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Test - 3 (Code E) (Answers) All India Aakash Test Series for Medical-2019

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1. (1)

2. (3)

3. (2)

4. (2)

5. (1)

6. (3)

7. (2)

8. (2)

9. (1)

10. (3)

11. (2)

12. (2)

13. (2)

14. (2)

15. (3)

16. (2)

17. (3)

18. (3)

19. (4)

20. (3)

21. (3)

22. (1)

23. (4)

24. (3)

25. (3)

26. (4)

27. (2)

28. (4)

29. (1)

30. (1)

31. (4)

32. (2)

33. (1)

34. (3)

35. (2)

36. (3)

Test Date : 20/01/2019

ANSWERS

TEST - 4 (Code E)

All India Aakash Test Series for Medical - 2019

37. (4)

38. (2)

39. (1)

40. (2)

41. (3)

42. (1)

43.Deleted

44. (3)

45. (4)

46. (2)

47. (3)

48. (2)

49. (3)

50. (4)

51. (3)

52. (3)

53. (3)

54. (1)

55. (4)

56. (4)

57. (1)

58. (3)

59. (3)

60. (4)

61. (3)

62. (1)

63. (2)

64. (4)

65. (3)

66. (3)

67. (2)

68. (1)

69. (4)

70. (3)

71. (2)

72. (4)

73. (2)

74. (2)

75. (1)

76. (4)

77. (1)

78. (4)

79. (2)

80. (2)

81. (2)

82. (2)

83. (4)

84. (2)

85. (2)

86. (4)

87. (4)

88. (3)

89. (3)

90. (2)

91. (1)

92. (3)

93. (4)

94. (2)

95. (1)

96. (2)

97. (4)

98. (4)

99. (1)

100. (3)

101. (2)

102. (2)

103. (4)

104. (1)

105. (2)

106. (3)

107. (4)

108. (3)

109. (4)

110. (2)

111. (4)

112. (3)

113. (2)

114. (4)

115. (4)

116. (1)

117. (2)

118. (3)

119. (3)

120. (3)

121. (1)

122. (2)

123. (3)

124. (1)

125. (4)

126. (2)

127. (3)

128. (3)

129. (2)

130. (2)

131. (4)

132. (3)

133. (2)

134. (3)

135. (4)

136. (1)

137. (1)

138. (4)

139. (3)

140. (3)

141. (1)

142. (1)

143. (2)

144. (3)

145. (2)

146. (3)

147. (2)

148. (2)

149. (3)

150. (4)

151. (3)

152. (2)

153. (1)

154. (3)

155. (2)

156. (3)

157. (2)

158. (4)

159. (1)

160. (3)

161. (1)

162. (2)

163. (2)

164. (4)

165. (1)

166. (2)

167. (2)

168. (1)

169. (3)

170. (4)

171. (2)

172. (4)

173. (2)

174. (3)

175. (1)

176. (3)

177. (4)

178. (4)

179. (1)

180. (3)

All India Aakash Test Series for Medical-2019 Test - 4 (Code E) (Hints and Solutions)

2/15

PHYSICS

1. Answer (1)

Hint: Wave transport energy and momentum.

Sol.: Transverse wave do not propagate inside a

fluid.

2. Answer (3)

Hint: RMS

3RTV

M

Sol.: RMS

3

Mass of gas

PVV

At constant volume RMS

V P

3. Answer (2)

Hint: avg2

fKTE for per molecule

Sol.: rms

3RTV

M

4. Answer (2)

Hint: PV = nRT

Sol.: PV = nRT

Mass( Molecular weight)PV RT M

M

MassPVR

T M

massPV R

T M

1Slope

M

5. Answer (1)

Hint: Sound RMS

3v v

Sol.: RMS

3RTv

M

HINTS AND SOLUTIONS

Sound

RTv

M

RMS av mp > > v v v

6. Answer (3)

Hint: PV = nRT

Sol.: PV = nRT

V nR

T P

Slopen

P

tan60A

A

n

P

tan30B

B

n

P

1A

B

P

P

7. Answer (2)

Hint: (mix)

mix

mix

P

V

C

C

Sol.: 1 21 2

(mix)

1 2

P P

P

nC n CC

n n

1 21 2

(mix)

1 2

V V

V

nC n CC

n n

(mix)

5 7

2 23

2P

R R

C R

CV (mix)

= 2R

2

3

V

P

C

C

8. Answer (2)

Hint: 0

1V

T V

Test - 4 (Code E) (Hints and Solutions) All India Aakash Test Series for Medical-2019

3/15

Sol.: PV = nRT

P V = nRT

V nR

T P

0

nR nR

PV nRT

1

T

n = –1

9. Answer (1)

Hint: PV = nRT

Sol.:nR

P TV

As, SlopenR

V is constant.

So, V = constant, W = 0 and = constant

10. Answer (3)

Hint: a = –2x

Sol.: 2 2 2

2 2–

–

aT x TT

x x

– 2 2constantT

11. Answer (2)

Hint: 2 2 21 1and KE ( – )

2 2U kx k A x

Sol.: As body moves from mean position, speed

decreases. So, KE decreases and PE increases.

12. Answer (2)

Hint: 0

2H

Tg

Sol.: At equilibrium condition

FB = weight

AHg = Ah0g

H = h0

[A is circular area of cylinder]

2h

Tg

13. Answer (2)

Hint: 2 2 2– –a x v A x

Sol.: a = v

2 2 2–x A x

2x2 = A2 – x2

At 21 2 second

2

Ax T

14. Answer (2)

Hint: 2R

Tg

Sol.: Time period of particle is independent of length

of chord.

15. Answer (3)

Hint: If y = f(ax ± bt)

then b

va

Sol.: 2

9

3 ( – 10 )y

x t

v = 10 m/s

16. Answer (2)

Hint: a = –2x

Sol.: a = –2x

2–B A B

A

2 B

T A

2A

TB

17. Answer (3)

Hint: 2 2

Resultant 1 2 1 22 cosA A A A A

Sol.: x = x1 + x

2

3 sin( )x A t

max3 ;A A

2 2 2

net2 cos60A A A A

∵

amax

= 2Amax

23 A


4/15

18. Answer (3)

Hint: Conservation of ME and linear momentum.

Sol.: 2 21 1( )

2 2M m v Kx

and mv0 = (M + m)v

x = Amplitude of oscillation

On solving

0mv M m

xM m K

19. Answer (4)

Hint: vparticle

= –(Slope) vwave

Sol.: Acceleration of particle always along their mean

position.

vA is upward and v

C is downward.

20. Answer (3)

Hint & Sol.: Two waves of same frequency & moves

in opposite direction.

21. Answer (3)

Hint: v = f

Sol.: v = f1

1 and v = f

2

2

1

1

vf

and

2

2

vf

1 2

1 2

1 1– –f f v

f1 – f

2 = n

1 2

2 1–

n

v

22. Answer (1)

Hint: 0

0

S

v vf f

v v

Sol.:

fref.

f0

v

Reflected frequency

ref. 0–

c vf f

c v

23. Answer (4)

Hint: In closed organ pipe only odd harmonics are

present.

Sol.: 50 × 1 = 50, 50 × 3 = 150, 50 × 5 = 250.

Odd harmonics are present. So, pipe is closed at

one end.

340 34m

50 5

v

f

34 171.7 m

4 20 10l

24. Answer (3)

Hint: 1 2

(2 – 1),

2 4o c

nv nf f v

l l

Sol.: 1 2

1 5;

2 4o c

vf f v

l l

fo = f

c

1 2

1 5

2 4l l

2 1

5

2l l

l1 = 10 cm; l

2 = 25 cm.

25. Answer (3)

Hint and Sol.: Doppler’s effect depends on relative

motion.

26. Answer (4)

Hint: 0

S.L 10logI

I

Sol.: 1

1

0

S.L 10logI

I

2

2

0

S.L 10logI

I

S.L = S.L2 – S.L

1 = 30

2 11000I I

27. Answer (2)

Hint and Sol.: All medium particles between two

successive nodes oscillate in same phase and

particles on one side of a node oscillate in opposite

phase with those on the other side of same node.


5/15

28. Answer (4)

Hint: 2L

Tg

Sol.: L = 2.45 m

g = 9.8 m/s2

2 1sL

Tg

29. Answer (1)

Hint: 2

nf v

l

Sol.: 3

3

2f v

l v = 48 m/s

f3 = 48 Hz l = 1.5 m

1 mv

f

30. Answer (1)

Hint: RT

vM

Sol.: 2 RTv

M

v2 T (T in Kelvin)

31. Answer (4)

Hint: = kx

Sol.: Path difference 3 5

–4 8 8

x

2 5

8

5

4

32. Answer (2)

Hint and Sol.: particle

=

y

vt

33. Answer (1)

Hint: max

waveand

Pv A v

k

Sol.: sin( )y A t kx

= 50; k = 5

10 m/svk

maxp

v = A = 10 × 50 = 500 m/s

max

50p

v

v

34. Answer (3)

Hint: P

v

Sol.: air

air

Pv

2

2

H

H

Pv

2air

4 4 332 1328 m/sH

v v

35. Answer (2)

Hint and Sol.: Phase difference between incident

and reflected wave is when reflected from rigid

boundary.

36. Answer (3)

Hint: Doppler’s effect depends on relative motion.

37. Answer (4)

Hint: fBeat

= |f1 – f

2|

Sol.:

fv

s

Frequency of reflected sound.

–

r

c vf f

c v

Beat

2–

–r

vff f f

c v


6/15

38. Answer (2)

Hint: fbeat

= |f1 – f

2|

Sol.: fB – f

A = 2

1 1

30 25A Bf f

25

30

A

B

f

f

On solving

fA = 10 Hz; f

B = 12 Hz

39. Answer (1)

Hint: 2 2

net 1 2 1 22 cosA A A A A .

Sol.: 2A

2A

60°

60°

A

Anet

= 2A – A = A

40. Answer (2)

Hint: x = A sin(t + )

Sol.:

45°

45°

t t =

A(1)

t = 0

(2)

(2)

x

mean

(t t) =

t = 0

x = Acos45° 2

A

41. Answer (3)

Hint: 2

AV

1KE

3kA

Sol.: ( )AV

0

1KE KE

A

xdx

A

On solving

2 2

AV

1(KE)

3m A

42. Answer (1)

Hint and Sol.: 2m

Tk

43. Deleted

44. Answer (3)

Hint: Damped and forced oscillation.

Sol.: In damped oscillation amplitude and energy

both decreases exponentially.

At resonance amplitude becomes maximum.

45. Answer (4)

Hint: U = nCvT

Sol.: Since5

2v

RC for diatomic

3

2v

RC for monoatomic.

Slopev

C

CHEMISTRY

46. Answer (2)

Hint: 32

%wt. 100 94.12%34

47. Answer (3)

Hint:

Zn + 2NaOH Na2

ZnO2 + H

2

2Al + 2NaOH + 6H2O 2Na[Al(OH)

4] + 3H

2

48. Answer (2)

Hint: At low temperature para form dominant.

Sol.: At absolute zero only para form of hydrogen

exists.


7/15

49. Answer (3)

Hint:

Reducing :

atomic >

nascent > dihydrogen.

ability hydrogen hydrogen

50. Answer (4)

Hint: Solubility order

CaCO3 < NaHCO

3 < KHCO

3

51. Answer (3)

Hint: Li is hardest alkali metal.

Sol. : Lithium being smallest in size shows

dominating covalent character.

52. Answer (3)

Hint: – O – O – linkage is peroxide linkage.

Sol.: BaO2 and Na

2O

2 are peroxides.

53. Answer (3)

Hint: Higher the lattice energy, higher the thermal

stability.

Sol.: Order of thermal stability : LiF > NaF > KF.

54. Answer (1)

Hint: Phosphorus give disproportionation reaction

with Ca(OH)2.

Sol.:

2P4 + 3Ca(OH)

2 + 6H

2O

3Ca(H

2PO

2)2

+ 2PH3

55. Answer (4)

Hint: Beryllium shows covalent nature in its

compounds.

Sol.: Be shows diagonal relationship with Al.

56. Answer (4)

Hint: NaH is ionic hydride.

Sol.: Li, Be and Mg form covalent hydrides.

57. Answer (1)

Hint: H4P

2O

5 contains only two P – OH bond.

HO – P – O – P – OH

OII

IH

OII

IH

58. Answer (3)

Hint: CNG consist of lower alkane as major

constituent.

59. Answer (3)

Hint: Higher the (charge/radius) value, higher is

hydration.

60. Answer (4)

Hint: Ammoniated electron gives blue colouration.

61. Answer (3)

Hint: Correct order of bond angle is

H2O > H

2S > H

2Te

62. Answer (1)


BeSO4 > MgSO

4 > BaSO

4

63. Answer (2)

Hint: CaO is a basic oxide

64. Answer (4)

Hint: 4 2 4 2

(gypsum) (Dead burnt plaster)

CaSO 2H O CaSO 2H O

65. Answer (3)

Hint: In carbon and oxygen d-orbitals are absent.

66. Answer (3)

Hint: 1H1,

1D2,

1T3 same electronic configuration

but different atomic masses.

Sol.: BE of D2 > BE of H

2

67. Answer (2)

Hint: Br in BrF5 is sp3d2 hybridised.

Br

F

F

F

F

F

68. Answer (1)

Hint: Higher the proton affinity higher is the

basicity.

Sol.: NH3 is most basic.

69. Answer (4)

Hint: Dielectric constant of H2O and D

2O

respectively are 78.39 and 78.06 C2 N–1 m–2.

70. Answer (3)

Hint: Solvay process

Sol.: NH3 + H

2O + CO

2 NH

4HCO

3

NH4HCO

3 + NaCl

4 3NH Cl + NaHCO


8/15

71. Answer (2)

Hint: Be(OH)2 is amphoteric in nature.

Sol.: Be(OH)2 + 2NaOH Na

2[Be(OH)

4].

72. Answer (4)

Hint: CaC2 + 2H

2O Ca(OH)

2 + C

2H

2

73. Answer (2)

Hint: NO2 is a brown coloured gas.

Sol.: 2NaNO3 2NaNO

2 + O

2

74. Answer (2)

Hint: Castner-Kellner cell is used to prepare NaOH.

Sol.: At cathode: HgNa e Na Hg

2Na – Hg + 2H2O 2NaOH + H

2 + 2Hg

At anode: 2

1Cl Cl e

2

75. Answer (1)

Hint: Be (OH)2 is amphoteric in nature.

Sol.: Basicity of second group hydroxides increases

down the group.

76. Answer (4)

Hint: Na2CO

3 + H

2O + CO

2 2NaHCO

3

77. Answer (1)

Hint: H, T, D are isotopes of hydrogen.

Sol.: H2, D

2, T

2, H–D, D–T, H–T

78. Answer (4)

Hint: Clark’s method for the removal of temporary

hardness.

79. Answer (2)

Hint: Formula of brown coloured complex is

[Fe(H2O)

5(NO)]2+

80. Answer (2)

Hint: 6.8 ml O2 at STP is obtained by 1 ml H

2O

2

solution by decomposition.

Sol.:

224 ml O2 at STP is obtained by

224ml

6.8H

2O

2.

= 32.94 ml H2O

2

81. Answer (2)

Hint: Cu + 8HNO3(dilute) 3Cu(NO

3)2 + 2NO + 4H

2O.

82. Answer (2)

Hint: CaCN2 is calcium cyanamide.

Sol.: CaCN2 + 3H

2O CaCO

3 + 2NH

3(g)

83. Answer (4)

Hint: NO2

is paramagnetic oxide of nitrogen.

N

O O

84. Answer (2)

Hint: H2O is neutral oxide.

Sol.: Oxides of alkali metals are basic.

85. Answer (2)

Hint: A suspension of magnesium hydroxide in water

is called milk of magnesia.

86. Answer (4)

Hint: 4Zn + 10HNO3(dilute) 4Zn(NO

3)2 + 5H

2O + N

2O.

87. Answer (4)

Hint: Spontaneous combustion of phosphine is used

in Holme’s signal.

88. Answer (3)

Hint: Cobalt is used as a catalyst in preparation of

methanol.

89. Answer (3)

Hint: H2O

2 oxidises PbS into PbSO

4.

90. Answer (2)

Hint: Zn reacts both with acid and base.

Sol.: Zn + 2HCl ZnCl2 + H

2

Zn + 2NaOH Na2ZnO

2 + H

2

BIOLOGY

91. Answer (1)

Hint : More the H+ ions, more acidic the condition

is and less will be the pH.

Sol. : During stomatal opening, due to the activity

of hydrogen-potassium ion-exchange pump, H+ from

guard cells are transported to neighbouring

subsidiary cells while K+ ions are transported into

the guard cells. This increases the pH of guard cells.

92. Answer (3)

Hint : Transpiration pull and root pressure causes

upliftment of water by pulling and pushing

respectively.


9/15

Sol. : Transpiration develops a negative water

potential in the xylem which creates a ‘pull’ for

translocation of water while root pressure is a

positive hydrostatic pressure responsible for pushing

of water.

93. Answer (4)

Hint : In girdling experiment, a ring of bark upto the

depth of the phloem layer is carefully removed.

Sol. : In a girdled plant, root cells die first than the

shoot cells, due to stoppage of translocation of

sugars and other materials to the roots.

94. Answer (2)

Sol. : In guard cells, cellulose microfibrils are

radially oriented.

95. Answer (1)

Hint : Water molecules move from an area of high

water potential (w) to an area of low water potential

(w).

Sol. : w of cell A =

s +

p = – 2 atm

w of cell B = – DPD = – 10 atm

w of cell C = OP – TP = – 3 atm

w of cell D = – 6 atm

So the correct direction of movement of water is

A B

C D

↗

96. Answer (2)

Hint : In this process, loss of water in the form of

liquid droplets occurs through special openings

called hydathodes.

Sol. : Herbaceous plants lose water in the form of

droplets due to high root pressure. This process is

called guttation.

97. Answer (4)

Hint : Movement of sucrose from mesophyll cells to

sieve tube elements via companion cells, requires

energy in the form of ATP.

Sol. : Loading of sucrose in sieve tube cells via

companion cells is an active process.

98. Answer (4)

Hint : Facilitated diffusion is a passive transport.

Sol. : In facilitated diffusion, transport of ions/

molecules occurs with the help of specific membrane

proteins called transporters. This process is highly

specific because transporter proteins are highly

selective. The rate of facilitated transport may

saturate.

99. Answer (1)

Hint : Intercellular movement of water occurs

through cytoplasmic connections between

neighbouring cells.

Sol. : These structures are called plasmodesmata.

100. Answer (3)

Hint : Water potential (w) is free energy of water

molecules which decreases by adding solutes.

Sol. : Water potential of pure water is zero at

atmospheric pressure. By increasing external

pressure on pure water, its water potential increases.

101. Answer (2)

Hint : Mg is the co-factor of RuBisCO enzyme.

Sol. : Zn – Synthesis of auxin

K – Maintains turgidity of cells

Ni – Component of urease

102. Answer (2)

Hint : For immobile minerals, deficiency symptoms

tend to appear first in younger tissues.

Sol. : Among the given minerals, calcium is an

immobile element.

103. Answer (4)

Sol. : Na is a beneficial element.

104. Answer (1)

Hint : Ammonification involves decomposition of

organic nitrogen into ammonia.

Sol. : Ammonifying bacteria convert organic nitrogen

of dead plants and animals into ammonia.

105. Answer (2)

Hint : In green plants, Mn and Cl are involved in

splitting of water to release oxygen during

photosynthesis.

Sol. : Due to Mn toxicity translocation of Ca to

shoot apex is inhibited. Mn deficiency causes grey

spots in oats. Little leaf symptom is caused by the

deficiency of Zn. Transport of carbohydrate in phloem

is facilitated by boron.

106. Answer (3)

Hint : Leghaemoglobin is an O2 scavenger pigment

present in the cytosol of root nodule cells.


10/15

Sol. : After the formation of root nodule, bacteroids

present in root nodule cells produce a reddish-pink

pigment called leghaemoglobin which protects

nitrogenase from oxygen. Nodule formation is

induced by bacterial and plant signals.

107. Answer (4)

Hint : Nitrifying bacteria have chemoautotrophic

mode of nutrition.

Sol. : Nitrosomonas is a nitrifying bacterium which

converts NH3 into NO

2–.

108. Answer (3)

Hint : During nitrogen fixation atmospheric nitrogen

is converted into ammonia.

Sol. : Nitrogenase enzyme is a Mo – Fe protein. It

requires ATPs, strong reducing agents like FADH2,

NADPH and anaerobic condition for fixation of

nitrogen into ammonia.

109. Answer (4)

Hint : In C3 plants, CO

2 acceptor is a 5-carbon

containing molecule.

Sol. : In C3 plants, RuBP is the primary CO

2

acceptor molecule.

110. Answer (2)

Hint : Non-cyclic photophosphorylation is associated

with formation of assimilatory power and splitting of

water.

Sol. : Non-cyclic phosphorylation occurs only in

grana thylakoids. In this process both photosystems

work in a series, where PS II operates first, followed

by PS I. Products of non-cyclic photophosphorylation

are ATP, NADPH and O2.

111. Answer (4)

Hint : Dictyosomes are unconnected units of Golgi

body found in plant cells.

Sol. : Golgi bodies are not associated with

photorespiration. Organelles involved in photo

respiration are chloroplast, peroxisome and

mitochondria.

112. Answer (3)

Hint : C4 plants lack the wasteful process

responsible for loss of fixed CO2.

Sol. : In C3 plants, when there is high

concentration of O2 in stroma, oxygenase activity of

RuBisCO leads to photorespiratory loss. On the

other hand C4 plants have mechanism to increase

the CO2 concentration in comparison to O

2

concentration at the enzyme site, because of which

C4 plants do not have photorespiration. Thus

productivity of C4 plant is better than C

3 plants.

113. Answer (2)

Hint : Electron transport of non-cyclic

photophosphorylation starts with photosystem II.

Sol. : A = photosystem II

B = Photosystem I

C = Stroma

D = Lumen

114. Answer (4)

Hint : C4 plants have Kranz anatomy.

Sol. : In C4 plant, first carboxylation occurs in

mesophyll cells by PEPcase enzyme which

eventually leads to malic acid formation. Malic acid

gets decarboxylated in bundle sheath cells where

second carboxylation takes place by RuBisCO

enzyme.

115. Answer (4)

Hint : Sorghum shows spatial difference in double

carboxylation.

Sol. : Sorghum is a C4 plant.

116. Answer (1)

Hint : Photorespiration is a process which involves

loss of fixed carbon as CO2 in plants.

Sol. : Carboxylation, reduction and regeneration are

three steps of Calvin cycle.

117. Answer (2)

Hint : CO2 is the major limiting factor, influencing

the rate of photosynthesis.

Sol. : C3 plants show saturation at 450 ppm of

CO2, while C

4 plants show saturation at 360 ppm of

CO2 concentration at high light intensities. C

3 plants

show CO2 fertilization effect as in the CO

2 enriched

atmosphere they show higher yield.

118. Answer (3)

Hint : First action spectrum of photosynthesis was

prepared by T.W. Engelmann.

Sol. : Engelmann used a green alga Cladophora to

describe action spectrum of photosynthesis.

119. Answer (3)

Hint : Wheat is a C3 plant while maize is a C

4

plant.


11/15

Sol. : For formation of 1 molecule of glucose

C3 plants require = 18 ATPs


So for each glucose molecule, C3 plants require 12

ATPs less than C4 plants.

120. Answer (3)

Hint : The process through which electrons pass

from one carrier to another is called electron

transport system (ETS).

Sol. : In mitochondria, ETS is linked to ATP

synthesis (oxidative phosphorylation). Electron

carriers of ETS are present in inner mitochondrial

membrane.

121. Answer (1)

Hint : Fermentation is a kind of anaerobic

respiration, carried out primarily by fungi and

anaerobic bacteria.

Sol. : In fermentation, pyruvic acid is partially

oxidized in the presence of external electron donor

(reducing agent), NADH.

In fermentation, no new ATP or NADH molecules are

produced apart from those produced during

glycolysis.

122. Answer (2)

Hint : Succinyl CoA is an intermediate of Krebs

cycle, used in the biosynthesis of chlorophyll.

Sol. : Terminal electron acceptor of mitochondrial

ETS is O2.

Cofactor of pyruvic acid decarboxylase enzyme = Mg+2

Cofactor of lactic acid dehydrogenase enzyme = Zn+2.

123. Answer (3)

Hint : In human kidney cells, NADH molecules

produced in glycolysis enter the mitochondria via

malate-aspartate shuttle.

Sol. : In aerobic respiration, ATP produced by

oxidative phosphorylation is coupled with ETS.

By ETS and oxidative phosphorylation

1 NADH produces 3 ATPs

and 1 FADH2 produces 2 ATPs

Total NADH molecules produced in aerobic

respiration = 10 (30 ATPs)

Total FADH2 molecules produced in aerobic

respiration = 2 (4 ATPs) so total ATP produced by

ETS and oxidative phosphorylation is 34.

124. Answer (1)

Hint : Respiratory quotient of any respiratory

substrate is the ratio of volume of CO2 released and

volume of O2 consumed during respiration.

Sol. : RQ of carbohydrate = 1

RQ of Proteins (Albumin) = 0.9

RQ of Fats (Palmitic acid) = 0.7

RQ of Organic acids >1 (for oxalic acid it is 4)

So descending order of RQ values for given

substrates is oxalic acid > glucose > albumin >

palmitic acid.

125. Answer (4)

Hint : Respiration mediated breakdown of different

substrates such as carbohydrate, fats and proteins

show interconnection as all produce energy through

Krebs cycle eventually.

Sol. : Acetyl CoA is a metabolite which is a

common product during respiratory breakdown of fats,

proteins and carbohydrates. It is also a raw material

for synthesis of carotenoids, terpenes and gibberellins.

126. Answer (2)

Sol. : Induction of -amylase in barley endosperms

is a bioassay of gibberellins.

127. Answer (3)

Hint : Initially zygotic division shows geometric

growth pattern.

Sol. : In geometric growth, every cell divides with

all the daughter cells growing and dividing again.

128. Answer (3)

Hint : Some synthetic auxins are used as

weedicides.

Sol. : 2, 4-D and 2, 4, 5-T are synthetic auxins

which act as weedicides.

129. Answer (2)

Hint : Abscisic acid is called stress hormone.

Sol. : Abscisic acid induces dormancy in buds and

seeds whereas gibberellin prevents it.

ABA acts as antagonist of gibberellins.

130. Answer (2)

Sol. : Garner and Allard first reported the

photoperiodic response of flowering in tobacco.

131. Answer (4)

Hint : Apical dominance is a phenomenon in which

apical buds do not allow growth of lateral buds.


12/15

Sol. : Auxin promotes apical dominance while

cytokinin counteracts it.

132. Answer (3)

Hint : Phytohormone that counteracts apical

dominance is best suitable for tea plantation.

Sol. : Promotion of parthenocarpy – Auxin.

Antitranspiratory action – Abscisic acid

Root development in various cuttings – Auxin.

Cytokinin counteracts with auxin and help in making

bushes of tea plants.

133. Answer (2)

Hint : These cells are present just next to the cells

of meristematic zone.

Sol. : Increased vacuolation is a feature of the cells

which are in phase of elongation.

134. Answer (3)

Hint : Cytokinins are modified purines.

Sol. : Apart from cytokinin, auxin, gibberellin and

abscisic acid are acidic in nature.

135. Answer (4)

Hint : Thigmonasty is a contact or touch stimulated

variation of movement in plants like Drosera.

Sol. : Phytochromes are chromoproteins or

photoreceptor pigments which are used to detect

light by plants.

136. Answer (1)

Hint: This animal is also called bookworm.

Sol.: Loligo and Laccifer are economically beneficial

mollusc and insect respectively. Limulus is a living

fossil.

137. Answer (1)

Hint: Identify an arthropod.

Sol.: Pheretima (an annelid), Pavo (a bird) and

Petromyzon (cyclostome) have closed circulatory

system. Capillaries are a feature of closed circulatory

system.

138. Answer (4)

Hint: Metamorphosis involves conversion of larva to

adult form.

Sol.: Direct development occurs in Hirudinaria.

Butterfly, Ascidia and Culex show metamorphosis

and their larval forms are caterpillar, tadpole &

wriggler respectively.

139. Answer (3)

Hint: This parameter is shared by larval

echinoderms with platyhelminthes.

Sol.: Adult echinoderms are radially symmetrical

while their larvae are bilaterally symmetrical. Anus

develops from blastopore in deuterostome body plan.

140. Answer (3)

Hint: These organisms occur in exclusively marine

conditions.

Sol.: Saccoglossus a hemichordate, has proboscis

gland as its excretory organ. Gills are meant for

respiration in hemichordates. In molluscs, gills serve

both the function of respiration and excretion.

141. Answer (1)

Hint: This organism forms gemmules to reproduce.

Sol.: Sponges lack distinct germ layers, has ostia

for entry of water and has collar cells in body.

142. Answer (1)

Hint: Metagenesis.

Sol.: Physalia exists in both polyp and medusa

form. Hydra, Meandrina and Adamsia exist as

polyps.

143. Answer (2)

Hint: Identify organisms with cnidocytes.

Sol.: Adamsia is sea anemone i.e., a cnidarian

whose tentacles are rich in stinging cells.

Euspongia has flagellated choanocytes.

Pleurobrachia does not show metagenesis and

Planaria is a free living worm, hence it lacks hooks.

144. Answer (3)

Hint: Parapodia are lateral appendages that help in

swimming and respiration.

Sol.: Malpighian tubules are found in insects

(arthropods) while parapodia are found in annelids

such as Nereis.

145. Answer (2)

Hint: Identify a roundworm which gives birth to young

worms.

Sol.: Ancylostoma and Ascaris are oviparous round

worms. Fasciola is a flatworm.

146. Answer (3)

Hint: Identify worms that are pseudocoelomate.


13/15

Sol.: Ringworm is a fungus; earthworm is an

annelid; bookworm is an arthropod; Liver fluke &

tapeworm are flatworms. Pinworm, filarial worm,

hookworm and roundworm belong to aschelminthes.

147. Answer (2)

Hint: Comb plates are locomotory structures in

members of this phylum.

Sol.: Annelids, arthropods and molluscs occupy

either aquatic or terrestrial habitats.

148. Answer (2)

Hint: This structure is composed of cellulose in

plants.

Sol.: Animal cells lack a cell wall. All animals are

heterotrophs and show division of labour. Poriferans

lack nervous system.

149. Answer (3)

Hint: Identify first triploblastic animals.

Sol.: Regeneration is a prominent feature of Porifera

Coelenterata, Platyhelminthes and Echinoderms.

150. Answer (4)

Hint: Select the organism that occurs exclusively in

marine water.

Sol.: Asterias has water vascular system and show

external fertilisation. Poriferans have water canal

system.

151. Answer (3)

Hint: Identify a true fish.

Sol.: Flying fish is a bony fish. Hagfish is a

cyclostome. Cuttlefish and devil fish belong to

phylum Mollusca.

152. Answer (2)

Hint: Identify a mollusc with scientific name

Dentalium.

Sol.: Tusk shell has a calcareous shell

(exoskeleton) while chitinous exoskeleton is a feature

of arthropods.

153. Answer (1)

Hint: Metamerism refers to presence of segments

and probable repeat of organs.

Sol.: Wuchereria is a filarial worm, where excretory

pore eliminates nitrogenous waste. Wastes present

in alimentary canal are eliminated through anus.

Pseudocoelom, absence of segmentation and

presence of bilateral symmetry are features of

Aschelminthes.

154. Answer (3)

Hint: This animal lives in loosely organised

community.

Sol.: Locusta is an economically harmful insect

(arthropod).

155. Answer (2)

Hint: Organisms that show bioluminescence.

Sol.: Flatworms like Schistosoma are dioecious but

digestion is only extracellular. In triploblastic animals

such as aschelminthes, extracellular digestion is

observed.

156. Answer (3)

Hint: Arthropods have open circulatory system.

Sol.: Pila is a mollusc with open circulatory system

and lacks segmentation. Anopheles has malpighian

tubules for excretion. Nereis is segmented and has

nephridia.

Presence of solid ventral nerve cord is a character of

non chordates.

157. Answer (2)

Hint: This organism is commonly called lac insect.

Sol.: Laccifer is a useful insect. Its secretion called

lac and shellac are used in bangle industry.

Anopheles, Culex and Aedes act as vectors to

spread diseases such as malaria, filariasis and

dengue respectively.

158. Answer (4)

Hint: Triploblastic organisms show extracellular

digestion.

Sol.: Adult echinoderms are radially symmetrical and

most poriferans are asymmetrical. Notochord is

absent in both Echinoderms and Poriferans. True

coelom is a feature of Echinoderms.

159. Answer (1)

Hint: Select an arthropod.

Sol.: Sea hare is Aplysia with unsegmented body.

Sea cucumber is dioecious. Tapeworm shows

internal fertilisation.

160. Answer (3)

Hint: Another name for this organism is Euspongia.

Sol.: Antedon and Sea urchin exhibit radial

symmetry in adult form. Salpa shows bilateral

symmetry.


14/15

161. Answer (1)

Hint: This structure divides body cavity into thoracic

and abdominal cavity.

Sol.: Mammals show direct development. Heterodont

dentition is absent in aquatic mammals such as

whale, sea cow, etc. Sloth and sea cow have 9 and

6 cervical vertebrae respectively.

162. Answer (2)

Hint: Select an avian adaptation.

Sol.: Feathers reduce weight and are found in birds.

163. Answer (2)

Hint: Pneumatic bones have air cavities to reduce

weight of flying birds.

Sol.: Neophron i.e. vulture is a flying bird that has

both pneumatic bones and preen/oil gland. Air sacs

in birds are avascular and meant for storage of air

but not exchange of gases.

164. Answer (4)

Hint: Snakes detect vibrations through jaw bones.

Sol.: Vipera has epidermal scales; 3 chambered

heart and is limbless in adult form. It has dry

cornified aglandular skin.

165. Answer (1)

Hint: Such an organism produces internal body heat

through metabolic activity to maintain body

temperature.

Sol.: Mammals are homeotherms and endotherms.

They exhibit internal fertilisation (Felis) as do birds.

Chelone inhabits water. Replites show shedding of

skin (ecdysis).

166. Answer (2)

Hint: Heart and blood vessels are present in open

and closed circulatory system.

Sol.: Heart is dorsal in position in non-chordates

usually while it is ventral in chordates. Nerve cord is

ventral in non-chordates but dorsal in position in

chordates. Post anal tail is a feature of chordates.

Gill slits are lateral in position in chordates.

167. Answer (2)

Hint: Identify a circular mouth scaleless fish like

organism.

Sol.: Myxine is hagfish, a cyclostome. Torpedo,

Pristis and Trygon belong to super class Pisces.

168. Answer (1)

Hint: Select a cartilaginous fish.

Sol.: Air/swim bladder helps to maintain buoyancy in

bony fish. Dog fish has to swim continuously to

avoid sinking.

169. Answer (3)

Hint: Dry, aglandular skin is a feature of reptiles.

Sol.: Ichthyophis is a limbless amphibian

Bangarus is a limbless reptile

Balaenoptera is a limbless mammal

They all have bony vertebral column. Skin is glandular

in mammals.

170. Answer (4)

Hint: Identify an animal where development of

embryo can occur on land.

Sol.: Pisces and amphibians are anamniotes. Their

development is linked to water directly. Reptiles

evolved prior to birds. Mammals, birds and reptiles

are all amniotes.

171. Answer (2)

Hint: Gill cover is absent in cartilaginous fish.

Sol.: Chondrichthyes are cartilaginous fish which

lack operculum generally; their males have pelvic

claspers and show internal fertilisation. Bony fishes

possess operculum, have cycloid/ctenoid scales and

show external fertilisation.

172. Answer (4)

Hint: It is another term for arthrodial membrane.

Sol.: Arthrodial membrane is flexible membrane that

connects tergite with pleurites and sternites.

173. Answer (2)

Hint: First pair of wings is mesothoracic.

Sol.: Tegmina/Elytra or wing cover is the alternative

name for first pair of wings in cockroach.

174. Answer (3)

Hint: Identify mouth parts that are paired in

cockroach.

Sol.: Hypopharynx acts as tongue. Maxillae are

paired structures that help bring food to mandibles.

175. Answer (1)

Hint: Glands whose ducts open into brood or genital

pouch.

Sol.: Seminal vesicles nourish the sperms. Phallic/

conglobate secrete substances that help to form a

protective cover of spermatophore.


15/15

176. Answer (3)

Hint: Sexual dimorphism in cockroach is based on

presence of unjointed filamentous structures attached

to 9th sternum.

Sol.: Anal/caudal styles are unjointed structures

attached to 9th sternum in male cockroach. Anal

cerci are jointed structures present in both male and

female cockroach.

177. Answer (4)

Hint: External segmentation is found in arthropods.

Sol.: Schizocoelom and presence of external

metamerism/segmentation is a feature of Periplaneta.

178. Answer (4)

Hint: Part of foregut lined by cuticle is not secretory

in nature.

Sol.: Crop part of foregut in cockroach does not

secrete enzymes. Haemoglobin is absent in blood of

cockroach. Hepatic caecae are present at the

junction of foregut and midgut.

179. Answer (1)

Hint: These paired structures help in respiration in

cockroach.

Sol.: 10 pairs of spiracles are present in body of

Periplaneta.

180. Answer (3)

Hint: Appositional image.

Sol.: Vision in cockroach has low resolution but high

sensitivity.

��

Test - 4 (Code F) (Answers) All India Aakash Test Series for Medical-2019

1/15

1. (4)

2. (3)

3. Deleted

4. (1)

5. (3)

6. (2)

7. (1)

8. (2)

9. (4)

10. (3)

11. (2)

12. (3)

13. (1)

14. (2)

15. (4)

16. (1)

17. (1)

18. (4)

19. (2)

20. (4)

21. (3)

22. (3)

23. (4)

24. (1)

25. (3)

26. (3)

27. (4)

28. (3)

29. (3)

30. (2)

31. (3)

32. (2)

33. (2)

34. (2)

35. (2)

36. (3)

Test Date : 20/01/2019

ANSWERS

TEST - 4 (Code F)

All India Aakash Test Series for Medical - 2019

37. (1)

38. (2)

39. (2)

40. (3)

41. (1)

42. (2)

43. (2)

44. (3)

45. (1)

46. (2)

47. (3)

48. (3)

49. (4)

50. (4)

51. (2)

52. (2)

53. (4)

54. (2)

55. (2)

56. (2)

57. (2)

58. (4)

59. (1)

60. (4)

61. (1)

62. (2)

63. (2)

64. (4)

65. (2)

66. (3)

67. (4)

68. (1)

69. (2)

70. (3)

71. (3)

72. (4)

73. (2)

74. (1)

75. (3)

76. (4)

77. (3)

78. (3)

79. (1)

80. (4)

81. (4)

82. (1)

83. (3)

84. (3)

85. (3)

86. (4)

87. (3)

88. (2)

89. (3)

90. (2)

91. (4)

92. (3)

93. (2)

94. (3)

95. (4)

96. (2)

97. (2)

98. (3)

99. (3)

100. (2)

101. (4)

102. (1)

103. (3)

104. (2)

105. (1)

106. (3)

107. (3)

108. (3)

109. (2)

110. (1)

111. (4)

112. (4)

113. (2)

114. (3)

115. (4)

116. (2)

117. (4)

118. (3)

119. (4)

120. (3)

121. (2)

122. (1)

123. (4)

124. (2)

125. (2)

126. (3)

127. (1)

128. (4)

129. (4)

130. (2)

131. (1)

132. (2)

133. (4)

134. (3)

135. (1)

136. (3)

137. (1)

138. (4)

139. (4)

140. (3)

141. (1)

142. (3)

143. (2)

144. (4)

145. (2)

146. (4)

147. (3)

148. (1)

149. (2)

150. (2)

151. (1)

152. (4)

153. (2)

154. (2)

155. (1)

156. (3)

157. (1)

158. (4)

159. (2)

160. (3)

161. (2)

162. (3)

163. (1)

164. (2)

165. (3)

166. (4)

167. (3)

168. (2)

169. (2)

170. (3)

171. (2)

172. (3)

173. (2)

174. (1)

175. (1)

176. (3)

177. (3)

178. (4)

179. (1)

180. (1)

All India Aakash Test Series for Medical-2019 Test - 4 (Code F) (Hints and Solutions)

2/15

PHYSICS

1. Answer (4)

Hint: U = nCvT

Sol.: Since5

2v

RC for diatomic

3

2v

RC for monoatomic.

Slopev

C

2. Answer (3)

Hint: Damped and forced oscillation.

Sol.: In damped oscillation amplitude and energy

both decreases exponentially.

At resonance amplitude becomes maximum.

3. Deleted

4. Answer (1)

Hint and Sol.: 2m

Tk

5. Answer (3)

Hint: 2

AV

1KE

3kA

Sol.: ( )AV

0

1KE KE

A

xdx

A

on solving, 2 2

AV

1(KE)

3m A

6. Answer (2)

Hint: x = A sin(t + )Sol.:

45°

45°

t t =

A(1)

t = 0

(2)

(2)

x

mean

(t t) =

t = 0

x = Acos45° 2

A

HINTS AND SOLUTIONS

7. Answer (1)

Hint: 2 2

net 1 2 1 22 cosA A A A A .

Sol.: 2A

2A

60°

60°

A

Anet

= 2A – A = A

8. Answer (2)

Hint: fbeat

= |f1 – f

2|

Sol.: fB – f

A = 2

1 1

30 25A Bf f

25

30

A

B

f

f

On solving

fA = 10 Hz; f

B = 12 Hz

9. Answer (4)

Hint: fBeat

= |f1 – f

2|

Sol.:

fv

s

Frequency of reflected sound.

–

r

c vf f

c v

Beat

2–

–r

vff f f

c v

10. Answer (3)

Hint: Doppler’s effect depends on relative motion.

11. Answer (2)

Hint and Sol.: Phase difference between incident

and reflected wave is when reflected from rigid

boundary.

Test - 4 (Code F) (Hints and Solutions) All India Aakash Test Series for Medical-2019

3/15

12. Answer (3)

Hint: P

v

Sol.:air

air

Pv

2

2

H

H

Pv

2air

4 4 332 1328 m/sH

v v

13. Answer (1)

Hint: max

waveand

Pv A v

k

Sol.: sin( )y A t kx

= 50; k = 5

10 m/svk

maxp

v = A = 10 × 50 = 500 m/s

max

50p

v

v

14. Answer (2)

Hint and Sol.: particle

=

y

vt

15. Answer (4)

Hint: = kx

Sol.: Path difference 3 5

–4 8 8

x

2 5

8

5

4

16. Answer (1)

Hint: RT

vM

Sol.: 2 RTv

M

v2 T (T in Kelvin)

17. Answer (1)

Hint: 2

nf v

l

Sol.: 3

3

2f v

l v = 48 m/s

f3 = 48 Hz l = 1.5 m

1 mv

f

18. Answer (4)

Hint: 2L

Tg

Sol.: L = 2.45 m

g = 9.8 m/s2

2 1sL

Tg

19. Answer (2)

Hint and Sol.: All medium particles between two

successive nodes oscillate in same phase and

particles on one side of a node oscillate in opposite

phase with those on the other side of same node.

20. Answer (4)

Hint: 0

S.L 10logI

I

Hint: 0

S.L 10logI

I

Sol.: 1

1

0

S.L 10logI

I

2

2

0

S.L 10logI

I

S.L = S.L2 – S.L

1 = 30

2 11000I I

21. Answer (3)

Hint and Sol.: Doppler’s effect depends on relative

motion.

22. Answer (3)

Hint: 1 2

(2 – 1),

2 4o c

nv nf f v

l l


4/15

Sol.: 1 2

1 5;

2 4o c

vf f v

l l

fo = f

c

1 2

1 5

2 4l l

2 1

5

2l l

l1 = 10 cm; l

2 = 25 cm.

23. Answer (4)

Hint: In closed organ pipe only odd harmonics are

present.

Sol.: 50 × 1 = 50, 50 × 3 = 150, 50 × 5 = 250.

Odd harmonics are present. So, pipe is closed at

one end.

340 34m

50 5

v

f

34 171.7 m

4 20 10l

24. Answer (1)

Hint: 0

0

S

v vf f

v v

Sol.:

fref.

f0

v

Reflected frequency

ref. 0–

c vf f

c v

25. Answer (3)

Hint: v = f

Sol.: v = f1

1 and v = f

2

2

1

1

vf

and

2

2

vf

1 2

1 2

1 1– –f f v

f1 – f

2 = n

1 2

2 1–

n

v

26. Answer (3)

Hint & Sol.: Two waves of same frequency & moves

in opposite direction.

27. Answer (4)

Hint: vparticle

= –(Slope) vwave

Sol.: Acceleration of particle always along their mean

position.

vA is upward and v

C is downward.

28. Answer (3)

Hint: Conservation of ME and linear momentum.

Sol.: 2 21 1( )

2 2M m v Kx

and mv0 = (M + m)v

x = Amplitude of oscillation

On solving

0mv M m

xM m K

29. Answer (3)

Hint: 2 2

Resultant 1 2 1 22 cosA A A A A

Sol.: x = x1 + x

2

3 sin( )x A t

max3 ;A A

2 2 2

net2 cos60A A A A

∵

amax

= 2Amax

23 A

30. Answer (2)

Hint: a = –2x

Sol.: a = –2x

2–B A B

A

2 B

T A

2A

TB

31. Answer (3)

Hint: If y = f(ax ± bt)

then b

va


5/15

Sol.: 2

9

3 ( – 10 )y

x t

v = 10 m/s

32. Answer (2)

Hint: 2R

Tg

Sol.: Time period of particle is independent of length

of chord.

33. Answer (2)

Hint: 2 2 2– –a x v A x

Sol.: a = v

2 2 2–x A x

2x2 = A2 – x2

At 21 2 second

2

Ax T

34. Answer (2)

Hint: 0

2H

Tg

Sol.: At equilibrium condition

FB = weight

AHg = Ah0g

H = h0

[A is circular area of cylinder]

2h

Tg

35. Answer (2)

Hint: 2 2 21 1and KE ( – )

2 2U kx k A x

Sol.: As body moves from mean position, speed

decreases. So, KE decreases and PE increases.

36. Answer (3)

Hint: a = –2x

Sol.: 2 2 2

2 2–

–

aT x TT

x x

– 2 2constantT

37. Answer (1)

Hint: PV = nRT

Sol.:nR

P TV

As, SlopenR

V is constant.

So, V = constant, W = 0 and = constant

38. Answer (2)

Hint: 0

1V

T V

Sol.: PV = nRT

P V = nRT

V nR

T P

0

nR nR

PV nRT

1

T

n = –1

39. Answer (2)

Hint: (mix)

mix

mix

P

V

C

C

Sol.: 1 21 2

(mix)

1 2

P P

P

nC n CC

n n

1 21 2

(mix)

1 2

V V

V

nC n CC

n n

(mix)

5 7

2 23

2P

R R

C R

CV (mix)

= 2R

2

3

V

P

C

C


6/15

40. Answer (3)

Hint: PV = nRT

Sol.: PV = nRT

V nR

T P

Slopen

P

tan60A

A

n

P

tan30B

B

n

P

1A

B

P

P

41. Answer (1)

Hint: Sound RMS

3v v

Sol.: RMS

3RTv

M

Sound

RTv

M

RMS av mp > > v v v

42. Answer (2)

Hint: PV = nRT

Sol.: PV = nRT

Mass( Molecular weight)PV RT M

M

MassPVR

T M

massPV R

T M

1Slope

M

43. Answer (2)

Hint: avg2

fKTE for per molecule

Sol.: rms

3RTV

M

44. Answer (3)

Hint: RMS

3RTV

M

Sol.: RMS

3

Mass of gas

PVV

At constant volume RMS

V P

45. Answer (1)

Hint: Wave transport energy and momentum.

Sol.: Transverse wave do not propagate inside a

fluid.

CHEMISTRY

46. Answer (2)

Hint: Zn reacts both with acid and base.

Sol.: Zn + 2HCl ZnCl2 + H

2

Zn + 2NaOH Na2ZnO

2 + H

2

47. Answer (3)

Hint: H2O

2 oxidises PbS into PbSO

4.

48. Answer (3)

Hint: Cobalt is used as a catalyst in preparation of

methanol.

49. Answer (4)

Hint: Spontaneous combustion of phosphine is used

in Holme’s signal.

50. Answer (4)

Hint: 4Zn + 10HNO3(dilute) 4Zn(NO

3)2 + 5H

2O + N

2O.

51. Answer (2)

Hint: A suspension of magnesium hydroxide in water

is called milk of magnesia.

52. Answer (2)

Hint: H2O is neutral oxide.

Sol.: Oxides of alkali metals are basic.


7/15

53. Answer (4)

Hint: NO2

is paramagnetic oxide of nitrogen.

N

O O

54. Answer (2)

Hint: CaCN2 is calcium cyanamide.

Sol.: CaCN2 + 3H

2O CaCO

3 + 2NH

3(g)

55. Answer (2)

Hint: Cu + 8HNO3(dilute) 3Cu(NO

3)2 + 2NO + 4H

2O.

56. Answer (2)

Hint: 6.8 ml O2 at STP is obtained by 1 ml H

2O

2

solution by decomposition.

Sol.:

224 ml O2 at STP is obtained by

224ml

6.8H

2O

2.

= 32.94 ml H2O

2

57. Answer (2)

Hint: Formula of brown coloured complex is

[Fe(H2O)

5(NO)]2+

58. Answer (4)

Hint: Clark’s method for the removal of temporary

hardness.

59. Answer (1)

Hint: H, T, D are isotopes of hydrogen.

Sol.: H2, D

2, T

2, H–D, D–T, H–T

60. Answer (4)

Hint: Na2CO

3 + H

2O + CO

2 2NaHCO

3

61. Answer (1)

Hint: Be (OH)2 is amphoteric in nature.

Sol.: Basicity of second group hydroxides increases

down the group.

62. Answer (2)

Hint: Castner-Kellner cell is used to prepare NaOH.

Sol.: At cathode: HgNa e Na Hg

2Na – Hg + 2H2O 2NaOH + H

2 + 2Hg

At anode: 2

1Cl Cl e

2

63. Answer (2)

Hint: NO2 is a brown coloured gas.

Sol.: 2NaNO3 2NaNO

2 + O

2

64. Answer (4)

Hint: CaC2 + 2H

2O Ca(OH)

2 + C

2H

2

65. Answer (2)

Hint: Be(OH)2 is amphoteric in nature.

Sol.: Be(OH)2 + 2NaOH Na

2[Be(OH)

4].

66. Answer (3)

Hint: Solvay process

Sol.: NH3 + H

2O + CO

2 NH

4HCO

3

NH4HCO

3 + NaCl

4 3NH Cl + NaHCO

67. Answer (4)

Hint: Dielectric constant of H2O and D

2O

respectively are 78.39 and 78.06 C2 N–1 m–2.

68. Answer (1)

Hint: Higher the proton affinity higher is the

basicity.

Sol.: NH3 is most basic.

69. Answer (2)

Hint: Br in BrF5 is sp3d 2 hybridised.

Br

F

F

F

F

F

70. Answer (3)

Hint: 1H1,

1D2,

1T3 same electronic configuration

but different atomic masses.

Sol.: BE of D2 > BE of H

2

71. Answer (3)

Hint: In carbon and oxygen d-orbitals are absent.

72. Answer (4)

Hint: 4 2 4 2

(gypsum) (Dead burnt plaster)

CaSO 2H O CaSO 2H O

73. Answer (2)

Hint: CaO is a basic oxide


8/15

74. Answer (1)


BeSO4 > MgSO

4 > BaSO

4

75. Answer (3)

Hint: Correct order of bond angle is

H2O > H

2S > H

2Te

76. Answer (4)

Hint: Ammoniated electron gives blue colouration.

77. Answer (3)

Hint: Higher the (charge/radius) value, higher is

hydration.

78. Answer (3)

Hint: CNG consist of lower alkane as major

constituent.

79. Answer (1)

Hint: H4P

2O

5 contains only two P – OH bond.

HO – P – O – P – OH

OII

IH

OII

IH

80. Answer (4)

Hint: NaH is ionic hydride.

Sol.: Li, Be and Mg form covalent hydrides.

81. Answer (4)

Hint: Beryllium shows covalent nature in its

compounds.

Sol.: Be shows diagonal relationship with Al.

82. Answer (1)

Hint: Phosphorus give disproportionation reaction

with Ca(OH)2.

Sol.:

2P4 + 3Ca(OH)

2 + 6H

2O

3Ca(H

2PO

2)2

+ 2PH3

83. Answer (3)

Hint: Higher the lattice energy, higher the thermal

stability.

Sol.: Order of thermal stability : LiF > NaF > KF.

84. Answer (3)

Hint: – O – O – linkage is peroxide linkage.

Sol.: BaO2 and Na

2O

2 are peroxides.

85. Answer (3)

Hint: Li is hardest alkali metal.

Sol. : Lithium being smallest in size shows

dominating covalent character.

86. Answer (4)


CaCO3 < NaHCO

3 < KHCO

3

87. Answer (3)

Hint:

Reducing :

atomic >

nascent > dihydrogen.

ability hydrogen hydrogen

88. Answer (2)

Hint: At low temperature para form dominant.

Sol.: At absolute zero only para form of hydrogen

exists.

89. Answer (3)

Hint:

Zn + 2NaOH Na2

ZnO2 + H

2

2Al + 2NaOH + 6H2O 2Na[Al(OH)

4] + 3H

2

90. Answer (2)

Hint: 32

%wt. 100 94.12%34

BIOLOGY

91. Answer (4)

Hint: Thigmonasty is a contact or touch stimulated

variation of movement in plants like Drosera.

Sol.: Phytochromes are chromoproteins or

photoreceptor pigments which are used to detect

light by plants.

92. Answer (3)

Hint: Cytokinins are modified purines.

Sol.: Apart from cytokinin, auxin, gibberellin and

abscisic acid are acidic in nature.

93. Answer (2)

Hint: These cells are present just next to the cells

of meristematic zone.

Sol.: Increased vacuolation is a feature of the cells

which are in phase of elongation.


9/15

94. Answer (3)

Hint: Phytohormone that counteracts apical

dominance is best suitable for tea plantation.

Sol.: Promotion of parthenocarpy – Auxin.

Antitranspiratory action – Abscisic acid

Root development in various cuttings – Auxin.

Cytokinin counteracts with auxin and help in making

bushes of tea plants.

95. Answer (4)

Hint: Apical dominance is a phenomenon in which

apical buds do not allow growth of lateral buds.

Sol.: Auxin promotes apical dominance while

cytokinin counteracts it.

96. Answer (2)

Sol.: Garner and Allard first reported the

photoperiodic response of flowering in tobacco.

97. Answer (2)

Hint: Abscisic acid is called stress hormone.

Sol.: Abscisic acid induces dormancy in buds and

seeds whereas gibberellin prevents it.

ABA acts as antagonist of gibberellins.

98. Answer (3)

Hint: Some synthetic auxins are used as

weedicides.

Sol.: 2, 4-D and 2, 4, 5-T are synthetic auxins

which act as weedicides.

99. Answer (3)

Hint: Initially zygotic division shows geometric growth

pattern.

Sol.: In geometric growth, every cell divides with all

the daughter cells growing and dividing again.

100. Answer (2)

Sol.: Induction of -amylase in barley endosperms

is a bioassay of gibberellins.

101. Answer (4)

Hint: Respiration mediated breakdown of different

substrates such as carbohydrate, fats and proteins

show interconnection as all produce energy through

Krebs cycle eventually.

Sol.: Acetyl CoA is a metabolite which is a common

product during respiratory breakdown of fats, proteins

and carbohydrates. It is also a raw material for

synthesis of carotenoids, terpenes and gibberellins.

102. Answer (1)

Hint: Respiratory quotient of any respiratory

substrate is the ratio of volume of CO2 released and

volume of O2 consumed during respiration.

Sol.: RQ of carbohydrate = 1

RQ of Proteins (Albumin) = 0.9

RQ of Fats (Palmitic acid) = 0.7

RQ of Organic acids >1 (for oxalic acid it is 4)

So descending order of RQ values for given

substrates is oxalic acid > glucose > albumin >

palmitic acid.

103. Answer (3)

Hint: In human kidney cells, NADH molecules

produced in glycolysis enter the mitochondria via

malate-aspartate shuttle.

Sol.: In aerobic respiration, ATP produced by

oxidative phosphorylation is coupled with ETS.

By ETS and oxidative phosphorylation

1 NADH produces 3 ATPs

and 1 FADH2 produces 2 ATPs

Total NADH molecules produced in aerobic

respiration = 10 (30 ATPs)

Total FADH2 molecules produced in aerobic

respiration = 2 (4 ATPs) so total ATP produced by

ETS and oxidative phosphorylation is 34.

104. Answer (2)

Hint: Succinyl CoA is an intermediate of Krebs

cycle, used in the biosynthesis of chlorophyll.

Sol.: Terminal electron acceptor of mitochondrial

ETS is O2.

Cofactor of pyruvic acid decarboxylase enzyme = Mg+2

Cofactor of lactic acid dehydrogenase enzyme = Zn+2.

105. Answer (1)

Hint: Fermentation is a kind of anaerobic respiration,

carried out primarily by fungi and anaerobic bacteria.

Sol.: In fermentation, pyruvic acid is partially oxidized

in the presence of external electron donor (reducing

agent), NADH.

In fermentation, no new ATP or NADH molecules are

produced apart from those produced during

glycolysis.


10/15

106. Answer (3)

Hint: The process through which electrons pass

from one carrier to another is called electron

transport system (ETS).

Sol.: In mitochondria, ETS is linked to ATP

synthesis (oxidative phosphorylation). Electron

carriers of ETS are present in inner mitochondrial

membrane.

107. Answer (3)

Hint: Wheat is a C3 plant while maize is a C

4 plant.

Sol.: For formation of 1 molecule of glucose



So for each glucose molecule, C3 plants require 12

ATPs less than C4 plants.

108. Answer (3)

Hint: First action spectrum of photosynthesis was

prepared by T.W. Engelmann.

Sol.: Engelmann used a green alga Cladophora to

describe action spectrum of photosynthesis.

109. Answer (2)

Hint: CO2 is the major limiting factor, influencing the

rate of photosynthesis.

Sol.: C3 plants show saturation at 450 ppm of CO

2,

while C4 plants show saturation at 360 ppm of CO

2

concentration at high light intensities. C3 plants

show CO2 fertilization effect as in the CO

2 enriched

atmosphere they show higher yield.

110. Answer (1)

Hint: Photorespiration is a process which involves

loss of fixed carbon as CO2 in plants.

Sol.: Carboxylation, reduction and regeneration are

three steps of Calvin cycle.

111. Answer (4)

Hint: Sorghum shows spatial difference in double

carboxylation.

Sol.: Sorghum is a C4 plant.

112. Answer (4)

Hint: C4 plants have Kranz anatomy.

Sol.: In C4 plant, first carboxylation occurs in

mesophyll cells by PEPcase enzyme which

eventually leads to malic acid formation. Malic acid

gets decarboxylated in bundle sheath cells where

second carboxylation takes place by RuBisCO

enzyme.

113. Answer (2)

Hint: Electron transport of non-cyclic

photophosphorylation starts with photosystem II.

Sol.: A = photosystem II

B = Photosystem I

C = Stroma

D = Lumen

114. Answer (3)

Hint: C4 plants lack the wasteful process

responsible for loss of fixed CO2.

Sol.: In C3 plants, when there is high concentration

of O2 in stroma, oxygenase activity of RuBisCO

leads to photorespiratory loss. On the other hand C4

plants have mechanism to increase the CO2

concentration in comparison to O2 concentration at

the enzyme site, because of which C4 plants do not

have photorespiration. Thus productivity of C4 plant is

better than C3 plants.

115. Answer (4)

Hint: Dictyosomes are unconnected units of Golgi

body found in plant cells.

Sol.: Golgi bodies are not associated with

photorespiration. Organelles involved in photo

respiration are chloroplast, peroxisome and

mitochondria.

116. Answer (2)

Hint: Non-cyclic photophosphorylation is associ-ated

with formation of assimilatory power and splitting of

water.

Sol.: Non-cyclic phosphorylation occurs only in

grana thylakoids. In this process both photosystems

work in a series, where PS II operates first, followed

by PS I. Products of non-cyclic photophosphorylation

are ATP, NADPH and O2.

117. Answer (4)

Hint: In C3 plants, CO

2 acceptor is a 5-carbon

containing molecule.

Sol.: In C3 plants, RuBP is the primary CO

2 acceptor

molecule.

118. Answer (3)

Hint: During nitrogen fixation atmospheric nitrogen is

converted into ammonia.

Sol.: Nitrogenase enzyme is a Mo – Fe protein. It

requires ATPs, strong reducing agents like FADH2,

NADPH and anaerobic condition for fixation of

nitrogen into ammonia.


11/15

119. Answer (4)

Hint: Nitrifying bacteria have chemoautotrophic mode

of nutrition.

Sol.: Nitrosomonas is a nitrifying bacterium which

converts NH3 into NO

2–.

120. Answer (3)

Hint: Leghaemoglobin is an O2 scavenger pigment

present in the cytosol of root nodule cells.

Sol.: After the formation of root nodule, bacteroids

present in root nodule cells produce a reddish-pink

pigment called leghaemoglobin which protects

nitrogenase from oxygen. Nodule formation is

induced by bacterial and plant signals.

121. Answer (2)

Hint: In green plants, Mn and Cl are involved in

splitting of water to release oxygen during

photosynthesis.

Sol.: Due to Mn toxicity translocation of Ca to shoot

apex is inhibited. Mn deficiency causes grey spots

in oats. Little leaf symptom is caused by the

deficiency of Zn. Transport of carbohydrate in phloem

is facilitated by boron.

122. Answer (1)

Hint: Ammonification involves decomposition of

organic nitrogen into ammonia.

Sol.: Ammonifying bacteria convert organic nitrogen

of dead plants and animals into ammonia.

123. Answer (4)

Sol.: Na is a beneficial element.

124. Answer (2)

Hint: For immobile minerals, deficiency symptoms

tend to appear first in younger tissues.

Sol.: Among the given minerals, calcium is an

immobile element.

125. Answer (2)

Hint: Mg is the co-factor of RuBisCO enzyme.

Sol.: Zn – Synthesis of auxin

K – Maintains turgidity of cells

Ni – Component of urease

126. Answer (3)

Hint: Water potential (w) is free energy of water

molecules which decreases by adding solutes.

Sol.: Water potential of pure water is zero at

atmospheric pressure. By increasing external

pressure on pure water, its water potential increases.

127. Answer (1)

Hint: Intercellular movement of water occurs through

cytoplasmic connections between neighbouring cells.

Sol.: These structures are called plasmodesmata.

128. Answer (4)

Hint: Facilitated diffusion is a passive transport.

Sol.: In facilitated diffusion, transport of ions/

molecules occurs with the help of specific membrane

proteins called transporters. This process is highly

specific because transporter proteins are highly

selective. The rate of facilitated transport may

saturate.

129. Answer (4)

Hint: Movement of sucrose from mesophyll cells to

sieve tube elements via companion cells, requires

energy in the form of ATP.

Sol.: Loading of sucrose in sieve tube cells via

companion cells is an active process.

130. Answer (2)

Hint: In this process, loss of water in the form of

liquid droplets occurs through special openings

called hydathodes.

Sol.: Herbaceous plants lose water in the form of

droplets due to high root pressure. This process is

called guttation.

131. Answer (1)

Hint: Water molecules move from an area of high

water potential (w) to an area of low water potential

(w).

Sol.: w of cell A =

s +

p = – 2 atm

w of cell B = – DPD = – 10 atm

w of cell C = OP – TP = – 3 atm

w of cell D = – 6 atm

So the correct direction of movement of water is

A B

C D

↗

132. Answer (2)

Sol.: In guard cells, cellulose microfibrils are radially

oriented.


12/15

133. Answer (4)

Hint: In girdling experiment, a ring of bark upto the

depth of the phloem layer is carefully removed.

Sol.: In a girdled plant, root cells die first than the

shoot cells, due to stoppage of translocation of

sugars and other materials to the roots.

134. Answer (3)

Hint: Transpiration pull and root pressure causes

upliftment of water by pulling and pushing

respectively.

Sol.: Transpiration develops a negative water

potential in the xylem which creates a ‘pull’ for

translocation of water while root pressure is a

positive hydrostatic pressure responsible for pushing

of water.

135. Answer (1)

Hint: More the H+ ions, more acidic the condition is

and less will be the pH.

Sol.: During stomatal opening, due to the activity of

hydrogen-potassium ion-exchange pump, H+ from

guard cells are transported to neighbouring

subsidiary cells while K+ ions are transported into

the guard cells. This increases the pH of guard cells.

136. Answer (3)

Hint: Appositional image.

Sol.: Vision in cockroach has low resolution but high

sensitivity.

137. Answer (1)

Hint: These paired structures help in respiration in

cockroach.

Sol.: 10 pairs of spiracles are present in body of

Periplaneta.

138. Answer (4)

Hint: Part of foregut lined by cuticle is not secretory

in nature.

Sol.: Crop part of foregut in cockroach does not

secrete enzymes. Haemoglobin is absent in blood of

cockroach. Hepatic caecae are present at the

junction of foregut and midgut.

139. Answer (4)

Hint: External segmentation is found in arthropods.

Sol.: Schizocoelom and presence of external

metamerism/segmentation is a feature of Periplaneta.

140. Answer (3)

Hint: Sexual dimorphism in cockroach is based on

presence of unjointed filamentous structures attached

to 9th sternum.

Sol.: Anal/caudal styles are unjointed structures

attached to 9th sternum in male cockroach. Anal

cerci are jointed structures present in both male and

female cockroach.

141. Answer (1)

Hint: Glands whose ducts open into brood or genital

pouch.

Sol.: Seminal vesicles nourish the sperms. Phallic/

conglobate secrete substances that help to form a

protective cover of spermatophore.

142. Answer (3)

Hint: Identify mouth parts that are paired in

cockroach.

Sol.: Hypopharynx acts as tongue. Maxillae are

paired structures that help bring food to mandibles.

143. Answer (2)

Hint: First pair of wings is mesothoracic.

Sol.: Tegmina/Elytra or wing cover is the alternative

name for first pair of wings in cockroach.

144. Answer (4)

Hint: It is another term for arthrodial membrane.

Sol.: Arthrodial membrane is flexible membrane that

connects tergite with pleurites and sternites.

145. Answer (2)

Hint: Gill cover is absent in cartilaginous fish.

Sol.: Chondrichthyes are cartilaginous fish which

lack operculum generally; their males have pelvic

claspers and show internal fertilisation. Bony fishes

possess operculum, have cycloid/ctenoid scales and

show external fertilisation.

146. Answer (4)

Hint: Identify an animal where development of

embryo can occur on land.

Sol.: Pisces and amphibians are anamniotes. Their

development is linked to water directly. Reptiles

evolved prior to birds. Mammals, birds and reptiles

are all amniotes.

147. Answer (3)

Hint: Dry, aglandular skin is a feature of reptiles.

Sol.: Ichthyophis is a limbless amphibian.

Bangarus is a limbless reptile

Balaenoptera is a limbless mammal


13/15

They all have bony vertebral column. Skin is glandular

in mammals.

148. Answer (1)

Hint: Select a cartilaginous fish.

Sol.: Air/swim bladder helps to maintain buoyancy in

bony fish. Dog fish has to swim continuously to

avoid sinking.

149. Answer (2)

Hint: Identify a circular mouth scaleless fish like

organism.

Sol.: Myxine is hagfish, a cyclostome. Torpedo,

Pristis and Trygon belong to super class Pisces.

150. Answer (2)

Hint: Heart and blood vessels are present in open

and closed circulatory system.

Sol.: Heart is dorsal in position in non-chordates

usually while it is ventral in chordates. Nerve cord is

ventral in non-chordates but dorsal in position in

chordates. Post anal tail is a feature of chordates.

Gill slits are lateral in position in chordates.

151. Answer (1)

Hint: Such an organism produces internal body heat

through metabolic activity to maintain body

temperature.

Sol.: Mammals are homeotherms and endotherms.

They exhibit internal fertilisation (Felis) as do birds.

Chelone inhabits water. Replites show shedding of

skin (ecdysis).

152. Answer (4)

Hint: Snakes detect vibrations through jaw bones.

Sol.: Vipera has epidermal scales; 3 chambered

heart and is limbless in adult form. It has dry

cornified aglandular skin.

153. Answer (2)

Hint: Pneumatic bones have air cavities to reduce

weight of flying birds.

Sol.: Neophron i.e. vulture is a flying bird that has

both pneumatic bones and preen/oil gland. Air sacs

in birds are avascular and meant for storage of air

but not exchange of gases.

154. Answer (2)

Hint: Select an avian adaptation.

Sol.: Feathers reduce weight and are found in birds.

155. Answer (1)

Hint: This structure divides body cavity into thoracic

and abdominal cavity.

Sol.: Mammals show direct development. Heterodont

dentition is absent in aquatic mammals such as

whale, sea cow, etc. Sloth and sea cow have 9 and

6 cervical vertebrae respectively.

156. Answer (3)

Hint: Another name for this organism is Euspongia.

Sol.: Antedon and Sea urchin exhibit radial

symmetry in adult form. Salpa shows bilateral

symmetry.

157. Answer (1)

Hint: Select an arthropod.

Sol.: Sea hare is Aplysia with unsegmented body.

Sea cucumber is dioecious. Tapeworm shows

internal fertilisation.

158. Answer (4)

Hint: Triploblastic organisms show extracellular

digestion.

Sol.: Adult echinoderms are radially symmetrical and

most poriferans are asymmetrical. Notochord is

absent in both Echinoderms and Poriferans. True

coelom is a feature of Echinoderms.

159. Answer (2)

Hint: This organism is commonly called lac insect.

Sol.: Laccifer is a useful insect. Its secretion called

lac and shellac are used in bangle industry.

Anopheles, Culex and Aedes act as vectors to

spread diseases such as malaria, filariasis and

dengue respectively.

160. Answer (3)

Hint: Arthropods have open circulatory system.

Sol.: Pila is a mollusc with open circulatory system

and lacks segmentation. Anopheles has malpighian

tubules for excretion. Nereis is segmented and has

nephridia.

Presence of solid ventral nerve cord is a character of

non chordates.

161. Answer (2)

Hint: Organisms that show bioluminescence.

Sol.: Flatworms like Schistosoma are dioecious but

digestion is only extracellular. In triploblastic animals

such as aschelminthes, extracellular digestion is

observed.


14/15

162. Answer (3)

Hint: This animal lives in loosely organised

community.

Sol.: Locusta is an economically harmful insect

(arthropod).

163. Answer (1)

Hint: Metamerism refers to presence of segments

and probable repeat of organs.

Sol.: Wuchereria is a filarial worm, where excretory

pore eliminates nitrogenous waste. Wastes present

in alimentary canal are eliminated through anus.

Pseudocoelom, absence of segmentation and

presence of bilateral symmetry are features of

Aschelminthes.

164. Answer (2)

Hint: Identify a mollusc with scientific name

Dentalium.

Sol.: Tusk shell has a calcareous shell

(exoskeleton) while chitinous exoskeleton is a feature

of arthropods.

165. Answer (3)

Hint: Identify a true fish.

Sol.: Flying fish is a bony fish. Hagfish is a

cyclostome. Cuttlefish and devil fish belong to

phylum Mollusca.

166. Answer (4)

Hint: Select the organism that occurs exclusively in

marine water.

Sol.: Asterias has water vascular system and show

external fertilisation. Poriferans have water canal

system.

167. Answer (3)

Hint: Identify first triploblastic animals.

Sol.: Regeneration is a prominent feature of Porifera

Coelenterata, Platyhelminthes and Echinoderms.

168. Answer (2)

Hint: This structure is composed of cellulose in

plants.

Sol.: Animal cells lack a cell wall. All animals are

heterotrophs and show division of labour. Poriferans

lack nervous system.

169. Answer (2)

Hint: Comb plates are locomotory structures in

members of this phylum.

Sol.: Annelids, arthropods and molluscs occupy

either aquatic or terrestrial habitats.

170. Answer (3)

Hint: Identify worms that are pseudocoelomate.

Sol.: Ringworm is a fungus; earthworm is an

annelid; bookworm is an arthropod; Liver fluke &

tapeworm are flatworms. Pinworm, filarial worm,

hookworm and roundworm belong to aschelminthes.

171. Answer (2)

Hint: Identify a roundworm which gives birth to young

worms.

Sol.: Ancylostoma and Ascaris are oviparous round

worms. Fasciola is a flatworm.

172. Answer (3)

Hint: Parapodia are lateral appendages that help in

swimming and respiration.

Sol.: Malpighian tubules are found in insects

(arthropods) while parapodia are found in annelids

such as Nereis.

173. Answer (2)

Hint: Identify organisms with cnidocytes.

Sol.: Adamsia is sea anemone i.e., a cnidarian

whose tentacles are rich in stinging cells.

Euspongia has flagellated choanocytes.

Pleurobrachia does not show metagenesis and

Planaria is a free living worm, hence it lacks hooks.

174. Answer (1)

Hint: Metagenesis.

Sol.: Physalia exists in both polyp and medusa

form. Hydra, Meandrina and Adamsia exist as

polyps.

175. Answer (1)

Hint: This organism forms gemmules to reproduce.

Sol.: Sponges lack distinct germ layers, has ostia

for entry of water and has collar cells in body.

176. Answer (3)

Hint: These organisms occur in exclusively marine

conditions.

Sol.: Saccoglossus a hemichordate, has proboscis

gland as its excretory organ. Gills are meant for

respiration in hemichordates. In molluscs, gills serve

both the function of respiration and excretion.


15/15

177. Answer (3)

Hint: This parameter is shared by larval

echinoderms with platyhelminthes.

Sol.: Adult echinoderms are radially symmetrical

while their larvae are bilaterally symmetrical. Anus

develops from blastopore in deuterostome body plan.

178. Answer (4)

Hint: Metamorphosis involves conversion of larva to

adult form.

Sol.: Direct development occurs in Hirudinaria.

Butterfly, Ascidia and Culex show metamorphosis

and their larval forms are caterpillar, tadpole &

wriggler respectively.

��

179. Answer (1)

Hint: Identify an arthropod.

Sol.: Pheretima (an annelid), Pavo (a bird) and

Petromyzon (cyclostome) have closed circulatory

system. Capillaries are a feature of closed circulatory

system.

180. Answer (1)

Hint: This animal is also called bookworm.

Sol.: Loligo and Laccifer are economically beneficial

mollusc and insect respectively. Limulus is a living

fossil.

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