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Test - 3 (Code-C) (Answers) All India Aakash Test Series for Medical-2020

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1. (3)

2. (2)

3. (3)

4. (2)

5. (2)

6. (2)

7. (1)

8. (1)

9. (3)

10. (4)

11. (3)

12. (3)

13. (2)

14. (1)

15. (1)

16. (2)

17. (1)

18. (2)

19. (2)

20. (1)

21. (2)

22. (2)

23. (1)

24. (2)

25. (3)

26. (3)

27. (1)

28. (2)

29. (4)

30. (2)

31. (3)

32. (1)

33. (1)

34. (1)

35. (2)

36. (2)

Test Date : 02/12/2018

ANSWERS

TEST - 3 (Code-C)

All India Aakash Test Series for Medical-2020

37. (2)

38. (2)

39. (4)

40. (1)

41. (3)

42. (4)

43. (1)

44. (1)

45. (4)

46. (2)

47. (3)

48. (3)

49. (1)

50. (3)

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52. (1)

53. (4)

54. (4)

55. (3)

56. (2)

57. (3)

58. (2)

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61. (1)

62. (1)

63. (2)

64. (4)

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69. (4)

70. (3)

71. (4)

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74. (1)

75. (2)

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77. (4)

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79. (3)

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81. (4)

82. (2)

83. (4)

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86. (3)

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88. (4)

89. (2)

90. (3)

91. (1)

92. (4)

93. (4)

94. (3)

95. (2)

96. (3)

97. (1)

98. (3)

99. (3)

100. (4)

101. (2)

102. (3)

103. (4)

104. (2)

105. (1)

106. (3)

107. (2)

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109. (3)

110. (2)

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119. (3)

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124. (4)

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126. (3)

127. (4)

128. (3)

129. (4)

130. (3)

131. (1)

132. (3)

133. (4)

134. (4)

135. (4)

136. (3)

137. (3)

138. (1)

139. (2)

140. (4)

141. (3)

142. (4)

143. (1)

144. (4)

145. (3)

146. (2)

147. (2)

148. (2)

149. (3)

150. (4)

151. (3)

152. (1)

153. (3)

154. (4)

155. (3)

156. (1)

157. (1)

158. (3)

159. (1)

160. (4)

161. (2)

162. (3)

163. (1)

164. (2)

165. (3)

166. (2)

167. (3)

168. (2)

169. (1)

170. (4)

171. (3)

172. (4)

173. (3)

174. (2)

175. (1)

176. (4)

177. (3)

178. (3)

179. (2)

180. (4)

All India Aakash Test Series for Medical-2020 Test - 3 (Code-C) (Answers & Hints)

2/19

ANSWERS & HINTS

1. Answer (3)

Hint: Change in momentum normal to wall.

Sol.:

30°

30°

vcos30°

v sin 30°

m

v cos30°

v sin 30°

m

Px

= –mvcos30° – mvcos30°

= –2 mv cos30° = 3

–22

mv

|Px| = 3mv

|Py| = 0

P = 2 2 2

4( 3)

xP P mv

3OP mv

2. Answer (2)

Hint:cm

ext

dPF

dt

�

�

Sol.:cm

ext

�

� dPF

dt

ext0

� �

F

or�

cmP = constant

�

cmMV = constant

�

cmV = constant

3. Answer (3)

Hint: Use, constraint relation to find relation in

acceleration of blocks.

Sol.: From constraint relation.

a = 2a1

...(i)

m

mg

T

N

a

Now from F.B.D. of pulley

T1 = 2T ...(ii)

[ PHYSICS]

From F.B.D. of blocks.

T = ma ...(iii)

M

T1

Mg

a1

Mg – 2T = Ma1

...(iv)

Mg – 2ma = 2

aM

Mg = 2

Ma M TT

T1

Mg = 3

2

Ma

2

3

ga

4. Answer (2)

Hint: Fnet

= ma

Sol.: Let upward force be F.

F – mg = ma ...(i)

ma

F

mg

F = m (g + a) = 2(10 + 4)= 28 N = 2.8 kgf.

5. Answer (2)

Hint: Tmax

= m(g + amax

)

Sol.: Tmax

= 70 × 9.8 N

Now, Tmax

= 50(g + amax

)

70 × 9.8 = 50 × 9.8 + 50amax

amax

= – 2

20 9.83.92ms

50

6. Answer (2)

Hint: Use friction and pseudo force.

Sol.: N = ma ...(i)

maN

W

fs

Test - 3 (Code-C) (Answers & Hints) All India Aakash Test Series for Medical-2020

3/19

fsmax

= Nfsmax

= ma ...(ii)

for vertical equilibrium W fs max

mg ma.

ga

7. Answer (1)

Hint: opt

tanv Rg

Sol.: vopt

= 4

tan 120 103

Rg

vopt

= 40 10 4 = 40 m/s

8. Answer (1)

Hint: sinm

mg F f

Sol.: 1.7

cos 0.2 100 17 N2m

f mg .

Driving force along plane is

FDrive

= 50 – 10 = 40 N, downward

So, Friction will act upward.

Now by Newton’s 2nd Law.

40 – 17 = 10a

22.3 m/sa downward to plane.

9. Answer (3)

Hint: min2

1

mgF

Sol.: min2

1

mgF

min2

12 10 3

43

14

F

Fmin

= 72 N

10. Answer (4)

Hint: fm =

sN

Sol.: Limiting frictional force on block depends on

nature of surfaces in contact and normal reaction.

11. Answer (3)

Hint: F = ma

Sol.: ˆ ˆ ˆ(6 8 – 10 )NF i j k �

| | 36 64 100 10 2F �

N.

–2| | 5 msa �

m = ?

| | 10 2

| | 5

Fm

a

�

�

2 2 kgm

12. Answer (3)

Hint:

P = F dt Area under F -t curve.

Sol.:

p = Area under force - time curve

p = 20 + 5 – 5 – 5 – 2.5

12.5 kN s p

13. Answer (2)

Hint: 1 1 2 2 3 3

cm

1 2 3

mv m v m vv

m m m

� � �

�

Sol.: cm

0v �

1 1 2 2 3 3

1 2 3

0mv m v m v

m m m

� � �

1 1 2 2 3 30mv m v m v

� � �

3

ˆ ˆ ˆ3 –4 2 –v i j k �

3

ˆ ˆ ˆ–4 2 – m/s

3

� i j kv

14. Answer (1)

Hint: Use the concept of variable acceleration to find

the final velocity.

Sol.:

1 2 t (s)

1

0.5

For time (t = 0 to t = 1 s), retardation

a = g = 10 ms–2

Speed at t = 1 s

v1 = 20 – 10 × 1 = 10 m/s


4/19

For 1 t 2 s

av

= 1 0.5

0.752

Speed at t = 2 s

v2 = 10 – 0.75 × 1 = 2.5 m/s

15. Answer (1)

Hint: Fnet

= ma

Sol.:

37° 30°

5 kg 4 kg

40

T

30

N1

20

T

20 3

N2

fr

fr = N

2 = .1 × 20 × 1.7 = 3.4 N

Now, Fnet

= ma.

30 – T = 5a

T – 20 – 3.4 = 4a

6.6 = 9a

a = 6.6 2.2

9 3 =

–2

11ms

15

16. Answer (2)

Hint: Use Lami’s theorem for equilibrium.

= 180 – (30 + )

Sol.: From F.B.D of block

T = 40 g ...(i)

30°

T

T2

T1

A

From F.B.D. of pulley.

T = 2T = 80 g ...(ii)

From F.B.D. of point A

1 2

sin(120) sin sin(90 )

T TT

T

T T

T1 =

80 sin(120)

sin(180 – (30 ))

g

T

40 g

T1 =

40 3

sin( 30)

g

T

1 = minimum sin( + 30)

max 1

= 90 – 30 = 60° 60

17. Answer (1)

Hint: For conical pendulum.

Tcos = mg

Sol.: l = 1.3 m

r = 0.5 m

mg

T sin

T cos

l = 1

.30 m

T

50 cmm = 200 g

Now sin = 0.5 5

1.3 13

Hence, cos = 12

13.

Now, T cos = mg

T = 0.2 10 2 13

13 N12cos 12 6

13

mg

13N

6T

18. Answer (2)

Hint:

Calculate frictional force and apply Newton’s 2nd Law.

Sol.:

m 2

m 1

T

mg1

sin

m 2

sin

g

N 2

m 2

cos

g0.25

0.75

mg1

cos

N 1

a

37°

1rf

2rf

N1 = m

1gcos

1rf =

1m

1gcos =

3 44 10 24 N

4 5

22 2

1 4cos37 2 10 4 N

4 5rf m g


5/19

Now, m2gsin + m

1gsin –

1rf –

1rf = (m

1 + m

2)a

12 + 24 – 24 – 4 = 6a

a = 28 4

m/s6 3

Now, 12 – T – 1rf = 2 a.

12 – 4 – T = 4 8

2 8 –3 3

T

16N

3T

19. Answer (2)

Hint:

For equilibrium, 0dU

dr

Sol.:


dr

U = 3 2

2–

a b

r r

4 3

3 2– 0

dU a b

dr r r

3

3 1– 2 0

ab

r r

r 0 3

– 2 0a

br

3

2

ar

b

20. Answer (1)

Sol.: Linear momentum is always conserved

because Fext.

= 0.

Total energy also remains always conserved in all

cases. But kinetic energy will not be conserved

during collision.

21. Answer (2)

Hint: W = Pt when P is constant

Sol.: P = constant

P t =

22

2

v Ptm v

m

2 202

10

tv t

20 10

2

PF

v t t

22. Answer (2)

Hint: W = F d� �

for constant force.

Sol.: 2 2

ˆ ˆ– (5 – 3 ) m �

� �

d r r i j

ˆ ˆ(10 – 2 ) N�

F i j

ˆ ˆ ˆ ˆ(10 – 2 ) (5 – 3 ) 50 6 � �

W F d i j i j

56 JW

23. Answer (1)

Hint: Work – Energy theorem

Sol.:

K = 100 N/m

2 kg

8 m/s

Smooth

1 m

= 0.7

x

2 21 11

2 2mv kx mg ...(i)

21 12 64 100 0.7 2 10 1

2 2x

64 – 14 = 50x2

50 = 50x2

1mx

24. Answer (2)

Hint:

Work – Energy theorem

Sol.:

m

m

10 m

m = 4 kg

W = 500 J

h = 10 m

Now, W = mgh + Wwork done against friction.

500 = 400 + Wwork done against friction.

Wwork done against friction

= 500 – 400

work done against friction100 JW


6/19

25. Answer (3)

Hint: 0F S ��

Sol.: ˆ ˆ= sin( ) cos( )F at i at j�

ˆ ˆ= sin cos3 3

at atS i j

�

If andF S

��

are perpendicular then

0F S ��

sin( )sin cos( ) cos 03 3

at atat at

cos – 03

atat

–3

atat = /2

2

3 2

at

3

4t

a

26. Answer (3)

Hint: Area under the F–x curve.

Sol.: W = area under F – x curve.

W = 1

(200 100) (100 2) – 1002

= 300

200 –1002

W = 250 J

250 JW

27. Answer (1)

Hint: P F v �

�

Sol.:

1–5F y

1–5

a y

1–5

dvv k ydy

vdv =

1–5k y dy

2 4/5

42

5

v yk

v2 = 4/55

2

ky

v y2/5

P =

1–

2/55F v P y v �

�

1

5P y

28. Answer (2)

Hint: Momentum conservation.

Sol.:

m

2m

m + m2

4 m/s4 m/s

v

Since Fext

= 0

Hence, by conservation of linear momentum.

ˆ ˆ(4 ) 2 (4 ) 3m i m j m v �

4 8ˆ ˆ m/s

3 3

v i j

29. Answer (4)

Hint: Conservation of momentum and projectile

motion.

Sol.:

0.05 kg

400 m/s

3.95 kg

4 kg

v

4V = 400 × .05 + 3.95 × 0

V = 5 m/s.

As horizontal range, 2 h

R vg

2 205 10 m

10

30. Answer (2)

Hint: Work energy theorem.

Sol.: By work energy theorem

21Loss of mechanical energy

2mv mgh


7/19

212 20 20 10 17 Loss of mechanical

2 energy

400 – 340 = Loss of mechanical energy

Loss of mechanical energy 60 J

31. Answer (3)

Hint: P.E. energy of spring is = 21

2kx

1 2

1and

Pk k k k

L

Sol.:

Given21

(1)2k U ...(i)

k = 2k

kP = 2k + 2k = 4k

Now, U = 1

(4 ) (4) 162 k U

16U U

32. Answer (1)

Hint: cosA B

AB

� �

Sol.: Displacement vector

ˆ ˆ ˆ(3 4 5 )mS i j k �

Force ˆ ˆ ˆ(5 4 – 3 )NF i j k �

| | 5 2S �

m

| | 5 2F �

N

15 16 – 15 16F S ��

J

Hence,

16 16 8cos

25 2 25| || | 5 2 5 2

F S

F S

��

��

–1

8cos

25

33. Answer (1)

Hint: W = KE

Sol.: m = 2 kg, F = 4t, 2F

a tm

2 2

2

0 0 0

2

v

adt adt t dt t

At t = 2 s, v = 4 ms–1

2 21( – )

2f i

w m v v 12 (16 – 0) 16 J

2

34. Answer (1)

Hint: Elastic head-on collision between two equal

masses.

Sol.: If m1 = m

2 = m

3 = m

4 and v

4 = u.

Then, for 4th and 3rd ball collision.

v3 = u, v

4 = 0

for 3rd and 2nd ball collision.

v3 = 0 and v

2 = u

for 2nd and 1st ball collision.

v2 = 0 and v

1 = u

Hence 1st ball will move with speed u and other will

remain at rest.

35. Answer (2)

Hint: 2

0

n

nh e h

Sol.: We know that

hn = e2n h

0

h10

= e2 ×10 h0

20

10 0h e h

36. Answer (2)

Hint: Conservation of linear momentum

Px = 0 and P

y = 0.

Sol.:

m1

m2

m1

m2

u

v

3

u

Rest

Px = 0 m

1u – m

2v cos = 0

m1u = m

2vcos ...(i)

Py = 0, 0 =

1 2– sin

3

um m v

1

2sin

3

mum v ...(ii)

Dividing equation (i) by (ii)

tan = 1

3

–1

1tan

3


8/19

37. Answer (2)

Hint: 2P mKSol.: K

i = K

Kf = K

0 + 8K

0 = 9K

0

02

iP mK

0 02 2 9 3

f fP mK m K P

0 0

0

3 –100 100 200%

P PP

P P

38. Answer (2)

Hint: VLowest Point

= 2 gR for a tube.

Sol.: For a tube

ulowers point

= 4gR

Hence by M.E. conservation.

1(4 )

2m gR mgh

2R = h

2h R

39. Answer (4)

Hint: mgcos – N =

2mv

R

Sol.:

N

N

mg

mgcos

2mv

R 2mv

R

vgR

Now, UTop

= gR

Hence at top

mg = N +

2mv

R

mg = 0m gR

N NR

0N at Top i.e. 0

40. Answer (1)

Hint: Stopping distance × retarding force

= Loss of kinetic energy.

Sol.: As is same,

Given, K = (mg)x1 = (2mg)x

2

Then, x1 = 2x

2

41. Answer (3)

Hint: Actual power delivered

Efficiency = Rated power

Sol.:

PDelivered

=

4373 10 16

4 373 W40

mgh

t

PDelivered

= 2hp.

PDelivered

= engine

80

100P

Prated

= 10 20

2 2.5hp8 8

hp

Rated2.5 hpP

42. Answer (4)

Hint: xcm

= 1 1 2 2 1 1 2 2

cm

1 2 1 2

– –and

– –

m x m x m y m yy

m m m m

Sol.:

x

y

y

R (0, 0)

3 /5R

W1 = 90 N

W = 65 N

W2 = W

1 – W = 25 N

Now, ycm

= 1 1 2 2 1 1 2 2

1 2 1 2

– –

( – ) –

m gy m gy W y W y

m m g W W

=

390(0) – 25

15 –35–

65 65 13

R

R R

ycm

= 3

–13

R

Hence, cm cm

–3, 0,

13

Rx y

43. Answer (1)

Hint: Concept of motion of C.O.M.

Sol.: In explosion centre of mass will move in original

path.


9/19

Hence displacement = R

R =

2 2sin2 10sin(2 30 )

10

u

g

= 3

10sin60 102

R = 5 3 m

Hence Displacement = 5 3 m

44. Answer (1)

Hint:

1 1 2 2

cm

1 2

m r m rr

m m

� �

�

Sol.:

1

ˆ ˆ ˆ2r i j k , m1 = 2 kg

2

ˆ ˆ ˆ3 2 –r i j k , m2 = 5 kg

Hencecmr

�

=

ˆ ˆ ˆ ˆ ˆ ˆ2( 2 ) 5(3 2 – )

7

i j k i j k

=

ˆ ˆ ˆ17 14 – 3

7

i j k

cm

17 3ˆ ˆ ˆ2 – m

7 7

�

r i j k

45. Answer (4)

Hint: 1 1 2 2

cm

1 2

ma m aa

m m

� �

�

Sol.:

4 kg

4 kg

1a

2a

In this case

| |a

�

= 2 2

mg g

m

i.e.210

5 m/s2

a

Now,1

ˆ ˆ5 �

a ai i

2a

�

= ˆ ˆ– –5aj j

cm

ˆ ˆ ˆ ˆ4(5) 4(–5 ) 5 – 5

4 4 2

� i j i ja

cm 2

mˆ ˆ(2 5 – 2 5 )s

a i j

46. Answer (2)

Hint: PM

d =RT

Sol.: 4 32

d = = 3.9 g/L0.082 400

47. Answer (3)

Hint: 1

rM

and

Rate of effusion (r) =

Volume of gas effused (V)

Time taken (t)

Sol.: 1

rM

, so 1 1 2

2 2 1

V /t M=

V /t M

Since V1 = V

2, so

2 2

1 1

t M=

t M

1 2

1

2t M=

t 16 M

2 = 64 g/mol

[ CHEMISTRY]

48. Answer (3)

Hint: Higher the polarity of the molecule, higher is

the attractive force.

Sol.: NH3 is polar molecule hence it will be

associated by attractive force and will be liquefied

easily.

49. Answer (1)

Hint: Average speed (u) = 8RT

M

Sol.: 2

2

SOHe

SO He

Mu 64= = 4 :1

u M 4

50. Answer (3)

Hint: 2 2

1CO(g) O (g) CO (g) 283.5 kJ / mol

2

Sol.: Mole of CO2 formed =

55g 5= mole

44 g / mole 4

Heat released = 5

4 × (283.5) = 354.4 kJ


10/19

51. Answer (3)

Hint: For a reaction to be spontaneous

G = H – TS = –ve

Sol.: For G to be –ve, TS > H

H 7100T 284 K = 11°C

S 25

52. Answer (1)

Hint: Process is adiabatic so q = 0.

Sol.:

w = –1.5(4.2 – 1.2) L-atm = – 4.5 L-atm

= –4.5 × 101.3 J = –455.85 J

U = q + w

∵ q = 0, so U = w = –455.85 J

53. Answer (4)

Hint: S = q

T

Sol.: T = 127 + 273 = 400 K

S =

–1

4800 Jmol

400 K= 12 JK–1mol–1

54. Answer (4)

Hint: urms

= 3RT

M

Sol.: urms

1

M

Lower is the molar mass, higher is the root mean

square speed.

55. Answer (3)

Hint: (V – nb), Volume available for the movement of

gas molecules.

Sol.: ‘nb’ is the volume occupied by n mole of the

gas molecules.

56. Answer (2)

Hint: Avg. Kinetic energy (E) = 3

n RT2

Sol.: 2

N

14 3E = R 300

28 2

2O

24 3E = R 500

32 2

2

2

N

O

E 14 32 300 2= =

E 28 24 500 5

57. Answer (3)

Hint: K.E. per atom

Energy absorbed Bond energy–

per molecule per molecule

2

Sol.:

Bond energy/Molecule =

3

–19

23

180.6 10 J= 3 10 J

6.02 10

K.E. per atom =

–19 –19

3.5 10 – 3 10 J

2

= 2.5 × 10–20 J

58. Answer (2)

Hint:

Average speed 8RT

M

Most probable speed 2RT

M

rms speed 3RT

M

Sol.:

8RT 2RT 3RTAverage : mps : rms : :

M M Mspeed speed

= 8: 2 : 3

59. Answer (2)

Hint: H = B.E(reactant)

– B.E(product)

Sol.: 2 2

1 1H I HI

2 2

H = 2 2

(H ) (I ) (HI)

1 1B.E. B.E. – B.E.

2 2

or BE(HI)

= 2 2

(H ) (I )

1B.E. B.E. – H

2

= 1435 150 – 26.5

2 = 266 kJ/mole

60. Answer (2)

Hint: q and w are path functions.

Sol.: U + PV = H (State function)

S(Entropy) (State function)

H – TS = G(State function)


11/19

61. Answer (1)

Hint: H = U + ngRT

Sol.: For the reaction

H2(g) + I

2(g) 2HI(g)

ng = 2 – 2 = 0

H = U

62. Answer (1)

Hint: U = q + w;

Sol.: w = –PV = 0

q + w = U q = U = 750 J

63. Answer (2)

Hint: S = nRln2

1

V

V

Sol.:

S = 5 × 2 × 2.303 log2

1

V

V= 5 × 2 × 2.303 log

40

5

= 5 × 2 × 2.303 log8 = 5 × 2 × 2.303 × 3 log2

= 5 × 2 × 2.303 × 3 × 0.301 = 20.8 cal K–1

64. Answer (4)

Hint: r f product f reactantH H – H

Sol.: C3H

8(g) + 5O

2(g) 3CO

2(g) + 4H

2O(l)

r f product f reactantH H – H

rH = {3 (–394) 4(–286) – (–104)}

rH = {–1182 – 1144 + 104} = –2222 kJ/mole

Mole of C3H

8 =

66=1.5

44

rH for 1.5 mole = –2222 × 1.5 = –3333 kJ

65. Answer (2)

Hint: for a gaseous mixture, 2 2

2 2

CO CO

N N

p n

p n

Sol.: Mole of CO2 =

220= 5

44

Mole of N2 =

280=10

28

Now, 2 2

2 2

CO CO

N N

p n

p n

2CO

p 5

1.5 10

2

COp = 0.75 atm

66. Answer (2)

Hint: PV = nRT

or V T

p

Sol.:

T1 = 273 + 17 = 290 K

T2 = 273 + 27 = 300 K

V1

290

1.8

V2

300

1

or2

1

V 300 1.8= =1.86

V 290

67. Answer (4)

Hint: ZC =

C C

C

P V

RT

Sol.: PC

= C C2

a 8a, V = 3b, T

27b 27Rb

ZC =

2

a 3b 27Rb 3

27b R 8a 8

68. Answer (1)

Hint: Rate of effusion (r) P

M

Sol.: 2

O

1.5r

32 and He

4.5r

4

2O

He

r 1.5 2= =1: 6 2

r 4.532

69. Answer (4)

Hint: Naturally occurring most stable form of an

element has zero standard enthalpy of formation.

Sol.: In standard state chlorine exist as Cl2(g)

70. Answer (3)

Hint: dw = –PdV

Sol.: dw = –PdV

dw = – d(PV) (∵ Pressure is constant)

dw = –d(nRT) (For ideal gas)

w = –nRT

w = –1 × R × 1 = –R

Hence, work done by the gas is ‘R’


12/19

71. Answer (4)

Hint: G = Hvap

– TSvap

At equilibrium, G = 0

Sol.: G = H – TS

At equilibrium, G = 0, so H = TS

H 60240T 400 K

S 150.6

72. Answer (3)

Hint: Entropy decreases when disorder decreases.

Sol.: During condensation of vapour, vapour is

converted into liquid/solid. More disordered vapour is

converted into more ordered liquid/solid.

73. Answer (3)

Hint: Surface tension is force acting per unit length.

Sol.: Unit of surface tension is N m–1.

74. Answer (1)

Hint: Heat of neutralization for strong acid and strong

base is 13.7 kcal/eq.

Sol.: meq of H2SO

4 = 200 × 0.2 = 40

meq of NaOH = 200 × 0.15 = 30

Heat evolve = 13.7 × 30 × 10–3 × 103 cal

= 411 cal

75. Answer (2)

Hint: Higher the molecular attraction, higher is the

boiling point.

Sol.: HF molecules are associated with

intermolecular hydrogen bonding hence its boiling

point is highest.

76. Answer (4)

Hint: Boyle’s law, PV = constant

Sol.:

PV

V

77. Answer (4)

Hint: PV = nRT

Sol.: PHe

V = nHe

RT

42.5 V 0.082 375

4

V = 12.3 L

78. Answer (2)

Hint: For spontaneous reaction,

G = H – TS < 0

Sol.: at all temperature G < 0,

if H < 0 and S > 0

79. Answer (3)

Hint: For 1 mole of gas, van der waals equation is

2

aP (V – b) RT

V

and at high pressure 2

a0

V

�

Sol.: 2

aP (V – b) RT

V

for 2

a0

V

�

P(V – b) = RT

PV Pb– 1

RT RT

PV PbZ 1

RT RT

80. Answer (3)

Hint: mps

2RTu

M and for ideal gas PV = nRT

Sol.: PM RT P

dRT M d

umps

= 2RT 2P

M d

umps

1

d

81. Answer (4)

Hint: At any temperature, different particles in the

gas have different speeds.

Sol.: Due to different speeds, the gas particles have

different kinetic energy at same temperature.

82. Answer (2)

Hint: H = U + ng RT

Sol.:

H – U = ngRT = (6 – 9) × 8.314 × 300 J

= –7.48 kJ


13/19

83. Answer (4)

Hint: H = U + (PV)

Sol.:

H = U + (P2V

2 – P

1V

1)

P1 = 3 atm, V

1 = 4 L

P2 = 5 atm, V

2 = 6 L

H = 45 + (6 × 5 – 3 × 4) = 63 L-atm

84. Answer (2)

Hint: Standard enthalpy of formation is defined for

formation of one mole of a product from its

constituent elements in their most stable states of

aggregation.

Sol.: For reaction H2(g) + Br

2(l) 2HBr(g), enthalpy

change is not standard enthalpy of formation. Here

two moles instead of one mole of the product is

formed.

85. Answer (2)

Hint: wrev

= –2.303 nRTlogf

i

V

V

H = nCpT

Sol.:

wrev

= –2.303 × 5 × 8.314 × 300 log4

wrev

= –2.303 × 5 × 8.314 × 300 × 2 log2

= –2.303 × 5 × 8.314 × 300 × 2 × 0.301

= –17289.87 J = –17.3 kJ

H = nCpT

H = 0 ∵ T = 0

86. Answer (3)

Hint: w = – PexV

Sol.:

w = – PexV = –1.0 atm (150 × 15) cm3

= –1.0 atm × 2250 cm3

= –1.0 × 2.25 L-atm

= –2.25 L-atm

87. Answer (2)

Hint: Q = mLfus

+ msT

Sol.:

Q = 75 × 1000 × 80 + 75 × 1000 × 1 × 10 cal

= 6750000 cal

= 6.75 × 103 kcal

88. Answer (4)

Hint: Intensive property does not depend on the

quantity of matter.

Sol.: Pressure = Force

Area

is an intensive property.

89. Answer (2)

Hint: Hydration energy of F– is high.

Sol.: During neutralization of HF with NaOH, extra

energy is released due to extensive hydration of

F–

.

90. Answer (3)

Hint: CCl4 is a non-polar molecule.

Sol.: In non-polar molecules (CCl4), only London

Forces are present.

[ BIOLOGY]

91. Answer (1)

Hint: This plant is used as fodder.

Sol.: Sesbania is used as fodder and it belongs to

family Fabaceae.

92. Answer (4)

Hint: This condition is seen in potato family.

Sol.: In the floral formula of family Solanaceae,

C A(5) 5 represents epipetalous condition.

93. Answer (4)

Hint: Lateral meristem is responsible for secondary

growth.

Sol.: Lateral meristem includes intrafascicular

cambium (primary), interfascicular cambium and cork

cambium (secondary).

94. Answer (3)

Hint: Mesophyll tissue is not differentiated into

palisade and spongy parenchyma in monocot leaves.

Sol.: Monocot leaves are isobilateral. These have

stomata on both surfaces and below the stoma of

abaxial epidermis sub-stomatal cavities are present.

95. Answer (2)

Hint: In racemose inflorescence, younger flowers are

present towards the apex whereas in cymose

inflorescence, younger flowers are present towards

base.

Sol.:

Cymose inflorescence – Basipetal succession of

flowers


14/19

Racemose inflorescence – Acropetal succession of

flowers

Radial symmetry – Actinomorphic flower

Symmetry by one – Zygomorphic flower

plane only

96. Answer (3)

Hint: Rhizome is a modified underground stem which

grows horizontal to the soil surface.

Sol.: Potato is a tuber i.e. modified underground

stem branch which get swollen on account of

accumulation of food.

97. Answer (1)

Hint: Pulvinate leaf is found in some leguminous

plants.

Sol.: Leaf having swollen base is called pulvinate

leaf.

98. Answer (3)

Hint: This aestivation is found in corolla of family

Fabaceae.

Sol.: In family Fabaceae corolla is papilionaceous,

in which smallest anterior petals are referred as keel.

99. Answer (3)

Hint: These roots are adventitious roots.

Sol.: Stilt roots arise from the lower nodes of stem

to support the main axis.

100. Answer (4)

Hint: The endodermal cells of dicot stem store some

carbohydrate grains.

Sol.: These cells store starch grains and hence

endodermis is known as starch sheath.

101. Answer (2)

Hint: Hypodermis is absent in roots.

Sol.:

- All tissues outside the vascular cambium

constitute bark

- In leaves, ground tissue is not well differentiated

& is known as mesophyll

- Casparian strips are formed by deposition of a

waxy material suberin on the radial and

tangential walls of endodermal cells.

102. Answer (3)

Hint: Coconut is a drupe type of fruit.

Sol.: Drupe type of fruits are mostly one seeded,

develop from monocarpellary superior ovary and have

hard, stony endocarp.

103. Answer (4)

Hint: In tetradynamous condition there are

6 stamens4 larger in ring2 smaller in ring

inner outer

Sol.: Tetradynamous condition is a feature of family

Brassicaceae. In pea, stamens exhibit diadelphous

condition.

104. Answer (2)

Hint: In litchi, seed is covered by an outgrowth of

funicle.

Sol.: This white, translucent, fleshy edible covering

of seed is called aril.

105. Answer (1)

Hint: A sterile stamen is called staminode.

Sol.:

- Obliquely placed ovary and swollen placenta are

the features of family Solanaceae.

- Versatile fixation of anthers is feature of family

Poaceae

- Female flowers are known as pistillate flowers.

106. Answer (3)

Hint: This is a monocot family.

Sol.: In family Liliaceae, gynoecium is tricarpellary,

syncarpous, superior, trilocular and with axile

placentation. It is commonly called lily family.

107. Answer (2)

Hint: Collenchyma is an elastic, living mechanical

tissue.

Sol.: Collenchyma cells have thickening of cellulose,

hemicellulose and pectin at their corners.

108. Answer (1)

Hint: In exarch type of primary xylem, protoxylem

lies towards the periphery and metaxylem lies

towards the centre.

Sol.: Exarch arrangement of primary xylem is a

feature of roots.

109. Answer (3)

Hint: Companion cell controls the function of sieve

tube.

Sol.: Companion cells retain nucleus throughout

their life and control functions of anucleated sieve

tubes.

110. Answer (2)

Hint: Coleoptile and coleorhiza are sheaths that

enclose plumule and radicle respectively.

Sol.: Proteinaceous layer of endosperm that

separates embryo is called aleurone layer.

111. Answer (1)

Sol.: Rice, coconut and wheat are endospermic

seeds but seeds of orchids are non-endospermic.


15/19

112. Answer (2)

Hint: In Fabaceae, stamens show diadelphous

condition.

Sol.: Floral formula of family Fabaceae is

K (5)C A1 + 2 + (2) (9) + 1 1G%

113. Answer (4)

Hint: Guard cells are generally bean-shaped except

grasses.

Sol.: In grasses, the guard cells are dumb-bell

shaped.

114. Answer (3)

Hint: Root hairs are epidermal appendages.

Sol.: Root hairs are unicellular elongations of the

epidermal cells i.e. they are exogenous in origin.

115. Answer (2)

Hint: Relative positions of primary xylem

(Protoxylem & metaxylem) decides the endarch or

exarch conditions.

Sol.: In endarch arrangement, protoxylem occurs

towards centre and metaxylem towards periphery.

116. Answer (2)

Hint: In hypogynous flower, ovary is superior.

Sol.:

nG – Superior ovary

nG – Inferior ovary (Epigynous flower)

117. Answer (4)

Hint: Replum is thin membranous structure, which is

formed in siliqua type of fruit.

Sol.: Replum is a pseudoseptum and it is a feature

of family Brassicaceae.

118. Answer (2)

Sol.: Some pepo fruits are bitter in taste due to

tetracyclic triterpenes.

119. Answer (3)

Hint: In cymose inflorescence, the main axis

terminates into a flower.

Sol.: In Tri t icum (Wheat) inf lorescence is

spikelet type (racemose).

120. Answer (2)

Hint: Cassia flower can be divided into two equal

halves only by one plane.

Sol.: It is known as bilateral symmetry (Zygomorphic

flower)

121. Answer (4)

Hint: With the passage of time, rings of the sap

wood are changed into heart wood.

Sol.: As a result of secondary growth amount of

heart wood increases and amount of sap wood

almost remains constant.

122. Answer (2)

Hint: Lenticels occur in most woody trees and

permit the exchange of gasses and water.

Sol.:

- Annual rings are distinct in plants growing in

temperate regions because climatic conditions

are not uniform over there.

- Stelar secondary growth is performed by vascular

cambium as a result central cylinder of wood is

formed which remains surrounded by secondary

phloem.

- Pericycle cells opposite to protoxylem, become

meristematic and give rise to vascular cambium

in dicot roots.

123. Answer (3)

Hint: Heartwood is called duramen and sapwood is

called alburnum.

Sol.: Heartwood is hard, durable and resistant to

attack of microorganisms.

124. Answer (4)

Hint: Axillary buds modify into tendrils in cucurbits.

Sol.: In cucurbits, stem modifies to provide support

to the plant.

125. Answer (2)

Hint: Tap roots directly arise from radicle.

Sol.: Adventitious roots arise from plant parts other

than radicle.

126. Answer (3)

Hint: Flower is a modified shoot which may have

floral appendages in multiple of three, four or five.

Sol.: Trimerous flowers are found in Liliaceae family

in which floral appendages are in multiples of three.

127. Answer (4)


Sol.: In soyabean (Fabaceae), brinjal, chilli

(Solanaceae) and tulip (Liliaceae) ovary is superior.

128. Answer (3)

Hint: Pericycle is absent in monocot stem.

Sol.: Pericycle may be single layered or

multilayered.

129. Answer (4)

Hint: Lateral roots develope from pericycle.

Sol.: Lateral roots are present in both dicot &

monocot roots.


16/19

130. Answer (3)

Hint: In monocot stem, vascular bundles are

scattered in whole ground tissue.

Sol.: In monocot stem, endodermis is absent but it

is well developed in roots.

131. Answer (1)

Hint: In caryopsis fruits, pericarp is completely fused

with seed coat.

Sol.: Caryopsis fruits are found in members of

Poaceae like wheat (Triticum aestivum), maize, rice

etc.

132. Answer (3)

Hint: In some seeds nucellus is not fully consumed

during development of embryo.

Sol.: Such persistent nucellus is called perisperm

and such seeds are known as perispermic seeds.

133. Answer (4)

Hint: In a compound leaf, incision reaches upto

midrib.

Sol.: Lamina is broken into leaflets and midrib forms

a common axis called rachis, but bud is present in

the axil of petiole of compound leaves.

134. Answer (4)

Hint: In dicot stem, cork cambium develops from cell

of cortex region.

Sol.: Cork cambium is also called phellogen,its

origin is outside of the stele i.e. extrastelar.

In dicot root, it develops from pericycle cells.

135. Answer (4)


growth.

Sol.: In monocots, secondary growth is absent.

Thus lateral meristem is also absent.

136. Answer (3)

Hint : Factors which are used up during coagulation

of blood.

Sol. : Plasma lacks formed elements and plasma

without clotting factors is known as serum.

137. Answer (3)

Hint : Vertebrates possess ventral heart.

Sol. : Blood groups are determined by presence of

surface antigen on RBCs. Conversion of fibrinogen

into fibrin is done by thrombin.

Chordates like urochordates have an open

circulatory system.

138. Answer (1)

Hint : Removal of this compound requires large

amount of water.

Sol. : Ammonia is the most toxic nitrogenous waste

and uric acid is the least toxic nitrogenous waste

which is excreted in form of pellets.

139. Answer (2)

Hint : Antibodies protect us from pathogens.

Sol. : Antibodies are immunoglobulins which are

produced by B-lymphocytes (plasma cells) and

protect body from pathogens. Albumin is mainly

concerned with maintaining osmotic balance of

blood. Heparin is an anticoagulant while fibrinogen is

one of the clotting factors.

140. Answer (4)

Hint : Obligatory reabsorption of water occurs in this

part of nephron.

Sol. : Obligatory reabsorption of water around

60-70% occurs in PCT and some occurs in loop of

Henle. Facultative reabsorption of about

10 - 15% of water occurs in DCT and CD under the

influence of ADH.

141. Answer (3)

Hint : Chordae tendinae are attached to papillary

muscles in ventricular wall & with flaps of AV valves.

Sol. : Chordae tendinae are attached to AV valves on

one end and papillary muscles on other end. These

cord like structures prevent the collapse of AV valves

during powerful ventricular contraction. They are not

attached to semilunar valves.

142. Answer (4)

Hint : Haemoglobin helps in transport of gases.

Sol. : Haemoglobin is an iron containing pigment

present inside RBCs. Basophils secrete histamine,

serotonin and heparin. Neutrophils have multilobed

nucleus.

143. Answer (1)

Hint : Cortex present between medullary pyramids

was named after Joseph Bertin.

Sol. : At a number of places renal cortex invaginates

within medulla and divides medulla into a number of

pyramidal structures known as medullary pyramids.

Invaginations of cortex into medulla are known as

columns of Bertini.

144. Answer (4)

Hint : Human heart is myogenic.

Sol. : SA node (pacemaker) is a modified cardiac

muscular tissue capable of generating impulses.

RBCs lack mitochondria to perform aerobic

breakdown of glucose. Single circulation of blood

occurs in heart of fishes.


17/19

145. Answer (3)

Hint : Antibodies bind to antigen present on RBCs

surface.

Sol. : Person having O-blood group lacks antigens

A & B on the RBC surface but have anti-A and anti-

B antibodies in their plasma.

146 Answer (2)

Hint : Osmolarity depends upon number of solute

particles/volume.

Sol. : Because PCT is permeable for both water and

electrolytes as well as other substances so

osmolarity in PCT remains unchanged and filtrate

remains isotonic in comparison to blood plasma.

147. Answer (2)

Hint : Various plasma proteins promote blood

coagulation.

Sol. : Heparin is an anticoagulant produced by

basophils and mast cells while thrombin, Ca2+ and

thrombokinase are involved in blood clotting.

148. Answer (2)

Hint : Left ventricle pumps blood into aorta from

where it is carried to different parts of the body.

Sol. : Different nodal tissues of heart are capable of

generating actions potential at different rates.

Lymphocytes are found in lymph. Volume of blood

pumped by each ventricle in one cardiac cycle is

stroke volume.

149. Answer (3)

Hint : It is a tuft of blood capillaries present in

Bowman’s capsule.

Sol. : Renal tubule involves components of tubular

parts of nephrons such as Bowman’s capsule, PCT,

DCT and loop of Henle.

150. Answer (4)

Hint : Rh–ve individuals will generate antibodies

against Rh+ve antigen in their plasma.

Sol. : Erythroblastosis foetalis occurs due to Rh

incompatibility between Rh –ve mother and Rh +ve

foetus. Antibodies against Rh antigen cross the

placenta and start destroying foetal RBCs. In order

to prevent this, anti-Rh antibodies are given to

mother.

151. Answer (3)

Hint : Maximum reabsorption of water & solutes

occurs in PCT.

Sol. : Reabsorption of maximum solutes including

glucose occurs in PCT. Reabsorption of water

occurs in PCT, descending limb of loop of Henle,

DCT and collecting duct.

152. Answer (1)

Hint : Members of superclass Pisces have venous

heart.

Sol. : Only deoxygenated blood is pumped by fish’s

heart; which means blood is circulated only once

through their heart i.e. single circulation. Most

reptiles show incomplete double circulation.

153. Answer (3)

Hint : Net filtration pressure (NFP)/Glomerular

filtration pressure (GFP) is the pressure being

exerted by glomerular blood for ultrafiltration.

Sol. : NFP GHP (BCOP CHP)Glomerular Blood CapsularHydrostatic Colloidal HydrostaticPressure Osmotic Pressure

Pressure

154. Answer (4)

Hint : Oxygen requirement of body increases during

strenuous exercise.

Sol. : During strenuous exercise, under sympathetic

stimulation heart rate increases due to increase in

number of action potentials generated by

pacemaker. Also, there is an increase in force of

contraction of ventricles leading to an increase in

stroke volume. However, the sequence of various

events occurring in a cardiac cycle i.e. atrial systole,

ventricular systole and joint diastole remain same.

155. Answer (3)

Hint : T-wave represents repolarisation of ventricles.

Sol. : T-wave represents end of ventricular systole

which is followed by joint-diastole. Electrocardiogram

represents electrical activity of heart and it is

recorded by electrocardiograph machine.

156. Answer (1)

Hint : Systemic circulation begins with left ventricle

and ends in right atrium.

Sol. : Aorta carries oxygenated blood from left

ventricle towards body tissues while vena cava

returns deoxygenated blood to right atrium.

157. Answer (1)

Hint : Damage to kidney causes edema.

Sol. : Damage to kidney disrupts its functions

leading to accumulation of urea in blood which is

known as uremia. Formation of large volume of urine

is known as polyuria. Cystinuria refers to increased

presence of cysteine in urine while in hematuria,

blood is found in urine.


18/19

158. Answer (3)

Hint : Sudden damage to heart muscles causes

myocardial infarction

Sol. : Heart attack is also known as myocardial

infarction. Damage to nodal tissue causes stoppage

of heart-beat which is cardiac arrest.

159. Answer (1)

Hint : Sympathetic nervous system regulates

various activities of body in emergency situation.

Sol. : Parasympathetic stimulation decreases heart

rate, stroke volume and cardiac output. P-wave in

ECG represents atrial depolarisation.

160. Answer (4)

Hint : Animals which live in water deficient conditions

are uricotelic.

Sol. : Uric acid is excreted by using minimum

amount of water e.g., Reptiles, Birds, Insects, Land

snails etc.

161. Answer (2)

Hint : These blood vessels carry deoxygenated

blood.

Sol. : Blood must pass through pulmonary

circulation in order to flow from right atrium to left

atrium.

162. Answer (3)

Hint : Micturition reflex is initiated by stretching of

wall of urinary bladder.

Sol. : Stretch receptors in the wall of urinary bladder

send signals to CNS, which passes on motor

messages to initiate the contraction of smooth

muscles of bladder and simultaneous relaxation of

urethral sphincter causing the release of urine.

163. Answer (1)

Hint : This is associated with skeletal structures of

body.

Sol.: Liver and spleen are involved in erythropoiesis

in embryonic life.

164. Answer (2)

Hint : Sympathetic neuronal endings release

adrenaline.

Sol. : Adrenaline causes an increase in heart rate

resulting in decrease in the duration of individual

cardiac cycle. However, adrenaline causes an

increase in cardiac output and stroke volume.

165. Answer (3)

Hint : Heart beats about 72 times in a minute.

Sol. : In an cardiac cycle, both systemic and

pulmonary circulation take place. So about 72

cardiac cycles take place in a minute, hence about

72 times double circulations are normally completed

in one minute.

166. Answer (2)

Hint : Counter current mechanism helps in producing

concentrated urine.

Sol. : If there is no loop of Henle, there will be no

counter current mechanism and medullary

concentration gradient will not be maintained. As a

result dilute urine will be formed.

167. Answer (3)

Hint : First heart sound is ‘Lubb’ and second heart

sound is ‘dub’.

Sol. : First heart sound is produced by closure of

A-V valves at the beginning of ventricular systole,

while second heart second ‘dub’ is produced by

closure of semilunar valves at the beginning of

ventricular diastole.

168. Answer (2)

Hint : In normal ECG, T-waves are positive wave

produced during ventricular repolarisation.

Sol. : End of T-wave represents end of ventricular

systole and initiation of joint diastole or diastasis.

169. Answer (1)

Hint : Urea cycle occurs is liver.

Sol. : Formation of urea occurs in liver and hepatic

vein carrying blood from liver has maximum quantity

of urea while renal vein carrying blood from kidney

after filtration has least amount of urea.

170. Answer (4)

Hint : Malpighian body of Juxtamedullary nephrons

are located close to medulla of kidney.

Sol. : Cortical nephrons are more abundant as

compared to juxtamedullary nephrons.

Juxtamedullary nephrons have longer loop of Henle

and well developed vasa recta as compared to

cortical nephron. Peritubular capillaries are present

around tubules of nephrons in both Cortical and

Juxtamedullary nephron.

171. Answer (3)

Hint : Bundle of His is a part of nodal tissue of heart.

Sol. : Nodal tissue of heart are modified muscle

fibres which exhibit autoexcitability. Nodal tissue

include SA node, AV node, Bundle of His and

Purkinje fibres.

172. Answer (4)

Hint : Lymph flows through separate lymphatic

capillaries and lymphatic vessels.

Sol. : Lymphatic vessels have valves to prevent

backflow of lymph. Left and right subclavian veins

drain lymph into superior vena cava. Thoracic duct

collects lymph from most of part of gastrointestinal

tract.


19/19

��

173. Answer (3)

Hint : Portal system involving kidneys.

Sol. : Hepatic & hypophyseal portal system are well

developed in humans.

174. Answer (2)

Hint : Renin is a part of RAAS pathway.

Sol. : In response to fall in blood pressure/GFR/

glomerular blood flow, renin is released by JGA to

initiate RAAS pathway which will eventually restore

the normal level.

175. Answer (1)

Hint : Involuntary muscles are present in tunica

media.

Sol. : Smooth muscles are present in tunica media

in both type of blood vessels.

176. Answer (4)

Hint : Both atria contract simultaneously during atrial

systole.

Sol. : All four chambers of heart do not contract

simultaneously. Increase in ventricular pressure

causes closure of AV valves. Major filling i.e., (2/3rd)

of ventricles occurs during joint diastole.

177. Answer (3)

Hint : Blood group mismatch leads to clumping of

RBC.

Sol. : O–

blood group is universal donor as it lacks

A, B and D antigens on the surface of RBC, but it

has both anti-A, anti-B and anti-D antibodies in

plasma. That’s why a person having O blood group

can accept blood from another person having O blood

group only.

178. Answer (3)

Hint : Pacemaker maintains rhythmic contractility of

heart.

Sol. : Artificial pacemaker is required when there is

a damage to nodal tissue of heart like in case of

heart attack. This device sends out small electrical

current to stimulate heart to contract thus

maintaining rhythm of cardiac activity.

179. Answer (2)

Hint : SA node generates cardiac impulses.

Sol. : From SA node impulse travels to AV node.

From AV node, Bundle of His carries impulse to

bundle branches and Purkinje fibres spread impulse

to entire ventricular musculature.

180. Answer (4)

Hint : JGA secretes renin in response to fall in GFR.

Sol. : Angiotensinogen is produced by liver and

released into blood stream.

Angiotensinogen Angiotensin-I

ACE

Renin

Angiotensin-II

(Angiotensinconverting enzyme)

Vasoconstriction Increase in heart rate

Increase in blood pressure

Adrenal cortex

Aldosterone(Mineralocorticoid)

ACE is produced by lung capillaries.

Test - 3 (Code-D) (Answers) All India Aakash Test Series for Medical-2020

1/19

1. (4)

2. (1)

3. (1)

4. (4)

5. (3)

6. (1)

7. (4)

8. (2)

9. (2)

10. (2)

11. (2)

12. (1)

13. (1)

14. (1)

15. (3)

16. (2)

17. (4)

18. (2)

19. (1)

20. (3)

21. (3)

22. (2)

23. (1)

24. (2)

25. (2)

26. (1)

27. (2)

28. (2)

29. (1)

30. (2)

31. (1)

32. (1)

33. (2)

34. (3)

35. (3)

36. (4)

Test Date : 02/12/2018

ANSWERS

TEST - 3 (Code-D)

All India Aakash Test Series for Medical-2020

37. (3)

38. (1)

39. (1)

40. (2)

41. (2)

42. (2)

43. (3)

44. (2)

45. (3)

46. (3)

47. (2)

48. (4)

49. (2)

50. (3)

51. (2)

52. (2)

53. (4)

54. (2)

55. (4)

56. (3)

57. (3)

58. (2)

59. (4)

60. (4)

61. (2)

62. (1)

63. (3)

64. (3)

65. (4)

66. (3)

67. (4)

68. (1)

69. (4)

70. (2)

71. (2)

72. (4)

73. (2)

74. (1)

75. (1)

76. (2)

77. (2)

78. (2)

79. (3)

80. (2)

81. (3)

82. (4)

83. (4)

84. (1)

85. (3)

86. (3)

87. (1)

88. (3)

89. (3)

90. (2)

91. (4)

92. (4)

93. (4)

94. (3)

95. (1)

96. (3)

97. (4)

98. (3)

99. (4)

100. (3)

101. (2)

102. (4)

103. (3)

104. (2)

105. (4)

106. (2)

107. (3)

108. (2)

109. (4)

110. (2)

111. (2)

112. (3)

113. (4)

114. (2)

115. (1)

116. (2)

117. (3)

118. (1)

119. (2)

120. (3)

121. (1)

122. (2)

123. (4)

124. (3)

125. (2)

126. (4)

127. (3)

128. (3)

129. (1)

130. (3)

131. (2)

132. (3)

133. (4)

134. (4)

135. (1)

136. (4)

137. (2)

138. (3)

139. (3)

140. (4)

141. (1)

142. (2)

143. (3)

144. (4)

145. (3)

146. (4)

147. (1)

148. (2)

149. (3)

150. (2)

151. (3)

152. (2)

153. (1)

154. (3)

155. (2)

156. (4)

157. (1)

158. (3)

159. (1)

160. (1)

161. (3)

162. (4)

163. (3)

164. (1)

165. (3)

166. (4)

167. (3)

168. (2)

169. (2)

170. (2)

171. (3)

172. (4)

173. (1)

174. (4)

175. (3)

176. (4)

177. (2)

178. (1)

179. (3)

180. (3)

All India Aakash Test Series for Medical-2020 Test - 3 (Code-D) (Answers & Hints)

2/19

ANSWERS & HINTS

1. Answer (4)

Hint: 1 1 2 2

cm

1 2

ma m aa

m m

� �

�

Sol.:

4 kg

4 kg

1a

2a

In this case

| |a

�

= 2 2

mg g

m

i.e.210

5 m/s2

a

Now,1

ˆ ˆ5 �

a ai i

2a

�

= ˆ ˆ– –5aj j

cm

ˆ ˆ ˆ ˆ4(5) 4(–5 ) 5 – 5

4 4 2

� i j i ja

cm 2

mˆ ˆ(2 5 – 2 5 )s

a i j

2. Answer (1)

Hint:

1 1 2 2

cm

1 2

m r m rr

m m

� �

�

Sol.:

1

ˆ ˆ ˆ2r i j k , m1 = 2 kg

2

ˆ ˆ ˆ3 2 –r i j k , m2 = 5 kg

Hencecmr

�

=

ˆ ˆ ˆ ˆ ˆ ˆ2( 2 ) 5(3 2 – )

7

i j k i j k

=

ˆ ˆ ˆ17 14 – 3

7

i j k

cm

17 3ˆ ˆ ˆ2 – m

7 7

�

r i j k

[ PHYSICS]

3. Answer (1)

Hint: Concept of motion of C.O.M.

Sol.: In explosion centre of mass will move in original

path.

Hence displacement = R

R =

2 2sin2 10sin(2 30 )

10

u

g

= 3

10sin60 102

R = 5 3 m

Hence Displacement = 5 3 m

4. Answer (4)

Hint: xcm

= 1 1 2 2 1 1 2 2

cm

1 2 1 2

– –and

– –

m x m x m y m yy

m m m m

Sol.:

x

y

y

R (0, 0)

3 /5R

W1 = 90 N

W = 65 N

W2 = W

1 – W = 25 N

Now, ycm

= 1 1 2 2 1 1 2 2

1 2 1 2

– –

( – ) –

m gy m gy W y W y

m m g W W

=

390(0) – 25

15 –35–

65 65 13

R

R R

ycm

= 3

–13

R

Hence, cm cm

–3, 0,

13

Rx y

Test - 3 (Code-D) (Answers & Hints) All India Aakash Test Series for Medical-2020

3/19

5. Answer (3)

Hint: Actual power delivered

Efficiency = Rated power

Sol.:

PDelivered

=

4373 10 16

4 373 W40

mgh

t

PDelivered

= 2hp.

PDelivered

= engine

80

100P

Prated

= 10 20

2 2.5hp8 8

hp

Rated2.5 hpP

6. Answer (1)

Hint: Stopping distance × retarding force

= Loss of kinetic energy.

Sol.: As is same,

Given, K = (mg)x1 = (2mg)x

2

Then, x1 = 2x

2

7. Answer (4)

Hint: mgcos – N =

2mv

R

Sol.:

N

N

mg

mgcos

2mv

R 2mv

R

vgR

Now, UTop

= gR

Hence at top

mg = N +

2mv

R

mg = 0m gR

N NR

0N at Top i.e. 0

8. Answer (2)

Hint: VLowest Point

= 2 gR for a tube.

Sol.: For a tube

ulowers point

= 4gR

Hence by M.E. conservation.

1(4 )

2m gR mgh

2R = h

2h R

9. Answer (2)

Hint: 2P mK

Sol.: Ki = K

Kf = K

0 + 8K

0 = 9K

0

02

iP mK

0 02 2 9 3

f fP mK m K P

0 0

0

3 –100 100 200%

P PP

P P

10. Answer (2)

Hint: Conservation of linear momentum

Px = 0 and P

y = 0.

Sol.:

m1

m2

m1

m2

u

v

3

u

Rest

Px = 0 m

1u – m

2v cos = 0

m1u = m

2vcos ...(i)

Py = 0, 0 =

1 2– sin

3

um m v

1

2sin

3

mum v ...(ii)

Dividing equation (i) by (ii)

tan = 1

3

–1

1tan

3

11. Answer (2)

Hint: 2

0

n

nh e h

Sol.: We know that

hn = e2n h

0

h10

= e2 ×10 h0

20

10 0h e h


4/19

12. Answer (1)

Hint: Elastic head-on collision between two equal

masses.

Sol.: If m1 = m

2 = m

3 = m

4 and v

4 = u.

Then, for 4th and 3rd ball collision.

v3 = u, v

4 = 0

for 3rd and 2nd ball collision.

v3 = 0 and v

2 = u

for 2nd and 1st ball collision.

v2 = 0 and v

1 = u

Hence 1st ball will move with speed u and other will

remain at rest.

13. Answer (1)

Hint: W = KE

Sol.: m = 2 kg, F = 4t, 2F

a tm

2 2

2

0 0 0

2

v

adt adt t dt t At t = 2 s, v = 4 ms–1

2 21( – )

2f i

w m v v 12 (16 – 0) 16 J

2

14. Answer (1)

Hint: cosA B

AB

� �

Sol.: Displacement vector

ˆ ˆ ˆ(3 4 5 )mS i j k �

Force ˆ ˆ ˆ(5 4 – 3 )NF i j k �

| | 5 2S �

m

| | 5 2F �

N

15 16 –15 16F S ��

J

Hence,

16 16 8cos

25 2 25| || | 5 2 5 2

F S

F S

��

��

–1

8cos

25

15. Answer (3)

Hint: P.E. energy of spring is = 21

2kx

1 2

1and

Pk k k k

L

Sol.:

Given21

(1)2k U ...(i)

k = 2k

kP = 2k + 2k = 4k

Now, U = 1

(4 ) (4) 162 k U

16U U

16. Answer (2)

Hint: Work energy theorem.

Sol.: By work energy theorem

21Loss of mechanical energy

2mv mgh

212 20 20 10 17 Loss of mechanical

2 energy

400 – 340 = Loss of mechanical energy

Loss of mechanical energy 60 J

17. Answer (4)

Hint: Conservation of momentum and projectile

motion.

Sol.:

0.05 kg

400 m/s

3.95 kg

4 kg

v

4V = 400 × .05 + 3.95 × 0

V = 5 m/s.

As horizontal range, 2 h

R vg

2 205 10 m

10

18. Answer (2)

Hint: Momentum conservation.

Sol.:

m

2m

m + m2

4 m/s4 m/s

v


5/19

Since Fext

= 0

Hence, by conservation of linear momentum.

ˆ ˆ(4 ) 2 (4 ) 3m i m j m v �

4 8ˆ ˆ m/s

3 3

v i j

19. Answer (1)

Hint: P F v �

�

Sol.:

1–5F y

1–5

a y

1–5

dvv k ydy

vdv =

1–5k y dy

2 4/5

42

5

v yk

v2 =

4/55

2

ky

v y2/5

P =

1–

2/55F v P y v �

�

1

5P y

20. Answer (3)

Hint: Area under the F–x curve.

Sol.: W = area under F – x curve.

W = 1

(200 100) (100 2) – 1002

= 300

200 –1002

W = 250 J

250 JW

21. Answer (3)

Hint: 0F S ��

Sol.: ˆ ˆ= sin( ) cos( )F at i at j�

ˆ ˆ= sin cos3 3

at atS i j

�

If andF S

��

are perpendicular then

0F S ��

sin( )sin cos( ) cos 03 3

at atat at

cos – 03

atat

–3

atat = /2

2

3 2

at

3

4t

a

22. Answer (2)

Hint:

Work – Energy theorem

Sol.:

m

m

10 m

m = 4 kg

W = 500 J

h = 10 m

Now, W = mgh + Wwork done against friction.

500 = 400 + Wwork done against friction.

Wwork done against friction

= 500 – 400

work done against friction100 JW

23. Answer (1)

Hint: Work – Energy theorem

Sol.: K = 100 N/m

2 kg

8 m/s

Smooth

1 m

= 0.7

x

2 21 11

2 2mv kx mg ...(i)

21 12 64 100 0.7 2 10 1

2 2x

64 – 14 = 50x2

50 = 50x2

1mx


6/19

24. Answer (2)

Hint: W = F d� �

for constant force.

Sol.: 2 2

ˆ ˆ– (5 – 3 ) m �

� �

d r r i j

ˆ ˆ(10 – 2 ) N�

F i j

ˆ ˆ ˆ ˆ(10 – 2 ) (5 – 3 ) 50 6 � �

W F d i j i j

56 JW

25. Answer (2)

Hint: W = Pt when P is constant

Sol.: P = constant

P t =

22

2

v Ptm v

m

2 202

10

tv t

20 10

2

PF

v t t

26. Answer (1)

Sol.: Linear momentum is always conserved

because Fext.

= 0.

Total energy also remains always conserved in all

cases. But kinetic energy will not be conserved

during collision.

27. Answer (2)

Hint:


dr

Sol.:


dr

U = 3 2

2–

a b

r r

4 3

3 2– 0

dU a b

dr r r

3

3 1– 2 0

ab

r r

r 0 3

– 2 0a

br

3

2

ar

b

28. Answer (2)

Hint:

Calculate frictional force and apply Newton’s 2nd Law.

Sol.:

m 2

m 1

T

mg1

sin

m 2

sin

g

N 2

m 2

cos

g0.25

0.75

mg1

cos

N 1

a

37°

1rf

2rf

N1 = m

1gcos

1rf =

1m

1gcos =

3 44 10 24 N

4 5

22 2

1 4cos37 2 10 4 N

4 5rf m g

Now, m2gsin + m

1gsin –

1rf –

1rf = (m

1 + m

2)a

12 + 24 – 24 – 4 = 6a

a = 28 4

m/s6 3

Now, 12 – T – 1rf = 2a.

12 – 4 – T = 4 8

2 8 –3 3

T

16N

3T

29. Answer (1)

Hint: For conical pendulum.

Tcos = mg

Sol.: l = 1.3 m

r = 0.5 m

mg

T sin

T cos

l = 1

.30 m

T

50 cmm = 200 g


7/19

Now sin = 0.5 5

1.3 13

Hence, cos = 12

13.

Now, T cos = mg

T = 0.2 10 2 13

13 N12cos 12 6

13

mg

13N

6T

30. Answer (2)

Hint: Use Lami’s theorem for equilibrium.

= 180 – (30 + )

Sol.: From F.B.D of block

T = 40 g ...(i)

30°

T

T2

T1

A

From F.B.D. of pulley.

T = 2T = 80 g ...(ii)

From F.B.D. of point A

1 2

sin(120) sin sin(90 )

T TT

T

T T

T1 =

80 sin(120)

sin(180 – (30 ))

g

T

40 g

T1 =

40 3

sin( 30)

g

T

1 = minimum sin( + 30)

max 1

= 90 – 30 = 60° 60

31. Answer (1)

Hint: Fnet

= ma

Sol.:

37° 30°

5 kg 4 kg

40

T

30

N1

20

T

20 3

N2

fr

fr = N

2 = .1 × 20 × 1.7 = 3.4 N

Now, Fnet

= ma.

30 – T = 5a

T – 20 – 3.4 = 4a

6.6 = 9a

a = 6.6 2.2

9 3 =

–2

11ms

15

32. Answer (1)

Hint: Use the concept of variable acceleration to find

the final velocity.

Sol.:

1 2 t (s)

1

0.5

For time (t = 0 to t = 1 s), retardation

a = g = 10 ms–2

Speed at t = 1 s

v1 = 20 – 10 × 1 = 10 m/s

For 1 t 2 s

av

= 1 0.5

0.752

Speed at t = 2 s

v2 = 10 – 0.75 × 1 = 2.5 m/s

33. Answer (2)

Hint: 1 1 2 2 3 3

cm

1 2 3

mv m v m vv

m m m

� � �

�

Sol.: cm

0v �

1 1 2 2 3 3

1 2 3

0mv m v m v

m m m

� � �

1 1 2 2 3 30mv m v m v

� � �

3

ˆ ˆ ˆ3 –4 2 –v i j k �

3

ˆ ˆ ˆ–4 2 – m/s

3

� i j kv

34. Answer (3)

Hint:

P = F dt Area under F -t curve.

Sol.:

p = Area under force - time curve

p = 20 + 5 – 5 – 5 – 2.5

12.5 kN s p


8/19

35. Answer (3)

Hint: F = ma

Sol.: ˆ ˆ ˆ(6 8 – 10 )NF i j k �

| | 36 64 100 10 2F �

N.

–2| | 5 msa �

m = ?

| | 10 2

| | 5

Fm

a

�

�

2 2 kgm

36. Answer (4)

Hint: fm =

sN

Sol.: Limiting frictional force on block depends on

nature of surfaces in contact and normal reaction.

37. Answer (3)

Hint: min2

1

mg

F

Sol.: min2

1

mgF

min2

12 10 3

43

14

F

Fmin

= 72 N

38. Answer (1)

Hint: sinm

mg F f

Sol.: 1.7

cos 0.2 100 17 N2m

f mg .

Driving force along plane is

FDrive

= 50 – 10 = 40 N, downward

So, Friction will act upward.

Now by Newton’s 2nd Law.

40 – 17 = 10a

22.3 m/sa downward to plane.

39. Answer (1)

Hint: opt

tanv Rg

Sol.: vopt

= 4

tan 120 103

Rg

vopt

= 40 10 4 = 40 m/s

40. Answer (2)

Hint: Use friction and pseudo force.

Sol.: N = ma ...(i)

maN

W

fs

fsmax

= N

fsmax

= ma ...(ii)

for vertical equilibrium W fs max

mg ma.

ga

41. Answer (2)

Hint: Tmax

= m(g + amax

)

Sol.: Tmax

= 70 × 9.8 N

Now, Tmax

= 50(g + amax

)

70 × 9.8 = 50 × 9.8 + 50amax

amax

= – 2

20 9.83.92ms

50

42. Answer (2)

Hint: Fnet

= ma

Sol.: Let upward force be F.

F – mg = ma ...(i)

ma

F

mg

F = m (g + a) = 2(10 + 4)= 28 N = 2.8 kgf.

43. Answer (3)

Hint: Use, constraint relation to find relation in

acceleration of blocks.

Sol.: From constraint relation.

a = 2a1

...(i) m

mg

T

N

a

Now from F.B.D. of pulley

T1 = 2T ...(ii)


9/19

46. Answer (3)

Hint: CCl4 is a non-polar molecule.

Sol.: In non-polar molecules (CCl4), only London

Forces are present.

47. Answer (2)

Hint: Hydration energy of F– is high.

Sol.: During neutralization of HF with NaOH, extra

energy is released due to extensive hydration of

F–

.

48. Answer (4)

Hint: Intensive property does not depend on the

quantity of matter.

Sol.: Pressure = Force

Area

is an intensive property.

[ CHEMISTRY]

From F.B.D. of blocks.

T = ma ...(iii)

M

T1

Mg

a1

Mg – 2T = Ma1

...(iv)

Mg – 2 ma = 2

aM

Mg = 2

Ma M TT

T1

Mg = 3

2

Ma

2

3

ga

44. Answer (2)

Hint:cm

ext

dPF

dt

�

�

Sol.:cm

ext

�

� dPF

dt

ext0

� �

F

or�

cmP = constant

�

cmMV = constant

�

cmV = constant

45. Answer (3)

Hint: Change in momentum normal to wall.

Sol.:

30°

30°

vcos30°

v sin 30°

m

v cos30°

v sin 30°

m

Px

= –mvcos30° – mvcos30°

= –2 mv cos30° = 3

–22

mv

|Px| = 3mv

|Py| = 0

P = 2 2 2

4( 3)

xP P mv

3OP mv

49. Answer (2)

Hint: Q = mLfus

+ msT

Sol.:

Q = 75 × 1000 × 80 + 75 × 1000 × 1 × 10 cal

= 6750000 cal

= 6.75 × 103 kcal

50. Answer (3)

Hint: w = – PexV

Sol.:

w = – PexV = –1.0 atm (150 × 15) cm3

= –1.0 atm × 2250 cm3

= –1.0 × 2.25 L-atm

= –2.25 L-atm


10/19

51. Answer (2)

Hint: wrev

= –2.303 nRTlogf

i

V

V

H = nCpT

Sol.:

wrev

= –2.303 × 5 × 8.314 × 300 log4

wrev

= –2.303 × 5 × 8.314 × 300 × 2 log2

= –2.303 × 5 × 8.314 × 300 × 2 × 0.301

= –17289.87 J = –17.3 kJ

H = nCpT

H = 0 ∵ T = 0

52. Answer (2)

Hint: Standard enthalpy of formation is defined for

formation of one mole of a product from its

constituent elements in their most stable states of

aggregation.

Sol.: For reaction H2(g) + Br

2(l) 2HBr(g), enthalpy

change is not standard enthalpy of formation. Here

two moles instead of one mole of the product is

formed.

53. Answer (4)

Hint: H = U + (PV)

Sol.:

H = U + (P2V

2 – P

1V

1)

P1 = 3 atm, V

1 = 4 L

P2 = 5 atm, V

2 = 6 L

H = 45 + (6 × 5 – 3 × 4) = 63 L-atm

54. Answer (2)

Hint: H = U + ng RT

Sol.:

H – U = ngRT = (6 – 9) × 8.314 × 300 J

= –7.48 kJ

55. Answer (4)

Hint: At any temperature, different particles in the

gas have different speeds.

Sol.: Due to different speeds, the gas particles have

different kinetic energy at same temperature.

56. Answer (3)

Hint: mps

2RTu

M and for ideal gas PV = nRT

Sol.: PM RT P

dRT M d

umps

= 2RT 2P

M d

umps

1

d

57. Answer (3)

Hint: For 1 mole of gas, van der waals equation is

2

aP (V – b) RT

V

and at high pressure 2

a0

V

�

Sol.: 2

aP (V – b) RT

V

for 2

a0

V

�

P(V – b) = RT

PV Pb– 1

RT RT

PV PbZ 1

RT RT

58. Answer (2)

Hint: For spontaneous reaction,

G = H – TS < 0

Sol.: at all temperature G < 0,

if H < 0 and S > 0

59. Answer (4)

Hint: PV = nRT

Sol.: PHe

V = nHe

RT

42.5 V 0.082 375

4

V = 12.3 L

60. Answer (4)

Hint: Boyle’s law, PV = constant

Sol.:

PV

V

61. Answer (2)

Hint: Higher the molecular attraction, higher is the

boiling point.

Sol.: HF molecules are associated with

intermolecular hydrogen bonding hence its boiling

point is highest.


11/19

62. Answer (1)

Hint: Heat of neutralization for strong acid and strong

base is 13.7 kcal/eq.

Sol.: meq of H2SO

4 = 200 × 0.2 = 40

meq of NaOH = 200 × 0.15 = 30

Heat evolve = 13.7 × 30 × 10–3 × 103 cal

= 411 cal

63. Answer (3)

Hint: Surface tension is force acting per unit length.

Sol.: Unit of surface tension is N m–1.

64. Answer (3)

Hint: Entropy decreases when disorder decreases.

Sol.: During condensation of vapour, vapour is

converted into liquid/solid. More disordered vapour is

converted into more ordered liquid/solid.

65. Answer (4)

Hint: G = Hvap

– TSvap

At equilibrium, G = 0

Sol.: G = H – TS

At equilibrium, G = 0, so H = TS

H 60240T 400 K

S 150.6

66. Answer (3)

Hint: dw = –PdV

Sol.: dw = –PdV

dw = – d(PV) (∵ Pressure is constant)

dw = –d(nRT) (For ideal gas)

w = –nRT

w = –1 × R × 1 = –R

Hence, work done by the gas is ‘R’

67. Answer (4)

Hint: Naturally occurring most stable form of an

element has zero standard enthalpy of formation.

Sol.: In standard state chlorine exist as Cl2(g)

68. Answer (1)

Hint: Rate of effusion (r) P

M

Sol.: 2

O

1.5r

32 and He

4.5r

4

2O

He

r 1.5 2= =1: 6 2

r 4.532

69. Answer (4)

Hint: ZC =

C C

C

P V

RT

Sol.: PC

= C C2

a 8a, V = 3b, T

27b 27Rb

ZC =

2

a 3b 27Rb 3

27b R 8a 8

70. Answer (2)

Hint: PV = nRT

or V T

p

Sol.:

T1 = 273 + 17 = 290 K

T2 = 273 + 27 = 300 K

V1

290

1.8

V2

300

1

or2

1

V 300 1.8= =1.86

V 290

71. Answer (2)

Hint: for a gaseous mixture, 2 2

2 2

CO CO

N N

p n

p n

Sol.: Mole of CO2 =

220= 5

44

Mole of N2 =

280=10

28

Now, 2 2

2 2

CO CO

N N

p n

p n

2CO

p 5

1.5 10

2

COp = 0.75 atm

72. Answer (4)

Hint: r f product f reactantH H – H

Sol.: C3H

8(g) + 5O

2(g) 3CO

2(g) + 4H

2O(l)

r f product f reactantH H – H

rH = {3 (–394) 4(–286) – (–104)}


12/19

rH = {–1182 – 1144 + 104} = –2222 kJ/mole

Mole of C3H

8 =

66=1.5

44

rH for 1.5 mole = –2222 × 1.5 = –3333 kJ

73. Answer (2)

Hint: S = nRln2

1

V

V

Sol.:

S = 5 × 2 × 2.303 log2

1

V

V= 5 × 2 × 2.303 log

40

5

= 5 × 2 × 2.303 log8 = 5 × 2 × 2.303 × 3 log2

= 5 × 2 × 2.303 × 3 × 0.301 = 20.8 cal K–1

74. Answer (1)

Hint: U = q + w;

Sol.: w = –PV = 0

q + w = U q = U = 750 J

75. Answer (1)

Hint: H = U + ngRT

Sol.: For the reaction

H2(g) + I

2(g) 2HI(g)

ng = 2 – 2 = 0

H = U

76. Answer (2)

Hint: q and w are path functions.

Sol.: U + PV = H (State function)

S(Entropy) (State function)

H – TS = G(State function)

77. Answer (2)

Hint: H = B.E(reactant)

– B.E(product)

Sol.: 2 2

1 1H I HI

2 2

H = 2 2

(H ) (I ) (HI)

1 1B.E. B.E. – B.E.

2 2

or BE(HI)

= 2 2

(H ) (I )

1B.E. B.E. – H

2

= 1435 150 – 26.5

2 = 266 kJ/mole

78. Answer (2)

Hint: Average speed 8RT

M

Most probable speed 2RT

M

rms speed 3RT

M

Sol.:

8RT 2RT 3RTAverage : mps : rms : :

M M Mspeed speed

= 8: 2 : 3

79. Answer (3)

Hint: K.E. per atom

Energy absorbed Bond energy–

per molecule per molecule

2

Sol.:

Bond energy/Molecule =

3

–19

23

180.6 10 J= 3 10 J

6.02 10

K.E. per atom =

–19 –19

3.5 10 – 3 10 J

2

= 2.5 × 10–20 J

80. Answer (2)

Hint: Avg. Kinetic energy (E) = 3

n RT2

Sol.: 2

N

14 3E = R 300

28 2

2O

24 3E = R 500

32 2

2

2

N

O

E 14 32 300 2= =

E 28 24 500 5

81. Answer (3)

Hint: (V – nb), Volume available for the movement of

gas molecules.

Sol.: ‘nb’ is the volume occupied by n mole of the

gas molecules.

82. Answer (4)

Hint: urms

= 3RT

M

Sol.: urms

1

M

Lower is the molar mass, higher is the root mean

square speed.


13/19

[ BIOLOGY]

91. Answer (4)


growth.

Sol.: In monocots, secondary growth is absent.

Thus lateral meristem is also absent.

92. Answer (4)

Hint: In dicot stem, cork cambium develops from cell

of cortex region.

Sol.: Cork cambium is also called phellogen,its

origin is outside of the stele i.e. extrastelar.

In dicot root, it develops from pericycle cells.

93. Answer (4)

Hint: In a compound leaf, incision reaches upto

midrib.

Sol.: Lamina is broken into leaflets and midrib forms

a common axis called rachis, but bud is present in

the axil of petiole of compound leaves.

94. Answer (3)

Hint: In some seeds nucellus is not fully consumed

during development of embryo.

Sol.: Such persistent nucellus is called perisperm

and such seeds are known as perispermic seeds.

95. Answer (1)

Hint: In caryopsis fruits, pericarp is completely fused

with seed coat.

Sol.: Caryopsis fruits are found in members of

Poaceae like wheat (Triticum aestivum), maize, rice

etc.

83. Answer (4)

Hint: S = q

T

Sol.: T = 127 + 273 = 400 K

S =

–1

4800 Jmol

400 K= 12 JK–1mol–1

84. Answer (1)

Hint: Process is adiabatic so q = 0.

Sol.:

w = –1.5(4.2 – 1.2) L-atm = – 4.5 L-atm

= –4.5 × 101.3 J = –455.85 J

U = q + w

∵ q = 0, so U = w = –455.85 J

85. Answer (3)

Hint: For a reaction to be spontaneous

G = H – TS = –ve

Sol.: For G to be –ve, TS > H

H 7100T 284 K = 11°C

S 25

86. Answer (3)

Hint: 2 2

1CO(g) O (g) CO (g) 283.5 kJ / mol

2

Sol.: Mole of CO2 formed =

55g 5= mole

44 g / mole 4

Heat released = 5

4 × (283.5) = 354.4 kJ

87. Answer (1)

Hint: Average speed (u) = 8RT

M

Sol.: 2

2

SOHe

SO He

Mu 64= = 4 :1

u M 4

88. Answer (3)

Hint: Higher the polarity of the molecule, higher is

the attractive force.

Sol.: NH3 is polar molecule hence it will be

associated by attractive force and will be liquefied

easily.

89. Answer (3)

Hint: 1

rM

and

Rate of effusion (r) =

Volume of gas effused (V)

Time taken (t)

Sol.: 1

rM

, so 1 1 2

2 2 1

V /t M=

V /t M

Since V1 = V

2, so

2 2

1 1

t M=

t M

1 2

1

2t M=

t 16 M

2 = 64 g/mol

90. Answer (2)

Hint: PM

d =RT

Sol.: 4 32

d = = 3.9 g/L0.082 400


14/19

96. Answer (3)

Hint: In monocot stem, vascular bundles are

scattered in whole ground tissue.

Sol.: In monocot stem, endodermis is absent but it

is well developed in roots.

97. Answer (4)

Hint: Lateral roots develope from pericycle.

Sol.: Lateral roots are present in both dicot &

monocot roots.

98. Answer (3)

Hint: Pericycle is absent in monocot stem.

Sol.: Pericycle may be single layered or

multilayered.

99. Answer (4)


Sol.: In soyabean (Fabaceae), brinjal, chilli

(Solanaceae) and tulip (Liliaceae) ovary is superior.

100. Answer (3)

Hint: Flower is a modified shoot which may have

floral appendages in multiple of three, four or five.

Sol.: Trimerous flowers are found in Liliaceae family

in which floral appendages are in multiples of three.

101. Answer (2)

Hint: Tap roots directly arise from radicle.

Sol.: Adventitious roots arise from plant parts other

than radicle.

102. Answer (4)

Hint: Axillary buds modify into tendrils in cucurbits.

Sol.: In cucurbits, stem modifies to provide support

to the plant.

103. Answer (3)

Hint: Heartwood is called duramen and sapwood is

called alburnum.

Sol.: Heartwood is hard, durable and resistant to

attack of microorganisms.

104. Answer (2)

Hint: Lenticels occur in most woody trees and

permit the exchange of gasses and water.

Sol.:

- Annual rings are distinct in plants growing in

temperate regions because climatic conditions

are not uniform over there.

- Stelar secondary growth is performed by vascular

cambium as a result central cylinder of wood is

formed which remains surrounded by secondary

phloem.

- Pericycle cells opposite to protoxylem, become

meristematic and give rise to vascular cambium

in dicot roots.

105. Answer (4)

Hint: With the passage of time, rings of the sap

wood are changed into heart wood.

Sol.: As a result of secondary growth amount of

heart wood increases and amount of sap wood

almost remains constant.

106. Answer (2)

Hint: Cassia flower can be divided into two equal

halves only by one plane.

Sol.: It is known as bilateral symmetry (Zygomorphic

flower)

107. Answer (3)

Hint: In cymose inflorescence, the main axis

terminates into a flower.

Sol.: In Tri t icum (Wheat) inf lorescence is

spikelet type (racemose).

108. Answer (2)

Sol.: Some pepo fruits are bitter in taste due to

tetracyclic triterpenes.

109. Answer (4)

Hint: Replum is thin membranous structure, which is

formed in siliqua type of fruit.

Sol.: Replum is a pseudoseptum and it is a feature

of family Brassicaceae.

110. Answer (2)


Sol.:

nG – Superior ovary

nG – Inferior ovary (Epigynous flower)

111. Answer (2)

Hint: Relative positions of primary xylem

(Protoxylem & metaxylem) decides the endarch or

exarch conditions.

Sol.: In endarch arrangement, protoxylem occurs

towards centre and metaxylem towards periphery.

112. Answer (3)

Hint: Root hairs are epidermal appendages.

Sol.: Root hairs are unicellular elongations of the

epidermal cells i.e. they are exogenous in origin.

113. Answer (4)

Hint: Guard cells are generally bean-shaped except

grasses.

Sol.: In grasses, the guard cells are dumb-bell

shaped.


15/19

114. Answer (2)

Hint: In Fabaceae, stamens show diadelphous

condition.

Sol.: Floral formula of family Fabaceae is

K (5)C A1 + 2 + (2) (9) + 1 1G%

115. Answer (1)

Sol.: Rice, coconut and wheat are endospermic

seeds but seeds of orchids are non-endospermic.

116. Answer (2)

Hint: Coleoptile and coleorhiza are sheaths that

enclose plumule and radicle respectively.

Sol.: Proteinaceous layer of endosperm that

separates embryo is called aleurone layer.

117. Answer (3)

Hint: Companion cell controls the function of sieve

tube.

Sol.: Companion cells retain nucleus throughout

their life and control functions of anucleated sieve

tubes.

118. Answer (1)

Hint: In exarch type of primary xylem, protoxylem

lies towards the periphery and metaxylem lies

towards the centre.

Sol.: Exarch arrangement of primary xylem is a

feature of roots.

119. Answer (2)

Hint: Collenchyma is an elastic, living mechanical

tissue.

Sol.: Collenchyma cells have thickening of cellulose,

hemicellulose and pectin at their corners.

120. Answer (3)

Hint: This is a monocot family.

Sol.: In family Liliaceae, gynoecium is tricarpellary,

syncarpous, superior, trilocular and with axile

placentation. It is commonly called lily family.

121. Answer (1)

Hint: A sterile stamen is called staminode.

Sol.:

- Obliquely placed ovary and swollen placenta are

the features of family Solanaceae.

- Versatile fixation of anthers is feature of family

Poaceae

- Female flowers are known as pistillate flowers.

122. Answer (2)

Hint: In litchi, seed is covered by an outgrowth of

funicle.

Sol.: This white, translucent, fleshy edible covering

of seed is called aril.

123. Answer (4)

Hint: In tetradynamous condition there are

6 stamens4 larger in ring2 smaller in ring

inner outer

Sol.: Tetradynamous condition is a feature of family

Brassicaceae. In pea, stamens exhibit diadelphous

condition.

124. Answer (3)

Hint: Coconut is a drupe type of fruit.

Sol.: Drupe type of fruits are mostly one seeded,

develop from monocarpellary superior ovary and have

hard, stony endocarp.

125. Answer (2)

Hint: Hypodermis is absent in roots.

Sol.:

- All tissues outside the vascular cambium

constitute bark

- In leaves, ground tissue is not well differentiated

& is known as mesophyll

- Casparian strips are formed by deposition of a

waxy material suberin on the radial and

tangential walls of endodermal cells.

126. Answer (4)

Hint: The endodermal cells of dicot stem store some

carbohydrate grains.

Sol.: These cells store starch grains and hence

endodermis is known as starch sheath.

127. Answer (3)

Hint: These roots are adventitious roots.

Sol.: Stilt roots arise from the lower nodes of stem

to support the main axis.

128. Answer (3)

Hint: This aestivation is found in corolla of family

Fabaceae.

Sol.: In family Fabaceae corolla is papilionaceous,

in which smallest anterior petals are referred as keel.

129. Answer (1)

Hint: Pulvinate leaf is found in some leguminous

plants.

Sol.: Leaf having swollen base is called pulvinate

leaf.

130. Answer (3)

Hint: Rhizome is a modified underground stem which

grows horizontal to the soil surface.

Sol.: Potato is a tuber i.e. modified underground

stem branch which get swollen on account of

accumulation of food.


16/19

131. Answer (2)

Hint: In racemose inflorescence, younger flowers are

present towards the apex whereas in cymose

inflorescence, younger flowers are present towards

base.

Sol.:

Cymose inflorescence – Basipetal succession of

flowers

Racemose inflorescence – Acropetal succession of

flowers

Radial symmetry – Actinomorphic flower

Symmetry by one – Zygomorphic flower

plane only

132. Answer (3)

Hint: Mesophyll tissue is not differentiated into

palisade and spongy parenchyma in monocot leaves.

Sol.: Monocot leaves are isobilateral. These have

stomata on both surfaces and below the stoma of

abaxial epidermis sub-stomatal cavities are present.

133. Answer (4)


growth.

Sol.: Lateral meristem includes intrafascicular

cambium (primary), interfascicular cambium and cork

cambium (secondary).

134. Answer (4)

Hint: This condition is seen in potato family.

Sol.: In the floral formula of family Solanaceae,

C A(5) 5 represents epipetalous condition.

135. Answer (1)

Hint: This plant is used as fodder.

Sol.: Sesbania is used as fodder and it belongs to

family Fabaceae.

136. Answer (4)

Hint : JGA secretes renin in response to fall in GFR.

Sol. : Angiotensinogen is produced by liver and

released into blood stream.

Angiotensinogen Angiotensin-I

ACE

Renin

Angiotensin-II

(Angiotensinconverting enzyme)

Vasoconstriction Increase in heart rate

Increase in blood pressure

Adrenal cortex

Aldosterone(Mineralocorticoid)

ACE is produced by lung capillaries.

137. Answer (2)

Hint : SA node generates cardiac impulses.

Sol. : From SA node impulse travels to AV node.

From AV node, Bundle of His carries impulse to

bundle branches and Purkinje fibres spread impulse

to entire ventricular musculature.

138. Answer (3)

Hint : Pacemaker maintains rhythmic contractility of

heart.

Sol. : Artificial pacemaker is required when there is

a damage to nodal tissue of heart like in case of

heart attack. This device sends out small electrical

current to stimulate heart to contract thus

maintaining rhythm of cardiac activity.

139. Answer (3)

Hint : Blood group mismatch leads to clumping of

RBC.

Sol. : O–

blood group is universal donor as it lacks

A, B and D antigens on the surface of RBC, but it

has both anti-A, anti-B and anti-D antibodies in

plasma. That’s why a person having O blood group

can accept blood from another person having O blood

group only.

140. Answer (4)

Hint : Both atria contract simultaneously during atrial

systole.

Sol. : All four chambers of heart do not contract

simultaneously. Increase in ventricular pressure

causes closure of AV valves. Major filling i.e., (2/3rd)

of ventricles occurs during joint diastole.

141. Answer (1)

Hint : Involuntary muscles are present in tunica

media.

Sol. : Smooth muscles are present in tunica media

in both type of blood vessels.

142. Answer (2)

Hint : Renin is a part of RAAS pathway.

Sol. : In response to fall in blood pressure/GFR/

glomerular blood flow, renin is released by JGA to

initiate RAAS pathway which will eventually restore

the normal level.

143. Answer (3)

Hint : Portal system involving kidneys.

Sol. : Hepatic & hypophyseal portal system are well

developed in humans.


17/19

144. Answer (4)

Hint : Lymph flows through separate lymphatic

capillaries and lymphatic vessels.

Sol. : Lymphatic vessels have valves to prevent

backflow of lymph. Left and right subclavian veins

drain lymph into superior vena cava. Thoracic duct

collects lymph from most of part of gastrointestinal

tract.

145. Answer (3)

Hint : Bundle of His is a part of nodal tissue of heart.

Sol. : Nodal tissue of heart are modified muscle

fibres which exhibit autoexcitability. Nodal tissue

include SA node, AV node, Bundle of His and

Purkinje fibres.

146. Answer (4)

Hint : Malpighian body of Juxtamedullary nephrons

are located close to medulla of kidney.

Sol. : Cortical nephrons are more abundant as

compared to juxtamedullary nephrons.

Juxtamedullary nephrons have longer loop of Henle

and well developed vasa recta as compared to

cortical nephron. Peritubular capillaries are present

around tubules of nephrons in both Cortical and

Juxtamedullary nephron.

147. Answer (1)

Hint : Urea cycle occurs is liver.

Sol. : Formation of urea occurs in liver and hepatic

vein carrying blood from liver has maximum quantity

of urea while renal vein carrying blood from kidney

after filtration has least amount of urea.

148. Answer (2)

Hint : In normal ECG, T-waves are positive wave

produced during ventricular repolarisation.

Sol. : End of T-wave represents end of ventricular

systole and initiation of joint diastole or diastasis.

149. Answer (3)

Hint : First heart sound is ‘Lubb’ and second heart

sound is ‘dub’.

Sol. : First heart sound is produced by closure of

A-V valves at the beginning of ventricular systole,

while second heart second ‘dub’ is produced by

closure of semilunar valves at the beginning of

ventricular diastole.

150. Answer (2)

Hint : Counter current mechanism helps in producing

concentrated urine.

Sol. : If there is no loop of Henle, there will be no

counter current mechanism and medullary

concentration gradient will not be maintained. As a

result dilute urine will be formed.

151. Answer (3)

Hint : Heart beats about 72 times in a minute.

Sol. : In an cardiac cycle, both systemic and

pulmonary circulation take place. So about 72

cardiac cycles take place in a minute, hence about

72 times double circulations are normally completed

in one minute.

152. Answer (2)

Hint : Sympathetic neuronal endings release

adrenaline.

Sol. : Adrenaline causes an increase in heart rate

resulting in decrease in the duration of individual

cardiac cycle. However, adrenaline causes an

increase in cardiac output and stroke volume.

153. Answer (1)

Hint : This is associated with skeletal structures of

body.

Sol.: Liver and spleen are involved in erythropoiesis

in embryonic life.

154. Answer (3)

Hint : Micturition reflex is initiated by stretching of

wall of urinary bladder.

Sol. : Stretch receptors in the wall of urinary bladder

send signals to CNS, which passes on motor

messages to initiate the contraction of smooth

muscles of bladder and simultaneous relaxation of

urethral sphincter causing the release of urine.

155. Answer (2)

Hint : These blood vessels carry deoxygenated

blood.

Sol. : Blood must pass through pulmonary

circulation in order to flow from right atrium to left

atrium.

156. Answer (4)

Hint : Animals which live in water deficient conditions

are uricotelic.

Sol. : Uric acid is excreted by using minimum

amount of water e.g., Reptiles, Birds, Insects, Land

snails etc.

157. Answer (1)

Hint : Sympathetic nervous system regulates

various activities of body in emergency situation.

Sol. : Parasympathetic stimulation decreases heart

rate, stroke volume and cardiac output. P-wave in

ECG represents atrial depolarisation.

158. Answer (3)

Hint : Sudden damage to heart muscles causes

myocardial infarction

Sol. : Heart attack is also known as myocardial

infarction. Damage to nodal tissue causes stoppage

of heart-beat which is cardiac arrest.


18/19

159. Answer (1)

Hint : Damage to kidney causes edema.

Sol. : Damage to kidney disrupts its functions

leading to accumulation of urea in blood which is

known as uremia. Formation of large volume of urine

is known as polyuria. Cystinuria refers to increased

presence of cysteine in urine while in hematuria,

blood is found in urine.

160. Answer (1)

Hint : Systemic circulation begins with left ventricle

and ends in right atrium.

Sol. : Aorta carries oxygenated blood from left

ventricle towards body tissues while vena cava

returns deoxygenated blood to right atrium.

161. Answer (3)

Hint : T-wave represents repolarisation of ventricles.

Sol. : T-wave represents end of ventricular systole

which is followed by joint-diastole. Electrocardiogram

represents electrical activity of heart and it is

recorded by electrocardiograph machine.

162. Answer (4)

Hint : Oxygen requirement of body increases during

strenuous exercise.

Sol. : During strenuous exercise, under sympathetic

stimulation heart rate increases due to increase in

number of action potentials generated by

pacemaker. Also, there is an increase in force of

contraction of ventricles leading to an increase in

stroke volume. However, the sequence of various

events occurring in a cardiac cycle i.e. atrial systole,

ventricular systole and joint diastole remain same.

163. Answer (3)

Hint : Net filtration pressure (NFP)/Glomerular

filtration pressure (GFP) is the pressure being

exerted by glomerular blood for ultrafiltration.

Sol. : NFP GHP (BCOP CHP)Glomerular Blood CapsularHydrostatic Colloidal HydrostaticPressure Osmotic Pressure

Pressure

164. Answer (1)

Hint : Members of superclass Pisces have venous

heart.

Sol. : Only deoxygenated blood is pumped by fish’s

heart; which means blood is circulated only once

through their heart i.e. single circulation. Most

reptiles show incomplete double circulation.

165. Answer (3)

Hint : Maximum reabsorption of water & solutes

occurs in PCT.

Sol. : Reabsorption of maximum solutes including

glucose occurs in PCT. Reabsorption of water

occurs in PCT, descending limb of loop of Henle,

DCT and collecting duct.

166. Answer (4)

Hint : Rh–ve individuals will generate antibodies

against Rh+ve antigen in their plasma.

Sol. : Erythroblastosis foetalis occurs due to Rh

incompatibility between Rh –ve mother and Rh +ve

foetus. Antibodies against Rh antigen cross the

placenta and start destroying foetal RBCs. In order

to prevent this, anti-Rh antibodies are given to

mother.

167. Answer (3)

Hint : It is a tuft of blood capillaries present in

Bowman’s capsule.

Sol. : Renal tubule involves components of tubular

parts of nephrons such as Bowman’s capsule, PCT,

DCT and loop of Henle.

168. Answer (2)

Hint : Left ventricle pumps blood into aorta from

where it is carried to different parts of the body.

Sol. : Different nodal tissues of heart are capable of

generating actions potential at different rates.

Lymphocytes are found in lymph. Volume of blood

pumped by each ventricle in one cardiac cycle is

stroke volume.

169. Answer (2)

Hint : Various plasma proteins promote blood

coagulation.

Sol. : Heparin is an anticoagulant produced by

basophils and mast cells while thrombin, Ca2+ and

thrombokinase are involved in blood clotting.

170 Answer (2)

Hint : Osmolarity depends upon number of solute

particles/volume.

Sol. : Because PCT is permeable for both water and

electrolytes as well as other substances so

osmolarity in PCT remains unchanged and filtrate

remains isotonic in comparison to blood plasma.

171. Answer (3)

Hint : Antibodies bind to antigen present on RBCs

surface.

Sol. : Person having O-blood group lacks antigens

A & B on the RBC surface but have anti-A and anti-

B antibodies in their plasma.


19/19

��

172. Answer (4)

Hint : Human heart is myogenic.

Sol. : SA node (pacemaker) is a modified cardiac

muscular tissue capable of generating impulses.

RBCs lack mitochondria to perform aerobic

breakdown of glucose. Single circulation of blood

occurs in heart of fishes.

173. Answer (1)

Hint : Cortex present between medullary pyramids

was named after Joseph Bertin.

Sol. : At a number of places renal cortex invaginates

within medulla and divides medulla into a number of

pyramidal structures known as medullary pyramids.

Invaginations of cortex into medulla are known as

columns of Bertini.

174. Answer (4)

Hint : Haemoglobin helps in transport of gases.

Sol. : Haemoglobin is an iron containing pigment

present inside RBCs. Basophils secrete histamine,

serotonin and heparin. Neutrophils have multilobed

nucleus.

175. Answer (3)

Hint : Chordae tendinae are attached to papillary

muscles in ventricular wall & with flaps of AV valves.

Sol. : Chordae tendinae are attached to AV valves on

one end and papillary muscles on other end. These

cord like structures prevent the collapse of AV valves

during powerful ventricular contraction. They are not

attached to semilunar valves.

176. Answer (4)

Hint : Obligatory reabsorption of water occurs in this

part of nephron.

Sol. : Obligatory reabsorption of water around

60-70% occurs in PCT and some occurs in loop of

Henle. Facultative reabsorption of about

10 - 15% of water occurs in DCT and CD under the

influence of ADH.

177. Answer (2)

Hint : Antibodies protect us from pathogens.

Sol. : Antibodies are immunoglobulins which are

produced by B-lymphocytes (plasma cells) and

protect body from pathogens. Albumin is mainly

concerned with maintaining osmotic balance of

blood. Heparin is an anticoagulant while fibrinogen is

one of the clotting factors.

178. Answer (1)

Hint : Removal of this compound requires large

amount of water.

Sol. : Ammonia is the most toxic nitrogenous waste

and uric acid is the least toxic nitrogenous waste

which is excreted in form of pellets.

179. Answer (3)

Hint : Vertebrates possess ventral heart.

Sol. : Blood groups are determined by presence of

surface antigen on RBCs. Conversion of fibrinogen

into fibrin is done by thrombin.

Chordates like urochordates have an open

circulatory system.

180. Answer (3)

Hint : Factors which are used up during coagulation

of blood.

Sol. : Plasma lacks formed elements and plasma

without clotting factors is known as serum.

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