All India Aakash Test Series for NEET - 2020 · 2020. 4. 7. · All India Aakash Test Series for...

Open Mock Test-4 (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 : 05/04/2020 ANSWERS

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

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

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

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

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

OPEN MOCK TEST - 4 (Code-A)

All India Aakash Test Series for NEET-2020 Open Mock Test-4 (Code-A)_(Hints & Solutions)


HINTS & SOLUTIONS

[PHYSICS] 1. Answer (2) Hint : If two vector are perpendicular then their dot

product should be zero.

Sol. : Let ˆ ˆr ai bj= +

ˆ ˆ ˆ ˆ( ) (3 4 ) 0ai bj i j+ ⋅ + =

34ab = −

2 2 10a b+ =

2

2 3 1004aa + − =

a = 8

3 84

b = − ×

b = –6

ˆ ˆ8 6r i j= −

2. Answer (2) Hint & Sol. : Density of electric field lines indicate

the relative magnitude of charge. From positive charge, electric field lines are radially outward, while for negatively charged particle, electric field lines are radially inwards.

3. Answer (4) Hint and Sol. : Inside a conducting plate, the electrostatic field is

zero. While outside conducting plate, electrostatic field would be uniform.

4. Answer (3)

Hint : −∂ ∂ ∂= − −

∂ ∂ ∂ˆ ˆ ˆ U U UF i j k

x y z.

Sol. :

( ) ( ) ( )2 2 22 2 2

ˆ ˆ ˆ−∂ + ∂ + ∂ +

= − −∂ ∂ ∂

xyz x xyz x xyz xF i j k

x y z

( ) ( ) ( )2 2ˆ ˆ ˆ2 1 2 4= − + − −

F yz i xz j xyz k

At origin ( )ˆ1 N= −

F i

5. Answer (2) Hint : Series and parallel combination.

Sol. : = + + =net 22 2C CC C C

6. Answer (1) Hint : Velocity of efflux = 2gh

Sol. : ( ) ( )2 10 2 40 m/sv = × × = .

Rate of flow of water = av

64 10 40−= × ×

68 10 10−= × × = 8 × 3.16 × 10–6 = 25 × 10–6 m3/s = 25.0 cc/s 7. Answer (3) Hint : Slope of (stress-strain) graph = Young’s

modulus.

Sol. : ( ) 4tan 533AY = ° =

( ) 3tan 374BY = ° =

16 9 169

AA B

B

Y Y YY

= ⇒ =

8. Answer (3) Hint : For adiabatic process γPV = constant Sol. : Since mass of gas remains constant and

=

m dV

1⇒ ∝d

V

∴ for adiabatic process

1 constantγ

=

Pd

Pd–γ = constant

−γ −γ′ ′=Pd P d

32γ

γ′ ′ = =

P dP d

for diatomic gas : 75

γ =

( )7/5 732 2 128PP

′= = =

9. Answer (2)

Hint : T = Time to attain maximum height = ug

Open Mock Test-4 (Code-A)_(Hints & Solutions) All India Aakash Test Series for NEET-2020


Sol. : 67 6.7 s10

T = =

Distance travelled in 7th second

( ) ( )2 21 10.7 0.3 2.9 m2 2

g g= + =

10. Answer (1) Hint : Lenz’s Law. Sol. : As current in loop A increases, the flux

through loop B would increase. As per the Lenz’s law, the loop B would tend to move away from loop A i.e. rightwards. Thus, a force towards left must be applied.

11. Answer (4)

Hint : Current ( ) dQidt

= and = ∫ 2H i Rdt

Sol. : 2 12= = −dQi tdt

( )22 2 12 2H i R dt t dt= = −∫ ∫

1

2

02 (4 144 48 )

= + −

∫ t t dt

1 1 1

2

0 0 02 4 144 48

= + −

∫ ∫ ∫dt t dt tdt

144 482 43 2

= + −

= 8 + 96 – 48 = 56 J 12. Answer (2)

Hint : ( )2 22 f i

Be R Rω= − .

Sol. : ( )2 202 2

B Be a aω ω= − =

13. Answer (4) Hint & Sol. : For A = 0, B = 1 and C = 1, Y = 0 14. Answer (1) Hint & Sol. : Electromagnetic spectrum in order of

increasing frequency and decreasing wavelength is

Radiowaves – Microwaves – Infrared – Visible – Ultravoilet – X ray – Gamma rays

15. Answer (4) Hint : Normal force by wall of tube provides

necessary centripetal force to a small element of water while normal force by water on the wall of tube leads to develop tension in it.

Sol. :

Centripetal force dN = 2Tsinθ 2( )

=dm v

R

2

2 sin= θv dm T

R

= π θ ρ

2

(2 )2ddm R

As θ → θ = θ0 sin

π θ ρ

= = πρθ

22

2 2(2 ) 14

2 sin 4

dv RT v d

R

16. Answer (1) Hint : Tangential acceleration will be zero if

0a v⋅ = .

Sol. : ( ) ( )ˆ ˆ ˆ ˆ2 2 2 2 0i j i j+ α ⋅ + = + α =

1α = − 17 Answer (2) Hint : Energy conservation law.

Sol. : ° = 21sin30 ( )2

mg K x

× × × = ×1 110 10 100 92 2

= 9 m

Before hitting spring distance travelled = 9 – 3 = 6 m 18. Answer (4) Hint & Sol. : Electrostatic field is always conservative in nature.

If lift is moving up or down with uniform velocity, the apparent weight is equal to true weight.

19. Answer (2)

Hint : Work done = change in kinetic energy

Sol. : 314

x t=

23Velocity4

dx tdt

= =

( )1s3 m/s4tv = =



( )3 s27 m/s4tv = =

Work done = ( )2 21 27 34 90 J

2 4 4

− =

20. Answer (3) Hint : Spring force (F) = kx

Potential energy (U) = 2

212 2

Fkxk

=

Sol. : 1 2

2 12U k

U k= =

21. Answer (2) Hint : T = m (g – a)

Sol. : 34 4g mgT m g = − =

Displacement = ( )21 903 m2 4 8

=

g

W = TScos180°

× × ×= − × = −

×3 90 3 6 10 90

4 8 4 8mgW

2025 506.25 J4

= − = −

22. Answer (4) Hint : In case of non-coherent sources I net I= ∑

Sol. : Let intensity from one source by I0 3I0 = 6I

⇒ I0 =2I

If one source is switched off Net intensity = 2I0 = 4I 23. Answer (1) Sol. : mg – T = ma

20 – T = 2 × 0.1 α …(i)

TR = Iα

T × 0.1 = 0.4α …(ii) On solving (i) and (ii)

α = 4.8 rad/s2

24. Answer (4)

Hint : Escape speed ( ) 22 83

= = πeGMv GR dR

Sol. : 283

= πev GR d

∝ev R for same density.

planet

earth

( )3

( )= =e p

e e

v Rv R

25. Answer (2) Hint & Sol. : Since time required for same

temperature increase is larger in liquid state than solid state, the specific heat capacity is higher in liquid state.

Since in phase transition, the heat required is less in fusion than evaporation, hence latent heat of fusion is less.

26. Answer (3)

Hint : 2eff

lTg

= π

Sol. : eff 4= − ⋅

m dmg mg gd

34 4effg gg g= − =

23

′⇒ =T T

27. Answer (2) Hint : apparent apparent actual actualλ = λf f , for a

stationary observer

Sol. : apparent340340 200380

= λ

apparent380 1.9 m200

λ = =

28. Answer (2)

Hint : g

g

I GS

I I=

−

Sol. : I = 6Ig

5GS⇒ =

S = 9.6 Ω 29. Answer (1)

Hint & Sol. : 2 ITMB

= π

When cut along axis and2 2I MI M′ ′= =

T T′∴ =



30. Answer (4) Hint & Sol. : The magnetic field due to circular

loop is zero. Also, magnetic field due to straight wires also cancel each other as they are opposite in direction.

31. Answer (1) Hint : Resonance in series L.C R A.C. circuit. Sol. : When L is removed, phase difference is 45°

thus Xc = R as on removal of C, phase difference is also 45°,

thus XL = R ∴ In original circuit, impedance

( )22c LZ R x x R= + − =

∴ Power factor cos 1φ = =RZ

32. Answer (4)

Hint : hc k= φ +λ

(φ : work function, k : kinetic energy of fastest electrons)

Sol. : ( )

2case I :/ 2

hc hck k= φ + ⇒ = φ +λ λ

( )

6case II : 6 6/ 6

hc hck k= φ + ⇒ = φ +λ λ

Solving the above two equations,

6 5hc= φ

λ

65

hc⇒ φ =

λ

33. Answer (2)

Hint : Band width ∆ω =RL

Sol. : 32 2 5 10−π

∆ = =π π × ×Rf

L

= 100 Hz 34. Answer (3) Hint & Sol. : At midpoint of line joining the wires,

magnetic field would be non-zero and minimum and near the wires, the magnetic field would be very large.

35. Answer (4)

Hint : realapp

dd =µ

Sol. : Apparent depth = 20205/3

+

= 32 cm 36. Answer (4) Hint : Dimensional analysis Sol. : [h] = [M1 L2 T–1] [l] = [L] [F] = [MLT–2] [Speed] = [h]a [l]b [F]c

[LT–1] = [ML2 T–1]a [L]b [MLT–2]c

[L T–1] = [Ma+c L2a+b+c T– a – 2c] a + c = 0 2a + b + c = 1 – a – 2c = – 1 ⇒ a + 2c = 1 ⇒ a = –1, b = 2, c = 1 37. Answer (2) Hint : Pulley – mass systems Sol. : acceleration (a) of blocks

( )( )

A B C

A B C

m m mg

m m m + − = + +

8 48 4

g− = +

412 3

= =ga g

3

mgmg T− =

2 2 2 10 40 N3 3 3

× ×= = =

mgT

38. Answer (4)

Hint : Power F v= ⋅

Sol. : ( ) ( )ˆ ˆ ˆ ˆ ˆ ˆ6 2 4 2i j k i j k= + + α ⋅ + +

6 = 2 + 4 + 2α ⇒ α = 0 39. Answer (4)

Hint : Moment of inertia = 2mr∑ (r : perpendicular distance of mass from axis) Sol. : I = 1 (2)2 + 2(1)2 = 6 kg m2



40. Answer (1) Hint : Areal velocity of planets remains constant

Sol. : ( )31 2 constant2 4 8

AA A k= = =

∴ A1 : A2 : A3 = 2 : 4 : 8 = 1 : 2 : 4 41. Answer (3)

Hint : Excess pressure = 4TR

T : Surface tensionR : Radius of bubble

Sol. : V = αt3

3 343

R t R tπ = α ⇒ ∝

∆ = ∝ ⇒2 2T TPR t

Rectangular hyperbola

42. Answer (3) Hint : Angular momentum about cmCOM = ωI Sol. : On reaching the bottom,

θ= = θ

+2

2

2 sin sin1

gv gKR

θ= ω = 2

cmsingL I maa

sin= θma g 43. Answer (4)

Hint : 2 ,Kq KqE Vrr

= =

Sol. : E0 = Zero (as electric field of one charge cancelled by electric field of charge at diagonally opposite end)

( ) ( ) ( ) ( )0

2 2k q k q k q k qkq kqVa a a a a a

− −= + + + + +

( )0

4k q qa a

= =πε

44. Answer (3)

Hint : ( )∝ −1/2 1/21 2t H H

Sol. : −=

−

1/21

1/2 1/22

( 0)((4 ) )

t Ht H H

=1/2

1/22

t Ht H

t2 = t

45. Answer (2)

Hint : 1 2

1 1 1( 1)f R R

= µ − −

Sol. : For double convex lens

1 2

1 1 1( 1)f R R

= µ − +

Let 1 2

1 2

R RxR R

+=

For power is maximum when x will be maximum.

[CHEMISTRY]

46. Answer (1)

Hint : Molar mass = Given massNumber of moles

Sol. : Suppose the molar masses of X and Y are x g mol–1 and y g mol–1 respectively

Molar mass of X2Y = 2x + y = 100g

Molar mass of X3Y4 = 3x + 4y = 300g

On solving above equations for x and y

x = 20 u, y = 60 u

47. Answer (2) Hint : Possible value of m = – l to + l.

Sol. : For l = 2 the value of m can not be –3.

48. Answer (2) Hint : For 2z

d and 2 2x yd

−the electron density lies

along the axes. 49. Answer (2) Hint : The electronic configuration of Au is

[Xe] 4f14 5d10 6s1. Sol. : Since last electron is present in 6s orbital,

so, period number : 6 Group number : 10 + 1 = 11. 50. Answer (4)

Hint : NH3 and both contains 10 electrons each.

Sol. : Shape of NH3 and are pyramidal.



51. Answer (3) Hint : In hybridisation, sigma bond pair and lone

pair are considered only.

Sol. : NH+4 : sp3 hybridised

NH–2 : sp3 hybridised

NO–2 : sp2 hybridised

52. Answer (4)

Hint : 4 4 2

2 2 4

CH CH SO

SO SO CH

r n Mr n M

=

Sol. : 4

2

CH

SO

1r 6416 4 2 81r 1664= = × =

53. Answer (4) Hint : For a spontaneous reaction ∆G < 0. Sol. : If ∆H < 0 and ∆S > 0, then reaction is

spontaneous at all temperature. 54. Answer (2)

Hint : ( ) ( )r Reactant ProductH BE BE∆ = −∑ ∑

Sol. :

∆rH = [6 BEC–H + BEC–C + BEC=C + BEH–H]

– [8BEC–H + 2BEC–C] = BEC=C + BEH–H – 2BEC–H – BEC–C = y + x – 2u – z 55. Answer (2) Hint : pOH = 14 – pH. Sol. : pH = 10 ⇒ pOH = 4 ⇒ [OH–] = 10–4 M

22B(OH) (s) B (aq) 2OH (aq)+ −+

Ksp = [B2+] [OH–]2 = 410

2

−

(10–4)2

=0.5 × 10–12 =5 × 10–13M3 56. Answer (1) Hint : Apply Le Chatelier’s principle. Sol. : High temperature favours endothermic

equilibrium, in forward direction. 57. Answer (1) Hint : H3PO4 is a tribasic acid.

58. Answer (3)

Hint : Structure of CrO5 :

59. Answer (2) Hint : Volume strength = 11.2 × Molarity Sol. : Volume strength = 11.2 × 1 = 11.2 V 60. Answer (3) Hint : Molar conductivity increases with charge.

Sol. : Ions λ° / (S cm2 mol–1) Na+ 50.1 K+ 73.5 Ca2+ 119.0 Mg2+ 106.0 61. Answer (1) Hint : Be(NO3)2 decomposes easily.

62. Answer (4) Hint : Borax is Na2B4O7 . 10H2O

Sol. :

63. Answer (3)

Hint : Silicone is

Sol. : The chain length of the silicone polymer can be controlled by adding (CH3)3SiCl which blocks the end.

64. Answer (2) Hint : Positive charge on nitrogen increases – I

effect. 65. Answer (3) Hint : The molecule which does not contain

enolizable proton will not show tautomerism.

Sol. :



66. Answer (3) Hint : Carbocation can rearrange to gain stability. Sol. :

67. Answer (4) Hint : FCC unit cell contains four atoms.

Sol. : d = 3A

Z.MN a

23 3

4 1801.26 10 a

×=

×

323

4 180a1.2 6 10

×=

× ×

a = 10–7 cm 68. Answer (4) Hint : Al2 (SO4)3 give five ions per molecule. 69. Answer (2)

Hint : Ecell = E°cell −0.0591 log Q

n

Sol. : 2

0.1M 0.1MA(s) 2B A 2B(s)+ ++ +

Q = ( )( )

2

2 2

A 0.110

0.1B

+

+

= =

5 = E°cell – 0.0591 log102

E°cell = 5.03 V 70. Answer (1) Hint :1 Faraday charge displaces 1 equivalent

substance.

Sol. : Equivalent of O2 = 5600 1eqv.5600

=

1 eqv. Of O2 = 0.25 mole Charge required for 5600 mL of O2 gas = 1 Faraday 71. Answer (2)

Hint : For 1st order reaction 12

0.693tk

= .

Sol. : Number of half lives from 8 g to 2 g = 2

Half life of reaction = 10 5sec2

=

Rate constant 10.693k sec5

−=

72. Answer (4) Hint : Adsorption is spontaneous process in which

entropy decreases. 73. Answer (1)

Hint : 74. Answer (2) Hint : Zone refining is basically fractional

crystallisation. 75. Answer (3) Hint : H2S2O8 is peroxodisulphuric acid.

Sol. :

76. Answer (2) Hint : Nitrogen dioxide is obtained. Sol. : Zn+ 4HNO3 (conc.) → Zn(NO3)2

+ 2H2O + 2NO2. 77. Answer (2) Hint : Along the period, effective nuclear charge of

lanthanoids increases gradually so basic character decreases.

78. Answer (2) Hint : Jahn–Teller effect is observed in

unsymmetrically filled d-orbitals. 79. Answer (4) Hint : 3° carbocation is more stable than 1°

carbocation.

Sol. :

80. Answer (4) Hint : Reaction of Toluene with CrO2Cl2/CS2

followed by H3O+is Etard oxidation.



Sol. :

81. Answer (2) Hint : Steric inhibition of resonance effect

increases the acidic strength of ortho derivatives of benzoic acid.

Sol. : Acidic strength :

82. Answer (4) Hint : 3° Amines do not react with Hinsberg’s

reagent. 83. Answer (2) Hint : Nylon-6 is obtained by heating caprolactam

with water at high temperature. Sol. :

Nylon-6 84. Answer (1) Hint : Veronal, Amytal and Nembutal are

derivatives of barbituric acid.

85. Answer (3) Hint : Deficiency of vitamin B12 causes pernicious

anaemia. 86. Answer (2) Hint : In SNAr reaction, Intermediate carbanion is

stabilised by -R effect causing group present at ortho / para positions.

Sol. :

87. Answer (4)

Hint : Compounds containing group or

those in reaction condition form group give positive iodoform test.

Sol. :

88. Answer (3) Hint : Apply Kohlrausch’s Law. mΛ° CH3COOH = mΛ° CH3COO– + mΛ° H+.

Sol. : mΛ° CH3COOH = mΛ° CH3COONa

+ mΛ° HCl – mΛ° NaCl = y + z – x 89. Answer (4) Hint : Excess of nitrate causes blue baby

syndrome. 90. Answer (3) Hint : Resonance energy of thiophene is highest

among the given compounds.

Sol. :

[BIOLOGY]91. Answer (3) Hint : Mules, sterile worker bees, infertile human

couples etc. don’t reproduce. Sol. : Reproduction is not a defining property as

it is absent in some organisms. 92. Answer (1)

Hint : Arthropoda is a phylum. Sol. : Housefly-Musca domestica Family-Muscidae Order-Diptera Class-Insecta.



93. Answer (2) Hint : Red tide is caused by dinoflagellates. Sol. : Gonyaulax causes red tides. 94. Answer (4) Sol. : Albugo (Parasitic fungus on mustard) Claviceps (Ergot disease of rye) Puccinia (Rust of wheat) Smut fungi (Ustilago) 95. Answer (3) Sol. : Ecotocarpus and Polysiphonia – Haplo-

diplontic life cycle. Chlamydomonas – haplontic life cycle. Fucus – diplontic life cycle 96. Answer (1) Hint : Coralloid roots are associated with

N2 – fixing cyanobacteria. Sol. : Coralloid roots found in Cycas, are

associated with N2-fixing cyanobacteria. 97. Answer (3) Sol. : Root hairs – Region of maturation,

Palmately compound leaf – Leaflets at a common point.

98. Answer (1) Hint : Mango and coconut both are one seeded

drupe fruits. Sol. : Both develop from monocarpellary superior

ovary. 99. Answer (2) Hint : Secondary growth does not occur in

leaves. Sol. : Vascular bundles in leaves are conjoint

and closed. 100. Answer (4) Sol. : a-Phellem, b-Cambium ring, c-Secondary

phloem, d-Secondary xylem. 101. Answer (2) Hint : Small bristle like fibres on bacteria are

called fimbriae. Sol. : Fimbriae are helpful in attachment to the

surface like host tissue or substratum. 102. Answer (3) Sol. : Vacuoles – Contain ions and excretory

products. Lysosomes – Rich in hydrolytic enzymes. Golgi apparatus – Formation of glycoproteins

and glycolipids. SER – Synthesis of steroidal

hormones.

103. Answer (2) Hint : Chromosomal condensation, mitotic

spindle formation and disappearance of endomembrane system begins in prophase.

Sol. : Spindle fibres attach to kinetochore of chromosomes during mitotic metaphase.

104. Answer (4) Hint : Bivalent is formed in zygotene stage. Sol. : Bivalent is stabilised by synaptonemal

complex. 105. Answer (2) Hint : Pressure potential is usually positive

though in plants it can be negative. Sol. : Negative water potential or tension in water

column in xylem plays a major role in water transport up a stem.

106. Answer (1) Sol. : Cohesion, adhesion and surface tension

are responsible for tensile strength and capillarity of water in xylem. Active absorption is not related with tensile strength and capillarity.

107. Answer (3) Sol. : Manganese toxicity results in deficiency

symptoms of Ca, Mg and Fe. Potassium deficiency is not directly affected by Mn

toxicity. 108. Answer (2) Hint : Rhizobium and Frankia are symbiotic N2-

fixers. Sol. : Anabaena, Nostoc – fix N2 in both free living

and symbiotic form. Rhodospirillum – Anaerobic nitrogen fixer

Nitrogenase – A Mo – Fe protein. 109. Answer (1) Hint : Electron gradient is not required for ATP

synthesis. Sol. : According to chemiosmotic theory, ATP

synthesis requires – Proton gradient, proton pump, ATP synthase.

110. Answer (3) Sol. : There is a linear relationship between

incident light and CO2 fixation rates at low light intensities.

111. Answer (2) Hint : This enzyme is called pacemaker enzyme

of glycolysis. Sol. : Phosphofructokinase converts fructose

6-phosphate to fructose 1, 6 bisphosphate.



112. Answer (1) Sol. : Correct sequence of electron carriers is FMN → UQ → Cyt b → Cyt C1 → Cyt c → Cyt

a-a3. 113. Answer (1) Hint : Fats require more oxygen for oxidation

than carbohydrate and protein. Sol. : RQ for tripalmitin is 0.7. 114. Answer (3) Hint : In geometric growth every daughter cell

divides. Sol. : Geometrical growth is expressed as

W1 = W0ert. 115. Answer (3) Sol. : Cytokinin → Promotes nutrient

mobilisation. Gibberellin → Stem elongation in sugarcane. Ethylene → Induce flowering in pineapple. ABA → Closure of stomata. 116. Answer (1) Hint : Rhizome, bulbil and offset are vegetative

propagules. Sol. : Conidia is asexual, non-motile spore in

Penicillium. 117. Answer (2) Sol. : Strobilanthus kunthiana flowers once in

every 12 years. 118. Answer (2) Hint : Hydrophily is restricted to 30 genera of

monocots only. Sol. : Aquatic plants like water hyacinth and

water lily are insect pollinated plants. 119. Answer (1) Hint : Legumes are non-albuminous. Sol. : Pea and groundnut have non-albuminous

seeds. 120. Answer (4) Sol. : Phenotypes in dihybrid test cross.

121. Answer (4) Hint : Given pedigree is not applicable for sex

linked recessive disorders. Sol. : Haemophilia being X-linked recessive

disorder, is not applicable in above pedigree. 122. Answer (2) Hint : RNA is less stable than DNA. Sol. : DNA is less reactive and more stable

hence preferred as genetic material. 123. Answer (2) Hint : In eukaryotes the monocistronic structural

genes have interrupted coding sequences. Sol. : Split genes are present in eukaryotes.

Expressed sequences are known as exons. 124. Answer (3) Sol. : Himgiri – Resistant to leaf and stripe rust,

hill bunt. Parbhani Kranti – Resistant to yellow mosaic virus. Pusa Shubhra – Resistant to black rot and curl. Pusa Komal - Resistant to bacterial blight. 125. Answer (4) Sol. : Totipotency is generation of whole plant

from a single cell/explant. 126. Answer (1) Sol. : Aspergillus niger → citric acid

Acetobacter → Acetic acid Clostridium → butyric acid Trichoderma → Cyclosporin A. 127. Answer (4) Hint : Ladybirds are used to control aphids. Sol. : Baculoviruses are species specific.

Trichoderma is a biocontrol agent . 128. Answer (1) Sol. At high altitude, binding affinity of ‘Hb to

oxygen decreases. 129. Answer (2) Hint : Visiting flamingos and resident fishes

requires the same food. Sol. : Both fight for the same resources thus they

show competition. 130. Answer (2) Hint : Stratification and anabolism do not occur

during decomposition. Sol. : Steps of decomposition – Fragmentation,

Leaching, Catabolism, Humification, Mineralisation.



131. Answer (3) Hint : Lichens are pioneers on bare rocks. Sol. : In hydrarch succession phytoplanktons are

the pioneers while trees form the climax community.

132. Answer (3) Sol. : Insects (invertebrates), Fishes

(Vertebrates), Fungi (plants) show maximum biodiversity.

133. Answer (4) Hint : ‘The Evil Quartet’ are habitat loss and

fragmentation, over exploitation, alien species invasion, co-extinction.

Sol. : Co-evolution is not an Evil Quartet rather it is related to mutualism.

134. Answer (2) Sol. : Environment (Protection) act was given in

1986. 135. Answer (1) Hint : CO2 has the maximum global warming

potential. Sol. : Correct sequence of green house gases

are N2O (6%) < CFCs (14%) < CH4 (20%) < CO2 (60%). 136. Answer (2) Hint : Mast cells secrete a variety of inflammatory

mediators. Sol. : Inflammation is triggered by release of

chemicals such as histamine and kinins by mast cells and damaged cells.

137. Answer (2) Hint : This was developed by Calmette and

Guerin. Sol. : BCG is a vaccine against tuberculosis that

is prepared from a strain of attenuated live bovine tuberculosis bacillus. It loses its virulence considerably by being subcultured in an artificial medium for years.

138. Answer (4) Hint : Polio drops provide artificial active immunity. Sol. : Attenuated (i.e. live organisms that are

weakened to eliminate their virulence) poliovirus was used by Albert Sabin to develop oral vaccine for polio.

139. Answer (4) Hint : Fatty fish are rich in omega 3 fatty acids. e.g.

sardines and herring. Sol. : Edible marine fishes are Hilsa, Anguilla,

Sardinella. Mystus, Magur and Mrigala are fresh water fishes.

140. Answer (4) Hint : Phosphodiester bonds are present in DNA. Sol. : Insulin consists of 51 amino acids and is one

of the smallest proteins in the human body. It contains two polypeptide chains A and B linked by disulfide bonds.

141. Answer (1) Hint : Nyctalopia is night blindness. Sol. : Golden rice contains beta carotene gene

from daffodil plants. Beta carotene is precursor of vitamin – A.

142. Answer (2) Hint : These cells are pyramidal in appearance. Sol. : Large non-dividing Sertoli cells function as

supportive cells for developing sperms. Leydig cells are present in connective tissue around seminiferous tubules.

143. Answer (3) Hint : This is an Australian marsupial. Sol. : Australian marsupials and placental

mammals show convergent evolution. 144. Answer (3) Hint : Fetal cells can be analysed by

amniocentesis. Sol. : Amniocentesis is based on chromosomal

pattern (Karyotype) to determine genetic makeup of the developing foetus.

145. Answer (3) Hint : ZIFT stands for zygote intrafallopian

transfer. Sol. : In ZIFT, the early embryo is transferred in

the 8 blastomere stage in the fallopian tube of the mother.

146. Answer (2) Hint : Cranial capacity of modern day man is 1450

cc. Sol. : Neanderthal man is considered to be very

close to modern day man in line of ancestry. 147. Answer (3) Hint : Interferons are produced against virus

particles. Sol. : Measles is caused by a single stranded

enveloped virus spread by coughing and sneezing.

148. Answer (3) Hint : Mycobacterium leprae and Mycobacterium

tuberculosis are the two different species of Mycobacterium.

Sol. : Mycobacterium tuberculosis causes tuberculosis and M. leprae causes leprosy.



149. Answer (2) Hint : Algae have cell wall similar to plants. Sol. : For lysing the cell wall of algae, cellulase and

pectinase can be used because cell wall of algae is similar to cell wall of plants.

150. Answer (2) Hint : Depolarisation will occur. Sol. : Application of threshold stimulus will cause

depolarisation by opening of voltage gated Na+ channels.

151. Answer (1) Hint : This acronym stands for genetic engineering

approval committee. Sol. : GEAC monitors genetic engineering

programs in India and makes decisions regarding the validity of GM research and the safety of introducing GM organisms in India.

152. Answer (1) Hint : This spot has only cones and is a

depression in retinal wall. Sol. : Yellow spot or macula lutea has only cone

cells filled with a yellow pigment. It is the area of most acute vision in the eye where sharp and bright vision is formed. Region ‘D’ is blind spot with neither rods nor cones.

153. Answer (4) Hint : Neuroendocrine organ of the body. Sol. : Hypothalamus is the centre for hunger,

thirst, sweating, sleep, fatigue, temperature (thermoregulation) etc.

154. Answer (4) Hint : This bacteria can cause food poisoning. Sol. : Clostridium tetani causes tetanus which is

also known as lock jaw. Clostridium botulinum causes botulism.

155. Answer (4) Hint : It stimulates excretion of sodium and water

by the kidneys. Sol. : ANF is a peptide hormone produced by the

right atrium of the heart. It is produced in response to increased blood pressure.

156. Answer (2) Hint : It is the chief nitrogenous waste product of

humans. Sol. : Urea is maximum in concentration in hepatic

vein. Hepatic vein drains blood from liver into post caval vein. Urea is minimum in renal vein.

157. Answer (2) Hint : Maximum ventricular filling occurs during

atrial diastole.

Sol. : Majority of ventricular filling occurs when both ventricles and atria are relaxing. In atrial systole about 30% filling occurs.

158. Answer (4) Hint : Members of this phylum have dorsoventrally

flattened body. Sol. : Flame cells are specialized excretory

structures found in simple freshwater invertebrates, flatworms and rotifers.

159. Answer (4) Hint : It functions as a hydrostatic skeleton in

some lower animals. Sol. : Mesoglea is a translucent, non-living jelly

like substance found between two epithelial cell layers in the bodies of cnidarians. It is comparable to mesohyl of sponges.

160. Answer (2) Hint : Eyes of many arthropods are compound

eyes with mosaic vision. Sol. : Ommatidia are units in the eye of a

cockroach and each of it functions as a separate visual receptor.

161. Answer (4) Hint : Function of hepatic caecae is to secrete

digestive juices. Sol. : In cockroach (Periplaneta) a ring of 6 – 8

blind tubules called hepatic caecae are present at the junction of foregut and midgut.

162. Answer (4) Hint : RNA molecules. Sol. : Almost all enzymes are proteins, however,

certain RNA molecules called ribozymes (e.g. ribonuclease – P) directly act as enzymes.

163. Answer (4) Hint : ATP is adenosine triphosphate – the energy

currency of the cell. Sol. : ATP consists of three phosphate groups

attached to adenosine. Energy transfer typically involves hydrolysis of the last phosphate bond of ATP.

164. Answer (3) Hint : Km is the Michaelis – Menten constant. Sol. : Km shows the concentration of the substrate

when the velocity of the reaction is equal to one half of the maximal velocity of the reaction.

165. Answer (2) Hint : Bile contains no digestive enzymes. Sol. : Hepatocytes use cholesterol to make bile

salts.



166. Answer (4) Hint : Laetrile is a partly man made vitamin. Sol. : Laetrile is a synthetic form of amygdalin

which is found naturally in raw nuts. Laetrile is called Vit B17 although it is not a true

vitamin. Vitamin D2 is ergocalciferol, Vitamin E is tocopherol. 167. Answer (2) Hint : FRC is the functional residual capacity. Sol. : FRC = ERV + RV = 1100 + 1200 = 2300 ml 168. Answer (2) Hint : Glomerular filtrate lacks proteins

responsible for maintaining osmolarity of blood. Sol. : Plasma fluid that filters out from glomerular

capillaries into Bowman’s capsule of nephrons is called glomerular filtrate.

169. Answer (3) Hint : Visceral wall of Bowman’s capsule. Sol. : Podocytes are found on the inner wall of

Bowman’s capsule and make slit pores. They prevent / restrict passage of colloids.

170. Answer (4) Hint : It is the region between two

Z – lines. Sol. : Z – line or Krause’s membrane divides the

myofibrils into functional units called sarcomeres. 171. Answer (2) Hint : ADH is also known as vasopressin. Sol. : Diabetes insipidus is characterized by

diuresis due to deficiency of antidiuretic hormone. Grave’s disease is also known as exopthalmic goitre.

172. Answer (4) Hint : It secretes progesterone. Sol. : Corpus luteum functions as a temporary

endocrine gland secreting progesterone. 173. Answer (3) Hint : Haploid cell stage is reached after meiosis.

Sol. : Primary spermatocytes undergo meiosis I producing haploid secondary spermatocytes.

174. Answer (3) Hint : Norplant is surgically implanted under the

skin. Sol. : Norplant contains progestin as an active

ingredient and slowly releases the hormone to block ovulation and implantation.

175. Answer (1) Hint : Gene flow causes arrival of new alleles. Sol. : Maintenance of Hardy Weinberg equilibrium

requires absence of genetic drift, gene flow or migrations.

176. Answer (4) Hint : Mother’s milk provides natural passive

immunity. Sol. : Nursing mothers transfer IgA antibodies

through milk (colostrum) to the infants. 177. Answer (3) Hint : cry genes are of various types. Sol. : cryIAc and cryIIAb controls cotton

bollworms, dipterans like flies and coleopterans such as beetles.

cryIAb controls corn borer. 178. Answer (3) Hint : Pseudomonas genus has been modified to

control oil spills. Sol. : Pseudomonas putida is used in scavenging

of oil spills by digesting hydrocarbons of crude oil. 179. Answer (3) Hint : Cortisol is considered a life saving hormone. Sol. : Cortisol inhibits the normal defense

mechanisms and mobilises help from all parts of body to sustain life. It works in stress and shock situations.

180. Answer (4) Hint : Embryonic care is greatest in viviparous

animals. Sol. : Due to proper embryonic care and

protection, the chances of survival of young ones are greatest in viviparous animals.

All India Aakash Test Series for NEET - 2020 · 2020. 4. 7. · All India Aakash Test Series for...

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