.
95. Answer (2)
Hint: Lactic acid (3C) is produced from pyruvic acid
which is also a 3C compound.
Sol.: In lactic acid fermentation pyruvic acid (3C) is
converted into lactic acid (3C). Hence there is no
release of CO2 which occurs in alcohol fermentation.
Both processes are hazardous and produce 2 ATP
as net gain.
96. Answer (4)
Hint: Krebs’ cycle starts with the condensation of
2C compound with 4C compound to yield a 6C
compound.
Sol.: Krebs’ cycle starts with condensation of Acetyl
CoA (2C) with oxaloacetic acid (4C) to yield citric
acid (6C).
97. Answer (2)
Hint: It is a 2C compound.
Sol.: Acetyl CoA is common intermediate among
fats, proteins and carbohydrates breakdown during
aerobic respiration. It enters into Krebs’ cycle to
yield energy.
98. Answer (4)
Hint: In aerobic respiration CO2 is released in link
reaction & during Krebs’ cycle.
Sol.: In link reaction CO2 is released at one step.
During Krebs’ cycle CO2 is released at two steps.
In case of anaerobic condition
2
2
2CORQ (infinite)
0O
86. Answer (4)
Hint : Order of deactivation is :
– NO2 > – CN > – COCH
3 > – F
Sol. :
87. Answer (3)
Hint : 4np electrons in a planar ring makes the ring
antiaromatic.
88. Answer (4)
Hint : 31.4×meq.ofNH
% of N =wt.of sample
Sol. : meq of NH3 evolved = meq. of H
2SO
4 used
= N × V(ml) = 0.05 × 40 = 2
31.4×meq.ofNH 1.4 2
% of N = 14%wt.of sample. 0.2
89. Answer (4)
Hint : More the stability of conjugate base formed,
more will be acidic nature of hydrocarbon.
Sol. :
+ H
(More acidic) (Conjugate base)(Aromatic)More stable
H –
+
90. Answer (1)
Hint : C – H bond dissociation energy in
CH = CH – CH2 2
H
is slightly less than bond
dissociation energy of C – H in CH2
H
Sol. : Stability order :
CH2
CH2
CH – – CH3 2
C
CH3
CH3
CH3
Test - 6 (Code-C) (Answers & Hints) All India Aakash Test Series for Medical-2020
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99. Answer (3)
Hint: Yeast performs alcoholic fermentation.
Sol.: In alcoholic fermentation alcohol is produced
along with CO2 & ATP.
100. Answer (1)
Hint: Complete oxidation of acetyl CoA occurs in
aerobic respiration through Krebs’ cycle.
Sol.: One turn of Krebs’ cycle involves complete
breakdown of acetyl CoA which produces 3NADH +
H+, 1 FADH2 & 1 GTP.
3 NADH + H+ = 3 × 3 = 9 ATP
1 FADH2 = 1 × 2 = 2 ATP
1 GTP = 1 × 1 = 1 ATP
12 ATP
NADH + H+ & FADH2 produce ATP through ETS.
101. Answer (2)
Hint: The phytohormone which was extracted from a
fungus Gibberella fujikuroi is responsible for bolting.
Sol.: Gibberellins are active substances which
induce stem elongation in rossette plants. This
phenomenon is called bolting.
102. Answer (4)
Hint: The plant hormone which is responsible for
closure of stomata is known as stress hormone.
Sol.: ABA is called stress hormone as its synthesis
is stimulated by drought, water logging & other
adverse environmental conditions.
103. Answer (3)
Sol.: Auxin helps in root formation on stem cuttings,
shows apical dominance & phototropism but does
not play any role to speed up the malting process
which is done by gibberellin.
104. Answer (1)
Hint: Ethylene is a volatile substance which
promotes senescence.
Sol.: Ethylene is the only gaseous hormone of
plants. It breaks the dormancy of seeds.
Bakanae disease is caused by gibberellins.
105. Answer (4)
Hint: The continued growth due to activity of
meristems occurs in plants.
Sol.: New cells are always being added due to
activity of meristem, this type of growth is open form
of growth and indeterminate.
Animals show definite & diffused type of growth.
106. Answer (4)
Hint: Tissues formed from de-differentiation of
parenchyma are secondary in origin.
Sol.: Wound cambium, cork cambium &
interfascicular cambium are secondary in origin. They
are formed by de-differentiation of parenchyma cells.
Intrafascicular cambium is primary in origin.
107. Answer (1)
Hint: This hormone is a plant growth inhibitor.
Sol.: ABA is a derivative of carotenoids.
108. Answer (2)
Hint: Relative growth can be calculated by the
formula i.e. Growth per unit time
100Initial size
.
Sol.: For Leaf A
5100 50%
10
For Leaf B
5100 20%
25
109. Answer (3)
Hint: Rice & maize show hypogeal seed
germination.
Sol.: In hypogeal seed germination, epicotyl grows
first & cotyledons remain underground.
110. Answer (3)
Hint: Short day plant flowers after receiving
photoperiod shorter than critical period.
Sol.: Tobacco is a short day plant. Wheat, radish &
sugarbeet are long day plants.
111. Answer (1)
Hint: Phytochrome is a photoreceptor.
Sol.: Phytochrome is a blue-green pigment,
regulates photoperiodic induction of flowering and
helps in seed germination too.
It occurs in two forms Pr & P
fr. Its P
r form becomes
active by absorbing far-red light. Pfr is physiologically
active form.
112. Answer (4)
Hint: Auxin occurs in growing apices.
Sol.: If apical bud is removed then auxin is removed,
due to which apical dominance is prevented.
The plant will show more lateral branches & axillary
buds.
All India Aakash Test Series for Medical-2020 Test - 6 (Code-C) (Answers & Hints)
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113. Answer (3)
Sol.: The stimulus of vernalisation is perceived by the
mature stem apex or by the embryo of the seed.
114. Answer (2)
Hint: Gibberellin was first isolated from a fungus
called Gibberella fujikuroi.
Sol.: It is acidic in nature, increases sugarcane yield
by increasing the length of internodes, delays
ripening of fruits and helps in seed germination.
115. Answer (3)
Sol.: Buttercup shows environmental (extrinsic)
heterophylly. Cotton, coriander & larkspur show
developmental heterophylly due to intrinsic factors.
116. Answer (3)
Hint: Inner membrane of mitochondria forms cristae.
Oxidative phosphorylation occurs during oxidation of
food.
Sol.: Cristae contains cytochromes which helps in
ETS.
During formation of bread, yeast if used performs
alcoholic fermentation which releases CO2.
117. Answer (2)
Hint: The F1 headpiece contains the site of the ATP
synthesis.
Sol.: F1 headpiece remains present towards the
matrix of the mitochondria. Hence ATP synthesis
takes place towards the matrix.
118. Answer (3)
Hint: Krebs’ cycle produces reduced coenzymes.
Sol.: NADH + H+ is produced during Krebs’ cycle
along with FADH2 & GTP.
It starts with condensation of acetyl CoA with OAA
to form citric acid.
119. Answer (3)
Hint: Cytochrome c is a mobile carrier protein.
Sol.: Cytochrome c is not the part of complex IV. It
is attached to the outer surface of the inner
membrane & transfers electrons between complex III
& IV.
120. Answer (4)
Hint: Tryptophan is precursor of auxin.
Sol.: Tryptophan is an amino acid. Final product of
transition/link reaction of aerobic respiration is acetyl
CoA which is precurssor of gibberellin.
A pigment violaxanthin is a precursor of ABA.
121. Answer (3)
Sol.: 2, 4-D is widely used as weedicide in cereal
crops.
122. Answer (1)
Hint: Cytokinin(CK) helps in production of new
leaves.
Sol.: It (CK) enhances chloroplast production in
leaves. It delays senescence by controlling protein
synthesis.
123. Answer (2)
Sol.: Cytokinin along with auxin are essential in
tissue culture. High cytokinin to auxin ratio causes
shoot formation in callus.
124. Answer (3)
Hint: Coconut milk has a phytohormone which helps
in cell division.
Sol.: It contains cytokinin.
125. Answer (3)
Hint: Antagonistic means showing opposite effects
towards something.
Sol.: When two hormones bring about opposite
effects on the same function, it means they are
antagonistic to each other, whereas when they act
together they are called synergistic.
126. Answer (4)
Hint: Short day plants require uninterrupted dark
period to flower.
Sol.: SDP will not flower if its dark period is
interuppted by a flash of light.
127. Answer (3)
Sol.: Gibberellin can replace vernalisation. As by
exogenous application of gibberellins many biennials
can be induced to behave as annuals.
128. Answer (3)
Hint: ABA is a plant growth inhibiting phytohormone.
Sol.: ABA induces dormancy of buds and seeds &
promotes abscission of flowers.
129. Answer (4)
Hint: It is a gaseous hormone.
Sol.: Ethylene is responsible for ripening of fruits. It
can be used artificially to ripen the fruits.
130. Answer (2)
Hint: Auxin gets removed along with apical buds of
a plant.
Sol.: Auxin was first time isolated from human urine.
It is an amino acid derivative.
131. Answer (4)
Sol.: Growth curve in plants is sigmoid, also called
S-shaped growth curve.
Test - 6 (Code-C) (Answers & Hints) All India Aakash Test Series for Medical-2020
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132. Answer (3)
Hint: Pr absorbs red light.
Sol.: 660 nm is red light.
Pfr absorbs far red light.
133. Answer (3)
Hint: During dormancy, growth & development of
seed is temporarily stopped.
Sol.: In dormancy, seeds cannot germinate inspite
of availability of all suitable environmental conditions.
134. Answer (3)
Hint: In link reaction pyruvic acid (3C) forms acetyl
CoA(2C) & CO2.
Sol.:
4C compound – Malic acid
3C compound – Pyruvic acid
2C compound – Acetyl CoA
5C compound – -ketoglutaric acid
135. Answer (2)
Hint: RQ of fat is less than 1 and of organic acids
it is more than 1.
Sol.:
Tripalmitin is a fatty acid (RQ = 0.7)
Proteins (RQ = 0.9)
Malic acid (RQ = 1.33).
136. Answer (3)
Hint : In sponges, the body cavity opens to exterior
through osculum.
Sol. : All the given characters are shown by
sponges. They are not considered as diploblastic
and they have cellular grade of body organisation.
The cavity inside body is known as spongocoel or
paragastric cavity.
137. Answer (1)
Hint : This animal shows internal metamerism.
Sol. : In annelids, both external and internal
segmentation is found in adult stage
In arthropods, only external segmentation is present
in adults. Chordates show internal segments in
certain body parts.
138. Answer (4)
Hint : Animal which is bilaterally symmetrical in larval
stage but becomes asymmetrical due to torsion in
adult life.
Sol. : Adult snails are asymmetrical due to torsion.
In larval stage, it is bilaterally symmetrical. It is a
type of retrograde metamorphosis.
139. Answer (3)
Hint : Obelia exhibits alternation of generation.
Sol. : In class Hydrozoa, all members except Hydra
show metagenesis. Metastasis is spread of cancer,
metamorphosis is the development of larval stage
into adult stage.
140. Answer (4)
Hint : The roundworm which is found attached with
intestinal wall.
Sol. : • Euspongia – Bath sponge
• Meandrina – Brain coral
• Fasciola – Liver fluke
141. Answer (4)
Hint : Chordates have hollow, dorsal nerve cord
Sol. : Arthropoda – Chitinous exoskeleton
Mollusca – Calcareous exoskeleton
Echinodermata – Spiny skinned, ventral
solid nerve cord
Chordata – Presence of notochord
142. Answer (4)
Hint : This process helps in replacement of lost
body parts.
Sol. : With the help of water vascular system
echinoderms perform locomotion, excretion,
respiration and capturing of food.
143. Answer (2)
Hint : Capillaries are absent in Hemichordates
Sol. : In leech and Balanoglossus, open type of
circulatory system is present.
144. Answer (3)
Hint : Platyhelminthes are acoelomates.
Sol. : Protostomic animals are schizocoelic
eucoelomates and deuterostomic animals are
enterocoelic eucoelomates. Only arthropods placed
in class Insecta are considered as insects.
145. Answer (3)
Hint : Identify common Indian bull frog.
Sol. : Rana tigrina is common Indian bull frog.
Alytes is commonly known as midwife toad because
male carries eggs wrapped around its hind limbs.
146 Answer (2)
Hint : Members of Class Osteichthyes.
Sol. : Air bladder/swim bladder is found in bony
fishes to provide buoyancy which is helpful in
swimming. They do not to have continuously swim
as in case of cartilaginous fishes which lack swim
bladder.
All India Aakash Test Series for Medical-2020 Test - 6 (Code-C) (Answers & Hints)
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147. Answer (4)
Hint : The first segment bears mouth and a lobe that
can act as a wedge.
Sol. : The first body segment of earthworm is known
as peristomium. Prostomium is a hood like structure
on peristomium that is helpful in burrowing. The fecal
deposits of earthworm are known as worm castings.
Vermicomposting is use of worm castings as
manure.
148. Answer (3)
Hint : Members of this class show aerial
adaptations.
Sol. : In birds, long bones of the body contain
cavities filled with air to decrease weight of the body.
Such type of bones found in birds are known as
pneumatic bones.
149. Answer (3)
Hint : Fish like organisms having circular mouth.
Sol. : Cyclostomes (Petromyzon) are animals
without scales having circular mouth. They have non
operculated gills and lack paired fins. Scoliodon
(cartilaginous fish) and Hippocampus and Labeo
(bony fishes) have scales on their body.
150. Answer (3)
Hint : Common name for Hemichordates.
Sol. : • Trygon – Poison sting (Sting ray)
• Torpedo – Muscle modified into
electric organ (Electric
eel)
• Tongue worm – Proboscis present not
placoid scales
• Hippocampus – Brood pouch in male to
carry fertilized eggs.
151. Answer (2)
Hint : Common name for Antedon.
Sol. :
Metridium – sea anemone – Coelenterata
Antedon – sea lily – Echinodermata
Hormiphora – sea walnut – Ctenophora
Pennatula – sea pen – Coelenterata
152. Answer (3)
Hint : Number of moults is equivalent to the number
of chambers in its heart.
Sol. : In cockroach, nymph is a young stage which
is similar to cockroach but smaller in size, light in
colour, with immature gonads and without wings. It
undergoes gradual metamorphosis known as
paurometabola in 6 months upto two years. During
this time, it moults 13 times, to become sexually
mature adult.
153. Answer (1)
Hint : Animals which are commonly known as
roundworms are classified in this phylum.
Sol. : Aschelminthes/Nematodes are considered as
first animals to have complete digestive tract with
distinct mouth and anus.
154. Answer (3)
Hint : Most vertebrates are poikilothermous except
birds and mammals generally.
Sol. : Testudo is a reptile and cannot maintain its
body temperature, hence considered as
poikilothermic/cold blooded animal.
Columba is a bird while Felis and Macropus are
mammals. So, all these are warm-blooded animals.
155. Answer (2)
Hint : Larval stage which causes taeniasis in
humans.
Sol. : In fecal matter of infected man, onchosphere
is present which is converted into hexacanth larva
and finally cysticercus larva/bladderworm in muscles
of pig. Human beings are infected when they feed on
undercooked measly pork and this larva in adult
develops into Taenia solium causing taeniasis.
156. Answer (4)
Hint : Identify a roundworm whose larvae are spirally
coiled in muscles.
Sol. : Trichinella is an endoparasitic, viviparous
roundworm in rodents, pig, humans etc. causing
disease trichinosis characterised by abdominal pain
and muscular inflammation.
157. Answer (4)
Hint : Most primitive lizard named as Tua-tara.
Sol. : Sphenodon is the most primitive lizard,
considered as living fossil but not as connecting link.
Peripatus is placed in taxon-Onychophora and
Neopilina is a connecting link between molluscs and
annelids.
158. Answer (3)
Hint : Mammal with avian characters.
Sol. : Ornithorhynchus – Duckbilled platypus
Pinctada – Pearl oyster
Pteropus – Flying fox
Balanoglossus – Tongue worm
159. Answer (4)
Hint : Flagellated cells of sponges are also named
as collar cells.
Sol. : Choanocytes or collar cells are found in inner
lining of spongocoel as well as flagellated chambers
of sponges. They continuously produce water current
for uptake of food, respiration, excretion and
reproduction.
Test - 6 (Code-C) (Answers & Hints) All India Aakash Test Series for Medical-2020
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160. Answer (1)
Hint : Syngamy occurs inside female’s body.
Sol. : Cartilaginous fishes show internal fertilization
and ovoviviparous condition.
161. Answer (4)
Hint : Fish like organism with cartilaginous cranium
and vertebral column.
Sol. : In cyclostomes, such as Myxine, scales and
tympanum are absent. Heart is two chambered.
Fishes have membranous labyrinth. All the given
characters belong to members of class Amphibia.
162. Answer (4)
Hint : These structures are sensitive to sound and
other vibrations.
Sol. : Anal cerci are jointed filamentous structures
present on 10th tergum. Anal style is unsegmented
spiny structure on 9th sternum present only in male
cockroach.
163. Answer (3)
Hint : Vocal cords are sound producing structures in
frogs.
Sol. : Male frog has sound producing vocal cords
and vocal sacs make the sound louder.
Cockroach can survive upto 7 days with amputated
head because most of its nervous tissue is ventrally
placed in belly region.
164. Answer (3)
Hint : This organism can live on both land and water.
Sol. : Bidder’s canal present in kidney is responsible
for conduction of sperms
Hirudinaria – nephridia as excretory organ
Locusta – jointed legs for locomotion
Ctenoplana – cnidoblasts are absent
165. Answer (3)
Hint : These structures are associated with
segmented worms.
Sol. : Parapodia are associated with annelid like
Nereis. They help in locomotion and respiration.
166. Answer (1)
Hint : Chemoreceptors which taste chemical nature
of water
Sol. : In molluscs, osphradium in mantle cavity
serves to test chemical nature of water.
167. Answer (3)
Hint : A – Fastest running mammal.
B – Bird exhibiting polygamy.
Sol. : Panthera tigris (tiger) is considered as national
terrestrial animal and Pavo cristatus (Peacock) is
national bird of India.
168. Answer (3)
Hint : Respiratory structures present on body from
2nd thoracic segment onwards.
Sol. : In cockroach, 10 pair of spiracles are present
in between sclerites for gaseous exchange during
respiration.
169. Answer (3)
Hint : Fighting fish is a bony fish.
Sol. :
Earthworm – Nephridia – Act as excretory
structures
Cockroach – Titillator – Helps in copulation
present in
males only
Ascaris – Pineal setae – In males for
copulation
170. Answer (2)
Hint : Such type of metamorphosis is found in silk
moth.
Sol. : Silk moth develops through larval and pupal
stages. Conversion of larva into pupa is known as
pupation, characterised by histolysis followed by
histogenesis.
171. Answer (2)
Hint : Flatworms have solid body without any cavity.
Sol. : Flatworms are bilaterally symmetrical,
triploblastic, dorsoventrally flattened and acoelomate.
Taenia solium has no alimentary canal. It absorbs
predigested nutrients from intestine of host.
172. Answer (1)
Hint : Wuchereria is a representative of these
worms.
Sol. : Being circular in cross-section, these animals
are commonly known as roundworms. Sexual
dimorphism evolved first in roundworms. They have
muscular pharynx and both free living and parasitic
forms are present.
173. Answer (4)
Hint : Identify an animal exhibiting epitoky
Sol. : Nereis is an aquatic unisexual annelid.
Parapodia are lateral unsegmented appendages for
swimming and respiration. Body is metamerically
segmented, development is indirect through
trochophore larval stage.
All India Aakash Test Series for Medical-2020 Test - 6 (Code-C) (Answers & Hints)
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174. Answer (3)
Hint : In molluscs, mantle cavity containing feathery
gills is present between mantle and visceral hump.
Sol. : Soft and spongy skin present just below shell
is known as mantle. Space between mantle and
visceral hump is known as mantle cavity. Feather like
gills responsible for both respiration and excretion
are present in mantle cavity.
175. Answer (3)
Hint : Shell is a calcareous exoskeleton found
around most molluscs.
Sol. : Calcareous spicules form endoskeleton in
sponges. Calcareous ossicles constitute
endoskeleton in sea stars and cartilaginous
endoskeleton is found in cartilaginous fishes.
Dentalium has shell as calcareous exoskeleton.
176. Answer (1)
Hint : Blood glands and tufts of pharyngeal nephridia
are present in same set of body segments.
Sol. :
Typhlosole is a fold in intestinal wall to increase
absorptive surface area. Ovaries with ovarioles are
situated between 2nd to 6th abdominal segments in
female cockroach.
177. Answer (4)
Hint : Brood pouch harbours mammary nipples.
Sol. : In Balanoglossus, indirect development is
seen due to presence of larval stage. In female
Macropus (kangaroo), a mammal, mammary nipples
are present in brood pouch (marsupium).
178. Answer (1)
Hint : Mouth parts which contain chitinous teeth
known as denticles.
Sol. : In cockroach, mandibulate type of mouth parts
are found which are responsible for grinding and
incising food.
179. Answer (4)
Hint : Commonly called star fish.
Sol. : Pheretima has nephridia for excretion and is
an annelid. Interstitial cells are present in the body
wall of coelenterates. Ctenoplana belongs to
Ctenophora.
180. Answer (3)
Hint : In non-chordates, heart beat is initiated by
hormone from brain.
Sol. : In non-chordates, heart is either absent or
present dorsally. If present, it is generally neurogenic
because heart beat is initiated by a ganglia situated
near heart. In myogenic heart, heart beat is initiated
by specialized muscular tissue situated in wall of
heart known as S.A node/pace maker of the heart.
Test - 6 (Code-D) (Answers) All India Aakash Test Series for Medical-2020
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Test Date : 17/02/2019
ANSWERS
TEST - 6 (Code-D)
All India Aakash Test Series for Medical-2020
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175. (4)
176. (4)
177. (3)
178. (4)
179. (1)
180. (3)
All India Aakash Test Series for Medical-2020 Test - 6 (Code-D) (Answers & Hints)
2/16
ANSWERS & HINTS
1. Answer (2)
Hint: 2m
Tk
Sol.: 2 kg
2300N/m
T
2s
150T
2. Answer (1)
Hint: F = –kx, k = m2.
Sol.: 2 2
2
4 2 8
k
m
2 2
2
2 2T
or, 4 2 sT
3. Answer (1)
Hint: y = f (x ± vt)
Sol.: y = f (x + vt) when wave is moving towards
negative x-axis
y = f (x – vt) when wave is moving towards
positive x-axis
4. Answer (4)
Hint: Pressure variation is maximum at displacement
nodes.
Sol.: For stationary waves, all particles oscillates in
same phase with in a loop. Stationary wave is
formed when two waves are travelling in opposite
direction.
5. Answer (3)
Hint: 1
v
Sol.: Density of moist air is less than that of dry air.
6. Answer (3)
Hint: 0
0
s
v vf f
v v
⎡ ⎤ ⎢ ⎥⎣ ⎦
[ PHYSICS]
Sol.:
Motorist Band
5 ms–1
Wall
10 ms–1
Frequency of reflected sound
1
330 3301000 1000
330 –10 320f
⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
As motorist is moving toward wall
0
2 1
330 3351000
320 330
v vf f
v
⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎣ ⎦
2
3351000
320f
⎡ ⎤ ⎢ ⎥⎣ ⎦
7. Answer (2)
Hint: 0
0
s
v vf f
v v
⎛ ⎞ ⎜ ⎟⎝ ⎠.
Sol.: When engine approaches man then frequency
is higher and constant, when engine passes man
then frequency heard lowered but remains constant.
8. Answer (2)
Hint: f1 – f
n = (n – 1) f
B
Sol.:1– 29 5 145
nf f
1
1– 1452
ff
f1 = 290 Hz 1
145 Hz2
f⇒ .
9. Answer (2)
Hint: nl = constant
Sol.: 2
1
20 4
25 5
A
B
n l
n l n
B > n
A.
So, nB – n
A = 5 on solving,
nA = 20
nB = 25
Test - 6 (Code-D) (Answers & Hints) All India Aakash Test Series for Medical-2020
3/16
10. Answer (1)
Hint: Use Newton’s formula P
v
Sol.: According to Laplace’s correction
v v Percentage error
v v 1.4 1100 100%
v 1.4
15%�
11. Answer (1)
Hint: 1 1
2 2
f v
f v
Sol.: 1 1 2
2 2 1
, If Constantf v m
Tf v m
1
2
600 2 1
32 4
f
f f
f = 2400 Hz
12. Answer (1)
Hint: v
f
Sol.:330 1
m 50 cm660 2
3 512.5, 37.5, 62.5
4 4 4
(Not possible)
Only 1st and 2nd resonance can be produced.
So minimum height of water
hmin
= 50 – 37.5 = 12.5 cm
13. Answer (4)
Hint: y = 2A sinkx cost
Sol.: y = 2A sinkx cost
2sin cos(90 )30
xy t
⎛ ⎞ ⎜ ⎟⎝ ⎠
2
30
= 60 cm
30 cm2
Number of loops = 10
Number of nodes = 10 + 1 = 11
Number of antinodes = 10
Amplitude of x = 5 cm
2sin 5 1cm30
14. Answer (1)
Hint: 1 2 3
1 2 3
1 1 1: : : :n n n
l l l
Sol.:1 2 3: : 1: 2 :1n n n
1 2 3: : 2 :1: 2l l l
l1 + l
2 + l
3 = 150 cm 5K = 150 K = 30.
So, l1 = 60 cm
l2 = 30 cm
l3 = 60 cm
so, x1 = 60 cm
x2 = 90 cm
15. Answer (2)
Hint & Sol.: Sound travel fastest in solids.
16. Answer (4)
Hint: Stress
v .
Sol.: f0 = 100 Hz, v = f
0
0 = 200 m/s
Now, 2Stress v .
Stress = (200)2 × 8 × 103 = 3.2 × 108 N/m2
17. Answer (1)
Hint: 0
Source–
vf f
v v
⎛ ⎞ ⎜ ⎟
⎝ ⎠
Sol.:350
5500 5000350 – v
⎛ ⎞ ⎜ ⎟⎝ ⎠
11 350
10 350 – v
350 – 11v = 0
350m/s
11v
18. Answer (3)
Hint: End correction, e = 0.6r
Sol.:1 2
3
4( ) 4( )
v vf
l e l e
3(l1 + e) = l
2 + e
2 1– 3
2
l le
72.6 – 720.3 0.6
2e r
3 1cm
6 2r
All India Aakash Test Series for Medical-2020 Test - 6 (Code-D) (Answers & Hints)
4/16
19. Answer (4)
Hint: m m
a a
v
v
Sol.:1650 5
330 1
m
a
v
v
5
1
m
a
5
m a
20. Answer (4)
Hint: fB = f
1 ~ f
2
Sol.: f = 4 5t = 1st 2nd
fB
4 5
21. Answer (1)
Hint: Only odd harmonic in closed organ pipe.
Sol.: f1 : f
2 : f
3 = 5 : 7 : 9
So pipe is closed
So, fundamental frequency 0
28557 Hz
5f
0
0
380 20m
57 3
v
f
22. Answer (4)
Hint: Y
v
Sol.: 9
3 –1
3
8 102 10 m s
2 10
Yv
L
5
4
O
In second overtone
5
4L
41.6 m
5
L
3
32 101.25 10 1250 Hz
1.6
vf
23. Answer (3)
Hint: Wavelength depends on medium.
Sol.: (1) Frequency does not change by changing
the medium of propagation.
(2) If wave is reflected from denser medium,
the phase change is .
24. Answer (1)
Hint: Beat frequency = |f1 – f
2|.
Sol.: 1 = 1020 f
1 = 510 Hz.
2 = 1004 f
2 = 502 Hz
fb = |f
1 – f
2| = 8 Hz
25. Answer (2)
Hint: Amplitude = 2Asinkx
Sol.: Amplitude = 4sinx ...(i)
22m
⇒
...(ii)
at2m,
3 3x
2Amplitude 4sin 2 3 m
3
26. Answer (2)
Hint: Resultant amplitude formula.
Sol.: When amplitude of component waves are
equal, amplitude of resultant, 2 cos2
R A⎛ ⎞ ⎜ ⎟
⎝ ⎠
Amplitude, 3
2 cos 2 36 2
R A A A
27. Answer (4)
Hint: T
v
Sol.: Stress
v
11 –5
3
1.6 10 1.6 10 20
8 10
Yv
4
–11.6 1.6 2 1080 m s
8
28. Answer (4)
Hint: Pitch is related to frequency of sound.
Sol.: Quality is the sensation of human ear due to
wave form of sound.
29. Answer (1)
Hint: P RT
vM
Sol.: If temperature is constant v will remain
constant.
Since pressure and density will change in the same
ratio.
Test - 6 (Code-D) (Answers & Hints) All India Aakash Test Series for Medical-2020
5/16
30. Answer (1)
Hint: B
v
Sol.: 2
B
v
or, 2 2 –6 –2 2
8 3
1 1
8 10 (2.5 10 )
2 10 kg/m
B
v kv
31. Answer (3)
Hint: Fundamental frequency, 0
4
vf
l
Sol.:0
330110 Hz
4 0.75f
only odd harmonic are produced in closed pipe.
So frequencies 330 Hz, 550 Hz, 770 Hz are
possible. 660 Hz is not possible.
32. Answer (3)
Hint: = kx
Sol.:2
x
0.52
k
[0.5 – 0.25]2 8
33. Answer (3)
Hint:p
yv
t
Sol.: y = 20 sin(100 t – 2x)
20 100 cos(100 – 2 )p
yv t x
t
3 12 10 cos 100 0 – 2
6p
v⎡ ⎤ ⎢ ⎥⎣ ⎦
3 312 10 10 cm/s
2p
v⎛ ⎞ ⎜ ⎟⎝ ⎠
34. Answer (3)
Hint: 0
1
2
Tf
l
Sol.:10
20
If 100 N100 1
169 N20 169 1.3
⎡ ⎤ ⎢ ⎥ ⎣ ⎦
Tf
Tf
0.3f0 = 20
0
200Hz
3f
35. Answer (1)
Hint and Sol.: Open end of a closed organ pipe is
displacement antinode and pressure node.
36. Answer (3)
Hint: (vp)max
= A
Sol.:1
2
p wv v
0
1(2 )
2y f f
= 4y0
37. Answer (4)
Hint: T
v xg
Sol.: v xg
1
22
dv ga v xg g
dx x (constant)
So a x°
38. Answer (1)
Hint: 2
20
md x dxb kxdtdt
Sol.:2
20
md x dxb kxdtdt
...(i)
2
22 0d x dx
xdt dt
This is equation of damped oscillation.
39. Answer (1)
Hint: dx
vdt
Sol.: 10 sin5
v t
x vdt ∫
–10 cos5
5
t C
x
–50cos5
x t C
at t = 0, v = 0, x = –50 so, C = 0
50cos5
x t⎛ ⎞ ⎜ ⎟
⎝ ⎠
All India Aakash Test Series for Medical-2020 Test - 6 (Code-D) (Answers & Hints)
6/16
46. Answer (1)
Hint : C – H bond dissociation energy in
CH = CH – CH2 2
H
is slightly less than bond
dissociation energy of C – H in CH2
H
Sol. : Stability order :
CH2
CH2
CH – – CH3 2
C
CH3
CH3
CH3
[ CHEMISTRY]
40. Answer (3)
Hint: 2m
Tk
Sol.: 3
2 22 2
3
m mT
kk
⎛ ⎞⎜ ⎟⎝ ⎠
41. Answer (2)
Hint: vmax
= ASol.: x = Asint
(10 cm)sin2
x t
–1 –1
max(10 cm) rad s 5 cm s
2v
42. Answer (1)
Sol.: Time taken from x = A to is .2 6
A Tx
x = Acost
2cos 3
2
AA
T
1 2cos 3
2 T
23
3 T
T = 18 s
43. Answer (4)
Hint: y = Asin(t + )
Sol.: The angle at time 2
( )4
t tT
.
The projection on y axis
2 2( ) sin sin
4 120 4
⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
y t R t R tT
sin60 4
y R t ⎡ ⎤ ⎢ ⎥
⎣ ⎦
44. Answer (4)
Hint: If function increases or decreases
monotonically with time, function is non-periodic.
Sol.: (1) y = cos3t = 13cos – cos3
4t t
it is periodic but not SHM
(2) 2 sin SHM4
⎛ ⎞ ⎜ ⎟⎝ ⎠
y t
(3) y = cos2t, it is SHM
(4) y = aet is non periodic
45. Answer (1)
Hint: The least time interval after which particle
repeats its motion is called time period.
Sol.: Least time for given combination is 2
47. Answer (4)
Hint : More the stability of conjugate base formed,
more will be acidic nature of hydrocarbon.
Sol. :
+ H
(More acidic) (Conjugate base)(Aromatic)More stable
H –
+
48. Answer (4)
Hint : 31.4×meq.ofNH
% of N =wt.of sample
Test - 6 (Code-D) (Answers & Hints) All India Aakash Test Series for Medical-2020
7/16
Sol. : meq of NH3 evolved = meq. of H
2SO
4 used
= N × V(ml) = 0.05 × 40 = 2
31.4×meq.ofNH 1.4 2
% of N = 14%wt.of sample. 0.2
49. Answer (3)
Hint : 4np electrons in a planar ring makes the ring
antiaromatic.
50. Answer (4)
Hint : Order of deactivation is :
– NO2 > – CN > – COCH
3 > – F
51. Answer (3)
Hint : The degree of unsaturation (DOU) for C4H
6 is
two.
Sol. : ⇒C H (DOU = 3)4 4
52. Answer (1)
Hint : More the s-character in carbanion, more will
be stability.
Sol. : In ethyne carbanion (CH C–), carbon is sp
hybridised which has more s-character (50%), in
other words can hold negative charge more better, so
more stable.
53. Answer (2)
Hint : Nitration takes place in that benzene ring
which has more electron density.
Sol. :
C – O
O
Conc. HNO3
Conc. H SO2 4
C – O
O
NO2
(Major)
(–M) (+M)
54. Answer (3)
Hint :
CH – C C – CH3 3
Lindlar’s catalyst
Na in liq. NH
3
C = C
CH3
HH
CH3
(Cis – but – 2 – ene)
C = CH
H
CH3
(trans – but – 2 – ene)
CH3
(A)
(B)
Sol. : For A, 0
For B, = 0
55. Answer (4)
Hint :
3 C H2 2
Red Hot
Fe tube
Ethyne 873 K
COONa + NaOHCaO
+ Na CO2 3
Sodium
benzoate
OH + Zn
Phenol
+ ZnO
56. Answer (4)
Hint : More the surface area of Isomeric alkanes,
more will be the boiling point.
Sol. : As the branching in isomeric alkane increases
their surface area decreases so boiling point
decreases
BP : > >
57. Answer (3)
Hint : Less stable alkenes are more reactive towards
catalytic hydrogenation.
Sol. :
CH = CH2
has only one alpha–H
atom. So, hyperconjugation effect is small and
alkene is least stable.
58. Answer (1)
Hint : Less stable carbocation can rearrange to
more stable carbocation.
Sol. :
OH
H+
OH2
+
–H O2
CH – – CH3 2
C
CH3
CH3
+ CH3
–
shiftCH – – CH – CH
3 3 C
CH3
+
H
CH3 – C = CH – CH
3
CH3
– H+
All India Aakash Test Series for Medical-2020 Test - 6 (Code-D) (Answers & Hints)
8/16
59. Answer (3)
Hint : Propyne contains acidic hydrogen.
Sol. : 4
3
NH OH3 3AgNO
whiteppt.
CH C CH CH C CAg
4
3
NH OH3 2 AgNO
CH C CH Noreaction
60. Answer (3)
Hint : Symmetrical alkene does not show
Markovnikov’s addition of HBr.
Sol. : is a symmetrical alkene.
61. Answer (2)
Hint : H atom attach to sp-hybridised carbon is
acidic in nature.
Sol. : CH C 3 – C– H acidic H atom.
Propyne
62. Answer (4)
Hint : By Kolbe’s electrolytic method alkane, alkene
and alkyne all can be prepared.
Sol. :
2 CH3 – COONa
+ H
2 O CH
3 – CH
3 + 2CO
2
(Sodium acetate)ethane
+ 2NaOH + H 2
CH2 – COONa + 2H
2 O CH
2 + 2CO
2 + 2NaOH + H
2
CH
ethene2
CH –2
COONa
(Sodium succinate)
CH – COONa + 2H
2 O CH + 2CO
2
+ 2NaOH + H2
CHethyne
CH –
COONa
(Sodium fumarate)
63. Answer (1)
Hint : During hydration of alkene in acidic medium
carbocation is formed as intermediate.
Sol. :
CH3
CH – CH = CH2
H2 O H
CH3
C
H
CH
+CH
3
H
shift
CH3
C
CH2 CH
3
+
H2 O
–H+
CH3
C
CH2
OH
CH3
+
–
(3° alcohol)
CH3
CH
CH CH
2
(i) B2 H THF
6,
CH3
CH CH2 CH
2
=
(ii) H2O ,NaOH
2
(1° alcohol)
OH
64. Answer (2)
Hint : R of R-MgX on reaction with acidic-H give
alkane.
Sol. :
CH3
CH – CH MgCI2
CH3
– +
CH –3 O – H
+–
CH3
CH3
– CH – CH
3 + Mg
CI
OCH3
65. Answer (4)
Hint : More the H-bonding and lesser the torsional
strain, more will be the stability of conformers.
Sol. : Stability order:
(1) Cyclohexane : Chair > Boat > Half chair
(2) Butane (around C2 and C
3) : Anti > Gauche >
Partially eclipsed
(3) Propane : Staggered > Skew > Eclipsed
(4) Ethane -1, 2-diol: Gauche > Anti > Eclipsed
66. Answer (3)
Hint : In presence of sunlight addition reaction and
in presence of FeCl3 electrophilic substitution
reaction take place.
Sol. :
+ 3CI2
Sunlight
H CI
CI H
H
CI
H
CI
CI
H
CI
H
( )C H Cl6 6 6
( ) (X)Benzene hexachloride
+ CI2
FeCI3 + HCI
CI
( ) (Y)Chlorobenzene
67. Answer (4)
Hint : 1-alkynes are acidic in nature.
Sol. :
Ch C CH3 –
NaNH2
CH3– C C Na
CH3 – CH – CI
2
CH3– C CH
2 3–C – CH
Pent–2–yne
– +
68. Answer (1)
Hint : 2, 3–dimethylbutane has two type of
H-atoms.
Test - 6 (Code-D) (Answers & Hints) All India Aakash Test Series for Medical-2020
9/16
Sol. :
CI2
h Cl+
Cl
Chiral carbon
69. Answer (2)
Hint : When C – CI bond acquires double bond
character then reactivity of C – Cl bond decreases.
Sol. : CH2 = CH
– CI
..CH
2 –CH CI
(vinyl chloride)
– +
70. Answer (1)
Hint : (i) O3
(ii) Zn+H O2
C=C C =O+O =C
Sol. :
2–oxo propanal
CH3 – C – CH = O + O = CH2
O+O = CH2
CH3 – C – CH = CH2
CH2
2-methylbuta-1, 3-diene
2 3 4
(C H5 8)
Methanal
Methanal
1
71. Answer (2)
Hint : Carbonium ion intermediate is formed.
Sol. :
CH = CH2
H+
H
+
H – +
CI–
shift
CI
A
72. Answer (1)
Hint : H2SO
4 + H
2SO
4 SO
3 + HSO
4
–
+ H
3O+
Sol. : SO3 is an electrophile.
73. Answer (4)
Hint : For highly polluted water, BOD value is more
than 17 ppm.
74. Answer (4)
Hint : N2O is a greenhouse gas.
75. Answer (3)
Hint : Photochemical smog is oxidising smog.
76. Answer (3)
Hint : Kjeldahl’s method is not applicable to
compounds containing
(i) Nitrogen in ring.
(ii) (–NO2) and azo (– N = N –) groups.
Sol. : Nitrogen present in CN , can be
quantitatively estimated by Kjeldahl’s method.
77. Answer (4)
Hint : Pent–2–ene and 1-chloro-2-bromopropene
show geometrical isomerism.
78. Answer (2)
Hint : Glycerol decomposes near its boiling point.
Sol. : Compounds which decompose near its boiling
point are purified by vacuum distillation method.
79. Answer (2)
Hint : Higher the availability of lone pair on nitrogen
atom higher is its basic nature and hence higher the
pKa
80. Answer (1)
Hint : Order of -I effect.
– CN > – COOH > – OH
81. Answer (1)
Hint : More the delocalization of +ve charge, more
will be the stability of carbocation.
Sol. : CH2
+ has maximum number of
resonating structures, so most stable.
82. Answer (4)
Hint : HNO3 converts NaCN and Na
2S into HCN and
H2S respectively.
83. Answer (1)
Hint. : CH3
CH2
– –C
CH3
CH3
––
–CH2– is neohexyl group.
84. Answer (3)
Hint : For tautomerism, acidic H-atom should be
present.
Sol. :
O
does not contain any acidic H-atom.
85. Answer (3)
Hint : Octet of N in 4
NH
is complete and no vacant
orbital is present on N atom.
All India Aakash Test Series for Medical-2020 Test - 6 (Code-D) (Answers & Hints)
10/16
[ BIOLOGY]
91. Answer (2)
Hint: RQ of fat is less than 1 and of organic acids
it is more than 1.
Sol.:
Tripalmitin is a fatty acid (RQ = 0.7)
Proteins (RQ = 0.9)
Malic acid (RQ = 1.33).
92. Answer (3)
Hint: In link reaction pyruvic acid (3C) forms acetyl
CoA(2C) & CO2.
Sol.:
4C compound – Malic acid
3C compound – Pyruvic acid
2C compound – Acetyl CoA
5C compound – -ketoglutaric acid
93. Answer (3)
Hint: During dormancy, growth & development of
seed is temporarily stopped.
Sol.: In dormancy, seeds cannot germinate inspite
of availability of all suitable environmental conditions.
94. Answer (3)
Hint: Pr absorbs red light.
Sol.: 660 nm is red light.
Pfr absorbs far red light.
86. Answer (2)
Hint : 2Volumeof N (atSTP)28
% of N 10022400 wt.of compound
Sol. : 1 1 2 2
1 2
P V P V
T T
(Experiment condition) (STP condition)
2760 V(725 – 25) 60
300 273
∵ Volume of N2 (at STP) = 50.29 ml
28 50.29% of N 100 22.45%
22400 0.28
87. Answer (4)
Hint : Alkyl group with –H atoms attach to
benzene, activates the ring for SE reactions.
Sol. : does not have –H atom.
88. Answer (1)
Hint : Cyclic, planar and conjugated compounds
containing (4n + 2)e– are aromatic in nature.
Sol. : HH is non planar because of steric
repulsion of inner hydrogen.
89. Answer (3)
Hint : More the s-character in hybrid orbitals, more
will be the electronegativity.
Sol. : C2H
2 : sp hybridised carbon
50% s-character.
90. Answer (2)
Hint : 1 2 3 4 5 6
2 2 3CH CH– CH – C C–CH
Hex–1–en–4–yne
95. Answer (4)
Sol.: Growth curve in plants is sigmoid, also called
S-shaped growth curve.
96. Answer (2)
Hint: Auxin gets removed along with apical buds of
a plant.
Sol.: Auxin was first time isolated from human urine.
It is an amino acid derivative.
97. Answer (4)
Hint: It is a gaseous hormone.
Sol.: Ethylene is responsible for ripening of fruits. It
can be used artificially to ripen the fruits.
98. Answer (3)
Hint: ABA is a plant growth inhibiting phytohormone.
Sol.: ABA induces dormancy of buds and seeds &
promotes abscission of flowers.
99. Answer (3)
Sol.: Gibberellin can replace vernalisation. As by
exogenous application of gibberellins many biennials
can be induced to behave as annuals.
100. Answer (4)
Hint: Short day plants require uninterrupted dark
period to flower.
Sol.: SDP will not flower if its dark period is
interuppted by a flash of light.
Test - 6 (Code-D) (Answers & Hints) All India Aakash Test Series for Medical-2020
11/16
101. Answer (3)
Hint: Antagonistic means showing opposite effects
towards something.
Sol.: When two hormones bring about opposite
effects on the same function, it means they are
antagonistic to each other, whereas when they act
together they are called synergistic.
102. Answer (3)
Hint: Coconut milk has a phytohormone which helps
in cell division.
Sol.: It contains cytokinin.
103. Answer (2)
Sol.: Cytokinin along with auxin are essential in
tissue culture. High cytokinin to auxin ratio causes
shoot formation in callus.
104. Answer (1)
Hint: Cytokinin(CK) helps in production of new
leaves.
Sol.: It (CK) enhances chloroplast production in
leaves. It delays senescence by controlling protein
synthesis.
105. Answer (3)
Sol.: 2, 4-D is widely used as weedicide in cereal
crops.
106. Answer (4)
Hint: Tryptophan is precursor of auxin.
Sol.: Tryptophan is an amino acid. Final product of
transition/link reaction of aerobic respiration is acetyl
CoA which is precurssor of gibberellin.
A pigment violaxanthin is a precursor of ABA.
107. Answer (3)
Hint: Cytochrome c is a mobile carrier protein.
Sol.: Cytochrome c is not the part of complex IV. It
is attached to the outer surface of the inner
membrane & transfers electrons between complex III
& IV.
108. Answer (3)
Hint: Krebs’ cycle produces reduced coenzymes.
Sol.: NADH + H+ is produced during Krebs’ cycle
along with FADH2 & GTP.
It starts with condensation of acetyl CoA with OAA
to form citric acid.
109. Answer (2)
Hint: The F1 headpiece contains the site of the ATP
synthesis.
Sol.: F1 headpiece remains present towards the
matrix of the mitochondria. Hence ATP synthesis
takes place towards the matrix.
110. Answer (3)
Hint: Inner membrane of mitochondria forms cristae.
Oxidative phosphorylation occurs during oxidation of
food.
Sol.: Cristae contains cytochromes which helps in
ETS.
During formation of bread, yeast if used performs
alcoholic fermentation which releases CO2.
111. Answer (3)
Sol.: Buttercup shows environmental (extrinsic)
heterophylly. Cotton, coriander & larkspur show
developmental heterophylly due to intrinsic factors.
112. Answer (2)
Hint: Gibberellin was first isolated from a fungus
called Gibberella fujikuroi.
Sol.: It is acidic in nature, increases sugarcane yield
by increasing the length of internodes, delays
ripening of fruits and helps in seed germination.
113. Answer (3)
Sol.: The stimulus of vernalisation is perceived by the
mature stem apex or by the embryo of the seed.
114. Answer (4)
Hint: Auxin occurs in growing apices.
Sol.: If apical bud is removed then auxin is removed,
due to which apical dominance is prevented.
The plant will show more lateral branches & axillary
buds.
115. Answer (1)
Hint: Phytochrome is a photoreceptor.
Sol.: Phytochrome is a blue-green pigment,
regulates photoperiodic induction of flowering and
helps in seed germination too.
It occurs in two forms Pr & P
fr. Its P
r form becomes
active by absorbing far-red light. Pfr is physiologically
active form.
116. Answer (3)
Hint: Short day plant flowers after receiving
photoperiod shorter than critical period.
Sol.: Tobacco is a short day plant. Wheat, radish &
sugarbeet are long day plants.
117. Answer (3)
Hint: Rice & maize show hypogeal seed
germination.
Sol.: In hypogeal seed germination, epicotyl grows
first & cotyledons remain underground.
118. Answer (2)
Hint: Relative growth can be calculated by the
formula i.e. Growth per unit time
100Initial size
.
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Sol.: For Leaf A
5100 50%
10
For Leaf B
5100 20%
25
119. Answer (1)
Hint: This hormone is a plant growth inhibitor.
Sol.: ABA is a derivative of carotenoids.
120. Answer (4)
Hint: Tissues formed from de-differentiation of
parenchyma are secondary in origin.
Sol.: Wound cambium, cork cambium &
interfascicular cambium are secondary in origin. They
are formed by de-differentiation of parenchyma cells.
Intrafascicular cambium is primary in origin.
121. Answer (4)
Hint: The continued growth due to activity of
meristems occurs in plants.
Sol.: New cells are always being added due to
activity of meristem, this type of growth is open form
of growth and indeterminate.
Animals show definite & diffused type of growth.
122. Answer (1)
Hint: Ethylene is a volatile substance which
promotes senescence.
Sol.: Ethylene is the only gaseous hormone of
plants. It breaks the dormancy of seeds.
Bakanae disease is caused by gibberellins.
123. Answer (3)
Sol.: Auxin helps in root formation on stem cuttings,
shows apical dominance & phototropism but does
not play any role to speed up the malting process
which is done by gibberellin.
124. Answer (4)
Hint: The plant hormone which is responsible for
closure of stomata is known as stress hormone.
Sol.: ABA is called stress hormone as its synthesis
is stimulated by drought, water logging & other
adverse environmental conditions.
125. Answer (2)
Hint: The phytohormone which was extracted from a
fungus Gibberella fujikuroi is responsible for bolting.
Sol.: Gibberellins are active substances which
induce stem elongation in rossette plants. This
phenomenon is called bolting.
126. Answer (1)
Hint: Complete oxidation of acetyl CoA occurs in
aerobic respiration through Krebs’ cycle.
Sol.: One turn of Krebs’ cycle involves complete
breakdown of acetyl CoA which produces 3NADH +
H+, 1 FADH2 & 1 GTP.
3 NADH + H+ = 3 × 3 = 9 ATP
1 FADH2 = 1 × 2 = 2 ATP
1 GTP = 1 × 1 = 1 ATP
12 ATP
NADH + H+ & FADH2 produce ATP through ETS.
127. Answer (3)
Hint: Yeast performs alcoholic fermentation.
Sol.: In alcoholic fermentation alcohol is produced
along with CO2 & ATP.
128. Answer (4)
Hint: In aerobic respiration CO2 is released in link
reaction & during Krebs’ cycle.
Sol.: In link reaction CO2 is released at one step.
During Krebs’ cycle CO2 is released at two steps.
In case of anaerobic condition
2
2
2CORQ (infinite)
0O
129. Answer (2)
Hint: It is a 2C compound.
Sol.: Acetyl CoA is common intermediate among
fats, proteins and carbohydrates breakdown during
aerobic respiration. It enters into Krebs’ cycle to
yield energy.
130. Answer (4)
Hint: Krebs’ cycle starts with the condensation of
2C compound with 4C compound to yield a 6C
compound.
Sol.: Krebs’ cycle starts with condensation of Acetyl
CoA (2C) with oxaloacetic acid (4C) to yield citric
acid (6C).
131. Answer (2)
Hint: Lactic acid (3C) is produced from pyruvic acid
which is also a 3C compound.
Sol.: In lactic acid fermentation pyruvic acid (3C) is
converted into lactic acid (3C). Hence there is no
release of CO2 which occurs in alcohol fermentation.
Both processes are hazardous and produce 2 ATP
as net gain.
132. Answer (1)
Sol.: The term glycolysis has originated from two
Greek words that are ‘glycon’ which means sugar
and ‘lysis’ which means splitting.
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133. Answer (3)
Hint: Partial oxidation of glucose without utilization
of oxygen is called glycolysis.
Sol.: Glycolysis takes place in all organisms in the
cytoplasm. Hence all organisms whether anaerobes
or aerobes retain enzymes for glycolysis.
134. Answer (4)
Hint: Plants have lesser demands for gaseous
exchange as compared to animals.
Sol.: Roots, stems and leaves all respire at lower
rates than animals. In plants, cells are closely
packed and located quite close to the surface of the
plant. Hence it is easy for gases to diffuse inside.
135. Answer (4)
Hint: ATP is synthesized during respiration.
Sol.: ATP acts as the energy currency of the cell.
The energy released by oxidation of food in
respiration is used to synthesise ATP and not used
directly.
136. Answer (3)
Hint : In non-chordates, heart beat is initiated by
hormone from brain.
Sol. : In non-chordates, heart is either absent or
present dorsally. If present, it is generally neurogenic
because heart beat is initiated by a ganglia situated
near heart. In myogenic heart, heart beat is initiated
by specialized muscular tissue situated in wall of
heart known as S.A node/pace maker of the heart.
137. Answer (4)
Hint : Commonly called star fish.
Sol. : Pheretima has nephridia for excretion and is
an annelid. Interstitial cells are present in the body
wall of coelenterates. Ctenoplana belongs to
Ctenophora.
138. Answer (1)
Hint : Mouth parts which contain chitinous teeth
known as denticles.
Sol. : In cockroach, mandibulate type of mouth parts
are found which are responsible for grinding and
incising food.
139. Answer (4)
Hint : Brood pouch harbours mammary nipples.
Sol. : In Balanoglossus, indirect development is
seen due to presence of larval stage. In female
Macropus (kangaroo), a mammal, mammary nipples
are present in brood pouch (marsupium).
140. Answer (1)
Hint : Blood glands and tufts of pharyngeal nephridia
are present in same set of body segments.
Sol. :
Typhlosole is a fold in intestinal wall to increase
absorptive surface area. Ovaries with ovarioles are
situated between 2nd to 6th abdominal segments in
female cockroach.
141. Answer (3)
Hint : Shell is a calcareous exoskeleton found
around most molluscs.
Sol. : Calcareous spicules form endoskeleton in
sponges. Calcareous ossicles constitute
endoskeleton in sea stars and cartilaginous
endoskeleton is found in cartilaginous fishes.
Dentalium has shell as calcareous exoskeleton.
142. Answer (3)
Hint : In molluscs, mantle cavity containing feathery
gills is present between mantle and visceral hump.
Sol. : Soft and spongy skin present just below shell
is known as mantle. Space between mantle and
visceral hump is known as mantle cavity. Feather like
gills responsible for both respiration and excretion
are present in mantle cavity.
143. Answer (4)
Hint : Identify an animal exhibiting epitoky
Sol. : Nereis is an aquatic unisexual annelid.
Parapodia are lateral unsegmented appendages for
swimming and respiration. Body is metamerically
segmented, development is indirect through
trochophore larval stage.
144. Answer (1)
Hint : Wuchereria is a representative of these
worms.
Sol. : Being circular in cross-section, these animals
are commonly known as roundworms. Sexual
dimorphism evolved first in roundworms. They have
muscular pharynx and both free living and parasitic
forms are present.
145. Answer (2)
Hint : Flatworms have solid body without any cavity.
Sol. : Flatworms are bilaterally symmetrical,
triploblastic, dorsoventrally flattened and acoelomate.
Taenia solium has no alimentary canal. It absorbs
predigested nutrients from intestine of host.
146. Answer (2)
Hint : Such type of metamorphosis is found in silk
moth.
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Sol. : Silk moth develops through larval and pupal
stages. Conversion of larva into pupa is known as
pupation, characterised by histolysis followed by
histogenesis.
147. Answer (3)
Hint : Fighting fish is a bony fish.
Sol. :
Earthworm – Nephridia – Act as excretory
structures
Cockroach – Titillator – Helps in copulation
present in
males only
Ascaris – Pineal setae – In males for
copulation
148. Answer (3)
Hint : Respiratory structures present on body from
2nd thoracic segment onwards.
Sol. : In cockroach, 10 pair of spiracles are present
in between sclerites for gaseous exchange during
respiration.
149. Answer (3)
Hint : A – Fastest running mammal.
B – Bird exhibiting polygamy.
Sol. : Panthera tigris (tiger) is considered as national
terrestrial animal and Pavo cristatus (Peacock) is
national bird of India.
150. Answer (1)
Hint : Chemoreceptors which taste chemical nature
of water
Sol. : In molluscs, osphradium in mantle cavity
serves to test chemical nature of water.
151. Answer (3)
Hint : These structures are associated with
segmented worms.
Sol. : Parapodia are associated with annelid like
Nereis. They help in locomotion and respiration.
152. Answer (3)
Hint : This organism can live on both land and water.
Sol. : Bidder’s canal present in kidney is responsible
for conduction of sperms
Hirudinaria – nephridia as excretory organ
Locusta – jointed legs for locomotion
Ctenoplana – cnidoblasts are absent
153. Answer (3)
Hint : Vocal cords are sound producing structures in
frogs.
Sol. : Male frog has sound producing vocal cords
and vocal sacs make the sound louder.
Cockroach can survive upto 7 days with amputated
head because most of its nervous tissue is ventrally
placed in belly region.
154. Answer (4)
Hint : These structures are sensitive to sound and
other vibrations.
Sol. : Anal cerci are jointed filamentous structures
present on 10th tergum. Anal style is unsegmented
spiny structure on 9th sternum present only in male
cockroach.
155. Answer (4)
Hint : Fish like organism with cartilaginous cranium
and vertebral column.
Sol. : In cyclostomes, such as Myxine, scales and
tympanum are absent. Heart is two chambered.
Fishes have membranous labyrinth. All the given
characters belong to members of class Amphibia.
156. Answer (1)
Hint : Syngamy occurs inside female’s body.
Sol. : Cartilaginous fishes show internal fertilization
and ovoviviparous condition.
157. Answer (4)
Hint : Flagellated cells of sponges are also named
as collar cells.
Sol. : Choanocytes or collar cells are found in inner
lining of spongocoel as well as flagellated chambers
of sponges. They continuously produce water current
for uptake of food, respiration, excretion and
reproduction.
158. Answer (3)
Hint : Mammal with avian characters.
Sol. : Ornithorhynchus – Duckbilled platypus
Pinctada – Pearl oyster
Pteropus – Flying fox
Balanoglossus – Tongue worm
159. Answer (4)
Hint : Most primitive lizard named as Tua-tara.
Sol. : Sphenodon is the most primitive lizard,
considered as living fossil but not as connecting link.
Peripatus is placed in taxon-Onychophora and
Neopilina is a connecting link between molluscs and
annelids.
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160. Answer (4)
Hint : Identify a roundworm whose larvae are spirally
coiled in muscles.
Sol. : Trichinella is an endoparasitic, viviparous
roundworm in rodents, pig, humans etc. causing
disease trichinosis characterised by abdominal pain
and muscular inflammation.
161. Answer (2)
Hint : Larval stage which causes taeniasis in
humans.
Sol. : In fecal matter of infected man, onchosphere
is present which is converted into hexacanth larva
and finally cysticercus larva/bladderworm in muscles
of pig. Human beings are infected when they feed on
undercooked measly pork and this larva in adult
develops into Taenia solium causing taeniasis.
162. Answer (3)
Hint : Most vertebrates are poikilothermous except
birds and mammals generally.
Sol. : Testudo is a reptile and cannot maintain its
body temperature, hence considered as
poikilothermic/cold blooded animal.
Columba is a bird while Felis and Macropus are
mammals. So, all these are warm-blooded animals.
163. Answer (1)
Hint : Animals which are commonly known as
roundworms are classified in this phylum.
Sol. : Aschelminthes/Nematodes are considered as
first animals to have complete digestive tract with
distinct mouth and anus.
164. Answer (3)
Hint : Number of moults is equivalent to the number
of chambers in its heart.
Sol. : In cockroach, nymph is a young stage which
is similar to cockroach but smaller in size, light in
colour, with immature gonads and without wings. It
undergoes gradual metamorphosis known as
paurometabola in 6 months upto two years. During
this time, it moults 13 times, to become sexually
mature adult.
165. Answer (2)
Hint : Common name for Antedon.
Sol. :
Metridium – sea anemone – Coelenterata
Antedon – sea lily – Echinodermata
Hormiphora – sea walnut – Ctenophora
Pennatula – sea pen – Coelenterata
166. Answer (3)
Hint : Common name for Hemichordates.
Sol. : • Trygon – Poison sting (Sting ray)
• Torpedo – Muscle modified into
electric organ (Electric
eel)
• Tongue worm – Proboscis present not
placoid scales
• Hippocampus – Brood pouch in male to
carry fertilized eggs.
167. Answer (3)
Hint : Fish like organisms having circular mouth.
Sol. : Cyclostomes (Petromyzon) are animals
without scales having circular mouth. They have non
operculated gills and lack paired fins. Scoliodon
(cartilaginous fish) and Hippocampus and Labeo
(bony fishes) have scales on their body.
168. Answer (3)
Hint : Members of this class show aerial
adaptations.
Sol. : In birds, long bones of the body contain
cavities filled with air to decrease weight of the body.
Such type of bones found in birds are known as
pneumatic bones.
169. Answer (4)
Hint : The first segment bears mouth and a lobe that
can act as a wedge.
Sol. : The first body segment of earthworm is known
as peristomium. Prostomium is a hood like structure
on peristomium that is helpful in burrowing. The fecal
deposits of earthworm are known as worm castings.
Vermicomposting is use of worm castings as
manure.
170 Answer (2)
Hint : Members of Class Osteichthyes.
Sol. : Air bladder/swim bladder is found in bony
fishes to provide buoyancy which is helpful in
swimming. They do not to have continuously swim
as in case of cartilaginous fishes which lack swim
bladder.
171. Answer (3)
Hint : Identify common Indian bull frog.
Sol. : Rana tigrina is common Indian bull frog.
Alytes is commonly known as midwife toad because
male carries eggs wrapped around its hind limbs.
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172. Answer (3)
Hint : Platyhelminthes are acoelomates.
Sol. : Protostomic animals are schizocoelic
eucoelomates and deuterostomic animals are
enterocoelic eucoelomates. Only arthropods placed
in class Insecta are considered as insects.
173. Answer (2)
Hint : Capillaries are absent in Hemichordates
Sol. : In leech and Balanoglossus, open type of
circulatory system is present.
174. Answer (4)
Hint : This process helps in replacement of lost
body parts.
Sol. : With the help of water vascular system
echinoderms perform locomotion, excretion,
respiration and capturing of food.
175. Answer (4)
Hint : Chordates have hollow, dorsal nerve cord
Sol. : Arthropoda – Chitinous exoskeleton
Mollusca – Calcareous exoskeleton
Echinodermata – Spiny skinned, ventral
solid nerve cord
Chordata – Presence of notochord
176. Answer (4)
Hint : The roundworm which is found attached with
intestinal wall.
Sol. : • Euspongia – Bath sponge
• Meandrina – Brain coral
• Fasciola – Liver fluke
177. Answer (3)
Hint : Obelia exhibits alternation of generation.
Sol. : In class Hydrozoa, all members except Hydra
show metagenesis. Metastasis is spread of cancer,
metamorphosis is the development of larval stage
into adult stage.
178. Answer (4)
Hint : Animal which is bilaterally symmetrical in larval
stage but becomes asymmetrical due to torsion in
adult life.
Sol. : Adult snails are asymmetrical due to torsion.
In larval stage, it is bilaterally symmetrical. It is a
type of retrograde metamorphosis.
179. Answer (1)
Hint : This animal shows internal metamerism.
Sol. : In annelids, both external and internal
segmentation is found in adult stage
In arthropods, only external segmentation is present
in adults. Chordates show internal segments in
certain body parts.
180. Answer (3)
Hint : In sponges, the body cavity opens to exterior
through osculum.
Sol. : All the given characters are shown by
sponges. They are not considered as diploblastic
and they have cellular grade of body organisation.
The cavity inside body is known as spongocoel or
paragastric cavity.