all india aakash test series for neet - 2020...2020/02/02 · test-8 (code-a)_(hints &...

Test-8 (Code-A)_(Answers) All India Aakash Test Series for NEET-2020

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All India Aakash Test Series for NEET - 2020

Test Date : 02/02/2020

ANSWERS1. (1)2. (3)3. (1)4. (3)5. (2)6. (1)7. (4)8. (3)9. (2)10. (1)11. (3)12. (4)13. (2)14. (3)15. (2)16. (3)17. (3)18. (2)19. (3)20. (1)21. (3)22. (2)23. (1)24. (4)25. (2)26. (3)27. (1)28. (4)29. (1)30. (3)31. (3)32. (4)33. (3)34. (2)35. (4)36. (1)

37. (3)38. (2)39. (4)40. (1)41. (3)42. (1)43. (1)44. (2)45. (3)46. (4)47. (2)48. (3)49. (2)50. (1)51. (3)52. (4)53. (4)54. (4)55. (1)56. (3)57. (4)58. (3)59. (3)60. (4)61. (3)62. (4)63. (4)64. (2)65. (2)66. (1)67. (1)68. (1)69. (3)70. (1)71. (4)72. (4)

73. (3)74. (1)75. (4)76. (3)77. (1)78. (3)79. (4)80. (4)81. (2)82. (2)83. (2)84. (2)85. (1)86. (2)87. (4)88. (1)89. (4)90. (2)91. (3)92. (3)93. (3)94. (4)95. (4)96. (3)97. (3)98. (1)99. (4)100. (2)101. (2)102. (3)103. (4)104. (2)105. (4)106. (4)107. (4)108. (1)

109. (2)110. (2)111. (2)112. (3)113. (1)114. (2)115. (2)116. (4)117. (3)118. (4)119. (2)120. (3)121. (3)122. (4)123. (2)124. (3)125. (4)126. (2)127. (3)128. (3)129. (3)130. (2)131. (2)132. (1)133. (2)134. (3)135. (2)136. (2)137. (3)138. (4)139. (3)140. Deleted141. (4)142. (2)143. (4)144. (4)

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TEST - 8 (Code-A)

All India Aakash Test Series for NEET-2020 Test-8 (Code-A)_(Hints & Solutions)

Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456 2/16

HINTS & SOLUTIONS

[PHYSICS] 1. Answer (1) Hint : Boyle’s law Sol. : PV = constant

P09V = P1(9V + V)

01

910

=PP

P1 × 9V = P2(9V + V)

12

910

=PP

2 09 9

10 10= ×P P

2 081

100=P P

2. Answer (3) Hint : Dalton’s law of partial pressure.

Sol. : 1 =PVnRT

2 =PVnRT

3 =PVnRT

N = n1 + n2 + n3

1 = + +PV PV PV PVRT RT RT RT

P1 = 3P

3. Answer (1) Hint and Sol. : Charle’s law that leads to absolute

scale of temperature. 4. Answer (3) Hint : PV = nRT

Sol. : 1 500=

×PVn

R

2432 500

=PV

R

2 2 400= ×

×P Vn

R

2 800=

PVnR

232 800 32 = =

x PV xnR

24 58×

=x

x = 15 g

5. Answer (2) Hint and Sol. :

. 1molecule- degree of freedom 2

K E kT=

Diatomic molecule have two rotational degree of freedom. Average rotational kinetic energy per molecule

122

= × kT

= kT 6. Answer (1) Hint : PV = nRT

Sol. : 1400

(27 )nRP

x A=

+

2500

(27 )nRP

x A=

−

P1 = P2

⇒ 400 500(27 ) (27 )nR nR

x A x A=

− +

⇒ 4 5(27 ) 27

=− +x x

⇒ 27 × 4 + 4x = 5 × 27 – 5x ⇒ 9x = 27 ⇒ x = 3 cm 7. Answer (4) Hint : V = V0(1 + αt)

Sol. : 0 1273

= +

tV V

0273

273+ =

tV V

0273

=V TV

Here T is in kelvin

0tan273

θ =V

Test-8 (Code-A)_(Hints & Solutions) All India Aakash Test Series for NEET-2020


8. Answer (3)

Hint and Sol. : 0

=RkN

R = N0k, therefore statement 3 is incorrect. 9. Answer (2)

Hint : ms3RTV

M=

Sol. : 1

2

ms 1

ms 2

V TV T

=

1

1

ms

ms 2

3002VV T

=

T2 = 600 K T2 = 327°C 10. Answer (1)

Hint : 2 2= ±ω −v A x

Sol. : 2 2= ±ω −v A x

2

2 22

= −

ω v A x

2

2 22 + =

ωv x A

2 2

2 2 1( )

+ =ω

v xA A

→ This is the equation of an

ellipse. 11. Answer (3)

Hint : 2 2 21 12 2

= = ωE kA m A

Sol. : 2 2 2 21 1 2 2

1 12 2

ω = ωm A m A

2 2 2 21 1 2 2ω = ωA A

⇒ 2 2 2 225 8 4× = ω

⇒ 2 25 4ω = × ⇒ 10 unitω =

12. Answer (4)

Hint : 2 2= ±ω −v A x

Sol. : 2 2

112 22 2

ω −=

ω −

A xvv A x

⇒ 2

2

16 912 16

−=

−

A

A

⇒ 2

2

4 93 16

−=

−

A

A

On solving A = 5 cm 13. Answer (2)

Hint : 2= πMTk

Sol. : M = 8m + m = 9 m mg = kx

=mgkx

92= πmT mgx

92= πxT

g

6= πxTg

14. Answer (3)

Hint : 2= πTg

Sol. : 2

1 24=

π

T g

2

2 244

=π

T g

1 21 2 +

= π T

g

On putting values of 1 2and

1 5T T=


6sin100 8sin 1002π = π + π +

x t t

⇒ 6sin100 8cos100x t t= π + π ⇒ 10sin(100 )= π + φx t

4tan3

φ =

⇒ 53φ = °

10sin(100 53 )= π + °x t

Particle execute S.H.M with amplitude 10.



16. Answer (3)

Hint : 2= πIT

mgd

Sol. :

2

23

2

= π× ×

mTmg

223

= πTg

2 ∝ T

17. Answer (3) Hint : cos= − ω ωV A t

Sol. : cos= − ω ωdx A tdt

cos= − ω ω∫ ∫dx A tdt

sinω= − ω

ωAx t

sin( )= ω + πx A t

18. Answer (2)

Hint : sin4π

=tx a

Sol. : x(0) = asin0 = 0 xi = 0

sin 64π = ×

fx a

3sin2π =

fx a

sin2π

= − ⇒ = −f fx a x a

Magnitude of displacement = a

19. Answer (3)

Hint : and ω= ω =Pv A v

k

Sol. : vP = 100 × 0.05 = 5 m/s

10050

=v

v = 2 m/s

5 : 2=Pvv

20. Answer (1) Hint : sin( )y A kx t= + ω

Sol. : A = 2 m

2π=

λk

11 m2

−=k

2ω = πf

122

ω = π ×π

= 1 rad/s

2sin2

= +

xy t

21. Answer (3)

Hint : 2

1 2

sintancos

∆φφ =

+ ∆φA

A A

Y = Rsin(ωt + φ)

Sol. : 10sin

2tan10 10cos

2

π

φ =π

+

tan 1φ =

4π

φ =

22. Answer (2) Hint : ∆φ = ∆k x

Sol. : 20.03 50 10−∆φ = π × ×

10.032

= π ×

3200

π=



23. Answer (1) Hint and Sol. : Propagation of mechanical wave

through a medium carries energy and momentum. 24. Answer (4) Hint : Time interval between two maximum

compression to appear at a place T.

Sol. : 1400

=T

25. Answer (2)

Hint : =ρTvA

Sol. : On increase in density of wire, speed of wave in wire decreases.

1100 1002

∆ ∆ρ× = ×

ρv

v

1 4%2

= ×

= 2% decrease 26. Answer (3) Hint and Sol. :

According to above diagram, velocity of particle at

point D will be along positive y direction. 27. Answer (1)

Hint : 12

=µ

Tf

f1 – f2 = 4

Sol. : 12

TfA

=ρ

12

∆ ∆=

f Tf T

4 1

400 2∆

=T

T

0.02∆=

TT

28. Answer (4)

Hint : γ=

RTvM

Sol. : r .m.s3

=RTvM

20

RTCM

γ=

2

3=

C RTM

2

20 3

CC γ=

2

20

43 3

CC = ×

023

C C=


Distance travelled by the sound in 10 second = 330 × 10 = 3300 m Distance travelled by the engine = 700 m v × 10 = 700 m v = 70 m/s 30. Answer (3) Hint and Sol. : In stationary wave, strain at the

node will be maximum. 31. Answer (3)

Hint : Fundamental frequency for open 0 2vf =

Fundamental frequency of closed pipe 4

=

v

Sol. : 0 2vf =

When dipped in water pipe behave as close organ pipe.

Frequency for third harmonic of a closed organ pipe

3 4 24 3 2

×= = =

×

v v vf

02=f f



32. Answer (4) Hint and Sol. : In a closed organ pipe only odd harmonics will be

present i.e. 3 5, ,

4 4 4⋅ ⋅ ⋅

v v v therefore 800 Hz

frequency cannot be obtained in it. 33. Answer (3) Hint and Sol. : Apparent change in wavelength takes place due to

motion of source while apparent change in frequency takes place due to motion of source and observer both.

34. Answer (2) Hint : Use trigonometric identity cos2θ = 1 – 2sin2θ 2cosAcosB = cos(A + B) + cos(A – B)

Sol. : 24sin cos(500 )2ty t =

1 cos4 cos(500 )2

ty t− =

y = 2cos500t – 2cost cos(500t) = 2cos500t – [cos(500t + t) + cos(500 – t)] = 2cos500t – cos501t – cos499t So the given expression is the resultant of three

independent harmonic oscillation. 35. Answer (4) Hint and Sol. :

sin cos(60 )3xy tπ

= π

3π

=k

⇒ 2

3π π

=λ

⇒ 6 mλ =

Distance between two consecutive node = 3 m2λ

=

Distance between two consecutive node and

antinode 1.5 m4λ

= =

2 60π = πf

30 Hz=f

60 180 m/s

3

ω π= = =

πv

k

36. Answer (1)

Hint : In string, 2vfL

= (for same T and µ)

Sol. : L = L1 + L2 + L3

0 1 2 32 2 2 2

= + +v v v vf f f f

0 1 2 3

1 1 1 1= + +

f f f f

37. Answer (3)

Hint : =µTv

Sol. : ω

=vk

5054

=v

v = 40 m/s

3404 10

T−=

×

T = 1600 × 4 × 10–3 T = 6.4 N 38. Answer (2) Hint : Distance = v × t

Sol. : 1 250T

=

1 s

250=T

50 1 s250 5

= =t

13405

d = × = 68 m

39. Answer (4)

Hint : 1 ′λ = + λ

svv

Sol. : 1 604

′λ = + × ×

vv

5 604

′λ = × = 75 cm

40. Answer (1) Hint and Sol. : Laplace’s correction in the formula for the speed

of sound in gases given by Newton, because sound wave propagates adiabatically, while Netwon assumed it isothermally.



41. Answer (3)

Hint : 00

− = v vf f

v

Sol. : v0 = u + at v0 = at

0− =

v atf f

v

00= −

f af f tv

42. Answer (1)

Hint : 1 35( ) ,

4 4λ λ

+ = + = e e

Sol. : 1 = 22.5

3 = 116.5

⇒ 1

3

15

+=

+

ee

⇒ 5 1 + 5e = 3 + e

⇒ 3 154−

= e

⇒ 116.5 112.54−

=e

= 1 cm

43. Answer (1)

Hint : 1 2 1 22 cosI I I I I= + + φ

Sol. : ( )2max 1 2I I I= + for lmax φ = 0°

( )2min 1 2I I I= − for lmin φ = 180°

( )

1 2max min

max min 1 2

42

I II II I I I

−=

+ +

45

=

44. Answer (2) Hint : Frequency of nth overtones of open pipe

0( 1)

2+

=

n vf

Frequency of nth overtones of closed organ pipe(2 1)

4+

=

cn vf

Sol. : 0

1)2

(2 1)4

( +

=+

c

n vf

n vf

⇒ 0 2( 1)(2 1)

+=

+c

f nf n

45. Answer (3) Hint and Sol. : Beat frequency is the number of times the

resultant intensity becomes maximum or minimum in one second.

[CHEMISTRY]

46. Answer (4) Hint : Urea is NH2CONH2

Sol. :

47. Answer (2) Hint : –OH group increases electron density on

ortho and para positions of benzene ring.

Sol. : and so on

–OH group activates the benzene ring for the attack of and deactivates for attack of 2NO .−

48. Answer (3) Hint : Carbonyl compound is obtained. Sol. :

49. Answer (2) Hint : Electron donating group present on

benzene increases the rate of SE reaction. Sol. :

2 3 2NMe CH Cl NO( M effect) ( H effect) ( I effect) ( M effect)

− − − −+ + − −

50. Answer (1) Hint : Maximum prescribed limit of Cd in drinking

water is 0.005 ppm.



51. Answer (3) Hint : Br2 decolourisation is test for unsaturation. Sol. :

52. Answer (4)

Hint :

Sol. : Hyperconjugation is a permanent effect. 53. Answer (4) Hint : Resonance increases the stability of

carbocation.

Sol. : effect

effect


55. Answer (1) Hint : Longest chain of carbon atoms containing

–COOH is considered. Sol. :

56. Answer (3) Hint : Cyclopentadienyl anion is an aromatic. 57. Answer (4) Hint : AgNO3 gives ppt. with NaCN and Na2S.

58. Answer (3)

Hint : Compound B. P.Chloroform 334 K

Aniline 457 K

Sol. : Liquids having sufficient difference in their boiling points are separated by distillation.

59. Answer (3)

Hint : 1 2 3 4

2CH CH C CH= − ≡

Sol. : C2 ⇒ sp2 hybridised C3 ⇒ sp hybridised 60. Answer (4) Hint : Soda lime is CaO + NaOH Sol. :

61. Answer (3) Hint : CH3OH is more acidic than hydrocarbons.

Sol. : Acidic behavior : CH3OH > C2H2 > C2H4 > C2H6 (i) (iii) (iv) (ii) 62. Answer (4) Hint : Uncatalysed oxidation of SO2 to SO3 is slow. 63. Answer (4) Hint : Photochemical smog has high

concentration of oxidising agents. Sol. : The main components of the

photochemical smog result from the action of sunlight on unsaturated hydrocarbons and nitrogen oxides produced by automobiles and factories.

64. Answer (2)

Hint : shows +R effect.

Sol. : +R effect : – NH2

–R effect : – COOH, = O, – CN

65. Answer (2) Hint : The given characteristic is associated with

mesomeric effect. 66. Answer (1) Hint : Metamerism arises due to different alkyl

chains on either side of the functional group in the molecule.



67. Answer (1)

Hint : Trans-But-2-ene is

Sol. :

68. Answer (1) Hint : Markovnikov addition of H2O take place.

Sol. :

69. Answer (3) Hint : HBr gives electrophilic addition to alkene.

Sol. :

70. Answer (1) Hint : Halogenation of alkanes is a free radical

substitution reaction.

Sol. :

71. Answer (4) Hint : Rotation about C – C single bond in

alkanes is almost free for all practical purposes. Sol. : Conformers of ethane are infinite. 72. Answer (4) Hint : Propane can be prepared by Kolbe’s

electrolytic method.

Sol. :

(1)

(2)

(3) (4) 73. Answer (3) Hint : H2(excess) Pd reduces C ≡ C as well as

C = C.

Sol. :

3, 4, 4, 5- tetramethylheptane. 74. Answer (1) Hint : Reaction of alkyl bromide with alc. KOH is

an example of β-elimination Sol. : Formation of more substituted alkene is

preferred in the case of β-elimination. Hence, rate of reaction is

75. Answer (4)

Hint :

Sol. : trans-but-2-ene is more stable than cis form.

76. Answer (3) Hint : Zn/H2O prevents oxidation of aldehyde to

carboxylic acid. Sol. :



77. Answer (1) Hint : HBr in presence of peroxides (C6H5CO)2O2

generates Br (free radical) Sol. :

78. Answer (3) Hint : BOD value measures how much oxygen is

required by bacteria to break down the organic matter present in water.

Sol. : Highly polluted water requires more oxygen by bacteria. Hence, its BOD value is more than 17 ppm.

79. Answer (4) Hint : DDT is a pesticide. Sol. : Organic toxins : Aldrin, Dieldrin Herbicides : Sodium arsinite (Na3AsO3)

80. Answer (4)

Hint :

81. Answer (2) Hint : Greenhouse gases contribute to global

warming and hence, climate change. Sol. : Greenhouse gases : Carbon dioxide, Methane, Ozone, Nitrous oxide,

Water vapour. 82. Answer (2) Hint : A non-benzenoid compound does not

contain any benzene ring.

Sol. :

83. Answer (2) Hint : Rain water is acidic in nature. Sol. : Normally it has a pH of 5.6

84. Answer (2) Hint : N2 is obtained.

Sol. : x y zyC H N 2x CuO2

+ + →

2 2 2y z yxCO H O N 2x Cu2 2 2

+ + + +

85. Answer (1) Hint : NaNH2 is a strong base, capable of

elimination. Sol. :

86. Answer (2) Hint : Precipitate of ammonium phosphomolybdate

is obtained.

Sol. :

87. Answer (4)

Hint : % of 31.4 meq. of NHNMass of compound

×=

Sol. : Meq. of NH3 = meq. of H2SO4

= (0.1 × 2) × 100 = 20

% of 1.4 20N 37.33%0.75

×= =

88. Answer (1) Hint : An aromatic compound should have

complete delocalisation of π-electrons in the ring. Sol. : Heterocyclic ring must comprise of different

atoms. 89. Answer (4) Hint : Degree of unsaturation of C4H8 = 1

Sol. : Number of π bond + ring = 1

Hence, 90. Answer (2) Hint : n-hexane in such conditions undergoes

aromatization. Sol. :



[BIOLOGY]91. Answer (3) Hint : Water moves from high to low water

potential. Sol. : Cell A Cell B

ψw = – 5 ψw = – 2

Cell B has higher water potential than cell A, hence cell A is more deficit in water. Osmotic potential (ψs) of cell B is – 8 atm, whereas osmotic pressure will be 8 atm as ψs and OP are opposite in sign.

92. Answer (3) Hint : Active absorption is absorption by root. Sol. : Force for active absorption develops in

root. 93. Answer (3) Hint : Addition of solutes increases the osmotic

pressure. Sol. : During phloem loading, sucrose moves into

the phloem which increases its OP and becomes hypertonic.

Phloem loading is an active process. At the sink, osmotic pressure in sieve tube decreases and water moves out of the phloem.

94. Answer (4) Hint : Girdling experiment showed food

translocation by a tissue. Sol. : Food translocation takes place by phloem

and it was identified by the girdling experiment. 95. Answer (4) Sol. : Xylem is associated with translocation of

mineral salts, water, organic nitrogen and hormones. Phloem transports photosynthates.

96. Answer (3) Hint : Facilitated diffusion involves movement of

substances through membrane proteins. Sol. : Hydrophilic substances cannot move

through the lipid bilayer hence these molecules are facilitated by some membrane proteins.

97. Answer (3) Hint : Osmotic concentration is directly

proportional to the amount of solutes. Sol. : If cell sap has high osmotic concentration,

it means it has more solutes and it is hypertonic as compared to the surrounding solution.

98. Answer (1) Hint : Chief sinks for mineral elements are those

parts of plants which are either storage organs or they are growing regions.

Sol. : Apical meristem requires minerals as the cells of it are in continuous state of division.

99. Answer (4) Hint : In roots, generally the concentration of

minerals is higher than the soil. Sol. : Most minerals are actively absorbed by the

roots. For carbon fixation, electrons and hydrogens are

given by H2O.

100. Answer (2) Hint : Solute potential is developed when solute

is added to water.

Sol. : ψw of pure water is zero and ψs is always negative for a solution.

101. Answer (2) Hint : Only prokaryotes can fix the atmospheric

nitrogen for a solution. Sol. : Only some bacteria and cyanobacteria can

fix the atmospheric nitrogen. 102. Answer (3) Hint : Chemoautotrophs do not require solar

energy to synthesise food. Sol. : Nitrosomonas, Nitrobacter } Chemoautotrophs Rhizobium, Frankia } Heterotrophs Nostoc, Anabaena } Photoautotrophs 103. Answer (4) Hint : Trace elements are micronutrients. Sol. : B, Cu and Mo are micronutrients. 104. Answer (2) Hint : Nitrate cannot be used directly by plants as

such. First, NO3– is converted into NO2

– and then NH3.

Sol. : Nitrate assimilation involves nitrate reduction into nitrite.

Conversion of atmospheric N2 to 3NO− occurs by non-biological nitrogen fixation.



105. Answer (4) Hint : Mobile elements readily move from older to

younger tissues. Sol. : Deficiency symptoms of mobile elements

are first seen in the older tissues. 106. Answer (4) Sol. : Calcium is required for the formation of

mitotic spindle. 107. Answer (4) Hint : Mn is a micronutrient. Slight excess of it

causes toxicity. Sol. : Mn toxicity results in reduction in uptake of

Fe and Mg. 108. Answer (1) Hint : Nitrite reductase reduces nitrite ions to

ammonia. Sol. : It is found in leaves and contain copper and

iron but not Mo. 109. Answer (2) Hint : Calvin cycle takes place in bundle sheath

cell of C4 plants.

Sol. : Sugar is produced in Calvin cycle. In mesophyll cells of C4 plants, primary fixation of CO2 takes place.

110. Answer (2) Hint : C4 pathway is called Hatch and Slack

pathway. Sol. : In C4 plants the primary acceptor is PEP

which is a 3C compound and found in mesophyll cells.

111. Answer (2) Sol. : Calvin cycle or C3 cycle is present in all

plants.

4Kranz anatomy

C plantsDouble carboxylation

Scotoactive stomata } CAM plants 112. Answer (3) Hint : Photorespiration takes place in C3 plants

not in C4 plants.

Sol. : OEC involves Mn and Cl for splitting of water.

113. Answer (1) Hint : C4 plants show double carboxylation.

Sol. : In C4 plants Calvin cycle operates in bundle sheath cells so concentration of CO2 increases at enzyme site.

PEPcase does not show oxygenase activity. 114. Answer (2) Hint : In C3 and C4 plants Calvin cycle is

common. Sol. : Same amount of NADPH are required in C3

and C4 plants as regeneration of PEP does not require NADPH.

115. Answer (2) Hint : CO2 concentration and light intensity both

should be optimum to increase the yield of plants. Sol. : At high light intensity and high CO2

concentration, both C3 and C4 plants show increase in rate of photosynthesis.

116. Answer (4) Sol. : Cyclic flow of electrons is seen in cyclic

photophosphorylation. PS II involves splitting of H2O.

117. Answer (3) Hint : Dimorphic chloroplasts are seen in C4

plants. Sol. : Sugarcane is a C4 plant.

Opuntia – CAM plant Pea and wheat } C3 plants

118. Answer (4) Hint : To develop the proton gradient more

H+ ions should be present in the lumen of thylakoids than stroma of chloroplast.

Sol. : When number of H+ ions increases in lumen, pH of lumen decrease.

119. Answer (2)

Sol. :

120. Answer (3) Hint : Acetyl CoA is a 2C compound. Sol. : End product of glycolysis is pyruvic acid

(3C). Acetyl CoA releases 2CO2 molecules upon

complete oxidation through TCA cycle. It is the product of link reaction.


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121. Answer (3)

Hint : Glycolysis is also known as EMP pathway.

Sol. : Glycolysis is common step in both aerobicand anaerobic respiration.

122. Answer (4)

Hint : 2

2

Volume of CO evolvedRQVolume of O consumed

=

Sol. : RQ value for oxalic acid is 4.

123. Answer (2)

Hint : Substrate level phosphorylation meansdirect ATP synthesis.

Sol. : Glycolysis and Krebs cycle or citric acidcycle produces direct ATP molecules throughsubstrate level phosphorylation.

124. Answer (3)

Hint : During fermentation, pyruvic acid isreduced to lactic acid.

Sol. : In EMP, hexose sugar splits into twomolecules of triose sugar with the help of aldolase.

125. Answer (4)

Hint : Reduced ubiquinone is ubiquinol.

Sol. : Cyt c is a small protein found attached tothe outer surface of inner mitochondrialmembrane.

126. Answer (2)

Hint : Pepper is a DNP.

Sol. : Day Neutral Plants (DNP) do not show anycorrelation between exposure to light duration andinduction of flowering response.

127. Answer (3)

Hint : Ethylene is a gaseous hormone.

Sol. : ABA is called stress hormone as stressstimulates its synthesis.

128. Answer (3)

Hint : Seed requires anchorage first.

Sol. : Radicle is the first developed structurewithin the seed.

129. Answer (3)

Hint : Ethylene is a gaseous phytohormone.

Sol. : Precusor of ethylene is methionine which isa sulphur containing amino acid.

130. Answer (2)Hint : Cytokinin was first isolated from unripe maize grains.Sol. : Cytokinin delays senescence, promotes cell division and is used in tissue culture.

131. Answer (2)Hint : GA induces synthesis of hydrolytic enzymes.Sol. : Aleurone layer has hydrolytic enzymes which move to starchy endosperm to convert complex carbohydrates to simpler ones.

132. Answer (1)Sol. : Viviparous seed germination is shown by Sonneratia

133. Answer (2)Sol. : Buttercup shows environmental heterophylly. Cotton, coriander and larkspur show developmental heterophylly.

134. Answer (3)Sol. : Gibberellins-composition is terpenes.

135. Answer (2)Hint : Apical dominance is shown by auxin. Sol. : Cytokinin counteracts apical dominance by increasing the supply of water and minerals to lateral buds.

136. Answer (2)Hint : They are shelled organisms.Sol. : Octopus, Aplysia and Chaetopleura belong to phylum Mollusca. Taenia is a flatworm. Sepia is a mollusc and Ancylostoma is a roundworm.

137. Answer (3)Hint : These organisms have pentamerous radial symmetry.Sol. : Radula is present in molluscs. Echinoderms lack excretory system.

138. Answer (4)Hint : They show true metamerism.Sol. : Annelids are characterised by presence of metameres, bilateral symmetry, presence of a true coelom and triploblasty.

139. Answer (3)Hint : They belong to phylum Coelenterata. Sol. : Sea pen is Pennatula, sea fan is Gorgonia, Sea hare is Aplysia (Mollusc), sea horse is fish named Hippocampus.

140. Deleted



141. Answer (4) Hint : Lepisma and scorpion are arthropods. Sol. : Dolphin is a mammal and it respires by

lungs. Dentalium is a mollusc and respires by feather like gills.

142. Answer (2) Hint : Wing muscles are anchored here. Sol. : Birds have a well developed keel

(sternum), the bone that anchors the muscles that move a bird’s wings.

143. Answer (4) Hint : Island country near Australia. Sol. : Apteryx (kiwi) is native to New Zealand. Pavo can fly to a limited extent. 144. Answer (4) Hint : Snake does not undergo metamorphosis. Sol. : Amphibians possess larval stages but the

same are absent in reptiles. 145. Answer (4) Hint : Scoliodon is shark. Sol. : Myxine, a Cyclostomate has a circular

mouth. Exocoetus is a bony fish and Scoliodon is a cartilaginous fish.

146. Answer (2) Hint : It is a tunicate. Sol. : Retrogressive metamorphosis is a

phenomenon wherein the larva is more advanced than adult. Progressive metamorphosis is shown by Rana, Amphioxus and Petromyzon.

147. Answer (3) Hint : Metagenesis is exhibited by Obelia. Sol. : Polyp and medusae both are diploid

stages. Comb plates are present in ctenophores. 148. Answer (1) Hint : Bony fishes have operculum. Sol. : Chondrichthyes are marine and possess

5-7 pairs of gills without operculum. Osteichthyes are freshwater and marine and possess 4 pairs of gills with operculum.

149. Answer (2) Hint : Cluster of relevant cells can develop into a

new organism. Sol. : They may have calcareous or siliceous

spicules. Spongilla is freshwater sponge. They are mostly asymmetrical.

150. Answer (1) Hint : It is called the boring sponge. Sol. : Cliona bores through the oyster shell and

prevents pearl formation. 151. Answer (1) Hint : Lepisma does not exhibit intermediate

stages in development. Sol. :

Ametabolous – Egg → Young → Adult

Hemimetabolous – Egg → Naiad → Adult

Paurometabolous – Egg → Nymph → Adult

Holometabolous – Egg → Larva → Pupa → Adult 152. Answer (1) Hint : They are edible fishes. Sol. : Labeo (Rohu), Catla (Katla) and Clarias

(Magur) are freshwater bony fishes. 153. Answer (2) Hint : Alternation of generation. Sol. : Most species of Hydra are dioecious. No

free larval stage is present in Hydra. Hydra does not show metagenesis.

154. Answer (3) Hint : Pseudocoelom is a feature of

Aschelminthes. Sol. : Protostomes are schizocoelomates

Ancylostoma and Enterobius are Aschelminthes. Blind sac body plan is found in Coelenterates, Ctenophores and Platyhelminthes.

155. Answer (4) Hint : They are also named as collar cells. Sol. : Archaeocytes are undifferentiated cells

Flagellated choanocytes are responsible for water current and receiving sperms during sexual reproduction.

156. Answer (4) Hint : Open circulatory system. Sol. : Nereis and Pheretima are annelids which

possess closed circulatory system. Among molluscs, cephalopods such as Octopus possess closed circulatory system.

157. Answer (3) Hint : It causes elephantiasis. Sol. : Hookworm and roundworm are

monogenetic and oviparous. Tapeworm is a digenetic platyhelminth but oviparous.



158. Answer (4) Hint : This phylum includes animals with flame

cells as excretory structures. Sol. : Cephalization, acoelomate condition and

bilateral symmetry are observed in phylum Platyhelminthes.

159. Answer (4) Hint : They are locomotory structures. Sol. : Parapodia and setae are present in Nereis

while only setae are present in Pheretima. 160. Answer (2) Hint : This is located in anterior part of body. Sol. : Body of Balanoglossus is divisible into

anterior proboscis, a collar and a long trunk. 161. Answer (4) Hint : Exoskeleton is present ventrally, laterally

and dorsally. Sol. : Dorsally present sclerites are called

tergites, ventrally present ones are called sternites and laterally present are called pleurites.

All are collectively known as sclerites. 162. Answer (4) Hint : Petromyzon is a cyclostome Sol. : The mouth of cyclostomes is suctorial,

circular and without jaws. Mouth of chondrichthyes is located ventrally. Terminal mouth is a feature of bony fishes.

163. Answer (1) Hint : This is also present in coelenterates. Sol. : Statocysts are statoreceptors in

coelenterates and arthropods. 164. Answer (2) Hint : Avian feature. Sol. : Aves are warm blooded, oviparous and

dioecious. 165. Answer (3) Hint : These excretory structures are most

numerous. Sol. : Hepatic caecae are present at junction of

foregut and midgut and secrete digestive juices. Malpighian tubules perform excretion. 10 pairs of spiracles are present in cockroach.

166. Answer (4) Hint : This structure is ventrally located. Sol. : Earthworm and cockroach both possess a

double, ventral, solid nerve cord with paired ganglia.

167. Answer (3) Hint : Identify a reptile. Sol. : Bangarus, commonly known as krait is a

snake belonging to Class Reptilia. 168. Answer (4) Hint : It is an echinoderm. Sol. : Asterias contains a calcareous

endoskeleton. 169. Answer (1) Hint : They originate from ventral side of 9th

segment. Sol. : Anal or caudal styles are the features of

sexual dimorphism. They are chitinous spine like unjointed structures.

170. Answer (2) Hint : It has pneumatic bones. Sol. : Corvus is a bird and a homeotherm. 171. Answer (3) Hint : Petromyzon belongs to Class Cyclostomata. Sol. : Animals belonging to Division Agnatha

possess mouth without jaws and have a single nostril. Notochord persists throughout life in primitive craniates such as Myxine.

172. Answer (1) Hint : Ecdysis is shedding of skin. Sol. : Shedding of scaly epidermis of skin

periodically is called ecdysis or moulting. It is shown by snakes and lizards among vertebrates.

173. Answer (2) Hint : It operates between intestine and liver. Sol. : The portal system of circulation bypasses

the heart. Hepatic portal and renal portal system are well developed in amphibians.

174. Answer (1) Hint : It includes true fishes. Sol. : Pisces and amphibians lack amnion so

considered as anamniotes. 175. Answer (4) Hint : It is an exception among members of Class

Reptilia. Sol. : Reptiles usually have 3 chambered heart;

but it is 4 chambered in crocodiles. 176. Answer (1) Hint : Identify a urochordate. Sol. : Ichthyophis : a limbless amphibian Pterophyllum : Angel fish Equus : Horse



177. Answer (2)

Hint : Club shaped gland in male cockroach.

Sol. : They reach anteriorly upto 5th abdominal segment. They secrete the outer layer of spermatophore.

178. Answer (1)

Hint : Unsegmented body.

Sol. : True metamerism was first shown by annelids. Tapeworm exhibits pseudometamerism.

179. Answer (4)

Hint : They absorb nitrogenous waste.

Sol. : Each tubule is lined by glandular ciliated epithelium which absorbs nitrogenous waste and converts it into uric acid. Hepatic caecae secrete digestive enzymes.

180. Answer (4)

Hint : It is unlike mammalian heart.

Sol. : Heart of cockroach is 13 chambered and neurogenic.

Test-8 (Code-B)_(Answers) All India Aakash Test Series for NEET-2020


All India Aakash Test Series for NEET - 2020

Test Date : 02/02/2020

ANSWERS

1. (3) 2. (2) 3. (1) 4. (1) 5. (3) 6. (1) 7. (4) 8. (2) 9. (3) 10. (1) 11. (4) 12. (2) 13. (3) 14. (4) 15. (3) 16. (3) 17. (1) 18. (4) 19. (1) 20. (3) 21. (2) 22. (4) 23. (1) 24. (2) 25. (3) 26. (1) 27. (3) 28. (2) 29. (3) 30. (3) 31. (2) 32. (3) 33. (2) 34. (4) 35. (3) 36. (1)

37. (2) 38. (3) 39. (4) 40. (1) 41. (2) 42. (3) 43. (1) 44. (3) 45. (1) 46. (2) 47. (4) 48. (1) 49. (4) 50. (2) 51. (1) 52. (2) 53. (2) 54. (2) 55. (2) 56. (4) 57. (4) 58. (3) 59. (1) 60. (3) 61. (4) 62. (1) 63. (3) 64. (4) 65. (4) 66. (1) 67. (3) 68. (1) 69. (1) 70. (1) 71. (2) 72. (2)

73. (4) 74. (4) 75. (3) 76. (4) 77. (3) 78. (3) 79. (4) 80. (3) 81. (1) 82. (4) 83. (4) 84. (4) 85. (3) 86. (1) 87. (2) 88. (3) 89. (2) 90. (4) 91. (2) 92. (3) 93. (2) 94. (1) 95. (2) 96. (2) 97. (3) 98. (3) 99. (3) 100. (2) 101. (4) 102. (3) 103. (2) 104. (4) 105. (3) 106. (3) 107. (2) 108. (4)

109. (3) 110. (4) 111. (2) 112. (2) 113. (1) 114. (3) 115. (2) 116. (2) 117. (2) 118. (1) 119. (4) 120. (4) 121. (4) 122. (2) 123. (4) 124. (3) 125. (2) 126. (2) 127. (4) 128. (1) 129. (3) 130. (3) 131. (4) 132. (4) 133. (3) 134. (3) 135. (3) 136. (4) 137. (4) 138. (1) 139. (2) 140. (1) 141. (4) 142. (1) 143. (2) 144. (1)

145. (3) 146. (2) 147. (1) 148. (4) 149. (3) 150. (4) 151. (3) 152. (2) 153. (1) 154. (4) 155. (4) 156. (2) 157. (4) 158. (4) 159. (3) 160. (4) 161. (4) 162. (3) 163. (2) 164. (1) 165. (1) 166. (1) 167. (2) 168. (1) 169. (3) 170. (2) 171. (4) 172. (4) 173. (4) 174. (2) 175. (4) 176. Deleted 177. (3) 178. (4) 179. (3) 180. (2)

TEST - 8 (Code-B)

All India Aakash Test Series for NEET-2020 Test-8 (Code-B)_(Hints & Solutions)


HINTS & SOLUTIONS

[PHYSICS] 1. Answer (3) Hint and Sol. : Beat frequency is the number of times the

resultant intensity becomes maximum or minimum in one second.

2. Answer (2) Hint : Frequency of nth overtones of open pipe

0( 1)

2+

=

n vf

Frequency of nth overtones of closed organ pipe(2 1)

4+

=

cn vf

Sol. : 0

1)2

(2 1)4

( +

=+

c

n vf

n vf

⇒ 0 2( 1)(2 1)

+=

+c

f nf n

3. Answer (1)

Hint : 1 2 1 22 cosI I I I I= + + φ

Sol. : ( )2max 1 2I I I= + for lmax φ = 0°

( )2min 1 2I I I= − for lmin φ = 180°

( )

1 2max min

max min 1 2

42

I II II I I I

−=

+ +

45

=

4. Answer (1)

Hint : 1 35( ) ,

4 4λ λ

+ = + = e e

Sol. : 1 = 22.5

3 = 116.5

⇒ 1

3

15

+=

+

ee

⇒ 5 1 + 5e = 3 + e

⇒ 3 154−

= e

⇒ 116.5 112.54−

=e

= 1cm 5. Answer (3)

Hint : 00

− = v vf f

v

Sol. : v0 = u + at v0 = at

0− =

v atf f

v

00= −

f af f tv

6. Answer (1) Hint and Sol. : Laplace’s correction in the formula for the speed

of sound in gases given by Newton, because sound wave propagates adiabatically, while Netwon assumed it isothermally.

7. Answer (4)

Hint : 1 ′λ = + λ

svv

Sol. : 1 604

′λ = + × ×

vv

5 604

′λ = × = 75 cm

8. Answer (2) Hint : Distance = v × t

Sol. : 1 250T

=

1 s

250=T

50 1 s250 5

= =t

13405

d = × = 68 m

Test-8 (Code-B)_(Hints & Solutions) All India Aakash Test Series for NEET-2020


9. Answer (3)

Hint : =µTv

Sol. : ω

=vk

5054

=v

v = 40 m/s

3404 10

T−=

×

T = 1600 × 4 × 10–3 T = 6.4 N 10. Answer (1)

Hint : In string, 2vfL

= (for same T and µ)

Sol. : L = L1 + L2 + L3

0 1 2 32 2 2 2

= + +v v v vf f f f

0 1 2 3

1 1 1 1= + +

f f f f


sin cos(60 )3xy tπ

= π

3π

=k

⇒ 2

3π π

=λ

⇒ 6 mλ =

Distance between two consecutive node = 3 m2λ

=

Distance between two consecutive node and

antinode 1.5 m4λ

= =

2 60π = πf

30 Hz=f

60 180 m/s

3

ω π= = =

πv

k

12. Answer (2) Hint : Use trigonometric identity

cos2θ = 1 – 2sin2θ 2cosAcosB = cos(A + B) + cos(A – B)

Sol. : 24sin cos(500 )2ty t =

1 cos4 cos(500 )

2ty t− =

y = 2cos500t – 2cost cos(500t) = 2cos500t – [cos(500t + t) + cos(500 – t)] = 2cos500t – cos501t – cos499t So the given expression is the resultant of three

independent harmonic oscillation. 13. Answer (3) Hint and Sol. : Apparent change in wavelength takes place due to

motion of source while apparent change in frequency takes place due to motion of source and observer both.

14. Answer (4) Hint and Sol. : In a closed organ pipe only odd harmonics will be

present i.e. 3 5, ,

4 4 4⋅ ⋅ ⋅

v v v therefore 800 Hz

frequency cannot be obtained in it. 15. Answer (3)

Hint : Fundamental frequency for open 0 2vf =

Fundamental frequency of closed pipe 4

=

v

Sol. : 0 2vf =

When dipped in water pipe behave as close organ pipe

Frequency for third harmonic of a closed organ pipe.

3 4 24 3 2

×= = =

×

v v vf

02=f f

16. Answer (3) Hint and Sol. : In stationary wave, strain at the

node will be maximum.




Distance travelled by the sound in 10 second = 330 × 10 = 3300 m Distance travelled by the engine = 700 m v × 10 = 700 m v = 70 m/s 18. Answer (4)

Hint : γ=

RTvM

Sol. : r .m.s3

=RTvM

20

RTCM

γ=

2

3=

C RTM

2

20 3

CC γ=

2

20

43 3

CC = ×

023

C C=

19. Answer (1)

Hint : 12

=µ

Tf

f1 – f2 = 4

Sol. : 12

TfA

=ρ

12

∆ ∆=

f Tf T

4 1

400 2∆

=T

T

0.02∆=

TT


According to above diagram, velocity of particle at

point D will be along positive y direction. 21. Answer (2)

Hint : =ρTvA

Sol. : On increase in density of wire, speed of wave in wire decreases.

1100 1002

∆ ∆ρ× = ×

ρv

v

1 4%2

= ×

= 2% decrease 22. Answer (4) Hint : Time interval between two maximum

compression to appear at a place T.

Sol. : 1400

=T

23. Answer (1) Hint and Sol. : Propagation of mechanical wave

through a medium carries energy and momentum. 24. Answer (2) Hint : ∆φ = ∆k x

Sol. : 20.03 50 10−∆φ = π × ×

10.032

= π ×

3200

π=

25. Answer (3)

Hint : 2

1 2

sintancos

∆φφ =

+ ∆φA

A A

Y = Rsin(ωt + φ)

Sol. : 10sin

2tan10 10cos

2

π

φ =π

+

tan 1φ =

4π

φ =



26. Answer (1) Hint : sin( )y A kx t= + ω

Sol. : A = 2 m

2π=

λk

11 m2

−=k

2ω = πf

122

ω = π ×π

= 1 rad/s

2sin2

= +

xy t

27. Answer (3)

Hint : and ω= ω =Pv A v

k

Sol. : vP = 100 × 0.05 = 5 m/s

10050

=v

v = 2 m/s

5 : 2=Pvv

28. Answer (2)

Hint : sin4π

=tx a

Sol. : x(0) = asin0 = 0 xi = 0

sin 64π = ×

fx a

3sin2π =

fx a

sin2π

= − ⇒ = −f fx a x a

Magnitude of displacement = a 29. Answer (3) Hint : cos= − ω ωV A t

Sol. : cos= − ω ωdx A tdt

cos= − ω ω∫ ∫dx A tdt

sinω= − ω

ωAx t

sin( )= ω + πx A t

30. Answer (3)

Hint : 2= πIT

mgd

Sol. :

2

23

2

= π× ×

mTmg

223

= πTg

2 ∝ T


6sin100 8sin 1002π = π + π +

x t t

⇒ 6sin100 8cos100x t t= π + π ⇒ 10sin(100 )= π + φx t

4tan3

φ =

⇒ 53φ = ° 10sin(100 53 )= π + °x t Particle execute S.H.M with amplitude 10. 32. Answer (3)

Hint : 2= πTg

Sol. : 2

1 24=

π

T g

2

2 244

=π

T g

1 21 2 +

= π T

g

On putting values of 1 2and

1 5T T=



33. Answer (2)

Hint : 2= πMTk

Sol. : M = 8m + m = 9 m mg = kx

=mgkx

92= πmT mgx

92= πxT

g

6= πxTg

34. Answer (4)

Hint : 2 2= ±ω −v A x

Sol. : 2 2

112 22 2

ω −=

ω −

A xvv A x

⇒ 2

2

16 912 16

−=

−

A

A

⇒ 2

2

4 93 16

−=

−

A

A

On solving A = 5 cm 35. Answer (3)

Hint : 2 2 21 12 2

= = ωE kA m A

Sol. : 2 2 2 21 1 2 2

1 12 2

ω = ωm A m A

2 2 2 21 1 2 2ω = ωA A

⇒ 2 2 2 225 8 4× = ω

⇒ 2 25 4ω = ×

⇒ 10 unitω =

36. Answer (1)

Hint : 2 2= ±ω −v A x

Sol. : 2 2= ±ω −v A x

2

2 22

= −

ω v A x

2

2 22 + =

ωv x A

2 2

2 2 1( )

+ =ω

v xA A

→ This is the equation of an

ellipse. 37. Answer (2)

Hint : ms3RTVM

=

Sol. : 1

2

ms 1

ms 2

V TV T

=

1

1

ms

ms 2

3002VV T

=

T2 = 600 K T2 = 327°C 38. Answer (3)

Hint and Sol. : 0

=RkN

R = N0k, therefore statement 3 is incorrect. 39. Answer (4)

Hint : V = V0(1 + αt)

Sol. : 0 1273

= +

tV V

0273

273+ =

tV V

0273

=V TV

Here T is in kelvin

0tan273

θ =V

40. Answer (1) Hint : PV = nRT



Sol. : 1400

(27 )nRP

x A=

+

2500

(27 )nRP

x A=

−

P1 = P2

⇒ 400 500(27 ) (27 )nR nR

x A x A=

− +

⇒ 4 5(27 ) 27

=− +x x

⇒ 27 × 4 + 4x = 5 × 27 – 5x ⇒ 9x = 27 ⇒ x = 3 cm 41. Answer (2) Hint and Sol. :

. 1molecule- degree of freedom 2

K E kT=

Diatomic molecule have two rotational degree of freedom. Average rotational kinetic energy per molecule

122

= × kT

= kT 42. Answer (3) Hint : PV = nRT

Sol. : 1 500=

×PVn

R

2432 500

=PV

R

2 2 400= ×

×P Vn

R

2 800=

PVnR

232 800 32 = =

x PV xnR

24 58×

=x

x = 15 g 43. Answer (1) Hint and Sol. : Charle’s law that leads to absolute

scale of temperature. 44. Answer (3) Hint : Dalton’s law of partial pressure.

Sol. : 1 =PVnRT

2 =PVnRT

3 =PVnRT

N = n1 + n2 + n3

1 = + +PV PV PV PVRT RT RT RT

P1 = 3P

45. Answer (1) Hint : Boyle’s law Sol. : PV = constant

P09V = P1(9V + V)

01

910

=PP

P1 × 9V = P2(9V + V)

12

910

=PP

2 09 9

10 10= ×P P

2 081

100=P P

[CHEMISTRY]

46. Answer (2)

Hint : n-hexane in such conditions undergoes aromatization.

Sol. :

47. Answer (4)

Hint : Degree of unsaturation of C4H8 = 1 Sol. : Number of π bond + ring = 1

Hence, 48. Answer (1) Hint : An aromatic compound should have

complete delocalisation of π-electrons in the ring. Sol. : Heterocyclic ring must comprise of different

atoms.



49. Answer (4)

Hint : % of 31.4 meq. of NHNMass of compound

×=

Sol. : Meq. of NH3 = meq. of H2SO4

= (0.1 × 2) × 100 = 20

% of 1.4 20N 37.33%0.75

×= =

50. Answer (2) Hint : Precipitate of ammonium phosphomolybdate

is obtained.

Sol. :

51. Answer (1) Hint : NaNH2 is a strong base, capable of

elimination. Sol. :

52. Answer (2) Hint : N2 is obtained.

Sol. : x y zyC H N 2x CuO2

+ + →

2 2 2y z yxCO H O N 2x Cu2 2 2

+ + + +

53. Answer (2) Hint : Rain water is acidic in nature. Sol. : Normally it has a pH of 5.6 54. Answer (2) Hint : A non-benzenoid compound does not

contain any benzene ring.

Sol. :

55. Answer (2) Hint : Greenhouse gases contribute to global

warming and hence, climate change.

Sol. : Greenhouse gases : Carbon dioxide, Methane, Ozone, Nitrous oxide,

Water vapour. 56. Answer (4)

Hint :

57. Answer (4) Hint : DDT is a pesticide. Sol. : Organic toxins : Aldrin, Dieldrin Herbicides : Sodium arsinite (Na3AsO3)

58. Answer (3) Hint : BOD value measures how much oxygen is

required by bacteria to break down the organic matter present in water.

Sol. : Highly polluted water requires more oxygen by bacteria. Hence, its BOD value is more than 17 ppm.

59. Answer (1) Hint : HBr in presence of peroxides (C6H5CO)2O2

generates Br (free radical) Sol. :

60. Answer (3) Hint : Zn/H2O prevents oxidation of aldehyde to

carboxylic acid. Sol. :

61. Answer (4)

Hint :

Sol. : trans-but-2-ene is more stable than cis form.



62. Answer (1) Hint : Reaction of alkyl bromide with alc. KOH is

an example of β-elimination Sol. : Formation of more substituted alkene is

preferred in the case of β-elimination. Hence, rate of reaction is

63. Answer (3) Hint : H2(excess) Pd reduces C ≡ C as well as

C = C.

Sol. :

3, 4, 4, 5- tetramethylheptane. 64. Answer (4) Hint : Propane can be prepared by Kolbe’s

electrolytic method. Sol. :

(1)

(2)

(3)

(4) 65. Answer (4) Hint : Rotation about C – C single bond in

alkanes is almost free for all practical purposes. Sol. : Conformers of ethane are infinite. 66. Answer (1) Hint : Halogenation of alkanes is a free radical

substitution reaction. Sol. :

67. Answer (3) Hint : HBr gives electrophilic addition to alkene.

Sol. :

68. Answer (1) Hint : Markovnikov addition of H2O take place. Sol. :

69. Answer (1)

Hint : Trans-But-2-ene is

Sol. :

70. Answer (1) Hint : Metamerism arises due to different alkyl

chains on either side of the functional group in the molecule.

71. Answer (2) Hint : The given characteristic is associated with

mesomeric effect. 72. Answer (2)

Hint : shows +R effect.

Sol. : +R effect : – NH2

–R effect : – COOH, = O, – CN

73. Answer (4) Hint : Photochemical smog has high

concentration of oxidising agents. Sol. : The main components of the

photochemical smog result from the action of sunlight on unsaturated hydrocarbons and nitrogen oxides produced by automobiles and factories.



74. Answer (4)

Hint : Uncatalysed oxidation of SO2 to SO3 is slow.

75. Answer (3)

Hint : CH3OH is more acidic than hydrocarbons.

Sol. : Acidic behavior :

CH3OH > C2H2 > C2H4 > C2H6

(i) (iii) (iv) (ii)

76. Answer (4)

Hint : Soda lime is CaO + NaOH

Sol. :

77. Answer (3)

Hint : 1 2 3 4

2CH CH C CH= − ≡

Sol. : C2 ⇒ sp2 hybridised

C3 ⇒ sp hybridised

78. Answer (3)

Hint : Compound B. P.Chloroform 334 K

Aniline 457 K

Sol. : Liquids having sufficient difference in their boiling points are separated by distillation.

79. Answer (4)

Hint : AgNO3 gives ppt. with NaCN and Na2S.

80. Answer (3)

Hint : Cyclopentadienyl anion is an aromatic.

81. Answer (1)

Hint : Longest chain of carbon atoms containing –COOH is considered.

Sol. :

82. Answer (4)

Hint : Carbonyl compound is obtained.

Sol. :

83. Answer (4) Hint : Resonance increases the stability of

carbocation.

Sol. : effect

effect

84. Answer (4)

Hint :

Sol. : Hyperconjugation is a permanent effect. 85. Answer (3) Hint : Br2 decolourisation is test for unsaturation. Sol. :

86. Answer (1) Hint : Maximum prescribed limit of Cd in drinking

water is 0.005 ppm. 87. Answer (2) Hint : Electron donating group present on

benzene increases the rate of SE reaction. Sol. :

2 3 2NMe CH Cl NO( M effect) ( H effect) ( I effect) ( M effect)

− − − −+ + − −


89. Answer (2) Hint : –OH group increases electron density on

ortho and para positions of benzene ring.



Sol. : and so on

–OH group activates the benzene ring for the attack of and deactivates for attack of 2NO .−

90. Answer (4) Hint : Urea is NH2CONH2

Sol. :

[BIOLOGY]91. Answer (2) Hint : Apical dominance is shown by auxin. Sol. : Cytokinin counteracts apical dominance by

increasing the supply of water and minerals to lateral buds.

92. Answer (3) Sol. : Gibberellins-composition is terpenes. 93. Answer (2) Sol. : Buttercup shows environmental

heterophylly. Cotton, coriander and larkspur show developmental heterophylly.

94. Answer (1) Sol. : Viviparous seed germination is shown by

Sonneratia 95. Answer (2) Hint : GA induces synthesis of hydrolytic

enzymes. Sol. : Aleurone layer has hydrolytic enzymes

which move to starchy endosperm to convert complex carbohydrates to simpler ones.

96. Answer (2) Hint : Cytokinin was first isolated from unripe

maize grains. Sol. : Cytokinin delays senescence, promotes

cell division and is used in tissue culture. 97. Answer (3) Hint : Ethylene is a gaseous phytohormone. Sol. : Precusor of ethylene is methionine which is

a sulphur containing amino acid. 98. Answer (3) Hint : Seed requires anchorage first. Sol. : Radicle is the first developed structure

within the seed. 99. Answer (3) Hint : Ethylene is a gaseous hormone. Sol. : ABA is called stress hormone as stress

stimulates its synthesis.

100. Answer (2) Hint : Pepper is a DNP. Sol. : Day Neutral Plants (DNP) do not show any

correlation between exposure to light duration and induction of flowering response.

101. Answer (4) Hint : Reduced ubiquinone is ubiquinol. Sol. : Cyt c is a small protein found attached to

the outer surface of inner mitochondrial membrane.

102. Answer (3) Hint : During fermentation, pyruvic acid is

reduced to lactic acid. Sol. : In EMP, hexose sugar splits into two

molecules of triose sugar with the help of aldolase. 103. Answer (2) Hint : Substrate level phosphorylation means

direct ATP synthesis. Sol. : Glycolysis and Krebs cycle or citric acid

cycle produces direct ATP molecules through substrate level phosphorylation.

104. Answer (4)

Hint : 2

2

Volume of CO evolvedRQVolume of O consumed

=

Sol. : RQ value for oxalic acid is 4. 105. Answer (3) Hint : Glycolysis is also known as EMP pathway. Sol. : Glycolysis is common step in both aerobic

and anaerobic respiration. 106. Answer (3) Hint : Acetyl CoA is a 2C compound. Sol. : End product of glycolysis is pyruvic acid

(3C). Acetyl CoA releases 2CO2 molecules upon

complete oxidation through TCA cycle. It is the product of link reaction.



107. Answer (2)

Sol. :

108. Answer (4) Hint : To develop the proton gradient more

H+ ions should be present in the lumen of thylakoids than stroma of chloroplast.

Sol. : When number of H+ ions increases in lumen, pH of lumen decrease.

109. Answer (3) Hint : Dimorphic chloroplasts are seen in C4

plants. Sol. : Sugarcane is a C4 plant.

Opuntia – CAM plant Pea and wheat } C3 plants

110. Answer (4) Sol. : Cyclic flow of electrons is seen in cyclic

photophosphorylation. PS II involves splitting of H2O.

111. Answer (2) Hint : CO2 concentration and light intensity both

should be optimum to increase the yield of plants. Sol. : At high light intensity and high CO2

concentration, both C3 and C4 plants show increase in rate of photosynthesis.

112. Answer (2) Hint : In C3 and C4 plants Calvin cycle is

common. Sol. : Same amount of NADPH are required in C3

and C4 plants as regeneration of PEP does not require NADPH.

113. Answer (1) Hint : C4 plants show double carboxylation.

Sol. : In C4 plants Calvin cycle operates in bundle sheath cells so concentration of CO2 increases at enzyme site.

PEPcase does not show oxygenase activity. 114. Answer (3) Hint : Photorespiration takes place in C3 plants

not in C4 plants.

Sol. : OEC involves Mn and Cl for splitting of water.

115. Answer (2) Sol. : Calvin cycle or C3 cycle is present in all

plants.

4Kranz anatomy

C plantsDouble carboxylation

Scotoactive stomata } CAM plants 116. Answer (2) Hint : C4 pathway is called Hatch and Slack

pathway. Sol. : In C4 plants the primary acceptor is PEP

which is a 3C compound and found in mesophyll cells.

117. Answer (2) Hint : Calvin cycle takes place in bundle sheath

cell of C4 plants.

Sol. : Sugar is produced in Calvin cycle. In mesophyll cells of C4 plants, primary fixation of CO2 takes place.

118. Answer (1) Hint : Nitrite reductase reduces nitrite ions to

ammonia. Sol. : It is found in leaves and contain copper and

iron but not Mo. 119. Answer (4) Hint : Mn is a micronutrient. Slight excess of it

causes toxicity. Sol. : Mn toxicity results in reduction in uptake of

Fe and Mg. 120. Answer (4) Sol. : Calcium is required for the formation of

mitotic spindle. 121. Answer (4) Hint : Mobile elements readily move from older to

younger tissues. Sol. : Deficiency symptoms of mobile elements

are first seen in the older tissues. 122. Answer (2) Hint : Nitrate cannot be used directly by plants as

such. First, NO3– is converted into NO2

– and then NH3.

Sol. : Nitrate assimilation involves nitrate reduction into nitrite.

Conversion of atmospheric N2 to 3NO− occurs by non-biological nitrogen fixation.



123. Answer (4) Hint : Trace elements are micronutrients. Sol. : B, Cu and Mo are micronutrients. 124. Answer (3) Hint : Chemoautotrophs do not require solar

energy to synthesise food. Sol. : Nitrosomonas, Nitrobacter } Chemoautotrophs Rhizobium, Frankia } Heterotrophs Nostoc, Anabaena } Photoautotrophs 125. Answer (2) Hint : Only prokaryotes can fix the atmospheric

nitrogen for a solution. Sol. : Only some bacteria and cyanobacteria can

fix the atmospheric nitrogen. 126. Answer (2) Hint : Solute potential is developed when solute

is added to water.

Sol. : ψw of pure water is zero and ψs is always negative for a solution.

127. Answer (4) Hint : In roots, generally the concentration of

minerals is higher than the soil. Sol. : Most minerals are actively absorbed by the

roots. For carbon fixation, electrons and hydrogens are

given by H2O.

128. Answer (1) Hint : Chief sinks for mineral elements are those

parts of plants which are either storage organs or they are growing regions.

Sol. : Apical meristem requires minerals as the cells of it are in continuous state of division.

129. Answer (3) Hint : Osmotic concentration is directly

proportional to the amount of solutes. Sol. : If cell sap has high osmotic concentration,

it means it has more solutes and it is hypertonic as compared to the surrounding solution.

130. Answer (3) Hint : Facilitated diffusion involves movement of

substances through membrane proteins. Sol. : Hydrophilic substances cannot move

through the lipid bilayer hence these molecules are facilitated by some membrane proteins.

131. Answer (4) Sol. : Xylem is associated with translocation of

mineral salts, water, organic nitrogen and hormones. Phloem transports photosynthates.

132. Answer (4) Hint : Girdling experiment showed food

translocation by a tissue. Sol. : Food translocation takes place by phloem

and it was identified by the girdling experiment. 133. Answer (3) Hint : Addition of solutes increases the osmotic

pressure. Sol. : During phloem loading, sucrose moves into

the phloem which increases its OP and becomes hypertonic.

Phloem loading is an active process. At the sink, osmotic pressure in sieve tube decreases and water moves out of the phloem.

134. Answer (3) Hint : Active absorption is absorption by root. Sol. : Force for active absorption develops in

root. 135. Answer (3) Hint : Water moves from high to low water

potential. Sol. : Cell A Cell B

ψw = – 5 ψw = – 2

Cell B has higher water potential than cell A, hence cell A is more deficit in water. Osmotic potential (ψs) of cell B is – 8 atm, whereas osmotic pressure will be 8 atm as ψs and OP are opposite in sign.

136. Answer (4) Hint : It is unlike mammalian heart. Sol. : Heart of cockroach is 13 chambered and

neurogenic. 137. Answer (4) Hint : They absorb nitrogenous waste. Sol. : Each tubule is lined by glandular ciliated

epithelium which absorbs nitrogenous waste and converts it into uric acid. Hepatic caecae secrete digestive enzymes.

138. Answer (1) Hint : Unsegmented body. Sol. : True metamerism was first shown by

annelids. Tapeworm exhibits pseudometamerism.



139. Answer (2) Hint : Club shaped gland in male cockroach. Sol. : They reach anteriorly upto 5th abdominal

segment. They secrete the outer layer of spermatophore.

140. Answer (1) Hint : Identify a urochordate. Sol. : Ichthyophis : a limbless amphibian Pterophyllum : Angel fish Equus : Horse 141. Answer (4) Hint : It is an exception among members of Class

Reptilia. Sol. : Reptiles usually have 3 chambered heart;

but it is 4 chambered in crocodiles. 142. Answer (1) Hint : It includes true fishes. Sol. : Pisces and amphibians lack amnion so

considered as anamniotes. 143. Answer (2) Hint : It operates between intestine and liver. Sol. : The portal system of circulation bypasses

the heart. Hepatic portal and renal portal system are well developed in amphibians.

144. Answer (1) Hint : Ecdysis is shedding of skin. Sol. : Shedding of scaly epidermis of skin

periodically is called ecdysis or moulting. It is shown by snakes and lizards among vertebrates.

145. Answer (3) Hint : Petromyzon belongs to Class Cyclostomata. Sol. : Animals belonging to Division Agnatha

possess mouth without jaws and have a single nostril. Notochord persists throughout life in primitive craniates such as Myxine.

146. Answer (2) Hint : It has pneumatic bones. Sol. : Corvus is a bird and a homeotherm. 147. Answer (1) Hint : They originate from ventral side of 9th

segment. Sol. : Anal or caudal styles are the features of

sexual dimorphism. They are chitinous spine like unjointed structures.

148. Answer (4) Hint : It is an echinoderm.

Sol. : Asterias contains a calcareous endoskeleton.

149. Answer (3) Hint : Identify a reptile. Sol. : Bangarus, commonly known as krait is a

snake belonging to Class Reptilia. 150. Answer (4) Hint : This structure is ventrally located. Sol. : Earthworm and cockroach both possess a

double, ventral, solid nerve cord with paired ganglia.

151. Answer (3) Hint : These excretory structures are most

numerous. Sol. : Hepatic caecae are present at junction of

foregut and midgut and secrete digestive juices. Malpighian tubules perform excretion. 10 pairs of spiracles are present in cockroach.

152. Answer (2) Hint : Avian feature. Sol. : Aves are warm blooded, oviparous and

dioecious. 153. Answer (1) Hint : This is also present in coelenterates. Sol. : Statocysts are statoreceptors in

coelenterates and arthropods. 154. Answer (4) Hint : Petromyzon is a cyclostome Sol. : The mouth of cyclostomes is suctorial,

circular and without jaws. Mouth of chondrichthyes is located ventrally. Terminal mouth is a feature of bony fishes.

155. Answer (4) Hint : Exoskeleton is present ventrally, laterally

and dorsally. Sol. : Dorsally present sclerites are called

tergites, ventrally present ones are called sternites and laterally present are called pleurites.

All are collectively known as sclerites. 156. Answer (2) Hint : This is located in anterior part of body. Sol. : Body of Balanoglossus is divisible into

anterior proboscis, a collar and a long trunk. 157. Answer (4) Hint : They are locomotory structures. Sol. : Parapodia and setae are present in Nereis

while only setae are present in Pheretima.



158. Answer (4) Hint : This phylum includes animals with flame

cells as excretory structures. Sol. : Cephalization, acoelomate condition and

bilateral symmetry are observed in phylum Platyhelminthes.

159. Answer (3) Hint : It causes elephantiasis. Sol. : Hookworm and roundworm are

monogenetic and oviparous. Tapeworm is a digenetic platyhelminth but oviparous.

160. Answer (4) Hint : Open circulatory system. Sol. : Nereis and Pheretima are annelids which

possess closed circulatory system. Among molluscs, cephalopods such as Octopus possess closed circulatory system.

161. Answer (4) Hint : They are also named as collar cells. Sol. : Archaeocytes are undifferentiated cells

Flagellated choanocytes are responsible for water current and receiving sperms during sexual reproduction.

162. Answer (3) Hint : Pseudocoelom is a feature of

Aschelminthes. Sol. : Protostomes are schizocoelomates

Ancylostoma and Enterobius are Aschelminthes. Blind sac body plan is found in Coelenterates, Ctenophores and Platyhelminthes.

163. Answer (2) Hint : Alternation of generation. Sol. : Most species of Hydra are dioecious. No

free larval stage is present in Hydra. Hydra does not show metagenesis.

164. Answer (1) Hint : They are edible fishes. Sol. : Labeo (Rohu), Catla (Katla) and Clarias

(Magur) are freshwater bony fishes. 165. Answer (1) Hint : Lepisma does not exhibit intermediate

stages in development. Sol. : Ametabolous – Egg → Young → Adult Hemimetabolous – Egg → Naiad → Adult Paurometabolous – Egg → Nymph → Adult

Holometabolous – Egg → Larva → Pupa → Adult

166. Answer (1)

Hint : It is called the boring sponge. Sol. : Cliona bores through the oyster shell and

prevents pearl formation.

167. Answer (2)

Hint : Cluster of relevant cells can develop into a new organism.

Sol. : They may have calcareous or siliceous spicules. Spongilla is freshwater sponge. They are mostly asymmetrical.

168. Answer (1) Hint : Bony fishes have operculum.

Sol. : Chondrichthyes are marine and possess 5-7 pairs of gills without operculum. Osteichthyes are freshwater and marine and possess 4 pairs of gills with operculum.

169. Answer (3)

Hint : Metagenesis is exhibited by Obelia. Sol. : Polyp and medusae both are diploid

stages. Comb plates are present in ctenophores.

170. Answer (2) Hint : It is a tunicate.

Sol. : Retrogressive metamorphosis is a phenomenon wherein the larva is more advanced than adult. Progressive metamorphosis is shown by Rana, Amphioxus and Petromyzon.

171. Answer (4)

Hint : Scoliodon is shark. Sol. : Myxine, a Cyclostomate has a circular

mouth. Exocoetus is a bony fish and Scoliodon is a cartilaginous fish.

172. Answer (4)

Hint : Snake does not undergo metamorphosis. Sol. : Amphibians possess larval stages but the

same are absent in reptiles.

173. Answer (4) Hint : Island country near Australia.

Sol. : Apteryx (kiwi) is native to New Zealand.

Pavo can fly to a limited extent. 174. Answer (2)

Hint : Wing muscles are anchored here. Sol. : Birds have a well developed keel

(sternum), the bone that anchors the muscles that move a bird’s wings.



175. Answer (4) Hint : Lepisma and scorpion are arthropods. Sol. : Dolphin is a mammal and it respires by

lungs. Dentalium is a mollusc and respires by feather like gills.

176. Deleted 177. Answer (3) Hint : They belong to phylum Coelenterata. Sol. : Sea pen is Pennatula, sea fan is Gorgonia,

Sea hare is Aplysia (Mollusc), sea horse is fish named Hippocampus.

178. Answer (4) Hint : They show true metamerism.

Sol. : Annelids are characterised by presence of metameres, bilateral symmetry, presence of a true coelom and triploblasty.

179. Answer (3) Hint : These organisms have pentamerous radial

symmetry. Sol. : Radula is present in molluscs.

Echinoderms lack excretory system. 180. Answer (2) Hint : They are shelled organisms. Sol. : Octopus, Aplysia and Chaetopleura belong

to phylum Mollusca. Taenia is a flatworm. Sepia is a mollusc and Ancylostoma is a roundworm.

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