All India Aakash Test Series for NEET - 2020 · Open Mock Test-6 (Code-A)_(Hints & Solutions) All...

Open Mock Test-6 (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 : 26/04/2020 ANSWERS

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

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

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

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

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

OPEN MOCK TEST - 6 (Code-A)

All India Aakash Test Series for NEET-2020 Open Mock Test-6 (Code-A)_(Hints & Solutions)


HINTS & SOLUTIONS

[PHYSICS] 1. Answer (2)

Hint : 2

24 LgTπ

=

Sol. : T tT t∆ ∆

=

hence, 2g L Tg L T

∆ ∆ ∆ = +

2L tL t

∆ ∆ = +

% error in 2 100 ∆ ∆ = + ×

L tgL t

0.1 1100 2 10010.0 80

= × + ×

= 3.5% 2. Answer (4)

Hint : .dva vds

=

Sol. : v.dv = a.ds

2

2 3/2

0

u x

u

v dv k x dx⇒ =∫ ∫

22 2 5/22 .

2 5

⇒ =

u

u

v k x

2 2 2

2 5/2 2 5/24 – 2 3 22 5 2 5

u u uk x k x⇒ = ⇒ =

4/52/52

215 154 2

u uxkk

⇒ = =

3. Answer (3)

Hint : ⊥ω =vr

Sol. :

vx sinα = rω...(i)

2 2(R/2) Hr = + ...(ii) Here R = 80 m and H = 20 m

On solving (i) and (ii) we get ω = 0.2 rads

4. Answer (4) Hint : Fs ≤ Fsmax = µs.N Sol. : During free fall of a lift, normal reaction on

block by lift floor will be zero. Hence no friction act on the block.

5. Answer (2)

Hint : 1 2

1 2

− = +

m ma gm m

Sol. : Acceleration of system (6 – 3)9 3

ga g= =

Now for block ‘C’

6. Answer (2) Hint : Use conservation of momentum

Sol. : 1 1 2 2 1 2( )+ = +

m u m u m m v

+ =.4 (–7) 2 .m m m v

= −32

v

= 1.5 m/sv 7. Answer (3) Hint : Use conservation of momentum and energy.

Sol. : Initial extension in spring is 0mgxk

=

After collision, speed of combined system is 2v

Applying conservation of energy.

2 21 1(2 ) 2

2 2 2v mg mgm k mg

k k + =

On solving, 26mgv

k=

Open Mock Test-6 (Code-A)_(Hints & Solutions) All India Aakash Test Series for NEET-2020


8. Answer (4) Hint : Kinetic energy of rolling body

2

22

1 12

kE mvR

= +

Sol. :

1/2

1/211

2 5 1012 7 715

+ × = = = +

s

c

vv

710

=c

s

vv

9. Answer (4) Hint and Sol. : Centre of mass of a body depends

on its mass distribution 10. Answer (2)

Hint : 2=e

GMvR

Sol. : 2 83

= = πρeGMv R GR

∴ ve ∝ R [ρ = constant]

here 1 1, planet , earth

2 2

1 22

ee e

e

v R v vv R

= = ⇒ =

= 22.4 km/s 11. Answer (4) Hint : at constant velocity Fnet = 0

Sol. : Volume of ball 1

=ρmV

Buoyant force on ball 2 21

= ρ = ρρmV g g

When ball attains constant velocity Fnet = 0 Fviscous = weight – upthrust

2 2

1 11mmg g mg

ρ ρ= − = − ρ ρ

1 2

1mg

ρ − ρ= ρ

12. Answer (4)

Hint : Angle of shear rLθ

φ =

Sol. : 33 10 30 0.0452

−× × °φ = = °

13. Answer (2) Hint : Principle of calorimetry Sol. : m × 540 + m × 1 × (100 – 80) =

30 × 1 × (80 – 30°) ⇒ (540 + 20) m = 30 × 50 m = 2.68 g Hence total mass of water present = 32.68 g 14. Answer (2) Hint : Stefan’s law E ∝ T4

Sol. : 44

1 1 14

2 22

E T TE TT

= =

4 4200 1 1

800 4 256 = = =

15. Answer (4)

Hint : Molar specific heat .

QCT n

∆=

∆

Sol. :

= + = +∆ ∆

5 Area under the curve2v

W R P –VC CT T0 0

0 0

85 5 42 14 2 7

P V RR R RP V

= + = +

4314

R=

16. Answer (1) Hint : Work done in an isothermal process

=

2

1ln VW nRT

V

Sol. : n = 3 mol, T = 300 K, V2 = 3V and V1 = V

∴ Work done W = 3R300 3In

VV

= 900 R ln[3] = 450 R ln[9] 17. Answer (4) Hint and Sol. : Molecules of an ideal gas moves

randomly with different speeds. 18. Answer (3)

Hint : 1 12

kT ff m

= ⇒ =π

Sol. : 1 1

2 2

2 2f k kf k k

= = =

12 22 2

f ff f∴ = ⇒ =



19. Answer (2) Hint and Sol. For any particle which is executing

SHM a = – ω2x ⇒ a ∝ (–x) 20. Answer (1)

Hint : Speed of string wave Tv =µ

Sol. :

T = µxg

T xgv gxµ= = =

µ µ

v x∴ ∝

21. Answer (3) Hint : fbeat = |f1 – f2|

Sol. : beat1 1–

2 2( ) 2v v vfl l x l l x

= − = + +

( )22=

+

vxl lx

22. Answer (2)

Hint :

−∂ ∂ ∂= − −

∂ ∂

V V VE i j kdx y z

F qE=

Sol. : −∂= = − −

∂2(3 4 )

xVE y ix

−∂= = − − − +

∂( 8 4 3 )

yVE xy z jy

−∂= = − −

∂(3 4 )

zVE y z kz

At origin ⇒ x = 0, y = 0, z = 0

3 4E i j∴ = − +

Hence 5 Force 3= ⇒ =

E F E = 15 N

23. Answer (1) Hint : Work done w = q∆V

Sol. : 0

14A

QVx

=πε

and 0

14B

QVy

=πε

( )0

1 14B A

QW q V V qy x

= − = − πε

04

Qq x yxy

−= πε

24. Answer (2)

Hint : 2kEr

λ=

Sol. : 0

24

Erλ

=πε

9

52

2 9 1018 1010 10−

× × × λ⇒ × =

×

− µ⇒ λ = × =6 C10 10 10

m

25. Answer (4) Hint : Redraw the circuit Sol. : Equivalent circuit will be

Ceq = 2 µF 26. Answer (1) Hint : Redraw the circuit after closing the switch. Sol. : Equivalent circuit will be

= Ω203eqR

10 3

20eq

VIR

×= =

1.5 A 27. Answer (2) Hint : Kirchoff’s voltage law Sol. : Equivalent circuit.



0

0 0 00

0

2 (2 )( 4 )

22

2

V R RVi R R R RR RRR

+= =

+

+ +

0 0

0

(2 )2 ( 4 )R

R R RV V i V VR R

+= − = −

+

0

(2 )( 4 )R

RVVR R

=+

28. Answer (3)

Hint : 2 ITmB

= π

Sol. : In vertical plane 1 2v

ITmB

= π

In Horizontal plane 2 2H

ITmB

= π

here T1 = T2 ⇒ Bv = BH

angle of dip tan 1v

H

BB

θ = =

∴ θ = 45° 29. Answer (4)

Hint : Magnetic field due to circular are 04

lR

µ θ=

π

Sol. : 0 0( / 2) ( / 2)4 2 4

l IBR R

µ µπ π= +

π π

0 0338 2 16

llR R

µ µ = =

30. Answer (4)

Hint : 0 1 22 N/m4

l lFd

µ=

π

Sol. : Net force on 1m2

length of wire C

net / /1| |2C A C BF F F= − ×

02 2

2 10 20 2 30 10 14 22 10 3 10− −

µ × × × × = − × π × ×

4 40 2 10 2 104µ = × − × π

= zero 31. Answer (3) Hint : ε = BVleff

Induced emf between lowest and highest point of

ring ε = 2B0vR Both rings are moving in opposite direction ∴ Potential difference required = 4B0vR 32. Answer (4)

Hint : emf dedt

φ= −

Sol. : φ = 3 tn – 2n

1 13 | | 3n nd dnt e ntdt dt

− −φ φ= − ⇒ = =

13 −⇒ = ne nt

33. Answer (4) Hint and Sol. :

For 1ω >

LC circuit will be inductive.

Hence current will lag behind the voltage.

For 1ω =

LCcircuit will have resistive nature.

Power factor, φ = ⇒ φ =cos cos 1RZ

if XL = XC

34. Answer (1) Hint and Sol. : λmicro > λIR > λuv > λgamma 35. Answer (4) Hint : δ = i + e – A Sol. : For minimum deviation i = e δmin = 2i – A 60° = 2 × 60° – A ⇒ A = 60°

sin

2Refractive index 3sin

2

mA

A

+ δ µ = =

δ v/s i curve is not parabolic 36. Answer (2) Hint and Sol. : Object ‘O’ is at ‘2f’ of lens. So

image by lens formed ‘2f’ behind it. Pole of mirror will reflect the incident rays

symmetrically, therefore image will be formed on the object itself



37. Answer (3) Hint : For 1000th maxima, dsinθ = 1000 λ

10

31000 1000 6000 10sin

1 10d

−

−λ × ×

⇒ θ = =×

1 36 10 375

−= × = ⇒ θ = °

∴ y = D tanθ = 1.tan37° 3 m4

=

38. Answer (1) Hint : kmax = hν – φ ⇒ hν = φ + kmax

2 1 2 8 3 33 4

φ += ⇒ φ + = φ +

φ +

⇒ φ = 5 eV 39. Answer (3)

Hint : de-Broglie wavelength 3

hmKT

λ =

Sol. : For same temperature 1m

λ ∝

as 2 2 20 N He Hm m m H> > >

2 2 20∴ λ < λ < λ < λ

eN H H

40. Answer (4) Hint and Sol. :

E1 = E2 + E3

hν1 = hν2 + hν3 = ⇒ ν1 = ν2 + ν3 41. Answer (2)

Hint : dU Fdr

− =

Sol. : = − = −2

432

dU keFdr r

But = =2 2

432

mv keFr r

Also =π2

nhmvr

On solving 2 2

2 26 ke mr

n hπ

=

42. Answer (4)

Hint : λ = =min12400 Å(in volt)

hceV V

=12400 Å18000

≈ 0.7 Å 43. Answer (4) Hint : After n-half life the radioactive nuelei

remaining is 0

2nN

Sol. : Number of nuclei disintegrated in n-half lives

is 00 2n

NN = −

1 1for , fraction disintegrated 12 2

n∴ = = −

2 1

2−

=

44. Answer (3) Hint Sol. : Only the light nuclei having sufficient

initial energy can overcome the coulomb potential barrier, hence only those can fuse.

45. Answer (3) Hint and Sol. : Boolean expression of the given

arrangement is

= + = ⋅ = .y A B A B A B ∴ This is Boolean expression for AND gate.

[CHEMISTRY]

46. Answer (1) Hint : Diene and yne stands for 2π bonds each

Sol. :

Number of σ bonds = 14 and number of π bonds = 4 47. Answer (2)

Hint : Preparation of phenol from cumene.

Sol. :



48. Answer (3) Hint : The structure of tribromooctaoxide is

Sol. :

The oxidation states of bromine are +6, +4. 49. Answer (1) Hint : For multielectronic system, greater is the

sum of (n + l), greater is the energy of orbitals. Sol. : In case, two subshell have same (n + l)

values then the subshell with greater n value will have more energy

Value of (n + l) : 4f = 4 + 3 = 7 5p = 5 + 1 = 6 4d = 4 + 2 = 6 4p = 4 + 1 = 5 ∴ The order of energy of orbital is 4f > 5p > 4d > 4p 50. Answer (4) Hint : A redox reaction, in which same element is

simultaneously oxidised and reduced in the same reaction is called disproportionation reaction.

Sol. : a. 2Cu+ → Cu2+ + Cu° (Disproportionation reaction)

b. 50 1

2 3 26NaOH 3Cl 5NaCl NaClO 3H O+−

+ → + + (Disproportionation reaction)

c. 30 1

4 2 3 2 2P 3NaOH 3H O PH 3NaH PO− +

+ + → + (Disproportionation reaction)

d. 6 47

4 2 4 2 22KMnO K MnO MnO O+ ++

∆→ + + (∴ Not a disproportionation reaction) 51. Answer (1)

Hint : W = – 2.303 nRT log f

i

VV

Sol. :

W = – 2.303 × 2 × 8.314 × 300 × log 5010

= – 2.303 × 2 × 8.314 × 300 × 0.7 = – 8.04 kJ 52. Answer (4) Hint : Common components of photochemical

smog are ozone, NO, acrolein, formaldehyde and PAN.

53. Answer (1) Hint :

cell cathode anode0.059 [Product]E E E log

n [Reactant]= ° − ° −

Sol. : anode0.059 0.010.58 0.34 E log

2 0.1= − ° −

1anode

0.059E 0.34 0.58 log102

−° = − −

= 0.34 – 0.58 + 0.0295 = – 0.21 V 2Ni /Ni

E 0.21 V+° = −

54. Answer (4) Hint : Drug – Enzyme interaction 55. Answer (2) Hint : Alkyne reacts with Na/Liq. NH3 and forms

trans alkene. 56. Answer (2) Hint : Bond angle of 15th group hydrides

decreases down the group.

Sol. : 3 3 3 3Hydrides NH PH AsH SbHBond angle 107.8 93.6 91.8 91.3° ° ° °

57. Answer (4) Hint : Boric acid is not a protonic acid but acts as

a lewis acid. 58. Answer (1) Hint : The structure of salicylic acid is

Sol. : a. Kolbe’s reaction

b. Stephen reaction R – CN + SnCl2 + HCl → RCH = NH

RCHO c. Hoffmann bromamide degradation reaction

+ Br2 + 4NaOH → R – NH2 + Na2CO3 + 2NaBr + 2H2O

d. Swarts reaction 3 3CH Br AgF CH F AgBr− + → − +



59. Answer (3) Hint : 2s – 2p mixing takes place in C2 molecule. Sol. : Molecular orbital configuration of C2 is σ1s2, σ*1s2, σ2s2, σ*2s2, π2px

2 = π2py2

Bond order 1 (8 4) 22

= − =

Both the bonds are π bonds 60. Answer (4) Hint : N has half filled p-subshell. Sol. : Correct order of ionization potential is C < O < N. 61. Answer (2) Hint : The molecular formula of Teflon is

Sol. :

Persulphate catalyst2 2 High pressurenCF CF= →

Tetrafluoro Teflon ethene 62. Answer (1) Hint : Common ion effect Sol. : M(OH)2 + H2O(l) M2+(aq) + 2OH–(aq)

At initial 0 0.2 At equilibrium state s 0.2+2s ≈ 0.2 Ksp = [M2+] [OH–]2

4.0 × 10–15 = (s) (0.2)2

15

24 10s4 10

−

−×

=×

= 10–13 mol/L 63. Answer (2) Hint : For first order reaction Integrated rate law is

0[R]2.303t logk [R]

=

Sol. : 0[R]2.303t logk [R]

=

03

0 0

[R]2.303 log[R] 0.6[R]3 10−=

−×

03

0

[R]2.303 log0.4[R]3 10−=

×

32.303 0.398

3 10−= ××

= 305 sec

64. Answer (4) Hint : During denaturation of proteins only 1°

structure remains intact. 65. Answer (2) Hint :

Sol. :

Electrophilic substitution reaction 66. Answer (1) Hint : Solution of ethanol and water show positive

deviation from Raoult’s law. Sol. : Solution showing positive deviation from

Raoult’s law form minimum boiling azeotropes.

Azeotropes have same composition in liquid and vapour phase.

67. Answer (2)

Hint : For the reaction, N2 + 3H2 2NH3

Rate of reaction 32 2 d[NH ]d[N ] d[H ]1 1dt 3 dt 2 dt

= − = − =

Sol. : 32 d[NH ]d[H ]1 13 dt 2 dt

− =

3 2d[NH ] d[H ]1 2 1 4g 2 mol;3 1 2 dt dt h h

− × = = =

3d[NH ] 4 moldt 3 h

=

Rate of change of NH3 in g/h

4 17 22.7 g/h3

= × =

68. Answer (1) Hint : 2KMnO4 + 16HCl → 2KCl + 2MnCl2 + 8H2O + 5Cl2 69. Answer (3) Hint : Stronger is the base, higher will be the

equilibrium constant for protonation. Sol. : Among the given compounds, only

pyridine, has localized lone pair of electron. ∴ It is most basic. As a result it has the maximum

value of equilibrium constant for protonation.



70. Answer (2)

Hint : ∆mixS is always greater than zero while ∆mixG is always less than zero.

Sol. : ∆mixS is always greater than zero also, ∆mixG is always less than zero. While, ∆mixV and ∆mixH both are zero for ideal solution.

71. Answer (1)

Hint : H2SO4 is a stronger acid than HNO3

Sol. : 2 4 3 4 2 3H SO HNO HSO H NO+

−+ +

72. Answer (1) Hint : Charge of colloidal particles will depend

upon adsorption of common ion present from species present in excess.

Sol. : When FeCl3 is added to the excess of hot water, a positively charged solution of hydrated ferric oxide is formed due to adsorption of Fe3+ ion. On the other hand, when ferric chloride is added to NaOH solution a negatively charged sol is obtained with adsorption of OH– ion.

73. Answer (4) Hint : The percentage of phenol in disinfectant is

more than its percentage in an antiseptic solution.

Sol. : 0.2 percent solution of phenol is an antiseptic solution while 1 percent solution of phenol is disinfectant.

74. Answer (1) Hint : In presence of organic peroxide, addition of

HBr to alkene takes place according to anti-Markovnikov rule.

Sol. :

75. Answer (1)

Hint : F– is a weak field ligand.

Sol. : [FeF6]4–

Fe2+ = [Ar]183d64s°

F– is a weak field ligand

∴ The configuration will be t2g4 eg

2 CFSE value = 4 x (– 4Dq) + 2 x (+ 6Dq) = – 16Dq + 12Dq = – 4Dq 76. Answer (4) Hint : On hydrolysis, the oxidation state of

phosphorus remains same. Sol. : PCl5 hydrolyses to POCl3 and finally gets

converted to phosphoric acid. PCl5 + H2O → POCl3 + 2HCl POCl3 + 3H2O → H3PO4 + 3HCl 77. Answer (3) Hint : Acidic buffer is a mixture of weak acid and

its salt with a strong base. Sol. : KOH + C6H5COOH → C6H5COOK + H2O Initially 500 mL×1M 500 mL×1M 0 = 500 mmol = 500 mmol Finally 0 mmol 0 mmol 500 mmol

∴ This is not an acidic buffer. 78. Answer (2) Hint : β-keto acid readily decarboxylates on

heating.

Sol. :

79. Answer (1) Hint : Use VSEPR theory. Sol. : a. The shape of XeF6 is distorted octahedral

b. The shape of I3

+ is V-shape



c. The shape of BrF3 is bent T-shape

d. The shape of ClO4

– is tetrahedral.

80. Answer (2) Hint : Manganate and permanganate ions show

colour due to charge transfer. Sol. : Solution of manganate ion shows green

colour. Whereas solution of permanganate ion shows purple colour.

81. Answer (4) Hint : Due to non-availability of d-orbital, Boron is

unable to expand its octet. Therefore, the maximum covalency of boron cannot exceed 4.

Sol. : PI5 and [SiCl6]2– is unstable due to steric crowding

[BF6]3– doesn’t exist due to unavailability of d-orbital

∴ [Be(OH)4]2– is the only stable species.

82. Answer (1)

Hint : cellnFElnKRT

°=

Sol. : cell cathode anodeE E E° = ° − °

= 0.80 – (– 0.25) = 1.05

cellnFElogK2.303RT

°=

2 96500 1.05logK2.303 8.314 298

× ×=

× ×

logK = 35.51

35.5K 10≈ 83. Answer (2) Hint : Generally metal oxides are basic in nature

while non-metallic oxides are acidic in nature. Sol. : (1) Be(OH)2 : Amphoteric

(2) N2O : Neutral oxide

(3) SiO2 : Acidic oxide

(4) BaO : Basic oxide

84. Answer (1) Hint : For most of the gases at low pressure

deviation is negative, while at high pressure deviation is positive.

Sol. : At low pressure, attractive forces dominates as a result Z < 1.

85. Answer (3) Hint : Number of tetrahedral voids = 2 × number

of close packed atoms. Sol. : Suppose ‘N’ A ions are forming FCC array

then, number of tetrahedral voids = 2N Number of B ion = 50% of tetrahedral voids = 50% of 2N = N The formula of the compound is ANBN or AB.

86. Answer (4) Hint : Boiling of water increases entropy. Sol. : During freezing of water entropy decreases

as the randomness of particle decreases. 87. Answer (3) Hint : Transition from higher shell to first shell lies

in Lyman series. Sol. : Lyman series lies in the ultraviolet region

and for this series. n1 = 1 and n2 = 2, 3,.....

88. Answer (3) Hint : CO + N2 is producer gas

Sol. : The mixture of CO and H2 is called water gas. As this mixture is used for the synthesis of methanol hence, it is also called synthesis gas or syn gas.

89. Answer (4) Hint : Malachite is an ore of copper

CuCO3. Cu(OH)2.

Sol. : Kaolinite is an ore of aluminium, [Al2(OH)4Si2O5] Calamine is an ore of Zinc, ZnCO3

Siderite is an ore of Iron, FeCO3

90. Answer (1) Hint : +I effect causing group decreases the

acidic strength of carboxylic acid.

Sol. : H – COOH > C6H5 – COOH >



[BIOLOGY]91. Answer (2) Sol. : Mesosome helps in DNA replication,

respiration and cell division of bacteria. 92. Answer (2) Hint : Shape of chromosomes during anaphase

depends upon position of centromere Sol. : Metacentric – V-shaped Sub-metacentric – L-shaped Acrocentric – J-shaped Telocentric – I-shaped 93. Answer (1) Hint : Megaspore and egg cell of angiosperm is

haploid (n = 10)

Sol. : Megaspore mother cell and archesporial cell are diploid. The chromosome number in MMC in S and G2 phase will be 20 because there is no change in chromosome number in S phase.

94. Answer (4) Hint : Chromatids separate from each other during

anaphase-II Sol. : At anaphase II, haploid set of chromosomes

are present in cells. Dipoloid set of chromosomes are present in cells at anaphase-I

95. Answer (1) Hint : At lower hierarchial level, similarities in

characters are more. Sol. : Specificity decreases when we go from

species to kingdom. When we go from kingdom to species the number of similar characteristics of organisms increases.

96. Answer (2) Sol. : Proliferation of some dinoflagellates like

Gonyaulax, causes red tide in the sea. 97. Answer (2) Sol. : Prions are abnormally folded protein

particles 98. Answer (3) Hint : Woese divided kingdom Monera into two

kingdoms. Sol. : Based on the sequence of 16S ribosomal

RNA genes, Woese categorised all organisms into three main categories. He called them domains of life. These domain were Bacteria, Archaea and Eukarya.

99. Answer (4) Hint : Roots developed from any parts of plant

other than the radical are called adventitious roots.

Sol. : Sweet potato is an adventitious root modified for storage of food.

100. Answer (3) Hint : Stamens show epipetalous condition in

Solanaceae. Sol. : Members of Solanaceae family have

bisexual, actinomorphic flowers with fused calyx as well as corolla. Stamens are five in number and epipetalous. Gynoecium is bicarpellary, syncarpous with superior ovary.

Floral formula ;

101. Answer (1) Hint : Collenchyma tissue is absent in

monocotyledonous plants. Sol. : Parenchymatous tissue having large air

cavity is called aerenchyma. Xylem is a complex permanent tissue. Sclerieds are present in leaves of tea.

102. Answer (2) Hint : Periderm is formed due to the activity of cork

cambium. Sol. : Cork cambium (phellogen) divides to form

phelloderm towards inner side and phellem towards outer side. Phellogen, phelloderm and phellem collectively called periderm.

103. Answer (3) Hint : In haplontic life cycle, major phase of life is

haploid. Sol. : Organisms showing haplontic life cycle

shows zygotic meiosis. Dominant phase of life is gametophyte, that is formed by germination of haploid spores. Fucus, a brown alga has diplontic type of life cycle.

104. Answer (2) Hint : Bryophytes need water for their sexual

reproduction Sol. : Bryophytes are called non-vascular

amphibians of plant kingdom as these plants live in soil but are dependent on water for sexual reproduction. Bryophytes lack vascular system.

105. Answer (3) Hint : Movement of water always takes place from

its high to low concentration. Sol. : Hypertonic solution has high DPD value as

compared to hypotonic solution. So, water moves from low DPD to high DPD.



106. Answer (4) Hint : Exosmosis from guard cells causes closing

of stomata. Sol. : Lower K+ and malate ion concentration in

guard cell will cause exosmosis from guard cells, that will result into closure of stomata.

107. Answer (2) Sol. : Total 18 ATPs are used to form one glucose

molecule through Calvin cycle 108. Answer (3) Hint : T.W Engelmann used a green alga. Sol. : He used Cladophora for his experiment and

described first action spectrum of photosynthesis. 109. Answer (3) Hint : Lactic acid is the product of anaerobic

respiration in muscle cells. Sol. : Net 2 ATP molecules are gained in

conversion of glucose into lactic acid during anaerobic conditions in muscle cells.

110. Answer (3) Sol. : Both types of plants (C3 and C4) show

increase in rate of photosynthesis at high light intensity.

111. Answer (2) Hint : Short day plants need light lesser than

critical day length for flowering Sol. : Photoperiod should never exceed critical

day length in case of short day plant. The long night should not be interrupted by flashes of light.

112. Answer (3) Hint : Gibberellins are growth promoters. Sol. : Gibberellin induces seed germination and

internodal stem elongation in rossette plants. It also increases stem length in sugarcane, which stores sugar as a result, increases yield of crop.

113. Answer (2) Hint : Interflowering period is the period between

two successive flowering phases. Sol. : Mango is a polycarpic plant, whereas wheat

is a monocarpic plant. Polycarpic plants show interflowering period.

114. Answer (3) Hint : Turmeric has rhizome. Sol. : Turmeric is a modified stem. Radish, Carrot

and sweet potato are storage roots. 115. Answer (1) Hint : Pericarp is fruit wall and perisperm is

persistent nucellus. Sol. : Cotyledon of monocots is called scutellum.

The lower end of embryonal axis has radical which is enclosed in an undifferentiated sheath called coleorhiza.

116. Answer (1) Sol. : Polar nuclei : (n + n), Male gamete : (n) 117. Answer (1) Hint : Mendelian dihybrid cross ratio is 9 : 3 : 3 : 1 Sol. : In a dihybrid cross, one out of 16 genotypes

is homozygous recessive for both the traits. Number of seeds with both homozygous recessive

trait = 1 800 5016

× =

118. Answer (2) Hint : Types of gametes formed are 2n, where “n”

is number of hybrid. Sol. : In PpQqRr × ppqqrr, probability of getting

1 1 1Pp ,qq and Rr2 2 2

= = =

So, probability of getting genotype PpqqRr is 1 1 1 12 2 2 8

= × × =

119. Answer (3) Hint : Peptidyl transferase is a ribozyme found in

larger subunit of ribosome. Sol. : 28 S rRNA of larger ribosomal subunit act as

peptidyl transferase in eukaryotic cell. 120. Answer (2) Hint : Permeability for lactose is increased by the

action of permease. Sol. : Gene y codes for enzyme permease. 121. Answer (3) Hint : Histone proteins are absent in eubacteria. Sol. : Nucleosome is octamer of histones over

which DNA is wrapped. Such DNA packaging is not present in prokaryotes. So, there is no nucleosome present in eubacteria.

122. Answer (4) Sol. : Pusa Sawani is resistant to shoot and fruit

borer. 123. Answer (2) Hint : Each of the plant produced through tissue

culture will be genetically identical to parent plant are called somaclones.

Sol. : Somaclonal variations are variations seen in plants that have been produced by tissue culture. Chromosomal rearrangements is an important source of such variations.

124. Answer (3) Sol. : Acetobacter aceti is used to produce acetic

acid.



125. Answer (1) Sol. : Dragonflies are useful to get rid of

mosquitoes and baculoviruses are in the genus NPV. These viruses have no negative impact on plants, mammals, birds, fish or even on non-target insects.

126. Answer (2) Hint : Logistic growth is described by the equation,

dN K – NrNdt K

=

Sol. : N = 400 K = 500 r = 0.01

So, dN 500 – 4000.01 400dt 500

= ×

14 0.85

= × =

dN 0.8dt

=

127. Answer (1) Sol. : Principle of competitive exclusion was given

by Gause. 128. Answer (4) Sol. : Sequence of communities in hydrosere is :

Phytoplankton → submerged plant stage → Submerged free floating → Reed swamp stage → Marsh-meadow stage → Forest.

129. Answer (1) Sol. : The phenomenon of incorporation of

nutrients in living microbes is called nutrient immobilisation.

130. Answer (3) Hint : β-diversity is a type of community dirversity. Sol. : Diversity between two communities referred

as β-diversity. 131. Answer (1) Sol. : Methane (CH4) contributes 20% of green

house gases to total global warming. 132. Answer (3) Hint : Hot spots are areas of high endemism and

very high level of species richness. Sol. : Hot spots are most threatend reservoir of

plants and animals life on earth. 133. Answer (3) Sol. : Montreal protocol aimed at protecting

stratospheric ozone.

134. Answer (2) Hint : Nitrogenase enzyme responsible for

nitrogen fixation functions in anaerobic conditions. Sol. : Leghaemoglobin is present in cells of

nodules to scavenge oxygen from those cells. 135. Answer (1) Sol. : Boron is essential for pollen grain

germination. 136. Answer (4) Hint. : Select a cell which forms myelin sheath in

PNS. Sol. : Astrocytes/Macrocytes are large glial cells

with a number of protoplasmic processes. They form maximum number of glial cells. They help in repair of nervous tissue and form blood brain barrier. Schwann cells form myelin sheath and neurilemma around an axon in PNS.

137. Answer (2) Hint. : The cells of absorptive surfaces often bear

microvilli Sol. : Proximal convoluted tubules of kidney are

lined by cuboidal epithelial cells. Microvilli greatly increase the area of free surface of the cells and thereby enhance reabsorption.

138. Answer (3) Hint. : Iron containing part of respiratory pigment. Sol. : Amino acids, sugars, chlorophyll, haem etc.

are few examples of primary metabolites. Carotenoids are pigments, Abrin is a toxin while concanavalin A is a lectin. These all are considered as secondary metabolites.

139. Answer (3) Sol. : Alcoholic amino acids have –OH group in

side chain. Hint. : Serine and threonine are examples of

alcoholic amino acids. Lysine is a basic amino acid. Glycine is simplest amino acid with ‘H’ as its side chain.

140. Answer (3) Hint. : Non-competitive inhibition cannot be

overcome by increasing the concentration of substrate.

Sol. : An example of allosteric modulation or feedback inhibition is inhibition of threonine deaminase by isoleucine. Amino acid isoleucine is formed in bacterium Escherichia coli through 5-step reactions from threonine. When isoleucine accumulates beyond a threshold value, its further production stops.



141. Answer (2) Hint. : They are also known as peptic cells. Sol. : Mucus neck cells secrete mucus. Peptic or

chief cells secrete the proenzyme pepsinogen. Parietal or oxyntic cells secrete HCl and intrinsic factor (factor essential for absorption of vitamin B12 in intestine).

142. Answer (4) Hint. : Pancreatic juice contains inactive enzymes. Sol. : Goblet cells secrete mucus. Pancreatic juice

contains – trypsinogen, chymotrypsinogen, procarboxypeptidase, amylase, lipase and nucleases.

143. Answer (1) Hint. : Each haem group of Hb molecule carries

centrally placed iron atom. Sol. : Each Hb molecule has four polypeptide

chains and each polypeptide chain carries a haem group containing iron in ferrous form.

144. Answer (4) Hint. : Select the antibody secreting cell. Sol. :

Human blood cell Lifespan Monocytes 10 to 20 hours

Eosinophils 4 to 8 hours in the blood. 4 to 5 days in the tissue

Neutrophils 4 to 8 hours in the blood and 4 to 5 days in the tissues

Lymphocytes Lifespan is about few days or months or even years.

145. Answer (1) Hint. : EC = TV + ERV Sol. : Vital capacity is ERV + TV + IRV. TLC = IC + FRC FRC = ERV + RV 146. Answer (4) Hint. : Ventricular diastole is the period of filling of

ventricles with blood. Sol. : Heart in ventricular diastole is relaxed and

AV valves are open. When ventricular pressure falls, it causes the closure of semilunar valves, thereby preventing the back flow of blood into the ventricles.

147. Answer (2) Hint. : Blood in veins flows under low pressure. Sol. : Veins have valves which prevent backward

flow of blood. Arteries in general have no valves.

148. Answer (2) Hint. : Network of capillaries. Sol. : Renal tubule is composed of Bowman’s

capsule, PCT, loop of Henle and DCT. 149. Answer (3) Hint. : Osmoreceptors sense changes in blood

osmolarity. Sol. : An excessive loss of fluids from the body can

activate osmoreceptors in hypothalamus receptors which stimulate the hypothalamus to release vasopressin from the neurohypophysis. ADH prevents diuresis. An increase in body fluid volume can “switch off” the osmoreceptors and suppress the ADH release to complete the feedback.

150. Answer (2) Hint. : Enzyme involved in cross bridge formation. Sol. : ATP + H2O ADP + Pi + Energy

Phosphocreatine + ADP ATP + Creatine 151. Answer (3) Hint. : This tissue/structure exhibits excitability. Sol. : Contractility, conductivity and extensibility

are properties of muscular tissue. Neurons show excitability but not contractibility. Hematopoietic tissue synthesizes blood.

152. Answer (2) Hint. : Muscle fibres which have slow rate of

contraction for long periods Sol. : Eye ball muscles have white muscle fibres.

Extensor muscle of the human back have red muscle fibres.

153. Answer (3) Hint. : Na+ ions are more in ECF. Sol. : On application of a stimulus, voltage-gated

Na+ channels suddenly open, allowing Na+ ions to enter into neuron because ions move down their concentration gradient.

154. Answer (2) Hint. : This structure is the anterior part of choroid. Sol. : Suspensory ligaments attach lens with

ciliary body. Contraction of the muscles of the ciliary body relaxes suspensory ligaments and permits the lens to take a more or less spherical shape. These changes in lens shape enable the eye to change focal length of lens (accommodate). The external layer of eye is called sclera or fibrous tunic, middle layer is called uvea or vascular tunic and third innermost layer of eyeball is called retina.



155. Answer (4) Hint. : Hypothalamus is ventral part of this relay

centre in forebrain. Sol. : Forebrain consists of telencephalon and

diencephalon. Mesencephalon is midbrain. Rhombencephalon is hindbrain.

156. Answer (3) Hint. : Estrogen shows a cyclic rise and fall during

menstrual cycle. Sol. : Estrogen is secreted by the ovary prior to as

well as after ovulation. Progesterone is mainly secreted by ovary after ovulation.

157. Answer (4) Hint. : Food materials directly diffuse through body

surface in Taenia. Sol. : Parasitic flatworms are specially adapted to

obtain nutrition from their hosts. Thick external tegument protects the parasitic worms from the digestive juices of host. Hooks and suckers help in anchorage of parasitic flatworm with host.

158. Answer (2) Hint. : Xenopsylla is known as rat flea. Sol. : Anopheles mosquito is the vector of

malaria. Xenopsylla/ rat flea spreads plague. Glossina (Tse-tse fly)spreads sleeping sickness. Aedes mosquito is the vector of dengue fever and chikungunya.

159. Answer (4) Hint. : Found in highly cold regions of Earth Sol. : Delphinus and Ornithorhynchus are

mammals with non pneumatic bones. Neophron is a flying bird with well developed keel.

160. Answer (1) Hint. : Organisms belonging to the subclass

prototheria are oviparous. Sol. : Ornithorhynchus (Platypus) is oviparous.

Egg laying mammals are called monotremes. In egg laying mammals, eggs are macrolecithal with meroblastic cleavage.

161. Answer (3) Hint. : Gizzard is followed by hepatic caecae Sol. : At the junction of midgut and gizzard (i.e.,

foregut) arise six to eight finger like structures called hepatic or gastric caecae that pour digestive enzymes. Malpighian tubules are present at junction of midgut and hindgut.

162. Answer (3) Hint. : Heart of cockroach is an elongated

muscular tube with more number of chambers than pairs of alary muscles

Sol. : Cockroach’s heart consists of 13 funnel-shaped contractile chambers. Alary muscles play an important role in circulation of blood. Collaterial glands and phallomeres are helpful in reproduction. Ommatidia are associated with compound eye.

163. Answer (4) Hint. : End branch of trachea closest to body cells Sol. : The trachea is divided into fine branches

known as tracheoles. They terminate in the tissues and contain a tissue fluid at the distal end which plays a significant role during the diffusion of gases.

164. Answer (3) Hint. : Amoeba secretes a three-layered hard

covering around itself. Sol. : Encystation of Amoeba is a protective

response in unfavourable environment. Gemmule formation is seen in sponges. Sporulation is seen when Amoeba divides by multiple fission and produces many minute amoebae upon arrival of favorable conditions.

165. Answer (3) Hint. : Oestrus cycle is seen in non-primate

mammals. Sol. : In primates (Humans, Ape and monkey)

female sexual cycle is called menstrual cycle. It occurs all around the year with generally one ovulation in a month. Non-primates such as sheep, cow, rat and dog reproduce seasonally (oestrus cycle) and are known as seasonal breeders.

166. Answer (3) Hint. : This structure is commonly called oviduct. Sol. : Fertilization occurs in ampulla of the

fallopian tube. Post fertilization, growing embryo passes through fallopian tube to enter uterine cavity in about 3 days. First cleavage is completed after about 30 hours of fertilization within fallopian tube.

167. Answer (4) Hint. : This type of placenta has three barriers. Sol. : Human placenta is haemochorial in nature

and has only three barriers with chorionic villi in contact with maternal blood. The maternal part of placenta is completely eroded in humans.

168. Answer (3) Hint. : Removal of foreskin. Sol. : Surgical removal of prepuce or foreskin from

penis is known as circumcision. Causes of infertility in males are alcoholism, radiation, cytotoxic drugs, cryptorchidism etc. but not circumcision.



169. Answer (2) Hint. : It is a method of assisted reproduction. Sol. : Surrogate female is also called gestational

carrier. IUI is Intrauterine insemination. AI is Artificial insemination while in GIFT, gamete

is transferred and not the embryo. 170. Answer (2) Hint. : Wasserman’s test can detect this disease. Sol. : Syphilis and gonorrhea are bacterial STIs

while Trichomoniasis is caused by protozoa called Trichomonas vaginalis.

171. Answer (1) Hint. : Placentals lack a pouch to protect young

ones. Sol. : Marsupials are currently geographically

located in Australia while placentals are located all across world except australian belt due to continental drift.

172. Answer (4) Hint. : Craniosynostosis is seen in birds. Sol. : An intimate fusion of the skull bones is

characteristically seen in the birds. Other avian characters of Archaeopteryx include (a) Presence of feathers on the body (b) Two jaws modified into beak (c) Forelimbs modified into wings (d) Hindlimbs built on the typical avian plan

173. Answer (4) Hint. : Artificial selection. Sol. : Artificial selection, as the name suggests, is

the selective breeding of plants or animals for desired traits. Examples of artificial selection are: 1. Generation of different breeds of dogs, cows,

sheep. 2. Generation of different vegetables like

broccoli, kale, cauliflower, etc. from wild mustard.

174. Answer (4) Hint. : Vector for this disease is Culex.

Sol. : Filariasis is caused by Wuchereria (W. bancrofti and W. malayi), aka filarial worm. Poliomyelitis, dengue fever and chikungunya are few examples of viral diseases.

175. Answer (1) Hint. : Salk’s polio vaccine is a first generation

vaccine. Sol. : Small pox, oral polio vaccine (OPV), BCG

(Bacillus-Calmette-Guerin), Influenza vaccines are attenuated.

176. Answer (3) Hint. : Select an organism associated with

sericulture. Sol. : Silk moth is reared for the production of silk.

The life history of the silk moth passes through four stages, eggs, larva, pupa and adult.

177. Answer (4) Hint. : Edible marine fish. Sol. : Bombay duck, Hilsa, Eel, Pomfret, Salmon

and Sardines are few examples of edible marine fishes.

178. Answer (3) Hint. : DNA synthesis occurs during primer

extension step. Sol. : In polymerization step of PCR, DNA

polymerase adds nucleotides to the primer, synthesizing a new DNA strand.

179. Answer (2) Hint. : Select a technique used to produce

genetically modified crops. Sol. : Gene addition is a technique in which

cloning is used to alter the characteristics of a plant by providing it with one or more new genes.

180. Answer (1) Hint. : Bioethics are a set of standards. Sol. : GEAC stands for Genetic Engineering

Approval Committee. It may be used to regulate human activities in relation to the biological world.

NACO : National AIDS control organisation WHO : World Health Organisation EFB : European Federation of Biotechnology.

All India Aakash Test Series for NEET - 2020 · Open Mock Test-6 (Code-A)_(Hints & Solutions) All...

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