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Test - 8 (Code-C) (Answers & Hints) All India Aakash Test Series for Medical-2018

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1. (4)

2. (3)

3. (3)

4. (4)

5. (2)

6. (3)

7. (1)

8. (4)

9. (3)

10. (4)

11. (1)

12. (4)

13. (2)

14. (2)

15. (4)

16. (2)

17. (1)

18. (4)

19. (3)

20. (3)

21. (3)

22. (3)

23. (2)

24. (1)

25. (4)

26. (1)

27. (4)

28. (1)

29. (4)

30. (1)

31. (3)

32. (3)

33. (2)

34. (2)

35. (3)

36. (3)

37. (4)

38. (2)

39. (4)

40. (3)

41. (2)

42. (4)

43. (1)

44. (2)

45. (4)

46. (3)

47. (4)

48. (2)

49. (3)

50. (4)

51. (2)

52. (1)

53. (2)

54. (4)

55. (1)

56. (2)

57. (2)

58. (3)

59. (3)

60. (1)

61. (4)

62. (4)

63. (3)

64. (4)

65. (2)

66. (2)

67. (4)

68. (3)

69. (2)

70. (4)

71. (2)

72. (2)

73. (2)

74. (4)

75. (1)

76. (3)

77. (1)

78. (2)

79. (1)

80. (4)

81. (1)

82. (3)

83. (4)

84. (2)

85. (1)

86. (2)

87. (1)

88. (4)

89. (1)

90. (2)

91. (3)

92. (2)

93. (1)

94. (2)

95. (3)

96. (1)

97. (3)

98. (2)

99. (3)

100. (2)

101. (2)

102. (2)

103. (3)

104. (3)

105. (4)

106. (3)

107. (1)

108. (1)

109. (2)

110. (2)

111. (2)

112. (3)

113. (4)

114. (4)

115. (2)

116. (4)

117. (1)

118. (2)

119. (2)

120. (1)

121. (1)

122. (2)

123. (2)

124. (2)

125. (1)

126. (3)

127. (4)

128. (3)

129. (3)

130. (1)

131. (3)

132. (4)

133. (4)

134. (1)

135. (4)

136. (4)

137. (3)

138. (3)

139. (3)

140. (3)

141. (3)

142. (3)

143. (3)

144. (4)

145. (2)

146. (3)

147. (4)

148. (2)

149. (3)

150. (3)

151. (3)

152. (3)

153. (3)

154. (2)

155. (4)

156. (1)

157. (1)

158. (3)

159. (1)

160. (1)

161. (3)

162. (3)

163. (4)

164. (2)

165. (2)

166. (1)

167. (2)

168. (3)

169. (3)

170. (3)

171. (2)

172. (3)

173. (3)

174. (4)

175. (3)

176. (2)

177. (2)

178. (3)

179. (3)

180. (1)

ANSWERS

TEST - 8 (Code-C)

All India Aakash Test Series for Medical-2018

Test Date : 19-03-2017

All India Aakash Test Series for Medical-2018 Test - 8 (Code-C) (Answers & Hints)

2/10

Hints to Selected Questions

[ PHYSICS]

1. Answer (4)

2. Answer (3)

3. Answer (3)

v2 = u2 + 2as

or v2 = u2 + 2 (Area under a–x graph)

v2 = 21 + 2 × 30

v = 9 m/s

4. Answer (4)

At v = 2, x = 1 m

2 8dv

v xdx

24 4 m/s

dva v x

dx

5. Answer (2)

T = 4 + 8 = 12 s

21

120 m2

Ts

av

12010 m/s

12v

6. Answer (3)

2(2 )2

8t c

ta a ⇒

t = 2 s

v = 2t = 4 m/s

7. Answer (1)

By comparison,

tan = 1

2 20.1

2 cos

g

u

u = 10 m/s

8. Answer (4)

T = 1 s

2 2(horizontal) 3 4 5 m/su

R = 5 m/s × 1 s = 5 m

9. Answer (3)

� � �

BA B Av v v

45°

45°

A

B

vB

–vA

vA

Cd

vBA

�

d

2

10. Answer (4)

1 1 2 2F m a m a �

� �

11. Answer (1)

Free body diagram :

F 2F

3 = 30 Ng

3F = 30 N

F = 10 N

12. Answer (4)

6 sin60 4 sin30

10

g ga

a 3.2 m/s2

13. Answer (2)

fL =

sN =

smg = 100 N

Body will remain static.

Static friction = f = F = 80 N


3/10

14. Answer (2)

2 2(20)tan 1

40 10

v

Rg

= 45°

15. Answer (4)

Tension = Weight of the lower part

16. Answer (2)

W F s ��

= ˆ ˆ ˆ ˆ(6 2 )(2 12 )i j i j

= 12 – 24

= –12 J

17. Answer (1)

Work energy theorem,

21

2Pt mv

2Ptv

m

18. Answer (4)

21(2 )

2mv mg R H

2 (2 )v g R H

19. Answer (3)

At equilibrium,

F = 0 x = ±1

20. Answer (3)

Centre of mass lies on the symmetric line.

21. Answer (3)

IZ

= IX + I

Y

=

2

212

ML

=

2

6

ML

22. Answer (3)

vP = 2v

C v

C = 3 m/s

23. Answer (2)

2 2

0

1 3 1

2 4 2mv mv

⎛ ⎞ ⎜ ⎟⎝ ⎠

2 2

0

3

4v v

0

3

2v v

0

3

2

ve

v

24. Answer (1)

28 2 8 1; 0.5 m/s

8 8 8 8

x y

a a

2 2 25m/s

2x y

a a a

25. Answer (4)

2R

Tg

26. Answer (1)

T R3/2

3

2

T R

T R

27. Answer (4)

28. Answer (1)

P

V

V

PV

V

V = VT

PV P

V T T ⇒

29. Answer (4)

1000r3 = R3

R = 10r

Now,

2

2

1

101000

R

r

30. Answer (1)

PA = ab = gb

PB = gh

PA – P

B = g(b – h)

31. Answer (3)

1 2

0

2( )

At H H

A g

Here, T1 = T

2


4/10

H H

H

4H

H

= 4

32. Answer (3)

22 2

2 2 2I MR R

TMgd MgR g

33. Answer (2)

34. Answer (2)

2

av max;

4

Aa a A

T

av

max

2a

a

35. Answer (3)

A = 2 2

1 2 1 22 cosA A A A

= 2 cos2

A

= 2A cos30°

= 3

2 32

A A

36. Answer (3)

2

2K

= 4 m

So, L can be 3

, , , 22 2

etc.

37. Answer (4)

vs = 20 cos60 = 10 m/s

340660 680 Hz

340 10

f

s

vf

v v

⎡ ⎤ ⎢ ⎥ ⎣ ⎦

38. Answer (2)

2

2 1 10

1

10 logI

SL SLI

⎛ ⎞ ⎜ ⎟⎝ ⎠

30 – 3 = 27 dB

39. Answer (4)

14

e l ...(i)

2

3

4e

l ...(ii)

Solving,

2 13

1.05 cm2

e

l l

40. Answer (3)

U = U1 + U

2

= 5 3

2 42 2

RT RT

= 11RT

41. Answer (2)

W = WAB

+ WBC

0 0 0 0

0 0

3 5 72

2 2 2

P P P VV V

42. Answer (4)

11K

43. Answer (1)

10 5

3 2 RT RT

W

5

6 RT

1

4

W

Q

4

QW

44. Answer (2)

1120 × 1 × (80 – 15) = m(540 + 20)

m = 130 g

45. Answer (4)

P R2T4

Pnew

= 4 × 440 = 1760 watt


5/10

[ CHEMISTRY]

46. Answer (3)

Number of moles of CaCO3

=

3100 10 g

100 g/mole

= 10–3 mole

Number of moles of O atoms = 3 × 10–3

Number of O atoms = 6.02 × 1023 × 3 × 10–3

= 18.06 × 1020

= 1.8 × 1021

47. Answer (4)

48. Answer (2)

4KClO3

3KClO + KCl4

+5 +7 –1

Reduction

Oxidation

49. Answer (3)

50. Answer (4)

51. Answer (2)

52. Answer (1)

Volume strength = 5.6 × N

=

8.5

175.6

560

1000

= 8.5 1000

5.617 560

= 5 V

53. Answer (2)

54. Answer (4)

PV = nRT (Ideal gas equation)

55. Answer (1)

Gas [X],

x x x

x x

x

P M d RTd or P

RT M

Gas [Y],

y y y

y y

y

P M d RTd or P

RT M

According to the problem,

dx = 3d

y

My = 1.5M

x

or Mx =

1

1.5My

y y yx x

y x y y y

M 3d MP d RT1.5 4.5

P M d RT M d

56. Answer (2)

Cr (Z = 24) = 1s2, 2s

2, 2p6, 3s

2, 3p6, 4s

1, 3d5

For s-orbital,

l = 0

Total number of electrons having l = 0 are 7.

57. Answer (2)

Na has low melting point and it spread over water.

58. Answer (3)

2

20.8 mole0.4 mole

MgCl Mg 2Cl

2 2

4 4

0.2 mole 0.2 mole

MgSO Mg SO

Number of moles of Mg2+ = (0.4 + 0.2) = 0.6

59. Answer (3)

60. Answer (1)

H3BO

3 accepts one OH– ion.

61. Answer (4)

Higher the value of standard reduction potential,

higher is the oxidising power.

62. Answer (4)

Hydrogen shows oxidation number of –1 in metal

hydrides.

63. Answer (3)

64. Answer (4)

65. Answer (2)

3° free radical is more stable.

66. Answer (2)

Addition follows Anti-Markownikoff’s rule due to

–I effect of –CF3.


6/10

67. Answer (4)

Trans but-2-ene is highly stable alkene amongst all

of these.

68. Answer (3)

More electron withdrawing groups increase the acidic

strength.

69. Answer (2)

n(n 2) B.M.

3(3 2)

15 B.M.

70. Answer (4)

71. Answer (2)

N2(g) + 2H

2(g) N

2H4(g)

Hf° N

2H4(g) = [BE of N N + 2BE of H – H]

– [BE of N – N + 4BE (N – H)]

= 941 + 2 × 436 – (159 + 4 × 398)

= 62 kJ/mol

72. Answer (2)

In nature, carbon exist as graphite allotrope.

73. Answer (2)

(High volume) (Low volume)Ice Water��⇀

↽��

Melting of ice will be favoured by increase in

pressure on system.

74. Answer (4)

AB(g) ��⇀↽�� A(g) + B(g)

Initially : 1 0 0

At equilibrium : (1 – 0.33) 0.33 0.33

Total moles at equilibrium = 1.33

p

0.33 0.33P P

1.33 1.33K 0.122P

0.67P

1.33

⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

p

1K P

8

P = 8Kp

75. Answer (1)

Kb = 10–10, K

a = 10–4

pKa = –log(10–4) = 4

a

[X ]pH pK log

[HX]

[X ]pH 4 log

[HX]

pH = 4

76. Answer (3)

77. Answer (1)

CA of base is formed by accepting the H+ ion

NH2

–H+ NH3

78. Answer (2)

2

2 4 4

S2S

Ag SO 2Ag SO ��⇀

↽��

Ksp

= [Ag+]2[SO4

2–]

= (2S)2(S)

= 4S3

79. Answer (1)

80. Answer (4)

81. Answer (1)

Br — Br — Br

O

O

O

O

O

O

O

O

+6 +4 +6

Average oxidation number is 16

3

or 3x – 16 = 0

3x = +16

x = 16

3

82. Answer (3)

83. Answer (4)

8Al 4Al2O3 + 24e–

84. Answer (2)

C2H5OH + 4I

2 + 6OH– CHI

3 + HCO

2

– + 5I– + 5H2O

85. Answer (1)

On 10 times dilution, pH is increased by 1 unit.

86. Answer (2)

Hard boiling of an egg produces disorderness in

structure of protene inside the shell of egg.


7/10

For isothermal process,

2

1

VS nR ln

V

Since P2 < P

1 and V

2 > V

1, hence entropy will be

increased.

87. Answer (1)

Since carbon and oxygen both have 2p1 and 2p

4

(n = 2) orbital electrons and can easily participate in

-bonding (parallel to plane of internuclear axis).

88. Answer (4)

C

Cl

Cl

ClCl

net

= 0

89. Answer (1)

1 amu = 1.66 × 10–24 g

12 amu = 1.99 × 10–23 g

90. Answer (2)

[ BIOLOGY

]

91. Answer (3)

92. Answer (2)

C – Higher is the category, greater is the difficulty of

determining the relationship to other texa at the

same level.

93. Answer (1)

IV. Nature of cell wall.

VIII. Mode of respiration.

94. Answer (2)

95. Answer (3)

2H+, 2e–, 4ATP

96. Answer (1)

97. Answer (3)

98. Answer (2)

99. Answer (3)

100. Answer (2)

101. Answer (2)

Natural system of classification given by Bentham

and Hooker.

It is based on external features

internal features

ultrastructure

anatomy

embryology and phytochemistry

102. Answer (2)

103. Answer (3)

104. Answer (3)

105. Answer (4)

The female gametophyte produces two or more

archegonia.

106. Answer (3)

107. Answer (1)

108. Answer (1)

Mimosa pudica – Leguminaceae (family)

characterised by pulvinus leaf base.

109. Answer (2)

110. Answer (2)

111. Answer (2)

112. Answer (3)

113. Answer (4)

114. Answer (4)

Male gamete (n) + Egg (n)

Male gamete (n) + Secondary nucleus (2n)

Syngamy + triple fusion = Double fertilisation

Zygote (2n)

PEN (3n)

Triple fusion

Syngamy

115. Answer (2)

116. Answer (4)

117. Answer (1)

III. Mechanical support to stem.

V. Resistant to the attacks of insects.

These features are related to heart wood.

118. Answer (2)

119. Answer (2)

A – 3-phosphoglyceric acid

B – Lactic acid

C – Ethyl alcohol (ethanol)


8/10

120. Answer (1)

121. Answer (1)

122. Answer (2)

B – Sequence of cytochromes for electron flow is

cyt b cyt c1 cyt c cyt a cyt a

3.

D – Cyt c is electron carrier between complex III

and IV.

123. Answer (2)

124. Answer (2)

Water potential for cell A,

w = DPD +

w = –8 bars

Water potential for cell B = s +

p =

w

So, w = –18 + 7 = –11 bars

Water potential for cell C,

DPD = OP – TP

= 6 – 3

= 3

w = –3 bars

So, direction of water flow

B w = –11

A

Cw = –3

w = –8

125. Answer (1)

126. Answer (3)

A B — V

127. Answer (4)

128. Answer (3)

129. Answer (3)

130. Answer (1)

In oocytes of some vertebrates lost for months or

years, true for diplotene stage.

131. Answer (3)

132. Answer (4)

a. Sugar moiety towards outer region.

b. Quasi-fluid nature of lipid enables lateral

movement of proteins within the overall bi-lipid

layer.

133. Answer (4)

Figure shown is lysosome.

134. Answer (1)

135. Answer (4)

Steroidal hormones secreted from smooth

endoplasmic reticulum (SER).

136. Answer (4)

A resting membrane, is said to be in polarised

state. In this, the outside of the membrane is

positively charged, and inside is negatively charged.

Outside has more Na+ ions and inside has more K+

ions. K+ ions leak/diffuse from inside to outside.

137. Answer (3)

7% of CO2 is transported as dissolved gas in

plasma. 70% of CO2 is transported as HCO

3

– in

RBC and plasma CO2 is mainly transported as

NaHCO3 in plasma. 20–25% is transported as

carbaminohemoglobin.

138. Answer (3)

Knee jerk reflex is simple reflex. It is monosynaptic

reflex, i.e., between the sensory and motor neuron

there is only one synapse, no interneuron are

present. Quadriceps femoris is extensor of thigh.

139. Answer (3)

A person with B blood group has B antigen and

anti-A antibodies. If he is B negative Rh antigen and

Rh antibodies are absent.

140. Answer (3)

Total fat consumption is 1 g, i.e., 1 × 9 = 9 K

calories are released.

Total calories from serving is 217 kcal.

Total fat 1 g × 9 = 9 kcal

Starch 43 g × 4 = 172

Sugar 3 g × 4 = 12

Protein 6 g × 4 = 24

= 217 kcal

Percentage of calories that come from fat

9100 4.1%

217

141. Answer (3)

If the level of cortisol rises in blood, then increase in

this level will exert a negative feedback control on

the hypothalamus, so less CRH is released Soless ACTH released So less cortisol till the

level is normalised.


9/10

142. Answer (3)

Estrogen hormone is a steroid hormone which

directly enters the cell, no second messenger is

required. It binds with intracellular receptor in

nucleus to form hormone receptor complex which

binds with a specific DNA segment to stimulate

gene expression.

143. Answer (3)

Mitral valve is bicuspid valve between left atrium and

left ventricle. If we cut the chordae tendinae of mitral

valve, the valve will close, but during ventricular

systole, they will regurgitate into atria along with some

blood. So, the flow of blood would decrease in aorta.

144. Answer (4)

Endothermal organisms are warm blooded with a

constant body temperature. They have a higher

metabolic rate, so they eat more than ectotherms.

145. Answer (2)

The presence of true body cavity poses a problem

for direct exchange of oxygen and carbon dioxide

with environment. But it provides space for the

intestine to become longer for better digestion and

absorption.

146. Answer (3)

The only phylum in the kingdom animalia is phylum

porifera.

147. Answer (4)

During ventricular systole, the blood from the left

ventricle will flow into aorta, and the blood from right

ventricle will enter into pulmonary artery.

148. Answer (2)

a. Reabsorption is minimum in DLH.

b. Selective secretion of H+ and K+ ions occur in

PCT, DCT and CD.

c. Conditional reabsorption of Na+ and water occur

in DCT.

d. Filtrate becomes hypotonic in ALH.

149. Answer (3)

Aldosterone is produced by adrenal cortex in

response to low levels of sodium ions in the blood.

150. Answer (3)

Neutrophils are 60–65%, they help in phagocytosis,

but they are not precursor of macrophages.

Monocytes are precursor of macrophages.

151. Answer (3)

If we hold our breath for some time CO2 will build up

in the body, pH will decrease and the respiratory

rate would increase.

152. Answer (3)

If the oxygen hemoglobin dissociation curve shifts

towards RHS, it means binding has decreased

under the same pO2 because the secondary factors

have increased, causing dissociation of HbO2.

Strenuous exercise would increase lactic acid in

muscles and blood, pH would decrease.

153. Answer (3)

Estrogen promotes osteoblast activity.

154. Answer (2)

Knee joint is hinge joint, is formed between femur

and tibia, fibula does not take part.

155. Answer (4)

Periplaneta americana is a protostome with

determinate cleavage.

156. Answer (1)

In tight junction, the two adjacent cell membranes

are fused together to prevent the leaking of any

substance between the cells.

157. Answer (1)

Parathyroid gland secretes parathormone, which

increases the level of calcium ions in blood plasma.

158. Answer (3)

Lyases breakdown the covalent bond with the

removal of the group leaving double bond behind.

159. Answer (1)

160. Answer (1)

Haem is the prosthetic group for both catalase and

peroxidase.

For digestive enzyme, carboxypeptidase Zn++ is the

cofactor.

For tyrosine, copper is the cofactor.

161. Answer (3)

Adenylic acid is a nucleotide, AMP.

A is adenine

B is cytocine

C is ribose

D is phosphoric acid

E is guanine


10/10

� � �

AMP is

O

H

C1

Adenine

H

C

HO – P – O – CH2

OH

C2

H

OH

C3

HOH

O

162. Answer (3)

A is serine amino acid, it is alcoholic.

B is cysteine, it is sulphur containing amino acid.

C is tyrosine, it is aromatic and non-essential amino

acid.

D is glutamic, acidic amino acid.

163. Answer (4)

Negative feedback inhibition is responsible to

maintain homeostasis in the body. If the level of

hormone rises, it is detected and immediately

reversed, it is negative feedback inhibition.

Positive feedback are rare, the level of hormone

keeps rising but is not immediately reversed.

164. Answer (2)

Melatonin hormone controls sleep-wake cycle.

165. Answer (2)

Epiglottis closes the glottis at the time of swallowing

of food.

166. Answer (1)

Appendicular skeleton is present on the lateral sides

of axial skeleton. It includes the bones of the girdle

and limbs.

167. Answer (2)

The various levels of organisation are

Cellular level Tissue level Organ level Organ

system

168. Answer (3)

In fishes, middle ear is absent. So, they have lateral

line sense organs to detect the vibrations and

currents in water. Cochlea is for hearing.

Semicircular canals have sensory spots cristae with

sensory hair cells to detect dynamic equilibrium or

when a person is in motion.

169. Answer (3)

Birds are endothermal and reptiles are ectotherms.

170. Answer (3)

Sub-phylum vertebrata is divided into two divisions –

Agnatha (jawless vertebrates) and Gnathostomata

(jawed vertebrates).

171. Answer (2)

Ascidians are non-vertebrate chordates. They are

urochordates. In them, notochord is present only in

the larval tail not in adult, nor it is replaced by

vertebrae column.

172. Answer (3)

In ctenophores, the sexes are not separate. They

reproduce only sexually. They are exclusively

marine.

173. Answer (3)

174. Answer (4)

In Bufo and Rana, ureters carry both sperms and

urine.

175. Answer (3)

In echinoderms the endoskeleton is made up of

calcareous ossicles which are mesodermal in origin.

176. Answer (2)

In male frog, the copulatory pads are present in the

1st digit of forelimbs.

177. Answer (2)

Planarian has two way digestive system, as it has

blind sac type of body plan.

178. Answer (3)

The second heart sound 'dub' associated with the

closure of semilunar valves.

179. Answer (3)

Dark adaptation : Rods adapt later, cones adapt first.

180. Answer (1)

Test - 8 (Code-D) (Answers & Hints) All India Aakash Test Series for Medical-2018

1/10

1. (2)

2. (4)

3. (3)

4. (2)

5. (4)

6. (1)

7. (2)

8. (4)

9. (2)

10. (1)

11. (1)

12. (4)

13. (4)

14. (1)

15. (1)

16. (3)

17. (2)

18. (3)

19. (2)

20. (3)

21. (2)

22. (3)

23. (4)

24. (1)

25. (1)

26. (1)

27. (3)

28. (2)

29. (3)

30. (4)

31. (2)

32. (4)

33. (4)

34. (2)

35. (3)

36. (2)

37. (1)

38. (2)

39. (3)

40. (1)

41. (4)

42. (2)

43. (1)

44. (1)

45. (2)

46. (4)

47. (3)

48. (2)

49. (3)

50. (4)

51. (3)

52. (4)

53. (2)

54. (1)

55. (3)

56. (2)

57. (3)

58. (4)

59. (3)

60. (1)

61. (3)

62. (2)

63. (4)

64. (4)

65. (4)

66. (2)

67. (4)

68. (1)

69. (2)

70. (4)

71. (4)

72. (2)

73. (1)

74. (2)

75. (4)

76. (3)

77. (1)

78. (1)

79. (4)

80. (4)

81. (3)

82. (4)

83. (4)

84. (3)

85. (4)

86. (2)

87. (1)

88. (4)

89. (4)

90. (1)

91. (2)

92. (3)

93. (4)

94. (2)

95. (1)

96. (3)

97. (1)

98. (1)

99. (2)

100. (1)

101. (1)

102. (4)

103. (4)

104. (4)

105. (3)

106. (3)

107. (4)

108. (4)

109. (3)

110. (2)

111. (2)

112. (2)

113. (2)

114. (1)

115. (4)

116. (4)

117. (4)

118. (3)

119. (3)

120. (1)

121. (4)

122. (1)

123. (1)

124. (2)

125. (4)

126. (2)

127. (1)

128. (4)

129. (3)

130. (3)

131. (1)

132. (4)

133. (3)

134. (4)

135. (3)

136. (3)

137. (3)

138. (1)

139. (4)

140. (4)

141. (3)

142. (4)

143. (1)

144. (1)

145. (4)

146. (1)

147. (1)

148. (1)

149. (4)

150. (3)

151. (4)

152. (4)

153. (2)

154. (1)

155. (1)

156. (3)

157. (3)

158. (1)

159. (3)

160. (3)

161. (2)

162. (2)

163. (1)

164. (1)

165. (1)

166. (3)

167. (1)

168. (4)

169. (2)

170. (1)

171. (4)

172. (2)

173. (1)

174. (1)

175. (1)

176. (1)

177. (1)

178. (1)

179. (1)

180. (2)

ANSWERS

TEST - 8 (Code-D)

All India Aakash Test Series for Medical-2018

Test Date : 19-03-2017

All India Aakash Test Series for Medical-2018 Test - 8 (Code-D) (Answers & Hints)

2/10

Hints to Selected Questions

[ PHYSICS]

1. Answer (2)

P R2T4

Pnew

= 4 × 440 = 1760 watt

2. Answer (4)

1120 × 1 × (80 – 15) = m(540 + 20)

m = 130 g

3. Answer (3)

10 5

3 2 RT RT

W

5

6 RT

1

4

W

Q

4

QW

4. Answer (2)

11K

5. Answer (4)

W = WAB

+ WBC

0 0 0 0

0 0

3 5 72

2 2 2

P P P VV V

6. Answer (1)

U = U1 + U

2

= 5 3

2 42 2

RT RT

= 11RT

7. Answer (2)

14

e l ...(i)

2

3

4e

l ...(ii)

Solving,

2 13

1.05 cm2

e

l l

8. Answer (4)

2

2 1 10

1

10 logI

SL SLI

⎛ ⎞ ⎜ ⎟⎝ ⎠

30 – 3 = 27 dB

9. Answer (2)

vs = 20 cos60 = 10 m/s

340660 680 Hz

340 10

f

s

vf

v v

⎡ ⎤ ⎢ ⎥ ⎣ ⎦

10. Answer (1)

2

2K

= 4 m

So, L can be 3

, , , 22 2

etc.

11. Answer (1)

A = 2 2

1 2 1 22 cosA A A A

= 2 cos2

A

= 2A cos30°

= 3

2 32

A A

12. Answer (4)

2

av max;

4

Aa a A

T

av

max

2a

a

13. Answer (4)


3/10

14. Answer (1)

22 2

2 2 2I MR R

TMgd MgR g

15. Answer (1)

1 2

0

2( )

At H H

A g

Here, T1 = T

2

H H

H

4H

H

= 4

16. Answer (3)

PA = ab = gb

PB = gh

PA – P

B = g(b – h)

17. Answer (2)

1000r3 = R3

R = 10r

Now,

2

2

1

101000

R

r

18. Answer (3)

P

V

V

PV

V

V = VT

PV P

V T T ⇒

19. Answer (2)

20. Answer (3)

T R3/2

3

2

T R

T R

21. Answer (2)

2R

Tg

22. Answer (3)

28 2 8 1; 0.5 m/s

8 8 8 8

x y

a a

2 2 25m/s

2x y

a a a

23. Answer (4)

2 2

0

1 3 1

2 4 2mv mv

⎛ ⎞ ⎜ ⎟⎝ ⎠

2 2

0

3

4v v

0

3

2v v

0

3

2

ve

v

24. Answer (1)

vP = 2v

C v

C = 3 m/s

25. Answer (1)

IZ

= IX + I

Y

=

2

212

ML

=

2

6

ML

26. Answer (1)

Centre of mass lies on the symmetric line.

27. Answer (3)

At equilibrium,

F = 0 x = ±1

28. Answer (2)

21(2 )

2mv mg R H

2 (2 )v g R H

29. Answer (3)

Work energy theorem,

21

2Pt mv

2Ptv

m


4/10

30. Answer (4)

W F s ��

= ˆ ˆ ˆ ˆ(6 2 )(2 12 )i j i j

= 12 – 24

= –12 J

31. Answer (2)

Tension = Weight of the lower part

32. Answer (4)

2 2(20)tan 1

40 10

v

Rg

= 45°

33. Answer (4)

fL =

sN =

smg = 100 N

Body will remain static.

Static friction = f = F = 80 N

34. Answer (2)

6 sin60 4 sin30

10

g ga

a 3.2 m/s2

35. Answer (3)

Free body diagram :

F 2F

3 = 30 Ng

3F = 30 N

F = 10 N

36. Answer (2)

1 1 2 2F m a m a �

� �

37. Answer (1)

� � �

BA B Av v v

45°

45°

A

B

vB

–vA

vA

Cd

vBA

�

d

2

38. Answer (2)

T = 1 s

2 2(horizontal) 3 4 5 m/su

R = 5 m/s × 1 s = 5 m

39. Answer (3)

By comparison,

tan = 1

2 20.1

2 cos

g

u

u = 10 m/s

40. Answer (1)

2(2 )2

8t c

ta a ⇒

t = 2 s

v = 2t = 4 m/s

41. Answer (4)

T = 4 + 8 = 12 s

21

120 m2

Ts

av

12010 m/s

12v

42. Answer (2)

At v = 2, x = 1 m

2 8dv

v xdx

24 4 m/s

dva v x

dx

43. Answer (1)

v2 = u2 + 2as

or v2 = u2 + 2 (Area under a–x graph)

v2 = 21 + 2 × 30

v = 9 m/s

44. Answer (1)

45. Answer (2)


5/10

[ CHEMISTRY]

46. Answer (4)

47. Answer (3)

1 amu = 1.66 × 10–24 g

12 amu = 1.99 × 10–23 g

48. Answer (2)

C

Cl

Cl

ClCl

net

= 0

49. Answer (3)

Since carbon and oxygen both have 2p1 and 2p4

(n = 2) orbital electrons and can easily participate in

-bonding (parallel to plane of internuclear axis).

50. Answer (4)

Hard boiling of an egg produces disorderness in

structure of protene inside the shell of egg.

For isothermal process,

2

1

VS nR ln

V

Since P2 < P

1 and V

2 > V

1, hence entropy will be

increased.

51. Answer (3)

On 10 times dilution, pH is increased by 1 unit.

52. Answer (4)

C2H5OH + 4I

2 + 6OH– CHI

3 + HCO

2

– + 5I– + 5H2O

53. Answer (2)

8Al 4Al2O3 + 24e–

54. Answer (1)

55. Answer (3)

Br — Br — Br

O

O

O

O

O

O

O

O

+6 +4 +6

Average oxidation number is 16

3

or 3x – 16 = 0

3x = +16

x = 16

3

56. Answer (2)

57. Answer (3)

58. Answer (4)

2

2 4 4

S2S

Ag SO 2Ag SO ��⇀

↽��

Ksp

= [Ag+]2[SO4

2–]

= (2S)2(S)

= 4S3

59. Answer (3)

CA of base is formed by accepting the H+ ion

NH2

–H+ NH3

60. Answer (1)

61. Answer (3)

Kb = 10–10, K

a = 10–4

pKa = –log(10–4) = 4

a

[X ]pH pK log

[HX]

[X ]pH 4 log

[HX]

pH = 4

62. Answer (2)

AB(g) ��⇀↽�� A(g) + B(g)

Initially : 1 0 0

At equilibrium : (1 – 0.33) 0.33 0.33

Total moles at equilibrium = 1.33

p

0.33 0.33P P

1.33 1.33K 0.122P

0.67P

1.33

⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

p

1K P

8

P = 8Kp


6/10

63. Answer (4)

(High volume) (Low volume)Ice Water��⇀

↽��

Melting of ice will be favoured by increase in

pressure on system.

64. Answer (4)

In nature, carbon exist as graphite allotrope.

65. Answer (4)

N2(g) + 2H

2(g) N

2H4(g)

Hf° N

2H4(g) = [BE of N N + 2BE of H – H]

– [BE of N – N + 4BE (N – H)]

= 941 + 2 × 436 – (159 + 4 × 398)

= 62 kJ/mol

66. Answer (2)

67. Answer (4)

n(n 2) B.M.

3(3 2)

15 B.M.

68. Answer (1)

More electron withdrawing groups increase the acidic

strength.

69. Answer (2)

Trans but-2-ene is highly stable alkene amongst all

of these.

70. Answer (4)

Addition follows Anti-Markownikoff’s rule due to

–I effect of –CF3.

71. Answer (4)

3° free radical is more stable.

72. Answer (2)

73. Answer (1)

74. Answer (2)

Hydrogen shows oxidation number of –1 in metal

hydrides.

75. Answer (4)

Higher the value of standard reduction potential,

higher is the oxidising power.

76. Answer (3)

H3BO

3 accepts one OH– ion.

77. Answer (1)

78. Answer (1)

2

20.8 mole0.4 mole

MgCl Mg 2Cl

2 2

4 4

0.2 mole 0.2 mole

MgSO Mg SO

Number of moles of Mg2+ = (0.4 + 0.2) = 0.6

79. Answer (4)

Na has low melting point and it spread over water.

80. Answer (4)

Cr (Z = 24) = 1s2, 2s2, 2p6, 3s2, 3p6, 4s1, 3d5

For s-orbital,

l = 0

Total number of electrons having l = 0 are 7.

81. Answer (3)

Gas [X],

x x x

x x

x

P M d RTd or P

RT M

Gas [Y],

y y y

y y

y

P M d RTd or P

RT M

According to the problem,

dx = 3d

y

My = 1.5M

x

or Mx =

1

1.5My

y y yx x

y x y y y

M 3d MP d RT1.5 4.5

P M d RT M d

82. Answer (4)

PV = nRT (Ideal gas equation)

83. Answer (4)

84. Answer (3)

Volume strength = 5.6 × N

=

8.5

175.6

560

1000

= 8.5 1000

5.617 560

= 5 V


7/10

[ BIOLOGY

]

91. Answer (2)

Steroidal hormones secreted from smooth

endoplasmic reticulum (SER).

92. Answer (3)

93. Answer (4)

Figure shown is lysosome.

94. Answer (2)

a. Sugar moiety towards outer region.

b. Quasi-fluid nature of lipid enables lateral

movement of proteins within the overall bi-lipid

layer.

95. Answer (1)

96. Answer (3)

In oocytes of some vertebrates lost for months or

years, true for diplotene stage.

97. Answer (1)

98. Answer (1)

99. Answer (2)

100. Answer (1)

A B — V

101. Answer (1)

102. Answer (4)

Water potential for cell A,

w = DPD +

w = –8 bars

Water potential for cell B = s +

p =

w

So, w = –18 + 7 = –11 bars

Water potential for cell C,

DPD = OP – TP

= 6 – 3

= 3

w = –3 bars

So, direction of water flow

B w = –11

A

Cw = –3

w = –8

103. Answer (4)

104. Answer (4)

B – Sequence of cytochromes for electron flow is

cyt b cyt c1 cyt c cyt a cyt a

3.

D – Cyt c is electron carrier between complex III

and IV.

105. Answer (3)

106. Answer (3)

107. Answer (4)

A – 3-phosphoglyceric acid

B – Lactic acid

C – Ethyl alcohol (ethanol)

108. Answer (4)

109. Answer (3)

III. Mechanical support to stem.

V. Resistant to the attacks of insects.

These features are related to heart wood.

110. Answer (2)

111. Answer (2)

85. Answer (4)

86. Answer (2)

87. Answer (1)

88. Answer (4)

4KClO3

3KClO + KCl4

+5 +7 –1

Reduction

Oxidation

89. Answer (4)

90. Answer (1)

Number of moles of CaCO3

=

3100 10 g

100 g/mole

= 10–3 mole

Number of moles of O atoms = 3 × 10–3

Number of O atoms = 6.02 × 1023 × 3 × 10–3

= 18.06 × 1020

= 1.8 × 1021


8/10

112. Answer (2)

Male gamete (n) + Egg (n)

Male gamete (n) + Secondary nucleus (2n)

Syngamy + triple fusion = Double fertilisation

Zygote (2n)

PEN (3n)

Triple fusion

Syngamy

113. Answer (2)

114. Answer (1)

115. Answer (4)

116. Answer (4)

117. Answer (4)

118. Answer (3)

Mimosa pudica – Leguminaceae (family)

characterised by pulvinus leaf base.

119. Answer (3)

120. Answer (1)

121. Answer (4)

The female gametophyte produces two or more

archegonia.

122. Answer (1)

123. Answer (1)

124. Answer (2)

125. Answer (4)

Natural system of classification given by Bentham

and Hooker.

It is based on external features

internal features

ultrastructure

anatomy

embryology and phytochemistry

126. Answer (2)

127. Answer (1)

128. Answer (4)

129. Answer (3)

130. Answer (3)

131. Answer (1)

2H+, 2e–, 4ATP

132. Answer (4)

133. Answer (3)

IV. Nature of cell wall.

VIII. Mode of respiration.

134. Answer (4)

C – Higher is the category, greater is the difficulty of

determining the relationship to other texa at the

same level.

135. Answer (3)

136. Answer (3)

137. Answer (3)

Dark adaptation : Rods adapt later, cones adapt first.

138. Answer (1)

The second heart sound 'dub' associated with the

closure of semilunar valves.

139. Answer (4)

Planarian has two way digestive system, as it has

blind sac type of body plan.

140. Answer (4)

In male frog, the copulatory pads are present in the

1st digit of forelimbs.

141. Answer (3)

In echinoderms the endoskeleton is made up of

calcareous ossicles which are mesodermal in origin.

142. Answer (4)

In Bufo and Rana, ureters carry both sperms and

urine.

143. Answer (1)

144. Answer (1)

In ctenophores, the sexes are not separate. They

reproduce only sexually. They are exclusively

marine.

145. Answer (4)

Ascidians are non-vertebrate chordates. They are

urochordates. In them, notochord is present only in

the larval tail not in adult, nor it is replaced by

vertebrae column.

146. Answer (1)

Sub-phylum vertebrata is divided into two divisions –

Agnatha (jawless vertebrates) and Gnathostomata

(jawed vertebrates).

147. Answer (1)

Birds are endothermal and reptiles are ectotherms.

148. Answer (1)

In fishes, middle ear is absent. So, they have lateral

line sense organs to detect the vibrations and

currents in water. Cochlea is for hearing.

Semicircular canals have sensory spots cristae with

sensory hair cells to detect dynamic equilibrium or

when a person is in motion.


9/10

149. Answer (4)

The various levels of organisation are

Cellular level Tissue level Organ level Organ

system

150. Answer (3)

Appendicular skeleton is present on the lateral sides

of axial skeleton. It includes the bones of the girdle

and limbs.

151. Answer (4)

Epiglottis closes the glottis at the time of swallowing

of food.

152. Answer (4)

Melatonin hormone controls sleep-wake cycle.

153. Answer (2)

Negative feedback inhibition is responsible to

maintain homeostasis in the body. If the level of

hormone rises, it is detected and immediately

reversed, it is negative feedback inhibition.

Positive feedback are rare, the level of hormone

keeps rising but is not immediately reversed.

154. Answer (1)

A is serine amino acid, it is alcoholic.

B is cysteine, it is sulphur containing amino acid.

C is tyrosine, it is aromatic and non-essential amino

acid.

D is glutamic, acidic amino acid.

155. Answer (1)

Adenylic acid is a nucleotide, AMP.

A is adenine

B is cytocine

C is ribose

D is phosphoric acid

E is guanine

AMP is

O

H

C1

Adenine

H

C

HO – P – O – CH2

OH

C2

H

OH

C3

HOH

O

156. Answer (3)

Haem is the prosthetic group for both catalase and

peroxidase.

For digestive enzyme, carboxypeptidase Zn++ is the

cofactor.

For tyrosine, copper is the cofactor.

157. Answer (3)

158. Answer (1)

Lyases breakdown the covalent bond with the

removal of the group leaving double bond behind.

159. Answer (3)

Parathyroid gland secretes parathormone, which

increases the level of calcium ions in blood plasma.

160. Answer (3)

In tight junction, the two adjacent cell membranes

are fused together to prevent the leaking of any

substance between the cells.

161. Answer (2)

Periplaneta americana is a protostome with

determinate cleavage.

162. Answer (2)

Knee joint is hinge joint, is formed between femur

and tibia, fibula does not take part.

163. Answer (1)

Estrogen promotes osteoblast activity.

164. Answer (1)

If the oxygen hemoglobin dissociation curve shifts

towards RHS, it means binding has decreased

under the same pO2 because the secondary factors

have increased, causing dissociation of HbO2.

Strenuous exercise would increase lactic acid in

muscles and blood, pH would decrease.

165. Answer (1)

If we hold our breath for some time CO2 will build up

in the body, pH will decrease and the respiratory

rate would increase.

166. Answer (3)

Neutrophils are 60–65%, they help in phagocytosis,

but they are not precursor of macrophages.

Monocytes are precursor of macrophages.


10/10

� � �

167. Answer (1)

Aldosterone is produced by adrenal cortex in

response to low levels of sodium ions in the blood.

168. Answer (4)

a. Reabsorption is minimum in DLH.

b. Selective secretion of H+ and K+ ions occur in

PCT, DCT and CD.

c. Conditional reabsorption of Na+ and water occur

in DCT.

d. Filtrate becomes hypotonic in ALH.

169. Answer (2)

During ventricular systole, the blood from the left

ventricle will flow into aorta, and the blood from right

ventricle will enter into pulmonary artery.

170. Answer (1)

The only phylum in the kingdom animalia is phylum

porifera.

171. Answer (4)

The presence of true body cavity poses a problem

for direct exchange of oxygen and carbon dioxide

with environment. But it provides space for the

intestine to become longer for better digestion and

absorption.

172. Answer (2)

Endothermal organisms are warm blooded with a

constant body temperature. They have a higher

metabolic rate, so they eat more than ectotherms.

173. Answer (1)

Mitral valve is bicuspid valve between left atrium and

left ventricle. If we cut the chordae tendinae of mitral

valve, the valve will close, but during ventricular

systole, they will regurgitate into atria along with some

blood. So, the flow of blood would decrease in aorta.

174. Answer (1)

Estrogen hormone is a steroid hormone which

directly enters the cell, no second messenger is

required. It binds with intracellular receptor in

nucleus to form hormone receptor complex which

binds with a specific DNA segment to stimulate

gene expression.

175. Answer (1)

If the level of cortisol rises in blood, then increase in

this level will exert a negative feedback control on

the hypothalamus, so less CRH is released Soless ACTH released So less cortisol till the

level is normalised.

176. Answer (1)

Total fat consumption is 1 g, i.e., 1 × 9 = 9 K

calories are released.

Total calories from serving is 217 kcal.

Total fat 1 g × 9 = 9 kcal

Starch 43 g × 4 = 172

Sugar 3 g × 4 = 12

Protein 6 g × 4 = 24

= 217 kcal

Percentage of calories that come from fat

9100 4.1%

217

177. Answer (1)

A person with B blood group has B antigen and

anti-A antibodies. If he is B negative Rh antigen and

Rh antibodies are absent.

178. Answer (1)

Knee jerk reflex is simple reflex. It is monosynaptic

reflex, i.e., between the sensory and motor neuron

there is only one synapse, no interneuron are

present. Quadriceps femoris is extensor of thigh.

179. Answer (1)

7% of CO2 is transported as dissolved gas in

plasma. 70% of CO2 is transported as HCO

3

– in

RBC and plasma CO2 is mainly transported as

NaHCO3 in plasma. 20–25% is transported as

carbaminohemoglobin.

180. Answer (2)

A resting membrane, is said to be in polarised

state. In this, the outside of the membrane is

positively charged, and inside is negatively charged.

Outside has more Na+ ions and inside has more K+

ions. K+ ions leak/diffuse from inside to outside.

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