All India Aakash Test Series for NEET - 2019 TEST - …...All India Aakash Test Series for NEET-2019...

Test - 2 (Code-A) (Answers) All India Aakash Test Series for NEET-2019

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1. (2)

2. (1)

3. (3)

4. (4)

5. (1)

6. (3)

7. (4)

8. (2)

9. (4)

10. (1)

11. (3)

12. (2)

13. (3)

14. (3)

15. (2)

16. (2)

17. (1)

18. (4)

19. (2)

20. (2)

21. (1)

22. (2)

23. (3)

24. (1)

25. (2)

26. (4)

27. (3)

28. (3)

29. (4)

30. (2)

31. (3)

32. (1)

33. (1)

34. (1)

35. (4)

36. (2)

Test Date : 07/10/2018

ANSWERS

TEST - 2 (Code-A)

All India Aakash Test Series for NEET - 2019

37. (2)

38. (2)

39. (2)

40. (2)

41. (2)

42. (1)

43. (3)

44. (4)

45. (1)

46. (4)

47. (1)

48. (3)

49. (3)

50. (4)

51. (3)

52. (1)

53. (4)

54. (4)

55. (1)

56. (1)

57. (2)

58. (2)

59. (3)

60. (1)

61. (2)

62. (1)

63. (4)

64. (4)

65. (2)

66. (3)

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68. (4)

69. (3)

70. (3)

71. (1)

72. (3)

73. (4)

74. (1)

75. (1)

76. (2)

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78. (3)

79. (4)

80. (3)

81. (1)

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86. (1)

87. (4)

88. (1)

89. (2)

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91. (3)

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96. (4)

97. (3)

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100. (2)

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105. (1)

106. (3)

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109. (4)

110. (3)

111. (1)

112. (4)

113. (2)

114. (4)

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118. (1)

119. (3)

120. (4)

121. (3)

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123. (1)

124. (1)

125. (2)

126. (4)

127. (2)

128. (1)

129. (3)

130. (2)

131. (4)

132. (2)

133. (4)

134. (3)

135. (3)

136. (1)

137. (3)

138. (4)

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140. (1)

141. (3)

142. (2)

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145. (2)

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147. (3)

148. (2)

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150. (1)

151. (4)

152. (3)

153. (2)

154. (1)

155. (4)

156. (4)

157. (2)

158. (4)

159. (1)

160. (3)

161. (1)

162. (1)

163. (4)

164. (4)

165. (2)

166. (4)

167. (3)

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170. (3)

171. (1)

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178. (2)

179. (1)

180. (3)

All India Aakash Test Series for NEET-2019 Test - 2 (Code-A) (Hints and Solutions)

2/21

PHYSICS

1. Answer (2)

Hint: mF q V B � � �

r

mF B� �

Sol.: r

mF B� �

therefore, r

a B�

�

a B 0 �

�

ˆ ˆ ˆ ˆ ˆ ˆ2 3 2 3 0i j k i bj k

2 + 2b – 3 = 0

1

2b

2. Answer (1)

Hint: Time period of charge particle in uniform

magnetic field 2 m

TqB

Sol.:2

qBf

m

P P

P

mf q

f q m

1 4

2 1 = 2

Pf f

3. Answer (3)

Hint: mF i l B � � �

effsin

mF i L B

Sol.:

Leff

= R

Fm

= IRB = 1

2 22

= 2 N

4. Answer (4)

Hint: mF I B � ��

l

M B � �

�

Sol.:effl sin

mF I B

0 sinm

F I B

0m

F

M B � �

�

= MB sin = Il2 B sin = 0

Both (1) and (2) are incorrect

5. Answer (1)

Hint:

0

34

dl rIB

r

��

�

�

Sol.:0

2 2arc

IB

R

�

0

2 2 2arc

IB

R

�

0

8arc

IB

R

�

0sin sin

4straight wire

IB

R

�

0I

sin90 sin04 R

I

R

0

4

net arc straight wireB B B � � �

0 0

8 4

I I

R R

⎛ ⎞ ⎜ ⎟⎝ ⎠

net

I

R

0B 1

4 2

⎛ ⎞ ⎜ ⎟ ⎝ ⎠

6. Answer (3)

Hint: Ampere circuital law

0 encB dl I ∫

��

�

Sol.:

IrB

R

0

22

when r R

for r > R

IB

r

0

2

Br

1

7. Answer (4)

Hint: M NIA� �

Sol.: M NIA� �

Magnetic dipole moment depends on

(1) Number of turn of the loop

(2) Current in the loop

(3) Area of the loop

HINTS & SOLUTIONS

Test - 2 (Code-A) (Hints and Solutions) All India Aakash Test Series for NEET-2019

3/21

8. Answer (2)

Hint:

2

0

3

2 2 22

axis

IRB

R X

Sol.:1

27axis centre

B B

2

0 0

3

2 2 2

1

27 22

IR I

RR X

3 3

2 2 2

1 1

2 272

RR X

2 23R X R

R2 + X

2 = 9R2

X

2 = 8R2

2 2X R9. Answer (4)

Hint: mF q v B � �

�

eF qE� �

Sol.: mF q v B � �

� ˆ ˆqvB i k ˆqvBj

eF qEjˆ�

As both magnetic and electric force are in opposite

direction, net force on the charge may be zero.

Therefore, particle may pass undeflected.

10. Answer (1)

Hint: M B � �

�

Sol.: MB sin �

M = NIA

qI

t

RI

2

2

RI

2

2

A = R2

22

2

RR B

= R3B

11. Answer (3)

Hint: Current sensitivity = I

Sol.:NAB

I K

Current sensitivity

So, current sensitivity depend on

(1) The number of turn in the coil

(2) The area of the coil

(3) The magnetic field applied

12. Answer (2)

Hint: 0 encB dl I ∫

��

�

Sol.:

0 encB dl I ∫

��

�

0

32

2

aB I

0

3P

IB

a

0 encB dl I ∫

��

�

0

5 32

2 2

a IB I

⎛ ⎞ ⎜ ⎟⎝ ⎠

0

10Q

IB

a

P

Q

B

B

10

3

⎛ ⎞ ⎜ ⎟⎝ ⎠

13. Answer (3)

Hint: 0

34

I dl r

B

r

��

�

�

Sol.: Magnetic field due to current carrying wire

0sin sin

4

IB

r

[ = 90°, = – 30°]

0 1

14 2

I

r

⎛ ⎞ ⎜ ⎟ ⎝ ⎠

0

8

IB

r

14. Answer (3)

Hint: Magnetic field due to bar magnet at its

equatorial and axial position respectively are M

r

0

34

and M

r

0

3

2

4

Sol.: N

S

S N

r

M

M

B1

B2


4/21

MB

r

0

1 3

2

4

M

Br

0

2 34

netB B B

2 2

1 2

M

r

0

3

5

4

15. Answer (2)

Hint: For paramagnetic material Curie's law

0

m

C

T

Sol.:0

m

C

T

m

TO

16. Answer (2)

Hint: B = H

Sol.:B

H

0.25

1000

= 2.5 × 10–4 TmA–1

0

r

r

4

7

2.5 10

4 10

= 0.2 × 103

r

22 10

17. Answer (1)

Hint: Wagent

= U + K

Sol.: Wagent

= U + K

Wagent

= Uf – U

i

Uf = – MB cos120°

MB

2

Ui = – MB cos 0

= – MB

Wagent

MBMB

2

agent

3

2

MBW

18. Answer (4)

Hint:V

H

B

Btan

Sol.:V

H

B

B

tan

sintan

cos cos

B

B

tantan

cos

tan60tan

cos30

3

3

2

1tan 2

19. Answer (2)

Hint: Tangent law

Sol.: H =

V

MBHsin = MB

Vsin(90 – )

tanV

H

B

B

= tan–1(3)

This from horizontal

1 1tan

3

⎛ ⎞ ⎜ ⎟⎝ ⎠

from vertical

20. Answer (2)

Hint:d

Ndt

Sol.: From 0 to t1 emf is zero because is

constant from t1 to t

2 emf is positive constant

because d

dt

is negative constant.

21. Answer (1)

Hint:d

Ndt

Sol.:d

dt

dt r

dt

2 2

04

= – 8tr0

2

|| = 16r2

0[at t = 2 s)

rI

R

2

016

22. Answer (2)

Hint: motional

� �

�

v B l

Sol.: = vBl cos60°

0 0

2

v B l

Force on positive charge will be toward end A

therefore, end A will be at higher potential.

23. Answer (3)

Hint: = MI


5/21

Sol.:

0sin45 sin45 4

4

⎡ ⎤ ⎢ ⎥⎣ ⎦

IB

d

IB

a

0 8

24

2

⎛ ⎞ ⎜ ⎟⎝ ⎠

IB

a

02 2

= BA

Ir

a

202 2

= MI

I rMI

a

2

02 2

rM

a

2

02 2

24. Answer (1)

Hint: rotation 0 0

0

l

B xdx ∫

Sol.:

2

3

0 0

0

l

PAB xdx ∫

2

2 3

0 0

02

l

PA

xB

⎡ ⎤ ⎢ ⎥

⎢ ⎥⎣ ⎦

2

0 04

9 2

B l

2

0 02

9PA

B l

3

0 0

0

l

PBB xdx ∫

2

0 0

18PB

B l

Both have opposite polarity

= PA

– PB

2 2

0 0 0 02

9 18

B l B l

2 2

0 0 0 04

18

B l B l

2

0 0

6

B l

25. Answer (2)

Hint: VP – V

Q = V

b + V

R + V

L

Sol.: VP – V

Q = 10 + 4 (2t + 3) + 2

dI

dt

VP – V

Q = 10 + 4 × 7 + 2 × 2

VP – V

Q = 10 + 28 + 4

= 42 V

26. Answer (4)

Hint:

tR

LE

I eR

1

⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠

Sol.:

R

SE

L

I

L

dIV L

dt

tR

LE

I eR

1⎛ ⎞

⎜ ⎟ ⎜ ⎟⎝ ⎠

⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠

tR

LdI E R

edt R L

tR

LdI E

edt L

L

dIV L

dt

tR

LE e

e

2

810

e

1

410

27. Answer (3)

Hint:d

dt

Sol.: Flux at any time (t)

= BAcos = BAcost

= B0a2 cost

d

dt

= – B0a2(– sint)

= B0a2sint

– B a0 2

B a0 2

tO


6/21

28. Answer (3)

Hint:dI

Ldt

Sol.:dI

Ldt

LdI

dt

0.080 5H

0.032 2 L

2AB

LL L

L3

2

3 5

2 2

15H

4

29. Answer (4)

Hint:L

R

V VI

Z V

0

0, tan

⎛ ⎞ ⎜ ⎟

⎝ ⎠

Sol.: Z2 2

4 3 Z = 5

= 53°

VR

VL

V

O

VI

Z

0

0

I0

20

5

I0 = 4A

LX I

RItan

4tan

3

⎛ ⎞ ⎜ ⎟⎝ ⎠

= 53°

In R – L circuit voltage leads the current

I = (4A)sin[(100rad/s)t – 53°]

30. Answer (2)

Hint:2 2 2

rms net dc rms acI I I

Sol.:2 2


. .

11Ar m sI

31. Answer (3)

Hint & Sol.:

In R – C circuit current lead the voltage

32. Answer (1)

Hint:T

t4

, here T is time period of the current.

Sol.: I = I0sin 50 t

= 50 rad/s

T2

T

T

4

t

I

1

25T s

Tt

4

1

100t s

t 10 ms33. Answer (1)

Hint: P = Vr.m.s

Ir.m.s

cosSol.: P = V

r.m.s Ir.m.s

cos

= Vr.m.s

Ir.m.s

R

Z

r m sV

. .

195 2

2

= 195 V

Z R

2

2

3

1

100 2 10

⎛ ⎞ ⎜ ⎟ ⎝ ⎠

Z2 2

12 5 Z = 13

r m sI. .

195

13

= 15 A

12195 15 2700 W

13 P

34. Answer (1)

Hint:avg

IdtI

dt

∫∫

Sol.:0

1

2 2avg

l bI

b

⎛ ⎞ ⎜ ⎟⎝ ⎠

0

4

I

35. Answer (4)

Hint: At resonance V

IR

1

2f

LC

And voltage across resistor, inductor and capacitor

will be in different phase.

Sol.: I200

20

I = 10 A

f

6

1

2 22 10


7/21

1000

4

= 250 Hz

36. Answer (2)

Hint: S S P PV I V I

80

100

Sol.: S S P PV I V I

80

100

SV

801 220 2

100

352 VS

V

37. Answer (2)

Hint:QQ

LIC C

222 01 1 1

2 2 2

Sol.:QQ

LIC C

222 01 1 1

2 2 2

QLI

C

221 3 1

2 4 2

QLI

C

221 4 1

2 3 2

⎛ ⎞ ⎜ ⎟⎜ ⎟

⎝ ⎠

22 2

01 4 1 1

2 3 2 2

QQ Q

C C C

⎛ ⎞ ⎜ ⎟⎜ ⎟

⎝ ⎠

QQ

22

0

7

3

Q Q0

3

7

⎛ ⎞ ⎜ ⎟⎜ ⎟⎝ ⎠

38. Answer (2)

Hint: Choke coil is device which is used to reduce

the current in ac circuit.

Sol.: Z R L22

To minimise the loss we keep the value of R less

and value of XL more.

39. Answer (2)

Hint: Direction of propagation is in the direction of

E B� �

Sol.: j n iˆ ˆˆ

n kˆ

ˆ

Ec

B

0

0

EB

c

8

6

3 10

= 2 × 10–8 T

8 8 ˆ2 10 cos 1.2 – 3.6 10 T

�

B x t k

40. Answer (2)

Hint: Poynting vector is defined as the flow of energy

in the direction of the propagation of a wave per unit

time through a unit cross sectional area

perpendicular to the propagation direction.

Sol.:E B

S

0

� �

�

41. Answer (2)

Hint: 0d

dEI A

dt

Sol.: 0d

dEI A

dt

0

dIdE

dt A

12

6

8.85 10 0.04

dE

dt

13 1 11.7 10 Vm s

dE

dt

42. Answer (1)

Hint:

2

0

20

1,

24

BPI I C

r

Sol.: P20

100100

P = 20 W

2

0

20

1

24

BPC

r

2

80

2 7

20 13 10

2 4 104 1

B

7

0

410 T

3

B

43. Answer (3)

Hint: 0ˆsinE E kx t i

�

0ˆsinB B kx t j

�

Sol.:

(1) Direction of propagation i j kˆ ˆ ˆ

(2) Speed of EM wave in vacuum

0 0

1

therefore

it will travel in vacuum.

(3)

2

2 0

0 0

0

1 1

2 2

BE


8/21

44. Answer (4)

Hint: Production of electromagnetic wave.

Sol.: Electromagnetic wave is combination of

electric and magnetic field only therefore, it can be

produced only by the accelerating charge.

45. Answer (1)

Hint: Wavelength 1

Frequency .

Sol.: Television and FM radio wave have low

frequency therefore wavelength will be longer.

CHEMISTRY

46. Answer (4)

Hint: Phosphinic acid is hypophosphorous acid.

Sol:

P OH

O

HH

(H PO )3 2

Oxidation state of phosphorus is +1.

Hypophosphorous acid is a good reducing agent

because it contains two P–H bonds and precipitates

Ag from AgNO3 solution

4AgNO3 + H

3PO

2 + 2H

2O 4Ag + H

3PO

4 + 4HNO

3

47. Answer (1)

Hint: More the number of unpaired electrons involved

in metallic bonding, higher will be the melting point.

Sol: In case of Cr, there are six unpaired electrons

in their ground state which provide strongest metallic

bonding whereas in case of Zn, the number of

unpaired electrons is zero and hence show weakest

metallic bonding.

48. Answer (3)

Hint: Spin only magnetic moment,

n(n 2) B.M. Sol: No. of unpaired electrons present in Fe2+ and

Cr2+ are equal to four.

4(4 2) 24 = 4.89 B.M.

49. Answer (3)

Hint: The coordination of EDTA ligand to the metal

atom takes place through two N atoms and four O

atoms forming five 5-membered rings.

Sol:

O

Fe

O

O

C

CH2

CH2

CH2

CH2

CH2

C

O

O

C

O

O

C

O

CH2N

N

50. Answer (4)

Hint:O

O

S

OHO H

(H SO )2 5

OOO

O

S

OHO

O

S

OH

(H S O )2 2 8

Sol:

O ,

S

HO

(H SO )2 3

O

O

S

O

(H S O )2 2 7

OH

O

S

OH OHO

O

S

O

O

S

O

OH , HO

(H S O )2 2 6

O

S

O

S OHHO

(H S O )2 2 4

51. Answer (3)

Hint:

4FeCr2O

4 + 8Na

2CO

3 + 7O

2

8Na2CrO

4 + 2Fe

2O

3 + 8CO

2

Sol: Yellow solution of Na2CrO

4 is filtered and

obtained as filtrate.

52. Answer (1)

Hint: Ba(N3)2

Ba + 3N2

Sol: 4 2 2 7 2 2 2 3

(NH ) Cr O N + 4H O Cr O

4 3NH Cl NH +HCl

3 2 2 22Pb (NO ) 4NO + 2PbO+O

4 3 2 2NH NO N O+2H O

53. Answer (4)

Hint: Due to stability of fully filled subshell 4f14 in D2+

Sol: Metal Third ionisation

enthalpy (kJ/mol)

A – Gd– 1990

B – Pr – 2086

C – Er – 2194

D – Yb – 2417

54. Answer (4)

Hint: CaCN2 + 3H

2O 2NH

3 + CaCO

3

Sol: NH3 is basic and hence can turn moist litmus

paper blue.

An alkaline solution of K2[HgI

4] is called Nessler’s

reagent. Ammonia is quantitatively estimated using

Nessler’s reagent.

55. Answer (1)

Hint: I2 is a weaker oxidising agent.


9/21

Sol: Cu2+ can oxidise I– to I2. That’s why CuI

2 does

not exist.

56. Answer (1)

Hint: XeF6 + 3H

2O XeO

3 + 6HF

Sol:

O

Xe

OO

(Pyramidal)

57. Answer (2)

Hint: Hypochlorous acid is HOCl

Sol:

1 2 1

HO Cl

58. Answer (2)

Hint: Malachite : CuCO3 · Cu(OH)

2

Azurite: 2CuCO3 Cu(OH)

2

Sol: Carnalite: KCl · MgCl2 · 6H

2O

Limonite: Fe2O

3 · 3H

2O

Calamine: ZnCO3

59. Answer (3)

Hint: Higher is the oxidation state of element, higher

will be the acidic nature of its oxide.

Sol: F2 is the strongest oxidising agent among all

halogens.

H2O H2S H2Se H2Te

273 188 208 222

Bond angle (°)

NH3 PH3 ASH3 S Hb 3

107.8 93.6 91.8 91.3

60. Answer (1)

Hint: 4 2 4 2 2513 K(Green) (Black)(A)

2KMnO K MnO MnO O

Sol: KMnO4 shows colour due to ligand to metal

charge transfer.

61. Answer (2)

Hint: I2 + 10HNO

3 2HIO

3 + 10NO

2 + 4H

2O

(conc.)

Sol: Let the oxidation state of I in HIO3 is x

1 + x + 3(–2) = 0 x = + 5

62. Answer (1)

Hint: With limited NH3, N

2 and HCl are formed.

Sol: On reaction with excess ammonia, chlorine

gas gives nitrogen and ammonium chloride.

8NH3 + 3Cl

2 6NH

4Cl + N

2

(Excess)

63. Answer (4)

Hint: Ce4+, Lu3+ and Yb2+ are diamagnetic.

Sol: Eu2+ : [Xe]4f 7

Ce4+ : [Xe] 4f 0

Lu3+ : [Xe] 4f 14

Yb2+ : [Xe] 4f 14

64. Answer (4)

Hint: O

O

P

P

O

P

O

O

P

O

OO O

O

Sol: Number of bridging oxygen atom is equal to 6.

65. Answer (2)

Hint: ClF5 is sp3d2 hybridised.

Sol:

Cl

F

FF

FF

Br

F

F

F , l F

F

F

F

F

F

F

66. Answer (3)

Hint: CrO3 – Acidic

CrO – Basic

Cr2O

3 – Amphoteric

Sol: V2O

5 – Acidic

V2O

3 – Basic

67. Answer (2)

Hint: KMnO4 has highest +7 oxidation state of Mn.

Sol: PbS + 4O3 PbSO

4 + 4O

2

2Hg + O3 Hg

2O + O

2

2K4[Fe(CN)

6] + H

2O + O

3 2K

3[Fe(CN)

6] + 2KOH + O

2

68. Answer (4)

Hint: Ce3+(4f

1) is colourless.

Sol: Eu2+ is a good reducing agent changing to

common +3 oxidation state actinoids show greater

range of oxidation states because 5f, 6d and 7s

levels are of comparable energies.


10/21

69. Answer (3)

Hint: XeF2 + PF

5 [XeF ]+ [PF

6]–

Sol:

F

FF

FFF

Hybridisation of

P : sp d3 2

d

–

70. Answer (3)

Hint: In neutral and faintly alkaline solutions, n

factor for MnO4

–

is 3.

Sol:

8MnO4

–

+3S2O

3

2– + H2O 8MnO

2 + 6SO

4

2–

+ 2OH

71. Answer (1)

Hint: K2[HgI

4] is a soluble complex.

Sol: HgCl2 + 2KI HgI

2 + 2KCl

(Red ppt.)

Hint: HgI2 + 2KI K

2 [HgI

4]

(Soluble)

72. Answer (3)

Hint: Wilkinson's catalyst is [(Ph3P)

3 RhCl].

Sol: Organometallic compounds contain at least one

chemical bond between metal and carbon atom of an

organic molecule.

Grignard reagent RMgBr

Tetracarbonyl nickel [Ni(CO)4]

Ferrocene [Fe(5 – C5H

5)2] Fe

73. Answer (4)

Hint: If complex is neutral, then the word ‘ate’ is not

used with the name of metal.

Sol: Triamminetrinitrito-N-cobalt (III)

74. Answer (1)

Hint: The oxides having high melting points are

difficult to reduce by carbon reduction method.

Sol: PbO + C Pb + CO

75. Answer (1)

Hint: [Co(en) (NH3)

2Cl

2]+ exists as cis-trans

isomers. Trans-isomer is optically inactive.

Sol: H N

3

enCo

ClCl

(cis)

Co

ClCl

en

NH3

NH3

NH3

optically active

en ,Co

Cl

Co

Cl

en

ClNH3

H N3

NH3

Cl

H N3

optically inactive

76. Answer (2)

Hint: In blast furnace at 900–1500 K (higher

temperature range)

C + CO2 2CO

FeO + CO Fe + CO2

Sol: In blast furnace at 500-800 K (lower

temperature range)

3Fe2O

3 + CO 2Fe

3O

4 + CO

2

Fe3O

4 + 4CO 3Fe + 4CO

2

Fe2O

3 + CO 2FeO + CO

2

77. Answer (2)

Hint: The effective nuclear charge of the metal ion

increases as we move down the group from 3d to 5d

series. Thus, 0 increases.

Sol: Higher the strength of the ligand, higher will be

the value of 0.

Higher the oxidation state of the metal, higher will be

the value of 0.

78. Answer (3)

Hint: Spectrochemical series

Sol: N < C O < NO < CN

3

– 2– – –

2 4 2

Ligand strengthfield

79. Answer (4)

Hint: Distillation is useful for low boiling metals like

zinc and mercury.

Sol: Cu is refined by electrolytic refining.

80. Answer (3)

Hint: H2O and C

2O

4

2– acts as strong field ligand for

Co3+.

Sol:

Complex No. of unpaired electrons

1. [Co(H2O)

6]3+ Zero

2. [Co(C2O

4)3]3– Zero

3. [FeF6]4– Four

4. [Fe(H2O)

5NO]2+ Three

81. Answer (1)

Hint: EAN of central atom = Atomic number of metal

atom – oxidation number of metal atom + No. of

electrons donated by ligands.

Sol: Both [Co(en)3]2+ and [Cr(H

2O)

6]2+ have three

unpaired electrons. The hybridisation of metal atom

in [Cu(NH3)4]2+ and [Ni(CN)

4]2– is dsp2.


11/21

EAN of [Mn(H2O)

6]2+ = 25 – 2 + 12 = 35

EAN of [Fe(CN)6]3– = 26 – 3 + 12 = 35

[Cr(gly)3] and [Pt(gly)

2] has cis and trans-isomers.

82. Answer (3)

Hint: Strong Jahn-Teller distortion is observed for d9

configuration 6 32g g(t e ) .

Sol: 2+ 3 2

2 6 2g g[Mn(H O) ] : t e

2+ 6 42 6 2g g[Zn(H O) ] : t e

2+ 6 33 6 2g g[Cu(NH ) ] : t e

3+ 3 03 6 2g g[Cr(NH ) ] : t e

83. Answer (2)

Hint: Primary valency is number of charges on the

central metal atom and secondary valency is equal

to coordination number.

Sol: Oxidation state of Co is +3. Coordination no.

of Co is 6(4NH3 and 2Cl–).

84. Answer (4)

Hint: In coordination compounds, colour arises due

to d – d transition or charge transfer.

Sol: Complexes of Ti (IV), Zn (II) and Sc (III) are

colourless.

[Co(CN)6]3– yellow, [CoF

6]3– Green

[Ni(NO2)6]4– Violet

85. Answer (2)

Hint: Wrought iron is the purest from of commercial

iron.

Sol: Extraction of gold and silver involves leaching

the metal with CN–. The metal is later recovered by

displacement method.

86. Answer (1)

Hint: Synergic bonding is least in [Mn(CO)6]+.

Sol: Due to the presence of positive charge,

synergic bonding from metal to carbonyl ligand will

decrease. Hence, M – C bond length will increase

and C – O bond length will decrease.

87. Answer (4)

Hint: Complex having unpaired electrons will be

paramagnetic.

Sol: Outer orbital complexes are having outer shell

d-orbitals in its hybridization.

88. Answer (1)

Hint: CSFE = [– 0.4 (nt2g

) + 0.6(neg

)] 0

Sol:

eg

+0.6 0

t2g

–0.40

d6

CFSE = – 6 × 0.4 0 + 0 = –2.4

0

89. Answer (2)

Hint: Froth stabilisers such as cresols, aniline

stabilise the froth.

Sol: Collectors such as xanthates enhance

non-wettability of mineral particles in water.

90. Answer (1)

Hint: CN– is an ambidentate ligand.

Sol: Linkage isomerism is shown by the ligands

which contain more than one atom which could

donate an electron pair.

BIOLOGY

91. Answer (3)

Hint: Polypeptide synthesis takes place from mRNA.

Sol.: mRNA is synthesized from DNA and has

information for protein synthesis.

Detailed Sol.: mRNA, tRNA and rRNA are directly

synthesized from DNA.

92. Answer (3)

Hint: Termination codons are stop codons.

Sol.: UAA, UAG, UGA are stop codons.

Detailed Sol.: All stop codons of universal genetic

codes begin with uracil.

93. Answer (1)

Hint: Nucleoside has sugar and nitrogenous base.

Sol.: A nitrogenous base is linked to a pentose

sugar by N-glycosidic linkage.

94. Answer (2)

Hint: Replication is copying of DNA. In DNA Uracil

is absent, it is present in RNA.

Sol.: Chargaff’s rule is not applicable for single

stranded DNA.

Detailed Sol.: Repetitive DNA is the part of DNA

which contains the same sequence of nitrogen bases

repeated more than once in a genome and the area

with long sequence of short repetitive DNA is called

satellite DNA.

95. Answer (1)

Hint: DNA with more G C content has high melting


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point.

Sol.: DNA with more A = T pairs, has low melting

point and get denatured more easily. DNA with more

G C pairs than A = T pairs has high melting point.

96. Answer (4)

Hint: Nucleosomes, the structure in chromatin are

seen as ‘beads-on-string’

Sol.: Nucleosomes have histone octamer, i.e.

organised form of H2A, H

2B, H

3 and H

4 histones. H

1

histone is not a part of histone octamer.

97. Answer (3)

Hint: In reverse transcription DNA is synthesised

from RNA.

Sol.: E.coli, Lambda phage and T4 bacteriophage have

DNA as genetic material.

Detailed Sol.: HIV is a retrovirus in which flow of

information takes place from RNA to DNA.

98. Answer (4)

Hint: hn RNA is converted into functional m-RNA by

post transcriptional process.

Sol.: Splicing occurs in nucleus in eukaryotes

whereas in prokaryotes it is absent as introns are

absent except in Archaebacteria.

Detailed Sol.: Post transcriptional modifications

involve splicing, capping and tailing which are absent

in prokaryotes but are found in eukaryotes.

99. Answer (3)

Hint: Nonsense codon is stop codon, i.e., UAG.

Sol.: Start codon is AUG. GUG is ambiguous

codon. UGG codes for amino acid Tryptophan.

100. Answer (2)

Hint: Cairns used tritiated thymidine in

autoradiography experiment.

Sol.: By using tritiated thymidine in auto radiography

experiment; Cairns proved semi-conservative mode of

DNA replication in E.coli.

101. Answer (4)

Hint: This enzyme is a ribozyme.

Sol.: Peptidyl transferase is an RNA enzyme rather

being proteinaceous. In case of prokaryotes 23 S

rRNA acts as ribozyme whereas in eukaryotes it is

28 S rRNA. It catalyses the peptide bond formation

between amino acids.

102. Answer (4)

Hint: Dystrophin is the largest gene.

Sol.: Regulator and operator are part of lac operon.

Detailed Sol.: SRY (sex determining region Y) is

the smallest gene codes for TDF (Testis determining

factor).

103. Answer (2)

Hint: It is the process of genetic recombination in

which donor and recipient do not come in contact.

Sol.: Griffith discovered transformation process in

bacteria Streptococcus pneumoniae. In which the

donor cell releases a piece of DNA which is actively

taken up by the recipient cell from the solution.

104. Answer (4)

Hint: First step of translation mechanism is

aminoacylation of tRNA.

Sol.: mRNA first binds to smaller subunit of

ribosome then initiator tRNA comes to P site of

ribosome.

Detailed Sol.: Formation of polypeptide is the final

step in protein synthesis.

105. Answer (1)

Hint: Satellite DNA is repetitive DNA.

Sol.: Repetitive DNA is heterochromatin region of

DNA.

In bacteria, larger subunit of ribosome has 23S rRNA

and 5S rRNA.

Detailed Sol.: During protein synthesis 23S rRNA

acts as ribozyme.

106. Answer (3)

Hint: The mRNA has complementary sequence of

nitrogenous bases to template strand.

Sol.: ATCTGGCGT is the sequence of DNA on

template strand then the sequence of mRNA will be

UAGACCGCA. In RNA uracil is present instead of

thymine.

107. Answer (3)

Hint: Lac operon is inducible operon system.

Sol.: Lactose acts as an inducer. Regulation of lac

operon by repressor protein is negative regulation.

Detailed Sol.: When lactose is absent repressor

binds to the operator and in the presence of lactose

repressor binds to lactose.

108. Answer (3)

Sol.: Direction of both RNA and DNA synthesis is 5'

to 3'.

109. Answer (4)

Hint: RNA polymerase I synthesizes 28S rRNA.

Sol.: tRNA and 5S rRNA are synthesized by RNA

polymerase III.

110. Answer (3)

Hint: It is used in analysis of a gene for disease, in

identifying organisms etc.

Sol.: RFLP is not used in determining the structures

of proteins.


13/21

111. Answer (1)

Hint: Griffith used Streptococcus or Diplococcus

bacteria for his experiment.

Sol.: QB Bacteriophage has RNA as genetic

material.

Taylor et. al. proved semiconservative mode of DNA

replication in Vicia faba.

Hershy and Chase experimented with bacteriophage.

112. Answer (4)

Hint: Minisatellites are repetitive DNA.

Sol.: Minisatellites are also known as VNTR and

surrounded by conserved restriction sites.

113. Answer (2)

Hint: In eukaryotes both introns and exons are

present.

Sol.: hnRNA is processed in the nucleus and after

processing it is called mRNA that contains exons

only.

Detailed Sol.: The intervening sequences in

between coding regions are called introns. Exons

are coding regions of genes. Therefore, eukaryotic

genes are split genes.

114. Answer (4)

Hint: Genetic variability increases chances of getting

desirable characters.

Sol.: Genetic variability is root of any breeding

programme and we can choose better characters.

Hybridisation increases hybrid vigours.

115. Answer (3)

Sol.: During green revolution, production of wheat

and rice led to increase in food production.

116. Answer (4)

Hint: Spirogyra is a filamentous green alga.

Sol.: Spirulina is used to produce SCP but Spirogyra

is not used for such production.

117. Answer (3)

Hint: Genetically identical offsprings are called

clones.

Sol.: A somatic cell produces identical clones called

somaclones during tissue culture.

Somatic hybrids are obtained by fusion of protoplast

of two different species, varieties.

Cybrid is produced during somatic hybridisation of

two different cytoplasms, if one of the nuclei get

degenerated.

118. Answer (1)

Hint: Downy mildew of grapes is a fungal disease.

Sol.: Bordeaux mixture is effective against fungal

diseases.

Use of disease resistant varieties protect plant from

infection of microbes.

119. Answer (3)

Hint: Jaya and Ratna are rice varieties.

Sol.: Himgiri is a wheat variety which is disease

resistant.

Atlas 66 has high protein content and has been

used as a donor for improving cultivated wheat.

120. Answer (4)

Hint: Triticale was first man-made crop produced by

hybridisation of wheat and rye.

Sol.: Parbhani Kranti is a variety of okra produced

by hybridisation of two varieties through conventional

breeding.

121. Answer (3)

Hint: Biofertilisers improve the soil fertility.

Sol.: Biofertilisers are organisms which enrich the

nutrient quality of soil by increasing the organic and

inorganic matter in the soil.

122. Answer (3)

Hint: Gobar gas is biogas.

Sol.: Xanthomonas campestris is a bacterial species

that causes a variety of plant disease including black

rot in crucifers .

Detailed Sol.: The sediment of settling tank is called

activated sludge. Which are of bacterial and fungal

flocs.

Lipases are used in detergent formulation.

123. Answer (1)

Hint: BOD Polluting potential of water.

Flocs are masses of fungi and bacteria.

Sol.: Biogas is inflammable hence used as source

of energy.

Heterotrophic microbes are naturally present in the

sewage and treat waste water.

124. Answer (1)

Hint: A yeast is used to produce statins.

Sol.: Monascus purpureus is a yeast which is used

to produce statins which is blood cholesterol

lowering agent.

125. Answer (2)

Hint: It is a bacterium, used to produced vinegar.

Sol.: Penicillium and Aspergillus are fungi.

Acetobacter aceti is used to produce vinegar

commercially.


14/21

126. Answer (4)

Hint: Methane is the major component of biogas

which is 50-70%.

Sol.: Biogas contains methane, carbon dioxide, H2,

H2S, etc.

127. Answer (2)

Hint: Rhizobium is used as a biofertiliser in

leguminous plants.

Sol.: Soyabean is a leguminous plant.

Organic farming uses naturally produced compost.

Anaerobic sludge digesters produce biogas due to

anaerobic digestion of organic material.

128. Answer (1)

Hint: Fruit juice is clarified due to the degradation of

some cell wall material.

Sol.: Pectinases and proteases help in clarifying fruit

juices.

Lipases are used in detergent formulations.

Amylases degrade starch, streptokinase is used as

clot buster.

129. Answer (3)

Sol.: To prevent the discharge of untreated sewage

into Ganga and Yamuna rivers, The Ministry of

Environment and Forests has initiated Ganga and

Yamuna action plan.

130. Answer (2)

Hint: Secondary treatment is biological treatment.

Sol.: Primary treatment of waste water is mainly a

physical process which involves sequential filtration

and sedimentation.

131. Answer (4)

Sol.: Biocontrol agents are non toxic, do not kill

useful organisms and keep pests at manageable

levels.

132. Answer (2)

Hint: They are species specific.

Sol.: Nucleopolyhedroviruses do not have side effect

on plants and animals.

They are narrow spectrum insecticides. They

conserve beneficial insects.

133. Answer (4)

Hint: These are oxygenic eubacteria.

Sol.: Cyanobacteria can fix atmospheric nitrogen

such as Nostoc and Oscillatoria.

Glomus – Fungi

134. Answer (3)

Hint: Antibiotics are effective against bacterial

diseases.

Sol.: Diphtheria, whooping cough and pneumonia

are bacterial diseases and can be cured by taking

antibiotics.

Curd is more nutritious than milk as it contains

vitamin B12

.

135. Answer (3)

Hint: Toddy is obtained from fermenting sap of

Caryota urens.

Sol.: It is a traditional drink of some parts of South

India.

Caryota urens is scientific name of a palm.

136. Answer (1)

Hint: Changes leading to evolution of organisms.

Sol.: Study of external features of organisms is

known as morphology and physiology is the science

of body functions that how the body parts work.

Biogenesis is formation of organisms from pre

existing organisms.

137. Answer (3)

Hint: Theory of panspermia.

Sol.: Life arises from non-living matter according to

theory of spontaneous generation.

138. Answer (4)

Hint: A single huge explosion formed Universe.

Sol.: Origin of species was explained by Charles

Darwin and Origin of life by A.I. Oparin. Origin of

Universe is explained by Big-Bang hypothesis.

139. Answer (4)

Hint: Equus is modern present day horse.

Sol.: Pliohippus was first one toed horse.

140. Answer (1)

Hint: Descent with modification.

Sol.: Key points of evolution i.e., descent with

modifications and natural selection were proposed by

Darwin. Darwin stressed on reproductive fitness.

Saltation was explained by De Vries and genetic drift

by Sewall & Wright.

141. Answer (3)

Hint: Origin of life occurred in water.

Sol.: Plants evolved prior to animals. Sauropsids

were ancestors of thecodonts.

142. Answer (2)

Hint: Given study is based on the genetic material

which is transmitted only by females.

Sol.: Older the organism more will be the

accumulation of variations, in its mitochondrial DNA.

Mitochondrial DNA is uniparental and does not

undergo recombination unlike nuclear DNA.


15/21

143. Answer (3)

Hint: Eyes of cephalopod molluscs are analogous to

vertebrate eyes.

Sol.: Nictating membrane is a vestigial organ and

presents anatomical evidence. In Miller’s experiment

of chemical evolution, CH4, NH

3, H

2 and water

vapour were used which on exposure to electric

discharge resulted in formation of amino acids.

Industrial melanism is an example of natural

selection.

144. Answer (1)

Hint: Tasmanian tiger cat is an Australian marsupial.

Sol.: Wombat, Bandicoot and Spotted cuscus are

Australian marsupials whereas Bobcat is a placental

mammal.

145. Answer (2)

Hint: Elongated neck of giraffe can be explained by

continuous stretching according to this theory.

Sol.: All acquired characters of a generation are

passed on to the next one according to Lamarck.

146. Answer (1)

Hint: Common ancestral finch was a herbivore.

Sol.: Parent finch was seed eating and other forms

arose with altered beaks enabling them to become

insectivorous.

147. Answer (3)

Hint: Bipedal gait evolved prior to verbal language.

Sol.: Increased cranial capacity led to natural

selection. Erect posture appeared in

Australopithecines.

148. Answer (2)

Hint: This is a case of stabilising selection.

Sol.: In case of stabilizing selection, average mean

value is selected so that average phenotype exhibits

higher and narrower peak. Two peaks are obtained in

disruptive selection.

149. Answer (2)

Hint: Simpler molecules give rise to complex

organisms.

Sol.: During course of origin of life following events

take place in given order.

1. Synthesis of organic monomers.

2. Synthesis of organic polymers.

3. Formation of protobionts with RNA

4. Formation of protobionts with DNA based

genetic systems.

150. Answer (1)

Hint: Adaptive radiation in ancestral stock of

Australian marsupials and placental mammals

separately.

Sol.: A number of marsupials, each different from the

other, evolved from an ancestral stock, but all within

Australian island. Divergent evolution, represented

homology while convergent evolution represents

analogy. Saltation refer to large mutation leading to

variation according to mutation theory.

151. Answer (4)

Hint: Presence of artificial selection alters Hardy-

Weinberg principle.

Sol.: The factors which affect genetic equilibrium in

a population are presence of (1) genetic drift (2) gene

migration/gene flow (3) mutation (4) genetic

recombination (5) natural selection and (6) lack of

random mating.

152. Answer (3)

Hint: Melanic forms have selective advantage and

inheritance of this character was naturally selected.

Sol.: Natural selection brings about evolution

Industrial melanism is an example of directional type

of natural selection. In a mixed population having

different variants of moths, those that are better

adapted will survive and will be naturally selected.

Such as after industrialization, melanic forms could

camouflage themselves in the background.

153. Answer (2)

Hint: Identify a case of Sewall-Wright effect.

Sol.: When gene migration occurs many time, there

will be gene flow. If the change in allele frequency

occurs by chance, it is called genetic drift. Following

two effects are ramifications of genetic drift i.e., (1)

Founder’s effect (2) Bottleneck effect.

154. Answer (1)

Hint: According to Ernst Haeckl, it is summarised as

biogenetic law.

Sol.: Ontogeny is development of embryo and

phylogeny is the ancestral sequence. Biogenetic law

states that ontogeny is recapitulation of phylogeny.

As per this law, the sequence of embryonic

development in different vertebrates show striking

similarities.

155. Answer (4)

Hint: Common ancestry in different organisms.

Sol.: Organs which have same origin but different

function are called homologous organs. e.g.

Forelimbs of human, horse and bat.


16/21

156. Answer (4)

Hint: Homo erectus appeared about 1.5 mya.

Sol.: Cranial capacities of Cro-Magnon man,

H.erectus, H.habilis and H.neanderthalensis are

1650, 900, 650-800 & 1400 cc respectively.

Neanderthal man buried his dead with flowers and

tools.

157. Answer (2)

Hint: Large population has constant allelic frequency

in absence of evolutionary factors.

Sol.:

According to HW law,

p + q = 1

p + 0.4 = 1

p = 0.6

frequency of carriers (heterozygous) = 2 pq

= 2 × 0.4 × 0.6 = 0.48

= 48%

So, number of carriers = 96 among 2000 individuals

2000 48960

100

⎛ ⎞ ⎜ ⎟⎝ ⎠

158. Answer (4)

Hint: Age of fossils based on position and pairing of

electrons.

Sol.: Electron spin resonance measures no. of

unpaired electrons in crystalline structures which

were previously exposed to radiation. Age of fossil is

determined by measuring dosage of radiations.

159. Answer (1)

Hint: Birds have shelled calcareous eggs.

Sol.: Presence of scales on hind limbs and shelled/

cleidoic eggs are considered as reptilian ancestry of

birds because both features together appeared first

in reptiles but are also found in birds. Birds lack

teeth and have a four chambered heart.

160. Answer (3)

Hint: Golden Age of reptiles.

Sol.: Mesozoic era is considered as golden age of

reptiles because reptiles were dominant. Jurassic

period is considered as golden age of Dinosaurs.

161. Answer (1)

Hint: Mutations are non directional in nature.

Sol.: Evolution is a non-directional, stochastic

process based on chance events and chance

mutations. Evolution is occurring on a fast pace due

to anthropogenic interference.

162. Answer (1)

Hint: Bond formed between two sulphur containing

amino acids.

Sol.: Heavy and light chains are connected to each

other by disulphide bonds between cysteines.

163. Answer (4)

Hint: Pathogen causes filariasis, a helminthic

disease.

Sol.: Ascaris causes ascariasis, E.histolytica cause

amoebiasis, Streptococcus pneumoniae causes

pneumonia. Wuchereria causes elephantiasis/

filariasis which is chronic inflammation of lymphatic

vessels of lower limb.

164. Answer (4)

Hint: This disorder is characterised by presence of

swollen, reddened, running eyes, nose and anti-

histamines are given to the patient.

Sol.: Myasthenia gravis, rheumatoid arthritis and

vitiligo are autoimmune disorders. Hay fever is a type

of allergy.

165. Answer (2)

Hint: Widal test is the diagnostic method of

diseases caused by bacteria S.typhi.

Sol.: Typhoid is caused by a bacterium Salmonella

typhi through contaminated food and water.

Wassermann test is used to detect syphilis.

166. Answer (4)

Hint: Identify a viral disease.

Sol.: Amoebic dysentery caused by E. histolytica,

ascariasis by Ascaris, typhoid fever by Salmonella

typhi, are transmitted through faeco-oral route.

Chikungunya is a vector borne disease transmitted

by Aedes mosquito.

167. Answer (3)

Hint: Disease characterised by sneezing and

watering from eyes.

Sol.: In common cold, only upper respiratory tract is

involved but lungs and alveoli remain unaffected.

168. Answer (3)

Hint: Anti-histamine drugs are used to counter

hypersensitivity of a person to foreign substances.

Sol.: Common cold reaction and dengue are viral

diseases. Common cold is caused by Rhino virus.

Dengue is transmitted by Aedes mosquitoes.

169. Answer (3)

Hint: Gambusia is an example of larvivorous fish.

Sol.: Cholera is a bacterial disease caused by Vibrio

cholerae. Larvae of vectors such as Anopheles

mosquito are kept under control by larvivorous fish

such as Gambusia.


17/21

170. Answer (3)

Hint: Asexual reproduction (schizogony) occurs in

erythrocytes & hepatocytes.

Sol.: Malarial parasites are sporozoans and

reproduce asexually in hepatocytes as well as in

RBCs of man. Sporozoite forms of Plasmodium are

stored in salivary glands of infected female

Anopheles.

Mature infective stage (sporozoite) escapes from gut

and migrates to the mosquito’s salivary glands.

171. Answer (1)

Hint: Fungus growing in ring shape on skin.

Sol.: Ringworm infection are caused by fungi such

as Trichophyton, Epidermophyton & Microsporum.

172. Answer (2)

Hint : Select antibodies chiefly responsible for

allergic response.

Sol.: IgE activates mast cells that release histamine

that acts as vasodilator & bronchoconstrictor.

173. Answer (3)

Hint: One of the disease in given pairs could be

confirmed by Widal test.

Sol.: Dysentery, diphtheria, typhoid and plague are

bacterial diseases. Dengue is a vector borne

protozoan disease. Polio is a viral disease and

Ascariasis is a helminthic disease.

174. Answer (3)

Hint: Organ which acts as “filter of blood” is known

as graveyard of RBCs.

Sol.:

(a) Spleen : Filter of the blood (graveyard of RBCs)

(b) Appendix : Secondary lymphoid organ

(vestigial in human)

(c) Thymus : Primary lymphoid organ where T

cells mature.

(d) Bone marrow : Production house of all types

of lymphocytes (i.e., B & T cells)

175. Answer (3)

Hint: Preformed antibodies are administered in

passive immunity.

Sol.: ATS against tetanus, antivenin against

snakebites contain ready made antibodies.

Injecting tetanus toxoid triggers active immunity.

176. Answer (4)

Hint: Macrophages like neutrophils, participate in

phagocytosis to destroy microbes.

Sol.: Skin and mucous are considered as physical

barrier, tear and saliva as physiological barrier.

Interferons as cellular barrier for uninfected cells and

macrophages as cellular barrier in tissues NK cells,

PMNL and monocytes as cellular barriers in blood.

177. Answer (4)

Hint: Cells that originate and mature in bone marrow

produce antibodies.

Sol.: Antibodies are proteinaceous in nature and are

produced by B-lymphocytes only.

T-lymphocytes facilitate B-lymphocytes to produce

antibodies.

178. Answer (2)

Hint: A primary lymphoid organ located near heart

and beneath the breastbone and provides the site for

T cell maturation.

Sol.: Thymus is quite large at the time of birth but

keeps reducing in size with age and by the time of

puberty reduces to a very small size (atrophy).

179. Answer (1)

Hint: Amino terminal of light chains of antibody

interacts with epitope of antigen.

Sol.: Variable part of N/amino terminal of both heavy

& light chains form antigen binding site (paratope) of

antibody that recognises epitope of antigen.

180. Answer (3)

Hint: Acquired immunity is pathogen specific.

Sol.: Innate immunity is non-specific and is not

characterised by memory.

Test - 2 (Code-B) (Answers) All India Aakash Test Series for NEET-2019

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1. (1)

2. (4)

3. (3)

4. (1)

5. (2)

6. (2)

7. (2)

8. (2)

9. (2)

10. (2)

11. (4)

12. (1)

13. (1)

14. (1)

15. (3)

16. (2)

17. (4)

18. (3)

19. (3)

20. (4)

21. (2)

22. (1)

23. (3)

24. (2)

25. (1)

26. (2)

27. (2)

28. (4)

29. (1)

30. (2)

31. (2)

32. (3)

33. (3)

34. (2)

35. (3)

36. (1)

Test Date : 07/10/2018

ANSWERS

TEST - 2 (Code-B)

All India Aakash Test Series for NEET - 2019

37. (4)

38. (2)

39. (4)

40. (3)

41. (1)

42. (4)

43. (3)

44. (1)

45. (2)

46. (1)

47. (2)

48. (1)

49. (4)

50. (1)

51. (2)

52. (4)

53. (2)

54. (3)

55. (1)

56. (3)

57. (4)

58. (3)

59. (2)

60. (2)

61. (1)

62. (1)

63. (4)

64. (3)

65. (1)

66. (3)

67. (3)

68. (4)

69. (2)

70. (3)

71. (2)

72. (4)

73. (4)

74. (1)

75. (2)

76. (1)

77. (3)

78. (2)

79. (2)

80. (1)

81. (1)

82. (4)

83. (4)

84. (1)

85. (3)

86. (4)

87. (3)

88. (3)

89. (1)

90. (4)

91. (3)

92. (3)

93. (4)

94. (2)

95. (4)

96. (2)

97. (3)

98. (1)

99. (2)

100. (4)

101. (2)

102. (1)

103. (1)

104. (3)

105. (3)

106. (4)

107. (3)

108. (1)

109. (3)

110. (4)

111. (3)

112. (4)

113. (2)

114. (4)

115. (1)

116. (3)

117. (4)

118. (3)

119. (3)

120. (3)

121. (1)

122. (4)

123. (2)

124. (4)

125. (4)

126. (2)

127. (3)

128. (4)

129. (3)

130. (4)

131. (1)

132. (2)

133. (1)

134. (3)

135. (3)

136. (3)

137. (1)

138. (2)

139. (4)

140. (4)

141. (3)

142. (3)

143. (3)

144. (2)

145. (1)

146. (3)

147. (3)

148. (3)

149. (3)

150. (4)

151. (2)

152. (4)

153. (4)

154. (1)

155. (1)

156. (3)

157. (1)

158. (4)

159. (2)

160. (4)

161. (4)

162. (1)

163. (2)

164. (3)

165. (4)

166. (1)

167. (2)

168. (2)

169. (3)

170. (1)

171. (2)

172. (1)

173. (3)

174. (2)

175. (3)

176. (1)

177. (4)

178. (4)

179. (3)

180. (1)

All India Aakash Test Series for NEET-2019 Test - 2 (Code-B) (Hints and Solutions)

2/21

PHYSICS

1. Answer (1)

Hint: Wavelength 1

Frequency .

Sol.: Television and FM radio wave have low

frequency therefore wavelength will be longer.

2. Answer (4)

Hint: Production of electromagnetic wave.

Sol.: Electromagnetic wave is combination of

electric and magnetic field only therefore, it can be

produced only by the accelerating charge.

3. Answer (3)

Hint: 0ˆsinE E kx t i

�

0ˆsinB B kx t j

�

Sol.:

(1) Direction of propagation i j kˆ ˆ ˆ

(2) Speed of EM wave in vacuum

0 0

1

therefore it

will travel in vacuum.

(3)

2

2 0

0 0

0

1 1

2 2

BE

4. Answer (1)

Hint:

2

0

20

1,

24

BPI I C

r

Sol.: P20

100100

P = 20 W

2

0

20

1

24

BPC

r

2

80

2 7

20 13 10

2 4 104 1

B

7

0

410 T

3

B

5. Answer (2)

Hint: 0d

dEI A

dt

Sol.: 0d

dEI A

dt

0

dIdE

dt A

12

6

8.85 10 0.04

dE

dt

13 1 11.7 10 Vm s

dE

dt

6. Answer (2)

Hint: Poynting vector is defined as the flow of

energy in the direction of the propagation of a wave

per unit time through a unit cross sectional area

perpendicular to the propagation direction.

Sol.:E B

S

0

� �

�

7. Answer (2)

Hint: Direction of propagation is in the direction of

E B� �

Sol.: j n iˆ ˆ

ˆ

n kˆ

ˆ

Ec

B

0

0

EB

c

8

6

3 10

= 2 × 10–8 T

8 8 ˆ2 10 cos 1.2 – 3.6 10 T

�

B x t k

8. Answer (2)

Hint: Choke coil is device which is used to reduce

the current in ac circuit.

Sol.: Z R L22

To minimise the loss we keep the value of R less

and value of XL

more.

HINTS & SOLUTIONS

Test - 2 (Code-B) (Hints and Solutions) All India Aakash Test Series for NEET-2019

3/21

9. Answer (2)

Hint:QQ

LIC C

222 01 1 1

2 2 2

Sol.:QQ

LIC C

222 01 1 1

2 2 2

QLI

C

221 3 1

2 4 2

QLI

C

221 4 1

2 3 2

⎛ ⎞ ⎜ ⎟⎜ ⎟

⎝ ⎠

22 2

01 4 1 1

2 3 2 2

QQ Q

C C C

⎛ ⎞ ⎜ ⎟⎜ ⎟

⎝ ⎠

QQ

22

0

7

3

Q Q0

3

7

⎛ ⎞ ⎜ ⎟⎜ ⎟⎝ ⎠

10. Answer (2)

Hint: S S P PV I V I

80

100

Sol.: S S P PV I V I

80

100

SV

801 220 2

100

352 VS

V

11. Answer (4)

Hint: At resonance V

IR

1

2f

LC

And voltage across resistor, inductor and capacitor

will be in different phase.

Sol.: I200

20

I = 10 A

f

6

1

2 22 10

1000

4

= 250 Hz

12. Answer (1)

Hint: avg

IdtI

dt

∫∫

Sol.:0

1

2 2avg

l bI

b

⎛ ⎞ ⎜ ⎟⎝ ⎠

0

4

I

13. Answer (1)

Hint: P = Vr.m.s

Ir.m.s

cosSol.: P = V

r.m.s

Ir.m.s

cos

= Vr.m.s

Ir.m.s

R

Z

r m sV

. .

195 2

2

= 195 V

Z R

2

2

3

1

100 2 10

⎛ ⎞ ⎜ ⎟ ⎝ ⎠

Z2 2

12 5

Z = 13

r m sI. .

195

13

= 15 A

12195 15 2700 W

13 P

14. Answer (1)

Hint:T

t4

, here T is time period of the current.

Sol.: I = I0

sin 50 t

= 50 rad/s

T2

1

25T s

T

T

4

t

I

Tt

4

1

100t s

t 10 ms

15. Answer (3)

Hint & Sol.:

In R – C circuit current lead the voltage


4/21

16. Answer (2)

Hint:2 2 2


Sol.:2 2


. .

11Ar m sI

17. Answer (4)

Hint:L

R

V VI

Z V

0

0, tan

⎛ ⎞ ⎜ ⎟

⎝ ⎠

Sol.: Z2 2

4 3

Z = 5

= 53°

VR

VL

V

O

VI

Z

0

0

I0

20

5

I0

= 4A

LX I

RItan

4tan

3

⎛ ⎞ ⎜ ⎟⎝ ⎠

= 53°

In R – L circuit voltage leads the current

I = (4A)sin[(100rad/s)t – 53°]

18. Answer (3)

Hint:dI

Ldt

Sol.:dI

Ldt

LdI

dt

0.080 5H

0.032 2 L

2AB

LL L

L3

2

3 5

2 2

15H

4

19. Answer (3)

Hint:d

dt

Sol.: Flux at any time (t)

= BAcos = BAcost

= B0

a2 cost

d

dt

= – B0

a2(– sint)

= B0

a2sint

– B a0 2

B a0 2

tO

20. Answer (4)

Hint:

tR

LE

I eR

1

⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠

Sol.:R

SE

L

I

L

dIV L

dt

tR

LE

I eR

1⎛ ⎞

⎜ ⎟ ⎜ ⎟⎝ ⎠

⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠

tR

LdI E R

edt R L

tR

LdI E

edt L

L

dIV L

dt

tR

LE e

e

2

810

e

1

410

21. Answer (2)

Hint: VP – V

Q = V

b + V

R + V

L

Sol.: VP – V

Q = 10 + 4 (2t + 3) + 2

dI

dt

VP – V

Q = 10 + 4 × 7 + 2 × 2

VP – V

Q = 10 + 28 + 4

= 42 V


5/21

22. Answer (1)

Hint: rotation 0 0

0

l

B xdx ∫

Sol.:

2

3

0 0

0

l

PAB xdx ∫

2

2 3

0 0

02

l

PA

xB

⎡ ⎤ ⎢ ⎥

⎢ ⎥⎣ ⎦

2

0 04

9 2

B l

2

0 02

9PA

B l

3

0 0

0

l

PBB xdx ∫

2

0 0

18PB

B l

Both have opposite polarity

= PA

– PB

2 2

0 0 0 02

9 18

B l B l

2 2

0 0 0 04

18

B l B l

2

0 0

6

B l

23. Answer (3)

Hint: = MI

Sol.:

0sin45 sin45 4

4

⎡ ⎤ ⎢ ⎥⎣ ⎦

IB

d

IB

a

0 8

24

2

⎛ ⎞ ⎜ ⎟⎝ ⎠

IB

a

02 2

= BA

Ir

a

202 2

= MI

I rMI

a

2

02 2

rM

a

2

02 2

24. Answer (2)

Hint: motional

� �

�

v B l

Sol.: = vBl cos60°

0 0

2

v B l

Force on positive charge will be toward end A

therefore, end A will be at higher potential.

25. Answer (1)

Hint:d

Ndt

Sol.:d

dt

dt r

dt

2 2

04

= – 8tr0

2

|| = 16r2

0

[at t = 2 s)

rI

R

2

016

26. Answer (2)

Hint:d

Ndt

Sol.: From 0 to t1

emf is zero because is

constant from t1

to t2

emf is positive constant

because d

dt

is negative constant.

27. Answer (2)

Hint: Tangent law

Sol.: H =

V

MBHsin = MB

Vsin(90 – )

tanV

H

B

B

= tan–1(3)

This from horizontal

1 1tan

3

⎛ ⎞ ⎜ ⎟⎝ ⎠

from vertical


6/21

28. Answer (4)

Hint:V

H

B

Btan

Sol.:

V

H

B

B

tan

sintan

cos cos

B

B

tantan

cos

tan60tan

cos30

3

3

2

1tan 2

29. Answer (1)

Hint: Wagent

= U + K

Sol.: Wagent

= U + K

Wagent

= Uf – U

i

Uf = – MB cos120°

MB

2

Ui

= – MB cos 0

= – MB

Wagent

MBMB

2

agent

3

2

MBW

30. Answer (2)

Hint: B = H

Sol.:B

H

0.25

1000

= 2.5 × 10–4 TmA–1

0

r

r

4

7

2.5 10

4 10

= 0.2 × 103

r

22 10

31. Answer (2)

Hint: For paramagnetic material Curie's law

0

m

C

T

Sol.:0

m

C

T

m

TO

32. Answer (3)

Hint: Magnetic field due to bar magnet at its

equatorial and axial position respectively are M

r

0

34

and M

r

0

3

2

4

Sol.:

MB

r

0

1 3

2

4

MB

r

0

2 34

netB B B

2 2

1 2

M

r

0

3

5

4

33. Answer (3)

Hint: 0

34

I dl r

B

r

��

�

�

Sol.: Magnetic field due to current carrying wire

0sin sin

4

IB

r

[ = 90°, = – 30°]

0 11

4 2

I

r

⎛ ⎞ ⎜ ⎟ ⎝ ⎠

0

8

IB

r


7/21

34. Answer (2)

Hint: 0 encB dl I ∫

��

�

Sol.:

0 encB dl I ∫

��

�

0

32

2

aB I

0

3P

IB

a

0 encB dl I ∫

��

�

0

5 32

2 2

a IB I

⎛ ⎞ ⎜ ⎟⎝ ⎠

0

10Q

IB

a

P

Q

B

B

10

3

⎛ ⎞ ⎜ ⎟⎝ ⎠

35. Answer (3)

Hint: Current sensitivity = I

Sol.:NAB

I K

Current sensitivity

So, current sensitivity depend on

(1) The number of turn in the coil

(2) The area of the coil

(3) The magnetic field applied

36. Answer (1)

Hint: M B � �

�

Sol.: MB sin �

M = NIA

qI

t

RI

2

2

RI

2

2

A = R2

22

2

RR B

= R3B

37. Answer (4)

Hint: mF q v B � �

�

eF qE� �

Sol.: mF q v B � �

�

ˆ ˆqvB i k

ˆqvBj

e

F qEjˆ�

As both magnetic and electric force are in opposite

direction, net force on the charge may be zero.

Therefore, particle may pass undeflected.

38. Answer (2)

Hint:

2

0

3

2 2 22

axis

IRB

R X

Sol.:1

27axis centre

B B

2

0 0

3

2 2 2

1

27 22

IR I

RR X

3 3

2 2 2

1 1

2 272

RR X

2 23R X R

R2 + X

2 = 9R2

X

2 = 8R2

2 2X R39. Answer (4)

Hint: M NIA� �

Sol.: M NIA� �

Magnetic dipole moment depends on

(1) Number of turn of the loop

(2) Current in the loop

(3) Area of the loop

40. Answer (3)

Hint: Ampere circuital law

0 encB dl I ∫

��

�


8/21

Sol.:

IrB

R

0

22

when r R

for r > R

IB

r

0

2

Br

1

41. Answer (1)

Hint:

0

34

dl rIB

r

��

�

�

Sol.:0

2 2arc

IB

R

�

0

2 2 2arc

IB

R

�

0

8arc

IB

R

�

0sin sin

4straight wire

IB

R

�

0I

sin90 sin04 R

I

R

0

4

net arc straight wireB B B � � �

0 0

8 4

I I

R R

⎛ ⎞ ⎜ ⎟⎝ ⎠

net

I

R

0B 1

4 2

⎛ ⎞ ⎜ ⎟ ⎝ ⎠

42. Answer (4)

Hint: mF I B � ��

l

M B � �

�

Sol.:effl sin

mF I B

0 sinm

F I B

0m

F

M B � �

�

= MB sin

= Il2 B sin = 0

Both (1) and (2) are incorrect

43. Answer (3)

Hint: mF i l B � � �

effsin

mF i L B

Sol.:

Leff

= R

Fm

= IRB

= 1

2 22

= 2 N

44. Answer (1)

Hint: Time period of charge particle in uniform

magnetic field 2 m

TqB

Sol.:2

qBf

m

P P

P

mf q

f q m

1 4

2 1

= 2

Pf f

45. Answer (2)

Hint: mF q V B � � �

r

mF B� �

Sol.:r

mF B� �

therefore, r

a B�

�

a B 0 �

�

ˆ ˆ ˆ ˆ ˆ ˆ2 3 2 3 0i j k i bj k

2 + 2b – 3 = 0

1

2b


9/21

CHEMISTRY

46. Answer (1)

Hint: CN– is an ambidentate ligand.

Sol: Linkage isomerism is shown by the ligands

which contain more than one atom which could

donate an electron pair.

47. Answer (2)

Hint: Froth stabilisers such as cresols, aniline

stabilise the froth.

Sol: Collectors such as xanthates enhance

non-wettability of mineral particles in water.

48. Answer (1)

Hint: CSFE = [– 0.4 (nt2g

) + 0.6(neg

)] 0

Sol:

eg

+0.6 0

t2g

–0.40

d6

CFSE = – 6 × 0.4 0

+ 0 = –2.4 0

49. Answer (4)

Hint: Complex having unpaired electrons will be

paramagnetic.

Sol: Outer orbital complexes are having outer shell

d-orbitals in its hybridization.

50. Answer (1)

Hint: Synergic bonding is least in [Mn(CO)6

]+.

Sol: Due to the presence of positive charge,

synergic bonding from metal to carbonyl ligand will

decrease. Hence, M – C bond length will increase

and C – O bond length will decrease.

51. Answer (2)

Hint: Wrought iron is the purest from of commercial

iron.

Sol: Extraction of gold and silver involves leaching

the metal with CN–. The metal is later recovered by

displacement method.

52. Answer (4)

Hint: In coordination compounds, colour arises due

to d – d transition or charge transfer.

Sol: Complexes of Ti (IV), Zn (II) and Sc (III) are

colourless.

[Co(CN)6

]3– yellow, [CoF6

]3– Green

[Ni(NO2

)6

]4– Violet

53. Answer (2)

Hint: Primary valency is number of charges on the

central metal atom and secondary valency is equal

to coordination number.

Sol: Oxidation state of Co is +3. Coordination no.

of Co is 6(4NH3

and 2Cl–).

54. Answer (3)

Hint: Strong Jahn-Teller distortion is observed for d9

configuration 6 32g g(t e ) .

Sol: 2+ 3 2

2 6 2g g[Mn(H O) ] : t e

2+ 6 42 6 2g g[Zn(H O) ] : t e

2+ 6 33 6 2g g[Cu(NH ) ] : t e

3+ 3 03 6 2g g[Cr(NH ) ] : t e

55. Answer (1)

Hint: EAN of central atom = Atomic number of metal

atom – oxidation number of metal atom + No. of

electrons donated by ligands.

Sol: Both [Co(en)3

]2+ and [Cr(H2

O)6

]2+ have three

unpaired electrons. The hybridisation of metal atom

in [Cu(NH3

)4

]2+ and [Ni(CN)4

]2– is dsp2.

EAN of [Mn(H2

O)6

]2+ = 25 – 2 + 12 = 35

EAN of [Fe(CN)6

]3– = 26 – 3 + 12 = 35

[Cr(gly)3

] and [Pt(gly)2

] has cis and trans-isomers.

56. Answer (3)

Hint: H2

O and C2

O4

2– acts as strong field ligand for

Co3+.

Sol:

Complex No. of unpaired electrons

1. [Co(H2

O)6

]3+ Zero

2. [Co(C2

O4

)3

]3– Zero

3. [FeF6

]4– Four

4. [Fe(H2

O)5

NO]2+ Three


10/21

57. Answer (4)

Hint: Distillation is useful for low boiling metals like

zinc and mercury.

Sol: Cu is refined by electrolytic refining.

58. Answer (3)

Hint: Spectrochemical series

Sol: N < C O < NO < CN

3

– 2– – –

2 4 2

Ligand strengthfield

59. Answer (2)

Hint: The effective nuclear charge of the metal ion

increases as we move down the group from 3d to 5d

series. Thus, 0

increases.

Sol: Higher the strength of the ligand, higher will be

the value of 0

.

Higher the oxidation state of the metal, higher will be

the value of 0

.

60. Answer (2)

Hint: In blast furnace at 900–1500 K (higher

temperature range)

C + CO2

2CO

FeO + CO Fe + CO2

Sol: In blast furnace at 500-800 K (lower

temperature range)

3Fe2

O3

+ CO 2Fe3

O4

+ CO2

Fe3

O4

+ 4CO 3Fe + 4CO2

Fe2

O3

+ CO 2FeO + CO2

61. Answer (1)

Hint: [Co(en) (NH3

)2

Cl2

]+ exists as cis-trans

isomers. Trans-isomer is optically inactive.

Sol: H N

3

enCo

ClCl

(cis)

Co

ClCl

en

NH3

NH3

NH3

optically active

en ,Co

Cl

Co

Cl

en

ClNH3

H N3

NH3

Cl

H N3

optically inactive

62. Answer (1)

Hint: The oxides having high melting points are

difficult to reduce by carbon reduction method.

Sol: PbO + C Pb + CO

63. Answer (4)

Hint: If complex is neutral, then the word ‘ate’ is not

used with the name of metal.

Sol: Triamminetrinitrito-N-cobalt (III)

64. Answer (3)

Hint: Wilkinson's catalyst is [(Ph3

P)3

RhCl].

Sol: Organometallic compounds contain at least one

chemical bond between metal and carbon atom of an

organic molecule.

Grignard reagent RMgBr

Tetracarbonyl nickel [Ni(CO)4

]

Ferrocene [Fe(5 – C5

H5

)2

] Fe

65. Answer (1)

Hint: K2

[HgI4

] is a soluble complex.

Sol: HgCl2

+ 2KI HgI2

+ 2KCl

(Red ppt.)

Hint: HgI2

+ 2KI K2

[HgI4

]

(Soluble)

66. Answer (3)

Hint: In neutral and faintly alkaline solutions, n

factor for MnO4

–

is 3.

Sol:

8MnO4

–

+3S2

O3

2– + H2

O 8MnO2

+ 6SO4

2–

+ 2OH

67. Answer (3)

Hint: XeF2

+ PF5

[XeF ]+ [PF6

]–

Sol:

F

FF

FFF

Hybridisation of

P : sp d3 2

d

–

68. Answer (4)

Hint: Ce3+(4f

1) is colourless.

Sol: Eu2+ is a good reducing agent changing to

common +3 oxidation state actinoids show greater

range of oxidation states because 5f, 6d and 7s

levels are of comparable energies.

69. Answer (2)

Hint: KMnO4

has highest +7 oxidation state of Mn.


11/21

Sol: PbS + 4O3

PbSO4

+ 4O2

2Hg + O3

Hg2

O + O2

2K4

[Fe(CN)6

] + H2

O + O3

2K3

[Fe(CN)6

] + 2KOH + O2

70. Answer (3)

Hint: CrO3

– Acidic

CrO – Basic

Cr2

O3

– Amphoteric

Sol: V2

O5

– Acidic

V2

O3

– Basic

71. Answer (2)

Hint: ClF5

is sp3d2 hybridised.

Sol: Cl

F

FF

FF

Br

F

F

F , l F

F

F

F

F

F

F

72. Answer (4)

Hint: O

O

P

P

O

P

O

O

P

O

OO O

O

Sol: Number of bridging oxygen atom is equal to 6.

73. Answer (4)

Hint: Ce4+, Lu3+ and Yb2+ are diamagnetic.

Sol: Eu2+ : [Xe]4f 7

Ce4+ : [Xe] 4f 0

Lu3+ : [Xe] 4f 14

Yb2+ : [Xe] 4f 14

74. Answer (1)

Hint: With limited NH3

, N2

and HCl are formed.

Sol: On reaction with excess ammonia, chlorine

gas gives nitrogen and ammonium chloride.

8NH3

+ 3Cl2

6NH4

Cl + N2

(Excess)

75. Answer (2)

Hint: I2

+ 10HNO3

2HIO3

+ 10NO2

+ 4H2

O

(conc.)

Sol: Let the oxidation state of I in HIO3

is x

1 + x + 3(–2) = 0 x = + 5

76. Answer (1)

Hint: 4 2 4 2 2513 K(Green) (Black)(A)

2KMnO K MnO MnO O

Sol: KMnO4

shows colour due to ligand to metal

charge transfer.

77. Answer (3)

Hint: Higher is the oxidation state of element, higher

will be the acidic nature of its oxide.

Sol: F2

is the strongest oxidising agent among all

halogens.

H2O H2S H2Se H2Te

273 188 208 222

Bond angle (°)

NH3 PH3 ASH3 S Hb 3

107.8 93.6 91.8 91.3

78. Answer (2)

Hint: Malachite : CuCO3

· Cu(OH)2

Azurite: 2CuCO3

Cu(OH)2

Sol: Carnalite: KCl · MgCl2

· 6H2

O

Limonite: Fe2

O3

· 3H2

O

Calamine: ZnCO3

79. Answer (2)

Hint: Hypochlorous acid is HOCl

Sol:

1 2 1

HO Cl

80. Answer (1)

Hint: XeF6

+ 3H2

O XeO3

+ 6HF

Sol: O

Xe

OO

(Pyramidal)

81. Answer (1)

Hint: I2

is a weaker oxidising agent.

Sol: Cu2+ can oxidise I– to I2

. That’s why CuI2

does

not exist.


12/21

82. Answer (4)

Hint: CaCN2

+ 3H2

O 2NH3

+ CaCO3

Sol: NH3

is basic and hence can turn moist litmus

paper blue.

An alkaline solution of K2

[HgI4

] is called Nessler’s

reagent. Ammonia is quantitatively estimated using

Nessler’s reagent.

83. Answer (4)

Hint: Due to stability of fully filled subshell 4f14 in D2+

Sol: Metal Third ionisation

enthalpy (kJ/mol)

A – Gd – 1990

B – Pr – 2086

C – Er – 2194

D – Yb – 2417

84. Answer (1)

Hint: Ba(N3

)2

Ba + 3N

2

Sol: 4 2 2 7 2 2 2 3

(NH ) Cr O N + 4H O Cr O

4 3NH Cl NH +HCl

3 2 2 22Pb (NO ) 4NO + 2PbO+O

4 3 2 2NH NO N O+2H O

85. Answer (3)

Hint:

4FeCr2

O4

+ 8Na2

CO3

+ 7O2

8Na2

CrO4

+ 2Fe2

O3

+ 8CO2

Sol: Yellow solution of Na2

CrO4

is filtered and

obtained as filtrate.

86. Answer (4)

Hint:O

O

S

OHO H

(H SO )2 5

OOO

O

S

OHO

O

S

OH

(H S O )2 2 8

Sol:

O ,

S

HO

(H SO )2 3

O

O

S

O

(H S O )2 2 7

OH

O

S

OH OHO

O

S

O

O

S

O

OH , HO

(H S O )2 2 6

O

S

O

S OHHO

(H S O )2 2 4

87. Answer (3)

Hint: The coordination of EDTA ligand to the metal

atom takes place through two N atoms and four O

atoms forming five 5-membered rings.

Sol:

O

Fe

O

O

C

CH2

CH2

CH2

CH2

CH2

C

O

O

C

O

O

C

O

CH2N

N

88. Answer (3)

Hint: Spin only magnetic moment,

n(n 2) B.M.

Sol: No. of unpaired electrons present in Fe2+ and

Cr2+ are equal to four.

4(4 2) 24 = 4.89 B.M.

89. Answer (1)

Hint: More the number of unpaired electrons involved

in metallic bonding, higher will be the melting point.

Sol: In case of Cr, there are six unpaired electrons

in their ground state which provide strongest metallic

bonding whereas in case of Zn, the number of

unpaired electrons is zero and hence show weakest

metallic bonding.

90. Answer (4)

Hint: Phosphinic acid is hypophosphorous acid.

Sol:

P OH

O

HH

(H PO )3 2


13/21

Oxidation state of phosphorus is +1.

Hypophosphorous acid is a good reducing agent

because it contains two P–H bonds and precipitates

Ag from AgNO3

solution

4AgNO3

+ H3

PO2

+ 2H2

O 4Ag + H3

PO4

+ 4HNO3

BIOLOGY

91. Answer (3)

Hint: Toddy is obtained from fermenting sap of

Caryota urens.

Sol.: It is a traditional drink of some parts of South

India.

Caryota urens is scientific name of a palm.

92. Answer (3)

Hint: Antibiotics are effective against bacterial

diseases.

Sol.: Diphtheria, whooping cough and pneumonia

are bacterial diseases and can be cured by taking

antibiotics.

Curd is more nutritious than milk as it contains

vitamin B12

.

93. Answer (4)

Hint: These are oxygenic eubacteria.

Sol.: Cyanobacteria can fix atmospheric nitrogen

such as Nostoc and Oscillatoria.

Glomus – Fungi

94. Answer (2)

Hint: They are species specific.

Sol.: Nucleopolyhedroviruses do not have side effect

on plants and animals.

They are narrow spectrum insecticides. They

conserve beneficial insects.

95. Answer (4)

Sol.: Biocontrol agents are non toxic, do not kill

useful organisms and keep pests at manageable

levels.

96. Answer (2)

Hint: Secondary treatment is biological treatment.

Sol.: Primary treatment of waste water is mainly a

physical process which involves sequential filtration

and sedimentation.

97. Answer (3)

Sol.: To prevent the discharge of untreated sewage

into Ganga and Yamuna rivers, The Ministry of

Environment and Forests has initiated Ganga and

Yamuna action plan.

98. Answer (1)

Hint: Fruit juice is clarified due to the degradation of

some cell wall material.

Sol.: Pectinases and proteases help in clarifying fruit

juices.

Lipases are used in detergent formulations.

Amylases degrade starch, streptokinase is used as

clot buster.

99. Answer (2)

Hint: Rhizobium is used as a biofertiliser in

leguminous plants.

Sol.: Soyabean is a leguminous plant.

Organic farming uses naturally produced compost.

Anaerobic sludge digesters produce biogas due to

anaerobic digestion of organic material.

100. Answer (4)

Hint: Methane is the major component of biogas

which is 50-70%.

Sol.: Biogas contains methane, carbon dioxide, H2

,

H2

S, etc.

101. Answer (2)

Hint: It is a bacterium, used to produced vinegar.

Sol.: Penicillium and Aspergillus are fungi.

Acetobacter aceti is used to produce vinegar

commercially.

102. Answer (1)

Hint: A yeast is used to produce statins.

Sol.: Monascus purpureus is a yeast which is used

to produce statins which is blood cholesterol

lowering agent.

103. Answer (1)

Hint: BOD Polluting potential of water.

Flocs are masses of fungi and bacteria.

Sol.: Biogas is inflammable hence used as source

of energy.

Heterotrophic microbes are naturally present in the

sewage and treat waste water.


14/21

104. Answer (3)

Hint: Gobar gas is biogas.

Sol.: Xanthomonas campestris is a bacterial species

that causes a variety of plant disease including black

rot in crucifers .

Detailed Sol.: The sediment of settling tank is called

activated sludge. Which are of bacterial and fungal

flocs.

Lipases are used in detergent formulation.

105. Answer (3)

Hint: Biofertilisers improve the soil fertility.

Sol.: Biofertilisers are organisms which enrich the

nutrient quality of soil by increasing the organic and

inorganic matter in the soil.

106. Answer (4)

Hint: Triticale was first man-made crop produced by

hybridisation of wheat and rye.

Sol.: Parbhani Kranti is a variety of okra produced

by hybridisation of two varieties through conventional

breeding.

107. Answer (3)

Hint: Jaya and Ratna are rice varieties.

Sol.: Himgiri is a wheat variety which is disease

resistant.

Atlas 66 has high protein content and has been

used as a donor for improving cultivated wheat.

108. Answer (1)

Hint: Downy mildew of grapes is a fungal disease.

Sol.: Bordeaux mixture is effective against fungal

diseases.

Use of disease resistant varieties protect plant from

infection of microbes.

109. Answer (3)

Hint: Genetically identical offsprings are called

clones.

Sol.: A somatic cell produces identical clones called

somaclones during tissue culture.

Somatic hybrids are obtained by fusion of protoplast

of two different species, varieties.

Cybrid is produced during somatic hybridisation of

two different cytoplasms, if one of the nuclei get

degenerated.

110. Answer (4)

Hint: Spirogyra is a filamentous green alga.

Sol.: Spirulina is used to produce SCP but Spirogyra

is not used for such production.

111. Answer (3)

Sol.: During green revolution, production of wheat

and rice led to increase in food production.

112. Answer (4)

Hint: Genetic variability increases chances of getting

desirable characters.

Sol.: Genetic variability is root of any breeding

programme and we can choose better characters.

Hybridisation increases hybrid vigours.

113. Answer (2)

Hint: In eukaryotes both introns and exons are

present.

Sol.: hnRNA is processed in the nucleus and after

processing it is called mRNA that contains exons

only.

Detailed Sol.: The intervening sequences in

between coding regions are called introns. Exons

are coding regions of genes. Therefore, eukaryotic

genes are split genes.

114. Answer (4)

Hint: Minisatellites are repetitive DNA.

Sol.: Minisatellites are also known as VNTR and

surrounded by conserved restriction sites.

115. Answer (1)

Hint: Griffith used Streptococcus or Diplococcus

bacteria for his experiment.

Sol.: QB Bacteriophage has RNA as genetic

material.

Taylor et. al. proved semiconservative mode of DNA

replication in Vicia faba.

Hershy and Chase experimented with bacteriophage.

116. Answer (3)

Hint: It is used in analysis of a gene for disease, in

identifying organisms etc.

Sol.: RFLP is not used in determining the structures

of proteins.

117. Answer (4)

Hint: RNA polymerase I synthesizes 28S rRNA.

Sol.: tRNA and 5S rRNA are synthesized by RNA

polymerase III.

118. Answer (3)

Sol.: Direction of both RNA and DNA synthesis is 5'

to 3'.


15/21

119. Answer (3)

Hint: Lac operon is inducible operon system.

Sol.: Lactose acts as an inducer. Regulation of lac

operon by repressor protein is negative regulation.

Detailed Sol.: When lactose is absent repressor

binds to the operator and in the presence of lactose

repressor binds to lactose.

120. Answer (3)

Hint: The mRNA has complementary sequence of

nitrogenous bases to template strand.

Sol.: ATCTGGCGT is the sequence of DNA on

template strand then the sequence of mRNA will be

UAGACCGCA. In RNA uracil is present instead of

thymine.

121. Answer (1)

Hint: Satellite DNA is repetitive DNA.

Sol.: Repetitive DNA is heterochromatin region of

DNA.

In bacteria, larger subunit of ribosome has 23S rRNA

and 5S rRNA.

Detailed Sol.: During protein synthesis 23S rRNA

acts as ribozyme.

122. Answer (4)

Hint: First step of translation mechanism is

aminoacylation of tRNA.

Sol.: mRNA first binds to smaller subunit of

ribosome then initiator tRNA comes to P site of

ribosome.

Detailed Sol.: Formation of polypeptide is the final

step in protein synthesis.

123. Answer (2)

Hint: It is the process of genetic recombination in

which donor and recipient do not come in contact.

Sol.: Griffith discovered transformation process in

bacteria Streptococcus pneumoniae. In which the

donor cell releases a piece of DNA which is actively

taken up by the recipient cell from the solution.

124. Answer (4)

Hint: Dystrophin is the largest gene.

Sol.: Regulator and operator are part of lac operon.

Detailed Sol.: SRY (sex determining region Y) is

the smallest gene codes for TDF (Testis determining

factor).

125. Answer (4)

Hint: This enzyme is a ribozyme.

Sol.: Peptidyl transferase is an RNA enzyme rather

being proteinaceous. In case of prokaryotes 23 S

rRNA acts as ribozyme whereas in eukaryotes it is

28 S rRNA. It catalyses the peptide bond formation

between amino acids.

126. Answer (2)

Hint: Cairns used tritiated thymidine in

autoradiography experiment.

Sol.: By using tritiated thymidine in auto radiography

experiment; Cairns proved semi-conservative mode of

DNA replication in E.coli.

127. Answer (3)

Hint: Nonsense codon is stop codon, i.e., UAG.

Sol.: Start codon is AUG. GUG is ambiguous

codon. UGG codes for amino acid Tryptophan.

128. Answer (4)

Hint: hn RNA is converted into functional m-RNA by

post transcriptional process.

Sol.: Splicing occurs in nucleus in eukaryotes

whereas in prokaryotes it is absent as introns are

absent except in Archaebacteria.

Detailed Sol.: Post transcriptional modifications

involve splicing, capping and tailing which are absent

in prokaryotes but are found in eukaryotes.

129. Answer (3)

Hint: In reverse transcription DNA is synthesised

from RNA.

Sol.: E.coli, Lambda phage and T4

bacteriophage have

DNA as genetic material.

Detailed Sol.: HIV is a retrovirus in which flow of

information takes place from RNA to DNA.

130. Answer (4)

Hint: Nucleosomes, the structure in chromatin are

seen as ‘beads-on-string’

Sol.: Nucleosomes have histone octamer, i.e.

organised form of H2

A, H2

B, H3

and H4

histones. H1

histone is not a part of histone octamer.

131. Answer (1)

Hint: DNA with more G C content has high melting

point.

Sol.: DNA with more A = T pairs, has low melting

point and get denatured more easily. DNA with more

G C pairs than A = T pairs has high melting point.


16/21

132. Answer (2)

Hint: Replication is copying of DNA. In DNA Uracil

is absent, it is present in RNA.

Sol.: Chargaff’s rule is not applicable for single

stranded DNA.

Detailed Sol.: Repetitive DNA is the part of DNA

which contains the same sequence of nitrogen bases

repeated more than once in a genome and the area

with long sequence of short repetitive DNA is called

satellite DNA.

133. Answer (1)

Hint: Nucleoside has sugar and nitrogenous base.

Sol.: A nitrogenous base is linked to a pentose

sugar by N-glycosidic linkage.

134. Answer (3)

Hint: Termination codons are stop codons.

Sol.: UAA, UAG, UGA are stop codons.

Detailed Sol.: All stop codons of universal genetic

codes begin with uracil.

135. Answer (3)

Hint: Polypeptide synthesis takes place from mRNA.

Sol.: mRNA is synthesized from DNA and has

information for protein synthesis.

Detailed Sol.: mRNA, tRNA and rRNA are directly

synthesized from DNA.

136. Answer (3)

Hint: Acquired immunity is pathogen specific.

Sol.: Innate immunity is non-specific and is not

characterised by memory.

137. Answer (1)

Hint: Amino terminal of light chains of antibody

interacts with epitope of antigen.

Sol.: Variable part of N/amino terminal of both heavy

& light chains form antigen binding site (paratope) of

antibody that recognises epitope of antigen.

138. Answer (2)

Hint: A primary lymphoid organ located near heart

and beneath the breastbone and provides the site for

T cell maturation.

Sol.: Thymus is quite large at the time of birth but

keeps reducing in size with age and by the time of

puberty reduces to a very small size (atrophy).

139. Answer (4)

Hint: Cells that originate and mature in bone marrow

produce antibodies.

Sol.: Antibodies are proteinaceous in nature and are

produced by B-lymphocytes only.

T-lymphocytes facilitate B-lymphocytes to produce

antibodies.

140. Answer (4)

Hint: Macrophages like neutrophils, participate in

phagocytosis to destroy microbes.

Sol.: Skin and mucous are considered as physical

barrier, tear and saliva as physiological barrier.

Interferons as cellular barrier for uninfected cells and

macrophages as cellular barrier in tissues NK cells,

PMNL and monocytes as cellular barriers in blood.

141. Answer (3)

Hint: Preformed antibodies are administered in

passive immunity.

Sol.: ATS against tetanus, antivenin against

snakebites contain ready made antibodies.

Injecting tetanus toxoid triggers active immunity.

142. Answer (3)

Hint: Organ which acts as “filter of blood” is known

as graveyard of RBCs.

Sol.:

(a) Spleen : Filter of the blood (graveyard of RBCs)

(b) Appendix : Secondary lymphoid organ (vestigial

in human)

(c) Thymus : Primary lymphoid organ where T cells

mature.

(d) Bone marrow : Production house of all types of

lymphocytes (i.e., B & T cells)

143. Answer (3)

Hint: One of the disease in given pairs could be

confirmed by Widal test.

Sol.: Dysentery, diphtheria, typhoid and plague are

bacterial diseases. Dengue is a vector borne

protozoan disease. Polio is a viral disease and

Ascariasis is a helminthic disease.

144. Answer (2)

Hint : Select antibodies chiefly responsible for

allergic response.

Sol.: IgE activates mast cells that release histamine

that acts as vasodilator & bronchoconstrictor.

145. Answer (1)

Hint: Fungus growing in ring shape on skin.

Sol.: Ringworm infection are caused by fungi such

as Trichophyton, Epidermophyton & Microsporum.


17/21

146. Answer (3)

Hint: Asexual reproduction (schizogony) occurs in

erythrocytes & hepatocytes.

Sol.: Malarial parasites are sporozoans and

reproduce asexually in hepatocytes as well as in

RBCs of man. Sporozoite forms of Plasmodium are

stored in salivary glands of infected female

Anopheles.

Mature infective stage (sporozoite) escapes from gut

and migrates to the mosquito’s salivary glands.

147. Answer (3)

Hint: Gambusia is an example of larvivorous fish.

Sol.: Cholera is a bacterial disease caused by Vibrio

cholerae. Larvae of vectors such as Anopheles

mosquito are kept under control by larvivorous fish

such as Gambusia.

148. Answer (3)

Hint: Anti-histamine drugs are used to counter

hypersensitivity of a person to foreign substances.

Sol.: Common cold reaction and dengue are viral

diseases. Common cold is caused by Rhino virus.

Dengue is transmitted by Aedes mosquitoes.

149. Answer (3)

Hint: Disease characterised by sneezing and

watering from eyes.

Sol.: In common cold, only upper respiratory tract is

involved but lungs and alveoli remain unaffected.

150. Answer (4)

Hint: Identify a viral disease.

Sol.: Amoebic dysentery caused by E. histolytica,

ascariasis by Ascaris, typhoid fever by Salmonella

typhi, are transmitted through faeco-oral route.

Chikungunya is a vector borne disease transmitted

by Aedes mosquito.

151. Answer (2)

Hint: Widal test is the diagnostic method of

diseases caused by bacteria S.typhi.

Sol.: Typhoid is caused by a bacterium Salmonella

typhi through contaminated food and water.

Wassermann test is used to detect syphilis.

152. Answer (4)

Hint: This disorder is characterised by presence of

swollen, reddened, running eyes, nose and anti-

histamines are given to the patient.

Sol.: Myasthenia gravis, rheumatoid arthritis and

vitiligo are autoimmune disorders. Hay fever is a type

of allergy.

153. Answer (4)

Hint: Pathogen causes filariasis, a helminthic

disease.

Sol.: Ascaris causes ascariasis, E.histolytica cause

amoebiasis, Streptococcus pneumoniae causes

pneumonia. Wuchereria causes elephantiasis/

filariasis which is chronic inflammation of lymphatic

vessels of lower limb.

154. Answer (1)

Hint: Bond formed between two sulphur containing

amino acids.

Sol.: Heavy and light chains are connected to each

other by disulphide bonds between cysteines.

155. Answer (1)

Hint: Mutations are non directional in nature.

Sol.: Evolution is a non-directional, stochastic

process based on chance events and chance

mutations. Evolution is occurring on a fast pace due

to anthropogenic interference.

156. Answer (3)

Hint: Golden Age of reptiles.

Sol.: Mesozoic era is considered as golden age of

reptiles because reptiles were dominant. Jurassic

period is considered as golden age of Dinosaurs.

157. Answer (1)

Hint: Birds have shelled calcareous eggs.

Sol.: Presence of scales on hind limbs and shelled/

cleidoic eggs are considered as reptilian ancestry of

birds because both features together appeared first

in reptiles but are also found in birds. Birds lack

teeth and have a four chambered heart.

158. Answer (4)

Hint: Age of fossils based on position and pairing of

electrons.

Sol.: Electron spin resonance measures no. of

unpaired electrons in crystalline structures which

were previously exposed to radiation. Age of fossil is

determined by measuring dosage of radiations.

159. Answer (2)

Hint: Large population has constant allelic frequency

in absence of evolutionary factors.


18/21

Sol.:

According to HW law,

p + q = 1

p + 0.4 = 1

p = 0.6

frequency of carriers (heterozygous) = 2 pq

= 2 × 0.4 × 0.6 = 0.48

= 48%

So, number of carriers = 96 among 2000 individuals

2000 48960

100

⎛ ⎞ ⎜ ⎟⎝ ⎠

160. Answer (4)

Hint: Homo erectus appeared about 1.5 mya.

Sol.: Cranial capacities of Cro-Magnon man,

H.erectus, H.habilis and H.neanderthalensis are

1650, 900, 650-800 & 1400 cc respectively.

Neanderthal man buried his dead with flowers and

tools.

161. Answer (4)

Hint: Common ancestry in different organisms.

Sol.: Organs which have same origin but different

function are called homologous organs. e.g.

Forelimbs of human, horse and bat.

162. Answer (1)

Hint: According to Ernst Haeckl, it is summarised as

biogenetic law.

Sol.: Ontogeny is development of embryo and

phylogeny is the ancestral sequence. Biogenetic law

states that ontogeny is recapitulation of phylogeny.

As per this law, the sequence of embryonic

development in different vertebrates show striking

similarities.

163. Answer (2)

Hint: Identify a case of Sewall-Wright effect.

Sol.: When gene migration occurs many time, there

will be gene flow. If the change in allele frequency

occurs by chance, it is called genetic drift. Following

two effects are ramifications of genetic drift i.e., (1)

Founder’s effect (2) Bottleneck effect.

164. Answer (3)

Hint: Melanic forms have selective advantage and

inheritance of this character was naturally selected.

Sol.: Natural selection brings about evolution.

Industrial melanism is an example of directional type

of natural selection. In a mixed population having

different variants of moths, those that are better

adapted will survive and will be naturally selected.

Such as after industrialization, melanic forms could

camouflage themselves in the background.

165. Answer (4)

Hint: Presence of artificial selection alters Hardy-

Weinberg principle.

Sol.: The factors which affect genetic equilibrium in

a population are presence of (1) genetic drift (2) gene

migration/gene flow (3) mutation (4) genetic

recombination (5) natural selection and (6) lack of

random mating.

166. Answer (1)

Hint: Adaptive radiation in ancestral stock of

Australian marsupials and placental mammals

separately.

Sol.: A number of marsupials, each different from the

other, evolved from an ancestral stock, but all within

Australian island. Divergent evolution, represented

homology while convergent evolution represents

analogy. Saltation refer to large mutation leading to

variation according to mutation theory.

167. Answer (2)

Hint: Simpler molecules give rise to complex

organisms.

Sol.: During course of origin of life following events

take place in given order.

1. Synthesis of organic monomers.

2. Synthesis of organic polymers.

3. Formation of protobionts with RNA

4. Formation of protobionts with DNA based genetic

systems.

168. Answer (2)

Hint: This is a case of stabilising selection.

Sol.: In case of stabilizing selection, average mean

value is selected so that average phenotype exhibits

higher and narrower peak. Two peaks are obtained in

disruptive selection.

169. Answer (3)

Hint: Bipedal gait evolved prior to verbal language.

Sol.: Increased cranial capacity led to natural

selection. Erect posture appeared in

Australopithecines.

170. Answer (1)

Hint: Common ancestral finch was a herbivore.

Sol.: Parent finch was seed eating and other forms

arose with altered beaks enabling them to become

insectivorous.

171. Answer (2)

Hint: Elongated neck of giraffe can be explained by

continuous stretching according to this theory.

Sol.: All acquired characters of a generation

are passed on to the next one according to

Lamarck.


19/21

172. Answer (1)

Hint: Tasmanian tiger cat is an Australian marsupial.

Sol.: Wombat, Bandicoot and Spotted cuscus are

Australian marsupials whereas Bobcat is a placental

mammal.

173. Answer (3)

Hint: Eyes of cephalopod molluscs are analogous to

vertebrate eyes.

Sol.: Nictating membrane is a vestigial organ and

presents anatomical evidence. In Miller’s experiment

of chemical evolution, CH4

, NH3

, H2

and water vapour

were used which on exposure to electric discharge

resulted in formation of amino acids. Industrial

melanism is an example of natural selection.

174. Answer (2)

Hint: Given study is based on the genetic material

which is transmitted only by females.

Sol.: Older the organism more will be the

accumulation of variations, in its mitochondrial DNA.

Mitochondrial DNA is uniparental and does not

undergo recombination unlike nuclear DNA.

175. Answer (3)

Hint: Origin of life occurred in water.

Sol.: Plants evolved prior to animals. Sauropsids

were ancestors of thecodonts.

176. Answer (1)

Hint: Descent with modification.

Sol.: Key points of evolution i.e., descent with

modifications and natural selection were proposed by

Darwin. Darwin stressed on reproductive fitness.

Saltation was explained by De Vries and genetic drift

by Sewall & Wright.

177. Answer (4)

Hint: Equus is modern present day horse.

Sol.: Pliohippus was first one toed horse.

178. Answer (4)

Hint: A single huge explosion formed Universe.

Sol.: Origin of species was explained by Charles

Darwin and Origin of life by A.I. Oparin. Origin of

Universe is explained by Big-Bang hypothesis.

179. Answer (3)

Hint: Theory of panspermia.

Sol.: Life arises from non-living matter according to

theory of spontaneous generation.

180. Answer (1)

Hint: Changes leading to evolution of organisms.

Sol.: Study of external features of organisms is

known as morphology and physiology is the science

of body functions that how the body parts work.

Biogenesis is formation of organisms from pre

existing organisms.

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