All India Aakash Test Series for NEET - 2019 TEST - 3 (Code-A)...All India Aakash Test Series for NEET-2019 Test - 3 (Code-A) (Hints and Solutions) 6/20 21. Answer (4) Hint: Photoelectric

Test - 3 (Code-A) (Answers) All India Aakash Test Series for NEET-2019

1/20

1. (4)

2. (4)

3. (2)

4. (3)

5. (2)

6. (3)

7. (2)

8. (1)

9. (2)

10. (1)

11. (3)

12. (1)

13. (3)

14. (2)

15. (3)

16. (4)

17. (1)

18. (2)

19. (1)

20. (2)

21. (4)

22. (1)

23. (3)

24. (2)

25. (3)

26. (4)

27. (1)

28. (4)

29. (1)

30. (3)

31. (2)

32. (2)

33. (3)

34. (1)

35. (2)

36. (1)

Test Date : 09/12/2018

ANSWERS

TEST - 3 (Code-A)

All India Aakash Test Series for NEET - 2019

37. (3)

38. (4)

39. (2)

40. (4)

41. (3)

42. (3)

43. (1)

44. (2)

45. (2)

46. (2)

47. (4)

48. (2)

49. (2)

50. (3)

51. (4)

52. (3)

53. (3)

54. (2)

55. (2)

56. (3)

57. (3)

58. (2)

59. (2)

60. (3)

61. (2)

62. (3)

63. (3)

64. (2)

65. (2)

66. (4)

67. (1)

68. (4)

69. (2)

70. (2)

71. (4)

72. (3)

73. (3)

74. (4)

75. (3)

76. (4)

77. (3)

78. (2)

79. (3)

80. (4)

81. (2)

82. (1)

83. (1)

84. (1)

85. (4)

86. (3)

87. (1)

88. (2)

89. (2)

90. (4)

91. (3)

92. (3)

93. (2)

94. (1)

95. (1)

96. (4)

97. (4)

98. (1)

99. (2)

100. (3)

101. (3)

102. (4)

103. (3)

104. (2)

105. (4)

106. (1)

107. (2)

108. (2)

109. (1)

110. (3)

111. (3)

112. (4)

113. (4)

114. (2)

115. (1)

116. (4)

117. (2)

118. (1)

119. (4)

120. (3)

121. (2)

122. (3)

123. (3)

124. (2)

125. (1)

126. (2)

127. (2)

128. (2)

129. (4)

130. (3)

131. (4)

132. (1)

133. (4)

134. (3)

135. (4)

136. (2)

137. (4)

138. (3)

139. (2)

140. (2)

141. (4)

142. (4)

143. (2)

144. (1)

145. (1)

146 (3)

147. (1)

148. (4)

149. (4)

150. (2)

151. (3)

152. (2)

153. (1)

154. (3)

155. (4)

156. (2)

157. (2)

158. (2)

159. (4)

160. (3)

161. (1)

162. (1)

163. (3)

164. (1)

165. (4)

166. (3)

167. (1)

168. (3)

169. (4)

170. (1)

171. (2)

172. (4)

173. (4)

174. (3)

175. (2)

176. (4)

177. (1)

178. (2)

179. (3)

180. (2)

All India Aakash Test Series for NEET-2019 Test - 3 (Code-A) (Hints and Solutions)

2/20

PHYSICS

1. Answer (4)

Hint: Light may enter from denser medium or from

rarer medium in glass slab.

Solution: Snell’s law 2

1

sin

sin

i

r

Light travels away from normal in rarer medium

(denser to rarer) and bend toward normal in denser

medium (rarer to denser)

2. Answer (4)

Hint: Reflection from convex mirror

Solution: v u f

1 1 1

f fO

I F

v = ?

u = – f

f = + f

v f f

1 1 1

1 2

v f

2

fv

3. Answer (2)

Hint: dreal

= dApp

Solution:

x2

x1

d

1

36 9 cm

2x

2

3 279 13.5 cm

2 3x

d = x1 + x2

= 22.5 cm

4. Answer (3)

Hint: f 2 = O1 O

2

Solution: 1 2

f O O

25 9

225

f 15 cm

5. Answer (2)

Hint: Refraction through curved surface.

Solution: For spherical surface

2 1 2 1

v u R

Put u =

4 3 4

3 3 2 3

22

9

v

HINTS & SOLUTIONS

Test - 3 (Code-A) (Hints and Solutions) All India Aakash Test Series for NEET-2019

3/20

1

3 1 60

22

9

v

3 1 3

2 4v

1 1

2v

v = 2 m

6. Answer (3)

Hint: Using Snell’s law

Solution: 2

1

sin

sin

i

r

37°i = 60°

2

1

37°

From geometry

r = A = 37°

i = 60°

Using Snell’s law 2

1

sin60

sin37

2

3

2

3

5

2

5 3 5

2 3 2 3

2

5

2 3

7. Answer (2)

Hint: For thin lens use 1 1 1

v u f and for mirror use

1 1 1

v u f

Solution:

f = 15 cm

0

30 cm

40

F

f = ?

For rays to be parallel image from lens must be on

focus mirror

For lens 1 1 1

v f u

1 1

15 30

30 cmv

Focal length of mirror f = 40 – 30 = 10 cm

8. Answer (1)

Hint: For longitudinal velocity VI = m2 × V

0.

for lateral velocity VI = m V

0.

Solution:1 1 1

v u f

1 1

20 30

v = 60 cm

602

30

vm

u

(VI)long

= (–2)2 × 8 = 32 cm/s

9. Answer (2)

Hint: L = V0 + u

e and

e

Dm

f

Solution:

O L E

v0

ueu = 2.4 cm

0

f0 = 2 cm f

e = 5 cm


4/20

from 0 0 0

1 1 1

v u f

0

0

1 1 112 cm

2 2.4v

v

⇒

125

2.4

vm

u

As image will form at infinity D

mf

ue = – 5 cm

255

5m

Now magnifying power M = m × m

M = – 5 × 5 = – 25

and length of tube L = | v0 | + | u

e |

= 12 cm + 5 cm

L = 17 cm

Hence M, L = – 25, 17 cm

10. Answer (1)

Hint: 1 2

1 1 11

f R R

⎛ ⎞ ⎜ ⎟

⎝ ⎠

Solution: 1

1 11.5 1

10f

⎛ ⎞ ⎜ ⎟⎝ ⎠

1

1 1

20f

Similarly3

1 1

20f

2

1 8 1 11

7 10 10f

⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

2

1 1 2 1

7 10 35f

1 1 1 1

20 20 35f

1 7 7 4

140f

1 10

140f

| f | = 14 cm

11. Answer (3)

Hint: Contrast means relative intensity

Solution: Intensity of fringes depends on intensity of

slits (sources)

2max 1 2I I I

2min 1 2I I I

12. Answer (1)

Hint: 1

Resolving power

Solution: 1

R.P

1

1

1R

2

2

1R

1

2

R 600

R 450

1

2

R 4

R 3

13. Answer (3)

Hint: I I2

04 cos2

Solution: Here 4I 0 = I

0 (given)

2

0cos

6I I

0

3

4I

03

4

II

14. Answer (2)

Hint: n

n Dy

d


5/20

Solution: 1

1 1

n

n Dy

d

2

2 2

n

n Dy

d

1 2n n

y y

n1

1 = n

2

2

1 2

2 1

800

500

n

n

1

2

8 16 24........

5 10 15

n

n

For minimum value of n1 such that

1 21

8;n n

y y n ⇒

n1 = 8

15. Answer (3)

Hint: 1st minima 1

1

Dy

a

2nd maxima 2

2

5

2

Dy

a

Solution: 1

1

Dy

a

2

2

5

2

Dy

a

Since y1 = y

2

1 25

2

D D

a a

1 2

5

2

1 = 2.5

2

16. Answer (4)

Hint: For polarisation tan p =

Solution: If i = p.

Then reflected ray is fully polarised and refracted ray

is partially polarised.

Also, both reflected and refracted rays are

perpendicular to each other.

17. Answer (1)

Hint: Intensity of unpolarised light used in Malus law

0

2

I.

Solution: I

I20

cos2

0 01616

2 25 50

I II

Fractional intensity transmitted.

0

0 0

16 16

50 50

II

I I

% I = 32%

18. Answer (2)

Hint: 2d sin = n (for crystal diffraction)

Solution: d = 2 × 10–8 m

2 sind

n

max

= 2d for max sin 1

1n

⎧ ⎫⎨ ⎬⎩ ⎭

max

= 2 × 2 × 10–8 m

max

= 4 × 10–8 m

19. Answer (1)

Hint: Property of work function.

Solution: Work function of a material depends on

nature of material.

20. Answer (2)

Hint: ( )E

Ve

Solution: 34 8

7

6.6 10 3 10

4.36 10

hcE

19

19

4.54 102.838 eV

1.6 10

E = 2.84 eV

K.Emax

= E – = (2.84 – 2.34) eV

= 0.5 eV

eV = K.Emax

0.5eVV

e

V = 0.5 V


6/20

21. Answer (4)

Hint: Photoelectric effect

Solution: Photocurrent Intensity

K.E = E – Since K.E does not depend on intensity of incident

light.

22. Answer (1)

Hint: 3

h

mKT (de-Broglie equation)

Solution: 2

He H H

H He He

M T

M T

2 500 5

4 300 6

He

H

2 5

2 6

a

a

2 5

2 6

a

a

a = 3

23. Answer (3)

Hint: p p

m vp

m v

Solution: m = 4 mp

p

p p

v m

v m

4 3 12

1 1

p p

p

v m

v m

vp : v = 12 : 1

24. Answer (2)

Hint: P = number of photon per second × energy of

photon

Solution: P = n E

P

nhc

3 7

34 8

18 10 4.4 10

6.6 10 3 10n

n = 4 × 1016 photons per second.

25. Answer (3)

Hint: nh

mvr2

Solution: for 2nd excited state n = 3

hmv

r

3

2

2 3

3 3 2

2

h h rr

hmv

r

⇒ ⎛ ⎞

⎜ ⎟⎝ ⎠

26. Answer (4)

Hint: Intensity power 2

1.

r

Solution: Let intensity at distance 1 m is I, then no.

of photoelectrons will be n0

0 0

I n

I n

0

0At 0.5 4

2

rr m I I ⇒

0

0 0

4I n

I n

n = 4 n0

27. Answer (1)

Hint : For Paschen.

2 2

2

1 1 1

3v R

n

⎡ ⎤ ⎢ ⎥ ⎢ ⎥⎣ ⎦

For Lyman

2 2

2

1 1 1

1v R

n

⎡ ⎤ ⎢ ⎥ ⎢ ⎥⎣ ⎦

Solutions : 1

1 1

9 9

Rv R

⎡ ⎤ ⎢ ⎥⎣ ⎦

2

11v R R⎡ ⎤ ⎢ ⎥⎣ ⎦

1

2

1

9 9

v R

v R


7/20

28. Answer (4)

Hint : For stable orbit, 2

nhL

Solutions : For n = 8, 8

8 4

2

h hL

For n = 6, 6

6 3

2

h hL

For n = 2, 6

2

2

h hL

29. Answer (1)

Hint: 2

2(13.6) eV

ZE

n

Solution: E

2

1 2

213.6 54.4 eV

1

2

2 2

213.6 13.6 eV

2E

Erequired

= E2 – E

1

= [–13.6– (–54.4)] eV

E = 40.8 eV

30. Answer (3)

Hint: Impact parameter 2 2

0

cot4 2

z eb

E

Solution: given b = 0

2 2

2

0

0 cot4 2

z e

E

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠⎝ ⎠

cot 02

⇒

902

= 180°

31. Answer (2)

Hint: Isotone have same number of neutron.

Solution: Isotone have same number of neutron but

different mass number.

32. Answer (2)

Hint :

1

3R A

Solution :

1

31

(64)R ...(i)

1

32

(125)R ...(ii)

Divide equation (ii) by (i)

2

1

5

4

R

R

2

4.8 5

4R

R2 = 6 fermi

33. Answer (3)

Hint: Sum of atomic number and mass number at

reactant and product remain constant.

Solution: 13 + 0 = 11 + Z

Z = 2

27 + 1 = A + 24

A = 4

particle is alpha.

34. Answer (1)

Hint: S

tv

Solution:3

2500 10

2500t

t = 1000 s

1/2

0.693

T

× 693 = 0.693

11s

1000

As N = N0 e–t

11000

10000

N N e

0N

Ne


8/20

35. Answer (2)

Hint: Activity 1/2

1/2

ln2and

R N T

Solution: At t = 10 h, R 0

2

R

e

Since R = N0.e–t

0

022

tRR e t

e

⇒

2 1

10 5 (hours)–1

Now for 0

1/2 02

Rt T R ⇒

Hence 1/2

25 2

�

�n

T n

1/25 2 hoursT n �

36. Answer (1)

Solution : In common emitter amplifier, the output

signal voltage is 180° out of phase with input signal

voltage.

37. Answer (3)

Hint: Apply KVL in input and output loop.

RC

10 V

10 V

E

RB

B

C

Solution: VBE

= 0 and VCE

= 6 V

RB = 400 k R

C = 2 k

BI

10 025 A

400k

10 6

2mA2k

C

I

IB, I

C 25 A and 2 m A.

38. Answer (4)

Hint : 1

Solution : 0.981

0.98 + 0.98 = 0.98 = 0.02

49

39. Answer (2)

Hint: Vac

= 0.3V

Solution: VAB

= 0.3V

Hence 10 – (– 2) = 0.3 + IR

IR = 11.7

V0 – (– 2) = IR

V0 = 11.7 – 2

= 9.7 V

40. Answer (4)

Hint : out

in

m

Ig

V

Solution : out

in

m

Ig

V

in

m

in

Ig

V

50in

in in

I

R I

3

50

2 10

= 25 × 10–3

225

1010

= 2.5 × 10–2 mho

41. Answer (3)

Hint : Half wave rectifier output current.

Solution :


9/20

vm for diode

210 221 2

10m

v V

As 21 2

70

m

m

F L

vI

R R

3 2

10

3 2 3

5 2 5 2A

42. Answer (3)

Hint: Y A B

Solution: De Morgan’s law

Y A A B A BB

Y A + B OR gate.

43. Answer (1)

Hint: X A B ⇒ for NOR gate and 1

Y X X for NAND gate.

Sol.: X A B for A = 0, B = 1

X = 1

for NAND gate.

1Y X X

for A = 0, B = 1

Y = 1

44. Answer (2)

Hint: Mass action law ne

nh = n

i2

Solution: ne.n

h = n

i2

2

i

e

h

nn

n

en

16 2 32

22 22

(2 10 ) 4 10

4 10 4 10

ne = 1010 m–3

45. Answer (2)

Hint: V2 = 6 V, voltage in parallel remains same.

Solution: V2k = 6 V I2k

63 mA

2

Current across 1 k resistor is 21 6

10 mA1.5

Iz = 10 – 3 = 7 mA

Hence Iz, I

1k, I2k

are 7mA, 10mA, 3mA respectively.

CHEMISTRY

46. Answer (2)

Hint: The presence of an electron withdrawing group

at ortho and para-positions increases the reactivity of

haloarenes towards nucleophilic substitution reaction.

Solution:

Cl

N

O O

Slow step

OH

OH

N

O O

Cl OH

N

O O

Cl OH

N

O O

Cl

OH

N

O O

Cl

Fast

Step

OH

NO2

+ Cl

47. Answer (4)

Hint: Compound which is more acidic than water will

react with aqueous alkali.

Solution:

OH

+ KOH

OK

+ H O2


10/20

48. Answer (2)

Hint: The ease of dehydration is decided by the

stability of carbonium ion and the product form.

Solution:

O

OHH

O

–H

O

Less

stable

O

OH

H

O

–H

O

More

stable

Product is

stabilised

due to

resonance

(i)

(ii)

O

OH

H

O

–H

O

O

OH

H

O

–H

O

(iii)

(iv)

49. Answer (2)

Hint: Hydrolysis of cyanide in acidic medium gives

carboxylic acid.

Solution:

PhCH — Cl2

KCNPhCH — CN

2

A

H O3

PhCH COOH2

BC H OH/H

2 5

PhCH — C — OC H2 2 5

O

(Ester) C

50. Answer (3)

Hint: Draw Fischer projections of different

stereoisomers of tartaric acid.

Solution:

COOH

COOH

H OH

HHO

A

COOH

COOH

HO H

OHH

COOH

COOH

H OH

OHH

Plane of symmetry B C

Opticallyactive

Opticallyactive

Opticallyinactive (meso)

Compounds A and B are enantiomers and each one

is optically active. Compound C is optically inactive

as it contains plane of symmetry within the

molecule.

51. Answer (4)

Hint: Lower the value of pKa, stronger is the acid

Solution:

Compound pKa

COOH

O N2

NO2

2.83

COOH

NO2

3.43

COOH

CH3

3.91

OH

NO2O N

2

NO2

1.02

(Picric acid)

Anion of picric acid is highly stabilised by resonance

of three nitro groups present at ortho and para

positions. Hence, picric acid is most acidic among

all.

52. Answer (3)

Hint: Reduction of ester by DIBAL-H gives aldehyde.

Solution:

CH CH — C — OC H3 2 2 5

DIBAL-HCH CH — C — H

3 2

OO


11/20

53. Answer (3)

Hint: Coupling reaction of phenol

Solution:

NH2

NaNO /HCl

0–5°C

2

N Cl2

PhOH/OHN = N OH

54. Answer (2)

Hint: Hoffmann bromamide degradation of benzamide

Solution:

NH2O

Br

KOH

2

NH2

aniline

55. Answer (2)

Hint: Nitrobenzene is converted to hydrazobenzene

in presence of Zinc dust and aqueous alkali.

Solution:

NO2

Zn/NaOHPh — NH — NH — Ph

56. Answer (3)

Hint: Lone pair is not taking part in resonance in

N

and

N|H

Solution:

Lone pair of N-atom is occupying sp3 hybrid orbital

in

N|H

and sp2 hybrid orbital in

N

57. Answer (3)

Hint: Compound which contains aldehyde group will

reduce Tollen’s reagent.

Solution: Sucrose is made up of -D-glucose and

-D-fructose by 1,2-glycosidic linkage. It does not

contain free aldehyde group. In alkaline medium

fructose converts into glucose and mannose hence

can be reduced by Tollen’s reagent.

58. Answer (2)

Hint: Vitamin ‘C’ is chemically an acid, called

Ascorbic acid.

Solution: Vitamin C is readily soluble in water.

59. Answer (2)

Hint: Monomer of teflon is tetrafluoroethylene

Solution: n CF = CF2 2 CF — CF

2 2

nTeflon

60. Answer (3)

Hint: In condensation polymerisation small

molecules like H2O or ammonia is removed.

Solution:

n CH CHCN2

Addition

Polymerisation — CH — CH —2

CN

n

Acrylonitrile

Orlon 61. Answer (2)

Hint: Phenolic hydrogen is easily abstracted by

aqueous alkali and phenoxide ion is formed.

Solution:

NaOH

OH

OH

O

OH

PH — CH — Br2

O

OH

Ph

62. Answer (3)

Hint: Electrophile formed in Reimer-Tiemann reaction

is :CCl2

Solution:

NaOH

CHCl3

OH ONa

CHCl2

NaOH

ONa

CHO

H

OH

CHO

Intermediate


12/20

63. Answer (3)

Hint: Cleavage of ether using HI

Solution:

OCH3

H

HO — CH3

I

OH

+ CH I3

S 2N

64. Answer (2)

Hint: Wolff-Kishner reduction

Solution:

O

(i) NH — NH2 2

(ii) NaOH +OH OH

65. Answer (2)

Hint: Formation of chiral cyanohydrin compound.

Solution:

Ph

H

OH

CN

Ph

HOH

CN

+

Racemic mixture

H O3

Ph

H

OH

CO H2

Ph

HOH

+

CO H2

Ph — C — H + HCN

O

66. Answer (4)

Hint: Alitame is the sweetest compound

Solution:

Compounds Sweetness value in comparison

to cane sugar

Saccharin 550

Aspartame 100

Sucralose 600

Alitame 2000

67. Answer (1)

Hint: Analgesics reduce pain without causing anydisturbance of nerve system.

Solution:

Drug Therapeutic action

Aspirin Analgesic

Equanil Tranquilizers

Ofloxacin Antibiotic

Salvarsan Antibiotic

68. Answer (4)

Hint: The compound which does not contain plane

of symmetry is optically active.

Solution: A properly substituted allene is chiral but

Me

H

C C C

Me

Me

contains same Me group on

one carbon leading to have POS hence achiral.

Rotation of bond keeps two phenyl ring in different

plane in

Br Me

Me Me

but due to presence of

same Me group at ortho position in phenyl group,

holds POS hence achiral.

CH3

H

H

OH

OH

CH3

POS

CH3

Br

H

H

Br(R)

CH3

(Chiral)

(R)

(Achiral)

69. Answer (2)

Hint: Tertiary alkyl halide undergoes nucleophilic

substitution reaction by SN1 mode.

Solution: Tert-butylchloride is tertiary alkyl halide

hence produces stable carbocation which further

gives substitution product via SN1 mechanism.

70. Answer (2)

Hint: Addition of water to alkene in presence of acid

gives stable rearranged alcohol.

Solution:

CH — CH — CH — CH3 2

CH3

Cl

H OH

CH CH — CH 3

CH2

CH3

H

CH — C — CH — CH3 2 3

CH3

H O2

CH — C — CH — CH3 2 3

OH

CH3

CH — C — CH — CH3 3

H

CH3

H

shift


13/20

71. Answer (4)

Hint: Tertiary amines do not react with Hinsberg’s

reagent.

Solution: Tertiary amine N

will not

react with Hinsberg’s reagent (C6H

5SO

2Cl) due to

absence of H-atom at Nitrogen.

72. Answer (3)

Hint: Carbylamine reaction.

Solution:

CH NH2 2

CHCl3

KOH

CH NC2

73. Answer (3)

Hint: Reaction of Grignard reagent with aldehyde.

Solution:

Cl

Mg

Dry ether

MgCl

A

(i) HCHO

(ii) H O3

CH OH2

B

74. Answer (4)

Hint: Compounds containing keto methyl group or

compounds which on oxidation generate keto methyl

group will give haloform reaction.

Solution:

(a) Ph — C — CH3

O

I2

NaOHPh — C — COO + CHI

3

O

Keto methyl group

(b) CH CH OH3 2

[O]CH CHO

3

I2

NaOHCHI + H — C — O

3

O

(c) CH CH CHCH3 2 3

OH

[O]CH CH — C — CH

3 2 3

O

I

NaOH

2

CHI + CH CH — C — O3 3 2

O

75. Answer (3)

Hint: Structural formula of Glyoxal is

H

O

H

O

Solution: PhCH CH — CH CHPh(i) O

(ii) Zn/H O

3

2

2 PhCHO + CH — CH

O O

76. Answer (4)

Hint: -D-glucose is a cyclic hemiacetal.

Solution:

O

CH OH2

H

HO

H

OH HOH

H

H OH

-D-glucose

In -D-glucose number of asymmetric carbon is 5.

77. Answer (3)

Hint: Cellulose is polysaccharide composed of only

-D-glucose units.

Solution:

Amylopectin is a branched chain polymer of

-D-glucose units.

78. Answer (2)

Hint: ,-unsaturated carbonyl compounds are

formed in aldol condensation reaction.

Solution:

O

H

H

OH

O

+ H O2

O

OOH

OO

H H

O

O

+ H O2


14/20

79. Answer (3)

Hint: Most electrophilic carbonyl carbon will be one,

in which benzene ring is attached with strong

electron withdrawing group at o/p-positions.

Solution:

OCOCH3

NaOH

NO2

OH

NO2

+ CH — C — O3

O

– NO2 is strong electron withdrawing group so the

above ester will be most easily hydrolysed in

alkaline medium.

80. Answer (4)

Hint: Grignard reagent adds to ethylene oxide to give

primary alcohol.

Solution: PhMgBr + OPh O MgBr

H O3

+

PhOH

81. Answer (2)

Hint: Ribose is present in RNA and deoxyribose in

DNA.

Solution: Sugar component in DNA is

-D-2-deoxyribose whereas in RNA molecule, it is

-D-ribose.

82. Answer (1)

Hint: Amino acid containing five membered ring.

Solution: Proline NIH

COOH is a cyclic

amino acid.

83. Answer (1)

Hint: Elastomers involve weakest van der Waals

interactions.

Solution: — CH — C CH — CH —2 2

Cl n

Neoprene is

an elastomer.

84. Answer (1)

Hint: Ziegler-Natta catalyst is triethylaluminium and

titanium tetrachloride.

Solution: It is a mixture of (C2H

5)3Al & TiCl

4.

85. Answer (4)

Hint: Polymer of Ethylene glycol and Phthalic acid.

Solution:

Glyptal — O — CH — CH OOC2 2

CO —

n

is

used in the manufacture of paints.

86. Answer (3)

Hint: Dettol contains chloroxylenol.

Solution: Dettol is a mixture of chloroxylenol and

terpineol.

87. Answer (1)

Hint: Chloramphenicol inhibit the growth of

microorganisms.

Solution: Chloramphenicol is a broad spectrum

antibiotics.

88. Answer (2)

Hint: Phenol formaldehyde resin.

Solution:

OH

+ H CO2

H

or

OH

Bakelite

89. Answer (2)

Hint: Rosenmund reduction.

Solution:

ClO

H2

Pd-BaSO4

CHO

90. Answer (4)

Hint: NaHCO3 reacts with carboxylic and sulphonic

acids or with an acid having acidic strength more

than carboxylic acid.

Solution: Phenol is less acidic than H2CO

3 so does

not react with NaHCO3.

BIOLOGY

91. Answer (3)

Hint: 1974 - Water (Prevention and control of

Pollution) Act.

Sol: Government of India has passed the

Environment (Protection) Act in 1986.

1987 - Montreal Protocol was signed.

1981 - Air (Prevention and Control of pollution) Act.


15/20

92. Answer (3)

Hint: Electrostatic precipitator is most widely used

way of removing particulate matter.

Sol: Electrostatic precipitator can remove over 99%

particulate matter present in the exhaust from

thermal power plant.

According to CPCB, particulate size 2.5 m or less

in diameter causes greatest harm to human health.

93. Answer (2)

Hint: Amrita Devi Bishnoi Wildlife Protection Award

is instituted for individuals or communities from rural

areas.

Sol: Catalytic converters have metals like platinum-

palladium and rhodium as catalyst. Lead in the petrol

inactivates these catalysts.

Reforestation may occur naturally.

94. Answer (1)

Hint: Tropical rainforests have the maximum mean

annual rainfall among all the biomes of India.

Sol: Tropical rainforests are rich in biodiversity

because mean annual rainfall is 2000 - 3500 mm and

mean annual temperature is 23 - 27°C.

95. Answer (1)

Hint: Triangular age pyramid is a graphic

representation of a young population.

Sol: Young population is growing or expanding

population because it has very high proportion of pre-

reproductive individuals as compared to others.

96. Answer (4)

Hint: Stenohalines are those which are restricted to

a narrow range of salinity.

Sol: Eurythermal - organisms can tolerate and thrive

in a wide range of temperature.

Hypersaline lagoons - Salinity > 100 (ppt)

UV - C - Wavelength 100 - 280 nm (0.1 - 0.28 m)

97. Answer (4)

Hint: Decomposition concerns with breakdown of

complex organic matter into inorganic nutrients.

Sol: Mineralisation is a part of catabolism. It is

release of inorganic nutrients from organic matter or

humus during the process of decomposition.

98. Answer (1)

Hint: It is expressed as biomass or numbers of

organisms per unit area.

Sol: Amount of living material present in different

trophic levels at a given time is called standing crop.

GPP - Rate of production of organic matter by

producers.

NPP = GPP – Respiratory loss.

99. Answer (2)

Sol: Term biodiversity was popularised by Edward

Wilson.

100. Answer (3)

Sol: Dodo was found in Mauritius.

101. Answer (3)

Hint: Mammals are able to maintain homeostasis.

Sol: Due to the capacity to maintain constant body

temperature, mammals are successful on earth but

majority of the animals and nearly all plants do not

have this ability.

102. Answer (4)

Hint: CO2 and CH

4 are the major cause of

greenhouse effect.

Sol: CO – 60%2

CH – 20%4

CFCs – 14%

N O – 6%2

Relative contributions to thetotal global warming

103. Answer (3)

Hint: Tropical Amazonian rainforest has the greatest

biodiversity on earth.

Sol: Amazon forest is estimated to produce 20% of

the total O2 of earth’s atmosphere, thus it is called

the lungs of the planet.

104. Answer (2)

Sol: A.G. Tansley coined the term ecosystem and

defined it as sum total of interactions between biotic

and abiotic components.

105. Answer (4)

Hint: Detritus is the substrate of decomposition.

Sol: Dead remains of plants and animals including

faecal matter constitute detritus and it is raw material

for decomposition.

106. Answer (1)

Hint: Sun is the only source of energy in aquatic

food chains.

Sol: Grazing food chain is the major conduit of

energy flow in aquatic ecosystems.

107. Answer (2)

Hint: Heat loss or gain is a function of surface area.

Sol: Mammals from colder climates have shorter

extremities.

Population size is not a static paramater.

Resources for growth of population are never

infinite.


16/20

108. Answer (2)

Hint: Natality and immigration are responsible for

increase in size of population of an area.

Sol: Population density will increase if the number

of births plus number of immigrants (B + I) is more

than the number of deaths plus the number of

emigrants (D + E) otherwise it will decrease.

Thus, t 1 tN N B I D E ⎡ ⎤ ⎣ ⎦

109. Answer (1)

Hint: It is an interaction between two organisms of

different species in which one species is harmed

whereas the other is unaffected.

Sol: Amensalism will be expressed as (–, 0).

110. Answer (3)

Hint: Ecological pyramid assumes a simple food

chain and does not accommodate a food web.

Sol: The entire series of communities occurring in

biotic succession is called sere.

The natural reservoir of phosphrous is rock.

111. Answer (3)

Hint: Upright pyramids indicate that given

parameters at lower trophic level are more than at a

higher level.

Sol: In most ecosystems all upright pyramids of

energy, number and biomass indicates that

producers are more in number and biomass than the

herbivores and energy at a lower trophic level is

always more than at a higher level.

112. Answer (4)

Hint: Primary pollutants are emitted directly in the

environment from some definite sources.

Sol: DDT, CO and Pollen are primary pollutants.

PAN (Peroxyacetyl nitrate) is a secondary pollutant.

113. Answer (4)

Hint: Cadmium causes painful skeletal deformities.

Sol: Painful skeletal deformities i.e, itai-itai disease

is caused by cadmium.

114. Answer (2)

Hint: It occurs in upper part of atmosphere.

Sol: Good ozone acts as a shield and absorb

harmful UV radiations from the sun. It is found in

upper part of atmosphere i.e, stratosphere.

115. Answer (1)

Hint: Biomass is the amount of living matter,

expressed as dry weight in kg / m².

Sol: If total biomass of producers in a specific area

is more than that of primary consumers and

gradually decreases at each successive trophic level,

forms upright pyramid.

116. Answer (4)

Hint: The species is able to secrete acids to

dissolve rock, helping in weathering and soil

formation.

Sol: Lichens are pioneer species on rock.

117. Answer (2)

Hint: Among various ecosystem services soil

formation accounts major percent.

Sol: Soil formation - 50%

Recreation < 10%

Nutrient cycling < 10%

118. Answer (1)

Hint: Doha Amendment (8th Dec. 2012) - Committed

to reduce GHG emission by at least 18% below

1990 level.

Sol: COP (Conference of Parties) - 13 - Bali Action

Plan.

COP - 15 - Copenhagen Accord

COP - 22 - Held at Marrakesh (Morocco)

119. Answer (4)

Hint: Speciation is a function of time and tropics had

more evolutionary time.

Sol: Tropical environments are less seasonal,

relatively more constant, predictable and receive

more solar energy over the year.

120. Answer (3)

Hint: High dose of UV-B radiation causes

inflammation of cornea.

Sol: Inflammation of cornea due to high dose of

UV–B, is called snow-blindness. It is an effect of

ozone layer depletion.

121. Answer (2)

Hint: This food chain begins with dead organic

matter.

Sol: Major fraction of energy flow in terrestrial

ecosystem, occurs through detritus food chain (DFC)

than grazing food chain

122. Answer (3)

Hint: Food web is more realistic form of transfer of

energy & food in nature than food chain.

Sol: The natural interconnection of food chains

make food web, which provides stability to

ecosystem. It operates because of taste preference

and unavailability of food.

123. Answer (3)

Sol: Red Data Book or Red list has eight categories

of species.


17/20

124. Answer (2)

Sol: Initially 25 hot spots were identified globally.

125. Answer (1)

Hint: National Forest policy of India came in theyear 1988.

Sol: This policy recommends 33% forest cover forthe plains and 67% for the hills.

126. Answer (2)

Hint: Chipko Movement was started at Gopeshwar inChamoli district.

Sol: At Present Chamoli district is situated inUttarakhand.

127. Answer (2)

Hint: The sum of environmental factors that limits thepopulation size is called environmental resistance.

Sol: Influence of environmental resistance over the

biotic potential is denoted by K N

K

⎛ ⎞⎜ ⎟⎝ ⎠

or N

1K

⎛ ⎞⎜ ⎟⎝ ⎠

.

128. Answer (2)

Hint: Reduced aeration slows down the process ofdecomposition.

Sol: Decomposition is largely an oxygen requiringprocess.

129. Answer (4)

Hint: High concentration of DDT causes thinning ofegg shells and their premature breaking.

Sol: Thinning of egg shell is result of disturbance incalcium metabolism. Finally, it results in decline ofbird population.

130. Answer (3)

Hint: Endemism is species confined to a particularregion and not found anywhere else.

Sol: High degree of endemism helps in in-situconservation, not in biodiversity loss.

131. Answer (4)

Hint: There are two basic strategies for biodiversityconservation, in-situ & ex-situ.

Sol: Zoological park – ex-situ conservation

Biosphere Reserve

Sacred Groves - conservation

National Park

⎫⎪⎬⎪⎭

in situ

132. Answer (1)

Hint: Convention of Biological diversity held in Rio deJaneiro in 1992.

Sol: Earth summit - Rio de Janeiro (1992)

World Summit - Johannesburg (2002) on sustainabledevelopment.

Biodiversity plays major role in many ecosystemservices - Broadly utilitarian.

Direct economic benefits from nature - Narrowlyutilitarian.

133. Answer (4)

Hint: More than 50% of incident solar radiation isabsorbed by gases, water vapours etc.

Sol: PAR is < 50% of the incident solar radiation.

134. Answer (3)

Hint: When nitrite in body combines withhaemoglobin it forms methaemoglobinemia.

Sol: Formation of methaemoglobinemia reducesoxygen carrying capacity of blood or causes bluebaby syndrome. It is an effect of water pollution.

135. Answer (4)

Hint: Hot spot is on-site conservation strategy.

Sol: Hot spots have high degree of habitat loss.

136. Answer (2)

Hint : AIDS is acquired during life time of anindividual.

Solution :AIDS is an immuno deficiency syndrome.HIV can pass from mother to her child but AIDSoccurs when the immune system of a person iscompromised i.e when T-lymphocyte count fallsbelow 200/mm3 of blood.

137. Answer (4)

Hint : Alien DNA is first transferred into retrovirus.

Solution : Retrovirus carrying alien DNA infects hostcell and transfers alien DNA to the host cell. So, itis an indirect method of gene transfer but the othermethods given are direct methods of transfer.

138. Answer (3)

Hint : Name of restriction enzyme is derived fromname of source organism.

Solution :

Name of restriction enzyme isolated fromHaemophilus influenzae is based on genus andspecies

H ind

Haemophilus

influenzae

139. Answer (2)

Hint : Anabolic steroids disturb hormone balancewhich affects adiposis in males.

Solution : Breast enlargement occurs in males, assteroids are converted into estrogen in body whichcause men to develop unwanted breast tissue(adipose). Breast atrophy occurs in females.

140. Answer (2)

Hint : Native E.coli lacks resistance to antibiotics.

Solution : Plasmid was isolated from Salmonellatyphimurium and antibiotic resistance gene wasintroduced in plasmid. Recombinant DNA wasintroduced into E.coli.


18/20

141. Answer (4)

Hint : Purelines are homozygous individuals.

Solution : Purelines are developed by mating relatedmembers of a species for many generations.Purelines are homozygous for different traits leadingto reduction in allele variety in a population. Theyare more prone to inbreeding depression. Inbreedingdepression can be overcome by outcrossing.

142. Answer (4)

Hint : This drug primarily affects cardiovascularsystem of the body.

Solution : Hallucinogen LSD (Lysergic acid diethylamide) is an extract of fruiting body of fungusClaviceps purpurea. Coke is another name forcocaine and is obtained from Erythroxylum coca.Marijuana is a hallucinogen obtained from plantCannabis sativa. Heroin is an opiate narcoticobtained from diacetylation of morphine.

143. Answer (2)

Hint : Metagenesis refers to alternation of generationin cnidarians.

Solution : Cancer cells show high telomeraseactivity. High telomerase activity in cancer cellsprevents senescence and ageing. Depletion of ozonelayer causes increased penetration by UV rayswhich causes DNA damage and neoplastictransformation of normal cells. Metastasis is shownby malignant tumors.

144. Answer (1)

Hint : Restriction enzymes recognise palindromicsequences.

Solution : Palindromic sequences of base pairsread the same on the two strands when orientationof reading is kept same i.e. 5 3 on both strands.

145. Answer (1)

Hint : Different sized DNA fragments are separatedby agarose gel electrophoresis.

Solution : DNA threads are isolated from solution byspooling. Extraction of DNA fragments from agarosegel is elution. Agglutination can refer to clumpingresulting from interaction between antigen andantibodies.

146 Answer (3)

Hint : In genetically modified (GM) plants, theirgenes have been altered or manipulated by RDT.

Solution : Genetic manipulations are aimed atimproving certain characteristics of plants.Genetically modified crops are usually nutrient rich,resistant to abiotic stress and prevent earlyexhaustion of fertility of soil.

147. Answer (1)

Hint : Probes are radiolabelled single stranded DNAor RNA molecules.

Solution : Probes used in autoradiography arecomplementary to a particular gene sequence. In amutated gene, sequence of bases is altered, soprobe is no longer complementary to mutated gene.As a result cell/colony having mutated gene will notappear on photographic film.

148. Answer (4)

Hint : Number of copies of desired/target DNA isdetermined by copy number of plasmid.

Solution :

Ori regulates copy number of a plasmid. Ori is asequence from where replication starts; so anyforeign DNA linked with Ori can replicate and multiplyitself in host organism.

149. Answer (4)

Hint : This sterile animal is produced by a crossbetween horse and donkey.

Solution :

True product of honey bee is beewax.

150. Answer (2)

Hint : Recombinants carry recombinant plasmidwhile non-recombinants carry native plasmid.

Solution :

Insertional inactivation of antibiotic resistance gene inpBR322 helps in selection of recombinants and non-recombinants. Hind III recognition site is locatedoutside the selectable marker, so both recombinantsand non-recombinants will show resistance toampicillin and tetracycline and we will not be able todifferentiate between them.

151. Answer (3)

Hint : Multiple Ovulation Embryo Transfer (MOET)

Solution : Embryos are transferred from geneticmother to surrogate mother. In MOET, fertilizationtakes place inside female’s body i.e in-vivofertilization.

152. Answer (2)

Hint : X-rays are used to detect cancers of internalorgans.

Solution :

In biopsy, a piece of suspected tissue cut into thinsections is stained and examined under microscope.X rays are ionising rays used for CT scan.

153. Answer (1)

Hint : RNAi prevents expression of specific genes.

Solution : RNAi prevents translation of specificmRNA. It is a post transcriptional mRNA silencingtechnique. Genes introduced in tobacco plantproduce both sense and antisense RNA. TheseRNA being complementary to each other formsdsRNA that initiates RNAi and thus silences specificmRNA of parasite. As a result parasite cannotsurvive in transgenic hosts.


19/20

154. Answer (3)

Hint : Plasmids replicate independent of host cell

division.

Solution : Most plasmids are extrachromosomal

DNA; they are not a part of chromosomal DNA and

they do not integrate with it generally. Histones are

associated with chromosomal DNA.

155. Answer (4)

Hint : Enzyme is used in accordance with

composition of cell wall/membrane to isolate DNA

from cell.

Solution : Different enzymes are used depending on

the composition of cell wall and cell membrane.

Animal cells lack cell wall and their cell membrane

can be digested by SDS and alkaline salts. RNase

hydrolysis the RNA in cell and not the cell membrane

constituents.

156. Answer (2)

Hint : Proinsulin is immature form of insulin.

Solution :During maturation of proinsulin, C-peptide

is removed. A and B chains are linked by disulphide

linkage but do not form a single polypeptide.

157. Answer (2)

Hint : Transformation is introduction of foreign DNA

in a cell.

Solution : Transformed cells can receive

recombinant plasmid or native plasmid. In blue/white

colony selection there is inactivation of -

galactosidase by introduction of foreign DNA,

producing white colonies.

158. Answer (2)

Hint : Toxin gets activated in gut of insect.

Solution : Alkaline pH in gut of insect converts

protoxin into active toxin. The active toxin damages

the epithelial lining of gut and creates pores that

cause swelling and lysis and eventually cause death

of insect.

159. Answer (4)

Hint : DNA is a negatively charged molecule.

Solution : Larger fragment, faces more resistance

while moving in matrix as compared to smaller

fragment. In gel electrophoresis, DNA fragments are

separated on the basis of size. Smaller fragments

move faster through agarose matrix towards

positively charged electrode i.e anode.

160. Answer (3)

Hint : Methanol is converted to formaldehyde.

Solution : Chronic intake of alcohol during

pregnancy cause FAS which includes facial

changes, poorly formed concha, defects in atria and

ventricles of heart.

161. Answer (1)

Hint : Identify the enzyme which is also known asmolecular glue.

Solution : DNA is cut using restriction endonucleaseand is linked by DNA ligase. DNA polymerase isresponsible for synthesis of DNA strand using DNAtemplate.

162. Answer (1)

Hint : A disarmed vector was used for transformationof lymphocytes.

Solution : cDNA was introduced into lymphocytesby retrovirus.

163. Answer (3)

Hint : Antibiotic resistance genes act as selectablemarkers in plasmids.

Solution : Ori determines copy number of aplasmid in host cell

A – ampR

B – tetR (recombinant will be sensitive totetracycline)

C – EcoRI

D – Rop (Supports Ori)

E – Ori

F – Hind III

164. Answer (1)

Hint : Plasmid pBR322 is extra chromosomal DNA.

Solution : Ti plasmid of Agrobacterium tumefaciensincorporates foreign DNA in chromosomal DNA ofplant cell. Bacteriophage infects bacterial cell.

165. Answer (4)

Hint : CNS depressants can induce sleep.

Solution : Person suffering from insomnia hasdifficulty in falling asleep at night. Drugs thatsuppress the activity of CNS are used to treatinsomnia. Cocaine is a stimulant, which makes aperson more wakeful and alert. Valium is trademarkfor diazepam.

166. Answer (3)

Hint : cry gene codes for crystalline proteins foundin Bacillus thuringiensis.

Solution :

cry I Ab encodes for toxin which protects corn plantsfrom corn borer. Toxin encoded by cry I Ac and cryII Ab control cotton bollworms.

167. Answer (1)

Hint : Competent cells are ready to take up theforeign DNA.

Solution : CaCl2 causes deposition of DNA on cell

surface thus increasing their chance for entry in cell.Heat shock creates transient pores in cell membranewhich facilitates the entry of DNA in cell.


20/20

168. Answer (3)

Hint : ELISA is a diagnostic test.

Solution : ELISA is based on antigen antibody

interaction.

169. Answer (4)

Hint : Anti-cancer drugs also target the rapidly

dividing cells in body.

Solution : Anti-cancer drugs inhibit the proliferation

of hair follicle cells leading to hair loss and anaemia.

Alcohol addicts are more prone to liver cirrhosis. UV

rays and X-rays are physical carcinogens. Cancer is

a non-infectious disease.

170. Answer (1)

Hint : Identify the acronym for Genetic Engineering

Approval Committee.

Solution : GMO - Genetically modified organisms.

ICMR – Indian Council of Medical Research

171. Answer (2)

Hint : Thermostable DNA polymerase is used during

PCR.

Solution : Taq polymerase used in PCR is isolated

from Thermus aquaticus, which can survive high

temperature (90°C).

172. Answer (4)

Hint : Foreign gene product is isolated from host cell

after biosynthetic stage.

Solution : After the product is formed at

biosynthetic stage, it is separated from reaction

mixture and further purified and preserved. These

processes of separation and purification is referred

as downstream processing.

173. Answer (4)

Hint : About 27 varieties of this type of rice are

grown in India.

Solution : Lerma rojo – wheat variety.

Sharbati sonora – wheat variety.

174. Answer (3)

Hint : Bubbles dramatically increase the oxygen

transfer area.

Solution : Sparger ensures oxygen availability

throughout the reaction mixture. Purification of

product occurs after biosynthetic stage.

175. Answer (2)

Hint : This phase is also called exponential phase.

Solution : Higher yields of desired products areobtained by this method of culture in bioreactors.

176. Answer (4)

Hint : Identify a stimulant of nervous system.

Solution : Crack or cocaine is obtained fromErythroxylum coca.

177. Answer (1)

Hint : A single cut in closed circular DNA produceslinear DNA.

Solution :

⇒6 fragments

Circular DNA

⇒7 fragments

Linear DNATotal : 13 Fragments

178. Answer (2)

Hint : Denaturation of DNA in PCR occurs at90°C - 94°C.

Solution : Annealing of primers to template strandoccurs on the basis of complementarity betweenbases. Extension of primer occurs by Taqpolymerase at 72°C.

179. Answer (3)

Hint : Crossbreed is developed by crossing animalsof different breeds.

Solution : Hisardale is a cross breed of sheepdeveloped in Punjab by crossing Bikaneri ewes andMerino rams.

180. Answer (2)

Hint : Restriction enzymes restrict the growth ofvirus inside bacteria.

Solution : Restriction enzymes digest the viral DNAthus inhibiting the growth of virus inside bacteria.Bacterial DNA is protected from action of restrictionenzymes as it is modified by the action of DNAmethylase, so that bacterial DNA is no longerrecognised by restriction enzymes.

� � �

Test - 3 (Code-B) (Answers) All India Aakash Test Series for NEET-2019

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Test Date : 09/12/2018

ANSWERS

TEST - 3 (Code-B)

All India Aakash Test Series for NEET - 2019

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147. (4)

148. (3)

149. (1)

150. (3)

151. (4)

152. (1)

153. (3)

154. (1)

155. (1)

156. (3)

157. (4)

158. (2)

159. (2)

160. (2)

161. (4)

162. (3)

163. (1)

164. (2)

165. (3)

166. (2)

167. (4)

168. (4)

169. (1)

170. (3)

171. (1)

172. (1)

173. (2)

174. (4)

175. (4)

176. (2)

177. (2)

178. (3)

179. (4)

180. (2)

All India Aakash Test Series for NEET-2019 Test - 3 (Code-B) (Hints and Solutions)

2/20

PHYSICS

1. Answer (2)

Hint: V2 = 6 V, voltage in parallel remains same.

Solution: V2k = 6 V I2k

63 mA

2

Current across 1 k resistor is 21 6

10 mA1.5

Iz = 10 – 3 = 7 mA

Hence Iz, I

1k, I2k

are 7mA, 10mA, 3mA respectively.

2. Answer (2)

Hint: Mass action law ne

nh = n

i2

Solution: ne.n

h = n

i2

2

i

e

h

nn

n

en

16 2 32

22 22

(2 10 ) 4 10

4 10 4 10

ne = 1010 m–3

3. Answer (1)

Hint: X A B ⇒ for NOR gate and 1

Y X X for NAND gate.

Sol.: X A B for A = 0, B = 1

X = 1

for NAND gate.

1Y X X

for A = 0, B = 1

Y = 1

4. Answer (3)

Hint: Y A B

Solution: De Morgan’s law

Y A A B A BB

Y A + B OR gate.

5. Answer (3)

Hint : Half wave rectifier output current.

Solution :

vm for diode

210 221 2

10m

v V

As 21 2

70

m

m

F L

vI

R R

3 2

10

3 2 3

5 2 5 2A

6. Answer (4)

Hint : out

in

m

Ig

V

Solution : out

in

m

Ig

V

in

m

in

Ig

V

50in

in in

I

R I

3

50

2 10

= 25 × 10–3

225

1010

= 2.5 × 10–2 mho

7. Answer (2)

Hint: Vac

= 0.3V

HINTS & SOLUTIONS

Test - 3 (Code-B) (Hints and Solutions) All India Aakash Test Series for NEET-2019

3/20

Solution: VAB

= 0.3V

Hence 10 – (– 2) = 0.3 + IR

IR = 11.7

V0 – (– 2) = IR

V0 = 11.7 – 2

= 9.7 V

8. Answer (4)

Hint : 1

Solution : 0.981

0.98 + 0.98 = 0.98 = 0.02

49

9. Answer (3)

Hint: Apply KVL in input and output loop.

RC

10 V

10 V

E

RB

B

C

Solution: VBE

= 0 and VCE

= 6 V

RB = 400 k R

C = 2 k

BI

10 025 A

400k

10 6

2mA2k

C

I

IB, I

C 25 A and 2 m A.

10. Answer (1)

Solution : In common emitter amplifier, the output

signal voltage is 180° out of phase with input signal

voltage.

11. Answer (2)

Hint: Activity 1/2

1/2

ln2and

R N T

Solution: At t = 10 h, R 0

2

R

e

Since R = N0.e–t

0

022

tRR e t

e

⇒

2 1

10 5 (hours)–1

Now for 0

1/2 02

Rt T R ⇒

Hence 1/2

25 2

�

�n

T n

1/25 2 hoursT n �

12. Answer (1)

Hint: S

tv

Solution:3

2500 10

2500t

t = 1000 s

1/2

0.693

T

× 693 = 0.693

11s

1000

As N = N0 e–t

11000

10000

N N e

0N

Ne

13. Answer (3)

Hint: Sum of atomic number and mass number at

reactant and product remain constant.

Solution: 13 + 0 = 11 + Z

Z = 2

27 + 1 = A + 24

A = 4

particle is alpha.


4/20

14. Answer (2)

Hint :

1

3R A

Solution :

1

31

(64)R ...(i)

1

32

(125)R ...(ii)

Divide equation (ii) by (i)

2

1

5

4

R

R

2

4.8 5

4R

R2 = 6 fermi

15. Answer (2)

Hint: Isotone have same number of neutron.

Solution: Isotone have same number of neutron but

different mass number.

16. Answer (3)

Hint: Impact parameter 2 2

0

cot4 2

z eb

E

Solution: given b = 0

2 2

2

0

0 cot4 2

z e

E

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠⎝ ⎠

cot 02

⇒

902

= 180°

17. Answer (1)

Hint: 2

2(13.6) eV

ZE

n

Solution: E

2

1 2

213.6 54.4 eV

1

2

2 2

213.6 13.6 eV

2E

Erequired

= E2 – E

1

= [–13.6– (–54.4)] eV

E = 40.8 eV

18. Answer (4)

Hint : For stable orbit, 2

nhL

Solutions : For n = 8, 8

8 4

2

h hL

For n = 6, 6

6 3

2

h hL

For n = 2, 6

2

2

h hL

19. Answer (1)

Hint : For Paschen.

2 2

2

1 1 1

3v R

n

⎡ ⎤ ⎢ ⎥ ⎢ ⎥⎣ ⎦

For Lyman

2 2

2

1 1 1

1v R

n

⎡ ⎤ ⎢ ⎥ ⎢ ⎥⎣ ⎦

Solutions : 1

1 1

9 9

Rv R

⎡ ⎤ ⎢ ⎥⎣ ⎦

2

11v R R⎡ ⎤ ⎢ ⎥⎣ ⎦

1

2

1

9 9

v R

v R

20. Answer (4)

Hint: Intensity power 2

1.

r

Solution: Let intensity at distance 1 m is I, then no.

of photoelectrons will be n0

0 0

I n

I n

0

0At 0.5 4

2

rr m I I ⇒

0

0 0

4I n

I n

n = 4 n0


5/20

21. Answer (3)

Hint: nh

mvr2

Solution: for 2nd excited state n = 3

hmv

r

3

2

2 3

3 3 2

2

h h rr

hmv

r

⇒ ⎛ ⎞

⎜ ⎟⎝ ⎠

22. Answer (2)

Hint: P = number of photon per second × energy of

photon

Solution: P = n E

P

nhc

3 7

34 8

18 10 4.4 10

6.6 10 3 10n

n = 4 × 1016 photons per second.

23. Answer (3)

Hint: p p

m vp

m v

Solution: m = 4 mp

p

p p

v m

v m

4 3 12

1 1

p p

p

v m

v m

vp : v = 12 : 1

24. Answer (1)

Hint: 3

h

mKT (de-Broglie equation)

Solution: 2

He H H

H He He

M T

M T

2 500 5

4 300 6

He

H

2 5

2 6

a

a

2 5

2 6

a

a

a = 3

25. Answer (4)

Hint: Photoelectric effect

Solution: Photocurrent Intensity

K.E = E – Since K.E does not depend on intensity of incident

light.

26. Answer (2)

Hint: ( )E

Ve

Solution: 34 8

7

6.6 10 3 10

4.36 10

hcE

19

19

4.54 102.838 eV

1.6 10

E = 2.84 eV

K.Emax

= E – = (2.84 – 2.34) eV

= 0.5 eV

eV = K.Emax

0.5eV

Ve

V = 0.5 V

27. Answer (1)

Hint: Property of work function.

Solution: Work function of a material depends on

nature of material.

28. Answer (2)

Hint: 2d sin = n (for crystal diffraction)

Solution: d = 2 × 10–8 m

2 sind

n

max = 2d for max sin 1

1n

⎧ ⎫⎨ ⎬⎩ ⎭

max

= 2 × 2 × 10–8 m

max

= 4 × 10–8 m


6/20

29. Answer (1)

Hint: Intensity of unpolarised light used in Malus law

0

2

I.

Solution: I

I20

cos2

0 01616

2 25 50

I II

Fractional intensity transmitted.

0

0 0

16 16

50 50

II

I I

% I = 32%

30. Answer (4)

Hint: For polarisation tan p =

Solution: If i = p.

Then reflected ray is fully polarised and refracted ray

is partially polarised.

Also, both reflected and refracted rays are

perpendicular to each other.

31. Answer (3)

Hint: 1st minima 1

1

Dy

a

2nd maxima 2

2

5

2

Dy

a

Solution: 1

1

Dy

a

2

2

5

2

Dy

a

Since y1 = y

2

1 25

2

D D

a a

1 2

5

2

1 = 2.5

2

32. Answer (2)

Hint: n

n Dy

d

Solution: 1

1 1

n

n Dy

d

2

2 2

n

n Dy

d

1 2n n

y y

n1

1 = n

2

2

1 2

2 1

800

500

n

n

1

2

8 16 24........

5 10 15

n

n

For minimum value of n1 such that

1 21

8;n n

y y n ⇒

n1 = 8

33. Answer (3)

Hint: I I2

04 cos2

Solution: Here 4I 0 = I

0 (given)

2

0cos

6I I

0

3

4I

03

4

II

34. Answer (1)

Hint: 1

Resolving power

Solution: 1

R.P

1

1

1R

2

2

1R

1

2

R 600

R 450

1

2

R 4

R 3

35. Answer (3)

Hint: Contrast means relative intensity

Solution: Intensity of fringes depends on intensity of

slits (sources)


7/20

2max 1 2I I I

2min 1 2I I I

36. Answer (1)

Hint: 1 2

1 1 11

f R R

⎛ ⎞ ⎜ ⎟

⎝ ⎠

Solution: 1

1 11.5 1

10f

⎛ ⎞ ⎜ ⎟⎝ ⎠

1

1 1

20f

Similarly3

1 1

20f

2

1 8 1 11

7 10 10f

⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

2

1 1 2 1

7 10 35f

1 1 1 1

20 20 35f

1 7 7 4

140f

1 10

140f

| f | = 14 cm

37. Answer (2)

Hint: L = V0 + u

e and

e

Dm

f

Solution:

O L E

v0

ueu = 2.4 cm

0

f0 = 2 cm f

e = 5 cm

from 0 0 0

1 1 1

v u f

0

0

1 1 112 cm

2 2.4v

v

⇒

125

2.4

vm

u

As image will form at infinity D

mf

ue = – 5 cm

255

5m

Now magnifying power M = m × m

M = – 5 × 5 = – 25

and length of tube L = | v0 | + | u

e |

= 12 cm + 5 cm

L = 17 cm

Hence M, L = – 25, 17 cm

38. Answer (1)

Hint: For longitudinal velocity VI = m2 × V

0.

for lateral velocity VI = m V

0.

Solution:1 1 1

v u f

1 1

20 30

v = 60 cm

602

30

vm

u

(VI)long

= (–2)2 × 8 = 32 cm/s

39. Answer (2)

Hint: For thin lens use 1 1 1

v u f and for mirror use

1 1 1

v u f

Solution:

f = 15 cm

0

30 cm

40

F

f = ?


8/20

For rays to be parallel image from lens must be on

focus mirror

For lens 1 1 1

v f u

1 1

15 30

30 cmv

Focal length of mirror f = 40 – 30 = 10 cm

40. Answer (3)

Hint: Using Snell’s law

Solution: 2

1

sin

sin

i

r

37°i = 60°

2

1

37°

From geometry

r = A = 37°

i = 60°

Using Snell’s law 2

1

sin60

sin37

2

3

2

3

5

2

5 3 5

2 3 2 3

2

5

2 3

41. Answer (2)

Hint: Refraction through curved surface.

Solution: For spherical surface

2 1 2 1

v u R

Put u =

4 3 4

3 3 2 3

22

9

v

1

3 1 60

22

9

v

3 1 3

2 4v

1 1

2v

v = 2 m

42. Answer (3)

Hint: f 2 = O1 O

2

Solution: 1 2

f O O

25 9

225

f 15 cm

43. Answer (2)

Hint: dreal

= dApp

Solution:

x2

x1

d

1

36 9 cm

2x

2

3 279 13.5 cm

2 3x

d = x1 + x

2

= 22.5 cm


9/20

44. Answer (4)

Hint: Reflection from convex mirror

Solution: v u f

1 1 1

f fO

I F

v = ?

u = – f

f = + f

v f f

1 1 1

1 2

v f

2

fv

45. Answer (4)

Hint: Light may enter from denser medium or fromrarer medium in glass slab.

Solution: Snell’s law 2

1

sin

sin

i

r

Light travels away from normal in rarer medium(denser to rarer) and bend toward normal in densermedium (rarer to denser)

CHEMISTRY

46. Answer (4)

Hint: NaHCO3 reacts with carboxylic and sulphonic

acids or with an acid having acidic strength morethan carboxylic acid.

Solution: Phenol is less acidic than H2CO

3 so does

not react with NaHCO3.

47. Answer (2)

Hint: Rosenmund reduction.

Solution:

ClO

H2

Pd-BaSO4

CHO

48. Answer (2)

Hint: Phenol formaldehyde resin.

Solution:

OH

+ H CO2

H

or

OH

Bakelite

49. Answer (1)

Hint: Chloramphenicol inhibit the growth of

microorganisms.

Solution: Chloramphenicol is a broad spectrum

antibiotics.

50. Answer (3)

Hint: Dettol contains chloroxylenol.

Solution: Dettol is a mixture of chloroxylenol and

terpineol.

51. Answer (4)

Hint: Polymer of Ethylene glycol and Phthalic acid.

Solution:

Glyptal — O — CH — CH OOC2 2

CO —

n

is

used in the manufacture of paints.

52. Answer (1)

Hint: Ziegler-Natta catalyst is triethylaluminium and

titanium tetrachloride.

Solution: It is a mixture of (C2H

5)3Al & TiCl

4.

53. Answer (1)

Hint: Elastomers involve weakest van der Waals

interactions.

Solution: — CH — C CH — CH —2 2

Cl n

Neoprene is

an elastomer.

54. Answer (1)

Hint: Amino acid containing five membered ring.

Solution: Proline NIH

COOH is a cyclic

amino acid.

55. Answer (2)

Hint: Ribose is present in RNA and deoxyribose in

DNA.


10/20

Solution: Sugar component in DNA is

-D-2-deoxyribose whereas in RNA molecule, it is

-D-ribose.

56. Answer (4)

Hint: Grignard reagent adds to ethylene oxide to give

primary alcohol.

Solution: PhMgBr + OPh O MgBr

H O3

+

PhOH

57. Answer (3)

Hint: Most electrophilic carbonyl carbon will be one,

in which benzene ring is attached with strong

electron withdrawing group at o/p-positions.

Solution:

OCOCH3

NaOH

NO2

OH

NO2

+ CH — C — O3

O

– NO2 is strong electron withdrawing group so the

above ester will be most easily hydrolysed in

alkaline medium.

58. Answer (2)

Hint: ,-unsaturated carbonyl compounds are

formed in aldol condensation reaction.

Solution:

O

H

H

OH

O

+ H O2

O

OOH

OO

H H

O

O

+ H O2

59. Answer (3)

Hint: Cellulose is polysaccharide composed of only

-D-glucose units.

Solution:

Amylopectin is a branched chain polymer of

-D-glucose units.

60. Answer (4)

Hint: -D-glucose is a cyclic hemiacetal.

Solution:

O

CH OH2

H

HO

H

OH HOH

H

H OH

-D-glucose

In -D-glucose number of asymmetric carbon is 5.

61. Answer (3)

Hint: Structural formula of Glyoxal is

H

O

H

O

Solution: PhCH CH — CH CHPh(i) O

(ii) Zn/H O

3

2

2 PhCHO + CH — CH

O O

62. Answer (4)

Hint: Compounds containing keto methyl group or

compounds which on oxidation generate keto methyl

group will give haloform reaction.

Solution:

(a) Ph — C — CH3

O

I2

NaOHPh — C — COO + CHI

3

O

Keto methyl group

(b) CH CH OH3 2

[O]CH CHO

3

I2

NaOHCHI + H — C — O

3

O

(c) CH CH CHCH3 2 3

OH

[O]CH CH — C — CH

3 2 3

O

I

NaOH

2

CHI + CH CH — C — O3 3 2

O


11/20

63. Answer (3)

Hint: Reaction of Grignard reagent with aldehyde.

Solution:

Cl

Mg

Dry ether

MgCl

A

(i) HCHO

(ii) H O3

CH OH2

B64. Answer (3)

Hint: Carbylamine reaction.

Solution:

CH NH2 2

CHCl3

KOH

CH NC2

65. Answer (4)

Hint: Tertiary amines do not react with Hinsberg’s

reagent.

Solution: Tertiary amine N

will not

react with Hinsberg’s reagent (C6H

5SO

2Cl) due to

absence of H-atom at Nitrogen.

66. Answer (2)

Hint: Addition of water to alkene in presence of acid

gives stable rearranged alcohol.

Solution:

CH — CH — CH — CH3 2

CH3

Cl

H OH

CH CH — CH 3

CH2

CH3

H

CH — C — CH — CH3 2 3

CH3

H O2

CH — C — CH — CH3 2 3

OH

CH3

CH — C — CH — CH3 3

H

CH3

H

shift

67. Answer (2)

Hint: Tertiary alkyl halide undergoes nucleophilic

substitution reaction by SN1 mode.

Solution: Tert-butylchloride is tertiary alkyl halide

hence produces stable carbocation which further

gives substitution product via SN1 mechanism.

68. Answer (4)

Hint: The compound which does not contain plane

of symmetry is optically active.

Solution: A properly substituted allene is chiral but

Me

H

C C C

Me

Me

contains same Me group on

one carbon leading to have POS hence achiral.

Rotation of bond keeps two phenyl ring in different

plane in

Br Me

Me Me

but due to presence of

same Me group at ortho position in phenyl group,

holds POS hence achiral.

CH3

H

H

OH

OH

CH3

POS

CH3

Br

H

H

Br(R)

CH3

(Chiral)

(R)

(Achiral)

69. Answer (1)

Hint: Analgesics reduce pain without causing any

disturbance of nerve system.

Solution:

Drug Therapeutic action

Aspirin Analgesic

Equanil Tranquilizers

Ofloxacin Antibiotic

Salvarsan Antibiotic

70. Answer (4)

Hint: Alitame is the sweetest compound

Solution:

Compounds Sweetness value in comparison

to cane sugar

Saccharin 550

Aspartame 100

Sucralose 600

Alitame 2000

71. Answer (2)

Hint: Formation of chiral cyanohydrin compound.


12/20

Solution:

Ph

H

OH

CN

Ph

HOH

CN

+

Racemic mixture

H O3

Ph

H

OH

CO H2

Ph

HOH

+

CO H2

Ph — C — H + HCN

O

72. Answer (2)

Hint: Wolff-Kishner reduction

Solution:

O

(i) NH — NH2 2

(ii) NaOH +OH OH

73. Answer (3)

Hint: Cleavage of ether using HI

Solution:

OCH3

H

HO — CH3

I

OH

+ CH I3

S 2N

74. Answer (3)

Hint: Electrophile formed in Reimer-Tiemann reaction

is :CCl2

Solution:

NaOH

CHCl3

OH ONa

CHCl2

NaOH

ONa

CHO

H

OH

CHO

Intermediate

75. Answer (2)

Hint: Phenolic hydrogen is easily abstracted by

aqueous alkali and phenoxide ion is formed.

Solution:

NaOH

OH

OH

O

OH

PH — CH — Br2

O

OH

Ph

76. Answer (3)

Hint: In condensation polymerisation small

molecules like H2O or ammonia is removed.

Solution:

n CH CHCN2

Addition

Polymerisation — CH — CH —2

CN

n

Acrylonitrile

Orlon 77. Answer (2)

Hint: Monomer of teflon is tetrafluoroethylene

Solution: n CF = CF2 2 CF — CF

2 2

nTeflon

78. Answer (2)

Hint: Vitamin ‘C’ is chemically an acid, called

Ascorbic acid.

Solution: Vitamin C is readily soluble in water.

79. Answer (3)

Hint: Compound which contains aldehyde group will

reduce Tollen’s reagent.

Solution: Sucrose is made up of -D-glucose and

-D-fructose by 1,2-glycosidic linkage. It does not

contain free aldehyde group. In alkaline medium

fructose converts into glucose and mannose hence

can be reduced by Tollen’s reagent.

80. Answer (3)

Hint: Lone pair is not taking part in resonance in

N

and

N|H

Solution:

Lone pair of N-atom is occupying sp3 hybrid orbital


13/20

in

N|H

and sp2 hybrid orbital in

N

81. Answer (2)

Hint: Nitrobenzene is converted to hydrazobenzene

in presence of Zinc dust and aqueous alkali.

Solution:

NO2

Zn/NaOHPh — NH — NH — Ph

82. Answer (2)

Hint: Hoffmann bromamide degradation of benzamide

Solution:

NH2O

Br

KOH

2

NH2

aniline

83. Answer (3)

Hint: Coupling reaction of phenol

Solution:

NH2

NaNO /HCl

0–5°C

2

N Cl2

PhOH/OHN = N OH

84. Answer (3)

Hint: Reduction of ester by DIBAL-H gives aldehyde.

Solution:

CH CH — C — OC H3 2 2 5

DIBAL-HCH CH — C — H

3 2

OO

85. Answer (4)

Hint: Lower the value of pKa, stronger is the acid

Solution:

Compound pKa

COOH

O N2

NO2

2.83

COOH

NO2

3.43

COOH

CH3

3.91

OH

NO2O N

2

NO2

1.02

(Picric acid)

Anion of picric acid is highly stabilised by resonance

of three nitro groups present at ortho and para

positions. Hence, picric acid is most acidic among

all.

86. Answer (3)

Hint: Draw Fischer projections of different

stereoisomers of tartaric acid.

Solution:

COOH

COOH

H OH

HHO

A

COOH

COOH

HO H

OHH

COOH

COOH

H OH

OHH

Plane of symmetry B C

Opticallyactive

Opticallyactive

Opticallyinactive (meso)

Compounds A and B are enantiomers and each one

is optically active. Compound C is optically inactive

as it contains plane of symmetry within the

molecule.

87. Answer (2)

Hint: Hydrolysis of cyanide in acidic medium gives

carboxylic acid.

Solution:

PhCH — Cl2

KCNPhCH — CN

2

A

H O3

PhCH COOH2

BC H OH/H

2 5

PhCH — C — OC H2 2 5

O

(Ester) C


14/20

88. Answer (2)

Hint: The ease of dehydration is decided by the

stability of carbonium ion and the product form.

Solution:

O

OHH

O

–H

O

Less

stable

O

OH

H

O

–H

O

More

stable

Product is

stabilised

due to

resonance

(i)

(ii)

O

OH

H

O

–H

O

O

OH

H

O

–H

O

(iii)

(iv)

89. Answer (4)

Hint: Compound which is more acidic than water will

react with aqueous alkali.

Solution:

OH

+ KOH

OK

+ H O2

90. Answer (2)

Hint: The presence of an electron withdrawing group

at ortho and para-positions increases the reactivity of

haloarenes towards nucleophilic substitution reaction.

Solution:

Cl

N

O O

Slow step

OH

OH

N

O O

Cl OH

N

O O

Cl OH

N

O O

Cl

OH

N

O O

Cl

Fast

Step

OH

NO2

+ Cl

BIOLOGY

91. Answer (4)

Hint: Hot spot is on-site conservation strategy.

Sol: Hot spots have high degree of habitat loss.

92. Answer (3)

Hint: When nitrite in body combines with

haemoglobin it forms methaemoglobinemia.

Sol: Formation of methaemoglobinemia reduces

oxygen carrying capacity of blood or causes blue

baby syndrome. It is an effect of water pollution.

93. Answer (4)

Hint: More than 50% of incident solar radiation is

absorbed by gases, water vapours etc.

Sol: PAR is < 50% of the incident solar radiation.

94. Answer (1)

Hint: Convention of Biological diversity held in Rio de

Janeiro in 1992.

Sol: Earth summit - Rio de Janeiro (1992)

World Summit - Johannesburg (2002) on sustainable

development.

Biodiversity plays major role in many ecosystem

services - Broadly utilitarian.

Direct economic benefits from nature - Narrowly

utilitarian.

95. Answer (4)

Hint: There are two basic strategies for biodiversity

conservation, in-situ & ex-situ.

Sol: Zoological park – ex-situ conservation

Biosphere Reserve

Sacred Groves - conservation

National Park

⎫⎪⎬⎪⎭

in situ


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96. Answer (3)

Hint: Endemism is species confined to a particularregion and not found anywhere else.

Sol: High degree of endemism helps in in-situconservation, not in biodiversity loss.

97. Answer (4)

Hint: High concentration of DDT causes thinning ofegg shells and their premature breaking.

Sol: Thinning of egg shell is result of disturbance incalcium metabolism. Finally, it results in decline ofbird population.

98. Answer (2)

Hint: Reduced aeration slows down the process ofdecomposition.

Sol: Decomposition is largely an oxygen requiringprocess.

99. Answer (2)

Hint: The sum of environmental factors that limits thepopulation size is called environmental resistance.

Sol: Influence of environmental resistance over the

biotic potential is denoted by K N

K

⎛ ⎞⎜ ⎟⎝ ⎠

or N

1K

⎛ ⎞⎜ ⎟⎝ ⎠

.

100. Answer (2)

Hint: Chipko Movement was started at Gopeshwar inChamoli district.

Sol: At Present Chamoli district is situated inUttarakhand.

101. Answer (1)

Hint: National Forest policy of India came in theyear 1988.

Sol: This policy recommends 33% forest cover forthe plains and 67% for the hills.

102. Answer (2)

Sol: Initially 25 hot spots were identified globally.

103. Answer (3)

Sol: Red Data Book or Red list has eight categories

of species.

104. Answer (3)

Hint: Food web is more realistic form of transfer of

energy & food in nature than food chain.

Sol: The natural interconnection of food chains

make food web, which provides stability to

ecosystem. It operates because of taste preference

and unavailability of food.

105. Answer (2)

Hint: This food chain begins with dead organic

matter.

Sol: Major fraction of energy flow in terrestrial

ecosystem, occurs through detritus food chain (DFC)

than grazing food chain

106. Answer (3)

Hint: High dose of UV-B radiation causes

inflammation of cornea.

Sol: Inflammation of cornea due to high dose of

UV–B, is called snow-blindness. It is an effect of

ozone layer depletion.

107. Answer (4)

Hint: Speciation is a function of time and tropics had

more evolutionary time.

Sol: Tropical environments are less seasonal,

relatively more constant, predictable and receive

more solar energy over the year.

108. Answer (1)

Hint: Doha Amendment (8th Dec. 2012) - Committed

to reduce GHG emission by at least 18% below

1990 level.

Sol: COP (Conference of Parties) - 13 - Bali Action

Plan.

COP - 15 - Copenhagen Accord

COP - 22 - Held at Marrakesh (Morocco)

109. Answer (2)

Hint: Among various ecosystem services soil

formation accounts major percent.

Sol: Soil formation - 50%

Recreation < 10%

Nutrient cycling < 10%

110. Answer (4)

Hint: The species is able to secrete acids to

dissolve rock, helping in weathering and soil

formation.

Sol: Lichens are pioneer species on rock.

111. Answer (1)

Hint: Biomass is the amount of living matter,

expressed as dry weight in kg / m².

Sol: If total biomass of producers in a specific area

is more than that of primary consumers and

gradually decreases at each successive trophic level,

forms upright pyramid.

112. Answer (2)

Hint: It occurs in upper part of atmosphere.

Sol: Good ozone acts as a shield and absorb

harmful UV radiations from the sun. It is found in

upper part of atmosphere i.e, stratosphere.

113. Answer (4)

Hint: Cadmium causes painful skeletal deformities.

Sol: Painful skeletal deformities i.e, itai-itai disease

is caused by cadmium.


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114. Answer (4)

Hint: Primary pollutants are emitted directly in the

environment from some definite sources.

Sol: DDT, CO and Pollen are primary pollutants.

PAN (Peroxyacetyl nitrate) is a secondary pollutant.

115. Answer (3)

Hint: Upright pyramids indicate that given

parameters at lower trophic level are more than at a

higher level.

Sol: In most ecosystems all upright pyramids of

energy, number and biomass indicates that

producers are more in number and biomass than the

herbivores and energy at a lower trophic level is

always more than at a higher level.

116. Answer (3)

Hint: Ecological pyramid assumes a simple food

chain and does not accommodate a food web.

Sol: The entire series of communities occurring in

biotic succession is called sere.

The natural reservoir of phosphrous is rock.

117. Answer (1)

Hint: It is an interaction between two organisms of

different species in which one species is harmed

whereas the other is unaffected.

Sol: Amensalism will be expressed as (–, 0).

118. Answer (2)

Hint: Natality and immigration are responsible for

increase in size of population of an area.

Sol: Population density will increase if the number

of births plus number of immigrants (B + I) is more

than the number of deaths plus the number of

emigrants (D + E) otherwise it will decrease.

Thus, t 1 tN N B I D E ⎡ ⎤ ⎣ ⎦

119. Answer (2)

Hint: Heat loss or gain is a function of surface area.

Sol: Mammals from colder climates have shorter

extremities.

Population size is not a static paramater.

Resources for growth of population are never

infinite.

120. Answer (1)

Hint: Sun is the only source of energy in aquatic

food chains.

Sol: Grazing food chain is the major conduit of

energy flow in aquatic ecosystems.

121. Answer (4)

Hint: Detritus is the substrate of decomposition.

Sol: Dead remains of plants and animals including

faecal matter constitute detritus and it is raw material

for decomposition.

122. Answer (2)

Sol: A.G. Tansley coined the term ecosystem and

defined it as sum total of interactions between biotic

and abiotic components.

123. Answer (3)

Hint: Tropical Amazonian rainforest has the greatest

biodiversity on earth.

Sol: Amazon forest is estimated to produce 20% of

the total O2 of earth’s atmosphere, thus it is called

the lungs of the planet.

124. Answer (4)

Hint: CO2 and CH

4 are the major cause of

greenhouse effect.

Sol: CO – 60%2

CH – 20%4

CFCs – 14%

N O – 6%2

Relative contributions to thetotal global warming

125. Answer (3)

Hint: Mammals are able to maintain homeostasis.

Sol: Due to the capacity to maintain constant body

temperature, mammals are successful on earth but

majority of the animals and nearly all plants do not

have this ability.

126. Answer (3)

Sol: Dodo was found in Mauritius.

127. Answer (2)

Sol: Term biodiversity was popularised by Edward

Wilson.

128. Answer (1)

Hint: It is expressed as biomass or numbers of

organisms per unit area.

Sol: Amount of living material present in different

trophic levels at a given time is called standing crop.

GPP - Rate of production of organic matter by

producers.

NPP = GPP – Respiratory loss.

129. Answer (4)

Hint: Decomposition concerns with breakdown of

complex organic matter into inorganic nutrients.

Sol: Mineralisation is a part of catabolism. It is

release of inorganic nutrients from organic matter or

humus during the process of decomposition.


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130. Answer (4)

Hint: Stenohalines are those which are restricted to

a narrow range of salinity.

Sol: Eurythermal - organisms can tolerate and thrive

in a wide range of temperature.

Hypersaline lagoons - Salinity > 100 (ppt)

UV - C - Wavelength 100 - 280 nm (0.1 - 0.28 m)

131. Answer (1)

Hint: Triangular age pyramid is a graphic

representation of a young population.

Sol: Young population is growing or expanding

population because it has very high proportion of pre-

reproductive individuals as compared to others.

132. Answer (1)

Hint: Tropical rainforests have the maximum mean

annual rainfall among all the biomes of India.

Sol: Tropical rainforests are rich in biodiversity

because mean annual rainfall is 2000 - 3500 mm and

mean annual temperature is 23 - 27°C.

133. Answer (2)

Hint: Amrita Devi Bishnoi Wildlife Protection Award

is instituted for individuals or communities from rural

areas.

Sol: Catalytic converters have metals like platinum-

palladium and rhodium as catalyst. Lead in the petrol

inactivates these catalysts.

Reforestation may occur naturally.

134. Answer (3)

Hint: Electrostatic precipitator is most widely used

way of removing particulate matter.

Sol: Electrostatic precipitator can remove over 99%

particulate matter present in the exhaust from

thermal power plant.

According to CPCB, particulate size 2.5 m or less

in diameter causes greatest harm to human health.

135. Answer (3)

Hint: 1974 - Water (Prevention and control of

Pollution) Act.

Sol: Government of India has passed the

Environment (Protection) Act in 1986.

1987 - Montreal Protocol was signed.

1981 - Air (Prevention and Control of pollution) Act.

136. Answer (2)

Hint : Restriction enzymes restrict the growth ofvirus inside bacteria.

Solution : Restriction enzymes digest the viral DNA

thus inhibiting the growth of virus inside bacteria.

Bacterial DNA is protected from action of restriction

enzymes as it is modified by the action of DNA

methylase, so that bacterial DNA is no longer

recognised by restriction enzymes.

137. Answer (3)

Hint : Crossbreed is developed by crossing animals

of different breeds.

Solution : Hisardale is a cross breed of sheep

developed in Punjab by crossing Bikaneri ewes and

Merino rams.

138. Answer (2)

Hint : Denaturation of DNA in PCR occurs at

90°C - 94°C.

Solution : Annealing of primers to template strand

occurs on the basis of complementarity between

bases. Extension of primer occurs by Taq

polymerase at 72°C.

139. Answer (1)

Hint : A single cut in closed circular DNA produces

linear DNA.

Solution :

⇒6 fragments

Circular DNA

⇒7 fragments

Linear DNATotal : 13 Fragments

140. Answer (4)

Hint : Identify a stimulant of nervous system.

Solution : Crack or cocaine is obtained from

Erythroxylum coca.

141. Answer (2)

Hint : This phase is also called exponential phase.

Solution : Higher yields of desired products are

obtained by this method of culture in bioreactors.

142. Answer (3)

Hint : Bubbles dramatically increase the oxygen

transfer area.

Solution : Sparger ensures oxygen availability

throughout the reaction mixture. Purification of

product occurs after biosynthetic stage.

143. Answer (4)

Hint : About 27 varieties of this type of rice are

grown in India.

Solution : Lerma rojo – wheat variety.

Sharbati sonora – wheat variety.


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144. Answer (4)

Hint : Foreign gene product is isolated from host cell

after biosynthetic stage.

Solution : After the product is formed at

biosynthetic stage, it is separated from reaction

mixture and further purified and preserved. These

processes of separation and purification is referred

as downstream processing.

145. Answer (2)

Hint : Thermostable DNA polymerase is used during

PCR.

Solution : Taq polymerase used in PCR is isolated

from Thermus aquaticus, which can survive high

temperature (90°C).

146. Answer (1)

Hint : Identify the acronym for Genetic Engineering

Approval Committee.

Solution : GMO - Genetically modified organisms.

ICMR – Indian Council of Medical Research

147. Answer (4)

Hint : Anti-cancer drugs also target the rapidly

dividing cells in body.

Solution : Anti-cancer drugs inhibit the proliferation

of hair follicle cells leading to hair loss and anaemia.

Alcohol addicts are more prone to liver cirrhosis. UV

rays and X-rays are physical carcinogens. Cancer is

a non-infectious disease.

148. Answer (3)

Hint : ELISA is a diagnostic test.

Solution : ELISA is based on antigen antibody

interaction.

149. Answer (1)

Hint : Competent cells are ready to take up theforeign DNA.

Solution : CaCl2 causes deposition of DNA on cell

surface thus increasing their chance for entry in cell.Heat shock creates transient pores in cell membranewhich facilitates the entry of DNA in cell.

150. Answer (3)

Hint : cry gene codes for crystalline proteins foundin Bacillus thuringiensis.

Solution :

cry I Ab encodes for toxin which protects corn plantsfrom corn borer. Toxin encoded by cry I Ac and cryII Ab control cotton bollworms.

151. Answer (4)

Hint : CNS depressants can induce sleep.

Solution : Person suffering from insomnia has

difficulty in falling asleep at night. Drugs that

suppress the activity of CNS are used to treat

insomnia. Cocaine is a stimulant, which makes a

person more wakeful and alert. Valium is trademark

for diazepam.

152. Answer (1)

Hint : Plasmid pBR322 is extra chromosomal DNA.

Solution : Ti plasmid of Agrobacterium tumefaciens

incorporates foreign DNA in chromosomal DNA of

plant cell. Bacteriophage infects bacterial cell.

153. Answer (3)

Hint : Antibiotic resistance genes act as selectable

markers in plasmids.

Solution : Ori determines copy number of a

plasmid in host cell

A – ampR

B – tetR (recombinant will be sensitive to

tetracycline)

C – EcoRI

D – Rop (Supports Ori)

E – Ori

F – Hind III

154. Answer (1)

Hint : A disarmed vector was used for transformation

of lymphocytes.

Solution : cDNA was introduced into lymphocytes

by retrovirus.

155. Answer (1)

Hint : Identify the enzyme which is also known as

molecular glue.

Solution : DNA is cut using restriction endonuclease

and is linked by DNA ligase. DNA polymerase is

responsible for synthesis of DNA strand using DNA

template.

156. Answer (3)

Hint : Methanol is converted to formaldehyde.

Solution : Chronic intake of alcohol during

pregnancy cause FAS which includes facial

changes, poorly formed concha, defects in atria and

ventricles of heart.

157. Answer (4)

Hint : DNA is a negatively charged molecule.

Solution : Larger fragment, faces more resistance

while moving in matrix as compared to smaller

fragment. In gel electrophoresis, DNA fragments are

separated on the basis of size. Smaller fragments

move faster through agarose matrix towards

positively charged electrode i.e anode.


19/20

158. Answer (2)

Hint : Toxin gets activated in gut of insect.

Solution : Alkaline pH in gut of insect converts

protoxin into active toxin. The active toxin damages

the epithelial lining of gut and creates pores that

cause swelling and lysis and eventually cause death

of insect.

159. Answer (2)

Hint : Transformation is introduction of foreign DNA

in a cell.

Solution : Transformed cells can receive

recombinant plasmid or native plasmid. In blue/white

colony selection there is inactivation of -

galactosidase by introduction of foreign DNA,

producing white colonies.

160. Answer (2)

Hint : Proinsulin is immature form of insulin.

Solution :During maturation of proinsulin, C-peptide

is removed. A and B chains are linked by disulphide

linkage but do not form a single polypeptide.

161. Answer (4)

Hint : Enzyme is used in accordance with

composition of cell wall/membrane to isolate DNA

from cell.

Solution : Different enzymes are used depending on

the composition of cell wall and cell membrane.

Animal cells lack cell wall and their cell membrane

can be digested by SDS and alkaline salts. RNase

hydrolysis the RNA in cell and not the cell membrane

constituents.

162. Answer (3)

Hint : Plasmids replicate independent of host cell

division.

Solution : Most plasmids are extrachromosomal

DNA; they are not a part of chromosomal DNA and

they do not integrate with it generally. Histones are

associated with chromosomal DNA.

163. Answer (1)

Hint : RNAi prevents expression of specific genes.

Solution : RNAi prevents translation of specific

mRNA. It is a post transcriptional mRNA silencing

technique. Genes introduced in tobacco plant

produce both sense and antisense RNA. These

RNA being complementary to each other forms

dsRNA that initiates RNAi and thus silences specific

mRNA of parasite. As a result parasite cannot

survive in transgenic hosts.

164. Answer (2)

Hint : X-rays are used to detect cancers of internal

organs.

Solution :

In biopsy, a piece of suspected tissue cut into thinsections is stained and examined under microscope.X rays are ionising rays used for CT scan.

165. Answer (3)

Hint : Multiple Ovulation Embryo Transfer (MOET)

Solution : Embryos are transferred from geneticmother to surrogate mother. In MOET, fertilizationtakes place inside female’s body i.e in-vivofertilization.

166. Answer (2)

Hint : Recombinants carry recombinant plasmidwhile non-recombinants carry native plasmid.

Solution :

Insertional inactivation of antibiotic resistance gene inpBR322 helps in selection of recombinants and non-recombinants. Hind III recognition site is locatedoutside the selectable marker, so both recombinantsand non-recombinants will show resistance toampicillin and tetracycline and we will not be able todifferentiate between them.

167. Answer (4)

Hint : This sterile animal is produced by a crossbetween horse and donkey.

Solution :

True product of honey bee is beewax.

168. Answer (4)

Hint : Number of copies of desired/target DNA isdetermined by copy number of plasmid.

Solution :

Ori regulates copy number of a plasmid. Ori is asequence from where replication starts; so anyforeign DNA linked with Ori can replicate and multiplyitself in host organism.

169. Answer (1)

Hint : Probes are radiolabelled single stranded DNAor RNA molecules.

Solution : Probes used in autoradiography arecomplementary to a particular gene sequence. In amutated gene, sequence of bases is altered, soprobe is no longer complementary to mutated gene.As a result cell/colony having mutated gene will notappear on photographic film.

170 Answer (3)

Hint : In genetically modified (GM) plants, theirgenes have been altered or manipulated by RDT.

Solution : Genetic manipulations are aimed atimproving certain characteristics of plants.Genetically modified crops are usually nutrient rich,resistant to abiotic stress and prevent earlyexhaustion of fertility of soil.


20/20

� � �

171. Answer (1)

Hint : Different sized DNA fragments are separatedby agarose gel electrophoresis.

Solution : DNA threads are isolated from solution byspooling. Extraction of DNA fragments from agarosegel is elution. Agglutination can refer to clumpingresulting from interaction between antigen andantibodies.

172. Answer (1)

Hint : Restriction enzymes recognise palindromicsequences.

Solution : Palindromic sequences of base pairsread the same on the two strands when orientationof reading is kept same i.e. 5 3 on both strands.

173. Answer (2)

Hint : Metagenesis refers to alternation of generationin cnidarians.

Solution : Cancer cells show high telomeraseactivity. High telomerase activity in cancer cellsprevents senescence and ageing. Depletion of ozonelayer causes increased penetration by UV rayswhich causes DNA damage and neoplastictransformation of normal cells. Metastasis is shownby malignant tumors.

174. Answer (4)

Hint : This drug primarily affects cardiovascularsystem of the body.

Solution : Hallucinogen LSD (Lysergic acid diethylamide) is an extract of fruiting body of fungusClaviceps purpurea. Coke is another name forcocaine and is obtained from Erythroxylum coca.Marijuana is a hallucinogen obtained from plantCannabis sativa. Heroin is an opiate narcoticobtained from diacetylation of morphine.

175. Answer (4)

Hint : Purelines are homozygous individuals.

Solution : Purelines are developed by mating relatedmembers of a species for many generations.Purelines are homozygous for different traits leadingto reduction in allele variety in a population. Theyare more prone to inbreeding depression. Inbreedingdepression can be overcome by outcrossing.

176. Answer (2)

Hint : Native E.coli lacks resistance to antibiotics.

Solution : Plasmid was isolated from Salmonella

typhimurium and antibiotic resistance gene was

introduced in plasmid. Recombinant DNA was

introduced into E.coli.

177. Answer (2)

Hint : Anabolic steroids disturb hormone balance

which affects adiposis in males.

Solution : Breast enlargement occurs in males, as

steroids are converted into estrogen in body which

cause men to develop unwanted breast tissue

(adipose). Breast atrophy occurs in females.

178. Answer (3)

Hint : Name of restriction enzyme is derived from

name of source organism.

Solution :

Name of restriction enzyme isolated from

Haemophilus influenzae is based on genus and

species

H ind

Haemophilus

influenzae

179. Answer (4)

Hint : Alien DNA is first transferred into retrovirus.

Solution : Retrovirus carrying alien DNA infects host

cell and transfers alien DNA to the host cell. So, it

is an indirect method of gene transfer but the other

methods given are direct methods of transfer.

180. Answer (2)

Hint : AIDS is acquired during life time of an

individual.

Solution :AIDS is an immuno deficiency syndrome.

HIV can pass from mother to her child but AIDS

occurs when the immune system of a person is

compromised i.e when T-lymphocyte count falls

below 200/mm3 of blood.

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