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Test - 8 (Code-C)_(Answers) All India Aakash Test Series for NEET-2021

Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456 1/16

All India Aakash Test Series for Medical - 2021

Test Date : 15/03/2020

ANSWERS

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

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

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

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

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

TEST - 8 (Code-C)

All India Aakash Test Series for NEET-2021 Test - 8 (Code-C)_(Hints & Solutions)

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

2/16

HINTS & SOLUTIONS

[PHYSICS]1. Answer (3) Hint: Principle of homogeneity.

Sol.: 2 2PxU

x b=

+

2(unit of energy)(m )(Unit of )

m=P

(Unit of P) → joule-metre 2. Answer (1)

Hint: ∆ ∆ ∆= +

A a bA a b

Sol.: a bA Aa b

∆ ∆ ∆ = + ×

0.01 0.01 62 3

= + ×

= 0.05 Area = (6.00 ± 0.05) cm 3. Answer (2) Hint & Sol.: The range of the strong nuclear force

is 10–15 m. 4. Answer (4) Hint: 1/2 1 2= −

v v v

Sol.: 2 2car/truck 80 60 100 m/s= + =

v

5. Answer (3)

Hint: dva vdx

=

Sol.: 3= −dvv vdx

3= −vdv dx

v

0

9 0

3= −∫ ∫x

vdv dx

0

3/2

9

332

= −

v x

3/22 (9) 33

x− = −

x = 6 m

6. Answer (1) Hint: and = =∫ y xy v dt x u t

Sol.: vy = 2t

t

dy tdt0

2=∫ ∫

y = t2

x = 3t

2

9=

xy

7. Answer (1) Hint: = −

bc bG cGv v v

Sol.: = +

bG bc cGv v v

ˆ ˆ3 4i i= +

3tan4

θ =

37θ = ° 8. Answer (3) Hint: cos= θxv u Sol.: v = 20 cos 53°

= 320 12 m/s5

× =

9. Answer (3) Hint: At equilibrium ΣF = 0 Sol.:

As string is inextensible T = mgsinθ N = mgcosθ

Test - 8 (Code-C)_(Hints & Solutions) All India Aakash Test Series for NEET-2021


3/16

10. Answer (4)

Hint & Sol.: 1

1

2−

=

F m ga

m

=

60 32 zero

3

−=

g

11. Answer (3) Hint: max = µf N and max =f ma

Sol.: N = F = 10 N fmax = 5 N

max−=

mg fam

220 5 7.5 m/s2−

= =

12. Answer (2) Hint: There will be loss of kinetic energy in an

inelastic collision. Sol.: (1) Work done in the motion of a body over any

closed loop will be zero in the case of conservative force.

(2) In an inelastic collision total momentum of the system will be conserved but there will be loss of kinetic energy.

(3) Total energy of the system may not be conserved in the presence of external force.

(4) In an elastic collision momentum and energy of the system will be conserved but momentum and energy of individual body will not conserved.

13. Answer (1)

Hint: = ⋅

P F v Sol.: At highest point in the given case, Tension (T)

= 0 ⇒ power = 0 14. Answer (3) Hint: Acceleration will be non-zero if F > fmax. Sol.: fmax = µmg = 0.3 × 5 × g = 15 N F = 3 × 10 = 30 N Fnet = 30 – 15 = 15 N

215 N 3 m/s5

= =a

15. Answer (4) Hint & Sol.: For no slipping between sphere and

plank, a1 should be equal to α.

To provide α in anticlockwise direction friction on

sphere should be along positive x direction. 16. Answer (2) Hint: I = mk2 Sol.:

( )m l

I

2

1

2 2

3=

2

18

3mlI =

2

22 (3 )

2mlI m l= +

mlI ml

22

2 92

= +

2

219

2mlI =

1 2I I I= +

2 28 19

3 2ml ml

= +

2 216 57

6ml ml+

=

273

6ml

=

=IkM

(M = 2m)

7312

=k l

17. Answer (3) Hint: Conservation of angular momentum about

centre of mass.



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Sol.:

i fL L=

2 22

0 0 4ˆ ˆ ˆ2 2 12 2 2

v l mv l ml l lm k k m m k + = + + ω

2 2

0 3 2ml mlmv l

= + ω

206 5mv l ml= ω

065vl

ω =

t π=

ω

0

56

ltvπ

=

18. Answer (2) Hint: f = µmg and Fnet = ma

F f ma− =

225

af R mRR

× =

52fF f− =

72fF =

722

mg mg= µ

47

µ =

19. Answer (1) Hint & Sol.: A very large attractive coulomb force

will be produced due to huge amounts of opposite charges on the sun and the planets. This force will add up with the gravitational force, as both are central forces. In this situation, all the three Kepler’s law would be valid.

20. Answer (2) Hint & Sol.: Inside the shell gravitational field is

zero.

21. Answer (3)

Hint: 2GGMEr

=

Sol.:

Enet = 0

22. Answer (3) Hint & Sol.:

∆=

h al g

∆ =alhg

23. Answer (4) Hint & Sol.: In streamline flow velocity of all fluid

particle crossing a given position is constant and no two stream lines can cross each other.

24. Answer (1) Hint: 0(1 )= + γ∆VV t

Sol.: 1 1 1(1 )V V t′ = + γ ∆

2 2 2(1 )V V t′ = + γ ∆

2 1 2 1 2 2 1 1V V V V V t V t′ ′− = − + γ ∆ − γ ∆

2 2 1 1γ = γV V

25. Answer (3) Hint: PV = nRT

Sol.: 1mP RT

MV=

1 300mP RMV

= ×

2 350−= ×

m xP RMV

1 2P P=



5/16

300 350− = ×

m M xR RMV MV

300m = 350m – 350x 350x = 50 × 21

50 21350

×=x

= 3 g 26. Answer (2) Hint & Sol.: The kinetic energy of gas molecules

will be stored as internal energy and temperature would increase.

27. Answer (3) Hint: Work done = area under PV graph with

volume axis. Sol.: W = area

0 012

= PV


Sol.: P RTMρ

=

A → B temperature is constant, pressure is increased, therefore density will increase.

B → C pressure is constant, temperature is

increasing, therefore density will decrease 1T

ρ ∝ .

C → D temperature is constant, pressure is decreasing, therefore density will decrease.

D → A pressure is constant, temperature is

decreasing, therefore density will increase 1T

ρ ∝ .

29. Answer (2)

Hint: 2 2 2

1 2 ..... nms

v v vvn

+ +=

Sol.: 2 2 2 2 21 2 3 4 5

5msv + + + +=

555

=

= 11 (m/s)2

30. Answer (2)

Hint: dH dTKAdt dx

=

Sol.: (100 25) (100 )3 1

KA KA T− −=

75 1003

T= −

T = 100 – 25 T 75 C= °

31. Answer (4) Hint: dQ mSdT=

Sol.: 3

1

Q SdT= ∫

3

2

1

bT dT= ∫

3313

b T =

263

bQ =

32. Answer (2) Hint & Sol.: Stress has a unit N/m2 and snow is

better insulator due to presence of air pockets. 33. Answer (1)

Hint: 20

btmA A e

−=

Sol.: 020 2

btm AA e

−=

2 2=btme

ln 22

=btm

2 ln 21

t =

t = 2 ln 2 34. Answer (1)

Hint: eff

2TKµ

= π

Sol.: 1 1 1m m

= +µ

2m

µ =

eff 2KK =

22

2

mT K= π×

2 mTK

= π

35. Answer (4)

Hint & Sol.: 212

( 10 31)=

− +y

x x

212

( 10 25 6)=

− + +y

x x



6/16

212

( 5) 6y

x=

− +

max126

y =

max 2 cmy =

36. Answer (3) Hint & Sol.: Light wave is transverse while sound

wave in air is longitudinal. 37. Answer (4)

Hint: 00

s

v vf fv v

−= −

Sol.: 0

0

v vff v

+=

0

0

1 vff v

= +

0 0

0

100 100f f vf v

−× = ×

025 100= ×vv

0 4=

vv

38. Answer (2) Hint: 5A Bf f− =

Sol.: 5 3 5100 100

c cC C

f ff f + − − =

5 3 500+ =c cf f

8 500cf =

62.5 Hz=cf

62.562.5 3100

= − ×Bf

= 60.625 Hz

39. Answer (1) Hint & Sol.: k = Yr

k = 2 × 1011 × 2 × 10–10 ⇒ k = 40 N/m 40. Answer (2) Hint & Sol.: Breaking stress remains constant. 41. Answer (1)

Hint: 212

s ut at= +

Sol.: t = 5 s to 5.8 s → upward motion t = 5.8 s to 6 s → downward motion

2 21 1(0.8) (0.2)2 2

= +d g g

1 0.68{0.64 0.04} 10 3.4 m2 2

= + = × =g

42. Answer (2) Hint: Fnet = ma Sol.: T – 4g = 4a T = 4(g + a) = 4 (10 + 2) = 48 N 43. Answer (2) Hint & Sol.: As radius of orbit increases the value

of gravitational potential energy GMm

R−

increases. 44. Answer (4) Hint: PVγ = constant

Sol.: 4

γγ =

i fVPV P

1.5 3/24 4 8= = =f

i

PP

45. Answer (1) Hint & Sol.: Time period of spring pendulum is

independent of acceleration due to gravity.

[CHEMISTRY]

46. Answer (2)

Hint: number of moles of soluteMolarityvolume of solution in litre

=

Sol.: number of moles of solute 22

2312.04 106.02 10

×=

×

12 10−= × = 0.2 mole

n 0.2M 1000 0.4 MV 500

= = × =

47. Answer (4) Hint: Isoelectronic means same number of

electrons and isostructural means same shape. Sol.: Number of electrons in 2

3CO − = 32e–

3NO 32e− −=

Shape: trigonal planar



7/16

48. Answer (1) Hint: Extensive property depends on the quantity

or size of matter present in the system. Sol.: Enthalpy, volume and internal energy are

extensive properties. Temperature is an intensive property. 49. Answer (2) Hint: At high temperature, the intermolecular force

of attraction between the gas molecules are negligible and at low pressure the volume of the molecules of the gas in comparison to the volume occupied by the gas is negligibly small.

50. Answer (1) Hint: 2Ca + O2 → 2CaO Sol.: 22Ca O 2CaO

80 g 32 g+ →

4 g of Ca requires 32 4 g 1.6 g80

× = of oxygen

∴ O2 is present in excess Amount of O2 present in excess = 2.4 g – 1.6 g = 0.8 g 51. Answer (2) Hint: Equilibrium constant for the reverse reaction

is the inverse of the equilibrium constant for the reaction in the forward direction.

Sol.: For, H2(g) + I2(g) 2HI(g);

Equilibrium constant = K ∴ For 2HI(g) I2(g) + H2(g);

Equilibrium constant 1K

=

For 2 21 1HI(g) I (g) H (g);2 2

+

Equilibrium constant1/2

1/2

1 1K K

= =

52. Answer (4) Hint: The structure of diborane is

Sol.: In Diborane, the four terminal B–H bonds are

two-centre-two electron bonds and the two bridge bonds are three-centre-two electron bonds.

53. Answer (3)

Hint: (Carboxylic group) is given highest priority.

Sol.:

3-Ethyl-4-methylocta-3, 5, 7-trien -1-oic acid 54. Answer (4) Hint: Down the group, the tendency to show

catenation decreases. Sol.: The correct order of catenation tendency of

elements of group-14 is C >> Si > Ge ≈ Sn Pb does not show catenation. 55. Answer (4) Hint: Pd/BaSO4 is Lindlar’s catalyst. Sol.: CH3 – C ≡ C – CH3 + H2

CH3 – C ≡ C – CH3

56. Answer (2) Hint: The conjugate bases should have one proton

less and conjugate acid should have one extra proton.

Sol.:



8/16

57. Answer (4) Hint: Single bond has one σ bond Double bond has one σ and one π bond Triple bond has one σ and two π bonds.

Sol.:

Total number of σ bonds = 18 Total number of π bonds = 4 58. Answer (1)

Hint: 182 2i f

1 1E 2.18 10 Jn n

− ∆ = × −

/atom

59. Answer (1)

Hint: Average atomic mass 1 1 2 2

1 2

M X M XX X

+=

+

Where M1 and M2 are the atomic masses and X1 and X2 are the % composition

Sol.: Average atomic mass of element A

(12 40) (15 60)100

× + ×=

480 900 13.8 u100

+= =

60. Answer (2)

Hint: ( )1 1(HCl) 1 1(NaOH)

H1 2

M V M VM

(V V ) +

−=

+

Sol.: 1 1(HCl) 2 2(NaOH)(H )

1 2

M V M VM

(V V )+

−=

+

200 0.2 200 0.02400

× − ×=

= 9.0 × 10–2 M

pH = –log[H+] = –log[9 × 10–2] = 2 – log 9 = 2 – 0.95 = 1.05 61. Answer (3) Hint: The hydration enthalpy of alkali metal ions

decreases with increase in ionic sizes. Sol.: The correct order of hydration enthalpy of

alkali metal ions is Li+ > Na+ > K+ > Rb+ 62. Answer (3) Hint: Ca(OH)2 is used to remove temporary

hardness of water. Sol.: Clark’s method is used to remove temporary

hardness rather than permanent hardness.

Ca(HCO3)2 + Ca(OH)2 → 2CaCO3↓ + 2H2O

Mg(HCO3)2 + 2Ca(OH)2→2CaCO3↓+ Mg(OH)2↓

+ 2H2O 63. Answer (4) Hint: n = 4 and l = 1 represent 4p orbital Sol.: Maximum number of electrons that can have

quantum number n = 4, l = 1 is 6. 64. Answer (3)

Hint: In acidic medium, 4MnO− converts into Mn2+. Sol.: Balanced chemical equation

5Fe2+ + 4MnO− + 8H+ → Mn2+ + 4H2O + 5Fe3+

For 5 moles of Fe2+, 1 mole of 4MnO− is required

For 2 mole of Fe2+, 25

moles of 4MnO− are

required 65. Answer (4) Hint: Higher the standard reduction potential value

of metal, more will be its oxidizing power. 66. Answer (2) Hint: A group which deactivates the benzene ring

through resonance is a meta directing group. 67. Answer (2)

Hint: Orbital angular momentum ( ) hl l 12

= +π

Sol.: For p-orbital; l = 1

∴ Angular momentum ( ) hl l 12

= +π

2h2

=π

68. Answer (3) Hint: α-H atom of carbocation bonded with sp3

carbon takes part in hyperconjugation. Sol.: Two α–H atoms are present in

69. Answer (1) Hint: On moving left to right in the modern periodic

table, the acidic character of oxides generally increases.

Sol.: Cl2O7 is most acidic among the given oxides. 70. Answer (2) Hint: ∆H = ∆U + ∆ngRT If ∆ng > 0, then, ∆H > ∆U Sol.: 2NH3(g) → N2(g) + 3H2(g) ∆ng = 4 – 2 = 2 ∴ ∆H > ∆U



9/16

71. Answer (4) Hint: The gas with higher critical temperature is

most readily liquified. Sol.: The correct order of critical temperature of the

given gases is He < H2 < CO2 < NH3

72. Answer (3) Hint: Energy of an electromagnetic radiation is

given by E = hν Sol.: E = hν E = 6.6 × 10–34 × 10 × 1014 = 6.6 × 10–19 J 73. Answer (4) Hint: ∆G = ∆H – T∆S, ∆G < 0

Sol.: 1

1 1

H 30,000 JmolTS 60 JK mol

−

− −

∆= =

∆ = 500 K

∴ The reaction is spontaneous above 500 K 74. Answer (3)

Hint:

1 2 3 4 5 6

2 3

3 2

Hybridisation CH C CH CH CH CHof carbonatom sp sp sp

≡ − − = −↑ ↑ ↑

75. Answer (2) Hint.: For isoelectronic ions, size decreases with

increase in atomic number. Sol.: The correct order of decreasing ionic radii is

N3– > O2– > F– > Na+ 76. Answer (4) Hint: An electrophile is a species which has the

ability to accept an electron pair. 77. Answer (2) Hint: Alkene on reductive ozonolysis gives

aldehyde or ketone. Sol.:

78. Answer (3) Hint: Generally, down the group, metallic character

increases. Sol.: When the electron is added to F, due to its

small size, the interelectronic repulsion increases, and hence the electron affinity decreases. On the other hand, in Cl, the electron-electron repulsion is much less due to its larger size, so the electron affinity of Cl is more than F.

79. Answer (2)

Hint: In trigonal bipyramidal geometry, the lone pair is most preferably placed at equatorial position.

Sol.:

80. Answer (2)

Hint: For a mixture of non-reacting gases, partial pressure is directly proportional to the number of moles.

Sol.: 2N He

1 1n : n :28 4

=

= 4 : 28

= 1 : 7

81. Answer (2)

Hint: N2 is not a green house gas. 82. Answer (1)

Hint: Element X is chlorine Sol.: Chlorine belongs to halogen family i.e. group

number 17th and period number 3.

83. Answer (2)

Hint: Oxidation state of S in H2SO4 is +6.

Sol.: In HNO3, 1 + x + 3(–2) = 0 ⇒ x = +5

In HClO4, 1 + x + 4(–2) = 0 ⇒ x = +7



10/16

84. Answer (4) Hint: ∆H is an additive property.

Sol.:

85. Answer (4)

Hint: ( )b a1B.O N N2

= −

Sol.: Species Bond O2 2

2O− 1.5

22O − 1

2O+ 2.5 86. Answer (4) Hint: Polar molecules have permanent dipole

moment.

Sol.:

87. Answer (3)

Hint: Volume changes with temperature.

Sol.: Normality has volume term, so it changes

with change in temperature.

88. Answer (2)

Hint: Electron rich hydrides have excess of

electrons which are present as lone pair(s).

Sol.: CH4 is an electron precise hydride.

89. Answer (3)

Hint: Dead burnt plaster is CaSO4.

Sol.: 2 3 2Na CO 10H O⋅ is washing soda

4 21CaSO H O2

⋅ is plaster of paris

NaOH is caustic soda

90. Answer (4)

Hint: gnp cK K (RT)∆=

Sol.: For 2 2 3N (g) 3H (g) 2NH (g)+

∆ng = 2 – 4 = –2

∴ Kp ≠ Kc

[BIOLOGY]91. Answer (2)

Hint: Chloroplast and mitochondria are semi-autonomous cell organelles.

Sol.: Several ribosomes attached to a single mRNA and form a chain called polyribosome or polysome. Therefore polysomes have mRNA and rRNA only.

92. Answer (1)

Hint: Amongst aleuroplast, tonoplast, amyloplast and elaioplast, tonoplast is odd one out.

Sol.: The vacuoles are bounded by a single, semi permeable membrane called tonoplast. It facilitates the transport of ions and other materials against the concentration gradient into the vacuole.

93. Answer (4) Hint: Kinetochores present around the centromere

forms the site of attachment of microtubules of the spindle fibres.

Sol.: The microtubules of spindle fibres attach to the kinetochores in metaphase. This process will occur if kinetochores are present on centromeres. Thus in the absence of kinetochores, metaphase and so, the other phases after metaphase will not occur.

94. Answer (3) Hint: During anaphase I, homologous chromosomes

separate. Sol.: Crossing over occurs during pachytene stage. Splitting of centromere occurs during anaphase II. Terminalisation of chiasmata occurs during

diakinesis.



11/16

95. Answer (2) Hint: Leopard belongs to the family Felidae. Sol.: Canis familiaris, Canis lupus and Canis

aureus belong to the family Canidae whereas Panthera pardus belongs to the family Felidae.

96. Answer (4) Hint: Diatoms are the chief producers in the ocean. Sol.: All protozoans are heterotrophic. Euglenoids

do not have cell walls but they are photosynthetic. Protozoans are believed to be primitive relatives of

animals. 97. Answer (3) Hint: Plasmogamy and karyogamy are the

processes in sexual reproduction. Sol.: Deuteromycetes reproduce only by asexual

spores known as conidia. The mycelium is septate and branched in these fungi.

98. Answer (1) Hint: In some plants growing in swampy areas,

many roots come out of the ground to get oxygen. Sol.: Stilt roots, prop roots and axillary buds arise

from stems. Pneumatophores are respiratory roots come out of the ground vertically upwards to get oxygen for respiration.

99. Answer (3) Hint: Seeds are formed after fertilisation. Sol.: Parthenocarpic fruits are formed without

fertilisation, therefore they lack seeds. 100. Answer (2) Hint: The companion cells help in maintaining the

pressure gradient in the sieve tubes. Sol.: Companion cells are closely associated with

sieve tube elements by pit fields. 101. Answer (2) Hint: Mesophyll is found in leaves. Sol.: Cambium is absent in monocots and leaves.

Therefore their vascular bundles are said to be closed.

102. Answer (1) Hint: The life-cycle shown in the figure is

Haplo-diplontic. Sol.: Along with bryophytes and pteridophytes,

some algae like Ectocarpus, Polysiphonia, kelps are haplo-diplontic. All these plants do not form seeds.

103. Answer (1) Hint: Gametophytes of homosporous species in

pteridophytes are monoecious, i.e., male and female sex organs are present on the same thallus.

Sol.: Dryopteris is homosporous whereas Selaginella, Marsilea and Salvinia are heterosporous.

104. Answer (4) Hint: Water potential (ψw) = solute potential (ψs) +

pressure potential (ψp) Sol.: Solute potential of a solution is always a

negative value whereas pressure potential is a positive value. If their magnitude is same, then the sum will be zero.

105. Answer (4) Hint: Phloem sap has water and other organic

molecules. Sol.: Phloem transports water, amino acids,

hormones, sucrose and other sugars. 106. Answer (4) Hint: The Mo-Fe protein in root-nodules of

leguminous plants is nitrogenase. Sol.: Nitrogenase catalyses the conversion of

atmospheric nitrogen to ammonia. 107. Answer (3) Hint: Critical enzymes involved in carbon fixation

during C3 and C4 cycles are RuBisCO and PEPcase respectively.

Sol.: Mg2+ is an activator of both RuBisCO and PEPcase.

108. Answer (2) Hint: In non-cyclic photophosphorylation NADPH is

formed. Sol.: NADP+ is reduced to NADPH + H+ during

non-cyclic photophosphorylation. Due to increase in concentration of H+ in thylakoid

lumen, pH decreases. 109. Answer (3) Hint: Along with FADH2 and GTP two molecules of

NADH + H+ are also formed after α-ketoglutaric acid in Krebs cycle.

Sol.:

2

2 NADH H 2 3 ATP1 FADH 2 ATP1 GTP 1 ATPTotal 9 ATP

++ = ×===

110. Answer (1) Hint: Heterophylly is an example of plasticity. Sol.: Transpiration does not occur from the

submerged part of the plant. 111. Answer (3) Hint.: Coleoptile bends towards unilateral light. Sol.: Phototropism is the directional movement of

growth in response to light.



12/16

112. Answer (4) Hint: Dormancy is due to internal factors. Sol.: Many seeds having endosperm also

germinate. 113. Answer (2) Hint.: α-ketoglutaric acid and succinyl CoA

respectively are 5 C and 4 C compounds. Sol.: 6 C isocitric acid is converted into 6 C

oxalosuccinic acid in the presence of isocitrate dehydrogenase.

114. Answer (2) Hint: Two redox equivalents are removed per

dehydrogenation reaction. Sol.: Total 5 dehydrogenation reactions occur in

complete oxidation of PGA forming 4 NADH2 and 1 FADH2 molecules. Thus 5 × 2 = 10 redox equivalents are removed.

115. Answer (3) Hint: Organic acids like oxalic acid and malic acid

have RQ value more than 1. Sol.: Malic acid – 1.33 Glucose – 1.0 Oxalic acid – 4.0 Tripalmitin – 0.7 116. Answer (3) Hint: In Monera, cyanobacteria show oxygenic

photosynthesis. Sol.: In five kingdom classification, chlorophyllous

photosynthetic organisms evolving oxygen belong to three kingdoms (Monera, Protista and Plantae).

117. Answer (4) Hint: Silica shells of dead diatoms are nearly

indestructible. Sol.: Chrysophytes include diatoms and desmids 118. Answer (1) Hint: Both archaebacteria and eubacteria are

prokaryotes. Sol.: Archaebacteria differ from other bacteria in

having different cell wall structure which is responsible for their survival in extreme conditions.

119. Answer (4) Hint: Zygospores are resting diploid spores. Sol.: Ascospore and basidiospores are haploid. 120. Answer (2) Sol.: In beans, seeds are non-endospermic. 121. Answer (2) Hint: Pulvinus is the swollen leaf base. Sol.: In some leguminous plants, the leaf base

may become swollen and is called pulvinus.

122. Answer (4) Sol.: Suckers - Chrysanthemum 123. Answer (3) Sol.: Mustard - Racemose inflorescence 124. Answer (4) Hint: In the flowers of Liliaceae, gynoecium is

tricarpellary. Sol.: Allium cepa (onion) belongs to the family

Liliaceae. 125. Answer (1) Sol.: In Alstonia whorled phyllotaxy is found. 126. Answer (2) Hint: Viruses are smaller than bacteria. Sol.: Virus take over biosynthetic machinery of

host cell and they are non-cellular obligate parasites.

127. Answer (4) Hint: Zygotic meiosis occurs in plants showing

haplontic life-cycle pattern. Sol.: Haplontic life-cycle pattern is shown by

Volvox and Spirogyra. 128. Answer (4) Hint.: In axile placentation, placenta is present in

the axial position in multilocular ovary. Sol.: In free central placentation, ovary is

unilocular, ovules are borne on central axis and septa are absent.

129. Answer (4) Sol.: Cell wall provides barrier to undesirable

macromolecules. 130. Answer (4) Hint.: Splitting of centromeres occurs just after

metaphase. Sol.: In anaphase, centromeres split and

chromatids are separated. This is disjunction of chromatids.

131. Answer (2) Hint.: Porins are found in plasma membrane. Sol.: Porins are the proteins that form huge pores

in outer membrane of plastids, mitochondria and some bacteria.

132. Answer (3) Hint.: Denitrification is the process through which

nitrates are converted into molecular nitrogen.

Sol.: 3NO N N− → ≡ [Denitrification]

3 2 3NH NO NO− −→ → [Nitrification]



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133. Answer (3) Sol.: Bulk transport of minerals, water and food

occurs in plant is called translocation. 134. Answer (4) Hint.: Annual plants require one season to

complete their life-cycle. Sol.: Sugarbeet, cabbages, carrots are some of

the common biennials which show vernalisation. 135. Answer (2) Hint: In C4 plants, Calvin cycle does not occur in

mesophyll cells. Sol.: In C4 plants, Calvin cycle occurs in bundle

sheath cell where RuBisCO is present. ATP for Calvin cycle comes from light reactions of

photosynthesis that occur in chloroplasts. 136. Answer (4) Hint: Sarcomere is present between two

Z-lines. Sol.: Z-line or Krause’s membrane is a dark

membrane which bisects I-band or isotropic band. 137. Answer (4) Hint: Select the fruit sugar. Sol.: Monosaccharides are modified variously to

form a number of different substances for eg., Deoxy sugar - Deoxyribose Amino sugar - Glucosamine and

galactosamine Fructose is also known as levulose (fruit sugar). It

is the sweetest among naturally occuring sugars. 138. Answer (4) Hint: Heat is released in an exothermic reaction. Sol.: In exothermic reactions, the energy content of

the product is lower than that of the substrate as heat is released.

139. Answer (4) Hint: It is the innermost layer lining the lumen of

alimentary canal. Sol.: Serosa is the outermost layer of alimentary

canal, muscularis layer is formed by smooth muscles usually arranged into an inner circular and an outer longitudinal layer. The submucosal layer is formed of loose connective tissue containing nerves, blood and lymph vessels.

140. Answer (3) Hint: Fatty acids are absorbed here. Sol.: Small intestine is the principal organ for

absorption of nutrients. Digestion is completed here and the final products of digestion such as glucose, fructose, fatty acids, glycerol and amino acids are absorbed through the mucosa into the blood stream and lymph.

141. Answer (3) Hint: Volume of thoracic cavity increases during

inspiration. Sol.: Statement (a), (c) & (d) are correct.

Contraction of external intercostal muscles lifts ribs and sternum up & outwards. Inspiration in mammals is an active process. It occurs due to contraction of external intercostal muscles and diaphragm.

142. Answer (2) Hint: Part of brainstem that forms ‘bridge’. Sol.: Pneumotaxic centre is present in the pons

region of hind brain. It moderates the function of respiratory rhythm centre. The neural signal from this centre can reduce the duration of inspiration and thereby alter the respiratory rate.

143. Answer (1) Hint: This vein starts and ends between capillary

beds. Sol.: The hepatic portal system involves portal vein

which carries blood from intestine to the liver before it is delivered to the heart.

144. Answer (3) Hint: It makes the lumen of vessel narrower. Sol.: Heart failure is the state of heart when it does

not pump blood effectively enough to meet the needs of the body. Heart attack is when the heart muscles are suddenly damaged by an inadequate blood supply. Cardiac arrest means complete stoppage of the heart beat.

Coronary artery diseases, often referred to as atherosclerosis, affects the vessels that supply blood to the heart muscle. It is caused due to the deposition of calcium, fat, cholesterol and fibrous tissues in the arteries supplying the heart musculature. These depositions make the lumen of arteries narrower.

145. Answer (2) Hint: Permeable to salts. Sol.: The descending limb of loop of Henle is

permeable to water but almost impermeable to electrolytes.

146. Answer (1) Hint: Almost all the glomerular filtrate is

reabsorbed by the renal tubules. Sol.: The tubular epithelial cells in different

segments of nephron carry out reabsorption either by active or passive mechanisms. For example, substances like glucose, amino acids, Na+, etc. in the filtrate are reabsorbed actively, whereas nitrogenous wastes are reabsorbed by passive transport.



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147. Answer (1) Hint: It is not usually inheritable. Sol.: In Myasthenia gravis, destruction of

acetylcholine receptors occurs due to autoimmune response. Muscular dystrophy is a genetic disorder leading to degeneration of skeletal muscles. Tetany occurs due to hypocalcemia.

148. Answer (3) Hint: Adrenaline increases heart rate. Sol.: The sympathetic neural system accelerates

the heart beat whereas the parasympathetic neural system slows down the heart beat. Parasympathetic neural system stimulates the secretion of saliva.

149. Answer (3) Hint: These ions act as a second messengers. Sol.: Ca++ ions diffuse into the axon terminal from

the surrounding fluid on opening of voltage-gated calcium channels.

150. Answer (2) Hint: Aldosterone is mineralocorticoid. Sol.: Principle actions of aldosterone is in

controlling electrolyte and water metabolism. Cortisol raises blood glucose level. Estrogen develops secondary sexual characteristics in females. Adrenaline constricts skin and visceral smooth muscle capillaries & dilates arterioles of heart and skeletal muscle.

151. Answer (3) Hint: It maintains circadian rhythm. Sol.: Melatonin is derived from amino acid

tryptophan. 152. Answer (2) Hint: Select a mollusc. Sol.: Aplysia belongs to phylum mollusca.

Segmentation is absent in body of molluscs. Nereis, Pheretima, Hirudinaria are annelids. The

body of annelids is metamerically segmented i.e. the external segmentations of the body correspond to the internal segmentations.

153. Answer (4) Hint: Echinoderms are free-living animals. Sol.: No parasitic form is found in echinoderms. All

echinoderms are exclusively marine and usually live at the bottom of sea hence they are called bottom dwellers.

154. Answer (1) Hint: Feature of aves and mammals. Sol.: The heart of lizard is 3 chambered. Among

reptiles, crocodiles possess 4 chambered heart.

155. Answer (4) Hint: Haemolymph does not contain respiratory

pigment. Sol.: Respiratory system of cockroach consists of

spiracles and a network of trachea and tracheoles. 156. Answer (3) Hint: Fluid connective tissue. Sol.: Examples of loose connective tissues are

areolar tissue and adipose tissue. Specialized connective tissue include skeletal connective tissue like cartilage, bone and fluid connective tissue like blood and lymph. Muscle tissue comprises muscles of body which enable movements of the body.

157. Answer (4) Hint: Compound epithelium. Sol.: Stratified keratinised squamous epithelium

covers the dry surface of skin. Heavy deposits of the insoluble protein keratin in the dead superficial cells make the epithelium impervious to water and highly resistant to mechanical abrasions. In contrast, non-keratinised stratified epithelium cannot prevent water loss and afford only moderate protection against abrasions.

158. Answer (1) Hint: Smooth muscle fibres are unstriated. Sol.: Each muscle fibre of cardiac muscle has a

single centrally located nucleus. Nuclei are peripheral in skeletal muscle fibres. They possess faint striations.

159. Answer (3) Hint: A carrier protein.

Sol.: Protein Functions

Collagen Intercellular ground substance

Trypsin Enzyme

Insulin Hormone

Antibody Fights infectious agents

GLUT-4 Enables glucose transport into cells

160. Answer (2) Hint: Component of lipids. Sol.: Substituted methane with amino and

carboxylic group. Amino acids have four substituent groups hydrogen, carboxyl group (–COOH), amino group (–NH2) and a variable group designated as R-group. Glycerol is trihydroxy propane and is a constituent of lipids.



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161. Answer (2) Hint: HCl converts pepsinogen to pepsin. Sol.: HCl provides acidic pH (pH 1.8) optimal for

pepsin.

Proteins Pepsin→ Proteoses + Peptones 162. Answer (1) Hint: Teeth are arranged in order I, C, PM, M.

Sol.: 21232123

represents arrangement of teeth in

each half of the upper and lower jaw in the order I, C, PM, M. Total number of teeth in adult human are I – 8, C – 4, PM – 8 and M – 12.

163. Answer (4) Hint: ERV + TV + IRV Sol.: Vital capacity (VC) is defined as the maximum

volume of air a person can breathe in, after a forceful expiration or the maximum volume of air a person can breathe out after a forceful inspiration.

Depending upon the age, sex and height of individual, its value varies from 4000 ml to 4600 ml.

Vital capacity (VC) = ERV + TV + IRV or IC + ERV = 4000 to 4600 ml.

164. Answer (4) Hint: Increased oxygen affinity of haemoglobin

causes left shift. Sol.: Shifting of curve towards right indicates the

dissociation of O2 from Hb. Conditions responsible for shifting oxygen-haemoglobin curve towards right are:

(1) Low pO2 (2) High pCO2 (3) High H+ ion concentration (decrease in pH) (4) High temperature 165. Answer (3) Hint: Heart activity at the beginning of ventricular

diastole. Sol.: Second heart sound (Dub) is produced by

closing of semilunar valves at the beginning of ventricular diastole. It is higher pitched and of shorter duration than first heart sound (Lub).

166. Answer (1) Hint: Repolarisation of the ventricles. Sol.: The contraction of the ventricles starts shortly

after Q and marks beginning of systole. T-wave represents the repolarisation of the ventricles. The end of T-wave marks the end of ventricular systole. P-wave represents depolarisation of the atria which leads to contraction of both the atria.

167. Answer (3) Hint: It opens into a straight tube called collecting

duct. Sol.: Conditional reabsorption of Na+ and water

takes place in DCT. DCT is also capable of reabsorption of 3HCO− and selective secretion of hydrogen and potassium ions and NH3 to maintain the pH and sodium-potassium balance in blood.

168. Answer (1) Hint: Diabetes mellitus. Sol.: Polyuria-amount of urine passed out is more.

Renal calculi are masses of crystallised salts within the kidney.

169. Answer (2) Hint: This joint allows movement only in one plane. Sol.: Saddle joint is present between carpals and

metacarpal of human thumb. Saddle joint is a joint in which the ball like or convex head of one bone fixes into saddle like depression of the other bone. It is a type of ball and socket joint and hence allows movement in many directions.

170. Answer (1) Hint: Hyoid is U-shaped. Sol.: Two occipital condyles are present on

occipital bone of skull. The skull articulates with the superior region of the vertebral column with the help of two occipital condyles (dicondylic skull). True ribs are connected to the sternum with the help of hyaline cartilage.

171. Answer (1) Hint: Grave’s disease occurs due to

hyperthyroidism. Sol.: Addison’s disease – Underproduction of

hormones by the adrenal cortex

Cretinism – Hypothyroidism in children

Acromegaly – Over-secretion of GH in adults

Diabetes mellitus – Insulin deficiency 172. Answer (3) Hint: Avian characters. Sol.: Ornithorhynchus and Camelus are mammals.

Chelone is a reptile which does not have pneumatic bones.

173. Answer (2) Hint: β-cells of Islets of Langerhans secrete this

hormone.



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Sol.: Insulin acts mainly on hepatocytes and adipocytes (cells of adipose tissue), and enhances cellular glucose uptake and utilisation. As a result, there is rapid movement of glucose from blood to hepatocytes and adipocytes resulting in hypoglycemia. Insulin also stimulates glycogenesis in the target cells. Glucagon reduces the cellular glucose uptake and utilisation & stimulates glycogenolysis resulting in hyperglycemia.

174. Answer (4) Hint: It is released by zona fasciculata. Sol.:

175. Answer (2) Hint: Collip’s hormone Sol.: When the calcium level in blood decreases,

parathyroid hormone is released in order to increase the calcium level in blood. Hence, PTH (Parathormone) is a hypercalcemic hormone. Calcitonin (TCT) is hypocalcemic and

hypophosphatemic. TSH stimulates the synthesis and secretion of thyroid hormones like thyroxine from the thyroid gland. Mineralocorticoids regulate the water and electrolyte balance in body.

176. Answer (3) Hint: They have one axon and one dendrite. Sol.: Bipolar neurons are found in retina of the eye,

inner ear and the olfactory area of brain. 177. Answer (3) Hint: A normal inspiration lasts for about 2 seconds

and expiration lasts for about 3 seconds. Sol.: A normal adult human breathes 12-16

times/min. An infant breathes about 44 times/min. 178. Answer (3) Hint: Female Anopheles mosquito. Sol.: Mosquitoes belong to class insecta of phylum

Arthropoda. 179. Answer (2) Hint: Removal of nitrogenous waste. Sol.: Proboscis pore is present at anterior region of

proboscis gland through which excretion occurs. 180. Answer (2) Hint: Shoulder joint. Sol.: Head of humerus articulates with a

depression in the scapula called glenoid cavity. The clavicle articulates with the acromion process.

Test - 8 (Code-D)_(Answers) All India Aakash Test Series for NEET-2021

Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456 1/16

All India Aakash Test Series for Medical - 2021

Test Date : 15/03/2020

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

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

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

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

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

TEST - 8 (Code-D)

All India Aakash Test Series for NEET-2021 Test - 8 (Code-D)_(Hints & Solutions)


2/16

HINTS & SOLUTIONS

[PHYSICS]1. Answer (1) Hint & Sol.: Time period of spring pendulum is

independent of acceleration due to gravity. 2. Answer (4) Hint: PVγ = constant

Sol.: 4

γγ =

i fVPV P

1.5 3/24 4 8= = =f

i

PP

3. Answer (2) Hint & Sol.: As radius of orbit increases the value

of gravitational potential energy GMm

R−

increases. 4. Answer (2) Hint: Fnet = ma Sol.: T – 4g = 4a T = 4(g + a) = 4 (10 + 2) = 48 N 5. Answer (1)

Hint: 212

s ut at= +

Sol.: t = 5 s to 5.8 s → upward motion t = 5.8 s to 6 s → downward motion

2 21 1(0.8) (0.2)2 2

= +d g g

1 0.68{0.64 0.04} 10 3.4 m2 2

= + = × =g

6. Answer (2) Hint & Sol.: Breaking stress remains constant. 7. Answer (1) Hint & Sol.: k = Yr

k = 2 × 1011 × 2 × 10–10 ⇒ k = 40 N/m 8. Answer (2) Hint: 5A Bf f− =

Sol.: 5 3 5100 100

c cC C

f ff f + − − =

5 3 500+ =c cf f

8 500cf =

62.5 Hz=cf

62.562.5 3100

= − ×Bf

= 60.625 Hz 9. Answer (4)

Hint: 00

s

v vf fv v

−= −

Sol.: 0

0

v vff v

+=

0

0

1 vff v

= +

0 0

0

100 100f f vf v

−× = ×

025 100= ×vv

0 4=

vv

10. Answer (3) Hint & Sol.: Light wave is transverse while sound

wave in air is longitudinal. 11. Answer (4)

Hint & Sol.: 212

( 10 31)=

− +y

x x

212

( 10 25 6)=

− + +y

x x

212

( 5) 6y

x=

− +

max126

y =

max 2 cmy =

12. Answer (1)

Hint: eff

2TKµ

= π

Sol.: 1 1 1m m

= +µ

2m

µ =

eff 2KK =

Test - 8 (Code-D)_(Hints & Solutions) All India Aakash Test Series for NEET-2021


3/16

22

2

mT K= π×

2 mTK

= π

13. Answer (1)

Hint: 20

btmA A e

−=

Sol.: 020 2

btm AA e

−=

2 2=btme

ln 22

=btm

2 ln 21

t =

t = 2 ln 2 14. Answer (2) Hint & Sol.: Stress has a unit N/m2 and snow is

better insulator due to presence of air pockets. 15. Answer (4) Hint: dQ mSdT=

Sol.: 3

1

Q SdT= ∫

3

2

1

bT dT= ∫

3313

b T =

263

bQ =

16. Answer (2)

Hint: dH dTKAdt dx

=

Sol.: (100 25) (100 )3 1

KA KA T− −=

75 1003

T= −

T = 100 – 25 T 75 C= ° 17. Answer (2)

Hint: 2 2 2

1 2 ..... nms

v v vvn

+ +=

Sol.: 2 2 2 2 21 2 3 4 5

5msv + + + +=

555

=

= 11 (m/s)2


Sol.: P RTMρ

=

A → B temperature is constant, pressure is increased, therefore density will increase.

B → C pressure is constant, temperature is

increasing, therefore density will decrease 1T

ρ ∝ .

C → D temperature is constant, pressure is decreasing, therefore density will decrease.

D → A pressure is constant, temperature is

decreasing, therefore density will increase 1T

ρ ∝ .

19. Answer (3) Hint: Work done = area under PV graph with

volume axis. Sol.: W = area

0 012

= PV

20. Answer (2) Hint & Sol.: The kinetic energy of gas molecules

will be stored as internal energy and temperature would increase.


Sol.: 1mP RT

MV=

1 300mP RMV

= ×

2 350−= ×

m xP RMV

1 2P P=

300 350− = ×

m M xR RMV MV

300m = 350m – 350x 350x = 50 × 21

50 21350

×=x

= 3 g 22. Answer (1) Hint: 0(1 )= + γ∆VV t

Sol.: 1 1 1(1 )V V t′ = + γ ∆

2 2 2(1 )V V t′ = + γ ∆

2 1 2 1 2 2 1 1V V V V V t V t′ ′− = − + γ ∆ − γ ∆

2 2 1 1γ = γV V



4/16

23. Answer (4) Hint & Sol.: In streamline flow velocity of all fluid

particle crossing a given position is constant and no two stream lines can cross each other.

24. Answer (3) Hint & Sol.:

∆=

h al g

∆ =alhg

25. Answer (3)

Hint: 2GGMEr

=

Sol.:

Enet = 0

26. Answer (2) Hint & Sol.: Inside the shell gravitational field is

zero. 27. Answer (1) Hint & Sol.: A very large attractive coulomb force

will be produced due to huge amounts of opposite charges on the sun and the planets. This force will add up with the gravitational force, as both are central forces. In this situation, all the three Kepler’s law would be valid.

28. Answer (2) Hint: f = µmg and Fnet = ma

F f ma− =

225

af R mRR

× =

52fF f− =

72fF =

722

mg mg= µ

47

µ =

29. Answer (3) Hint: Conservation of angular momentum about

centre of mass. Sol.:

i fL L=

2 22

0 0 4ˆ ˆ ˆ2 2 12 2 2

v l mv l ml l lm k k m m k + = + + ω

2 2

0 3 2ml mlmv l

= + ω

206 5mv l ml= ω

065vl

ω =

t π=

ω

0

56

ltvπ

=

30. Answer (2) Hint: I = mk2 Sol.:

( )m l

I

2

1

2 2

3=



5/16

2

18

3mlI =

2

22 (3 )

2mlI m l= +

mlI ml

22

2 92

= +

2

219

2mlI =

1 2I I I= +

2 28 19

3 2ml ml

= +

2 216 57

6ml ml+

=

273

6ml

=

=IkM

(M = 2m)

7312

=k l

31. Answer (4) Hint & Sol.: For no slipping between sphere and

plank, a1 should be equal to α.

To provide α in anticlockwise direction friction on

sphere should be along positive x direction. 32. Answer (3) Hint: Acceleration will be non-zero if F > fmax. Sol.: fmax = µmg = 0.3 × 5 × g = 15 N F = 3 × 10 = 30 N Fnet = 30 – 15 = 15 N

215 N 3 m/s5

= =a

33. Answer (1)

Hint: = ⋅

P F v Sol.: At highest point in the given case, Tension (T)

= 0 ⇒ power = 0

34. Answer (2) Hint: There will be loss of kinetic energy in an

inelastic collision. Sol.: (1) Work done in the motion of a body over any

closed loop will be zero in the case of conservative force.

(2) In an inelastic collision total momentum of the system will be conserved but there will be loss of kinetic energy.

(3) Total energy of the system may not be conserved in the presence of external force.

(4) In an elastic collision momentum and energy of the system will be conserved but momentum and energy of individual body will not conserved.

35. Answer (3) Hint: max = µf N and max =f ma

Sol.: N = F = 10 N fmax = 5 N

max−=

mg fam

220 5 7.5 m/s2−

= =

36. Answer (4)

Hint & Sol.: 1

1

2−

=

F m ga

m

=

60 32 zero

3

−=

g

37. Answer (3) Hint: At equilibrium ΣF = 0

Sol.:

As string is inextensible T = mgsinθ N = mgcosθ



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38. Answer (3) Hint: cos= θxv u

Sol.: v = 20 cos 53°

= 320 12 m/s5

× =

39. Answer (1) Hint: = −

bc bG cGv v v

Sol.: = +

bG bc cGv v v

ˆ ˆ3 4i i= +

3tan4

θ =

37θ = ° 40. Answer (1) Hint: and = =∫ y xy v dt x u t

Sol.: vy = 2t

t

dy tdt0

2=∫ ∫

y = t2

x = 3t

2

9=

xy

41. Answer (3)

Hint: dva vdx

=

Sol.: 3= −dvv vdx

3= −vdv dx

v

0

9 0

3= −∫ ∫x

vdv dx

0

3/2

9

332

= −

v x

3/22 (9) 33

x− = −

x = 6 m 42. Answer (4) Hint: 1/2 1 2= −

v v v

Sol.: 2 2car/truck 80 60 100 m/s= + =

v

43. Answer (2) Hint & Sol.: The range of the strong nuclear force

is 10–15 m. 44. Answer (1)

Hint: ∆ ∆ ∆= +

A a bA a b

Sol.: a bA Aa b

∆ ∆ ∆ = + ×

0.01 0.01 62 3

= + ×

= 0.05 Area = (6.00 ± 0.05) cm 45. Answer (3) Hint: Principle of homogeneity.

Sol.: 2 2PxU

x b=

+

2(unit of energy)(m )(Unit of )

m=P

(Unit of P) → joule-metre

[CHEMISTRY]

46. Answer (4)

Hint: gnp cK K (RT)∆=

Sol.: For 2 2 3N (g) 3H (g) 2NH (g)+

∆ng = 2 – 4 = –2

∴ Kp ≠ Kc

47. Answer (3)

Hint: Dead burnt plaster is CaSO4.

Sol.: 2 3 2Na CO 10H O⋅ is washing soda

4 21CaSO H O2

⋅ is plaster of paris

NaOH is caustic soda



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48. Answer (2) Hint: Electron rich hydrides have excess of

electrons which are present as lone pair(s). Sol.: CH4 is an electron precise hydride.

49. Answer (3) Hint: Volume changes with temperature. Sol.: Normality has volume term, so it changes

with change in temperature. 50. Answer (4) Hint: Polar molecules have permanent dipole

moment.

Sol.:

51. Answer (4)

Hint: ( )b a1B.O N N2

= −

Sol.: Species Bond O2 2

2O− 1.5

22O − 1

2O+ 2.5

52. Answer (4) Hint: ∆H is an additive property.

Sol.:

53. Answer (2) Hint: Oxidation state of S in H2SO4 is +6. Sol.: In HNO3, 1 + x + 3(–2) = 0 ⇒ x = +5 In HClO4, 1 + x + 4(–2) = 0 ⇒ x = +7 54. Answer (1) Hint: Element X is chlorine Sol.: Chlorine belongs to halogen family i.e. group

number 17th and period number 3.

55. Answer (2) Hint: N2 is not a green house gas. 56. Answer (2) Hint: For a mixture of non-reacting gases, partial

pressure is directly proportional to the number of moles.

Sol.: 2N He

1 1n : n :28 4

=

= 4 : 28 = 1 : 7 57. Answer (2) Hint: In trigonal bipyramidal geometry, the lone

pair is most preferably placed at equatorial position.

Sol.:

58. Answer (3) Hint: Generally, down the group, metallic character

increases. Sol.: When the electron is added to F, due to its

small size, the interelectronic repulsion increases, and hence the electron affinity decreases. On the other hand, in Cl, the electron-electron repulsion is much less due to its larger size, so the electron affinity of Cl is more than F.

59. Answer (2) Hint: Alkene on reductive ozonolysis gives

aldehyde or ketone.



8/16

Sol.:

60. Answer (4) Hint: An electrophile is a species which has the

ability to accept an electron pair. 61. Answer (2) Hint.: For isoelectronic ions, size decreases with

increase in atomic number. Sol.: The correct order of decreasing ionic radii is

N3– > O2– > F– > Na+ 62. Answer (3)

Hint:

1 2 3 4 5 6

2 3

3 2

Hybridisation CH C CH CH CH CHof carbonatom sp sp sp

≡ − − = −↑ ↑ ↑

63. Answer (4) Hint: ∆G = ∆H – T∆S, ∆G < 0

Sol.: 1

1 1

H 30,000 JmolTS 60 JK mol

−

− −

∆= =

∆ = 500 K

∴ The reaction is spontaneous above 500 K 64. Answer (3) Hint: Energy of an electromagnetic radiation is

given by E = hν Sol.: E = hν E = 6.6 × 10–34 × 10 × 1014 = 6.6 × 10–19 J 65. Answer (4) Hint: The gas with higher critical temperature is

most readily liquified. Sol.: The correct order of critical temperature of the

given gases is He < H2 < CO2 < NH3

66. Answer (2) Hint: ∆H = ∆U + ∆ngRT If ∆ng > 0, then, ∆H > ∆U Sol.: 2NH3(g) → N2(g) + 3H2(g) ∆ng = 4 – 2 = 2 ∴ ∆H > ∆U 67. Answer (1) Hint: On moving left to right in the modern periodic

table, the acidic character of oxides generally increases.

Sol.: Cl2O7 is most acidic among the given oxides.

68. Answer (3) Hint: α-H atom of carbocation bonded with sp3

carbon takes part in hyperconjugation. Sol.: Two α–H atoms are present in

69. Answer (2)

Hint: Orbital angular momentum ( ) hl l 12

= +π

Sol.: For p-orbital; l = 1

∴ Angular momentum ( ) hl l 12

= +π

2h2

=π

70. Answer (2) Hint: A group which deactivates the benzene ring

through resonance is a meta directing group. 71. Answer (4) Hint: Higher the standard reduction potential value

of metal, more will be its oxidizing power. 72. Answer (3)

Hint: In acidic medium, 4MnO− converts into Mn2+. Sol.: Balanced chemical equation

5Fe2+ + 4MnO− + 8H+ → Mn2+ + 4H2O + 5Fe3+

For 5 moles of Fe2+, 1 mole of 4MnO− is required

For 2 mole of Fe2+, 25

moles of 4MnO− are

required 73. Answer (4) Hint: n = 4 and l = 1 represent 4p orbital Sol.: Maximum number of electrons that can have

quantum number n = 4, l = 1 is 6. 74. Answer (3) Hint: Ca(OH)2 is used to remove temporary

hardness of water. Sol.: Clark’s method is used to remove temporary

hardness rather than permanent hardness. Ca(HCO3)2 + Ca(OH)2 → 2CaCO3↓ + 2H2O Mg(HCO3)2 + 2Ca(OH)2→2CaCO3↓+ Mg(OH)2↓

+ 2H2O 75. Answer (3) Hint: The hydration enthalpy of alkali metal ions

decreases with increase in ionic sizes. Sol.: The correct order of hydration enthalpy of

alkali metal ions is Li+ > Na+ > K+ > Rb+



9/16

76. Answer (2)

Hint: ( )1 1(HCl) 1 1(NaOH)

H1 2

M V M VM

(V V ) +

−=

+

Sol.: 1 1(HCl) 2 2(NaOH)(H )

1 2

M V M VM

(V V )+

−=

+

200 0.2 200 0.02400

× − ×=

= 9.0 × 10–2 M

pH = –log[H+] = –log[9 × 10–2] = 2 – log 9 = 2 – 0.95 = 1.05 77. Answer (1)

Hint: Average atomic mass 1 1 2 2

1 2

M X M XX X

+=

+

Where M1 and M2 are the atomic masses and X1 and X2 are the % composition

Sol.: Average atomic mass of element A

(12 40) (15 60)100

× + ×=

480 900 13.8 u100

+= =

78. Answer (1)

Hint: 182 2i f

1 1E 2.18 10 Jn n

− ∆ = × −

/atom

79. Answer (4) Hint: Single bond has one σ bond Double bond has one σ and one π bond Triple bond has one σ and two π bonds.

Sol.:

Total number of σ bonds = 18 Total number of π bonds = 4 80. Answer (2) Hint: The conjugate bases should have one proton

less and conjugate acid should have one extra proton.

Sol.:

81. Answer (4) Hint: Pd/BaSO4 is Lindlar’s catalyst. Sol.: CH3 – C ≡ C – CH3 + H2

CH3 – C ≡ C – CH3

82. Answer (4) Hint: Down the group, the tendency to show

catenation decreases. Sol.: The correct order of catenation tendency of

elements of group-14 is C >> Si > Ge ≈ Sn Pb does not show catenation. 83. Answer (3)

Hint: (Carboxylic group) is given highest priority.

Sol.:

3-Ethyl-4-methylocta-3, 5, 7-trien -1-oic acid 84. Answer (4) Hint: The structure of diborane is

Sol.: In Diborane, the four terminal B–H bonds are

two-centre-two electron bonds and the two bridge bonds are three-centre-two electron bonds.

85. Answer (2) Hint: Equilibrium constant for the reverse reaction

is the inverse of the equilibrium constant for the reaction in the forward direction.

Sol.: For, H2(g) + I2(g) 2HI(g);



10/16

Equilibrium constant = K ∴ For 2HI(g) I2(g) + H2(g);

Equilibrium constant 1K

=

For 2 21 1HI(g) I (g) H (g);2 2

+

Equilibrium constant1/2

1/2

1 1K K

= =

86. Answer (1)

Hint: 2Ca + O2 → 2CaO Sol.: 22Ca O 2CaO

80 g 32 g+ →

4 g of Ca requires 32 4 g 1.6 g80

× = of oxygen

∴ O2 is present in excess Amount of O2 present in excess = 2.4 g – 1.6 g = 0.8 g 87. Answer (2) Hint: At high temperature, the intermolecular force

of attraction between the gas molecules are negligible and at low pressure the volume of the molecules of the gas in comparison to the volume occupied by the gas is negligibly small.

88. Answer (1) Hint: Extensive property depends on the quantity

or size of matter present in the system. Sol.: Enthalpy, volume and internal energy are

extensive properties. Temperature is an intensive property. 89. Answer (4) Hint: Isoelectronic means same number of

electrons and isostructural means same shape. Sol.: Number of electrons in 2

3CO − = 32e–

3NO 32e− −=

Shape: trigonal planar

90. Answer (2)

Hint: number of moles of soluteMolarityvolume of solution in litre

=

Sol.: number of moles of solute 22

2312.04 106.02 10

×=

×

12 10−= × = 0.2 mole

n 0.2M 1000 0.4 MV 500

= = × =

[BIOLOGY]91. Answer (2) Hint: In C4 plants, Calvin cycle does not occur in

mesophyll cells. Sol.: In C4 plants, Calvin cycle occurs in bundle

sheath cell where RuBisCO is present. ATP for Calvin cycle comes from light reactions of

photosynthesis that occur in chloroplasts. 92. Answer (4) Hint.: Annual plants require one season to

complete their life-cycle. Sol.: Sugarbeet, cabbages, carrots are some of

the common biennials which show vernalisation. 93. Answer (3) Sol.: Bulk transport of minerals, water and food

occurs in plant is called translocation. 94. Answer (3) Hint.: Denitrification is the process through which

nitrates are converted into molecular nitrogen.

Sol.: 3NO N N− → ≡ [Denitrification]

3 2 3NH NO NO− −→ → [Nitrification]

95. Answer (2)

Hint.: Porins are found in plasma membrane.

Sol.: Porins are the proteins that form huge pores in outer membrane of plastids, mitochondria and some bacteria.

96. Answer (4)

Hint.: Splitting of centromeres occurs just after metaphase.

Sol.: In anaphase, centromeres split and chromatids are separated. This is disjunction of chromatids.

97. Answer (4)

Sol.: Cell wall provides barrier to undesirable macromolecules.



11/16

98. Answer (4) Hint.: In axile placentation, placenta is present in

the axial position in multilocular ovary. Sol.: In free central placentation, ovary is

unilocular, ovules are borne on central axis and septa are absent.

99. Answer (4) Hint: Zygotic meiosis occurs in plants showing

haplontic life-cycle pattern. Sol.: Haplontic life-cycle pattern is shown by

Volvox and Spirogyra. 100. Answer (2) Hint: Viruses are smaller than bacteria. Sol.: Virus take over biosynthetic machinery of

host cell and they are non-cellular obligate parasites.

101. Answer (1) Sol.: In Alstonia whorled phyllotaxy is found. 102. Answer (4) Hint: In the flowers of Liliaceae, gynoecium is

tricarpellary. Sol.: Allium cepa (onion) belongs to the family

Liliaceae. 103. Answer (3) Sol.: Mustard - Racemose inflorescence 104. Answer (4) Sol.: Suckers - Chrysanthemum 105. Answer (2) Hint: Pulvinus is the swollen leaf base. Sol.: In some leguminous plants, the leaf base

may become swollen and is called pulvinus. 106. Answer (2) Sol.: In beans, seeds are non-endospermic. 107. Answer (4) Hint: Zygospores are resting diploid spores. Sol.: Ascospore and basidiospores are haploid. 108. Answer (1) Hint: Both archaebacteria and eubacteria are

prokaryotes. Sol.: Archaebacteria differ from other bacteria in

having different cell wall structure which is responsible for their survival in extreme conditions.

109. Answer (4) Hint: Silica shells of dead diatoms are nearly

indestructible. Sol.: Chrysophytes include diatoms and desmids

110. Answer (3) Hint: In Monera, cyanobacteria show oxygenic

photosynthesis. Sol.: In five kingdom classification, chlorophyllous

photosynthetic organisms evolving oxygen belong to three kingdoms (Monera, Protista and Plantae).

111. Answer (3) Hint: Organic acids like oxalic acid and malic acid

have RQ value more than 1. Sol.: Malic acid – 1.33 Glucose – 1.0 Oxalic acid – 4.0 Tripalmitin – 0.7 112. Answer (2) Hint: Two redox equivalents are removed per

dehydrogenation reaction. Sol.: Total 5 dehydrogenation reactions occur in

complete oxidation of PGA forming 4 NADH2 and 1 FADH2 molecules. Thus 5 × 2 = 10 redox equivalents are removed.

113. Answer (2) Hint.: α-ketoglutaric acid and succinyl CoA

respectively are 5 C and 4 C compounds. Sol.: 6 C isocitric acid is converted into 6 C

oxalosuccinic acid in the presence of isocitrate dehydrogenase.

114. Answer (4) Hint: Dormancy is due to internal factors. Sol.: Many seeds having endosperm also

germinate. 115. Answer (3) Hint.: Coleoptile bends towards unilateral light. Sol.: Phototropism is the directional movement of

growth in response to light. 116. Answer (1) Hint: Heterophylly is an example of plasticity. Sol.: Transpiration does not occur from the

submerged part of the plant. 117. Answer (3) Hint: Along with FADH2 and GTP two molecules of

NADH + H+ are also formed after α-ketoglutaric acid in Krebs cycle.

Sol.:

2

2 NADH H 2 3 ATP1 FADH 2 ATP1 GTP 1 ATPTotal 9 ATP

++ = ×===



12/16

118. Answer (2) Hint: In non-cyclic photophosphorylation NADPH is

formed. Sol.: NADP+ is reduced to NADPH + H+ during

non-cyclic photophosphorylation. Due to increase in concentration of H+ in thylakoid

lumen, pH decreases. 119. Answer (3) Hint: Critical enzymes involved in carbon fixation

during C3 and C4 cycles are RuBisCO and PEPcase respectively.

Sol.: Mg2+ is an activator of both RuBisCO and PEPcase.

120. Answer (4) Hint: The Mo-Fe protein in root-nodules of

leguminous plants is nitrogenase. Sol.: Nitrogenase catalyses the conversion of

atmospheric nitrogen to ammonia. 121. Answer (4) Hint: Phloem sap has water and other organic

molecules. Sol.: Phloem transports water, amino acids,

hormones, sucrose and other sugars. 122. Answer (4) Hint: Water potential (ψw) = solute potential (ψs) +

pressure potential (ψp) Sol.: Solute potential of a solution is always a

negative value whereas pressure potential is a positive value. If their magnitude is same, then the sum will be zero.

123. Answer (1) Hint: Gametophytes of homosporous species in

pteridophytes are monoecious, i.e., male and female sex organs are present on the same thallus.

Sol.: Dryopteris is homosporous whereas Selaginella, Marsilea and Salvinia are heterosporous.

124. Answer (1) Hint: The life-cycle shown in the figure is

Haplo-diplontic. Sol.: Along with bryophytes and pteridophytes,

some algae like Ectocarpus, Polysiphonia, kelps are haplo-diplontic. All these plants do not form seeds.

125. Answer (2) Hint: Mesophyll is found in leaves. Sol.: Cambium is absent in monocots and leaves.

Therefore their vascular bundles are said to be closed.

126. Answer (2) Hint: The companion cells help in maintaining the

pressure gradient in the sieve tubes. Sol.: Companion cells are closely associated with

sieve tube elements by pit fields. 127. Answer (3) Hint: Seeds are formed after fertilisation. Sol.: Parthenocarpic fruits are formed without

fertilisation, therefore they lack seeds. 128. Answer (1) Hint: In some plants growing in swampy areas,

many roots come out of the ground to get oxygen. Sol.: Stilt roots, prop roots and axillary buds arise

from stems. Pneumatophores are respiratory roots come out of the ground vertically upwards to get oxygen for respiration.

129. Answer (3) Hint: Plasmogamy and karyogamy are the

processes in sexual reproduction. Sol.: Deuteromycetes reproduce only by asexual

spores known as conidia. The mycelium is septate and branched in these fungi.

130. Answer (4) Hint: Diatoms are the chief producers in the ocean. Sol.: All protozoans are heterotrophic. Euglenoids

do not have cell walls but they are photosynthetic. Protozoans are believed to be primitive relatives of

animals. 131. Answer (2) Hint: Leopard belongs to the family Felidae. Sol.: Canis familiaris, Canis lupus and Canis

aureus belong to the family Canidae whereas Panthera pardus belongs to the family Felidae.

132. Answer (3) Hint: During anaphase I, homologous chromosomes

separate. Sol.: Crossing over occurs during pachytene stage. Splitting of centromere occurs during anaphase II. Terminalisation of chiasmata occurs during

diakinesis. 133. Answer (4) Hint: Kinetochores present around the centromere

forms the site of attachment of microtubules of the spindle fibres.

Sol.: The microtubules of spindle fibres attach to the kinetochores in metaphase. This process will occur if kinetochores are present on centromeres. Thus in the absence of kinetochores, metaphase and so, the other phases after metaphase will not occur.



13/16

134. Answer (1) Hint: Amongst aleuroplast, tonoplast, amyloplast

and elaioplast, tonoplast is odd one out.

Sol.: The vacuoles are bounded by a single, semi permeable membrane called tonoplast. It facilitates the transport of ions and other materials against the concentration gradient into the vacuole.

135. Answer (2) Hint: Chloroplast and mitochondria are semi-

autonomous cell organelles.

Sol.: Several ribosomes attached to a single mRNA and form a chain called polyribosome or polysome. Therefore polysomes have mRNA and rRNA only.

136. Answer (2) Hint: Shoulder joint. Sol.: Head of humerus articulates with a

depression in the scapula called glenoid cavity. The clavicle articulates with the acromion process.

137. Answer (2) Hint: Removal of nitrogenous waste. Sol.: Proboscis pore is present at anterior region of

proboscis gland through which excretion occurs. 138. Answer (3) Hint: Female Anopheles mosquito. Sol.: Mosquitoes belong to class insecta of phylum

Arthropoda. 139. Answer (3) Hint: A normal inspiration lasts for about 2 seconds

and expiration lasts for about 3 seconds. Sol.: A normal adult human breathes 12-16

times/min. An infant breathes about 44 times/min. 140. Answer (3) Hint: They have one axon and one dendrite. Sol.: Bipolar neurons are found in retina of the eye,

inner ear and the olfactory area of brain. 141. Answer (2) Hint: Collip’s hormone Sol.: When the calcium level in blood decreases,

parathyroid hormone is released in order to increase the calcium level in blood. Hence, PTH (Parathormone) is a hypercalcemic hormone. Calcitonin (TCT) is hypocalcemic and hypophosphatemic. TSH stimulates the synthesis and secretion of thyroid hormones like thyroxine from the thyroid gland. Mineralocorticoids regulate the water and electrolyte balance in body.

142. Answer (4) Hint: It is released by zona fasciculata. Sol.:

143. Answer (2)

Hint: β-cells of Islets of Langerhans secrete this hormone.

Sol.: Insulin acts mainly on hepatocytes and adipocytes (cells of adipose tissue), and enhances cellular glucose uptake and utilisation. As a result, there is rapid movement of glucose from blood to hepatocytes and adipocytes resulting in hypoglycemia. Insulin also stimulates glycogenesis in the target cells. Glucagon reduces the cellular glucose uptake and utilisation & stimulates glycogenolysis resulting in hyperglycemia.

144. Answer (3) Hint: Avian characters. Sol.: Ornithorhynchus and Camelus are mammals.

Chelone is a reptile which does not have pneumatic bones.

145. Answer (1) Hint: Grave’s disease occurs due to

hyperthyroidism. Sol.: Addison’s disease – Underproduction of

hormones by the adrenal cortex

Cretinism – Hypothyroidism in children

Acromegaly – Over-secretion of GH in adults

Diabetes mellitus – Insulin deficiency 146. Answer (1) Hint: Hyoid is U-shaped. Sol.: Two occipital condyles are present on

occipital bone of skull. The skull articulates with the superior region of the vertebral column with the help of two occipital condyles (dicondylic skull). True ribs are connected to the sternum with the help of hyaline cartilage.



14/16

147. Answer (2) Hint: This joint allows movement only in one plane. Sol.: Saddle joint is present between carpals and

metacarpal of human thumb. Saddle joint is a joint in which the ball like or convex head of one bone fixes into saddle like depression of the other bone. It is a type of ball and socket joint and hence allows movement in many directions.

148. Answer (1) Hint: Diabetes mellitus. Sol.: Polyuria-amount of urine passed out is more.

Renal calculi are masses of crystallised salts within the kidney.

149. Answer (3) Hint: It opens into a straight tube called collecting

duct. Sol.: Conditional reabsorption of Na+ and water

takes place in DCT. DCT is also capable of reabsorption of 3HCO− and selective secretion of hydrogen and potassium ions and NH3 to maintain the pH and sodium-potassium balance in blood.

150. Answer (1) Hint: Repolarisation of the ventricles. Sol.: The contraction of the ventricles starts shortly

after Q and marks beginning of systole. T-wave represents the repolarisation of the ventricles. The end of T-wave marks the end of ventricular systole. P-wave represents depolarisation of the atria which leads to contraction of both the atria.

151. Answer (3) Hint: Heart activity at the beginning of ventricular

diastole. Sol.: Second heart sound (Dub) is produced by

closing of semilunar valves at the beginning of ventricular diastole. It is higher pitched and of shorter duration than first heart sound (Lub).

152. Answer (4) Hint: Increased oxygen affinity of haemoglobin

causes left shift. Sol.: Shifting of curve towards right indicates the

dissociation of O2 from Hb. Conditions responsible for shifting oxygen-haemoglobin curve towards right are:

(1) Low pO2 (2) High pCO2 (3) High H+ ion concentration (decrease in pH) (4) High temperature

153. Answer (4) Hint: ERV + TV + IRV Sol.: Vital capacity (VC) is defined as the maximum

volume of air a person can breathe in, after a forceful expiration or the maximum volume of air a person can breathe out after a forceful inspiration.

Depending upon the age, sex and height of individual, its value varies from 4000 ml to 4600 ml.

Vital capacity (VC) = ERV + TV + IRV or IC + ERV = 4000 to 4600 ml.

154. Answer (1) Hint: Teeth are arranged in order I, C, PM, M.

Sol.: 21232123

represents arrangement of teeth in

each half of the upper and lower jaw in the order I, C, PM, M. Total number of teeth in adult human are I – 8, C – 4, PM – 8 and M – 12.

155. Answer (2) Hint: HCl converts pepsinogen to pepsin. Sol.: HCl provides acidic pH (pH 1.8) optimal for

pepsin.

Proteins Pepsin→ Proteoses + Peptones 156. Answer (2) Hint: Component of lipids. Sol.: Substituted methane with amino and

carboxylic group. Amino acids have four substituent groups hydrogen, carboxyl group (–COOH), amino group (–NH2) and a variable group designated as R-group. Glycerol is trihydroxy propane and is a constituent of lipids.

157. Answer (3) Hint: A carrier protein.

Sol.: Protein Functions

Collagen Intercellular ground substance

Trypsin Enzyme

Insulin Hormone

Antibody Fights infectious agents

GLUT-4 Enables glucose transport into cells

158. Answer (1) Hint: Smooth muscle fibres are unstriated. Sol.: Each muscle fibre of cardiac muscle has a

single centrally located nucleus. Nuclei are peripheral in skeletal muscle fibres. They possess faint striations.



15/16

159. Answer (4) Hint: Compound epithelium. Sol.: Stratified keratinised squamous epithelium

covers the dry surface of skin. Heavy deposits of the insoluble protein keratin in the dead superficial cells make the epithelium impervious to water and highly resistant to mechanical abrasions. In contrast, non-keratinised stratified epithelium cannot prevent water loss and afford only moderate protection against abrasions.

160. Answer (3) Hint: Fluid connective tissue. Sol.: Examples of loose connective tissues are

areolar tissue and adipose tissue. Specialized connective tissue include skeletal connective tissue like cartilage, bone and fluid connective tissue like blood and lymph. Muscle tissue comprises muscles of body which enable movements of the body.

161. Answer (4) Hint: Haemolymph does not contain respiratory

pigment. Sol.: Respiratory system of cockroach consists of

spiracles and a network of trachea and tracheoles. 162. Answer (1) Hint: Feature of aves and mammals. Sol.: The heart of lizard is 3 chambered. Among

reptiles, crocodiles possess 4 chambered heart. 163. Answer (4) Hint: Echinoderms are free-living animals. Sol.: No parasitic form is found in echinoderms. All

echinoderms are exclusively marine and usually live at the bottom of sea hence they are called bottom dwellers.

164. Answer (2) Hint: Select a mollusc. Sol.: Aplysia belongs to phylum mollusca.

Segmentation is absent in body of molluscs. Nereis, Pheretima, Hirudinaria are annelids. The

body of annelids is metamerically segmented i.e. the external segmentations of the body correspond to the internal segmentations.

165. Answer (3) Hint: It maintains circadian rhythm. Sol.: Melatonin is derived from amino acid

tryptophan. 166. Answer (2) Hint: Aldosterone is mineralocorticoid. Sol.: Principle actions of aldosterone is in

controlling electrolyte and water metabolism. Cortisol raises blood glucose level. Estrogen develops secondary sexual characteristics in

females. Adrenaline constricts skin and visceral smooth muscle capillaries & dilates arterioles of heart and skeletal muscle.

167. Answer (3) Hint: These ions act as a second messengers. Sol.: Ca++ ions diffuse into the axon terminal from

the surrounding fluid on opening of voltage-gated calcium channels.

168. Answer (3) Hint: Adrenaline increases heart rate. Sol.: The sympathetic neural system accelerates

the heart beat whereas the parasympathetic neural system slows down the heart beat. Parasympathetic neural system stimulates the secretion of saliva.

169. Answer (1) Hint: It is not usually inheritable. Sol.: In Myasthenia gravis, destruction of

acetylcholine receptors occurs due to autoimmune response. Muscular dystrophy is a genetic disorder leading to degeneration of skeletal muscles. Tetany occurs due to hypocalcemia.

170. Answer (1) Hint: Almost all the glomerular filtrate is

reabsorbed by the renal tubules. Sol.: The tubular epithelial cells in different

segments of nephron carry out reabsorption either by active or passive mechanisms. For example, substances like glucose, amino acids, Na+, etc. in the filtrate are reabsorbed actively, whereas nitrogenous wastes are reabsorbed by passive transport.

171. Answer (2) Hint: Permeable to salts. Sol.: The descending limb of loop of Henle is

permeable to water but almost impermeable to electrolytes.

172. Answer (3) Hint: It makes the lumen of vessel narrower. Sol.: Heart failure is the state of heart when it does

not pump blood effectively enough to meet the needs of the body. Heart attack is when the heart muscles are suddenly damaged by an inadequate blood supply. Cardiac arrest means complete stoppage of the heart beat.

Coronary artery diseases, often referred to as atherosclerosis, affects the vessels that supply blood to the heart muscle. It is caused due to the deposition of calcium, fat, cholesterol and fibrous tissues in the arteries supplying the heart musculature. These depositions make the lumen of arteries narrower.



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173. Answer (1) Hint: This vein starts and ends between capillary

beds. Sol.: The hepatic portal system involves portal vein

which carries blood from intestine to the liver before it is delivered to the heart.

174. Answer (2) Hint: Part of brainstem that forms ‘bridge’. Sol.: Pneumotaxic centre is present in the pons

region of hind brain. It moderates the function of respiratory rhythm centre. The neural signal from this centre can reduce the duration of inspiration and thereby alter the respiratory rate.

175. Answer (3) Hint: Volume of thoracic cavity increases during

inspiration. Sol.: Statement (a), (c) & (d) are correct.

Contraction of external intercostal muscles lifts ribs and sternum up & outwards. Inspiration in mammals is an active process. It occurs due to contraction of external intercostal muscles and diaphragm.

176. Answer (3) Hint: Fatty acids are absorbed here. Sol.: Small intestine is the principal organ for

absorption of nutrients. Digestion is completed here and the final products of digestion such as glucose, fructose, fatty acids, glycerol and amino acids are absorbed through the mucosa into the blood stream and lymph.

177. Answer (4) Hint: It is the innermost layer lining the lumen of

alimentary canal. Sol.: Serosa is the outermost layer of alimentary

canal, muscularis layer is formed by smooth muscles usually arranged into an inner circular and an outer longitudinal layer. The submucosal layer is formed of loose connective tissue containing nerves, blood and lymph vessels.

178. Answer (4) Hint: Heat is released in an exothermic reaction. Sol.: In exothermic reactions, the energy content of

the product is lower than that of the substrate as heat is released.

179. Answer (4) Hint: Select the fruit sugar. Sol.: Monosaccharides are modified variously to

form a number of different substances for eg., Deoxy sugar - Deoxyribose Amino sugar - Glucosamine and

galactosamine Fructose is also known as levulose (fruit sugar). It

is the sweetest among naturally occuring sugars. 180. Answer (4) Hint: Sarcomere is present between two

Z-lines. Sol.: Z-line or Krause’s membrane is a dark

membrane which bisects I-band or isotropic band.

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