All India Aakash Test Series for NEET - 2021...Test-3 (Code-A)_(Hints & Solutions) All India Aakash...

Test-3 (Code-A)_(Answers) All India Aakash Test Series for NEET-2021

Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456 1/19

All India Aakash Test Series for NEET - 2021

Test Date : 08/11/2020

ANSWERS

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

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

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

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

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

TEST - 3 (Code-A)

All India Aakash Test Series for NEET-2021 Test-3 (Code-A)_(Hints & Solutions)


[PHYSICS] 1. Answer (3)

Hint : irms = 20

1 i dt

Sol. : irms = 22 0

1 49

t dt

3

30

439t

2 A3 3

2. Answer (4)

Hint : ieffective = irms = 0

2i

Sol. : 1.4A 1.0 A2

i

3. Answer (1)

Hint : and p p

s s

V N dVV N dt

Sol. : 20VpdVdt

4000. 20 800V100

ss p

p

NV V

N

4. Answer (4)

Hint : Q = Resonant frequencyBandwidth

Sol. : r r r LQR RL

Also rL = 1 1

r rQ

C RC

And r = 1 1 LQR CLC

5. Answer (2) Hint : Pavg = Vrms irms cos

Sol. : Pavg = 100 100 cos62 2

2500 3 W

6. Answer (1) Hint & Sol. : X = XL ~ XC

At resonance, X 1 0LC

7. Answer (3) Hint : 0 2 rmsI I

Sol. : R = 2V

P = 220 220

100

= 484

220 0.4545A484

rmsrms

VI

R

0 2 0.64ArmsI I

8. Answer (1) Hint : Pavg = Vrms Irms cos

Sol. : x = 1 LC

36

1 100 20 102 50 100 10

= 25.5

2 250 (25.5) 56.12 Z

2

2144 50. 2.286 W

(3150.25)

rmsavg

VP R

Z

Energy dissipated per second = Pavg × 1 2.3 J

9. Answer (3)

Hint : 2

2 1Z RC

Sol. : For 1st term i1 = 1

10Z

For 2nd term i2 = 2

10Z

where 2

21

1100

Z R

C

2

22

1500

Z R

C

As Z1 > Z2 i1 < i2

HINTS & SOLUTIONS

Test-3 (Code-A)_(Hints & Solutions) All India Aakash Test Series for NEET-2021


10. Answer (2)

Hint : Average energy density = 20 0

12

E

Sol. : U = uavg × V

20 0

1 .2

E A l

121 18.85 10 50 50 12 4

= 2.77 nJ 11. Answer (4) Hint : Use concept of displacement current

Sol. : For loop M, 0. dB dl I

And for loop N, 0. cB dl I

As Id and Ic are non zero during charging of capacitor.

Hence .B dl will be non zero for both loop.

12. Answer (3) Hint and Sol. : As magnetic and electric field

components of an EM wave are sinusoidal and hence they have zero average value. The magnetic and electric energy is square of sine or cosine function and hence they have non zero average value.

13. Answer (2)

Hint : I = 20 0

12

E C

Sol. : 0 80 0

322 23 10

IEC

4

00

10 N/CE

14. Answer (2) Hint and Sol. : Since the frequencies of gamma

rays, X-rays and infrared rays are related as > x > IR, hence IR > x > 15. Answer (1)

Hint : 2 1 2 1v u R

Sol. : 1 1.5 1 1.56 10v

1 1 1 1 520 4 20v

v = –5 cm

16. Answer (4)

Hint : 1 1 1v u f

Sol. : 1 1 140 40v

1 1 140 40v

1 120v

v = 20 cm Image will be behind the mirror at 20 cm i.e.

between pole and focus. 17. Answer (4)

Hint : 2 1 2 1v u R

Sol. : As for upper portion and lower portion on right side of spherical surface have different refractive index, hence they will have different focal length also. Hence there will be two images one at I1 and other at I2.

18. Answer (3) Hint and Sol. : For erect image by mirror, the

magnification should be positive i.e. possible if object or image, any one should be virtual.

19. Answer (4) Hint : P = P1 + P2

Sol. : 1 21 P Pf

1 2

1fP P

20. Answer (2)

Hint : 1 1 1v u f

Sol. : 1 1 1v u f

1 1 120 10 f

f = –20 cm



21. Answer (2)

Hint : Apparent shift = 11t

Sol. : Apparent shift = 6 213

= 2 cm

Microscope should shift 2 cm upwards. 22. Answer (4)

Hint : 2

1 1 2

1 1 11f R R

Sol. : Here (2 < 1) i.e. focal length will be negative now. Hence it will behave like diverging lens.

23. Answer (4)

Hint : 1 1 1v u f

Sol. : If SP1 = SP2 = 2f Then d = 4f

If SP1 = SP2 = f Then P1P2 = 2f d = 2f

24. Answer (1)

Hint : sin c = 1

Sol. : For TIR, i > c sin(90 – ) > sin c

cos > 1

cos > 32

cos > cos 30° < 30° max = 30°

25. Answer (2)

Hint : dapparent = reald

Sol. : Since distance between object O and image I is 30 cm

Then 6 cm + happ = 15 cm happ = 9 cm

12 cm

= 9 cm

12 49 3

= 1.33 26. Answer (3)

Hint : 1 2 3

1 1 1 1f f f f

Sol. : 1 2 3

1 1 1 115 f f f

…(1)

1 2

7 1 160 f f

…(2)

From equation (1) and (2)

3

1 7 115 60 f

3

4 7 160 f

f3 = –20 cm 27. Answer (1)

Hint : sin c = 1



Sol. : sin c = 14/3

sin c = 34

tan c = 37

32 7

r

67

r

2 222 36. 16.16 m7 7

A r

28. Answer (4)

Hint : sin c = 1 vc

Sol. : sin c =

61230

0.6 182

ns

ns

sin c = 23

c = 1 2sin3

29. Answer (2)

Hint : 1 1 1f v u

Sol. :

From the symmetry For 1st position of lens u = –45 cm, v = 90 cm

1 1 1 390 45 90f

f = 30 cm

30. Answer (1)

Hint : 1 2

1 1 1 ...f f f

Sol. : 1

1 1 11f R

1 1 12 3 60

2

1 1 11.8 130 30f

1.630

3

1 1 1(1.5 1)30f

160

1 2 3

1 1 1 1 1 1.6 160 30 60f f f f

1 3.2 1 1.260 60

f = 50 cm 31. Answer (3) Hint : For normal adjustment

0

0

e

v Dmu f

and 0 eL v f

Sol. : 0 0 0

1 1 1v u f

0

1 1 12.4 2v

0

1 1 1 6 52 2.4 12v

v0 = 12 cm

0

0

12 25 252.4 5

e

v Dmu f

L = v0 + fe = 12 + 5 = 17 cm



32. Answer (1)

Hint : 1 1 1v u f

Sol. : As far point is at infinity Hence power of the eye-lens will be minimum.

i.e. maxmin

1 1 2.5 cm40

fP

i.e. Distance of retina from eye-lens is 2.5 cm. Now if eye is focused at near point the power will

be maximum i.e.

1 560 cm.60 3

P D fD

Now 1 1 1v u f

1 1 152.53

u

1 2 35 5u

u = –5 cm i.e. near point is at 5 cm 33. Answer (1)

Hint : m = o

e

ff

for normal adjustment

Sol. : m = 1002

= –50

34. Answer (2) Hint and Sol. : Since prisms are identical so any

two can produce zero deviation and hence minimum deviation produced will be

min = ( – ) + ( – ) + = 35. Answer (4) Hint and Sol. : Huygen’s principle of secondary

wavelets is used to explain wave nature of light, Snell’s law and also to find the new position of wavefront.

36. Answer (3) Hint : Wavefront is perpendicular to direction of

propagation of light. Sol. : As light is moving along z-axis, hence wave

front will be in xy plane. The general equation of xy plane is z = c

37. Answer (1) Hint : Most of light is diffracted in central maxima. Sol. : a sin = +

sin = 7

42 2 5 10

2 10a

sin 5 × 10–3 rad.

38. Answer (2)

Hint : = 1.22d

Sol. : Limit of resolution = 1.22d

= 7

71.22 5.5 10 2.684 102.5

2.7 × 10–7 rad

39. Answer (4)

Hint : brightn Dy

d

and dark2 1

2n Dy

d

Sol. : 4,72D

Dyd

7,7 7

BD Dy

d d

Now 7 72

D Dd d

= 2.0 40. Answer (3)

Hint : I = 20 cos2I

Sol. : It = 2 20 cos 60 cos 302I

0 031 32 4 4 32I I

41. Answer (2)

Hint : 2t cos = (2n – 1)2

Sol. : 2t = (2n – 1)2 [cos = 1]

64 4 1.5 0.5 10

(2 1) (2 1)t

n n



63 10

(2 1)n

For n = 1, = 3000 nm

n = 2, = 1000 nm

n = 3, = 600 nm

n = 4, = 429 nm

n = 5, = 333 nm So, within the visible range 400 nm – 700 nm light

of wavelength 429 nm and 600 nm will be strongly reflected.

42. Answer (1)

Hint : 2

max 1 22

min 1 2

( )( )

I a aI a a

Sol. : 1 23a a

2 2 2

max 22 2 2

min 2

( 3 1) ( 3 1)( 3 1) ( 3 1)

I aI a

14 43. Answer (4) Hint : ( – 1)t = n Sol. : For minimum thickness n = 1 i.e. ( – 1)t =

9

9600 10 1200 100.5

t

t = 1.2 m 44. Answer (1)

Hint : arc = R

Sol. :

Distance between two dots

x = 2.54500

cm

R = 42.54

500 5 10x

R = 10.16 cm 45. Answer (3)

Hint : vc

Sol. : From Doppler’s effect in light if galaxy will approach wavelength of light will decrease and hence it is called blue shift

[CHEMISTRY]

46. Answer (1) Hint : Species which involve d-d transition

generally show colour. Sol. : Sc3+ : [Ar] 3d0, no d-d transition Mn3+ : Violet, Cr2+ : Blue, Fe2+ : green 47. Answer (3) Hint : Due to presence of stable half filled

d-subshell configuration Mn has weak metallic bonding hence melting point is low.

48. Answer (3) Hint : On moving left to right in 3d series density

of element generally increase. Sol. : Sc < Ti < V < Ni 49. Answer (3) Hint : Tb : [Xe]4f 9 6s2 50. Answer (4)

Hint : 4MnO can act as oxidizing agent and converted into MnO2 in neutral or faintly alkaline medium.

Sol. : 2 2

4 2 2 3 2 48MnO H O 3S O 8MnO 6SO 2OH

51. Answer (4)

Hint : Magnetic moment = n(n 2) BM

(Where n : number of unpaired electrons)

Sol. : Mn2+ : 3d5

Magnetic moment = 5 7 35 BM

52. Answer (2) Hint : K2Cr2O7 oxidize H2S into S, in acidic

medium and converts itself into Cr3+ Sol. : K2Cr2O7 + H2S Cr3+ + S

Equivalent(s) of K2 Cr2O7 = Equivalent(s) of H2S Mole of K2Cr2O7× 6 = 0.2 × 2

Mole of 2 2 70.2 2 1K Cr O 0.067 mol3 30 15



53. Answer (4) Hint : Ac, Cf and No only show +3 Oxidation state

Th shows only +4 and Np show +3,+4,+5,+6 and +7 Oxidation states 54. Answer (1) Hint : Ce : [Xe] 4f1 5d1 6s2, Eu : [Xe] 4f 7 6s2

Tm: [Xe] 4f13 6s2, Yb : [Xe] 4f 14 6s2, Sm : [Xe]4f 6 6s2

Sol. : Tm2+ :[Xe]4f 13 1 unpaired electron Ce3+ : [Xe]4f 1 1 unpaired electron 55. Answer (3) Hint : Mischmetal contains around 95% lanthanoid

metal and around 5% iron and traces of S, C, Ca and Al

56. Answer (1) Hint : Zn2+ = [Ar]3d10 4s0 and Br– is a weak field

ligand.

57. Answer (3) Hint : bpy is a bidented neutral ligand and ox is a

bidented anionic ligand. Sol. : x + 2(0) + 0 + 1(–2) + (–2) = 0 x = +4 58. Answer (1)

Hint :

59. Answer (1) Hint : en (ethylene diamine) is a neutral bidentate

ligand 60. Answer (4) Hint : Order of field strength of ligand : Cl– < H2O

< NH3 < CN–

Sol. :

61. Answer (3) Hint : Cr3+ : [Ar]3d3, Co3+ : [Ar]3d6

Sol. : Both [Cr(H2O)6]3+ and [Cr(CN)6]3– contain three unpaired electrons

[Co(H2O)6]3+ contains zero unpaired electron [CoF6]3– contains four unpaired electrons 62. Answer (2) Hint : Ni2+ : [Ar] 3d8 and magnetic moment =

n(n 2) BM

Sol. : Ni2+ ion contains two unpaired electrons Magnetic moment = 2(2 2) 8 BM

63. Answer (2) Hint : Higher is the negative charge on metal,

more is the back donation, larger is the bond length and lesser will be the bond order of C – O bond

Sol. : Bond order of C–O will be maximum in [Co(CO)4]+

64. Answer (3) Hint : [Ni(en)2Cl2] shows G. I. In which cis is

optically active while trans is optically inactive Sol. : [Pt (NH3)2Cl2] shows G. I. [Pd(NH3)3Cl]Br does not show G. I. or O. I. [Co(NH3)2(H2O)2ClBr] I shows both G. I. and O.

I. 65. Answer (3) Hint : Ionization isomers can be formed by

exchange of anions from one sphere to another. Sol. :

66. Answer (2) Hint : Compounds which form most stable

carbocation will undergo fastest SN1 reaction.

Sol. : is most stable carbocation



67. Answer (1) Hint : Anions are hydrated in polar protic solvent Sol. : Order of nucleophilicity : I– > Br– > Cl– > F– 68. Answer (4) Hint : Compound which does not contain plane of

symmetry is optically active.

Sol. :

69. Answer (3) Hint :

70. Answer (3) Hint : Electron withdrawing group present at

p-position to halobenzene facilitates nucleophilic substitution reaction.

71. Answer (3) Hint : Due to partial double bond character of

‘C – Cl’ bond in haloarenes, it shows slowest rate towards nucleophilic substitution reaction.

72. Answer (4) Hint : Compound having only one type of

hydrogen atoms will give only one product on mono chlorination

73. Answer (2) Hint : Reaction follows benzyne mechanism. Sol. :

74. Answer (4) Hint : Rearrangement of carbocation takes place

to generate more stable carbocation. Sol. :

75. Answer (1) Hint : Resultant dipole moment is the vector sum

of all individual dipole and dipole moment of Ph – CI is more than Ph – F

Sol. :

76. Answer (2) NBS (N-Bromosuccinamide) is a brominating

reagent used for allylic and benzylic substitution Sol. :

77. Answer (3) Hint : The reaction follows SN1 mechanism. Sol. : SN1 mechanism is accelerated in polar protic

solvent such as ethanol. 78. Answer (4) Hint : Lone pair of oxygen is more delocalized in

Ph – O – Ph, so difficult to protonate 79. Answer (3) Hint : –OH group attached with primary (or 1°)

Carbon will be 1° alcohol 80. Answer (1) Hint : In HBO (Hydroboration oxidation) reaction

addition of water take place as per anti Markovnikov’s rule.



Sol. :

81. Answer (3)

Hint : Phenol and its derivatives are generally more acidic than aliphatic alcohol

Sol. : CH3CH2OH (pka = 15.9), Phenol (pka = 10), o – cresol (pka = 10.2)

m-nitrophenol (pka = 8.3)

82. Answer (3)

Hint : In Tertiary alcohol turbidity is produced immediately on reaction with Lucas reagent.

Sol. :

83. Answer (1)

Hint : H2S2O7 (oleum) is a sulfonating reagent.

Sol. :

84. Answer (3)

Hint : In Denaturation of alcohol, it is made unfit for drinking by mixing in it some CuSO4 (to give colour to it) and pyridine (a foul smelling liquid)

85. Answer (3)

Hint : Carbocation intermediate is formed in dehydration of alcohol.

Sol. :

86. Answer (3)

Hint : Anisole forms para isomer as major product on electrophilic substitution reaction.

Sol. : (major)

87. Answer (2)

Hint :

Electrophile is CO2

88. Answer (4)

Hint : Electron donating groups activate ring towards electrophilic substitution reaction.

Sol. : +M effect of –OCH3 and –OH groups will make the benzene ring electron rich which facilitates electrophilic substitution reaction.

89. Answer (3)

Hint : First step is protonation of ether oxygen which leads to carbocation formation.

Sol. :

90. Answer (2)

Hint : Alkylhalide on reaction with Mg in presence of ether forms Grignard reagent.

Sol. :



[BIOLOGY]91. Answer (1)

Hint : Recovery of healthy plants from the diseased plants is possible by meristem culture.

Sol. : Virus free clones of plants can be obtained through meristem culture because meristem (apical and axillary) is free of viruses due to high concentration of auxin and rapid rate of cell division.

92. Answer (3)

Hint : Micro-organisms which can be explored for obtaining SCP are fungi, cyanobacteria as well as bacteria.

Sol. : SCP is not only obtained from unicellular micro-organisms but also from multicellular organisms.

93. Answer (2)

Hint : Somatic hybridisation is the fusion of protoplast of two plants belonging to different varieties, species and even genera.

Sol. : During somatic hybridisation, the cells are first treated with pectinase and cellulase for protoplast preparation then the naked protoplasts are fused by electrofusion or chemofusion (through sodium nitrate or PEG). It results in somatic hybrid. IAA and kinetin are phytohormones.

94. Answer (1)

Hint : Micro-organism are explored for obtaining SCP and, SCP is alternate source of proteins for animals and human nutrition.

Sol. : Blue green algae (Cyanobacteria) Spirulina are grown in large quantities and these serve as food rich in protein, minerals, fats, carbohydrate and vitamins.

95. Answer (1)

Hint : Cross-hybridisation among selected parents is one of the main steps required in plant breeding programme.

Sol. : Plant breeding programme is employed to develop new varieties of crop which includes cross

hybridisation among selected parents. In this step, a cross is made between two genetically diverse parents to obtain a progeny with desired superior traits.

96. Answer (2)

Hint : By application of tissue culture it is possible to achieve propagation of large number of plants in very short duration.

Sol. : The method of producing thousands of plants through tissue culture is called micropropagation.

97. Answer (4)

Sol. : Wheat variety Atlas-66 with high protein content has been used as a donor for improving cultivated wheat.

98. Answer (2)

Hint : Mutation breeding induces mutations in plants to develop improved varieties.

Sol. : High-yielding Mexican wheat were originally red-grained

Mexican wheat Improved variety

1. Sonora-64 Sharbati Sonora

2. Lerma Rojo-64 Pusa Lerma

99. Answer (4)

Hint : In conventional breeding, crop-varieties are breed by hybridisation and selection for resistance to several fungi, bacteria and viral agents.

Sol. : Pusa Swarnim (Karan Rai) is a variety of Brassica and is resistant to white rust.

100. Answer (1)

Sol. : Pusa Sawani bred by hybridisation and selection for shoot and fruit borer is a variety of Okra (Bhindi). While, Pusa sem 2 and Pusa sem 3 bred by hybridisation and selection for Jassids, aphids and fruit borer are varieties of flat bean.

101. Answer (4)

Hint : Indian Agricultural Research Institute (IARI), New Delhi, has developed many vegetable crops that are rich in mineral and vitamins.



Sol. :

Vitamin-A enriched crops :- Carrot, Pumpkin, Spinach.

Vitamin-C enriched crops :- Bitter gourd, Bathua, Tomato, Mustard.

Calcium and Iron enriched vegetable crop :- Spinach, Bathua.

Protein enriched vegetable crop :- Beans (Broad, Lablab, French), and Garden peas.

102. Answer (1)

Hint : Breeding for disease resistance is carried out by the conventional breeding technique.

Sol. : The various sequential steps w.r.t. conventional breeding are :-

i. Screening germplasm for resistance sources.

ii. Hybridisation of selected parents.

iii. Selection and evaluation of hybrids.

iv. Testing and release of new varieties.

103. Answer (2)

Sol. : Saccharum officinarum (noble sugarcane) had higher sugar content. Rest of other options are correct.

104. Answer (1)

Hint : Development of several high yielding varieties of wheat and rice in mid 1960’s increased the yield per unit area.

Sol. : From 1960 to 2000 wheat production increased from 11 million tonnes to 75 million tonnes. This was due to the development of semi dwarf varieties of wheat.

105. Answer (2)

Sol. : Gene norin-10, responsible for dwarfing in wheat, was first reported in Japan.

106. Answer (3)

Hint : Molecules that are functional in living system or can interact with their component, are bioactive molecules.

Sol. : Cyclosporin-A is used as an immunosuppressive agent in organ-transplant patients, is produced by the fungus, Trichoderma polysporum.

107. Answer (4)

Hint : Fungal hyphae penetrate into cortex cells of root e.g., orchids, coffee, and woody plants. These are called as endomycorrhizae.

Sol. : VAM refers to endomycorrhizae. Rest of other options are correctly matched.

108. Answer (4)

Hint : Cheese is one of the oldest food items in which microbes e.g., bacteria and fungi were used.

Sol. : Large holed Swiss cheese is ripened with the help of bacterium called Propionibacterium sharmanii whereas, Roquefort and camembert cheese are ripened by fungus Penicillium roquefortii and Penicillium camembertii respectively.

109. Answer (1)

Hint : Dosa and idli are the fermented preparations of rice and black gram, prepared by using bacteria.

Sol. : Microbes used in such preparations are – Leuconostoc and Streptococcus species of bacteria.

110. Answer (2)

Hint : Bread is prepared from dough, which is fermented using Baker’s yeast.

Sol. : Puffed up appearance of dough is due to production of CO2 during fermentation. CO2 gas along with ethyl alcohol is formed during baking, making bread porous and soft.

111. Answer (3)

Hint : Certain microbes have ability to convert carbohydrate into organic acid.

Sol. : Lactic acid, butyric acid and acetic acid are produced by different bacteria. Whereas, citric acid is commercially produced by Aspergillus niger (fungus)

112. Answer (2)

Sol. : Toddy is traditional drink of some parts of South India and is made by fermenting sap from palm known as Caryota urens.

Wine, beer and rum are alcoholic beverages.



113. Answer (2)

Hint : Microbes are used for commercial and industrial production of certain chemicals (bioactive molecules).

Sol. :

Bioactive molecule Source (Microbes)

1. Lipase Candida lipolytica

2. Streptokinase Streptococcus

3. Amylase Aspergillus

4. Statins Monascus purpureus

114. Answer (2)

Hint : Primary treatment is the initial step of sewage treatment.

Sol. : Primary treatment is a physical process which involves removal of large and small particles from sewage through filtration and sedimentation.

115. Answer (1)

Hint : Biogas is a mixture of gases produced by microbial activity that can be used as fuel.

Sol. : Major component of biogas is methane (50-70%) which is highly inflammable, rest are CO2 (30-40%) and mixture of other gases H2, H2S etc.

116. Answer (3)

Hint : Antibiotics are chemical substances, which are produced by some microbes and kill or retard the growth of other (disease-causing) microbes.

Sol. : The first antibiotic penicillin was discovered by Alexander Fleming.

117. Answer (4)

Sol. : The technology of biogas production was developed in India mainly due to the efforts of Indian Agricultural Research Institute (IARI) and Khadi and Village Industries commission (KVIC)

118. Answer (2)

Sol. : Methanogens grow anaerobically on cellulosic material, produce large amount of methane along with CO2, H2S and H2. Leuconostoc and Streptococcus species of bacteria are used in fermented preparation of rice and black gram.

119. Answer (2)

Hint : Biofertilisers are organisms that enrich the nutrient quality of soil.

Sol. : Anabaena, Azotobacter and Azospirillum are employed as biofertiliser.

Bacillus thuringiensis is a microbial biocontrol agent that can be introduced to control butterfly caterpillars.

120. Answer (4)

Hint : Biocontrol refers to the use of biological methods for controlling plant diseases and pests.

Sol. : Fungus, Trichoderma is a biological control agent developed for use in treatment of plant diseases Trichoderma species are free living fungi that are very common in the root ecosystems and are effective against several plant pathogens.

121. Answer (1)

Sol. : Natality refers to birth rate during a given period in the population.

Mortality refers to death rate during a given period in the population.

Immigration is the number of individuals of the same species that have come into the habitat from elsewhere.

122. Answer (1)

Hint : Amensalism is –, 0 relationship.

Sol. : Amensalism is an interaction between two organisms of different species in which one species inhibit the growth of other species by secreting certain chemicals.

Competition, Mutualism and Parasitism are also different types of population interactions and these can be represented as (–, –), (+, +) and (+, –) respectively

123. Answer (4)

Hint : An age pyramid is a graphic representation of proportion of various age group of population.

Sol. : An age pyramid can be urn-shaped when a population has small number of pre-reproductive individuals followed by a large number of reproductive individual.



124. Answer (1)

Hint : Birth rate can be calculated by following formula:

Birth rate = NN t

Sol. : In the given question,

N = 50, N = 10, t = 1 year

Birth rate 10 10 0.250 1 50

offspring per fish

per year.

125. Answer (2)

Hint : Altitude sickness is the physiological attribute of an organism that enables the organism to survive in its habitat.

Sol. : During altitude sickness, body increases RBCs production.

126. Answer (3)

Hint : Based upon thermal tolerance, organisms are classified into stenothermal and eurythermal.

Sol. : Stenothermal organisms cannot tolerate large temperature variation e.g., polar bears, lizards, Abies.

Eurythermal organisms can tolerate large changes in temperature e.g., most of the mammals and birds.

127. Answer (1)

Hint : Conformers cannot maintain a constant internal environment. Their body temperature changes with the ambient temperature

Sol. : Graphical representation of organismic response w.r.t. conformers is

128. Answer (1)

Hint : The shape of the pyramids reflect the growth status of the population.

Sol. : The age pyramid appears like a triangle in which the population has a very high proportion of pre-reproductive individuals, the population size will expand with time. Such population shows a positive growth.

129. Answer (1)

Hint : The salt concentration (measured as salinity in parts per thousand) for water bodies.

Sol. :

Water bodies

Salinity (Parts per thousand)

Inland water < 5

Sea 30-35

Hypersaline lagoons

> 100

130. Answer (1)

Hint : A-Conformers, B-Regulators, C-Partial regulators.

Sol. : Regulators are organisms like all birds and mammals and a very few lower vertebrate and invertebrate species, and able to maintain homeostasis by physiological as well as behavioural means.

131. Answer (1)

Sol. : Kangaroo rat, in North American desert is capable of meeting all of its water requirements through its internal fat oxidation where water is released as by-product.

132. Answer (2)

Hint : Logistic growth model is considered as the most realistic one.

Sol. : Logistic growth model is represented by sigmoid curve and is mathematically described by

the equation dN K NrNdt K



133. Answer (1)

Hint : Population interaction is an interaction between two different species where interaction can be ‘+’ → beneficial, ‘–‘ → detrimental and ‘0’ → neutral.

Sol. : Competition can be represented by (–, –). Hence, it is a type of interaction in which both of the species are negatively affected.

134. Answer (3)

Hint : Lichens show mutualism.

Sol. : Lichens is a mutualistic relationship between a fungus and photosynthesizing green algae or cyanobacteria. Algae (Phycobiont) produces food through photosynthesis and fungi (mycobiont) absorbs nutrient from soil.

135. Answer (1)

Hint : Some organisms avoid the stress by over-wintering.

Sol. : Bear escapes in time during winter by a process called hibernation (over-wintering) whereas, snails, and fishes escape in time by another mechanism called aestivation (over summer).

136. Answer (4)

Hint : BCG vaccine is used for this disease.

Sol. : Tuberculosis is a bacterial disease caused by Mycobacterium tuberculosis. Measles, chicken pox and rabies are viral diseases.

137. Answer (1)

Hint : Discoverer of blood circulation.

Sol. : William Harvey discovered blood circulation and demonstrated normal body temperature in person with black bile using thermometer disproved the ‘Good humor’ hypothesis of health. Karl Landsteiner discovered blood groups and Louis Pasteur disproved theory of abiogenesis. Mary Mallon is related with typhoid.

138. Answer (2)

Hint : AIDS is an example to it.

Sol. : Diseases which are not easily transmitted from one person to another are called non-infectious disease. Diseases caused due to helminthic infections are called infestational diseases and due to deficiency of vitamins and minerals are called deficiency diseases.

139. Delete

140. Answer (3)

Hint : Cyanosis is present in pneumonia due to accumulation of CO2 in blood.

Sol. : In severe cases of pneumonia, the lips and finger nails may turn gray to bluish in colour.

141. Answer (4)

Hint : Aedes mosquito is a vector.

Sol. : Both chikungunya and dengue are viral diseases transmitted by female Aedes agypti mosquito. Elephantiasis is caused by a roundworm and malaria is caused by a protozoan. Yellow fever is also a viral disease transmitted by Aedes mosquito.

142. Answer (2)

Hint : Occurrence of disease is called pathogenesis

Sol. : Allergy causing agents are termed as allergens. Any foreign substance which can activate immune response of the body is called antigen. Interferons are antiviral proteins which prevent multiplication of viruses.

143. Answer (1)

Hint : Disease characterised by loss of memory

Sol. : Parkinson’s disease is caused due to degeneration of dopaminergic neurons in basal ganglia and is characterised by muscular tremors. Huntington chorea is due to degeneration of GABA secreting neurons in corpus striatum and cholinergic neurons of other parts. Myasthenia gravis is autoimmune disease caused due to damaging of acetylcholine receptors at neuromuscular junction.



144. Answer (1)

Hint : Disease caused by Salmonella

Sol. : Typhoid – Widal test

Malaria – Blood smear test

Diphtheria – Schick test

Dengue – Tourniquet test

145. Answer (2)

Hint : It is an autoimmune disease

Sol. : Autoimmunity is an abnormal response in which the immune system of the body starts rejecting its own body cells and molecules. Sometimes body lose its ability to differentiate between pathogen and foreign molecules from self cells and attack self cells. This results in damage to the body. Autoimmune response is against autoantigens. Antibodies are produced and interact with autoantigens in the body in case of autoimmunity.

146. Answer (3)

Hint : Infective stage of parasite are stored in mosquito.

Sol. : Sporozoites are produced in gut wall of female Anopheles during sporulation and reach into salivary glands through haemolymph for their storage. When mosquito bites they reach in human blood along with its saliva.

147. Answer (2)

Hint : Haemolymph does not contain R.B.Cs.

Sol. : Human RBCs rupture to release malarial toxin hemozoin responsible for paroxysm of malaria. P. falciparum cause malignant tertian malaria also called cerebral malaria. Sporozoites reproduce asexually in human body to produce gametocytes within RBCs. Male and female gametocytes are considered as sexual stages which enter in body of mosquito through bite and undergo gametogenesis.

148. Answer (1)

Hint : Cutaneous disease causing permanent scar marks.

Sol. : Smallpox is a viral disease caused by Variola virus leaving a permanent scar marks on skin. This disease is completely eradicated now. HIV, hepatitis–B and genital herpes are not curable if diagnosed at early stages and treated properly

149. Answer (3)

Hint: Common cold is a viral disease.

Sol.: Streptococcus pneumoniae and Haemophilus influenzae are responsible for disease pneumonia in human which infect alveoli of lungs. Microsporum, Trichophyton and Epidermophyton are responsible for ringworms. Filariasis is caused by Wuchereria bancrofti and W. malayi. Amoebic dysentery is caused by Entamoeba histolytica.

150. Answer (2)

Hint : Interferons protect non-infected cells.

Sol. : Mucus coating present at inner lining of digestive, respiratory and urinogenital tracts are included in category of physical barriers. NK cells act as cellular barriers like PMNL and monocytes. Immunity present from birth is called innate immunity.

151. Answer (2)

Hint : Colostrum contains IgA antibodies.

Sol. : Antibodies attack self cells it is autoimmunity. Immunity which develops in the body against a particular antigen is called active immunity. Immunity, present at the time of birth is called innate immunity.

152. Answer (1)

Hint : Cells which are formed to recognize antigen/pathogen

Sol. : Memory cells for sensitized B-lymphocytes and T-lymphocytes are formed during primary immune response. IgM antibodies are produced during primary immune response whereas Ig G type of antibodies are produced during secondary immune response.



153. Answer (4)

Hint : FC fragment of antibody is situated towards C-terminal.

Sol. : Antigen binding site is formed by both light and heavy chains and present towards N-terminal.

154. Answer (3)

Hint : Smallest antibody

Sol. : IgG are smallest antibodies present in our body in maximum amount and are responsible for secondary immune response of the body. Ig M is largest antibody. Being smallest in size, Ig G antibodies easily cross placenta to reach foetus during pregnancy.

155. Answer (2)

Hint : Absorption of water is facilitated with absorption of Na+.

Sol. : Cholera disease is characterised by loss of water and minerals due to frequent diarrhoea. Na+ absorption increases osmotic pressure and creates osmotic gradient for absorption of water as well as glucose and amino acids to maintain blood volume and blood pressure.

156. Answer (1)

Hint : Autoantigens are RBCs.

Sol. : Pernicious anaemia is an autoimmune disease. Genital warts is an STI. Swine flu and rabies are viral diseases.

157. Answer (4)

Hint : Histamine is a potent vasodilator.

Sol. : The exaggerated response of immune system to certain antigens present in atmosphere is called allergy. Eosinophils and IgE antibodies are increased during allergy. Histamine is not used as a drug to prevent allergy.

158. Answer (3)

Hint : The primary lymphoid organs.

Sol. : The organs where immature lymphocytes differentiate into antigen sensitive lymphocytes are known as primary lymphoid organs eg. Thymus and bone marrow. After maturation the

lymphocytes migrate to secondary lymphoid organs like spleen, lymph nodes, tonsils, appendix and peyer’s patches of small intestine

159. Answer (2)

Hint: Connecting link between endocrine and immune system

Sol.: Thymus is a bilobed endocrine gland as well as lymphoid organ of the body situated dorsal to breast bone/sternum. Lymph nodes trap any foreign harmful substances e.g Antigens.

160. Answer (2)

Hint : Immunosuppressive drugs are used to prevent graft rejection.

Sol. : Cyclosporin is an immunosuppressive drug which is used to prevent graft rejection. Grafted tissues contain antigens which activates immune system of the body, specially T-lymphocytes which are responsible for damaging of grafted tissue

161. Answer (3)

Hint : These cells are also known as platelets

Sol. : Platelets synthesize certain factor for blood clotting but they are not involved in immune system of the body.

162. Answer (1)

Hint : It is prepared by recombinant DNA technology

Sol. : Hepatitis B vaccine is a second generation vaccine prepared by recombinant DNA technology using yeast. It induces active immunity and is given to persons above age of 14 years.

163. Answer (2)

Hint : Tissue present in defence organs of the body.

Sol. : MALT (mucosa associated lymphoid tissue) is associated with mucosa of digestive, respiratory and excretory tracts and constitute about 50% of total lymphoid tissue. Blood is fluid connective tissue. MALT is not associated with neural tissue. Connective tissue is specialized to form lymphoid tissue.



164. Answer (2)

Hint : Malaria is a disease caused by protozoans

Sol. : Cholera, tetanus, leprosy and diphtheria are bacterial disease whereas smallpox and measles are viral diseases.

165. Answer (3)

Hint : Telomerase activity is high in cancer patients

Sol. : Normal cells regulate their multiplication through property of contact inhibition which is lost in cancer patients. Spread of cancer cells from one place to another place of the body is called metastasis. Alternation of generation in members of phylum coelenterata is metagenesis.

166. Answer (1)

Hint : Shell is related with molluscs.

Sol. : Shellfish is a mollusc and used in fishery industry for catching and selling. Common carp is freshwater and Hilsa and pomfrets are marine edible fishes.

167. Answer (4)

Hint : Breeding in between two different species

Sol. : Breeding between two different related species is called hybridization eg. mule, formed by mating male ass with female horse. Mating between closely related individual within same breed is called inbreeding and within same breed but having no common ancestors is called out crossing. Mating between superior male of one breed with superior female of other breed is called cross breeding

168. Answer (2)

Hint : Apiary is where beehives kept.

Sol. : Rearing of domesticated birds for meat and their eggs is called poultry. Rearing of silk moth for silk is sericulture and production of fishes is called pisciculture.

169. Answer (4)

Hint : This disease is also known as New Castle’s disease.

Sol. : New castle’s disease of poultry is a viral disease. Pasteurellosis, salmonellosis and spirochetosis are bacterial diseases in poultry.

170. Answer (1)

Hint : Under fur of Kashmiri goats.

Sol. : Pashmina wool is obtained from under fur of Kashmiri goat. Shahtoosh wool is obtained from Tibetan antelope and Angoora wool is obtained from rabbit.

171. Answer (4)

Hint : HIV has two RNA molecules each having one enzyme reverse transcriptase.

Sol. :

172. Answer (3)

Hint : X-rays are ionizing radiation.

Sol. : Ionizing radiations such as x-rays and gamma rays and non-ionizing radiations such as UV light are responsible for oncogenic transformation. Visible spectrum of light has no carcinogenic effect.

173. Answer (1)

Hint : FNAC is an invasive technique

Sol. : MRI is safest technique for detection of cancer as it uses strong magnetic field and non-invasive. X-rays are responsible for oncogenic transformation and may be harmful. Histopathological studies and FNAC are invasive techniques.



174. Answer (3)

Hint : Cannabis sativa is not a fungus.

Sol. : Amanita muscaria is a poisonous fungus having hallucinogens. Morchella is edible sac fungi. Datura, Atropa are hallucinogenic angiosperms.

175. Answer (1)

Hint : It is also known as heroin.

Sol. : Morphine and its derivatives such as codeine and heroin has receptors in our CNS and GIT. Cannabinoids has receptors mainly in the brain. Cocaine interferes with the transport of dopamine.

176. Answer (1)

Hint : Catecholamines are pressure agents

Sol. : Adrenal medulla synthesizes adrenaline and non-adrenaline to increase both heart rate and blood pressure. Nicotine activates adrenal medulla. Pituitary gland and parathyroid glands do not show direct effect on blood pressure.

177. Answer (2)

Hint : Period which starts after puberty.

Sol. : Adolescence period is bridge linking between childhood and adulthood and comes between 12-18 years of age.

178. Answer (2)

Hint : Morphine is sedative and painkiller.

Sol. : Heroin is a depressant and slow down body function. Datura, Atropa and cannabinoids have hallucinogenic property.

179. Answer (2)

Hint : Receptors of this drug are mainly present in brain.

Sol. : This structure represents cannabinoid molecule. Receptor for cannabinoids are mainly present in brain.

180. Answer (1)

Hint : Female Mammals are characterised by this feature.

Sol. : Breast enlargement is an ill effect of use of anabolic steroids by males and not by females.

All India Aakash Test Series for NEET - 2021...Test-3 (Code-A)_(Hints & Solutions) All India Aakash...

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