All India Aakash Test Series for NEET-2022...Test - 2 (Code-C)_(Hints & Solutions) All India Aakash...

Test - 2 (Code-C)_(Answers) All India Aakash Test Series for NEET-2022

Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456 1/15

All India Aakash Test Series for NEET-2022

Test Date : 29/11/2020

ANSWERS

1. (4) 2. (2) 3. (2)

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

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

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

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

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

TEST - 2 (Code-C)

All India Aakash Test Series for NEET-2022 Test - 2 (Code-C)_(Hints & Solutions)

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

2/15

HINTS & SOLUTIONS

[PHYSICS]1. Answer (4)

Hint : x yv v v2 2= +

Sol. : x ydx dyv t vdt dt

5 – 6 and 5= = = =

at t = 1 s, v 2 2(–1) 5 26 m/s= + =

2. Answer (2)

Hint : R A B AB2 2sum 2 cos= + + θ

R A B AB2 2difference 2 cos= + + θ

Sol. : A2+B2+2AB cosθ = A2 + B2 – 2ABcosθ

4ABcosθ = 0, A and B ≠ 0

cosθ = 0 ⇒ θ = 90° 3. Answer (2)

Hint : rel = −

A Bv v v

Sol. : 2 2 2rel| | 2 cos90v v v v= + − °

v 2= 4. Answer (3)

Hint : ∆ = −

f iv v u Sol. : cos ( cos sin )v u i u i u j∆ = θ − θ + θ

u jsin= − θ

v u| | sin∆ = θ

5. Answer (4)

Hint : R ug

2 sin2θ=

Sol. : uR Rg

22

max maxArea(max) ( )= ⇒ = π

ug

4

2π

=

6. Answer (1) Hint : Relative motion Sol. : Relative acceleration of projectile is zero. Hence their relative velocity remains constant Srel = vrel t 7. Answer (1)

Hint : u uR Hg g

2 2 22 sin cos 2. sin22

θ θ θ= ⇒ =

Sol. : 2 1tan 2 sin and cos5 5

θ = ⇒ θ = θ =

Now Range u vRg g

2 22 sin cos 45

θ θ= =

8. Answer (2)

Hint : / = −

A B A Bv v v

Sol. : B A A B A B Av v v v v/ /= − = +

= ˆ ˆ ˆ ˆ(5 12 ) (3 4 )+ + −i j i j

= ˆ ˆ(8 8 ) m/s+i j 9. Answer (1) Hint : Concepts of River swimmer problems

Sol. : wtu v

1 2 2

2=

−

v → velocity of river u → velocity of swimmer in still water

2 2 22w w uwt

u v u v u v= + =

+ − −

wtu3

2= hence t t t2

1 2 3=

10. Answer (3) Hint : Umbrella should be in the direction of

velocity of rain with respect to boy

Sol. :

air

r

vv

12tan35

θ = =

1 12tan35

− θ =

11. Answer (2) Hints and Sol. : In uniform circular motion velocity

and acceleration are perpendicular to each other. 12. Answer (2)

Hint : 2 2net = +c ta a a

Test - 2 (Code-C)_(Hints & Solutions) All India Aakash Test Series for NEET-2022


3/15

Sol. : va rR

22 2

c3600 3 m/s1200

= ω = = =

at = 4 m/s2

hence 2 2 23 4 5 m/sneta = + =

13. Answer (1) Hint : v = rω Sol. : v = rω = r. 2πf = 0.44 × 12 = 5.3 cm/s 14. Answer (2) Hint : ∆ = −

f ir r r

Sol. : 2 2 22 cos( )∆ = + − ω

r r r r t

2 2 22 [1 cos( ) 2 .2sin2tr t r ω = − ω =

2 sin2

ω = tr

15. Answer (4) Hint and Sol. : Angular velocity is always directed

perpendicular to the plane of the circular path 16. Answer (1) Hint : v = rω Sol. : T1 = T2 ⇒ ω1 = ω2

1 1 1

2 2 2

v r Rv r r

ω= =

ω

17. Answer (3) Hint and Sol. : Displacement velocity and

acceleration changes continuously while speed remains constant.

18. Answer (1)

Hint : Range on ground = ug

2 sin2θ

Sol. : For maximum range ⇒ θ = 45°

uRg

2

max 10 m= =

19. Answer (3) Hint : displacement = shortest distance between

initial and final point Sol. :

Time to reach the ground

t 2 20 2 second10×

= =

x = 30 × 2 = 60 m

displacement d h x2 2= +

= + =400 3600 4000 m = 20 10 m

20. Answer (3) Hint and Sol. : Car can not negotiate a circular

turn, in the absence of sufficient centripetal force. 21. Answer (4)

Hint : angular speed T2π

ω =

Sol. : 1 2

2 1

3600 12 123600 1

TT

ω ×= = =

ω

22. Answer (4)

Hint and Sol. : 2 240260

π ×ω = π =f

= 8π rad/s 23. Answer (2) Hint and Sol. : a = ω2r

2

22 1200 0.6 1600 0.660

π × = × = × π

= 9600 ms–2 24. Answer (1) Hint and Sol. : Both the bullets have zero initial

velocity along vertical direction, hence they will hit the ground simultaneously.

25. Answer (1) Hint and Sol. : Both bomb and aeroplane has

same horizontal velocity at all times so bomb and aeroplane will remain in same vertical line

26. Answer (3) Hint : For a man on ground coin has same

horizontal velocity as that of train Sol. :

Path of coin will be a parabola.



4/15

27. Answer (2)

Hint : Horizontal range uRg

2 sin2θ=

Sol. : 2 2

1sin(2 15 ) 1.5 km

2× °

= = =u uR

g g

u uRg g

2 2

2sin(2 45 ) 3 km× °

= = =

28. Answer (3) Hint and Sol. : At highest point on trajectory

velocity is positive while slope is zero. 29. Answer (3) Hint and Sol. : Relative acceleration of bullet and

target is zero. So velocity of bullet w.r.t target remains constant and Hence it will hit the target.

30. Answer (1) Hint : Addition of vectors

Sol. :

Displacement along (E – W) 1 15 15 0S = − =

Displacement along (N – S) 2 15 8 7S = − =

km So net displacement is 7 km South 31. Answer (4)

Hint : 2 sin2θ

=uR

g

Sol. : Maximum range,2

1uRg

=

at 15°, 2 2 2

2sin 30 3 km

2= = =

u uRg g

hence R1 = 6 km i.e shell can’t hit the target. 32. Answer (1) Hint : Equation of trajectory y = x

2 2

2sectan

2θ

θ −gx

u

Sol. : x = 8 m, θ = 37°, u = 20 m/s

3 1 10 64 25 198 4.75m4 2 400 16 4

h y × ×= = × − × = =

×

33. Answer (2)

Hint : 2 2x yv v v= +

Sol. : at t = 2 second vx = ux

= 140 2 40 m/s2

× =

vy = uy – gt = 140 2 – 10 2 20 m/s2

× × =

hence 2 2 –140 20 20 5 ms= + =v

34. Answer (3)

Hint : rel /= = −

a b a bV v v v

Sol. : / /= = − ⇒ = +

s r s r s s r rv v v v v v

2 2 –16 8 10 ms= + =

sv

6tan 378

θ = = °

Now velocity of car relative to boat

2 2/ 2 cos(127 )c b c b c bv v v v v= + − °

3121 100 2 11 105

− = + − × × ×

–1353 ms=

35. Answer (4) Hint : Particle crosses x-axis if y = 0 Sol. : y = 0 (3t2 – 6t) = 0 ⇒ t = 0 and t = 2s 36. Answer (2) Hint : Range is equal for complementary angles

2 2

21 1 112 2 2

2 1 1

sin sin2 tan2 sin (90 – ) cos

h gh g u

θ θ= × = = θ

θ θ

But 11

2

1 303

hh

= ⇒ θ = °

2 2 2sin2 sin60 3

2u u uR

g g gθ °

= = =



5/15

37. Answer (4)

Hint and Sol. : 2

2 22t r tR vv a a a

Rπ

= × ⇒ = = π ×

r

t

aa

⇒ = π

38. Answer (3)

Hint : For y to be maximum 0dydx

=

Sol. : 2ky x xa

= −

21– 02

dy kx axdx a k

= = ⇒ =

2 4 4a a ayk k k

∴ = − =

39. Answer (3) Hint and Sol. : At maximum height vertical

component of velocity is zero but acceleration is ‘g’.

40. Answer (1)

Hint : Centripetal acceleration = 2

2 vrr

ω =

Sol. : ω2r = 4ω2r1 1 4

⇒ =rr

1 1 1 24 2r vv r= ω = ω =

41. Answer (2) Hint : at = αr ; ac = ω2r

Sol. :

When θ = 45° ⇒ ac = at

⇒ ω2 = α ⇒ α2 t2 = α , (u = 0)

1 1 second2

t = =α

42. Answer (4)

Hint : Average velocity, avgdisplacement

time takenV =

Sol. : ( )avg

2 4 22/4

r rVrT

v

= =π

2 2v=

π

43. Answer (3) Hint : If ball doesn’t hit the inclined plane, It should

travel a vertical distance H and horizontal distance

tanH

θ.

Sol. : Time of flight 2Htg

= , Horizotal range =

02Hvg

Minimum value of range so that it does not hit the

incline is tan

Hθ

0

0

1tan22

H gHvHv

g

⇒ θ = =

1

0

1tan2

gHv

− θ =

44. Answer (1) Hint : /r m r mv v v= −

Sol. :

/| | 25 9 4 m/sr mv = − =

Now when man reverse its direction /r m r mv v v= −

/| | 52 m/sr mv =

45. Answer (4)

Hint : Time period, T = 2πω

Sol. : at = rα = 20π ⇒ α = 10π rad s–2 [r = 2 m] Now, ω = ω0 + αt ⇒ ω = 0 + 10π × (4) ω = 40 π

2 2 1

40 20π π

∴ = = =ω π

T

T = 0.05 second



6/15

[CHEMISTRY]

46. Answer (2) Hint: For isoelectronic species ionic size depends

on effective nuclear charge. Solution: For isoelectronic species with increase

in number of protons (Nuclear charge) Ionic size decreases.

47. Answer (3) Hint: In case of d-block elements, group number =

no. of electrons in [(n – 1)d subshell + ns subshell]. Sol.: E.C. represents that it is ‘d’ block element. So

group no. = 6 (no. of e–s in 3d + 4s orbitals) 48. Answer (4) Hint: On moving left to right in a period of p-block,

negative electron gain enthalpy generally increases.

Sol.: ∆egH F > S > O > C 49. Answer (4) Hint: Non metals with high electronegativity and

higher oxidation states form strongly acidic oxides. Sol.: Cl2O7 is most acidic among given oxides. But

Al2O3 is amphoteric in nature. 50. Answer (4) Hint: Comparable sized atoms of highest effective

nuclear charge will require highest ionization energy.

51. Answer (2) Hint: Electronegativity decreases with increasing

atomic/ionic size. Sol. Fluorine is most electronegative among given

species with electronegativity value of 4.0. 52. Answer (4) Hint: In BeCl2, octet of Be is incomplete so it will

be electron deficient. 53. Answer (1)

Hint: CO2 has two π-bonds 1 1

1 1O C O

σ σ

π π

= =

54. Answer (3) Hint: Hydrogen bonding within a molecule is

known as intramolecular hydrogen bonding.

Sol. :

55. Answer (1) Hint: If terminal atoms are different then bond

angles are different. Sol. In CH4, all the bond angles are same as all

atoms surrounding carbon are same. 56. Answer (3) Hint: There is no lone pair on ‘Xe’ but there is two

lone pairs on each ‘O’ of XeO4.

Sol. :

No. of lone pairs of electrons = 8, in XeO4 57. Answer (2) Hint: Dipole moment of non-polar molecules is

zero. Solution: PCl5 (µNet = 0) due to its symmetrical

geometry. 58. Answer (4) Hint: Species having unpaired electron is

paramagnetic in nature. Sol.: P15 is [Ne]3s23p3 59. Answer (2) Hint: Hybridization (i.e., No. of hybrid orbitals)

= No. of σ bonds + No. of lone pairs. Solution:

2NH− :

CH4 :

60. Answer (4) Hint: Species of bond order of zero, does not exist.

Solution: 2 –2Li (4e )+ : σ1s2 σ*1s2 , 2 2B.O 0

2−

= = .

61. Answer (1)

Hint: Formal charge = 1V L S2

− − .

Solution: Formal charge of Nitrogen atom

= 5 – 0 – 82

= 5 – 4 = +1



7/15

62. Answer (4) Hint: Large size atoms form molecule of larger

bond length Solution: H2 (H – H : 74 pm) , F2 (F – F : 144 pm) Cl2(Cl – Cl : 199 pm) , Br2(Br – Br : 228 pm). 63. Answer (4) Hint: π bond is formed when the atomic orbitals

overlap parallel to each other and perpendicular to the internuclear axis.

Sol.:

64. Answer (3) Hint: Both I and Cl atoms involve in p–p

overlapping 65. Answer (4) Hint: PH3 with H+ forms co-ordinate bond. Solution:

66. Answer (3) Hint: Chlorine atom has one unpaired electron in

its ground state as E.C. is [Ne]3s23p5

Sol. : In HClO3 chlorine forms 5

covalent bonds, which is possible in second excited state.

67. Answer (4) Hint: Orbitals involved in dsp2 hybridization are

2 2x yd − , s, px and py.

68. Answer (3) Hint: Due to the presence of lone pairs in

T-shaped BrF3, no bond angle is perfect 90°.

Sol.:

69. Answer (1) Hint: sp hybridization has highest % s character.

(i.e., 50%)

Sol.: C2H2 ⇒ H – C C H≡ −sp sp

70. Answer (3) Hint: Species containing two bond pairs and 3 lone

pairs has linear shape.

Sol.: 2ICl−

71. Answer (4)

Hint: 2 * 2 2 * 2 2 22O : 1 , 1 , 2 , 2 , 2 , 2z xs s s s p p− σ σ σ σ σ π

2 2 12 , *2 *2= π π = πy x yp p p

72. Answer (4)

Hint : 2N+ [13e–] ⇒ Bond order = 2.5

2N− [15e–] ⇒ Bond order = 2.5

73. Answer (1) Hint: On increasing of size of cation, covalent

character decreases (Fajan’s rule) Solution: Order of covalent character is BeO > MgO > CaO > SrO > BaO 74. Answer (4) Hint: Hybridization (i.e. No. of hybrid orbitals) = No. of σ bonds + no. of lone pairs. Sol.:

75. Answer (2) Hint: Due to resonance –1 charge is distributed

equally at all four oxygen atoms.



8/15

Solution: 4ClO− :

Bond order 7 1.754

= =

76. Answer (2) Hint: Sulphur atom has two unpaired electrons in

ground state.

Solution: SO3 :

77. Answer (4) Hint: Generally a double bond has 1σ and 1π

bond. Solution:

78. Answer (1) Hint: Direction of dipole moment is towards more

electronegative atom. Solution: H2O(1.85D), NH3(1.47D), NF3(0.23D),

SiCl4(0D) 79. Answer (2)

Hint: 3I− contains 2bp and 3lp around centre atom

and has linear shape. Solution: Lone pair has more % s character than

bond pairs. 80. Answer (2) Hint: More the number of unpaired electrons, more

will be the paramagnetic nature. Sol.: 2 2 2 2 2 2

2 x yN : 1 , *1 , 2 , * 2 2 2 ,s s s s , p pσ σ σ σ π = π

2z x y z2 , * 2 * 2 , * 2p p p pσ π = π σ

N2 : No of unpaired electron = 0

2N− : No of unpaired electron = 1

22N − : No of unpaired electrons = 2

81. Answer (4) Hint: In positive overlap similar sign lobes are

overlapped.

Sol.:

82. Answer (2) Hint: Antibonding molecular orbitals always have

nodal planes. Sol.:

83. Answer (3) Hint:

2 0x x2 2 2 2

z2 0y y

2 *2NO : 1 *1 2 *2 2

2 *22 p p

s s s s pp p

+ π π σ σ σ σ σ

π π

Sol. : Lowest unoccupied orbital is π*2p 84. Answer (3) Hint: Increase in number of hybrid orbitals result in

decrease of % s character of hybrid orbital. Sol.: % s character in sp3d2 hybridization is

1 1006

= × = 16.67 %

85. Answer (2) Hint: Dipole moment (µ) = q × d and % ionic

character = obs

theo100µ

×µ

.

Sol.: µobs = 1.3 × 10–18 esu-cm µtheo = q × d = 4.8 ×10–10 × 2.6 ×10–8 esu-cm = 4.8 × 2.6 × 10–18 esu-cm % Ionic character

= 18

181.3 10 100 10.4%

4.8 2.6 10

−

−

×× =

× ×

86. Answer (3) Hint: Geometry of sp3d2 hybridization is octahedral

which has 12(90°) and 3(180°) bond angles. 87. Answer (4) Hint: If lone pair of an atom involved in back

bonding then it is not a part of hybrid orbital



9/15

Sol.:

88. Answer (3)

Hint:

89. Answer (4) Hint: PBr5 has axial and equatorial bond of

different length

Sol.:

90. Answer (3) Hint: H2O (B.A = 104.5°), CH4 (B.A = 109.5°)

Sol.: ψ1+ ψ2 ⇒ Bonding molecular orbital ψ1– ψ2 ⇒ Antibonding molecular orbital [Ni(CN)4]2– has dsp2 hybridization.

[BIOLOGY]91. Answer (4) Hint : Systematics is taxonomy along with

phylogeny. Sol. : Taxonomy and systematics commonly

include characterization, identification, nomenclature and classification of organism. Systematic includes evolutionary relationship among organisms also.

92. Answer (4) Hint : Category represents a rank in taxonomic

hierarchy. Sol. : Specificity decreases when we go from

species to kingdom, i.e. the higher the category, lesser will be the number of similar characteristics. Hence, numbers of similar characteristics at kingdom will be least as compared to other rank in hierarchy.

93. Answer (4) Hint : The scientific (botanical) name of brinjal is

Solanum melongena. Sol. : Solanum is genus (or generic name) while,

melongena is specific epithet (species). 94. Answer (2) Sol. : The rules of scientific naming of animals is

assigned in ICZN (International Code for Zoological Nomenclature).

95. Answer (3) Hint : Some universal rules of nomenclature

framed under codes of ICZN, ICBN, etc are the rules of binomial nomenclature.

Sol. : The scientific name is printed in italics or underlined separately when handwritten to indicate their Latin origin w.r.t rules of binomial nomenclature.

96. Answer (3) Hint : Linnaeus classified all living organisms into

two kingdoms – plantae and animalia. Sol. : The criteria for classification used by him

includes, cell wall, locomotion, mode of nutrition, response to external stimuli and contractile vacuole.

Cell structure (either prokaryotic or eukaryotic) was the basis of classification in R. H. Whittaker’s five kingdom classification.

97. Answer (1) Sol. : Five kingdom classification was proposed by

R. H. Whittaker in 1969. 98. Answer (1) Hint : Bacteria are grouped under four categories

on the basis of their shape : Cocci, Bacilli, Vibrio and Spirilla.

Sol. : Spherical shaped bacterium is Coccus. Rod shaped bacteria are called bacillus, Comma

shaped vibrium, Spiral shaped bacteria are called spirillum

99. Answer (1) Sol. : All unicellular eukaryotic organisms were

placed in kingdom Protista. 100. Answer (2) Hint : In bacteria when photosynthesis is

anoxygenic means there is no release of O2. Sol. : H2S acts as electron donor in bacterial

photosynthesis. 101. Answer (2) Hint : Purple sulphur bacteria and green sulphur

bacteria contain pigments bacteriochlorophyll, bacteriopurpurin and bacterioviridin.



10/15

Sol. : Nitrococcus is an example of chemosynthetic autotrophic bacteria while, Chlorobium is an example of green sulphur bacteria. Rest of other options are correctly matched.

102. Answer (4)

Hint : Chitin is fungal cellulose. Sol. : Cell wall of fungal hyphae is made up of

chitin.

103. Answer (3)

Hint : Bacteria exhibit a primitive form of sexual reproduction which differs from eukaryotic sexual reproduction.

Sol. : Sexual reproduction in bacteria takes place by genetic recombination.

104. Answer (4)

Hint : N2 - fixing bacteria maintain soil fertility and can be employed as biofertilizer.

Sol. : Frankia is symbiotic N2 – fixing bacteria. Rest of the other N2 – fixing bacteria in option are free living.

105. Answer (4)

Sol. : Bacillus subtilis is employed in production of antibiotics (i.e. subtilin).

Lactobacillus and Streptococcus lactis are used in production of curd, cheese, yoghurt.

106. Answer (1)

Sol. : Micrococcus candidans improves the flavour and taste in tea.

107. Answer (1)

Hint : Cell membrane of archaebacteria contains branched chain lipids which decreases membrane fluidity.

Sol. : Methanogens are the archaebacteria which are of marshy habitats and are capable of converting CO2, methanol and formic acid into methane.

108. Answer (2)

Hint : The given diagram is of a filamentous blue green algae called, Nostoc.

Sol. : ‘A’ → Heterocyst, ‘B’ → Mucilaginous sheath.

Heterocyst is a large, specialized cell in Nostoc in which N2 – fixation occurs under anaerobic condition. It also lacks PS – II activities.

109. Answer (1)

Hint : Cyanobacteria occur in symbiotic association with almost every group of eukaryotes.

Sol. : Anabaena cycadae is associated with coralloid roots of Cycas.

110. Answer (3)

Sol. : Cyanobacteria are gram negative photosynthetic prokaryotes. They are characterized by the absence of flagellum throughout life cycle.

111. Answer (1)

Hint : Lichens are composite organisms. Sol. : Lichens are symbiotic association of a fungal

partner (mycobiont) and an algal partner (phycobiont).

112. Answer (1)

Hint : Some cyanobacteria serve as food to several aquatic animals.

Sol. : Spirulina is edible, non-toxic, fast growing cyanobacterium. It is cultivated in tanks as source of protein rich animal food (SCP).

113. Answer (2)

Sol. : (A) Aulosira – Fixes N2 non-symbiotically in rice fields.

(B) Mycoplasma – are called “Bacteria with their coats off” due to many similarities with bacteria.

(C) Anabaena – Used for reclaiming usar soil.

114. Answer (2)

Hint : Protista forms a link between kingdom monera and other three kingdom plantae, fungi and animalia.

Sol. : Protistans are ancestors of all multicellular eukaryotes (plants, fungi and animals).

115. Answer (1)

Hint : Protists are photosynthetic as well as heterotrophic.

Sol. : Diatoms, Euglena and Gymnodinium are photosynthetic protists.

116. Answer (1)

Sol. : In diatoms, the cell wall is made up of two halves; one half covering the other (epitheca over hypotheca) resembling a ‘soap box’.



11/15

117. Answer (1) Hint : Dinoflagellates proliferate in large number in

sea. Sol. : Gonyaulax; a dinoflagellate causes red tide

of the sea. 118. Answer (3) Hint : By taking example of slime mould, now it

can be justified that protista forms a connecting link with plants, animals and fungi.

Sol. : Fungi like features :- Formation of fruiting bodies.

Plant like features :- cell wall around spores Animal like features :- Plasmodium is without cell

wall. 119. Answer (2) Sol. : Mycoplasma are smallest living organisms. 120. Answer (1) Sol. : Diatoms are chief producers in ocean. 121. Answer (3) Hint : Some dinoflagellates like Gonyaulax

catenella produce a toxin called saxitoxin into the sea water which is highly poisonous to vertebrates.

Sol. : Marine shell fishes consume dinoflagellates and accumulate the poison which is not harmful to the shell fish (mussel) but upon being consumed causes severe illness in man called paralytic shell fish poisoning (PSP) and even prove fatal.

122. Answer (1) Hint : The parasitic fungi on mustard is an algal

fungi. Sol. : Albugo candida is the parasitic fungi on

mustard and causes white rust of leaves. 123. Answer (2) Hint : Asexual reproduction occur through

zoospores in phycomycetes. Sol. : In oomycetes, asexual reproduction is by

zoospores. 124. Answer (4) Sol. : Claviceps causes ergot disease while,

Aspergillus is called as weed of laboratory. 125. Answer (3) Hint : Neurospora is extensively used in

biochemical and genetic work. Sol. : Neurospora is called as Drosophila of plant

kingdom.

126. Answer (1)

Hint : Mycelium of fungus can be septate as well as aseptate.

Sol. : Aseptate or coenocytic mycelium is seen in phycomycetes.

127. Answer (1)

Hint : Zoospore (planospores), sporangiospores (aplanospores), conidia are asexual spores, which can be produced in different classes of fungi.

Sol. : Conidia is produced in both ascomycetes as well as deuteromycetes.

128. Answer (1)

Hint : Deuteromycetes is the class of fungi in which fruiting bodies and sexual spores are absent.

Sol. : Colletotrichum is an example of deuteromycetes.

129. Answer (3)

Hint : Bovine spongiform encephalopathy (BSE) is also called mad cow disease.

Sol. : BSE is caused by prions. 130. Answer (4)

Sol. : Name of disease Causal agent (A) Tobacco mosaic – TMV

Disease

(B) Kuru disease – Prions

(C) Potato spindle tuber – Viroid

Disease

(D) Mumps – Paramyxovirus

131. Answer (1)

Hint : In TMV, 2130 capsomeres are arranged and its ssRNA consists of 6400 nucleotides.

Sol. : In TMV, the ratio of capsomeres : nucleotides = 1 : 3

132. Answer (3)

Hint : DNA containing viruses are called deoxyribovirus.

Sol. : dsDNA virus :- Pox virus

ssDNA virus :- φ × 174, M13 phage. 133. Answer (1) Hint : Virus means venom or poisonous fluid. Sol. : Term virus was coined by Pasteur (1880).



12/15

134. Answer (4) Hint : Viruses are connecting link between living

and non-living entities. Sol. : Living nature of virus is that they take over

biosynthetic machinery of host cells and produce chemicals required for their multiplication.

135. Answer (4) Hint : According to Whittaker’s classification

system, organisms are classified into monera, protista, fungi, plantae and animalia.

Sol. : In kingdom plantae, cells are eukaryotic with presence of chloroplast and cell wall. In plants, life-cycle follows alternation of generation.

136. Answer (3) Hint: This is a composite digestive gland. Solution: Nucleases are secreted by exocrine part

of pancreas. Duodenal cells secrete nucleosidase and nucleotidase. Endocrine part of pancreas releases hormones. Stomach does not secrete any nucleic acid digesting enzyme.

137. Answer (1) Hint: Phrenic muscle Solution: Forced expiration results in removal of

CO2 rich air by relaxation of diaphragm and contraction of abdominal muscles. Inspiration involves contraction of both the diaphragm and the external intercostal muscles.

138. Answer (3) Hint: This is the volume of air remaining in lungs

even after forceful expiration. Solution: Tidal volume, IRV and ERV when

measured by spirometer reflect a value of 500 ml, 2500-3000 ml and 1000-1100 ml respectively. Residual volume when evaluated by other methods equals 1100-1200 ml.

139. Answer (3) Hint: HCl is secreted by cells of gastric mucosa. Solution: Rennin, a milk protein (casein) digesting

enzyme is secreted by cells forming mucosal lining of stomach. Oesophagus does not secrete any enzyme.

140. Answer (1) Hint : Constituent of succus entericus. Solution: Enterokinase is a secretion of duodenal

mucosal cells that acts on trypsinogen. Trypsin, chymotrypsin and carboxypeptidase are secreted in pancreatic juice in zymogen form.

141. Answer (2) Hint: These enzyme act on N and C terminal of

polypeptides. Solution: Trypsin, pepsin and chymotrypsin are

endopeptidases and they generate peptide fragments. Carboxypeptidase acts on ‘C’ terminal while aminopeptidase acts on ‘N’ terminal. They are both called exopeptidases.

142. Answer (4) Hint: This type of transport requires carrier

proteins. Solution Lipids after emulsification and digestion

enter intestinal cells. They are absorbed as protein coated structures called chylomicrons into the lacteals.

Glucose and amino acids are absorbed along with sodium. Fructose uptake is not dependent on sodium, instead, its transport through GLUT proteins takes place.

143. Answer (2) Hint: This organ produces bile. Solution: Bile lacks mucus. Secretions of gastric

glands, salivary glands and intestinal glands contain mucus.

144. Answer (4) Hint: Oral-aboral axis. Solution: The egestion of faeces to the outside

through the anal opening of gut in man is called defaecation. Vomiting occurs through oral opening.

145. Answer (3) Hint: Identify enteroparasites. Solution: Tapeworm is a flatworm affecting small

intestine. Round worm, threadworm, hookworm and pinworm are nematodes/helminths infecting the human intestine.

146. Answer (2) Hint: These cells secrete HCl. Solution: Parietal/oxyntic cells secrete both HCl

and castle’s intrinsic factor. Vitamin B12 is not secreted but absorbed with help of castle’s intrinsic factor.

147. Answer (1) Hint: These are commonly called sugars. Solution: French fries, burger buns and ice-cream

are all digested initially by salivary amylase in the oral cavity



13/15

148. Answer (1) Hint: This includes milk sugar digesting enzyme. Solution: Succus entericus comprises brush

border enzymes and mucus. Nuclease, lipase and amylase are pancreatic secretions while rennin is a gastric secretion.

149. Answer (4) Hint: Duodenum is ‘C’ shaped. Solution: Appendix is attached to caecum part of

alimentary canal. Stomach is J shaped sac where digestion of proteins is initiated.

150. Answer (1) Hint: Succus entericus contains enzymes that

produce absorbable forms. Solution: Bile helps in emulsification of fats,

thereby increasing surface area for enzyme action. Enzymes of pancreatic juice allow partial digestion of food while intestinal enzymes act on substrates to produce simple absorbable forms.

151. Answer (4) Hint: This process involves diffusion or active

transport of nutrients. Solution: Assimilation makes nutrients available

for utilisation by cells. Ingestion refers to the process of taking in food. Deglutition refers to swallowing of food.

152. Answer (2) Hint: Gross calorific value is higher than net

calorific value. Solution:

Gross calorific value (Kcal/mol)

Net calorific value (Kcal/mol)

Carbohydrates 4.1 4.0

Proteins 5.65 4.0

Fats 9.45 9.0

153. Answer (3) Hint: In this undigested, unabsorbed substances

accumulate in large intestine. Solution: Food is not properly digested leading to

a feeling of fullness in indigestion. Abnormal frequency of bowel movement and increased liquidity of faecal discharge is known as diarrhoea. Obesity could be a case of overnutrition.

154. Answer (1) Hint: This is one case of PEM Solution: Marasmus is a form of malnutrition. It

occurs in infants i.e less than one year old. Simultaneous deficiency of both proteins and calories is observed in marasmus. Obesity is seen as a consequence of overnutrition while night blindness results from deficiency of vitamin A.

155. Answer (3) Hint: They have jointed appendages and are found

in water. Solution: Aquatic arthropods respire through

special vascularised structure called gills. 156. Answer (3) Hint: Lower invertebrates Solution: Flatworms respire through body surface

by simple diffusion in free living forms. 157. Answer (1) Hint: Accessory glands are digestive glands

associated with alimentary canal. Solution: Intestinal glands are part of alimentary

canal. They secrete digestive enzymes and mucus. Accessory glands pour their secretions into oral and duodenal lumen/cavity.

158. Answer (1) Hint: Monomers of nutrients can be absorbed. Solution: Polysaccharides are polymers. They

need to be digested and are digested in small intestine into absorbable forms. Large intestine does not secrete digestive enzymes. Water, drugs and minerals do not require digestion for absorption.

159. Answer (1) Hint: These erupt only once in lifetime. Solution: Premolars in humans are monophyodont

teeth which are present in both upper and lower jaw. They are actively involved in grinding of food.

160. Answer (4) Hint: This is the first part of small intestine in man. Solution: Colon is part of large intestine. Small

intestine is divisible into three parts duodenum, jejunum and ileum.

161. Answer (3) Hint: This is a J-shaped structure

https://www.google.com/search?q=special+vascularized+structure+called+gills.&spell=1&sa=X&ved=2ahUKEwiisODL0cToAhXUXSsKHQ_0D2YQBSgAegQIAhAl



14/15

Solution: Study of histology of human alimentary canal in T.S reveals that a layer of outer longitudinal and inner circular muscle fibres is present in oesophagus, colon and caecum. Additional layer of oblique muscle fibres is present in stomach.

162. Answer (3) Hint: Tiny raised/elevated structure located on

skeletal muscle organ in mouth. Solution: Lingual frenulum attaches tongue to

lower surface in mouth. Enamel constitutes outer layer of teeth. Epiglottis guards the opening of windpipe.

163. Answer (1) Hint: Site of exchange of gas in a lung. Solution ‘X’ is alveolar wall which is composed of

single layer of endothelial cells. The sequence of layers at diffusion membrane is alveolar membrane basement substance and blood capillary endothelium.

164. Answer (3) Hint: Disease related to chronic smokers. Solution: Occupational respiratory disorders

involve asbestosis, pneumoconiosis, and silicosis. In emphysema, patient’s residual volume increases.

165. Answer (2) Hint: Nearly a similar percent of carbohydrates are

digested by pancreatic amylase. Solution: 7% of CO2 is transported in dissolved

form. Solubility of CO2 is more than that of O2 in water/plasma. Total amount of CO2 being transported in plasma is nearly 77% (7 + 70%). Sodium bicarbonate form is chief form of transport of total CO2.

166. Answer (3) Hint: Minute volume. Solution: Volume of air a heathy man can inhale

or exhale effortlessly is minute volume Minute volume = Tidal volume × Respiratory rate = 500 ml × 14 = 6000 ml. 167. Answer (1) Hint: pO2 of 40 mmHg

Solution: pO2 and pCO2, both are 40 mm Hg in deoxygenated blood and oxygenated blood respectively.

Atmospheric air Alveoli Tissue

pO2 159 104 40

pCO2 0.3 40 45

168. Answer (2)

Hint: Food poisoning

Solution: Clostridium botulinum, a bacteria is responsible for causing food poisoning. It is not an occupational respiratory disorder.

169. Answer (3)

Hint: Fall in pH.

Solution: During hyperventilation, pO2 increases significantly. Hence apnea occurs for a short while. Urge to breathe is promoted by rise in pCO2 after a few seconds.

170. Answer (4)

Hint: This is a combination of ERV and RV

Solution: FRC = ERV +RV

= 1100 + 100 ml

= 2200 ml.

171. Answer (1)

Hint: Nasal chamber opens into this region of respiratory tract.

Solution: Pharynx opens through the larynx region into the trachea. Epiglottis is thin, elastic flap guarding the windpipe. Pleural fluid reduces friction on lung surface

172. Answer (2)

Hint: These vertebrae lie below neck and are associated with chest region.

Solution: Trachea divides at level of 5th thoracic vertebra.

173. Answer (4)

Hint: These can collapse under certain conditions.

Solution: Alveoli are region of actual exchange/diffusion of gases in human lungs. Incomplete rings of hyaline cartilage prevent collapse of primary, secondary, tertiary bronchi and initial bronchioles



15/15

174. Answer (3) Hint: Function performed by thin, irregular walled

vascularised bags. Solution: Diffusion of O2 and CO2 across alveolar

surface is not a feature of conducting part of respiratory tract.

175. Answer (2) Hint: Movement of particles from a region of high

to low concentration. Solution: Partial pressure of oxygen occurs

downhill/down the concentration gradient from atmospheric air to alveoli. Facilitated diffusion requires certain proteins. Active transport requires ATP expense.

176. Answer (2) Hint: Fluid component of blood. Solution: Maximum amount of carbonic anhydrase

is present in RBCs while minute quantities are present in fluid part of blood named plasma.

177. Answer (4) Hint: Receptors in carotid arch respond to pCO2

and H+ concentration. Solution: Pneumotaxic centre is located in pons

part of brain. Changes in pCO2 and H+ are recognised by receptors in aortic and carotid arch.

Respiratory centre in medulla oblongata is chiefly responsible for maintaining respiratory rhythm.

178. Answer (1)

Hint: Erythrocytes.

Solution: Affinity of haemoglobin for O2 does not decline rather it remain unaffected in tourists (Short-term travellers). Number and size of alveoli remains unchanged.

179. Answer (2)

Hint: Definition of partial pressure.

Solution Pressure contributed by an individual gas in a mixture of gases.

Total pressure at sea level 760 mmHg × 21% = 159 mmHg

At summit total pressure is 238 mmHg.

21% of 238 mmHg = 50 mmHg

180. Answer (2)

Hint: ‘Fe’ centres in haemoglobin.

Solution: 3% of oxygen is transported in dissolved form while 97% is transported bound to iron centres in haemoglobin. 23% of CO2 is transported in carbaminohaemoglobin form.

CO is transported bound to haemoglobin as carboxyhaemoglobin.

All India Aakash Test Series for NEET-2022...Test - 2 (Code-C)_(Hints & Solutions) All India Aakash...

Documents

Transcript of All India Aakash Test Series for NEET-2022...Test - 2 (Code-C)_(Hints & Solutions) All India Aakash...