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Test - 1 (Code-C) (Answers) All India Aakash Test Series for Medical-2020
1/17
1. (3)
2. (4)
3. (1)
4. (1)
5. (2)
6. (1)
7. (2)
8. (2)
9. (2)
10. (3)
11. (1)
12. (2)
13. (3)
14. (2)
15. (4)
16. (2)
17. (1)
18. (1)
19. (3)
20. (2)
21. (2)
22. (3)
23. (3)
24. (3)
25. (2)
26. (4)
27. (4)
28. (1)
29. (4)
30. (3)
31. (4)
32. (4)
33. (1)
34. (1)
35. (2)
36. (3)
Test Date : 14/10/2018
ANSWERS
TEST - 1 (Code-C)
All India Aakash Test Series for Medical-2020
37. (1)
38. (4)
39. (4)
40. (1)
41. (3)
42. (1)
43. (4)
44. (2)
45. (3)
46. (1)
47. (3)
48. (4)
49. (4)
50. (2)
51. (3)
52. (4)
53. (1)
54. (1)
55. (2)
56. (3)
57. (4)
58. (1)
59. (3)
60. (2)
61. (2)
62. (3)
63. (1)
64. (1)
65. (4)
66. (2)
67. (4)
68. (3)
69. (3)
70. (1)
71. (1)
72. (2)
73. (4)
74. (2)
75. (1)
76. (2)
77. (2)
78. (4)
79. (4)
80. (1)
81. (2)
82. (1)
83. (4)
84. (3)
85. (2)
86. (2)
87. (4)
88. (1)
89. (4)
90. (3)
91. (2)
92. (3)
93. (4)
94. (2)
95. (3)
96. (4)
97. (2)
98. (3)
99. (3)
100. (4)
101. (3)
102. (3)
103. (2)
104. (2)
105. (3)
106. (4)
107. (4)
108. (2)
109. (4)
110. (3)
111. (4)
112. (1)
113. (2)
114. (4)
115. (3)
116. (3)
117. (2)
118. (4)
119. (2)
120. (3)
121. (2)
122. (4)
123. (3)
124. (3)
125. (1)
126. (3)
127. (2)
128. (2)
129. (2)
130. (1)
131. (2)
132. (2)
133. (4)
134. (2)
135. (3)
136. (2)
137. (3)
138. (4)
139. (2)
140. (4)
141. (1)
142. (3)
143. (1)
144. (1)
145. (1)
146. (3)
147. (1)
148. (2)
149. (3)
150. (1)
151. (4)
152. (3)
153. (1)
154. (4)
155. (4)
156. (3)
157. (2)
158. (4)
159. (2)
160. (3)
161. (3)
162. (2)
163. (3)
164. (4)
165. (1)
166. (2)
167. (4)
168. (2)
169. (3)
170. (2)
171. (2)
172. (1)
173. (4)
174. (2)
175. (1)
176. (4)
177. (1)
178. (2)
179. (1)
180. (4)
All India Aakash Test Series for Medical-2020 Test - 1 (Code-C) (Answers & Hints)
2/17
ANSWERS & HINTS
1. Answer (3)
Hint : The interval in which displacement is zero.
Solution : For the curve, ACB displacement is zero
AB
C
Tt
x
0
0
����
�
av
SV
t
2. Answer (4)
Hint :
Displacement and velocity may be positive or
negative.
2nd
4th
0
+ve
s
v
Solution : Direction of displacement is vertically
downward.
S < 0
Direction of velocity is downward
So, v < 0
3. Answer (1)
Hint : max
2 v S
Solution :
u = 0 vmax v = 0
S2
S2
S1
S1
2
2 max
max 1 12
2
v
v S S ...(i)
[ PHYSICS]
2
2 max
max 2 20 2
2
vv S S
...(ii)
2
max
1 2
1 1
2
vS S S
max
2 2 2 3 100
5
v S
4 15 m/s
4. Answer (1)
Hint : 1 2
1 2
2
av
v vv
v v
Solution : v1 = 36 km/h = 10 ms–1
v2 = 54 km/h = 15 ms–1
1 2
1 2
av
S Sv
t t
1 1 1
210
2 10
S
SS t t
2 2 2
215
2 15
S
SS t t
112 ms
1 1
2 10 15
av
Sv
S
5. Answer (2)
Hint : v
at
Solution : v = t2 + 4t
At t = 1 s, v = 5 ms–1
t = 2 s, v = 12 ms–1
212 57 ms
2 1
a
Test - 1 (Code-C) (Answers & Hints) All India Aakash Test Series for Medical-2020
3/17
6. Answer (1)
Hint : dv
a vdx
Solution:
0 0
1v x
v x
...(i)
0
0
0
v
v x v
x
...(ii)
0
0
vdv
dx x ...(iii)
0 0
0
0 0
v vdva v x v
dx x x
2 2
0 0
200
v v
x
xx
7. Answer (2)
Hint : slope of graphdv
a v tdt
Solution : a = slope of v – t graph
= –ve value (constant)
t
a
0
8. Answer (2)
Hint : Average speed = Total distance travelled
Total time taken
Detailed Solution : From t = 0 to t = 4 s x1
= 4 m
From t = 4 s to t = 6 s x2 = 0
From t = 6 s to t = 10 s x3 = 4 m
Average speed = 14 0 4 8 4
ms10 10 5
9. Answer (2)
Hint : 2
0
1
2S S ut at
Solution:
v
tt0
v0
0
0
0
(constant)
va
t
2
0
1
2x x ut at
20
0
0
10
2
vx x t
t
20
0
0
1
2
vx x t
t
10. Answer (3)
Hint : | Average velocity | = |Total displacement travelled|
Total time taken
Solution:
ABO
| Displacement | = 10 m
Time taken = 5
10 2
s
| Average velocity | = 110 20
ms
2
11. Answer (1)
Hint : Reaction time
Solution:
Distance travelled during reaction time
S1 = 10 × 1 = 10 m ...(i)
Distance travelled during retarded motion
2
2
10050 m
2 2 1
vS
a
...(ii)
Total travelled distance = S1 + S
2 = 60 m
Distance of car from red signal
= 70 – 60 = 10 m
12. Answer (2)
Hint : 2
0
1
2h v t gt
Solution :
7
8
5
4
h
t = 6
All India Aakash Test Series for Medical-2020 Test - 1 (Code-C) (Answers & Hints)
4/17
Particle will be at maximum height at t = 6 s.
2110 6 180 m.
2 h
13. Answer (3)
Hint : 2 2
2 1 02v v a l
Detailed solution : 2 2
2 1 02v v a l
2 2
2 1
02
v vl
a
l
2
2 2
1 02
2
lv v a
2 2 2 2
2 2 2 1 1 2
1 0
02 2
v v v vv v a
a
2 2
1 2
2
v vv
14. Answer (2)
Hint : d + x = v0t ...(i)
21
2x at ...(ii)
Solution : 32 + x = 10t ...(i)
211
2x t ...(ii)
32 m
10 ms–1
a = 1 ms–2
x
2132 10 0
2 t t
t2 – 20t + 64 = 0
20 400 256 20 12
2 2t
t = 4, 16
t = 4 s
15. Answer (4)
Hint : Stone will acquire velocity of frame of
reference when it leave frame of reference.
Solution : 2
0
1
2h u t gt
h
u0
2
0
1
2h u t gt
–3.2 = 6t – 5t2
5t2 – 6t – 3.2 = 0
t = 1.6 s
16. Answer (2)
Hint : Average speed = Total travelled distance
Total time taken
Solution : Time taken by cyclist A
362 h
18
Time taken by B cyclist from cycling
1 32 h
2 2
Average speed of B for actual riding
3624 km/h
3
2
17. Answer (1)
Hint : S = S1 + S
2 + S
3
Solution : 2
1
11 4 8 m
2 S
S2 = v
0t2 = (a
1t1)t
2 = 1 × 4 × 10 = 40 m
2
0 2 30 2v a S
3
164 m
2 2
S
S = 40 + 8 + 4 = 52 m
18. Answer (1)
Hint : A A
B B
a u
a u
Detailed solution : tA = t
B = t
0
0 = 20 – aAt0
Test - 1 (Code-C) (Answers & Hints) All India Aakash Test Series for Medical-2020
5/17
0
20A
at
Similarly
0
10
Ba
t
20 2
10 1
A
B
a
a
19. Answer (3)
Hint : Total time (n – 1) T0 time interval
Solution : 21
2h gt
1
2
3
4
5
6
7
Total time = (7 – 1) T0
= 6 T0
21
2h gt
2
0
1{36 }
2h g T
2
018h gT
2
0180 18 10 T
T0 = 1 s
20. Answer (2)
Hint : dv
a vdx
Solution :
v(ms )–1
x(m)3 70
10
5
A
B
5
4
dv
dx
Velocity at t = 5 s
v = 7.5 ms–1
25 37.5(7.5) 9.4 ms
4 4
a
21. Answer (2)
Hint :
2
1
2
1
t
t
t
t
vdt
v
dt
Solution :
52
0
5
0
(30 4 2)t t dt
v
dt
53 2
0
30 42
3 2
(5 0)
t tt
3 2 5
0
1(10 2 2 )
5t t t
11[10 125 50 10] 262 ms
5
22. Answer (3)
Hint : Smaller least count.
Solution : Least count is minimum in = 0.001 m, so
it is most precise.
23. Answer (3)
Hint : In multiplication, product has number of
significant figures equal to least significant figure
present in calculation.
Solution :
V = IR
V = 1.25 × 25.425
= 31.78125
= 31.8 (3 significant figures)
24. Answer (3)
Hint : Limitations of dimensional analysis.
Solution :
1. Two and more physical quantities may have
same dimensional expression.
2. Numerical constant has no dimensions.
3. Method of dimensions can be used only for
product of physical quantities.
All India Aakash Test Series for Medical-2020 Test - 1 (Code-C) (Answers & Hints)
6/17
25. Answer (2)
Hint : n1u
1 = n
2u
2
Solution : 1 1 1
2 1
2 2 2
a b c
M L Tn n
M L T
h = 6.6 × 10–34 J-s
[h] = [M1L2T–1]
a = 1, b = 2, c = –1
1 2 1
34
2
kg s6.6 10
10 g 5m 100 s
mn
34 2 216.6 10 10 10
25
= 2.64 × 10–31
26. Answer (4)
Hint : Planck’s constant and angular momentum
have same dimensions.
Solution :
1. Energy density and pressure have same
dimensions.
2. Relative density and plane angle have no
dimension.
1 2 2[M L T ][ ]
[MK]
Q
Cm Q
0 2 2 1[M L T K ]
0 2 2[ ] [M L T ] QL
m
27. Answer (4)
Hint : n1u
1 = n
2u
2
Solution : [E] = [M1L2T–2]
2 kgm2s–2 = [2 kg]1 [2 m]2 [n s]–2
2
2
22 2
n
n2 = 4
n = 2, similarly we can find the relation for
momentum and power.
28. Answer (1)
Hint : Trignometric ratios have no dimensions
Solution : [LHS] = [RHS]
[y] = [Ax2]
1
2[ ] [ ]
yA L
x
[] = [M0L0T0]
[] = [M0L1T0]
[A2] = [M0L1T0]
29. Answer (4)
Hint : [L] = [pavbmc]
Solution : [L] = [M1L2T–1]
[p] = [M1L2T–3]
[v] = [M0L1T–1]
[m] = [M1L0T0]
[M1L2T–1] = k [M1L2T–3]a [M0L1T–1]b [M1L0T–0]c
1 = a + 0 + c
2 = 2a + b + 0
–1 = –3a – b
a = –1, b = 4, c = 2
30. Answer (3)
Hint :
2
21 1
2 2
lk mv m
t
Solution :
2
21 1
2 2
lk mv m
t
100 100 100 2 100 k m l t
k m l t
= 1% + 4% + 2% = 7%
31. Answer (4)
Hint : It has no dimension.
Solution : Solid angle 2
cos dsd
r
0 0 0
2
cos[ ] [M L T ]
dsd
r
32. Answer (4)
Hint : Fundamental forces.
Solution : Gravitational force is the weakest force.
Gravitational forces are central forces, electrostatic
forces are central forces, strong nuclear forces are
not central forces.
33. Answer (1)
34. Answer (1)
Hint : Mass and energy are inter-convertable.
Solution : Conservation laws in nature.
1. Law of conservation of energy
2. Law of conservation of linear momentum
3. Law of conservation of angular momentum
35. Answer (2)
Hint : Raman effect.
Solution : CV Raman won noble prize for scattering
of light by molecules.
Test - 1 (Code-C) (Answers & Hints) All India Aakash Test Series for Medical-2020
7/17
36. Answer (3)
Hint : Non-zero digits are significant.
Solution : Trailing zeros in a number without
decimal are insignificant.
37. Answer (1)
Hint : z = xy
Z X Y
Z X Y
Solution : z = xy
Z X Y
Z X Y
0.4 0.1 1
4 1 5
Z
Z
1
4 5
Z
Z = 0.8 m
Z = (4 ± 0.8) m
38. Answer (4)
Hint : If there is no digit after 5 then last digit should
be even.
Solution : 2.765 rounding off (3 significant)
= 2.76
5.735 = 5.74
39. Answer (4)
Hint : Minimum percentage error.
Solution : Percentage error in 2.000 m
0.001100 0.05%
2.000
40. Answer (1)
Hint : The difference with average value should be
least.
Solution : t1 = |10.00 – 9.95| = 0.05 s
t2 = |10.00 – 10.10| = 0.10 s
41. Answer (3)
Hint : � � �
rel A Ba a a
Solution : � � �
rel A Ba a a
= g – g = 0
42. Answer (1)
Hint : L.C = 1MSD – 1VSD
Solution : L.C = 1MSD – 1VSD
VSDMSD 1
MSD
1 MSD = 1 mm
19 MSD = 20 VSD
1 VSD = 19
VSD20
L.C = 19
1 1 0.05 mm20
43. Answer (4)
44. Answer (2)
Hint : R = R1 + R
2
Solution : Req
= R1 + R
2
1 2
eq 1 2
0.1 0.1
30
R RR
R R R
eq
0.2 2100 100 %
30 3
R
R
45. Answer (3)
Hint : Mean absolute error
1 2| | | | | |
nx x x
n
Solution : xm
= 330 ms–1
x1 = |342 – 330| = 12
x2 = |338 – 330| = 8
x3 = |318 – 330| = 12
x4 = |322 – 330| = 8
112 8 12 810 ms
4
46. Answer (1)
Hint: Orbital angular momentum = hl l 1
2
Solution: For f - orbital, I = 3
Orbital angular momentum = h 3h3 3 1
2
[ CHEMISTRY]
47. Answer (3)
Hint: Chromium has half-filled d-subshell.
Solution: Cr(24) : [Ar]4s13d5
48. Answer (4)
Hint: n-factor for acid is number of replaceable H+
ions
Solution: n-factor = 1
Mol. wt 120Eq. wt. 120
n-factor 1
All India Aakash Test Series for Medical-2020 Test - 1 (Code-C) (Answers & Hints)
8/17
49. Answer (4)
Hint: No. of particles = No. of moles × NA
Solution:
Moles of oxygen atom = 0.3 × 14 = 4.2
No. of electrons of oxygen atoms = 4.2 × 8 × NA
= 33.6 NA
50. Answer (2)
Hint: x y 2 2 2
y yC H x O xCO H O
4 2
2
2
Moles of CO obtainedx
y 2 Moles of H O obtained
Solution:
8.8
x 144
5.4y 32
18
Empirical formula = CH3
Molecular wt. 2 30n 4
Empirical formulawt. 15
Molecular formula = n × empirical formula
= 4 (CH3) = C
4H
12
51. Answer (3)
Hint: Only four lines of Balmer series of hydrogen
atom lie in visible region
Solution: For n = 6 (5th excited state) transition to
n = 1
* Maximum possible transitions
= 6 1 5 + 4 + 3 + 2 + 1 =15
* Maximum Paschen transitions
= 3(6 3, 5 3, 4 3)
52. Answer (4)
Hint: For one electron species, energy of subshell
depends only on the value of n.
Solution: For 3rd shell, No. of degenerates orbitals = n2
= 32 = 9.
53. Answer (1)
Hint: Frequency 2
32
z
v znf
r nn
z
Solution:
2
2
2
3
He Li
2
Li He
3
3
T f 2 9
T f 322
1
54. Answer (1)
Hint: Dilution equation, M1V
1 = M
2V
2
Solution: For stock solution,
1
49 1.5 10M 7.5
98
Now, M1V
1 = M
2V
2
7.5 × V1 = 0.1 × 2.5 × 1000
V1 = 33.33 ml
55. Answer (2)
Hint: Normality = Molarity × n-factor
Solution: Molarity of
22
23
3
9.033 10
6.022 10OH 0.3 M
500 10
Molarity of Ca(OH)2
= 0.3
M2
Normality of Ca(OH)2
=
0.32 0.3 M
2
56. Answer (3)
Hint: No. of atoms = No. of molecules × atomicity
Solution: Remaining molecules of CO2
=
323 20440 10
6.022 10 1044
No. of remaining atoms = 59.22 × 1020 × 3
= 1.77 × 1022
57. Answer (4)
Hint: 2 2 3
1 1FeO O Fe O
4 2
Solution:
(72g)
(8g)
2 2 31 mole
0.25 mole
1 1FeO O Fe O
4 2
% increase in wt. 8
100 11.11%72
58. Answer (1)
Hint: h
mv
Solution: H2 has minimum molar mass so have
longest de-Broglie wavelength.
59. Answer (3)
Hint: s, p and d-subshells have 1, 3 and 5 orbitals
respectively.
Solution:
In l subshell, number of orbitals = 2l + 1
Test - 1 (Code-C) (Answers & Hints) All India Aakash Test Series for Medical-2020
9/17
60. Answer (2)
Hint: Electrons in half filled p-subshell have different
orientations.
Solution: Orientation of orbitals is given by magnetic
quantum number.
61. Answer (2)
Hint: 2
2 2
1 2
1 1 1RZ
n n
Solution: For second line of Lyman series :
2
2 2
1 1 1R (1)
x 1 3
1 8R
x 9
9R
8x
For 3rd line of Lyman series : 2
2 2
1 1 1R (1)
1 4
15 9 15 135R
16 8x 16 128x
128x
135
62. Answer (3)
Hint: Ni2+ = [Ar] 3d8
Solution: 29
Cu = 1s2 2s2 2p6 3s2 3p6 4s1 3d10
s-electrons in Cu atom = 7
63. Answer (1)
Hint: 6
n
zv 2.18 10 m/sec
n
Solution: 6
4 He
8
22.18 10v 14
c 2753 10
64. Answer (1)
Hint: For minimum molar mass, one molecule
should contain at least one atom of oxygen
Solution:% of oxygen =wt.of oxygen
100wt.of compound
164 100
Mol.wt.
Minimum mol. wt. = 400 amu
65. Answer (4)
Hint: Molarity (M) = solute
n
V L
Solution: Mass of 1 ml D2O = 1 g
3
1
20M 50 M
1 10
66. Answer (2)
Hint: Law of multiple proportions is illustrated for
compounds which have two same elements.
Solution: CH4 : for 12 g C, wt. of H = 4 g = x
C2H
4 : for 12 g C, wt. of H = 2 g = y
x 4 2
y 2 1 = Simple whole no. ratio
67. Answer (4)
Hint: M1
V1 + M
2 V
2 = M
3 V
3
Solution: 0.5 × 2 + 1 × 1= M3 × 3
3
2M 0.67 M
3
68. Answer (3)
Hint: 2 2 3
32AI O Al O
2
Solution: 2 2 3
32Al O Al O
2
3 2
3KClO KCl O
2
From stoichiometry, 2 mole Al require 3
2 mole O
2
and 3
2 mole O
2 is obtained by 1 mole KCIO
3.
69. Answer (3)
Hint: Divalent metal chloride should be MCl2
Solution: Mol. wt.= 2 × vapour density = 60 × 2 = 120
x + 2 × 35.5 = 120
x = 49
Eq. wt. of metal 49
24.52
70. Answer (1)
Hint: Average atomic mass =
% abundance of isotope atomic mass
100
Solution: 13 x 12 (100 x)
12.011100
x = 1.1%
All India Aakash Test Series for Medical-2020 Test - 1 (Code-C) (Answers & Hints)
10/17
71. Answer (1)
Hint: Mass of electron is 9.1 × 10–31 kg
Solution: Positron is the particle having mass equal
to electron but having a unit positive charge
72. Answer (2)
Hint: -particle is He nucleus (He2+) which contains
2 protons and 2 neutrons.
Solution: For neutron, e
0m
For other particles, e
m order is
electron > proton > -particle
73. Answer (4)
Hint: l = 2 means d-subshell
Solution: d-subshell can have maximum 10
electrons.
74. Answer (2)
Hint: -rays, X-rays , UV, visible, IR, microwaves
Wavelength increases
Frequency decreases
Solution: Visible waves have higher frequency than
IR.
75. Answer (1)
Hint:
solvent
solute
solvent
X 1000molality(m) =
X MW
Solution: 0.2 1000
m 13.890.8 18
76. Answer (2)
Hint: Exchange energy of electrons is defined for
degenerate orbitals.
Solution: Exchange energy of electrons is defined
as energy released when an electron exchanges its
position with electron having same spin present in
degenerate orbitals.
77. Answer (2)
Hint: 21
KE mv2
, h
mv , v
Solution: 2v v mv 2 1 2
mv KEh h 2 h
27 7 1
34
23.313 10 10 sec
6.626 10
78. Answer (4)
Hint: h
x p4
Solution:h
x m v4
34
3 24
h 6.626 10x
1004 m v4 10 6.626 10
10x 2.5 10 m 2.5Å
79. Answer (4)
Hint: For principal quantum number n, l can be 0, 1,
2, ... (n – 1)
Solution: for n = 2,
l 2, it will have value 0 and 1.
80. Answer (1)
Hint:
2
n
nr 0.53 Å
Z
Solution: n = 2, Z = 4
2
4
2r 0.53 0.53 Å
4
81. Answer (2)
Hint: Ti : [Ar] 4s2 3d2
Solution:
Ti2+ : [Ar] 3d2
82. Answer (1)
Hint: 2
2 2
1 2
1 1E 13.6Z
n n
eV/atom
Solution: For, n1 = 4 and n
2 = 5
9E 13.6 eV/atom
16 25
83. Answer (4)
Hint: Electron density of the p-orbitals lie along the
axes.
Solution: dxy
orbital does not have electron density
along the axis.
84. Answer (3)
Hint: One AVOGRAM =
A
1
N
Test - 1 (Code-C) (Answers & Hints) All India Aakash Test Series for Medical-2020
11/17
85. Answer (2)
Hint: % of an element wt.of element
100wt.of compound
Solution: For YO: % of oxygen =16
100M 16
160040
M 16
M 24
For second oxide: Let YOx
% of oxygen =
16x100
16x 24
1600x20
16x 24
3x
8
x 3 8 3
8
YO YO Y O
86. Answer (2)
Hint: No. of molecules = No. of moles × NA
Solution: H2
: No. of molecules A A
10N 5N
2
O2 : No. of molecules = 10 N
A
CO2
: No. of molecules = 8 N
A
SO2
: No. of molecules = A A
64N N
64 .
87. Answer (4)
Hint: Isoelectronic species have same number of
electrons.
Solution: Species Total electrons
N2
14
CO 14
14
17
2
2O
2O
CN–
14
88. Answer (1)
Hint: Ionisation energy 2
2
Z13.6 eV
n
Solution: Z = 2, n = 1
Ionisation energy
2
2
13.6 254.4eV
1
89. Answer (4)
Hint: Bohr’s model is valid only for one electron
species.
Solution: According to Bohr’s theory, angular
momentum of an electron is quantised.
90. Answer (3)
Hint: nhc
E
Solution: Energy released = 66.26 × 60
J100
34 8
9
60 n 6.626 10 3 1066.26
100 100 10
n = 2 × 1019
[ BIOLOGY]
91. Answer (2)
Hint: Nuclear membrane is found in all eukaryotes.
Solution: Mycoplasma, BGA, Bacillus, purple
photosynthetic bacteria & E.coli are prokaryotes.
92. Answer (3)
Hint: Body of plants and animals are composed of
cells & product of cells was proposed by Theodor
Schwann.
Solution: Theodor Schwann was a British zoologist.
93. Answer (4)
Hint: Chloroplast is the site of photosynthesis and
contain pigments in eukaryotes.
Solution: Chromatophores are membranous
extension in the cytoplasm of a cyanobacterial cell
which have photosynthetic pigments
94. Answer (2)
Hint: This inclusion body provides buoyancy to the
bacteria
Solution: Ribosomes are found in all prokaryotes
and eukaryotes. Cyanophycean granules are found in
cyanobacteria only while gas vacuoles are found in
blue-green algae (BGA) as well as in purple & green
photosynthetic bacteria
95. Answer (3)
Hint: Lipid molecules of plasma membrane have
polar head and non-polar tail.
Solution: Polar head is hydrophilic in nature and it
interacts with water. Rest all the statements
regarding plasma membranes are true.
All India Aakash Test Series for Medical-2020 Test - 1 (Code-C) (Answers & Hints)
12/17
96. Answer (4)
Hint: Movement of molecules across the membrane
without help of carrier proteins and ATP is called
simple diffusion
Solution: Simple diffusion occurs only for those
molecules which are neutral or non-polar, while
transport of polar and hydrophilic substances need
carrier proteins.
97. Answer (2)
Hint: Smallest cell organelle is known as organelle
within organelle.
Solution: Ribosome is known as organelle within
organelle and its r-RNA part is synthesized inside the
nucleolus.
98. Answer (3)
Hint: These structures are present in pits.
Solution: Symplast of two adjacent cells are
connected via cytoplasmic strands or
plasmodesmata. They are lined by plasma
membrane.
99. Answer (3)
Hint: The given figure is of golgi bodies
Solution: Enzymatic precursors for lysosomes are
synthesized in ER. Golgi does processing and
packaging of materials for intra as well as extra-cellular
targets.
Rest all the features of golgi are true.
100. Answer (4)
Hint: ER, Golgi, lysosomes & vacuoles function in
a coordinated manner.
Solution: Oxidation of fatty acids, proteins and
carbohydrate occurs inside the mitochondria.
Rest all functions are performed by organelles of
endomembrane system.
101. Answer (3)
Hint: This organelle has hydrolytic enzymes which
become functional at acidic pH.
Solution: Lysosomes have hydrolytic enzymes for
digestion of almost all types of macromolecules
which are functional at acidic pH.
102. Answer (3)
Hint: This organelle is found in almost all eukaryotic
cells and is site of ATP synthesis.
Solution: Mitochondria are sausage shaped, have
their own 70S ribosomes i.e, palade particles and
they are viewed after staining with Janus green.
Usually their number is high in those cells which
have high metabolism.
103. Answer (2)
Hint: ‘S’ is a unit
Solution: ‘S’ stands for Svedberg coefficient or
sedimentation coefficient and it is indirect measure of
density and size of ribosomal sub units.
104. Answer (2)
Hint: These structures are absent in prokaryotes.
Solution: Cytoskeleton are proteinaceous
filamentous structures which provide mechanical
strength & support to the cell.
105. Answer (3)
Hint: Microfilaments are solid unbranched rod like
fibrils.
Solution: Microtubules have diameter of 25 nm.
Intermediate filaments are involved in formation of
scaffolds of chromatin.
106. Answer (4)
Hint: This structure is absent in higher plants.
Solution: Centrioles have 9 peripheral fibrils of
tubulin and these are absent in the centre therefore
the arrangement is 9 + 0.
107. Answer (4)
Hint: In leucoplast granum remains absent.
Solution: Protoplast is a cell without cell wall.
108. Answer (2)
Hint: Both mitochondria and chloroplast are semi
autonomous structures.
Solution: Mitochondria, chloroplast and bacteria all
have ds circular DNA, 70S ribosomes porins on
outer membrane and self duplication ability.
Mitochondria and chloroplast are partially dependent
on nucleus.
109. Answer (4)
Hint: Polyribosomes are formed in cytoplasm
Solution: Polysomes are not formed with the help of
RER
110. Answer (3)
Hint: Nucleolus is found inside the nucleus
Solution: Nucleolus is non-membrane bound
structure found in nucleoplasm of nucleus. They are
larger in cells involved in protein synthesis.
111. Answer (4)
Hint: Chromatin is packed DNA
Solution: Packed DNA has RNA, histones and
some non-histone proteins.
Test - 1 (Code-C) (Answers & Hints) All India Aakash Test Series for Medical-2020
13/17
112. Answer (1)
Hint: Chromatids of chromosomes are held together
at primary constriction.
Solution: Primary constriction is known as
centromere which holds the two halves of a
chromosome.
113. Answer (2)
Hint: Such chromosomes appear L-shaped during
anaphase.
Solution: A chromosome with centromere slightly
away from the centre is submetacentric chromosome.
114. Answer (4)
Hint: Loops of lampbrush chromosomes have hair
like structure and with protein that are known as
informosomes.
Solution: Lampbrush chromosomes form loops
which have hair like structures. These hairs are
bound to protein. Some of which are called
informosome i.e, having mRNA + protein. These
loops participate in transcription and form m-RNA
115. Answer (3)
Hint: These structures help in photorespiration
along with chloroplast and mitochondria.
Solution: Peroxisomes are involved in photorespiration
because they have enzymes for formation as well
as destruction of peroxide.
116. Answer (3)
Hint: These structures are tubular and help bacteria
to attach with rocks also.
Solution: Fimbriae are bristle like structures which
help bacteria to attach with rocks or host cell.
117. Answer (2)
Hint: The cell also has cell wall with plasmodesmata.
Solution: Plant cells have cell wall with
plasmodesmata. Their 90% volume is occupied by
vacuoles.
118. Answer (4)
Hint: Bacterial type ribosomes are also found in
some cell organelles of eukaryotes.
Solution: Both ribosomes and plasma membrane
are similar in prokaryotes and eukaryotes.
119. Answer (2)
Hint: Crossing over occurs in pachytene stage.
Solution: Point at which crossing over occurs forms
the recombination nodule.
120. Answer (3)
Hint: Spindle fibres are not directly attached to the
centromere.
Solution: A disc-shaped structure is found over
centromere through which spindle fibres are attached
called kinetochore.
121. Answer (2)
Hint: Meiosis involves two sequential karyokinesis
and cytokinesis.
Solution: Except DNA duplication and histone
protein synthesis, rest all phenomenon occur twice
in meiosis.
122. Answer (4)
Hint: Tubulin protein synthesis occurs in the stage
where duplication of mitochondria & chloroplast
occurs.
Solution: Most of the cell organelles get duplicated
in G1. But Golgi, chloroplast and mitochondria are
doubled at G2 phase along with tubulin protein
synthesis.
123. Answer (3)
Hint: Interphase is known as the most active stage
of cell cycle.
Solution: Interphase constitute more than 95%
duration of cell cycle of a human cell. Rest all the
features regarding the interphase are true.
124. Answer (3)
Hint: Last phase of karyokinesis in mitosis involves
reappearance of cell organelles and nuclear
membrane.
Solution: At the end of mitosis, the chromosomes
decondense into chromatin which occurs in
telophase.
125. Answer (1)
Hint: Major restriction point inhibits cells to go for
DNA synthesis.
Solution: Major control on cell cycle can be
imposed on 1
G S transition step.
126. Answer (3)
Hint: Number of mitotic divisions to form ‘n’ number
of cells are = n – 1
Solution: Number of generations (n) required to form
‘x’ number of cells are = 2n
For 32 cells
Mitotic divisions = 32 – 1 = 31
Number of generations = 25 = 32, x = 5
All India Aakash Test Series for Medical-2020 Test - 1 (Code-C) (Answers & Hints)
14/17
127. Answer (2)
Hint: The given stage is the best stage to study
morphology of chromosomes.
Solution: Given stage is Metaphase I. where,
Bivalent chromosomes align at equator forming
two metaphasic plates
Alignment of bivalents is totally random process.
Spindle fibre attaches to each chromosome of
homologous pair
Separation of chromosomes occur at anaphase I
128. Answer (2)
Hint: Each chromosome has 2 chromatids
Solution: 1 Bivalent = 2 chromosomes
20 Bivalent = 20 × 2 = 40 chromosomes
Since 1 chromosome = 2 chromatids
40 chromosomes = 80 chromatids
129. Answer (2)
Hint: Recombinase enzyme catalyses the process
occurring in pachytene stage.
Solution: Crossing over is enzyme mediated
process and it produces recombinants.
130. Answer (1)
Hint: Dyad of cells appear in last stage of heterotypic
division
Solution: Dyad of cells are formed at the end of
telophase I.
131. Answer (2)
Hint: Cell plate represents the future middle lamella
of adjacent cells.
Solution: Cell plate help in cytokinesis of plant cells
which represent the middle lamella of adjacent cells.
132. Answer (2)
Hint: Treatment of colchicine produces polyploid
condition.
Solution: Colchicine inhibits formation of tubulin and
therefore microtubules. Lack of microtubules results
in inhibition of spindle fibres formation and its affect
will be visible first at metaphase.
133. Answer (4)
Hint: Mitosis does not involve crossing over.
Solution: Mitosis helps in reproduction in unicellular
organisms. It helps in repair and regeneration but
cannot produce recombinants.
134. Answer (2)
Hint: Interkinesis is resting phase between meiosis
I and II.
Solution: Meiosis I and II involves into sequential
karyokinesis. Chromosomes do not decondense upto
DNA level in interkinesis because after telophase I
chromatin recondense in prophase II.
135. Answer (3)
Hint: Meiosis involves crossing over as well as
reduction of chromosome number.
Solution: Four cells produced after telophase II are
genetically dissimilar to each other as well as to
their parents.
136. Answer (2)
Hint : Compounds whose role or function we do not
understand at the moment.
Solution : Primary metabolites are essential for
physiological processes and have identifiable
functions. Secondary metabolities include alkaloids,
antibiotics, rubber, essential oils, spices etc.
137. Answer (3)
Hint : Order of percentage of biomolecules in cells.
Solution : Water > proteins > nucleic acids >
carbohydrates > lipids > ions
70 90% 10 15% 5 7% 3% 2% 1%
A C B D E
138. Answer (4)
Hint : Polymers are found in retentate/acid insoluble
fraction.
Solution : Cysteine, calcium & cytosine are
micromolecules that are obtained in filtrate/acid
soluble fraction. Cellulose is a polymer of glucose
and due to its large size it cannot cross the filtration
membrane.
139. Answer (2)
Hint : Bones are reservoirs of some elements.
Solution : Matrix of bone is largely composed of
hydroxyapatite crystals of calcium phosphate.
Essential fatty acids & essential amino acids are
supplied in diet.
140. Answer (4)
Hint : Identify a molecule which is a monomer.
Solution : Inulin is a polymer of fructose, while
insulin is a polymer of amino acids. Lactose is a
disaccharide. Polymers can be broken into
monomers by addition of water (hydrolysis).
141. Answer (1)
Hint : Dehydration is a condensation reaction
involving loss of water molecule.
Solution : Cystine formation requires removal of H2
and disulfide bonds are formed between two amino
acids. Formation of lecithin, adenosine and collagen
requires formation of ester & glycosidic bonds
respectively.
Test - 1 (Code-C) (Answers & Hints) All India Aakash Test Series for Medical-2020
15/17
142. Answer (3)
Hint : Structural sugar found in RNA.
Solution : Cellulose is unbranched structural
homopolymer of glucose comprising of (1-4)
glycosidic linkages. Ribose is a monomer formed
without glycosidic bond. Cytidylic acid is a
nucleotide found in RNA.
143. Answer (1)
Hint : Reducing sugar.
Solution : Sugars with free aldehyde or ketone
groups are detected by Benedict’s and Fehling’s
solution. Glucose and galactose have free aldehyde
group while fructose has free keto group. Sucrose
has no free keto or aldehyde group.
144. Answer (1)
Hint : Isomerism in hexose sugars.
Solution : Galactose & glucose are aldose sugars
while fructose is ketohexose. Ribose is a pentose
sugar.
145. Answer (1)
Hint : Pentose sugar derived from C5H
10O
5.
Solution : 5 carbon sugar in RNA (ribonucleic acid)
is ribose. DNA has deoxyribose with molecular
formula C5H
10O
4.
146. Answer (3)
Hint : Identify a nucleotide. Ribozyme is RNA acting
as enzyme.
Solution : ATP is adenosine triphosphate.
147. Answer (1)
Hint : Formation of polymers and disaccharides
requires glycosidic bonds.
Solution : Glucose is a monomer of cellulose but
cellulose is unbranched homopolymer glucose.
148. Answer (2)
Hint : Unsaturated fatty acids have low melting
points.
Solution : Gingelly oil/sesame oil and arachidonic
acid are unsaturated fatty acids, hence will have
lower melting points in comparison to other saturated
lipids given.
149. Answer (3)
Hint : Double bonds cannot be introduced beyond 9th
carbon position in most animals usually.
Solution : Linoleic acid has 2 double bonds
(9 & 12th carbon). Plants are able to synthesize
essential fatty acids by introducing double bonds at
12th & 15th carbon. Oleic acid has a single double
bond at 9th position. In most mammals, arachidonic
acid can be formed from linoleic acid.
150. Answer (1)
Hint : Identify a homopolymer.
Solution : Chitin comprises of repeating units of (1
4) N-acetylglucosamine. Nucleic acids (DNA),
proteins i.e. RuBisCO and insulin are
heteropolymers.
151. Answer (4)
Hint : Organic cofactor loosely attached to
apoenzyme.
Solution : Niacin and riboflavin are vitamins requires
for formation of coenzymes such as NAD, FAD &
FMN.
152. Answer (3)
Hint : First alphabet of amino acid is considered as
the basis of single alphabet code.
Solution : Tyrosine is coded by Y, phenylalanine by
‘F’ and glutamic acid by E. This is because
threonine, proline & glycine have already been named
as T, P & G respectively.
153. Answer (1)
Hint : Two or more amino acids can be linked by
peptide bonds.
Solution : Collagen is a heteropolymer of amino
acids. Chitin & cellulose are polysaccharides whose
monomers are linked by glycosidic bonds. Choline is
nitrogenous compound found in lecithin.
154. Answer (4)
Hint : All components given are sugars in raffinose.
Solution : Tripeptide has 3 amino acids &
triglyceride has 3 fatty acids linked to glycerol.
155. Answer (4)
Hint : Active site are formed at tertiary level of
protein organisation. Hydrogen bonds stabilise
predominantly protein structure formed at secondary
level of organisation.
Solution : Ionic bonds, Van der Waals forces,
hydrophobic interactions along with hydrogen bonds
stabilise quaternary structure.
156. Answer (3)
Hint : DNA of a eukaryote comprises of A, T, G, C
as nitrogenous base.
Solution : Cytosine is a nitrogenous base while
cysteine is an amino acid.
157. Answer (2)
Hint : Amino acids are monomers of proteins.
Solution : Amino acids are substituted methanes.
They exist as zwitterions at isoelectric pH.
Trihydroxypropane is glycerol. DNA & RNA are
negatively charged molecules.
All India Aakash Test Series for Medical-2020 Test - 1 (Code-C) (Answers & Hints)
16/17
158. Answer (4)
Hint : Identify odd one out w.r.t. macromolecules.
Solution : Lipids are neither polymer nor
macromolecules.
159. Answer (2)
Hint : Thymine is exclusive to DNA.
Solution : Thiamine is vitamin B1 while thymine is
nitrogenous base in DNA.
160. Answer (3)
Hint : Hydrolytic enzymes work by addition of water
to break the bond between biomolecules.
Solution : Hydrolases belong to class III according
to nomenclature given by IUB. Phosphodiesterases
also belong to this class of enzymes.
161. Answer (3)
Hint : While overcoming competitive inhibition Km
value increases.
Solution : Succinate is the substrate for enzyme
succinate dehydrogenase. Malonate is its
competitive inhibitor.
162. Answer (2)
Hint : Positional information of amino acids decides
specificity of enzyme action.
Solution : Apoenzymes are proteinaceous part of
holoenzymes. Enzymes never alter the equilibrium of
the reaction, they act by lowering activation energy
barrier.
163. Answer (3)
Hint : White fat is unilocular and adipocytes are
signet ring shaped cells.
Solution : Brown fat in newly born prevents shivering
in them. White fat acts as a reservior of energy and
the triglycerides are hydrolysed upon need.
164. Answer (4)
Hint : Intercalated discs are interdigitations between
adjacent cells that act as boosters of cardiac
impulse.
Solution : Cardiac fibres are striated and the edges
of sarcomeres are ‘Z’ lines. Connexons are part of
gap junctions present at intercalated discs. Visceral
muscles are uninucleated.
165. Answer (1)
Hint : Muscle fibres associated with limbs and heart
are striped in appearance.
Solution : Striated muscles include both cardiac &
skeletal muscles. The alternate light and dark bands
refer to the horizontal arrangement of actin & myosin
filaments.
166. Answer (2)
Hint : Muscles under the control of our will are
voluntary.
Solution : Smooth and cardiac muscles are
involuntary in nature i.e. they are not under the
control of our will. Smooth muscles contain
unbranched, unstriped, uninucleate fibres.
167. Answer (4)
Hint : Concentric arrangement of calcium salts
surrounding osteocytes.
Solution : The lamellae refer to concentric
arrangement of calcium phosphate salts surrounding
osteocytes in diaphysis of long bones of mammals.
168. Answer (2)
Hint : Exocrine glands release/pour their secretions
through ducts.
Solution : Insulin is a secretion of ductless part of
gland pancreas. Saliva, earwax and milk are
secretions of exocrine glands.
169. Answer (3)
Hint : Endocrine glands lack ducts.
Solution : Endocrine glands do not have ducts and
release their secretion into blood capillaries.
170. Answer (2)
Hint : Multiple layer in epithelial tissue are effective
in protection.
Solution : At surfaces prone to stress (wear & tear)
or abrasion compound epithelium is effective.
171. Answer (2)
Hint : Junctions involved in cementing neighbouring
cells.
Solution : Tight junctions help to stop substances
from leaking across a tissue.
172. Answer (1)
Hint : Stratified i.e. compound epithelium lines
surfaces that face mechanical stress.
Solution : Stratum germinativum comprises of cells
that divide regularly and forms the bottommost layer
of straitified epithelium. Diffusion surfaces and tubular
parts of nephron are lined by single layered simple
epithelium.
173. Answer (4)
Hint : Wine flask shaped cells of alimentary canal.
Solution : Goblet cells are mucus secreting
unicellular glands of alimentary canal. Cerumen and
sweat are secreted by sebaceous & sudorific glands
respectively. Hormones are secretions of endocrine
glands.
Test - 1 (Code-C) (Answers & Hints) All India Aakash Test Series for Medical-2020
17/17
174. Answer (2)
Hint : Neurilemma is a contribution of Schwann
cells.
Solution : Schwann cells form myelin sheath &
neurilemma in neurons of PNS. Oligodendrocytes
form myelin sheath in CNS. Neurilemma is absent in
CNS. Oligodendrocytes are branched cells while
Schwann cells are not branched.
175. Answer (1)
Hint : Nissl's granules are clusters of RNA &
ribosomes.
Solution : Nissl's granules are site of protein
synthesis located in cell body of neuron/soma & in
dendrites (afferent processes). They are absent in
afferent processes also called axons.
176. Answer (4)
Hint : Response to stimulus is a feature of structural
& functional units of neural system.
Solution : Extensibility is a property of muscle fibres
while neurons exhibit both excitability & conductivity.
177. Answer (1)
Hint : I band is composed of actin filaments while A
band comprises of both actin & myosin filaments.
Solution : Length of both actin & myosin filaments
remain constant. Size of A band also remains
contant. Only the extent of overlap between actin &
myosin filaments increases thereby reducing size of
Henson’s zone.
178. Answer (2)
Hint : Brush border membrane (BBM) due to
microvilli increase surface area for reabsorption &
absorption.
Solution : Cilia in trachea push mucus with
entrapped particles in specific (upward) direction.
BBM cuboidal is a feature of tubular part of nephrons
while BBM columnar epithelium dominants in
intestine.
179. Answer (1)
Hint : Keratinised epithelium occurs on water
resistant surfaces in living beings.
Solution : Stratum corneum is the topmost layer of
stratified keratinised epithelium. Pharynx, tongue &
cornea are non-keratinised epithelial surfaces.
Microvilli in alimentary canal support absorption of
nutrients not reabsorption. Transitional epithelium is
distensible/stretchable. Little intercellular matrix is
present in epithelial tissue.
180. Answer (4)
Hint : ‘Osteo’ refers to bone & clasts refers to
destruction/demineralisation.
Solution : Osteocytes & osteoblasts are bone
forming cells while osteoclasts are bone dissolving/
dissolution cells that decrease bone calcium.
�����
Test - 1 (Code-D) (Answers) All India Aakash Test Series for Medical-2020
1/17
1. (3)
2. (2)
3. (4)
4. (1)
5. (3)
6. (1)
7. (4)
8. (4)
9. (1)
10. (3)
11. (2)
12. (1)
13. (1)
14. (4)
15. (4)
16. (3)
17. (4)
18. (1)
19. (4)
20. (4)
21. (2)
22. (3)
23. (3)
24. (3)
25. (2)
26. (2)
27. (3)
28. (1)
29. (1)
30. (2)
31. (4)
32. (2)
33. (3)
34. (2)
35. (1)
36. (3)
Test Date : 14/10/2018
ANSWERS
TEST - 1 (Code-D)
All India Aakash Test Series for Medical-2020
37. (2)
38. (2)
39. (2)
40. (1)
41. (2)
42. (1)
43. (1)
44. (4)
45. (3)
46. (3)
47. (4)
48. (1)
49. (4)
50. (2)
51. (2)
52. (3)
53. (4)
54. (1)
55. (2)
56. (1)
57. (4)
58. (4)
59. (2)
60. (2)
61. (1)
62. (2)
63. (4)
64. (2)
65. (1)
66. (1)
67. (3)
68. (3)
69. (4)
70. (2)
71. (4)
72. (1)
73. (1)
74. (3)
75. (2)
76. (2)
77. (3)
78. (1)
79. (4)
80. (3)
81. (2)
82. (1)
83. (1)
84. (4)
85. (3)
86. (2)
87. (4)
88. (4)
89. (3)
90. (1)
91. (3)
92. (2)
93. (4)
94. (2)
95. (2)
96. (1)
97. (2)
98. (2)
99. (2)
100. (3)
101. (1)
102. (3)
103. (3)
104. (4)
105. (2)
106. (3)
107. (2)
108. (4)
109. (2)
110. (3)
111. (3)
112. (4)
113. (2)
114. (1)
115. (4)
116. (3)
117. (4)
118. (2)
119. (4)
120. (4)
121. (3)
122. (2)
123. (2)
124. (3)
125. (3)
126. (4)
127. (3)
128. (3)
129. (2)
130. (4)
131. (3)
132. (2)
133. (4)
134. (3)
135. (2)
136. (4)
137. (1)
138. (2)
139. (1)
140. (4)
141. (1)
142. (2)
143. (4)
144. (1)
145. (2)
146. (2)
147. (3)
148. (2)
149. (4)
150. (2)
151. (1)
152. (4)
153. (3)
154. (2)
155. (3)
156. (3)
157. (2)
158. (4)
159. (2)
160. (3)
161. (4)
162. (4)
163. (1)
164. (3)
165. (4)
166. (1)
167. (3)
168. (2)
169. (1)
170. (3)
171. (1)
172. (1)
173. (1)
174. (3)
175. (1)
176. (4)
177. (2)
178. (4)
179. (3)
180. (2)
All India Aakash Test Series for Medical-2020 Test - 1 (Code-D) (Answers & Hints)
2/17
ANSWERS & HINTS
1. Answer (3)
Hint : Mean absolute error
1 2| | | | | |
nx x x
n
Solution : xm
= 330 ms–1
x1
= |342 – 330| = 12
x2
= |338 – 330| = 8
x3
= |318 – 330| = 12
x4
= |322 – 330| = 8
112 8 12 810 ms
4
2. Answer (2)
Hint : R = R1
+ R2
Solution : Req
= R1
+ R2
1 2
eq 1 2
0.1 0.1
30
R RR
R R R
eq
0.2 2100 100 %
30 3
R
R
3. Answer (4)
4. Answer (1)
Hint : L.C = 1MSD – 1VSD
Solution : L.C = 1MSD – 1VSD
VSDMSD 1
MSD
1 MSD = 1 mm
19 MSD = 20 VSD
1 VSD = 19
VSD20
L.C = 19
1 1 0.05 mm20
5. Answer (3)
Hint : � � �
rel A Ba a a
Solution : � � �
rel A Ba a a
= g – g = 0
6. Answer (1)
Hint : The difference with average value should be
least.
Solution : t1
= |10.00 – 9.95| = 0.05 s
t2
= |10.00 – 10.10| = 0.10 s
[ PHYSICS]
7. Answer (4)
Hint : Minimum percentage error.
Solution : Percentage error in 2.000 m
0.001100 0.05%
2.000
8. Answer (4)
Hint : If there is no digit after 5 then last digit should
be even.
Solution : 2.765 rounding off (3 significant)
= 2.76
5.735 = 5.74
9. Answer (1)
Hint : z = xy
Z X Y
Z X Y
Solution : z = xy
Z X Y
Z X Y
0.4 0.1 1
4 1 5
Z
Z
1
4 5
Z
Z = 0.8 m
Z = (4 ± 0.8) m
10. Answer (3)
Hint : Non-zero digits are significant.
Solution : Trailing zeros in a number without
decimal are insignificant.
11. Answer (2)
Hint : Raman effect.
Solution : CV Raman won noble prize for scattering
of light by molecules.
12. Answer (1)
Hint : Mass and energy are inter-convertable.
Solution : Conservation laws in nature.
1. Law of conservation of energy
2. Law of conservation of linear momentum
3. Law of conservation of angular momentum
13. Answer (1)
Test - 1 (Code-D) (Answers & Hints) All India Aakash Test Series for Medical-2020
3/17
14. Answer (4)
Hint : Fundamental forces.
Solution : Gravitational force is the weakest force.
Gravitational forces are central forces, electrostatic
forces are central forces, strong nuclear forces are
not central forces.
15. Answer (4)
Hint : It has no dimension.
Solution : Solid angle 2
cos dsd
r
0 0 0
2
cos[ ] [M L T ]
dsd
r
16. Answer (3)
Hint :
2
21 1
2 2
lk mv m
t
Solution :
2
21 1
2 2
lk mv m
t
100 100 100 2 100 k m l t
k m l t
= 1% + 4% + 2% = 7%
17. Answer (4)
Hint : [L] = [pav
bm
c]
Solution : [L] = [M1L2T–1]
[p] = [M1L2T–3]
[v] = [M0L1T–1]
[m] = [M1L0T0]
[M1L2T–1] = k [M1L2T–3]a [M0L1T–1]b [M1L0T–0]c
1 = a + 0 + c
2 = 2a + b + 0
–1 = –3a – b
a = –1, b = 4, c = 2
18. Answer (1)
Hint : Trignometric ratios have no dimensions
Solution : [LHS] = [RHS]
[y] = [Ax2]
1
2[ ] [ ]
yA L
x
[] = [M0L0T0]
[] = [M0L1T0]
[A2] = [M0L1T0]
19. Answer (4)
Hint : n1
u1
= n2
u2
Solution : [E] = [M1L2T–2]
2 kgm2s–2 = [2 kg]1 [2 m]2 [n s]–2
2
2
22 2
n
n2 = 4
n = 2, similarly we can find the relation for
momentum and power.
20. Answer (4)
Hint : Planck’s constant and angular momentum
have same dimensions.
Solution :
1. Energy density and pressure have same
dimensions.
2. Relative density and plane angle have no
dimension.
1 2 2[M L T ][ ]
[MK]
Q
Cm Q
0 2 2 1[M L T K ]
0 2 2[ ] [M L T ] QL
m
21. Answer (2)
Hint : n1
u1
= n2
u2
Solution : 1 1 1
2 1
2 2 2
a b c
M L Tn n
M L T
h = 6.6 × 10–34 J-s
[h] = [M1L2T–1]
a = 1, b = 2, c = –1
1 2 1
34
2
kg s6.6 10
10 g 5m 100 s
mn
34 2 216.6 10 10 10
25
= 2.64 × 10–31
22. Answer (3)
Hint : Limitations of dimensional analysis.
Solution :
1. Two and more physical quantities may have
same dimensional expression.
2. Numerical constant has no dimensions.
3. Method of dimensions can be used only for
product of physical quantities.
23. Answer (3)
Hint : In multiplication, product has number of
significant figures equal to least significant figure
present in calculation.
All India Aakash Test Series for Medical-2020 Test - 1 (Code-D) (Answers & Hints)
4/17
Solution :
V = IR
V = 1.25 × 25.425
= 31.78125
= 31.8 (3 significant figures)
24. Answer (3)
Hint : Smaller least count.
Solution : Least count is minimum in = 0.001 m, so
it is most precise.
25. Answer (2)
Hint :
2
1
2
1
t
t
t
t
vdt
v
dt
Solution :
52
0
5
0
(30 4 2)t t dt
v
dt
53 2
0
30 42
3 2
(5 0)
t tt
3 2 5
0
1(10 2 2 )
5t t t
11[10 125 50 10] 262 ms
5
26. Answer (2)
Hint : dv
a vdx
Solution :
v(ms )–1
x(m)3 70
10
5
A
B
5
4
dv
dx
Velocity at t = 5 s
v = 7.5 ms–1
25 37.5(7.5) 9.4 ms
4 4
a
27. Answer (3)
Hint : Total time (n – 1) T0
time interval
Solution : 21
2h gt
1
2
3
4
5
6
7
Total time = (7 – 1) T0
= 6 T0
21
2h gt
2
0
1{36 }
2h g T
2
018h gT
2
0180 18 10 T
T0
= 1 s
28. Answer (1)
Hint : A A
B B
a u
a u
Detailed solution : tA = t
B = t
0
0 = 20 – aAt0
0
20
Aa
t
Similarly
0
10B
at
20 2
10 1
A
B
a
a
29. Answer (1)
Hint : S = S1
+ S2
+ S3
Solution : 2
1
11 4 8 m
2 S
S2
= v0
t2
= (a1
t1
)t2
= 1 × 4 × 10 = 40 m
2
0 2 30 2v a S
3
164 m
2 2
S
S = 40 + 8 + 4 = 52 m
Test - 1 (Code-D) (Answers & Hints) All India Aakash Test Series for Medical-2020
5/17
30. Answer (2)
Hint : Average speed = Total travelled distance
Total time taken
Solution : Time taken by cyclist A
362 h
18
Time taken by B cyclist from cycling
1 32 h
2 2
Average speed of B for actual riding
3624 km/h
3
2
31. Answer (4)
Hint : Stone will acquire velocity of frame of
reference when it leave frame of reference.
Solution : 2
0
1
2h u t gt
h
u0
2
0
1
2h u t gt
–3.2 = 6t – 5t2
5t2 – 6t – 3.2 = 0
t = 1.6 s
32. Answer (2)
Hint : d + x = v0
t ...(i)
21
2x at ...(ii)
Solution : 32 + x = 10t ...(i)
211
2x t ...(ii)
32 m
10 ms–1
a = 1 ms–2
x
2132 10 0
2 t t
t2 – 20t + 64 = 0
20 400 256 20 12
2 2t
t = 4, 16
t = 4 s
33. Answer (3)
Hint : 2 2
2 1 02v v a l
Detailed solution : 2 2
2 1 02v v a l
2 2
2 1
02
v vl
a
l
2
2 2
1 02
2
lv v a
2 2 2 2
2 2 2 1 1 2
1 0
02 2
v v v vv v a
a
2 2
1 2
2
v vv
34. Answer (2)
Hint : 2
0
1
2h v t gt
Solution :
7
8
5
4
h
t = 6
Particle will be at maximum height at t = 6 s.
2110 6 180 m.
2 h
35. Answer (1)
Hint : Reaction time
Solution:
Distance travelled during reaction time
S1
= 10 × 1 = 10 m ...(i)
Distance travelled during retarded motion
2
2
10050 m
2 2 1
vS
a
...(ii)
All India Aakash Test Series for Medical-2020 Test - 1 (Code-D) (Answers & Hints)
6/17
Total travelled distance = S1
+ S2
= 60 m
Distance of car from red signal
= 70 – 60 = 10 m
36. Answer (3)
Hint : | Average velocity | = |Total displacement travelled|
Total time taken
Solution:
ABO
| Displacement | = 10 m
Time taken = 5
10 2
s
| Average velocity | = 110 20
ms
2
37. Answer (2)
Hint : 2
0
1
2S S ut at
Solution:
v
tt0
v0
0
0
0
(constant)
va
t
2
0
1
2x x ut at
20
0
0
10
2
vx x t
t
20
0
0
1
2
vx x t
t
38. Answer (2)
Hint : Average speed = Total distance travelled
Total time taken
Detailed Solution : From t = 0 to t = 4 s x1
= 4 m
From t = 4 s to t = 6 s x2
= 0
From t = 6 s to t = 10 s x3
= 4 m
Average speed = 14 0 4 8 4
ms10 10 5
39. Answer (2)
Hint : slope of graphdv
a v tdt
Solution : a = slope of v – t graph
= –ve value (constant)
t
a
0
40. Answer (1)
Hint : dv
a vdx
Solution:
0 0
1v x
v x
...(i)
0
0
0
v
v x v
x
...(ii)
0
0
vdv
dx x ...(iii)
0 0
0
0 0
v vdva v x v
dx x x
2 2
0 0
200
v v
x
xx
41. Answer (2)
Hint : v
at
Solution : v = t2 + 4t
At t = 1 s, v = 5 ms–1
t = 2 s, v = 12 ms–1
212 57 ms
2 1
a
42. Answer (1)
Hint : 1 2
1 2
2
av
v vv
v v
Solution : v1
= 36 km/h = 10 ms–1
v2
= 54 km/h = 15 ms–1
1 2
1 2
av
S Sv
t t
1 1 1
210
2 10
S
SS t t
Test - 1 (Code-D) (Answers & Hints) All India Aakash Test Series for Medical-2020
7/17
46. Answer (3)
Hint: nhc
E
Solution: Energy released = 66.26 × 60
J100
34 8
9
60 n 6.626 10 3 1066.26
100 100 10
n = 2 × 1019
47. Answer (4)
Hint: Bohr’s model is valid only for one electron
species.
Solution: According to Bohr’s theory, angular
momentum of an electron is quantised.
[ CHEMISTRY]
2 2 2
215
2 15
S
SS t t
112 ms
1 1
2 10 15
av
Sv
S
43. Answer (1)
Hint : max
2 v S
Solution :
u = 0 vmax v = 0
S2
S2
S1
S1
2
2 max
max 1 12
2
v
v S S ...(i)
2
2 max
max 2 20 2
2
vv S S
...(ii)
2
max
1 2
1 1
2
vS S S
max
2 2 2 3 100
5
v S
4 15 m/s
44. Answer (4)
Hint : Displacement and velocity may be positive or
negative.
2nd
4th
0
+ve
s
v
Solution : Direction of displacement is vertically
downward.
S < 0
Direction of velocity is downward
So, v < 0
45. Answer (3)
Hint : The interval in which displacement is zero.
Solution : For the curve, ACB displacement is zero
AB
C
Tt
x
0
0
����
�
av
SV
t
48. Answer (1)
Hint: Ionisation energy 2
2
Z13.6 eV
n
Solution: Z = 2, n = 1
Ionisation energy
2
2
13.6 254.4eV
1
49. Answer (4)
Hint: Isoelectronic species have same number of
electrons.
Solution: Species Total electrons
N2
14
CO 14
14
17
2
2O
2O
CN–
14
All India Aakash Test Series for Medical-2020 Test - 1 (Code-D) (Answers & Hints)
8/17
50. Answer (2)
Hint: No. of molecules = No. of moles × NA
Solution: H2
: No. of molecules A A
10N 5N
2
O2
: No. of molecules = 10 NA
CO2
:
No. of molecules = 8 NA
SO2
: No. of molecules = A A
64N N
64 .
51. Answer (2)
Hint: % of an element wt.of element
100wt.of compound
Solution: For YO: % of oxygen =16
100M 16
160040
M 16
M 24
For second oxide: Let YOx
% of oxygen =
16x100
16x 24
1600x20
16x 24
3x
8
x 3 8 3
8
YO YO Y O
52. Answer (3)
Hint: One AVOGRAM =
A
1
N
53. Answer (4)
Hint: Electron density of the p-orbitals lie along the
axes.
Solution: dxy
orbital does not have electron density
along the axis.
54. Answer (1)
Hint: 2
2 2
1 2
1 1E 13.6Z
n n
eV/atom
Solution: For, n1
= 4 and n2
= 5
9E 13.6 eV/atom
16 25
55. Answer (2)
Hint: Ti : [Ar] 4s2 3d
2
Solution:
Ti2+ : [Ar] 3d2
56. Answer (1)
Hint:
2
n
nr 0.53 Å
Z
Solution: n = 2, Z = 4
2
4
2r 0.53 0.53Å
4
57. Answer (4)
Hint: For principal quantum number n, l can be 0, 1,
2, ... (n – 1)
Solution: for n = 2,
l 2, it will have value 0 and 1.
58. Answer (4)
Hint: h
x p4
Solution:h
x m v4
34
3 24
h 6.626 10x
1004 m v4 10 6.626 10
10x 2.5 10 m 2.5 Å
59. Answer (2)
Hint: 21
KE mv2
, h
mv , v
Solution: 2v v mv 2 1 2
mv KEh h 2 h
27 7 1
34
23.313 10 10 sec
6.626 10
60. Answer (2)
Hint: Exchange energy of electrons is defined for
degenerate orbitals.
Solution: Exchange energy of electrons is defined
as energy released when an electron exchanges its
position with electron having same spin present in
degenerate orbitals.
Test - 1 (Code-D) (Answers & Hints) All India Aakash Test Series for Medical-2020
9/17
61. Answer (1)
Hint:
solvent
solute
solvent
X 1000molality(m) =
X MW
Solution: 0.2 1000
m 13.890.8 18
62. Answer (2)
Hint: -rays, X-rays , UV, visible, IR, microwaves
Wavelength increases
Frequency decreases
Solution: Visible waves have higher frequency than
IR.
63. Answer (4)
Hint: l = 2 means d-subshell
Solution: d-subshell can have maximum 10
electrons.
64. Answer (2)
Hint: -particle is He nucleus (He2+) which contains
2 protons and 2 neutrons.
Solution: For neutron, e
0m
For other particles, e
m order is
electron > proton > -particle
65. Answer (1)
Hint: Mass of electron is 9.1 × 10–31 kg
Solution: Positron is the particle having mass equal
to electron but having a unit positive charge
66. Answer (1)
Hint: Average atomic mass =
% abundance of isotope atomic mass
100
Solution: 13 x 12 (100 x)
12.011100
x = 1.1%
67. Answer (3)
Hint: Divalent metal chloride should be MCl2
Solution: Mol. wt.= 2 × vapour density = 60 × 2 = 120
x + 2 × 35.5 = 120
x = 49
Eq. wt. of metal 49
24.52
68. Answer (3)
Hint: 2 2 3
32AI O Al O
2
Solution: 2 2 3
32Al O Al O
2
3 2
3KClO KCl O
2
From stoichiometry, 2 mole Al require 3
2 mole O
2
and 3
2 mole O
2
is obtained by 1 mole KCIO3
.
69. Answer (4)
Hint: M1
V1
+ M2
V2
= M3
V3
Solution: 0.5 × 2 + 1 × 1= M3
× 3
3
2M 0.67 M
3
70. Answer (2)
Hint: Law of multiple proportions is illustrated for
compounds which have two same elements.
Solution: CH4
: for 12 g C, wt. of H = 4 g = x
C2
H4
: for 12 g C, wt. of H = 2 g = y
x 4 2
y 2 1 = Simple whole no. ratio
71. Answer (4)
Hint: Molarity (M) = solute
n
V L
Solution: Mass of 1 ml D2
O = 1 g
3
1
20M 50 M
1 10
72. Answer (1)
Hint: For minimum molar mass, one molecule
should contain at least one atom of oxygen
Solution:% of oxygen =wt.of oxygen
100wt.of compound
164 100
Mol.wt.
Minimum mol. wt. = 400 amu
All India Aakash Test Series for Medical-2020 Test - 1 (Code-D) (Answers & Hints)
10/17
73. Answer (1)
Hint: 6
n
zv 2.18 10 m/sec
n
Solution: 6
4 He
8
22.18 10v 14
c 2753 10
74. Answer (3)
Hint: Ni2+ = [Ar] 3d8
Solution: 29
Cu = 1s2 2s
2 2p6 3s
2 3p6 4s
1 3d10
s-electrons in Cu atom = 7
75. Answer (2)
Hint: 2
2 2
1 2
1 1 1RZ
n n
Solution: For second line of Lyman series :
2
2 2
1 1 1R (1)
x 1 3
1 8R
x 9
9R
8x
For 3rd line of Lyman series : 2
2 2
1 1 1R (1)
1 4
15 9 15 135R
16 8x 16 128x
128x
135
76. Answer (2)
Hint: Electrons in half filled p-subshell have different
orientations.
Solution: Orientation of orbitals is given by magnetic
quantum number.
77. Answer (3)
Hint: s, p and d-subshells have 1, 3 and 5 orbitals
respectively.
Solution:
In l subshell, number of orbitals = 2l + 1
78. Answer (1)
Hint: h
mv
Solution: H2
has minimum molar mass so have
longest de-Broglie wavelength.
79. Answer (4)
Hint: 2 2 3
1 1FeO O Fe O
4 2
Solution:
(72g)
(8g)
2 2 31 mole
0.25 mole
1 1FeO O Fe O
4 2
% increase in wt. 8
100 11.11%72
80. Answer (3)
Hint: No. of atoms = No. of molecules × atomicity
Solution: Remaining molecules of CO2
=
323 20440 10
6.022 10 1044
No. of remaining atoms = 59.22 × 1020 × 3
= 1.77 × 1022
81. Answer (2)
Hint: Normality = Molarity × n-factor
Solution: Molarity of
22
23
3
9.033 10
6.022 10OH 0.3 M
500 10
Molarity of Ca(OH)2
= 0.3
M2
Normality of Ca(OH)2
=
0.32 0.3 M
2
82. Answer (1)
Hint: Dilution equation, M1
V1
= M2
V2
Solution: For stock solution,
1
49 1.5 10M 7.5
98
Now, M1
V1
= M2
V2
7.5 × V1
= 0.1 × 2.5 × 1000
V1
= 33.33 ml
83. Answer (1)
Hint: Frequency 2
32
z
v znf
r nn
z
Solution:
2
2
2
3
He Li
2
Li He
3
3
T f 2 9
T f 322
1
Test - 1 (Code-D) (Answers & Hints) All India Aakash Test Series for Medical-2020
11/17
[ BIOLOGY
]
91. Answer (3)
Hint: Meiosis involves crossing over as well as
reduction of chromosome number.
Solution: Four cells produced after telophase II are
genetically dissimilar to each other as well as to
their parents.
92. Answer (2)
Hint: Interkinesis is resting phase between meiosis
I and II.
Solution: Meiosis I and II involves into sequential
karyokinesis. Chromosomes do not decondense upto
DNA level in interkinesis because after telophase I
chromatin recondense in prophase II.
93. Answer (4)
Hint: Mitosis does not involve crossing over.
Solution: Mitosis helps in reproduction in unicellular
organisms. It helps in repair and regeneration but
cannot produce recombinants.
84. Answer (4)
Hint: For one electron species, energy of subshell
depends only on the value of n.
Solution: For 3rd shell, No. of degenerates orbitals = n2
= 32 = 9.
85. Answer (3)
Hint: Only four lines of Balmer series of hydrogen
atom lie in visible region
Solution: For n = 6 (5th excited state) transition to
n = 1
* Maximum possible transitions
= 6 1 5 + 4 + 3 + 2 + 1 =15
* Maximum Paschen transitions
= 3(6 3, 5 3, 4 3)
86. Answer (2)
Hint: x y 2 2 2
y yC H x O xCO H O
4 2
2
2
Moles of CO obtainedx
y 2 Moles of H O obtained
Solution:
8.8
x 144
5.4y 32
18
Empirical formula = CH3
Molecular wt. 2 30n 4
Empirical formulawt. 15
Molecular formula = n × empirical formula
= 4 (CH3
) = C4
H12
87. Answer (4)
Hint: No. of particles = No. of moles × NA
Solution:
Moles of oxygen atom = 0.3 × 14 = 4.2
No. of electrons of oxygen atoms = 4.2 × 8 × NA
= 33.6 NA
88. Answer (4)
Hint: n-factor for acid is number of replaceable H+
ions
Solution: n-factor = 1
Mol. wt 120Eq. wt. 120
n-factor 1
89. Answer (3)
Hint: Chromium has half-filled d-subshell.
Solution: Cr(24) : [Ar]4s13d
5
90. Answer (1)
Hint: Orbital angular momentum = hl l 1
2
Solution: For f - orbital, I = 3
Orbital angular momentum = h 3h3 3 1
2
94. Answer (2)
Hint: Treatment of colchicine produces polyploid
condition.
Solution: Colchicine inhibits formation of tubulin and
therefore microtubules. Lack of microtubules results
in inhibition of spindle fibres formation and its affect
will be visible first at metaphase.
95. Answer (2)
Hint: Cell plate represents the future middle lamella
of adjacent cells.
Solution: Cell plate help in cytokinesis of plant cells
which represent the middle lamella of adjacent cells.
96. Answer (1)
Hint: Dyad of cells appear in last stage of heterotypic
division
Solution: Dyad of cells are formed at the end of
telophase I.
All India Aakash Test Series for Medical-2020 Test - 1 (Code-D) (Answers & Hints)
12/17
97. Answer (2)
Hint: Recombinase enzyme catalyses the process
occurring in pachytene stage.
Solution: Crossing over is enzyme mediated
process and it produces recombinants.
98. Answer (2)
Hint: Each chromosome has 2 chromatids
Solution: 1 Bivalent = 2 chromosomes
20 Bivalent = 20 × 2 = 40 chromosomes
Since 1 chromosome = 2 chromatids
40 chromosomes = 80 chromatids
99. Answer (2)
Hint: The given stage is the best stage to study
morphology of chromosomes.
Solution: Given stage is Metaphase I. where,
Bivalent chromosomes align at equator forming
two metaphasic plates
Alignment of bivalents is totally random process.
Spindle fibre attaches to each chromosome of
homologous pair
Separation of chromosomes occur at anaphase I
100. Answer (3)
Hint: Number of mitotic divisions to form ‘n’ number
of cells are = n – 1
Solution: Number of generations (n) required to form
‘x’ number of cells are = 2n
For 32 cells
Mitotic divisions = 32 – 1 = 31
Number of generations = 25 = 32, x = 5
101. Answer (1)
Hint: Major restriction point inhibits cells to go for
DNA synthesis.
Solution: Major control on cell cycle can be
imposed on 1
G S transition step.
102. Answer (3)
Hint: Last phase of karyokinesis in mitosis involves
reappearance of cell organelles and nuclear
membrane.
Solution: At the end of mitosis, the chromosomes
decondense into chromatin which occurs in telophase.
103. Answer (3)
Hint: Interphase is known as the most active stage
of cell cycle.
Solution: Interphase constitute more than 95%
duration of cell cycle of a human cell. Rest all the
features regarding the interphase are true.
104. Answer (4)
Hint: Tubulin protein synthesis occurs in the stage
where duplication of mitochondria & chloroplast
occurs.
Solution: Most of the cell organelles get duplicated
in G1
. But Golgi, chloroplast and mitochondria are
doubled at G2
phase along with tubulin protein
synthesis.
105. Answer (2)
Hint: Meiosis involves two sequential karyokinesis
and cytokinesis.
Solution: Except DNA duplication and histone
protein synthesis, rest all phenomenon occur twice
in meiosis.
106. Answer (3)
Hint: Spindle fibres are not directly attached to the
centromere.
Solution: A disc-shaped structure is found over
centromere through which spindle fibres are attached
called kinetochore.
107. Answer (2)
Hint: Crossing over occurs in pachytene stage.
Solution: Point at which crossing over occurs forms
the recombination nodule.
108. Answer (4)
Hint: Bacterial type ribosomes are also found in
some cell organelles of eukaryotes.
Solution: Both ribosomes and plasma membrane
are similar in prokaryotes and eukaryotes.
109. Answer (2)
Hint: The cell also has cell wall with plasmodesmata.
Solution: Plant cells have cell wall with
plasmodesmata. Their 90% volume is occupied by
vacuoles.
110. Answer (3)
Hint: These structures are tubular and help bacteria
to attach with rocks also.
Solution: Fimbriae are bristle like structures which
help bacteria to attach with rocks or host cell.
111. Answer (3)
Hint: These structures help in photorespiration
along with chloroplast and mitochondria.
Solution: Peroxisomes are involved in photorespiration
because they have enzymes for formation as well
as destruction of peroxide.
112. Answer (4)
Hint: Loops of lampbrush chromosomes have hair
like structure and with protein that are known as
informosomes.
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Solution: Lampbrush chromosomes form loops
which have hair like structures. These hairs are
bound to protein. Some of which are called
informosome i.e, having mRNA + protein. These
loops participate in transcription and form m-RNA
113. Answer (2)
Hint: Such chromosomes appear L-shaped during
anaphase.
Solution: A chromosome with centromere slightly
away from the centre is submetacentric chromosome.
114. Answer (1)
Hint: Chromatids of chromosomes are held together
at primary constriction.
Solution: Primary constriction is known as
centromere which holds the two halves of a
chromosome.
115. Answer (4)
Hint: Chromatin is packed DNA
Solution: Packed DNA has RNA, histones and
some non-histone proteins.
116. Answer (3)
Hint: Nucleolus is found inside the nucleus
Solution: Nucleolus is non-membrane bound
structure found in nucleoplasm of nucleus. They are
larger in cells involved in protein synthesis.
117. Answer (4)
Hint: Polyribosomes are formed in cytoplasm
Solution: Polysomes are not formed with the help of
RER
118. Answer (2)
Hint: Both mitochondria and chloroplast are semi
autonomous structures.
Solution: Mitochondria, chloroplast and bacteria all
have ds circular DNA, 70S ribosomes porins on
outer membrane and self duplication ability.
Mitochondria and chloroplast are partially dependent
on nucleus.
119. Answer (4)
Hint: In leucoplast granum remains absent.
Solution: Protoplast is a cell without cell wall.
120. Answer (4)
Hint: This structure is absent in higher plants.
Solution: Centrioles have 9 peripheral fibrils of
tubulin and these are absent in the centre therefore
the arrangement is 9 + 0.
121. Answer (3)
Hint: Microfilaments are solid unbranched rod like
fibrils.
Solution: Microtubules have diameter of 25 nm.
Intermediate filaments are involved in formation of
scaffolds of chromatin.
122. Answer (2)
Hint: These structures are absent in prokaryotes.
Solution: Cytoskeleton are proteinaceous
filamentous structures which provide mechanical
strength & support to the cell.
123. Answer (2)
Hint: ‘S’ is a unit
Solution: ‘S’ stands for Svedberg coefficient or
sedimentation coefficient and it is indirect measure of
density and size of ribosomal sub units.
124. Answer (3)
Hint: This organelle is found in almost all eukaryotic
cells and is site of ATP synthesis.
Solution: Mitochondria are sausage shaped, have
their own 70S ribosomes i.e, palade particles and
they are viewed after staining with Janus green.
Usually their number is high in those cells which
have high metabolism.
125. Answer (3)
Hint: This organelle has hydrolytic enzymes which
become functional at acidic pH.
Solution: Lysosomes have hydrolytic enzymes for
digestion of almost all types of macromolecules
which are functional at acidic pH.
126. Answer (4)
Hint: ER, Golgi, lysosomes & vacuoles function in
a coordinated manner.
Solution: Oxidation of fatty acids, proteins and
carbohydrate occurs inside the mitochondria.
Rest all functions are performed by organelles of
endomembrane system.
127. Answer (3)
Hint: The given figure is of golgi bodies
Solution: Enzymatic precursors for lysosomes are
synthesized in ER. Golgi does processing and
packaging of materials for intra as well as extra-cellular
targets.
Rest all the features of golgi are true.
128. Answer (3)
Hint: These structures are present in pits.
Solution: Symplast of two adjacent cells are
connected via cytoplasmic strands or
plasmodesmata. They are lined by plasma
membrane.
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129. Answer (2)
Hint: Smallest cell organelle is known as organelle
within organelle.
Solution: Ribosome is known as organelle within
organelle and its r-RNA part is synthesized inside the
nucleolus.
130. Answer (4)
Hint: Movement of molecules across the membrane
without help of carrier proteins and ATP is called
simple diffusion
Solution: Simple diffusion occurs only for those
molecules which are neutral or non-polar, while
transport of polar and hydrophilic substances need
carrier proteins.
131. Answer (3)
Hint: Lipid molecules of plasma membrane have
polar head and non-polar tail.
Solution: Polar head is hydrophilic in nature and it
interacts with water. Rest all the statements
regarding plasma membranes are true.
132. Answer (2)
Hint: This inclusion body provides buoyancy to the
bacteria
Solution: Ribosomes are found in all prokaryotes
and eukaryotes. Cyanophycean granules are found in
cyanobacteria only while gas vacuoles are found in
blue-green algae (BGA) as well as in purple & green
photosynthetic bacteria
133. Answer (4)
Hint: Chloroplast is the site of photosynthesis and
contain pigments in eukaryotes.
Solution: Chromatophores are membranous
extension in the cytoplasm of a cyanobacterial cell
which have photosynthetic pigments
134. Answer (3)
Hint: Body of plants and animals are composed of
cells & product of cells was proposed by Theodor
Schwann.
Solution: Theodor Schwann was a British zoologist.
135. Answer (2)
Hint: Nuclear membrane is found in all eukaryotes.
Solution: Mycoplasma, BGA, Bacillus, purple
photosynthetic bacteria & E.coli are prokaryotes.
136. Answer (4)
Hint : ‘Osteo’ refers to bone & clasts refers to
destruction/demineralisation.
Solution : Osteocytes & osteoblasts are bone
forming cells while osteoclasts are bone dissolving/
dissolution cells that decrease bone calcium.
137. Answer (1)
Hint : Keratinised epithelium occurs on water
resistant surfaces in living beings.
Solution : Stratum corneum is the topmost layer of
stratified keratinised epithelium. Pharynx, tongue &
cornea are non-keratinised epithelial surfaces.
Microvilli in alimentary canal support absorption of
nutrients not reabsorption. Transitional epithelium is
distensible/stretchable. Little intercellular matrix is
present in epithelial tissue.
138. Answer (2)
Hint : Brush border membrane (BBM) due to
microvilli increase surface area for reabsorption &
absorption.
Solution : Cilia in trachea push mucus with
entrapped particles in specific (upward) direction.
BBM cuboidal is a feature of tubular part of nephrons
while BBM columnar epithelium dominants in
intestine.
139. Answer (1)
Hint : I band is composed of actin filaments while A
band comprises of both actin & myosin filaments.
Solution : Length of both actin & myosin filaments
remain constant. Size of A band also remains
contant. Only the extent of overlap between actin &
myosin filaments increases thereby reducing size of
Henson’s zone.
140. Answer (4)
Hint : Response to stimulus is a feature of structural
& functional units of neural system.
Solution : Extensibility is a property of muscle fibres
while neurons exhibit both excitability & conductivity.
141. Answer (1)
Hint : Nissl's granules are clusters of RNA &
ribosomes.
Solution : Nissl's granules are site of protein
synthesis located in cell body of neuron/soma & in
dendrites (afferent processes). They are absent in
afferent processes also called axons.
142. Answer (2)
Hint : Neurilemma is a contribution of Schwann
cells.
Solution : Schwann cells form myelin sheath &
neurilemma in neurons of PNS. Oligodendrocytes
form myelin sheath in CNS. Neurilemma is absent in
CNS. Oligodendrocytes are branched cells while
Schwann cells are not branched.
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143. Answer (4)
Hint : Wine flask shaped cells of alimentary canal.
Solution : Goblet cells are mucus secreting
unicellular glands of alimentary canal. Cerumen and
sweat are secreted by sebaceous & sudorific glands
respectively. Hormones are secretions of endocrine
glands.
144. Answer (1)
Hint : Stratified i.e. compound epithelium lines
surfaces that face mechanical stress.
Solution : Stratum germinativum comprises of cells
that divide regularly and forms the bottommost layer
of straitified epithelium. Diffusion surfaces and tubular
parts of nephron are lined by single layered simple
epithelium.
145. Answer (2)
Hint : Junctions involved in cementing neighbouring
cells.
Solution : Tight junctions help to stop substances
from leaking across a tissue.
146. Answer (2)
Hint : Multiple layer in epithelial tissue are effective
in protection.
Solution : At surfaces prone to stress (wear & tear)
or abrasion compound epithelium is effective.
147. Answer (3)
Hint : Endocrine glands lack ducts.
Solution : Endocrine glands do not have ducts and
release their secretion into blood capillaries.
148. Answer (2)
Hint : Exocrine glands release/pour their secretions
through ducts.
Solution : Insulin is a secretion of ductless part of
gland pancreas. Saliva, earwax and milk are
secretions of exocrine glands.
149. Answer (4)
Hint : Concentric arrangement of calcium salts
surrounding osteocytes.
Solution : The lamellae refer to concentric
arrangement of calcium phosphate salts surrounding
osteocytes in diaphysis of long bones of mammals.
150. Answer (2)
Hint : Muscles under the control of our will are
voluntary.
Solution : Smooth and cardiac muscles are
involuntary in nature i.e. they are not under the
control of our will. Smooth muscles contain
unbranched, unstriped, uninucleate fibres.
151. Answer (1)
Hint : Muscle fibres associated with limbs and heart
are striped in appearance.
Solution : Striated muscles include both cardiac &
skeletal muscles. The alternate light and dark bands
refer to the horizontal arrangement of actin & myosin
filaments.
152. Answer (4)
Hint : Intercalated discs are interdigitations between
adjacent cells that act as boosters of cardiac
impulse.
Solution : Cardiac fibres are striated and the edges
of sarcomeres are ‘Z’ lines. Connexons are part of
gap junctions present at intercalated discs. Visceral
muscles are uninucleated.
153. Answer (3)
Hint : White fat is unilocular and adipocytes are
signet ring shaped cells.
Solution : Brown fat in newly born prevents shivering
in them. White fat acts as a reservior of energy and
the triglycerides are hydrolysed upon need.
154. Answer (2)
Hint : Positional information of amino acids decides
specificity of enzyme action.
Solution : Apoenzymes are proteinaceous part of
holoenzymes. Enzymes never alter the equilibrium of
the reaction, they act by lowering activation energy
barrier.
155. Answer (3)
Hint : While overcoming competitive inhibition Km
value increases.
Solution : Succinate is the substrate for enzyme
succinate dehydrogenase. Malonate is its
competitive inhibitor.
156. Answer (3)
Hint : Hydrolytic enzymes work by addition of water
to break the bond between biomolecules.
Solution : Hydrolases belong to class III according
to nomenclature given by IUB. Phosphodiesterases
also belong to this class of enzymes.
157. Answer (2)
Hint : Thymine is exclusive to DNA.
Solution : Thiamine is vitamin B1
while thymine is
nitrogenous base in DNA.
158. Answer (4)
Hint : Identify odd one out w.r.t. macromolecules.
Solution : Lipids are neither polymer nor
macromolecules.
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159. Answer (2)
Hint : Amino acids are monomers of proteins.
Solution : Amino acids are substituted methanes.
They exist as zwitterions at isoelectric pH.
Trihydroxypropane is glycerol. DNA & RNA are
negatively charged molecules.
160. Answer (3)
Hint : DNA of a eukaryote comprises of A, T, G, C
as nitrogenous base.
Solution : Cytosine is a nitrogenous base while
cysteine is an amino acid.
161. Answer (4)
Hint : Active site are formed at tertiary level of
protein organisation. Hydrogen bonds stabilise
predominantly protein structure formed at secondary
level of organisation.
Solution : Ionic bonds, Van der Waals forces,
hydrophobic interactions along with hydrogen bonds
stabilise quaternary structure.
162. Answer (4)
Hint : All components given are sugars in raffinose.
Solution : Tripeptide has 3 amino acids &
triglyceride has 3 fatty acids linked to glycerol.
163. Answer (1)
Hint : Two or more amino acids can be linked by
peptide bonds.
Solution : Collagen is a heteropolymer of amino
acids. Chitin & cellulose are polysaccharides whose
monomers are linked by glycosidic bonds. Choline is
nitrogenous compound found in lecithin.
164. Answer (3)
Hint : First alphabet of amino acid is considered as
the basis of single alphabet code.
Solution : Tyrosine is coded by Y, phenylalanine by
‘F’ and glutamic acid by E. This is because
threonine, proline & glycine have already been named
as T, P & G respectively.
165. Answer (4)
Hint : Organic cofactor loosely attached to
apoenzyme.
Solution : Niacin and riboflavin are vitamins requires
for formation of coenzymes such as NAD, FAD &
FMN.
166. Answer (1)
Hint : Identify a homopolymer.
Solution : Chitin comprises of repeating units of (1
4) N-acetylglucosamine. Nucleic acids (DNA),
proteins i.e. RuBisCO and insulin are
heteropolymers.
167. Answer (3)
Hint : Double bonds cannot be introduced beyond 9th
carbon position in most animals usually.
Solution : Linoleic acid has 2 double bonds
(9 & 12th carbon). Plants are able to synthesize
essential fatty acids by introducing double bonds at
12th & 15th carbon. Oleic acid has a single double
bond at 9th position. In most mammals, arachidonic
acid can be formed from linoleic acid.
168. Answer (2)
Hint : Unsaturated fatty acids have low melting
points.
Solution : Gingelly oil/sesame oil and arachidonic
acid are unsaturated fatty acids, hence will have
lower melting points in comparison to other saturated
lipids given.
169. Answer (1)
Hint : Formation of polymers and disaccharides
requires glycosidic bonds.
Solution : Glucose is a monomer of cellulose but
cellulose is unbranched homopolymer glucose.
170. Answer (3)
Hint : Identify a nucleotide. Ribozyme is RNA acting
as enzyme.
Solution : ATP is adenosine triphosphate.
171. Answer (1)
Hint : Pentose sugar derived from C5
H10
O5
.
Solution : 5 carbon sugar in RNA (ribonucleic acid)
is ribose. DNA has deoxyribose with molecular
formula C5
H10
O4
.
172. Answer (1)
Hint : Isomerism in hexose sugars.
Solution : Galactose & glucose are aldose sugars
while fructose is ketohexose. Ribose is a pentose
sugar.
173. Answer (1)
Hint : Reducing sugar.
Solution : Sugars with free aldehyde or ketone
groups are detected by Benedict’s and Fehling’s
solution. Glucose and galactose have free aldehyde
group while fructose has free keto group. Sucrose
has no free keto or aldehyde group.
174. Answer (3)
Hint : Structural sugar found in RNA.
Solution : Cellulose is unbranched structural
homopolymer of glucose comprising of (1-4)
glycosidic linkages. Ribose is a monomer formed
without glycosidic bond. Cytidylic acid is a
nucleotide found in RNA.
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175. Answer (1)
Hint : Dehydration is a condensation reaction
involving loss of water molecule.
Solution : Cystine formation requires removal of H2
and disulfide bonds are formed between two amino
acids. Formation of lecithin, adenosine and collagen
requires formation of ester & glycosidic bonds
respectively.
176. Answer (4)
Hint : Identify a molecule which is a monomer.
Solution : Inulin is a polymer of fructose, while
insulin is a polymer of amino acids. Lactose is a
disaccharide. Polymers can be broken into
monomers by addition of water (hydrolysis).
177. Answer (2)
Hint : Bones are reservoirs of some elements.
Solution : Matrix of bone is largely composed of
hydroxyapatite crystals of calcium phosphate.
Essential fatty acids & essential amino acids are
supplied in diet.
178. Answer (4)
Hint : Polymers are found in retentate/acid insoluble
fraction.
Solution : Cysteine, calcium & cytosine are
micromolecules that are obtained in filtrate/acid
soluble fraction. Cellulose is a polymer of glucose
and due to its large size it cannot cross the filtration
membrane.
179. Answer (3)
Hint : Order of percentage of biomolecules in cells.
Solution : Water > proteins > nucleic acids >
carbohydrates > lipids > ions
70 90% 10 15% 5 7% 3% 2% 1%
A C B D E
180. Answer (2)
Hint : Compounds whose role or function we do not
understand at the moment.
Solution : Primary metabolites are essential for
physiological processes and have identifiable
functions. Secondary metabolities include alkaloids,
antibiotics, rubber, essential oils, spices etc.