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Test - 5 (Code-C) (Answers) All India Aakash Test Series for Medical-2020

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1. (2)

2. (4)

3. (2)

4. (3)

5. (1)

6. (3)

7. (1)

8. (4)

9. (4)

10. (2)

11. (1)

12. (2)

13. (2)

14. (1)

15. (3)

16. (2)

17. (1)

18. (4)

19. (1)

20. (2)

21. (3)

22. (4)

23. (2)

24. (1)

25. (2)

26. (2)

27. (4)

28. (2)

29. (2)

30. (4)

31. (4)

32. (1)

33. (4)

34. (2)

35. (3)

36. (2)

Test Date : 27/01/2019

ANSWERS

TEST - 5 (Code-C)

All India Aakash Test Series for Medical-2020

37. (4)

38. (1)

39. (3)

40. (1)

41. (2)

42. (2)

43. (2)

44. (2)

45. (3)

46. (3)

47. (4)

48. (1)

49. (4)

50. (2)

51. (2)

52. (3)

53. (1)

54. (2)

55. (4)

56. (1)

57. (3)

58. (4)

59. (2)

60. (3)

61. (1)

62. (3)

63. (2)

64. (4)

65. (3)

66. (1)

67. (4)

68. (4)

69. (3)

70. (2)

71. (3)

72. (2)

73. (1)

74. (3)

75. (2)

76. (2)

77. (3)

78. (4)

79. (4)

80. (3)

81. (3)

82. (4)

83. (3)

84. (1)

85. (4)

86. (1)

87. (2)

88. (3)

89. (1)

90. (2)

91. (4)

92. (4)

93. (3)

94. (4)

95. (3)

96. (1)

97. (4)

98. (4)

99. (3)

100. (4)

101. (2)

102. (2)

103. (3)

104. (3)

105. (2)

106. (4)

107. (1)

108. (3)

109. (1)

110. (1)

111. (4)

112. (4)

113. (1)

114. (1)

115. (4)

116. (3)

117. (4)

118. (2)

119. (4)

120. (3)

121. (3)

122. (4)

123. (2)

124. (1)

125. (2)

126. (4)

127. (1)

128. (2)

129. (2)

130. (4)

131. (4)

132. (4)

133. (2)

134. (2)

135. (3)

136. (4)

137. (4)

138. (4)

139. (3)

140. (2)

141. (3)

142. (3)

143. (1)

144. (3)

145. (1)

146. (1)

147. (4)

148. (4)

149. (2)

150. (1)

151. (4)

152. (1)

153. (3)

154. (1)

155. (4)

156. (2)

157. (2)

158. (3)

159. (4)

160. (4)

161. (3)

162. (4)

163. (4)

164. (2)

165. (4)

166. (4)

167. (4)

168. (4)

169. (4)

170. (4)

171. (2)

172. (3)

173. (3)

174. (2)

175. (4)

176. (3)

177. (2)

178. (4)

179. (3)

180. (4)

All India Aakash Test Series for Medical-2020 Test - 5 (Code-C) (Answers & Hints)

2/14

ANSWERS & HINTS

1. Answer (2)

Hint: CO molecule is diatomic.

Sol.: CO molecule is diatomic. At moderate

temperature it will be having 7 degrees of freedom,

as vibrational mode is also excited.

2. Answer (4)

Hint: rms

v T

Sol.: As temperature increases, speed also

increases. Hence number of collision per second will

increase as well as momentum change per collision

will also increase. Hence force to the container wall

will increase so pressure will increase.

3. Answer (2)

Hint: Dalton’s law of partial pressure.

Sol.: 1 2

1 10.082 300

( ) 4 4

5

⎛ ⎞ ⎜ ⎟ ⎝ ⎠ RT

PV

= 2.5 × 105 = 2.5 atm

4. Answer (3)

Hint: Q = CvT

Sol.: Q = CvT

2 330

20 2R 5

9K C F

⎡ ⎤ ⎢ ⎥⎣ ⎦∵

= 9 cal

5. Answer (1)

Hint: PV = constant

Sol.: As vessel is open pressure will be constant

and equal to the atmospheric pressure and

neglecting expansion of vessel so volume is also

constant

PV = RT

1T

1 =

2T

2

1

× 300 = 2

× 400

1

2

3

4

1Fraction of mass that will escape

4⇒

6. Answer (3)

Hint: Properties of solids

Sol.: Fluids have only Bulk modulus.

[ PHYSICS]

7. Answer (1)

Hint: Poisson's ratio = –�

�

dr

r

d

Sol.: V = r2 × �

2⇒ �

�

dV dr d

V r

–2×0.3 0.4⇒ � � �

� � �

dV d d d

V

100 0.4 100 0.4 1% 0.4%⇒ �

�

dV d

V

8. Answer (4)

Hint: Breaking stress (B.S.) F

A

Sol.: B.S. �

�F Mg A g

gA A A

B.S is independent of area of cross-section.

Hence, maximum length to be hung is �.

9. Answer (4)

Hint: Elongation in the wire is proportional to the

tension in the wire.

Sol.: In case A, T = Mg

2 2 3In case B, 2.4

2 3

M M g T Mg

M M

T �T �

2.42.4

Mg

Mg⇒ ⇒

�

� ��

10. Answer (2)

Hint: Air pressure decreases with increase in volume.

Sol.: P0 = hg + P

air

When tube will be pulled slightly the volume of air

above mercury column will increase hence pressure

will decrease.

So, height of mercury column will increase.

Hence slightly more than 70 cm.

Test - 5 (Code-C) (Answers & Hints) All India Aakash Test Series for Medical-2020

3/14

11. Answer (1)

Hint: Excess pressure inside soap bubble = T

r

4

Sol.: 4T

h gr

r

–3 34 0.033 10 0.9 10 10

2

24 3 10 410 m

3 9 9

r

Surface area = 4r2 × 2 = 496.2 × 10–6 m2

12. Answer (2)

Hint: By equation of continuity.

Sol.: A1v

1 = A

2v

2 + A

3v

3

4 × A = 1 × 2A + v × 3A

2v m/s

3

13. Answer (2)

Hint: Use Bernoulli's equation.

Sol.: 2

0 0

14 4

2P gh P v

2v gh

2 10 5

v = 10 m/s

14. Answer (1)

Hint: When anything lighter is unloaded no change

in the liquid level.

Sol.: Let mass of cork ball is M.

Mg = upthrust when ball is inside the boat.

Mg = Vin�g

Liquid displaced (Vin) =

M

�

...(i)

When ball is dropped in water, still Mg = upthrust

in

M

V�

...(ii)

Boat is floating in both cases. So total volume of

water displaced in both cases are same. Hence no

change in liquid level.

15. Answer (3)

Hint: PowerW P V

t t

Sol.: P = hg = 1.5 × 103 × 10 N/m2

–650 10 1.25

V

t

m3/s

Power = 15 × 103 × 50 × 1.25 × 10–6 = 0.94 W

16. Answer (2)

Hint: 1 2

1 2

mix

m md

V V

Sol.: 1 2 1 2 1 2

1

1 22

m m dV d V d d

V V V V

1 2 1 2

2

1 2 1 2

1 2

2m m d dm m

m mV V d d

d d

Now,

2

1 2 1 2 1 2 1 2

1 2

1 2 1 2

2 ( ) 4

2 2( )

d d d d d d d d

d d d d

2

1 2

1 2

( )0

2( )

d d

d d

So, 1 >

2

17. Answer (1)

Hint: vD

R

Sol.: 2

4

D vQ Av

Or, 2

4Qv

D

So,2

4 4 D Q Q

RD D

18. Answer (4)

Hint: = 2

Sol.: 100 2% ⇒ L

L LL

100 2% ⇒ r

r rr

Area = r2

2100 100 4%

⇒ dA r

A r


4/14

19. Answer (1)

Hint: H = (m + w)c Sol.: H = (m + w)c = 150 × 1 × 50 = 7500 cal

= 7.5 kcal

20. Answer (2)

Hint: Heat current will be same as connected in

series.

Sol.: 0 0( – 100)2 (200 – )3

2 3

KA KAH

� �

0 – 100 = 200 –

0

20 = 300

0 = 150°C

21. Answer (3)

Hint: Q = mL

Sol.: Heat given by steam = Heat taken by ice

mLv + mc= mL

1

m × 540 + m × 1 × 100 = 80 × 80

80 80

10 g640

m

Mass of water present = 80 + 10 = 90 g

22. Answer (4)

Hint: For linear scale Y = mx + c

Sol.: ( C)º

�(cm)50 cm

– 10º C

80º C

0

9– 10

5 �

When � = 70 cm

970 –10

5

= 116ºC

23. Answer (2)

Hint: If CP & C

V are molar specific heats then

CP – C

V = R

Sol.: Molar specific heat = M × specific heat

McP – Mc

V = R

–P V

Rc c

M

For helium = 4

R p

For oxygen 32

R q

p = 8q

24. Answer (1)

Hint: V V

Sol.: – V P

V B

And V

V

P

B⇒

P

B⇒

25. Answer (2)

Hint: P = eAT4

Sol.: P = eAT4

Now as radius is doubled. Hence surface area will

become 4 times i.e. 4A.

P A

P 4A

P = 4P = 4 × 300 = 1200 W

26. Answer (2)

Hint: th

Temperature differenceHeat flow =

Thermal resistance

R

Sol.: Temperature difference remains same 30ºC,

but thermal resistance eq

1 1 1 R R R

eq2

RR

Hence heat flow will double i.e. 20 J/s

27. Answer (4)

Hint: Apply Newton’s law of cooling.

Sol.: 1 2 1 20

––

2

⎡ ⎤ ⎢ ⎥⎣ ⎦

Kt

60 – 50 60 50– 20

9 2

⎡ ⎤⇒ ⎢ ⎥

⎣ ⎦K ...(i)


5/14

50 – 50– 20

9 2

⎡ ⎤ ⎢ ⎥⎣ ⎦

K ...(ii)

10 35

50 –25 – 20

2

= 42.5ºC

28. Answer (2)

Hint: For fixed mass of gas at constant pressure

V T.

Sol.: V

T

For fixed mass of gas at constant pressure V T.

29. Answer (2)

Hint: AB is Lsothermal process.

Sol.: Work done = i

F

PRT

P102.303 log

= 2.303 × 2 × 8.3 × 300 log1

2

= – 2.303 × 2 × 8.3 × 300 × 0.3010

= – 3452 J

Negative sign indicates that work is done on the gas.

30. Answer (4)

Hint: Change in internal energy is the function of

state only U1 = U

2 = U

3

Sol.: Q = U + W

W = area under P – V diagram

Hence W1 > W

2 > W

3

Q1 > Q

2 > Q

3

31. Answer (4)

Hint: Change in internal energy U = CVT

Sol.:

Since A A B B

A B

P V P V

T T

3 33 10 1 10 3

A BT T

TA = T

B

T = TB – T

A = 0

U = 0

32. Answer (1)

Hint: Q = U + W

Sol.: Q = U + W

–100 = U + 20

U = –120 J

33. Answer (4)

Hint and solution : If two bodies are in thermal

equilibrium then both the bodies are at same

temperature.

34. Answer (2)

Hint: V

V T

Sol.: PT = constant = K

Gas equation PV = RT

KV RT

T

V T2

dV 2T dT

dV dT

V T

2⇒

dV

VdT T

2

35. Answer (3)

Hint: For adiabatic process Q = 0

Sol.: 0 Q

ST

S will remain same.

36. Answer (2)

Hint: Conversion of P-V graph into V-T graph.

Sol.: A Carnot cycle consist of isothermal

expansion, adiabatic expansion, isothermal

compression and adiabatic compression respectively.

37. Answer (4)

Hint: Q = U + W

Sol.: Isobaric process

Q = Cv R

Q = 3RRQ = 4RW = R

4 :1Q

W⇒


6/14

46. Answer (3)

Hint: Basic character of oxide increases down the

group.

47. Answer (4)

Hint: For solubility, hydration energy > lattice energy

Sol.: The size of Ba2+ and 2

4SO

ions are very large

which leads to higher lattice enthalpy than that of

hydration enthalpy.

48. Answer (1)

Hint: Lattice enthalpy decreases with increase in the

size of cation.

[ CHEMISTRY]

Sol.: The order of thermal stability of hydrides is

LiH > NaH > KH > RbH > CsH

49. Answer (4)

Hint: In Castner-Kellner cell mercury cathode and

graphite anode are used.

Sol.: Cathode: Hg

Na e Na Hg

Anode: 2

1Cl Cl e

2

Cl2 gas evolved at anode.

50. Answer (2)

Hint: Ga: [Ar] 3d10 4s2 4p1

38. Answer (1)

Hint: PVN = constant

Sol.: PVN = constant

Molar heat capacity of the gas.

5 9

11– 2 21–

2

v

R R R RC C

N

39. Answer (3)

Hint: Coefficient of performance 2QK

W

Sol.: 2

–

L

H L

T QK

T T W

P

273 2000

45

2000 45330 W

273

�P

40. Answer (1)

Hint: PV = nRT

Sol.: nR

P TV

⎛ ⎞ ⎜ ⎟⎝ ⎠

P

T

V2

V1

P2

P1

0

For a given temperature, P2 > P

1

V2 < V

1

41. Answer (2)

Hint and solution : Isobaric process is at constant

pressure, isochoric process is at constant volume

and isothermal process is at constant temperature.

42. Answer (2)

Hint: For adiabatic Q = 0

Sol.: Q = U + W

U = –W

As in expansion W = Positive. Hence U =

Negative so temperature of the gas will decrease.

43. Answer (2)

Hint: Carnot

Sol.: Carnot

1–⎛ ⎞

⎜ ⎟⎝ ⎠

L

H

T

T

max

3031– 0.284

423

⎛ ⎞ ⎜ ⎟⎝ ⎠

= 28.4% maximum possible efficiency.

Efficiency may be 25%

44. Answer (2)

Hint: Boyle’s law.

Sol.: PV = constant.

P0V

0 = P × 2V

0

0

2

PP

45. Answer (3)

Hint: Isochoric process, i.e, V = 0

Sol.: From Ist law

Q = U + W

Q = U

All heat absorbed will be used in increasing the

internal energy of the gas.


7/14

Sol.: In ‘Ga’ atom, 10 electrons are filled in d orbitals

hence size decreases due to poor shielding of d

electrons.

51. Answer (2)

Hint: Amphoteric oxides means which will react with

acid and as well as base.

Sol.: SnO2 + 2NaOH Na

2SnO

3 + H

2O

SnO2 + 4HCl SnCl

4 + 2H

2O

52. Answer (3)

Hint.: Blue colour turns to bronze colour by increasing

the metal concentration leading to the cluster formation

of metal.

53. Answer (1)

Hint: BeH2 becomes polymer due to its electron

deficient nature through multicentre bridge bonds.

Sol.:

3c - 2e bond

BeH

H

BeH

H

BeH

H

BeH

H

54. Answer (2)

Hint: 2 2

wt. of H O%strength= 100

V(ml) of solution

Sol.: 2H2O

2 2H

2O + O

2

22.4 L O2 from 2 34 g H

2O

2 (at NTP)

100 L O2 from

2 34100

22.4

= 303.57 g

= 30.35% H2O

2 solution.

55. Answer (4)

Hint: Galena is PbS.

Sol.: Mica, zeolite and feldspar are silicates.

56. Answer (1)

Hint: CH SiCl CH Si(OH) + HCl3 3 3 3

H O2

.

Sol.:

CH Si(OH) 3 3

polymeriseCH

3

– O – Si – O – Si – O

CH3

O O

– O – Si – O – Si – O

CH3

CH3

(cross link polymer)

57. Answer (3)

Hint: All reactive metal releases H2 gas with dil acid

or water.

Sol.: 2NaHCO3 + 2HCl 2NaCl + 2H

2O + 2CO

2

58. Answer (4)

Hint: Thermal stability order : MgCO3 < CaCO

3 < BaCO

3

59. Answer (2)

Hint: Slaked lime is Ca(OH)2

Sol.: Slaked lime is prepared by adding water to quick

lime (CaO)

CaO + H2O Ca(OH)

2

60. Answer (3)

Hint: Cl

Cl

Cl

Cl

Al

Cl

Cl

101°Al

61. Answer (1)

Hint: Al forms a protective layer of Al2O

3 with conc.

HNO3.

62. Answer (3)

Hint: Boron nitride has 2 dimensional sheet like

structure.

63. Answer (2)

Hint:

F

B

F F

Sol.: Acidic strength : BI3 > BBr

3 >BCl

3 > BF

3

64. Answer (4)

Hint: Mobility decreases with increase in the extent

of solvation.

Sol.: Order of size of hydrated ion

Na+(aq) > K+(aq) > Rb+(aq) > Cs+(aq)

65. Answer (3)

Hint: Half filled and completely filled subshells are

more stable.

Sol.: IE1 order: Li < Be > B

IE2 order: Li >> Be > Mg

66. Answer (1)

Hint: Solubility order :

LiF < NaF < KF

LiCl > NaCl > KCl

67. Answer (4)

Hint: High hydration of Li+ compensates its high

ionisation enthalpy hence best reducing agent.


8/14

68. Answer (4)

Hint: LiNO3 decomposes into oxide on heating.

Sol.: All group-2 nitrates decompose in their oxides

2M(NO3)

2 2MO + 4NO

2 + O

2

69. Answer (3)

Hint: Ca(OCl)2 is calcium hypochlorite

70. Answer (2)

Hint: Average composition of Portland cement

CaO : 50–60%

SiO2 : 20–25%

Al2O

3 : 5–10%

71. Answer (3)

Hint: NH3 + H

2O + CO

2 NH

4HCO

3

72. Answer (2)

Hint: Be can show four co-ordination number.

Sol.: Be(OH)2 + 2OH– [Be(OH)

4]2–

73. Answer (1)

Hint: For alkaline earth metal hydroxides,

decrease in lattice energy is more than decrease in

hydration energy down the group.

74. Answer (3)

Hint: Large size anion is more stabilized by large size

cation.

Sol.: K and Rb form stable superoxides.

75. Answer (2)

Hint: Order of density is

Rb > Na > K > Li

76. Answer (2)

Hint: All alkali metals impart colour in flame test.

Sol.: Due to small size and high nuclear charge, Be

and Mg do not show colour to the flame.

77. Answer (3)

Hint: Cs : [Xe] 6s1

Sol.: Cs has least ionisation energy in its group.

78. Answer (4)

Hint: 4H3BO

3 + Na

2CO

3 Na

2B

4O

7 + 6H

2O + CO

2

Sol.: Na B O 2 4 7

750°C

heat2NaBO + B O

2 2 3

Glassy bead

79. Answer (4)

Hint: Borax is Na2B

4O

7 10 H

2O

Sol.: In borax, two boron atoms are sp2 and two boron

atoms are sp3 hybridised.

80. Answer (3)

Hint: 4BCl + 3LiAlH 2B H + 3LiCl + 3AlCl3 4 2 6 3

[A] [B] [C]

ether

81. Answer (3)

Hint: Electrovalency of alkali metals and phosphide

ion respectively are +1 and –3.

Sol.: Formula of phosphide of alkali metal is M3P.

82. Answer (4)

Hint: H3BO

3 is a Lewis acid.

Sol.: HO B

HO

HO

B

OH

OH

–

OH

H+ H

+

HO

HO

+

83. Answer (3)

Hint: Inert pair effect

Sol.: Stability of +2 oxidation state (Si < Ge < Sn < Pb)

84. Answer (1)

Hint: 2n

3 n(SiO ) represent cyclic and chain silicates.

85. Answer (4)

Hint: Mixture of CO and N2 is called producer gas.

86. Answer (1)

Hint: Generally beryllium compounds are covalent in

nature.

87. Answer (2)

Hint: Graphite has 2D sheet like structure and has

unpaired electrons.

Sol.: Graphite is thermodynamically more stable than

diamond.

88. Answer (3)

Hint: SiCl4 has tetrahedral structure

Sol.:

H

H

H HN

+

Tetrahedral

Cl

Cl

B,Cl

O–

Al–

Tetrahedral

Si

O–

O–

–

O ClCl

Cl

Cl

89. Answer (1)

Hint: Lewis acids are electron deficient species.

Sol.: B2H

6 acts as a Lewis acid

90. Answer (2)

Hint: 2Na + 2NH3

2NaNH

2 + H

2


9/14

[ BIOLOGY]

91. Answer (4)

Hint: Nitrogen is not absorbed by the plants in the

gaseous form

Sol.: Plants absorb nitrogen in the form of 3

NO ,

2NO

and 4

NH only. The major absorption occurs in

the form of 3

NO .

92. Answer (4)

Hint: Macronutrients are found in plants in excess of

10 mmole kg–1 of dry matter.

Sol.: Macronutrients are nine in number (C, H, O, N,

P, K, Ca, Mg, S).

93. Answer (3)

Hint: Sulphur is not found in nucleosides and nucleic

acids.

Sol.: Deoxyadenosine is a nucleoside which does

not have sulphur.

94. Answer (4)

Hint: Element which activates carboxylases is also

required for synthesis of a growth hormone i.e.

auxin.

Sol.: Zn activates various enzymes especially

carboxylases. Nitrite reductase contains copper and

iron.

95. Answer (3)

Hint: Deficiency symptoms of immobile elements

appear first in young leaves.

Sol.: Ca is an immobile element which is required

for the formation of mitotic spindles.

96. Answer (1)

Hint: This element is an important component of cell

membrane.

Sol.: Phosphorus (P) does not help in maintaining

cation - anion balance in cells.

97. Answer (4)

Hint: Toxicity of Mn affects uptake of some other

elements & causes their deficiencies.

Sol.: Excess of Mn reduces uptake of Mg and Fe.

Its excess also inhibits translocation of Ca to shoot

apex and produces brown spots surrounded by

chlorotic veins.

98. Answer (4)

Hint: Symplast comprises cytoplasm and vacuole

Sol.: Metabolic phase involves movement of ions into

symplast. It is an active process and transmembrane

proteins are required for this phase of mineral

absorption.

99. Answer (3)

Hint: It is the first step of nitrification.

Sol.: Nitrification is carried out by a group of nitrifying

bacteria which are free living and chemoautotrophs.

100. Answer (4)

Hint: Nitrogenase is found in all those bacteria which

convert N2 into NH

3.

Sol.: Nitrococcus is a nitrifying bacterium whereas

rest of all are N2 fixing bacteria.

101. Answer (2)

Hint: Rhizobium is involved in nitrogen fixation for

which nitrogenase is required.

Sol.: Regions of nodule where nitrogen fixation

occurs have leghaemoglobin, nitrogenase, ATP and

lack molecular oxygen. Nitrogenase is highly

sensitive to molecular oxygen.

102. Answer (2)

Hint: Plants can assimilate both 4

NH and

3NO

.

Sol.: Rhizobium is aerobic and free living in soil but

fixes N2 in anaerobic condition symbiotically. Nodule

is formed by division of cells of inner cortex and

pericycle.

103. Answer (3)

Hint: Given conversion shows reductive amination.

Sol.: A - - ketoglutaric acid

B - NADPH

C - Glutamate dehydrogenase

D - NADP

104. Answer (3)

Hint: Amides have more nitrogen than amino acids.

Sol.: Hydroxyl part of amino acids is replaced by

another amino group to form amides. Asparagine &

Glutamine are two important amides.

105. Answer (2)

Hint: Some plants transport nitrogen in other forms

which have high nitrogen to carbon ratio.

Sol.: Ureides have high N/C value thus high amount

of nitrogen is transported in this form.

106. Answer (4)

Sol.: Hydroponics is very useful technique but its

set up requires high cost to maintain.

107. Answer (1)

Hint: Death of tissue is known as necrosis.

Sol.: Necrosis is caused by deficiency of elements

such as Ca, Cu, Mg, K.


10/14

108. Answer (3)

Sol.: Bacillus vulgaris converts organic nitrogen into

ammonia, called ammonification.

109. Answer (1)

Hint: These elements are absorbed in the form of

cation and anion respectively.

Sol.: Fe and S are absorbed as Fe+3 and 2

4SO

respectively. These are constituents of ferredoxin.

110. Answer (1)

Hint: These elements are either components of

chlorophyll or required for its synthesis.

Sol.: Mg is an important component of chlorophyll

while K & Fe are required for chlorophyll synthesis.

111. Answer (4)

Hint: First action spectrum was prepared by

measuring O2 evolution.

Sol.: T.W. Engelmann used a green alga Cladophora

and aerobic bacteria to measure O2

evolution and

prepared first action spectrum of photosynthesis

112. Answer (4)

Hint: Carotenoids are yellow-orange pigments found

in green plants.

Sol.: Carotenoids along with xanthophyll &

chlorophyll b are called accessory pigments. They

absorb wavelength between 400 - 700 nm (visible

spectrum) only and prevent photo-oxidation.

113. Answer (1)

Hint: PS II is found in grana lamellae only.

Sol.: PS I is found on the outer surface while PS II

towards inner surface of thylakoids.

Both are involved in non cyclic photophosphorylation

but only PS II is associated with release of O2.

114. Answer (1)

Hint: Reaction centre of PS I absorbs light at wave

length 700 nm.

Sol.: Cyclic photophosphorylation occurs at

1. Light wavelength beyond 680 nm.

2. Anaerobic condition

3. Low CO2 availability

115. Answer (4)

Hint: Chemiosmosis theory involves ATP synthesis

via ATP synthase.

Sol.: Only protons are diffused back to stroma to

break the proton gradient between lumen and stroma.

116. Answer (3)

Hint: Molecules which provide energy for CO2 fixation

are called assimilatory power

Sol.: ATP and NADPH are assimilatory power for

dark reaction.

117. Answer (4)

Hint: Calvin cycle involves three steps of CO2

fixation.

Sol.: Deamination is the process which involves

removal of amine group. It does not occur during C3

cycle.

118. Answer (2)

Hint: C4 plants can perform photosynthesis at high

temperature also.

Sol.: C4 plants are adapted to dry tropical regions.

119. Answer (4)

Hint: C3 plants can show dual activity of RuBisCO.

Sol.: Maize, Sorghum, Sugarcane are C4 plants

Wheat is a C3 plant which can show carboxylation

as well as oxygenation (during photorespiration) by

RuBisCO.

120. Answer (3)

Hint: Potato is a C3 plant, while Amaranthus is a C

4

plant

Sol.: C3 cycle in potato - Mesophyll cells

C3 cycle in Amaranthus - Bundle sheath cells.

121. Answer (3)

Hint: RuBisCO is the most abundant enzyme of the

world.

Sol.: RuBisCO is heat sensitive but has greater

affinity for CO2 than O

2 when CO

2 : O

2 is nearly

equal.

122. Answer (4)

Sol.: For each molecule of CO2 to be fixed via Calvin

cycle 3 ATP and 2 NADPH molecules are required.

123. Answer (2)

Hint: PGA to PGAL conversion is a reduction

process.

Sol.: Reduction of PGA to PGAL requires ATP as

well as NADPH.

124. Answer (1)

Hint: C4 plants lose less water through transpiration

as compared to C3 plants.

Sol.: C4 plants have

High yield and productivity

C3 plants perform better at high CO

2

concentration as compared to C4 plants.

C4 plants show better water utilisation.

C4 plants lack photorespiration.

125. Answer (2)

Hint: Bundle sheath cells are specialized cells in C4

plants, where RuBisCO is found.


11/14

Sol.: Bundle sheath cells & mesophyll cells both

show CO2 fixation process.

Grana are absent in bundle sheath cells and primary

CO2 fixation product is different in both the cells.

126. Answer (4)

Hint: CAM pathway is also known as diurnal acid

cycle.

Sol.: CAM pathway is less efficient than C4

pathway. In CAM plants stomata open during night

(scotoactive). All chloroplasts are of same type.

127. Answer (1)

Hint: Photorespiration operates during light when

concentration of O2 > CO

2

Sol.: It consumes a part of light energy and prevents

plants from photooxidative damage.

128. Answer (2)

Hint: In C2 cycle, O

2 is used by RuBisCO in

chloroplast.

Sol.: Decarboxylation or release of CO2 occurs in

mitochondria.

129. Answer (2)


temperature.

Sol.: C4 plants have PEP synthetase which is cold

sensitive.

130. Answer (4)

Sol.: Pyruvic acid is a 3-carbon acid while rest all

are 4-carbon acids.

131. Answer (4)

Hint: Light rarely becomes a limiting factor except

plants found in shade or dense forest.

Sol.: Above 10% light intensity of the total sunlight,

light becomes damaging and there is no increase in

photosynthesis.

132. Answer (4)

Hint: At low light intensity photosynthesis rate does

not change.

Sol.: At low light intensity, no plant (C3 or C

4)

respond to high CO2 concentration.

133. Answer (2)

Hint: Optimum external and internal factors will

support maximum photosynthesis.

Sol.:

For C3 plants For C

4 plants

Green leaves Green leaves

Low temperature High temperature

CO2 = 450 IL–1 CO

2 = 360 IL–1

High light intensity High light intensity

134. Answer (2)

Hint: Cyclic photophosphorylation involves PS I only.

Sol.: During cyclic photophosphorylation only ATP is

formed while non-cyclic photophosphorylation yields

ATP, NADPH as well as O2

135. Answer (3)

Hint: To form two molecules of hexose sugar Calvin

cycle operates 12 times.

Sol.: In maize both C3 and C

4 cycles operate. Thus

here 12 C3 & C

4 cycles will occur.

Since 1 C3 cycle requires = 3 ATP

12 C3 cycles require = 12 × 3 = 36 ATP

1 C4 cycle requires = 2 ATP


136. Answer (4)

Hint: Endocrine glands release their secretion in

blood so that they can reach every cell.

Sol.: Neural system transmits messages through

impulses. This involves faster transmission of

information than endocrine system. Nervous system

is not connected directly to every cell while

endocrine glands release hormones in blood that

reach every cell of the body through blood. Neural

response is rapid and lasts for shorter duration as

compared to endocrine system.

137. Answer (4)

Hint: This hormone is also known as somatotrophin.

Sol.: GH hormone in muscles promotes protein

anabolism. Parathormone is related with Ca2+ level in

blood and thymosin helps in maturation of

T-lymphocytes. Cortisol is a stress hormone which

promotes protein catabolism in muscle.

138. Answer (4)

Hint: Somatotrophin is also known as growth

hormone.

Sol.: It regulates body growth by promoting protein

anabolism. Somatotrophin is released from

adenohypophysis or anterior pituitary gland which is

under the control of GHRH from hypothalamus.

139. Answer (3)

Hint: This hormone requires iodine for its synthesis.

Sol.: Deficiency of thyroxine during intrauterine life

causes cretinism characterised by stunted growth,

mental retardation and deaf mutism.

140. Answer (2)

Hint: Endocrine gland present in neck region.

Sol.: Thyroid is the largest endocrine gland in

humans. It secretes mainly three hormones

thyroxine or T4, T

3 and thyrocalcitonin(TCT).


12/14

141. Answer (3)

Hint: Hormone which increases Ca2+ level in blood.

Sol.: Hyposecretion of parathormone decreases the

Ca2+ level in blood that leads to tetany.

Hyposecretion of estrogen causes osteoporosis in

females. Hyposecretion of thyroxine causes

Myxedema in adults. Deficiency of vasopressin/ADH

causes diabetes insipidus.

142. Answer (3)

Hint: Zona reticularis is the innermost layer of

adrenal cortex.

Sol.: Zona reticularis layer secretes mainly

androgenic corticoids i.e., androstenedione,

dehydroepiandrosterone. Aldosterone is secreted from

zona glomerulosa and cortisol from zona fasciculata

of adrenal cortex. Epinephrine is secreted from

adrenal medulla.

143. Answer (1)

Hint: Hyposecretion of glucocorticoids and

mineralocorticoids causes this disease.

Sol.: Addison’s disease occurs due to deficiency of

aldosterone and cortisol. Cushing’s disease is

caused due to hypersecretion from adrenal cortex.

Exopthalmic goitre occurs due to hypersecretion

from thyroid gland. Conn’s disease occurs due to

hypersecretion of aldosterone.

144. Answer (3)

Hint: Deficiency of this hormone causes diabetes

mellitus.

Sol.: Insulin is a hypoglycemic hormone in humans

which increases uptake of glucose into cells from

blood. Glucagon, thyroxine and growth hormone can

result in hyperglycemia.

145. Answer (1)

Hint: These cells are also known as interstitial cells.

Sol.: Sertoli cells are also known as sustentacular

cells. They release ABP and inhibin. Spermatozoa

formation requires testosterone.

146 Answer (1)

Hint: Hormone secreted by thymus gland.

Sol.: Thymosin is a peptide hormone secreted by

thymus gland. This hormone helps in development of

immunity. T3, T

4 and TCT are hormones of thyroid

gland. Catecholamines are released during

emergency from adrenal medulla.

147. Answer (4)

Hint: Amylase is a carbohydrate digesting enzyme.

Sol.: Digestion of carbohydrates starts in mouth but

does not occur in stomach due to absence of gastric

amylase.

148. Answer (4)

Hint: These hormones are released by adrenal

medulla.

Sol.: Adrenal medulla is a sympathetic ganglia that

releases stress hormones. These hormones

produced by post ganglionic neurons are transported

through blood and are known as neurohormones. eg;

epinephrine and norepinephrine.

149. Answer (2)

Hint: Identify the disease characterised by hirsutism

and masculinization in females.

Sol.: ACTH is a pituitary hormone which acts on

adrenal gland and increases secretion from adrenal

cortex that results in Cushing’s disease. Deficiency

of insulin results in Diabetes mellitus. Deficiency of

hormones of adrenal cortex causes Addison’s

disease while Grave’s disease occurs due to

hypersecretion of thyroid gland.

150. Answer (1)

Hint: This hormone is secreted by -cells of

pancreas.

Sol.: Glucagon is a peptide hormone that requires

membrane bound receptors, whereas estrogen,

iodothyronines and cortisol require intra-cellular

receptors.

151. Answer (4)

Hint: Identify a neurohormone from hypothalamus.

Sol.: ADH is produced by hypothalamus and stored

in posterior pituitary. Pars distalis also known as

anterior pituitary secretes PRL, TSH and ACTH.

152. Answer (1)

Hint: This hormone is mainly responsible for

ovulation.

Sol.: After ovulation, remaining part of Graafian

follicle that changes into corpus luteum is maintained

by LH in a non-pregnant female. FSH is responsible

for growth and development of ovarian follicles.

Corpus luteum secretes progesterone followed by

estrogen during luteal phase.

153. Answer (3)

Hint: This gland is attached to epithalamus.

Sol.: Pineal gland is present on the roof of third

ventricle. Hypothalamus is present at the base of

thalamus.

154. Answer (1)

Hint: Both gustatoreceptors and olfactory receptors

are chemoreceptors.

Sol.: The receptors which receive chemical stimuli are

known as chemoreceptors and mechanoreceptors

receive mechanical stimuli. Algesireceptors receive

stimulus that is perceived as pain.


13/14

155. Answer (4)

Hint: This hormone is released from adrenal medulla.

Sol.: This hormone increases alertness and causes

pupillary dilation.

156. Answer (2)

Hint: This hormone is released by outer most layer

of adrenal cortex.

Sol.: Outer most layer of adrenal cortex called zona

glomerulosa releases aldosterone. This hormone

increases the reabsorption of Na+ in distal part of

nephron. Water is also reabsorbed following Na+

reabsorption. ANP (atrial natriuretic peptide) released

from atrial wall of heart affects water and electrolyte

balance but is not a corticoid. Melatonin and

testosterone are also not corticoids.

157. Answer (2)

Hint: This hormone is released in response to

increase in blood pressure.

Sol.: Atrial natriuretic factor causes vasodilation and

increases natriuresis (increased excretion of Na+ in

urine), thereby lowering blood volume and blood

pressure.

158. Answer (3)

Hint: This hormone increases thickness of

endometrium during proliferative phase of menstrual

cycle.

Sol.: Estrogen in females and testosterone in males

control secondary sexual characters. Progesterone

maintains pregnancy. FSH promotes gametogenesis.

159. Answer (4)

Hint: Testosterone increases muscular growth during

adolescence.

Sol.: Testosterone in males controls secondary

sexual characters and has anabolic effect on

proteins and carbohydrates.

160. Answer (4)

Hint: Disc shaped receptor sensitive to touch.

Sol.: Merkel’s disc is a disc shaped receptor

present in epidermis of our skin. Receptors such as

Pacinian corpuscle, Meissner’s corpuscle and

Ruffini’s organ are found below epidermis.

161. Answer (3)

Hint: This hormone is synthesised and released by

anterior pituitary.

Sol.: Prolactin is responsible for milk synthesis and

secretion from cells of alveoli of mammary gland.

Oxytocin is responsible for ejection of milk (Milk let

down hormone).

162. Answer (4)

Hint: Identify eye related defects.

Sol.: In myopia, image is formed in front of retina.

This defect is corrected by concave lens.

163. Answer (4)

Hint: Identify the gland that regulates diurnal rhythm.

Sol.: Melatonin released by pineal gland regulates

biological clock.

164. Answer (2)

Hint: Identify a vasoconstrictor.

Sol.: ANF is released in response to increase in

blood pressure while vasopressin is released in

response to low blood pressure. Insulin increases

glucose absorption into cells and aldosterone

increases Na+ reabsorption in nephrons.

165. Answer (4)

Hint: Identify hypothalamic hormone called GHIH.

Sol.: Pineal gland secretes melatonin and prolactin

is secreted by anterior pituitary. Erythropoietin is

secreted by JG cells of kidney. Somatostatin is also

called GHIH.

166. Answer (4)

Hint: This hormone is known as stress hormone.

Sol.: Adrenaline is derived from tyrosine and

functions during emergency conditions such as fear,

fight and flight

167. Answer (4)

Hint: Identify a ductless gland.

Sol.: Parathyroid gland is situated on dorsal side of

thyroid gland while salivary, lacrimal and mammary

glands are exocrine glands.

168. Answer (4)

Hint: This molecule is called energy currency of the

cell.

Sol.: Hormones which are hydrophilic in nature bind

to extracytoplasmic receptors present on plasma

membrane and generates a second messenger

which leads to activation of enzyme cascade.

Common secondary messengers are Ca2+, cAMP

and cGMP. ATP is not a secondary messenger.

169. Answer (4)

Hint: Sensory cells in ear

Sol.: Hair cells are present in organ of corti are

sensory in nature.

170. Answer (4)

Hint: Cochlea is composed of both bony and

membranous labyrinth.

Sol.: Scala media is filled with endolymph while

scala vestibuli and scala tympani are filled with

perilymph.


14/14

� � �

171. Answer (2)

Hint: This part of ear is involved in hearing.

Sol.: Utricle, saccule and cristae help in maintaining

body equilibrium.

172. Answer (3)

Hint: This membrane is found in scala media.

Sol.: Sensory hair cells are pressed against tectorial

membrane upon receiving stimulus. Reissner’s and

Basilar membrane form roof and floor of scala media

respectively.

173. Answer (3)

Hint: This pigment is also called rhodopsin.

Sol.: Rods are sensitive to dim light due to

presence of rhodopsin

174. Answer (2)

Hint: Part that gives white appearance to eye.

Sol.: Sclera is outermost layer, choroid is middle

layer and retina is innermost layer of eye.

175. Answer (4)

Hint: This structure connects middle ear to pharynx.

Sol.: Eustachian tube connects middle ear to

pharynx so that atmospheric pressure remains equal

on both sides of ear drum.

176. Answer (3)

Hint: This is the swollen part of a semicircular canal

Sol.: Cristae are associated with ampulla in inner ear

and help in maintaining equilibrium.

177. Answer (2)

Hint: Identify a steroid hormone.

Sol.: Norepinephrine, oxytocin and TCT are lipid

insoluble hormones hence, they cannot cross the

lipid bilayer. They execute their action through

extracellular receptors. Estrogen is lipid soluble,

steroid hormone which acts through intracellular

receptors.

178. Answer (4)

Hint: Some hormones are steroidal in nature or

derivative of an amino acid.

Sol.: Estrogen, progesterone, cortisol, etc are

steroidal in nature while catecholamines and

thyroxine are amino acid derivatives.

179. Answer (3)

Hint: This organ is associated with sense of smell

Sol.: Modified pseudostratified epithelium/

Schneidarian membrane is associated with roof of

nasal cavity

180. Answer (4)

Hint: This disease results from hyperthyroidism.

Sol.: Exopthalmic goitre occurs due to hyper

secretion of thyroxine hormone. High secretion of

GH after puberty and before puberty causes

acromegaly and gigantism respectively.

Test - 5 (Code-D) (Answers) All India Aakash Test Series for Medical-2020

1/14

1. (3)

2. (2)

3. (2)

4. (2)

5. (2)

6. (1)

7. (3)

8. (1)

9. (4)

10. (2)

11. (3)

12. (2)

13. (4)

14. (1)

15. (4)

16. (4)

17. (2)

18. (2)

19. (4)

20. (2)

21. (2)

22. (1)

23. (2)

24. (4)

25. (3)

26. (2)

27. (1)

28. (4)

29. (1)

30. (2)

31. (3)

32. (1)

33. (2)

34. (2)

35. (1)

36. (2)

Test Date : 27/01/2019

ANSWERS

TEST - 5 (Code-D)

All India Aakash Test Series for Medical-2020

37. (4)

38. (4)

39. (1)

40. (3)

41. (1)

42. (3)

43. (2)

44. (4)

45. (2)

46. (2)

47. (1)

48. (3)

49. (2)

50. (1)

51. (4)

52. (1)

53. (3)

54. (4)

55. (3)

56. (3)

57. (4)

58. (4)

59. (3)

60. (2)

61. (2)

62. (3)

63. (1)

64. (2)

65. (3)

66. (2)

67. (3)

68. (4)

69. (4)

70. (1)

71. (3)

72. (4)

73. (2)

74. (3)

75. (1)

76. (3)

77. (2)

78. (4)

79. (3)

80. (1)

81. (4)

82. (2)

83. (1)

84. (3)

85. (2)

86. (2)

87. (4)

88. (1)

89. (4)

90. (3)

91. (3)

92. (2)

93. (2)

94. (4)

95. (4)

96. (4)

97. (2)

98. (2)

99. (1)

100. (4)

101. (2)

102. (1)

103. (2)

104. (4)

105. (3)

106. (3)

107. (4)

108. (2)

109. (4)

110. (3)

111. (4)

112. (1)

113. (1)

114. (4)

115. (4)

116. (1)

117. (1)

118. (3)

119. (1)

120. (4)

121. (2)

122. (3)

123. (3)

124. (2)

125. (2)

126. (4)

127. (3)

128. (4)

129. (4)

130. (1)

131. (3)

132. (4)

133. (3)

134. (4)

135. (4)

136. (4)

137. (3)

138. (4)

139. (2)

140. (3)

141. (4)

142. (2)

143. (3)

144. (3)

145. (2)

146. (4)

147. (4)

148. (4)

149. (4)

150. (4)

151. (4)

152. (2)

153. (4)

154. (4)

155. (3)

156. (4)

157. (4)

158. (3)

159. (2)

160. (2)

161. (4)

162. (1)

163. (3)

164. (1)

165. (4)

166. (1)

167. (2)

168. (4)

169. (4)

170. (1)

171. (1)

172. (3)

173. (1)

174. (3)

175. (3)

176. (2)

177. (3)

178. (4)

179. (4)

180. (4)

All India Aakash Test Series for Medical-2020 Test - 5 (Code-D) (Answers & Hints)

2/14

ANSWERS & HINTS

1. Answer (3)

Hint: Isochoric process, i.e, V = 0

Sol.: From Ist law

Q = U + W

Q = U

All heat absorbed will be used in increasing the

internal energy of the gas.

2. Answer (2)

Hint: Boyle’s law.

Sol.: PV = constant.

P0V

0 = P × 2V

0

0

2

PP

3. Answer (2)

Hint: Carnot

Sol.: Carnot

1–⎛ ⎞

⎜ ⎟⎝ ⎠

L

H

T

T

max

3031– 0.284

423

⎛ ⎞ ⎜ ⎟⎝ ⎠

= 28.4% maximum possible efficiency.

Efficiency may be 25%

4. Answer (2)

Hint: For adiabatic Q = 0

Sol.: Q = U + W

U = –W

As in expansion W = Positive. Hence U =

Negative so temperature of the gas will decrease.

5. Answer (2)

Hint and solution : Isobaric process is at constant

pressure, isochoric process is at constant volume

and isothermal process is at constant temperature.

6. Answer (1)

Hint: PV = nRT

Sol.: nR

P TV

⎛ ⎞ ⎜ ⎟⎝ ⎠

P

T

V2

V1

P2

P1

0

For a given temperature, P2 > P

1

V2 < V

1

[ PHYSICS]

7. Answer (3)

Hint: Coefficient of performance 2QK

W

Sol.: 2

–

L

H L

T QK

T T W

P

273 2000

45

2000 45330 W

273

�P

8. Answer (1)

Hint: PVN = constant

Sol.: PVN = constant

Molar heat capacity of the gas.

5 9

11– 2 21–

2

v

R R R RC C

N

9. Answer (4)

Hint: Q = U + W

Sol.: Isobaric process

Q = Cv R

Q = 3RRQ = 4RW = R

4 :1Q

W⇒

10. Answer (2)

Hint: Conversion of P-V graph into V-T graph.

Sol.: A Carnot cycle consist of isothermal

expansion, adiabatic expansion, isothermal

compression and adiabatic compression respectively.

11. Answer (3)

Hint: For adiabatic process Q = 0

Sol.: 0 Q

ST

S will remain same.

12. Answer (2)

Hint: V

V T

Sol.: PT = constant = K

Test - 5 (Code-D) (Answers & Hints) All India Aakash Test Series for Medical-2020

3/14

Gas equation PV = RT

KV RT

T

V T2

dV 2T dT

dV dT

V T

2⇒

dV

VdT T

2

13. Answer (4)

Hint and solution : If two bodies are in thermal

equilibrium then both the bodies are at same

temperature.

14. Answer (1)

Hint: Q = U + W

Sol.: Q = U + W

–100 = U + 20

U = –120 J

15. Answer (4)

Hint: Change in internal energy U = CVT

Sol.:

Since A A B B

A B

P V P V

T T

3 33 10 1 10 3

A BT T

TA = T

B

T = TB – T

A = 0

U = 0

16. Answer (4)

Hint: Change in internal energy is the function of

state only U1 = U

2 = U

3

Sol.: Q = U + W

W = area under P – V diagram

Hence W1 > W

2 > W

3

Q1 > Q

2 > Q

3

17. Answer (2)

Hint: AB is Lsothermal process.

Sol.: Work done = i

F

PRT

P102.303 log

= 2.303 × 2 × 8.3 × 300 log1

2

= – 2.303 × 2 × 8.3 × 300 × 0.3010

= – 3452 J

Negative sign indicates that work is done on the gas.

18. Answer (2)

Hint: For fixed mass of gas at constant pressure

V T.

Sol.: V

T

For fixed mass of gas at constant pressure V T.

19. Answer (4)

Hint: Apply Newton’s law of cooling.

Sol.: 1 2 1 20

––

2

⎡ ⎤ ⎢ ⎥⎣ ⎦

Kt

60 – 50 60 50– 20

9 2

⎡ ⎤⇒ ⎢ ⎥

⎣ ⎦K ...(i)

50 – 50– 20

9 2

⎡ ⎤ ⎢ ⎥⎣ ⎦

K ...(ii)

10 35

50 –25 – 20

2

= 42.5ºC

20. Answer (2)

Hint: th

Temperature differenceHeat flow =

Thermal resistance

R

Sol.: Temperature difference remains same 30ºC,

but thermal resistance eq

1 1 1 R R R

eq2

RR

Hence heat flow will double i.e. 20 J/s

21. Answer (2)

Hint: P = eAT4

Sol.: P = eAT4

Now as radius is doubled. Hence surface area will

become 4 times i.e. 4A.

P A

P 4A

P = 4P = 4 × 300 = 1200 W


4/14

22. Answer (1)

Hint: V V

Sol.: – V P

V B

And V

V

P

B⇒

P

B⇒

23. Answer (2)

Hint: If CP & C

V are molar specific heats then

CP – C

V = R

Sol.: Molar specific heat = M × specific heat

McP – Mc

V = R

–P V

Rc c

M

For helium = 4

R p

For oxygen 32

R q

p = 8q

24. Answer (4)

Hint: For linear scale Y = mx + c

Sol.: ( C)º

�(cm)50 cm

– 10º C

80º C

0

9– 10

5 �

When � = 70 cm

970 –10

5

= 116ºC

25. Answer (3)

Hint: Q = mL

Sol.: Heat given by steam = Heat taken by ice

mLv + mc= mL

1

m × 540 + m × 1 × 100 = 80 × 80

80 80

10 g640

m

Mass of water present = 80 + 10 = 90 g

26. Answer (2)

Hint: Heat current will be same as connected in

series.

Sol.: 0 0( – 100)2 (200 – )3

2 3

KA KAH

� �

0 – 100 = 200 –

0

20 = 300

0 = 150°C

27. Answer (1)

Hint: H = (m + w)c Sol.: H = (m + w)c = 150 × 1 × 50 = 7500 cal

= 7.5 kcal

28. Answer (4)

Hint: = 2

Sol.: 100 2% ⇒ L

L LL

100 2% ⇒ r

r rr

Area = r2

2100 100 4%

⇒ dA r

A r

29. Answer (1)

Hint: vD

R

Sol.: 2

4

D vQ Av

Or, 2

4Qv

D

So,2

4 4 D Q Q

RD D

30. Answer (2)

Hint: 1 2

1 2

mix

m md

V V

Sol.: 1 2 1 2 1 2

1

1 22

m m dV d V d d

V V V V


5/14

1 2 1 2

2

1 2 1 2

1 2

2m m d dm m

m mV V d d

d d

Now,

2

1 2 1 2 1 2 1 2

1 2

1 2 1 2

2 ( ) 4

2 2( )

d d d d d d d d

d d d d

2

1 2

1 2

( )0

2( )

d d

d d

So, 1 >

2

31. Answer (3)

Hint: PowerW P V

t t

Sol.: P = hg = 1.5 × 103 × 10 N/m2

–650 10 1.25

V

t

m3/s

Power = 15 × 103 × 50 × 1.25 × 10–6 = 0.94 W

32. Answer (1)

Hint: When anything lighter is unloaded no change

in the liquid level.

Sol.: Let mass of cork ball is M.

Mg = upthrust when ball is inside the boat.

Mg = Vin�g

Liquid displaced (Vin) =

M

�

...(i)

When ball is dropped in water, still Mg = upthrust

in

M

V�

...(ii)

Boat is floating in both cases. So total volume of

water displaced in both cases are same. Hence no

change in liquid level.

33. Answer (2)

Hint: Use Bernoulli's equation.

Sol.: 2

0 0

14 4

2P gh P v

2v gh

2 10 5

v = 10 m/s

34. Answer (2)

Hint: By equation of continuity.

Sol.: A1v

1 = A

2v

2 + A

3v

3

4 × A = 1 × 2A + v × 3A

2v m/s

3

35. Answer (1)

Hint: Excess pressure inside soap bubble = T

r

4

Sol.: 4T

h gr

r

–3 34 0.033 10 0.9 10 10

2

24 3 10 410 m

3 9 9

r

Surface area = 4r2 × 2 = 496.2 × 10–6 m2

36. Answer (2)

Hint: Air pressure decreases with increase in volume.

Sol.: P0 = hg + P

air

When tube will be pulled slightly the volume of air

above mercury column will increase hence pressure

will decrease.

So, height of mercury column will increase.

Hence slightly more than 70 cm.

37. Answer (4)

Hint: Elongation in the wire is proportional to the

tension in the wire.

Sol.: In case A, T = Mg

2 2 3In case B, 2.4

2 3

M M g T Mg

M M

T �

T �

2.42.4

Mg

Mg⇒ ⇒

�

� ��

38. Answer (4)

Hint: Breaking stress (B.S.) F

A

Sol.: B.S. �

�F Mg A g

gA A A

B.S is independent of area of cross-section.

Hence, maximum length to be hung is �.


6/14

46. Answer (2)

Hint: 2Na + 2NH3

2NaNH

2 + H

2

47. Answer (1)

Hint: Lewis acids are electron deficient species.

Sol.: B2H

6 acts as a Lewis acid

48. Answer (3)

Hint: SiCl4 has tetrahedral structure

Sol.:

H

H

H HN

+

Tetrahedral

Cl

Cl

B,Cl

[ CHEMISTRY]

39. Answer (1)

Hint: Poisson's ratio = –�

�

dr

r

d

Sol.: V = r2 × �

2⇒ �

�

dV dr d

V r

–2×0.3 0.4⇒ � � �

� � �

dV d d d

V

100 0.4 100 0.4 1% 0.4%⇒ �

�

dV d

V

40. Answer (3)

Hint: Properties of solids

Sol.: Fluids have only Bulk modulus.

41. Answer (1)

Hint: PV = constant

Sol.: As vessel is open pressure will be constant

and equal to the atmospheric pressure and

neglecting expansion of vessel so volume is also

constant

PV = RT

1T

1 =

2T

2

1

× 300 = 2

× 400

1

2

3

4

1Fraction of mass that will escape

4⇒

42. Answer (3)

Hint: Q = CvT

Sol.: Q = CvT

2 330

20 2R 5

9K C F

⎡ ⎤ ⎢ ⎥⎣ ⎦∵

= 9 cal

43. Answer (2)

Hint: Dalton’s law of partial pressure.

Sol.: 1 2

1 10.082 300

( ) 4 4

5

⎛ ⎞ ⎜ ⎟ ⎝ ⎠ RT

PV

= 2.5 × 105 = 2.5 atm

44. Answer (4)

Hint: rms

v T

Sol.: As temperature increases, speed also

increases. Hence number of collision per second will

increase as well as momentum change per collision

will also increase. Hence force to the container wall

will increase so pressure will increase.

45. Answer (2)

Hint: CO molecule is diatomic.

Sol.: CO molecule is diatomic. At moderate

temperature it will be having 7 degrees of freedom,

as vibrational mode is also excited.

O–

Al–

Tetrahedral

Si

O–

O–

–

O ClCl

Cl

Cl

49. Answer (2)

Hint: Graphite has 2D sheet like structure and has

unpaired electrons.

Sol.: Graphite is thermodynamically more stable than

diamond.

50. Answer (1)

Hint: Generally beryllium compounds are covalent in

nature.


7/14

51. Answer (4)

Hint: Mixture of CO and N2 is called producer gas.

52. Answer (1)

Hint: 2n

3 n(SiO ) represent cyclic and chain silicates.

53. Answer (3)

Hint: Inert pair effect

Sol.: Stability of +2 oxidation state (Si < Ge < Sn < Pb)

54. Answer (4)

Hint: H3BO

3 is a Lewis acid.

Sol.: HO B

HO

HO

B

OH

OH

–

OH

H+ H

+

HO

HO

+

55. Answer (3)

Hint: Electrovalency of alkali metals and phosphide

ion respectively are +1 and –3.

Sol.: Formula of phosphide of alkali metal is M3P.

56. Answer (3)

Hint: 4BCl + 3LiAlH 2B H + 3LiCl + 3AlCl3 4 2 6 3

[A] [B] [C]

ether

57. Answer (4)

Hint: Borax is Na2B

4O

7 10 H

2O

Sol.: In borax, two boron atoms are sp2 and two boron

atoms are sp3 hybridised.

58. Answer (4)

Hint: 4H3BO

3 + Na

2CO

3 Na

2B

4O

7 + 6H

2O + CO

2

Sol.: Na B O 2 4 7

750°C

heat2NaBO + B O

2 2 3

Glassy bead

59. Answer (3)

Hint: Cs : [Xe] 6s1

Sol.: Cs has least ionisation energy in its group.

60. Answer (2)

Hint: All alkali metals impart colour in flame test.

Sol.: Due to small size and high nuclear charge, Be

and Mg do not show colour to the flame.

61. Answer (2)

Hint: Order of density is

Rb > Na > K > Li

62. Answer (3)

Hint: Large size anion is more stabilized by large size

cation.

Sol.: K and Rb form stable superoxides.

63. Answer (1)

Hint: For alkaline earth metal hydroxides,

decrease in lattice energy is more than decrease in

hydration energy down the group.

64. Answer (2)

Hint: Be can show four co-ordination number.

Sol.: Be(OH)2 + 2OH– [Be(OH)

4]2–

65. Answer (3)

Hint: NH3 + H

2O + CO

2 NH

4HCO

3

66. Answer (2)

Hint: Average composition of Portland cement

CaO : 50–60%

SiO2 : 20–25%

Al2O

3 : 5–10%

67. Answer (3)

Hint: Ca(OCl)2 is calcium hypochlorite

68. Answer (4)

Hint: LiNO3 decomposes into oxide on heating.

Sol.: All group-2 nitrates decompose in their oxides

2M(NO3)

2 2MO + 4NO

2 + O

2

69. Answer (4)

Hint: High hydration of Li+ compensates its high

ionisation enthalpy hence best reducing agent.

70. Answer (1)

Hint: Solubility order :

LiF < NaF < KF

LiCl > NaCl > KCl

71. Answer (3)

Hint: Half filled and completely filled subshells are

more stable.

Sol.: IE1 order: Li < Be > B

IE2 order: Li >> Be > Mg

72. Answer (4)

Hint: Mobility decreases with increase in the extent

of solvation.

Sol.: Order of size of hydrated ion

Na+(aq) > K+(aq) > Rb+(aq) > Cs+(aq)

73. Answer (2)

Hint:

F

B

F F

Sol.: Acidic strength : BI3 > BBr

3 >BCl

3 > BF

3


8/14

74. Answer (3)

Hint: Boron nitride has 2 dimensional sheet like

structure.

75. Answer (1)

Hint: Al forms a protective layer of Al2O

3 with conc.

HNO3.

76. Answer (3)

Hint: Cl

Cl

Cl

Cl

Al

Cl

Cl

101°Al

77. Answer (2)

Hint: Slaked lime is Ca(OH)2

Sol.: Slaked lime is prepared by adding water to quick

lime (CaO)

CaO + H2O Ca(OH)

2

78. Answer (4)

Hint: Thermal stability order : MgCO3 < CaCO

3 < BaCO

3

79. Answer (3)

Hint: All reactive metal releases H2 gas with dil acid

or water.

Sol.: 2NaHCO3 + 2HCl 2NaCl + 2H

2O + 2CO

2

80. Answer (1)

Hint: CH SiCl CH Si(OH) + HCl3 3 3 3

H O2

.

Sol.:

CH Si(OH) 3 3

polymeriseCH

3

– O – Si – O – Si – O

CH3

O O

– O – Si – O – Si – O

CH3

CH3

(cross link polymer)

81. Answer (4)

Hint: Galena is PbS.

Sol.: Mica, zeolite and feldspar are silicates.

82. Answer (2)

Hint: 2 2

wt. of H O%strength= 100

V(ml) of solution

Sol.: 2H2O

2 2H

2O + O

2

22.4 L O2 from 2 34 g H

2O

2 (at NTP)

100 L O2 from

2 34100

22.4

= 303.57 g

= 30.35% H2O

2 solution.

83. Answer (1)

Hint: BeH2 becomes polymer due to its electron

deficient nature through multicentre bridge bonds.

Sol.:

3c - 2e bond

BeH

H

BeH

H

BeH

H

BeH

H

84. Answer (3)

Hint.: Blue colour turns to bronze colour by increasing

the metal concentration leading to the cluster formation

of metal.

85. Answer (2)

Hint: Amphoteric oxides means which will react with

acid and as well as base.

Sol.: SnO2 + 2NaOH Na

2SnO

3 + H

2O

SnO2 + 4HCl SnCl

4 + 2H

2O

86. Answer (2)

Hint: Ga: [Ar] 3d10 4s2 4p1

Sol.: In ‘Ga’ atom, 10 electrons are filled in d orbitals

hence size decreases due to poor shielding of d

electrons.

87. Answer (4)

Hint: In Castner-Kellner cell mercury cathode and

graphite anode are used.

Sol.: Cathode: Hg

Na e Na Hg

Anode: 2

1Cl Cl e

2

Cl2 gas evolved at anode.

88. Answer (1)

Hint: Lattice enthalpy decreases with increase in the

size of cation.

Sol.: The order of thermal stability of hydrides is

LiH > NaH > KH > RbH > CsH

89. Answer (4)

Hint: For solubility, hydration energy > lattice energy

Sol.: The size of Ba2+ and 2

4SO

ions are very large

which leads to higher lattice enthalpy than that of

hydration enthalpy.

90. Answer (3)

Hint: Basic character of oxide increases down the

group.


9/14

[ BIOLOGY]

91. Answer (3)

Hint: To form two molecules of hexose sugar Calvin

cycle operates 12 times.

Sol.: In maize both C3 and C

4 cycles operate. Thus

here 12 C3 & C

4 cycles will occur.

Since 1 C3 cycle requires = 3 ATP


1 C4 cycle requires = 2 ATP


92. Answer (2)

Hint: Cyclic photophosphorylation involves PS I only.

Sol.: During cyclic photophosphorylation only ATP is

formed while non-cyclic photophosphorylation yields

ATP, NADPH as well as O2

93. Answer (2)

Hint: Optimum external and internal factors will

support maximum photosynthesis.

Sol.:

For C3 plants For C

4 plants

Green leaves Green leaves

Low temperature High temperature

CO2 = 450 IL–1 CO

2 = 360 IL–1

High light intensity High light intensity

94. Answer (4)

Hint: At low light intensity photosynthesis rate does

not change.

Sol.: At low light intensity, no plant (C3 or C

4)

respond to high CO2 concentration.

95. Answer (4)

Hint: Light rarely becomes a limiting factor except

plants found in shade or dense forest.

Sol.: Above 10% light intensity of the total sunlight,

light becomes damaging and there is no increase in

photosynthesis.

96. Answer (4)

Sol.: Pyruvic acid is a 3-carbon acid while rest all

are 4-carbon acids.

97. Answer (2)


temperature.

Sol.: C4 plants have PEP synthetase which is cold

sensitive.

98. Answer (2)

Hint: In C2 cycle, O

2 is used by RuBisCO in

chloroplast.

Sol.: Decarboxylation or release of CO2 occurs in

mitochondria.

99. Answer (1)

Hint: Photorespiration operates during light when

concentration of O2 > CO

2

Sol.: It consumes a part of light energy and prevents

plants from photooxidative damage.

100. Answer (4)

Hint: CAM pathway is also known as diurnal acid

cycle.

Sol.: CAM pathway is less efficient than C4

pathway. In CAM plants stomata open during night

(scotoactive). All chloroplasts are of same type.

101. Answer (2)

Hint: Bundle sheath cells are specialized cells in C4

plants, where RuBisCO is found.

Sol.: Bundle sheath cells & mesophyll cells both

show CO2 fixation process.

Grana are absent in bundle sheath cells and primary

CO2 fixation product is different in both the cells.

102. Answer (1)

Hint: C4 plants lose less water through transpiration

as compared to C3 plants.

Sol.: C4 plants have

High yield and productivity

C3 plants perform better at high CO

2

concentration as compared to C4 plants.

C4 plants show better water utilisation.

C4 plants lack photorespiration.

103. Answer (2)

Hint: PGA to PGAL conversion is a reduction

process.

Sol.: Reduction of PGA to PGAL requires ATP as

well as NADPH.

104. Answer (4)

Sol.: For each molecule of CO2 to be fixed via Calvin

cycle 3 ATP and 2 NADPH molecules are required.

105. Answer (3)

Hint: RuBisCO is the most abundant enzyme of the

world.

Sol.: RuBisCO is heat sensitive but has greater

affinity for CO2 than O

2 when CO

2 : O

2 is nearly

equal.


10/14

106. Answer (3)

Hint: Potato is a C3 plant, while Amaranthus is a C

4

plant

Sol.: C3 cycle in potato - Mesophyll cells

C3 cycle in Amaranthus - Bundle sheath cells.

107. Answer (4)

Hint: C3 plants can show dual activity of RuBisCO.

Sol.: Maize, Sorghum, Sugarcane are C4 plants

Wheat is a C3 plant which can show carboxylation

as well as oxygenation (during photorespiration) by

RuBisCO.

108. Answer (2)


temperature also.

Sol.: C4 plants are adapted to dry tropical regions.

109. Answer (4)

Hint: Calvin cycle involves three steps of CO2

fixation.

Sol.: Deamination is the process which involves

removal of amine group. It does not occur during C3

cycle.

110. Answer (3)

Hint: Molecules which provide energy for CO2 fixation

are called assimilatory power

Sol.: ATP and NADPH are assimilatory power for

dark reaction.

111. Answer (4)

Hint: Chemiosmosis theory involves ATP synthesis

via ATP synthase.

Sol.: Only protons are diffused back to stroma to

break the proton gradient between lumen and stroma.

112. Answer (1)

Hint: Reaction centre of PS I absorbs light at wave

length 700 nm.

Sol.: Cyclic photophosphorylation occurs at

1. Light wavelength beyond 680 nm.

2. Anaerobic condition

3. Low CO2 availability

113. Answer (1)

Hint: PS II is found in grana lamellae only.

Sol.: PS I is found on the outer surface while PS II

towards inner surface of thylakoids.

Both are involved in non cyclic photophosphorylation

but only PS II is associated with release of O2.

114. Answer (4)

Hint: Carotenoids are yellow-orange pigments found

in green plants.

Sol.: Carotenoids along with xanthophyll &

chlorophyll b are called accessory pigments. They

absorb wavelength between 400 - 700 nm (visible

spectrum) only and prevent photo-oxidation.

115. Answer (4)

Hint: First action spectrum was prepared by

measuring O2 evolution.

Sol.: T.W. Engelmann used a green alga Cladophora

and aerobic bacteria to measure O2

evolution and

prepared first action spectrum of photosynthesis

116. Answer (1)

Hint: These elements are either components of

chlorophyll or required for its synthesis.

Sol.: Mg is an important component of chlorophyll

while K & Fe are required for chlorophyll synthesis.

117. Answer (1)

Hint: These elements are absorbed in the form of

cation and anion respectively.

Sol.: Fe and S are absorbed as Fe+3 and 2

4SO

respectively. These are constituents of ferredoxin.

118. Answer (3)

Sol.: Bacillus vulgaris converts organic nitrogen into

ammonia, called ammonification.

119. Answer (1)

Hint: Death of tissue is known as necrosis.

Sol.: Necrosis is caused by deficiency of elements

such as Ca, Cu, Mg, K.

120. Answer (4)

Sol.: Hydroponics is very useful technique but its

set up requires high cost to maintain.

121. Answer (2)

Hint: Some plants transport nitrogen in other forms

which have high nitrogen to carbon ratio.

Sol.: Ureides have high N/C value thus high amount

of nitrogen is transported in this form.

122. Answer (3)

Hint: Amides have more nitrogen than amino acids.

Sol.: Hydroxyl part of amino acids is replaced by

another amino group to form amides. Asparagine &

Glutamine are two important amides.

123. Answer (3)

Hint: Given conversion shows reductive amination.

Sol.: A - - ketoglutaric acid

B - NADPH

C - Glutamate dehydrogenase

D - NADP


11/14

124. Answer (2)

Hint: Plants can assimilate both 4

NH and

3NO

.

Sol.: Rhizobium is aerobic and free living in soil but

fixes N2 in anaerobic condition symbiotically. Nodule

is formed by division of cells of inner cortex and

pericycle.

125. Answer (2)

Hint: Rhizobium is involved in nitrogen fixation for

which nitrogenase is required.

Sol.: Regions of nodule where nitrogen fixation

occurs have leghaemoglobin, nitrogenase, ATP and

lack molecular oxygen. Nitrogenase is highly

sensitive to molecular oxygen.

126. Answer (4)

Hint: Nitrogenase is found in all those bacteria which

convert N2 into NH

3.

Sol.: Nitrococcus is a nitrifying bacterium whereas

rest of all are N2 fixing bacteria.

127. Answer (3)

Hint: It is the first step of nitrification.

Sol.: Nitrification is carried out by a group of nitrifying

bacteria which are free living and chemoautotrophs.

128. Answer (4)

Hint: Symplast comprises cytoplasm and vacuole

Sol.: Metabolic phase involves movement of ions into

symplast. It is an active process and transmembrane

proteins are required for this phase of mineral

absorption.

129. Answer (4)

Hint: Toxicity of Mn affects uptake of some other

elements & causes their deficiencies.

Sol.: Excess of Mn reduces uptake of Mg and Fe.

Its excess also inhibits translocation of Ca to shoot

apex and produces brown spots surrounded by

chlorotic veins.

130. Answer (1)

Hint: This element is an important component of cell

membrane.

Sol.: Phosphorus (P) does not help in maintaining

cation - anion balance in cells.

131. Answer (3)

Hint: Deficiency symptoms of immobile elements

appear first in young leaves.

Sol.: Ca is an immobile element which is required

for the formation of mitotic spindles.

132. Answer (4)

Hint: Element which activates carboxylases is also

required for synthesis of a growth hormone i.e.

auxin.

Sol.: Zn activates various enzymes especially

carboxylases. Nitrite reductase contains copper and

iron.

133. Answer (3)

Hint: Sulphur is not found in nucleosides and nucleic

acids.

Sol.: Deoxyadenosine is a nucleoside which does

not have sulphur.

134. Answer (4)

Hint: Macronutrients are found in plants in excess of

10 mmole kg–1 of dry matter.

Sol.: Macronutrients are nine in number (C, H, O, N,

P, K, Ca, Mg, S).

135. Answer (4)

Hint: Nitrogen is not absorbed by the plants in the

gaseous form

Sol.: Plants absorb nitrogen in the form of 3

NO ,

2NO

and 4

NH only. The major absorption occurs in

the form of 3

NO .

136. Answer (4)

Hint: This disease results from hyperthyroidism.

Sol.: Exopthalmic goitre occurs due to hyper

secretion of thyroxine hormone. High secretion of

GH after puberty and before puberty causes

acromegaly and gigantism respectively.

137. Answer (3)

Hint: This organ is associated with sense of smell

Sol.: Modified pseudostratified epithelium/

Schneidarian membrane is associated with roof of

nasal cavity

138. Answer (4)

Hint: Some hormones are steroidal in nature or

derivative of an amino acid.

Sol.: Estrogen, progesterone, cortisol, etc are

steroidal in nature while catecholamines and

thyroxine are amino acid derivatives.

139. Answer (2)

Hint: Identify a steroid hormone.

Sol.: Norepinephrine, oxytocin and TCT are lipid

insoluble hormones hence, they cannot cross the

lipid bilayer. They execute their action through

extracellular receptors. Estrogen is lipid soluble,

steroid hormone which acts through intracellular

receptors.


12/14

140. Answer (3)

Hint: This is the swollen part of a semicircular canal

Sol.: Cristae are associated with ampulla in inner ear

and help in maintaining equilibrium.

141. Answer (4)

Hint: This structure connects middle ear to pharynx.

Sol.: Eustachian tube connects middle ear to

pharynx so that atmospheric pressure remains equal

on both sides of ear drum.

142. Answer (2)

Hint: Part that gives white appearance to eye.

Sol.: Sclera is outermost layer, choroid is middle

layer and retina is innermost layer of eye.

143. Answer (3)

Hint: This pigment is also called rhodopsin.

Sol.: Rods are sensitive to dim light due to

presence of rhodopsin

144. Answer (3)

Hint: This membrane is found in scala media.

Sol.: Sensory hair cells are pressed against tectorial

membrane upon receiving stimulus. Reissner’s and

Basilar membrane form roof and floor of scala media

respectively.

145. Answer (2)

Hint: This part of ear is involved in hearing.

Sol.: Utricle, saccule and cristae help in maintaining

body equilibrium.

146. Answer (4)

Hint: Cochlea is composed of both bony and

membranous labyrinth.

Sol.: Scala media is filled with endolymph while

scala vestibuli and scala tympani are filled with

perilymph.

147. Answer (4)

Hint: Sensory cells in ear

Sol.: Hair cells are present in organ of corti are

sensory in nature.

148. Answer (4)

Hint: This molecule is called energy currency of the

cell.

Sol.: Hormones which are hydrophilic in nature bind

to extracytoplasmic receptors present on plasma

membrane and generates a second messenger

which leads to activation of enzyme cascade.

Common secondary messengers are Ca2+, cAMP

and cGMP. ATP is not a secondary messenger.

149. Answer (4)

Hint: Identify a ductless gland.

Sol.: Parathyroid gland is situated on dorsal side of

thyroid gland while salivary, lacrimal and mammary

glands are exocrine glands.

150. Answer (4)

Hint: This hormone is known as stress hormone.

Sol.: Adrenaline is derived from tyrosine and

functions during emergency conditions such as fear,

fight and flight

151. Answer (4)

Hint: Identify hypothalamic hormone called GHIH.

Sol.: Pineal gland secretes melatonin and prolactin

is secreted by anterior pituitary. Erythropoietin is

secreted by JG cells of kidney. Somatostatin is also

called GHIH.

152. Answer (2)

Hint: Identify a vasoconstrictor.

Sol.: ANF is released in response to increase in

blood pressure while vasopressin is released in

response to low blood pressure. Insulin increases

glucose absorption into cells and aldosterone

increases Na+ reabsorption in nephrons.

153. Answer (4)

Hint: Identify the gland that regulates diurnal rhythm.

Sol.: Melatonin released by pineal gland regulates

biological clock.

154. Answer (4)

Hint: Identify eye related defects.

Sol.: In myopia, image is formed in front of retina.

This defect is corrected by concave lens.

155. Answer (3)

Hint: This hormone is synthesised and released by

anterior pituitary.

Sol.: Prolactin is responsible for milk synthesis and

secretion from cells of alveoli of mammary gland.

Oxytocin is responsible for ejection of milk (Milk let

down hormone).

156. Answer (4)

Hint: Disc shaped receptor sensitive to touch.

Sol.: Merkel’s disc is a disc shaped receptor

present in epidermis of our skin. Receptors such as

Pacinian corpuscle, Meissner’s corpuscle and

Ruffini’s organ are found below epidermis.

157. Answer (4)

Hint: Testosterone increases muscular growth during

adolescence.

Sol.: Testosterone in males controls secondary

sexual characters and has anabolic effect on

proteins and carbohydrates.


13/14

158. Answer (3)

Hint: This hormone increases thickness of

endometrium during proliferative phase of menstrual

cycle.

Sol.: Estrogen in females and testosterone in males

control secondary sexual characters. Progesterone

maintains pregnancy. FSH promotes gametogenesis.

159. Answer (2)

Hint: This hormone is released in response to

increase in blood pressure.

Sol.: Atrial natriuretic factor causes vasodilation and

increases natriuresis (increased excretion of Na+ in

urine), thereby lowering blood volume and blood

pressure.

160. Answer (2)

Hint: This hormone is released by outer most layer

of adrenal cortex.

Sol.: Outer most layer of adrenal cortex called zona

glomerulosa releases aldosterone. This hormone

increases the reabsorption of Na+ in distal part of

nephron. Water is also reabsorbed following Na+

reabsorption. ANP (atrial natriuretic peptide) released

from atrial wall of heart affects water and electrolyte

balance but is not a corticoid. Melatonin and

testosterone are also not corticoids.

161. Answer (4)

Hint: This hormone is released from adrenal medulla.

Sol.: This hormone increases alertness and causes

pupillary dilation.

162. Answer (1)

Hint: Both gustatoreceptors and olfactory receptors

are chemoreceptors.

Sol.: The receptors which receive chemical stimuli are

known as chemoreceptors and mechanoreceptors

receive mechanical stimuli. Algesireceptors receive

stimulus that is perceived as pain.

163. Answer (3)

Hint: This gland is attached to epithalamus.

Sol.: Pineal gland is present on the roof of third

ventricle. Hypothalamus is present at the base of

thalamus.

164. Answer (1)

Hint: This hormone is mainly responsible for

ovulation.

Sol.: After ovulation, remaining part of Graafian

follicle that changes into corpus luteum is maintained

by LH in a non-pregnant female. FSH is responsible

for growth and development of ovarian follicles.

Corpus luteum secretes progesterone followed by

estrogen during luteal phase.

165. Answer (4)

Hint: Identify a neurohormone from hypothalamus.

Sol.: ADH is produced by hypothalamus and stored

in posterior pituitary. Pars distalis also known as

anterior pituitary secretes PRL, TSH and ACTH.

166. Answer (1)

Hint: This hormone is secreted by -cells of

pancreas.

Sol.: Glucagon is a peptide hormone that requires

membrane bound receptors, whereas estrogen,

iodothyronines and cortisol require intra-cellular

receptors.

167. Answer (2)

Hint: Identify the disease characterised by hirsutism

and masculinization in females.

Sol.: ACTH is a pituitary hormone which acts on

adrenal gland and increases secretion from adrenal

cortex that results in Cushing’s disease. Deficiency

of insulin results in Diabetes mellitus. Deficiency of

hormones of adrenal cortex causes Addison’s

disease while Grave’s disease occurs due to

hypersecretion of thyroid gland.

168. Answer (4)

Hint: These hormones are released by adrenal

medulla.

Sol.: Adrenal medulla is a sympathetic ganglia that

releases stress hormones. These hormones

produced by post ganglionic neurons are transported

through blood and are known as neurohormones. eg;

epinephrine and norepinephrine.

169. Answer (4)

Hint: Amylase is a carbohydrate digesting enzyme.

Sol.: Digestion of carbohydrates starts in mouth but

does not occur in stomach due to absence of gastric

amylase.

170 Answer (1)

Hint: Hormone secreted by thymus gland.

Sol.: Thymosin is a peptide hormone secreted by

thymus gland. This hormone helps in development of

immunity. T3, T

4 and TCT are hormones of thyroid

gland. Catecholamines are released during

emergency from adrenal medulla.

171. Answer (1)

Hint: These cells are also known as interstitial cells.

Sol.: Sertoli cells are also known as sustentacular

cells. They release ABP and inhibin. Spermatozoa

formation requires testosterone.


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172. Answer (3)

Hint: Deficiency of this hormone causes diabetes

mellitus.

Sol.: Insulin is a hypoglycemic hormone in humans

which increases uptake of glucose into cells from

blood. Glucagon, thyroxine and growth hormone can

result in hyperglycemia.

173. Answer (1)

Hint: Hyposecretion of glucocorticoids and

mineralocorticoids causes this disease.

Sol.: Addison’s disease occurs due to deficiency of

aldosterone and cortisol. Cushing’s disease is

caused due to hypersecretion from adrenal cortex.

Exopthalmic goitre occurs due to hypersecretion

from thyroid gland. Conn’s disease occurs due to

hypersecretion of aldosterone.

174. Answer (3)

Hint: Zona reticularis is the innermost layer of

adrenal cortex.

Sol.: Zona reticularis layer secretes mainly

androgenic corticoids i.e., androstenedione,

dehydroepiandrosterone. Aldosterone is secreted from

zona glomerulosa and cortisol from zona fasciculata

of adrenal cortex. Epinephrine is secreted from

adrenal medulla.

175. Answer (3)

Hint: Hormone which increases Ca2+ level in blood.

Sol.: Hyposecretion of parathormone decreases the

Ca2+ level in blood that leads to tetany.

Hyposecretion of estrogen causes osteoporosis in

females. Hyposecretion of thyroxine causes

Myxedema in adults. Deficiency of vasopressin/ADH

causes diabetes insipidus.

176. Answer (2)

Hint: Endocrine gland present in neck region.

Sol.: Thyroid is the largest endocrine gland in

humans. It secretes mainly three hormones

thyroxine or T4, T

3 and thyrocalcitonin(TCT).

177. Answer (3)

Hint: This hormone requires iodine for its synthesis.

Sol.: Deficiency of thyroxine during intrauterine life

causes cretinism characterised by stunted growth,

mental retardation and deaf mutism.

178. Answer (4)

Hint: Somatotrophin is also known as growth

hormone.

Sol.: It regulates body growth by promoting protein

anabolism. Somatotrophin is released from

adenohypophysis or anterior pituitary gland which is

under the control of GHRH from hypothalamus.

179. Answer (4)

Hint: This hormone is also known as somatotrophin.

Sol.: GH hormone in muscles promotes protein

anabolism. Parathormone is related with Ca2+ level in

blood and thymosin helps in maturation of

T-lymphocytes. Cortisol is a stress hormone which

promotes protein catabolism in muscle.

180. Answer (4)

Hint: Endocrine glands release their secretion in

blood so that they can reach every cell.

Sol.: Neural system transmits messages through

impulses. This involves faster transmission of

information than endocrine system. Nervous system

is not connected directly to every cell while

endocrine glands release hormones in blood that

reach every cell of the body through blood. Neural

response is rapid and lasts for shorter duration as

compared to endocrine system.

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