Test - 2 (Code-C)_(Answers) All India Aakash Test Series for NEET-2022
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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)
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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
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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.
All India Aakash Test Series for NEET-2022 Test - 2 (Code-C)_(Hints & Solutions)
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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 °
= = =
Test - 2 (Code-C)_(Hints & Solutions) All India Aakash Test Series for NEET-2022
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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
All India Aakash Test Series for NEET-2022 Test - 2 (Code-C)_(Hints & Solutions)
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[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
Test - 2 (Code-C)_(Hints & Solutions) All India Aakash Test Series for NEET-2022
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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.
All India Aakash Test Series for NEET-2022 Test - 2 (Code-C)_(Hints & Solutions)
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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
Test - 2 (Code-C)_(Hints & Solutions) All India Aakash Test Series for NEET-2022
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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.
All India Aakash Test Series for NEET-2022 Test - 2 (Code-C)_(Hints & Solutions)
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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’.
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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).
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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
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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
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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
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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.