for Medical-2020 All India Aakash Test Series for …...All India Aakash Test Series for Medical-2020 Test - 6 (Code-C) (Answers & Hints) 4/16 20. Answer (2) Hint: Resultant amplitude

Test - 6 (Code-C) (Answers) All India Aakash Test Series for Medical-2020

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1. (1)

2. (4)

3. (4)

4. (1)

5. (2)

6. (3)

7. (1)

8. (1)

9. (4)

10. (3)

11. (1)

12. (3)

13. (3)

14. (3)

15. (3)

16. (1)

17. (1)

18. (4)

19. (4)

20. (2)

21. (2)

22. (1)

23. (3)

24. (4)

25. (1)

26. (4)

27. (4)

28. (3)

29. (1)

30. (4)

31. (2)

32. (1)

33. (4)

34. (1)

35. (1)

36. (1)

Test Date : 17/02/2019

ANSWERS

TEST - 6 (Code-C)

All India Aakash Test Series for Medical-2020

37. (2)

38. (2)

39. (2)

40. (3)

41. (3)

42. (4)

43. (1)

44. (1)

45. (2)

46. (2)

47. (3)

48. (1)

49. (4)

50. (2)

51. (3)

52. (3)

53. (1)

54. (4)

55. (1)

56. (1)

57. (2)

58. (2)

59. (4)

60. (3)

61. (3)

62. (4)

63. (4)

64. (1)

65. (2)

66. (1)

67. (2)

68. (1)

69. (4)

70. (3)

71. (4)

72. (2)

73. (1)

74. (4)

75. (2)

76. (3)

77. (3)

78. (1)

79. (3)

80. (4)

81. (4)

82. (3)

83. (2)

84. (1)

85. (3)

86. (4)

87. (3)

88. (4)

89. (4)

90. (1)

91. (4)

92. (4)

93. (3)

94. (1)

95. (2)

96. (4)

97. (2)

98. (4)

99. (3)

100. (1)

101. (2)

102. (4)

103. (3)

104. (1)

105. (4)

106. (4)

107. (1)

108. (2)

109. (3)

110. (3)

111. (1)

112. (4)

113. (3)

114. (2)

115. (3)

116. (3)

117. (2)

118. (3)

119. (3)

120. (4)

121. (3)

122. (1)

123. (2)

124. (3)

125. (3)

126. (4)

127. (3)

128. (3)

129. (4)

130. (2)

131. (4)

132. (3)

133. (3)

134. (3)

135. (2)

136. (3)

137. (1)

138. (4)

139. (3)

140. (4)

141. (4)

142. (4)

143. (2)

144. (3)

145. (3)

146. (2)

147. (4)

148. (3)

149. (3)

150. (3)

151. (2)

152. (3)

153. (1)

154. (3)

155. (2)

156. (4)

157. (4)

158. (3)

159. (4)

160. (1)

161. (4)

162. (4)

163. (3)

164. (3)

165. (3)

166. (1)

167. (3)

168. (3)

169. (3)

170. (2)

171. (2)

172. (1)

173. (4)

174. (3)

175. (3)

176. (1)

177. (4)

178. (1)

179. (4)

180. (3)

All India Aakash Test Series for Medical-2020 Test - 6 (Code-C) (Answers & Hints)

2/16

ANSWERS & HINTS

1. Answer (1)

Hint: The least time interval after which particle

repeats its motion is called time period.

Sol.: Least time for given combination is 2

2. Answer (4)

Hint: If function increases or decreases

monotonically with time, function is non-periodic.

Sol.: (1) y = cos3t = 13cos – cos3

4t t

it is periodic but not SHM

(2) 2 sin SHM4

⎛ ⎞ ⎜ ⎟⎝ ⎠

y t

(3) y = cos2t, it is SHM

(4) y = aet is non periodic

3. Answer (4)

Hint: y = Asin(t + )

Sol.: The angle at time 2

( )4

t tT

.

The projection on y axis

2 2( ) sin sin

4 120 4

⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

y t R t R tT

sin60 4

y R t ⎡ ⎤ ⎢ ⎥

⎣ ⎦

4. Answer (1)

Sol.: Time taken from x = A to is .2 6

A Tx

x = Acost

2cos 3

2

AA

T

1 2cos 3

2 T

23

3 T

T = 18 s

[ PHYSICS]

5. Answer (2)

Hint: vmax

= ASol.: x = Asint

(10 cm)sin2

x t

–1 –1

max(10 cm) rad s 5 cm s

2v

6. Answer (3)

Hint: 2m

Tk

Sol.: 3

2 22 2

3

m mT

kk

⎛ ⎞⎜ ⎟⎝ ⎠

7. Answer (1)

Hint: dx

vdt

Sol.: 10 sin5

v t

x vdt ∫

–10 cos5

5

t C

x

–50cos5

x t C

at t = 0, v = 0, x = –50 so, C = 0

50cos5

x t⎛ ⎞ ⎜ ⎟

⎝ ⎠

8. Answer (1)

Hint: 2

20

md x dxb kxdtdt

Sol.:2

20

md x dxb kxdtdt

...(i)

2

22 0d x dx

xdt dt

This is equation of damped oscillation.

Test - 6 (Code-C) (Answers & Hints) All India Aakash Test Series for Medical-2020

3/16

9. Answer (4)

Hint: T

v xg

Sol.: v xg

1

22

dv ga v xg g

dx x (constant)

So a x°

10. Answer (3)

Hint: (vp)max

= A

Sol.:1

2

p wv v

0

1(2 )

2y f f

= 4y0

11. Answer (1)

Hint and Sol.: Open end of a closed organ pipe is

displacement antinode and pressure node.

12. Answer (3)

Hint: 0

1

2

Tf

l

Sol.:10

20

If 100 N100 1

169 N20 169 1.3

⎡ ⎤ ⎢ ⎥ ⎣ ⎦

Tf

Tf

0.3f0 = 20

0

200Hz

3f

13. Answer (3)

Hint:p

yv

t

Sol.: y = 20 sin(100 t – 2x)

20 100 cos(100 – 2 )p

yv t x

t

3 12 10 cos 100 0 – 2

6p

v⎡ ⎤ ⎢ ⎥⎣ ⎦

3 312 10 10 cm/s

2p

v⎛ ⎞ ⎜ ⎟⎝ ⎠

14. Answer (3)

Hint: = kx

Sol.:2

x

0.52

k

[0.5 – 0.25]2 8

15. Answer (3)

Hint: Fundamental frequency, 0

4

vf

l

Sol.:0

330110 Hz

4 0.75f

only odd harmonic are produced in closed pipe.

So frequencies 330 Hz, 550 Hz, 770 Hz are

possible. 660 Hz is not possible.

16. Answer (1)

Hint: B

v

Sol.: 2

B

v

or, 2 2 –6 –2 2

8 3

1 1

8 10 (2.5 10 )

2 10 kg/m

B

v kv

17. Answer (1)

Hint: P RT

vM

Sol.: If temperature is constant v will remain

constant.

Since pressure and density will change in the same

ratio.

18. Answer (4)

Hint: Pitch is related to frequency of sound.

Sol.: Quality is the sensation of human ear due to

wave form of sound.

19. Answer (4)

Hint: T

v

Sol.: Stress

v

11 –5

3

1.6 10 1.6 10 20

8 10

Yv

4

–11.6 1.6 2 1080 m s

8


4/16

20. Answer (2)

Hint: Resultant amplitude formula.

Sol.: When amplitude of component waves are

equal, amplitude of resultant, 2 cos2

R A⎛ ⎞ ⎜ ⎟

⎝ ⎠

Amplitude, 3

2 cos 2 36 2

R A A A

21. Answer (2)

Hint: Amplitude = 2Asinkx

Sol.: Amplitude = 4sinx ...(i)

22m

⇒

...(ii)

at2m,

3 3x

2Amplitude 4sin 2 3 m

3

22. Answer (1)

Hint: Beat frequency = |f1 – f

2|.

Sol.: 1 = 1020 f

1 = 510 Hz.

2 = 1004 f

2 = 502 Hz

fb = |f

1 – f

2| = 8 Hz

23. Answer (3)

Hint: Wavelength depends on medium.

Sol.: (1) Frequency does not change by changing

the medium of propagation.

(2) If wave is reflected from denser medium,

the phase change is .

24. Answer (4)

Hint: Y

v

Sol.: 9

3 –1

3

8 102 10 m s

2 10

Yv

L

5

4

O

In second overtone

5

4L

41.6 m

5

L

3

32 101.25 10 1250 Hz

1.6

vf

25. Answer (1)

Hint: Only odd harmonic in closed organ pipe.

Sol.: f1 : f

2 : f

3 = 5 : 7 : 9

So pipe is closed

So, fundamental frequency 0

28557 Hz

5f

0

0

380 20m

57 3

v

f

26. Answer (4)

Hint: fB = f

1 ~ f

2

Sol.: f = 4 5t = 1st 2nd

fB

4 5

27. Answer (4)

Hint: m m

a a

v

v

Sol.:1650 5

330 1

m

a

v

v

5

1

m

a

5m a

28. Answer (3)

Hint: End correction, e = 0.6r

Sol.:1 2

3

4( ) 4( )

v vf

l e l e

3(l1 + e) = l

2 + e

2 1– 3

2

l le

72.6 – 720.3 0.6

2e r

3 1cm

6 2r

29. Answer (1)

Hint: 0

Source–

vf f

v v

⎛ ⎞ ⎜ ⎟

⎝ ⎠

Sol.:350

5500 5000350 – v

⎛ ⎞ ⎜ ⎟

⎝ ⎠

11 350

10 350 – v

350 – 11v = 0

350m/s

11v


5/16

30. Answer (4)

Hint: Stress

v .

Sol.: f0 = 100 Hz, v = f

0

0 = 200 m/s

Now, 2Stress v .

Stress = (200)2 × 8 × 103 = 3.2 × 108 N/m2

31. Answer (2)

Hint & Sol.: Sound travel fastest in solids.

32. Answer (1)

Hint: 1 2 3

1 2 3

1 1 1: : : :n n n

l l l

Sol.:1 2 3: : 1: 2 :1n n n

1 2 3: : 2 :1: 2l l l

l1 + l

2 + l

3 = 150 cm 5K = 150 K = 30.

So, l1 = 60 cm

l2 = 30 cm

l3 = 60 cm

so, x1 = 60 cm

x2 = 90 cm

33. Answer (4)

Hint: y = 2A sinkx cost

Sol.: y = 2A sinkx cost

2sin cos(90 )30

xy t

⎛ ⎞ ⎜ ⎟⎝ ⎠

2

30

= 60 cm

30 cm2

Number of loops = 10

Number of nodes = 10 + 1 = 11

Number of antinodes = 10

Amplitude of x = 5 cm

2sin 5 1cm30

34. Answer (1)

Hint: v

f

Sol.:330 1

m 50 cm660 2

3 512.5, 37.5, 62.5

4 4 4

(Not possible)

Only 1st and 2nd resonance can be produced.

So minimum height of water

hmin

= 50 – 37.5 = 12.5 cm

35. Answer (1)

Hint: 1 1

2 2

f v

f v

Sol.: 1 1 2

2 2 1

, If Constantf v m

Tf v m

1

2

600 2 1

32 4

f

f f

f = 2400 Hz

36. Answer (1)

Hint: Use Newton’s formula P

v

Sol.: According to Laplace’s correction

v v Percentage error

v v 1.4 1100 100%

v 1.4

15%�

37. Answer (2)

Hint: nl = constant

Sol.: 2

1

20 4

25 5

A

B

n l

n l n

B > n

A.

So, nB – n

A = 5 on solving,

nA = 20

nB = 25

38. Answer (2)

Hint: f1 – f

n = (n – 1) f

B

Sol.:1– 29 5 145

nf f

1

1– 1452

ff

f1 = 290 Hz 1

145 Hz2

f⇒ .

39. Answer (2)

Hint: 0

0

s

v vf f

v v

⎛ ⎞ ⎜ ⎟⎝ ⎠.

Sol.: When engine approaches man then frequency

is higher and constant, when engine passes man

then frequency heard lowered but remains constant.


6/16

46. Answer (2)

Hint : 1 2 3 4 5 6

2 2 3CH CH– CH – C C–CH

Hex–1–en–4–yne

47. Answer (3)

Hint : More the s-character in hybrid orbitals, more

will be the electronegativity.

Sol. : C2H

2 : sp hybridised carbon

50% s-character.

48. Answer (1)

Hint : Cyclic, planar and conjugated compounds

containing (4n + 2)e– are aromatic in nature.

[ CHEMISTRY]

Sol. : HH is non planar because of steric

repulsion of inner hydrogen.

49. Answer (4)

Hint : Alkyl group with –H atoms attach to

benzene, activates the ring for SE reactions.

Sol. : does not have –H atom.

50. Answer (2)

Hint :

2Volumeof N (atSTP)28

% of N 10022400 wt.of compound

40. Answer (3)

Hint: 0

0

s

v vf f

v v

⎡ ⎤ ⎢ ⎥⎣ ⎦

Sol.:

Motorist Band

5 ms–1

Wall

10 ms–1

Frequency of reflected sound

1

330 3301000 1000

330 –10 320f

⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

As motorist is moving toward wall

0

2 1

330 3351000

320 330

v vf f

v

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎣ ⎦

2

3351000

320f

⎡ ⎤ ⎢ ⎥⎣ ⎦

41. Answer (3)

Hint: 1

v

Sol.: Density of moist air is less than that of dry air.

42. Answer (4)

Hint: Pressure variation is maximum at displacement

nodes.

Sol.: For stationary waves, all particles oscillates in

same phase with in a loop. Stationary wave is

formed when two waves are travelling in opposite

direction.

43. Answer (1)

Hint: y = f (x ± vt)

Sol.: y = f (x + vt) when wave is moving towards

negative x-axis

y = f (x – vt) when wave is moving towards

positive x-axis

44. Answer (1)

Hint: F = –kx, k = m2.

Sol.: 2 2

2

4 2 8

k

m

2 2

2

2 2T

or, 4 2 sT

45. Answer (2)

Hint: 2m

Tk

Sol.: 2 kg

2300N/m

T

2s

150T


7/16

Sol. : 1 1 2 2

1 2

P V P V

T T

(Experiment condition) (STP condition)

2760 V(725 – 25) 60

300 273

∵ Volume of N2 (at STP) = 50.29 ml

28 50.29% of N 100 22.45%

22400 0.28

51. Answer (3)

Hint : Octet of N in 4

NH

is complete and no vacant

orbital is present on N atom.

52. Answer (3)

Hint : For tautomerism, acidic H-atom should be

present.

Sol. :

O

does not contain any acidic H-atom.

53. Answer (1)

Hint. : CH3

CH2

– –C

CH3

CH3

––

–CH2– is neohexyl group.

54. Answer (4)

Hint : HNO3 converts NaCN and Na

2S into HCN and

H2S respectively.

55. Answer (1)

Hint : More the delocalization of +ve charge, more

will be the stability of carbocation.

Sol. : CH2

+ has maximum number of

resonating structures, so most stable.

56. Answer (1)

Hint : Order of -I effect.

– CN > – COOH > – OH

57. Answer (2)

Hint : Higher the availability of lone pair on nitrogen

atom higher is its basic nature and hence higher the

pKa

58. Answer (2)

Hint : Glycerol decomposes near its boiling point.

Sol. : Compounds which decompose near its boiling

point are purified by vacuum distillation method.

59. Answer (4)

Hint : Pent–2–ene and 1-chloro-2-bromopropene

show geometrical isomerism.

60. Answer (3)

Hint : Kjeldahl’s method is not applicable to

compounds containing

(i) Nitrogen in ring.

(ii) (–NO2) and azo (– N = N –) groups.

Sol. : Nitrogen present in CN , can be

quantitatively estimated by Kjeldahl’s method.

61. Answer (3)

Hint : Photochemical smog is oxidising smog.

62. Answer (4)

Hint : N2O is a greenhouse gas.

63. Answer (4)

Hint : For highly polluted water, BOD value is more

than 17 ppm.

64. Answer (1)

Hint : H2SO

4 + H

2SO

4 SO

3 + HSO

4

–

+ H

3O+

Sol. : SO3 is an electrophile.

65. Answer (2)

Hint : Carbonium ion intermediate is formed.

Sol. :

CH = CH2

H+

H

+

H – +

CI–

shift

CI

A

66. Answer (1)

Hint : (i) O3

(ii) Zn+H O2

C=C C =O+O =C

Sol. :

2–oxo propanal

CH3 – C – CH = O + O = CH2

O+O = CH2

CH3 – C – CH = CH2

CH2

2-methylbuta-1, 3-diene

2 3 4

(C H5 8)

Methanal

Methanal

1

67. Answer (2)

Hint : When C – CI bond acquires double bond

character then reactivity of C – Cl bond decreases.

Sol. : CH2 = CH

– CI

..CH

2 –CH CI

(vinyl chloride)

– +


8/16

68. Answer (1)

Hint : 2, 3–dimethylbutane has two type of

H-atoms.

Sol. :

CI

2

h Cl+

Cl

Chiral carbon

69. Answer (4)

Hint : 1-alkynes are acidic in nature.

Sol. :

Ch C CH3 –

NaNH2

CH3– C C Na

CH3 – CH – CI

2

CH3– C CH

2 3–C – CH

Pent–2–yne

– +

70. Answer (3)

Hint : In presence of sunlight addition reaction and

in presence of FeCl3 electrophilic substitution

reaction take place.

Sol. :

+ 3CI2

Sunlight

H CI

CI H

H

CI

H

CI

CI

H

CI

H

( )C H Cl6 6 6

( ) (X)Benzene hexachloride

+ CI2

FeCI3 + HCI

CI

( ) (Y)Chlorobenzene

71. Answer (4)

Hint : More the H-bonding and lesser the torsional

strain, more will be the stability of conformers.

Sol. : Stability order:

(1) Cyclohexane : Chair > Boat > Half chair

(2) Butane (around C2 and C

3) : Anti > Gauche >

Partially eclipsed

(3) Propane : Staggered > Skew > Eclipsed

(4) Ethane -1, 2-diol: Gauche > Anti > Eclipsed

72. Answer (2)

Hint : R of R-MgX on reaction with acidic-H give

alkane.

Sol. :

CH3

CH – CH MgCI2

CH3

– +

CH –3 O – H

+–

CH3

CH3

– CH – CH

3 + Mg

CI

OCH3

73. Answer (1)

Hint : During hydration of alkene in acidic medium

carbocation is formed as intermediate.

Sol. :

CH3

CH – CH = CH2

H2 O H

CH3

C

H

CH

+CH

3

H

shift

CH3

C

CH2 CH

3

+

H2 O

–H+

CH3

C

CH2

OH

CH3

+

–

(3° alcohol)

CH3

CH

CH CH

2

(i) B2 H THF

6,

CH3

CH CH2 CH

2

=

(ii) H2O ,NaOH

2

(1° alcohol)

OH

74. Answer (4)

Hint : By Kolbe’s electrolytic method alkane, alkene

and alkyne all can be prepared.

Sol. :

2 CH3 – COONa

+ H

2 O CH

3 – CH

3 + 2CO

2

(Sodium acetate)ethane

+ 2NaOH + H 2

CH2 – COONa + 2H

2 O CH

2 + 2CO

2 + 2NaOH + H

2

CH

ethene2

CH –2

COONa

(Sodium succinate)

CH – COONa + 2H

2 O CH + 2CO

2

+ 2NaOH + H2

CHethyne

CH –

COONa

(Sodium fumarate)

75. Answer (2)

Hint : H atom attach to sp-hybridised carbon is

acidic in nature.

Sol. : CH C 3 – C– H acidic H atom.

Propyne

76. Answer (3)

Hint : Symmetrical alkene does not show

Markovnikov’s addition of HBr.


9/16

Sol. : is a symmetrical alkene.

77. Answer (3)

Hint : Propyne contains acidic hydrogen.

Sol. : 4

3

NH OH3 3AgNO

whiteppt.

CH C CH CH C CAg

4

3

NH OH3 2 AgNO

CH C CH Noreaction

78. Answer (1)

Hint : Less stable carbocation can rearrange to

more stable carbocation.

Sol. :

OH

H+

OH2

+

–H O2

CH – – CH3 2

C

CH3

CH3

+ CH3

–

shiftCH – – CH – CH

3 3 C

CH3

+

H

CH3 – C = CH – CH

3

CH3

– H+

79. Answer (3)

Hint : Less stable alkenes are more reactive towards

catalytic hydrogenation.

Sol. :

CH = CH2

has only one alpha–H

atom. So, hyperconjugation effect is small and

alkene is least stable.

80. Answer (4)

Hint : More the surface area of Isomeric alkanes,

more will be the boiling point.

Sol. : As the branching in isomeric alkane increases

their surface area decreases so boiling point

decreases

BP : > >

81. Answer (4)

Hint :

3 C H2 2

Red Hot

Fe tube

Ethyne 873 K

COONa + NaOHCaO

+ Na CO2 3

Sodium

benzoate

OH + Zn

Phenol

+ ZnO

82. Answer (3)

Hint :

CH – C C – CH3 3

Lindlar’s catalyst

Na in liq. NH

3

C = C

CH3

HH

CH3

(Cis – but – 2 – ene)

C = CH

H

CH3

(trans – but – 2 – ene)

CH3

(A)

(B)

Sol. : For A, 0

For B, = 0

83. Answer (2)

Hint : Nitration takes place in that benzene ring

which has more electron density.

Sol. :

C – O

O

Conc. HNO3

Conc. H SO2 4

C – O

O

NO2

(Major)

(–M) (+M)

84. Answer (1)

Hint : More the s-character in carbanion, more will

be stability.

Sol. : In ethyne carbanion (CH C–), carbon is sp

hybridised which has more s-character (50%), in

other words can hold negative charge more better, so

more stable.

85. Answer (3)

Hint : The degree of unsaturation (DOU) for C4H

6 is

two.

Sol. : ⇒C H (DOU = 3)4 4


10/16

[ BIOLOGY]

91. Answer (4)

Hint: ATP is synthesized during respiration.

Sol.: ATP acts as the energy currency of the cell.

The energy released by oxidation of food in

respiration is used to synthesise ATP and not used

directly.

92. Answer (4)

Hint: Plants have lesser demands for gaseous

exchange as compared to animals.

Sol.: Roots, stems and leaves all respire at lower

rates than animals. In plants, cells are closely

packed and located quite close to the surface of the

plant. Hence it is easy for gases to diffuse inside.

93. Answer (3)

Hint: Partial oxidation of glucose without utilization

of oxygen is called glycolysis.

Sol.: Glycolysis takes place in all organisms in the

cytoplasm. Hence all organisms whether anaerobes

or aerobes retain enzymes for glycolysis.

94. Answer (1)

Sol.: The term glycolysis has originated from two

Greek words that are ‘glycon’ which means sugar

and ‘lysis’ which means splitting.

95. Answer (2)

Hint: Lactic acid (3C) is produced from pyruvic acid

which is also a 3C compound.

Sol.: In lactic acid fermentation pyruvic acid (3C) is

converted into lactic acid (3C). Hence there is no

release of CO2 which occurs in alcohol fermentation.

Both processes are hazardous and produce 2 ATP

as net gain.

96. Answer (4)

Hint: Krebs’ cycle starts with the condensation of

2C compound with 4C compound to yield a 6C

compound.

Sol.: Krebs’ cycle starts with condensation of Acetyl

CoA (2C) with oxaloacetic acid (4C) to yield citric

acid (6C).

97. Answer (2)

Hint: It is a 2C compound.

Sol.: Acetyl CoA is common intermediate among

fats, proteins and carbohydrates breakdown during

aerobic respiration. It enters into Krebs’ cycle to

yield energy.

98. Answer (4)

Hint: In aerobic respiration CO2 is released in link

reaction & during Krebs’ cycle.

Sol.: In link reaction CO2 is released at one step.

During Krebs’ cycle CO2 is released at two steps.

In case of anaerobic condition

2

2

2CORQ (infinite)

0O

86. Answer (4)

Hint : Order of deactivation is :

– NO2 > – CN > – COCH

3 > – F

Sol. :

87. Answer (3)

Hint : 4np electrons in a planar ring makes the ring

antiaromatic.

88. Answer (4)

Hint : 31.4×meq.ofNH

% of N =wt.of sample

Sol. : meq of NH3 evolved = meq. of H

2SO

4 used

= N × V(ml) = 0.05 × 40 = 2

31.4×meq.ofNH 1.4 2

% of N = 14%wt.of sample. 0.2

89. Answer (4)

Hint : More the stability of conjugate base formed,

more will be acidic nature of hydrocarbon.

Sol. :

+ H

(More acidic) (Conjugate base)(Aromatic)More stable

H –

+

90. Answer (1)

Hint : C – H bond dissociation energy in

CH = CH – CH2 2

H

is slightly less than bond

dissociation energy of C – H in CH2

H

Sol. : Stability order :

CH2

CH2

CH – – CH3 2

C

CH3

CH3

CH3


11/16

99. Answer (3)

Hint: Yeast performs alcoholic fermentation.

Sol.: In alcoholic fermentation alcohol is produced

along with CO2 & ATP.

100. Answer (1)

Hint: Complete oxidation of acetyl CoA occurs in

aerobic respiration through Krebs’ cycle.

Sol.: One turn of Krebs’ cycle involves complete

breakdown of acetyl CoA which produces 3NADH +

H+, 1 FADH2 & 1 GTP.

3 NADH + H+ = 3 × 3 = 9 ATP

1 FADH2 = 1 × 2 = 2 ATP

1 GTP = 1 × 1 = 1 ATP

12 ATP

NADH + H+ & FADH2 produce ATP through ETS.

101. Answer (2)

Hint: The phytohormone which was extracted from a

fungus Gibberella fujikuroi is responsible for bolting.

Sol.: Gibberellins are active substances which

induce stem elongation in rossette plants. This

phenomenon is called bolting.

102. Answer (4)

Hint: The plant hormone which is responsible for

closure of stomata is known as stress hormone.

Sol.: ABA is called stress hormone as its synthesis

is stimulated by drought, water logging & other

adverse environmental conditions.

103. Answer (3)

Sol.: Auxin helps in root formation on stem cuttings,

shows apical dominance & phototropism but does

not play any role to speed up the malting process

which is done by gibberellin.

104. Answer (1)

Hint: Ethylene is a volatile substance which

promotes senescence.

Sol.: Ethylene is the only gaseous hormone of

plants. It breaks the dormancy of seeds.

Bakanae disease is caused by gibberellins.

105. Answer (4)

Hint: The continued growth due to activity of

meristems occurs in plants.

Sol.: New cells are always being added due to

activity of meristem, this type of growth is open form

of growth and indeterminate.

Animals show definite & diffused type of growth.

106. Answer (4)

Hint: Tissues formed from de-differentiation of

parenchyma are secondary in origin.

Sol.: Wound cambium, cork cambium &

interfascicular cambium are secondary in origin. They

are formed by de-differentiation of parenchyma cells.

Intrafascicular cambium is primary in origin.

107. Answer (1)

Hint: This hormone is a plant growth inhibitor.

Sol.: ABA is a derivative of carotenoids.

108. Answer (2)

Hint: Relative growth can be calculated by the

formula i.e. Growth per unit time

100Initial size

.

Sol.: For Leaf A

5100 50%

10

For Leaf B

5100 20%

25

109. Answer (3)

Hint: Rice & maize show hypogeal seed

germination.

Sol.: In hypogeal seed germination, epicotyl grows

first & cotyledons remain underground.

110. Answer (3)

Hint: Short day plant flowers after receiving

photoperiod shorter than critical period.

Sol.: Tobacco is a short day plant. Wheat, radish &

sugarbeet are long day plants.

111. Answer (1)

Hint: Phytochrome is a photoreceptor.

Sol.: Phytochrome is a blue-green pigment,

regulates photoperiodic induction of flowering and

helps in seed germination too.

It occurs in two forms Pr & P

fr. Its P

r form becomes

active by absorbing far-red light. Pfr is physiologically

active form.

112. Answer (4)

Hint: Auxin occurs in growing apices.

Sol.: If apical bud is removed then auxin is removed,

due to which apical dominance is prevented.

The plant will show more lateral branches & axillary

buds.


12/16

113. Answer (3)

Sol.: The stimulus of vernalisation is perceived by the

mature stem apex or by the embryo of the seed.

114. Answer (2)

Hint: Gibberellin was first isolated from a fungus

called Gibberella fujikuroi.

Sol.: It is acidic in nature, increases sugarcane yield

by increasing the length of internodes, delays

ripening of fruits and helps in seed germination.

115. Answer (3)

Sol.: Buttercup shows environmental (extrinsic)

heterophylly. Cotton, coriander & larkspur show

developmental heterophylly due to intrinsic factors.

116. Answer (3)

Hint: Inner membrane of mitochondria forms cristae.

Oxidative phosphorylation occurs during oxidation of

food.

Sol.: Cristae contains cytochromes which helps in

ETS.

During formation of bread, yeast if used performs

alcoholic fermentation which releases CO2.

117. Answer (2)

Hint: The F1 headpiece contains the site of the ATP

synthesis.

Sol.: F1 headpiece remains present towards the

matrix of the mitochondria. Hence ATP synthesis

takes place towards the matrix.

118. Answer (3)

Hint: Krebs’ cycle produces reduced coenzymes.

Sol.: NADH + H+ is produced during Krebs’ cycle

along with FADH2 & GTP.

It starts with condensation of acetyl CoA with OAA

to form citric acid.

119. Answer (3)

Hint: Cytochrome c is a mobile carrier protein.

Sol.: Cytochrome c is not the part of complex IV. It

is attached to the outer surface of the inner

membrane & transfers electrons between complex III

& IV.

120. Answer (4)

Hint: Tryptophan is precursor of auxin.

Sol.: Tryptophan is an amino acid. Final product of

transition/link reaction of aerobic respiration is acetyl

CoA which is precurssor of gibberellin.

A pigment violaxanthin is a precursor of ABA.

121. Answer (3)

Sol.: 2, 4-D is widely used as weedicide in cereal

crops.

122. Answer (1)

Hint: Cytokinin(CK) helps in production of new

leaves.

Sol.: It (CK) enhances chloroplast production in

leaves. It delays senescence by controlling protein

synthesis.

123. Answer (2)

Sol.: Cytokinin along with auxin are essential in

tissue culture. High cytokinin to auxin ratio causes

shoot formation in callus.

124. Answer (3)

Hint: Coconut milk has a phytohormone which helps

in cell division.

Sol.: It contains cytokinin.

125. Answer (3)

Hint: Antagonistic means showing opposite effects

towards something.

Sol.: When two hormones bring about opposite

effects on the same function, it means they are

antagonistic to each other, whereas when they act

together they are called synergistic.

126. Answer (4)

Hint: Short day plants require uninterrupted dark

period to flower.

Sol.: SDP will not flower if its dark period is

interuppted by a flash of light.

127. Answer (3)

Sol.: Gibberellin can replace vernalisation. As by

exogenous application of gibberellins many biennials

can be induced to behave as annuals.

128. Answer (3)

Hint: ABA is a plant growth inhibiting phytohormone.

Sol.: ABA induces dormancy of buds and seeds &

promotes abscission of flowers.

129. Answer (4)

Hint: It is a gaseous hormone.

Sol.: Ethylene is responsible for ripening of fruits. It

can be used artificially to ripen the fruits.

130. Answer (2)

Hint: Auxin gets removed along with apical buds of

a plant.

Sol.: Auxin was first time isolated from human urine.

It is an amino acid derivative.

131. Answer (4)

Sol.: Growth curve in plants is sigmoid, also called

S-shaped growth curve.


13/16

132. Answer (3)

Hint: Pr absorbs red light.

Sol.: 660 nm is red light.

Pfr absorbs far red light.

133. Answer (3)

Hint: During dormancy, growth & development of

seed is temporarily stopped.

Sol.: In dormancy, seeds cannot germinate inspite

of availability of all suitable environmental conditions.

134. Answer (3)

Hint: In link reaction pyruvic acid (3C) forms acetyl

CoA(2C) & CO2.

Sol.:

4C compound – Malic acid

3C compound – Pyruvic acid

2C compound – Acetyl CoA

5C compound – -ketoglutaric acid

135. Answer (2)

Hint: RQ of fat is less than 1 and of organic acids

it is more than 1.

Sol.:

Tripalmitin is a fatty acid (RQ = 0.7)

Proteins (RQ = 0.9)

Malic acid (RQ = 1.33).

136. Answer (3)

Hint : In sponges, the body cavity opens to exterior

through osculum.

Sol. : All the given characters are shown by

sponges. They are not considered as diploblastic

and they have cellular grade of body organisation.

The cavity inside body is known as spongocoel or

paragastric cavity.

137. Answer (1)

Hint : This animal shows internal metamerism.

Sol. : In annelids, both external and internal

segmentation is found in adult stage

In arthropods, only external segmentation is present

in adults. Chordates show internal segments in

certain body parts.

138. Answer (4)

Hint : Animal which is bilaterally symmetrical in larval

stage but becomes asymmetrical due to torsion in

adult life.

Sol. : Adult snails are asymmetrical due to torsion.

In larval stage, it is bilaterally symmetrical. It is a

type of retrograde metamorphosis.

139. Answer (3)

Hint : Obelia exhibits alternation of generation.

Sol. : In class Hydrozoa, all members except Hydra

show metagenesis. Metastasis is spread of cancer,

metamorphosis is the development of larval stage

into adult stage.

140. Answer (4)

Hint : The roundworm which is found attached with

intestinal wall.

Sol. : • Euspongia – Bath sponge

• Meandrina – Brain coral

• Fasciola – Liver fluke

141. Answer (4)

Hint : Chordates have hollow, dorsal nerve cord

Sol. : Arthropoda – Chitinous exoskeleton

Mollusca – Calcareous exoskeleton

Echinodermata – Spiny skinned, ventral

solid nerve cord

Chordata – Presence of notochord

142. Answer (4)

Hint : This process helps in replacement of lost

body parts.

Sol. : With the help of water vascular system

echinoderms perform locomotion, excretion,

respiration and capturing of food.

143. Answer (2)

Hint : Capillaries are absent in Hemichordates

Sol. : In leech and Balanoglossus, open type of

circulatory system is present.

144. Answer (3)

Hint : Platyhelminthes are acoelomates.

Sol. : Protostomic animals are schizocoelic

eucoelomates and deuterostomic animals are

enterocoelic eucoelomates. Only arthropods placed

in class Insecta are considered as insects.

145. Answer (3)

Hint : Identify common Indian bull frog.

Sol. : Rana tigrina is common Indian bull frog.

Alytes is commonly known as midwife toad because

male carries eggs wrapped around its hind limbs.

146 Answer (2)

Hint : Members of Class Osteichthyes.

Sol. : Air bladder/swim bladder is found in bony

fishes to provide buoyancy which is helpful in

swimming. They do not to have continuously swim

as in case of cartilaginous fishes which lack swim

bladder.


14/16

147. Answer (4)

Hint : The first segment bears mouth and a lobe that

can act as a wedge.

Sol. : The first body segment of earthworm is known

as peristomium. Prostomium is a hood like structure

on peristomium that is helpful in burrowing. The fecal

deposits of earthworm are known as worm castings.

Vermicomposting is use of worm castings as

manure.

148. Answer (3)

Hint : Members of this class show aerial

adaptations.

Sol. : In birds, long bones of the body contain

cavities filled with air to decrease weight of the body.

Such type of bones found in birds are known as

pneumatic bones.

149. Answer (3)

Hint : Fish like organisms having circular mouth.

Sol. : Cyclostomes (Petromyzon) are animals

without scales having circular mouth. They have non

operculated gills and lack paired fins. Scoliodon

(cartilaginous fish) and Hippocampus and Labeo

(bony fishes) have scales on their body.

150. Answer (3)

Hint : Common name for Hemichordates.

Sol. : • Trygon – Poison sting (Sting ray)

• Torpedo – Muscle modified into

electric organ (Electric

eel)

• Tongue worm – Proboscis present not

placoid scales

• Hippocampus – Brood pouch in male to

carry fertilized eggs.

151. Answer (2)

Hint : Common name for Antedon.

Sol. :

Metridium – sea anemone – Coelenterata

Antedon – sea lily – Echinodermata

Hormiphora – sea walnut – Ctenophora

Pennatula – sea pen – Coelenterata

152. Answer (3)

Hint : Number of moults is equivalent to the number

of chambers in its heart.

Sol. : In cockroach, nymph is a young stage which

is similar to cockroach but smaller in size, light in

colour, with immature gonads and without wings. It

undergoes gradual metamorphosis known as

paurometabola in 6 months upto two years. During

this time, it moults 13 times, to become sexually

mature adult.

153. Answer (1)

Hint : Animals which are commonly known as

roundworms are classified in this phylum.

Sol. : Aschelminthes/Nematodes are considered as

first animals to have complete digestive tract with

distinct mouth and anus.

154. Answer (3)

Hint : Most vertebrates are poikilothermous except

birds and mammals generally.

Sol. : Testudo is a reptile and cannot maintain its

body temperature, hence considered as

poikilothermic/cold blooded animal.

Columba is a bird while Felis and Macropus are

mammals. So, all these are warm-blooded animals.

155. Answer (2)

Hint : Larval stage which causes taeniasis in

humans.

Sol. : In fecal matter of infected man, onchosphere

is present which is converted into hexacanth larva

and finally cysticercus larva/bladderworm in muscles

of pig. Human beings are infected when they feed on

undercooked measly pork and this larva in adult

develops into Taenia solium causing taeniasis.

156. Answer (4)

Hint : Identify a roundworm whose larvae are spirally

coiled in muscles.

Sol. : Trichinella is an endoparasitic, viviparous

roundworm in rodents, pig, humans etc. causing

disease trichinosis characterised by abdominal pain

and muscular inflammation.

157. Answer (4)

Hint : Most primitive lizard named as Tua-tara.

Sol. : Sphenodon is the most primitive lizard,

considered as living fossil but not as connecting link.

Peripatus is placed in taxon-Onychophora and

Neopilina is a connecting link between molluscs and

annelids.

158. Answer (3)

Hint : Mammal with avian characters.

Sol. : Ornithorhynchus – Duckbilled platypus

Pinctada – Pearl oyster

Pteropus – Flying fox

Balanoglossus – Tongue worm

159. Answer (4)

Hint : Flagellated cells of sponges are also named

as collar cells.

Sol. : Choanocytes or collar cells are found in inner

lining of spongocoel as well as flagellated chambers

of sponges. They continuously produce water current

for uptake of food, respiration, excretion and

reproduction.


15/16

160. Answer (1)

Hint : Syngamy occurs inside female’s body.

Sol. : Cartilaginous fishes show internal fertilization

and ovoviviparous condition.

161. Answer (4)

Hint : Fish like organism with cartilaginous cranium

and vertebral column.

Sol. : In cyclostomes, such as Myxine, scales and

tympanum are absent. Heart is two chambered.

Fishes have membranous labyrinth. All the given

characters belong to members of class Amphibia.

162. Answer (4)

Hint : These structures are sensitive to sound and

other vibrations.

Sol. : Anal cerci are jointed filamentous structures

present on 10th tergum. Anal style is unsegmented

spiny structure on 9th sternum present only in male

cockroach.

163. Answer (3)

Hint : Vocal cords are sound producing structures in

frogs.

Sol. : Male frog has sound producing vocal cords

and vocal sacs make the sound louder.

Cockroach can survive upto 7 days with amputated

head because most of its nervous tissue is ventrally

placed in belly region.

164. Answer (3)

Hint : This organism can live on both land and water.

Sol. : Bidder’s canal present in kidney is responsible

for conduction of sperms

Hirudinaria – nephridia as excretory organ

Locusta – jointed legs for locomotion

Ctenoplana – cnidoblasts are absent

165. Answer (3)

Hint : These structures are associated with

segmented worms.

Sol. : Parapodia are associated with annelid like

Nereis. They help in locomotion and respiration.

166. Answer (1)

Hint : Chemoreceptors which taste chemical nature

of water

Sol. : In molluscs, osphradium in mantle cavity

serves to test chemical nature of water.

167. Answer (3)

Hint : A – Fastest running mammal.

B – Bird exhibiting polygamy.

Sol. : Panthera tigris (tiger) is considered as national

terrestrial animal and Pavo cristatus (Peacock) is

national bird of India.

168. Answer (3)

Hint : Respiratory structures present on body from

2nd thoracic segment onwards.

Sol. : In cockroach, 10 pair of spiracles are present

in between sclerites for gaseous exchange during

respiration.

169. Answer (3)

Hint : Fighting fish is a bony fish.

Sol. :

Earthworm – Nephridia – Act as excretory

structures

Cockroach – Titillator – Helps in copulation

present in

males only

Ascaris – Pineal setae – In males for

copulation

170. Answer (2)

Hint : Such type of metamorphosis is found in silk

moth.

Sol. : Silk moth develops through larval and pupal

stages. Conversion of larva into pupa is known as

pupation, characterised by histolysis followed by

histogenesis.

171. Answer (2)

Hint : Flatworms have solid body without any cavity.

Sol. : Flatworms are bilaterally symmetrical,

triploblastic, dorsoventrally flattened and acoelomate.

Taenia solium has no alimentary canal. It absorbs

predigested nutrients from intestine of host.

172. Answer (1)

Hint : Wuchereria is a representative of these

worms.

Sol. : Being circular in cross-section, these animals

are commonly known as roundworms. Sexual

dimorphism evolved first in roundworms. They have

muscular pharynx and both free living and parasitic

forms are present.

173. Answer (4)

Hint : Identify an animal exhibiting epitoky

Sol. : Nereis is an aquatic unisexual annelid.

Parapodia are lateral unsegmented appendages for

swimming and respiration. Body is metamerically

segmented, development is indirect through

trochophore larval stage.


16/16

� � �

174. Answer (3)

Hint : In molluscs, mantle cavity containing feathery

gills is present between mantle and visceral hump.

Sol. : Soft and spongy skin present just below shell

is known as mantle. Space between mantle and

visceral hump is known as mantle cavity. Feather like

gills responsible for both respiration and excretion

are present in mantle cavity.

175. Answer (3)

Hint : Shell is a calcareous exoskeleton found

around most molluscs.

Sol. : Calcareous spicules form endoskeleton in

sponges. Calcareous ossicles constitute

endoskeleton in sea stars and cartilaginous

endoskeleton is found in cartilaginous fishes.

Dentalium has shell as calcareous exoskeleton.

176. Answer (1)

Hint : Blood glands and tufts of pharyngeal nephridia

are present in same set of body segments.

Sol. :

Typhlosole is a fold in intestinal wall to increase

absorptive surface area. Ovaries with ovarioles are

situated between 2nd to 6th abdominal segments in

female cockroach.

177. Answer (4)

Hint : Brood pouch harbours mammary nipples.

Sol. : In Balanoglossus, indirect development is

seen due to presence of larval stage. In female

Macropus (kangaroo), a mammal, mammary nipples

are present in brood pouch (marsupium).

178. Answer (1)

Hint : Mouth parts which contain chitinous teeth

known as denticles.

Sol. : In cockroach, mandibulate type of mouth parts

are found which are responsible for grinding and

incising food.

179. Answer (4)

Hint : Commonly called star fish.

Sol. : Pheretima has nephridia for excretion and is

an annelid. Interstitial cells are present in the body

wall of coelenterates. Ctenoplana belongs to

Ctenophora.

180. Answer (3)

Hint : In non-chordates, heart beat is initiated by

hormone from brain.

Sol. : In non-chordates, heart is either absent or

present dorsally. If present, it is generally neurogenic

because heart beat is initiated by a ganglia situated

near heart. In myogenic heart, heart beat is initiated

by specialized muscular tissue situated in wall of

heart known as S.A node/pace maker of the heart.

Test - 6 (Code-D) (Answers) All India Aakash Test Series for Medical-2020

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1. (2)

2. (1)

3. (1)

4. (4)

5. (3)

6. (3)

7. (2)

8. (2)

9. (2)

10. (1)

11. (1)

12. (1)

13. (4)

14. (1)

15. (2)

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28. (4)

29. (1)

30. (1)

31. (3)

32. (3)

33. (3)

34. (3)

35. (1)

36. (3)

Test Date : 17/02/2019

ANSWERS

TEST - 6 (Code-D)

All India Aakash Test Series for Medical-2020

37. (4)

38. (1)

39. (1)

40. (3)

41. (2)

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173. (2)

174. (4)

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177. (3)

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180. (3)

All India Aakash Test Series for Medical-2020 Test - 6 (Code-D) (Answers & Hints)

2/16

ANSWERS & HINTS

1. Answer (2)

Hint: 2m

Tk

Sol.: 2 kg

2300N/m

T

2s

150T

2. Answer (1)

Hint: F = –kx, k = m2.

Sol.: 2 2

2

4 2 8

k

m

2 2

2

2 2T

or, 4 2 sT

3. Answer (1)

Hint: y = f (x ± vt)

Sol.: y = f (x + vt) when wave is moving towards

negative x-axis

y = f (x – vt) when wave is moving towards

positive x-axis

4. Answer (4)

Hint: Pressure variation is maximum at displacement

nodes.

Sol.: For stationary waves, all particles oscillates in

same phase with in a loop. Stationary wave is

formed when two waves are travelling in opposite

direction.

5. Answer (3)

Hint: 1

v

Sol.: Density of moist air is less than that of dry air.

6. Answer (3)

Hint: 0

0

s

v vf f

v v

⎡ ⎤ ⎢ ⎥⎣ ⎦

[ PHYSICS]

Sol.:

Motorist Band

5 ms–1

Wall

10 ms–1

Frequency of reflected sound

1

330 3301000 1000

330 –10 320f

⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

As motorist is moving toward wall

0

2 1

330 3351000

320 330

v vf f

v

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎣ ⎦

2

3351000

320f

⎡ ⎤ ⎢ ⎥⎣ ⎦

7. Answer (2)

Hint: 0

0

s

v vf f

v v

⎛ ⎞ ⎜ ⎟⎝ ⎠.

Sol.: When engine approaches man then frequency

is higher and constant, when engine passes man

then frequency heard lowered but remains constant.

8. Answer (2)

Hint: f1 – f

n = (n – 1) f

B

Sol.:1– 29 5 145

nf f

1

1– 1452

ff

f1 = 290 Hz 1

145 Hz2

f⇒ .

9. Answer (2)

Hint: nl = constant

Sol.: 2

1

20 4

25 5

A

B

n l

n l n

B > n

A.

So, nB – n

A = 5 on solving,

nA = 20

nB = 25

Test - 6 (Code-D) (Answers & Hints) All India Aakash Test Series for Medical-2020

3/16

10. Answer (1)

Hint: Use Newton’s formula P

v

Sol.: According to Laplace’s correction

v v Percentage error

v v 1.4 1100 100%

v 1.4

15%�

11. Answer (1)

Hint: 1 1

2 2

f v

f v

Sol.: 1 1 2

2 2 1

, If Constantf v m

Tf v m

1

2

600 2 1

32 4

f

f f

f = 2400 Hz

12. Answer (1)

Hint: v

f

Sol.:330 1

m 50 cm660 2

3 512.5, 37.5, 62.5

4 4 4

(Not possible)

Only 1st and 2nd resonance can be produced.

So minimum height of water

hmin

= 50 – 37.5 = 12.5 cm

13. Answer (4)

Hint: y = 2A sinkx cost

Sol.: y = 2A sinkx cost

2sin cos(90 )30

xy t

⎛ ⎞ ⎜ ⎟⎝ ⎠

2

30

= 60 cm

30 cm2

Number of loops = 10

Number of nodes = 10 + 1 = 11

Number of antinodes = 10

Amplitude of x = 5 cm

2sin 5 1cm30

14. Answer (1)

Hint: 1 2 3

1 2 3

1 1 1: : : :n n n

l l l

Sol.:1 2 3: : 1: 2 :1n n n

1 2 3: : 2 :1: 2l l l

l1 + l

2 + l

3 = 150 cm 5K = 150 K = 30.

So, l1 = 60 cm

l2 = 30 cm

l3 = 60 cm

so, x1 = 60 cm

x2 = 90 cm

15. Answer (2)

Hint & Sol.: Sound travel fastest in solids.

16. Answer (4)

Hint: Stress

v .

Sol.: f0 = 100 Hz, v = f

0

0 = 200 m/s

Now, 2Stress v .

Stress = (200)2 × 8 × 103 = 3.2 × 108 N/m2

17. Answer (1)

Hint: 0

Source–

vf f

v v

⎛ ⎞ ⎜ ⎟

⎝ ⎠

Sol.:350

5500 5000350 – v

⎛ ⎞ ⎜ ⎟⎝ ⎠

11 350

10 350 – v

350 – 11v = 0

350m/s

11v

18. Answer (3)

Hint: End correction, e = 0.6r

Sol.:1 2

3

4( ) 4( )

v vf

l e l e

3(l1 + e) = l

2 + e

2 1– 3

2

l le

72.6 – 720.3 0.6

2e r

3 1cm

6 2r


4/16

19. Answer (4)

Hint: m m

a a

v

v

Sol.:1650 5

330 1

m

a

v

v

5

1

m

a

5

m a

20. Answer (4)

Hint: fB = f

1 ~ f

2

Sol.: f = 4 5t = 1st 2nd

fB

4 5

21. Answer (1)

Hint: Only odd harmonic in closed organ pipe.

Sol.: f1 : f

2 : f

3 = 5 : 7 : 9

So pipe is closed

So, fundamental frequency 0

28557 Hz

5f

0

0

380 20m

57 3

v

f

22. Answer (4)

Hint: Y

v

Sol.: 9

3 –1

3

8 102 10 m s

2 10

Yv

L

5

4

O

In second overtone

5

4L

41.6 m

5

L

3

32 101.25 10 1250 Hz

1.6

vf

23. Answer (3)

Hint: Wavelength depends on medium.

Sol.: (1) Frequency does not change by changing

the medium of propagation.

(2) If wave is reflected from denser medium,

the phase change is .

24. Answer (1)

Hint: Beat frequency = |f1 – f

2|.

Sol.: 1 = 1020 f

1 = 510 Hz.

2 = 1004 f

2 = 502 Hz

fb = |f

1 – f

2| = 8 Hz

25. Answer (2)

Hint: Amplitude = 2Asinkx

Sol.: Amplitude = 4sinx ...(i)

22m

⇒

...(ii)

at2m,

3 3x

2Amplitude 4sin 2 3 m

3

26. Answer (2)

Hint: Resultant amplitude formula.

Sol.: When amplitude of component waves are

equal, amplitude of resultant, 2 cos2

R A⎛ ⎞ ⎜ ⎟

⎝ ⎠

Amplitude, 3

2 cos 2 36 2

R A A A

27. Answer (4)

Hint: T

v

Sol.: Stress

v

11 –5

3

1.6 10 1.6 10 20

8 10

Yv

4

–11.6 1.6 2 1080 m s

8

28. Answer (4)

Hint: Pitch is related to frequency of sound.

Sol.: Quality is the sensation of human ear due to

wave form of sound.

29. Answer (1)

Hint: P RT

vM

Sol.: If temperature is constant v will remain

constant.

Since pressure and density will change in the same

ratio.


5/16

30. Answer (1)

Hint: B

v

Sol.: 2

B

v

or, 2 2 –6 –2 2

8 3

1 1

8 10 (2.5 10 )

2 10 kg/m

B

v kv

31. Answer (3)

Hint: Fundamental frequency, 0

4

vf

l

Sol.:0

330110 Hz

4 0.75f

only odd harmonic are produced in closed pipe.

So frequencies 330 Hz, 550 Hz, 770 Hz are

possible. 660 Hz is not possible.

32. Answer (3)

Hint: = kx

Sol.:2

x

0.52

k

[0.5 – 0.25]2 8

33. Answer (3)

Hint:p

yv

t

Sol.: y = 20 sin(100 t – 2x)

20 100 cos(100 – 2 )p

yv t x

t

3 12 10 cos 100 0 – 2

6p

v⎡ ⎤ ⎢ ⎥⎣ ⎦

3 312 10 10 cm/s

2p

v⎛ ⎞ ⎜ ⎟⎝ ⎠

34. Answer (3)

Hint: 0

1

2

Tf

l

Sol.:10

20

If 100 N100 1

169 N20 169 1.3

⎡ ⎤ ⎢ ⎥ ⎣ ⎦

Tf

Tf

0.3f0 = 20

0

200Hz

3f

35. Answer (1)

Hint and Sol.: Open end of a closed organ pipe is

displacement antinode and pressure node.

36. Answer (3)

Hint: (vp)max

= A

Sol.:1

2

p wv v

0

1(2 )

2y f f

= 4y0

37. Answer (4)

Hint: T

v xg

Sol.: v xg

1

22

dv ga v xg g

dx x (constant)

So a x°

38. Answer (1)

Hint: 2

20

md x dxb kxdtdt

Sol.:2

20

md x dxb kxdtdt

...(i)

2

22 0d x dx

xdt dt

This is equation of damped oscillation.

39. Answer (1)

Hint: dx

vdt

Sol.: 10 sin5

v t

x vdt ∫

–10 cos5

5

t C

x

–50cos5

x t C

at t = 0, v = 0, x = –50 so, C = 0

50cos5

x t⎛ ⎞ ⎜ ⎟

⎝ ⎠


6/16

46. Answer (1)

Hint : C – H bond dissociation energy in

CH = CH – CH2 2

H

is slightly less than bond

dissociation energy of C – H in CH2

H

Sol. : Stability order :

CH2

CH2

CH – – CH3 2

C

CH3

CH3

CH3

[ CHEMISTRY]

40. Answer (3)

Hint: 2m

Tk

Sol.: 3

2 22 2

3

m mT

kk

⎛ ⎞⎜ ⎟⎝ ⎠

41. Answer (2)

Hint: vmax

= ASol.: x = Asint

(10 cm)sin2

x t

–1 –1

max(10 cm) rad s 5 cm s

2v

42. Answer (1)

Sol.: Time taken from x = A to is .2 6

A Tx

x = Acost

2cos 3

2

AA

T

1 2cos 3

2 T

23

3 T

T = 18 s

43. Answer (4)

Hint: y = Asin(t + )

Sol.: The angle at time 2

( )4

t tT

.

The projection on y axis

2 2( ) sin sin

4 120 4

⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

y t R t R tT

sin60 4

y R t ⎡ ⎤ ⎢ ⎥

⎣ ⎦

44. Answer (4)

Hint: If function increases or decreases

monotonically with time, function is non-periodic.

Sol.: (1) y = cos3t = 13cos – cos3

4t t

it is periodic but not SHM

(2) 2 sin SHM4

⎛ ⎞ ⎜ ⎟⎝ ⎠

y t

(3) y = cos2t, it is SHM

(4) y = aet is non periodic

45. Answer (1)

Hint: The least time interval after which particle

repeats its motion is called time period.

Sol.: Least time for given combination is 2

47. Answer (4)

Hint : More the stability of conjugate base formed,

more will be acidic nature of hydrocarbon.

Sol. :

+ H

(More acidic) (Conjugate base)(Aromatic)More stable

H –

+

48. Answer (4)

Hint : 31.4×meq.ofNH

% of N =wt.of sample


7/16

Sol. : meq of NH3 evolved = meq. of H

2SO

4 used

= N × V(ml) = 0.05 × 40 = 2

31.4×meq.ofNH 1.4 2

% of N = 14%wt.of sample. 0.2

49. Answer (3)

Hint : 4np electrons in a planar ring makes the ring

antiaromatic.

50. Answer (4)

Hint : Order of deactivation is :

– NO2 > – CN > – COCH

3 > – F

51. Answer (3)

Hint : The degree of unsaturation (DOU) for C4H

6 is

two.

Sol. : ⇒C H (DOU = 3)4 4

52. Answer (1)

Hint : More the s-character in carbanion, more will

be stability.

Sol. : In ethyne carbanion (CH C–), carbon is sp

hybridised which has more s-character (50%), in

other words can hold negative charge more better, so

more stable.

53. Answer (2)

Hint : Nitration takes place in that benzene ring

which has more electron density.

Sol. :

C – O

O

Conc. HNO3

Conc. H SO2 4

C – O

O

NO2

(Major)

(–M) (+M)

54. Answer (3)

Hint :

CH – C C – CH3 3

Lindlar’s catalyst

Na in liq. NH

3

C = C

CH3

HH

CH3

(Cis – but – 2 – ene)

C = CH

H

CH3

(trans – but – 2 – ene)

CH3

(A)

(B)

Sol. : For A, 0

For B, = 0

55. Answer (4)

Hint :

3 C H2 2

Red Hot

Fe tube

Ethyne 873 K

COONa + NaOHCaO

+ Na CO2 3

Sodium

benzoate

OH + Zn

Phenol

+ ZnO

56. Answer (4)

Hint : More the surface area of Isomeric alkanes,

more will be the boiling point.

Sol. : As the branching in isomeric alkane increases

their surface area decreases so boiling point

decreases

BP : > >

57. Answer (3)

Hint : Less stable alkenes are more reactive towards

catalytic hydrogenation.

Sol. :

CH = CH2

has only one alpha–H

atom. So, hyperconjugation effect is small and

alkene is least stable.

58. Answer (1)

Hint : Less stable carbocation can rearrange to

more stable carbocation.

Sol. :

OH

H+

OH2

+

–H O2

CH – – CH3 2

C

CH3

CH3

+ CH3

–

shiftCH – – CH – CH

3 3 C

CH3

+

H

CH3 – C = CH – CH

3

CH3

– H+


8/16

59. Answer (3)

Hint : Propyne contains acidic hydrogen.

Sol. : 4

3

NH OH3 3AgNO

whiteppt.

CH C CH CH C CAg

4

3

NH OH3 2 AgNO

CH C CH Noreaction

60. Answer (3)

Hint : Symmetrical alkene does not show

Markovnikov’s addition of HBr.

Sol. : is a symmetrical alkene.

61. Answer (2)

Hint : H atom attach to sp-hybridised carbon is

acidic in nature.

Sol. : CH C 3 – C– H acidic H atom.

Propyne

62. Answer (4)

Hint : By Kolbe’s electrolytic method alkane, alkene

and alkyne all can be prepared.

Sol. :

2 CH3 – COONa

+ H

2 O CH

3 – CH

3 + 2CO

2

(Sodium acetate)ethane

+ 2NaOH + H 2

CH2 – COONa + 2H

2 O CH

2 + 2CO

2 + 2NaOH + H

2

CH

ethene2

CH –2

COONa

(Sodium succinate)

CH – COONa + 2H

2 O CH + 2CO

2

+ 2NaOH + H2

CHethyne

CH –

COONa

(Sodium fumarate)

63. Answer (1)

Hint : During hydration of alkene in acidic medium

carbocation is formed as intermediate.

Sol. :

CH3

CH – CH = CH2

H2 O H

CH3

C

H

CH

+CH

3

H

shift

CH3

C

CH2 CH

3

+

H2 O

–H+

CH3

C

CH2

OH

CH3

+

–

(3° alcohol)

CH3

CH

CH CH

2

(i) B2 H THF

6,

CH3

CH CH2 CH

2

=

(ii) H2O ,NaOH

2

(1° alcohol)

OH

64. Answer (2)

Hint : R of R-MgX on reaction with acidic-H give

alkane.

Sol. :

CH3

CH – CH MgCI2

CH3

– +

CH –3 O – H

+–

CH3

CH3

– CH – CH

3 + Mg

CI

OCH3

65. Answer (4)

Hint : More the H-bonding and lesser the torsional

strain, more will be the stability of conformers.

Sol. : Stability order:

(1) Cyclohexane : Chair > Boat > Half chair

(2) Butane (around C2 and C

3) : Anti > Gauche >

Partially eclipsed

(3) Propane : Staggered > Skew > Eclipsed

(4) Ethane -1, 2-diol: Gauche > Anti > Eclipsed

66. Answer (3)

Hint : In presence of sunlight addition reaction and

in presence of FeCl3 electrophilic substitution

reaction take place.

Sol. :

+ 3CI2

Sunlight

H CI

CI H

H

CI

H

CI

CI

H

CI

H

( )C H Cl6 6 6

( ) (X)Benzene hexachloride

+ CI2

FeCI3 + HCI

CI

( ) (Y)Chlorobenzene

67. Answer (4)

Hint : 1-alkynes are acidic in nature.

Sol. :

Ch C CH3 –

NaNH2

CH3– C C Na

CH3 – CH – CI

2

CH3– C CH

2 3–C – CH

Pent–2–yne

– +

68. Answer (1)

Hint : 2, 3–dimethylbutane has two type of

H-atoms.


9/16

Sol. :

CI2

h Cl+

Cl

Chiral carbon

69. Answer (2)

Hint : When C – CI bond acquires double bond

character then reactivity of C – Cl bond decreases.

Sol. : CH2 = CH

– CI

..CH

2 –CH CI

(vinyl chloride)

– +

70. Answer (1)

Hint : (i) O3

(ii) Zn+H O2

C=C C =O+O =C

Sol. :

2–oxo propanal

CH3 – C – CH = O + O = CH2

O+O = CH2

CH3 – C – CH = CH2

CH2

2-methylbuta-1, 3-diene

2 3 4

(C H5 8)

Methanal

Methanal

1

71. Answer (2)

Hint : Carbonium ion intermediate is formed.

Sol. :

CH = CH2

H+

H

+

H – +

CI–

shift

CI

A

72. Answer (1)

Hint : H2SO

4 + H

2SO

4 SO

3 + HSO

4

–

+ H

3O+

Sol. : SO3 is an electrophile.

73. Answer (4)

Hint : For highly polluted water, BOD value is more

than 17 ppm.

74. Answer (4)

Hint : N2O is a greenhouse gas.

75. Answer (3)

Hint : Photochemical smog is oxidising smog.

76. Answer (3)

Hint : Kjeldahl’s method is not applicable to

compounds containing

(i) Nitrogen in ring.

(ii) (–NO2) and azo (– N = N –) groups.

Sol. : Nitrogen present in CN , can be

quantitatively estimated by Kjeldahl’s method.

77. Answer (4)

Hint : Pent–2–ene and 1-chloro-2-bromopropene

show geometrical isomerism.

78. Answer (2)

Hint : Glycerol decomposes near its boiling point.

Sol. : Compounds which decompose near its boiling

point are purified by vacuum distillation method.

79. Answer (2)

Hint : Higher the availability of lone pair on nitrogen

atom higher is its basic nature and hence higher the

pKa

80. Answer (1)

Hint : Order of -I effect.

– CN > – COOH > – OH

81. Answer (1)

Hint : More the delocalization of +ve charge, more

will be the stability of carbocation.

Sol. : CH2

+ has maximum number of

resonating structures, so most stable.

82. Answer (4)

Hint : HNO3 converts NaCN and Na

2S into HCN and

H2S respectively.

83. Answer (1)

Hint. : CH3

CH2

– –C

CH3

CH3

––

–CH2– is neohexyl group.

84. Answer (3)

Hint : For tautomerism, acidic H-atom should be

present.

Sol. :

O

does not contain any acidic H-atom.

85. Answer (3)

Hint : Octet of N in 4

NH

is complete and no vacant

orbital is present on N atom.


10/16

[ BIOLOGY]

91. Answer (2)

Hint: RQ of fat is less than 1 and of organic acids

it is more than 1.

Sol.:

Tripalmitin is a fatty acid (RQ = 0.7)

Proteins (RQ = 0.9)

Malic acid (RQ = 1.33).

92. Answer (3)

Hint: In link reaction pyruvic acid (3C) forms acetyl

CoA(2C) & CO2.

Sol.:

4C compound – Malic acid

3C compound – Pyruvic acid

2C compound – Acetyl CoA

5C compound – -ketoglutaric acid

93. Answer (3)

Hint: During dormancy, growth & development of

seed is temporarily stopped.

Sol.: In dormancy, seeds cannot germinate inspite

of availability of all suitable environmental conditions.

94. Answer (3)

Hint: Pr absorbs red light.

Sol.: 660 nm is red light.

Pfr absorbs far red light.

86. Answer (2)

Hint : 2Volumeof N (atSTP)28

% of N 10022400 wt.of compound

Sol. : 1 1 2 2

1 2

P V P V

T T

(Experiment condition) (STP condition)

2760 V(725 – 25) 60

300 273

∵ Volume of N2 (at STP) = 50.29 ml

28 50.29% of N 100 22.45%

22400 0.28

87. Answer (4)

Hint : Alkyl group with –H atoms attach to

benzene, activates the ring for SE reactions.

Sol. : does not have –H atom.

88. Answer (1)

Hint : Cyclic, planar and conjugated compounds

containing (4n + 2)e– are aromatic in nature.

Sol. : HH is non planar because of steric

repulsion of inner hydrogen.

89. Answer (3)

Hint : More the s-character in hybrid orbitals, more

will be the electronegativity.

Sol. : C2H

2 : sp hybridised carbon

50% s-character.

90. Answer (2)

Hint : 1 2 3 4 5 6

2 2 3CH CH– CH – C C–CH

Hex–1–en–4–yne

95. Answer (4)

Sol.: Growth curve in plants is sigmoid, also called

S-shaped growth curve.

96. Answer (2)

Hint: Auxin gets removed along with apical buds of

a plant.

Sol.: Auxin was first time isolated from human urine.

It is an amino acid derivative.

97. Answer (4)

Hint: It is a gaseous hormone.

Sol.: Ethylene is responsible for ripening of fruits. It

can be used artificially to ripen the fruits.

98. Answer (3)

Hint: ABA is a plant growth inhibiting phytohormone.

Sol.: ABA induces dormancy of buds and seeds &

promotes abscission of flowers.

99. Answer (3)

Sol.: Gibberellin can replace vernalisation. As by

exogenous application of gibberellins many biennials

can be induced to behave as annuals.

100. Answer (4)

Hint: Short day plants require uninterrupted dark

period to flower.

Sol.: SDP will not flower if its dark period is

interuppted by a flash of light.


11/16

101. Answer (3)

Hint: Antagonistic means showing opposite effects

towards something.

Sol.: When two hormones bring about opposite

effects on the same function, it means they are

antagonistic to each other, whereas when they act

together they are called synergistic.

102. Answer (3)

Hint: Coconut milk has a phytohormone which helps

in cell division.

Sol.: It contains cytokinin.

103. Answer (2)

Sol.: Cytokinin along with auxin are essential in

tissue culture. High cytokinin to auxin ratio causes

shoot formation in callus.

104. Answer (1)

Hint: Cytokinin(CK) helps in production of new

leaves.

Sol.: It (CK) enhances chloroplast production in

leaves. It delays senescence by controlling protein

synthesis.

105. Answer (3)

Sol.: 2, 4-D is widely used as weedicide in cereal

crops.

106. Answer (4)

Hint: Tryptophan is precursor of auxin.

Sol.: Tryptophan is an amino acid. Final product of

transition/link reaction of aerobic respiration is acetyl

CoA which is precurssor of gibberellin.

A pigment violaxanthin is a precursor of ABA.

107. Answer (3)

Hint: Cytochrome c is a mobile carrier protein.

Sol.: Cytochrome c is not the part of complex IV. It

is attached to the outer surface of the inner

membrane & transfers electrons between complex III

& IV.

108. Answer (3)

Hint: Krebs’ cycle produces reduced coenzymes.

Sol.: NADH + H+ is produced during Krebs’ cycle

along with FADH2 & GTP.

It starts with condensation of acetyl CoA with OAA

to form citric acid.

109. Answer (2)

Hint: The F1 headpiece contains the site of the ATP

synthesis.

Sol.: F1 headpiece remains present towards the

matrix of the mitochondria. Hence ATP synthesis

takes place towards the matrix.

110. Answer (3)

Hint: Inner membrane of mitochondria forms cristae.

Oxidative phosphorylation occurs during oxidation of

food.

Sol.: Cristae contains cytochromes which helps in

ETS.

During formation of bread, yeast if used performs

alcoholic fermentation which releases CO2.

111. Answer (3)

Sol.: Buttercup shows environmental (extrinsic)

heterophylly. Cotton, coriander & larkspur show

developmental heterophylly due to intrinsic factors.

112. Answer (2)

Hint: Gibberellin was first isolated from a fungus

called Gibberella fujikuroi.

Sol.: It is acidic in nature, increases sugarcane yield

by increasing the length of internodes, delays

ripening of fruits and helps in seed germination.

113. Answer (3)

Sol.: The stimulus of vernalisation is perceived by the

mature stem apex or by the embryo of the seed.

114. Answer (4)

Hint: Auxin occurs in growing apices.

Sol.: If apical bud is removed then auxin is removed,

due to which apical dominance is prevented.

The plant will show more lateral branches & axillary

buds.

115. Answer (1)

Hint: Phytochrome is a photoreceptor.

Sol.: Phytochrome is a blue-green pigment,

regulates photoperiodic induction of flowering and

helps in seed germination too.

It occurs in two forms Pr & P

fr. Its P

r form becomes

active by absorbing far-red light. Pfr is physiologically

active form.

116. Answer (3)

Hint: Short day plant flowers after receiving

photoperiod shorter than critical period.

Sol.: Tobacco is a short day plant. Wheat, radish &

sugarbeet are long day plants.

117. Answer (3)

Hint: Rice & maize show hypogeal seed

germination.

Sol.: In hypogeal seed germination, epicotyl grows

first & cotyledons remain underground.

118. Answer (2)

Hint: Relative growth can be calculated by the

formula i.e. Growth per unit time

100Initial size

.


12/16

Sol.: For Leaf A

5100 50%

10

For Leaf B

5100 20%

25

119. Answer (1)

Hint: This hormone is a plant growth inhibitor.

Sol.: ABA is a derivative of carotenoids.

120. Answer (4)

Hint: Tissues formed from de-differentiation of

parenchyma are secondary in origin.

Sol.: Wound cambium, cork cambium &

interfascicular cambium are secondary in origin. They

are formed by de-differentiation of parenchyma cells.

Intrafascicular cambium is primary in origin.

121. Answer (4)

Hint: The continued growth due to activity of

meristems occurs in plants.

Sol.: New cells are always being added due to

activity of meristem, this type of growth is open form

of growth and indeterminate.

Animals show definite & diffused type of growth.

122. Answer (1)

Hint: Ethylene is a volatile substance which

promotes senescence.

Sol.: Ethylene is the only gaseous hormone of

plants. It breaks the dormancy of seeds.

Bakanae disease is caused by gibberellins.

123. Answer (3)

Sol.: Auxin helps in root formation on stem cuttings,

shows apical dominance & phototropism but does

not play any role to speed up the malting process

which is done by gibberellin.

124. Answer (4)

Hint: The plant hormone which is responsible for

closure of stomata is known as stress hormone.

Sol.: ABA is called stress hormone as its synthesis

is stimulated by drought, water logging & other

adverse environmental conditions.

125. Answer (2)

Hint: The phytohormone which was extracted from a

fungus Gibberella fujikuroi is responsible for bolting.

Sol.: Gibberellins are active substances which

induce stem elongation in rossette plants. This

phenomenon is called bolting.

126. Answer (1)

Hint: Complete oxidation of acetyl CoA occurs in

aerobic respiration through Krebs’ cycle.

Sol.: One turn of Krebs’ cycle involves complete

breakdown of acetyl CoA which produces 3NADH +

H+, 1 FADH2 & 1 GTP.

3 NADH + H+ = 3 × 3 = 9 ATP

1 FADH2 = 1 × 2 = 2 ATP

1 GTP = 1 × 1 = 1 ATP

12 ATP

NADH + H+ & FADH2 produce ATP through ETS.

127. Answer (3)

Hint: Yeast performs alcoholic fermentation.

Sol.: In alcoholic fermentation alcohol is produced

along with CO2 & ATP.

128. Answer (4)

Hint: In aerobic respiration CO2 is released in link

reaction & during Krebs’ cycle.

Sol.: In link reaction CO2 is released at one step.

During Krebs’ cycle CO2 is released at two steps.

In case of anaerobic condition

2

2

2CORQ (infinite)

0O

129. Answer (2)

Hint: It is a 2C compound.

Sol.: Acetyl CoA is common intermediate among

fats, proteins and carbohydrates breakdown during

aerobic respiration. It enters into Krebs’ cycle to

yield energy.

130. Answer (4)

Hint: Krebs’ cycle starts with the condensation of

2C compound with 4C compound to yield a 6C

compound.

Sol.: Krebs’ cycle starts with condensation of Acetyl

CoA (2C) with oxaloacetic acid (4C) to yield citric

acid (6C).

131. Answer (2)

Hint: Lactic acid (3C) is produced from pyruvic acid

which is also a 3C compound.

Sol.: In lactic acid fermentation pyruvic acid (3C) is

converted into lactic acid (3C). Hence there is no

release of CO2 which occurs in alcohol fermentation.

Both processes are hazardous and produce 2 ATP

as net gain.

132. Answer (1)

Sol.: The term glycolysis has originated from two

Greek words that are ‘glycon’ which means sugar

and ‘lysis’ which means splitting.


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133. Answer (3)

Hint: Partial oxidation of glucose without utilization

of oxygen is called glycolysis.

Sol.: Glycolysis takes place in all organisms in the

cytoplasm. Hence all organisms whether anaerobes

or aerobes retain enzymes for glycolysis.

134. Answer (4)

Hint: Plants have lesser demands for gaseous

exchange as compared to animals.

Sol.: Roots, stems and leaves all respire at lower

rates than animals. In plants, cells are closely

packed and located quite close to the surface of the

plant. Hence it is easy for gases to diffuse inside.

135. Answer (4)

Hint: ATP is synthesized during respiration.

Sol.: ATP acts as the energy currency of the cell.

The energy released by oxidation of food in

respiration is used to synthesise ATP and not used

directly.

136. Answer (3)

Hint : In non-chordates, heart beat is initiated by

hormone from brain.

Sol. : In non-chordates, heart is either absent or

present dorsally. If present, it is generally neurogenic

because heart beat is initiated by a ganglia situated

near heart. In myogenic heart, heart beat is initiated

by specialized muscular tissue situated in wall of

heart known as S.A node/pace maker of the heart.

137. Answer (4)

Hint : Commonly called star fish.

Sol. : Pheretima has nephridia for excretion and is

an annelid. Interstitial cells are present in the body

wall of coelenterates. Ctenoplana belongs to

Ctenophora.

138. Answer (1)

Hint : Mouth parts which contain chitinous teeth

known as denticles.

Sol. : In cockroach, mandibulate type of mouth parts

are found which are responsible for grinding and

incising food.

139. Answer (4)

Hint : Brood pouch harbours mammary nipples.

Sol. : In Balanoglossus, indirect development is

seen due to presence of larval stage. In female

Macropus (kangaroo), a mammal, mammary nipples

are present in brood pouch (marsupium).

140. Answer (1)

Hint : Blood glands and tufts of pharyngeal nephridia

are present in same set of body segments.

Sol. :

Typhlosole is a fold in intestinal wall to increase

absorptive surface area. Ovaries with ovarioles are

situated between 2nd to 6th abdominal segments in

female cockroach.

141. Answer (3)

Hint : Shell is a calcareous exoskeleton found

around most molluscs.

Sol. : Calcareous spicules form endoskeleton in

sponges. Calcareous ossicles constitute

endoskeleton in sea stars and cartilaginous

endoskeleton is found in cartilaginous fishes.

Dentalium has shell as calcareous exoskeleton.

142. Answer (3)

Hint : In molluscs, mantle cavity containing feathery

gills is present between mantle and visceral hump.

Sol. : Soft and spongy skin present just below shell

is known as mantle. Space between mantle and

visceral hump is known as mantle cavity. Feather like

gills responsible for both respiration and excretion

are present in mantle cavity.

143. Answer (4)

Hint : Identify an animal exhibiting epitoky

Sol. : Nereis is an aquatic unisexual annelid.

Parapodia are lateral unsegmented appendages for

swimming and respiration. Body is metamerically

segmented, development is indirect through

trochophore larval stage.

144. Answer (1)

Hint : Wuchereria is a representative of these

worms.

Sol. : Being circular in cross-section, these animals

are commonly known as roundworms. Sexual

dimorphism evolved first in roundworms. They have

muscular pharynx and both free living and parasitic

forms are present.

145. Answer (2)

Hint : Flatworms have solid body without any cavity.

Sol. : Flatworms are bilaterally symmetrical,

triploblastic, dorsoventrally flattened and acoelomate.

Taenia solium has no alimentary canal. It absorbs

predigested nutrients from intestine of host.

146. Answer (2)

Hint : Such type of metamorphosis is found in silk

moth.


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Sol. : Silk moth develops through larval and pupal

stages. Conversion of larva into pupa is known as

pupation, characterised by histolysis followed by

histogenesis.

147. Answer (3)

Hint : Fighting fish is a bony fish.

Sol. :

Earthworm – Nephridia – Act as excretory

structures

Cockroach – Titillator – Helps in copulation

present in

males only

Ascaris – Pineal setae – In males for

copulation

148. Answer (3)

Hint : Respiratory structures present on body from

2nd thoracic segment onwards.

Sol. : In cockroach, 10 pair of spiracles are present

in between sclerites for gaseous exchange during

respiration.

149. Answer (3)

Hint : A – Fastest running mammal.

B – Bird exhibiting polygamy.

Sol. : Panthera tigris (tiger) is considered as national

terrestrial animal and Pavo cristatus (Peacock) is

national bird of India.

150. Answer (1)

Hint : Chemoreceptors which taste chemical nature

of water

Sol. : In molluscs, osphradium in mantle cavity

serves to test chemical nature of water.

151. Answer (3)

Hint : These structures are associated with

segmented worms.

Sol. : Parapodia are associated with annelid like

Nereis. They help in locomotion and respiration.

152. Answer (3)

Hint : This organism can live on both land and water.

Sol. : Bidder’s canal present in kidney is responsible

for conduction of sperms

Hirudinaria – nephridia as excretory organ

Locusta – jointed legs for locomotion

Ctenoplana – cnidoblasts are absent

153. Answer (3)

Hint : Vocal cords are sound producing structures in

frogs.

Sol. : Male frog has sound producing vocal cords

and vocal sacs make the sound louder.

Cockroach can survive upto 7 days with amputated

head because most of its nervous tissue is ventrally

placed in belly region.

154. Answer (4)

Hint : These structures are sensitive to sound and

other vibrations.

Sol. : Anal cerci are jointed filamentous structures

present on 10th tergum. Anal style is unsegmented

spiny structure on 9th sternum present only in male

cockroach.

155. Answer (4)

Hint : Fish like organism with cartilaginous cranium

and vertebral column.

Sol. : In cyclostomes, such as Myxine, scales and

tympanum are absent. Heart is two chambered.

Fishes have membranous labyrinth. All the given

characters belong to members of class Amphibia.

156. Answer (1)

Hint : Syngamy occurs inside female’s body.

Sol. : Cartilaginous fishes show internal fertilization

and ovoviviparous condition.

157. Answer (4)

Hint : Flagellated cells of sponges are also named

as collar cells.

Sol. : Choanocytes or collar cells are found in inner

lining of spongocoel as well as flagellated chambers

of sponges. They continuously produce water current

for uptake of food, respiration, excretion and

reproduction.

158. Answer (3)

Hint : Mammal with avian characters.

Sol. : Ornithorhynchus – Duckbilled platypus

Pinctada – Pearl oyster

Pteropus – Flying fox

Balanoglossus – Tongue worm

159. Answer (4)

Hint : Most primitive lizard named as Tua-tara.

Sol. : Sphenodon is the most primitive lizard,

considered as living fossil but not as connecting link.

Peripatus is placed in taxon-Onychophora and

Neopilina is a connecting link between molluscs and

annelids.


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160. Answer (4)

Hint : Identify a roundworm whose larvae are spirally

coiled in muscles.

Sol. : Trichinella is an endoparasitic, viviparous

roundworm in rodents, pig, humans etc. causing

disease trichinosis characterised by abdominal pain

and muscular inflammation.

161. Answer (2)

Hint : Larval stage which causes taeniasis in

humans.

Sol. : In fecal matter of infected man, onchosphere

is present which is converted into hexacanth larva

and finally cysticercus larva/bladderworm in muscles

of pig. Human beings are infected when they feed on

undercooked measly pork and this larva in adult

develops into Taenia solium causing taeniasis.

162. Answer (3)

Hint : Most vertebrates are poikilothermous except

birds and mammals generally.

Sol. : Testudo is a reptile and cannot maintain its

body temperature, hence considered as

poikilothermic/cold blooded animal.

Columba is a bird while Felis and Macropus are

mammals. So, all these are warm-blooded animals.

163. Answer (1)

Hint : Animals which are commonly known as

roundworms are classified in this phylum.

Sol. : Aschelminthes/Nematodes are considered as

first animals to have complete digestive tract with

distinct mouth and anus.

164. Answer (3)

Hint : Number of moults is equivalent to the number

of chambers in its heart.

Sol. : In cockroach, nymph is a young stage which

is similar to cockroach but smaller in size, light in

colour, with immature gonads and without wings. It

undergoes gradual metamorphosis known as

paurometabola in 6 months upto two years. During

this time, it moults 13 times, to become sexually

mature adult.

165. Answer (2)

Hint : Common name for Antedon.

Sol. :

Metridium – sea anemone – Coelenterata

Antedon – sea lily – Echinodermata

Hormiphora – sea walnut – Ctenophora

Pennatula – sea pen – Coelenterata

166. Answer (3)

Hint : Common name for Hemichordates.

Sol. : • Trygon – Poison sting (Sting ray)

• Torpedo – Muscle modified into

electric organ (Electric

eel)

• Tongue worm – Proboscis present not

placoid scales

• Hippocampus – Brood pouch in male to

carry fertilized eggs.

167. Answer (3)

Hint : Fish like organisms having circular mouth.

Sol. : Cyclostomes (Petromyzon) are animals

without scales having circular mouth. They have non

operculated gills and lack paired fins. Scoliodon

(cartilaginous fish) and Hippocampus and Labeo

(bony fishes) have scales on their body.

168. Answer (3)

Hint : Members of this class show aerial

adaptations.

Sol. : In birds, long bones of the body contain

cavities filled with air to decrease weight of the body.

Such type of bones found in birds are known as

pneumatic bones.

169. Answer (4)

Hint : The first segment bears mouth and a lobe that

can act as a wedge.

Sol. : The first body segment of earthworm is known

as peristomium. Prostomium is a hood like structure

on peristomium that is helpful in burrowing. The fecal

deposits of earthworm are known as worm castings.

Vermicomposting is use of worm castings as

manure.

170 Answer (2)

Hint : Members of Class Osteichthyes.

Sol. : Air bladder/swim bladder is found in bony

fishes to provide buoyancy which is helpful in

swimming. They do not to have continuously swim

as in case of cartilaginous fishes which lack swim

bladder.

171. Answer (3)

Hint : Identify common Indian bull frog.

Sol. : Rana tigrina is common Indian bull frog.

Alytes is commonly known as midwife toad because

male carries eggs wrapped around its hind limbs.


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172. Answer (3)

Hint : Platyhelminthes are acoelomates.

Sol. : Protostomic animals are schizocoelic

eucoelomates and deuterostomic animals are

enterocoelic eucoelomates. Only arthropods placed

in class Insecta are considered as insects.

173. Answer (2)

Hint : Capillaries are absent in Hemichordates

Sol. : In leech and Balanoglossus, open type of

circulatory system is present.

174. Answer (4)

Hint : This process helps in replacement of lost

body parts.

Sol. : With the help of water vascular system

echinoderms perform locomotion, excretion,

respiration and capturing of food.

175. Answer (4)

Hint : Chordates have hollow, dorsal nerve cord

Sol. : Arthropoda – Chitinous exoskeleton

Mollusca – Calcareous exoskeleton

Echinodermata – Spiny skinned, ventral

solid nerve cord

Chordata – Presence of notochord

176. Answer (4)

Hint : The roundworm which is found attached with

intestinal wall.

Sol. : • Euspongia – Bath sponge

• Meandrina – Brain coral

• Fasciola – Liver fluke

177. Answer (3)

Hint : Obelia exhibits alternation of generation.

Sol. : In class Hydrozoa, all members except Hydra

show metagenesis. Metastasis is spread of cancer,

metamorphosis is the development of larval stage

into adult stage.

178. Answer (4)

Hint : Animal which is bilaterally symmetrical in larval

stage but becomes asymmetrical due to torsion in

adult life.

Sol. : Adult snails are asymmetrical due to torsion.

In larval stage, it is bilaterally symmetrical. It is a

type of retrograde metamorphosis.

179. Answer (1)

Hint : This animal shows internal metamerism.

Sol. : In annelids, both external and internal

segmentation is found in adult stage

In arthropods, only external segmentation is present

in adults. Chordates show internal segments in

certain body parts.

180. Answer (3)

Hint : In sponges, the body cavity opens to exterior

through osculum.

Sol. : All the given characters are shown by

sponges. They are not considered as diploblastic

and they have cellular grade of body organisation.

The cavity inside body is known as spongocoel or

paragastric cavity.

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