All India Aakash Test Series for NEET-2020 TEST - 2 - Code-A · All India Aakash Test Series for NEET-2020 Test – 2 (Code-A)_(Answers) Aakash Educational Services Limited - Regd

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)


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


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



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



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



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



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.



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



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.



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


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.



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.



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.



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.



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%



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.



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


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.



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.



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)


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

All India Aakash Test Series for NEET-2020 Test - 2 (Code-B)_(Hints & Solutions)


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

Test - 2 (Code-B)_(Hints & Solutions) All India Aakash Test Series for NEET-2020


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



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



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



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



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)



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 :



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.



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.



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.



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.



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.



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


Sol. : Terminal position of flowers and wrinkled

seed shape in pea plant are recessive traits.

So, they appear phenotypically only in


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.



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.



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


Chimaera : Connecting link between

cartilaginous and bony fishes.

Peripatus : Connecting link between


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.



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.



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.

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