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ANSWERS

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170. (3)

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172. (1)

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174. (2)

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176. (3)

177. (4)

178. (4)

179. (3)

180. (4)

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

MM : 720 Mock Test for NEET - 2020 Time : 3 Hrs.

Home Assignment – 8

08/07/2020

(2)

Home Assignment-8 (Answers and Hints) Mock Test for NEET - 2020

Answers and Hints

1. Answer (1)

t sec

–4

6v

3 5

1 12 1 4 2

2 2 S

S = 5m

2. Answer (4)

x

dxv t

dt

3

2

y

dy dy dx tv

dt dx dt

At t = 2

2 4 � � �v i j m/s

3. Answer (3)

Time of flight = 3 + 7

= 10 sec

90 = ucos × 4 ...(i)

R = ucos × 10 ...(ii)

90 4

10

R

R = 225 m

4. Answer (4)

For minimum time swimmer should swinperpendicular to river.

5. Answer (2)

Mg = Fsin + Fcos

sin cos

Mg

F

6. Answer (4)

Area of F – t curve = p

O = p2 – p

1

p2 = p

1 = 20 kg m/s

7. Answer (4)

Work done by gravity is zero during total motion ofprojectile so it average power will also zero.

8. Answer (1)

22

0

1 11.2 25

2 2 K I t

2

0

11500 1.2 625

2 t

t0 = 2 sec

9. Answer (2)

4y – 3x = 2

3 1

4 2 y x

= 37°

37°y

x

5 m/s

13 5cos37

2 L

= 4 1

3 55 2

= 6 kgm2s–1

10. Answer (4)

W = Uf – U

i

=

2 23

32

Gm Gm

a a

=

23

2

Gm

a

11. Answer (2)

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

MM : 720 Mock Test for NEET - 2020 Time : 3 Hrs.

Home Assignment – 8

08/07/2020

(3)

Mock Test for NEET - 2020 Home Assignment-8 (Answers and Hints)

12. Answer (1)

P

v

CQ

U C

70 7

5

U U = 50 cal

13. Answer (1)

14. Answer (3)

80 60 80 6030

10 2

K

60 6030

10 2

T TK

T = 48°C

15. Answer (2)

1 s l

n l n l

n × 10 = (n – 1)12

10n = 12n – 12

n = 6

16. Answer (4)

y = acos(kx – t) – acos(kx + t)

y = 2asinkxsint

17. Answer (2)

1.21Å2 2

= 1.21 Å

18. Answer (2)

(3f)P (3f)P (3f)S(2f)

19. Answer (2)

1

12

6 I = 2Amp

2

12

12I = 1Amp

2 4AI1

I2

B

39

12 v

VA – 0 = 8

VB – 0 = 3

VA – V

B = 8 – 3 = 5 volt

20. Answer (4)

Current in B4 bulb is maximum so it glow more

brightly.

21. Answer (2)

4 41

100 100

IG I S

24 G

S ohm

22. Answer (2)

2. . �

dBE dl r

dt

E r

23. Answer (1)

110cos100

600 v

10cos6

v

= 5 3 volt

24. Answer (3)

2 2

1 1 1

1 3

R

8

9 h

P Rh

25. Answer (3)

2

2 2

1 132.4 13.6

3 4

z

z = 7

26. Answer (2)

BE = m x 931 Mev

27. Answer (3)

28. Answer (3)

29. Answer (2)

30. Answer (4)

0 0 0aEM L T

t

2 2

0 0 0

a ML T

M L TT

a = (M–1L–2T3) = b

31. Answer (3)

mu

2m mv1

=

2mv2

mu = mv1 + 2mv

2

u = v1 + 2v

2...(i)

v2 – v

1 = u ...(ii)

(i) & (ii)

3v2 = 2u

2 2

2 1,

3 3 v u v u

% k =

2

2

1 8

2 9100 88.8%

1

2

m u

mu

(4)


32. Answer (1)

v0

v

mv0r = mvr +

22

5

vmr

r

0

5

7v v

33. Answer (4)

2 2

2

3 3 ml ml

I ml

25

3I ml

34. Answer (1)

80 20 ( 60)

2 3

l l l

KA KA KA

2(80 –) = ( – 20) + 3( – 60)

160 – 2 = – 20 + 3 – 180

= 60°

35. Answer (2)

8 dvE x

dx

8 �

�E i

36. Answer (3)

8 3 11 net

R

4 123

11 11

ABV

1250 0.5

11 100

E eV

37. Answer (1)

0 02.

4 4

IIB

R R

38. Answer (4)

5' 5

1

I mA

k

5 = 5 × 10–3 × 5

S = 1 k

39. Answer (2)

7

x

x + 7 = 25, x = 18 cm

40. Answer (2)

sin2

cot2

cos2

mA

A

A

m = (180 – 2A)

41. Answer (3)

10 cm

v = 60 cm

x

1 1 1 f v u

1 1 1

20 30 v

v = 60 cm

x = 60 – 10 = 50 cm

42. Answer (2)

10 –(10)4500 10 6000 10

hc hc

KE

= 0.63 eV

43. Answer (4)

E = 1.5 – (–13.6)eV

44. Answer (4)

It is equal to time of half life.

45. Answer (4)

80 c

b

I

I

Ie

– Ib + I

c

Ib = 0.1 mA

46. Answer (3)

(CH ) – C – OH3 3

(CH ) C– O Na3 3

+

(CH ) – C – O – CH CH 3 3 2 3

Na

(A)

C H Br2 5

(B

47. Answer (3)

Fact

48. Answer (2)

In gas phase, +I effect decides the basicity while inaqueous solution, it is the combined effect of +Ieffect, hydration and steric hindrance.

(5)


49. Answer (4)

Epimess differ in the configuration about anyunsymmetrical carbon other than anomeric carbon.

50. Answer (3)

In anionic detergent, the negative part of themolecule is responsible for cleansing action.

51. Answer (4)

P1 = 715 – 15 = 700 mm Hg, V

1 = 30 ml,

T1 = 300K

STP STP 1 1

STP 1

P V P V

T T

STP

700 30 273V 25.14 ml

300 760

2

25.14W(N ) 28 0.0314g

22400

0.0314

% nitrogen 100 6.28%0.5

52. Answer (4)

In (1), it is syn-addition to form glycol.

In Answer (2), oxidation takes place to formdicarboxylic acid.

53. Answer (1)

2 2(excess) Potassium superoxide

K O KO

Superoxide ion is paramagnetic and coloured.

54. Answer (4)

Orthoboric acid (H3BO

3) is formed which is

polymeric due to hydrogen bonding.

55. Answer (1)

3KE nRT

2

= 0

3 WRT

2 M

0 3 W R TM

2 KE

3 8 2 30016 g (Approx)

2 447

The gas is CH4

56. Answer (1)

Since a bomb calorimeter has constant volume,

So w = 0, (w = – PV)

Zn(s) + 2HCl(aq) ZnCl2(aq) + H

2(g)

Since heat is evolved in the above reaction,

so U < O.

57. Answer (2)

PCl3Br

2 is unsymmetrical with net dipole moment

while PBr3Cl

2 is symmetrical with zero dipole

moment.

58. Answer (2)

Fact

59. Answer (4)

2

2paschen

1 1R 3 0

3

paschen

1

R

2

2 2visible

1 1 1 5RR 2

92 3

visible

9

5R

paschen

visible

5

9

60. Answer (3)

2 2H (g) Br (g) 2HBr(g) ��⇀

↽��

Initial : 15ml 20ml 0

Final : 5ml 10ml 20

20 20K 8

5 10

for the reaction :

2 22HBr(g) H (g) Br (g)��⇀

↽��

Equilibrium constant = 1

0.1258

61. Answer (3)

Because H3PO

4 does not have P–H bond.

62. Answer (3)

Fact

63. Answer (2)

The corresponding elements of second and thirdtransition series have almost similar atomic radii.

64. Answer (3)

Fact

65. Answer (4)

Fact

66. Answer (1)

Fact

67. Answer (1)

Higher the effective concentration, higher is theboiling point.

(6)


68. Answer (3)

m

K1000

C

232.3 0.05K

1000

= 1.162 ×10–2 –1 cm–1

Conductance = K

cell cons tant

= 2

1.162 10

0.367

= 0.0316 –1

69. Answer (4)

For the same oxidation number, the strength ofoxoacids of halogen decreases with the increase inthe size of the halogen.

70. Answer (3)

[Co(NH3)3(NO

2)3]

The complex doesnot show optional isomerism.

71. Answer (2)

Cr2O

7

2– + 3Sn+2 + 14H+ 2Cr+3 + 3Sn+4 + 7H2O

3 moles of Sn+2 requires = 1 mole of Cr2O

7

2–

0.03 moles of Sn+2 requires = 0.03 moles of Cr2O

7

2–

72. Answer (2)

Fact

73. Answer (3)

x 1log logP logK (n 1)

m n

1

Slopen

1

O 1n

74. Answer (4)

Fact

75. Answer (3)

For bcc metallic structure,

a 3R

4

4 1.86a

1.732

= 4.3 Å

d = 2R

= 2 × 1.86

= 3.72Å

76. Answer (3)

Ag Ag

Al Al

w E

w E

Ag

108w 4.5

9

= 54g

77. Answer (4)

Molecularity is defined for only elementary reactions

78. Answer (3)

(B) is B2O

3

(c) is Ni(BO2)2 brown

79. Answer (4)

Fact

80. Answer (3)

Perhydrol is H2O

2

81. Answer (1)

Due to maximum hyperconjugation.

82. Answer (3)

Eclipsed has higher energy than staggered due totorsional strain.

83. Answer (3)

Among Rx, C–Cl bond has highest dipole momentwhile R–I has maximum boiling point for the givenalkyl group.

84. Answer (3)

(CH ) CH – OH3 2

(CH ) CH – I3 2

(CH ) CH – NO3 2 2

(CH ) C – NO3 2 2

P + I2

AgNO2

HNO2

N = 0

85. Answer (2)

(A) is CH3COOC

2H

5

(B) is CH – C – OH3

CH3

CH3

86. Answer (2)

N = N

Azobenzene

(7)


87. Answer (2)

Electron given enthalpy of fluorine is negative.

88. Answer (3)

Due to resonance, bond order is 4/3.

N

O

OO

]–

89. Answer (4)

Common ion effect decreases the solubility. Due toreaction with ammonia, solubility in it is enhanced.

90. Answer (3)

N + 3H 2NH2 2 3

�

1 mol 3 mol 0

1

2–3

23

1–1 mole

Initial

Final

Moles of H2 =

3

2

Weight of H2 =

3

2× 2 = 3g

91. Answer (3)

Used for plants and animals. Each statement is lead.

92. Answer (4)

Both (1) & Answer (3)

93. Answer (2)

Shorter dikaryophase and crozier are feature ofascomycetes.

94. Answer (2)

Diatoms dont have flagella on somatic body.Chrysophytes include diatoms and golden algae

95. Answer (3)

96. Answer (1)

Used preferably 100’s of characters.

97. Answer (1)

98. Answer (3)

Sphagnum is peat or bog or cotton moss.

99. Answer (2)

Opposite phyllotary.

100. Answer (3)

101. Answer (3)

Fascicular vascular cambium - primary lateralmeristem. Interfascicular cambium and corkcambium - secondary lateral meristem.

102. Answer (2)

Physiologically inactive - heart wood

103. Answer (4)

RBC 52% protein, 40% lipid

104. Answer (4)

9 doublet fibrils at periphery.

105. Answer (2)

106. Answer (3)

107. Answer (2)

Simple diffusion does not show transport saturation.

108. Answer (2)

KGA + NH4+ Glutamate DH glutamic acid + H

2O

109. Answer (1)

110. Answer (2)

Doesnot show photorespiration

111. Answer (1)

Breakdown of pyruvate molecule produces 4 NADH& 1 FADH

2. For each NADH & FADH

2 oxidation one

oxygen atom in required

112. Answer (3)

(C) Pyruvic acid

113. Answer (3)

Phellogen, interfascicular cambium - Dedefferentiation

114. Answer (4)

Promote root formation

115. Answer (2)

116. Answer (1)

117. Answer (4)

Shape of cell in spindle like.

118. Answer (4)

Aplospory produces haploid embryo

119. Answer (3)

Gamete (n)

PEN (3n)

120. Answer (1)

121. Answer (2)

122. Answer (4)

Sterile males

123. Answer (4)

Two chains are antiparallel.

Backbone is sugar & PO4.

124. Answer (4)

Translocase is involved in protein synthesis.

125. Answer (1)

3 structural genes (z, y, a)

126. Answer (2)

Meristem is free from viruses.

(8)


127. Answer (4)

Produced in IRRI, Philippines

128. Answer (2)

Cyclosporin A - Trichoderma

129. Answer (2)

BOD in reduced to 15% of its original value

130. Answer (4)

Competition (– –)

Predation (+ –)

Mutualism (+ +)

131. Answer (2)

132. Answer (4)

Size of organisms increases with trophic level

133. Answer (2)

134. Answer (1)

Diversity decreases from lower to higher latitude.Indigenous catfishes are eliminated by exotic speciesClarias

135. Answer (2)

National park : In situ conservation stragegy.

136. Answer (4)

Decrease in mortality rate is the main reason forrapid increase in population.

137. Answer (3)

138. Answer (3)

Aspirin and disprin cause erosion of gastric mucosa,damaging it

139. Answer (2)

140. Answer (4)

141. Answer (3)

Archaeocytes and interstitial cells are totipotent cellsof porifera and coelenterata respectively.

142. Answer (3)

Annelids and arthropods have protostomic tube withinbody plan, bilateral symmetry and schizocoelom.

143. Answer (1)

Mean arterial pressure

= diastolic pressure + 1

3(Pulse pressure)

144. Answer (3)

Urea enters into the thin segment of loop of henle.

145. Answer (4)

146. Answer (4)

‘Lubb’ sound is produced at the end of atrial systoleor begining of ventricular systole.

147. Answer (4)

148. Answer (4)

149. Answer (2)

Inbreeding increases the homozygosity.

150. Answer (3)

151. Answer (4)

152. Answer (2)

Activation of proto-oncogenes leads to cancers.

153. Answer (2)

Cyclostomata are jawless vertebrates.

154. Answer (2)

155. Answer (4)

156. Answer (2)

Lipids get extracted in acid insoluble fraction.

157. Answer (4)

158. Answer (4)

159. Answer (4)

160. Answer (1)

161. Answer (3)

‘Ori’ site controls the origin of replication of plasmidsand their copy number.

162. Answer (3)

Cross breeding is mating of animals of differentbreeds.

163. Answer (3)

164. Answer (4)

165. Answer (3)

166. Answer (4)

167. Answer (2)

IUDs are the most accepted methods ofcontraception in India.

168. Answer (4)

‘Neries’ has both setae and parapodia while leechlacks both of them.

169. Answer (4)

Fertilization is internal in aves.

170. Answer (3)

The given characteristics are of smooth muscles.

(9)


171. Answer (3)

Microglial cells are mesodermal in origin.

172. Answer (1)

Tight junctions prevent the leakage/exchange ofmaterial between adjacent cells.

173. Answer (3)

174. Answer (2)

Maltose and straight chains of glycogen have (1 4) glycosidic bonds.

175. Answer (2)

Ptyalin or salivary amylase is starch digestingenzyme.

176. Answer (3)

177. Answer (4)

178. Answer (4)

Glucocorticoids promote/stimulate gluconeogenesislipolysis and proteolysis.

179. Answer (3)

Interneurons are absent in monosynaptic reflexes.

180. Answer (4)

After forced expiration, expiratory reserve volume,tidal volume and inspiratory reserve volume can beinhaled during maximum forceful inhalation.

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Edition: 2020-21

Regd. OfficeAakash Tower, 8, Pusa Road, New Delhi …...Home Assignment – 8 08/07/2020 (2) Home...

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