for Medical-2019 All India Aakash Test Series for Medical ... · Test - 6 (Code C) (Hints) All...

Test - 6 (Code C) (Answers) All India Aakash Test Series for Medical-2019

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Test Date : 18/02/2018

ANSWERS

TEST - 6 (Code C)

All India Aakash Test Series for Medical - 2019

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All India Aakash Test Series for Medical-2019 Test - 6 (Code C) (Hints)

2/8

HINTS

PHYSICS

2. Fact.

3. avg

2 3 .....v v v nvv

n

( 1)

2

nv

3 3 3 3

rmc

(2 ) (3 ) ..... ( )v v v nvv k

n

2

1 1

2

nkv n v

n

⎡ ⎤⎛ ⎞ ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

1

2

vk nn v

n

⎛ ⎞ ⎜ ⎟⎝ ⎠

Ratio = 1

k nv.

4. Both the pistons move to the right because of equal

pressure but greater force on right cylinder (greater

area).

5. At V = 3 m3, P = 22

1 Nm1 1

rms 3

3 3 3 1 3

32 10 2

RT PVV

M M N

⎛ ⎞ ⎜ ⎟⎝ ⎠

11510 ms

4.

6. U = 2 + 3PV nCVdT = dU = 3nRdT [PV = nRT]

CV = 3R

7. P6 V5 = const.

5

6const.PV

Now3 15

51 2 21

6

v

R R RC C R

x

Heat supplied, Q = nCT 15

(5)2

Rn⎛ ⎞ ⎜ ⎟⎝ ⎠

= 37.5 nR.

8. Clearly at state A, gas is an ideal one but in state

B it is real gas.

So PA < P

B and T

A > T

B

9. Ar

3(2) 3

2E nR nR

2H

5(3) 7.5

2E nR nR

2N

5(5) 12.5

2E nR nR

3O

(3 )(2) 6E n R nR .

11. A = A0e–kt

(20 )00

2

k TAA e

(T = Time period)

Also, (80 )

0k T

A A e

So, 0 15

cm16 16

AA .

12. Time taken will be minimum when it travels from

2

A to

2

A because of more average speed.

So, 2 s.12 12 6

T T Tt

13. O

N

R

mg mgsin

Taking moment about ‘O’

mgsin × R = I

2Rmg mR �

g

R

2 2 s.g R

TR g

14. Maximum tension will be at mean position and

minimum tension will be at extreme position

v

mgcosmg

T1

T2

mg

Test - 6 (Code C) (Hints) All India Aakash Test Series for Medical-2019

3/8

According to problem, 2T2 = T

1

2

2( cos )( )

mvmg mg

R

l...(i)

From conservation of energy

21(1 cos )

2mg mv l ...(ii)

Using (i) & (ii)

2 (1 cos )2 cos

mgmg mg

l

l

3cos

4

15. 100cos 20 03

v t⎛ ⎞ ⎜ ⎟

⎝ ⎠

For the second time, 3

203 2

t

7second.

120t

16.

T

k

x

Equilibrium position

2x

Let T is additional tension after shifting the block by

x.

Tension in spring (additional) = 2kx

T = 4

kx = restoring force on block

4

ka x

m

4 k

m

42 m

Tk

17.

eff.

2 Tg

l( g

eff. = g)

19. Fact.

21.1

100 1000 10 50200

T Mg B N

0

1 1 50125 Hz

2 2 0.2 0.02

Tf

L

22.1f k T

2

19 9

100 10f k T T k

⎛ ⎞ ⎜ ⎟⎝ ⎠

% decrease 2 1

1

100 10%f f

f

23.

L

L/4

2200

2

T

L

For second case

L

L/4

6

2

Tf

L

f = 3 × 200 = 600 Hz.

24.0

50 10logI

I

⎛ ⎞ ⎜ ⎟⎝ ⎠

Also

4

0

1010log

IL

I

⎛ ⎞ ⎜ ⎟

⎝ ⎠

L – 50 = 10log(104) = 40

L = 90 dB

25. Frequency heard directlly

1 0 0

340

339s

vf f f

v v

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠⎝ ⎠

Frequency heard after reflection

2 0 0

340

341s

vf f f

v v

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠⎝ ⎠

According to problem, f1 – f

2 = 4

Solving f0 � 680 Hz.

26. ⎛ ⎞ ⎛ ⎞ ⇒ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠1 2

340 3400 338 342

340 340

v vf f

v = 2 ms–1

27. At, t = 0, x = 0; 3.

2y

Also phase constant will be .3

Also slope at t = 0 is positive. That leads to second

option.


4/8

30. Using phasor diagram,

60º

A

A

2 2( cos60º ) ( cos30º )r

A A A A = A

31. y = 2Asinkx cost

= (hsinkx) cost

So, a = hsinkx

From mode of vibration, 3 2

2 3

LL

3K

L

1

3sin 0.5

18 2

L hy h h

L

⎛ ⎞ ⎜ ⎟⎝ ⎠

33.1

2 360 2x and 2

2 360 4x

Solving, 1 2

1080x x m.

34. p

v A ab

w

bv

k c

1w

p

v

v ac

35.T xg

v xg

constant dvv adx

.

36. ⇒ ⇒ 2 5

2 2

0.4625 10 W

8 1

PP A P

39. ∵string

,f T we can conclude from question that

ftuning

– fstring

= 6

So fstring

= 512 – 6 = 506 Hz

40. 0

02

vf

l and

0

0

42

c c

vf so f f

l

⎛ ⎞⎜ ⎟⎝ ⎠

42.0

0

11 4 8

4 2 11c c

lv v

l l l ⇒ .

43. Basic relations.

44.350 7

300 6

v

f m and

3

Using 2

x

45.1 1,k

vk

CHEMISTRY

46.

O

OH1

2

3

4

6

5 6-Hydroxycyclohex-2-enone

47.

OHO

~99.9%

Aromatic

~0.1%

48. Acidic strength stability of conjugate base.

After removal of acidic hydrogen various ions formed

are

Antiaromatic

Aromatic(most stable) Antiaromatic

CH CH2

49. Tolerable limit for nitrates, fluorides, sulphates and

lead ions for drinking water are: 50 ppm, 1 ppm, 500

ppm and 50 ppb respectively.

51. Meq of excess acid = meq of NaOH 1

100 205

Total meq of acid = 1

2 150 605

Meq of H2SO

4 reacted = 60 – 20 = 40

= meq of NH3

Mole of NH3 × 1 × 103 = 40

Mole of NH3 = 40 × 10–3 = mole of N

Mass of N = 40 × 10–3 × 14 g = 0.56 g

% of N = 0.56

100 80%0.7

53. CH CH CH COOH3 2 2

CH CH COOCH3 2 3

CH CH CH3 3

COOH

CH COOCH CH3 2 3

HCOOCH CH CH2 2 3

HCOOCHCH3

CH3


5/8

54. In hydroboration oxidation, product obtained is similar

to that by anti-Markovnikov addition of water.

CH3

H C3

C CH2

1. B H2 6

2. H O /OH2 2

–

CH3

H C3

CH CH OH2

55. Electrophilic substitution reaction of type F.C.

Alkylation on benzene.

H – F

–F EAS+ HF

57. CH CH CH CH CH3 3

CH3

Number of -H = number of hyperconjugative

structures

59. CH CH C CH CH3 2 2 3

CH3

60. Higher the extent of +M/+R more is stability of

carbocation.

61.

O /V O2 2 5

773 KC H

66

CH C

CH CO

O

OMaleic anhydride

62. CH CH CH3 2

CH CH CH3 2

HOCl

Markovnikov

addition

– +

OH Cl63. CH

3COOH

(A) (B) (C)

64. Higher the number of -H of alkyl group directly

attach on benzene, more is reactivity towards

electrophilic substitution reaction.

67. Option (1) is least stable as like charge on adjacent

atoms.

69.Cl

2

UV

H Cl

H

ClH

Cl

Cl

HCl

H

Cl HGammexane

(666)

71. Sulphonation is a reversible process in which

generation of electrophile take place as:

2 4 3 3 42H SO H O SO HSO

��⇀↽��

On addition of KHSO4, concentration of

4HSO

increases so, backward reaction is favoured.

72. Structure of product formed will be:

C C

H

CH3

H

C C

H Z H

C C

H

CH3EHE

73.

O OO

CH3

O

H C3

O O

74. CH3

CCl3

(A) (B) (C)

CCl3

Cl

75. Intermediate formed during anti-Markovnikov addition

is CH CH CH Br3 2

which may undergo

following changes.

CH CH CH Br3 2

BrCH CH CH Br

3 2

Br

CH CH CH Br3 2 2

CH CH CH Br3 2

CH CH CH CH3 3

CH Br2

CH Br2

H

76. CH CH CH CH CH CH CH CH CH CH CH CH3 2 2 2 2 2 2 3

CH3

CH3

CH3

CH CH3

CH CH3

CH3

1 2 3 4 5 6 7 8 9 10 11 12

77.Cl

COOH1

23

4

5

COOH

Cl

1

23

4

3-chloro-2-methylpentanoic acid

3-chloro-2-ethylbutanoic acid

78.CH CH CH CH Cl

3 2 2

Cl

CH CH C CH3 2 3

Cl

Cl

CH CH CH CH3 2 2

ClCl

CH CH CH CH3 2 2 2

ClCl

CH CH CH CH2 2 2 2

ClCl

CH CH CH CH3 3

ClCl


6/8

79. CH CH CH CHO3

CHCl CHO2

(A) (B) (C)

80. Bromination of alkane is highly selective. Reactivity

of 1°H : 2°H : 3°H of alkane is 1 : 82 : 1600 for

bromination by free radical mechanism.

81.4 3 2

CH CH NO

(B) (C)

85.

1. O3

2. Zn/H O2 OOO

86. W = 0.30 g V1 = 50 ml

P1 = (P – a) = 715 – 15 = 700 mm of Hg

T1 = 300 K

Where a = aqueous tension

Now, 1 1 2 2

1 2

PV P V

T T 1 1 2

2

1 2

PV T 700 50 273V

T P 300 760

V2 = 41.9 ml

% of nitrogen = 28 41.9

100 17.46%22400 0.3

87.C C

H C3

CH3

CH3

CH3

Proceed through more stable carbocation

88. For reduction of each mole of alcohol, ketone &

acids, number of moles of HI used are 2, 4 & 6

respectively.

89. (A) CH – CH – CH – Br3 2

Br

NaNH (2 eq)2

CH – C CH3

NaNH2

CH – C C : Na3

H O2

CH – C CH3

90. Ammonium ion does not act as electrophile as N

cannot form 5 bonds.

BIOLOGY

91. Translocation in phloem is explained by pressure

flow hypothesis.

Diffusion is not dependent on a ‘‘living system.’’

98. Cell A w = –18 + 8 = –10

Cell B w = –14 + 2 = –12

Cell C w = –12 + 2 = –10

So flow of water is A = –10w

w = –12 B

w = –10 C

101. Etiolation occurs when plants are grown in darkness.

105. A – Denitrification

B – Denitrifying bacteria – Pseudomonas,

Thiobacillus

106. NitriteNitrite reductase

6H + 6e+ –

PSI

NH (Ammonia)3

110. (i) Transmembrane channel

(ii) Facilitated diffusion

(iii) Outer

(iv) Thylakoid membrane

(v) Stroma

112. Cyclic ETS – P700

, Hydrogen carrier, PSI

114. Sugarcane and maize are C4 plants

122. 38 ATP × 40 = 1520 ATP

124. CO2 is produced at 3 steps

(i) During acetyl Co-A formation

(ii) During formation of -ketoglutaric acid

(iii) During succinic acid formation

136. Actin filaments slide over myosin filaments towards

H-zone but their length remains the same.

137. The Ca2+ ions bind with troponin-C, this binding will

shift the tropomyosin-troponin complex, so that the

myosin binding sites of actin protein become

exposed to bind with actin binding sites of myosin.

138. ‘‘All or none law’’ is applicable to muscle fibres. If we

apply a threshold stimulus, muscle fibre will contract

with maximum force. If the stimulus is of sub-

threshold value the fibre will not contract at all.

139. Skeletal muscles at rest obtain most of their energy

from the aerobic respiration of fatty acids. During

exercise, muscle glycogen and blood glucose are

also used as energy sources.

140. In this disorder dystrophin protein is not formed. It

adds strength to the muscle by providing scaffolding

for the fibrils which is essential for muscle

contraction. It helps to reinforce the sarcolemma

and thus may help to transmit tension generated by

sarcomeres to the tendons.


7/8

141. Humerus is the part of appendicular endoskeleton.

142. (a) Tarsals in 1 hind limb = 7

Bones in cranium = 8

(b) Cranial bones = 8

Carpals in each forelimb = 8

(c) Thoracic cage of human adult = 25

( 12 × 2 ribs + 1 sternum)

Each forelimb has 30 bones.

(d) Facial region has = 14 bones

Tarsals of both hindlimbs = 2 × 7 = 14 bones

143. Nodding movement of head occurs due to the

atlanto-occipital / condyloid joint between atlas and

occipital.

144. Between sternum and ribs, there is present

amphiarthrose / cartilaginous joint.

145. Rheumatoid arthritis is diagnosed by the presence of

rheumatoid factor antibodies which resemble IgM. In

this disease, finally the joint becomes immovable.

146. Through filtration membrane all blood cells and

proteins cannot be filtered, so the glomerular filterate

is blood without formed elements and proteins. It is

isotonic to blood plasma.

147. Fresh water fishes excrete hypotonic urine through

which there occurs loss of ions also. So to

compensate it and to maintain their osmolarity

higher, they actively uptake monovalent ions from

surrounding water.

148. As their loop of Henle is very short they are not used

in concentration of urine by counter-current

mechanism.

149. Due to the presence of liver on right side, right

kidney is slightly lower than the left kidney.

150. In diabetes mellitus, both glucose and ketone

bodies are passed out in urine.

151. % Filtration fraction (F.F) = GFR

RPF

Renal plasma flow (RPF) = GFR 125 125

FF 20% 0.2

= 625 ml/min

If we have to calculate renal blood flow

Renal plasma constitutes only 55% of blood volume

so renal blood flow is

RPF 625100 1136.36364 ml./min

55% 55

Nearly = 1137 ml/min

153. Collip's hormone / parathormone / PTH secretion

causes the rise in blood Ca+2 level, so due to this

there occurs loss of Ca+ minerals from bones and in

the hypersecretion, cavities are formed in bones

filled with fibres. TCT is hypocalcemic hormone.

154. cAMP acts as secondary messenger because

adrenaline cannot cross lipid bilayer of cell

membrane.

155. Neurohypophysis, i.e. the posterior pituitary is only

a storing and releasing centre of neurohormones

which are synthesized and secreted from

hypothalamus and reach the neurohypophysis via

neuronal axons.

156. Knee jerk reflex is monosynaptic/simple reflex.

Between sensory and motor neurons there are no

interneurons.

157. 3P's is diabetes mellitus which express the

followings–

Polyuria – frequent urination – large volume of urine

along with glucose.

Polyphagia – frequent eating (hunger).

Polydipsia – frequent thirst.

158. Toxic goitre is due to nodules in thyroid gland

leading to hyperthyroidism.

159. Kidney does not secrete angiotensinogen

Angiotensinogen Angiotensin-I

Liver

Renin

Kidney

Secretes

160. Cortisol, a glucocorticoid increases blood glucose

level, so it will inhibit the cellular uptake of glucose

and synthesis of glycogen from glucose

(= glycogenesis)

162. In blind girls and boys puberty is early as priming of

pineal gland hasn't been achieved.

163. Bronze like skin pigmentation is feature of Addison's

disease due to hyposecretion of adrenal cortex, rest

all are the cases of hypersecretion.

164. Progesterone, a lipid soluble hormone can cross

lipid bilayer of target organ, so it does not require a

secondary messenger. It binds with intracellular

receptors.

167. From inside to outside, meninges are arranged

as–piamater–arachnoid–duramater.


8/8

168. Cavities of cerebrum i.e. lateral ventricles or paracoel

open in cavity of diencephalon i.e. diocoel or III

ventricle through foramen of Monro.

170. Cerebellum is a part of hind brain

171. Medulla controls respiration cardiovascular reflexes

and gastric secretions.

172. Parasympathetic nervous system constitutes cranio-

sacral outflow i.e. III, VII, IX, X are cranial autonomics

= 4 pairs

173. Hypermetropia can be due to shortening of eyeball

on decrease in convexity of the lens.

174. When canal of Schlemm is blocked, aqueous humor

is not drained in blood so there occurs increase in

intraocular pressure and retina is damaged along

with optic nerve.

175. Tympanum is ear drum, separates outer & middle

ear.

Ear ossicles are present in middle ear.

Membranous labyrinth is internal ear.

176. Photoreceptors are hyperpolarised during light and

depolarised in dark human eye.

177. This is the case of near vision, so–

• Pressure tension / develops in ciliary muscles

i.e. they contract.

• Ligament – Relax.

• Lens become thick i.e. more convex so

convergence of diverging light rays increased.

178. Tapetum Lucidium cellulosum layer has refractive

crystals which reflects the light rays when

illuminated during night.

180. Both pitressin and vasopressin are the other names

of ADH (Anti diuretic hormone), which is released to

reabsorb more H2O from DCT of nephron.

� � �

Test - 6 (Code D) (Answers) All India Aakash Test Series for Medical-2019

1/8

1. (4)

2. (3)

3. (3)

4. (4)

5. (4)

6. (3)

7. (1)

8. (1)

9. (3)

10. (3)

11. (4)

12. (1)

13. (3)

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15. (4)

16. (1)

17. (3)

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20. (2)

21. (4)

22. (3)

23. (4)

24. (2)

25. (2)

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27. (1)

28. (4)

29. (4)

30. (3)

31. (3)

32. (3)

33. (3)

34. (2)

35. (2)

36. (4)

Test Date : 18/02/2018

ANSWERS

TEST - 6 (Code D)

All India Aakash Test Series for Medical - 2019

37. (1)

38. (1)

39. (4)

40. (3)

41. (2)

42. (1)

43. (3)

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47. (2)

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49. (4)

50. (2)

51. (2)

52. (3)

53. (2)

54. (1)

55. (3)

56. (1)

57. (4)

58. (4)

59. (2)

60. (2)

61. (4)

62. (3)

63. (4)

64. (3)

65. (2)

66. (1)

67. (1)

68. (2)

69. (1)

70. (3)

71. (2)

72. (1)

73. (2)

74. (3)

75. (4)

76. (4)

77. (2)

78. (3)

79. (1)

80. (2)

81. (2)

82. (2)

83. (3)

84. (3)

85. (3)

86. (2)

87. (3)

88. (2)

89. (4)

90. (1)

91. (2)

92. (1)

93. (4)

94. (1)

95. (1)

96. (3)

97. (2)

98. (1)

99. (3)

100. (3)

101. (2)

102. (3)

103. (4)

104. (3)

105. (4)

106. (3)

107. (3)

108. (1)

109. (1)

110. (3)

111. (3)

112. (1)

113. (1)

114. (1)

115. (3)

116. (2)

117. (2)

118. (3)

119. (3)

120. (3)

121. (3)

122. (1)

123. (2)

124. (3)

125. (3)

126. (4)

127. (2)

128. (2)

129. (4)

130. (3)

131. (3)

132. (1)

133. (1)

134. (3)

135. (2)

136. (3)

137. (2)

138. (2)

139. (3)

140. (4)

141. (1)

142. (1)

143. (1)

144. (2)

145. (4)

146. (1)

147. (3)

148. (4)

149. (3)

150. (1)

151. (3)

152. (4)

153. (2)

154. (3)

155. (1)

156. (1)

157. (3)

158. (3)

159. (1)

160. (4)

161. (4)

162. (2)

163. (1)

164. (3)

165. (2)

166. (2)

167. (3)

168. (4)

169. (2)

170. (1)

171. (2)

172. (3)

173. (2)

174. (3)

175. (3)

176. (4)

177. (4)

178. (1)

179. (2)

180. (2)

All India Aakash Test Series for Medical-2019 Test - 6 (Code D) (Hints)

2/8

HINTS

PHYSICS

1.1 1,k

vk

2.350 7

300 6

v

f m and

3

Using 2

x

3. Basic relations.

4.0

0

11 4 8

4 2 11c c

lv v

l l l ⇒ .

6. 0

02

vf

l and

0

0

42

c c

vf so f f

l

⎛ ⎞⎜ ⎟⎝ ⎠

7. ∵string

,f T we can conclude from question that

ftuning

– fstring

= 6

So fstring

= 512 – 6 = 506 Hz

10. ⇒ ⇒ 2 5

2 2

0.4625 10 W

8 1

PP A P

11.T xg

v xg

constant dvv adx

.

12. p

v A ab

w

bv

k c

1w

p

v

v ac

13.1

2 360 2x and 2

2 360 4x

Solving, 1 2

1080x x m.

15. y = 2Asinkx cost

= (hsinkx) cost

So, a = hsinkx

From mode of vibration, 3 2

2 3

LL

3K

L

1

3sin 0.5

18 2

L hy h h

L

⎛ ⎞ ⎜ ⎟⎝ ⎠

16. Using phasor diagram,

60º

A

A

2 2( cos60º ) ( cos30º )r

A A A A = A

19. At, t = 0, x = 0; 3.

2y

Also phase constant will be .3

Also slope at t = 0 is positive. That leads to second

option.

20. ⎛ ⎞ ⎛ ⎞ ⇒ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠1 2

340 3400 338 342

340 340

v vf f

v = 2 ms–1

21. Frequency heard directlly

1 0 0

340

339s

vf f f

v v

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠⎝ ⎠

Frequency heard after reflection

2 0 0

340

341s

vf f f

v v

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠⎝ ⎠

According to problem, f1 – f

2 = 4

Solving f0 � 680 Hz.

22.0

50 10logI

I

⎛ ⎞ ⎜ ⎟⎝ ⎠

Also

4

0

1010log

IL

I

⎛ ⎞ ⎜ ⎟

⎝ ⎠

L – 50 = 10log(104) = 40

L = 90 dB

Test - 6 (Code D) (Hints) All India Aakash Test Series for Medical-2019

3/8

23.

L

L/4

2200

2

T

L

For second case

L

L/4

6

2

Tf

L

f = 3 × 200 = 600 Hz.

24.1f k T

2

19 9

100 10f k T T k

⎛ ⎞ ⎜ ⎟⎝ ⎠

% decrease 2 1

1

100 10%f f

f

25.1

100 1000 10 50200

T Mg B N

0

1 1 50125 Hz

2 2 0.2 0.02

Tf

L

27. Fact.

29.

eff.

2 Tg

l( g

eff. = g)

30.

T

k

x

Equilibrium position

2x

Let T is additional tension after shifting the block by

x.

Tension in spring (additional) = 2kx

T = 4

kx = restoring force on block

4

ka x

m

4 k

m

42 m

Tk

31. 100cos 20 03

v t⎛ ⎞ ⎜ ⎟

⎝ ⎠

For the second time, 3

203 2

t

7second.

120t

32. Maximum tension will be at mean position and

minimum tension will be at extreme position

v

mgcosmg

T1

T2

mg

According to problem, 2T2 = T

1

2

2( cos )( )

mvmg mg

R

l...(i)

From conservation of energy

21(1 cos )

2mg mv l ...(ii)

Using (i) & (ii)

2 (1 cos )2 cos

mgmg mg

l

l

3cos

4

33. O

N

R

mg mgsin

Taking moment about ‘O’

mgsin × R = I

2Rmg mR �


4/8

g

R

2 2 s.g R

TR g

34. Time taken will be minimum when it travels from

2

A to

2

A because of more average speed.

So, 2 s.12 12 6

T T Tt

35. A = A0e–kt

(20 )00

2

k TAA e

(T = Time period)

Also, (80 )

0k T

A A e

So, 0 15

cm16 16

AA .

37. Ar

3(2) 3

2E nR nR

2H

5(3) 7.5

2E nR nR

2N

5(5) 12.5

2E nR nR

3O

(3 )(2) 6E n R nR .

38. Clearly at state A, gas is an ideal one but in state

B it is real gas.

So PA < P

B and T

A > T

B

39. P6 V5 = const.

5

6const.PV

Now3 15

51 2 21

6

v

R R RC C R

x

Heat supplied, Q = nCT 15

(5)2

Rn⎛ ⎞ ⎜ ⎟⎝ ⎠

= 37.5 nR.

40. U = 2 + 3PV nCVdT = dU = 3nRdT [PV = nRT]

CV = 3R

41. At V = 3 m3, P = 22

1 Nm1 1

rms 3

3 3 3 1 3

32 10 2

RT PVV

M M N

⎛ ⎞ ⎜ ⎟⎝ ⎠

11510 ms

4.

42. Both the pistons move to the right because of equal

pressure but greater force on right cylinder (greater

area).

43. avg

2 3 .....v v v nvv

n

( 1)

2

nv

3 3 3 3

rmc

(2 ) (3 ) ..... ( )v v v nvv k

n

2

1 1

2

nkv n v

n

⎡ ⎤⎛ ⎞ ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

1

2

vk nn v

n

⎛ ⎞ ⎜ ⎟⎝ ⎠

Ratio = 1

k nv.

44. Fact.

CHEMISTRY

46. Ammonium ion does not act as electrophile as N

cannot form 5 bonds.

47. (A) CH – CH – CH – Br3 2

Br

NaNH (2 eq)2

CH – C CH3

NaNH2

CH – C C : Na3

H O2

CH – C CH3

48. For reduction of each mole of alcohol, ketone &

acids, number of moles of HI used are 2, 4 & 6

respectively.

49.C C

H C3

CH3

CH3

CH3

Proceed through more stable carbocation

50. W = 0.30 g V1 = 50 ml

P1 = (P – a) = 715 – 15 = 700 mm of Hg

T1 = 300 K

Where a = aqueous tension

Now, 1 1 2 2

1 2

PV P V

T T 1 1 2

2

1 2

PV T 700 50 273V

T P 300 760


5/8

V2 = 41.9 ml

% of nitrogen = 28 41.9

100 17.46%22400 0.3

51.

1. O3

2. Zn/H O2 OOO

55.4 3 2

CH CH NO

(B) (C)

56. Bromination of alkane is highly selective. Reactivity

of 1°H : 2°H : 3°H of alkane is 1 : 82 : 1600 for

bromination by free radical mechanism.

57. CH CH CH CHO3

CHCl CHO2

(A) (B) (C)

58.CH CH CH CH Cl

3 2 2

Cl

CH CH C CH3 2 3

Cl

Cl

CH CH CH CH3 2 2

ClCl

CH CH CH CH3 2 2 2

ClCl

CH CH CH CH2 2 2 2

ClCl

CH CH CH CH3 3

ClCl

59.Cl

COOH1

23

4

5

COOH

Cl

1

23

4

3-chloro-2-methylpentanoic acid

3-chloro-2-ethylbutanoic acid

60. CH CH CH CH CH CH CH CH CH CH CH CH3 2 2 2 2 2 2 3

CH3

CH3

CH3

CH CH3

CH CH3

CH3

1 2 3 4 5 6 7 8 9 10 11 12

61. Intermediate formed during anti-Markovnikov addition

is CH CH CH Br3 2

which may undergo

following changes.

CH CH CH Br3 2

BrCH CH CH Br

3 2

Br

CH CH CH Br3 2 2

CH CH CH Br3 2

CH CH CH CH3 3

CH Br2

CH Br2

H

62. CH3

CCl3

(A) (B) (C)

CCl3

Cl

63.

O OO

CH3

O

H C3

O O

64. Structure of product formed will be:

C C

H

CH3

H

C C

H Z H

C C

H

CH3EHE

65. Sulphonation is a reversible process in which

generation of electrophile take place as:

2 4 3 3 42H SO H O SO HSO

��⇀↽��

On addition of KHSO4, concentration of

4HSO

increases so, backward reaction is favoured.

67.Cl

2

UV

H Cl

H

ClH

Cl

Cl

HCl

H

Cl HGammexane

(666)

69. Option (1) is least stable as like charge on adjacent

atoms.

72. Higher the number of -H of alkyl group directly

attach on benzene, more is reactivity towards

electrophilic substitution reaction.

73. CH3

COOH

(A) (B) (C)

74. CH CH CH3 2

CH CH CH3 2

HOCl

Markovnikov

addition

– +

OH Cl

75.

O /V O2 2 5

773 KC H

66

CH C

CH CO

O

OMaleic anhydride

76. Higher the extent of +M/+R more is stability of

carbocation.


6/8

77. CH CH C CH CH3 2 2 3

CH3

79. CH CH CH CH CH3 3

CH3

Number of -H = number of hyperconjugative

structures

81. Electrophilic substitution reaction of type F.C.

Alkylation on benzene.

H – F

–F EAS+ HF

82. In hydroboration oxidation, product obtained is similar

to that by anti-Markovnikov addition of water.

CH3

H C3

C CH2

1. B H2 6

2. H O /OH2 2

–

CH3

H C3

CH CH OH2

83. CH CH CH COOH3 2 2

CH CH COOCH3 2 3

CH CH CH3 3

COOH

CH COOCH CH3 2 3

HCOOCH CH CH2 2 3

HCOOCHCH3

CH3

85. Meq of excess acid = meq of NaOH 1

100 205

Total meq of acid = 1

2 150 605

Meq of H2SO

4 reacted = 60 – 20 = 40

= meq of NH3

Mole of NH3 × 1 × 103 = 40

Mole of NH3 = 40 × 10–3 = mole of N

Mass of N = 40 × 10–3 × 14 g = 0.56 g

% of N = 0.56

100 80%0.7

87. Tolerable limit for nitrates, fluorides, sulphates and

lead ions for drinking water are: 50 ppm, 1 ppm, 500

ppm and 50 ppb respectively.

88. Acidic strength stability of conjugate base.

After removal of acidic hydrogen various ions formed

are

Antiaromatic

Aromatic(most stable) Antiaromatic

CH CH2

89.

OHO

~99.9%

Aromatic

~0.1%

90.

O

OH1

2

3

4

6

5 6-Hydroxycyclohex-2-enone

BIOLOGY

102. CO2 is produced at 3 steps

(i) During acetyl Co-A formation

(ii) During formation of -ketoglutaric acid

(iii) During succinic acid formation

104. 38 ATP × 40 = 1520 ATP

112. Sugarcane and maize are C4 plants

114. Cyclic ETS – P700

, Hydrogen carrier, PSI

116. (i) Transmembrane channel

(ii) Facilitated diffusion

(iii) Outer

(iv) Thylakoid membrane

(v) Stroma

120. NitriteNitrite reductase

6H + 6e+ –

PSI

NH (Ammonia)3

121. A – Denitrification

B – Denitrifying bacteria – Pseudomonas,

Thiobacillus

125. Etiolation occurs when plants are grown in darkness.

128. Cell A w = –18 + 8 = –10

Cell B w = –14 + 2 = –12

Cell C w = –12 + 2 = –10

So flow of water is A = –10w

w = –12 B

w = –10 C

135. Translocation in phloem is explained by pressure

flow hypothesis.

Diffusion is not dependent on a ‘‘living system.’’

136. Both pitressin and vasopressin are the other names

of ADH (Anti diuretic hormone), which is released to

reabsorb more H2O from DCT of nephron.


7/8

138. Tapetum Lucidium cellulosum layer has refractive

crystals which reflects the light rays when

illuminated during night.

139. This is the case of near vision, so–

• Pressure tension / develops in ciliary muscles

i.e. they contract.

• Ligament – Relax.

• Lens become thick i.e. more convex so

convergence of diverging light rays increased.

140. Photoreceptors are hyperpolarised during light and

depolarised in dark human eye.

141. Tympanum is ear drum, separates outer & middle

ear.

Ear ossicles are present in middle ear.

Membranous labyrinth is internal ear.

142. When canal of Schlemm is blocked, aqueous humor

is not drained in blood so there occurs increase in

intraocular pressure and retina is damaged along

with optic nerve.

143. Hypermetropia can be due to shortening of eyeball

on decrease in convexity of the lens.

144. Parasympathetic nervous system constitutes cranio-

sacral outflow i.e. III, VII, IX, X are cranial autonomics

= 4 pairs

145. Medulla controls respiration cardiovascular reflexes

and gastric secretions.

146. Cerebellum is a part of hind brain

148. Cavities of cerebrum i.e. lateral ventricles or paracoel

open in cavity of diencephalon i.e. diocoel or III

ventricle through foramen of Monro.

149. From inside to outside, meninges are arranged

as–piamater–arachnoid–duramater.

152. Progesterone, a lipid soluble hormone can cross

lipid bilayer of target organ, so it does not require a

secondary messenger. It binds with intracellular

receptors.

153. Bronze like skin pigmentation is feature of Addison's

disease due to hyposecretion of adrenal cortex, rest

all are the cases of hypersecretion.

154. In blind girls and boys puberty is early as priming of

pineal gland hasn't been achieved.

156. Cortisol, a glucocorticoid increases blood glucose

level, so it will inhibit the cellular uptake of glucose

and synthesis of glycogen from glucose

(= glycogenesis)

157. Kidney does not secrete angiotensinogen

Angiotensinogen Angiotensin-I

Liver

Renin

Kidney

Secretes

158. Toxic goitre is due to nodules in thyroid gland

leading to hyperthyroidism.

159. 3P's is diabetes mellitus which express the

followings–

Polyuria – frequent urination – large volume of urine

along with glucose.

Polyphagia – frequent eating (hunger).

Polydipsia – frequent thirst.

160. Knee jerk reflex is monosynaptic/simple reflex.

Between sensory and motor neurons there are no

interneurons.

161. Neurohypophysis, i.e. the posterior pituitary is only

a storing and releasing centre of neurohormones

which are synthesized and secreted from

hypothalamus and reach the neurohypophysis via

neuronal axons.

162. cAMP acts as secondary messenger because

adrenaline cannot cross lipid bilayer of cell

membrane.

163. Collip's hormone / parathormone / PTH secretion

causes the rise in blood Ca+2 level, so due to this

there occurs loss of Ca+ minerals from bones and in

the hypersecretion, cavities are formed in bones

filled with fibres. TCT is hypocalcemic hormone.

165. % Filtration fraction (F.F) = GFR

RPF

Renal plasma flow (RPF) = GFR 125 125

FF 20% 0.2

= 625 ml/min

If we have to calculate renal blood flow

Renal plasma constitutes only 55% of blood volume

so renal blood flow is

RPF 625100 1136.36364 ml./min

55% 55

Nearly = 1137 ml/min

166. In diabetes mellitus, both glucose and ketone

bodies are passed out in urine.


8/8

� � �

167. Due to the presence of liver on right side, right

kidney is slightly lower than the left kidney.

168. As their loop of Henle is very short they are not used

in concentration of urine by counter-current

mechanism.

169. Fresh water fishes excrete hypotonic urine through

which there occurs loss of ions also. So to

compensate it and to maintain their osmolarity

higher, they actively uptake monovalent ions from

surrounding water.

170. Through filtration membrane all blood cells and

proteins cannot be filtered, so the glomerular filterate

is blood without formed elements and proteins. It is

isotonic to blood plasma.

171. Rheumatoid arthritis is diagnosed by the presence of

rheumatoid factor antibodies which resemble IgM. In

this disease, finally the joint becomes immovable.

172. Between sternum and ribs, there is present

amphiarthrose / cartilaginous joint.

173. Nodding movement of head occurs due to the

atlanto-occipital / condyloid joint between atlas and

occipital.

174. (a) Tarsals in 1 hind limb = 7

Bones in cranium = 8

(b) Cranial bones = 8

Carpals in each forelimb = 8

(c) Thoracic cage of human adult = 25

( 12 × 2 ribs + 1 sternum)

Each forelimb has 30 bones.

(d) Facial region has = 14 bones

Tarsals of both hindlimbs = 2 × 7 = 14 bones

175. Humerus is the part of appendicular endoskeleton.

176. In this disorder dystrophin protein is not formed. It

adds strength to the muscle by providing scaffolding

for the fibrils which is essential for muscle

contraction. It helps to reinforce the sarcolemma

and thus may help to transmit tension generated by

sarcomeres to the tendons.

177. Skeletal muscles at rest obtain most of their energy

from the aerobic respiration of fatty acids. During

exercise, muscle glycogen and blood glucose are

also used as energy sources.

178. ‘‘All or none law’’ is applicable to muscle fibres. If we

apply a threshold stimulus, muscle fibre will contract

with maximum force. If the stimulus is of sub-

threshold value the fibre will not contract at all.

179. The Ca2+ ions bind with troponin-C, this binding will

shift the tropomyosin-troponin complex, so that the

myosin binding sites of actin protein become

exposed to bind with actin binding sites of myosin.

180. Actin filaments slide over myosin filaments towards

H-zone but their length remains the same.

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