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Test - 5 (Code-C) (Answers & Hints) All India Aakash Test Series for Medical-2018

1/7

1. (3)

2. (2)

3. (4)

4. (2)

5. (3)

6. (3)

7. (2)

8. (2)

9. (1)

10. (1)

11. (2)

12. (2)

13. (3)

14. (1)

15. (2)

16. (2)

17. (4)

18. (1)

19. (4)

20. (1)

21. (4)

22. (2)

23. (1)

24. (2)

25. (3)

26. (1)

27. (3)

28. (3)

29. (1)

30. (3)

31. (1)

32. (2)

33. (3)

34. (1)

35. (3)

36. (2)

37. (3)

38. (3)

39. (1)

40. (3)

41. (3)

42. (1)

43. (2)

44. (4)

45. (2)

46. (4)

47. (4)

48. (4)

49. (3)

50. (2)

51. (4)

52. (1)

53. (4)

54. (1)

55. (1)

56. (4)

57. (3)

58. (3)

59. (3)

60. (3)

61. (2)

62. (4)

63. (4)

64. (3)

65. (3)

66. (1)

67. (2)

68. (1)

69. (1)

70. (2)

71. (3)

72. (1)

73. (2)

74. (1)

75. (1)

76. (4)

77. (3)

78. (3)

79. (4)

80. (4)

81. (1)

82. (3)

83. (2)

84. (1)

85. (3)

86. (2)

87. (2)

88. (4)

89. (1)

90. (2)

91. (4)

92. (3)

93. (3)

94. (2)

95. (3)

96. (1)

97. (4)

98. (3)

99. (3)

100. (2)

101. (4)

102. (3)

103. (4)

104. (2)

105. (3)

106. (4)

107. (4)

108. (1)

109. (2)

110. (3)

111. (4)

112. (2)

113. (2)

114. (2)

115. (3)

116. (1)

117. (1)

118. (4)

119. (2)

120. (4)

121. (4)

122. (3)

123. (3)

124. (4)

125. (2)

126. (4)

127. (3)

128. (2)

129. (3)

130. (4)

131. (3)

132. (4)

133. (3)

134. (4)

135. (1)

136. (4)

137. (4)

138. (2)

139. (3)

140. (4)

141. (3)

142. (3)

143. (1)

144. (2)

145. (4)

146. (3)

147. (2)

148. (2)

149. (1)

150. (3)

151. (3)

152. (3)

153. (1)

154. (1)

155. (3)

156. (2)

157. (3)

158. (3)

159. (1)

160. (2)

161. (3)

162. (2)

163. (2)

164. (2)

165. (1)

166. (1)

167. (2)

168. (4)

169. (2)

170. (2)

171. (3)

172. (4)

173. (4)

174. (2)

175. (3)

176. (3)

177. (2)

178. (2)

179. (4)

180. (1)

ANSWERS

TEST - 5 (Code-C)

All India Aakash Test Series for Medical-2018

Test Date : 22-01-2017

All India Aakash Test Series for Medical-2018 Test - 5 (Code-C) (Answers & Hints)

2/7

Hints to Selected Questions

[ PHYSICS]

1. Answer (3)

4

3g RG

2. Answer (2)

m 4mr

A x c B( – )r x

2 2

(4 )

( )

Gm G m

x r x

3

rx

3GmV

r

3. Answer (4)

4. Answer (2)

21

21

mghmv

h

R

⎛ ⎞⎜ ⎟⎝ ⎠

5. Answer (3)

6. Answer (3)

2 2 2

ev v v

7. Answer (2)

5/2

kF

r

2

5/2

mv k

r r

2 rT

v

2 7/2T r

8. Answer (2)

E = –KE

9. Answer (1)

dvI

dr

10. Answer (1)

1 1 1 2 2

2 2 2 1 1

l F l A Y

l F l A Y

11. Answer (2)

PVB

V

12. Answer (2)

l

l/2

2

2 2

mgl gll l

AY AY

⇒

13. Answer (3)

2 1y

14. Answer (1)

15. Answer (2)

16. Answer (2)

A1V1 = na

2v2

17. Answer (4)

18. Answer (1)

19. Answer (4)

20. Answer (1)

2/3

Tv n v

21. Answer (4)

22. Answer (2)

2

12

mlI

2l l

l l

= 2IT

2

3

I T

23. Answer (1)

F Ay T

24. Answer (2)

1

2Tt


3/7

25. Answer (3)

26. Answer (1)

Insufficient heat at MP.

27. Answer (3)

28. Answer (3)

29. Answer (1)

nR T w 30. Answer (3)

1 2 1 2( ) Q Q W W

31. Answer (1)

W P dV ∫0

0

3

4 ∫V

V

V dV

W = 16V0

2

32. Answer (2)

33. Answer (3)

P

V

CQ

U C

34. Answer (1)

1 1

2 2 1 1

r rT V TV

35. Answer (3)

2

1 1

1T W

T Q

36. Answer (2)

Polytropic process (n = –1)

,1

n

v

RPV K C C

n

37. Answer (3)

VPr = Constant

0V P

rV P

⇒

P P

V r

V

⇒

⎛ ⎞⎜ ⎟⎝ ⎠

38. Answer (3)

39. Answer (1)

1 2

eq2

K K

K

40. Answer (3)

L

B

LA

2L

D

C

100°C 40°CH1

H2

HH H1 2 + = H

100 40 20

2

2

T T T

l L L

kA kA kA

41. Answer (3)

Use Newton's law of cooling

1 2 1 2

02

T T T Tk T

t

⎛ ⎞ ⎜ ⎟⎝ ⎠

42. Answer (1)

43. Answer (2)

m

T = b = Wein's constant

44. Answer (4)

45. Answer (2)

[ CHEMISTRY]

46. Answer (4)

3 2CaCO CaO(s) CO (g)

There is no change in oxidation state.

47. Answer (4)

2PbO + 4HCl 2PbCl2 + 2H

2O

(Acid-base reaction)

PbO2 + 4HCl PbCl

2 + Cl

2 + 2H

2O

(Redox reaction)

48. Answer (4)

H2SO

5 : O has –1 and –2 O.S.

H2S2O3 : S has 0 and +4 O.S.

CaOCl2 : Cl has +1 and –1 O.S.

49. Answer (3)

CrO2

– : +3; ClO3

– : +5

CrO4

2– : +6; MnO4

– : +7


4/7

50. Answer (2)

Fact.

51. Answer (4)

Fact.

52. Answer (1)

Average oxidation state of oxygen =2( 1 2 2 2)

8

= –1.75

53. Answer (4)

o

CellEMF =

o oCathode (RP) Anode (RP)E E

= 0.80 – 0.34

= 0.46 V

54. Answer (1)

H – O – S – O – O – H; S has O.S. + 6

O

O

55. Answer (1)

Acidic medium :

KMnO4 Mn2+

ME

5

Basic medium :

KMnO4 K

2MnO

4

ME

1

56. Answer (4)

57. Answer (3)

Fact.

58. Answer (3)

'B' has incomplete octet.

59. Answer (3)

Fact.

60. Answer (3)

Group 7 to 9 is known as hydride gap because

elements of these group do not form hydrides.

61. Answer (2)

Fact.

62. Answer (4)

63. Answer (4)

2H2O2 2H

2O + O

2(g)

68 g 22.7 L at STP

105 g 500 mL at STP

64. Answer (3)

Fact.

65. Answer (3)

H2SO

4 is a stronger acid and can act as a catalyst

and this decomposes H2O2, thus weak acid such

as H3PO

4 is preferred over H

2SO

4.

66. Answer (1)

1 100 5.6V 56 mL

10

67. Answer (2)

Fact.

68. Answer (1)

Fact.

69. Answer (1)

70. Answer (2)

71. Answer (3)

Fact.

72. Answer (1)

The stability of peroxide and superoxide of alkali

metals increases as we go down the group.

73. Answer (2)

2NH4Cl + Ca(OH)

2 2NH

3 + CaCl

2 + 2H

2O

74. Answer (1)

Fact.

75. Answer (1)

76. Answer (4)

77. Answer (3)

AlCl3 exist as dimer in solid state (sp3).

Al Al

Cl

Cl

Cl Cl

Cl Cl

But in aqueous medium, it produces [Al(H2O)

6]3+

and Cl–(aq).

78. Answer (3)


5/7

79. Answer (4)

BF3 + 3LiAlH

4 B

2H6 + 3LiF + 3AlF

3

B2H6 + 6H

2O 2H

3BO

3 + 6H

2

B2H6 + 3O

2 B

2O3 + 3H

2O

80. Answer (4)

Fact.

81. Answer (1)

82. Answer (3)

83. Answer (2)

84. Answer (1)

Fact.

85. Answer (3)

Fact.

86. Answer (2)

87. Answer (2)

Fact.

88. Answer (4)

Due to least extent of back-bonding.

89. Answer (1)

Due to back bonding in BF3, B – F bond length in

BF3 is shorter than that in BF

4

–.

90. Answer (2)

Borax contains only 8H2O molecules.

[ BIOLOGY

]

91. Answer (4)

a – Elaioplast

b – Polysome

92. Answer (3)

Centriole makes its own copy.

93. Answer (3)

Both mitochondria and plastid have 70 S ribosome,

double membrane and circular DNA.

94. Answer (2)

95. Answer (3)

Mitochondria is not included in endomembrane

system.

96. Answer (1)

Non-polar molecules can pass through the lipid

bilayer.

97. Answer (4)

98. Answer (3)

99. Answer (3)

RER has a continuous connection with outer

membrane of nucleus.

100. Answer (2)

101. Answer (4)

Prokaryotic cells have 70 S ribosome.

102. Answer (3)

103. Answer (4)

Ribosome and centriole are non-membranous

organelles.

104. Answer (2)

105. Answer (3)

Secondary wall in plant cell is located inside the

primary wall.

106. Answer (4)

107. Answer (4)

108. Answer (1)

109. Answer (2)

Acrocentric

110. Answer (3)

A – Inner membrane

B – Outer membrane

C – Matrix

D – Inner membrane space

111. Answer (4)

Cytoskeleton are involved in many function such as

mechanical support, motility, maintenance of shape

of cell.

112. Answer (2)

113. Answer (2)

114. Answer (2)

Ribosome is present in both prokaryotic and

eukaryotic.


6/7

115. Answer (3)

116. Answer (1)

Equational division also occurs in some haploid cell

like in case of some lower plants and in some

social insects.

117. Answer (1)

118. Answer (4)

In G0-phase, cell is metabolically active.

119. Answer (2)

120. Answer (4)

121. Answer (4)

122. Answer (3)

2 picogram

123. Answer (3)

Mitogen induce mitosis.

124. Answer (4)

125. Answer (2)

Diakinesis represents transition of metaphase.

126. Answer (4)

127. Answer (3)

128. Answer (2)

129. Answer (3)

130. Answer (4)

DNA duplication occurs in nucleus.

131. Answer (3)

132. Answer (4)

133. Answer (3)

134. Answer (4)

Telophase I is called a dyad of cells.

135. Answer (1)

136. Answer (4)

Residual volume cannot be measured by spirometry

as it cannot be expired out.

VC = ERV + TV + IRV

It does not included RV.

137. Answer (4)

138. Answer (2)

139. Answer (3)

Finer dust particles are trapped in mucus. Dust cells

are located in alveoli. Cilia propel the mucus trapped

particles upwards.

140. Answer (4)

Epiglottis is made up of elastic cartilage.

141. Answer (3)

142. Answer (3)

Abdominal breathing is normal breathing at rest

mostly involving contractions of diaphragm.

Abdominal muscles and internal intercostal muscles

are involved in forceful thoracic breathing.

143. Answer (1)

[Tidal volume – Respiratory dead space volume]

reaches into the alveoli (350 mL).

Alveolar ventilation is

Alveolar volume × Breathing rate

144. Answer (2)

The pO2 in exhaled air is 116 mmHg, while pO

2 in

alveolar air is 104 mmHg.

145. Answer (4)

Hamburger phenomenon (chloride shift) occurs in

transport of CO2 to the lungs.

146. Answer (3)

100 mL oxygenated blood transports 15 mL O2 to

the tissues during exercise.

147. Answer (2)

70% CO2 is transported in the form of HCO

3

– ions

through blood plasma.

148. Answer (2)

Presence of high amount of diphosphoglycerate

(DPG) increases the P50

value of haemoglobin.

149. Answer (1)

The chemoreceptors are stimulated by increase in

CO2 concentration in blood. The role of oxygen in

regulation of respiratory rhythm is insignificant.

150. Answer (3)

Inhalation of coal/carbon dust is pneumoconiosis.

151. Answer (3)

152. Answer (3)

Thrombocytopenia (lack of platelets) is the reason for

lack of clotting.


7/7

153. Answer (1)

PMNL are neutrophils (granulocytes).

154. Answer (1)

Pericardial cavity is enclosed between the parietal

and visceral pericardial layers, both of which

constitute the serous pericardium.

155. Answer (3)

156. Answer (2)

Stroke volume of both ventricles is same. Pulmonary

trunk has lesser resistance than aorta, so right

ventricle contracts with lesser force to pump out the

same quantity of stroke volume.

157. Answer (3)

158. Answer (3)

159. Answer (1)

Isovolumetric relaxation is followed by opening of

tricuspid and bicuspid valves, causing first rapid

filling of ventricles.

160. Answer (2)

161. Answer (3)

Both statements are correct according to

Frank Starling law of heart muscles.

162. Answer (2)

163. Answer (2)

Tunica media of artery is thicker while its lumen is

narrower.

164. Answer (2)

165. Answer (1)

166. Answer (1)

167. Answer (2)

168. Answer (4)

All premolars and last molars are arise only in

permanent dentition.

169. Answer (2)

GALT is gut associated lymphoid tissue.

170. Answer (2)

171. Answer (3)

172. Answer (4)

173. Answer (4)

Enterogastrone will inhibit gastric motility and

release of gastric juice.

174. Answer (2)

175. Answer (3)

176. Answer (3)

177. Answer (2)

178. Answer (2)

179. Answer (4)

Haemorrhoids is characterised by leakage of blood

out of varicose veins of rectum; the cause being

prolonged constipation / low fibre diet.

180. Answer (1)

� � �

Test - 5 (Code-D) (Answers & Hints) All India Aakash Test Series for Medical-2018

1/7

1. (4)

2. (2)

3. (4)

4. (1)

5. (1)

6. (1)

7. (3)

8. (3)

9. (1)

10. (4)

11. (1)

12. (3)

13. (1)

14. (4)

15. (3)

16. (1)

17. (3)

18. (3)

19. (3)

20. (3)

21. (1)

22. (4)

23. (3)

24. (4)

25. (4)

26. (3)

27. (4)

28. (3)

29. (2)

30. (4)

31. (4)

32. (3)

33. (1)

34. (4)

35. (2)

36. (3)

37. (3)

38. (4)

39. (4)

40. (1)

41. (1)

42. (4)

43. (2)

44. (4)

45. (1)

46. (4)

47. (3)

48. (2)

49. (4)

50. (4)

51. (1)

52. (3)

53. (4)

54. (1)

55. (3)

56. (2)

57. (2)

58. (1)

59. (1)

60. (2)

61. (3)

62. (3)

63. (4)

64. (3)

65. (1)

66. (4)

67. (3)

68. (3)

69. (4)

70. (3)

71. (1)

72. (1)

73. (2)

74. (2)

75. (4)

76. (1)

77. (1)

78. (1)

79. (1)

80. (4)

81. (3)

82. (3)

83. (2)

84. (3)

85. (4)

86. (4)

87. (1)

88. (4)

89. (2)

90. (2)

91. (3)

92. (2)

93. (1)

94. (4)

95. (1)

96. (2)

97. (1)

98. (4)

99. (1)

100. (2)

101. (4)

102. (2)

103. (1)

104. (1)

105. (2)

106. (2)

107. (4)

108. (2)

109. (3)

110. (3)

111. (3)

112. (4)

113. (4)

114. (4)

115. (2)

116. (1)

117. (4)

118. (3)

119. (2)

120. (2)

121. (1)

122. (4)

123. (4)

124. (1)

125. (2)

126. (4)

127. (1)

128. (1)

129. (2)

130. (3)

131. (1)

132. (4)

133. (1)

134. (1)

135. (2)

136. (3)

137. (4)

138. (4)

139. (4)

140. (1)

141. (1)

142. (4)

143. (2)

144. (4)

145. (1)

146. (4)

147. (4)

148. (4)

149. (4)

150. (3)

151. (3)

152. (4)

153. (4)

154. (4)

155. (1)

156. (4)

157. (3)

158. (1)

159. (1)

160. (4)

161. (1)

162. (3)

163. (3)

164. (1)

165. (3)

166. (1)

167. (3)

168. (4)

169. (4)

170. (1)

171. (2)

172. (4)

173. (3)

174. (1)

175. (1)

176. (2)

177. (3)

178. (4)

179. (2)

180. (2)

ANSWERS

TEST - 5 (Code-D)

All India Aakash Test Series for Medical-2018

Test Date : 22-01-2017

All India Aakash Test Series for Medical-2018 Test - 5 (Code-D) (Answers & Hints)

2/7

Hints to Selected Questions

[ PHYSICS]

1. Answer (4)

2. Answer (2)

3. Answer (4)

m

T = b = Wein's constant

4. Answer (1)

5. Answer (1)

Use Newton's law of cooling

1 2 1 2

02

T T T Tk T

t

⎛ ⎞ ⎜ ⎟⎝ ⎠

6. Answer (1)

L

B

LA

2L

D

C

100°C 40°CH1

H2

HH H1 2 + = H

100 40 20

2

2

T T T

l L L

kA kA kA

7. Answer (3)

1 2

eq2

K K

K

8. Answer (3)

9. Answer (1)

VPr = Constant

0V P

rV P

⇒

P P

V r

V

⇒

⎛ ⎞⎜ ⎟⎝ ⎠

10. Answer (4)

Polytropic process (n = –1)

,1

n

v

RPV K C C

n

11. Answer (1)

2

1 1

1T W

T Q

12. Answer (3)

1 1

2 2 1 1

r rT V TV

13. Answer (1)

P

V

CQ

U C

14. Answer (4)

15. Answer (3)

W P dV ∫0

0

3

4 ∫V

V

V dV

W = 16V0

2

16. Answer (1)

1 2 1 2( ) Q Q W W

17. Answer (3)

nR T w 18. Answer (3)

19. Answer (3)

20. Answer (3)

Insufficient heat at MP.

21. Answer (1)

22. Answer (4)

1

2Tt

23. Answer (3)

F Ay T

24. Answer (4)

2

12

mlI

2l l

l l

= 2IT

2

3

I T

25. Answer (4)


3/7

26. Answer (3)

2/3

Tv n v

27. Answer (4)

28. Answer (3)

29. Answer (2)

30. Answer (4)

A1V1 = na

2v2

31. Answer (4)

32. Answer (3)

33. Answer (1)

2 1y

34. Answer (4)

l

l/2

2

2 2

mgl gll l

AY AY

⇒

35. Answer (2)

PVB

V

36. Answer (3)

1 1 1 2 2

2 2 2 1 1

l F l A Y

l F l A Y

37. Answer (3)

dvI

dr

[ CHEMISTRY]

46. Answer (4)

Borax contains only 8H2O molecules.

47. Answer (3)

Due to back bonding in BF3, B – F bond length in

BF3 is shorter than that in BF

4

–.

48. Answer (2)

Due to least extent of back-bonding.

49. Answer (4)

Fact.

50. Answer (4)

51. Answer (1)

Fact.

52. Answer (3)

Fact.

53. Answer (4)

54. Answer (1)

55. Answer (3)

38. Answer (4)

E = –KE

39. Answer (4)

5/2

kF

r

2

5/2

mv k

r r

2 rT

v

2 7/2T r

40. Answer (1)

2 2 2

ev v v

41. Answer (1)

42. Answer (4)

21

21

mghmv

h

R

⎛ ⎞⎜ ⎟⎝ ⎠

43. Answer (2)

44. Answer (4)

m 4mr

A x c B( – )r x

2 2

(4 )

( )

Gm G m

x r x

3

rx

3GmV

r

45. Answer (1)

4

3g RG


4/7

56. Answer (2)

Fact.

57. Answer (2)

BF3 + 3LiAlH

4 B

2H6 + 3LiF + 3AlF

3

B2H6 + 6H

2O 2H

3BO

3 + 6H

2

B2H6 + 3O

2 B

2O3 + 3H

2O

58. Answer (1)

59. Answer (1)

AlCl3 exist as dimer in solid state (sp3).

Al Al

Cl

Cl

Cl Cl

Cl Cl

But in aqueous medium, it produces [Al(H2O)

6]3+

and Cl–(aq).

60. Answer (2)

61. Answer (3)

62. Answer (3)

Fact.

63. Answer (4)

2NH4Cl + Ca(OH)

2 2NH

3 + CaCl

2 + 2H

2O

64. Answer (3)

The stability of peroxide and superoxide of alkali

metals increases as we go down the group.

65. Answer (1)

Fact.

66. Answer (4)

67. Answer (3)

68. Answer (3)

Fact.

69. Answer (4)

Fact.

70. Answer (3)

1 100 5.6V 56 mL

10

71. Answer (1)

H2SO

4 is a stronger acid and can act as a catalyst

and this decomposes H2O2, thus weak acid such

as H3PO

4 is preferred over H

2SO

4.

72. Answer (1)

Fact.

73. Answer (2)

2H2O2 2H

2O + O

2(g)

68 g 22.7 L at STP

105 g 500 mL at STP

74. Answer (2)

75. Answer (4)

Fact.

76. Answer (1)

Group 7 to 9 is known as hydride gap because

elements of these group do not form hydrides.

77. Answer (1)

Fact.

78. Answer (1)

'B' has incomplete octet.

79. Answer (1)

Fact.

80. Answer (4)

81. Answer (3)

Acidic medium :

KMnO4 Mn2+

ME

5

Basic medium :

KMnO4 K

2MnO

4

ME

1

82. Answer (3)

H – O – S – O – O – H; S has O.S. + 6

O

O

83. Answer (2)

o

CellEMF =

o oCathode (RP) Anode (RP)E E

= 0.80 – 0.34

= 0.46 V


5/7

84. Answer (3)

Average oxidation state of oxygen =2( 1 2 2 2)

8

= –1.75

85. Answer (4)

Fact.

86. Answer (4)

Fact.

87. Answer (1)

CrO2

– : +3; ClO3

– : +5

CrO4

2– : +6; MnO4

– : +7

[ BIOLOGY

]

91. Answer (3)

92. Answer (2)

Telophase I is called a dyad of cells.

93. Answer (1)

94. Answer (4)

95. Answer (1)

96. Answer (2)

DNA duplication occurs in nucleus.

97. Answer (1)

98. Answer (4)

99. Answer (1)

100. Answer (2)

101. Answer (4)

Diakinesis represents transition of metaphase.

102. Answer (2)

103. Answer (1)

Mitogen induce mitosis.

104. Answer (1)

2 picogram

105. Answer (2)

106. Answer (2)

107. Answer (4)

108. Answer (2)

In G0-phase, cell is metabolically active.

109. Answer (3)

110. Answer (3)

Equational division also occurs in some haploid cell

like in case of some lower plants and in some

social insects.

111. Answer (3)

112. Answer (4)

Ribosome is present in both prokaryotic and

eukaryotic.

113. Answer (4)

114. Answer (4)

115. Answer (2)

Cytoskeleton are involved in many function such as

mechanical support, motility, maintenance of shape

of cell.

116. Answer (1)

A – Inner membrane

B – Outer membrane

C – Matrix

D – Inner membrane space

117. Answer (4)

Acrocentric

88. Answer (4)

H2SO

5 : O has –1 and –2 O.S.

H2S2O3 : S has 0 and +4 O.S.

CaOCl2 : Cl has +1 and –1 O.S.

89. Answer (2)

2PbO + 4HCl 2PbCl2 + 2H

2O

(Acid-base reaction)

PbO2 + 4HCl PbCl

2 + Cl

2 + 2H

2O

(Redox reaction)

90. Answer (2)

3 2CaCO CaO(s) CO (g)

There is no change in oxidation state.


6/7

118. Answer (3)

119. Answer (2)

120. Answer (2)

121. Answer (1)

Secondary wall in plant cell is located inside the

primary wall.

122. Answer (4)

123. Answer (4)

Ribosome and centriole are non-membranous

organelles.

124. Answer (1)

125. Answer (2)

Prokaryotic cells have 70 S ribosome.

126. Answer (4)

127. Answer (1)

RER has a continuous connection with outer

membrane of nucleus.

128. Answer (1)

129. Answer (2)

130. Answer (3)

Non-polar molecules can pass through the lipid

bilayer.

131. Answer (1)

Mitochondria is not included in endomembrane

system.

132. Answer (4)

133. Answer (1)

Both mitochondria and plastid have 70 S ribosome,

double membrane and circular DNA.

134. Answer (1)

Centriole makes its own copy.

135. Answer (2)

a – Elaioplast

b – Polysome

136. Answer (3)

137. Answer (4)

Haemorrhoids is characterised by leakage of blood

out of varicose veins of rectum; the cause being

prolonged constipation / low fibre diet.

138. Answer (4)

139. Answer (4)

140. Answer (1)

141. Answer (1)

142. Answer (4)

143. Answer (2)

Enterogastrone will inhibit gastric motility and

release of gastric juice.

144. Answer (4)

145. Answer (1)

146. Answer (4)

147. Answer (4)

GALT is gut associated lymphoid tissue.

148. Answer (4)

All premolars and last molars are arise only in

permanent dentition.

149. Answer (4)

150. Answer (3)

151. Answer (3)

152. Answer (4)

153. Answer (4)

Tunica media of artery is thicker while its lumen is

narrower.

154. Answer (4)

155. Answer (1)

Both statements are correct according to

Frank Starling law of heart muscles.

156. Answer (4)

157. Answer (3)

Isovolumetric relaxation is followed by opening of

tricuspid and bicuspid valves, causing first rapid

filling of ventricles.

158. Answer (1)

159. Answer (1)

160. Answer (4)

Stroke volume of both ventricles is same. Pulmonary

trunk has lesser resistance than aorta, so right

ventricle contracts with lesser force to pump out the

same quantity of stroke volume.

161. Answer (1)


7/7

162. Answer (3)

Pericardial cavity is enclosed between the parietal

and visceral pericardial layers, both of which

constitute the serous pericardium.

163. Answer (3)

PMNL are neutrophils (granulocytes).

164. Answer (1)

Thrombocytopenia (lack of platelets) is the reason for

lack of clotting.

165. Answer (3)

166. Answer (1)

Inhalation of coal/carbon dust is pneumoconiosis.

167. Answer (3)

The chemoreceptors are stimulated by increase in

CO2 concentration in blood. The role of oxygen in

regulation of respiratory rhythm is insignificant.

168. Answer (4)

Presence of high amount of diphosphoglycerate

(DPG) increases the P50

value of haemoglobin.

169. Answer (4)

70% CO2 is transported in the form of HCO

3

– ions

through blood plasma.

170. Answer (1)

100 mL oxygenated blood transports 15 mL O2 to

the tissues during exercise.

171. Answer (2)

Hamburger phenomenon (chloride shift) occurs in

transport of CO2 to the lungs.

172. Answer (4)

The pO2 in exhaled air is 116 mmHg, while pO

2 in

alveolar air is 104 mmHg.

173. Answer (3)

[Tidal volume – Respiratory dead space volume]

reaches into the alveoli (350 mL).

Alveolar ventilation is

Alveolar volume × Breathing rate

174. Answer (1)

Abdominal breathing is normal breathing at rest

mostly involving contractions of diaphragm.

Abdominal muscles and internal intercostal muscles

are involved in forceful thoracic breathing.

175. Answer (1)

176. Answer (2)

Epiglottis is made up of elastic cartilage.

177. Answer (3)

Finer dust particles are trapped in mucus. Dust cells

are located in alveoli. Cilia propel the mucus trapped

particles upwards.

178. Answer (4)

179. Answer (2)

180. Answer (2)

Residual volume cannot be measured by spirometry

as it cannot be expired out.

VC = ERV + TV + IRV

It does not included RV.

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