All India Aakash Test Series for NEET-2020 Test – 2 (Code-A)_(Answers)
Aakash Educational Services Limited - Regd. Office: Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456 1/18
All India Aakash Test Series for NEET-2020
Test Date : 15/09/2019
ANSWERS
1. (4)
2. (1)
3. (2)
4. (1)
5. (3)
6. (1)
7. (3)
8. (2)
9. (3)
10. (1)
11. (2)
12. (1)
13. (1)
14. (3)
15. (4)
16. (1)
17. (2)
18. (2)
19. (2)
20. (4)
21. (1)
22. (2)
23. (4)
24. (3)
25. (4)
26. (3)
27. (2)
28. (2)
29. (2)
30. (4)
31. (2)
32. (4)
33. (1)
34. (3)
35. (1)
36. (4)
37. (2)
38. (2)
39. (3)
40. (4)
41. (1)
42. (2)
43. (2)
44. (2)
45. (1)
46. (2)
47. (1)
48. (4)
49. (2)
50. (4)
51. (3)
52. (3)
53. (1)
54. (2)
55. (2)
56. (1)
57. (3)
58. (1)
59. (1)
60. (4)
61. (2)
62. (2)
63. (1)
64. (3)
65. (3)
66. (4)
67. (1)
68. (4)
69. (4)
70. (4)
71. (1)
72. (2)
73. (1)
74. (1)
75. (3)
76. (3)
77. (2)
78. (2)
79. (3)
80. (2)
81. (1)
82. (2)
83. (1)
84. (2)
85. (3)
86. (3)
87. (4)
88. (2)
89. (4)
90. (4)
91. (2)
92. (4)
93. (3)
94. (1)
95. (2)
96. (3)
97. (3)
98. (2)
99. (1)
100. (4)
101. (4)
102. (3)
103. (1)
104. (2)
105. (1)
106. (3)
107. (4)
108. (2)
109. (1)
110. (3)
111. (4)
112. (4)
113. (1)
114. (1)
115. (4)
116. (2)
117. (3)
118. (2)
119. (3)
120. (1)
121. (2)
122. (4)
123. (1)
124. (3)
125. (4)
126. (1)
127. (2)
128. (1)
129. (4)
130. (3)
131. (2)
132. (1)
133. (2)
134. (3)
135. (4)
136. (2)
137. (3)
138. (2)
139. (4)
140. (4)
141. (1)
142. (2)
143. (3)
144. (2)
145. (1)
146. (1)
147. (3)
148. (2)
149. (1)
150. (3)
151. (3)
152. (2)
153. (4)
154. (1)
155. (2)
156. (4)
157. (3)
158. (4)
159. (2)
160. (4)
161. (3)
162. (1)
163. (3)
164. (4)
165. (3)
166. (3)
167. (1)
168. (4)
169. (1)
170. (1)
171. (4)
172. (2)
173. (4)
174. (2)
175. (4)
176. (3)
177. (1)
178. (1)
179. (4)
180. (2)
TEST - 2 - Code-A
All India Aakash Test Series for NEET-2020 Test - 2 (Code-A)_(Hints & Solutions)
Aakash Educational Services Limited - Regd. Office: Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456 2/18
HINTS & SOLUTIONS
PHYSICS
1. Answer (4)
Hint : mF q v B
Sol. : As magnetic force is normal to velocity of
charged particle. Hence kinetic energy of
particle remains constant but momentum
changes.
2. Answer (1)
Hint : Application of Biot-Savart‟s law.
Sol. :
i
Bd
00 sin sin
4
B B 0net 0
3 .3 33 sin60 sin60
34
2
= 61.8 10 T
3. Answer (2)
Hint and Sol. : 0long wire
2
iB
r
4. Answer (1)
Hint : Two current carrying wires exert a force
on each other.
Sol. : The net magnetic force on loop is
repulsive. Hence loop will move away from the
wire.
5. Answer (3)
Hint : mF q v B
Sol. : 6 6 ˆ ˆ ˆ1 10 10 2 2mF i j j
ˆ2 Nk
6. Answer (1)
Hint : Magnetic field due to long current
carrying wire.
Sol. : At any points on line y = x the point will
be equidistant from both wires and magnetic
fields are in opposite direction.
7. Answer (3)
Hint : Use Ampere‟s circuital law.
Sol. : 0 0. 2aB dl l l l
0. .2bB dl l
0. 2 2 0
cB dl l l
0 0. 6 2 4
d B dl l l l
Hence d, b, a, c
8. Answer (2)
Hint : lv
SS
R
Sol. : Current sensitivity NBA
I
Voltage sensitivity NBA
V R
1
NBA
V RNBA R
I
9. Answer (3)
Hint : Use Biot-Savart‟s law
Sol. : B0 = B1 + B2
0 0 .4 4
i i
a a
0 00
4 4 2
i iB
a a
0 0
4 8
i i
a a
0 . 14 2
i
a
10. Answer (1)
Hint : eff
mF i B
Test - 2 (Code-A)_(Hints & Solutions) All India Aakash Test Series for NEET-2020
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Sol. : i = 1 A
ˆ2B k
y2 = 4x
2y
effˆ4 j
ˆ ˆ1 4 2mF j k
ˆ ˆ8 j k
ˆ8i
11. Answer (2)
Hint : 0 0F q v B
Sol. : Arc length AB = distance travelled in
magnetic field.
Time t = 32 6
T
T
03
m
qB
12. Answer (1)
Hint : mF i dL B
Sol. : 0ˆ1
yB B k
Now for line AD
1 0 0
0 ˆ ˆ1
F iB j iB j
l
For line BC
2 0 0ˆ ˆ1 2F iB j iB j
The forces due to line AB and CD are equal
and opposite. Hence net force will be
net 2 1 0 0 0ˆ ˆF F F iB iB j iB j
net 0ˆF iB j
13. Answer (1)
Hint : Use Biot-Savart‟s law.
Sol. :
0
2 2
lB
a
3
2
0 3
2 2 2
lB
a
03
8
lB
a
14. Answer (3)
Hint : Use mv
rqB
Sol. : 2mqVmv
rqB qB
m
rq
2 pr r
2 1 2 10 2 cmr r
15. Answer (4)
Hint and Sol. :
C
INBA
i
16. Answer (1)
Hint : Use 2
M q
mL
Sol. : 2 2
.2 4 4
q mr qrM
m
6 2
8 24 10 10 22 10 Am
4M
17. Answer (2)
Hint : Use and M B M NIA
All India Aakash Test Series for NEET-2020 Test - 2 (Code-A)_(Hints & Solutions)
Aakash Educational Services Limited - Regd. Office: Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456 4/18
Sol. : 2 ˆM ia k
0 0ˆ ˆcos45 sin45B B i B j
2
0 ˆ ˆ2
B iaM B j i
20
ˆ ˆB a i j
18. Answer (2)
Hint :
20
axis, Ring 32 2 22
iaB
x a
Sol. :
20
32 2 2
0
2
2
P
C
ia
x aB
iB
a
3
32 2 2
1
27
a
x a
1
2 2 2 3x a a
x2+ a
2 = 9a
2
x
2 = 8a
2
2 2x a
19. Answer (2)
Hint : 0loop, centre
2
niB
a
Sol. : 0 ,2
iB
a
where
1 and 1
2a n
0 0
2.
2
i iB
l I
0. 3 1
, where 2. 3. 2
iB a
a
0 0.39
2.
6
i i
l l
9B B
20. Answer (4)
Hint and Sol. : The current i is divided equally
in both branches, hence at the centre the
magnetic field due to both branch is equal and
opposite. Hence Bnet at centre of loop is zero.
21. Answer (1)
Hint : 0
34
q v rB
r
Sol. :
02
sin90 ˆ4
qvB k
r
7 9 610 2 10 2 10
4
10 ˆ10 TB k
22. Answer (2)
Hint : Use vector addition rule.
Sol. : 2 2 22 cos60netM M M M
23 3M M
23. Answer (4)
Hint : Use BH = Becos
Sol. : 1
seccos
e
H
B
B
24. Answer (3)
Hint and Sol. : Since diamagnetic substances
are weakly repelled by the field so if they are to
move, they tend to move away from stronger
field.
25. Answer (4)
Hint : Use 1 and m r
B
H
Sol. : 0 0
r
B
H
0
1m
B
H
26. Answer (3)
Hint : Use cos and tan vH H
H
BB B
B
Test - 2 (Code-A)_(Hints & Solutions) All India Aakash Test Series for NEET-2020
Aakash Educational Services Limited - Regd. Office: Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456 5/18
Sol. : tan
tancos
3tan37 34
1cos60 2
2
1 3tan
2
27. Answer (2)
Hint : Use 2H
lT
MB
Sol. :
1
602
30 cos30
e
l
MB
1
602
20 cos60
e
l
MB
2
1
cos602
3 cos30
e
e
B
B
2
1
4 1
9 3
e
e
B
B
1
2
9 3 3
44 3
e
e
B
B
28. Answer (2)
Hint : Use and .M B U M B
Sol. : 1 2cos cosW MB
3
5 15
MB
5 5
2MB
Now 5 5 4
sin 2 5 N m2 5
MB
29. Answer (2)
Hint : Use 034
MB
r
Sol. :
M HB B
034
M
r
5 7
3
2.403 10 10
r
3
1000 1
8 r
10 1
2 r
r = 0.2 m
30. Answer (4)
Hint : Use 0axial 3
2
4
MB
r
Sol. : Since 2eqM M
Now point P will be on axial position
Hence 0axial 3
2
4
eqMB
r
0axial 3
0
2 2
4
MB
x
31. Answer (2)
Hint : Use tan
tancos
Sol. : tan
tancos
Then, tan tan
tancos 90 sin
2 2sin cos 1
2 2
2 2
tan tan1
tan tan
2 2 2
1 1 1
tan tan tan
2 2 2cot cot cot
32. Answer (4)
Hint : Use .B B S
Sol. : ˆ ˆ ˆ. 2 4 . 1 0B B S i j k
33. Answer (1)
Hint :
dAB
dtiR
All India Aakash Test Series for NEET-2020 Test - 2 (Code-A)_(Hints & Solutions)
Aakash Educational Services Limited - Regd. Office: Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456 6/18
Sol. : l vt bB
d
dt
vbB
vbB
iR
34. Answer (3)
Hint : Use 2
0
and
TE d
H dt ER dt
Sol. : 2t T t
Induced emf 2 2d
E T tdt
22
0 0
4 2T T T tEH dt dt
R R
34
3
T
R
34
3
TH
R
35. Answer (1)
Hint : Use 1 21 2
1 2
and P S
L LL L L L
L L
Sol. : 1 2P
S
L LL
L 1 2 4.8 20 96L L
(L1 – L2)2 = (L1 + L2)
2 – 4L1L2
= 400 – 4 × 96 = 16
L1 – L2 = 4 mH
Hence, L1 = 12 mH and L2 = 8 mH
36. Answer (4)
Hint : Use 0 1
t
i i e
Sol. : 0 and E L
iR R
0
0 0
1
t
q idt i e dt
0i
e
.
E L
R Re
2
ELq
eR
37. Answer (2)
Hint : di
e Ldt
Sol. : For time 0 to 3
T, i t
Hence, di
cdt
e = – constant
For time 2
,3 3
T Tt i = constant
i.e. e = 0
For time 2
,3
Tt T i a bt
i.e. e = + constant
38. Answer (2)
Hint : Use KVL and eL di
Ldt
Sol. : VP – 4 × 4 – 10 – 5 × 10–3
× 104– VQ = 0
VP – VQ = 76 V
39. Answer (3)
Hint : 1 d
iR dt
Sol. : 2
0.5
13 8 2
10 t
di t t
dt
0.5
16 8
10 tt
5
0.5 A10
Test - 2 (Code-A)_(Hints & Solutions) All India Aakash Test Series for NEET-2020
Aakash Educational Services Limited - Regd. Office: Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456 7/18
40. Answer (4)
Hint : Use and .M B dAi
Sol. : Since 0
2
a b
B
a
iB cdr cdr
r
0 In 2
a b
a
icr
0 In2
ic a b
a
0 In 12
c bM
i a
41. Answer (1)
Hint : .
v B I
Sol. :
2 4 3AC
24 V
4 2 4CB
32 V
56 V AB AC CB
42. Answer (2)
Hint : di
Ldt
Sol. : 3 sin d
L t tdt
3sin 3 cosL t t t
2 3 sin 3 cos2 2 2
6 V
6 V
43. Answer (2)
Hint and Sol. : Among the given options,
Nickel has minimum Curie temperature (631 K).
44. Answer (2)
Hint : Use eq.
and e Blv iR
Sol. :
1 1 1 0.1 4 0.4 VBlv
2 2 1 0.1 5Blv
0.5 V
eq 1 2
eq 1 2
r r r
eq
0.4 0.50.45 V
2 2
0.45
0.045 V9 1
i
45 mAi
45. Answer (1)
Hint : Use .dB
d Adt
Sol. : B = 40 t
2.2 .dB
E r rdt
2
r dBE
dt
–22 10
402
= 0.4 N/C
CHEMISTRY
46. Answer (2)
Hint : Physisorption involves weak van der
Waals forces so reversible in nature.
47. Answer (1)
Hint : Increase in surface area, increases the
rate of adsorption of gases on solid surface.
All India Aakash Test Series for NEET-2020 Test - 2 (Code-A)_(Hints & Solutions)
Aakash Educational Services Limited - Regd. Office: Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456 8/18
48. Answer (4)
Hint : Gases having high critical temperature
will easily liquify and more easily adsorb.
Sol. : Tc : NH3 > CO2 > CH4 > H2
49. Answer (2)
Hint : If reactants and catalyst are in same
phase then process is known as homogenous
catalysis.
50. Answer (4)
Hint : Au, gelatin and Sb2S3 sols are negatively
charged sols.
51. Answer (3)
Hint : Peptization is the method used for the
preparation of colloids.
52. Answer (3)
Hint : Size of colloidal particles ranges from
1 nm to 1000 nm (or 10 Å to 10000 Å).
53. Answer (1)
Hint : Electrolyte of highest coagulating power
has lowest flocculating value.
Sol. : Fe(OH)3 is a positively charged sol and
for its coagulation, order of flocculating value of
given electrolytes is KCl > K2SO4 > K3[Fe(CN)6]
54. Answer (2)
Hint : Emulsions show Tyndall effect.
55. Answer (2)
Hint : Coagulation and electrophoresis depend
on charge present on colloids.
Sol. : Tyndall effect depends on the size of
colloidal particles.
56. Answer (1)
Hint : As2S3 sol is formed by double
decomposition method.
Double
decomposition2 3 2 2 3 2As O 3H S As S sol 3H O
57. Answer (3)
Hint : Foam : Gas dispersed in liquid.
Sol. : Cell fluids is an example of sol.
58. Answer (1)
Hint : Copper glance : Cu2S
Sol. : Zincite : ZnO
Malachite : CuCO3Cu(OH)2
Copper pyrites : CuFeS2
59. Answer (1)
Hint : Leaching of bauxite ore is done by
Baeyer‟s process using NaOH as solvent.
60. Answer (4)
Hint : Mg with BaO2 is used as ignition mixture
in Goldschmidt alumino thermite process.
61. Answer (2)
Hint : Iron has impurity of SiO2 which combines
with CaO and forms slag of CaSiO3.
62. Answer (2)
Hint : Poling process is used to refine Cu and
Sn containing impurities of their metal oxides.
63. Answer (1)
Hint : van Arkel process is used for refining of
Ti, Zr, etc.
64. Answer (3)
Hint : In metallurgy of Ag and Au, complexes
formed are [Ag(CN)2]–, [Au(CN)2]
– and
[Zn(CN)4]2–
65. Answer (3)
Hint : After roasting of copper pyrites, copper is
obtained in the form of copper matte.
Sol. : Copper matte : (Cu2S and FeS)
66. Answer (4)
Hint : Cast iron has impurity of carbon, sulphur,
silicon and phosphorus.
Sol. : Impurities present in cast iron are
oxidised by haematite (oxidizing agent).
2 3Fe O 3C 2Fe 3CO
67. Answer (1)
Hint : In liquation impurities of high melting
points are removed from metal.
Sol. : Refining is based on difference in melting
points of metal and impurities.
68. Answer (4)
Hint : Concentration of white bauxite is done by
Serpeck‟s method in the presence of coke and
N2(g)
Sol. :
Al2O3 . 2H2O + N2 + 3C 2AIN + 3CO + 2H2O
AIN + 3H2O Al(OH)3 + NH3
2 3 232Al OH Al O 3H O
Test - 2 (Code-A)_(Hints & Solutions) All India Aakash Test Series for NEET-2020
Aakash Educational Services Limited - Regd. Office: Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456 9/18
69. Answer (4)
Hint : CuFeS2 has impurity of FeS, which is
converted into FeSiO3 as slag.
Sol. : Froth floatation process is generally used
for the concentration of sulphide ores.
70. Answer (4)
Hint :
Element N P As Bi
Density (g/cm3) 0.879 1.823 5.778 9.808
71. Answer (1)
Hint : More the electron density on central
atom, more will be the basic nature.
Sol. : NH3 > PH3 > AsH3 > SbH3 > BiH3
(Basicity order)
72. Answer (2)
Hint : 4 2 7 2 2 3 22NH Cr O N Cr O 4H O
Sol. : • 3 22NaN 3N g 2Na
• 4 2 4 2NH Cl aq. NaNO aq. NH NO aq.
NaCl aq.
• 4 3 2 222NH Cl Ca OH 2NH g 2H O CaCl
73. Answer (1)
Hint : NO2 and N2O5 are acidic oxides.
Sol. : NO and N2O are neutral oxides
NO is paramagnetic
N2O is diamagnetic
74. Answer (1)
Hint : Laboratory grade nitric acid contains ~
68% HNO3 by mass.
75. Answer (3)
Hint : PH3 becomes inflammable due to
presence of impurity of P2H4 or P4 vapours.
76. Answer (3)
Hint : 5 2 3 4PCl 4H O H PO 5HCl
77. Answer (2)
Hint :
Number of S – S bonds is 8
78. Answer (2)
Hint : linkage is known as peroxy
linkage.
Sol. :
79. Answer (3)
Hint : Due to H-bonding present in „HF‟, it has
highest boiling point among the hydrides of
group 17 elements.
Sol. : HF > HI > HBr > HCl (Boiling point).
80. Answer (2)
Hint : Au dissolves in aqua regia (HCl + HNO3)
Sol. : Au + 4H+ + NO3
– + 4Cl
– AuCl4
– + NO
+ 2H2O
81. Answer (1)
Hint : Xe made similar type of compound as
prepared by O2 with PtF6.
Sol. :
Species O2 Xe N2 I F
Ionization energy
(in kJ/mol)
1175 1170 1503 1008 1680
82. Answer (2)
Hint : XeF6 on partial hydrolysis forms XeO2F2
and XeOF4.
All India Aakash Test Series for NEET-2020 Test - 2 (Code-A)_(Hints & Solutions)
Aakash Educational Services Limited - Regd. Office: Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456 10/18
Sol. : XeF6 + H2O XeOF4 + 2HF
XeOF4 + H2O XeO2F2 + 2HF
XeO2F2 + H2O XeO3 + 2HF
83. Answer (1)
Hint : Phosphinic acid : H3PO2
Phosphonic acid : H3PO3
Sol. :
84. Answer (2)
Hint : Due to presence of two P – H bonds,
H3PO2 is a strong reducing agent.
Sol. : 4AgNO3 + 2H2O + H3PO2 4Ag + 4HNO3
+ H3PO4
85. Answer (3)
Hint : PH3 is slightly soluble in water.
Sol. : The solution of PH3 in water decomposes
in light and gives red phosphorus and H2(g).
86. Answer (3)
Hint : Minimum and maximum covalency of
halogen is 1 and 7 respectively.
Sol. : XX′ : Minimum halogen atoms = 2
XX′7 : Maximum halogen atoms = 8
87. Answer (4)
Hint : HBrO2 is least likely to exist among the
given compounds
88. Answer (2)
Hint : Monoclinic sulphur is -sulphur.
Sol. : Monoclinic sulphur has m.p. 393 K and
m.p. of rhombic sulphur is 385.8 K.
89. Answer (4)
Hint : CIF3 is a colourless gas
Sol. : CIF3 : Bent-T-shaped
ClF3 + 2H2O HClO2 + 3HF
90. Answer (4)
Hint : H2O > H2S > H2Se > H2Te : Bond angle
Sol. :
H2O > H2Te > H2Se > H2S : (Melting point
and boiling point)
H2Te > H2Se > H2S > H2O : (Dissociation
constant)
BIOLOGY
91. Answer (2)
Hint : Recombination of genetic material
occurs in sexual reproduction.
Sol. : Reproduction through leaf buds is a type
of vegetative reproduction. In such type of
reproduction, there is neither meiosis nor fusion
of gametes and thus least or no genetic
variation will occur in the next generations.
92. Answer (4)
Hint : Recessive phenotypes appear only in
homozygous condition.
Sol. : Terminal position of flowers and wrinkled
seed shape in pea plant are recessive traits.
So, they appear phenotypically only in
homozygous condition.
93. Answer (3)
Hint : All prokaryotes and eukaryotes have
both RNA and DNA.
Sol. : Some viruses do not have DNA so in
these viruses, RNA acts as genetic material.
94. Answer (1)
Hint : The length of DNA in viruses is much
smaller than that of prokaryotes and
eukaryotes.
Sol. : × 174 bacteriophage – 5386
nucleotides Bacteriophage lambda – 48502 bp
E. coli – 4.6 × 106 bp
Human genome – 3.3 × 109 bp
95. Answer (2)
Hint : Yellow seed colour, violet flower colour
and inflated pod shape are the dominant traits
in pea plant
Sol. : Only solitary flowers are present in pea
plant.
Test - 2 (Code-A)_(Hints & Solutions) All India Aakash Test Series for NEET-2020
Aakash Educational Services Limited - Regd. Office: Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456 11/18
96. Answer (3)
Sol. : Punnett square was developed by a
British geneticist, Reginald C. Punnett.
97. Answer (3)
Hint : On the basis of X-ray diffraction pictures
of DNA, the earlier scientists suggested that
DNA was sort of helix with 3.4 Å periodicity.
Sol. : On the basis of X-ray diffraction pictures
of DNA, the Watson and Crick proposed a very
simple but famous double helix model for the
structure of DNA.
98. Answer (2)
Hint : In a nucleotide, a phosphate group is
linked to OH of 5C of nucleoside.
Sol. : When a phosphate group is linked to OH
of 5C of a nucleoside through phosphoester
linkage, a corresponding nucleotide is formed.
99. Answer (1)
Hint : A tall pea plant may be homozygous
dominant or heterozygous for alleles
responsible for stem height.
Sol. :
100. Answer (4)
Hint : In heterozygous condition for the genes
which show incomplete dominance, the
intermediate phenotype is produced.
Sol. : Test cross is done to know the genotype
of individual that shows dominant phenotype of
the trait which follows Law of Dominance.
101. Answer (4)
Hint : In viruses, the genetic material is either
DNA or RNA.
Sol. : 5-methyl uracil is another name of
thymine. RNA lacks thymine and DNA lacks
uracil.
102. Answer (3)
Hint : In dsDNA, purines of one strand are
paired with pyrimidines of corresponding strand
by formation of hydrogen bonds.
Sol. : Adenine forms two H-bonds with thymine
and guanine is bonded with cytosine with three
H-bonds.
103. Answer (1)
Hint : Some viruses produce an enzyme
reverse transcriptase which can synthesize
DNA over RNA template.
Sol. :
104. Answer (2)
Hint : The phenotype is effected if the
responsible allele produces non-functional
enzyme or no enzyme.
Sol. : If the modified allele produces normal or
less efficient enzyme then it is said to be
equivalent to the unmodified allele.
105. Answer (1)
Hint : Regarding ABO blood groups, total
number of genotypes in human is six.
Sol. : Regarding ABO blood types in human
beings
Total number of
genotypes
: Total number of
phenotypes
= 6 : 4
= 3 : 2
106. Answer (3)
Sol. : The negatively charged DNA is wrapped
around the positively charged histone octamer
to form a structure called nucleosome.
107. Answer (4)
Hint : S-strain of Pneumococcus is virulent and
cause pneumonia.
Sol. : In the experiment conducted by Griffith,
the R-strain bacteria had been transformed by
the heat killed S-strain bacteria.
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108. Answer (2)
Hint : A single gene product can produce more
than one effect when there is interrelationship
between the metabolic pathways that contribute
towards different phenotypes.
Sol. : In pleiotropy, a single gene product may
produce more than one effect or control several
phenotypes depending on its position.
109. Answer (1)
Hint : True breeding lines show homozygous
condition for the characters
Sol. : In Mendel‟s dihybrid cross, the true
breeding lines in F2 generation are RRYY,
RRyy, rrYY and rryy.
110. Answer (3)
Hint : Sulphur is present in proteins whereas
phosphorus is present in genetic materials.
Sol. : Only genetic material of bacteriophage
enters the bacteria and the protein coat of the
virus is synthesised inside the bacteria will not
contain radioactive sulphur. But the genetic
material of bacteriophage formed inside the
bacteria will have radioactive phosphorus.
111. Answer (4)
Hint : Two terminal phosphates in a
deoxyribonucleoside triphosphates are high
energy phosphates.
Sol. : In addition to acting as substrate
deoxyribonucleoside triphosphates also provide
energy for polymerisation.
112. Answer (4)
Hint : For being suitable experimental material
in genetics, the specimen should have smaller
number of chromosomes which should be
morphologically distinct.
Sol. :Breeding throughout the year, production
of large number of offsprings in single mating
and easily visible hereditary variations in
organism are some of the properties which
make it suitable as experimental material in
genetics.
113. Answer (1)
Hint : For flower colour, snapdragon plant
shows incomplete dominance.
Sol. :
Out of eight individuals, two produce white
flowers, i.e., 25%
114. Answer (1)
Sol. : On the template strand of DNA with
polarity 5 3 DNA synthesis is discontinuous
in the form of Okazaki fragments. These
fragments are joined by the enzyme DNA
ligase.
115. Answer (4)
Hint : DNA of E. coli has 4.6 × 106 bp.
Sol. : The rate of polymerisation in E. coli
will be 64.6 10
bp/second38 60
= 2000 bp/second
116. Answer (2)
Hint : The probability of beings first child a girl
is 1
.2
Sol. :
Probability of blood group A is 1
.2
Therefore, the probability of being first child a
girl with blood group 1 1 1
A2 2 4
117. Answer (3)
Hint : The human female with Turner‟s
syndrome has only one X-chromosome in her
cells.
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Sol. : The females inflicted with Turner‟s
syndrome are sterile as ovaries are
rudimentary including lack of other secondary
sexual characters.
118. Answer (2)
Hint : The DNA dependent DNA polymerase
catalyse polymerisation only in one direction
that is 5 3.
Sol. : The replication is continuous on one
template strand with polarity 3 5and is
known as leading daughter strand.
119. Answer (3)
Hint : To maintain the ploidy of the cell, the
replication of DNA and cell division cycle
should be highly coordinated.
Sol. : A failure in cell division after DNA
replication results into polyploidy.
120. Answer (1)
Hint : According to law of independent
assortment, the genes controlling different
characters get assorted independent to each
other.
Sol. : Law of independent assortment is correct
if the genes are present on two different
chromosomes, may also by their segregation
through crossing over if they are present on the
same chromosome.
121. Answer (2)
Hint : RNA transcription starts from 3 end of
the template strand of the transcription unit.
Sol. : Promoter sequences are present towards
5 end of the structural gene, i. e., with respect
to the polarity of coding strand.
122. Answer (4)
Hint : If we switch the position of promoter with
terminator in the transcription unit, the template
strand becomes coding strand and the coding
strand becomes template strand.
Sol. : By switching the positions of promoter
with terminator, the template strand will be
5-G C C T A T A G G T T A-3 template strand
3-C G G A U A U C C A A U-5- mRNA
123. Answer (1)
Hint : Brown body, red eyes and normal wings
are the wild type traits in Drosophila.
Sol. : In an experiment conducted by Morgan
red eyed and miniature winged and white eyed
and brown bodied Drosophila are recombinant
type.
124. Answer (3)
Hint : Polygenic traits are controlled by two or
more genes.
Sol. : There are 3 pairs of genes may be
involved in controlling the skin colour in human
beings. Therefore it is an example of polygenic
inheritance
125. Answer (4)
Hint : Transcription is copying genetic
information from one strand of the DNA into
RNA.
Sol. : The DNA sequence coding for tRNA or
rRNA molecules also define a gene.
126. Answer (1)
Hint : rRNAs synthesised in eukaryotes are 5S,
5.8S, 18S and 28S.
Sol. : RNA polymerase I – 5.8S, 18S, and 28S
rRNAs
RNA polymerase II – hnRNA
RNA polymerase III – tRNA, ScRNA, 5S rRNA
and SnRNA.
127. Answer (2)
Hint : Male honey bees have only one set of
chromosomes.
Sol. : Unfertilized egg of honey bee develops
into male bee.
128. Answer (1)
Hint : ZW-ZZ type of sex determination is
found in birds.
Sol. : Hens are heterogametic and thus they
produce two types of eggs, i.e., (A + Z) and
(A + W). Therefore, females are responsible to
determine the sex of the chicks.
129. Answer (4)
Hint : Amino acid gets attached to the 3 end of
tRNA.
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Sol. : At the 3 end of tRNA, unpaired –CCA
sequence is present. Amino acid gets attached
at this end only.
130. Answer (3)
Hint : In lac operon, operator gene interacts
with a protein molecule or regulator molecule,
which prevents the transcription of structural
genes.
Sol. :
Operator gene – Interacts with regulator
molecule
Promoter gene – Provides attachment site for
RNA polymerase
Structural gene – Transcribe mRNA for
polypeptide synthesis
Regulator gene – Controls the activity of
operator gene
131. Answer (2)
Hint : Haemophilia is X-linked recessive trait.
Sol. :
132. Answer (1)
Hint : Individuals inflicted with Klinefelter‟s
syndrome are sterile males with overall
masculine development and some female
characteristics.
Sol. : Klinefelter‟s syndrome is caused due
to chromosome complement 44 + XXY. This
results by the union of an abnormal egg
(22 + XX) and a normal sperm (22 + Y) or normal
egg (22 + X) and abnormal sperm (22 + XY).
133. Answer (2)
Hint : Minisatellites are VNTRs.
Sol. : VNTRs show very high degree of
polymorphism. The size of VNTR varies from
0.1 to 20 kb.
134. Answer (3)
Hint : In phenylketonuria, accumulation of
phenylpyruvic acid and other derivatives of
phenylalanine occur in brain.
Sol. :Phenylketonuria is an autosomal
recessive trait while Down‟s syndrome
develops due to aneuploidy. Mental retardation
occurs in both the disorders.
135. Answer (4)
Hint : Sex-linked traits include X-linked traits.
Sol. : Haemophilia and colour blindness are
sex linked disorders.
Thalassemia, sickle-cell anaemia and myotonic
distrophy are autosomal disorders.
Turner‟s syndrome is due to aneuploidy.
136. Answer (2)
Hint : Sterilisation procedure is the terminal
method of family planning.
Sol. : Saheli is a non-steroidal contraceptive
drug.
Condoms are used to prevent the meeting of
egg and sperm. They also provide protection
against STIs.
Cu7 is copper releasing IUD which suppresses
sperm motility and fertilising capacity of
sperms.
137. Answer (3)
Hint : Teenagers are more vulnerable to STIs.
Sol. : Persons with 15-24 years of age group
are more vulnerable to STIs.
138. Answer (2)
Hint : The average failure rate of natural
methods of contraception is 20-30%.
Sol. :
Contraceptive method Average failure
rate
Rhythm (natural) method – 20-30%
Oral contraceptives – 2-3%
Barrier methods – 10-15%
Coitus interruptus – 20%
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139. Answer (4)
Hint : Syphilis is a STI caused by bacterium
Treponema pallidum.
Sol. :
STI Causative agent
Genital herpes Herpes simplex virus (HSV)
Gonorrhoea Neisseria gonorrhoeae
AIDS HIV
140. Answer (4)
Hint : IUI is Intra uterine insemination.
Sol. : ICSI = Intra cytoplasmic sperm injection
In ICSI, the partner‟s sperm is placed inside the
egg with a microscopic needle.
141. Answer (1)
Hint. : The population growth rate was
20/1000/year.
Sol. : According to the 2011 census report, the
population growth rate was less than 2%, while
it was around 1.7% as per 2001 census.
142. Answer (2)
Hint : During lactation period, the level of FSH
is low.
Sol. : In actively lactating mothers, the level of
prolactin is high, which suppresses the release
of gonadotrophin i.e. FSH and LH due to
inhibitory effect on GnRH.
143. Answer (3)
Hint : Select an ART.
Sol. : GIFT stands for Gamete intra fallopian
transfer.
144. Answer (2)
Hint : Select the STI caused by bacteria.
Sol. : Except for hepatitis B, genital herpes and
HIV infections, most STIs are completely
curable if detected early and treated properly.
145. Answer (1)
Hint : Technique used to diagnose genetic
disorders in the foetus.
Sol. : Amniocentesis is used to detect genetic
disorders like Down‟s syndrome.
A rapid decrease in MMR and IMR are possible
reasons for population explosion.
Saheli was discovered at CDRI, Lucknow, UP.
Family planning programmes were initiated in
1951 in India.
146. Answer (1)
Hint : Failure of testes to descend into scrotum.
Sol. : Infertility in a female can occur due to
various reasons such as:
(i) Anovulation
(ii) Oligoovulation
(iii) Inadequate growth of corpus luteum
(iv) Fibroid uterus
(v) Defective vaginal growth etc.
147. Answer (3)
Hint : Shedding of endometrial wall.
Sol. : The possible ill effects of using
contraceptive devices are nausea, abdominal
pain, breakthrough bleeding, irregular
menstrual bleeding or even breast cancer.
148. Answer (2)
Hint : ZIFT is Zygote Intra Fallopian Transfer.
Sol. : In GIFT i.e. Gamete Intra Fallopian
Transfer, fertilization of sperm and egg occurs
inside the female body, i.e. in vivo, while in
ZIFT, the fertilization of sperm and egg occurs
in a petridish in a laboratory i.e. in vitro.
149. Answer (1)
Hint : Identify a marsupial.
Sol. : Flying phalanger, marsupial mole and
Tasmanian wolf are Australian marsupials.
150. Answer (3)
Hint : Extra terrestrial origin of life.
Sol. : According to Panspermia, life was
transferred from one planet to other in the form
of small units called spores or seeds or sperms.
According to the theory of Biogenesis, life
arose from some pre-existing life.
According to the theory of Abiogenesis, life
originated from non-living matter.
151. Answer (3)
Hint : Age of Earth is estimated to be 4.5
billion years.
Sol. : The “Big bang theory” attempts to explain
the origin of universe. Origin of Earth occurred
around 4.5 billion years ago.
152. Answer (2)
Hint : Radioactive dating is used to determine
the age of rocks on Earth.
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Sol. : According to the theory of special
creation, given by Father Suarez, Earth is
about 4000 years old.
153. Answer (4)
Hint : This region comprises present day
Indonesia, Philippines and East Timor.
Sol. : Naturalist “Alfred Wallace”, worked in
Malay Archipelago. Like Darwin, he also talked
about the fitness of organisms. He also came to
the similar conclusion that those organisms
which can adapt better in their environment are
selected by nature.
154. Answer (1)
Hint : Its function is to collect the sound waves.
Sol. : Vestigial organs are those organs which
are non-functional in humans for eg. Nictitating
membrane, wisdom tooth, vermiform appendix,
nipples and dense body hair in males etc.
155. Answer (2)
Hint : Archaeopteryx is a missing link between
birds and reptiles.
Sol. : Forelimbs of whale, bat, cheetah and
human show homology, hence exemplify
divergent evolution.
Sweet potato is modified underground root,
while potato is modified underground stem.
Their origin is different, but function is same i.e.
storage of food.
156. Answer (4)
Hint : Select the basic amino acid.
Sol. : Lysine is a basic amino acid, with extra
amino group and was not obtained during
Miller‟s initial experiment.
157. Answer (3)
Hint : Epoch of tertiary period.
Sol. :
Period Epoch Age
(million of years)
Pliocene 5
Tertiary Miocene 23
Oligocene 34
Eocene 57
Paleocene 65
158. Answer (4)
Hint : Animals which live on land.
Sol. : Mesozoic era is the age of reptiles and in
Jurassic period dinosaurs became dominant.
159. Answer (2)
Hint : This horse evolved in Pliocene epoch.
Sol. : Pliohippus had one complete finger and
one complete toe and two splints hidden
beneath the skin, hence was considered to be
the first one toed horse.
160. Answer (4)
Hint : Hardy Weinberg principle.
Sol. : p2 + q
2 + 2pq = 1
Frequency of MM individual is p2 = (0.6)
2 =
0.36 or 36%
Frequency of NN individual is q2 = (0.4)
2 = 0.16
or 16%
Frequency of MN individual is 2pq = 2 × 0.6 ×
0.4 0.48 or 48%
161. Answer (3)
Hint : Sugar glider and Bandicoot have
common ancestor.
Sol. : As they have a common ancestor, they
exhibit divergent evolution.
162. Answer (1)
Hint : Coelacanth is called lobe finned fish.
Sol. :
Neopilina : Connecting link between
annelids and arthropods.
Chimaera : Connecting link between
cartilaginous and bony fishes.
Peripatus : Connecting link between
annelids and arthropods.
Latimeria : Connecting link between fishes
(Coelacanth) and amphibians.
163. Answer (3)
Hint : Biogenetic law.
Sol. : According to Biogenetic law, Ontogeny
recapitulates Phylogeny.
Presence of gills in tadpole of frog indicate that
frogs have evolved from gilled ancestors i.e.
fishes.
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164. Answer (4)
Hint : Single step large mutations are
saltations.
Sol. : According to Darwin, variations or
changes are slow, continuous and occur in a
directional manner.
165. Answer (3)
Hint : It is also called genetic drift.
Sol. : Sewall Wright effect/genetic drift is the
change in gene frequency by chance in a small
population.
Gene migration/gene flow : When genes are
exchanged between two different populations
of a species frequently.
Gene pool : The total collection of genes and
their alleles in a population.
166. Answer (3)
Hint : It is also known as diversifying selection.
Sol. : Disruptive selection is observed when
selection does not favour the mean character
value, rather favours both the peripheral
character values.
Directional selection is seen when selection
acts to eliminate one extreme form and
supports the other extreme, the peak shifts in
the direction which is selected by nature.
167. Answer (1)
Hint : Sulphur containing compound.
Sol. : S. L. Miller created electric discharge in a
closed flask containing CH4, H2, NH3 in 2 : 2 : 1
ratio and water vapour at 800°C.
168. Answer (4)
Hint : Neanderthal man.
Sol. : Neanderthal man buried his dead with
flowers, not Dryopithecus.
169. Answer (1)
Hint : Substances released by them inhibit
implantation.
Sol. : Intra uterine device is one of the most
widely accepted method of contraception in
India.
CuT Copper releasing IUD
Femidom Female condom (Barrier method)
Diaphragm Barrier method.
170. Answer (1)
Hint : Levonorgestrel
Sol. : The hormone releasing IUDs make the
uterus unsuitable for implantation and cervix
hostile to sperms.
171. Answer (4)
Hint : Cro-Magnon man.
Sol. : The cranial capacity of Homo sapiens
fossilis i.e. Cro-Magnon is 1650 cc.
172. Answer (2)
Hint : Lamarck‟s theory is also known as theory
of use and disuse of organs.
Sol. : It is based upon the inheritance of
acquired characters, hence Lamarck‟s theory is
often called theory of inheritance of acquired
characters.
173. Answer (4)
Hint : 1000 million years form a billion.
Sol. : Origin of Earth occurred 4.5 billion years
ago. Coacervates as the model of protocells
were presented by Oparin. Oparin and Haldane
gave the theory of chemical evolution of life.
174. Answer (2)
Hint : Darwin gave the theory of Natural
Selection.
Sol. : The phrase “Survival of fittest” was first
used by Herbert Spencer. The same context
was asserted by Darwin as “Natural selection”.
175. Answer (4)
Hint : Brachiosaurus had long neck and long
tail.
Sol. : Statement B is incorrect because
Stegosaurus had big kite like plates on its back
for protection.
176. Answer (3)
Hint : Saltations occur suddenly.
Sol. : According to mutation theory, mutations
are large, discontinuous changes and can
appear suddenly in any direction.
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177. Answer (1)
Hint : Chlorophyte ancestors were the aquatic
green algae that possibly gave rise to all green
plants.
Sol. :
178. Answer (1)
Hint : Hybrid inviability and hybrid breakdown
are post zygotic mechanisms of reproductive
isolation.
Sol. : Mechanical isolation : The structural
differences in genitalia of individuals belonging
to different animal species interfere with
mating.
179. Answer (4)
Hint : Biogeographical evidence.
Sol. : Restriction distribution of pouched
mammals in Australia support biogeographical
evidences, while remaining options support
embryological evidences.
180. Answer (2)
Hint : Artificial selection.
Sol. : Artificial selection is the selective
breeding of plants and animals for desired traits
by humans.
e.g Evolution of wild mustard
Variation among breeds of domestic
pigeon.
All India Aakash Test Series for NEET-2020 Test – 2 (Code-B)_(Answers)
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All India Aakash Test Series for NEET-2020
Test Date : 15/09/2019
ANSWERS
1. (1)
2. (2)
3. (2)
4. (2)
5. (1)
6. (4)
7. (3)
8. (2)
9. (2)
10. (4)
11. (1)
12. (3)
13. (1)
14. (4)
15. (2)
16. (4)
17. (2)
18. (2)
19. (2)
20. (3)
21. (4)
22. (3)
23. (4)
24. (2)
25. (1)
26. (4)
27. (2)
28. (2)
29. (2)
30. (1)
31. (4)
32. (3)
33. (1)
34. (1)
35. (2)
36. (1)
37. (3)
38. (2)
39. (3)
40. (1)
41. (3)
42. (1)
43. (2)
44. (1)
45. (4)
46. (4)
47. (4)
48. (2)
49. (4)
50. (3)
51. (3)
52. (2)
53. (1)
54. (2)
55. (1)
56. (2)
57. (3)
58. (2)
59. (2)
60. (3)
61. (3)
62. (1)
63. (1)
64. (2)
65. (1)
66. (4)
67. (4)
68. (4)
69. (1)
70. (4)
71. (3)
72. (3)
73. (1)
74. (2)
75. (2)
76. (4)
77. (1)
78. (1)
79. (3)
80. (1)
81. (2)
82. (2)
83. (1)
84. (3)
85. (3)
86. (4)
87. (2)
88. (4)
89. (1)
90. (2)
91. (4)
92. (3)
93. (2)
94. (1)
95. (2)
96. (3)
97. (4)
98. (1)
99. (2)
100. (1)
101. (4)
102. (3)
103. (1)
104. (4)
105. (2)
106. (1)
107. (3)
108. (2)
109. (3)
110. (2)
111. (4)
112. (1)
113. (1)
114. (4)
115. (4)
116. (3)
117. (1)
118. (2)
119. (4)
120. (3)
121. (1)
122. (2)
123. (1)
124. (3)
125. (4)
126. (4)
127. (1)
128. (2)
129. (3)
130. (3)
131. (2)
132. (1)
133. (3)
134. (4)
135. (2)
136. (2)
137. (4)
138. (1)
139. (1)
140. (3)
141. (4)
142. (2)
143. (4)
144. (2)
145. (4)
146. (1)
147. (1)
148. (4)
149. (1)
150. (3)
151. (3)
152. (4)
153. (3)
154. (1)
155. (3)
156. (4)
157. (2)
158. (4)
159. (3)
160. (4)
161. (2)
162. (1)
163. (4)
164. (2)
165. (3)
166. (3)
167. (1)
168. (2)
169. (3)
170. (1)
171. (1)
172. (2)
173. (3)
174. (2)
175. (1)
176. (4)
177. (4)
178. (2)
179. (3)
180. (2)
TEST - 2 - Code-B
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HINTS & SOLUTIONS
PHYSICS
1. Answer (1)
Hint : Use .dB
d Adt
Sol. : B = 40 t
2.2 .dB
E r rdt
2
r dBE
dt
–22 10
402
= 0.4 N/C
2. Answer (2)
Hint : Use eq.
and e Blv iR
Sol. :
1 1 1 0.1 4 0.4 VBlv
2 2 1 0.1 5Blv
0.5 V
eq 1 2
eq 1 2
r r r
eq
0.4 0.50.45 V
2 2
0.45
0.045 V9 1
i
45 mAi
3. Answer (2)
Hint and Sol. : Among the given options,
Nickel has minimum Curie temperature (631 K).
4. Answer (2)
Hint : di
Ldt
Sol. : 3 sin d
L t tdt
3sin 3 cosL t t t
2 3 sin 3 cos2 2 2
6 V
6 V
5. Answer (1)
Hint : .
v B I
Sol. :
2 4 3AC
24 V
4 2 4CB
32 V
56 V AB AC CB
6. Answer (4)
Hint : Use and .M B dAi
Sol. : Since 0
2
a b
B
a
iB cdr cdr
r
0 In 2
a b
a
icr
0 In2
ic a b
a
0 In 12
c bM
i a
7. Answer (3)
Hint : 1 d
iR dt
Sol. : 2
0.5
13 8 2
10 t
di t t
dt
0.5
16 8
10 tt
5
0.5 A10
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8. Answer (2)
Hint : Use KVL and eL di
Ldt
Sol. : VP – 4 × 4 – 10 – 5 × 10–3
× 104– VQ = 0
VP – VQ = 76 V
9. Answer (2)
Hint : di
e Ldt
Sol. : For time 0 to 3
T, i t
Hence, di
cdt
e = – constant
For time 2
,3 3
T Tt i = constant
i.e. e = 0
For time 2
,3
Tt T i a bt
i.e. e = + constant
10. Answer (4)
Hint : Use 0 1
t
i i e
Sol. : 0 and E L
iR R
0
0 0
1
t
q idt i e dt
0i
e
.
E L
R Re
2
ELq
eR
11. Answer (1)
Hint : Use 1 21 2
1 2
and P S
L LL L L L
L L
Sol. : 1 2P
S
L LL
L 1 2 4.8 20 96L L
(L1 – L2)2 = (L1 + L2)
2 – 4L1L2
= 400 – 4 × 96 = 16
L1 – L2 = 4 mH
Hence, L1 = 12 mH and L2 = 8 mH
12. Answer (3)
Hint : Use 2
0
and
TE d
H dt ER dt
Sol. : 2t T t
Induced emf 2 2d
E T tdt
22
0 0
4 2T T T tEH dt dt
R R
34
3
T
R
34
3
TH
R
13. Answer (1)
Hint :
dAB
dtiR
Sol. : l vt bB
d
dt
vbB
vbB
iR
14. Answer (4)
Hint : Use .B B S
Sol. : ˆ ˆ ˆ. 2 4 . 1 0B B S i j k
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15. Answer (2)
Hint : Use tan
tancos
Sol. : tan
tancos
Then, tan tan
tancos 90 sin
2 2sin cos 1
2 2
2 2
tan tan1
tan tan
2 2 2
1 1 1
tan tan tan
2 2 2cot cot cot
16. Answer (4)
Hint : Use 0axial 3
2
4
MB
r
Sol. : Since 2eqM M
Now point P will be on axial position
Hence 0axial 3
2
4
eqMB
r
0axial 3
0
2 2
4
MB
x
17. Answer (2)
Hint : Use 034
MB
r
Sol. :
M HB B
034
M
r
5 7
3
2.403 10 10
r
3
1000 1
8 r
10 1
2 r
r = 0.2 m
18. Answer (2)
Hint : Use and .M B U M B
Sol. : 1 2cos cosW MB
3
5 15
MB
5 5
2MB
Now 5 5 4
sin 2 5 N m2 5
MB
19. Answer (2)
Hint : Use 2H
lT
MB
Sol. :
1
602
30 cos30
e
l
MB
1
602
20 cos60
e
l
MB
2
1
cos602
3 cos30
e
e
B
B
2
1
4 1
9 3
e
e
B
B
1
2
9 3 3
44 3
e
e
B
B
20. Answer (3)
Hint : Use cos and tan vH H
H
BB B
B
Sol. : tan
tancos
3tan37 34
1cos60 2
2
1 3tan
2
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21. Answer (4)
Hint : Use 1 and m r
B
H
Sol. : 0 0
r
B
H
0
1m
B
H
22. Answer (3)
Hint and Sol. : Since diamagnetic substances
are weakly repelled by the field so if they are to
move, they tend to move away from stronger
field.
23. Answer (4)
Hint : Use BH = Becos
Sol. : 1
seccos
e
H
B
B
24. Answer (2)
Hint : Use vector addition rule.
Sol. : 2 2 22 cos60netM M M M
23 3M M
25. Answer (1)
Hint : 0
34
q v rB
r
Sol. :
02
sin90 ˆ4
qvB k
r
7 9 610 2 10 2 10
4
10 ˆ10 TB k
26. Answer (4)
Hint and Sol. : The current i is divided equally
in both branches, hence at the centre the
magnetic field due to both branch is equal and
opposite. Hence Bnet at centre of loop is zero.
27. Answer (2)
Hint : 0loop, centre
2
niB
a
Sol. : 0 ,2
iB
a
where
1 and 1
2a n
0 0
2.
2
i iB
l I
0. 3 1
, where 2. 3. 2
iB a
a
0 0.39
2.
6
i i
l l
9B B
28. Answer (2)
Hint :
20
axis, Ring 32 2 22
iaB
x a
Sol. :
20
32 2 2
0
2
2
P
C
ia
x aB
iB
a
3
32 2 2
1
27
a
x a
1
2 2 2 3x a a
x2+ a
2 = 9a
2
x
2 = 8a
2
2 2x a
29. Answer (2)
Hint : Use and M B M NIA
Sol. : 2 ˆM ia k
0 0ˆ ˆcos45 sin45B B i B j
2
0 ˆ ˆ2
B iaM B j i
20
ˆ ˆB a i j
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30. Answer (1)
Hint : Use 2
M q
mL
Sol. : 2 2
.2 4 4
q mr qrM
m
6 2
8 24 10 10 22 10 Am
4M
31. Answer (4)
Hint and Sol. :
C
INBA
i
32. Answer (3)
Hint : Use mv
rqB
Sol. : 2mqVmv
rqB qB
m
rq
2 pr r
2 1 2 10 2 cmr r
33. Answer (1)
Hint : Use Biot-Savart‟s law.
Sol. :
0
2 2
lB
a
3
2
0 3
2 2 2
lB
a
03
8
lB
a
34. Answer (1)
Hint : mF i dL B
Sol. : 0ˆ1
yB B k
Now for line AD
1 0 0
0 ˆ ˆ1
F iB j iB j
l
For line BC
2 0 0ˆ ˆ1 2F iB j iB j
The forces due to line AB and CD are equal
and opposite. Hence net force will be
net 2 1 0 0 0ˆ ˆF F F iB iB j iB j
net 0ˆF iB j
35. Answer (2)
Hint : 0 0F q v B
Sol. : Arc length AB = distance travelled in
magnetic field.
Time t = 32 6
T
T
03
m
qB
36. Answer (1)
Hint : eff
mF i B
Sol. : i = 1 A
ˆ2B k
y2 = 4x
2y
effˆ4 j
ˆ ˆ1 4 2mF j k
ˆ ˆ8 j k
ˆ8i
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37. Answer (3)
Hint : Use Biot-Savart‟s law
Sol. : B0 = B1 + B2
0 0 .4 4
i i
a a
0 00
4 4 2
i iB
a a
0 0
4 8
i i
a a
0 . 14 2
i
a
38. Answer (2)
Hint : lv
SS
R
Sol. : Current sensitivity NBA
I
Voltage sensitivity NBA
V R
1
NBA
V RNBA R
I
39. Answer (3)
Hint : Use Ampere‟s circuital law.
Sol. : 0 0. 2aB dl l l l
0. .2bB dl l
0. 2 2 0
cB dl l l
0 0. 6 2 4
d B dl l l l
Hence d, b, a, c
40. Answer (1)
Hint : Magnetic field due to long current
carrying wire.
Sol. : At any points on line y = x the point will
be equidistant from both wires and magnetic
fields are in opposite direction.
41. Answer (3)
Hint : mF q v B
Sol. : 6 6 ˆ ˆ ˆ1 10 10 2 2mF i j j
ˆ2 Nk
42. Answer (1)
Hint : Two current carrying wires exert a force
on each other.
Sol. : The net magnetic force on loop is
repulsive. Hence loop will move away from the
wire.
43. Answer (2)
Hint and Sol. : 0long wire
2
iB
r
44. Answer (1)
Hint : Application of Biot-Savart‟s law.
Sol. :
i
Bd
00 sin sin
4
B B 0net 0
3 .3 33 sin60 sin60
34
2
= 61.8 10 T
45. Answer (4)
Hint : mF q v B
Sol. : As magnetic force is normal to velocity of
charged particle. Hence kinetic energy of
particle remains constant but momentum
changes.
CHEMISTRY
46. Answer (4)
Hint : H2O > H2S > H2Se > H2Te : Bond angle
Sol. :
H2O > H2Te > H2Se > H2S : (Melting point
and boiling point)
H2Te > H2Se > H2S > H2O : (Dissociation
constant)
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47. Answer (4)
Hint : CIF3 is a colourless gas
Sol. : CIF3 : Bent-T-shaped
ClF3 + 2H2O HClO2 + 3HF
48. Answer (2)
Hint : Monoclinic sulphur is -sulphur.
Sol. : Monoclinic sulphur has m.p. 393 K and
m.p. of rhombic sulphur is 385.8 K.
49. Answer (4)
Hint : HBrO2 is least likely to exist among the
given compounds
50. Answer (3)
Hint : Minimum and maximum covalency of
halogen is 1 and 7 respectively.
Sol. : XX′ : Minimum halogen atoms = 2
XX′7 : Maximum halogen atoms = 8
51. Answer (3)
Hint : PH3 is slightly soluble in water.
Sol. : The solution of PH3 in water decomposes
in light and gives red phosphorus and H2(g).
52. Answer (2)
Hint : Due to presence of two P – H bonds,
H3PO2 is a strong reducing agent.
Sol. : 4AgNO3 + 2H2O + H3PO2 4Ag + 4HNO3
+ H3PO4
53. Answer (1)
Hint : Phosphinic acid : H3PO2
Phosphonic acid : H3PO3
Sol. :
54. Answer (2)
Hint : XeF6 on partial hydrolysis forms XeO2F2
and XeOF4.
Sol. : XeF6 + H2O XeOF4 + 2HF
XeOF4 + H2O XeO2F2 + 2HF
XeO2F2 + H2O XeO3 + 2HF
55. Answer (1)
Hint : Xe made similar type of compound as
prepared by O2 with PtF6.
Sol. :
Species O2 Xe N2 I F
Ionization energy
(in kJ/mol)
1175 1170 1503 1008 1680
56. Answer (2)
Hint : Au dissolves in aqua regia (HCl + HNO3)
Sol. : Au + 4H+ + NO3
– + 4Cl
– AuCl4
– + NO
+ 2H2O
57. Answer (3)
Hint : Due to H-bonding present in „HF‟, it has
highest boiling point among the hydrides of
group 17 elements.
Sol. : HF > HI > HBr > HCl (Boiling point).
58. Answer (2)
Hint : linkage is known as peroxy
linkage.
Sol. :
59. Answer (2)
Hint :
Test - 2 (Code-B)_(Hints & Solutions) All India Aakash Test Series for NEET-2020
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Number of S – S bonds is 8
60. Answer (3)
Hint : 5 2 3 4PCl 4H O H PO 5HCl
61. Answer (3)
Hint : PH3 becomes inflammable due to
presence of impurity of P2H4 or P4 vapours.
62. Answer (1)
Hint : Laboratory grade nitric acid contains
~ 68% HNO3 by mass.
63. Answer (1)
Hint : NO2 and N2O5 are acidic oxides.
Sol. : NO and N2O are neutral oxides
NO is paramagnetic
N2O is diamagnetic
64. Answer (2)
Hint : 4 2 7 2 2 3 22NH Cr O N Cr O 4H O
Sol. : • 3 22NaN 3N g 2Na
• 4 2 4 2NH Cl aq. NaNO aq. NH NO aq.
NaCl aq.
• 4 3 2 222NH Cl Ca OH 2NH g 2H O CaCl
65. Answer (1)
Hint : More the electron density on central
atom, more will be the basic nature.
Sol. : NH3 > PH3 > AsH3 > SbH3 > BiH3
(Basicity order)
66. Answer (4)
Hint :
Element N P As Bi
Density (g/cm3) 0.879 1.823 5.778 9.808
67. Answer (4)
Hint : CuFeS2 has impurity of FeS, which is
converted into FeSiO3 as slag.
Sol. : Froth floatation process is generally used
for the concentration of sulphide ores.
68. Answer (4)
Hint : Concentration of white bauxite is done by
Serpeck‟s method in the presence of coke and
N2(g)
Sol. :
Al2O3 . 2H2O + N2 + 3C 2AIN + 3CO + 2H2O
AIN + 3H2O Al(OH)3 + NH3
2 3 232Al OH Al O 3H O
69. Answer (1)
Hint : In liquation impurities of high melting
points are removed from metal.
Sol. : Refining is based on difference in melting
points of metal and impurities.
70. Answer (4)
Hint : Cast iron has impurity of carbon, sulphur,
silicon and phosphorus.
Sol. : Impurities present in cast iron are
oxidised by haematite (oxidizing agent).
2 3Fe O 3C 2Fe 3CO
71. Answer (3)
Hint : After roasting of copper pyrites, copper is
obtained in the form of copper matte.
Sol. : Copper matte : (Cu2S and FeS)
72. Answer (3)
Hint : In metallurgy of Ag and Au, complexes
formed are [Ag(CN)2]–, [Au(CN)2]
– and
[Zn(CN)4]2–
73. Answer (1)
Hint : van Arkel process is used for refining of
Ti, Zr, etc.
74. Answer (2)
Hint : Poling process is used to refine Cu and
Sn containing impurities of their metal oxides.
75. Answer (2)
Hint : Iron has impurity of SiO2 which combines
with CaO and forms slag of CaSiO3.
76. Answer (4)
Hint : Mg with BaO2 is used as ignition mixture
in Goldschmidt alumino thermite process.
77. Answer (1)
Hint : Leaching of bauxite ore is done by
Baeyer‟s process using NaOH as solvent.
78. Answer (1)
Hint : Copper glance : Cu2S
Sol. : Zincite : ZnO
Malachite : CuCO3Cu(OH)2
Copper pyrites : CuFeS2
79. Answer (3)
Hint : Foam : Gas dispersed in liquid.
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Sol. : Cell fluids is an example of sol.
80. Answer (1)
Hint : As2S3 sol is formed by double
decomposition method.
Double
decomposition2 3 2 2 3 2As O 3H S As S sol 3H O
81. Answer (2)
Hint : Coagulation and electrophoresis depend
on charge present on colloids.
Sol. : Tyndall effect depends on the size of
colloidal particles.
82. Answer (2)
Hint : Emulsions show Tyndall effect.
83. Answer (1)
Hint : Electrolyte of highest coagulating power
has lowest flocculating value.
Sol. : Fe(OH)3 is a positively charged sol and
for its coagulation, order of flocculating value of
given electrolytes is KCl > K2SO4 > K3[Fe(CN)6]
84. Answer (3)
Hint : Size of colloidal particles ranges from
1 nm to 1000 nm (or 10 Å to 10000 Å).
85. Answer (3)
Hint : Peptization is the method used for the
preparation of colloids.
86. Answer (4)
Hint : Au, gelatin and Sb2S3 sols are negatively
charged sols.
87. Answer (2)
Hint : If reactants and catalyst are in same
phase then process is known as homogenous
catalysis.
88. Answer (4)
Hint : Gases having high critical temperature
will easily liquify and more easily adsorb.
Sol. : Tc : NH3 > CO2 > CH4 > H2
89. Answer (1)
Hint : Increase in surface area, increases the
rate of adsorption of gases on solid surface.
90. Answer (2)
Hint : Physisorption involves weak van der
Waals forces so reversible in nature.
BIOLOGY
91. Answer (4)
Hint : Sex-linked traits include X-linked traits.
Sol. : Haemophilia and colour blindness are
sex linked disorders.
Thalassemia, sickle-cell anaemia and myotonic
distrophy are autosomal disorders.
Turner‟s syndrome is due to aneuploidy.
92. Answer (3)
Hint : In phenylketonuria, accumulation of
phenylpyruvic acid and other derivatives of
phenylalanine occur in brain.
Sol. : Phenylketonuria is an autosomal
recessive trait while Down‟s syndrome
develops due to aneuploidy. Mental retardation
occurs in both the disorders.
93. Answer (2)
Hint : Minisatellites are VNTRs.
Sol. : VNTRs show very high degree of
polymorphism. The size of VNTR varies from
0.1 to 20 kb.
94. Answer (1)
Hint : Individuals inflicted with Klinefelter‟s
syndrome are sterile males with overall
masculine development and some female
characteristics.
Sol. : Klinefelter‟s syndrome is caused due to
chromosome complement 44 + XXY. This
results by the union of an abnormal egg
(22 + XX) and a normal sperm (22 + Y) or
normal egg (22 + X) and abnormal sperm (22 +
XY).
95. Answer (2)
Hint : Haemophilia is X-linked recessive trait.
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Sol. :
96. Answer (3)
Hint : In lac operon, operator gene interacts
with a protein molecule or regulator molecule,
which prevents the transcription of structural
genes.
Sol. :
Operator gene – Interacts with regulator
molecule
Promoter gene – Provides attachment site for
RNA polymerase
Structural gene – Transcribe mRNA for
polypeptide synthesis
Regulator gene – Controls the activity of
operator gene
97. Answer (4)
Hint : Amino acid gets attached to the 3 end of
tRNA.
Sol. : At the 3 end of tRNA, unpaired –CCA
sequence is present. Amino acid gets attached
at this end only.
98. Answer (1)
Hint : ZW-ZZ type of sex determination is
found in birds.
Sol. : Hens are heterogametic and thus they
produce two types of eggs, i.e., (A + Z) and
(A + W). Therefore, females are responsible to
determine the sex of the chicks.
99. Answer (2)
Hint : Male honey bees have only one set of
chromosomes.
Sol. : Unfertilized egg of honey bee develops
into male bee.
100. Answer (1)
Hint : rRNAs synthesised in eukaryotes are 5S,
5.8S, 18S and 28S.
Sol. : RNA polymerase I – 5.8S, 18S, and 28S
rRNAs
RNA polymerase II – hnRNA
RNA polymerase III – tRNA, ScRNA, 5S rRNA
and SnRNA.
101. Answer (4)
Hint : Transcription is copying genetic
information from one strand of the DNA into
RNA.
Sol. : The DNA sequence coding for tRNA or
rRNA molecules also define a gene.
102. Answer (3)
Hint : Polygenic traits are controlled by two or
more genes.
Sol. : There are 3 pairs of genes may be
involved in controlling the skin colour in human
beings. Therefore it is an example of polygenic
inheritance
103. Answer (1)
Hint : Brown body, red eyes and normal wings
are the wild type traits in Drosophila.
Sol. : In an experiment conducted by Morgan
red eyed and miniature winged and white eyed
and brown bodied Drosophila are recombinant
type.
104. Answer (4)
Hint : If we switch the position of promoter with
terminator in the transcription unit, the template
strand becomes coding strand and the coding
strand becomes template strand.
Sol. : By switching the positions of promoter
with terminator, the template strand will be
5-G C C T A T A G G T T A-3 template strand
3-C G G A U A U C C A A U-5- mRNA
105. Answer (2)
Hint : RNA transcription starts from 3 end of
the template strand of the transcription unit.
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Sol. : Promoter sequences are present towards
5 end of the structural gene, i. e., with respect
to the polarity of coding strand.
106. Answer (1)
Hint : According to law of independent
assortment, the genes controlling different
characters get assorted independent to each
other.
Sol. : Law of independent assortment is correct
if the genes are present on two different
chromosomes, may also by their segregation
through crossing over if they are present on the
same chromosome.
107. Answer (3)
Hint : To maintain the ploidy of the cell, the
replication of DNA and cell division cycle
should be highly coordinated.
Sol. : A failure in cell division after DNA
replication results into polyploidy.
108. Answer (2)
Hint : The DNA dependent DNA polymerase
catalyse polymerisation only in one direction
that is 5 3.
Sol. : The replication is continuous on one
template strand with polarity 3 5and is
known as leading daughter strand.
109. Answer (3)
Hint : The human female with Turner‟s
syndrome has only one X-chromosome in her
cells.
Sol. : The females inflicted with Turner‟s
syndrome are sterile as ovaries are
rudimentary including lack of other secondary
sexual characters.
110. Answer (2)
Hint : The probability of beings first child a girl
is 1
.2
Sol. :
Probability of blood group A is 1
.2
Therefore, the probability of being first child a
girl with blood group 1 1 1
A2 2 4
111. Answer (4)
Hint : DNA of E. coli has 4.6 × 106 bp.
Sol. : The rate of polymerisation in E. coli
will be 64.6 10
bp/second38 60
= 2000 bp/second
112. Answer (1)
Sol. : On the template strand of DNA with
polarity 5 3 DNA synthesis is discontinuous
in the form of Okazaki fragments. These
fragments are joined by the enzyme DNA
ligase.
113. Answer (1)
Hint : For flower colour, snapdragon plant
shows incomplete dominance.
Sol. :
Out of eight individuals, two produce white
flowers, i.e., 25%
114. Answer (4)
Hint : For being suitable experimental material
in genetics, the specimen should have smaller
number of chromosomes which should be
morphologically distinct.
Sol. :Breeding throughout the year, production
of large number of offsprings in single mating
and easily visible hereditary variations in
organism are some of the properties which
make it suitable as experimental material in
genetics.
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115. Answer (4)
Hint : Two terminal phosphates in a
deoxyribonucleoside triphosphates are high
energy phosphates.
Sol. : In addition to acting as substrate
deoxyribonucleoside triphosphates also provide
energy for polymerisation.
116. Answer (3)
Hint : Sulphur is present in proteins whereas
phosphorus is present in genetic materials.
Sol. : Only genetic material of bacteriophage
enters the bacteria and the protein coat of the
virus is synthesised inside the bacteria will not
contain radioactive sulphur. But the genetic
material of bacteriophage formed inside the
bacteria will have radioactive phosphorus.
117. Answer (1)
Hint : True breeding lines show homozygous
condition for the characters
Sol. : In Mendel‟s dihybrid cross, the true
breeding lines in F2 generation are RRYY,
RRyy, rrYY and rryy.
118. Answer (2)
Hint : A single gene product can produce more
than one effect when there is interrelationship
between the metabolic pathways that contribute
towards different phenotypes.
Sol. : In pleiotropy, a single gene product may
produce more than one effect or control several
phenotypes depending on its position.
119. Answer (4)
Hint : S-strain of Pneumococcus is virulent and
cause pneumonia.
Sol. : In the experiment conducted by Griffith,
the R-strain bacteria had been transformed by
the heat killed S-strain bacteria.
120. Answer (3)
Sol. : The negatively charged DNA is wrapped
around the positively charged histone octamer
to form a structure called nucleosome.
121. Answer (1)
Hint : Regarding ABO blood groups, total
number of genotypes in human is six.
Sol. : Regarding ABO blood types in human
beings
Total number of
genotypes
: Total number of
phenotypes
= 6 : 4
= 3 : 2
122. Answer (2)
Hint : The phenotype is effected if the
responsible allele produces non-functional
enzyme or no enzyme.
Sol. : If the modified allele produces normal or
less efficient enzyme then it is said to be
equivalent to the unmodified allele.
123. Answer (1)
Hint : Some viruses produce an enzyme
reverse transcriptase which can synthesize
DNA over RNA template.
Sol. :
124. Answer (3)
Hint : In dsDNA, purines of one strand are
paired with pyrimidines of corresponding strand
by formation of hydrogen bonds.
Sol. : Adenine forms two H-bonds with thymine
and guanine is bonded with cytosine with three
H-bonds.
125. Answer (4)
Hint : In viruses, the genetic material is either
DNA or RNA.
Sol. : 5-methyl uracil is another name of
thymine. RNA lacks thymine and DNA lacks
uracil.
126. Answer (4)
Hint : In heterozygous condition for the genes
which show incomplete dominance, the
intermediate phenotype is produced.
Sol.: Test cross is done to know the genotype
of individual that shows dominant phenotype of
the trait which follows Law of Dominance.
127. Answer (1)
Hint : A tall pea plant may be homozygous
dominant or heterozygous for alleles
responsible for stem height.
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Sol. :
128. Answer (2)
Hint : In a nucleotide, a phosphate group is
linked to OH of 5C of nucleoside.
Sol. : When a phosphate group is linked to OH
of 5C of a nucleoside through phosphoester
linkage, a corresponding nucleotide is formed.
129. Answer (3)
Hint : On the basis of X-ray diffraction pictures
of DNA, the earlier scientists suggested that
DNA was sort of helix with 3.4 Å periodicity.
Sol. : On the basis of X-ray diffraction pictures
of DNA, the Watson and Crick proposed a very
simple but famous double helix model for the
structure of DNA.
130. Answer (3)
Sol. : Punnett square was developed by a
British geneticist, Reginald C. Punnett.
131. Answer (2)
Hint : Yellow seed colour, violet flower colour
and inflated pod shape are the dominant traits
in pea plant
Sol. : Only solitary flowers are present in pea
plant.
132. Answer (1)
Hint : The length of DNA in viruses is much
smaller than that of prokaryotes and
eukaryotes.
Sol. : × 174 bacteriophage – 5386
nucleotides Bacteriophage lambda – 48502 bp
E. coli – 4.6 × 106 bp
Human genome – 3.3 × 109 bp
133. Answer (3)
Hint : All prokaryotes and eukaryotes have
both RNA and DNA.
Sol. : Some viruses do not have DNA so in
these viruses, RNA acts as genetic material.
134. Answer (4)
Hint : Recessive phenotypes appear only in
homozygous condition.
Sol. : Terminal position of flowers and wrinkled
seed shape in pea plant are recessive traits.
So, they appear phenotypically only in
homozygous condition.
135. Answer (2)
Hint : Recombination of genetic material
occurs in sexual reproduction.
Sol. : Reproduction through leaf buds is a type
of vegetative reproduction. In such type of
reproduction, there is neither meiosis nor fusion
of gametes and thus least or no genetic
variation will occur in the next generations.
136. Answer (2)
Hint : Artificial selection.
Sol. : Artificial selection is the selective
breeding of plants and animals for desired traits
by humans.
e.g Evolution of wild mustard
Variation among breeds of domestic
pigeon.
137. Answer (4)
Hint : Biogeographical evidence.
Sol. : Restriction distribution of pouched
mammals in Australia support biogeographical
evidences, while remaining options support
embryological evidences.
138. Answer (1)
Hint : Hybrid inviability and hybrid breakdown
are post zygotic mechanisms of reproductive
isolation.
Sol. : Mechanical isolation : The structural
differences in genitalia of individuals belonging
to different animal species interfere with
mating.
Test - 2 (Code-B)_(Hints & Solutions) All India Aakash Test Series for NEET-2020
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139. Answer (1)
Hint : Chlorophyte ancestors were the aquatic
green algae that possibly gave rise to all green
plants.
Sol. :
140. Answer (3)
Hint : Saltations occur suddenly.
Sol. : According to mutation theory, mutations
are large, discontinuous changes and can
appear suddenly in any direction.
141. Answer (4)
Hint : Brachiosaurus had long neck and long
tail.
Sol. : Statement B is incorrect because
Stegosaurus had big kite like plates on its back
for protection.
142. Answer (2)
Hint : Darwin gave the theory of Natural
Selection.
Sol. : The phrase “Survival of fittest” was first
used by Herbert Spencer. The same context
was asserted by Darwin as “Natural selection”.
143. Answer (4)
Hint : 1000 million years form a billion.
Sol. : Origin of Earth occurred 4.5 billion years
ago. Coacervates as the model of protocells
were presented by Oparin. Oparin and Haldane
gave the theory of chemical evolution of life.
144. Answer (2)
Hint : Lamarck‟s theory is also known as theory
of use and disuse of organs.
Sol. : It is based upon the inheritance of
acquired characters, hence Lamarck‟s theory is
often called theory of inheritance of acquired
characters.
145. Answer (4)
Hint : Cro-Magnon man.
Sol. : The cranial capacity of Homo sapiens
fossilis i.e. Cro-Magnon is 1650 cc.
146. Answer (1)
Hint : Levonorgestrel
Sol. : The hormone releasing IUDs make the
uterus unsuitable for implantation and cervix
hostile to sperms.
147. Answer (1)
Hint : Substances released by them inhibit
implantation.
Sol. : Intra uterine device is one of the most
widely accepted method of contraception in
India.
CuT Copper releasing IUD
Femidom Female condom (Barrier method)
Diaphragm Barrier method.
148. Answer (4)
Hint : Neanderthal man.
Sol. : Neanderthal man buried his dead with
flowers, not Dryopithecus.
149. Answer (1)
Hint : Sulphur containing compound.
Sol. : S. L. Miller created electric discharge in a
closed flask containing CH4, H2, NH3 in 2 : 2 : 1
ratio and water vapour at 800°C.
150. Answer (3)
Hint : It is also known as diversifying selection.
Sol. : Disruptive selection is observed when
selection does not favour the mean character
value, rather favours both the peripheral
character values.
Directional selection is seen when selection
acts to eliminate one extreme form and
supports the other extreme, the peak shifts in
the direction which is selected by nature.
All India Aakash Test Series for NEET-2020 Test - 2 (Code-B)_(Hints & Solutions)
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151. Answer (3)
Hint : It is also called genetic drift.
Sol. : Sewall Wright effect/genetic drift is the
change in gene frequency by chance in a small
population.
Gene migration/gene flow : When genes are
exchanged between two different populations
of a species frequently.
Gene pool : The total collection of genes and
their alleles in a population.
152. Answer (4)
Hint : Single step large mutations are
saltations.
Sol. : According to Darwin, variations or
changes are slow, continuous and occur in a
directional manner.
153. Answer (3)
Hint : Biogenetic law.
Sol. : According to Biogenetic law, Ontogeny
recapitulates Phylogeny.
Presence of gills in tadpole of frog indicate that
frogs have evolved from gilled ancestors i.e.
fishes.
154. Answer (1)
Hint : Coelacanth is called lobe finned fish.
Sol. :
Neopilina : Connecting link between
annelids and arthropods.
Chimaera : Connecting link between
cartilaginous and bony fishes.
Peripatus : Connecting link between
annelids and arthropods.
Latimeria : Connecting link between fishes
(Coelacanth) and amphibians.
155. Answer (3)
Hint : Sugar glider and Bandicoot have
common ancestor.
Sol. : As they have a common ancestor, they
exhibit divergent evolution.
156. Answer (4)
Hint : Hardy Weinberg principle.
Sol. : p2 + q
2 + 2pq = 1
Frequency of MM individual is p2 = (0.6)
2 =
0.36 or 36%
Frequency of NN individual is q2 = (0.4)
2 = 0.16
or 16%
Frequency of MN individual is 2pq = 2 × 0.6 ×
0.4 0.48 or 48%
157. Answer (2)
Hint : This horse evolved in Pliocene epoch.
Sol. : Pliohippus had one complete finger and
one complete toe and two splints hidden
beneath the skin, hence was considered to be
the first one toed horse.
158. Answer (4)
Hint : Animals which live on land.
Sol. : Mesozoic era is the age of reptiles and in
Jurassic period dinosaurs became dominant.
159. Answer (3)
Hint : Epoch of tertiary period.
Sol. :
Period Epoch Age
(million of years)
Pliocene 5
Tertiary Miocene 23
Oligocene 34
Eocene 57
Paleocene 65
160. Answer (4)
Hint : Select the basic amino acid.
Sol. : Lysine is a basic amino acid, with extra
amino group and was not obtained during
Miller‟s initial experiment.
161. Answer (2)
Hint : Archaeopteryx is a missing link between
birds and reptiles.
Sol. : Forelimbs of whale, bat, cheetah and
human show homology, hence exemplify
divergent evolution.
Sweet potato is modified underground root,
while potato is modified underground stem.
Their origin is different, but function is same i.e.
storage of food.
Test - 2 (Code-B)_(Hints & Solutions) All India Aakash Test Series for NEET-2020
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162. Answer (1)
Hint : Its function is to collect the sound waves.
Sol. : Vestigial organs are those organs which
are non-functional in humans for eg. Nictitating
membrane, wisdom tooth, vermiform appendix,
nipples and dense body hair in males etc.
163. Answer (4)
Hint : This region comprises present day
Indonesia, Philippines and East Timor.
Sol. : Naturalist “Alfred Wallace”, worked in
Malay Archipelago. Like Darwin, he also talked
about the fitness of organisms. He also came to
the similar conclusion that those organisms
which can adapt better in their environment are
selected by nature.
164. Answer (2)
Hint : Radioactive dating is used to determine
the age of rocks on Earth.
Sol. : According to the theory of special
creation, given by Father Suarez, Earth is
about 4000 years old.
165. Answer (3)
Hint : Age of Earth is estimated to be 4.5
billion years.
Sol. : The “Big bang theory” attempts to explain
the origin of universe. Origin of Earth occurred
around 4.5 billion years ago.
166. Answer (3)
Hint : Extra terrestrial origin of life.
Sol. : According to Panspermia, life was
transferred from one planet to other in the form
of small units called spores or seeds or sperms.
According to the theory of Biogenesis, life
arose from some pre-existing life.
According to the theory of Abiogenesis, life
originated from non-living matter.
167. Answer (1)
Hint : Identify a marsupial.
Sol. : Flying phalanger, marsupial mole and
Tasmanian wolf are Australian marsupials.
168. Answer (2)
Hint : ZIFT is Zygote Intra Fallopian Transfer.
Sol. : In GIFT i.e. Gamete Intra Fallopian
Transfer, fertilization of sperm and egg occurs
inside the female body, i.e. in vivo, while in
ZIFT, the fertilization of sperm and egg occurs
in a petridish in a laboratory i.e. in vitro.
169. Answer (3)
Hint : Shedding of endometrial wall.
Sol. : The possible ill effects of using
contraceptive devices are nausea, abdominal
pain, breakthrough bleeding, irregular
menstrual bleeding or even breast cancer.
170. Answer (1)
Hint : Failure of testes to descend into scrotum.
Sol. : Infertility in a female can occur due to
various reasons such as:
(i) Anovulation
(ii) Oligoovulation
(iii) Inadequate growth of corpus luteum
(iv) Fibroid uterus
(v) Defective vaginal growth etc.
171. Answer (1)
Hint : Technique used to diagnose genetic
disorders in the foetus.
Sol. : Amniocentesis is used to detect genetic
disorders like Down‟s syndrome.
A rapid decrease in MMR and IMR are possible
reasons for population explosion.
Saheli was discovered at CDRI, Lucknow, UP.
Family planning programmes were initiated in
1951 in India.
172. Answer (2)
Hint : Select the STI caused by bacteria.
Sol. : Except for hepatitis B, genital herpes and
HIV infections, most STIs are completely
curable if detected early and treated properly.
173. Answer (3)
Hint : Select an ART.
Sol. : GIFT stands for Gamete intra fallopian
transfer.
174. Answer (2)
Hint : During lactation period, the level of FSH
is low.
All India Aakash Test Series for NEET-2020 Test - 2 (Code-B)_(Hints & Solutions)
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Sol. : In actively lactating mothers, the level of
prolactin is high, which suppresses the release
of gonadotrophin i.e. FSH and LH due to
inhibitory effect on GnRH.
175. Answer (1)
Hint. : The population growth rate was
20/1000/year.
Sol. : According to the 2011 census report, the
population growth rate was less than 2%, while
it was around 1.7% as per 2001 census.
176. Answer (4)
Hint : IUI is Intra uterine insemination.
Sol. : ICSI = Intra cytoplasmic sperm injection
In ICSI, the partner‟s sperm is placed inside the
egg with a microscopic needle.
177. Answer (4)
Hint : Syphilis is a STI caused by bacterium
Treponema pallidum.
Sol. :
STI Causative agent
Genital herpes Herpes simplex virus (HSV)
Gonorrhoea Neisseria gonorrhoeae
AIDS HIV
178. Answer (2)
Hint : The average failure rate of natural
methods of contraception is 20-30%.
Sol. :
Contraceptive method Average failure
rate
Rhythm (natural) method – 20-30%
Oral contraceptives – 2-3%
Barrier methods – 10-15%
Coitus interruptus – 20%
179. Answer (3)
Hint : Teenagers are more vulnerable to STIs.
Sol. : Persons with 15-24 years of age group
are more vulnerable to STIs.
180. Answer (2)
Hint : Sterilisation procedure is the terminal
method of family planning.
Sol. : Saheli is a non-steroidal contraceptive
drug.
Condoms are used to prevent the meeting of
egg and sperm. They also provide protection
against STIs.
Cu7 is copper releasing IUD which suppresses
sperm motility and fertilising capacity of
sperms.