Test - 3 (Code C) (Answers & Hints) All India Aakash Test ... · All India Aakash Test Series for...

Test - 3 (Code C) (Answers & Hints) All India Aakash Test Series for Medical-2019

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1. (3)

2. (3)

3. (2)

4. (2)

5. (4)

6. (2)

7. (2)

8. (4)

9. (1)

10. (2)

11. (2)

12. (3)

13. (2)

14. (2)

15. (4)

16. (1)

17. (4)

18. (3)

19. (2)

20. (1)

21. (4)

22. (3)

23. (1)

24. (1)

25. (3)

26. (1)

27. (4)

28. (3)

29. (1)

30. (4)

31. (1)

32. (1)

33. (3)

34. (3)

35. (2)

36. (2)

Test Date : 03/12/2017

ANSWERS

TEST - 3 (Code C)

All India Aakash Test Series for Medical - 2019

37. (2)

38. (2)

39. (2)

40. (3)

41. (1)

42. (1)

43. (2)

44. (2)

45. (4)

46. (3)

47. (1)

48. (4)

49. (1)

50. (4)

51. (1)

52. (1)

53. (1)

54. (3)

55. (1)

56. (4)

57. (4)

58. (2)

59. (4)

60. (1)

61. (1)

62. (3)

63. (Deleted)

64. (3)

65. (4)

66. (3)

67. (4)

68. (1)

69. (2)

70. (3)

71. (3)

72. (4)

73. (4)

74. (1)

75. (1)

76. (2)

77. (4)

78. (4)

79. (2)

80. (2)

81. (3)

82. (3)

83. (4)

84. (2)

85. (3)

86. (1)

87. (2)

88. (2)

89. (3)

90. (2)

91. (3)

92. (4)

93. (4)

94. (2)

95. (2)

96. (3)

97. (4)

98. (4)

99. (1)

100. (2)

101. (4)

102. (1)

103. (4)

104. (4)

105. (4)

106. (4)

107. (1)

108. (3)

109. (3)

110. (2)

111. (3)

112. (1)

113. (4)

114. (4)

115. (3)

116. (4)

117. (1)

118. (4)

119. (2)

120. (2)

121. (2)

122. (2)

123. (4)

124. (1)

125. (1)

126. (3)

127. (3)

128. (4)

129. (2)

130. (4)

131. (3)

132. (1)

133. (4)

134. (3)

135. (3)

136. (2)

137. (1)

138. (3)

139. (3)

140. (3)

141. (4)

142. (2)

143. (4)

144. (4)

145. (3)

146. (3)

147. (2)

148. (1)

149. (2)

150. (3)

151. (2)

152. (2)

153. (3)

154. (2)

155. (3)

156. (3)

157. (3)

158. (2)

159. (3)

160. (2)

161. (4)

162. (3)

163. (1)

164. (4)

165. (2)

166. (3)

167. (3)

168. (4)

169. (2)

170. (4)

171. (2)

172. (4)

173. (4)

174. (2)

175. (1)

176. (4)

177. (3)

178. (3)

179. (3)

180. (4)

All India Aakash Test Series for Medical-2019 Test - 3 (Code C) (Answers & Hints)

2/12

ANSWERS & HINTS

1. Answer (3)

As downward acceleration of elevator a > g, block

loses the contact with the elevator and its

acceleration is g downward. So speed of block at

t = 1 s,

v = u + gt

= 10 × 1 = 10 m/s

2. Answer (3)

When trolley is sliding with acceleration gsin, string

becomes perpendicular to the ceiling.

mgsin

mgcosmg

T

T = mgcos

3. Answer (2)

In case (I), acceleration of block in horizontal

direction

aI = (gsin30°) cos30° ...(1)

In case (II), acceleration of block in horizontal

direction is acceleration of wedge.

aII = gtan30° ...(2)

Ratio 2I

II

3cos 30

4

⎛ ⎞ ⎜ ⎟⎝ ⎠

a

a

4. Answer (2)

Acceleration 21 2

305 m/s

6

F F

aM

Mass of 10 cm part of uniform rod.

610 1.2 kg

50 m

Now,

10 cm

TF2

a

[ PHYSICS]

T – F2 = ma

T = F2 + ma

= 10 + 1.2 × 5 = 16 N

5. Answer (4)

Just before t = 1 s

Velocity v1 = +2 m/s

Just after t = 1 s

Velocity, v2 = –4 m/s

Change in velocity, v = v2 – v

1 = –6 m/s

Impulse = change in linear momentum.

= 2 × (–6) = –12 Ns

6. Answer (2)

Maximum common acceleration of two blocks, when

force is applied on the upper block.

21

max

2

0.5 20 1010 m/s

10

m g

a

m

Fmax

= (m1 + m

2) a

max = 30 × 10 = 300 N

As F < Fmax

, both blocks move together with

common acceleration.

21505 m/s

30 a

Fs = m

2a = 10 × 5 = 50 N

7. Answer (2)

Limiting force of friction.

For m1 ; (f

1)max

= 1m1g = 25 N

m2 ; (f

2)max

= 2m2g = 40 N

Maximum frictional force fmax

= 65 N

60 N

m1

T

( ) = 25 Nf1 max

As F = 60 N, both blocks will be at rest. Frictional

force on block of mass m1 be (f

1)max

. Tension in the

string.

T = 60 – 25 = 35 N


3/12

8. Answer (4)

2

2 2( )2

x

mT L x

L

T(x = 0) 21

2 m L

2

2 2

( 0) 4

( ) 1( )

T x L

T x x L x

4L2 – 4x

2 = L2 4x2 = 3L

2

3

2 L

x

9. Answer (1)

mgcosmg

FcosN

F

Fsin

mgsin

N

Block begins to slide when

Fcos = mgsin + N

Fcos = mgsin + (mgcos + Fsin)

F(cos – sin) = mg(sin + cos)

(sin cos )

(cos sin )

mgF

10. Answer (2)

2

12 sin 2 sin45 v g S g S ...(1)

2

22 (sin45 cos45 )

kv g S ...(2)

2

1

2

2

1 116

1511

16

k

v

v

1

2

4 :1v

v

11. Answer (2)

As 1 >

2, both blocks will start sliding together,

when

(m1 + m

2)gsin =

1m1gcos +

2m2gcos

Or (m1 + m

2)gsin = gcos(

1m1 +

2m2)

1 1 2 2

1 2

tan

m m

m m

12. Answer (3)

60°

u

v

From conservation of linear momentum

m(v cos60° + u) + Mu = 0

13. Answer (2)

mgcosmg

N

v

As block is at rest w.r.t. lift

N = mgcos ...(1)

Displacement of block w.r.t. ground

S = vt vertically upward ...(2)

Work done by normal contact force

W = N × S × cos

= (mgcos)(vt)cos

= mgvt cos2

14. Answer (2)

10 kg

kxm

100 N

N

Normal contact force is minimum when spring force

is maximum.

4 × 10 × xmax

= 21

2m

kx [K = 0]

kxm

= 80 N

Nmin

= 100 – 80 = 20 N


4/12

15. Answer (4)

m1

m2

a1 a

2

2T

T T2T

(2T + T) a1 = Ta

2

16. Answer (1)

17. Answer (4)

e p2

ep

ep

p

+ve

Momentum of ball after first collision = +ep

Change in momentum of ball after first collision.

(p)1 = ep – (–p) = p(1 + e)

Similarly after second collision

(p)2 = e2p – (–ep) = ep(1 + e)

So total momentum imparted

1 2( ) ( )p p p

2(1 ) (1 ) (1 ) terms p e ep e e p e

(1 )

(1 )

p e

e

18. Answer (3)

v

90 -

u

Line of impact

ucos

Let particle strikes with speed u and moves

horizontally with speed v after the collision.

Momentum is conserved along the plane.

So vcos = usin ...(1)

and vcos(90 – ) = eucos

or, vsin = eucos ...(2)

Dividing equation (2) by equation (1)

2 2 1tan cot tan tan 30

3e e ⇒

19. Answer (2)

After collision velocity of heavy block remains the

same.

8 m/sv1

Now,

Relative velocity of separation = Relative velocity of

approach

8 – v1 = (10 – 8)

v1 = +6 m/s (in positive x direction)

20. Answer (1)

At the highest point, speed

v = 100 cos60° = 50 m/s horizontally

Conservation of linear momentum.

2ˆ ˆ2 50 1 100 1i j v

�

2

ˆ ˆ100 100v i j �

2100 2 m/sv

�

So kinetic energy of this part.

2 411 100 2 10 J

2

21. Answer (4)

A

ˆui

B2

4u g i ��

Velocity at point A.

ˆui ...(1)

Speed at the highest point.

2 24 �

Bv u g

24 �

Bv u g

Velocity 2 ˆ4 �

�B

v u g i ...(2)

Change in velocity

2 ˆ4 4 � � �

�B A

v v v u g i i

24

��v u u g


5/12

22. Answer (3)

F kxs =

mg

From work-energy theorem.

21

2mgx kx

2mgx

k ...(1)

Spring force at this instant

Fs = kx = 2mg

s2

upward

F mg mg mga g

m m

23. Answer (1)

Instantaneous velocity.

23 8 dx

v t tdt

...(1)

At t = 0, v1 = 3 m/s

t = 9 s, v2 = 12 m/s

W = K

2 2

2 1

1

2 m v v

2 21 1[(12) (3) ]

2 2

135J

4

24. Answer (1)

s

dvm mg

dt

25. Answer (3)

mg

hO

v

4u g �

Let string becomes slack when it makes angle

with upward vertical. Then

2

cosmv

mg �

2cosv g � ...(1)

Also,

2 2 22 2 (1 cos )v u gh u g �

or, cos 4 2 2 cosg g g g � � � �

or,2

3 cos 2 cos3

g g ⇒ � �

Tangential acceleration

5sin

3g g

26. Answer (1)

Fv at t

m

⎛ ⎞ ⎜ ⎟⎝ ⎠ ...(1)

2F

P Fv tm

P t

27. Answer (4)

m

m v/2

v

Conservation of mechanical energy

2

21 1

2 2 2

vmv m

⎛ ⎞ ⎜ ⎟⎝ ⎠

2mg mg �

�

2

5 3 12

8 2 5

mv mg gv ⇒ � �

28. Answer (3)

vx = u

x = 3 m/s

vy = eu

y =

34 3 m/s

4

Speed 2 2

3 2 m/sx y

v v v


6/12

29. Answer (1)

3x

uF

x

23 m/s

x

x

Fa

m

4y

uF

y

ay = –4 m/s2

vx = a

xt = –3t; v

y = a

yt = –4t

2 25

x yv v v t

At t = 2 s, v = 5 × 2 = 10 m/s

30. Answer (4)

B

A

ar

vB

g

O

vA

2

, B

r t

va a g

R

2 2( ) ( ) r t

a a a

2 22

B Av v gR

31. Answer (1)

2

cos60 mvN mg

R

2 2(4)cos60 1 5

2

vN m g

R

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= 13 newton

fk =

kN = 6.5 N

N

mgmgsin60°

60°

fk

mgcos60°

Tangential acceleration sin60

kmg f

m

25 3 6.5 2.1 m/s

32. Answer (1)

O

Y

X

50,

3

⎛ ⎞⎜ ⎟⎝ ⎠

5, 0

4

⎛ ⎞⎜ ⎟⎝ ⎠

4 5

3 3Y X

5 5ˆ ˆ0 0

4 3r i j

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

�

= 5 5ˆ ˆ

4 3i j

0r F �

�

33. Answer (3)

Area under F - t graph

= change in linear momentum

1(2 3) 10

2

mv – mu = 25

v = 25 m/s

Kinetic energy 21 625

J2 2

k mv

34. Answer (3)

Speed of ball just before the collision.

2(10) 2 10 2.2

= 12 m/s

Speed of ball just after the collision.

2 10 5 10 m/s

Coefficient of restitution

10 5

12 6e

35. Answer (2)

For equilibrium

F = 0

2x2 – x = 0 2(2x – 1) = 0


7/12

Either x = 0 or1

2x

4 1dF

xdx

At x = 0, 1 0dF

dx

So equilibrium is stable at x = 0.

36. Answer (2)

Kinetic energy is maximum when potential energy is

minimum.

U = 2x2 – 2 ...(1)

4dU

xdx

...(2)

For minimum, , 0

dUU

dx

x = 0

2

24

d U

dx U is minimum at x = 0

Umin

= –2 J

Now,

Kmax

+ Umin

= Emech

Kmax

– 2 = 34 2

max

136

2mv

vmax

= 6 m/s

37. Answer (2)

20 W �

�

P F v

38. Answer (2)

W = Pt = (m)gh

8000 60( ) 3200 kg

10 15m

39. Answer (2)

(REF LEVEL)

L/10

GL/5

Mass of hanging part = 5

M

5 10

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠i

M LU g

50

MgL

Uf = 0

50 MgL

W U

40. Answer (3)

10dp

F tdt

At t = 2 s F = 20 N

41. Answer (1)

( 0)

(1)

p n muF mnu

t

⎛ ⎞ ⎜ ⎟⎝ ⎠

42. Answer (1)

fs

Maximum acceleration upto which block does not

slide on the floor

amax

= g = 0.5 × 10 = 5 m/s2

Given, a = 4 m/s2

So, fs = 2 × 4 = 8 N

Displacement of block in t = 2 s

214 (2) 8 m

2 S

wf = 8 × 8 = + 64 J

43. Answer (2)

Angle of repose R = 45°

So block does not slide

fs = mgsin37° =

310 6 N

5

44. Answer (2)

2 2

mg gmg ma a ⇒

45. Answer (4)

At the highest point �

�

F v

So 0 �

�

P F v


8/12

46. Answer (3)

2 2 2 2SO Cl SO Cl

Hence this is not applicable to Dalton's law of partial

pressure.

47. Answer (1)

Charge on ionLattice energy

Size of ion

48. Answer (4)

Due to resonance in 2 2

CH CH CH , all C atoms

undergo sp2 hybridization.

49. Answer (1)

dRT dRTP M

M P ⇒

3d 2.64 gm / dm

3 1 1 3R 0.0821dm atm mole k ( 1 dm 1 litre)

∵

775P atm

760

�T 310 273 583 K

2.64 0.0821 583M 124

775 / 760

No. of P atoms in a molecule =124 124

At. Wt of P 31

= 4

50. Answer (4)

In 3 3

N(SiH )�� , N-atom donate its lone pair to vacant

d-orbital of Si to form back bonding.

51. Answer- (1)

52. Answer (1)

Due to presence of lone pair on Br in BrF5, bond

angle is not equal to 90°.

53. Answer (1)

%ionic character = 2

A B A B16(X X ) 3.5(X X )

(Hannay – Smith Equation)

=2

16(2.5 2.1) 3.5(2.5 2.1)

= 6.96%.

[ CHEMISTRY]

54. Answer (3)

Fact

55. Answer (1)

O

O O

O

H

HCH3

CH3( 0) ( 0)

56. Answer (4)

Fact

57. Answer (4)

58. Answer (2)

59. Answer (4)

PV PV 1 1Z n 0.0247

nRT ZRT 1.8 0.0821 273

⇒

No. of H2 molecules

23 220.0247 6.022 10 1.487 10

60. Answer (1)

Fact.

61. Answer (1)

2 4 2N O 2NO��⇀

↽��

No. of moles initially 1 0

At equilibrium 1 – 0.4 0.8

0.6 92 0.8 46Molar mass ofmixture

1.4

∵

= 65.71

W PV RT (760 torr 1 atm)

M ∵

W PM 65.71 1d 2.67g / litre

V RT 0.0821 300

62. Answer (3)

Due to presence of 3 lone pairs on Xe in XeF2,

shape is linear.

63. Deleted


9/12

64. Answer (3)

Lewis dot structure of 4

NH

is H N H

H+

H

65. Answer (4)

In BF3, B is sp

2 hybridized

In PCl5, P is sp

3d hybridized

In 4

BF , B

is sp3 hybridized

In 6

PCl , P

is sp3d2 hybridized

66. Answer (3)

Cl Cl

Cl Cl

( = 0)

67. Answer (4)

68. Answer (1)

CH3

CN

has high dipole moment than

CH3

C N due to

presence of Electron with drawing (CN–) at para

position.

69. Answer (2)

Greater the value of 'a' i.e greater the intermolecular

forces of attraction for a gas, then the gas more

easily will be liquified.

70. Answer (3)

AlCl3 has more covalent character but more soluble

in H2O. It is due to high hydration enthalpy.

71. Answer (3)

O N O O N O

Bond order = 1 2

1.52

72. Answer (4)

3 3 3 3 3

* * 2 1

2 21 1 2 2 2 x y

z

p ps s s s p

73. Answer (4)

Fact

74. Answer (1)

Fact

75. Answer (1)

Due to formation of intramolecular H-bonding

76. Answer (2)

Fact.

77. Answer (4)

Fact.

78. Answer (4)

a s

Tv

M

4

2

CH

SO

v300 64 3

v 16 400 1

79. Answer (2)

Xe can form compounds with most electronegative

elements XeF2, XeF

4, XeF

6

80. Answer (2)

Fact

81. Answer (3)

Bond order Bond strength

82. Answer (3)

PVZ 1; 1

nRT ,

m V 22.4 L.∵

83. Answer (4)

84. Answer (2)

Density of a gas = PM

RT; density P; density

1

T

85. Answer (3)

In cyclooctatetraene all are p – p bonds.

86. Answer (1)

4CH

T

P

P =

4CH

1

116X

1 1 9

2 16


10/12

[ BIOLOGY]

91. Answer (3)

92. Answer (4)

Natural system of classification is based on

phytochemistry also.

93. Answer (4)

Iodine is obtained from brown algae (Phaeophyceae)

94. Answer (2)

Non-flagellated male gametes are found in Spirogyra

and flagellated male gametes are found in Fucus.

95. Answer (2)

Sex organs are non-Jacketed

96. Answer (3)

Generally in algae haplontic life cycle is found but

Fucus shows diplontic life cycle.

97. Answer (4)

Both Ectocarpus and Gelidium have chl-a and

cellulosic cell wall

98. Answer (4)

Majority of the red algae are marine and generally

found in warmer areas.

99. Answer (1)

Kelps are large sized brown algae

100. Answer (2)

101. Answer (4)

102. Answer (1)

Gemmae, haploid asexual bud which helps in

asexual reproduction of Marchantia.

103. Answer (4)

Archegonia are found in Cycas

Female cone is not found in Cycas

104. Answer (4)

Gametophytes in gymnosperms are dependent on

sporophytes.

105. Answer (4)

106. Answer (4)

In gymnosperm, veins and veinlets are absent in

leaves.

107. Answer (1)

108. Answer (3)

Pinus and Polytrichum both reproduce by oogamous

type sexual reproduction.

109. Answer (3)

All gymnosperms are heterosporous and produce

haploid microspores and megaspores.

110. Answer (2)

In bryophytes, gametophyte is dominant.

111. Answer (3)

Rhizoids of mosses are multicellular and branched

Prothallus is free-living inconspicuous and

photosynthetic gametophyte

112. Answer (1)

113. Answer (4)

In mosses, spores are formed after meiosis

Synergids and antipodals degenerate after

fertilisation in most of the flowering plant

Gemmae are asexual buds

114. Answer (4)

Zygote does not undergo reduction division

immediately

Sporophyte is dependent on gametophyte

87. Answer (2)

Polarising power of cation charge on cation

88. Answer (2)

dRTP

M

d M

P RT

d 1

P T ⇒

1 2

1 2

d P

P d

1 373

273 1

2 1

2 1

d d

P P

273

373

2

2

d

P

273 273x

373 373x

89. Answer (3)

In any resonance structure atom should obey octet

rule and should have symmetrical charge separation.

90. Answer (2)

In SF4, 'S' atom undergoes sp

3d hybridisation.


11/12

115. Answer (3)

In bryophyta zygote does not undergo reduction

division immediately

eg. Polytrichum and Sphagnum

116. Answer (4)

Double fertilisation is found in only angiosperms only

but Cedrus, Pinus are Gymnosperm

117. Answer (1)

'a' and 'c' features are correct.

118. Answer (4)

Only 'b' statement is correct.

'a', 'c' and 'd' statements are incorrect.

Gymnospermic root are generally tap root.

Egg apparatus is group of three cell (two synergid

and one female gamete)

Generally algae show haplontic life cycle

119. Answer (2)

Ploidy of PEN is triploid in nature (3n), e.g. maize.

120. Answer (2)

121. Answer (2)

Synergids(n), Antipodal(n), Pollen grain(n)

Embryo (2n) Integument (2n)

122. Answer (2)

123. Answer (4)

124. Answer (1)

Club moss - Lycopodium (pteridophytes)

Cord moss - Funaria (Bryophytes)

Hair cap moss - Polytrichum (Bryophytes)

Peat moss - Sphagnum (Bryophytes)

125. Answer (1)

126. Answer (3)

In numerical taxonomy, all characters are given equal

importance

127. Answer (3)

128. Answer (4)

The sporophyte in mosses is more elaborate than

that in liverwort.

Spores are formed after meiosis.

129. Answer (2)

Vascular tissue is present in vascular cryptogams

and some are heterosporous.

130. Answer (4)

Naked ovule is the important feature of gymnosperm

131. Answer (3)

‘‘Meiosis in the zygote results in the formation of

haploid spores in algae’’

132. Answer (1)

Diplontic life cycle found in all flowering and

gymnosperm and these both are seed bearing

plants.

133. Answer (4)

During double fertilisation one male gamete fuses

with diploid secondary nucleus.


fertilisation.

134. Answer (3)

Thalloid body organisation is found in Marchantia and

algae.

135. Answer (3)

Brown algae possess the photosynthetic pigments

chl-a and chl-c and fucoxanthin.

136. Answer (2)

Dorsal blood vessel is both collecting & distributing

blood vessel with valves.

137. Answer (1)

Lumbricus has a closed circulatory system and

nucleated leucocytes.

138. Answer (3)

Respiratory exchange occurs through moist body

surface.

139. Answer (3)

Blood glands are present in 4th, 5th & 6th segment.

Typhlosole starts from 27th segment of earthworm.

140. Answer (3)

Lateral hearts are present in segment 7 & 9.

141. Answer (4)

142. Answer (2)

143. Answer (4)

144. Answer (4)

Setae & dorsal pores are absent in last segment of

earthworm while integumentary nephridiopores are

present.

145. Answer (3)

Spermathecae are present in segment 6, 7, 8 & 9.


12/12

146. Answer (3)

Fertilization occurs in cocoon of earthworm,

therefore, it is external. Calciferous glands neutralize

humic acid. Chloragogen cells are analogous to

human liver.

147. Answer (2)

148. Answer (1)

Integumentary nephridia are exonephric and play no

role in osmoregulation.

149. Answer (2)

150. Answer (3)

Blood glands are present in segment 4th, 5th & 6th.

151. Answer (2)

Antennae are cephalic appendages.

152. Answer (2)

Fused, segmentally arranged ganglia are present

ventrally in cockroach.

153. Answer (3)

Cockroaches are active at night hence, nocturnal.

154. Answer (2)

Midgut/mesenteron is lined by peritrophic membrane

while foregut is lined by chitinous cuticle.

155. Answer (3)

Gradual metamorphosis i.e. paurometabolous

condition is observed in Periplaneta.

156. Answer (3)

Titillator is associated with left phallomere spiracles

are present on lateral sides of the body.

157. Answer (3)

Maxilla & mandible are paired appendages in

cockroach.

158. Answer (2)

159. Answer (3)

Ejaculatory duct, pharynx, conglobate gland &

phallic duct are four unpaired structures in

cockroach.

160. Answer (2)

161. Answer (4)

In cockroach, brain is represented by supra-

oesophageal ganglia; mosaic vision has more

sensitivity & sperms are glued together to form a

spermatophore.

162. Answer (3)

163. Answer (1)

Abdomen is long & narrow in males and caudal

styles are present in 9th segment of a male only.

164. Answer (4)

Alveolar sacs are absent in cockroach.

165. Answer (2)

Each ovary contains 8 ovarioles.

166. Answer (3)

Corpora allata is source of juvenile hormone. Its

absence leads to moulting.

167. Answer (3)

Mushroom gland is an unpaired structure in male

cockroach only.

168. Answer (4)

Anal styles are present in male cockroach only.

169. Answer (2)

Uricose gland are excretory structures found in

males of some species. Conglobate gland are

reproductive gland found in male cockroach only.

170. Answer (4)

Anal styles are unjointed structures.

171. Answer (2)

In frog RBCs are nucleated.

172. Answer (4)

173. Answer (4)

In frog teeth are not meant for chewing, digestion of

food begins in stomach & bile juices help in

emulsification.

174. Answer (2)

175. Answer (1)

Corpus callosum, ribs & salivary glands are absent

in frog.

176. Answer (4)

177. Answer (3)

Tadpole is a herbivore while frog is a carnivore.

Tadpole respire by gills while frog shifts to

pulmonary, cutaneous etc. form of respiration.

178. Answer (3)

179. Answer (3)

180. Answer (4)

� � �

Test - 3 (Code D) (Answers & Hints) All India Aakash Test Series for Medical-2019

1/12

1. (4)

2. (2)

3. (2)

4. (1)

5. (1)

6. (3)

7. (2)

8. (2)

9. (2)

10. (2)

11. (2)

12. (3)

13. (3)

14. (1)

15. (1)

16. (4)

17. (1)

18. (3)

19. (4)

20. (1)

21. (3)

22. (1)

23. (1)

24. (3)

25. (4)

26. (1)

27. (2)

28. (3)

29. (4)

30. (1)

31. (4)

32. (2)

33. (2)

34. (3)

35. (2)

36. (2)

Test Date : 03/12/2017

ANSWERS

TEST - 3 (Code D)

All India Aakash Test Series for Medical - 2019

37. (1)

38. (4)

39. (2)

40. (2)

41. (4)

42. (2)

43. (2)

44. (3)

45. (3)

46. (2)

47. (3)

48. (2)

49. (2)

50. (1)

51. (3)

52. (2)

53. (4)

54. (3)

55. (3)

56. (2)

57. (2)

58. (4)

59. (4)

60. (2)

61. (1)

62. (1)

63. (4)

64. (4)

65. (3)

66. (3)

67. (2)

68. (1)

69. (4)

70. (3)

71. (4)

72. (3)

73. (Deleted)

74. (3)

75. (1)

76. (1)

77. (4)

78. (2)

79. (4)

80. (4)

81. (1)

82. (3)

83. (1)

84. (1)

85. (1)

86. (4)

87. (1)

88. (4)

89. (1)

90. (3)

91. (3)

92. (3)

93. (4)

94. (1)

95. (3)

96. (4)

97. (2)

98. (4)

99. (3)

100. (3)

101. (1)

102. (1)

103. (4)

104. (2)

105. (2)

106. (2)

107. (2)

108. (4)

109. (1)

110. (4)

111. (3)

112. (4)

113. (4)

114. (1)

115. (3)

116. (2)

117. (3)

118. (3)

119. (1)

120. (4)

121. (4)

122. (4)

123. (4)

124. (1)

125. (4)

126. (2)

127. (1)

128. (4)

129. (4)

130. (3)

131. (2)

132. (2)

133. (4)

134. (4)

135. (3)

136. (4)

137. (3)

138. (3)

139. (3)

140. (4)

141. (1)

142. (2)

143. (4)

144. (4)

145. (2)

146. (4)

147. (2)

148. (4)

149. (3)

150. (3)

151. (2)

152. (4)

153. (1)

154. (3)

155. (4)

156. (2)

157. (3)

158. (2)

159. (3)

160. (3)

161. (3)

162. (2)

163. (3)

164. (2)

165. (2)

166. (3)

167. (2)

168. (1)

169. (2)

170. (3)

171. (3)

172. (4)

173. (4)

174. (2)

175. (4)

176. (3)

177. (3)

178. (3)

179. (1)

180. (2)

All India Aakash Test Series for Medical-2019 Test - 3 (Code D) (Answers & Hints)

2/12

ANSWERS & HINTS

1. Answer (4)

At the highest point �

�

F v

So 0 �

�

P F v

2. Answer (2)

2 2

mg gmg ma a ⇒

3. Answer (2)

Angle of repose R = 45°

So block does not slide

fs = mgsin37° =

310 6 N

5

4. Answer (1)

fs

Maximum acceleration upto which block does not

slide on the floor

amax

= g = 0.5 × 10 = 5 m/s2

Given, a = 4 m/s2

So, fs = 2 × 4 = 8 N

Displacement of block in t = 2 s

214 (2) 8 m

2 S

wf = 8 × 8 = + 64 J

5. Answer (1)

( 0)

(1)

p n muF mnu

t

⎛ ⎞ ⎜ ⎟⎝ ⎠

6. Answer (3)

10dp

F tdt

At t = 2 s F = 20 N

7. Answer (2)

(REF LEVEL)

L/10

GL/5

[ PHYSICS]

Mass of hanging part = 5

M

5 10

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠i

M LU g

50

MgL

Uf = 0

50 MgL

W U

8. Answer (2)

W = Pt = (m)gh

8000 60( ) 3200 kg

10 15m

9. Answer (2)

20 W �

�

P F v

10. Answer (2)

Kinetic energy is maximum when potential energy is

minimum.

U = 2x2 – 2 ...(1)

4dU

xdx

...(2)

For minimum, , 0

dUU

dx

x = 0

2

24

d U

dx U is minimum at x = 0

Umin

= –2 J

Now,

Kmax

+ Umin

= Emech

Kmax

– 2 = 34 2

max

136

2mv

vmax

= 6 m/s


3/12

11. Answer (2)

For equilibrium

F = 0

2x2 – x = 0 2(2x – 1) = 0

Either x = 0 or1

2x

4 1dF

xdx

At x = 0, 1 0dF

dx

So equilibrium is stable at x = 0.

12. Answer (3)

Speed of ball just before the collision.

2(10) 2 10 2.2

= 12 m/s

Speed of ball just after the collision.

2 10 5 10 m/s

Coefficient of restitution

10 5

12 6e

13. Answer (3)

Area under F - t graph

= change in linear momentum

1(2 3) 10

2

mv – mu = 25

v = 25 m/s

Kinetic energy 21 625

J2 2

k mv

14. Answer (1)

O

Y

X

50,

3

⎛ ⎞⎜ ⎟⎝ ⎠

5, 0

4

⎛ ⎞⎜ ⎟⎝ ⎠

4 5

3 3Y X

5 5ˆ ˆ0 0

4 3r i j

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

�

= 5 5ˆ ˆ

4 3i j

0r F �

�

15. Answer (1)

2

cos60 mvN mg

R

2 2(4)cos60 1 5

2

vN m g

R

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= 13 newton

fk =

kN = 6.5 N

N

mgmgsin60°

60°

fk

mgcos60°

Tangential acceleration sin60

kmg f

m

25 3 6.5 2.1 m/s

16. Answer (4)

B

A

ar

vB

g

O

vA

2

, B

r t

va a g

R

2 2( ) ( ) r t

a a a

2 22

B Av v gR

17. Answer (1)

3x

uF

x

23 m/s

x

x

Fa

m

4y

uF

y


4/12

ay = –4 m/s2

vx = a

xt = –3t; v

y = a

yt = –4t

2 25

x yv v v t

At t = 2 s, v = 5 × 2 = 10 m/s

18. Answer (3)

vx = u

x = 3 m/s

vy = eu

y =

34 3 m/s

4

Speed 2 2

3 2 m/sx y

v v v

19. Answer (4)

m

m v/2

v

Conservation of mechanical energy

2

21 1

2 2 2

vmv m

⎛ ⎞ ⎜ ⎟⎝ ⎠

2mg mg �

�

2

5 3 12

8 2 5

mv mg gv ⇒ � �

20. Answer (1)

Fv at t

m

⎛ ⎞ ⎜ ⎟⎝ ⎠ ...(1)

2F

P Fv tm

P t

21. Answer (3)

mg

hO

v

4u g �

Let string becomes slack when it makes angle

with upward vertical. Then

2

cosmv

mg �

2cosv g � ...(1)

Also,

2 2 22 2 (1 cos )v u gh u g �

or, cos 4 2 2 cosg g g g � � � �

or,2

3 cos 2 cos3

g g ⇒ � �

Tangential acceleration

5sin

3g g

22. Answer (1)

s

dvm mg

dt

23. Answer (1)

Instantaneous velocity.

23 8 dx

v t tdt

...(1)

At t = 0, v1 = 3 m/s

t = 9 s, v2 = 12 m/s

W = K

2 2

2 1

1

2 m v v

2 21 1[(12) (3) ]

2 2

135J

4

24. Answer (3)

F kxs =

mg


5/12

From work-energy theorem.

21

2mgx kx

2mgx

k ...(1)

Spring force at this instant

Fs = kx = 2mg

s2

upward

F mg mg mga g

m m

25. Answer (4)

A

ˆui

B2

4u g i ��

Velocity at point A.

ˆui ...(1)

Speed at the highest point.

2 24 �

Bv u g

24 �

Bv u g

Velocity 2 ˆ4 �

�B

v u g i ...(2)

Change in velocity

2 ˆ4 4 � � �

�B A

v v v u g i i

24

��v u u g

26. Answer (1)

At the highest point, speed

v = 100 cos60° = 50 m/s horizontally

Conservation of linear momentum.

2ˆ ˆ2 50 1 100 1i j v

�

2

ˆ ˆ100 100v i j �

2100 2 m/sv

�

So kinetic energy of this part.

2 411 100 2 10 J

2

27. Answer (2)

After collision velocity of heavy block remains the

same.

8 m/sv1

Now,

Relative velocity of separation = Relative velocity of

approach

8 – v1 = (10 – 8)

v1 = +6 m/s (in positive x direction)

28. Answer (3)

v

90 -

u

Line of impact

ucos

Let particle strikes with speed u and moves

horizontally with speed v after the collision.

Momentum is conserved along the plane.

So vcos = usin ...(1)

and vcos(90 – ) = eucos

or, vsin = eucos ...(2)

Dividing equation (2) by equation (1)

2 2 1tan cot tan tan 30

3e e ⇒

29. Answer (4)

e p2

ep

ep

p

+ve

Momentum of ball after first collision = +ep

Change in momentum of ball after first collision.

(p)1 = ep – (–p) = p(1 + e)

Similarly after second collision

(p)2 = e2p – (–ep) = ep(1 + e)

So total momentum imparted

1 2( ) ( )p p p

2(1 ) (1 ) (1 ) terms p e ep e e p e

(1 )

(1 )

p e

e

30. Answer (1)


6/12

31. Answer (4)

m1

m2

a1 a

2

2T

T T2T

(2T + T) a1 = Ta

2

32. Answer (2)

10 kg

kxm

100 N

N

Normal contact force is minimum when spring force

is maximum.

4 × 10 × xmax

= 21

2m

kx [K = 0]

kxm

= 80 N

Nmin

= 100 – 80 = 20 N

33. Answer (2)

mgcosmg

N

v

As block is at rest w.r.t. lift

N = mgcos ...(1)

Displacement of block w.r.t. ground

S = vt vertically upward ...(2)

Work done by normal contact force

W = N × S × cos

= (mgcos)(vt)cos

= mgvt cos2

34. Answer (3)

60°

u

v

From conservation of linear momentum

m(v cos60° + u) + Mu = 0

35. Answer (2)

As 1 >

2, both blocks will start sliding together,

when

(m1 + m

2)gsin =

1m1gcos +

2m2gcos

Or (m1 + m

2)gsin = gcos(

1m1 +

2m2)

1 1 2 2

1 2

tan

m m

m m

36. Answer (2)

2

12 sin 2 sin45 v g S g S ...(1)

2

22 (sin45 cos45 )

kv g S ...(2)

2

1

2

2

1 116

1511

16

k

v

v

1

2

4 :1v

v

37. Answer (1)

mgcosmg

FcosN

F

Fsin

mgsin

N

Block begins to slide when

Fcos = mgsin + N

Fcos = mgsin + (mgcos + Fsin)

F(cos – sin) = mg(sin + cos)

(sin cos )

(cos sin )

mgF

38. Answer (4)

2

2 2( )2

x

mT L x

L


7/12

T(x = 0) 21

2 m L

2

2 2

( 0) 4

( ) 1( )

T x L

T x x L x

4L2 – 4x2 = L2 4x2 = 3L2

3

2 L

x

39. Answer (2)

Limiting force of friction.

For m1 ; (f

1)max

= 1m1g = 25 N

m2 ; (f

2)max

= 2m2g = 40 N

Maximum frictional force fmax

= 65 N

60 N

m1

T

( ) = 25 Nf1 max

As F = 60 N, both blocks will be at rest. Frictional

force on block of mass m1 be (f

1)max

. Tension in the

string.

T = 60 – 25 = 35 N

40. Answer (2)

Maximum common acceleration of two blocks, when

force is applied on the upper block.

21

max

2

0.5 20 1010 m/s

10

m g

a

m

Fmax

= (m1 + m

2) a

max = 30 × 10 = 300 N

As F < Fmax

, both blocks move together with

common acceleration.

21505 m/s

30 a

Fs = m

2a = 10 × 5 = 50 N

41. Answer (4)

Just before t = 1 s

Velocity v1 = +2 m/s

Just after t = 1 s

Velocity, v2 = –4 m/s

Change in velocity, v = v2 – v

1 = –6 m/s

Impulse = change in linear momentum.

= 2 × (–6) = –12 Ns

42. Answer (2)

Acceleration 21 2

305 m/s

6

F F

aM

Mass of 10 cm part of uniform rod.

610 1.2 kg

50 m

Now,

10 cm

TF2

a

T – F2 = ma

T = F2 + ma

= 10 + 1.2 × 5 = 16 N

43. Answer (2)

In case (I), acceleration of block in horizontal

direction

aI = (gsin30°) cos30° ...(1)

In case (II), acceleration of block in horizontal

direction is acceleration of wedge.

aII = gtan30° ...(2)

Ratio 2I

II

3cos 30

4

⎛ ⎞ ⎜ ⎟⎝ ⎠

a

a

44. Answer (3)

When trolley is sliding with acceleration gsin, string

becomes perpendicular to the ceiling.

mgsin

mgcosmg

T

T = mgcos

45. Answer (3)

As downward acceleration of elevator a > g, block

loses the contact with the elevator and its

acceleration is g downward. So speed of block at

t = 1 s,

v = u + gt

= 10 × 1 = 10 m/s


8/12

46. Answer (2)

In SF4, 'S' atom undergoes sp3d hybridisation.

47. Answer (3)

In any resonance structure atom should obey octet

rule and should have symmetrical charge separation.

48. Answer (2)

dRTP

M

d M

P RT

d 1

P T ⇒

1 2

1 2

d P

P d

1 373

273 1

2 1

2 1

d d

P P

273

373

2

2

d

P

273 273x

373 373x

49. Answer (2)

Polarising power of cation charge on cation

50. Answer (1)

4CH

T

P

P =

4CH

1

116X

1 1 9

2 16

51. Answer (3)

In cyclooctatetraene all are p – p bonds.

52. Answer (2)

Density of a gas = PM

RT; density P; density

1

T

53. Answer (4)

54. Answer (3)

PVZ 1; 1

nRT ,

m V 22.4 L.∵

55. Answer (3)

Bond order Bond strength

56. Answer (2)

Fact

[ CHEMISTRY]

57. Answer (2)

Xe can form compounds with most electronegative

elements XeF2, XeF

4, XeF

6

58. Answer (4)

a s

Tv

M

4

2

CH

SO

v300 64 3

v 16 400 1

59. Answer (4)

Fact.

60. Answer (2)

Fact.

61. Answer (1)

Due to formation of intramolecular H-bonding

62. Answer (1)

Fact

63. Answer (4)

Fact

64. Answer (4)

3 3 3 3 3

* * 2 1

2 21 1 2 2 2 x y

z

p ps s s s p

65. Answer (3)

O N O O N O

Bond order = 1 2

1.52

66. Answer (3)

AlCl3 has more covalent character but more soluble

in H2O. It is due to high hydration enthalpy.

67. Answer (2)

Greater the value of 'a' i.e greater the intermolecular

forces of attraction for a gas, then the gas more

easily will be liquified.

68. Answer (1)

CH3

CN

has high dipole moment than

CH3

C N due to

presence of Electron with drawing (CN–) at para

position.


9/12

69. Answer (4)

70. Answer (3)

Cl Cl

Cl Cl

( = 0)

71. Answer (4)

In BF3, B is sp2 hybridized

In PCl5, P is sp3d hybridized

In 4

BF , B

is sp3 hybridized

In 6

PCl , P

is sp3d2 hybridized

72. Answer (3)

Lewis dot structure of 4

NH

is H N H

H+

H

73. Deleted

74. Answer (3)

Due to presence of 3 lone pairs on Xe in XeF2,

shape is linear.

75. Answer (1)

2 4 2N O 2NO��⇀

↽��

No. of moles initially 1 0

At equilibrium 1 – 0.4 0.8

0.6 92 0.8 46Molar mass ofmixture

1.4

∵

= 65.71

W PV RT (760 torr 1 atm)

M ∵

W PM 65.71 1d 2.67g / litre

V RT 0.0821 300

76. Answer (1)

Fact.

77. Answer (4)

PV PV 1 1Z n 0.0247

nRT ZRT 1.8 0.0821 273

⇒

No. of H2 molecules

23 220.0247 6.022 10 1.487 10

78. Answer (2)

79. Answer (4)

80. Answer (4)

Fact

81. Answer (1)

O

O O

O

H

HCH3

CH3( 0) ( 0)

82. Answer (3)

Fact

83. Answer (1)

%ionic character = 2

A B A B16(X X ) 3.5(X X )

(Hannay – Smith Equation)

=2

16(2.5 2.1) 3.5(2.5 2.1)

= 6.96%.

84. Answer (1)

Due to presence of lone pair on Br in BrF5, bond

angle is not equal to 90°.

85. Answer- (1)

86. Answer (4)

In 3 3

N(SiH )�� , N-atom donate its lone pair to vacant

d-orbital of Si to form back bonding.

87. Answer (1)

dRT dRTP M

M P ⇒

3d 2.64 gm / dm

3 1 1 3R 0.0821dm atm mole k ( 1 dm 1 litre)

∵


10/12

[ BIOLOGY]

91. Answer (3)

Brown algae possess the photosynthetic pigments

chl-a and chl-c and fucoxanthin.

92. Answer (3)

Thalloid body organisation is found in Marchantia and

algae.

93. Answer (4)

During double fertilisation one male gamete fuses

with diploid secondary nucleus.


fertilisation.

94. Answer (1)

Diplontic life cycle found in all flowering and

gymnosperm and these both are seed bearing

plants.

95. Answer (3)

‘‘Meiosis in the zygote results in the formation of

haploid spores in algae’’

96. Answer (4)

Naked ovule is the important feature of gymnosperm

97. Answer (2)

Vascular tissue is present in vascular cryptogams

and some are heterosporous.

98. Answer (4)

The sporophyte in mosses is more elaborate than

that in liverwort.

Spores are formed after meiosis.

99. Answer (3)

100. Answer (3)

In numerical taxonomy, all characters are given equal

importance

101. Answer (1)

102. Answer (1)

Club moss - Lycopodium (pteridophytes)

Cord moss - Funaria (Bryophytes)

Hair cap moss - Polytrichum (Bryophytes)

Peat moss - Sphagnum (Bryophytes)

103. Answer (4)

104. Answer (2)

105. Answer (2)

Synergids(n), Antipodal(n), Pollen grain(n)

Embryo (2n) Integument (2n)

106. Answer (2)

107. Answer (2)

Ploidy of PEN is triploid in nature (3n), e.g. maize.

108. Answer (4)

Only 'b' statement is correct.

'a', 'c' and 'd' statements are incorrect.

Gymnospermic root are generally tap root.

Egg apparatus is group of three cell (two synergid

and one female gamete)

Generally algae show haplontic life cycle

109. Answer (1)

'a' and 'c' features are correct.

110. Answer (4)

Double fertilisation is found in only angiosperms only

but Cedrus, Pinus are Gymnosperm

775P atm

760

�T 310 273 583 K

2.64 0.0821 583M 124

775 / 760

No. of P atoms in a molecule =124 124

At. Wt of P 31

= 4

88. Answer (4)

Due to resonance in 2 2

CH CH CH , all C atoms

undergo sp2 hybridization.

89. Answer (1)

Charge on ionLattice energy

Size of ion

90. Answer (3)

2 2 2 2SO Cl SO Cl

Hence this is not applicable to Dalton's law of partial

pressure.


11/12

111. Answer (3)

In bryophyta zygote does not undergo reduction

division immediately

eg. Polytrichum and Sphagnum

112. Answer (4)

Zygote does not undergo reduction division

immediately

Sporophyte is dependent on gametophyte

113. Answer (4)

In mosses, spores are formed after meiosis


fertilisation in most of the flowering plant

Gemmae are asexual buds

114. Answer (1)

115. Answer (3)

Rhizoids of mosses are multicellular and branched

Prothallus is free-living inconspicuous and

photosynthetic gametophyte

116. Answer (2)

In bryophytes, gametophyte is dominant.

117. Answer (3)

All gymnosperms are heterosporous and produce

haploid microspores and megaspores.

118. Answer (3)

Pinus and Polytrichum both reproduce by oogamous

type sexual reproduction.

119. Answer (1)

120. Answer (4)

In gymnosperm, veins and veinlets are absent in

leaves.

121. Answer (4)

122. Answer (4)

Gametophytes in gymnosperms are dependent on

sporophytes.

123. Answer (4)

Archegonia are found in Cycas

Female cone is not found in Cycas

124. Answer (1)

Gemmae, haploid asexual bud which helps in

asexual reproduction of Marchantia.

125. Answer (4)

126. Answer (2)

127. Answer (1)

Kelps are large sized brown algae

128. Answer (4)

Majority of the red algae are marine and generally

found in warmer areas.

129. Answer (4)

Both Ectocarpus and Gelidium have chl-a and

cellulosic cell wall

130. Answer (3)

Generally in algae haplontic life cycle is found but

Fucus shows diplontic life cycle.

131. Answer (2)

Sex organs are non-Jacketed

132. Answer (2)

Non-flagellated male gametes are found in Spirogyra

and flagellated male gametes are found in Fucus.

133. Answer (4)

Iodine is obtained from brown algae (Phaeophyceae)

134. Answer (4)

Natural system of classification is based on

phytochemistry also.

135. Answer (3)

136. Answer (4)

137. Answer (3)

138. Answer (3)

139. Answer (3)

Tadpole is a herbivore while frog is a carnivore.

Tadpole respire by gills while frog shifts to

pulmonary, cutaneous etc. form of respiration.

140. Answer (4)

141. Answer (1)

Corpus callosum, ribs & salivary glands are absent

in frog.

142. Answer (2)

143. Answer (4)

In frog teeth are not meant for chewing, digestion of

food begins in stomach & bile juices help in

emulsification.

144. Answer (4)

145. Answer (2)

In frog RBCs are nucleated.

146. Answer (4)

Anal styles are unjointed structures.


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� � �

147. Answer (2)

Uricose gland are excretory structures found in

males of some species. Conglobate gland are

reproductive gland found in male cockroach only.

148. Answer (4)

Anal styles are present in male cockroach only.

149. Answer (3)

Mushroom gland is an unpaired structure in male

cockroach only.

150. Answer (3)

Corpora allata is source of juvenile hormone. Its

absence leads to moulting.

151. Answer (2)

Each ovary contains 8 ovarioles.

152. Answer (4)

Alveolar sacs are absent in cockroach.

153. Answer (1)

Abdomen is long & narrow in males and caudal

styles are present in 9th segment of a male only.

154. Answer (3)

155. Answer (4)

In cockroach, brain is represented by supra-

oesophageal ganglia; mosaic vision has more

sensitivity & sperms are glued together to form a

spermatophore.

156. Answer (2)

157. Answer (3)

Ejaculatory duct, pharynx, conglobate gland &

phallic duct are four unpaired structures in

cockroach.

158. Answer (2)

159. Answer (3)

Maxilla & mandible are paired appendages in

cockroach.

160. Answer (3)

Titillator is associated with left phallomere spiracles

are present on lateral sides of the body.

161. Answer (3)

Gradual metamorphosis i.e. paurometabolous

condition is observed in Periplaneta.

162. Answer (2)

Midgut/mesenteron is lined by peritrophic membrane

while foregut is lined by chitinous cuticle.

163. Answer (3)

Cockroaches are active at night hence, nocturnal.

164. Answer (2)

Fused, segmentally arranged ganglia are present

ventrally in cockroach.

165. Answer (2)

Antennae are cephalic appendages.

166. Answer (3)

Blood glands are present in segment 4th, 5th & 6th.

167. Answer (2)

168. Answer (1)

Integumentary nephridia are exonephric and play no

role in osmoregulation.

169. Answer (2)

170. Answer (3)

Fertilization occurs in cocoon of earthworm,

therefore, it is external. Calciferous glands neutralize

humic acid. Chloragogen cells are analogous to

human liver.

171. Answer (3)

Spermathecae are present in segment 6, 7, 8 & 9.

172. Answer (4)

Setae & dorsal pores are absent in last segment of

earthworm while integumentary nephridiopores are

present.

173. Answer (4)

174. Answer (2)

175. Answer (4)

176. Answer (3)

Lateral hearts are present in segment 7 & 9.

177. Answer (3)

Blood glands are present in 4th, 5th & 6th segment.

Typhlosole starts from 27th segment of earthworm.

178. Answer (3)

Respiratory exchange occurs through moist body

surface.

179. Answer (1)

Lumbricus has a closed circulatory system and

nucleated leucocytes.

180. Answer (2)

Dorsal blood vessel is both collecting & distributing

blood vessel with valves.

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