for Medical-2019 All India Aakash Test Series for Medical ... · Test - 2 (Code E) (Hints and Solutions) All India Aakash Test Series for Medical-2019 5/22 Sol.: 37° 3 m 4 m 5 m

Test - 2 (Code E) (Answers) All India Aakash Test Series for Medical-2019

1/22

1. (2)

2. (3)

3. (1)

4. (1)

5. (1)

6. (2)

7. (3)

8. (2)

9. (1)

10. (4)

11. (1)

12. (3)

13. (3)

14. (1)

15. (4)

16. (1)

17. (4)

18. (1)

19. (4)

20. (3)

21. (3)

22. (1)

23. (1)

24. (2)

25. (2)

26. (1)

27. (4)

28. (2)

29. (1)

30. (2)

31. (1)

32. (2)

33. (3)

34. (2)

35. (1)

36. (3)

Test Date : 25/11/2018

ANSWERS

TEST - 2 (Code E)

All India Aakash Test Series for Medical - 2019

37. (2)

38. (1)

39. (2)

40. (3)

41. (4)

42. (1)

43. (4)

44. (1)

45. (2)

46. (3)

47. (2)

48. (4)

49. (3)

50. (3)

51. (4)

52. (4)

53. (2)

54. (4)

55. (3)

56. (4)

57. (1)

58. (1)

59. (3)

60. (4)

61. (3)

62. (2)

63. (4)

64. (2)

65. (3)

66. (2)

67. (1)

68. (4)

69. (4)

70. (1)

71. (2)

72. (1)

73. (1)

74. (4)

75. (2)

76. (3)

77. (2)

78. (4)

79. (1)

80. (3)

81. (2)

82. (4)

83. (4)

84. (1)

85. (4)

86. (1)

87. (2)

88. (3)

89. (2)

90. (4)

91. (4)

92. (4)

93. (2)

94. (3)

95. (4)

96. (4)

97. (3)

98. (2)

99. (4)

100. (2)

101. (4)

102. (1)

103. (4)

104. (4)

105. (3)

106. (2)

107. (4)

108. (1)

109. (2)

110. (3)

111. (1)

112. (2)

113. (4)

114. (4)

115. (3)

116. (3)

117. (4)

118. (3)

119. (3)

120. (1)

121. (2)

122. (2)

123. (1)

124. (4)

125. (1)

126. (4)

127. (2)

128. (3)

129. (3)

130. (4)

131. (4)

132. (4)

133. (1)

134. (4)

135. (3)

136. (3)

137. (1)

138. (3)

139. (1)

140. (4)

141. (3)

142. (3)

143. (3)

144. (4)

145. (3)

146. (1)

147. (4)

148. (2)

149. (2)

150. (2)

151. (1)

152. (4)

153. (3)

154. (1)

155. (4)

156. (4)

157. (3)

158. (4)

159. (2)

160. (3)

161. (2)

162. (3)

163. (1)

164. (3)

165. (2)

166. (4)

167. (2)

168. (2)

169. (3)

170. (2)

171. (3)

172. (4)

173. (4)

174. (2)

175. (3)

176. (2)

177. (2)

178. (3)

179. (1)

180. (3)

All India Aakash Test Series for Medical-2019 Test - 2 (Code E) (Hints and Solutions)

2/22

PHYSICS

1. Answer (2)

Hint: Apply F = ma

Sol.:

30°

10 kg

5 kg

T1

T2

T2

5 kg

T1

aa

a

100 – T1 = 10a

T1 – T

2 – 50 sin30° = 5a

T1 – 50 sin30° = 5a

20 a = 50

a = 5

2 ms–2

T1 = 100 –

50

2

T1 =

150

2N

T2 =

55 25

2

T2 =

75 N

2

1

2

150= = 2

2

75

2

T

T

2. Answer (3)

Hint: Translational equilibrium

F = 0

Sol.:

k

20 kg

T

2

T

2

T

4

T

4

T

HINTS AND SOLUTIONS

T = 200 N

4

T= k x

1000x = 200

4

x = 1

m20

= 5 cmx

3. Answer (1)

Hint: Use concept of pseudo force

Sol.:

A B

F.B.D. of B w.r.t. cart

NAB

m aB B

NAB

= 5 × 2

NAB

= 10 N

F.B.D. of A w.r.t. cart

NAB

NWA

m aA

NWA

= mAa = N

AB = 10 × 2 + 10 = 30 N

30= = 3 :110

WA

AB

N

N

4. Answer (1)

Hint: Use concept of constraint equation.

Sol.:

9 kg

30°

10 kg

60°

T

T

a1

a2

Test - 2 (Code E) (Hints and Solutions) All India Aakash Test Series for Medical-2019

3/22

T cos60° = 10 a2

T = 20 a2

...(i)

9g – T = 9a1

90 – T = 9a1

...(ii)

a2cos60° = a

1

a2 = 2a

1

T = 40a1

...(iii)

From equation (ii) & (iii),

49a1 = 90

or, a1 � 1.8 m/s2

5. Answer (1)

Hint: Properties of the friction force.

Sol.: fs Static friction

0 fs f

s max

0 Ns sf

flim

= s

N

6. Answer (2)

Hint: Concept of static and kinetic friction.

Sol.:

F = 25 N

QP

1maxsf

2maxsf

1 1max 1

= = 0.4 5 10s sf m g

1max

= 20Nsf

2 2max 2

= = 0.5 10 10 = 50Ns sf m g

1 2

max max appAs

s sf f F

So, system will be at rest

F.B.D of P

25 N

N

1maxs

fP

20 + N = 25

N = 5 N

F.B.D. of Q

Nfs Q

N = fs

= 5 Nsf

7. Answer (3)

Hint: Use F = ma

Sol.:

mB P A

10 N

x = 5 m x = 2 m x = 0

T T

Total mass of rope

M = 5 × 2 = 10 kg

2101 m/s

10 a

Mass of length 3 m

m = 3 × 2 = 6 kg

T = ma = 6 N

8. Answer (2)

Hint: Fnet

= 0, fk = N

Sol.:

37°

A

B

f

f

Net force on A will be zero

Wsin 37° = fk

3= cos37

5

BW m g

3 4= 3

5 5W W

1=4


4/22

9. Answer (1)

Hint: Fmin

= 2

1

mg

; when = tan–1()

Sol.: Fmin

= 2

1

mg

=

2

4 40010 10

3 3=

54

1 33

⎛ ⎞ ⎜ ⎟⎝ ⎠

Fmin

= 80 N

10. Answer (4)

Hint: For equilibrium Fnet

= 0

Sol.:

200 N

T1

T2

T4

T2

100 N

P

T1cos

T1sin

T1cos

T1sin T

2

T3

10 kg

100 N

T3

20 kg

200 N

T2

(F.B.D. ofblock 20 kg)


(F.B.D. ofpoint )P

P

T1cos = 100 N

T1sin = 200 N

tan = 2

–1

= tan 2

2 2 2

1= (100) (200)T

1=100 5 NT

11. Answer (1)

Hint: fmax

= s(m

1 + m

2)g

Sol.:

100 N10 kg

5 kg fk

fk

Maximum force for common acceleration

fmax

= 0.4 × 15 × 10 = 60 N

Force applied is greater than 60 N. So friction will be

kinetic.

fk =

kN = 0.4 × 50 = 20 N

2100 – 208 m/s

10a

12. Answer (3)

Hint: Concept of constraint equation.

Sol.:

P

T

aP

T/2

T/2

Q

T

Let movable pulley displaces by 'x' and block P by 'y '

l1 + l

2 = l

1 + x + l

2 + x – y

y = 2x

aP = 2a

Q

13. Answer (3)

Hint: Use dW = F dr��

Sol.: dW = ˆ ˆ ˆ ˆ( ) ( ) x y

F i F i dx i dyi

=x y

dW F dx F dy∫ ∫ ∫

W =

3 3

1 0

2xdx y dy∫ ∫

W =

32

32

0

1

9 1= – 9

2 2 2

xy

⎡ ⎤ ⎡ ⎤⎡ ⎤ ⎢ ⎥ ⎢ ⎥⎣ ⎦

⎣ ⎦⎣ ⎦= 4 + 9 = 13 J

14. Answer (1)

Hint: Wnet

= Wmg

+ Wf + W

N


5/22

Sol.:

37°

3 m

4 m

5 m10 k

g

Wnet

= Wmg

+ Wf + W

N

Wmg

= mg × 3 cos0° = 10 × 10 × 3 = 300 J

WN

= N × 5 cos 0°

WN

= 0

Wf

= N (5cos180°) = mgcos37°(–1) × 5

= –1 4

10 10 54 5 = –100 J

Wnet

= 300 – 100 net

= 200 JW

15. Answer (4)

Hint: W = 0 = 0F dr��

Sol.: ˆ ˆ ˆ= ( 3 – 4 ) N�

F i i bk

ˆ ˆ ˆ= (2 – 0) (2 – (–1) (3 – 1) �

r i j k

ˆ ˆ ˆ= (2 3 2 ) m �

r i i k

= 0F r�

�

ˆ ˆ ˆ ˆ ˆ ˆ(3 – 4 ) (2 3 2 )i j bk i j k = 0

6 – 12 + 2b = 0

= 3b

16. Answer (1)

Hint: Apply work–energy theorem

Wnet

= K

Sol.:

x m( )2

–5

10

6 8 10

F (N)

0

Wnet

= 1

10 2 10 4 – 2 52

Wnet

= 40 J

Wnet

= Kf – K

i

2 21 12 – 2 (2) = 40

2 2v

v2 – 4 = 40

v2 = 44

= 2 11m/sv

17. Answer (4)

Hint: Work = F ds��

= F(ds) cos

Sol.: W = F dscosWork done by static friction force may either be

positive, negative or zero. It depends on angle

between direction of force and displacement.

18. Answer (1)

Hint: Use ˆ ˆ= – –

U UF i j

x y

�

Sol.: U = –6x – 8y

= –6

U

x

= –8

U

y

ˆ ˆ(6 8 ) N �

F i j

2 2| | = 6 8 = 10 N�

F

2ˆ ˆ= = (3 4 ) m/s⇒ �

� �

F ma a i j

a = 10

2

2= 5 m/sa

x

y

(6, 4) m


6/22

Particle will move in the direction of acceleration

therefore, it will never cross y-axis and x-axis.

19. Answer (4)

Hint: =P F v� �

�

Sol.: F = constant

0 0

∫ ∫v x

vdv dx

2 v x

½v x

P F v �

�

½P x

20. Answer (3)

Hint: Use concept of coefficient of restitution.

Sol.: e = (Relative velocity of separation)

(Relative velocity of approach)

e = 2

1

v

v

v1

v2

e < 1

0 1e

21. Answer (3)

Hint: Conservation of energy and conservation of

linear momentum.

Sol.:

2mv1

4

m

0

2

v

4

m

v0

By conservation of linear momentum of system

0

0 1

ˆ ˆ ˆ= 24 4 2

vm mv i mv i i

0

116

vv

By conservation energy of the block.

2

1

12 = 2

2m v mgh

h =

2

0

2

1

2 (16)

v

g

2

0=512

vh

g

22. Answer (1)

Hint: Conservation of energy.

Sol.:

v0

R

h

(90° – )

B

RefA

v

2 2

0

1 1= ( )

2 2mv mg R h mv

At point B

mgcos(90° – ) – N =

2mv

R

mgsin =

2mv

R[∵ N = 0]

and h = R sin

2 mgR = 1

sin sin2

mgR mgR mgR

3 2sin = 1 sin =

2 3 ⇒

H = h + R

H = R + R sin

H = R + 2

3

R

5=

3

RH


7/22

23. Answer (1)

Hint: Conservation of mechanical energy

Sol.: At maximum extension of spring, speed of

block will be zero.

x

K

m

MEi = M.E

f

0 = 21

2kx mg

1

2kx2 = mgx

2=

mgx

k

24. Answer (2)

Hint: =W F d� �

Sol.:

T

mg

=2

ga

T – mg = ma

T = 3

2

mg

W = Td cos0°

3=

2

mgdW

25. Answer (2)

Hint: Condition for stable and unstable equilibrium.

Sol.: For equilibrium,

0dU

dx

3 23

– 2 03 2

d x xx

dx

⎛ ⎞ ⎜ ⎟

⎝ ⎠

(x2 – 3x + 2) = 0

x = 1, 2

Now,

2

22 – 3

d Ux

dx

For stable equilibrium,

2

20

d U

dx

For unstable equilibrium,

2

20

d U

dx

at x = 1

2

2–1 0

d U

dx

So unstable equilibrium,

at x = 2

2

21 0

d U

dx

So stable equilibrium.

26. Answer (1)

Hint: W = F S��

Sol.:

B

A

32 N

Let both block move together

32 = 8 a

a = 4 m/s2

fmax

= mg

fmax

= 1

3 10 =15 N2

Pseudo force on block A = ma = 4 × 3 = 12 N

There will be no slipping between the block and

12 N friction force will act on block A in forward

direction.

S = ut + 21

2at


8/22

S = 21

4 (2) = 8 m2

W = fs

× S

W = 12 × 8

= 96 JW

27. Answer (4)

Hint: Properties of center of mass.

Sol.: Center of mass relative to the body will not

depend on reference of frame. And it may lie inside

or outside the body.

28. Answer (2)

Hint: I = Mk2 k Radius of gyration.

Sol.:

2

2 =6

MaI I

I =

2

12

Ma

2

12

Ma = Mk2

=2 3

ak

29. Answer (1)

Hint: A1r1 = A

2r2

Sol.:

r1

r2

C1C C

2

2

2 2,

4 2

R RA r

A1 = R2 – A

2 = R2 –

2 23

4 4

R R

2 2

1

3

4 4 2

R R Rr

1

6

Rr

30. Answer (2)

Hint: =

dl

dt

�

�

= r F �

��

Sol.:

y

x

z

(2, 0, 0)

y

0

S = 21

2ut at

21– 10(2) = – 20 m2

y

ˆ= – 20 N�

F j

ˆ ˆ= (2 – 20 ) m�

r i j

ˆ ˆ ˆ

= 2 –20 0

0 –20 0

i j k

�

ˆ ˆ ˆ= (0 – 0) – (0 – 0) (–40 – 0)i j k �

ˆ= – 40 N-m� k

31. Answer (1)

Hint: Apply conservation of angular momentum about

the hinged point and conservation of M.E.

Sol.: =i f

L L

2

2

03

⎛ ⎞⎜ ⎟⎝ ⎠

��

mmv = m +

03

=4

v�


9/22

Conservation of M E

Ref

21 3= 2

2 2 2

mg mgI mg � �

�

22

0

2

14

92= 3

3 16

mv

mg�

��

2

0= 8v g�

v0 = 2 2g�

32. Answer (2)

Hint: Apply parallel axis theorem.

Sol.:

R

5R

I = 2 2

2(2 )5

3 2

⎡ ⎤ ⎢ ⎥⎣ ⎦

m R mRm R

I =

2 24 11

3 2

⎡ ⎤ ⎢ ⎥⎣ ⎦

mR mR

I =

2 2

28 33 41=

6 6

mR mRmR

33. Answer (3)

Hint: C.M. = 0S��

Sol.:

C.M.

C.M.

x

Initialposition

Final position

0

0

ˆ=

CPS li�

ˆ= –

PGS xi�

SCG

= ˆ= ( – )CP PG

S S l x i� �

ˆ ˆ( – ) – 2=

3

�

CM

l x im x miS

m

ml – 3xm = 0

x = 3

l

2– =

3

ll x

34. Answer (2)

Hint: net 1 2 3

= � � � �

Sol.:

45°

F

F 32 =F F

F

F F2 =

F F1 =

O

net

ˆ ˆ ˆ= (– ) (– ) ( ) 0 �

FR k FR k FR k

net

ˆ= –FRk�

35. Answer (1)

Hint: Use = andF ma = I �

��

.

Sol.:

A

B

T2

T2

T1

T1


10/22

100 – T1 = 10a ...(i)

T2 = 5a ...(ii)

T1R – T

2R = I

T1R – T

2R =

2

2mR a

R

T1 – T

2 =

2

ma...(iii)

T1 – T

2 = a

From equation (i), (ii) & (iii)

16 a = 100

a = 225

m/s4

= a

R

=

25

4

1

4

2= 25 rad/s

36. Answer (3)

Hint: C.M. C.M.

= �

� � �

L M r v I

Sol.:

x

zL

yM R,

v0

0

0

2 0 0

0

–72ˆ ˆ ˆ= (– ) (– ) =5 5

� v Mv RL Mv R k MR k k

R

37. Answer (2)

Hint: Use conservation of angular momentum and

v2 = u2 + 2as

Sol.:

v0 v

= 0

Conservation of angular momentum about the point

of contact.

mv0R (– ˆk ) =

2

ˆ(– )2

mRmvR k

⎛ ⎞ ⎜ ⎟

⎝ ⎠

2

02

mR vmv R mvR

R

02

=3

v

v

a = g

v2 = u2 + 2as

2

20

0

4 1= – 2

9 2

vv gs

2

05

=9

vs

g

38. Answer (1)

Hint: 2 21 1

=2 2

K E I mv

Sol.: KE1 = KE

2

2 2 2 2

1 1 2

1 1 1=

2 2 2mR mv mv

2 2 2

1 1 2

1 1 1=

2 2 2mv mv mv

2

1

2

1=2

v

v

⎛ ⎞⎜ ⎟⎝ ⎠

1

2

1=

2

v

v

39. Answer (2)

Sol.:

v

Velocity of center of mass decreases. To maintain

pure rolling angular velocity will also decrease. Here

direction of is clockwise.

Therefore, torque due to static friction will be

anticlockwise therefore, it will act upwards.


11/22

40. Answer (3)

Hint: C.M.

= ��

� �

�

Pv v OP

Sol.:

P

v1

vC.M.

60°30°

60°

vC.M.

O

0

C.M. 1| | = | |� � �

Pv v v

2 2

C.M. 1 1 C.M.| | = 2 cos60 �

Pv v v v v

2 2

0 1 0 1

�

Pv v v v v

As v1 =

02

R

v1 =

0

2

v

2 2

2 0 0 0

0

7| | = =

4 2 2

�

P

v v v

v v

41. Answer (4)

Hint: couple

mg

Sol.: fs = F, so they constitute a couple

F3

4

a

mgf = Fs

a/2

P

N

Torque about point P.

3 2

4 2 3

a a mgF mg F ⇒

maxs sf f

mg F

2

3

mgmg

2

3

42. Answer (1)

Hint: =L r P� �

�

Sol.: =L r P� �

�

andr r

L r L P � � �

�

= 0 = 0L r L P � � �

�

43. Answer (4)

Hint: Conservation of angular momentum about Q.

Sol.: =i f

L L� �

Taking anticlockwise direction positive

2

– =2 12

mmv I � �

v

Q

P

3 22

– =3 2 2 3

m m m

l

� � � �

– =3 12 3

4 –=

3 12

3=

4

44. Answer (1)

Hint: 2

0

1=

2t t

Sol.: 2

1

1= (1)2

1=2

21= (3)2

9=

2

2 1

9= – = – = 4

2 2

2

1

= 8


12/22

45. Answer (2)

Hint: 2

2

2=

1

ghv

k

R

⎛ ⎞⎜ ⎟

⎝ ⎠

Sol.: ring

2=

2

ghv ...(i)

Hollow, Sph.

2 2= =

2 51

3 3

gh ghv

...(ii)

ring

H,sp

2

52= =

62

5

6

gh

v

v gh

CHEMISTRY

46. Answer (3)

Hint: KE T

Sol.: Average molar kinetic energy is 3RT

2

⎛ ⎞⎜ ⎟⎝ ⎠

47. Answer (2)

Hint: Density of ideal PM

gas(d) =RT

Sol.:

d = PM 1.2 atm × 32 (g/mol.)

=RT L-atm

0.0821 600KK-mol.

⎛ ⎞⎜ ⎟⎝ ⎠

= 0.779 � 0.78 g/L

48. Answer (4)

Hint:H = U + ngRT

H > U (if ng > 0)

Sol.:ng of (i) Reaction = –5

ng of (ii) Reaction = 0

ng of (iiii) Reaction = 0

ng of (iv) Reaction = +1

Hence, for reaction (iv) H > U

49. Answer (3)

Hint: G = H – TS

For non-spontaneous reaction, G > 0

Sol.: If H > 0 and S < 0

G > 0 (For all temperature)

50. Answer (3)

Hint: In an isolated system, mass as well as energy

do not exchange with surrounding.

Sol.: Universe is an example of isolated system.

51. Answer (4)

Hint: Active mass of solid is unity.

Sol.: Calcium oxide is solid so its active mass is

unity.

52. Answer (4)

Hint: Keq

=

Coefficient

Coefficient

[Product]

[Reactant]

Sol.: 2 5 2 2 eq

1N O (g) 2NO (g) O (g), K = 0.5

2�

2 2 2 54NO (g) O (g) 2N O (g), � K

eq=

2

eq

1

K

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

= 1

0.5 × 0.5 = 4

53. Answer (2)

Hint: Acidic buffer is a mixture of weak acid and its

salt with strong base.

Sol.: (CH3COOH + CH

3COONa) Acidic buffer

54. Answer (4)

Hint: pH = – log[H+]

Sol.:[OH–] obtained from 0.1 N NaOH = 10–1 M

pOH = 1

pH = 13

55. Answer (3)

Hint: pH = a

17 (pK – pK )

2b

Sol.:pH of 0.1 M CH3COONH

4

pH = 1

7 (pKa – pK )2

b

pH = 7 (∵ pKa = pK

b)

56. Answer (4)

Hint: A3B

2(s) + H

2O � 3A2+(aq) + 2B3–(aq)

K = [A2+]3 [B3–]2


13/22

Sol.: K = (3s)3(2s)2 = 27s3 × 4s2 = 108 s5

s =

1/5

K

108

⎛ ⎞⎜ ⎟⎝ ⎠

57. Answer (1)

Hint: Unit of Kp : g

n

(atm) .

Sol.: For reaction PCl5(g) � PCl

3(g) + Cl

2(g),

ng = 1

So, unit of Kp

is atm

58. Answer (1)

Hint: Conjugate acid is formed by addition of one H+

ion to the given species.

Sol.: Conjugate acid of 2– –

4 2 4HPO is H PO

59. Answer (3)

Hint: Interaction energy of dipole-dipole force in

stationary polar molecule 3

1

r

60. Answer (4)

Rate of Volume effused=

effusion Time taken

⎛ ⎞⎜ ⎟⎝ ⎠

1

Molar mass

Sol.:

2

gas

H

r

r

= 2 2 2

2

2

gas

H H Hgas

H gas gas gas

H

V

M t Mt= = =

V M t M

t

=

gas

1 2=

8 M

Mgas

= 64 × 2 = 128 u

61. Answer (3)

Hint: Standard boiling point of water is 99.6°C

62. Answer (2)

Hint: Greater the atomicity of gas, greater is the

entropy.

Sol.: Correct order of entropy is

He < O2 < CO

2 < PCl

3

63. Answer (4)

Hint: SReaction

= (S)Product

– (S)Reactants

(:coefficients of reactants and products)

Sol.:Sreaction

= 6 × 100 – (2 × 30 + 5 × 60)

= 600 – (60 + 300) = 600 – 360 = 240

GReaction

= HReaction

– TSReaction

For spontaneous reaction

GReaction

< 0

HReaction

< TSReaction

300 × 103 < T × 240

T >

3300 10

240

T > 1250 K

64. Answer (2)

Hint: KC = [CO

2]

Sol.: CaCO3(s) � CaO(s) + CO

2(g)

KC = [CO

2]

0.5 2

mol.= [CO ]

⎛ ⎞⎜ ⎟⎝ ⎠�

moles of CO2 = 0.5 × 5 = 2.5

65. Answer (3)

Hint: Salt having minimum value of Ksp

will give ppt

first.

Sol.: Since, ppt appearance order is

AgI AgBr AgCl

Hence, Ksp

(AgI) < Ksp

(AgBr) < Ksp

(AgCI)

So, z < y < x

66. Answer (2)

Hint: Ksp

of Al(OH)3 = [Al3+] [OH–]3

Sol.: Al(OH)3(s) � Al3+(aq) + 3OH–(aq)

S 3S

∵ pH = 9, Hence, pOH = 5 [OH–] = 10–5 M

3S = 10–5 S =

–5

10

3 M

Ksp

(Al(OH)3) = S(3S)3 = 27S4 =

4–5

1027

3

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

=

–20 –20

27 10 10=

27 3 3

⎛ ⎞⎜ ⎟ ⎝ ⎠

67. Answer (1)

Hint: [H+]Resulting

(Normality) = 1 1 2 2

1 2

N V +N V

V V

Sol.: [H+] = 0.1 2 V + 0.1×1× V

V + V

= 0.2 + 0.1 0.3

= = 0.152 2


14/22

pH = – log[H+] = – log(15 × 10–2)

= 2 – log(15) = 0.824 0.82�

68. Answer (4)

Hint: Salt of weak acid and strong base will undergo

anionic hydrolysis.

Sol.: CH3COONa + H

2O CH

3COO– (aq) + Na+(aq)

CH COO3 + H

2O � CH

3COOH + OH (aq)

69. Answer (4)

Hint: Due to common ion effect, main source of Cl–

in solution is strong electrolyte NaCl.

Sol.: Ksp

(AgCl) = [Ag+] [Cl–]

1.6 × 10–10 = [Ag+] × 10–2

[Ag+] = 1.6 × 10–8

70. Answer (1)

Hint: For reaction 2A(g) � B(g) + C(g),

Equilibrium constant KC =

2

[B][C]

[A]

Sol.: 2A(g) � B(g) + C(g)

0.2 0 0

0.2 – 2x x x

KC =

2

2= 0.09

(0.2 – 2 )

x

x

= 0.3(0.2 – 2 )

x

x

x = 0.06 – 0.6x

1.6x = 0.06

x = 0.06

1.6

Concentration of B is 0.06

1.6 10

⎛ ⎞⎜ ⎟⎝ ⎠

= 3.75 × 10–3 M.

71. Answer (2)

Hint: For reaction

A(g) � B(g), KC =

[B] 0.02= = 2

[A] 0.01

Sol.:

A(g) � B(g)

At new eq. 0.02 – x 0.02 + x

KC =

0.02 + x= 2

0.02 – x

0.02 + x = 0.02 × 2 – 2x

3x = 0.02

x = 0.02

= 0.006673

[B] = 0.02 + 0.0067 = 0.0267

72. Answer (1)

Hint: Pure water is neutral and in pure water,

[H+] = [OH–] = w

K

Sol.: For acidic solution, [H+] > w

K

For basic solution, [OH–] > w

K

73. Answer (1)

Hint: Kp = K

c(RT)ng

Sol.: If ng = 0

Kp = K

c

For reaction N2(g) + O

2(g) � 2NO(g), n

g = 0

74. Answer (4)

Hint: pH of salt of weak base and strong acid is

b

17 – (pK log C)

2

⎧ ⎫⎨ ⎬⎩ ⎭

.

Sol.:

4 4 2NH OH HCl NH Cl H O

Initial mol, 0.1 V 0.1 V

Final mol, 0 0 0.1 V

⎧ ⎫⎨ ⎬⎩ ⎭

Concentration of NH4Cl =

–2

0.1= 5 ×10

2

pH = –2

17 – 4.75 log(5 ×10 )

2

= 17 – 4.75 – 2 log5

2

= 5.2755 � 5.28

75. Answer (2)

Hint: Partial pressure of a gas

= Mole fraction × Total pressure

Sol.: Moles of 2

xO

32

(Assume, X is mass of each gas mixed together)


15/22

Moles of x

He4

Total moles x x 9

x32 4 32

pHe

x

84P P

9 9x

32

Fraction of total pressure exerted by He = 8

9.

76. Answer (3)

Hint:

P

L G(L + G)

V

Sol.: At temp. TC and pressure P

2, gas phase

exists.

77. Answer (2)

Hint: Number of mole = Volume at STP(ml)

22400

Sol.: For A vapours, mole = 112 1

mol22400 200

Molar mass of A = 0.48 × 200 = 96 g

2

A(g)

H

r 2 1 1= = =

r 96 48 4 3

78. Answer (4)

Hint: For H2 gas, value of a is negligible.

Sol.: m2

m

aP + (V – b) = RT [ a is negligible]

V

⎛ ⎞⎜ ⎟⎝ ⎠

∵

P(Vm

– b) = RT

PbZ =1+

RT

79. Answer (1)

Hint: Partial pressure = (Total pressure)

× Mole fraction of gas

Sol.: 2

O

20n = = 0.625

32

CO

2.8n = = 0.1

28

Mole fraction oxygen gas = 0.625

0.1 0.625

= 0.625 625

=0.725 725

= 25

29

⎛ ⎞⎜ ⎟⎝ ⎠

pO2

= 25

29

⎛ ⎞⎜ ⎟⎝ ⎠

× 2 = 50

atm29

Partial volume of oxygen gas = 25

1029

250

29 L

80. Answer (3)

Hint: Fusion

2 2

Fusion

n HH O(s) H O(l), S =

T

⎛ ⎞⎜ ⎟

⎝ ⎠�

Sol.: S =

3

–1 –16 10= 21.98 Jmol K

273

At equilibrium, G = 0

81. Answer (2)

Hint: H2O(l) � H

2O(v) (1 atm, 100°C)

U = q + W

Sol.: q = nHVapourization

q = 1 × 40.66 × 103 J = 40.66 × 103 J

w = – ngRT

= –1 × 8.314 × 373 = –3101.12 J

U = q + w = (40.66 × 103 – 3101.12) J

= 37558.878 J = 37.56 × 103 J

82. Answer (4)

Hint: U = q + w and w = –2.303 nRT log2

1

V

V

⎛ ⎞⎜ ⎟⎝ ⎠

Sol.: w = –2.303 × 2 × 8.314 × 300 log20

5

⎛ ⎞⎜ ⎟⎝ ⎠

= –2.303 × 2 × 8.314 × 300 × log4

= –2.303 × 2 × 8.314 × 300 × 2 × 0.30

= – 6892.9 J � –6.9 kJ

U = q + w

q = –w (for isothermal process)

q = +6.9 kJ


16/22

83. Answer (4)

Hint: S = 2 1

p

1 2

T PnC ln + nRln

T P

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Sol.: S = 1

2

PnR ln

P

⎛ ⎞⎜ ⎟⎝ ⎠

(At constant temperature)

S = 1

2

PnR ln

P

⎛ ⎞⎜ ⎟⎝ ⎠

= 1 × R × 2

ln0.5

⎛ ⎞⎜ ⎟⎝ ⎠

= R ln 4 = 2R ln2

84. Answer (1)

Hint: Overall enthalpy of a process, is the sum of

the enthalpies of all steps involved.

Sol.:

(i) C(s) + O2(g) CO

2(g) + 94 kcal/mol

(ii) H2(g) +

1

2O

2(g) H

2O(l) + 68 kcal/mol

(iii) CH4(g) + 2O

2(g) CO

2(g) + 2H

2O(l) + 213 Kcal/mol.

On applying, (i) + 2 × (ii) – (iii)

Reaction obtained is

C(s) + 2H2(g) CH

4(g)

H = (–94) + 2(–68) –(–213) = –17 kcal.

85. Answer (4)

Hint: Vapour pressure of a liquid depends on

temperature.

86. Answer (1)

Hint:

Enthalpy of

= 3 ×

Enthalpy of hydrogenation –

Resonance hydrogenation of benzene of cyclohexane

energy

Sol.:

+ H2 ; H = –x kJ mol–1

+ 3H2

Enthalpy of hydrogenation of benzene

= 3H – Resonance energy.

= 3(–x) – (–y)

= (y – 3x) kJ mol–1

87. Answer (2)

Hint: pH of acidic buffer pH = pKa + log

Salt

Acid

⎡ ⎤⎢ ⎥⎣ ⎦

Sol.:

CH3COOH + NaOH CH

3COONa + H

2O

(mmol)initial

10 6

(mmol)final

4 0 6

Resulting solution is an acidic buffer

pH = 4.74 + 6

log4

⎛ ⎞⎜ ⎟⎝ ⎠

= 4.74 + 0.18 = 4.92

88. Answer (3)

Hint: Salt having minimum value of 'solubility (mole/l)'

will be least soluble.

Sol.: For binary electrolyte, lower the value of Ksp

,

lower will be its solubility.

89. Answer (2)

Hint: Lewis acids are electron pair acceptors.

Sol.: B2H

6 is an electron deficient species.

90. Answer (4)

Hint: G = G0 + 2.303RT log Qp

Sol.: QP =

2 5

2 2

2 2N O

2 5 2

N O

P 2= =1

P P 2 1

G = G0 + 2.303 RT log10

(1) = G0 = –43 kJ/mol

BIOLOGY

91. Answer (4)

Hint: The body of fungi is not differentiated into root,

stem & leaves.

Sol.: The body of fungi is thalloid & haploid (n).

Each cell contains one set of chromosomes.

92. Answer (4)

Hint: On basidium exogenously produced spores are

basidiospores.

Sol.: Basidiospores are produced by members of

basidiomycetes. These are haploid and sexual

spores.

93. Answer (2)

Hint: Claviceps purpurea causes ergot of rye.

Sol.: Claviceps belongs to the class Ascomycetes.

It is a sac fungus which possesses septate

mycelium and chitinous cell wall.

It reproduces by conidia which are non-motile

asexual spores.


17/22

94. Answer (3)

Hint: Deuteromycetes are called imperfect fungi.

Sol.:Morels

AscomycetesAspergillus

⎤⎥⎦

Bracket fungi ] Basidiomycetes

Deuteromycetes

Alternaria

Colletotrichum

Trichoderma

⎤⎥⎥⎥⎦

95. Answer (4)

Hint: Rhizopus is a conjugation fungus, also known

as bread mould.

Sol.: It reproduces by non-motile asexual spores and

non-motile zygospores.

It is a saprophytic fungus with chitinous cell wall.

96. Answer (4)

Sol.: Bladderwort is a partially heterotrophic

insectivorous plant.

97. Answer (3)

Hint: Phycobiont is algal partner in lichen

association. They are either green algae or blue-

green algae.

Sol.: Algal partner synthesizes food for fungal

partner. Lichens do not grow in air polluted areas.

Animals store their food in form of glycogen or fat.

98. Answer (2)

Hint: A virus is larger than sub viral agents/particles.

Sol.: Viroids are infectious RNA particles. They lack

protein. They cause diseases in plants only.

Viruses have proteinaceous capsid.

99. Answer (4)

Hint: Prions lack genetic material.

Sol.: Prions are proteinaceous infectious particles. It

causes mad cow disease in cattles & scrapie

disease in sheep.

100. Answer (2)

Hint: Viruses are obligate intracellular parasites.

Sol.: Viruses can multiply within the host cell only.

They cannot respire, divide and grow externally.

101. Answer (4)

Hint: All viruses have capsid.

Sol.: Bacterial viruses are known as bacteriophages,

they have capsid (protein coat) & genetic material,

usually dsDNA.

102. Answer (1)

Hint: Mycorrhiza is a mutually beneficial association.

Sol.: Mycorrhiza is a symbiotic association between

roots of higher plants & fungi.

103. Answer (4)

Hint: Dinoflagellates are flagellated, mostly marine

photosynthetic protist.

Sol.: Dinoflagellates have two flagella, one is

transverse and another is longitudinal

104. Answer (4)

Hint: Paramoecium has two nuclei.

Sol.: Paramoecium is a ciliated protozoan

105. Answer (3)

Hint: Cell wall of diatoms is indestructible due to

presence of silica.

Sol.: Due to presence of silica, the fossilized

diatoms get deposited at the sea bed to form

diatomaceous earth.

Diatoms are microscopic, unicellular protists which

float passively in water current.

106. Answer (2)

Hint: Euglenoids lack cell wall.

Sol.: Slime moulds are heterotrophic (Saprophytic).

Protozoans are predators & parasites. Chrysophytes

include diatoms and desmids which are mostly

photosynthetic.

107. Answer (4)

Hint: Protists are unicellular eukaryotes.

Sol.: Eukaryotes have well defined nucleus and

other membrane bound cell organelles.

108. Answer (1)

Hint: Heterocyst is related to N2 fixation which takes

place in anaerobic condition.

Sol.: In heterocyst cell wall is impermeable to

oxygen but permeable to nitrogen. It lacks PS-II but

PS-I remains active and produces ATP required for

N2-fixation.

109. Answer (2)

Hint: Archaebacteria contain pseudomurein in their

cell wall.

Sol.: The cell wall of eubacteria is made up of

mainly peptidoglycan. They are prokaryotes hence

they lack membrane bounded cell organelles.

Archaebacteria include halophiles, methanogens and

thermoacidophiles.

Archaebacteria have introns in their genetic material.


18/22

110. Answer (3)

Hint: Mycoplasma are smallest living organisms.

Sol.: They lack cell wall and can survive without

oxygen.

Methanogens are responsible for production of

biogas.

111. Answer (1)

Hint: Chemoautotrophic bacteria lack photosynthetic

pigments.

Sol.: ChemoautotrophicNitrosomonas

Nitrobacter

⎤⎥⎦

PhotosyntheticNostoc

Chlorobium

⎤⎥⎦

Lactobacillus] Heterotrophic.

112. Answer (2)

Hint: Prokaryotes lack nuclear membrane &

membrane bound cell organelles.

Sol.: Streptococcus is a prokaryote. Remaining

organisms are eukaryotes.

113. Answer (4)

Hint:

Cyanobacteria evolve O2 during photosynthesis.

Sol.: Cyanobacteria have enzyme nitrogenase which

fixes atmospheric nitrogen. Some cyanobacteria

form symbiotic associations with ferns. Such as

Anabaena with Azolla.

Nostoc is a filamentous BGA.

114. Answer (4)

Sol.: Bordeaux mixture is first fungicide discovered

by RMA Millardet. It is commonly known as holy

water of plant pathology.

115. Answer (3)

Hint: Tap root is the primary root.

Sol.: Tap roots (Primary root) arise from radicle

whereas fibrous roots arise from the base of the

stem.

116. Answer (3)

Hint: Carrot & Turnip are modifications of tap root.

Sol.: Ginger is a modified underground stem. Edible

part of sweet potato is a modified adventitious root.

117. Answer (4)

Hint: Pneumatophores help a plant to get oxygen

from the atmosphere.

Sol.: They grow vertically upward hence negatively

geotropic & can be seen in Rhizophora which grow

in marshy/swampy areas.

118. Answer (3)

Hint: Stem bears nodes, internodes and buds.

Sol.: Potato is a tuber which is a modified

underground stem because it bears buds.

119. Answer (3)

Hint: Phylloclade is a modified stem.

Sol.: In Opuntia, phylloclade is flattened and

photosynthetic.

120. Answer (1)

Hint: Albuminous seed contains endosperm.

Sol.: Coconut has endospermic seed where seed

coat is thick & non-membranous.

Groundnut, pea, and soyabean have ex-albuminous

(non-endospermic) seeds.

121. Answer (2)

Hint: Axile placentation has placenta in the axial

position and ovules are attached to it.

Sol.: Tomato & lemon have axile placentation. Pea

has marginal, mustard has parietal & sunflower and

marigold have basal placentation.

122. Answer (2)

Hint: In some seeds nucellus persists in the form of

a layer.

Sol.: Perisperm is persistent nucellus. Seeds having

perisperm are called perispermic seeds.

123. Answer (1)

Hint: Given floral formula belongs to family

Solanaceae.

Sol.: Solanaceae members have persistent calyx,

superior ovary with many ovules. Seeds are

endospermic. Flowers are bisexual and

actinomorphic.

124. Answer (4)

Hint: In monadelphous condition, all stamens are

united in a single bundle but anthers remain free. It

is found in family Malvaceae.

Sol.: In pea flower the stamens are united in two

bundles so it is diadelphous.

125. Answer (1)

Hint: In opposite phyllotaxy, a pair of leaves arise at

each node opposite to each other.

Sol.: Guava, Calotropis have opposite phyllotaxy.


19/22

Mustard, China roseAlternate phyllotaxy

Sunflower

⎤⎥⎦

Alstonia, Nerium ] Whorled phyllotaxy.

126. Answer (4)

Hint: Banana is a triploid plant hence cannot form

gametes.

Sol.: Banana fruit develops without fertilisation of

gametes, hence it is a parthenocarpic fruit. Banana

has suckers.

127. Answer (2)

Hint: In racemose inflorescence the terminal flower

is absent.

Sol.: It (main axis) has unlimited growth and flowers

ever borne in acropetal manner.

128. Answer (3)

Hint: Fused carpels— syncarpous

Sol.: Staminode is sterile anther. Twisted aestivation

is seen in petals of cotton. Unisexual flowers are

seen in maize.

129. Answer (3)

Hint: Papilionaceous corolla shows vexillary

aestivation.

Sol.: Mango is a drupe. Vexillum is largest posterior

petal in pea also called standard petal.

130. Answer (4)

Hint: In members of family Poaceae, the seed coat

is fused with fruit wall (Caryopsis type of fruit).

Sol.: Gram is a dicot. Wheat and maize are

monocots and belong to family Poaceae.

131. Answer (4)

Hint: This plant belongs to the family Fabaceae.

Sol.: It is Indigofera.

132. Answer (4)

Hint: Thorns can be seen in Citrus or Bougainvillea.

Sol.: Spines are modification of leaves.

133. Answer (1)

Sol.: Root hairs arise from epidermal cells of

maturation zone.

134. Answer (4)

Hint: Aleurone layer is a part of endosperm.

Sol.: It is triploid, provides nourishment and seen in

maize grains.

135. Answer (3)

Sol.: Strawberry is an aggregate fruit.

Mulberry & jackfruits are composite fruits or multiple

fruits.

Pomegranate is a simple fleshy or succulent fruit.

136. Answer (3)

Hint: The primary organ involved in this process is

kidney.

Sol.: Removal of metabolic waste products from the

body is known as excretion. Main role of sweating

is to facilitate thermoregulation. Defaecation involves

removal of undigested and unabsorbed food.

137. Answer (1)

Hint: Part of nephron exhibiting brush border

appearance.

Sol.: ADH facilitates facultative/conditional

reabsorption of water from DCT & collecting tubule.

PCT is responsible for obligate reabsorption and is

internally lined by simple cuboidal epithelium

containing microvilli giving brush border appearance

which increase surface area for reabsorption.

138. Answer (3)

Hint: These cells are granular, phagocytic and most

abundant type of WBCs.

Sol.: Differential leukocyte count i.e., DLC reveals

percentage of type of WBCs in blood.

Neutrophils – 60–65%

Lymphocytes – 20–25%

Monocytes – 6–8%

Eosinophils – 2–3%

139. Answer (1)

Hint: Excretory organ in annelids.

Sol.: Respiration in earthworm (annelids) occurs

through moist cuticle.

140. Answer (4)

Hint: Squamous epithelium is found at this surface.

Sol.: Gaseous exchange takes place in the alveoli

in lungs.

141. Answer (3)

Hint: Fluid filled cavity is present around lungs.

Sol.: Pleura is divided into two layers. Outer pleural

membrane is in close contact with thoracic lining

whereas inner pleural membrane is in contact with

lung surface.


20/22

142. Answer (3)

Hint: TV + IRV + ERV = TLC – RV

Sol.: TLC – RV = VC,

EC = ERV + TV = 1100 + 500 = 1600 ml

143. Answer (3)

Hint: Lungs are situated in cavity above diaphragm.

Sol.: Trachea divides into two primary bronchi at the

level of 5th thoracic vertebra and enters lungs.

144. Answer (4)

Hint: The nitrogenous waste product which requires

minimum loss of water in its elimination.

Sol.: Land snails excrete uric acid whereas most

terrestrial organisms excrete either urea (humans,

amphibians) or uric acid (insects, birds, reptiles).

Aminotelic organisms excrete amino acid (Unio and

echinoderms)

145. Answer (3)

Hint: These animals live on both land and water.

Sol.: Crocodiles are reptiles with completely divided

four chambered heart. Amphibians have three

chambered heart.

146. Answer (1)

Hint: Dialysis helps to counter uremia.

Sol.: Dialysing fluid is isotonic to blood plasma.

147. Answer (4)

Hint: This membrane divides body cavity into

thoracic and abdominal cavity.

Sol.: Pneumotaxic centre is called "switch-off" point

of inspiration. It stimulates neurons of expiratory

centre located in medulla. Diaphragm is stimulated

by respiratory rhythm centre in medulla.

148. Answer (2)

Hint: Enzyme which is released in body injured site

in presence of Ca+2.

Sol.: Prothrombinase/thrombokinase is also

responsible for conversion of inactive plasma protein

prothrombin into an active enzyme thrombin

responsible for conversion of fibrinogen into fibrin.

149. Answer (2)

Hint: This hormone secreted by JG cells stimulates

RAAS.

Sol.: Renin secreted by JG cells in response to low

B.P. stimulates RAAS to maintain GFR. Angiotensin-

II is a vasoconstrictor. Rennin digests casein in

infant stomach.

150. Answer (2)

Hint: These cells give rise to enucleated platelets.

Sol.: Megakaryocytes are specialised large cells in

red bone marrow which divide to form cellular

fragments lacking nucleus known as blood platelets/

thrombocytes.

151. Answer (1)

Hint: WBCs which participate in allergic reaction

during worm infestation.

Sol.: Eosinophils are WBCs that stain with acidic

dye eosin and have bilobed nucleus. They participate

in a number of allergic reactions and fight against

worm infestation.

152. Answer (4)

Hint: Identify the blood group of universal donors.

Sol.: Universal donors (O–) and individuals with

blood group B– and O+ can donate blood to the

patient.

153. Answer (3)

Hint: This heart sound is produced during ventricular

systole.

Sol.: First heart sound 'lub' is produced during

ventricular systole. It has low frequency of about

25–45 Hz and is of comparatively longer duration

(0.15 sec) in comparison to second heart sound

which has higher frequency than 50 Hz but duration

is only 0.12 sec.

154. Answer (1)

Hint: Lymphatic fluid is poured into venous blood.

Sol.: Lymph from right thoracic duct drains into right

subclavian vein whereas left thoracic duct drains

lymph into left subclavian vein

155. Answer (4)

Hint: Identify a layer absent in alveolar capillaries.

Sol.: Capillaries lack tunica media formed by

smooth muscle fibres. So, diffusion membrane does

not contain muscular tissue.

156. Answer (4)

Hint: Affinity of Hb for O2 is inversely proportional to

P50

value.

Sol.: Shift of curve B towards right indicates

increase in P50

value and thus decreased affinity of

haemoglobin for oxygen, leading to greater

dissociation of oxygen from hemoglobin.


21/22

157. Answer (3)

Hint: Maximum percentage of CO2 is transported in

plasma in bicarbonate form.

Sol.: 77% CO2 is transported through blood plasma.

Dissolvedin plasma – 7%

In the form ofbicarbontates

in plasma – 70%

Plasma(77%)

158. Answer (4)

Hint: ADH acts on this region to prevent diuresis.

Sol.: The collecting duct is responsible for secretion

of urea not for absorption. Various nephrons open

into collecting ducts. Collecting duct secretes urea

into interstitium to increase osmolarity for

reabsorption of water.

159. Answer (2)

Hint: Stretch receptors are activated upon filling of

urinary bladder beyond threshold.

Sol.: Hemodialysis is needed when kidneys

malfunction. Urine formation is independent of stretch

reflex.

160. Answer (3)

Hint: Volume of ventricle decreases during their

contraction.

Sol.: During ventricular systole, closure of cuspid/AV

valves occurs. Blood enters ventricles during atrial

systole.

161. Answer (2)

Hint: Volume of blood pumped by each ventricle in

0.8 seconds.

Sol.: Cardiac output = heart rate x stroke volume.

Hypertension is observed if blood pressure exceeds

140/90 or more repeatedly. Opening of pulmonary

artery is guarded by semilunar valves.

162. Answer (3)

Hint: During this phase, blood is not pumped from

ventricles to aorta.

Sol.: During isovolumetric contraction, both

auriculoventricular valves and semilunar valves remain

closed. Pressure within ventricles increases but

amount of blood remains unchanged in isovolumetric

contraction.

163. Answer (1)

Hint: 2/3rd filling of ventricles occurs when all

chambers of heart are relaxing.

Sol.: When auricles are in diastole auriculoventricular

valves remain open but semilunar valves present at

opening of aorta are closed.

164. Answer (3)

Hint: Duration of each auricular cycle is equal to

duration of the cardiac cycle.

Sol.: During each cardiac cycle, an auricular cycle

as well as a ventricular cycle is completed. Each

cycle has duration of 0.8 sec. So time of auricular

cycle is 0.8 sec not 0.6 second.

165. Answer (2)

Hint: Identify the machine use to measure an ECG.

Sol.: Electrical activities of heart is measured by an

instrument known as electrocardiograph and graph

obtained is known as electrocardiogram.

Stethoscope is used to hear heart sounds.

166. Answer (4)

Hint: Well developed region of JG nephrons

extending into medulla in JG nephrons.

Sol.: Loop of Henle of both cortical and juxtamedullary

nephron lie in medullary pyramid.

167. Answer (2)

Hint: This hormone is released by atrial walls.

Sol.: Vasa recta acts as countercurrent exchanger

because they exchange water for ions. ANF

opposes RAAS.

168. Answer (2)

Hint: These are region between medullary pyramids

in kidney.

Sol.: Invagination of cortex into medulla forms

columns of Bertini which divide medullary region of

kidney into renal pyramidals.

169. Answer (3)

Hint: Ascending limb of loop of Henle is permeable

to salt.

Sol.: Descending limb is called concentrating

segment of loop of Henle as it is permeable to water

and impermeable to salts.

170. Answer (2)

Hint: High threshold substance like glucose is

completely absorbed in kidney in a normal healthy

person.


22/22

Sol.: Diabetes insipidus occurs due to deficiency of

ADH (vasopressin) from hypothalamus. All types of

nephrons have peritubular capillary networks. GFR in

healthy individual is 125 ml/minute or 180 L/day.

171. Answer (3)

Hint: Maximum amount of air which can be inspired

or expired forcefully.

Sol.: Vital capacity is measured for lung function

test by an instrument known as spirometer. With

exception of RV, TLC, FRC. All other lung capacities

and volumes can be measured by spirometer.

172. Answer (4)

Hint: Innermost lining of wall of blood vessels is

constituted by squamous cells.

Sol.: Arteries have narrower lumen than veins and

mostly carry oxygenated blood.

173. Answer (4)

Hint: Sympathetic stimulation alters heart rate.

Sol.: Sympathetic neural signals increase cardiac

output and stroke volume but decrease duration of

cardiac cycle. Cardiac muscles are not under the

control of our will.

174. Answer (2)

Hint: Pressure difference on both sides of filtration

membrane.

Sol.: NFP = (GHP) – (BCOP + CHP)

= 60 – (30 + 20)

= 10 mmHg

175. Answer (3)

Hint: Blood vessel exiting from major osmoregulatory

organ of human body.

Sol.: Largest amount of urea is present in hepatic

vein.

176. Answer (2)

Hint: This hormone is released in response to

increase in blood pressure than normal.

Sol.: During dehydration (profuse sweating)

osmolarity of body fluid increases, increased

secretion of ADH from hypothalamus occurs. GFR

decreases during profuse sweating.

177. Answer (2)

Hint: Ischemic conditions reflect oxygen deficiency

in cardiac muscle fibres.

Sol.: When heart muscle is suddenly damaged by

inadequate blood supply, the condition is called heart

attack.

178. Answer (3)

Hint: This wave indicates ventricular repolarisation.

Sol.: T-wave represents the return of ventricles from

excited to normal state. QRS complex represents

depolarisation of ventricles while P-wave represents

atrial depolarisation.

179. Answer (1)

Hint: pCO2 in alveoli is 40 mmHg.

Sol.:

Atmospheric air

Alveoli Deoxy-generated

blood

Oxy-genated

blood

Tissue

pO2 159 104 40 95 40

pCO2 0.3 40 45 40 45

180. Answer (3)

Hint: Disorder associated with accumulation of dust

particles.

Sol.: Pneumoconiosis is occupational disease of

lungs due to inhalation of dust, characterised by

inflammation, coughing and fibrosis.

� � �

Test - 2 (Code F) (Answers) All India Aakash Test Series for Medical-2019

1/22

Test Date : 25/11/2018

ANSWERS

TEST - 2 (Code F)

All India Aakash Test Series for Medical - 2019

1. (2)

2. (1)

3. (4)

4. (1)

5. (4)

6. (3)

7. (2)

8. (1)

9. (2)

10. (3)

11. (1)

12. (2)

13. (3)

14. (2)

15. (1)

16. (2)

17. (1)

18. (2)

19. (4)

20. (1)

21. (2)

22. (2)

23. (1)

24. (1)

25. (3)

26. (3)

27. (4)

28. (1)

29. (4)

30. (1)

31. (4)

32. (1)

33. (3)

34. (3)

35. (1)

36. (4)

37. (1)

38. (2)

39. (3)

40. (2)

41. (1)

42. (1)

43. (1)

44. (3)

45. (2)

46. (4)

47. (2)

48. (3)

49. (2)

50. (1)

51. (4)

52. (1)

53. (4)

54. (4)

55. (2)

56. (3)

57. (1)

58. (4)

59. (2)

60. (3)

61. (2)

62. (4)

63. (1)

64. (1)

65. (2)

66. (1)

67. (4)

68. (4)

69. (1)

70. (2)

71. (3)

72. (2)

73. (4)

74. (2)

75. (3)

76. (4)

77. (3)

78. (1)

79. (1)

80. (4)

81. (3)

82. (4)

83. (2)

84. (4)

85. (4)

86. (3)

87. (3)

88. (4)

89. (2)

90. (3)

91. (3)

92. (4)

93. (1)

94 (4)

95. (4)

96. (4)

97. (3)

98. (3)

99. (2)

100. (4)

101. (1)

102. (4)

103. (1)

104. (2)

105. (2)

106. (1)

107. (3)

108. (3)

109. (4)

110. (3)

111. (3)

112. (4)

113. (4)

114. (2)

115. (1)

116. (3)

117. (2)

118. (1)

119. (4)

120. (2)

121. (3)

122. (4)

123. (4)

124. (1)

125. (4)

126. (2)

127. (4)

128. (2)

129. (3)

130. (4)

131. (4)

132. (3)

133. (2)

134. (4)

135. (4)

136. (3)

137. (1)

138. (3)

139. (2)

140. (2)

141. (3)

142. (2)

143. (4)

144. (4)

145. (3)

146. (2)

147. (3)

148. (2)

149. (2)

150. (4)

151. (2)

152. (3)

153. (1)

154. (3)

155. (2)

156. (3)

157. (2)

158. (4)

159. (3)

160. (4)

161. (4)

162. (1)

163. (3)

164. (4)

165. (1)

166. (2)

167. (2)

168. (2)

169. (4)

170. (1)

171. (3)

172. (4)

173. (3)

174. (3)

175. (3)

176. (4)

177. (1)

178. (3)

179. (1)

180. (3)

All India Aakash Test Series for Medical-2019 Test - 2 (Code F) (Hints and Solutions)

2/22

PHYSICS

1. Answer (2)

Hint: 2

2

2=

1

ghv

k

R

⎛ ⎞⎜ ⎟

⎝ ⎠

Sol.: ring

2=

2

ghv ...(i)

Hollow, Sph.

2 2= =

2 51

3 3

gh ghv

...(ii)

ring

H,sp

2

52= =

62

5

6

gh

v

v gh

2. Answer (1)

Hint: 2

0

1=

2t t

Sol.: 2

1

1= (1)2

1=2

21= (3)2

9=

2

2 1

9= – = – = 4

2 2

2

1

= 8

3. Answer (4)

Hint: Conservation of angular momentum about Q.

Sol.: =i f

L L� �

HINTS AND SOLUTIONS

v

Q

P

Taking anticlockwise direction positive

2

– =2 12

mmv I � �

3 22

– =3 2 2 3

m m m

l

� � � �

– =3 12 3

4 –=

3 12

3=

4

4. Answer (1)

Hint: =L r P� �

�

Sol.: =L r P� �

�

andr r

L r L P � � �

�

= 0 = 0L r L P � � �

�

5. Answer (4)

Hint: couple

mg

Sol.: fs = F, so they constitute a couple

F3

4

a

mgf = Fs

a/2

P

N

Torque about point P.

3 2

4 2 3

a a mgF mg F ⇒

Test - 2 (Code F) (Hints and Solutions) All India Aakash Test Series for Medical-2019

3/22

maxs sf f

mg F

2

3

mgmg

2

3

6. Answer (3)

Hint: C.M.

= ��

� � �

Pv v OP

Sol.:

P

v1

vC.M.

60°30°

60°

vC.M.

O

0

C.M. 1| | = | |� � �

Pv v v

2 2

C.M. 1 1 C.M.| | = 2 cos60 �

Pv v v v v

2 2

0 1 0 1

�

Pv v v v v

As v1 =

02

R

v1 =

0

2

v

2 2

2 0 0 0

0

7| | = =

4 2 2

�

P

v v v

v v

7. Answer (2)

Sol.:

v

Velocity of center of mass decreases. To maintain

pure rolling angular velocity will also decrease. Here

direction of is clockwise.

Therefore, torque due to static friction will be

anticlockwise therefore, it will act upwards.

8. Answer (1)

Hint: 2 21 1

=2 2

K E I mv

Sol.: KE1 = KE

2

2 2 2 2

1 1 2

1 1 1=

2 2 2mR mv mv

2 2 2

1 1 2

1 1 1=

2 2 2mv mv mv

2

1

2

1=2

v

v

⎛ ⎞⎜ ⎟⎝ ⎠

1

2

1=

2

v

v

9. Answer (2)

Hint: Use conservation of angular momentum and

v2 = u2 + 2as

Sol.:

v0 v

= 0

Conservation of angular momentum about the point

of contact.

mv0R (– ˆk ) =

2

ˆ(– )2

mRmvR k

⎛ ⎞ ⎜ ⎟

⎝ ⎠

2

02

mR vmv R mvR

R

02

=3

v

v

a = g

v2 = u2 + 2as

2

20

0

4 1= – 2

9 2

vv gs

2

05

=9

vs

g

10. Answer (3)

Hint: C.M. C.M.

= �

� � �

L M r v I


4/22

Sol.:

x

zL

yM R,

v0

0

0

2 0 0

0

–72ˆ ˆ ˆ= (– ) (– ) =5 5

� v Mv RL Mv R k MR k k

R

11. Answer (1)

Hint: Use = andF ma = I �

��

.

Sol.:

A

B

T2

T2

T1

T1

100 – T1 = 10a ...(i)

T2 = 5a ...(ii)

T1R – T

2R = I

T1R – T

2R =

2

2mR a

R

T1 – T

2 =

2

ma...(iii)

T1 – T

2 = a

From equation (i), (ii) & (iii)

16 a = 100

a = 225

m/s4

= a

R

=

25

4

1

4

2= 25 rad/s

12. Answer (2)

Hint: net 1 2 3

= � � � �

Sol.:

45°

F

F 32 =F F

F

F F2 =

F F1 =

O

net

ˆ ˆ ˆ= (– ) (– ) ( ) 0 �

FR k FR k FR k

net

ˆ= –FRk�

13. Answer (3)

Hint: C.M. = 0S��

Sol.:

C.M.

C.M.

x

Initialposition

Final position

0

0

ˆ=

CPS li�

ˆ= –

PGS xi�

SCG

= ˆ= ( – )CP PG

S S l x i� �

ˆ ˆ( – ) – 2=

3

�

CM

l x im x miS

m

ml – 3xm = 0

x = 3

l

2– =

3

ll x


5/22

14. Answer (2)

Hint: Apply parallel axis theorem.

Sol.:

R

5R

I = 2 2

2(2 )5

3 2

⎡ ⎤ ⎢ ⎥⎣ ⎦

m R mRm R

I =

2 24 11

3 2

⎡ ⎤ ⎢ ⎥⎣ ⎦

mR mR

I =

2 2

28 33 41=

6 6

mR mRmR

15. Answer (1)

Hint: Apply conservation of angular momentum about

the hinged point and conservation of M.E.

Sol.: =i f

L L

2

2

03

⎛ ⎞⎜ ⎟⎝ ⎠

��

mmv = m +

03

=4

v�

Conservation of M E

Ref

21 3= 2

2 2 2

mg mgI mg � �

�

22

0

2

14

92= 3

3 16

mv

mg�

��

2

0= 8v g�

v0 = 2 2g�

16. Answer (2)

Hint: =

dl

dt

�

�

= r F �

��

Sol.:

y

x

z

(2, 0, 0)

y

0

S = 21

2ut at

21– 10(2) = – 20 m2

y

ˆ= – 20 N�

F j

ˆ ˆ= (2 – 20 ) m�

r i j

ˆ ˆ ˆ

= 2 –20 0

0 –20 0

i j k

�

ˆ ˆ ˆ= (0 – 0) – (0 – 0) (–40 – 0)i j k �

ˆ= – 40 N-m� k

17. Answer (1)

Hint: A1r1 = A

2r2

Sol.:

r1

r2

C1C C

2


6/22

2

2 2,

4 2

R RA r

A1 = R2 – A

2 = R2 –

2 23

4 4

R R

2 2

1

3

4 4 2

R R Rr

1

6

Rr

18. Answer (2)

Hint: I = Mk2 k Radius of gyration.

Sol.:

2

2 =6

MaI I

I =

2

12

Ma

2

12

Ma = Mk2

=2 3

ak

19. Answer (4)

Hint: Properties of center of mass.

Sol.: Center of mass relative to the body will not

depend on reference of frame. And it may lie inside

or outside the body.

20. Answer (1)

Hint: W = F S��

Sol.:

B

A

32 N

Let both block move together

32 = 8 a

a = 4 m/s2

fmax

= mg

fmax

= 1

3 10 =15 N2

Pseudo force on block A = ma = 4 × 3 = 12 N

There will be no slipping between the block and

12 N friction force will act on block A in forward

direction.

S = ut + 21

2at

S = 21

4 (2) = 8 m2

W = fs

× S

W = 12 × 8

= 96 JW

21. Answer (2)

Hint: Condition for stable and unstable equilibrium.

Sol.: For equilibrium,

0dU

dx

3 23

– 2 03 2

d x xx

dx

⎛ ⎞ ⎜ ⎟

⎝ ⎠

(x2 – 3x + 2) = 0

x = 1, 2

Now,

2

22 – 3

d Ux

dx

For stable equilibrium,

2

20

d U

dx

For unstable equilibrium,

2

20

d U

dx

at x = 1

2

2–1 0

d U

dx

So unstable equilibrium,

at x = 2

2

21 0

d U

dx

So stable equilibrium.


7/22

22. Answer (2)

Hint: =W F d� �

Sol.: T

mg

=2

ga

T – mg = ma

T = 3

2

mg

W = Td cos0°

3=

2

mgdW

23. Answer (1)

Hint: Conservation of mechanical energy

Sol.: At maximum extension of spring, speed of

block will be zero.

x

K

m

MEi = M.E

f

0 = 21

2kx mg

1

2kx2 = mgx

2=

mgx

k

24. Answer (1)

Hint: Conservation of energy.

Sol.:

v0

R

h

(90° – )

B

RefA

v

2 2

0

1 1= ( )

2 2mv mg R h mv

At point B

mgcos(90° – ) – N =

2mv

R

mgsin =

2mv

R[∵ N = 0]

and h = R sin

2 mgR = 1

sin sin2

mgR mgR mgR

3 2sin = 1 sin =

2 3 ⇒

H = h + R

H = R + R sin

H = R + 2

3

R

5=

3

RH

25. Answer (3)

Hint: Conservation of energy and conservation of

linear momentum.

Sol.:

2mv1

4

m

0

2

v

4

m

v0

By conservation of linear momentum of system

0

0 1

ˆ ˆ ˆ= 24 4 2

vm mv i mv i i

0

116

vv

By conservation energy of the block.

2

1

12 = 2

2m v mgh


8/22

h =

2

0

2

1

2 (16)

v

g

2

0=512

vh

g

26. Answer (3)

Hint: Use concept of coefficient of restitution.

Sol.: e = (Relative velocity of separation)

(Relative velocity of approach)

e = 2

1

v

v

v1

v2

e < 1

0 1e

27. Answer (4)

Hint: =P F v� �

�

Sol.: F = constant

0 0

∫ ∫v x

vdv dx

2 v x

½v x

P F v �

�

½P x

28. Answer (1)

Hint: Use ˆ ˆ= – –

U UF i j

x y

�

Sol.: U = –6x – 8y

= –6

U

x

= –8

U

y

ˆ ˆ(6 8 ) N �

F i j

2 2| | = 6 8 = 10 N�

F

2ˆ ˆ= = (3 4 ) m/s⇒ �

� �

F ma a i j

a = 10

2

2= 5 m/sa

x

y

(6, 4) m

Particle will move in the direction of acceleration

therefore, it will never cross y-axis and x-axis.

29. Answer (4)

Hint: Work = F ds��

= F(ds) cos

Sol.: W = F dscosWork done by static friction force may either be

positive, negative or zero. It depends on angle

between direction of force and displacement.

30. Answer (1)

Hint: Apply work–energy theorem

Wnet

= K

Sol.:

x m( )2

–5

10

6 8 10

F (N)

0

Wnet

= 1

10 2 10 4 – 2 52

Wnet

= 40 J

Wnet

= Kf – K

i

2 21 12 – 2 (2) = 40

2 2v

v2 – 4 = 40

v2 = 44

= 2 11m/sv


9/22

31. Answer (4)

Hint: W = 0 = 0F dr��

Sol.: ˆ ˆ ˆ= ( 3 – 4 ) N�

F i i bk

ˆ ˆ ˆ= (2 – 0) (2 – (–1) (3 – 1) �

r i j k

ˆ ˆ ˆ= (2 3 2 ) m �

r i i k

= 0F r�

�

ˆ ˆ ˆ ˆ ˆ ˆ(3 – 4 ) (2 3 2 )i j bk i j k = 0

6 – 12 + 2b = 0

= 3b

32. Answer (1)

Hint: Wnet

= Wmg

+ Wf + W

N

Sol.:

37°

3 m

4 m

5 m10 k

g

Wnet

= Wmg

+ Wf + W

N

Wmg

= mg × 3 cos0° = 10 × 10 × 3 = 300 J

WN

= N × 5 cos 0°

WN

= 0

Wf

= N (5cos180°) = mgcos37°(–1) × 5

= –1 4

10 10 54 5 = –100 J

Wnet

= 300 – 100 net

= 200 JW

33. Answer (3)

Hint: Use dW = F dr��

Sol.: dW = ˆ ˆ ˆ ˆ( ) ( ) x y

F i F i dx i dyi

=x y

dW F dx F dy∫ ∫ ∫

W =

3 3

1 0

2xdx y dy∫ ∫

W =

32

32

0

1

9 1= – 9

2 2 2

xy

⎡ ⎤ ⎡ ⎤⎡ ⎤ ⎢ ⎥ ⎢ ⎥⎣ ⎦

⎣ ⎦⎣ ⎦= 4 + 9 = 13 J

34. Answer (3)

Hint: Concept of constraint equation.

Sol.:

P

T

aP

T/2

T/2

Q

T

Let movable pulley displaces by 'x' and block P by 'y'

l1 + l

2 = l

1 + x + l

2 + x – y

y = 2x

aP = 2a

Q

35. Answer (1)

Hint: fmax

= s(m

1 + m

2)g

Sol.:

100 N10 kg

5 kg fk

fk

Maximum force for common acceleration

fmax

= 0.4 × 15 × 10 = 60 N

Force applied is greater than 60 N. So friction will be

kinetic.

fk =

kN = 0.4 × 50 = 20 N

2100 – 208 m/s

10a

36. Answer (4)

Hint: For equilibrium Fnet

= 0

Sol.:

200 N

T1

T2

T4

T2

100 N

P

T1cos

T1sin

T1cos

T1sin T

2

T3

10 kg

100 N

T3

20 kg

200 N

T2



(F.B.D. ofpoint )P

P


10/22

T1cos = 100 N

T1sin = 200 N

tan = 2

–1

= tan 2

2 2 2

1= (100) (200)T

1=100 5 NT

37. Answer (1)

Hint: Fmin

= 2

1

mg

; when = tan–1()

Sol.: Fmin

= 2

1

mg

=

2

4 40010 10

3 3=

54

1 33

⎛ ⎞ ⎜ ⎟⎝ ⎠

Fmin

= 80 N

38. Answer (2)

Hint: Fnet

= 0, fk = N

Sol.:

37°

A

B

f

f

Net force on A will be zero

Wsin 37° = fk

3= cos37

5

BW m g

3 4= 3

5 5W W

1=4

39. Answer (3)

Hint: Use F = ma

Sol.:

mB P A

10 N

x = 5 m x = 2 m x = 0

T T

Total mass of rope

M = 5 × 2 = 10 kg

2101m/s

10 a

Mass of length 3 m

m = 3 × 2 = 6 kg

T = ma = 6 N

40. Answer (2)

Hint: Concept of static and kinetic friction.

Sol.:

F = 25 N

QP

1maxsf

2maxsf

1 1max 1

= = 0.4 5 10s sf m g

1max

= 20Nsf

2 2max 2

= = 0.5 10 10 = 50Ns sf m g

1 2

max max appAs

s sf f F

So, system will be at rest

F.B.D of P

25 N

N

1maxs

fP

20 + N = 25

N = 5 N

F.B.D. of Q

Nfs Q

N = fs

= 5 Nsf

41. Answer (1)

Hint: Properties of the friction force.

Sol.: fs Static friction

0 fs f

s max

0 Ns sf

flim

= s

N


11/22

42. Answer (1)

Hint: Use concept of constraint equation.

Sol.:

9 kg

30°

10 kg

60°

T

T

a1

a2

T cos60° = 10 a2

T = 20 a2

...(i)

9g – T = 9a1

90 – T = 9a1

...(ii)

a2cos60° = a

1

a2 = 2a

1

T = 40a1

...(iii)

From equation (ii) & (iii),

49a1 = 90

or, a1 � 1.8 m/s2

43. Answer (1)

Hint: Use concept of pseudo force

Sol.:

A B

F.B.D. of B w.r.t. cart

NAB

m aB B

NAB

= 5 × 2

NAB

= 10 N

F.B.D. of A w.r.t. cart

NAB

NWA

m aA

NWA

= mAa = N

AB = 10 × 2 + 10 = 30 N

30= = 3 :110

WA

AB

N

N

44. Answer (3)

Hint: Translational equilibrium

F = 0

Sol.:

k

20 kg

T

2

T

2

T

4

T

4

T

T = 200 N

4

T= k x

1000x = 200

4

x = 1

m20

= 5 cmx

45. Answer (2)

Hint: Apply F = ma

Sol.:

30°

10 kg

5 kg

T1

T2

T2

5 kg

T1

aa

a

100 – T1 = 10a

T1 – T

2 – 50 sin30° = 5a

T1 – 50 sin30° = 5a

20 a = 50

a = 5

2 ms–2

T1 = 100 –

50

2

T1 =

150

2N

T2 =

55 25

2

T2 =

75 N

2

1

2

150= = 2

2

75

2

T

T


12/22

CHEMISTRY

46. Answer (4)

Hint: G = G0 + 2.303RT log Qp

Sol.: QP =

2 5

2 2

2 2N O

2 5 2

N O

P 2= =1

P P 2 1

G = G0 + 2.303 RT log10

(1) = G0 = –43 kJ/mol

47. Answer (2)

Hint: Lewis acids are electron pair acceptors.

Sol.: B2H

6 is an electron deficient species.

48. Answer (3)

Hint: Salt having minimum value of 'solubility (mole/l)'

will be least soluble.

Sol.: For binary electrolyte, lower the value of Ksp

,

lower will be its solubility.

49. Answer (2)

Hint: pH of acidic buffer pH = pKa + log

Salt

Acid

⎡ ⎤⎢ ⎥⎣ ⎦

Sol.:

CH3COOH + NaOH CH

3COONa + H

2O

(mmol)initial

10 6

(mmol)final

4 0 6

Resulting solution is an acidic buffer

pH = 4.74 + 6

log4

⎛ ⎞⎜ ⎟⎝ ⎠

= 4.74 + 0.18 = 4.92

50. Answer (1)

Hint:

Enthalpy of

= 3 ×

Enthalpy of hydrogenation –

Resonance hydrogenation of benzene of cyclohexane

energy

Sol.:

+ H2 ; H = –x kJ mol–1

+ 3H2

Enthalpy of hydrogenation of benzene

= 3H – Resonance energy.

= 3(–x) – (–y)

= (y – 3x) kJ mol–1

51. Answer (4)

Hint: Vapour pressure of a liquid depends on

temperature.

52. Answer (1)

Hint: Overall enthalpy of a process, is the sum of

the enthalpies of all steps involved.

Sol.:

(i) C(s) + O2(g) CO

2(g) + 94 kcal/mol

(ii) H2(g) +

1

2O

2(g) H

2O(l) + 68 kcal/mol

(iii) CH4(g) + 2O

2(g) CO

2(g) + 2H

2O(l) + 213 Kcal/mol.

On applying, (i) + 2 × (ii) – (iii)

Reaction obtained is

C(s) + 2H2(g) CH

4(g)

H = (–94) + 2(–68) –(–213) = –17 kcal.

53. Answer (4)

Hint: S = 2 1

p

1 2

T PnC ln + nRln

T P

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Sol.: S = 1

2

PnR ln

P

⎛ ⎞⎜ ⎟⎝ ⎠

(At constant temperature)

S = 1

2

PnR ln

P

⎛ ⎞⎜ ⎟⎝ ⎠

= 1 × R × 2

ln0.5

⎛ ⎞⎜ ⎟⎝ ⎠

= R ln 4 = 2R ln2

54. Answer (4)

Hint: U = q + w and w = –2.303 nRT log2

1

V

V

⎛ ⎞⎜ ⎟⎝ ⎠

Sol.: w = –2.303 × 2 × 8.314 × 300 log20

5

⎛ ⎞⎜ ⎟⎝ ⎠

= –2.303 × 2 × 8.314 × 300 × log4

= –2.303 × 2 × 8.314 × 300 × 2 × 0.30

= – 6892.9 J � –6.9 kJ

U = q + w

q = –w (for isothermal process)

q = +6.9 kJ

55. Answer (2)

Hint: H2O(l) � H

2O(v) (1 atm, 100°C)

U = q + W

Sol.: q = nHVapourization

q = 1 × 40.66 × 103 J = 40.66 × 103 J


13/22

w = – ngRT

= –1 × 8.314 × 373 = –3101.12 J

U = q + w = (40.66 × 103 – 3101.12) J

= 37558.878 J = 37.56 × 103 J

56. Answer (3)

Hint: Fusion

2 2

Fusion

n HH O(s) H O(l), S =

T

⎛ ⎞⎜ ⎟

⎝ ⎠�

Sol.: S =

3

–1 –16 10= 21.98 Jmol K

273

At equilibrium, G = 0

57. Answer (1)

Hint:

Partial pressure = (Total pressure) × Mole fraction of gas

Sol.:

2O

20n = = 0.625

32

CO

2.8n = = 0.1

28

Mole fraction oxygen gas = 0.625

0.1 0.625

= 0.625 625

=0.725 725

= 25

29

⎛ ⎞⎜ ⎟⎝ ⎠

pO2

= 25

29

⎛ ⎞⎜ ⎟⎝ ⎠

× 2 = 50

atm29

Partial volume of oxygen gas = 25

1029

250

29 L

58. Answer (4)

Hint: For H2 gas, value of a is negligible.

Sol.:

m2

m

aP + (V – b) = RT [ a is negligible]

V

⎛ ⎞⎜ ⎟⎝ ⎠

∵

P(Vm

– b) = RT

PbZ =1+

RT

59. Answer (2)

Hint: Number of mole = Volume at STP(ml)

22400

Sol.: For A vapours, mole = 112 1

mol22400 200

Molar mass of A = 0.48 × 200 = 96 g

2

A(g)

H

r 2 1 1= = =

r 96 48 4 3

60. Answer (3)

Hint:

P

L G(L + G)

V

Sol.: At temperature TC and pressure P

2, gas phase

exists.

61. Answer (2)

Hint: Partial pressure of a gas

= Mole fraction × Total pressure

Sol.: Moles of 2

xO

32 (Assume, X is mass of

each gas mixed together)

Moles of x

He4

Total moles x x 9

x32 4 32

pHe

x

84P P

9 9x

32

Fraction of total pressure exerted by He = 8

9.

62. Answer (4)

Hint: pH of salt of weak base and strong acid is

b

17 – (pK log C)

2

⎧ ⎫⎨ ⎬⎩ ⎭

.


14/22

Sol.:

4 4 2NH OH HCl NH Cl H O

Initial mol, 0.1 V 0.1 V

Final mol, 0 0 0.1 V

⎧ ⎫⎨ ⎬⎩ ⎭

Concentration of NH4Cl =

–2

0.1= 5 ×10

2

pH = –2

17 – 4.75 log(5 ×10 )

2

= 17 – 4.75 – 2 log5

2

= 5.2755 � 5.28

63. Answer (1)

Hint: Kp = K

c(RT)ng.

Sol.: If ng = 0

Kp = K

c

For reaction N2(g) + O

2(g) � 2NO(g), n

g = 0

64. Answer (1)

Hint: Pure water is neutral and in pure water,

[H+] = [OH–] = w

K

Sol.: For acidic solution, [H+] > w

K

For basic solution, [OH–] > w

K

65. Answer (2)

Hint: For reaction

A(g) � B(g), KC =

[B] 0.02= = 2

[A] 0.01

Sol.:

A(g) � B(g)

At new eq. 0.02 – x 0.02 + x

KC =

0.02 + x= 2

0.02 – x

0.02 + x = 0.02 × 2 – 2x

3x = 0.02

x = 0.02

= 0.006673

[B] = 0.02 + 0.0067 = 0.0267

66. Answer (1)

Hint: For reaction 2A(g) � B(g) + C(g),

Equilibrium constant KC =

2

[B][C]

[A]

Sol.: 2A(g) � B(g) + C(g)

0.2 0 0

0.2 – 2x x x

KC =

2

2= 0.09

(0.2 – 2 )

x

x

= 0.3(0.2 – 2 )

x

x

x = 0.06 – 0.6x

1.6x = 0.06

x = 0.06

1.6

Concentration of B is 0.06

1.6 10

⎛ ⎞⎜ ⎟⎝ ⎠

= 3.75 × 10–3 M.

67. Answer (4)

Hint: Due to common ion effect, main source of Cl–

in solution is strong electrolyte NaCl.

Sol.: Ksp

(AgCl) = [Ag+] [Cl–]

1.6 × 10–10 = [Ag+] × 10–2

[Ag+] = 1.6 × 10–8

68. Answer (4)

Hint: Salt of weak acid and strong base will undergo

anionic hydrolysis.

Sol.: CH3COONa + H

2O CH

3COO– (aq) + Na+(aq)

CH COO3

+ H2O � CH

3COOH + OH (aq)

69. Answer (1)

Hint: [H+]Resulting

(Normality) = 1 1 2 2

1 2

N V +N V

V V

Sol.: [H+] = 0.1 2 V + 0.1×1× V

V + V

= 0.2 + 0.1 0.3

= = 0.152 2

pH = – log[H+] = – log(15 × 10–2)

= 2 – log(15) = 0.824 0.82�


15/22

70. Answer (2)

Hint: Ksp

of Al(OH)3 = [Al3+] [OH–]3

Sol.: Al(OH)3(s) � Al3+(aq) + 3OH–(aq)

S 3S

∵ pH = 9, Hence, pOH = 5 [OH–] = 10–5 M

3S = 10–5 S =

–5

10

3 M

Ksp

(Al(OH)3)= S(3S)3 = 27S4 =

4–5

1027

3

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

=

–20 –20

27 10 10=

27 3 3

⎛ ⎞⎜ ⎟ ⎝ ⎠

71. Answer (3)

Hint: Salt having minimum value of Ksp

will give ppt

first.

Sol.: Since, ppt appearance order is

AgI AgBr AgCl

Hence, Ksp

(AgI) < Ksp

(AgBr) < Ksp

(AgCI)

So, z < y < x

72. Answer (2)

Hint: KC = [CO

2]

Sol.: CaCO3(s) � CaO(s) + CO

2(g)

KC = [CO

2]

0.5 2

mol.= [CO ]

⎛ ⎞⎜ ⎟⎝ ⎠�

moles of CO2 = 0.5 × 5 = 2.5

73. Answer (4)

Hint: SReaction

= (S)Product

– (S)Reactants

(:coefficients of reactants and products)

Sol.:Sreaction

= 6 × 100 – (2 × 30 + 5 × 60)

= 600 – (60 + 300) = 600 – 360 = 240

GReaction

= HReaction

– TSReaction

for spontaneous reaction

GReaction

< 0

HReaction

< TSReaction

300 × 103 < T × 240

T >

3300 10

240

T > 1250 K

74. Answer (2)

Hint: Greater the atomicity of gas, greater is the

entropy.

Sol.: Correct order of entropy is

He < O2 < CO

2 < PCl

3

75. Answer (3)

Hint: Standard boiling point of water is 99.6°C

76. Answer (4)

Rate of Volume effused=

effusion Time taken

⎛ ⎞⎜ ⎟⎝ ⎠

1

Molar mass

Sol.:

2

gas

H

r

r

= 2 2 2

2

2

gas

H H Hgas

H gas gas gas

H

V

M t Mt= = =

V M t M

t

=

gas

1 2=

8 M

Mgas

= 64 × 2 = 128 u

77. Answer (3)

Hint: Interaction energy of dipole-dipole force in

stationary polar molecule 3

1

r

78. Answer (1)

Hint: Conjugate acid is formed by addition of one H+

ion to the given species.

Sol.: Conjugate acid of 2– –

4 2 4HPO is H PO

79. Answer (1)

Hint: Unit of Kp : g

n

(atm) .

Sol.: For reaction PCl5(g) � PCl

3(g) + Cl

2(g),

ng = 1

So, unit of Kp

is atm

80. Answer (4)

Hint: A3B

2(s) + H

2O � 3A2+(aq) + 2B3–(aq)

K = [A2+]3 [B3–]2

Sol.: K = (3s)3(2s)2 = 27s3 × 4s2 = 108 s5

s =

1/5

K

108

⎛ ⎞⎜ ⎟⎝ ⎠

81. Answer (3)

Hint: pH = a

17 (pK – pK )

2b

Sol.:pH of 0.1 M CH3COONH

4

pH = 1

7 (pKa – pK )2

b

pH = 7 (∵ pKa = pK

b)


16/22

82. Answer (4)

Hint: pH = – log[H+]

Sol.: [OH–] obtained from 0.1 N NaOH = 10–1 M

pOH = 1

pH = 13

83. Answer (2)

Hint: Acidic buffer is a mixture of weak acid and its

salt with strong base.

Sol.: (CH3COOH + CH

3COONa) Acidic buffer

84. Answer (4)

Hint: Keq

=

Coefficient

Coefficient

[Product]

[Reactant]

Sol.: 2 5 2 2 eq

1N O (g) 2NO (g) O (g), K = 0.5

2�

2 2 2 54NO (g) O (g) 2N O (g), � K

eq=

2

eq

1

K

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

= 1

0.5 × 0.5 = 4

85. Answer (4)

Hint: Active mass of solid is unity.

Sol.: Calcium oxide is solid so its active mass is

unity.

86. Answer (3)

Hint: In an isolated system, mass as well as energy

do not exchange with surrounding.

Sol.: Universe is an example of isolated system.

87. Answer (3)

Hint: G = H – TS

For non-spontaneous reaction, G > 0

Sol.: If H > 0 and S < 0

G > 0 (For all temperature)

88. Answer (4)

Hint:H = U + ngRT

H > U (if ng > 0)

Sol.:ng of (i) Reaction = –5

ng of (ii) Reaction = 0

ng of (iiii) Reaction = 0

ng of (iv) Reaction = +1

Hence, for reaction (iv) H > U

89. Answer (2)

Hint: Density of ideal PM

gas(d) =RT

Sol.: d = PM 1.2 atm × 32 (g/mol.)

=RT L-atm

0.0821 600KK-mol.

⎛ ⎞⎜ ⎟⎝ ⎠

= 0.779 � 0.78 g/L

90. Answer (3)

Hint: KE T

Sol.: Average molar kinetic energy is 3RT

2

⎛ ⎞⎜ ⎟⎝ ⎠

BIOLOGY

91. Answer (3)

Sol.: Strawberry is an aggregate fruit.

Mulberry & jackfruits are composite fruits or multiple

fruits.

Pomegranate is a simple fleshy or succulent fruit.

92. Answer (4)

Hint: Aleurone layer is a part of endosperm.

Sol.: It is triploid, provides nourishment and seen in

maize grains.

93. Answer (1)

Sol.: Root hairs arise from epidermal cells of

maturation zone.

94. Answer (4)

Hint: Thorns can be seen in Citrus or Bougainvillea.

Sol.: Spines are modification of leaves.

95. Answer (4)

Hint: This plant belongs to the family Fabaceae.

Sol.: It is Indigofera.

96. Answer (4)

Hint: In members of family Poaceae, the seed coat

is fused with fruit wall (Caryopsis type of fruit).

Sol.: Gram is a dicot. Wheat and maize are

monocots and belong to family Poaceae.

97. Answer (3)

Hint: Papilionaceous corolla shows vexillary

aestivation.

Sol.: Mango is a drupe. Vexillum is largest posterior

petal in pea also called standard petal.


17/22

98. Answer (3)

Hint: Fused carpels— syncarpous

Sol.: Staminode is sterile anther. Twisted aestivation

is seen in petals of cotton. Unisexual flowers are

seen in maize.

99. Answer (2)

Hint: In racemose inflorescence the terminal flower

is absent.

Sol.: It (main axis) has unlimited growth and flowers

ever borne in acropetal manner.

100. Answer (4)

Hint: Banana is a triploid plant hence cannot form

gametes.

Sol.: Banana fruit develops without fertilisation of

gametes, hence it is a parthenocarpic fruit. Banana

has suckers.

101. Answer (1)

Hint: In opposite phyllotaxy, a pair of leaves arise at

each node opposite to each other.

Sol.: Guava, Calotropis have opposite phyllotaxy.

Mustard, China roseAlternate phyllotaxy

Sunflower

⎤⎥⎦

Alstonia, Nerium ] Whorled phyllotaxy.

102. Answer (4)

Hint: In monadelphous condition, all stamens are

united in a single bundle but anthers remain free. It

is found in family Malvaceae.

Sol.: In pea flower the stamens are united in two

bundles so it is diadelphous.

103. Answer (1)

Hint: Given floral formula belongs to family

Solanaceae.

Sol.: Solanaceae members have persistent calyx,

superior ovary with many ovules. Seeds are

endospermic. Flowers are bisexual and

actinomorphic.

104. Answer (2)

Hint: In some seeds nucellus persists in the form of

a layer.

Sol.: Perisperm is persistent nucellus. Seeds having

perisperm are called perispermic seeds.

105. Answer (2)

Hint: Axile placentation has placenta in the axial

position and ovules are attached to it.

Sol.: Tomato & lemon have axile placentation. Pea

has marginal, mustard has parietal & sunflower and

marigold have basal placentation.

106. Answer (1)

Hint: Albuminous seed contains endosperm.

Sol.: Coconut has endospermic seed where seed

coat is thick & non-membranous.

Groundnut, pea, and soyabean have ex-albuminous

(non-endospermic) seeds.

107. Answer (3)

Hint: Phylloclade is a modified stem.

Sol.: In Opuntia, phylloclade is flattened and

photosynthetic.

108. Answer (3)

Hint: Stem bears nodes, internodes and buds.

Sol.: Potato is a tuber which is a modified

underground stem because it bears buds.

109. Answer (4)

Hint: Pneumatophores help a plant to get oxygen

from the atmosphere.

Sol.: They grow vertically upward hence negatively

geotropic & can be seen in Rhizophora which grow

in marshy/swampy areas.

110. Answer (3)

Hint: Carrot & Turnip are modifications of tap root.

Sol.: Ginger is a modified underground stem. Edible

part of sweet potato is a modified adventitious root.

111. Answer (3)

Hint: Tap root is the primary root.

Sol.: Tap roots (Primary root) arise from radicle

whereas fibrous roots arise from the base of the

stem.

112. Answer (4)

Sol.: Bordeaux mixture is first fungicide discovered

by RMA Millardet. It is commonly known as holy

water of plant pathology.

113. Answer (4)

Hint: Cyanobacteria evolve O2 during photosynthesis.

Sol.: Cyanobacteria have enzyme nitrogenase which

fixes atmospheric nitrogen. Some cyanobacteria

form symbiotic associations with ferns. Such as

Anabaena with Azolla.

Nostoc is a filamentous BGA.


18/22

114. Answer (2)

Hint: Prokaryotes lack nuclear membrane &

membrane bound cell organelles.

Sol.: Streptococcus is a prokaryote. Remaining

organisms are eukaryotes.

115. Answer (1)

Hint: Chemoautotrophic bacteria lack photosynthetic

pigments.

Sol.: ChemoautotrophicNitrosomonas

Nitrobacter

⎤⎥⎦

PhotosyntheticNostoc

Chlorobium

⎤⎥⎦

Lactobacillus] Heterotrophic.

116. Answer (3)

Hint: Mycoplasma are smallest living organisms.

Sol.: They lack cell wall and can survive without

oxygen.

Methanogens are responsible for production of

biogas.

117. Answer (2)

Hint: Archaebacteria contain pseudomurein in their

cell wall.

Sol.: The cell wall of eubacteria is made up of

mainly peptidoglycan. They are prokaryotes hence

they lack membrane bounded cell organelles.

Archaebacteria include halophiles, methanogens and

thermoacidophiles.

Archaebacteria have introns in their genetic material.

118. Answer (1)

Hint: Heterocyst is related to N2 fixation which takes

place in anaerobic condition.

Sol.: In heterocyst cell wall is impermeable to

oxygen but permeable to nitrogen. It lacks PS-II but

PS-I remains active and produces ATP required for

N2-fixation.

119. Answer (4)

Hint: Protists are unicellular eukaryotes.

Sol.: Eukaryotes have well defined nucleus and

other membrane bound cell organelles.

120. Answer (2)

Hint: Euglenoids lack cell wall.

Sol.: Slime moulds are heterotrophic (Saprophytic).

Protozoans are predators & parasites. Chrysophytes

include diatoms and desmids which are mostly

photosynthetic.

121. Answer (3)

Hint: Cell wall of diatoms is indestructible due to

presence of silica.

Sol.: Due to presence of silica, the fossilized

diatoms get deposited at the sea bed to form

diatomaceous earth.

Diatoms are microscopic, unicellular protists which

float passively in water current.

122. Answer (4)

Hint: Paramoecium has two nuclei.

Sol.: Paramoecium is a ciliated protozoan

123. Answer (4)

Hint: Dinoflagellates are flagellated, mostly marine

photosynthetic protist.

Sol.: Dinoflagellates have two flagella, one is

transverse and another is longitudinal.

124. Answer (1)

Hint: Mycorrhiza is a mutually beneficial association.

Sol.: Mycorrhiza is a symbiotic association between

roots of higher plants & fungi.

125. Answer (4)

Hint: All viruses have capsid.

Sol.: Bacterial viruses are known as bacteriophages,

they have capsid (protein coat) & genetic material,

usually dsDNA.

126. Answer (2)

Hint: Viruses are obligate intracellular parasites.

Sol.: Viruses can multiply within the host cell only.

They cannot respire, divide and grow externally.

127. Answer (4)

Hint: Prions lack genetic material.

Sol.: Prions are proteinaceous infectious particles. It

causes mad cow disease in cattles & scrapie

disease in sheep.

128. Answer (2)

Hint: A virus is larger than sub viral agents/particles.

Sol.: Viroids are infectious RNA particles. They lack

protein. They cause diseases in plants only.

Viruses have proteinaceous capsid.

129. Answer (3)

Hint: Phycobiont is algal partner in lichen

association. They are either green algae or blue-

green algae.

Sol.: Algal partner synthesizes food for fungal

partner. Lichens do not grow in air polluted areas.

Animals store their food in form of glycogen or fat.


19/22

130. Answer (4)

Sol.: Bladderwort is a partially heterotrophic

insectivorous plant.

131. Answer (4)

Hint: Rhizopus is a conjugation fungus, also known

as bread mould.

Sol.: It reproduces by non-motile asexual spores and

non-motile zygospores.

It is a saprophytic fungus with chitinous cell wall.

132. Answer (3)

Hint: Deuteromycetes are called imperfect fungi.

Sol.:Morels

AscomycetesAspergillus

⎤⎥⎦

Bracket fungi] Basidiomycetes

Deuteromycetes

Alternaria

Colletotrichum

Trichoderma

⎤⎥⎥⎥⎦

133. Answer (2)

Hint: Claviceps purpurea causes ergot of rye.

Sol.: Claviceps belongs to the class Ascomycetes.

It is a sac fungus which possesses septate

mycelium and chitinous cell wall.

It reproduces by conidia which are non-motile

asexual spores.

134. Answer (4)

Hint: On basidium exogenously produced spores are

basidiospores.

Sol.: Basidiospores are produced by members of

basidiomycetes. These are haploid and sexual

spores.

135. Answer (4)

Hint: The body of fungi is not differentiated into root,

stem & leaves.

Sol.: The body of fungi is thalloid & haploid (n).

Each cell contains one set of chromosomes.

136. Answer (3)

Hint: Disorder associated with accumulation of dust

particles.

Sol.: Pneumoconiosis is occupational disease of

lungs due to inhalation of dust, characterised by

inflammation, coughing and fibrosis.

137. Answer (1)

Hint: pCO2 in alveoli is 40 mmHg.

Sol.:

Atmospheric air

Alveoli Deoxy-generated

blood

Oxy-genated

blood

Tissue

pO2 159 104 40 95 40

pCO2 0.3 40 45 40 45

138. Answer (3)

Hint: This wave indicates ventricular repolarisation.

Sol.: T-wave represents the return of ventricles from

excited to normal state. QRS complex represents

depolarisation of ventricles while P-wave represents

atrial depolarisation.

139. Answer (2)

Hint: Ischemic conditions reflect oxygen deficiency

in cardiac muscle fibres.

Sol.: When heart muscle is suddenly damaged by

inadequate blood supply, the condition is called heart

attack.

140. Answer (2)

Hint: This hormone is released in response to

increase in blood pressure than normal.

Sol.: During dehydration (profuse sweating)

osmolarity of body fluid increases, increased

secretion of ADH from hypothalamus occurs. GFR

decreases during profuse sweating.

141. Answer (3)

Hint: Blood vessel exiting from major osmoregulatory

organ of human body.

Sol.: Largest amount of urea is present in hepatic

vein.

142. Answer (2)

Hint: Pressure difference on both sides of filtration

membrane.

Sol.: NFP = (GHP) – (BCOP + CHP)

= 60 – (30 + 20)

= 10 mmHg

143. Answer (4)

Hint: Sympathetic stimulation alters heart rate.

Sol.: Sympathetic neural signals increase cardiac

output and stroke volume but decrease duration of

cardiac cycle. Cardiac muscles are not under the

control of our will.


20/22

144. Answer (4)

Hint: Innermost lining of wall of blood vessels is

constituted by squamous cells.

Sol.: Arteries have narrower lumen than veins and

mostly carry oxygenated blood.

145. Answer (3)

Hint: Maximum amount of air which can be inspired

or expired forcefully.

Sol.: Vital capacity is measured for lung function

test by an instrument known as spirometer. With

exception of RV, TLC, FRC. All other lung capacities

and volumes can be measured by spirometer.

146. Answer (2)

Hint: High threshold substance like glucose is

completely absorbed in kidney in a normal healthy

person.

Sol.: Diabetes insipidus occurs due to deficiency of

ADH (vasopressin) from hypothalamus. All types of

nephrons have peritubular capillary networks. GFR in

healthy individual is 125 ml/minute or 180 L/day.

147. Answer (3)

Hint: Ascending limb of loop of Henle is permeable

to salt.

Sol.: Descending limb is called concentrating

segment of loop of Henle as it is permeable to water

and impermeable to salts.

148. Answer (2)

Hint: These are region between medullary pyramids

in kidney.

Sol.: Invagination of cortex into medulla forms

columns of Bertini which divide medullary region of

kidney into renal pyramidals.

149. Answer (2)

Hint: This hormone is released by atrial walls.

Sol.: Vasa recta acts as countercurrent exchanger

because they exchange water for ions. ANF

opposes RAAS.

150. Answer (4)

Hint: Well developed region of JG nephrons

extending into medulla in JG nephrons.

Sol.: Loop of Henle of both cortical and

juxtamedullary nephron lie in medullary pyramid.

151. Answer (2)

Hint: Identify the machine use to measure an ECG.

Sol.: Electrical activities of heart is measured by an

instrument known as electrocardiograph and graph

obtained is known as electrocardiogram.

Stethoscope is used to hear heart sounds.

152. Answer (3)

Hint: Duration of each auricular cycle is equal to

duration of the cardiac cycle.

Sol.: During each cardiac cycle, an auricular cycle

as well as a ventricular cycle is completed. Each

cycle has duration of 0.8 sec. So time of auricular

cycle is 0.8 sec not 0.6 second.

153. Answer (1)

Hint: 2/3rd filling of ventricles occurs when all

chambers of heart are relaxing.

Sol.: When auricles are in diastole auriculoventricular

valves remain open but semilunar valves present at

opening of aorta are closed.

154. Answer (3)

Hint: During this phase, blood is not pumped from

ventricles to aorta.

Sol.: During isovolumetric contraction, both

auriculoventricular valves and semilunar valves remain

closed. Pressure within ventricles increases but

amount of blood remains unchanged in isovolumetric

contraction.

155. Answer (2)

Hint: Volume of blood pumped by each ventricle in

0.8 seconds.

Sol.: Cardiac output = heart rate x stroke volume.

Hypertension is observed if blood pressure exceeds

140/90 or more repeatedly. Opening of pulmonary

artery is guarded by semilunar valves.

156. Answer (3)

Hint: Volume of ventricle decreases during their

contraction.

Sol.: During ventricular systole, closure of cuspid/AV

valves occurs. Blood enters ventricles during atrial

systole.

157. Answer (2)

Hint: Stretch receptors are activated upon filling of

urinary bladder beyond threshold.

Sol.: Hemodialysis is needed when kidneys

malfunction. Urine formation is independent of stretch

reflex.

158. Answer (4)

Hint: ADH acts on this region to prevent diuresis.

Sol.: The collecting duct is responsible for secretion

of urea not for absorption. Various nephrons open

into collecting ducts. Collecting duct secretes urea

into interstitium to increase osmolarity for

reabsorption of water.


21/22

159. Answer (3)

Hint: Maximum percentage of CO2 is transported in

plasma in bicarbonate form.

Sol.: 77% CO2 is transported through blood plasma.

Dissolvedin plasma – 7%

In the form ofbicarbontates

in plasma – 70%

Plasma(77%)

160. Answer (4)

Hint: Affinity of Hb for O2 is inversely proportional to

P50

value.

Sol.: Shift of curve B towards right indicates

increase in P50

value and thus decreased affinity of

haemoglobin for oxygen, leading to greater

dissociation of oxygen from hemoglobin.

161. Answer (4)

Hint: Identify a layer absent in alveolar capillaries.

Sol.: Capillaries lack tunica media formed by

smooth muscle fibres. So, diffusion membrane does

not contain muscular tissue.

162. Answer (1)

Hint: Lymphatic fluid is poured into venous blood.

Sol.: Lymph from right thoracic duct drains into right

subclavian vein whereas left thoracic duct drains

lymph into left subclavian vein.

163. Answer (3)

Hint: This heart sound is produced during ventricular

systole.

Sol.: First heart sound 'lub' is produced during

ventricular systole. It has low frequency of about

25–45 Hz and is of comparatively longer duration

(0.15 sec) in comparison to second heart sound

which has higher frequency than 50 Hz but duration

is only 0.12 sec.

164. Answer (4)

Hint: Identify the blood group of universal donors.

Sol.: Universal donors (O–) and individuals with

blood group B– and O+ can donate blood to the

patient.

165. Answer (1)

Hint: WBCs which participate in allergic reaction

during worm infestation.

Sol.: Eosinophils are WBCs that stain with acidic

dye eosin and have bilobed nucleus. They participate

in a number of allergic reactions and fight against

worm infestation.

166. Answer (2)

Hint: These cells give rise to enucleated platelets.

Sol.: Megakaryocytes are specialised large cells in

red bone marrow which divide to form cellular

fragments lacking nucleus known as blood platelets/

thrombocytes.

167. Answer (2)

Hint: This hormone secreted by JG cells stimulates

RAAS.

Sol.: Renin secreted by JG cells in response to low

B.P. stimulates RAAS to maintain GFR. Angiotensin-

II is a vasoconstrictor. Rennin digests casein in

infant stomach.

168. Answer (2)

Hint: Enzyme which is released in body injured site

in presence of Ca+2.

Sol.: Prothrombinase/thrombokinase is also

responsible for conversion of inactive plasma protein

prothrombin into an active enzyme thrombin

responsible for conversion of fibrinogen into fibrin.

169. Answer (4)

Hint: This membrane divides body cavity into

thoracic and abdominal cavity.

Sol.: Pneumotaxic centre is called "switch-off" point

of inspiration. It stimulates neurons of expiratory

centre located in medulla. Diaphragm is stimulated

by respiratory rhythm centre in medulla.

170. Answer (1)

Hint: Dialysis helps to counter uremia.

Sol.: Dialysing fluid is isotonic to blood plasma.

171. Answer (3)

Hint: These animals live on both land and water.

Sol.: Crocodiles are reptiles with completely divided

four chambered heart. Amphibians have three

chambered heart.

172. Answer (4)

Hint: The nitrogenous waste product which requires

minimum loss of water in its elimination.

Sol.: Land snails excrete uric acid whereas most

terrestrial organisms excrete either urea (humans,

amphibians) or uric acid (insects, birds, reptiles).

Aminotelic organisms excrete amino acid (Unio and

echinoderms)


22/22

173. Answer (3)

Hint: Lungs are situated in cavity above diaphragm.

Sol.: Trachea divides into two primary bronchi at the

level of 5th thoracic vertebra and enters lungs.

174. Answer (3)

Hint: TV + IRV + ERV = TLC – RV

Sol.: TLC – RV = VC,

EC = ERV + TV = 1100 + 500 = 1600 ml

175. Answer (3)

Hint: Fluid filled cavity is present around lungs.

Sol.: Pleura is divided into two layers. Outer pleural

membrane is in close contact with thoracic lining

whereas inner pleural membrane is in contact with

lung surface.

176. Answer (4)

Hint: Squamous epithelium is found at this surface.

Sol.: Gaseous exchange takes place in the alveoli

in lungs.

177. Answer (1)

Hint: Excretory organ in annelids.

Sol.: Respiration in earthworm (annelids) occurs

through moist cuticle.

178. Answer (3)

Hint: These cells are granular, phagocytic and most

abundant type of WBCs.

Sol.: Differential leukocyte count i.e., DLC reveals

percentage of type of WBCs in blood.

Neutrophils – 60–65%

Lymphocytes – 20–25%

Monocytes – 6–8%

Eosinophils – 2–3%

179. Answer (1)

Hint: Part of nephron exhibiting brush border

appearance.

Sol.: ADH facilitates facultative/conditional

reabsorption of water from DCT & collecting tubule.

PCT is responsible for obligate reabsorption and is

internally lined by simple cuboidal epithelium

containing microvilli giving brush border appearance

which increase surface area for reabsorption.

180. Answer (3)

Hint: The primary organ involved in this process is

kidney.

Sol.: Removal of metabolic waste products from the

body is known as excretion. Main role of sweating

is to facilitate thermoregulation. Defaecation involves

removal of undigested and unabsorbed food.

� � �

Documents

for Medical-2019 All India Aakash Test Series for Medical ... · Test - 2 (Code E) (Hints and Solutions) All India Aakash Test Series for Medical-2019 5/22 Sol.: 37° 3 m 4 m 5 m