for Medical-2020 All India Aakash Test Series for Medical ... · Test - 4 (Code-C) (Answers & Hints) All India Aakash Test Series for Medical-2020 5/16 31. Answer (4) Hint : Both

Test - 4 (Code-C) (Answers) All India Aakash Test Series for Medical-2020

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1. (4)

2. (2)

3. (4)

4. (3)

5. (3)

6. (2)

7. (2)

8. (4)

9. (4)

10. (3)

11. (4)

12. (2)

13. (3)

14. (1)

15. (2)

16. (2)

17. (2)

18. (2)

19. (3)

20. (4)

21. (2)

22. (2)

23. (Delete)

24. (3)

25. (3)

26. (4)

27. (2)

28. (4)

29. (4)

30. (2)

31. (4)

32. (4)

33. (3)

34. (1)

35. (4)

36. (3)

Test Date : 30/12/2018

ANSWERS

TEST - 4 (Code-C)

All India Aakash Test Series for Medical-2020

37. (2)

38. (3)

39. (3)

40. (2)

41. (2)

42. (4)

43. (4)

44. (2)

45. (3)

46. (2)

47. (4)

48. (2)

49. (3)

50. (4)

51. (2)

52. (4)

53. (4)

54. (2)

55. (2)

56. (4)

57. (4)

58. (3)

59. (4)

60. (3)

61. (2)

62. (4)

63. (3)

64. (3)

65. (3)

66. (2)

67. (2)

68. (3)

69. (2)

70. (3)

71. (4)

72. (3)

73. (3)

74. (2)

75. (2)

76. (1)

77. (3)

78. (2)

79. (4)

80. (4)

81. (3)

82. (3)

83. (3)

84. (1)

85. (3)

86. (1)

87. (1)

88. (4)

89. (4)

90. (1)

91. (3)

92. (3)

93. (4)

94. (4)

95. (1)

96. (4)

97. (3)

98. (3)

99. (4)

100. (3)

101. (3)

102. (3)

103. (4)

104. (3)

105. (3)

106. (4)

107. (2)

108. (4)

109. (2)

110. (4)

111. (3)

112. (2)

113. (3)

114. (3)

115. (2)

116. (2)

117. (3)

118. (4)

119. (3)

120. (3)

121. (3)

122. (2)

123. (3)

124. (1)

125. (2)

126. (4)

127. (3)

128. (3)

129. (2)

130. (3)

131. (3)

132. (2)

133. (3)

134. (3)

135. (4)

136. (2)

137. (4)

138. (4)

139. (3)

140. (3)

141. (4)

142. (2)

143. (2)

144. (2)

145. (4)

146. (2)

147. (3)

148. (3)

149. (4)

150. (4)

151. (2)

152. (3)

153. (1)

154. (3)

155. (3)

156. (4)

157. (2)

158. (1)

159. (3)

160. (2)

161. (2)

162. (4)

163. (3)

164. (1)

165. (1)

166. (2)

167. (3)

168. (3)

169. (4)

170. (3)

171. (1)

172. (4)

173. (4)

174. (2)

175. (1)

176. (2)

177. (4)

178. (2)

179. (3)

180. (1)

All India Aakash Test Series for Medical-2020 Test - 4 (Code-C) (Answers & Hints)

2/16

ANSWERS & HINTS

1. Answer (4)

Hint : = mgx

Sol. :

60°

u

x

y

xomg

At t = 2 s

1cos 20 2 20 m

2 x u t

= mgx = 2 × 10 × 20 = 400 Nm

2. Answer (2)

Hint : I = mr 2

Sol. : Moment of inertia depends on axis of rotation,

mass of the body and distribution of mass about

axis of rotation.

3. Answer (4)

Hint : F + f = macm

Sol. :

f

R

F

Ov

F + f = macm

...(i)

cm

= Icm

(F – f)R = mR2

or F – f = mR = macm

...(ii) (∵ acm

= R)

From (i) & (ii)

f = 0

4. Answer (3)

Hint : L r p �

� �

Sol. : L�

about O is constant as r⊥ and v both are

constant.

[ PHYSICS]

5. Answer (3)

Hint : r F �

��

Sol. :

P(1, 2)

O(0, 0)

r

ˆ4F k

ˆ ˆ(– – 2 ) mr i j�

ˆ4 �

F k

�

��

r F

ˆ ˆ4( 2 ) �

i j Nm

6. Answer (2)

Hint : If ext

= 0 then L = constant

If ext

F

��

= 0 the cm

0a �

Sol. : For a rigid body rotating about a fixed axis L

is always parallel to .

Torque produced by central force is always zero.

7. Answer (2)

Hint : vp = 2v

0 cos (/2).

Sol. :

O60° v

0

r v = 0

120°

Here = 120°

vnet

= 2v0cos60°

vnet

= v0

8. Answer (4)

Hint : Formula for moment of inertia of any

symmetrical part of a body = Formula for moment of

inertia of body.

Sol. : 2

Semidisc Disc2

MRI I

9. Answer (4)

Hint : = ISol. : = II = mk 2

= mk 2

Test - 4 (Code-C) (Answers & Hints) All India Aakash Test Series for Medical-2020

3/16

10. Answer (3)

Hint : Mechanical energy is conserved.

Sol. : In pure rolling on inclined plane potential

energy is converted into translational and rotational

kinetic energy. Net work done by friction and normal

force is zero.

11. Answer (4)

Hint : �

� �

L r p

Sol. : �

L is always prependicular to �

r and �

p .

12. Answer (2)

Hint : In pure rolling on fixed surface net tangential

acceleration of point of contact is zero.

Sol. :

Ov

R2

Rv

For pure rolling with constant angular velocity.

v = R

Radial acceleration of point is R2.

13. Answer (3)

Hint : 2

21

kt

R

⎛ ⎞ ⎜ ⎟

⎝ ⎠

Sol. : 2

2

k

Ris same for disc and solid cylinder.

14. Answer (1)

Hint : Conservation of angular momentum.

Sol. : Li = Lf

Li = mr2

Lf = (mr2 + 2mr2)

2 23mr mr

3

15. Answer (2)

Hint : Conservation of angular momentum about the

point of contact.

Sol. :

Pure rolling

0

v

Applying conservation of angular momentum about

the point of contact,

Li = L

f

mR20 = mR2 + mvR

2 0

02

2

RmR mvR v

⇒

16. Answer (2)

Hint and Sol. : Kepler’s law of area is based on

conservation of angular momentum.

17. Answer (2)

Hint : and � �

� � ��

L r p r F

Sol. : Net torque about C is non zero and net

torque about O is zero.

∴ L is conserve about O but variable about C.

18. Answer (2)

Hint : L r p �

� �

Sol. :

v

y

x

v cos

r H =

L = rp

r = H

L = H mv cos

vL mv

g

2 2

sincos

2

mv

g

3 2

sin cos

2

19. Answer (3)

Hint : .P ��

Sol. : ˆ ˆ ˆ( – 3 2 ) Nmi j k �

ˆ ˆ ˆ(4 2 5 ) rad/si j k �

ˆ ˆ ˆ ˆ ˆ ˆ( – 3 2 ) (4 2 5 )P i j k i j k P = 4 – 6 + 10

P = 8 W

20. Answer (4)

Hint : Use parallel axis theorem.

Sol. : L/2

rL

O


4/16

2 2

2

3 12

ML MLI Mr

2

2 5

4

Lr

2

5

3 ML

I

21. Answer (2)

Hint : Fraction of kinetic energy which is rotational

2

2 2.

k

k R

Sol. :

2

2

2

21

k

Rf

k

R

for solid sphere 2

2

2

5

k

R .

2 / 5% 100 28%

21

5

22. Answer (2)

Hint and Sol. : I = mk2.

23. Answer (Delete)

24. Answer (3)

Hint : net r t

a a a � � �

Sol. :

ar

v

at

anet

25. Answer (3)

Hint : , d d

dt dt

Sol. : = 3t2 – 4t + 3

6 – 4d

tdt

= 6 rad/s2

at = R, a

r = R2

At t = 1 s = 2 rad/s

at = 6 m/s2 a

r = 4 m/s2

2 2 2

net6 4 52 m/sa

26. Answer (4)

Hint : Use Newton’s law of gravitation.

Sol.: To make M in equilibrium resultant of F1 and

F2 should be equal in magnitude and opposite in

direction to T.

∴ From the figure,

1

2

tan30F

F 30°

30°

F1

F2

m1

T

m2

∴

1

2

1

3

m

m

27. Answer (2)


Sol. :

1 = 0

U1 + K

1 = U

2 + K

2

210 cos

2 2 2

L LMg mg I

3(1– cos )

g

L

3

2 g

L

28. Answer (4)

Hint and Sol. : I = mr2.

29. Answer (4)

Hint and Sol. : MI is same for any symmetrical part

of body.

I1 = I

2 = I

3

30. Answer (2)

Hint : rot total

1for a disc.

3K KE

Sol. : Total KE = mg h = mgl sin.

So, rot

1sin

3K mgl .


5/16

31. Answer (4)

Hint : Both particles rotate about their centre of mass.

Sol. : Both particles are always opposite to each

other along the line joining them so 1 = 2.

32. Answer (4)

Hint : Conservation of mechanical energy.

Sol. :

H

v = 0

u

U1 + K1 = U2 + K2

20 0

0

– –10

2

GMm GMmm u

R R H

Now, 2 2 0

0 0

1 1

2 4 2E

GMmm u m v

R

H = R

33. Answer (3)

Hint : 2

E

GMv

R

Sol. : 32 4

3

E

Gv R

R

28

3

Ev GR

E

v R

1

2 2

E E E

p p p

v R

v R

34. Answer (1)

Hint : Angular momentum is conserved.

Sol. : r is minimum at point A and maximum at

point C.

So, KA > K

B > K

C

35. Answer (4)

Hint and Sol.: Total energy of bound system is

negative.

36. Answer (3)

Hint : g = g – R2 cos 2Sol. : = 90° for pole g

pole = g(always) if = 0,

ge = g i.e., increases.

37. Answer (2)

Hint : orbital

1V

r .

Sol. : Orbital speed is independent of mass of

satellite.

38. Answer (3)


Sol. : 2 2

0

1 1– –

2 2 ( )

GMm GMm

mv mvR R h

2 2

0

1 1–

2 21

mghmv mv

h

R

2

0

2–

1

⎛ ⎞⎜ ⎟⎝ ⎠

ghv v

h

R

39. Answer (3)

Hint : 2 2 2

interstellar–1

E Ev v v v n

Sol. : v = nvE

n = 3

infinity 9 – 1 8 2 2 E E E

v v v v

40. Answer (2)

Hint : –f i

E E E

Sol. : –

2

GMmE

r

1 1 1 1– –

2 2 8 14i f E E

E GMm GMmr r R R

⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦

7 – 4 3

56 56

⎡ ⎤ ⎢ ⎥

⎣ ⎦E E

GMmGMm

R R

41. Answer (2)

Hint : L r p �

� �

Sol. : L = m0vr

r = R + h

0

GMv

r

0 0( ) ( )

GML m R h m GM R h

R h


6/16

42. Answer (4)

Hint : 2

1

⎛ ⎞⎜ ⎟⎝ ⎠

h

gg

h

R

Sol. : gh = g /2

2

1 1

21

h

R

⎛ ⎞⎜ ⎟⎝ ⎠

2 – 1h R

43. Answer (4)

Hint : 1n

Fr

Sol. : 2

k mvF

r r

v r°

44. Answer (2)

Hint : vara = v

prp

Sol. : vara = v

prp

2 2

a a p pr r

2

pa

p a

r

r

⎛ ⎞ ⎜ ⎟ ⎝ ⎠

ra = a + ae r

p = a – ae

2

1

1–

p

a

e

e

⎛ ⎞ ⎜ ⎟ ⎝ ⎠

45. Answer (3)

Hint : Mechanical energy and angular momentum

are conserved.

Sol. : Wgravity

= KE

0– GMm

PEr

.

46. Answer (2)

Hint: In a compound algebraic sum of oxidation

numbers of all atoms is zero.

Sol. : NH3 x + 3(+1) = 0 x = – 3

HNO3 (+1) + x + 3(–2) = 0 x = + 5

N2 x = 0

N2H

4 2(x) + 4(+1) = 0 x = – 2

N3H 3x + 1 = 0 x = –1/3

–3 –1/3 0 5

2 33 3NH < N H < N < HNO

47. Answer (4)

Hint: Balance the reaction by ion electron method.

Sol. : 2+ 2– + 3+ 3

2 7 26Fe + Cr O + 14H 6Fe 2Cr 7H O

Sum of coefficient of reactants = 6 + 1 + 14 = 21

48. Answer (2)

Hint: Ozone is better oxidising agent than H2O

2.

Sol.: O3 O

2 + [O]

H2O

2 + [O] H

2O + O

2

H2O

2 is oxidised here.

49. Answer (3)

Hint: When O.N. of an element in a compound is

intermediate, then compound can act as oxidant and

reductant both.

[ CHEMISTRY]

Sol. : HNO2 Oxidation number of N = +3

H2O

2 Oxidation number of O = –1

H2S Oxidation number of S = –2

SO2

Oxidation number of S = +4

50. Answer (4)

Hint: 2– 2– + 3+

2 7 2 4 2 2Cr O +3C O + 14H 2Cr +7H O + 6CO

Sol. :

1 mole Cr2O

72– requires = 3 mole C

2O

42–

So, 1 mole C2O

42–

requires 2–

2 7

1mole Cr O

3

51. Answer (2)

Hint: Sum of the oxidation numbers of all atoms in

a compound is zero.

Sol. : K4[Fe(CN)

6] 4(+1) + x + 6(–1) = 0

x = +2

Fe (CO)5

x + 5(0) = 0 x = 0

[Fe(H2O)

6]Cl

3 x + 6 x (0) + 3 x (–1) = 0

x = +3

52. Answer (4)

Hint: Electrode with higher o

redE value will act as

cathode.


7/16

53. Answer (4)

Hint: o o ocell cathode anodeE E E

Sol. : o o ocell cathode anodeE E E = 2 2

o o

Cu /Cu Zn /ZnE E

= +0.34 – (–0.76) = 1.10 V

54. Answer (2)

Hint: Sum of O.N. of all atoms in a compound is

zero.

Sol. : A2 (BC

4)3

[2(+3) + 3 [(+6) + 4(–2)] = 0

55. Answer (2)

Hint: Ksp

= s2

Sol. : –3

0.07175 10s

143.5

= 5 × 10–7 mol L–1

–

ss

AgCl Ag Cl �

Ksp

= s × s = s²

= (5 × 10–7)2 = 2.5 × 10–13 mol2 L–2

56. Answer (4)

Hint: For sparingly soluble salt, Ksp

<< 1.

Sol. : 1

2spAB s= K

1

3sp2

KAB s=

4

⎛ ⎞ ⎜ ⎟

⎝ ⎠

1

4sp3

KAB s=

27

⎛ ⎞ ⎜ ⎟

⎝ ⎠

1

5sp2 3

KA B s=

108

⎛ ⎞ ⎜ ⎟

⎝ ⎠

57. Answer (4)

Hint: With increase in OH concentration, pH will

increase.

Sol. :

(i)

1 12 100 1 100

10 10N

200

= 0.05 [H+]

So, pH < 7

(ii)

1 12 100 1 100

10 5N200

= 0 [Neutral]

So, pH = 7

(iii)

1 12 100 1 100

5 5N

200

+200.1(H )

200

So, pH 7

(iv)

1 11 100 2 100

2 10N

200

–

300.15(OH )

200

So, pH > 7

58. Answer (3)

Hint: When equations are added, then their

equilibrium constants are multiplied

Sol. : A � 2B K1 ...(i)

B � C+D K2 ...(ii)

2C + 2D � E K3 ...(iii)

2 (ii) + (i) + (iii)

A � E, Keq

= K1 K

22 K

3,

On reversing, E � A

eq'K =

2

1 2 3

1

K K K

59. Answer (4)

Hint: wh

a b

KK =

K K

Sol. : Salt of (WA + WB) will have higher KH value.

60. Answer (3)

Hint: Acidic buffer contains weak acid & its salt with

strong base.

Sol. : HClO4 is a strong acid.

61. Answer (2)

Hint: Larger is the Keq

, greater is the extent of

reaction.

Sol. : If Keq

<< 1, then (Reactant) >> (Product)

62. Answer (4)

Hint:

2w

h

a

K chK =

K 1 h

,

Sol. : On increasing the concentration of salt, ‘h’

decreases. On adding base, common ion effect is

applied therefore, h decreases and on increasing

temperature Kw and K

a changes.

63. Answer (3)

Hint: a b

1 1pH 7 pK pK

2 2

Sol. : pH of salt of weak acid and weak base is

concentration independent.


8/16

64. Answer (3)

Hint: In presence of common ion solubility decreases

while in complex formation solubility increases.

Sol. : More the concentration of common ion, lesser

will be the solubility.

3 4 1 2 3 3

(Complex formation) No common ion

0.1 M NH (s ) Water (s ) 0.1 M HCl (s ) 0.2 M AgNO (s )

65. Answer (3)

Hint : G° = –2.303RTlogKeq

Sol.: G° = –2.303 × R × 298 × log10

= –2.303 × 298 × R

66. Answer (2)

Hint:

½ ½2 2

c

(H ) (I )K =

(HI)

Sol. : 2 2

1 IHI H I

2 2

1 0 0

12 2

��⇀↽��

½ ½

2 2

C

(H ) (I )K =

(HI)=

½ ½

2 2

(1– )

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

2(1 )

0.5 1

2(1 0.5) 2

67. Answer (2)

Hint: Ksp

= [Ba2+][OH–]2

Sol. : 2

2Ba(OH) Ba 2OH

s 2s

��⇀↽��

Ksp

= (s) (2s)2 5.0 × 10–7 = 4s3

3 –75s = 10

4

= 1.25 × 10–7

s = 5 × 10–3

[OH–] = 2s = 2 × 5 × 10–3 = 10–2

pH = 14 – pOH = 12

68. Answer (3)

Hint: Thermoneutral reactions (H = 0) are

temperature independent.

Sol. : On increasing pressure the equilibrium is

shifted towards the lesser number of gaseous moles.

69. Answer (2)

Hint: Eq. mass = molar mass

n-factor

Sol. : 3

2

–6eN 2N

2

14 2 14Eq. mass

6 3(N )

70. Answer (3)

Hint: [H+]total

= [H+]acid

+ [H+]H2O

Sol. : pH = 5 [H+] = 10–5 M

N1 V

1 = N

2 V

2

10–5 × 1 = N2 × 1000

N2 = 10–8

Total H+ = H+ ion from acid + H+ ion from H2O

[H+] = 10–8 + 10–7 = 1.1 × 10–7

pH = –log(1.1 × 10–7) = 6.95

71. Answer (4)

Hint: [H+][OH–] = 10–14

Sol. :

[OH–] = 1

–14+ –1410

[H ] = 101

pH = 14

72. Answer (3)

Hint: Lewis acid is a substance which can accept

a pair of electrons.

Sol. : Al in AlCl3 and Sn in SnCl

4 have vacant d-

orbitals

In CO2, oxygen atoms are more

electronegative than the carbon atom so

carbon atom become electron deficient.

73. Answer (3)

Hint: Keq

depends only on temperature.

Sol. : On changing concentration Keq

remains same.

74. Answer (2)

Hint: For an exothermic reaction, on increasing

temperature Keq

decreases.

Sol. : Increasing temperature in exothermic reaction,

rate of forward & backward both reaction increases,

but rate of backward reaction increases more.


9/16

75. Answer (2)

Hint: Active mass of solid substance remains

constant.

Sol. : 3 2

p NH H SK (P )(P )

On adding 3

3 NHNH ,P increases and

2H S

P decreases.

76. Answer (1)

Hint: Conjugate base is obtained by removing a

proton of the species.

Sol. : 2

4SO

is conjugate base of 4

HSO .

77. Answer (3)

Hint: PB = P

C and P

E = P

F.

Sol. : A(s) B(g) C(g)��⇀↽��

PB = P

C P

B + P

C = 80 atm

PB = P

C = 40 atm

1P

K 40 40 = 1600 atm2

D(s) E(g) F(g)�

PE = P

F P

E + P

F

= 40

atm

PE = P

F = 20 atm

2P

K 20 20 400 atm2

1

2

P

P

K 1600= = 4

K 400

78. Answer (2)

Hint: Electron deficient species are Lewis acid.

Sol. : B2H

6 is a Lewis acid.

79. Answer (4)

Hint: w

1pH = pK

2

Sol. : 14

w

1 1pH = pK log(2 10 ) 6.85

2 2

80. Answer (4)

Hint. : 4

Acidic Salt Salt Salt

(SA WB) (SB WA)

pH 7 pH 7 pH 7 pH 7

HCl NH Cl NaCl NaCN

81. Answer (3)

Hint: Applicable only for conjugate acid & base pair.

Sol. : HCOOH – Acid

HCOO– – Conjugate base

82. Answer (3)

Hint: In presence of common ion solubility of salt

decreases.

Sol. : + –

+ –

Ag CN Ag + CN

s s

KCN K + CN

0.02 M 0.02 M

�

Ksp

= [Ag+] [CN–]

1.0×10–16 = (s) (s+0.02) s(0.02)�

–161.0 10

s =0.02

= 5 × 10–15 M

83. Answer (3)

Hint: For precipitation, Ksp

< Ionic product (IP)

Sol. :

(i)

–4 –6 –10–102 10 10 10

I.P 0.5 102 2 2

⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

(ii)

–5 –5–1010 2 10

I.P 0.5 102 2

⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

(iii)

–4 –4–1010 10

I.P 25 102 2

⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

84. Answer (1)

Hint: For an acidic buffer solution a weak acid its

salt with strong base is taken.

Sol. : (H3PO

4 + NaH

2PO

4), (NaH

2PO

4 + Na

2HPO

4)

and (Na2HPO

4 + Na

3PO

4)

85. Answer (3)

Hint: Salt of weak acid & strong base undergoes

anionic hydrolysis.

Sol. : HCOOH, CH3COOH and HCN are weak acid.

86. Answer (1)

Hint: pH of acidic buffer is given by

a

saltpH pK log

acid

⎛ ⎞ ⎜ ⎟⎝ ⎠

Sol. :

Initial, meq,

final, meq, = 100

==


10/16

[ BIOLOGY]

91. Answer (3)

Sol. : DNA sequence, chemical nature of proteins,

crystals & aromatic compounds are used in

chemotaxonomy by scientists to resolve confusions

in classification.

92. Answer (3)

Hint : Natural system of classification is based on

morphology, anatomy, embryology & phytochemistry.

Sol. : Phylogenetic system of classification includes

evolutionary relationships between organisms, as a

basis of classification.

93. Answer (4)

Hint : Oogamous type of sexual reproduction

involves fusion between non-motile female gamete

and motile male gamete in some green and brown

algae.

Sol. : Ulothrix and Spirogyra have isogamous type of

reproduction whereas Volvox and Polysiphonia have

oogamous reproduction but Polysiphonia has non-

motile male gametes.

94. Answer (4)

Hint : Porphyra is a red alga.

Sol. : The stored food of red algae is floridean starch.

95. Answer (1)

Hint : Peptidoglycan is constituent of bacterial cell

wall.

Sol. : In chlorophyceae outer layer of cell wall is

made up of pectose, while inner layer is constituted

by cellulose.

96. Answer (4)

Hint : Pear shaped zoospores & pyriform motile

gametes are found in brown algae.

Sol. : Brown algae are found mostly in marine

habitats.

a

SaltpH pK log

Acid

⎛ ⎞ ⎜ ⎟⎝ ⎠

–3

3

100 10

800= 4.74 log

200 10

800

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

= 4.74 – 0.30 = 4.44

87. Answer (1)

Hint: KP = K

C (RT)ng

Sol. : KP = K

C (RT)ng

– nC

P

K(RT) g

K

(i)1C

P

K(RT)

K

(ii)C

P

K1

K

(iii)–1C

P

K= (RT)

K

(iv)–2C

P

K= (RT)

K

88. Answer (4)

Hint: Adding inert gas at constant volume has no

effect on equilibrium.

89. Answer (4)

Hint: eq

H llogK logA

2.303R T

⎛ ⎞ ⎜ ⎟⎝ ⎠

Sol. : For exothermic reaction.

Hslope ( )ve

2.303R

log Keq

1/T

90. Answer (1)

Hint: 1 1 2 2

1 2

N V + N VN =

V + V

Sol. : pH = 2 pH = 3

N1 = 10–2 N

2 = 10–3

–2 –310 × V +10 × V

N =V + V

0.011=

2= 55 × 10–4

4pH = – log(55 10 ) 2.26


11/16

97. Answer (3)

Hint : Funaria is a moss.

Sol. : In mosses, leafy gametophytes have

multicellular and branched rhizoids.

98. Answer (3)

Hint : In liverworts, sporophytes are diploid

structures which develop from embryo and are

dependent on gametophytes.

Sol. : All bryophytes, including Marchantia are

homosporous.

99. Answer (4)

Sol. : A - Dictyota (Brown alga)

B - Porphyra (Red alga)

C - Selaginella (Pteridophyte)

D - Salvinia (Aquatic fern)

100. Answer (3)

Sol. : Bryophytes have gametophytic main plant body.

Pteridophytes, gymnosperms and angiosperms have

sporophytic main plant body.

101. Answer (3)

Hint : In bryophytes, the sporophyte is parasitic over

gametophyte.

Sol. : Members of pteridophytes may be

homosporous (one kind of spores) or heterosporous

(two kinds of spores).

102. Answer (3)

Hint : Adiantum is a fern which belongs to the class

Pteropsida.

Sol. : Psilopsida – Psilotum

Lycopsida – Selaginella

Sphenopsida – Equisetum

103. Answer (4)

Hint : Angiosperms are called flowering plants.

Sol. : Occurrence of triploid endosperm, double

fertilisation and presence of fruits are exclusive

features of angiosperms.

104. Answer (3)

Hint : Bryophytes have haplo-diplontic life cycle.

Sol. : Fucus, Cycas and mango all have diplontic life

cycle. Funaria being a bryophyte has haplo-diplontic

life cycle.

105. Answer (3)

Hint : It is group of antiparasitic drugs that expel

parasitic worms, i.e., helminths hence called

anthelmintics or vermifuges.

Sol. : Dryopteris is used to obtain anthelmintic drug

to treat helmenthiasis.

106. Answer (4)

Hint : Red wood tree (Sequoia) – tallest gymnosperm

Sol. :

Polysiphonia shows haplo-diplontic life cycle

pattern whereas Volvox has haplontic life cycle

pattern.

Statements (a) & (d) are correct.

107. Answer (2)

Hint : Stems are usually branched in orders

Coniferales and Gnetales of gymnosperms.

Sol. : Cedrus – branched stem.

Cycas – unbranched stem.

108. Answer (4)

Hint : Pteridophytes are vascular plants with

dominant diploid sporophyte.

Sol. : In pteridophytes, the gametophyte is short-

lived, haploid and independent.

109. Answer (2)

Hint : Gymnosperms have archegonia in their ovule.

Sol. : Pinus has 2-8 archegonia inside its ovule,

which is absent in the ovule of papaya as it is an

angiosperm.

110. Answer (4)

Hint :

Cycas

Eucalyptus

(gymnosperm) = nEndosperm in

(angiosperm) = 3n

⎡⎢⎣

Sol. : Wolfia is an angiosperm, within ovules of

which are present highly reduced female

gametophytes i.e., embryo sacs.

111. Answer (3)

Hint : All bryophytes are called amphibians of plant

kingdom.

Sol. : Mosses like Funaria, have sporophyte which

is partially dependent on gametophyte.

112. Answer (2)

Sol. : Primary protonema is a filamentous structure

develops from spore germination in Funaria.

113. Answer (3)

Sol. : Antherozoids of bryophytes are biflagellated.

114. Answer (3)

Hint : In red algae, the major pigments are chl a and

d, carotenoids and xanthophylls.

Sol. : Porphyra is a red alga, while Fucus, Dictyota

& Laminaria are brown algae.


12/16

115. Answer (2)

Hint : Carrageen is sulphated polysaccharide.

Sol. : Sulphated polysaccharides or hydrocolloids

are found in the cell wall of rhodophyceae.

116. Answer (2)

Sol. : Brown alga – Sargassum,

Green alga – Chara,

Red alga - Gracilaria

117. Answer (3)

Hint : Pteridophytes are the first vascular

embryophytes.

Sol. : Dryopteris – Pteridophyte

Funaria – Bryophyte

Cycas – Gymnosperm

Mangifera - Angiosperm

118. Answer (4)

Sol. : Gametophytes of mosses have multicellular

and branched rhizoids.

119. Answer (3)

Hint : Homosporous pteridophytes have monoecious

gametophyte.

Sol. : Heterospory in pteridophytes is precursor to

the seed habit.

120. Answer (3)

Hint : Coralloid roots of Cycas are irregular & do not

possess root hairs.

Sol. : Coralloid roots of Cycas are symbiotically

associated to Anabaena (cyanobacteria).

121. Answer (3)

Hint : Gymnosperms have integumented ovules.

Sol. : In ovules of gymnosperms, nucellus is

protected by single integument.

122. Answer (2)

Hint : Endosperm represents female gametophyte in

gymnospermic seed.

Sol. :

Plumule, radicle, suspensor & cotyledons

represent future sporophyte.

Testa, tegmen & perisperm represent parental

sporophyte.

123. Answer (3)

Hint : Ephedra, Gnetum & Welwitschia are placed in

the most advanced order of gymnosperms.

Sol. : Ephedra is placed in order Gnetales.

124. Answer (1)

Sol. : Tonoplast, membrane of sap vacuole is a

selectively permeable membrane.

125. Answer (2)

Hint : Water/solvent always moves from lower DPD

to higher DPD of solution.

Sol. : OP = 13TP = 5DPD = 8

AOP = 6TP = 2DPD = 4

B

Low DPDHigh DPD

126. Answer (4)

Hint : Most of the absorbed water is lost through

stomata of the leaves.

Sol. : Less than 1 percent of water, reaching the

leaves is used in photosynthesis & plant growth.

127. Answer (3)

Hint : Elements most readily mobilised are

phosphorus, nitrogen & potassium.

Sol. :

Mobile elements are frequently remobilised from

older parts to younger parts.

A C4 plant loses only half as much water as a

C3 plant for the same amount of CO

2 fixed.

128. Answer (3)

Hint : Water in the adjacent xylem moves into the

phloem by osmosis.

Sol. : Sucrose transport in the living phloem sieve

tube cells occurs through osmotic pressure gradient.

Loading of sucrose into sieve tube cells is an active

process.

129. Answer (2)

Hint : Water potential is chemical potential of water

molecules which regulates water movement as water

always moves from high water potenitial to low water

potential.

Sol. : Water potential of a cell = solute potential +

pressure potential.

Water potential of pure water is zero at atmospheric

pressure but by adding solutes it decreases and

become negative.

130. Answer (3)

Hint : In both facilitated diffusion and active transport,

movement of molecules occur with the help of

membrane proteins.

Sol. : Both facilitated diffusion and active transport,

show saturation of the transport and response to

protein inhibitors.


13/16

Facilitated diffusion is a downhill process as it does

not require ATP while active transport is an uphill

process.

131. Answer (3)

Hint : Osmotic pressure of an electrolyte is higher

than a non-electrolyte.

Sol. : A cell which does not show any change in

0.7 M sugar solution because it is isotonic, would

show decreased volume of cell due to 0.7 M NaCl

solution by exosmosis. As 0.7 M NaCl solution due

to dissociation has double osmotic pressure than

0.7 M sugar solution, hence become hypertonic in

comparison to cytoplasm of cell.

132. Answer (2)

Hint : In dry atmosphere when the relative humidity

is low, a plant has higher rate of transpiration.

Sol. : In humid atmosphere i.e., at high relative

humidity, the rate of transpiration decreases.

133. Answer (3)

Hint : Opening of stomata starts with incomplete

oxidation of carbohydrate.

Sol. :

Step I – In light, starch in guard cells is

incompletely oxidised into phosphoenol pyruvate

(PEP).

Step II – H+ from guard cells are transported to the

epidermal cells.

Step III – K+ ions are absorbed into guard cell.

Step IV – Increased K+ & malate ion form

potassium malate & store it in vacuoles.

134. Answer (3)

Hint : The chief sinks for the mineral elements are

the growing regions of the plants.

Sol. : Mature leaves are the chief source of

mobilised minerals while young leaves, developing

flowers, fruits, seeds, apical & lateral meristems are

the chief sinks for the mineral elements.

135. Answer (4)

Hint : Guttation is the process which occurs in early

morning and night while transpiration occurs in day

time.

Sol. : Due to transpiration a negative pressure is

generated in xylem vessels while guttation occurs

due to positive root pressure.

136. Answer (2)

Hint : The structure is responsible for propulsion/

whip like movement in tail of male gamete.

Sol. : Male gamete called sperm in humans is

flagellated. Flagella is the propulsion equipment that

pushes the sperm towards the ovum. Amoeboid

movement shown by leucocytes involves

pseudopodia formation.

137. Answer (4)

Hint : Filaments of this protein run close to the

F-actin throughout its length.

Sol. : Troponin - Tropomyosin complex masks the

myosin binding site on actin. Troponin binds to

calcium which results in unmasking of myosin

binding site on actin.

138. Answer (4)

Hint : Leaky channels are always open.

Sol. : Efflux of K+ is essential for repolarisation,

which occurs through opening of K+ voltage gated

channels.

139. Answer (3)

Hint : All muscle fibres contain actin and myosin.

Sol. : Smooth muscle fibres appear non-striated as

actin and myosin filaments are not regularly arrayed

along the length of the cell.

140. Answer (3)

Hint : Identify an enzyme.

Sol. : Acetylcholinesterase breaks acetylcholine in

synaptic cleft into acetate and choline.

141. Answer (4)

Hint : Actin filaments slide towards the M-line when

a sarcomere contracts.

Sol. : Thin and thick filaments in a myofibre are

composed of actin and myosin respectively. Thin

filaments slide over thick filaments during muscle

contraction.

142. Answer (2)

Hint : Myosin head has more affinity for ATP than for

actin.

Sol. : Cross bridge between actin and myosin will

not break until a new ATP attaches to myosin head.

143. Answer (2)

Hint : Dark meat is rich in myoglobin.

Sol. : White meat has fast contracting, glycolytic/

anaerobic white muscle fibres. They have more

developed sarcoplasmic reticulum than red muscle

fibres. Red muscle fibres rich in myoglobin have more

mitochondria than white muscle fibres.


14/16

144. Answer (2)

Hint : Extension is observed in contractile fibres.

Sol. : Nerve fibres do not exhibit contractility and

elasticity.

145. Answer (4)

Hint : A common collagenous connective tissue

layer holds together a number of muscle bundles.

Sol. : Each muscle bundle is called fascicle. Many

fascicles are together wrapped around by fascia.

Sarcolemma is plasma membrane of a muscle fibre.

Endomysium is connective tissue covering of

individual muscle fibre.

146 Answer (2)

Hint : Deficiency of estrogen adds to the increased

chances of fracture in this case.

Sol. : Bones become weak due to demineralisation

caused by enhanced activity of bone dissolving cells

called osteoclasts.

Rickets is observed in children in case of calcium

deficiency. Accumulation of uric acid crystals in

joints is seen in gouty arthritis while rheumatoid

arthritis is an autoimmune disorder.

147. Answer (3)

Hint : In humans, the 11th and 12th pair of ribs are

not attached ventrally to sternum.

Sol. : Some mammals like sloth and manatee have

9 and 6 cervical vertebrae respectively. Cranium in

humans is composed of eight bones. There are

usually 14 phalanges in a limb of a human.

148. Answer (3)

Hint : Muscles associated with heart are striped.

Sol. : All types of muscles have actin and myosin.

Both skeletal and cardiac muscle fibres are striated/

striped. Longer refractory period belongs to cardiac

muscle fibres.

149. Answer (4)

Hint : These are most common type of synovial

joints in humans.

Sol. : In humans, carpo-carpal and tarso-tarsal joints

are gliding joints; atlas-axis joint is a pivot joint; saddle

joint is present between carpal and metacarpal of

thumb; cartilaginous joints are found in between

vertebrae.

150. Answer (4)

Hint : Sternum is placed on same side as human

heart.

Sol. : Sternum is placed ventrally in the human body.

Myasthenia gravis affects skeletal muscles of the

human body.

151. Answer (2)

Hint : Select the triangular flat bone that is also

called shoulder bone.

Sol. : Scapula and clavicle of both side together form

pectoral girdle. Pelvic girdle involves innominate/coxal

bones.

152. Answer (3)

Hint : 8 carpals are arranged equally in two rows in

one wrist.

Sol.: Cranium has 8 bones in humans fixed together

by sutures. Carpals interact with each other through

gliding joints. Carpals are 16 while tarsals are 14 in

number. 2 coxal bones form pelvic girdle while 4

bones form pectoral girdle. 26 vertebrae occur in

vertebral column of an adult and 30 bones are

present in an upper limb of man.

153. Answer (1)

Hint : It is a supporting bone of lower limb.

Sol. : Fibula does not interact with femur at knee

joint. Fibula interacts with tibia at two points.

154. Answer (3)

Hint : Hammer, anvil and stirrup shaped bones are

part of middle ear in man.

Sol. : Incus is anvil shaped bone of middle ear that

interacts with stapes and malleus. Human skull has

two occipital condyles through which it interacts with

atlas vertebra.

155. Answer (3)

Hint : Coordination is the process through which two

or more organs interact and complement the

functions of one another.

Sol. : The neural system provides an organised

network of point to point connections for a quick

coordination. The endocrine system provides

chemical coordination through hormones.

156. Answer (4)

Hint : The neural organisation is very simple in lower

invertebrates.

Sol. : Poriferans, simplest invertebrates lack neural

system. In coelenterates such as Hydra, it appeared

for the first time as a network of neurons.

157. Answer (2)

Hint : Neurons respond to threshold stimulus

generated externally or internally in body.

Sol. : Neurons do not produce stimulus. They detect,

receive and transmit the stimulus.


15/16

158. Answer (1)

Hint : These neuroglial cells form myelin sheath in

PNS.

Sol. : Neurilemma is formed by Schwann cells by

wrapping only one time around unmyelinated

neurons. Oligodendrocytes form myelin sheath in

CNS. Microglial cells are macrophages of neural

system.

159. Answer (3)

Hint : These neurons contain one afferent and one

efferent process.

Sol. : Bipolar neurons contain one dendrite (afferent)

and one axon (one efferent) process. Unipolar

neurons are found in embryonic stage, while

multipolar neurons are found in CNS of adult human.

Dorsal root ganglion of spinal cord has

pseudounipolar neurons.

160. Answer (2)

Hint : Unconditioned reflex is inborn and not an

acquired reaction.

Sol. : Watering of mouth on seeing food is

conditioned reflex as prior exposure has already

occurred and memory of taste has been formed.

161. Answer (2)

Hint : Extensor muscle ‘Quadriceps’ is involved.

Sol. : Bending of knee involves the flexor muscle

Hamstring. This is a stretch reflex where the lower

leg moves outwards and upwards. Interneurons are

absent. Patella is the knee bone/cap involved.

162. Answer (4)

Hint : It appears like sea horse in a cross section

of forebrain.

Sol. : Hippocampus is a part of a complex structure

called limbic system of forebrain. Hippocampus

converts short term memory to long term memory in

man. Amygdala is also a part of limbic system

involved in emotional reactions such as anger and

rage.

163. Answer (3)

Hint : Canal of midbrain. This duct is also known as

‘‘Aqueduct of Sylvius’’.

Sol. : Iter/cerebral aqueduct canal is a part of the

midbrain through which CSF flows.

164. Answer (1)

Hint : These ions are essential for impulse generation

and transmission.

Sol. : Voltage gated Na+ channels opens upon

reaching threshold stimulus. Opening of calcium ion

channels promotes exocytosis of neurotransmitter

rich synaptic vesicles in synaptic cleft.

165. Answer (1)

Hint : Communication channels are present in

synapses where size of synaptic cleft is smaller

than 6 nm.

Sol. : Acetylcholine is a neurotransmitter in a

chemical synapse. In case of electrical synapse,

impulse jumps from pre-synaptic membrane to

postsynaptic membrane rapidly through gap

junctions.

166. Answer (2)

Hint : This centre is also responsible for

cardiovascular reflex and gastric secretions.

Sol. : Medulla is responsible for setting respiratory

rhythm. Thirst, hunger, satiety and body temperature

are regulated by hypothalamus.

167. Answer (3)

Hint : This tract of nerve fibres is a part of forebrain.

Sol. : Cerebral hemispheres are connected by

corpus callosum. Round swellings in midbrain form

corpora quadrigemina. Cerebellum is also divided into

two hemispheres.

168. Answer (3)

Hint : Gray matter of brain lies closer to pia mater

Sol. : Orientation of white and gray matter is

reversed in spinal cord in comparision to brain.

Hence, white matter is closest to pia mater followed

by arachnoid and outermost dura mater.

169. Answer (4)

Hint : Total change in potential difference across a

neuronal surface upon receiving threshold stimulus.

Sol. : Resting potential of neurons is around -70 mV.

If the overshoot after depolarisation is +40 mV, spike

potential = (70 + 40) = 110 mV.

170. Answer (3)

Hint : ECF contains 30 times less K+ than

axoplasm while ECF contains 10 times more Na+

than ICF.

Sol. : Potential difference around -70 mV exists

across axolemma due to differential distribution of

ions. Neural membrane is impermeable to movement

of negatively charged proteins. The exterior surface

of axolemma is positively charged.

171. Answer (1)

Hint : These granules are associated with non

dividing cells of neural tissue.

Sol. : Nissl’s granules are absent in axon of neurons

and glial cells.


16/16

� � �

172. Answer (4)

Hint : Presence of nodes of Ranvier promote

saltatory and consequently faster conduction of

impulse.

Sol. : Myelin sheath is an effective insulator around

axons. Impulse jumps from one node to another at

node of Ranvier. Threshold stimulus triggers the

impulse but is not directly responsible for its

conduction in a nerve fibre. Increase in temperature

increases speed of impulse conduction.

173. Answer (4)

Hint : This part acts as relay center.

Sol. : Amygdala in forebrain is responsible for anger

and rage.

174. Answer (2)

Hint : These are the apertures in wall of 4th ventricle.

Sol. : Each lateral ventricle is connected to third

ventricle by an interventricular foramen (Foramen of

Monro). The third ventricle is connected by iter to the

4th ventricle.

175. Answer (1)

Hint : This lobe is recognised as seat of intelligence

and creativity.

Sol. : Damage to motor speech area i.e, Broca's

area can result in aphasia.

176. Answer (2)

Hint : The neurotransmitter involved is a catecholamine

and derivative of tyrosine.

Sol. : Schizophrenia is a personality disorder

resulting from excessive secretion of dopamine.

177. Answer (4)

Hint : Function of autonomic nervous system is to

control and coordinate the activities of visceral

organs.

Sol. : Parasympathetic branch of nervous system

promotes secretion of intestinal juice.

178. Answer (2)

Hint : In conditions of fear, flight and fight

sympathetic system is activated.

Sol. : In emergencies, stimulation of sympathetic

branch of nervous system promotes relaxation of

breathing pathways. Parasympathetic system

hinders breathing and narrows respiratory pathways

by promoting bronchoconstriction. Therefore,

inhibition of bronchoconstrictions occurs as effect of

sympathetic branch.

179. Answer (3)

Hint : This part is associated with maintaining body

equilibrium.

Sol. : Purkinje cells are a feature of cerebellum part

of hindbrain.

180. Answer (1)

Hint : Transport of 3Na+ and 2K+ occurs in different

directions at the same time by the pump.

Sol. : One ATP is spent by Na+/K+ ATPase.

Test - 4 (Code-D) (Answers) All India Aakash Test Series for Medical-2020

1/16

1. (3)

2. (2)

3. (4)

4. (4)

5. (2)

6. (2)

7. (3)

8. (3)

9. (2)

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11. (4)

12. (1)

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14. (4)

15. (4)

16. (2)

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18. (4)

19. (2)

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21. (3)

22. (3)

23. (Delete)

24. (2)

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28. (2)

29. (2)

30. (2)

31. (2)

32. (1)

33. (3)

34. (2)

35. (4)

36. (3)

Test Date : 30/12/2018

ANSWERS

TEST - 4 (Code-D)

All India Aakash Test Series for Medical-2020

37. (4)

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179. (4)

180. (2)

All India Aakash Test Series for Medical-2020 Test - 4 (Code-D) (Answers & Hints)

2/16

ANSWERS & HINTS

1. Answer (3)

Hint : Mechanical energy and angular momentum

are conserved.

Sol. : Wgravity

= KE

0– GMm

PEr

.

2. Answer (2)

Hint : vara = v

prp

Sol. : vara = v

prp

2 2

a a p pr r

2

pa

p a

r

r

⎛ ⎞ ⎜ ⎟ ⎝ ⎠

ra = a + ae r

p = a – ae

2

1

1–

p

a

e

e

⎛ ⎞ ⎜ ⎟ ⎝ ⎠

3. Answer (4)

Hint : 1n

Fr

Sol. : 2

k mvF

r r

v r°

4. Answer (4)

Hint : 2

1

⎛ ⎞⎜ ⎟⎝ ⎠

h

gg

h

R

Sol. : gh = g /2

2

1 1

21

h

R

⎛ ⎞⎜ ⎟⎝ ⎠

2 – 1h R

5. Answer (2)

Hint : L r p �

� �

Sol. : L = m0vr

r = R + h

[ PHYSICS]

0

GMv

r

0 0( ) ( )

GML m R h m GM R h

R h

6. Answer (2)

Hint : –f i

E E E

Sol. : –

2

GMmE

r

1 1 1 1– –

2 2 8 14i f E E

E GMm GMmr r R R

⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦

7 – 4 3

56 56

⎡ ⎤ ⎢ ⎥

⎣ ⎦E E

GMmGMm

R R

7. Answer (3)

Hint : 2 2 2

interstellar–1

E Ev v v v n

Sol. : v = nvE

n = 3

infinity 9 –1 8 2 2 E E E

v v v v

8. Answer (3)


Sol. : 2 2

0

1 1– –

2 2 ( )

GMm GMm

mv mvR R h

2 2

0

1 1–

2 21

mghmv mv

h

R

2

0

2–

1

⎛ ⎞⎜ ⎟⎝ ⎠

ghv v

h

R

9. Answer (2)

Hint : orbital

1V

r .

Sol. : Orbital speed is independent of mass of

satellite.

10. Answer (3)

Hint : g = g – R2 cos 2Sol. : = 90° for pole g

pole = g(always) if = 0,

ge = g i.e., increases.

Test - 4 (Code-D) (Answers & Hints) All India Aakash Test Series for Medical-2020

3/16

11. Answer (4)

Hint and Sol.: Total energy of bound system is

negative.

12. Answer (1)

Hint : Angular momentum is conserved.

Sol. : r is minimum at point A and maximum at

point C.

So, KA > K

B > K

C

13. Answer (3)

Hint : 2

E

GMv

R

Sol. : 32 4

3

E

Gv R

R

28

3

Ev GR

E

v R

1

2 2

E E E

p p p

v R

v R

14. Answer (4)


Sol. :

H

v = 0

u

U1 + K

1 = U

2 + K

2

20 0

0

– –10

2

GMm GMmm u

R R H

Now, 2 2 0

0 0

1 1

2 4 2E

GMmm u m v

R

H = R

15. Answer (4)

Hint : Both particles rotate about their centre of mass.

Sol. : Both particles are always opposite to each

other along the line joining them so 1 =

2.

16. Answer (2)

Hint : rot total

1for a disc.

3K KE

Sol. : Total KE = mg h = mgl sin.

So, rot

1sin

3K mgl .

17. Answer (4)

Hint and Sol. : MI is same for any symmetrical part

of body.

I1 = I2 = I318. Answer (4)

Hint and Sol. : I = mr2.

19. Answer (2)


Sol. :

1 = 0

U1 + K

1 = U

2 + K

2

210 cos

2 2 2

L LMg mg I

3(1– cos )

g

L

3

2 g

L

20. Answer (4)

Hint : Use Newton’s law of gravitation.

Sol.: To make M in equilibrium resultant of F1 and

F2 should be equal in magnitude and opposite in

direction to T.

∴ From the figure,

1

2

tan30F

F 30°

30°

F1

F2

m1

T

m2

∴

1

2

1

3

m

m

21. Answer (3)

Hint : , d d

dt dt

Sol. : = 3t 2 – 4t + 3

6 – 4d

tdt

= 6 rad/s2

at = R, a

r = R2

At t = 1 s = 2 rad/s

at = 6 m/s2 a

r = 4 m/s2

2 2 2

net6 4 52 m/sa


4/16

22. Answer (3)

Hint : net r t

a a a � � �

Sol. :

ar

v

at

anet

23. Answer (Delete)

24. Answer (2)

Hint and Sol. : I = mk2.

25. Answer (2)

Hint : Fraction of kinetic energy which is rotational

2

2 2.

k

k R

Sol. :

2

2

2

21

k

Rf

k

R

for solid sphere 2

2

2

5

k

R .

2 / 5% 100 28%

21

5

26. Answer (4)

Hint : Use parallel axis theorem.

Sol. : L/2

rL

O

2 2

2

3 12

ML MLI Mr

2

2 5

4

Lr

2

5

3 ML

I

27. Answer (3)

Hint : .P ��

Sol. : ˆ ˆ ˆ( – 3 2 ) Nmi j k �

ˆ ˆ ˆ(4 2 5 ) rad/si j k �

ˆ ˆ ˆ ˆ ˆ ˆ( – 3 2 ) (4 2 5 )P i j k i j k P = 4 – 6 + 10

P = 8 W

28. Answer (2)

Hint : L r p �

� �

Sol. :

v

y

x

v cos

r H =

L = rp

r = H

L = H mv cos

vL mv

g

2 2

sincos

2

mv

g

3 2

sin cos

2

29. Answer (2)

Hint : and � �

� � ��

L r p r F

Sol. : Net torque about C is non zero and net

torque about O is zero.

∴ L is conserve about O but variable about C.

30. Answer (2)

Hint and Sol. : Kepler’s law of area is based on

conservation of angular momentum.

31. Answer (2)

Hint : Conservation of angular momentum about the

point of contact.

Sol. :

Pure rolling

0

v

Applying conservation of angular momentum about

the point of contact,

Li = L

f

mR20 = mR2 + mvR

2 0

02

2

RmR mvR v

⇒


5/16

32. Answer (1)

Hint : Conservation of angular momentum.

Sol. : Li = L

f

Li = mr2

Lf = (mr2 + 2mr2)

2 23mr mr

3

33. Answer (3)

Hint : 2

21

kt

R

⎛ ⎞ ⎜ ⎟

⎝ ⎠

Sol. : 2

2

k

Ris same for disc and solid cylinder.

34. Answer (2)

Hint : In pure rolling on fixed surface net tangential

acceleration of point of contact is zero.

Sol. :

Ov

R2

Rv

For pure rolling with constant angular velocity.

v = R

Radial acceleration of point is R2.

35. Answer (4)

Hint : �

� �

L r p

Sol. : �

L is always prependicular to �r and

�

p .

36. Answer (3)


Sol. : In pure rolling on inclined plane potential

energy is converted into translational and rotational

kinetic energy. Net work done by friction and normal

force is zero.

37. Answer (4)

Hint : = I

Sol. : = I

I = mk 2

= mk 2

38. Answer (4)

Hint : Formula for moment of inertia of any

symmetrical part of a body = Formula for moment of

inertia of body.

Sol. : 2

Semidisc Disc2

MRI I

39. Answer (2)

Hint : vp = 2v

0 cos (/2).

Sol. :

O60° v

0

r v = 0

120°

Here = 120°

vnet

= 2v0cos60°

vnet

= v0

40. Answer (2)

Hint : If ext

= 0 then L = constant

If ext

F��

= 0 the cm

0a �

Sol. : For a rigid body rotating about a fixed axis L

is always parallel to .

Torque produced by central force is always zero.

41. Answer (3)

Hint : r F �

��

Sol. :

P(1, 2)

O(0, 0)

r

ˆ4F k

ˆ ˆ(– – 2 ) mr i j�

ˆ4 �

F k

�

��

r F

ˆ ˆ4( 2 ) �

i j Nm

42. Answer (3)

Hint : L r p �

� �

Sol. : L�

about O is constant as r⊥ and v both are

constant.


6/16

46. Answer (1)

Hint: 1 1 2 2

1 2

N V + N VN =

V + V

Sol. : pH = 2 pH = 3

N1 = 10–2 N

2 = 10–3

–2 –310 × V +10 × V

N =V + V

0.011=

2= 55 × 10–4

4pH = – log(55 10 ) 2.26 47. Answer (4)

Hint: eq

H llogK logA

2.303R T

⎛ ⎞ ⎜ ⎟⎝ ⎠

Sol. : For exothermic reaction.

Hslope ( )ve

2.303R

log Keq

1/T

48. Answer (4)

Hint: Adding inert gas at constant volume has no

effect on equilibrium.

[ CHEMISTRY]

49. Answer (1)

Hint: KP = K

C (RT)ng

Sol. : KP = K

C (RT)ng

– nC

P

K(RT) g

K

(i)1C

P

K(RT)

K

(ii)C

P

K1

K

(iii)–1C

P

K= (RT)

K

(iv)–2C

P

K= (RT)

K

50. Answer (1)

Hint: pH of acidic buffer is given by

a

saltpH pK log

acid

⎛ ⎞ ⎜ ⎟⎝ ⎠

Sol. :

Initial, meq,

final, meq, = 100

==

43. Answer (4)

Hint : F + f = macm

Sol. :

f

R

F

Ov

F + f = macm

...(i)

cm

= Icm

(F – f)R = mR2

or F – f = mR = macm

...(ii) (∵ acm

= R)

From (i) & (ii)

f = 0

44. Answer (2)

Hint : I = mr2

Sol. : Moment of inertia depends on axis of rotation,

mass of the body and distribution of mass about

axis of rotation.

45. Answer (4)

Hint : = mgx

Sol. :

60°

u

x

y

xomg

At t = 2 s

1cos 20 2 20 m

2 x u t

= mgx = 2 × 10 × 20 = 400 Nm


7/16

a

SaltpH pK log

Acid

⎛ ⎞ ⎜ ⎟⎝ ⎠

–3

3

100 10

800= 4.74 log

200 10

800

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

= 4.74 – 0.30 = 4.44

51. Answer (3)

Hint: Salt of weak acid & strong base undergoes

anionic hydrolysis.

Sol. : HCOOH, CH3COOH and HCN are weak acid.

52. Answer (1)

Hint: For an acidic buffer solution a weak acid its

salt with strong base is taken.

Sol. : (H3PO

4 + NaH

2PO

4), (NaH

2PO

4 + Na

2HPO

4)

and (Na2HPO

4 + Na

3PO

4)

53. Answer (3)

Hint: For precipitation, Ksp

< Ionic product (IP)

Sol. :

(i)–4 –6 –10

–102 10 10 10I.P 0.5 10

2 2 2

⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

(ii)–5 –5

–1010 2 10I.P 0.5 10

2 2

⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

(iii)–4 –4

–1010 10I.P 25 10

2 2

⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

54. Answer (3)

Hint: In presence of common ion solubility of salt

decreases.

Sol. : + –

+ –

Ag CN Ag + CN

s s

KCN K + CN

0.02 M 0.02 M

�

Ksp

= [Ag+] [CN–]

1.0×10–16 = (s) (s+0.02) s(0.02)�

–161.0 10

s =0.02

= 5 × 10–15 M

55. Answer (3)

Hint: Applicable only for conjugate acid & base pair.

Sol. : HCOOH – Acid

HCOO– – Conjugate base

56. Answer (4)

Hint. : 4

Acidic Salt Salt Salt

(SA WB) (SB WA)

pH 7 pH 7 pH 7 pH 7

HCl NH Cl NaCl NaCN

57. Answer (4)

Hint: w

1pH = pK

2

Sol. : 14

w

1 1pH = pK log(2 10 ) 6.85

2 2

58. Answer (2)

Hint: Electron deficient species are Lewis acid.

Sol. : B2H

6 is a Lewis acid.

59. Answer (3)

Hint: PB = P

C and P

E = P

F.

Sol. : A(s) B(g) C(g)��⇀↽��

PB = P

C P

B + P

C = 80 atm

PB = P

C = 40 atm

1P

K 40 40 = 1600 atm2

D(s) E(g) F(g)�

PE = P

F P

E + P

F

= 40

atm

PE = P

F = 20 atm

2P

K 20 20 400 atm2

1

2

P

P

K 1600= = 4

K 400

60. Answer (1)

Hint: Conjugate base is obtained by removing a

proton of the species.

Sol. : 2

4SO

is conjugate base of 4

HSO .

61. Answer (2)

Hint: Active mass of solid substance remains

constant.

Sol. : 3 2

p NH H SK (P )(P )

On adding 3

3 NHNH , P increases and

2H S

P decreases.

62. Answer (2)

Hint: For an exothermic reaction, on increasing

temperature Keq

decreases.

Sol. : Increasing temperature in exothermic reaction,

rate of forward & backward both reaction increases,

but rate of backward reaction increases more.


8/16

63. Answer (3)

Hint: Keq

depends only on temperature.

Sol. : On changing concentration Keq

remains same.

64. Answer (3)

Hint: Lewis acid is a substance which can accept

a pair of electrons.

Sol. : Al in AlCl3 and Sn in SnCl

4 have vacant d-

orbitals

In CO2, oxygen atoms are more

electronegative than the carbon atom so

carbon atom become electron deficient.

65. Answer (4)

Hint: [H+][OH–] = 10–14

Sol. :

[OH–] = 1

–14+ –1410

[H ] = 101

pH = 14

66. Answer (3)

Hint: [H+]total

= [H+]acid

+ [H+]H2O

Sol. : pH = 5 [H+] = 10–5 M

N1 V

1 = N

2 V

2

10–5 × 1 = N2 × 1000

N2 = 10–8

Total H+ = H+ ion from acid + H+ ion from H2O

[H+] = 10–8 + 10–7 = 1.1 × 10–7

pH = –log(1.1 × 10–7) = 6.95

67. Answer (2)

Hint: Eq. mass = molar mass

n-factor

Sol. : 3

2

–6eN 2N

2

14 2 14Eq. mass

6 3(N )

68. Answer (3)

Hint: Thermoneutral reactions (H = 0) are

temperature independent.

Sol. : On increasing pressure the equilibrium is

shifted towards the lesser number of gaseous moles.

69. Answer (2)

Hint: Ksp

= [Ba2+][OH–]2

Sol. : 22Ba(OH) Ba 2OH

s 2s

��⇀↽��

Ksp

= (s) (2s)2 5.0 × 10–7 = 4s3

3 –75s = 10

4

= 1.25 × 10–7

s = 5 × 10–3

[OH–] = 2s = 2 × 5 × 10–3 = 10–2

pH = 14 – pOH = 12

70. Answer (2)

Hint:

½ ½2 2

c

(H ) (I )K =

(HI)

Sol. : 2 2

1 IHI H I

2 2

1 0 0

12 2

��⇀↽��

½ ½

2 2

C

(H ) (I )K =

(HI)=

½ ½

2 2

(1– )

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

2(1 )

0.5 1

2(1 0.5) 2

71. Answer (3)

Hint : G° = –2.303RTlogKeq

Sol.: G° = –2.303 × R × 298 × log10

= –2.303 × 298 × R

72. Answer (3)

Hint: In presence of common ion solubility decreases

while in complex formation solubility increases.

Sol. : More the concentration of common ion, lesser

will be the solubility.

3 4 1 2 3 3

(Complex formation) No common ion

0.1 M NH (s ) Water (s ) 0.1 M HCl (s ) 0.2 M AgNO (s )

73. Answer (3)

Hint: a b

1 1pH 7 pK pK

2 2

Sol. : pH of salt of weak acid and weak base is

concentration independent.


9/16

74. Answer (4)

Hint:

2w

h

a

K chK =

K 1 h

,

Sol. : On increasing the concentration of salt, ‘h’

decreases. On adding base, common ion effect is

applied therefore, h decreases and on increasing

temperature Kw and K

a changes.

75. Answer (2)

Hint: Larger is the Keq

, greater is the extent of

reaction.

Sol. : If Keq

<< 1, then (Reactant) >> (Product)

76. Answer (3)

Hint: Acidic buffer contains weak acid & its salt with

strong base.

Sol. : HClO4 is a strong acid.

77. Answer (4)

Hint: wh

a b

KK =

K K

Sol. : Salt of (WA + WB) will have higher KH value.

78. Answer (3)

Hint: When equations are added, then their

equilibrium constants are multiplied

Sol. : A � 2B K1 ...(i)

B � C+D K2 ...(ii)

2C + 2D � E K3 ...(iii)

2 (ii) + (i) + (iii)

A � E, Keq

= K1 K

22 K

3,

On reversing, E � A

eq'K =

2

1 2 3

1

K K K

79. Answer (4)

Hint: With increase in OH concentration, pH will

increase.

Sol. :

(i)

1 12 100 1 100

10 10N

200

= 0.05 [H+]

So, pH < 7

(ii)

1 12 100 1 100

10 5N200

= 0 [Neutral]

So, pH = 7

(iii)

1 12 100 1 100

5 5N

200

+200.1(H )

200

So, pH 7

(iv)

1 11 100 2 100

2 10N

200

–

300.15(OH )

200

So, pH > 7

80. Answer (4)

Hint: For sparingly soluble salt, Ksp

<< 1.

Sol. : 1

2spAB s= K

1

3sp2

KAB s=

4

⎛ ⎞ ⎜ ⎟

⎝ ⎠

1

4sp3

KAB s=

27

⎛ ⎞ ⎜ ⎟

⎝ ⎠

1

5sp2 3

KA B s=

108

⎛ ⎞ ⎜ ⎟

⎝ ⎠

81. Answer (2)

Hint: Ksp

= s2

Sol. : –3

0.07175 10s

143.5

= 5 × 10–7 mol L–1

–

ss

AgCl Ag Cl �

Ksp

= s × s = s²

= (5 × 10–7)2 = 2.5 × 10–13 mol2 L–2

82. Answer (2)

Hint: Sum of O.N. of all atoms in a compound is

zero.

Sol. : A2 (BC

4)3

[2(+3) + 3 [(+6) + 4(–2)] = 0

83. Answer (4)

Hint: o o ocell cathode anodeE E E

Sol. : o o ocell cathode anodeE E E = 2 2

o o

Cu /Cu Zn /ZnE E

= +0.34 – (–0.76) = 1.10 V

84. Answer (4)

Hint: Electrode with higher o

redE value will act as

cathode.


10/16

[ BIOLOGY]

91. Answer (4)

Hint : Guttation is the process which occurs in early

morning and night while transpiration occurs in day

time.

Sol. : Due to transpiration a negative pressure is

generated in xylem vessels while guttation occurs

due to positive root pressure.

92. Answer (3)

Hint : The chief sinks for the mineral elements are

the growing regions of the plants.

Sol. : Mature leaves are the chief source of

mobilised minerals while young leaves, developing

flowers, fruits, seeds, apical & lateral meristems are

the chief sinks for the mineral elements.

93. Answer (3)

Hint : Opening of stomata starts with incomplete

oxidation of carbohydrate.

Sol. :

Step I – In light, starch in guard cells is

incompletely oxidised into phosphoenol pyruvate

(PEP).

Step II – H+ from guard cells are transported to the

epidermal cells.

Step III – K+ ions are absorbed into guard cell.

Step IV – Increased K+ & malate ion form

potassium malate & store it in vacuoles.

94. Answer (2)

Hint : In dry atmosphere when the relative humidity

is low, a plant has higher rate of transpiration.

Sol. : In humid atmosphere i.e., at high relative

humidity, the rate of transpiration decreases.

95. Answer (3)

Hint : Osmotic pressure of an electrolyte is higher

than a non-electrolyte.

Sol. : A cell which does not show any change in

0.7 M sugar solution because it is isotonic, would

show decreased volume of cell due to 0.7 M NaCl

solution by exosmosis. As 0.7 M NaCl solution due

to dissociation has double osmotic pressure than

0.7 M sugar solution, hence become hypertonic in

comparison to cytoplasm of cell.

85. Answer (2)

Hint: Sum of the oxidation numbers of all atoms in

a compound is zero.

Sol. : K4[Fe(CN)

6] 4(+1) + x + 6(–1) = 0

x = +2

Fe (CO)5

x + 5(0) = 0 x = 0

[Fe(H2O)

6]Cl

3 x + 6 x (0) + 3 x (–1) = 0

x = +3

86. Answer (4)

Hint: 2– 2– + 3+

2 7 2 4 2 2Cr O +3C O + 14H 2Cr +7H O + 6CO

Sol. :

1 mole Cr2O

72– requires = 3 mole C

2O

42–

So, 1 mole C2O

42–

requires 2–

2 7

1mole Cr O

3

87. Answer (3)

Hint: When O.N. of an element in a compound is

intermediate, then compound can act as oxidant and

reductant both.

Sol. : HNO2 Oxidation number of N = +3

H2O

2 Oxidation number of O = –1

H2S Oxidation number of S = –2

SO2

Oxidation number of S = +4

88. Answer (2)

Hint: Ozone is better oxidising agent than H2O

2.

Sol.: O3 O

2 + [O]

H2O

2 + [O] H

2O + O

2

H2O

2 is oxidised here.

89. Answer (4)

Hint: Balance the reaction by ion electron method.

Sol. : 2+ 2– + 3+ 3

2 7 26Fe + Cr O + 14H 6Fe 2Cr 7H O

Sum of coefficient of reactants = 6 + 1 + 14 = 21

90. Answer (2)

Hint: In a compound algebraic sum of oxidation

numbers of all atoms is zero.

Sol. : NH3 x + 3(+1) = 0 x = – 3

HNO3 (+1) + x + 3(–2) = 0 x = + 5

N2 x = 0

N2H

4 2(x) + 4(+1) = 0 x = – 2

N3H 3x + 1 = 0 x = –1/3

–3 –1/3 0 5

2 33 3NH < N H < N < HNO


11/16

96. Answer (3)

Hint : In both facilitated diffusion and active transport,

movement of molecules occur with the help of

membrane proteins.

Sol. : Both facilitated diffusion and active transport,

show saturation of the transport and response to

protein inhibitors.

Facilitated diffusion is a downhill process as it does

not require ATP while active transport is an uphill

process.

97. Answer (2)

Hint : Water potential is chemical potential of water

molecules which regulates water movement as water

always moves from high water potenitial to low water

potential.

Sol. : Water potential of a cell = solute potential +

pressure potential.

Water potential of pure water is zero at atmospheric

pressure but by adding solutes it decreases and

become negative.

98. Answer (3)

Hint : Water in the adjacent xylem moves into the

phloem by osmosis.

Sol. : Sucrose transport in the living phloem sieve

tube cells occurs through osmotic pressure gradient.

Loading of sucrose into sieve tube cells is an active

process.

99. Answer (3)

Hint : Elements most readily mobilised are

phosphorus, nitrogen & potassium.

Sol. :

Mobile elements are frequently remobilised from

older parts to younger parts.

A C4 plant loses only half as much water as a

C3 plant for the same amount of CO

2 fixed.

100. Answer (4)

Hint : Most of the absorbed water is lost through

stomata of the leaves.

Sol. : Less than 1 percent of water, reaching the

leaves is used in photosynthesis & plant growth.

101. Answer (2)

Hint : Water/solvent always moves from lower DPD

to higher DPD of solution.

Sol. : OP = 13TP = 5DPD = 8

AOP = 6TP = 2DPD = 4

B

Low DPDHigh DPD

102. Answer (1)

Sol. : Tonoplast, membrane of sap vacuole is a

selectively permeable membrane.

103. Answer (3)

Hint : Ephedra, Gnetum & Welwitschia are placed in

the most advanced order of gymnosperms.

Sol. : Ephedra is placed in order Gnetales.

104. Answer (2)

Hint : Endosperm represents female gametophyte in

gymnospermic seed.

Sol. :

Plumule, radicle, suspensor & cotyledons

represent future sporophyte.

Testa, tegmen & perisperm represent parental

sporophyte.

105. Answer (3)

Hint : Gymnosperms have integumented ovules.

Sol. : In ovules of gymnosperms, nucellus is

protected by single integument.

106. Answer (3)

Hint : Coralloid roots of Cycas are irregular & do not

possess root hairs.

Sol. : Coralloid roots of Cycas are symbiotically

associated to Anabaena (cyanobacteria).

107. Answer (3)

Hint : Homosporous pteridophytes have monoecious

gametophyte.

Sol. : Heterospory in pteridophytes is precursor to

the seed habit.

108. Answer (4)

Sol. : Gametophytes of mosses have multicellular

and branched rhizoids.

109. Answer (3)

Hint : Pteridophytes are the first vascular

embryophytes.

Sol. : Dryopteris – Pteridophyte

Funaria – Bryophyte

Cycas – Gymnosperm

Mangifera - Angiosperm

110. Answer (2)

Sol. : Brown alga – Sargassum,

Green alga – Chara,

Red alga - Gracilaria

111. Answer (2)

Hint : Carrageen is sulphated polysaccharide.

Sol. : Sulphated polysaccharides or hydrocolloids

are found in the cell wall of rhodophyceae.


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112. Answer (3)

Hint : In red algae, the major pigments are chl a and

d, carotenoids and xanthophylls.

Sol. : Porphyra is a red alga, while Fucus, Dictyota

& Laminaria are brown algae.

113. Answer (3)

Sol. : Antherozoids of bryophytes are biflagellated.

114. Answer (2)

Sol. : Primary protonema is a filamentous structure

develops from spore germination in Funaria.

115. Answer (3)

Hint : All bryophytes are called amphibians of plant

kingdom.

Sol. : Mosses like Funaria, have sporophyte which

is partially dependent on gametophyte.

116. Answer (4)

Hint :

Cycas

Eucalyptus

(gymnosperm) = nEndosperm in

(angiosperm) = 3n

⎡⎢⎣

Sol. : Wolfia is an angiosperm, within ovules of

which are present highly reduced female

gametophytes i.e., embryo sacs.

117. Answer (2)

Hint : Gymnosperms have archegonia in their ovule.

Sol. : Pinus has 2-8 archegonia inside its ovule,

which is absent in the ovule of papaya as it is an

angiosperm.

118. Answer (4)

Hint : Pteridophytes are vascular plants with

dominant diploid sporophyte.

Sol. : In pteridophytes, the gametophyte is short-

lived, haploid and independent.

119. Answer (2)

Hint : Stems are usually branched in orders

Coniferales and Gnetales of gymnosperms.

Sol. : Cedrus – branched stem.

Cycas – unbranched stem.

120. Answer (4)

Hint : Red wood tree (Sequoia) – tallest gymnosperm

Sol. :

Polysiphonia shows haplo-diplontic life cycle

pattern whereas Volvox has haplontic life cycle

pattern.

Statements (a) & (d) are correct.

121. Answer (3)

Hint : It is group of antiparasitic drugs that expel

parasitic worms, i.e., helminths hence called

anthelmintics or vermifuges.

Sol. : Dryopteris is used to obtain anthelmintic drug

to treat helmenthiasis.

122. Answer (3)

Hint : Bryophytes have haplo-diplontic life cycle.

Sol. : Fucus, Cycas and mango all have diplontic life

cycle. Funaria being a bryophyte has haplo-diplontic

life cycle.

123. Answer (4)

Hint : Angiosperms are called flowering plants.

Sol. : Occurrence of triploid endosperm, double

fertilisation and presence of fruits are exclusive

features of angiosperms.

124. Answer (3)

Hint : Adiantum is a fern which belongs to the class

Pteropsida.

Sol. : Psilopsida – Psilotum

Lycopsida – Selaginella

Sphenopsida – Equisetum

125. Answer (3)

Hint : In bryophytes, the sporophyte is parasitic over

gametophyte.

Sol. : Members of pteridophytes may be

homosporous (one kind of spores) or heterosporous

(two kinds of spores).

126. Answer (3)

Sol. : Bryophytes have gametophytic main plant body.

Pteridophytes, gymnosperms and angiosperms have

sporophytic main plant body.

127. Answer (4)

Sol. : A - Dictyota (Brown alga)

B - Porphyra (Red alga)

C - Selaginella (Pteridophyte)

D - Salvinia (Aquatic fern)

128. Answer (3)

Hint : In liverworts, sporophytes are diploid

structures which develop from embryo and are

dependent on gametophytes.

Sol. : All bryophytes, including Marchantia are

homosporous.

129. Answer (3)

Hint : Funaria is a moss.

Sol. : In mosses, leafy gametophytes have

multicellular and branched rhizoids.


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130. Answer (4)

Hint : Pear shaped zoospores & pyriform motile

gametes are found in brown algae.

Sol. : Brown algae are found mostly in marine

habitats.

131. Answer (1)

Hint : Peptidoglycan is constituent of bacterial cell

wall.

Sol. : In chlorophyceae outer layer of cell wall is

made up of pectose, while inner layer is constituted

by cellulose.

132. Answer (4)

Hint : Porphyra is a red alga.

Sol. : The stored food of red algae is floridean starch.

133. Answer (4)

Hint : Oogamous type of sexual reproduction

involves fusion between non-motile female gamete

and motile male gamete in some green and brown

algae.

Sol. : Ulothrix and Spirogyra have isogamous type of

reproduction whereas Volvox and Polysiphonia have

oogamous reproduction but Polysiphonia has non-

motile male gametes.

134. Answer (3)

Hint : Natural system of classification is based on

morphology, anatomy, embryology & phytochemistry.

Sol. : Phylogenetic system of classification includes

evolutionary relationships between organisms, as a

basis of classification.

135. Answer (3)

Sol. : DNA sequence, chemical nature of proteins,

crystals & aromatic compounds are used in

chemotaxonomy by scientists to resolve confusions

in classification.

136. Answer (1)

Hint : Transport of 3Na+ and 2K+ occurs in different

directions at the same time by the pump.

Sol. : One ATP is spent by Na+/K+ ATPase.

137. Answer (3)

Hint : This part is associated with maintaining body

equilibrium.

Sol. : Purkinje cells are a feature of cerebellum part

of hindbrain.

138. Answer (2)

Hint : In conditions of fear, flight and fight

sympathetic system is activated.

Sol. : In emergencies, stimulation of sympathetic

branch of nervous system promotes relaxation of

breathing pathways. Parasympathetic system

hinders breathing and narrows respiratory pathways

by promoting bronchoconstriction. Therefore,

inhibition of bronchoconstrictions occurs as effect of

sympathetic branch.

139. Answer (4)

Hint : Function of autonomic nervous system is to

control and coordinate the activities of visceral

organs.

Sol. : Parasympathetic branch of nervous system

promotes secretion of intestinal juice.

140. Answer (2)

Hint : The neurotransmitter involved is a catecholamine

and derivative of tyrosine.

Sol. : Schizophrenia is a personality disorder

resulting from excessive secretion of dopamine.

141. Answer (1)

Hint : This lobe is recognised as seat of intelligence

and creativity.

Sol. : Damage to motor speech area i.e, Broca's

area can result in aphasia.

142. Answer (2)

Hint : These are the apertures in wall of 4th ventricle.

Sol. : Each lateral ventricle is connected to third

ventricle by an interventricular foramen (Foramen of

Monro). The third ventricle is connected by iter to the

4th ventricle.

143. Answer (4)

Hint : This part acts as relay center.

Sol. : Amygdala in forebrain is responsible for anger

and rage.

144. Answer (4)

Hint : Presence of nodes of Ranvier promote

saltatory and consequently faster conduction of

impulse.

Sol. : Myelin sheath is an effective insulator around

axons. Impulse jumps from one node to another at

node of Ranvier. Threshold stimulus triggers the

impulse but is not directly responsible for its

conduction in a nerve fibre. Increase in temperature

increases speed of impulse conduction.

145. Answer (1)

Hint : These granules are associated with non

dividing cells of neural tissue.

Sol. : Nissl’s granules are absent in axon of neurons

and glial cells.

146. Answer (3)

Hint : ECF contains 30 times less K+ than

axoplasm while ECF contains 10 times more Na+

than ICF.


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Sol. : Potential difference around -70 mV exists

across axolemma due to differential distribution of

ions. Neural membrane is impermeable to movement

of negatively charged proteins. The exterior surface

of axolemma is positively charged.

147. Answer (4)

Hint : Total change in potential difference across a

neuronal surface upon receiving threshold stimulus.

Sol. : Resting potential of neurons is around -70 mV.

If the overshoot after depolarisation is +40 mV, spike

potential = (70 + 40) = 110 mV.

148. Answer (3)

Hint : Gray matter of brain lies closer to pia mater

Sol. : Orientation of white and gray matter is

reversed in spinal cord in comparision to brain.

Hence, white matter is closest to pia mater followed

by arachnoid and outermost dura mater.

149. Answer (3)

Hint : This tract of nerve fibres is a part of forebrain.

Sol. : Cerebral hemispheres are connected by

corpus callosum. Round swellings in midbrain form

corpora quadrigemina. Cerebellum is also divided into

two hemispheres.

150. Answer (2)

Hint : This centre is also responsible for

cardiovascular reflex and gastric secretions.

Sol. : Medulla is responsible for setting respiratory

rhythm. Thirst, hunger, satiety and body temperature

are regulated by hypothalamus.

151. Answer (1)

Hint : Communication channels are present in

synapses where size of synaptic cleft is smaller

than 6 nm.

Sol. : Acetylcholine is a neurotransmitter in a

chemical synapse. In case of electrical synapse,

impulse jumps from pre-synaptic membrane to

postsynaptic membrane rapidly through gap

junctions.

152. Answer (1)

Hint : These ions are essential for impulse generation

and transmission.

Sol. : Voltage gated Na+ channels opens upon

reaching threshold stimulus. Opening of calcium ion

channels promotes exocytosis of neurotransmitter

rich synaptic vesicles in synaptic cleft.

153. Answer (3)

Hint : Canal of midbrain. This duct is also known as

‘‘Aqueduct of Sylvius’’.

Sol. : Iter/cerebral aqueduct canal is a part of the

midbrain through which CSF flows.

154. Answer (4)

Hint : It appears like sea horse in a cross section

of forebrain.

Sol. : Hippocampus is a part of a complex structure

called limbic system of forebrain. Hippocampus

converts short term memory to long term memory in

man. Amygdala is also a part of limbic system

involved in emotional reactions such as anger and

rage.

155. Answer (2)

Hint : Extensor muscle ‘Quadriceps’ is involved.

Sol. : Bending of knee involves the flexor muscle

Hamstring. This is a stretch reflex where the lower

leg moves outwards and upwards. Interneurons are

absent. Patella is the knee bone/cap involved.

156. Answer (2)

Hint : Unconditioned reflex is inborn and not an

acquired reaction.

Sol. : Watering of mouth on seeing food is

conditioned reflex as prior exposure has already

occurred and memory of taste has been formed.

157. Answer (3)

Hint : These neurons contain one afferent and one

efferent process.

Sol. : Bipolar neurons contain one dendrite (afferent)

and one axon (one efferent) process. Unipolar

neurons are found in embryonic stage, while

multipolar neurons are found in CNS of adult human.

Dorsal root ganglion of spinal cord has

pseudounipolar neurons.

158. Answer (1)

Hint : These neuroglial cells form myelin sheath in

PNS.

Sol. : Neurilemma is formed by Schwann cells by

wrapping only one time around unmyelinated

neurons. Oligodendrocytes form myelin sheath in

CNS. Microglial cells are macrophages of neural

system.

159. Answer (2)

Hint : Neurons respond to threshold stimulus

generated externally or internally in body.

Sol. : Neurons do not produce stimulus. They detect,

receive and transmit the stimulus.

160. Answer (4)

Hint : The neural organisation is very simple in lower

invertebrates.

Sol. : Poriferans, simplest invertebrates lack neural

system. In coelenterates such as Hydra, it appeared

for the first time as a network of neurons.


15/16

161. Answer (3)

Hint : Coordination is the process through which two

or more organs interact and complement the

functions of one another.

Sol. : The neural system provides an organised

network of point to point connections for a quick

coordination. The endocrine system provides

chemical coordination through hormones.

162. Answer (3)

Hint : Hammer, anvil and stirrup shaped bones are

part of middle ear in man.

Sol. : Incus is anvil shaped bone of middle ear that

interacts with stapes and malleus. Human skull has

two occipital condyles through which it interacts with

atlas vertebra.

163. Answer (1)

Hint : It is a supporting bone of lower limb.

Sol. : Fibula does not interact with femur at knee

joint. Fibula interacts with tibia at two points.

164. Answer (3)

Hint : 8 carpals are arranged equally in two rows in

one wrist.

Sol.: Cranium has 8 bones in humans fixed together

by sutures. Carpals interact with each other through

gliding joints. Carpals are 16 while tarsals are 14 in

number. 2 coxal bones form pelvic girdle while 4

bones form pectoral girdle. 26 vertebrae occur in

vertebral column of an adult and 30 bones are

present in an upper limb of man.

165. Answer (2)

Hint : Select the triangular flat bone that is also

called shoulder bone.

Sol. : Scapula and clavicle of both side together form

pectoral girdle. Pelvic girdle involves innominate/coxal

bones.

166. Answer (4)

Hint : Sternum is placed on same side as human

heart.

Sol. : Sternum is placed ventrally in the human body.

Myasthenia gravis affects skeletal muscles of the

human body.

167. Answer (4)

Hint : These are most common type of synovial

joints in humans.

Sol. : In humans, carpo-carpal and tarso-tarsal joints

are gliding joints; atlas-axis joint is a pivot joint; saddle

joint is present between carpal and metacarpal of

thumb; cartilaginous joints are found in between

vertebrae.

168. Answer (3)

Hint : Muscles associated with heart are striped.

Sol. : All types of muscles have actin and myosin.

Both skeletal and cardiac muscle fibres are striated/

striped. Longer refractory period belongs to cardiac

muscle fibres.

169. Answer (3)

Hint : In humans, the 11th and 12th pair of ribs are

not attached ventrally to sternum.

Sol. : Some mammals like sloth and manatee have

9 and 6 cervical vertebrae respectively. Cranium in

humans is composed of eight bones. There are

usually 14 phalanges in a limb of a human.

170 Answer (2)

Hint : Deficiency of estrogen adds to the increased

chances of fracture in this case.

Sol. : Bones become weak due to demineralisation

caused by enhanced activity of bone dissolving cells

called osteoclasts.

Rickets is observed in children in case of calcium

deficiency. Accumulation of uric acid crystals in

joints is seen in gouty arthritis while rheumatoid

arthritis is an autoimmune disorder.

171. Answer (4)

Hint : A common collagenous connective tissue

layer holds together a number of muscle bundles.

Sol. : Each muscle bundle is called fascicle. Many

fascicles are together wrapped around by fascia.

Sarcolemma is plasma membrane of a muscle fibre.

Endomysium is connective tissue covering of

individual muscle fibre.

172. Answer (2)

Hint : Extension is observed in contractile fibres.

Sol. : Nerve fibres do not exhibit contractility and

elasticity.

173. Answer (2)

Hint : Dark meat is rich in myoglobin.

Sol. : White meat has fast contracting, glycolytic/

anaerobic white muscle fibres. They have more

developed sarcoplasmic reticulum than red muscle

fibres. Red muscle fibres rich in myoglobin have more

mitochondria than white muscle fibres.

174. Answer (2)

Hint : Myosin head has more affinity for ATP than for

actin.

Sol. : Cross bridge between actin and myosin will

not break until a new ATP attaches to myosin head.


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� � �

175. Answer (4)

Hint : Actin filaments slide towards the M-line when

a sarcomere contracts.

Sol. : Thin and thick filaments in a myofibre are

composed of actin and myosin respectively. Thin

filaments slide over thick filaments during muscle

contraction.

176. Answer (3)

Hint : Identify an enzyme.

Sol. : Acetylcholinesterase breaks acetylcholine in

synaptic cleft into acetate and choline.

177. Answer (3)

Hint : All muscle fibres contain actin and myosin.

Sol. : Smooth muscle fibres appear non-striated as

actin and myosin filaments are not regularly arrayed

along the length of the cell.

178. Answer (4)

Hint : Leaky channels are always open.

Sol. : Efflux of K+ is essential for repolarisation,

which occurs through opening of K+ voltage gated

channels.

179. Answer (4)

Hint : Filaments of this protein run close to the

F-actin throughout its length.

Sol. : Troponin - Tropomyosin complex masks the

myosin binding site on actin. Troponin binds to

calcium which results in unmasking of myosin

binding site on actin.

180. Answer (2)

Hint : The structure is responsible for propulsion/

whip like movement in tail of male gamete.

Sol. : Male gamete called sperm in humans is

flagellated. Flagella is the propulsion equipment that

pushes the sperm towards the ovum. Amoeboid

movement shown by leucocytes involves

pseudopodia formation.

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