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Test-2 (Code-A)_(Answers) All India Aakash Test Series for NEET-2021
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456 1/16
All India Aakash Test Series for NEET - 2021
Test Date : 20/09/2020
ANSWERS
1. (1) 2. (4) 3. (2) 4. (1) 5. (1) 6. (2) 7. (2) 8. (4) 9. (1) 10. (2) 11. (2) 12. (1) 13. (4) 14. (3) 15. (4) 16. (3) 17. (4) 18. (3) 19. (2) 20. (4) 21. (3) 22. (2) 23. (1) 24. (1) 25. (3) 26. (3) 27. (3) 28. (2) 29. (1) 30. (3) 31. (1) 32. (3) 33. (3) 34. (1) 35. (1) 36. (3)
37. (2) 38. (1) 39. (1) 40. (3) 41. (2) 42. (4) 43. (3) 44. (2) 45. (4) 46. (3) 47. (3) 48. (2) 49. (3) 50. (1) 51. (3) 52. (1) 53. (4) 54. (4) 55. (3) 56. (4) 57. (4) 58. (4) 59. (4) 60. (1) 61. (3) 62. (1) 63. (4) 64. (4) 65. (2) 66. (4) 67. (2) 68. (3) 69. (4) 70. (4) 71. (2) 72. (2)
73. (1) 74. (4) 75. (4) 76. (4) 77. (1) 78. (4) 79. (1) 80. (3) 81. (2) 82. (4) 83. (1) 84. (3) 85. (4) 86. (3) 87. (4) 88. (4) 89. (4) 90. (3) 91. (1) 92. (4) 93. (1) 94. (2) 95. (3) 96. (1) 97. (4) 98. (2) 99. (4) 100. (2) 101. (2) 102. (4) 103. (1) 104. (3) 105. (3) 106. (4) 107. (4) 108. (3)
109. (1) 110. (3) 111. (1) 112. (2) 113. (1) 114. (2) 115. (4) 116. (3) 117. (2) 118. (4) 119. (1) 120. (3) 121. (3) 122. (4) 123. (1) 124. (3) 125. (3) 126. (4) 127. (3) 128. (3) 129. (2) 130. (1) 131. (1) 132. (Deleted) 133. (3) 134. (1) 135. (2) 136. (4) 137. (2) 138. (1) 139. (3) 140. (2) 141. (4) 142. (4) 143. (3) 144. (3)
145. (3) 146. (3) 147. (4) 148. (2) 149. (3) 150. (4) 151. (3) 152. (2) 153. (2) 154. (2) 155. (1) 156. (4) 157. (2) 158. (3) 159. (3) 160. (2) 161. (1) 162. (3) 163. (2) 164. (3) 165. (2) 166. (2) 167. (3) 168. (1) 169. (2) 170. (4) 171. (4) 172. (4) 173. (4) 174. (2) 175. (2) 176. (3) 177. (3) 178. (2) 179. (4) 180. (3)
TEST - 2 (Code-A)
All India Aakash Test Series for NEET-2021 Test-2 (Code-A)_(Hints & Solutions)
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456 2/16
[PHYSICS] 1. Answer (1)
Hint: mvrqB
Sol. 212
eV mv
122
eVvm
122
m eVrqB m
12
22
mVreB
.
2. Answer (4)
Hint: . F i l B
Sol. . sin F i lB
F will be maximum when = 90°. 3. Answer (2)
Hint: M B
Sol. M = r2I
= MBsin60°
2 3
2
r IB
23
2
r IB .
4. Answer (1) Hint & Sol.: Moving charge can produce both
electric and magnetic field 5. Answer (1)
Hint & sol.: 0
2
I
Br
1B
r
6. Answer (2) Hint & sol.:
20
32 2 22
InrB
r x
20
32 2 22 8
InrB
r r
2
032 27
InrB
r
054
InBr
7. Answer (2) Hint: At neutral point magnetic field due to
conductor, balance the horizontal magnetic field of earth.
Sol.:
02
IBr
02
H
IB
r
772 10 10 2 10
r
r = 10 m. 8. Answer (4) Hint: Magnetic field of the centre of the circular
loop having current I and number of turn n will be 0 I2
nr
Sol.:
HINTS & SOLUTIONS
Test-2 (Code-A)_(Hints & Solutions) All India Aakash Test Series for NEET-2021
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456 3/16
B = B1 – B2
0 1 1 0 2 2
1 22 2
I n I nr r
0 02 2
10 24 15 182 25 10 2 15 10
= 240 × 20 – 9000
= 4800 – 9000
= 4200 T
9. Answer (1)
Hint : Current sensitivity = NBAk
Sol.:
A
B
Current senstivityCurrent senstivity
A
B
NN
200100
(current sensitivity)A = 2(Current sensitivity)B 10. Answer (2)
Hint :
G G
sG
R IRI I
Sol.: 60020
GR
= 30
3
330 20 1010 20 10
SR
60010000 20
SR
Rs 0.06 11. Answer (2) Hint: To not enter in the region x > 3a deflection of
particle when it will reach at x = 3a, should be greater than 90°.
Sol.:
Rsin d Rsin90 2a
2mv aqB
2aqBvm
max2aqBV
m
12. Answer (1) Hint : magnetic field due to semi-circular arc at its
center is 02 I
r.
Magnetic field due to semi-infinite wire at distance
d is 04
Id
.
Sol.:
B = B1 + B2 + B3
0 0 0ˆ ˆ ˆ4 4 4
I I I
k k kr r r
0 ˆ4
I
kr
.
13. Answer (4) Hint: Magnetic field due to current carrying long
wire at a distance B is 02
Ir
.
Sol.: 1 2
2 1
B rB r
2
93
BB
2 3
BB
14. Answer (3) Hint: Magnetic field at the centre of circular wire is
02 I
R and magnetic moment of the coil is IR2 .
Sol.: 0
22
IB RM I R
032
R
All India Aakash Test Series for NEET-2021 Test-2 (Code-A)_(Hints & Solutions)
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456 4/16
15. Answer (4) Hint and Sol. : When charged particle moves in
uniform magnetic field the kinetic energy always remains constant because magnetic field does no work.
16. Answer (3) Hint: Use Ampere’s circuital law. Sol.: For r < a
2
1 2 3
I aI xa
1 9
II
0 1.
B dl I
01 2
3 9
IaB
01 6
IBa
For r > a I2 = I 0.
B dl I
2 0322aB I
02 3
I
Ba
0
1
02
6
3
IB a
IBa
12
17. Answer (4) Hint: magnetic field inside the current carrying
toroid 02
NIBr
.
Sol.: 7
24 10 3000 10
2 25 10B
B = 24 mT.
18. Answer (3)
Sol.: 2
qBf
m
19
271.6 10 1.4
2 3.14 1.67 10
= 2.2 × 107 s–1 19. Answer (2) Hint: Use Ampere circuital law
Sol.: 0 enc.B dI I
enc 4 1I = 3 A
0. 3
B dI .
20. Answer (4) Hint: Work done by magnetic field be zero. Sol.: By work energy theorem
net W K
22 21 6 82E BW W m
Q x0 E0 = 50m
00
50mxqE
00
50xE
.
21. Answer (3) Hint and sol.: The direction of magnetic field lines
of force produced by passing a direct current in conductor is given by Right hand palm rule.
22. Answer (2) Hint: Magnetic field due to bar magnetic at its axial
position is 03
24
Md
and on its equatorial position
034
Md
.
Sol.:
Test-2 (Code-A)_(Hints & Solutions) All India Aakash Test Series for NEET-2021
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456 5/16
|B1| = |B3| = 2|B2|
|B2| = 102
= 5 T
|B2| = 10 T 23. Answer (1) Hint: B = 0(H + I) Sol.:
Intensity of magnetisation Magnetic moment Volume
66
30 10
I
6
510 2 105
I
7 550004 10 2 104
B
B = 5 × 10–4 + 8 × 10–2 = 0.2517 T. 24. Answer (1)
Hint: 2H
ITMB
Sol.: 2H
ITMB
0 2H
ITMB
2 IT
MB
cos
HBB
cos2H
ITMB
122 cos
H
ITMB
120 cosT T
2 20 cos T T .
25. Answer (3) Hint & sol.: r = (1 + m)
= (1 + 0.725) = 1.725. 26. Answer (3) Hint and sol.: A magnet attract ferromagnetic
substance strongly and paramagnetic substance weakly. But it repel diamagnetic substance.
27. Answer (3) Hint: Resultant magnetic moment will be equal to
vector sum of magnetic moment of all the magnets.
Sol.:
Net ˆ ˆ ˆ ˆ2 cos 60 3 cos 60 2 sin60 M Mi M i M i M j
ˆ3 sin 60 M j
net3ˆ ˆ ˆsin602
MM M M i i M j
net3 3ˆ ˆ2 2
M MM i j
net| | 9 32
MM
= 3 M .
28. Answer (2) Hint and sol. When all the molecule in a magnet arrange
themselves in the direction of magnetic field the condition is called saturation.
29. Answer (1) Hint: Use work energy theorem. Sol.: Wnet = k –U = k –(Uf – Ui) = kf – ki –(–MB cos)= kf kf = 80 J.
All India Aakash Test Series for NEET-2021 Test-2 (Code-A)_(Hints & Solutions)
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456 6/16
30. Answer (3) Hint: In stable equilibrium position angle between
net magnetic moment and magnetic field will be zero.
Sol.:
3tan MM
= 60. 31. Answer (1)
Hint : At a position where angle of dip is . The horizontal component of earth’s magnetic field BH = Bcos and vertical component of earth’s magnetic field BV = Bsin.
Sol. :
sin30cos60
v A
H B
B BB B
= 1 : 1.
32. Answer (3)
Hint : Rate of heat dissipation dHdt
= I2R = 2E
R.
Sol.: E = Bvl
2 2 2dH B v
dt R
2dH vdt
2 14dH dHdt dt
33. Answer (3)
Hint: ddt
.
Sol.: = BAcos
= Br2
2d B rdt
2 drB rdt
4 Ba
34. Answer (1)
Hint: . B A
Sol.: 4 4ˆ ˆ ˆ20 9 10 .10 10 i k k
= –90 × 10–8 = –900 ×10–9 || = 900 nWb
35. Answer (1)
Hint: . B A .
Sol.: Magnetic field produced by coil having current will be perpendicular to area vector of other coil = BAcos90°
= 0. Change in flux in other coil will be zero therefore
no current will flow in other coil. 36. Answer (3)
Hint: sPdIM
dt
Sol.: 10( 0)2000 mI
t
Im = 200 × 10–3 Im = 0.2 A. 37. Answer (2) Hint: Heat developed in R1 will be equal to energy
stored in the inductor
Sol.: 212
H U LI
2
EIR
2
22
12
LEHR
.
38. Answer (1)
Hint: . effv B l
Sol.: leff = 2lcos45
Test-2 (Code-A)_(Hints & Solutions) All India Aakash Test Series for NEET-2021
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456 7/16
= 2l 2Blv By right hand palm rule P will be at higher
potential. 39. Answer (1) Hint and sol.: By Lenz’s law induce current
produced in ring oppose its cause. 40. Answer (3) Hint and Sol.: The mutual inductance of the pair
of coil depends upon position and orientation of the coil.
41. Answer (2) Hint.: Rotational emf. induced in a rod of length l
inside the uniform perpendicular magnetic field B
rotating about one end is 2
2B l
.
Sol. :
22(i)
2O Cl B
V V
= 2l2B
2
(ii)2O Al BV V
From (i) and (ii)
2
222
A C
l BV V l B
23
2
l B .
42. Answer (4) Hint : Use KVL Sol. : Case–I
a bdIV V L IRdt
12 = L + 3R …(i) Case–II
a bdIV V L IRdt
3 2 (ii) L R 15 = 5R R = 3
3 = – L + 6 L = 3 H. 43. Answer (3) Hint: Voltage across the inductor will be equal to
voltage across the 10 resistor. Sol.: At t = 0 current in the inductor will be zero but
rate of change of current will be maximum
315
I
1 A5
I
dIL I Rdt
1 105
dILdt
2 A/s5
dIdt
44. Answer (2)
Hint: ddt
Sol.: For 0 < t < 2T , d
dtis constant and negative,
ddt
which will be positive constant.
For 2T < t < T d
dt is constant and positive,
ddt
which will be negative constant.
45. Answer (4)
Hint: andd idt R
Sol.: 24 12 7t t
8 12d tdt
I will be zero, when ddt will be zero
Hence 8 12 0d tdt
12 1.5 s8
t
All India Aakash Test Series for NEET-2021 Test-2 (Code-A)_(Hints & Solutions)
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456 8/16
[CHEMISTRY]
46. Answer (3) Hint.: Chemisorption is specific in nature Sol.: Physical adsorption is not specific in nature. 47. Answer (3) Hint: Adsorption is exothermic in nature Sol. Higher is the temperature, less will be the
adsorption 48. Answer (2) Hint: Gas of highest critical temperature will be
adsorbed most easily Sol. CO2(g) has highest critical temperature
among the given gases. So, it is most easily adsorbed.
49. Answer (3) Hint: When physical state of catalyst is different
from physical state of reactants and products then it is an example of heterogeneous catalysis.
Sol.: 4NH3(g) + 5O2(g) Pt(s) 4NO(g) + 6 H2O(g) is an example of heterogeneous catalysis
50. Answer (1)
Hint: Sucrose Invertase Glucose + fructose 51. Answer (3) Hint: Paint is an example of sol Sol.: Paints contain solid as dispersed phase and
liquid as dispersion medium 52. Answer (1) Hint: For soaps the CMC value is 10–4 to 10–3 mol
L–1 53. Answer (4) Hint: CdS and charcoal are negatively charged sol Sol.: Al2O3 xH2O is a positively charged sol 54. Answer (4) Hint: The emulsifying agent forms an interfacial
film between suspended particles and the medium.
Sol.: The principal emulsifying agents for oil in water emulsion are proteins, gums, natural and synthetic soaps etc.
55. Answer (3) Hint: mmol of electrolyte required for coagulation
of 1000 mL of sol is called coagulation value. Sol. For 20 mL, mmol of NaCl required = 1 Hence, for 1000 ml, mmol of NaCl required = 50
56. Answer (4) Hint: Lyophilic sols are more stable than lyophobic
sols Sol. Smaller is the size and lesser the viscosity,
faster is the Brownian movement 57. Answer (4) Hint: Boiling, persistant dialysis and mixing of two
oppositely charged sols resulted in coagulation Sol. Peptization is the process of converting a
precipitate into colloidal sol by shaking it with dispersion medium in the presence of a small amount of electrolyte.
58. Answer (4) Hint.: Magnetite is Fe3O4 59. Answer (4) Hint: Removal of unwanted material from the ore
is known as concentration of an ore Sol.: Calcination is not a method of concentration. 60. Answer (1) Hint: Roasting is carried out for sulphides ores to
convert into oxide Sol.: 2ZnS + 3O2 2ZnO + 2SO2 61. Answer (3)
Hint: Reaction in which metal oxide on reaction with metal sulphide forms metal is self reduction
62. Answer (1) Hint: Vapour phase refining is used for refining of
Nickel.
Sol: Ni + 4CO 330 – 350K Ni(CO)4
Ni(CO)4 450– 470K Ni + 4CO
63. Answer (4) Hint: Silver and gold is concentrated by leaching
using aqueous NaCN solution. Sol.:
64. Answer (4) Hint: Zone refining method is very useful for
producing semiconductor and other metals of very high purity
Test-2 (Code-A)_(Hints & Solutions) All India Aakash Test Series for NEET-2021
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456 9/16
65. Answer (2) Hint: Depressant, selectively prevents ZnS from
coming to the froth but allows PbS to come with the froth.
Sol.: NaCN is used to convert ZnS into water soluble complex Na2[Zn(CN)4]
66. Answer (4) Hint: Bauxite ore contains impurities of SiO2, Iron
oxides and titanium oxide (TiO2) 67. Answer (2) Hint: Ellingham diagram is based only on the
thermodynamics concepts. Sol.: Ellingham diagram does not say about the
kinetics of the reduction processes. 68. Answer (3) Hint: In lower temperature range (500 K – 800 K) in
blast furnace, Fe2O3 is mainly reduced by CO. Sol.: At 900–1500 K (higher temperature range in
the blast furnace) reactions are C + CO2 2CO FeO + CO Fe + CO2 69. Answer (4) Hint: Cu2S : copper glance 70. Answer (4) Hint: NaNO3 is called as chile saltpetre Sol.: KNO3 is known as Indian saltpetre 71. Answer (2) Hint: AgCl(s) is white compound Sol.:
Zn(OH)2(s) is white compound Fe2O3 · xH2O(s) is brown compound [Cu(NH3)4]2+ is deep blue compound 72. Answer (2) Hint: NO is neutral oxide N2O3, N2O4 and N2O5 are acidic in nature Sol.: N2O3 is blue coloured solid which is acidic in
nature 73. Answer (1)
Hint: Hypophosphorous acid is also known as phosphinic acid Sol.: Phosphinic acid is H3PO2
74. Answer (4) Hint: In solid state, PCl5 exists as an ionic solid [PCl4]+ [PCl6]– Sol.: The hybridization of P in [PCl4]+ is sp3
75. Answer (4) Hint: Species with central atom having 1 lone pair
and 4 bond pair of electrons acquire see saw shape
Sol.: SF6(g) has octahedral shape. 76. Answer (4) Hint: Generally larger is the bond length, smaller
is the bond enthalpy Sol.: I2 has least bond enthalpy among the
halogen molecules. 77. Answer (1) Hint: Weakest acid has highest pKa value. Sol.: Order of pKa; HF > HCl > HBr > HI 78. Answer (4) Hint: 3Cu + 8HNO3(dilute) 3Cu(NO3)2 + 2NO + 4H2O 79. Answer (1) Hint: Ionic character of metal halides decreases
as the size of halide increases Sol.: Ionic Character MF > MCl > MBr > MI (Where M is a monovalent metal) 80. Answer (3) Hint: Compound containing hydrogen on reaction
with Cl2, forms HCl. Sol. 2 Ca(OH)2 + 2 Cl2 Ca(OCl)2 + CaCl2
+ 2H2O 81. Answer (2) Hint: Halous acid of only chlorine exist. Sol.: HOBrO is least likely to exist 82. Answer (4) Hint: 6XeF4 + 12H2O 4Xe + 2XeO3 + 24HF + 3O2
83. Answer (1) Hint: Molecule with 1 lone pair and 6 bond pairs of
electrons around central atom has distorted octahedral shape.
Sol.
XeF6 :
All India Aakash Test Series for NEET-2021 Test-2 (Code-A)_(Hints & Solutions)
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456 10/16
84. Answer (3) Hint: XeF2 and XeF4 are fluorinating reagent. Sol.: 143K
4 2 2 6 2XeF + O F XeF + O
2Xe:F = 1: 202 6573 K, 60 – 70 barXe(g) + 3F (g) XeF (s)
85. Answer (4) Hint: Depending upon the relative excess amount
of NH3 and Cl2, different products are formed. Sol.: NH3 + 3Cl2 (excess) NCl3 + 3HCl 8 NH3 (excess) + 3Cl2 6NH4Cl + N2 86. Answer (3) Hint: Catalyst used in Deacon’s process is CuCl2 Sol. Catalyst used in contact process is V2O5
Catalyst used in Haber’s process is Fe2O3 with K2O and Al2O3. Catalyst used in ostwald’s process is platinized asbestos.
87. Answer (4) Hint: Due to presence of H-bonding, Melting point
of NH3 is exceptionally high Sol.: Order of melting point of NH3 > SbH3 > AsH3 > PH3
88. Answer (4) Hint: The bond enthalpy of N2 is high Sol.: 3 22 NaN 2 Na + 3N (pure)
At high temperature, with some metals it predominantly form ionic nitrides e.g.
2 36Li + N 2Li N
2 3 23Mg N Mg N
89. Answer (4) Hint: H2SO4 gives charring action with
carbohydrates Sol.: Ca3P2 + 6H2O 2PH3(g) + 3Ca(OH)2
(A)
2PH3(g) + 3CuSO4 Cu3P2 + 3H2SO4 (B) (C)
PH3 is a colorless gas with rotten fish smell. It is slightly soluble in water. It explodes in contact of trace amount of HNO3.
90. Answer (3) Hint: In all oxoacids of phosphorous one P = O
bond is present for every phosphorous atom. Sol.: Formula Structure
Phosphonic Acid H3PO3
Orthophosphoric Acid H3PO4
Hypophosphoric acid H4P2O6
Pyrophosphoric acid H4P2O7
[BIOLOGY]91. Answer (1) Hint : Recessive traits are expressed in
homozygous condition only
Sol. : Terminal flower position, constricted pod
– Recessive traits
Yellow seeds, Violet flower
– Dominant traits
92. Answer (4) Hint : Yellow pod is recessive trait of pea plants. Sol. : To produce recessive feature in F1
generation, both the parents must have at least a recessive allele.
Hence YY × yy
Yy all green pods Will not have any progeny with yellow pods. 93. Answer (1) Hint : Formula to calculate number of different
types of gametes produced is = 2n. Sol. : Number of heterozygous locus ‘n’ = 3. Hence 2n = 23 = 8 types of gametes. 94. Answer (2) Hint : Flower colour in snapdragon plant is an
example of incomplete dominance.
Test-2 (Code-A)_(Hints & Solutions) All India Aakash Test Series for NEET-2021
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456 11/16
Sol. : In the given cross
F2 phenotypic ratio is 1 : 2 : 1 95. Answer (3) Hint : To have child with O blood group both the
parent must have one IO allele. Sol. :
96. Answer (1) Hint : A modified allele can produce normal or less
efficient enzyme. Sol. : If modified allele is produced due to silent
mutation it will have same phenotype and it can be a dominant allele.
97. Answer (4) Sol. : Punnett square is the graphical
representation of possible genotypes of offsprings of a genetic cross given by R. C. Punnett.
98. Answer (2) Hint : In Mendelian dihybrid cross 8 offsprings
have genotype heterozygous for single trait.
Sol. : Total number of offsprings heterozygous for both the traits = 4.
99. Answer (4) Hint : In test cross an individual with recessive
phenotype is crossed with individual with dominant phenotype.
Sol. : In 4 O’ clock plant, flower colour show incomplete dominance. So for this plant, if plant is showing red coloured flowers it means plant is homozygous dominant for red colour.
100. Answer (2) Sol. : Term linkage was coined by T. H. Morgan. 101. Answer (2) Hint : Pleiotropic gene controls multiple
phenotypes. Sol. : In pea starch synthesis gene shows
incomplete dominance. 102. Answer (4) Hint : Kernel colour in wheat show polygenic
inheritance. Sol. : Intermediate red kerneled wheat = AaBb White kerneled wheat = aabb AaBb × aabb
ab
AB AaBb Intermediate red wheat
Ab Aabb Light red wheat
aB aaBb Light red wheat
ab aabb White kernel wheat
Phenotypic ratio = 1 : 2 : 1 Genotypic ratio = 1 : 1 : 1 : 1 Thus only statement (ii) is correct 103. Answer (1) Sol. : Chromosomal theory of inheritance was
given by Sutton and Boveri. 104. Answer (3) Hint : An experimental material must be easily
grown and have short life span. Sol. : Both Drosophila and garden pea have short
life span. But only Drosophila can be grown in laboratory
and have distinguishable opposite sexes. 105. Answer (3) Hint : Recombination frequency (RF) is directly
proportional to distance between genes.
All India Aakash Test Series for NEET-2021 Test-2 (Code-A)_(Hints & Solutions)
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Sol. : Recombination or crossing over occurs between homologous chromosomes. Higher RF value indicates that genes are distantly situated on a chromosome not on separate chromosomes.
106. Answer (4) Sol. : In moth, females produce two types of
gametes i.e. heterogametic. 107. Answer (4) Hint : Infra red radiations due to high wavelength
have very low energy. Sol. : Being low energy radiations, infra red
radiations are not capable to directly damage the DNA.
108. Answer (3) Hint : Male cannot be carrier for X-linked recessive
disorders. Sol. : For haemophilia male can either be
diseased or normal
109. Answer (1) Sol. : Myotonic dystrophy is an autosomal
dominant disorders. 110. Answer (3) Hint : Colour blindness is X-linked recessive
disorder. Sol. : If a female is suffering from Colour blindness
then she must have genotype 44 + XCXC. This female can pass Colour blindness gene to her daughters as well as sons. Female ‘X’ must have received Colour blindness gene from both of her parents.
111. Answer (1) Hint : In sickle cell anaemia HbS peptide have
valine in place of glutamic acid. Sol. : m-RNA HbA peptide – 5 – GAG – 3 HbS peptide – 5 – GUG – 3 112. Answer (2) Hint : Phenylketonuria gene controls multiple
phenotypic expressions. Sol. : Gene responsible for phenylketonuria is a
pleiotropic gene. 113. Answer (1) Sol. : Four phenotypes A, B, AB and O are
possible for blood group in human population. 114. Answer (2) Sol. : Turner female have 44 + XO genetic
complement.
115. Answer (4) Hint : Cytidine is a nucleoside of RNA molecule. Sol. : A cytidine contains a ribose sugar a
pyrimidine nitrogenous base (6 membered) and N-glycosidic bond.
But since it’s a nucleoside it lacks phosphoester bond.
116. Answer (3) Sol. : Guanine is a purine base, rest three are
pyrimidine nitrogenous bases. 117. Answer (2) Hint : Watson and Crick were the first to propose
the idea of base complimentarity in DNA. Sol. : Watson and Crick were the first to propose
the base pairing between two strands of DNA. 118. Answer (4) Sol. : DNA packaging in prokaryotes involves non-
histone basic protein polyamines. 119. Answer (1) Hint : Reverse transcriptase catalyse the
synthesis of DNA over RNA. Sol. : Reverse transcriptase is found in retro
viruses and is RNA dependent DNA polymerase. It follows principle of complementarity. Viruses have very few enzymes such as reverse transcriptase, neuraminidase etc.
120. Answer (3) Sol. : Hershey and Chase proved that DNA act as
genetic material. 121. Answer (3) Hint : RNA transmits genetic information from
DNA to protein. Sol. : RNA is better material for transmission of
genetic information than DNA. 122. Answer (4) Hint : DNA polymerase has both exonuclease and
polymerase activity. Sol. : DNA polymerase is capable to synthesise
DNA as well as remove RNA primer. Joining of Okazaki fragments is a function of ligase enzyme.
123. Answer (1) Hint : m-RNA have bases sequences
complimentary to the non-coding DNA strand. Sol. : m-RNA transcribed from the given DNA
strand will be.
5th codon is stop codon i.e. UGA, hence only four
amino acids will be coded by this m-RNA.
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124. Answer (3) Hint : Translation occurs inside cytoplasm. Sol. : Charging of t-RNA a step of translation occur
inside cytoplasm. 125. Answer (3) Hint : Most of the rRNA are synthesised by RNA
polymerase I. Sol. : RNA polymerase I : 5.8 S, 18 S, 28 S rRNA RNA polymerase III : 5 S rRNA RNA polymerase II does not catalyse synthesis of
any rRNA. 126. Answer (4) Hint : VNTR is polymorphic DNA. Sol. : VNTR are/have Size 0.1 to 20kb Found in eukaryotes Non expressed part of DNA 127. Answer (3) Sol. : Semiconservative mode of DNA replication
was experimentally verified by Meselson and Stahl using heavy nitrogen i.e. N15.
128. Answer (3) Hint : Snurp catalyse splicing process. Sol. : Splicing i.e. removal of introns is not found
in prokaryotes. 129. Answer (2) Hint : is the initiation factor of transcription. Sol. : (2) is holoenzyme. Only 2 is
considered as core enzyme. 130. Answer (1) Hint : There are no t-RNAs for stop codons. Sol. : UAG is a stop codon for which there is no t-
RNA. 131. Answer (1) Hint : -galactosidase hydrolyses lactose into
glucose and galactose. Sol. : The gene lac z codes for enzymes
-galactosidase. 132. Deleted 133. Answer (3) Hint : Repeated sequences are non-coding
sequences. Sol. : A very large portion of human genome
contain non coding sequences or repetitive DNA.
134. Answer (1) Sol. : Chromosome 1 has the most genes i.e.
2968. 135. Answer (2) Hint : UTRs are found towards 5 and 3 end. Sol. : UTRs are found toward 5 end before start
codon and 3 end after stop codon in m-RNA. 136. Answer (4) Hint : Identify a bacterial disease. Sol. : Gonorrhoea is caused by bacteria Neisseria
gonorrhoeae. Genital herpes – Herpes simplex virus Genital warts – Human papilloma virus Hepatitis B – Hepatitis B virus 137. Answer (2) Hint : It is called fertile period. Sol. : During periodic abstinence, couples abstain
from coitus from day 10-17 of menstrual cycle, when ovulation is expected.
138. Answer (1) Hint : Identify a copper releasing IUD. Sol. : Cu ions present in copper releasing IUDs
suppress sperm motility and fertilizing capacity of sperms. Progestasert and LNG-20 are hormone releasing IUDs.
139. Answer (3) Hint : Identify the hormone released by growing
follicles. Sol. : Progestogens prevent ovulation and
implantation and alter the quality of cervical mucus which does not let sperms meet the ovum. Estrogen alone is not effective to carry out these functions.
140. Answer (2) Hint : Identify a technique. Sol. : Infertile childless couples could be assisted
to have children through special techniques known as assisted reproductive technologies (ART). A popular method is test tube baby programme.
MTP – Medical Termination of Pregnancy RCH – Reproductive and Child Health Care STI – Sexually Transmitted Infection 141. Answer (4) Hint : Amniotic fluid contains foetal cells. Sol. : In amniocentesis, amniotic fluid around the
developing fetus is taken to analyse fetal cells to check for genetic disorders.
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142. Answer (4) Hint : Not a goal of RCH. Sol. : Unprotected sexual co-habitation will lead to
more pregnancies, hence it encourages population growth.
143. Answer (3) Hint : Natural methods involve lactational
amenorrhea. Sol. : Natural methods of contraception work on
the principle of avoiding chances of meeting between ova and sperms. IUDs are one of the most widely used contraceptives in India.
144. Answer (3) Hint : Ovum donor is required. Sol. : Gamete Intra Fallopian Transfer involves
transfer of an ovum collected from a donor into the fallopian tube of another female who cannot produce one, but can provide suitable environment for fertilisation and further development.
145. Answer (3) Hint : MTP Sol. : Complications of reproductive tract
infections, if not treated early include pelvic inflammatory disease, abortions, still births, ectopic pregnancies, infertility or even cancer of reproductive tract. Haemophilia is a genetic disease and is not related with RTIs.
146. Answer (3) Hint : Viral STIs are generally incurable. Sol. : Hepatitis B, HIV and genital herpes are
incurable. Family planning programmes in India were launched at national level to attain total reproductive health as social goal.
147. Answer (4) Hint : Non-steroidal pill. Sol. : ‘Saheli’ the new oral contraceptive for
females contains a non-steroidal preparation – centchroman. It was developed by CDRI, Lucknow. It’s a once a week pill with very few side effects and high contraceptive value.
148. Answer (2) Hint : Chemical contraceptives include foams,
jellies etc. Sol. : Femidoms, Nirodh and chemical
contraceptives can be placed by the user itself. IUDs and implants are inserted by doctors or expert nurses. Injections are also administered by healthcare providers.
149. Answer (3) Hint : First trimester. Sol. : MTP is considered relatively safe during the
first trimester i.e. upto 12 weeks of pregnancy, second trimester abortions are much more riskier.
150. Answer (4) Hint : Zygote is single celled. Sol. : The zygote or early embryos formed after in-
vitro fertilisation are transferred into fallopian tube (ZIFT) and embryos with more than 8 blastomeres into the uterus (IUT), to complete its further development.
151. Answer (3) Hint : Vasectomy and tubectomy Sol. : Surgical methods also called sterilisation are
generally considered as a terminal methods to prevent any more pregnancies. These techniques are highly effective but their reversibility is very poor.
152. Answer (2) Hint : 2 percent. Sol. : According to 2011 census report, the
population growth rate was less than 2 percent i.e., 20/1000/year, a rate at which our population could increase rapidly.
153. Answer (2) Hint : Reducing atmosphere. Sol. : The early atmosphere was reducing in
nature containing CH4, NH3, H2, etc. 154. Answer (2) Hint : Swan neck flask experiment. Sol. : Louis Pasteur disproved the theory of
spontaneous generation which stated that life arose from decaying and rotting matter like straw, mud etc.
155. Answer (1) Hint : Selection by nature. Sol. : According to Darwin, fitness is the end result
of the ability to adapt and get selected by nature. 156. Answer (4) Hint : Hypothesis on chemical origin of first life. Sol. : Experimental evidence of chemical evolution
was given by Urey and Miller. This hypothesis was given by Oparin and Haldane who proposed that first form of life could have come from pre-existing non-living organic molecules.
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157. Answer (2) Hint : Biogenesis. Sol. : Water vapour condensed to form water
bodies on earth. Thus, first forms of life were probably unicellular and originated in water environment only.
158. Answer (3) Hint : They evolved oxygen. Sol. : First mammals were like shrews. Plants
invaded land first and released oxygen which was followed by arrival of animals.
159. Answer (3) Hint : Reproductive fitness. Sol. : The pre-existing advantageous mutations
when selected by nature lead to appearance of new phenotypes. This leads to speciation. Variations occurring due to various reasons result in changed frequency of genes and alleles in future generation.
160. Answer (2) Hint : Its height was 20 feet. Sol. : Dinosaurs were land reptiles.
Tyrannosaurus rex was about 20 feet in height and had huge fearsome dagger like teeth.
161. Answer (1) Hint : Ichthyosaurs. Sol. : Most dinosaurs probably evolved into birds.
Coelacanth which was earlier thought to be extinct was the ancestor of modern day frogs and salamanders.
162. Answer (3) Hint : Convergent evolution. Sol. : The given example shows convergent
evolution between placental mammals and Australian marsupials. Spotted cuscus (Australian marsupial) and Lemur (placental mammal) show convergent evolution.
163. Answer (2) Hint : Abrupt evolutionary change. Sol. : Hugo de Vries worked on evening primrose
and brought forth the idea of mutation – large differences arising suddenly in a population. According to him, mutations cause evolution.
Concept of branching descent was given by Darwin. Genetic drift involves a sudden changes.
164. Answer (3) Hint : Directional selection. Sol. : Selection of one type of moth (melanised)
against non-melanised moth due to camouflage represents selection of one type of species while the other is eliminated. Thus it represents directional selection
165. Answer (2) Hint : Modification of axillary branch. Sol. : Divergent evolution is based on homology.
Thorns of Bougainvillea and tendrils of Cucurbita are modification of axillary branches.
Tendrils of pea (Pisum) and spines of Cactus are modified leaves.
166. Answer (2) Hint : Homology indicates common ancestry. Sol. : When the same structure develops along
different direction due to adaptations to different needs, it results in divergent evolution and these structures are homologous.
167. Answer (3) Hint : Common ancestor of cabbage and broccoli. Sol. : Artificial selection of different parts of wild
mustard was done to make broccoli, kohlrabi, cauliflower etc.
Biston carbonaria was selected due to camouflage during industrial melanism, Darwin’s finches exemplify adaptation according to needs.
168. Answer (1) Hint : Sauropsids evolved into thecodonts. Sol. : Crocodiles and birds evolved from
thecodonts. Turtles, Lizards, Snakes and Tuataras evolved
from sauropsids. 169. Answer (2) Hint : Variations are small. Sol. : According to Darwin, evolution is not a
directionless process. It is a gradual process based on chance events in nature. According to Hugo de Vries, mutations cause evolution which are sudden and directionless.
170. Answer (4) Hint : Adaptive convergence. Sol. : Flying phalanger shows adaptive
convergence with flying squirrel. Wombat, Bandicoot and sugar glider exhibit adaptive radiation.
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171. Answer (4)
Hint : Adaptive radiation.
Sol. : Darwin’s finches evolved due to adaptive radiation on Galapagos island. Others are examples of evolution by anthropogenic action.
172. Answer (4)
Hint : Heterozygotes are 2 pq.
Sol. : A population in Hardy Weinberg equilibrium is represented by binomial expansion of (p + q)2 i.e. p2 + 2pq + q2 where p2 represents homozygous dominant, 2pq represents heterozygous and q2 represents homozygous recessive individuals.
173. Answer (4)
Hint : Embryology.
Sol. : Ernst von Baer disapproved the theory of Ernst Haeckel which stated that certain features during embryonic stage are common to all vertebrates, that may be absent in adult. Reproductive fitness was a concept given by Darwin.
174. Answer (2)
Hint : Connecting link between ape and man.
Sol. : Australopithecines probably lived in East-African grasslands and evidence suggests that they ate fruits. Hominids were first human like beings. Chimpanzees were apes.
175. Answer (2)
Hint : Hominids
Sol. : Cranial capacities.
Homo erectus – 900 cc
Homo habilis – 650-800 cc
Neanderthal man – 1400 cc
Cro magnon man – 1650 cc
176. Answer (3) Hint : Raisen district Sol. : Cave paintings by pre historic humans can
be seen at Bhimbetka rock shelter in Raisen district of Madhya Pradesh.
177. Answer (3) Hint : Bryophytes lack vessels. Sol. : Bryophytes evolved from chlorophyte
ancestors during Paleozoic era. Tracheophyte ancestors gave rise to Zosterophyllum in Silurian period.
178. Answer (2) Hint : Convergent evolution. Sol. : Marsupials in Australia have eutherian look
alikes with superficially similar adaptation. Eutherians complete their embryonic development in uterus whereas marsupials are born in under develop form and complete their development in an external pouch. Superficial similar adaptations are due to convergent evolution. Others are examples of divergent evolution.
179. Answer (4) Hint : Unit of evolution. Sol. : Micro evolution is the evolution on smallest
scale that occurs due to change in allele frequencies in a population over generation. Genetic variations refer to differences among individuals in the composition of their genes or other DNA sequences.
180. Answer (3) Hint : Chance event. Sol. : Genetic drift operates mostly on small
populations. It is a chance event that can cause allele frequencies to fluctuate unpredictably from one generation to next.