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Test-5 (Code-A)_(Answers) All India Aakash Test Series for NEET-2021

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All India Aakash Test Series for NEET - 2021

Test Date : 20/12/2020

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

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

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

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

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

TEST - 5 (Code-A)

All India Aakash Test Series for NEET-2021 Test-5 (Code-A)_(Hints & Solutions)


[PHYSICS] 1. Answer (3) Hint and Sol. : The working principle of steam

engine is based on laws of thermodynamics. 2. Answer (1) Hint and Sol. : Neutron was discovered by James

Chadwick. 3. Answer (4) Hint and Sol. : Pseudo forces are not fundamental

forces in nature. 4. Answer (1) Hint : L = Mvr Sol. : L = ML2T–1 a = 1, b = 2, c = –1 5. Answer (1) Hint and Sol. :

Relative error aa

= dimensionless quantity.

6. Answer (4) Hint and Sol. : 2.2 + 4.08 + 3.125 + 6.3755 = 15.7805 = 15.8 cm 7. Answer (3) Hint : L. C. = 1 M.S.D – 1 V.S.D Sol. : 25 V.S.D = 20 M.S.D

41V.S.D M.S.D5

L. C = 1 M.S.D – 45

M.S.D

1 1M.S.D 2 mm5 5

2 mm5

= 0.4 mm 8. Answer (2)

Hint : 100 100 100 v S tv S t

Sol. : 0.3 0.2100 100 0.4% 3% 3.4%10 50

vv

= 0.4% + 3% = 3.4%

9. Answer (4) Hint : 1 1 2 2n u n u

Sol. : 2 3 2 31 1 1 1 2 2 2 2 n M L T n M L T

2 3

1 1 12 1

2 2 2

M L Tn n

M L T

32

1 1 12

12 13200

2 22

M L Tn TM L

1 1 132002 4 8

= 50 unit 10. Answer (2) Hint and Sol. : In 200.00 mm, the uncertainty in

the measured value is smaller, that’s why accuracy will be greater.

11. Answer (1)

Hint : a b cv P L g

Sol. : 1 1 2 2LT ML T L LT a cb 1 2 2LT M L T a a b c a c a = 0 –a + b + c = 1 –2a – 2c = – 1

12

c

1 12

a b

1 12

a b

12

b

12. Answer (3) Hint : Use principle of homogeneity of dimensions. Sol. : CT = LT–2

C = LT–3 bT2 = LT–2 b = LT–4 a × b = L2T–7

HINTS & SOLUTIONS

Test-5 (Code-A)_(Hints & Solutions) All India Aakash Test Series for NEET-2021


13. Answer (2) Hint and Sol. : At maximum height, velocity

becomes zero due to retardation but acceleration due to gravity remains unchanged.

14. Answer (2)

Hint : dvadt

Sol. : dv = adt

2 2

1 1

v t

v t

dv adt

2

1

2 1 t

t

v v adt

v = Area under acceleration time graph. 15. Answer (3)

Hint : Use thn1 (2 1)2

S u a n

Sol. : Time taken by particle to reach the ground

from maximum height utg

4010

= 4 s

thn1 (2 1)2

S u a n

1 10(2 4 1)2

= 35 m 16. Answer (4)

Hint : dsvdt

Sol. : 2 4 3 dv t tdt v = 2t – 4 v = 0 t = 2 s At t = 2 s particle comes to rest momentarily. St = 0 = 3 St = 2 = 22 – 4 × 2 + 3 = 4 + 3 – 8 = – 1

S = |St = 2 – St = 0| = |–1–3| = 4 m 17. Answer (1) Hint and Sol. :

A B

v = u + at

1 uta

B C v = u + at

2 uta

1 2 u ut ta a

24

ua

2 104

a = 5 m/s2

18. Answer (3) Hint : Area under the velocity-time graph is

displacement.

Sol. :

1 18 2 8 8 8 22 2

h

= 64 + 16 = 80 m 19. Answer (2) Hint : Speed of the particle for motion under

gravity will be same at same position.



Sol. :

A B v2 = u2 + 2aS (3v)2 = v2 + 2gh

28

2vhg

24vh

g

20. Answer (4)

Hint : dvadt

Sol. : 2 3dv tdt

2

0

(2 3)dv t dt

2 2

0 0

2 3 v tdt dt

2 22

003t t

= 4 + 6 v = 10 m/s 21. Answer (1)

Hint : (2 1)2

naS u n

Sol. : 1 20 (2 1 1)2

aS u

202

au …(i)

2 18 (2 1 1)2

aS u

3182

au …(ii)

Solving equation (i) and (ii) u = 21 m/s and a = – 2 m/s2

Time taken by the particle to come to rest is

212

t

Now distance travelled in 21 2 s2

is

21 172

2 2

S

289 m4

22. Answer (3) Hint : Relative acceleration of the both bodies is

zero. Sol. : rel 1 2v v v (v1 and v2 in opposite direction)

= (40 – 10 × t) + 10t = 40 – 10 + 10 = 40 m/s 23. Answer (2) Hint and Sol. : |F1| – |F2| |R| |F1| + |F2| 24. Answer (4) Hint and Sol. : Centripetal acceleration is a

direction changing vector having constant magnitude.

25. Answer (2) Hint and Sol. :

3tan4

= 37° After rotating 74° clockwise

x = 5cos(–37°)



x = 4 y = 5sin(–37°) = –3

r xi y j

4 3 r i j

26. Answer (4) Hint and Sol. :

sin53 RSR

vv

45 R

SR

vv

54

SR

R

vv

27. Answer (2) Hint : When both particle collides then position

vector of the both particle will be same.

Sol. : 1 1 1 r r v t

1 4 6 2 2 2r i j i j 8 2 mi j 2 2 2

r r v t

3 4 2i j ai j 3 2 2 a i j

1 2r r

(For collision)

8 = –3 + 2a

112

a

28. Answer (4)

Hint and Sol. : 2 2 2 cos R a b ab

2 213 (3 ) 2 3 cos P P P P P

2 2 213 10 6 cos P P P

2 26 cos 3P P

1cos2

= 60° 29. Answer (2)

Hint : r xi y j

Sol. : 212

S ut at

21(4)(2)2

x

x = 8 (m)

215 2 (2)(2)2

y

= 10 + 4 y = 14 (m)

8 14 mr i j 30. Answer (1)

Hint and Sol. : 20 m/siv

20cos60fv

= 10 m/s

20 10i fv v

= 10 m/s 31. Answer (4)

Hint : 2 sin2uR

g

Sol. : 2(30) sin9010

R

R = 90 m

R = 60 m

R – R = 90 – 60 = 30 m 32. Answer (1)

Hint : 2 sin uTg

Sol. : 112 sin sin

2gTuT

g u

…(i)



22 sin(90 )uT

g

222 cos cos

2

gTuT

g u …(ii)

Squaring and adding

2 2

2 2 1 2sin cos2 2gT gT

u u

2

2 2 21 24

gu T T

2 21 22

gu T T

33. Answer (1)

Hint : 2 2sin

2uH

g

Sol. :

t2 = 5 s

12 sinut

g

(till level of projection)

12 20sin30

10t

t1 = 2 s t = t2 – t1

= 3 s (For next motion)

2 2

1(20) sin 30

2 10h

h1 = 5 m

221sin2

h u t gt

2110 3 10(3)2

= 75 m

h1 + h2 = 75 + 5 = 80 m

34. Answer (4)

Hint : sintancos

v gtv

Sol. : sin53tan45cos53

v gt

v

451 3

5

v gt

v

4 35 5 5

v v vgt

5

vtg

35. Answer (2) Hint and Sol. : Case-I for shortest time

b

dtv

10

dt

Case-II for shortest path

1 2 2b R

dtv v

1 2 210 6dt

1 8dt

t1 = t + 4

48 10d d

48 10d d

5 4 440

d d

d = 160 m



36. Answer (3)

Hint : drvdt

Sol. : 23.0 2.0 5.0dv ti t j kdt

3.0 4.0v i t j 1s 3.0 4.0 tv i j

4tan3

= 53° 37. Answer (1)

Hint : 2a r

Sol. : 2T

72100

= 0.44 rad/s a = 2r

a = (0.44)2 × 12 = 2.3 cm/s2

38. Answer (3)

Hint : 2 sin2

uR

g

Sol. : 12

sin(2 30)sin2

RR

5 33 2sin2

3 3sin210

11 3 3sin2 10

39. Answer (2)

Hint and Sol. : 25204

y x x

250 204

x x

R = 16 m

520 8 644

H (For man height 8 m2

Rx

= 80 m

40. Answer (2)

Hint : 2htg

Sol. : 12

22 9

t ht h

2

13

tt

t2 = 3t 41. Answer (1)

Hint : 2

x yu u

Rg

Sol. : ux = 4 m/s, uy = 5 m/s,

2 4 5 4 m10

R

42. Answer (3) Hint : v = u + at along vertical Sol. : vy = –10 × 1 (vy = 0) vy = –10 m/s

x yv v i v j

20 10 m/s v i j 43. Answer (3) Hint and Sol. : In horizontal projectile time of flight

of the object will be independent of horizontal velocity therefore they will reach the ground at same time with different velocities.

44. Answer (2)

Hint : tan

y

x

vv

Sol. :

y y yv u a t

= 0 + 9.8 × 2 = 19.6 m/s

tan45 yx

vv

19.61xv

vx = 19.6 m/s



45. Answer (1)

Hint and Sol. :

Component in y – z plane 2 2y z

2 2( 3) (4)

= 5

Component in z – x plane 2 2x z

2 23 4 = 5

Component in plane 5Component in plane 5

y zz x

= 1 : 1

[CHEMISTRY] 46. Answer (4) Hint : 1 mol gas at STP occupies 22.4 L volume. Sol : C3H8 + 5O2 3CO2 + 4H2O

3 8C H

n = 3 83 8

C H

C H

w 22.0=m 44

= 0.5 mol

2O

n = 3 8C H

5×n = 5 × 0.5 = 2.5 mol

2O

V = 2.5 × 22.4 = 56 L

47. Answer (4) Hint : M1 V1 = M2 V2 Sol : M1 = 0.2 M2 = ? V1 = 100 V2 = 100 + 400 = 500 ml

0.2 × 100 = M2 × 500

M2 = 0.2 100500

= 0.04 M

48. Answer (2)

Hint : x = nn+N

Sol : Mass of H2O = 1 kg = 1000 g

Moles of H2O = 100018

= 55.5

xurea = 2

urea

urea H O

n 2 2n +n 2 55.5 57.5

49. Answer (2) Hint : Apply limiting reagent concept. Sol : 2A + B 3C (i) 4 8 0 (f) 0 6 6 Limiting 6 mol of C will be formed. 50. Answer (4) Hint : Atoms are conserved in chemical reactions.

51. Answer (4) Hint : 1u = 1.66 × 10–24 g Sol : Mass of 4 He atoms = 4 × 4u = 16u = 16 × 1.66 × 10–24g 52. Answer (4) Hint : 1 g atoms = 1 mol atom 1 g molecule = 1 mol molecule Sol. : 12 g – atoms O = 12 × 16 g = 192 g 6 g – molecules O2 = 6 × 32 g = 192 g 5 mole O3 = 5 × 48 g = 240 g 224 L O2 = 10 mol O2 = 10 × 32 g = 320 g 53. Answer (2)

Hint : m = solutesolute solvent

W 1000m W

Sol : m = 49 100098 1500

= 0.33 m

54. Answer (3) Hint : Cathode rays particles are electrons.

Sol : em

of cathode ray is constant and equal to

1.76 × 1011C kg–1 55. Answer (4)

Hint : Number of spectral lines = –1

2n n

Sol : Number of spectral lines = 4 4 – 1

2 = 6

56. Answer (1)

Hint : hmv

Sol : m = 6.6 mg = 6.6 × 10–6 kg v = 100 m/s

= –34

–66.6 10

6.6 10 100

= 10–30 m



57. Answer (4)

Hint : 2

nnrz

Sol : He4r2

, 3Be9r4

3

He

Be

4r 82

9r 94

58. Answer (1)

Hint : En = 2

2Z–13.6 eV/atomn

Sol : E4(He+) = –3.4 eV/atom

59. Answer (1)

Hint : 2 21 1= R –2 n

Sol : For 1st line n = 3

2 21 1 5R= R –2 3 36

For last line n

2 21 1 R= R –2 4

60. Answer (4) Hint : Total number of nodes = n – 1 Sol : For 4d total nodes = 4 – 1 = 3 61. Answer (2)

Hint : hx. p4

Sol : 1 hv2m 2

2(mv)2 = h4

2p2 = h4

2p.p = h4

x.p = h4

(Heisenberg uncertainly principle) x = 2P

x 21p

62. Answer (1) Hint : K(19) : 1s2 2s2 2p6 3s2 3p6 4s1 Sol : Last electron of K is in 4s

n = 4, l = 0, m = 0, s = 1 –1or2 2

63. Answer (3) Hint : For H – atom energy of orbital depends only

on value of n. 64. Answer (4)

Hint : 2 r =n

Sol : Number of standing wave = shell number = 4.

65. Answer (4) Hint : Power = rate of consumption of energy Sol : 10 W = 10 Js–1

Energy of photons = –34 8

–9hc 6.626 10 3 10=

400 10

= 4.9 × 10–19 Rate of emission of photon

= 19–1910 2.04 10

4.9 10

66. Answer (4) Hint : KE Energy of photons. 67. Answer (2) Hint : More is the ease of adding electron in an

atom, more negative is egH (kJ mol–1)

Sol. :

68. Answer (1) Hint : For isoelectronic species, with increase of

number of protons, size decrease. Sol : Ionic radii 2– – 2+O > F > Mg

69. Answer (2) Hint : Uno is unniloctium. Sol : Un nil octium 1 0 8 70. Answer (2) Hint : Be has fully filled 2s subshell.

Sol : 1Li B Be

(in kJ mol ) 520 801 899



71. Answer (1)

Hint : Cu+ has pseudo inert gas configuration.

72. Answer (4)

Hint and Sol. :

73. Answer (2)

Hint : In SF4, S has 4 bond pairs and 1 lone pair.

Sol :

74. Answer (2)

Hint : Electronegative atom acquire Axial position.

Sol :

75. Answer (3)

Hint : Propadiene is H2C = C = CH2

Sol :

Maximum 5 atoms are coplanar.

76. Answer (2)

Hint : Only bond and lone pair e– take part in hybridisation.

Sol :

77. Answer (3)

Hint : Carbon cannot expand its octet.

78. Answer (3)

Hint : 2x2 2 2 2

2 2y

2pC 12 : 1s *1s 2s * 2s

2p

79. Answer (2) Hint : N2+ has lesser number of antibonding

electrons than N2– Sol : Stability order : N2+ > N2– > O2 > O2–

Bond order : 2.5 2.5 2 1.5 N2– is less stable than N2+ due to more number of

antibonding electrons. 80. Answer (2) Hint : Chloral hydrate is CCl3CH(OH)2 Sol :

81. Answer (2)

Hint :

Sol : Hybridisation of B in BF3 is sp2 Hybridisation of N in NH3 is sp3. Hybridisation of B in BH3 NH3 is sp3

Hybridisation of N in NH3 BF3 is sp3 82. Answer (3) Hint : FC =

Sol : FC of central O = 16 – 2 – 6 12

83. Answer (4) Hint : bond is due to axial overlap of atomic

orbitals. Sol :



84. Answer (4)

Hint : Naphthalene contains 5 bonds. Sol :

19 bonds and 5 bonds.

85. Answer (4)

Hint : Odd electron species are always paramagnetic.

Sol :

2 2

2– 2 2 2 2 22 z 2 2

2px *2pxO 18 : 1s *1s 2s * 2s 2p2py *2py

In O22– there is no unpaired electrons therefore diamagnetic in nature.

86. Answer (1)

Hint : 1s

87. Answer (4) Hint : In XeF2, Xe has 2 bond pairs and 3 lone

pairs.

Sol :

88. Answer (2) Hint : 1 Debye = 3.33564 × 10–30 C.m 89. Answer (3) Hint : +O – C O O = C = O +O C – O 90. Answer (2) Hint : Hybridisation of Ni in [Ni(CN)4]2– is dsp2

[BIOLOGY]91. Answer (3)

Hint : In non-living things growth is external. Sol. : Growth is not a defining property of living

beings but in living, growth is intrinsic and can be taken as defining property.

92. Answer (2)

Sol. : a. Binomial – Given by Linnaeus

nomenclature

b. Brinjal – Order polymoniales

c. Wheat – Family Poaceae

d. Taxonomy – Derived from Greek word

93. Answer (3)

Hint : Double equatorial plats are formed during meiosis I.

Sol. : Bivalents arrange in two rows at equator by formation of double metaphasic plate during meiosis I.

94. Answer (4)

Hint : During anaphase II centromere splits and chromatids separate. After separation each chromatid is considered as chromosome.

Sol. :

95. Answer (2) Hint : Diakinesis is the last phase of prophase I. It

is equivalent to late prophase of mitosis. Sol. : During diakinesis nucleolus disappear and it

represents transition to metaphase I. Chiasmata formation takes place during diplotene.

96. Answer (3) Hint : Chromosome pairs separate during

anaphase I. Sol. : Separation of chromosome pairs lead to

reduction in chromosome number in daughter cells.

97. Answer (4) Sol. : “Omnis cellula-e cellula” explains that all

cells arise from pre-existing cells.



98. Answer (2) Sol. : T. Schwann was a zoologist. He worked on

different types of animal cells. 99. Answer (2) Hint : DNA acts as genetic material in all cells

except some viruses. Ribosome is the only cell organelle found in both prokaryotes and eukaryotes.

Sol. : In prokaryotes, cell wall is not cellulosic, made up of peptidoglycan whereas in eukaryotes cellulose is the chief component of cell wall.

100. Answer (1) Sol. : In all cells, cytoplasm is the main arena of

cellular activities. 101. Answer (1) Hint : Prokaryotic flagella is made up of basal

body, hook and filament. Sol. : Prokaryotic flagella is made up of flagellin

protein whereas eukaryotic flagella is mainly made up of tubulin protein.

102. Answer (3) Sol. : In cells of meristem and parenchyma, cell

wall consists of only primary cell wall. Uphill transport requires energy.

103. Answer (4) Sol. : The middle lamella is made up of Ca++ and

Mg++ pectate. Rest of the statements are correct. 104. Answer (3) Hint : SER in muscle cells is known as

sarcoplasmic reticulum. Sol. : Sarcroplasmic reticulum (SER) is associated

with muscle contraction. 105. Answer (1) Sol. : Key – Analytical in nature. Botanical gardens – ex-situ conservation strategy Monograph – Includes information of any

one taxon Herberium – Store house of collected

plant species. 106. Answer (4) Hint : Human being is only organism who is aware

of himself. Sol. : Human being has self-consciousness too

along with awareness towards their surroundings.

107. Answer (1) Hint : Plants and animals are sensitive to external

stimuli. Sol. : Sensitivity towards surrounding and

response to it is called consciousness. 108. Answer (2) Hint : Disintegration of nuclear envelope starts

during late prophase. Sol. : Complete disintegration of the nuclear

envelope marks the start of metaphase. 109. Answer (1) Sol. : Anaphase is the best stage to study shapes

of chromosomes. 110. Answer (2) Sol. : a. Metacentric chromosome – V-shape b. Sub-metacentric chromosome – L-shape c. Acrocentric chromosome – J-shape d. Telocentric chromosome – I-shape 111. Answer (2) Sol. : G1 S transition is the major check point. 112. Answer (3) Hint : Cytoskeleton occurs only in eukaryotic cells. Sol. : Prokaryotic cells lack microtubules. 113. Answer (3) Hint : Spindle apparatus is mainly made up of

microtubules. It is meant for separation of chromatids/chromosomes.

Sol. : Pseudopodia formation, furrow formation and cytoplasmic streaming is carried out of microfilaments. For anaphasic movement of chromosomes microtubules are responsible.

114. Answer (1) Hint : In cell cycle from S-phase to anaphase,

DNA amount in a cell does not change. Sol. : G1 phase is the post mitotic gap phase

hence it has DNA amount just half to the G2 phase. 115. Answer (3) Hint : Kinetochore provides site for attachment of

spindle fibres. Sol. : Kinetochores are proteinaceous disc like

structures in primary constriction region.



116. Answer (2) Hint : This structure is not bound by any

membrane and is found in nucleoplasm. Sol. : It is nucleolus, which is the site of

r-RNA synthesis. 117. Answer (3) Hint : Gametes are haploid and fuse to form

zygote, which is diploid. Gametes are product of meiosis.

Sol. : Restorage of original diploid chromosome number in zygote is due to meiosis.

118. Answer (3) Hint : In a cell cycle karyokinesis is followed by

cytokinesis. Sol. : Cytokinesis marks the end of cell division. 119. Answer (1) Sol. : Number of generations(n) of mitosis for

producing ‘X’ cells is X = (2)n 128 cells = (2)7

256 cells = (2)8

Hence for 129 cells 8 generations of mitosis are required.

120. Answer (1) Hint : The complex formed by a pair of synapsed

chromosomes is called bivalent. Sol. : Bivalent is a pair of two homologous

chromosomes. 121. Answer (4) Sol. : When diplotene lasts for months or years in

oocyte of some vertebrates, called dictyotene stage.

122. Answer (3) Sol. : In human there are 5 pairs of SAT

chromosomes. These are chromosome number 13, 14, 15, 21 and 22.

123. Answer (1) Hint : During protein synthesis, generally in

prokaryotes many ribosomes form a chain on a common m-RNA called polysome or polyribosome.

Sol. : Fat storage – Elaioplast Tonoplast – Vacuole Colourless plastid – Leucoplast

124. Answer (2) Hint : In cilia and flagella, in centre of axoneme

two microtubules are found. Sol. : In cilia and flagella, microtubules exhibit 9 +

2 organisation. 125. Answer (3) Hint : ETS takes place in inner membrane of

mitochondria. Sol. : In mitochondria, for ATP synthesis electron

carriers and enzyme complexes are required, which is present in its inner membrane.

126. Answer (2) Sol. : In chloroplast, thylakoids are site of light

reaction. 127. Answer (3) Hint : Lysosomes are rich in hydrolytic enzymes,

which are active at acidic pH. Sol. : In plant cells vacuoles are larger and can

occupy 90% volume of the cells. 128. Answer (3) Hint : Mitochondrial DNA resembles to bacterial

DNA. Sol. : Mitochondrial DNA is double stranded and

circular. 129. Answer (2) Hint : In prokaryotes, cell wall is not cellulosic. Sol. : Bacterial cell wall is mainly made up of

peptidoglycan. 130. Answer (4) Sol. : Capsule is a layer of cell envelope of

bacteria. It gives gummy and sticky character to the cell and allows bacterium to hide from host’s immune system.

131. Answer (1) Sol. : Membrane of RBCs in human being has

approximately 52% protein and 40% lipid. 132. Answer (4) Sol. : ‘Fluid mosaic model’ about cell membrane

was given by Singer and Nicolson. 133. Answer (3) Hint : In mitochondria and chloroplast, ribosomes

are of 70S type. Sol. : In prokaryotes ribosomes are of only 70S

type found in cytoplasm. In eukaryotes ribosomes are of both 80S type (in cytoplasm) and 70S type.



134. Answer (2) Sol : Interphase is preparatory phase. During this

phase DNA replication takes place that’s why DNA is not in condensed form. DNA is in highly extended and elaborated form.

135. Answer (4) Sol. : Microtubules are found in cytoplasm. 136. Answer (4) Hint : Gastrectomy is surgical removal of

stomach. Sol. : Achlorhydria i.e. non secretion of HCl

impairs conversion of Fe3+ to Fe2+ which affects iron absorption. This results in iron - deficiency anemia. Vitamin B12 deficiency occurs due to lack of castle’s intrinsic factor. Protein digestion is only slightly delayed as it continues in intestine. Fat digestion is not completely absent as it primarily occurs in intestine.

137. Answer (3) Hint : Wisdom teeth are the third molars. Sol. : Each half of the upper or lower jaw contains

2 incisors, 1 canine, 2 premolars and 3 molars (including the 3rd molar).

138. Answer (2) Hint : Bile salts help in formation of chylomicrons. Sol. : Obstruction of bile duct affects release of

bile which contain bile salts. Bile salts help in emulsification of fats by incorporating fatty acids and glycerol into chylomicrons.

139. Answer (1) Hint : Ptyalin woks at optimum pH of 6.8. Sol. : Lemon has a pH of 2 to 3 which directly

lowers the pH of oral cavity to about 4.0 and impairs digestion of starch as salivary amylase function best at a pH of 6.8.

140. Answer (1) Hint : Deficiency of vitamins can cause certain

disorders. Sol. : Pepsin is a proteolytic enzyme. Liver

secretes bile which lacks digestive enzymes. Parotid, sublingual and submandibular glands are salivary glands. Deficiency of vitamin A causes night blindness.

141. Answer (4) Hint : Byproducts of bilirubin are responsible for

pigmentation of urine and faecal matter.

Sol. : The yellowish brown colour of stools is due to pigment derived from bilirubin and biliverdin which is present in bile juice secreted by liver. So, malfunction of liver is mostly cause of this symptom.

142. Answer (4) Hint : Posterior end of stomach is called the

pyloric antrum. Sol. : Oesophageal (cardiac) sphincter – Between

oesophageous and anterior stomach. Ileo-caecal valve – Between small and large

intestine. 143. Answer (1) Hint : Brunner’s glands secrete mucin. Sol. : Succus entericus (intestinal juice) contains

enzymes secreted by crypts of Lieberkuhn whereas Brunner’s glands secrete mucus.

144. Answer (2) Hint : Oxyntic cells secrete HCl. Sol. : Goblet cells are mucus secreting cells of

GIT. Acinar cells are exocrine cells of pancreas. Oxyntic cells are the parietal cells of stomach. Kupffer cells are phagocytic cells of liver.

145. Answer (2) Hint : Parietal cells are also called oxyntic cells. Sol. : Parotid gland is salivary gland. Pancreas

secretes certain hormones and digestive enzymes. Parietal cells secrete HCl and castle’s intrinsic factor. Paneth cells are intestinal cells which play a role in immunity.

146. Answer (1) Hint : These units contain hepatic cells. Sol. : Hepatic lobules are the structural and

functional units of liver 147. Answer (2) Hint : Prorennin is secreted by young mammals. Sol. : HCl converts prorennin to rennin which is a

milk digesting enzymes. Pepsin digests proteins in the stomach whereas

trypsin digests proteins in small intestine. 148. Answer (4) Hint : Chyle is basic in nature. Sol. : Masticated food from the oral cavity which

enters stomach is called bolus. Faeces is the end product after digestion, expelled as indigested waste.



149. Answer (4) Hint : Cilia are hair like structures. Sol. :

a. Squamous epithelium b. Cuboidal epithelium c. Columnar epithelium d. Ciliated epithelium 150. Answer (4) Hint : Carrier ions like Na+ help in active transport. Sol. : Glucose and amino acids are absorbed by

simple diffusion and also by facilitated transport with the help of carrier proteins. They can also be absorbed by active transport with electrolytes like Na+. Fatty acids and glycerol are absorbed as chylomicrons.

151. Answer (3) Hint : The condition in which food is not properly

digested. Sol. : The abnormal frequency of bowel

movements and increased liquidity of faecal discharge is known as diarrhoea. In constipation, the faeces are retained within the colon. Jaundice is yellowish discoloration caused due to deposit of bile pigments as the liver is affected.

152. Answer (1) Hint : Gross calorific value is greater than

physiologic value. Sol. :

Carbohydrates Proteins Fats

Gross calorific value (kcal/g)

4.1 5.65 9.45

Physiologic value (kcal/g)

4.0 4.0 9.0

153. Answer (4) Hint : Duct which transports bile. Sol. :

154. Answer (1) Hint : Areolar tissue Sol. :

155. Answer (4) Hint : Innermost layer of the alimentary canal. Sol. :

156. Answer (3) Hint : Systems in equilibrium cannot perform

work. Sol. : Living process is a constant effort to prevent

falling into equilibrium. This is achieved by energy input which is provided by metabolism.

157. Answer (1) Hint : Metallic ions can also be cofactors. Sol. : NAD and NADP contain the vitamin niacin.

Haem is a prosthetic group of peroxidase and catalase. Zinc is a cofactor for carboxypeptidase.



158. Answer (4) Hint :

Sol. : Hydrolases are enzymes which catalyse hydrolysis of esters, ethers etc. Transferases catalysis transfer of a group and ligases catalyse the linking together of 2 compounds.

159. Answer (4) Hint : Variable group in amino acids. 160. Answer (2) Hint : One turn occupies a distance of about

3.4nm. Sol. : In B-DNA, one turn of the helix has about 10

nucleotides on each stand of DNA. A turn occupies a distance of about 3.4nm –934Å or 3.4 10 m so that adjacent nucleotides or their bases are separated by about 0.34nm –90.34 10 m or 3.4Å .

161. Answer (2) Hint : Fatty acid has carboxyl group attached to a

R group. Sol. : Palmitic acid has 16 carbon atoms including

the carboxyl carbon whereas arachidonic acid has 20 carbon atoms.

162. Answer (2)

Hint :

Sol. : Most enzymes show maximum activity at a particular temperature or/and pH. The peak in the graph refers to optimum pH or temperature.

163. Answer (4) Hint : Malonate closely resembles the substrate in

this reaction. Sol. : Enzyme succinate dehydrogenase is a

classical example of competitive inhibition with succinic acid as its substrate. The compounds, namely, malonic acid and oxalic acid also have structural similarity with substrate for binding at the active site of succinate dehydrogenase enzyme.

164. Answer (3) Hint : Chitin is a complex polysaccharide. Sol. : Chitin is a complex polysaccharide made up

of N-acetyl glucosamine and forms the exoskeleton of arthropods.

165. Answer (4) Hint : Element with atomic number 14 present in

silica Sol. :

Element % weight in living matter

Carbon Nitrogen Calcium

18.5 3.3 1.5

166. Answer (3) Hint : Lemon grass oils is an essential oil. Sol. : Concanavalin A is a lectin and lemon grass

oil is an essential oil. 167. Answer (1)

Hint :

Sol. : Km is defined as the substrate concentration at which half of the maximum velocity in an enzyme catalysed reaction is achieved.

168. Answer (1) Hint : RuBisCO is a protein. Sol. : Keratin is a protein present in hair, nails and

claws. 169. Answer (4) Hint : - pleated sheet is a secondary structure. Sol. : In - helix, the polypeptide chain is coiled

spirally. In - pleated secondary structure, two or more polypeptide chains get interconnected by hydrogen bonds.

170. Answer (1) Hint : Reducing sugars reduce Cu2+ ions to Cu+

state. Sol. : Sucrose is non - reducing because the

aldose group of glucose and ketose group of fructose is lost due to formation of glycosidic bond. Reducing sugars have aldehyde group or ketone group.



171. Answer (2) Hint : Unused nutrition is stored as fat. Sol. : The cells of adipose tissue are specialized

to store fats. Areolar tissue serves as a support framework for epithelium. Skeletal tissue provides structural framework to the body and neural tissue is involved in control and coordination.

172. Answer (4)

Hint : Gaps help in movement of molecules. Sol. : There are three major junctions in

vertebrates:- (1) Anchoring junctions i.e. Adherens junction,

desmosome and hemidesmosome.

(2) Communicating junction i.e. gap junction. (3) Occluding junction i.e. tight junction. 173. Answer (3)

Hint : Compound epithelium is multi-layered. Sol. : Compound epithelium present at dry surface

of skin, moist surface of buccal cavity, pharynx, inner lining of ducts of salivary glands and pancreatic ducts.

174. Answer (1) Hint : Hair, nail and claws are composed of

keratin. Sol. : Excessive secretion or activity of

collagenase causes breakdown of structures rich in collagen like tendons, cartilage and bones.

175. Answer (1)

Hint : Tight junctions perform cementing.

Sol. : Communication junctions (intercalated discs) at some fusion points allow the cells to contract as a unit i.e. when one cell receives a signal the other neighbours are also stimulated to contract.

176. Answer (1) Hint : White fibrous connective tissue. Sol. : Tendon are cords of white fibrous

connective tissue that connect muscles to bone ligament connects bone to bone.

177. Answer (3) Hint : Schwann cells secrete myelin sheath. Sol. : Neurons are responsible for conduction of

nerve impulse whereas neuroglial cells help in providing support to the tissue.

178. Answer (4) Hint : Tissue which links and supports other

tissues and organs. Sol. : Connective tissue is the most abundant and

widely distributed tissue in the human body. It links and supports other tissues or organs of the body.

179. Answer (1) Hint : ‘Osteo’ refers to bone. Sol. : Osteoblasts are bone forming cells whereas

osteoclasts are bone resorbing cells. Fibroblasts secrete fibres.

180. Answer (4) Hint : Epiglottis is elastic in nature. Sol. : Hyaline cartilage is commonly found in ribs,

larynx and trachea. Fibrous cartilage is found in intervertebral disc and pubic symphysis.

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