Test - 4 (Code-A) (Answers) All India Aakash Test …...Test - 4 (Code-A) (Hints and Solutions) All India Aakash Test Series for NEET-2019 3/18 or, 3v b – 3v R = vb + vR or, v b

Test - 4 (Code-A) (Answers) All India Aakash Test Series for NEET-2019

1/18

1. (2)

2. (2)

3. (4)

4. (1)

5. (2)

6. (3)

7. (4)

8. (3)

9. (2)

10. (4)

11. (2)

12. (4)

13. (3)

14. (4)

15. (4)

16. (2)

17. (4)

18. (1)

19. (2)

20. (3)

21. (4)

22. (3)

23. (3)

24. (2)

25. (3)

26. (3)

27. (2)

28. (4)

29. (2)

30. (2)

31. (4)

32. (2)

33. (1)

34. (2)

35. (4)

36. (3)

Test Date : 23/12/2018

ANSWERS

TEST - 4 (Code-A)

All India Aakash Test Series for NEET - 2019

37. (4)

38. (2)

39. (3)

40. (3)

41. (3)

42. (2)

43. (2)

44. (2)

45. (3)

46. (4)

47. (3)

48. (3)

49. (4)

50. (1)

51. (2)

52. (1)

53. (1)

54. (4)

55. (1)

56. (1)

57. (4)

58. (2)

59. (2)

60. (3)

61. (2)

62. (1)

63. (2)

64. (1)

65. (2)

66. (4)

67. (3)

68. (3)

69. (1)

70. (4)

71. (2)

72. (3)

73. (3)

74. (4)

75. (3)

76. (2)

77. (1)

78. (4)

79. (1)

80. (2)

81. (1)

82. (2)

83. (1)

84. (1)

85. (4)

86. (3)

87. (4)

88. (2)

89. (2)

90. (1)

91. (3)

92. (4)

93. (2)

94. (3)

95. (2)

96. (4)

97. (1)

98. (3)

99. (2)

100. (4)

101. (3)

102. (1)

103. (3)

104. (4)

105. (4)

106. (1)

107. (2)

108. (4)

109. (4)

110. (3)

111. (2)

112. (1)

113. (3)

114. (4)

115. (3)

116. (2)

117. (4)

118. (3)

119. (2)

120. (2)

121. (1)

122. (4)

123. (1)

124. (3)

125. (2)

126. (1)

127. (2)

128. (2)

129. (3)

130. (1)

131. (3)

132. (2)

133. (1)

134. (2)

135. (4)

136. (2)

137. (3)

138. (2)

139. (1)

140. (3)

141. (4)

142. (2)

143. (3)

144. (1)

145. (3)

146. (1)

147. (1)

148. (1)

149. (2)

150. (1)

151. (2)

152. (2)

153. (3)

154. (2)

155. (1)

156. (4)

157. (2)

158. (3)

159. (2)

160. (2)

161. (1)

162. (1)

163. (3)

164. (3)

165. (3)

166. (1)

167. (3)

168. (1)

169. (2)

170. (3)

171. (1)

172. (3)

173. (3)

174. (2)

175. (1)

176. (4)

177. (4)

178. (2)

179. (2)

180. (2)

All India Aakash Test Series for NEET-2019 Test - 4 (Code-A) (Hints and Solutions)

2/18

PHYSICS1. Answer (2)

Hint :2 21 2

midpoint 2v v

v+

=

Sol :2 2

13 m/s2

P QR

v vv

+= =

P R Q

7 m/sPv = Rv 17 m/sQv =

13 172 2

+ += =R Qav

v vv

302

=

= 15 m/s2. Answer (2)

Hint : Area under velocity-time graph givesdisplacement.Sol : Average value of sine curve for positive halfcycle.

02 v=π

2 10×=π

= 6.37 m/s3. Answer (4)

Hint : If speed is varying, velocity must be variable.4. Answer (1)

Hint : If graph between physical quantities is astraight line then those quantities are directlyproportional.Sol : From graph v2 ∝ x ...(i)Now we know v2 – u2 = 2ax

0u =v2 = 2ax ...(ii)From (i) and (ii)a = constant∴ a is uniform

5. Answer (2)Hint : Principle of homogeneity.

Sol : 212

P v h g C+ ρ + ρ =

By principle of homogeneity

Dimensional formula of C should be of pressure[C] = [P] = [ML–1T–2 ]

Modulus of rigidity = Shear stressShear strain

FAx

x

=Δ

= [ML–1T–2]∴ [Modulus of rigidity] = [P] = [C]

6. Answer (3)

Hint : For expression z = m p

q

a b

c

Fractional error z m a p b q c

z a b cΔ Δ Δ Δ= + +

Sol : 2l

Tg

= π

2 24l

Tg

= π

22

4l

gT

= π

∴ 2g l Tg l T

Δ Δ Δ= +

2g

p qg

Δ = +

7. Answer (4)

Hint : avgDisplacement

Total timev =

Sol : Displacement = 9 cm.Time = 12.5 hours = 12.5 × 3600 s

avg9

12.5 3600v =

× = 2 × 10–4 cm/s

8. Answer (3)

Hint : still2 u d

u d

t tt

t t=

+

Sol : d = (vb – vR)tu ...(i)d = (vb + vR)td ...(ii)From (i) and (ii)tu(vb – vR) = td(vb + vR)120 (vb – vR) = 40 (vb + vR)

HINTS & SOLUTIONS

Test - 4 (Code-A) (Hints and Solutions) All India Aakash Test Series for NEET-2019

3/18

or, 3vb – 3vR = vb + vR

or, vb = 2 vR

Now, 332

R d

b R

v tdt

v v= =

3

40 60 minutes2

= × =

9. Answer (2)Hint : Time of ascent < Time of descentSol : Time of ascent

v2 – u2 = 2as0 – u2 = 2 (g + a) H

( )2

2( )2

uH u g a H

g a= ⇒ = +

+v = u + at

0 2( ) ( )g a H g a t= + − +

2( )( )g a H

tg a

+=+

up2

( )H

tg a

=+

Time of descent

210

2H ut at u= + =

21( )

2H g a t= −

down2

( )H

tg a

=−

∴

2

2d

a

Ht g a

t Hg a

−=

+

d

a

t g at g a

+=−

10. Answer (4)

Hint : Rm RG mGv v v= −��

Sol : Let us assume that if rain is falling at 30° itwould have horizontal component as vx and verticalcomponent vy

Now man starts running with 10 m/svRm → vertical

So,

Also 10

tan30 tan30 10 3xy

y y

vv

v v= ° ⇒ = ° ⇒ =

∴ When man starts running he will see onlyvertical component.

10 3 m/sRmv =��

11. Answer (2)

Sol :

vmax

T

α β

v

t

avgDisplacement

Timev =

Area under - graphtime

v t=

max12

2max

avg

v T vv

T

× ×= =

∴ max max

maxavg

21

2

v vvv

= = = 2 : 1

12. Answer (4)

Hint : net T Ca a a= +��

Sol : In case of increasing speed aT is in thedirection of ‘v’.

In case of decreasing speed aT is opposite directionspeed

aT

aCanet

v v

anet

=Decreasingspeed


4/18

13. Answer (3)

Hint :2 2sin 2 sin

,2

u uH T

g gθ θ= =

Sol : As usinθ = uy

2( ) 2,

2y yu u

H Tg g

= =

2

2

2 4

y

y

y

u

g uHuT

g

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠= =⎛ ⎞⎜ ⎟⎝ ⎠

yH

uT

∝

14. Answer (4)

Hint : If both magnitude as well as direction ofacceleration remains constant then it is uniformlyaccelerated motion.

Sol : In projectile motion direction of acceleration isconstant.

∴ It is uniformly accelerated motion

In uniform circular motion the direction ofcentripetal acceleration changes, although magnitudeis constant.

∴ It is non-uniformly accelerated motion.

15. Answer (4)

Hint :2 2

,y x yv v uR T

g g= =

Sol :v1

v2

1 2 12 2,

v v vR T

g g= =

16. Answer (2)

Hint : Principle of homogeneity.

Sol : If dimension of A and B are different.

→ In logAB

⎛ ⎞⎜ ⎟⎝ ⎠

and ABe

−,

AB

⎛ ⎞⎜ ⎟⎝ ⎠

should be dimensionless

which is not possible according to given condition

→ In A + B principle of homogeneity violates so

correct answer AB

⎛ ⎞⎜ ⎟⎝ ⎠

17. Answer (4)Hint: n1u1 = n2u2

Sol : F = [MLT–2] E = [ML2T–2]F ′ = [2M2LT–2] E ′ = [2M(2L)2T–2]F ′ = 4[MLT–2] = [8ML2T–2]

F ′ = 4F 8E E′ =

P = [ML2T–3]FA

σ =

P ′ = [2M(2L)2T–3]2

2

[MLT ]

[L ]

−=

= [8ML2T–3]

P ′ = 8P 2 2

2 2

2[M2LT ] [MLT ]

[2L] [L ]

− −′σ = =

′σ = σ18. Answer (1)

Hint : Dimensions are powers raised to fundamentalphysical quantity.

Sol : Energy density = EnergyVolume

2 2

3

[ML T ]

[L ]

−=

= [ML–1T–2]19. Answer (2)

Hint : Concept of significant figure used in additionSol : 5.0 m + 6.0 m = 11.0 m

20. Answer (3)

Hint : If v�

is constant, then speed is constant.

Sol : x = 4 sinωt y = 4 cosωt

⇒ sin4x

t= ω ⇒ cos4y

t= ω

sin2ωt + cos2ωt = 2 2

4 4x y+

2 2

14

x y+=

2 2 4x y+ =

Particle is moving in circular path.

4 cos , 4 sinx ydx

v t v tdt

= = ω ω = − ω ω

2 2x yv v v= +

| v | =4 ω


5/18

21. Answer (4)

Hint : 2 2net x yv v v= +

Sol : 2 2v u gh= + , irrespective of θ

22. Answer (3)Hint : LC = 1 MSD – 1 VSDSol : N(V.S.D.) = (N –1) M.S.D.

11( . . ) . . .

NV S D M S D

N−⎛ ⎞= ⎜ ⎟

⎝ ⎠

Least count:→ 1M.S.D. – 1V.S.D.

11 . . . . .

NM S D M S D

N−⎛ ⎞= − ⎜ ⎟

⎝ ⎠

1. .

N NM S D

N− +⎛ ⎞= ⎜ ⎟

⎝ ⎠

1 . . .M S DN

=

∴ Least count 1 . . . 1M S D mm

N N= =

L.C. = 1

cm10N

23. Answer (3)Hint : Derived quantities may have dimensions andsolution in terms of base quantities.

24. Answer (2)Hint : Z = a + b, ΔZ = Δa + Δb

Sol : Z = P + QΔZ = ΔP + ΔQ

% error 100Z

ZΔ= ×

100P QP Q

Δ + Δ= ×+

100P Q

P Q P QΔ Δ⎛ ⎞= + ×⎜ ⎟+ +⎝ ⎠

25. Answer (3)

Hint : Vector addition

Sol :

A−

B R

R′

βθ

A

26. Answer (3)

Hint : avgf i

f i

v va

t t

−=

−

� �

��

Sol :v′u

t = 0 t t t = + 1 2t2t1

�

a1

�

v�

a2

�

v = u + a1t1 v ′ = v + a2t2

avg1 2

v ua

t t

′ −=+

2 2

1 2

v a t ut t

+ −=+

1 1 2 2

1 2

u a t a t ut t

+ + −=+

1 1 2 2avg

1 2

a t a ta

t t+=+

27. Answer (2)

Hint : avgTotal distance

Total timev =

Sol :

From B → C 4.5 7.8 12

6 m/s2 2BCv+= = =

From A → C, 2 AB BC

ACAB BC

v vv

v v=

+

2 AB BCAC

AB BC

v vv

v v=

+

2 3 6 2 3 6

3 6 9× × × ×= =

+

4 m/sACv =

28. Answer (4)

Hint : 4

tanHR

θ =


6/18

Sol :4

tanHR

θ =

or, tanθ = 4 [∵ H = R]

4 1sin ; cos

17 17θ = θ =

2 22 sin cos 2 (20) 4 1 320m

10 17 17v

Rgθ θ × × ×= = =

×29. Answer (2)

Hint : y = ax – bx2, where y = 0, x = 0 or x = RSol : y = ax – bx2

where y = 0, x = 0 or x = Rax – bx2 = 0x(a – bx) = 0

a ax R

b b= =

Hmax at 2a

xb

=

Hmax = 2

2 2a a

a bb b

⎛ ⎞ ⎛ ⎞−⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

2

4ab

=

Also tan aθ =30. Answer (2)


Total timev =

Sol :

cos2

2r r

θΔ =

= 2 rcos45° = 2 2r =

22 14

4 2 2 2

rr

t sv

ππΔ = = =

× π

av21

2 2

v =

⇒ vav = 4 m/s

31. Answer (4)

Hint :2

2

dx d xv a

dt dt= =

Sol : If n > 22

2

d xa

dt=

= n(n – 1)tn – 2

Would depend on x and would be increasingwith t

If n < 0, a = ktn – 2( 1)

ve

k n n= −⎧ ⎫⎨ ⎬= +⎩ ⎭

∴ a would decrease

32. Answer (2)

Hint : t = drel/vrel

Sol :24

38

t s= =

33. Answer (1)

Hint : speed, velocityv v= =� �

Sol : | | 0d

vdt

=�

but 0dvdt

≠�

Example – circular motion.34. Answer (2)

Hint :4

tanHR

θ =

Sol :2

24 4 1tan

8 2H gT

gTR R R

θ = = × =

35. Answer (4)

Hint : At highest point v = u cos θ

Sol :

Rate of change of speed = 0


7/18

∵| |

0d v

dt=

�

| |v�

= constant = u cos θ

2 2cosug

Rθ =

2 2cosuR

gθ=

36. Answer (3)

Hint : tan 1x

y xR

⎛ ⎞= θ −⎜ ⎟⎝ ⎠

Sol : When θ = 45°R = 4H = 400 m

y = x tan θ 300

1 300 1400

xR

⎛ ⎞ ⎛ ⎞− = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= 75 m37. Answer (4)

Hint : ,x ydx dy

v vdt dt

= =

Sol : ˆ ˆ2 4v i xj= +�

2dxdt

= 4dy

xdt

=

x = 2t 4 2dy

x tdt

=

dy = 8 t dt

28

2t

y =

= 4 t2

2y x=

38. Answer (2)Hint : In projectile, radial acceleration ar = gcosαSol :

α reduces as particle goes from A → B, so cosαincreases∴ aradial → increases

From B → C, α → increases, cosα → decreases∴ aradial → decreases

39. Answer (3)

Hint : [ ]g eF F⎡ ⎤ =⎣ ⎦

Sol : [ ]g eF F⎡ ⎤ =⎣ ⎦

2 2

2 20

14

Gm e

r r

⎡ ⎤ ⎡ ⎤=⎢ ⎥ ⎢ ⎥πε⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

22

0

18

em

G

⎡ ⎤⎡ ⎤ = ⎢ ⎥⎣ ⎦ πε⎢ ⎥⎣ ⎦

[ ]1

2 2

0

18

em

G

⎡ ⎤= ⎢ ⎥πε⎢ ⎥⎣ ⎦

12 2

0

0

18

em C

G

⎡ ⎤= ⎢ ⎥πε⎢ ⎥⎣ ⎦

40. Answer (3)Hint : Reading = main scale reading + n × leastcountSol : x = M.S.R + n × L.C

= 1 cm + 25 × 0.001 cm = 1 cm + 0.025 =1.025 cm

41. Answer (3)

Hint :2 2

2 2x yd x d y

a adt dt

= =

Sol : x = at – bt2 y = ct vx = a – 2bt vy = cax = – 2b ay = 0

∴ |a| = 2b42. Answer (2)

Hint :2

radialTv

aR

=

Sol :2v

aR

=

2

55

v=

v2 = 25v = 5 m/s

43. Answer (2)

Hint : 2 1A A AΔ = −��

Sol :

2 sin2

A aθΔ =

��


8/18

44. Answer (2)Hint : In horizontal projectile R = uxT

Sol :

ux = 54 km/h = 5

54 15 m/s18

× =

uy = 0

215

2gt=

215 10

2t= × ×

t = 1 s

15 1 15 mR = × =

45. Answer (3)Hint : Concept of relative velocity.

Sol :

20 m

B

θ

1 1tan

2− ⎛ ⎞⎜ ⎟⎝ ⎠

ABv

40 m

'd'

The particle will be passing from below B

1 1tan

2− ⎛ ⎞⎜ ⎟⎝ ⎠

⇒ 1tan

2θ =

⇒1

sin5

θ =

d = 20 sinθ

120

5= × 20 5

5= 4 5 m=

CHEMISTRY46. Answer (4)

Hint: 6.022 × 1023 (NA) = 1 mole

Solution: Change in quantity of 1 mole will lead tothe change in weight of 1 g-atom of correspondingspecies.

47. Answer (3)

Hint: Molarity = No.of molesof solute

Volume(litre)

Solution:w 1000

MolarityM V(ml)

= ×

w 10001

98 500= × ⇒ w = 49 g

4980% 100

W= ×

49W 100 61.25 g

80= × =

48. Answer (3)

Hint: A tetra-atomic gas molecule contains 4 atoms

Solution: 1 mole tetra-atomic gas = 4 moles ofatoms

0.25 mole tetra-atomic gas = 4 × 0.25 = 1 mole atom

= NA atoms

49. Answer (4)

Hint: Na2CO3 + 2AgNO3 → 2NaNO3 + Ag2CO3 ↓

Solution: 2 3

21.2 50For Na CO : w 10.6 g

100×= =

2 3Na CO

10.6n 0.1mole

106= =

For AgNO3 : 17 100

w 17 g100×= =

3AgNO

17n 0.1mole

170= =

Limiting reagent is AgNO3

2 mole AgNO3 gives 1 mole Ag2CO3 so 0.1 moleAgNO3 gives 0.05 mole Ag2CO3

2 3Ag COw 276 0.05 13.8 g= × =


9/18

50. Answer (1)

Hint: m

dv

=

Solution: Density of ethanol = 0.8 g/cm3

Mass of 1 molecule of ethanol = 23

46g

6.022 10×

23 323

m 46 1v 9.55 10 cm

d 6.022 10 0.8−= = × = ×

×

51. Answer (2)

Hint: 1 mole CaCO3 gives 1 mole CO2 on completedecomposition.

Solution: CaCO3 → CaO + CO2

2 3CO CaCO

5.6n n 0.25 mole

22.4= = =

3CaCOw 0.25 100 25 g= × =

% purity = 25

100150

× = 16.67%

52. Answer (1)

Hint: Half filled and fully filled orbitals are morestable.

Solution: For one electron species, in its groundstate, all the orbitals of same shells remaindegenerated due to unavailability of electrons in theseshells.

2s = 2p < 3s = 3p = 3d (energy order).

• The uncertainty principle is given by

hE t

4Δ × Δ ≥

π

•2

2

ZE 2.178

n= − × gives energy of nth orbit.

53. Answer (1)

Hint: Maximum two electrons can beaccommodated in an orbital.

Solution: According to Pauli’s exclusion principle“No two electrons in an atom can have the same setof four quantum numbers”

54. Answer (4)

Hint: Orbital angular momentum = h

( 1)2

+π

� �

Solution: ∵ = 0� for s-orbital

∴ Orbital angular momentum for s-electron = 0

55. Answer (1)

Hint: dxy, dyz and dzx are non-axial orbitals.

Solution: 2 2Z x –y,p d and 2Z

d orbitals contain lobes

along the axes hence electron density is distributedalong the axes.

pZ dZ2 2 2x –yd

z z

y

x

56. Answer (1)

Hint: 2

H 2 21 2

1 1 1R Z

n n

⎡ ⎤= × −⎢ ⎥λ ⎣ ⎦

Solution: 21Z∝

λ ; higher the value of ‘Z’, smaller

will be the wavelength.

57. Answer (4)

Hint: h

mvλ =

Solution: 34

363 3

6.6 105 10 m

132 10 10

−−

−

×λ = = ×× ×

58. Answer (2)

Hint: Write electronic configuration of each species.

Solution:

Ni3+ : [Ar] 3d7 4s0 ;

Cr : [Ar] 3d5 4s1 ;

Fe : [Ar] 3d6 4s2 ;

Mn3+ : [Ar] 3d4 4s0 ;

59. Answer (2)

Hint: Electronic configuration and nuclear chargegoverns most of the properties.


10/18

Solution: 2 1Ca K Ar Radii

ve charge+ + ⎛ ⎞

< < ∝⎜ ⎟+⎝ ⎠

B < C < N [(I.E1)C > (I.E1)B]

I < Br < Cl [increasing order of ΔegH]

Li < Na < K [Radii ∝ n]

60. Answer (3)

Hint: Aufbau and (n + l) rule majorly decideselectronic configuration of element.

Solution:

Group-15 and period-6 : [ ] 14 10 2 3

54Xe 4 5 6 6f d s p

Group-17 and period-7 : [Rn]86 5f 14 6d10 7s2 7p5

Group-10 and period-5 : [Kr]36 4d10 5s0 [Pd]

Group-12 and period-6 : [ ] 14 10 2

54Xe 4 5 6f d s

61. Answer (2)

Hint: Electronic repulsion can make a processendothermic

Solution: H+ ⎯→ H (exothermic)

O– ⎯→ O2– (endothermic)

C ⎯→ C– (exothermic)

S ⎯→ S– (exothermic)

62. Answer (1)

Hint: Oxides which reacts with both acid and base.

Solution:

Amphoteric oxides : ZnO, PbO, SnO, Al2O3, Sb2O3

Acidic oxides : Cl2O7, SO3, NO2

63. Answer (2)

Hint:

Total energy = Sum of energies of eachrequired per mole ionisation.

Solution: 4 1 2 3 4Si/SiE (x x x x ) kJ/mole+ = + + +

Si

280n 10 moles

28= =

Energy required = 10(x1 + x2 + x3 + x4) kJ

64. Answer (1)

Hint: s-block and p-block elements are also calledrepresentative elements

Solution: Ni(Z = 28) is a d-block element.

65. Answer (2)

Hint: Chlorine has maximum negative electron gainenthalpy.

Solution: N is more electronegative than P.

66. Answer (4)

Hint: Species having same number of valenceelectrons are isoelectronic

Solution:

SF4 : Hybridised state of central atom = sp3d

(See-Saw shape)

CIF4

–: Hybridised state of central atom = sp3d2

(Square planar shape)

SO3 : Hybridised state of central atom = sp2

(Triangular planar shape)

ClO3 : Hybridised state of central atom = sp3

(Shallow pyramidal shape)

XeO4 : Hybridised state of central atom = sp3

(Tetrahedral shape)

34PO − : Hybridised state of central atom = sp3

(Tetrahedral shape)

ICl2– : Hybridised state of central atom = sp3d

(Linear shape)

XeF2 : Hybridised state of central atom = sp3d

(Linear shape)

67. Answer (3)

Hint: Molecule having sp3d2 and sp3d hybridizedcentral atom may have 90° bond angle.

Solution:

Species Hybridisation Bond angle(s)

BrF5 sp3d2 No bond angleis of 90° due topresence of lonepair

CO2 sp Bond angle is 180°

SF6 sp3d2 90° bond angles arepresent

[ ]4PBr⊕

sp3 Bond angles are of

109° 28′


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68. Answer (3)

Hint: Five or six membered rings are formed inintramolecular H-bonding.

Solution:

O

HO

NO

o-Nitrophenol

CCl C H

ClH

O

Cl OH

Chloral Hydrate

69. Answer (1)

Hint: Higher the bond order of molecule, higher willbe the stability of molecule.

Solution:

N more stable than Ndue to less electronsin ABMO

2 2⇒ N N N2 2 2

B.O = 3 2.5 2.5

⇒ O O O2 2 2

B.O = 2 1.5 2.5

⇒ B B B2 2 2

B.O = 1 1.5 0.5

⇒ C C C2 2 2

B.O = 2 1.5 2.5

+ – + –

+

+

+

–

–

–

70. Answer (4)

Hint: The number of sigma and pi bond formeddepends on electronic configuration of the species.

Solution: C2 and B2 contain only π-bonds

22 2 2N , C and O− contain both σ and π-bonds

71. Answer (2)

Hint: Generally higher electronegativity differencecauses higher polarity in a bond.

Solution:

2F O 0.3 D,μ = 2H O 1.85 Dμ = ,

3NH 1.47 Dμ =

3NF 0.23 Dμ =

72. Answer (3)

Hint: In C2H3Cl, C – Cl bond has double bondcharacter.

Solution:

• In PH5 the overlapping is poor due large variationin size.

• Nitrogen cannot expand its octet.

• LiF possess higher lattice energy than CsI dueto smaller ionic radii of Li+ and F–.

73. Answer (3)

Hint: Sulphur in SF4 is sp3d hybridised.

Solution:

S

F

F

F

F

see-saw structure

74. Answer (4)

Hint: Total no. of electrons in a molecule must beeven.

Solution: NO, NO2 and ClO2 contain odd electrons.

O

Cl Cl does not contain odd electrons

75. Answer (3)

Hint: For solubility hydration enthalpy should begreater than lattice energy.

Solution: Solubility order is

RbF > KF > NaF > LiF

76. Answer (2)

Hint: Higher is the molecular association, higher willbe the boiling point.

Solution: Due to two –OH groups, ethylene glycolmolecules will be more associated by intermolecularH–bonding.

77. Answer (1)

Hint: Nitrogen can form back bonding.

Solution: N(SiH3)3 contains pπ-dπ back bond.


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78. Answer (4)

Hint: CH4 has tetrahedral shape

Solution:

C

H

HH

H

So 6 angles are of 109°28′

79. Answer (1)

Hint: Lower the energy more stable is the ioniccrystal.

Solution: In the formation of ionic crystal followingsteps are involved

0iM(g) M (g) e ; H ve+ −⎯⎯→ + Δ = +

e –X(g) X (g)−+⎯⎯⎯→ ; eg

oH =–veΔ

M+(g) + X–(g) ⎯→ MX; ΔHo

Lattice = –ve

Net effect of above 3 steps must be negative for theformation of stable ionic bond.

80. Answer (2)

Hint: Xe is central atom in XeOF2.

81. Answer (1)

Hint: Valency of atoms to be satisfied

Solution: 2 22 2CaCN : Ca CN+ −

22CN − ≡ –

N = C = N–

2σ and 2π bonds

CO32– C

O O

O

:

NO3N

O O

O

:

NO : B.O. 3+

=

82. Answer (2)

Hint: Methane is a covalent compound.

Solution: Most electropositive element is bariumamong all which gives most ionic species

CH4 < CaH2 < BaH2

83. Answer (1)Hint: Acetone is a polar moleculeSolution: Acetone possess sufficient dipole momentwhich gives rise to sufficient attractive forces betweenmolecules and responsible for liquid state of acetone.

84. Answer (1)Hint: In XeF4, Xe has two lone pair and four bond pair.Solution: Electronic arrangement is octahedral

85. Answer (4)Hint: px and py orbitals are mutually perpendicular toeach other.Solution: px and py orbitals are of differentsymmetry hence no mixing occurs resulting information of no molecular orbitals.

86. Answer (3)Hint: Aufbau principle/(n + l) ruleSolution:

(n + �) :

Energy

87. Answer (4)Hint: Pauli’s exclusion principle.Solution: An orbital can have maximum twoelectrons with opposite spin.

88. Answer (2)

Hint: 2 2

ee 2

m vnh KZem vr and

2 r r= =

π

Solution: Energy of electron = 2KZe

2r−

Velocity 1

r∝

2πr = nλline spectra are characteristics of atoms

89. Answer (2)Hint: Azimuthal quantum number (l) decides theshape of the orbital.

90. Answer (1)

Hint: 2n

r 0.529 ÅZ

= ×

Solution:

2

1 1 2

2 2 1

r n Z

r n Z

⎛ ⎞= ×⎜ ⎟⎝ ⎠


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2

2

He

Li

r 1 3r 1 2+

+ ⎛ ⎞= ×⎜ ⎟⎝ ⎠

2Li

2xr Å

3+ =

BIOLOGY91. Answer (3)

Hint : Extensive compartmentalisation of cytoplasmis present in eukaryotic cells.Sol. : Presence of plastids, flagella (have 9 + 2arrangement of microtubules) and nuclear envelopeare eukaryotic features.Mesosomes are infoldings of cell membrane and arecharacterstic of prokaryotes.

92. Answer (4)Hint : It is a semi-fluid matrix present in all cells.Sol. : Main arena of cellular activities where allmetabolic reactions take place is cytoplasm in boththe plant and animal cells.

93. Answer (2)Hint : An elaborate network of proteinaceousfilaments present in eukaryotic cells is calledcytoskeleton.Sol. : The cytoskeleton helps in providingmechanical support to the cell, in formation offlagella and maintain the shape of the cell. Whereas,chloroplasts are involved in carbon assimilation.

94. Answer (3)Hint : Ribosomes are membrane-less organelles.Sol. : Ribosomes were discovered by GeorgePalade. Two ribosomal subunits remain attached toeach other at optimum concentration of Mg2+ ions.ER is absent in prokaryotes.

95. Answer (2)Hint : ER and Golgi apparatus both are parts ofendomembrane system.Sol. : Both ER and Golgi are structurally composedof cisternae, vesicles and tubules. Synthesis of lipidsand steroid hormones, is an exclusive feature ofSER while synthesis of plasma membrane duringcytokinesis is a function associated with Golgi.

96. Answer (4)Hint : Digestive vacuoles are secondary lysosomesalso called hetrophagosomes.Sol. : Secondary lysosomes are formed by fusion ofprimary lysosomes with phagosomes. Digestivevacuoles contain active enzymes against the materialto be digested.

97. Answer (1)Sol. : Movement of molecules against their

concentration gradient is called active transport. Inthis process energy is required in the form of ATP.

98. Answer (3)Hint : Electron transport system is present in innermitochondrial membrane and in thylakoid membraneof chloroplast.Sol. : Both mitochondria and chloroplasts have ATPsynthesising machinery and presence of porinproteins in their outer membrane, but machinery foraerobic respiration is found only in mitochondria.

99. Answer (2)Hint : This giant chromosome was discovered byRuckert.Sol. : Lampbrush chromosomes are diplotenebivalents and are found in the primary oocyte nucleiof some vertebrates and invertebrates. They haveloops which participate in transcription & RNAproduced from them could be stored in the form ofinformosomes(mRNA + proteins).

100. Answer (4)Hint : It is the site of synthesis of one of thecomponents of ribosomes.Sol. : Nucleolus is non-membrane bound structureand is the site for rRNA synthesis.

101. Answer (3)Hint : At interphase, nucleus contains looseextended & diffused network of nucleoprotein fibres,called chromatin.Sol. : Term chromatin was given by Flemming andit can be stained by basic dyes. It is composed ofDNA and histone proteins. Euchromatin is looselypacked part of chromatin which is transcriptionallyactive.

102. Answer (1)Hint : One form of plastid can be converted intoanother form according to the requirement.Sol. : During the ripening of tomato its colourchanges from green to reddish due to transformationof chloroplasts into chromoplasts.

103. Answer (3)Hint : Eukaryotic flagella have microtubule doubletsat periphery and a pair of microtubules in center.Sol. : Central tubules are covered by central sheathand the entire axoneme is covered by plasmamembrane.Formation of spindle fibres during cell division is afunction of centrioles.

104. Answer (4)Sol. : Eukaryotic genomic DNA is double strandedand linear.

105. Answer (4)Sol. : Small fragments which are formed beyondthe secondary constriction are called satellite.


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106. Answer (1)Hint : Quiescent stage is resting stage of nondividing cells.Sol. : Non-dividing cells enter the quiescent phaseor G0 phase of cell cycle where they remainmetabolically active but do not proliferate.

107. Answer (2)Hint : In this stage, splitting of centromere andseparation of sister chromatids occurs.Sol. : This is anaphase stage of mitosis. Atanaphase, chromatids move towards the poles andthe arms of chromosomes trail behind. This is thebest stage to study shape of chromosomes.

108. Answer (4)Hint : Chiasmata disappear and homologouschromosomes separate at diakinesis stage.Sol. : In prophase I, pairing of homologouschromosomes occurs at zygotene, formation ofrecombination nodule occurs at pachytene,separation of homologous chromosomes(desynapsis) occur at diplotene and terminalisationof chiasmata occurs at diakinesis stage.

109. Answer (4)Hint : Mitogens are substances which inducemitosis.Sol. : Insulin is a mitogen, while cyanide, mustardgas and chalones are mitotic poisons which inhibitmitosis.

110. Answer (3)Hint : Meiocyte is a diploid cell that undergoesmeiosis to produce haploid cells.Sol. : Due to DNA replication the amount of DNA isdoubled at G2 phase. A meiocyte at G2 phase hasfour times higher amount of DNA than its gametes.

111. Answer (2)Hint : Initiation of reduction in the number ofchromosomes occurs at anaphase I.Sol. : In Anaphase-I, the homologous chromosomesget separated and move towards the opposite polesof the cell, but their sister chromatids remainattached to each other.

112. Answer (1)Hint : Crossing over occurs only once duringreductional division.Sol. : Recombination or crossing over betweenhomologous chromosomes occurs once in meiosis.Duplication of centrioles, karyokinesis andmetaphasic plate formation occur twice in the wholemeiosis.

113. Answer (3)Hint : A bivalent is a pair of homologouschromosomes.

12 bivalents = 24 chromosomes. Hence in gameteit will be 12 chromosomes.Sol. : One chromosome has one chromatid ingamete. So, total 12 chromosomes & 12chromatids.

114. Answer (4)Hint : Mitosis occurs in somatic cells and fewhaploid cells of lower organisms.Sol. : Growth, repair and regeneration of body partsoccur through mitosis. But conservation ormaintenance of number of chromosomes throughoutthe generations occur via meiosis.

115. Answer (3)Hint : Class is an obligate taxonomic categorywhich is lower than phylum or division in hierarchy.Sol. :Mammalia, Dicotyledonae, Insecta - ClassesFelis, Mangifera - GeneraPlantae - KingdomPoales, Polymoniales, Diptera - OrderArthropoda - PhylumHominidae, Solanaceae - Families

116. Answer (2)Hint : A scientific name has two words in binomialnomenclature system.Sol. : In binomial nomenclature, first word of thename is genus and second word is specific epithetScientific names are derived from Latin language andprinted in italics.

117. Answer (4)Hint : Taxonomy is the branch of science whichdeals with study of principles and procedures ofclassification.Sol. : Modern taxonomy is based upon study ofboth external and internal features of organism alongwith cell structure, developmental process andecological informations of organisms. But phylogenyis included in systematics.

118. Answer (3)Hint : ER plays a major role in origin of thestructure which connects the cytoplasm ofneighbouring plant cells.Sol. : A - Plasmodesmata

B - Plasma membraneC - Desmotubule

119. Answer (2)Hint : These are rich in hydrolytic enzymes whichget activated at acidic pH.

Sol.: Lysosomes are the single membrane bound,polymorphic cell organelles which are associatedwith intracellular digestion of food materials. Lysis of


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old or dead organelles and ephagy of undigestedsubstances are also carried out by lysosomes. Butconversion of fats into carbohydrates is a function ofglyoxysomes.

120. Answer (2)

Hint : This mitochondrial structure is rich incardiolipins.

Sol. : Inner membrane of mitochondria containsenzymes and electron carriers for the formation ofATP.

121. Answer (1)

Hint : These bacteria appear purple in colour afterstaining.

Sol. : Gram positive bacteria have thick cell wall,retain the stain and due to which they appear purplein colour.

122. Answer (4)

Hint : A membraneless organelle which is present inanimal cell, but absent in higher plants is centriole.

Sol.: Centrioles have 9 + 0 arrangement ofmicrotubules and they are involved in spindleformation during animal cell division.

123. Answer (1)

Hint : ER which is free from ribosomes, is abundantin muscle cells.

Sol. : Smooth ER is associated with the functionslike detoxification of drugs, uptake and release ofCa2+ ions during muscle contraction and passing ofproducts of RER to golgi apparatus.

Synthesis of enzyme precursors for lysosomes, is afunction of RER.

124. Answer (3)

Hint : Chromosomes which have centromere veryclose to its one end, are J-shaped chromosomes.

Sol.: Chromosomes which have their centromerevery close to one end are called acrocentricchromosomes.

125. Answer (2)

Hint : It is a major component of the cell wall offungi.

Sol.: Chitin is not found in algal cell walls.

126. Answer (1)

Hint : These cells form ground tissue of leaves ofgreen plants.

Sol.: Chloroplasts are mainly present in themesophyll cells of leaves of green plants.

127. Answer (2)

Hint : Endocytosis is intake of materials in the formof carrier vesicles formed by invagination of smallregions of plasma membrane.

Sol.: Functions like cell growth, formation ofintercellular junctions, secretion etc. are due to fluidnature of plasma membrane.

128. Answer (2)Hint : Gap2 phase is followed by S-phase.Sol.: Synthesis of RNA & proteins occur in both G1& G2 stage. Duplication of single membrane boundcell organelles also take place in both G1 & G2. Butduplication of double membrane bound cellorganelles like mitochondria and chloroplast occur inG2 phase.

129. Answer (3)Hint : Syncytium shows multinucleate condition.Sol.: If karyokinesis is not followed by cytokinesisit leads to formation of a single cell with large numberof nuclei.

130. Answer (1)Hint : This stage includes spireme stage.Sol.: At early prophase, ends of chromosome arenot visible and they appear like a ‘ball of wool’.

131. Answer (3)Hint : Synaptonemal complex stabilises the pair ofsynapsed chromosomes or bivalent.Sol.: This complex is formed at zygotene stage anddissolves at diplotene stage. At diplotene thebivalent start separating.

132. Answer (2)Hint : Direct cell divison mostly occurs in bacteria,protozoan etc.Sol.: Amitosis or direct cell division lacks spindleformation. In amitosis, karyokinesis occurs by asimple constriction.

133. Answer (1)Hint : (a) Chiasmata formation occurs at diplotenestage of prophase-I.Sol.:(b) Alignment of univalents - Metaphase-II

on equatorial plate(c) Chromatids reach - Telophase-II

the pole(d) Initiation of reduction - Anaphase-I

in number ofchromosomes

Thus, correct sequence is : a → d → b → c.134. Answer (2)

Hint : It is an index of plant species found in aparticular area.Sol.: Flora is a book containing information aboutplants found in particular area. It gives actualaccount of habitat and distribution.


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135. Answer (4)Hint : Defining features are those features which arenecessarily present in all living organisms.Sol.: Sexual reproduction is shown by manyorganisms but it is absent in certain lower organismsand infertile higher organisms so it cannot be adefining feature of living organisms.

136. Answer (2)Hint : Alcohol is already a simple absorbable form.Sol. : Chips, burger and fish are complex foods thatneed to be digested in alimentary canal and aremaximally absorbed in small intestine.Wine contains alcohol which gets absorbed acrossthe gastric mucosa.

137. Answer (3)Hint : Middle part of small intestine.Sol. : Nearly 5500 ml of water each day isabsorbed across jejunum in a man. 1300 ml of wateris absorbed across colon and 200 ml across ileum.

138. Answer (2)Hint : Carbohydrases digest carbohydrates such asstarch.Sol. : Glandular cells of stomach secrete proteasesand acid. Carbohydrases are secreted in mouth,pancreas and small intestine.

139. Answer (1)Hint : Myenteric plexus is located in muscularisexterna.Sol. : Neurons in alimentary canal form Myentericplexus, a motor plexus responsible for peristalsis.Meissner’s plexus located in submucosa isresponsible for neural activity resulting in release ofenzymes and mucus.

140. Answer (3)Hint : Cells involved in defence activity in alimentarycanal.Sol. : Parietal cells in gastric mucosa secrete HCland Castle’s intrinsic factor. Kupffer cells aremacrophages of liver and argentaffin cells secreteprecursor molecules like 5-OH-tryptamine.

141. Answer (4)Hint : Sphincter that guards opening ofhepatopancreatic duct.Sol. : Sphincter of Boyden guards opening ofcommon bile duct. Ileocaecal valve prevents backflow of digested food from caecum to ileum. Pyloricsphincter is present at the junction of pyloric end ofstomach and duodenum

142. Answer (2)Hint : Identify the dental formula of milk teeth.

Sol. : At age 10, premolars and wisdom teeth (last

molar) are absent. Diphyodont dentition is seen.

21232123

is dental formula of adult human

143. Answer (3)

Hint : Action of these enzymes mostly results inproduction of simple absorbable forms.

Sol. : Enterokinase converts trypsinogen totrypsin. Pepsin and rennin are secreted aszymogens in stomach. Enterocrinin is a hormone.Carboxypeptidase, lipase (steapsin) and elastase aresecreted by pancreas.

144. Answer (1)

Hint : Bile does not contain any enzyme.

Sol. : Hepatocytes secrete bile that has bilirubin,biliverdin (bile pigments) and bile salts but nodigestive enzymes. Enterocytes secrete digestiveenzymes and parietal cells (oxyntic cells) releasehydrochloric acid to absorb iron.

145. Answer (3)

Hint : Competitive inhibitor competes with thesubstrate for the active site.

Sol. : Non-competitive inhibitor binds to the enzymeirreversibly, decreases the Vmax of the reactionwithout affecting the Km for a given substrate.

146. Answer (1)

Hint : Identify the disorder in which both protein andenergy malnutrition occur in infants less than1 year in age.

Sol. : Kwashiorkor results from deficiency ofproteins only in children above 1 year. Scurvy resultsfrom vitamin C deficiency. Anaemia does not occurdue to deficiency of riboflavin and thiamine.

147. Answer (1)Hint : Hormone whose secretion promotes gastricjuices

Sol. : Enterogastrone, secretin & duocrinin arereleased by cells in small intestine while gastrin isa secretion of G cells in stomach. Enterogastroneand secretin both decrease gastric secretions.Duocrinin stimulates Brunner’s glands.

148. Answer (1)

Hint : Identify a food source rich in cellulose.

Sol. : Lettuce is rich in cellulose which cannot bedigested by humans. Hence its calorific value isnearly ‘zero’. Egg white and meat produce grosscalorific value around 5.65 kcal/g while bread yields4 kcal/g.

149. Answer (2)Hint : Deficiency of HCl in human body isachlorhydria that would imply impaired functioning ofparietal cells.


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Sol. : Deficiency of HCl will affect iron absorptionleading to anaemia. Rennin is a hydrolytic enzyme,therefore, belongs to hydrolase category of enzymes.Parasympathetic stimulus promotes activity ofBrunner’s glands.

150. Answer (1)Hint : Aggregates of bile salts in enteric lumenpromote fat absorption by emulsification.Sol. : Chylomicrons are formed inside enterocytes.

151. Answer (2)Hint : Enzymes are absent in bile juice.Sol. : Bile juice is stored temporarily in gall bladder.Lipase is a secretion mainly from pancreas. Cysticduct is the term for duct arising from gall bladder.

152. Answer (2)Hint : Muscularis layer is found inner to serosa.Sol. : Innermost lining is mucosa followed by sub-mucosa, inner circular and outer longitudinal musclelayer and outermost serosa layer.

153. Answer (3)Hint : Vomiting centre is located in brain stem.Sol. : Vomiting is a reflex action controlled by thevomiting centre located in medulla oblongata.

154. Answer (2)Hint : pH 7.8 is present in small intestine.Sol. : Proteases hydrolysing peptide bonds ofproteins acting at pH 7.8 include trypsin,carboxypeptidase and chymotrypsin whosezymogenic form are secretions of pancreas.

155. Answer (1)Hint : Catalase a, holoenzyme comprises ofapoenzyme and cofactor.Sol. : Apoenzyme is proteinaceous in nature whileits cofactor is organic & non proteinaceous prostheticgroup called haem which is tightly attached to it.

156. Answer (4)Hint : Structural similarity between inhibitor andsubstrate is not the criteria for non competitiveinhibition enzyme inhibition.Sol. : Ethanol, malonate and sulfonamides arecompetitive inhibitors of methanol, succinate andPABA respectively.

157. Answer (2)Hint : Some nucleic acids can also act as enzymes.Sol. : Ribonuclease P is RNA acting as enzymewhose substrate is also RNA. It was discovered byAltmann from E.coli.

158. Answer (3)Hint : The activity of enzyme reaches a plateauonce all the active sites of enzyme are saturatedwith the substrate.

Sol. : A rectangular hyperbola curve is attainedwhen [S] is plotted against enzyme activity.

159. Answer (2)Hint : Deoxyribose sugar is found in DNA.Sol. : Niacin i.e. vitamin B3 forms NAD and NADPwhich act as coenzymes.

160. Answer (2)Hint : Uracil is a nitrogenous base exclusively foundin RNA.Sol. : Thiamine is vitamin B1. Cysteine is an aminoacid while cytosine is a base.

161. Answer (1)Hint : Hydrogen bonds link purines and pyrimidineson opposite strands.Sol. : Guanine and cytosine are linked through 3hydrogen bonds. AT base pair will have 2 hydrogenbonds.

162. Answer (1)Hint : 10 bp occupy a distance of about 34 Å inB-DNA.Sol. : Base pairs in DNA are stacked 3.4 Å apart.In B-DNA, diameter of helix is 20 Å and pitch of thehelix is 34 Å. Hydrogen bonds are absent in a singlenucleotide. A and G of one strand base pair with Tand C of other strand respectively.

163. Answer (3)Hint : Chargaff’s rule.Sol. : [A] = [T] and [G] = [C]

Therefore, [A] = 120 and [G] = 120 [T] = 120 and [C] = 120

Sum total of nucleotides is 480.164. Answer (3)

Hint : Formation of glycosidic, peptide and esterbond requires loss of water molecules.Sol. : Glycosidic bond formed by loss of water is adehydration/condensation reaction. Alphabet ‘P’represents amino acid Proline. A protein attains itsfunction by formation of its active site at tertiary levelof organisation.

165. Answer (3)Hint : Proteins are polymers of different repeatingunits.Sol. : Structural homopolymers include celluloseand chitin. Inulin and glycogen are storagehomopolymers whereas collagen and RuBisCO areheteropolymers.

166. Answer (1)Hint : Glycerol is trihydroxypropane.Sol. : Adenine is derived from substitution of purinering at 6th carbon while cytosine is a pyrimidineCholesterol is a sterol.


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167. Answer (3)Hint : Functional groups in a sugar are hydroxyland aldehyde.Sol. : Amino acid is identified by the position of ‘R’chain, amino and carboxyl group w.r.t. C∝ position.Sugar molecules have aldehyde or ketone groups.

168. Answer (1)Hint : Macromolecules are obtained in acidinsoluble fraction.Sol. : Insulin is a heteropolymer while cholesterolforms aggregates in vesicles. Both are obtained inacid insoluble fraction/retentate fraction. Proteins arelarge in size hence they are macromolecules andlipid like cholesterol does not dissolve in acid.

169. Answer (2)Hint : Porifera show cell aggregate body plan andtissues are lacking.Sol. : Members of Cnidaria/Coelenterata exhibittissue level of organisation for the first time inkingdom Animalia. Platyhelminthes andAschelminthes exhibit, organ and organ system levelof organisation respectively.

170. Answer (3)Hint : Identify epithelium with tall cells and basalnuclei.Sol. : Squamous epithelium is suitable for surfaceswhere filtration and diffusion are necessary. Ciliapush the ova towards ampulla of fallopian tube whilemucus is pushed upwards in trachea. Keratinisedepithelium features on surfaces that requireprotection from water.

171. Answer (1)Hint : Cell membrane of axon is called axolemma.Sol. : Nodes of Ranvier in neurons of PNS haveneurilemma. Oligodendrocytes form myelin sheath inCNS. Saltatory conduction occurs at nodes ofRanvier.

172. Answer (3)Hint : All cells of neural tissue are ectodermal inorigin except microglial cells.Sol. : Astrocytes are ectodermally derivedPerineurium forms a protective white sheath arounda single fasciculus. Centrosome is absent in neuronshence they cannot divide.

173. Answer (3)Hint : Property to decrease in size is associatedwith muscular tissue.Sol. : Property of excitability and conductivity iscommon to muscular and neural tissue.

174. Answer (2)

Hint : Uninucleated muscle fibres lines the wall ofstomach.

Sol. : Wall of internal organ stomach is lined byvisceral/smooth unstriped muscles. Actin and myosinfibrils occur in all type of muscle cells. Both skeletaland smooth muscle fibres are unbranched.

175. Answer (1)

Hint : Striped muscles are associated with skeletalsystem.

Sol. : Limb bones such as humerus and femur areassociated with voluntary, cylindrical muscle fibrescalled skeletal muscle fibres. Smooth, non-striatedmuscle fibres are present in wall of intestine whereasstriated, cardiac muscle fibres are associated withheart.

176. Answer (4)

Hint : Elastic cartilage is associated with tip ofnose.

Sol. : Haversian lamellae occur in diaphysis part oflong bones of mammals. Pliable matrix is a featureof cartilage. Chondroblasts and chondrocytes arelocated within lacuna.

177. Answer (4)

Hint : Muscles and bones are vascular tissues.

Sol. : Epithelial tissue is avascular as it lacks bloodsupply usually. Biceps are part of muscular tissue.

178. Answer (2)

Hint : This bone is also called knee cap.

Sol. : Patella is a sesamoid bone while other boneslisted are cartilaginous.

179. Answer (2)

Hint : Brown fat is thermogenic.

Sol. : Brown fat is known for producing heat. Whitefat is used as storehouse of energy, food reserve,insulation etc. Mast cells release heparin, serotoninand histamine upon activation. Collagen fibres arewhite and unbranched.

180. Answer (2)

Hint : Adhering junctions help to keep adjacent cellstogether during stress.

Sol. : Macula adherens/desmosomes prevent cellsfrom being pulled apart. Tight junctions preventleakage of substances and are also called zonulaoccludens. Intermediate junctions are also known aszonula adherens, usually occur below tight junctions.

� � �

Test - 4 (Code-B) (Answers) All India Aakash Test Series for NEET-2019

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Test Date : 23/12/2018

ANSWERS

TEST - 4 (Code-B)

All India Aakash Test Series for NEET - 2019

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113. (3)

114. (1)

115. (2)

116. (3)

117. (4)

118. (4)

119. (2)

120. (1)

121. (4)

122. (4)

123. (3)

124. (1)

125. (3)

126. (4)

127. (2)

128. (3)

129. (1)

130. (4)

131. (2)

132. (3)

133. (2)

134. (4)

135. (3)

136. (2)

137. (2)

138. (2)

139. (4)

140. (4)

141. (1)

142. (2)

143. (3)

144. (3)

145. (1)

146. (3)

147. (2)

148. (1)

149. (3)

150. (1)

151. (3)

152. (3)

153. (3)

154. (1)

155. (1)

156. (2)

157. (2)

158. (3)

159. (2)

160. (4)

161. (1)

162. (2)

163. (3)

164. (2)

165. (2)

166. (1)

167. (2)

168. (1)

169. (1)

170. (1)

171. (3)

172. (1)

173. (3)

174. (2)

175. (4)

176. (3)

177. (1)

178. (2)

179. (3)

180. (2)

All India Aakash Test Series for NEET-2019 Test - 4 (Code-B) (Hints and Solutions)

2/18

PHYSICS1. Answer (3)

Hint : Concept of relative velocity.

Sol :

20 m

B

θ

1 1tan

2− ⎛ ⎞⎜ ⎟⎝ ⎠

ABv

40 m

'd'

The particle will be passing from below B

1 1tan

2− ⎛ ⎞⎜ ⎟⎝ ⎠

⇒ 1tan

2θ =

⇒ 1sin

5θ =

d = 20 sinθ

120

5= × 20 5

5= 4 5 m=

2. Answer (2)Hint : In horizontal projectile R = uxT

Sol :

ux = 54 km/h = 5

54 15 m/s18

× =

uy = 0

215

2gt=

215 10

2t= × ×

t = 1 s

15 1 15 mR = × =

3. Answer (2)

Hint : 2 1A A AΔ = −��

Sol :

2 sin2

A aθΔ =

��

4. Answer (2)

Hint :2

radialTv

aR

=

Sol :2v

aR

=

2

55

v=

v2 = 25v = 5 m/s

5. Answer (3)

Hint :2 2

2 2x yd x d y

a adt dt

= =

Sol : x = at – bt2 y = ctvx = a – 2bt vy = cax = – 2b ay = 0

∴ |a| = 2b6. Answer (3)

Hint : Reading = main scale reading + n × leastcountSol : x = M.S.R + n × L.C

= 1 cm + 25 × 0.001 cm = 1 cm + 0.025 =1.025 cm

7. Answer (3)

Hint : [ ]g eF F⎡ ⎤ =⎣ ⎦

Sol : [ ]g eF F⎡ ⎤ =⎣ ⎦

HINTS & SOLUTIONS

Test - 4 (Code-B) (Hints and Solutions) All India Aakash Test Series for NEET-2019

3/18

2 2

2 20

14

Gm e

r r

⎡ ⎤ ⎡ ⎤=⎢ ⎥ ⎢ ⎥πε⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

22

0

18

em

G

⎡ ⎤⎡ ⎤ = ⎢ ⎥⎣ ⎦ πε⎢ ⎥⎣ ⎦

[ ]1

2 2

0

18

em

G

⎡ ⎤= ⎢ ⎥πε⎢ ⎥⎣ ⎦

12 2

0

0

18

em C

G

⎡ ⎤= ⎢ ⎥πε⎢ ⎥⎣ ⎦

8. Answer (2)Hint : In projectile, radial acceleration ar = gcosα

Sol :

α reduces as particle goes from A → B, so cosαincreases∴ aradial → increases

From B → C, α → increases, cosα → decreases∴ aradial → decreases

9. Answer (4)

Hint : ,x ydx dy

v vdt dt

= =

Sol : ˆ ˆ2 4v i xj= +�

2dxdt

= 4dy

xdt

=

x = 2t 4 2dy

x tdt

=

dy = 8 t dt

28

2t

y =

= 4 t2

2y x=

10. Answer (3)

Hint : tan 1x

y xR

⎛ ⎞= θ −⎜ ⎟⎝ ⎠

Sol : When θ = 45°R = 4H = 400 m

y = x tan θ 300

1 300 1400

xR

⎛ ⎞ ⎛ ⎞− = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= 75 m11. Answer (4)

Hint : At highest point v = u cos θ

Sol :

Rate of change of speed = 0

∵| |

0d v

dt=

�

| |v�

= constant = u cos θ

2 2cosug

Rθ =

2 2cosuR

gθ=

12. Answer (2)

Hint :4

tanHR

θ =

Sol :2

24 4 1tan

8 2H gT

gTR R R

θ = = × =

13. Answer (1)

Hint : speed, velocityv v= =� �

Sol : | | 0d

vdt

=�

but 0dvdt

≠�

Example – circular motion.14. Answer (2)

Hint : t = drel/vrel

Sol :24

38

t s= =

15. Answer (4)

Hint :2

2

dx d xv a

dt dt= =

Sol : If n > 22

2

d xa

dt=

= n(n – 1)tn – 2

Would depend on x and would be increasingwith t

If n < 0, a = ktn – 2( 1)

ve

k n n= −⎧ ⎫⎨ ⎬= +⎩ ⎭

∴ a would decrease


4/18

16. Answer (2)


Total timev =

Sol :

cos2

2r r

θΔ =

= 2 rcos45° = 2 2r =

22 14

4 2 2 2

rr

t sv

ππΔ = = =

× π

av21

2 2

v =

⇒ vav = 4 m/s17. Answer (2)

Hint : y = ax – bx2, where y = 0, x = 0 or x = RSol : y = ax – bx2

where y = 0, x = 0 or x = Rax – bx2 = 0x(a – bx) = 0

a ax R

b b= =

Hmax at 2a

xb

=

Hmax = 2

2 2a a

a bb b

⎛ ⎞ ⎛ ⎞−⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

2

4ab

=

Also tan aθ =

18. Answer (4)

Hint : 4

tanHR

θ =

Sol :4

tanHR

θ =

or, tanθ = 4 [∵ H = R]

4 1sin ; cos

17 17θ = θ =

2 22 sin cos 2 (20) 4 1 320m

10 17 17v

Rgθ θ × × ×= = =

×

19. Answer (2)

Hint : avgTotal distance

Total timev =

Sol :

From B → C 4.5 7.8 12

6 m/s2 2BCv+= = =

From A → C, 2 AB BC

ACAB BC

v vv

v v=

+

2 AB BCAC

AB BC

v vv

v v=

+

2 3 6 2 3 6

3 6 9× × × ×= =

+

4 m/sACv =

20. Answer (3)

Hint : avgf i

f i

v va

t t

−=

−

� �

��

Sol :v′u

t = 0 t t t = + 1 2t2t1

�

a1

�

v�

a2

�

v = u + a1t1 v ′ = v + a2t2

avg1 2

v ua

t t

′ −=+

2 2

1 2

v a t ut t

+ −=+


5/18

1 1 2 2

1 2

u a t a t ut t

+ + −=+

1 1 2 2avg

1 2

a t a ta

t t+=+

21. Answer (3)

Hint : Vector addition

Sol :

A−

B R

R′

βθ

A

22. Answer (2)

Hint : Z = a + b, ΔZ = Δa + Δb

Sol : Z = P + Q

ΔZ = ΔP + ΔQ

% error 100Z

ZΔ= ×

100P QP Q

Δ + Δ= ×+

100P Q

P Q P QΔ Δ⎛ ⎞= + ×⎜ ⎟+ +⎝ ⎠

23. Answer (3)

Hint : Derived quantities may have dimensions andsolution in terms of base quantities.

24. Answer (3)

Hint : LC = 1 MSD – 1 VSD

Sol : N(V.S.D.) = (N –1) M.S.D.

11( . . ) . . .

NV S D M S D

N−⎛ ⎞= ⎜ ⎟

⎝ ⎠

Least count:→ 1M.S.D. – 1V.S.D.

11 . . . . .

NM S D M S D

N−⎛ ⎞= − ⎜ ⎟

⎝ ⎠

1. .

N NM S D

N− +⎛ ⎞= ⎜ ⎟

⎝ ⎠

1 . . .M S DN

=

∴ Least count 1 . . . 1M S D mm

N N= =

L.C. = 1

cm10N

25. Answer (4)

Hint : 2 2net x yv v v= +

Sol : 2 2v u gh= + , irrespective of θ

26. Answer (3)

Hint : If v�

is constant, then speed is constant.

Sol : x = 4 sinωt y = 4 cosωt

⇒ sin4x

t= ω ⇒ cos4y

t= ω

sin2ωt + cos2ωt = 2 2

4 4x y+

2 2

14

x y+=

2 2 4x y+ =

Particle is moving in circular path.

4 cos , 4 sinx ydx

v t v tdt

= = ω ω = − ω ω

2 2x yv v v= +

| v | =4 ω27. Answer (2)

Hint : Concept of significant figure used in additionSol : 5.0 m + 6.0 m = 11.0 m

28. Answer (1)Hint : Dimensions are powers raised to fundamentalphysical quantity.

Sol : Energy density = EnergyVolume

2 2

3

[ML T ]

[L ]

−=

= [ML–1T–2]


6/18

29. Answer (4)

Hint: n1u1 = n2u2

Sol : F = [MLT–2] E = [ML2T–2]

F ′ = [2M2LT–2] E ′ = [2M(2L)2T–2]

F ′ = 4[MLT–2] = [8ML2T–2]

F ′ = 4F 8E E′ =

P = [ML2T–3]FA

σ =

P ′ = [2M(2L)2T–3]2

2

[MLT ]

[L ]

−=

= [8ML2T–3]

P ′ = 8P 2 2

2 2

2[M2LT ] [MLT ]

[2L] [L ]

− −′σ = =

′σ = σ

30. Answer (2)

Hint : Principle of homogeneity.

Sol : If dimension of A and B are different.

→ In logAB

⎛ ⎞⎜ ⎟⎝ ⎠

and ABe

−,

AB

⎛ ⎞⎜ ⎟⎝ ⎠

should be dimensionless

which is not possible according to given condition

→ In A + B principle of homogeneity violates so

correct answer AB

⎛ ⎞⎜ ⎟⎝ ⎠

31. Answer (4)

Hint :2 2

,y x yv v uR T

g g= =

Sol :v1

v2

1 2 12 2,

v v vR T

g g= =

32. Answer (4)

Hint : If both magnitude as well as direction ofacceleration remains constant then it is uniformlyaccelerated motion.

Sol : In projectile motion direction of acceleration isconstant.

∴ It is uniformly accelerated motion

In uniform circular motion the direction ofcentripetal acceleration changes, although magnitudeis constant.

∴ It is non-uniformly accelerated motion.

33. Answer (3)

Hint :2 2sin 2 sin

,2

u uH T

g gθ θ= =

Sol : As usinθ = uy

2( ) 2,

2y yu u

H Tg g

= =

2

2

2 4

y

y

y

u

g uHuT

g

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠= =⎛ ⎞⎜ ⎟⎝ ⎠

yH

uT

∝

34. Answer (4)

Hint : net T Ca a a= +��

Sol : In case of increasing speed aT is in thedirection of ‘v’.

In case of decreasing speed aT is opposite directionspeed

aT

aCanet

v v

anet

=Decreasingspeed

35. Answer (2)

Sol :

vmax

T

α β

v

t

avgDisplacement

Timev =

Area under - graphtime

v t=

max12

2max

avg

v T vv

T

× ×= =

∴ max max

maxavg

21

2

v vvv

= = = 2 : 1


7/18

36. Answer (4)

Hint : Rm RG mGv v v= −��

Sol : Let us assume that if rain is falling at 30° itwould have horizontal component as vx and verticalcomponent vy

Now man starts running with 10 m/svRm → vertical

So,

Also 10

tan30 tan30 10 3xy

y y

vv

v v= ° ⇒ = ° ⇒ =

∴ When man starts running he will see onlyvertical component.

10 3 m/sRmv =��

37. Answer (2)Hint : Time of ascent < Time of descentSol : Time of ascent

v2 – u2 = 2as0 – u2 = 2 (g + a) H

( )2

2( )2

uH u g a H

g a= ⇒ = +

+v = u + at

0 2( ) ( )g a H g a t= + − +

2( )( )g a H

tg a

+=+

up2

( )H

tg a

=+

Time of descent

210

2H ut at u= + =

21( )

2H g a t= −

down2

( )H

tg a

=−

∴

2

2d

a

Ht g a

t Hg a

−=

+

d

a

t g at g a

+=−

38. Answer (3)

Hint : still2 u d

u d

t tt

t t=

+

Sol : d = (vb – vR)tu ...(i)d = (vb + vR)td ...(ii)From (i) and (ii)tu(vb – vR) = td(vb + vR)120 (vb – vR) = 40 (vb + vR)

or, 3vb – 3vR = vb + vR

or, vb = 2 vR

Now, 332

R d

b R

v tdt

v v= =

3

40 60 minutes2

= × =

39. Answer (4)


Total timev =

Sol : Displacement = 9 cm.Time = 12.5 hours = 12.5 × 3600 s

avg9

12.5 3600v =

× = 2 × 10–4 cm/s

40. Answer (3)

Hint : For expression z = m p

q

a b

c

Fractional error z m a p b q c

z a b cΔ Δ Δ Δ= + +

Sol : 2l

Tg

= π

2 24l

Tg

= π

22

4l

gT

= π

∴ 2g l Tg l T

Δ Δ Δ= +


8/18

2g

p qg

Δ = +

41. Answer (2)Hint : Principle of homogeneity.

Sol : 212

P v h g C+ ρ + ρ =

By principle of homogeneityDimensional formula of C should be of pressure[C] = [P] = [ML–1T–2 ]

Modulus of rigidity = Shear stressShear strain

FAx

x

=Δ

= [ML–1T–2]∴ [Modulus of rigidity] = [P] = [C]

42. Answer (1)Hint : If graph between physical quantities is astraight line then those quantities are directlyproportional.Sol : From graph v2 ∝ x ...(i)Now we know v2 – u2 = 2ax

0u =v2 = 2ax ...(ii)From (i) and (ii)a = constant∴ a is uniform

43. Answer (4)Hint : If speed is varying, velocity must be variable.

44. Answer (2)Hint : Area under velocity-time graph givesdisplacement.Sol : Average value of sine curve for positive halfcycle.

02 v=π

2 10×=π

= 6.37 m/s45. Answer (2)

Hint :2 21 2

midpoint 2v v

v+=

Sol :2 2

13 m/s2

P QR

v vv

+= =

P R Q

7 m/sPv = Rv 17 m/sQv =

13 172 2

+ += =R Qav

v vv

302

=

= 15 m/s

CHEMISTRY46. Answer (1)

Hint: 2n

r 0.529 ÅZ

= ×

Solution:

2

1 1 2

2 2 1

r n Z

r n Z

⎛ ⎞= ×⎜ ⎟⎝ ⎠

2

2

He

Li

r 1 3r 1 2+

+ ⎛ ⎞= ×⎜ ⎟⎝ ⎠

2Li

2xr Å

3+ =

47. Answer (2)Hint: Azimuthal quantum number (l) decides theshape of the orbital.

48. Answer (2)

Hint: 2 2

ee 2

m vnh KZem vr and

2 r r= =

π

Solution: Energy of electron = 2KZe

2r−

Velocity 1

r∝

2πr = nλline spectra are characteristics of atoms

49. Answer (4)Hint: Pauli’s exclusion principle.Solution: An orbital can have maximum twoelectrons with opposite spin.

50. Answer (3)Hint: Aufbau principle/(n + l) ruleSolution:

(n + �) :

Energy

51. Answer (4)Hint: px and py orbitals are mutually perpendicular toeach other.Solution: px and py orbitals are of differentsymmetry hence no mixing occurs resulting information of no molecular orbitals.


9/18

52. Answer (1)

Hint: In XeF4, Xe has two lone pair and four bond pair.

Solution: Electronic arrangement is octahedral

53. Answer (1)

Hint: Acetone is a polar molecule

Solution: Acetone possess sufficient dipole momentwhich gives rise to sufficient attractive forces betweenmolecules and responsible for liquid state of acetone.

54. Answer (2)

Hint: Methane is a covalent compound.

Solution: Most electropositive element is bariumamong all which gives most ionic species

CH4 < CaH2 < BaH2

55. Answer (1)

Hint: Valency of atoms to be satisfied

Solution: 2 22 2CaCN : Ca CN+ −

22CN − ≡ –

N = C = N–

2σ and 2π bonds

CO32– C

O O

O

:

NO3N

O O

O

:

NO : B.O. 3+

=

56. Answer (2)

Hint: Xe is central atom in XeOF2.

57. Answer (1)

Hint: Lower the energy more stable is the ioniccrystal.

Solution: In the formation of ionic crystal followingsteps are involved

0iM(g) M (g) e ; H ve+ −⎯⎯→ + Δ = +

e –X(g) X (g)−+⎯⎯⎯→ ; eg

oH =–veΔ

M+(g) + X–(g) ⎯→ MX; ΔHo

Lattice = –ve

Net effect of above 3 steps must be negative for theformation of stable ionic bond.

58. Answer (4)

Hint: CH4 has tetrahedral shape

Solution: C

H

HH

H

So 6 angles are of 109°28′59. Answer (1)

Hint: Nitrogen can form back bonding.

Solution: N(SiH3)3 contains pπ-dπ back bond.

60. Answer (2)

Hint: Higher is the molecular association, higher willbe the boiling point.

Solution: Due to two –OH groups, ethylene glycolmolecules will be more associated by intermolecularH–bonding.

61. Answer (3)

Hint: For solubility hydration enthalpy should begreater than lattice energy.

Solution: Solubility order is

RbF > KF > NaF > LiF

62. Answer (4)

Hint: Total no. of electrons in a molecule must beeven.

Solution: NO, NO2 and ClO2 contain odd electrons.

O

Cl Cl does not contain odd electrons

63. Answer (3)

Hint: Sulphur in SF4 is sp3d hybridised.

Solution:

S

F

F

F

F

see-saw structure


10/18

64. Answer (3)

Hint: In C2H3Cl, C – Cl bond has double bondcharacter.

Solution:

• In PH5 the overlapping is poor due large variationin size.

• Nitrogen cannot expand its octet.

• LiF possess higher lattice energy than CsI dueto smaller ionic radii of Li+ and F–.

65. Answer (2)

Hint: Generally higher electronegativity differencecauses higher polarity in a bond.

Solution:

2F O 0.3 D,μ = 2H O 1.85 Dμ = ,

3NH 1.47 Dμ =

3NF 0.23 Dμ =

66. Answer (4)

Hint: The number of sigma and pi bond formeddepends on electronic configuration of the species.

Solution: C2 and B2 contain only π-bonds

22 2 2N , C and O− contain both σ and π-bonds

67. Answer (1)

Hint: Higher the bond order of molecule, higher willbe the stability of molecule.

Solution:

N more stable than Ndue to less electronsin ABMO

2 2⇒ N N N2 2 2

B.O = 3 2.5 2.5

⇒ O O O2 2 2

B.O = 2 1.5 2.5

⇒ B B B2 2 2

B.O = 1 1.5 0.5

⇒ C C C2 2 2

B.O = 2 1.5 2.5

+ – + –

+

+

+

–

–

–

68. Answer (3)

Hint: Five or six membered rings are formed inintramolecular H-bonding.

Solution:

O

HO

NO

o-Nitrophenol

CCl C H

ClH

O

Cl OH

Chloral Hydrate

69. Answer (3)

Hint: Molecule having sp3d2 and sp3d hybridizedcentral atom may have 90° bond angle.

Solution:

Species Hybridisation Bond angle(s)

BrF5 sp3d2 No bond angleis of 90° due topresence of lonepair

CO2 sp Bond angle is 180°

SF6 sp3d2 90° bond angles arepresent

[ ]4PBr⊕

sp3 Bond angles are of

109° 28′

70. Answer (4)

Hint: Species having same number of valenceelectrons are isoelectronic

Solution:

SF4 : Hybridised state of central atom = sp3d

(See-Saw shape)

CIF4

–: Hybridised state of central atom = sp3d2

(Square planar shape)

SO3 : Hybridised state of central atom = sp2

(Triangular planar shape)

ClO3 : Hybridised state of central atom = sp3

(Shallow pyramidal shape)


11/18

XeO4 : Hybridised state of central atom = sp3

(Tetrahedral shape)

34PO − : Hybridised state of central atom = sp3

(Tetrahedral shape)

ICl2– : Hybridised state of central atom = sp3d

(Linear shape)

XeF2 : Hybridised state of central atom = sp3d

(Linear shape)

71. Answer (2)

Hint: Chlorine has maximum negative electron gainenthalpy.

Solution: N is more electronegative than P.

72. Answer (1)

Hint: s-block and p-block elements are also calledrepresentative elements

Solution: Ni(Z = 28) is a d-block element.

73. Answer (2)

Hint:

Total energy = Sum of energies of eachrequired per mole ionisation.

Solution: 4 1 2 3 4Si/SiE (x x x x ) kJ/mole+ = + + +

Si

280n 10 moles

28= =

Energy required = 10(x1 + x2 + x3 + x4) kJ

74. Answer (1)

Hint: Oxides which reacts with both acid and base.

Solution:

Amphoteric oxides : ZnO, PbO, SnO, Al2O3, Sb2O3

Acidic oxides : Cl2O7, SO3, NO2

75. Answer (2)

Hint: Electronic repulsion can make a processendothermic

Solution: H+ ⎯→ H (exothermic)

O– ⎯→ O2– (endothermic)

C ⎯→ C– (exothermic)

S ⎯→ S– (exothermic)

76. Answer (3)

Hint: Aufbau and (n + l) rule majorly decideselectronic configuration of element.

Solution:

Group-15 and period-6 : [ ] 14 10 2 3

54Xe 4 5 6 6f d s p

Group-17 and period-7 : [Rn]86 5f 14 6d10 7s2 7p5

Group-10 and period-5 : [Kr]36 4d10 5s0 [Pd]

Group-12 and period-6 : [ ] 14 10 2

54Xe 4 5 6f d s

77. Answer (2)

Hint: Electronic configuration and nuclear chargegoverns most of the properties.

Solution: 2 1Ca K Ar Radii

ve charge+ + ⎛ ⎞

< < ∝⎜ ⎟+⎝ ⎠

B < C < N [(I.E1)C > (I.E1)B]

I < Br < Cl [increasing order of ΔegH]

Li < Na < K [Radii ∝ n]

78. Answer (2)

Hint: Write electronic configuration of each species.

Solution:

Ni3+ : [Ar] 3d7 4s0 ;

Cr : [Ar] 3d5 4s1 ;

Fe : [Ar] 3d6 4s2 ;

Mn3+ : [Ar] 3d4 4s0 ;

79. Answer (4)

Hint: h

mvλ =

Solution: 34

363 3

6.6 105 10 m

132 10 10

−−

−

×λ = = ×× ×

80. Answer (1)

Hint: 2

H 2 21 2

1 1 1R Z

n n

⎡ ⎤= × −⎢ ⎥λ ⎣ ⎦

Solution: 21Z∝

λ ; higher the value of ‘Z’, smaller

will be the wavelength.


12/18

81. Answer (1)

Hint: dxy, dyz and dzx are non-axial orbitals.

Solution: 2 2Z x –y,p d and 2Z

d orbitals contain lobes

along the axes hence electron density is distributedalong the axes.

pZ dZ2 2 2x –yd

z z

y

x

82. Answer (4)

Hint: Orbital angular momentum = h

( 1)2

+π

� �

Solution: ∵ = 0� for s-orbital

∴ Orbital angular momentum for s-electron = 0

83. Answer (1)

Hint: Maximum two electrons can beaccommodated in an orbital.

Solution: According to Pauli’s exclusion principle“No two electrons in an atom can have the same setof four quantum numbers”

84. Answer (1)

Hint: Half filled and fully filled orbitals are morestable.

Solution: For one electron species, in its groundstate, all the orbitals of same shells remaindegenerated due to unavailability of electrons in theseshells.

2s = 2p < 3s = 3p = 3d (energy order).

• The uncertainty principle is given by

hE t

4Δ × Δ ≥

π

•2

2

ZE 2.178

n= − × gives energy of nth orbit.

85. Answer (2)

Hint: 1 mole CaCO3 gives 1 mole CO2 on completedecomposition.

Solution: CaCO3 → CaO + CO2

2 3CO CaCO

5.6n n 0.25 mole

22.4= = =

3CaCOw 0.25 100 25 g= × =

% purity = 25

100150

× = 16.67%

86. Answer (1)

Hint: m

dv

=

Solution: Density of ethanol = 0.8 g/cm3

Mass of 1 molecule of ethanol = 23

46g

6.022 10×

23 323

m 46 1v 9.55 10 cm

d 6.022 10 0.8−= = × = ×

×

87. Answer (4)

Hint: Na2CO3 + 2AgNO3 → 2NaNO3 + Ag2CO3 ↓

Solution: 2 3

21.2 50For Na CO : w 10.6 g

100×= =

2 3Na CO

10.6n 0.1mole

106= =

For AgNO3 : 17 100

w 17 g100×= =

3AgNO

17n 0.1mole

170= =

Limiting reagent is AgNO3

2 mole AgNO3 gives 1 mole Ag2CO3 so 0.1 moleAgNO3 gives 0.05 mole Ag2CO3

2 3Ag COw 276 0.05 13.8 g= × =

88. Answer (3)

Hint: A tetra-atomic gas molecule contains 4 atoms

Solution: 1 mole tetra-atomic gas = 4 moles ofatoms

0.25 mole tetra-atomic gas = 4 × 0.25 = 1 mole atom

= NA atoms

89. Answer (3)

Hint: Molarity = No.of molesof solute

Volume(litre)

Solution:w 1000

MolarityM V(ml)

= ×


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w 10001

98 500= × ⇒ w = 49 g

4980% 100

W= ×

49W 100 61.25 g

80= × =

90. Answer (4)

Hint: 6.022 × 1023 (NA) = 1 mole

Solution: Change in quantity of 1 mole will lead tothe change in weight of 1 g-atom of correspondingspecies.

BIOLOGY91. Answer (4)

Hint : Defining features are those features which arenecessarily present in all living organisms.Sol.: Sexual reproduction is shown by manyorganisms but it is absent in certain lower organismsand infertile higher organisms so it cannot be adefining feature of living organisms.

92. Answer (2)Hint : It is an index of plant species found in aparticular area.Sol.: Flora is a book containing information aboutplants found in particular area. It gives actualaccount of habitat and distribution.

93. Answer (1)Hint : (a) Chiasmata formation occurs at diplotenestage of prophase-I.Sol.:(b) Alignment of univalents - Metaphase-II

on equatorial plate(c) Chromatids reach - Telophase-II

the pole(d) Initiation of reduction - Anaphase-I

in number ofchromosomes

Thus, correct sequence is : a → d → b → c.94. Answer (2)

Hint : Direct cell divison mostly occurs in bacteria,protozoan etc.Sol.: Amitosis or direct cell division lacks spindleformation. In amitosis, karyokinesis occurs by asimple constriction.

95. Answer (3)Hint : Synaptonemal complex stabilises the pair ofsynapsed chromosomes or bivalent.

Sol.: This complex is formed at zygotene stage anddissolves at diplotene stage. At diplotene thebivalent start separating.

96. Answer (1)

Hint : This stage includes spireme stage.Sol.: At early prophase, ends of chromosome arenot visible and they appear like a ‘ball of wool’.

97. Answer (3)

Hint : Syncytium shows multinucleate condition.

Sol.: If karyokinesis is not followed by cytokinesisit leads to formation of a single cell with large numberof nuclei.

98. Answer (2)

Hint : Gap2 phase is followed by S-phase.Sol.: Synthesis of RNA & proteins occur in both G1& G2 stage. Duplication of single membrane boundcell organelles also take place in both G1 & G2. Butduplication of double membrane bound cellorganelles like mitochondria and chloroplast occur inG2 phase.

99. Answer (2)Hint : Endocytosis is intake of materials in the formof carrier vesicles formed by invagination of smallregions of plasma membrane.

Sol.: Functions like cell growth, formation ofintercellular junctions, secretion etc. are due to fluidnature of plasma membrane.

100. Answer (1)

Hint : These cells form ground tissue of leaves ofgreen plants.

Sol.: Chloroplasts are mainly present in themesophyll cells of leaves of green plants.

101. Answer (2)Hint : It is a major component of the cell wall offungi.

Sol.: Chitin is not found in algal cell walls.

102. Answer (3)

Hint : Chromosomes which have centromere veryclose to its one end, are J-shaped chromosomes.

Sol.: Chromosomes which have their centromerevery close to one end are called acrocentricchromosomes.

103. Answer (1)

Hint : ER which is free from ribosomes, is abundantin muscle cells.

Sol. : Smooth ER is associated with the functionslike detoxification of drugs, uptake and release ofCa2+ ions during muscle contraction and passing ofproducts of RER to golgi apparatus.Synthesis of enzyme precursors for lysosomes, is afunction of RER.


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104. Answer (4)

Hint : A membraneless organelle which is present inanimal cell, but absent in higher plants is centriole.

Sol.: Centrioles have 9 + 0 arrangement ofmicrotubules and they are involved in spindleformation during animal cell division.

105. Answer (1)

Hint : These bacteria appear purple in colour afterstaining.

Sol. : Gram positive bacteria have thick cell wall,retain the stain and due to which they appear purplein colour.

106. Answer (2)

Hint : This mitochondrial structure is rich incardiolipins.

Sol. : Inner membrane of mitochondria containsenzymes and electron carriers for the formation ofATP.

107. Answer (2)

Hint : These are rich in hydrolytic enzymes whichget activated at acidic pH.

Sol.: Lysosomes are the single membrane bound,polymorphic cell organelles which are associatedwith intracellular digestion of food materials. Lysis ofold or dead organelles and ephagy of undigestedsubstances are also carried out by lysosomes. Butconversion of fats into carbohydrates is a function ofglyoxysomes.

108. Answer (3)

Hint : ER plays a major role in origin of thestructure which connects the cytoplasm ofneighbouring plant cells.

Sol. : A - Plasmodesmata

B - Plasma membrane

C - Desmotubule

109. Answer (4)

Hint : Taxonomy is the branch of science whichdeals with study of principles and procedures ofclassification.

Sol. : Modern taxonomy is based upon study ofboth external and internal features of organism alongwith cell structure, developmental process andecological informations of organisms. But phylogenyis included in systematics.

110. Answer (2)

Hint : A scientific name has two words in binomialnomenclature system.

Sol. : In binomial nomenclature, first word of thename is genus and second word is specific epithetScientific names are derived from Latin language andprinted in italics.

111. Answer (3)Hint : Class is an obligate taxonomic categorywhich is lower than phylum or division in hierarchy.Sol. :Mammalia, Dicotyledonae, Insecta - ClassesFelis, Mangifera - GeneraPlantae - KingdomPoales, Polymoniales, Diptera - OrderArthropoda - PhylumHominidae, Solanaceae - Families

112. Answer (4)Hint : Mitosis occurs in somatic cells and fewhaploid cells of lower organisms.Sol. : Growth, repair and regeneration of body partsoccur through mitosis. But conservation ormaintenance of number of chromosomes throughoutthe generations occur via meiosis.

113. Answer (3)Hint : A bivalent is a pair of homologouschromosomes.12 bivalents = 24 chromosomes. Hence in gameteit will be 12 chromosomes.Sol. : One chromosome has one chromatid ingamete. So, total 12 chromosomes & 12chromatids.

114. Answer (1)Hint : Crossing over occurs only once duringreductional division.Sol. : Recombination or crossing over betweenhomologous chromosomes occurs once in meiosis.Duplication of centrioles, karyokinesis andmetaphasic plate formation occur twice in the wholemeiosis.

115. Answer (2)Hint : Initiation of reduction in the number ofchromosomes occurs at anaphase I.Sol. : In Anaphase-I, the homologous chromosomesget separated and move towards the opposite polesof the cell, but their sister chromatids remainattached to each other.

116. Answer (3)Hint : Meiocyte is a diploid cell that undergoesmeiosis to produce haploid cells.Sol. : Due to DNA replication the amount of DNA isdoubled at G2 phase. A meiocyte at G2 phase hasfour times higher amount of DNA than its gametes.

117. Answer (4)Hint : Mitogens are substances which inducemitosis.Sol. : Insulin is a mitogen, while cyanide, mustardgas and chalones are mitotic poisons which inhibitmitosis.


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118. Answer (4)

Hint : Chiasmata disappear and homologouschromosomes separate at diakinesis stage.

Sol. : In prophase I, pairing of homologouschromosomes occurs at zygotene, formation ofrecombination nodule occurs at pachytene,separation of homologous chromosomes(desynapsis) occur at diplotene and terminalisationof chiasmata occurs at diakinesis stage.

119. Answer (2)

Hint : In this stage, splitting of centromere andseparation of sister chromatids occurs.

Sol. : This is anaphase stage of mitosis. Atanaphase, chromatids move towards the poles andthe arms of chromosomes trail behind. This is thebest stage to study shape of chromosomes.

120. Answer (1)

Hint : Quiescent stage is resting stage of nondividing cells.

Sol. : Non-dividing cells enter the quiescent phaseor G0 phase of cell cycle where they remainmetabolically active but do not proliferate.

121. Answer (4)Sol. : Small fragments which are formed beyondthe secondary constriction are called satellite.

122. Answer (4)Sol. : Eukaryotic genomic DNA is double strandedand linear.

123. Answer (3)Hint : Eukaryotic flagella have microtubule doubletsat periphery and a pair of microtubules in center.Sol. : Central tubules are covered by central sheathand the entire axoneme is covered by plasmamembrane.Formation of spindle fibres during cell division is afunction of centrioles.

124. Answer (1)Hint : One form of plastid can be converted intoanother form according to the requirement.Sol. : During the ripening of tomato its colourchanges from green to reddish due to transformationof chloroplasts into chromoplasts.

125. Answer (3)Hint : At interphase, nucleus contains looseextended & diffused network of nucleoprotein fibres,called chromatin.Sol. : Term chromatin was given by Flemming andit can be stained by basic dyes. It is composed ofDNA and histone proteins. Euchromatin is looselypacked part of chromatin which is transcriptionallyactive.

126. Answer (4)Hint : It is the site of synthesis of one of thecomponents of ribosomes.Sol. : Nucleolus is non-membrane bound structureand is the site for rRNA synthesis.

127. Answer (2)Hint : This giant chromosome was discovered byRuckert.Sol. : Lampbrush chromosomes are diplotenebivalents and are found in the primary oocyte nucleiof some vertebrates and invertebrates. They haveloops which participate in transcription & RNAproduced from them could be stored in the form ofinformosomes(mRNA + proteins).

128. Answer (3)Hint : Electron transport system is present in innermitochondrial membrane and in thylakoid membraneof chloroplast.Sol. : Both mitochondria and chloroplasts have ATPsynthesising machinery and presence of porinproteins in their outer membrane, but machinery foraerobic respiration is found only in mitochondria.

129. Answer (1)Sol. : Movement of molecules against theirconcentration gradient is called active transport. Inthis process energy is required in the form of ATP.

130. Answer (4)

Hint : Digestive vacuoles are secondary lysosomesalso called hetrophagosomes.

Sol. : Secondary lysosomes are formed by fusion ofprimary lysosomes with phagosomes. Digestivevacuoles contain active enzymes against the materialto be digested.

131. Answer (2)

Hint : ER and Golgi apparatus both are parts ofendomembrane system.

Sol. : Both ER and Golgi are structurally composedof cisternae, vesicles and tubules. Synthesis of lipidsand steroid hormones, is an exclusive feature ofSER while synthesis of plasma membrane duringcytokinesis is a function associated with Golgi.

132. Answer (3)

Hint : Ribosomes are membrane-less organelles.

Sol. : Ribosomes were discovered by GeorgePalade. Two ribosomal subunits remain attached toeach other at optimum concentration of Mg2+ ions.ER is absent in prokaryotes.

133. Answer (2)

Hint : An elaborate network of proteinaceousfilaments present in eukaryotic cells is calledcytoskeleton.


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Sol. : The cytoskeleton helps in providingmechanical support to the cell, in formation offlagella and maintain the shape of the cell. Whereas,chloroplasts are involved in carbon assimilation.

134. Answer (4)

Hint : It is a semi-fluid matrix present in all cells.

Sol. : Main arena of cellular activities where allmetabolic reactions take place is cytoplasm in boththe plant and animal cells.

135. Answer (3)

Hint : Extensive compartmentalisation of cytoplasmis present in eukaryotic cells.

Sol. : Presence of plastids, flagella (have 9 + 2arrangement of microtubules) and nuclear envelopeare eukaryotic features.

Mesosomes are infoldings of cell membrane and arecharacterstic of prokaryotes.

136. Answer (2)

Hint : Adhering junctions help to keep adjacent cellstogether during stress.

Sol. : Macula adherens/desmosomes prevent cellsfrom being pulled apart. Tight junctions preventleakage of substances and are also called zonulaoccludens. Intermediate junctions are also known aszonula adherens, usually occur below tight junctions.

137. Answer (2)

Hint : Brown fat is thermogenic.

Sol. : Brown fat is known for producing heat. Whitefat is used as storehouse of energy, food reserve,insulation etc. Mast cells release heparin, serotoninand histamine upon activation. Collagen fibres arewhite and unbranched.

138. Answer (2)

Hint : This bone is also called knee cap.

Sol. : Patella is a sesamoid bone while other boneslisted are cartilaginous.

139. Answer (4)

Hint : Muscles and bones are vascular tissues.

Sol. : Epithelial tissue is avascular as it lacks bloodsupply usually. Biceps are part of muscular tissue.

140. Answer (4)

Hint : Elastic cartilage is associated with tip ofnose.

Sol. : Haversian lamellae occur in diaphysis part oflong bones of mammals. Pliable matrix is a featureof cartilage. Chondroblasts and chondrocytes arelocated within lacuna.

141. Answer (1)

Hint : Striped muscles are associated with skeletalsystem.

Sol. : Limb bones such as humerus and femur areassociated with voluntary, cylindrical muscle fibrescalled skeletal muscle fibres. Smooth, non-striatedmuscle fibres are present in wall of intestine whereasstriated, cardiac muscle fibres are associated withheart.

142. Answer (2)

Hint : Uninucleated muscle fibres lines the wall ofstomach.

Sol. : Wall of internal organ stomach is lined byvisceral/smooth unstriped muscles. Actin and myosinfibrils occur in all type of muscle cells. Both skeletaland smooth muscle fibres are unbranched.

143. Answer (3)

Hint : Property to decrease in size is associatedwith muscular tissue.

Sol. : Property of excitability and conductivity iscommon to muscular and neural tissue.

144. Answer (3)

Hint : All cells of neural tissue are ectodermal inorigin except microglial cells.

Sol. : Astrocytes are ectodermally derivedPerineurium forms a protective white sheath arounda single fasciculus. Centrosome is absent in neuronshence they cannot divide.

145. Answer (1)

Hint : Cell membrane of axon is called axolemma.

Sol. : Nodes of Ranvier in neurons of PNS haveneurilemma. Oligodendrocytes form myelin sheath inCNS. Saltatory conduction occurs at nodes ofRanvier.

146. Answer (3)

Hint : Identify epithelium with tall cells and basalnuclei.

Sol. : Squamous epithelium is suitable for surfaceswhere filtration and diffusion are necessary. Ciliapush the ova towards ampulla of fallopian tube whilemucus is pushed upwards in trachea. Keratinisedepithelium features on surfaces that requireprotection from water.

147. Answer (2)

Hint : Porifera show cell aggregate body plan andtissues are lacking.

Sol. : Members of Cnidaria/Coelenterata exhibittissue level of organisation for the first time inkingdom Animalia. Platyhelminthes andAschelminthes exhibit, organ and organ system levelof organisation respectively.

148. Answer (1)

Hint : Macromolecules are obtained in acidinsoluble fraction.


17/18

Sol. : Insulin is a heteropolymer while cholesterolforms aggregates in vesicles. Both are obtained inacid insoluble fraction/retentate fraction. Proteins arelarge in size hence they are macromolecules andlipid like cholesterol does not dissolve in acid.

149. Answer (3)

Hint : Functional groups in a sugar are hydroxyland aldehyde.Sol. : Amino acid is identified by the position of ‘R’chain, amino and carboxyl group w.r.t. C∝ position.Sugar molecules have aldehyde or ketone groups.

150. Answer (1)Hint : Glycerol is trihydroxypropane.

Sol. : Adenine is derived from substitution of purinering at 6th carbon while cytosine is a pyrimidineCholesterol is a sterol.

151. Answer (3)

Hint : Proteins are polymers of different repeatingunits.

Sol. : Structural homopolymers include celluloseand chitin. Inulin and glycogen are storagehomopolymers whereas collagen and RuBisCO areheteropolymers.

152. Answer (3)Hint : Formation of glycosidic, peptide and esterbond requires loss of water molecules.

Sol. : Glycosidic bond formed by loss of water is adehydration/condensation reaction. Alphabet ‘P’represents amino acid Proline. A protein attains itsfunction by formation of its active site at tertiary levelof organisation.

153. Answer (3)

Hint : Chargaff’s rule.

Sol. : [A] = [T] and [G] = [C]

Therefore, [A] = 120 and [G] = 120

[T] = 120 and [C] = 120

Sum total of nucleotides is 480.154. Answer (1)

Hint : 10 bp occupy a distance of about 34 Å inB-DNA.

Sol. : Base pairs in DNA are stacked 3.4 Å apart.In B-DNA, diameter of helix is 20 Å and pitch of thehelix is 34 Å. Hydrogen bonds are absent in a singlenucleotide. A and G of one strand base pair with Tand C of other strand respectively.

155. Answer (1)

Hint : Hydrogen bonds link purines and pyrimidineson opposite strands.

Sol. : Guanine and cytosine are linked through 3hydrogen bonds. AT base pair will have 2 hydrogenbonds.

156. Answer (2)Hint : Uracil is a nitrogenous base exclusively foundin RNA.Sol. : Thiamine is vitamin B1. Cysteine is an aminoacid while cytosine is a base.

157. Answer (2)Hint : Deoxyribose sugar is found in DNA.Sol. : Niacin i.e. vitamin B3 forms NAD and NADPwhich act as coenzymes.

158. Answer (3)Hint : The activity of enzyme reaches a plateauonce all the active sites of enzyme are saturatedwith the substrate.Sol. : A rectangular hyperbola curve is attainedwhen [S] is plotted against enzyme activity.

159. Answer (2)Hint : Some nucleic acids can also act as enzymes.Sol. : Ribonuclease P is RNA acting as enzymewhose substrate is also RNA. It was discovered byAltmann from E.coli.

160. Answer (4)Hint : Structural similarity between inhibitor andsubstrate is not the criteria for non competitiveinhibition enzyme inhibition.Sol. : Ethanol, malonate and sulfonamides arecompetitive inhibitors of methanol, succinate andPABA respectively.

161. Answer (1)Hint : Catalase a, holoenzyme comprises ofapoenzyme and cofactor.Sol. : Apoenzyme is proteinaceous in nature whileits cofactor is organic & non proteinaceous prostheticgroup called haem which is tightly attached to it.

162. Answer (2)Hint : pH 7.8 is present in small intestine.Sol. : Proteases hydrolysing peptide bonds ofproteins acting at pH 7.8 include trypsin,carboxypeptidase and chymotrypsin whosezymogenic form are secretions of pancreas.

163. Answer (3)Hint : Vomiting centre is located in brain stem.Sol. : Vomiting is a reflex action controlled by thevomiting centre located in medulla oblongata.

164. Answer (2)Hint : Muscularis layer is found inner to serosa.Sol. : Innermost lining is mucosa followed by sub-mucosa, inner circular and outer longitudinal musclelayer and outermost serosa layer.

165. Answer (2)Hint : Enzymes are absent in bile juice.Sol. : Bile juice is stored temporarily in gall bladder.Lipase is a secretion mainly from pancreas. Cysticduct is the term for duct arising from gall bladder.


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� � �

166. Answer (1)Hint : Aggregates of bile salts in enteric lumenpromote fat absorption by emulsification.Sol. : Chylomicrons are formed inside enterocytes.

167. Answer (2)Hint : Deficiency of HCl in human body isachlorhydria that would imply impaired functioning ofparietal cells.Sol. : Deficiency of HCl will affect iron absorptionleading to anaemia. Rennin is a hydrolytic enzyme,therefore, belongs to hydrolase category of enzymes.Parasympathetic stimulus promotes activity ofBrunner’s glands.

168. Answer (1)Hint : Identify a food source rich in cellulose.Sol. : Lettuce is rich in cellulose which cannot bedigested by humans. Hence its calorific value isnearly ‘zero’. Egg white and meat produce grosscalorific value around 5.65 kcal/g while bread yields4 kcal/g.

169. Answer (1)Hint : Hormone whose secretion promotes gastricjuicesSol. : Enterogastrone, secretin & duocrinin arereleased by cells in small intestine while gastrin isa secretion of G cells in stomach. Enterogastroneand secretin both decrease gastric secretions.Duocrinin stimulates Brunner’s glands.

170. Answer (1)Hint : Identify the disorder in which both protein andenergy malnutrition occur in infants less than1 year in age.Sol. : Kwashiorkor results from deficiency ofproteins only in children above 1 year. Scurvy resultsfrom vitamin C deficiency. Anaemia does not occurdue to deficiency of riboflavin and thiamine.

171. Answer (3)Hint : Competitive inhibitor competes with thesubstrate for the active site.Sol. : Non-competitive inhibitor binds to the enzymeirreversibly, decreases the Vmax of the reactionwithout affecting the Km for a given substrate.

172. Answer (1)Hint : Bile does not contain any enzyme.Sol. : Hepatocytes secrete bile that has bilirubin,biliverdin (bile pigments) and bile salts but nodigestive enzymes. Enterocytes secrete digestiveenzymes and parietal cells (oxyntic cells) releasehydrochloric acid to absorb iron.

173. Answer (3)Hint : Action of these enzymes mostly results inproduction of simple absorbable forms.

Sol. : Enterokinase converts trypsinogen totrypsin. Pepsin and rennin are secreted aszymogens in stomach. Enterocrinin is a hormone.Carboxypeptidase, lipase (steapsin) and elastase aresecreted by pancreas.

174. Answer (2)Hint : Identify the dental formula of milk teeth.Sol. : At age 10, premolars and wisdom teeth (lastmolar) are absent. Diphyodont dentition is seen.

21232123

is dental formula of adult human

175. Answer (4)Hint : Sphincter that guards opening ofhepatopancreatic duct.Sol. : Sphincter of Boyden guards opening ofcommon bile duct. Ileocaecal valve prevents backflow of digested food from caecum to ileum. Pyloricsphincter is present at the junction of pyloric end ofstomach and duodenum

176. Answer (3)Hint : Cells involved in defence activity in alimentarycanal.Sol. : Parietal cells in gastric mucosa secrete HCland Castle’s intrinsic factor. Kupffer cells aremacrophages of liver and argentaffin cells secreteprecursor molecules like 5-OH-tryptamine.

177. Answer (1)Hint : Myenteric plexus is located in muscularisexterna.Sol. : Neurons in alimentary canal form Myentericplexus, a motor plexus responsible for peristalsis.Meissner’s plexus located in submucosa isresponsible for neural activity resulting in release ofenzymes and mucus.

178. Answer (2)Hint : Carbohydrases digest carbohydrates such asstarch.Sol. : Glandular cells of stomach secrete proteasesand acid. Carbohydrases are secreted in mouth,pancreas and small intestine.

179. Answer (3)Hint : Middle part of small intestine.Sol. : Nearly 5500 ml of water each day isabsorbed across jejunum in a man. 1300 ml of wateris absorbed across colon and 200 ml across ileum.

180. Answer (2)Hint : Alcohol is already a simple absorbable form.Sol. : Chips, burger and fish are complex foods thatneed to be digested in alimentary canal and aremaximally absorbed in small intestine.Wine contains alcohol which gets absorbed acrossthe gastric mucosa.

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