home assignment-6 sol · 2020-06-30 · (4) home assignment-6 (answers and hints) mock test for...

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ANSWERS

1. (1)

2. (1)

3. (4)

4. (3)

5. (1)

6. (3)

7. (3)

8. (3)

9. (4)

10. (2)

11. (1)

12. (2)

13. (3)

14. (1)

15. (2)

16. (2)

17. (2)

18. (3)

19. (3)

20. (3)

21. (1)

22. (1)

23. (1)

24. (3)

25. (1)

26. (4)

27. (2)

28. (2)

29. (3)

30. (1)

31. (2)

32. (1)

33. (4)

34. (1)

35. (3)

36. (3)

37. (2)

38. (3)

39. (2)

40. (3)

41. (2)

42. (2)

43. (3)

44. (1)

45. (4)

46. (3)

47. (3)

48. (2)

49. (3)

50. (4)

51. (1)

52. (1)

53. (2)

54. (3)

55. (4)

56. (3)

57. (2)

58. (2)

59. (3)

60. (3)

61. (1)

62. (3)

63. (1)

64. (1)

65. (3)

66. (3)

67. (4)

68. (1)

69. (3)

70. (2)

71. (1)

72. (2)

73. (2)

74. (1)

75. (4)

76. (1)

77. (3)

78. (3)

79. (2)

80. (3)

81. (4)

82. (4)

83. (3)

84. (1)

85. (1)

86. (3)

87. (2)

88. (4)

89. (2)

90. (3)

91. (3)

92. (1)

93. (4)

94. (2)

95. (2)

96. (3)

97. (4)

98. (3)

99. (1)

100. (1)

101. (2)

102. (4)

103. (2)

104. (4)

105. (3)

106. (2)

107. (3)

108. (4)

109. (1)

110. (1)

111. (3)

112. (1)

113. (3)

114. (1)

115. (2)

116. (2)

117. (1)

118. (3)

119. (3)

120. (2)

121. (4)

122. (1)

123. (1)

124. (2)

125. (4)

126. (1)

127. (3)

128. (3)

129. (4)

130. (2)

131. (3)

132. (4)

133. (3)

134. (4)

135. (3)

136. (1)

137. (2)

138. (2)

139. (4)

140. (4)

141. (3)

142. (1)

143. (2)

144. (4)

145. (3)

146. (3)

147. (2)

148. (3)

149. (4)

150. (2)

151. (2)

152. (4)

153. (2)

154. (4)

155. (2)

156. (3)

157. (4)

158. (1)

159. (4)

160. (3)

161. (4)

162. (2)

163. (2)

164. (3)

165. (2)

166. (3)

167. (4)

168. (2)

169. (4)

170. (4)

171. (4)

172. (1)

173. (3)

174. (2)

175. (3)

176. (4)

177. (3)

178. (3)

179. (4)

180. (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 – 6

30/06/2020

(2)

Home Assignment-6 (Answers and Hints) Mock Test for NEET - 2020

Answers and Hints

1. Answer (1)

At point of projection

= 90°

2. Answer (1)

Q = mL

Pt = mL

L = Pt

m

3. Answer (4)

F r4

4. Answer (3)

5. Answer (1)

A = R

32 = 40

= 5

4

6. Answer (3)

L1 = 10 log

0

I

I

L2 = 10 log

0

2I

I

L2 – L

1 = 10 log

2I

I

L2 – L

1 = 3 dB

7. Answer (3)

VC =

2

V

8. Answer (3)

a = 1 2 3

– sinF

gm m m

T = 3

1 2 3

m F

m m m

9. Answer (4)

10. Answer (2)

15

12 =

10

10

4

x

x

5 = 10

10

x

x50 + 5x = 10x

50 = 5x

x = 10 11. Answer (1)

ac =

16

2 = 8m/s2

at = 4m/s2

a = 2 2

c ta a

a = 80

12. Answer (2)

v = e2u

2

2

gh =

22e gh

e = 1

2

13. Answer (3)

N = m(g – a)

N = 60 × 8

R = 48 kg wt10

N

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 – 6

30/06/2020

(3)

Mock Test for NEET - 2020 Home Assignment-6 (Answers and Hints)

14. Answer (1)

15. Answer (2)

�2 = 3�

1

�2

= 3 × 20

= 60 cm

16. Answer (2)

C C

CEQ

= 0

2

Ad

17. Answer (2)

18. Answer (3)

19. Answer (3)

20. Answer (3)

Inward flux will decrease, to increase it induced

current is clockwise

21. Answer (1)

For melting ice 100 cal heat is left 100 = m80

m = 1.25 g

m1 = 5 – 1.25

m1 = 3.75 g

22. Answer (1)

U = – .P E

��

Umx

when = 180°

23. Answer (1)

fp

= fq – 4

= 250 – 4

= 246 Hz

24. Answer (3)

EC

B

EB

C

25. Answer (1)

26. Answer (4)

= g

L

x = a cos t

27. Answer (2)

F1 200

F2 25

F1/F

2 =

8

1

28. Answer (2)

29. Answer (3)

30. Answer (1)

I = 2 2

2 2 2219

4 4 2

mR mRmR mR mR

31. Answer (2)

1 2 1 2

min 1 2–1 –1

n n n n

32. Answer (1)

33. Answer (4)

Y0 2n = 4

2

2

n

= 0

2

Y

34. Answer (1)

35. Answer (3)

V = di

Ldt

5 2di

dt

2.5 A/sdi

dt

36. Answer (3)

i = V

R

= 6

1.5

= 4 A

37. Answer (2)

x = kt

y = kt – kt2

y = k x

k – k

2

2

x

k

y = x – k

x2

38. Answer (3)

F = –d

dx (ax2 – bx)

F = –2ax + b

0 = –2a × +b

x = 2

b

a

39. Answer (2)

40. Answer (3)

41. Answer (2)

(4)


42. Answer (2)

B = 0

2

i

r

B = 0

2r

100 1.6 10

1

43. Answer (3)

44. Answer (1)

Conductor C will repell with force 3 times force by

which conductor A will repell

45. Answer (4)

E = E2 – E

1

= 2

–13.6

2 – 2

13.6– eV

1

E = 10.2 eV

46. Answer (3)

3Fe(s) + 4H2O(g) Fe

3O

4 + 4H

2(g)

3 × 56 g 4mole

∵ 4 mole H2O 3 × 56g Fe

2 mole H2O

3 56 2

4

g Fe = 84 g

47. Answer (3)

1 1

1

PV

T =

2 2

2

P V

T[∵ V

1 = V

2]

1

1

P

T =

2

2

P

T

760

760 400 =

2

2

T

T2 = 800 K i.e. 527°C

48. Answer (2)

NOH

H+

N

H

C = O

49. Answer (3)

Addition of inert gas at constant volume to the

system at equilibrium, will not disturb the equilibrium

50. Answer (4)

Fact

51. Answer (1)

Fact

52. Answer (1)

Fact

53. Answer (2)

Ecell

= 0 – 0.591 × log –10

1

10 = 0.59 V

54. Answer (3)

CH3COOH + NaOH CH

3COONa + H

2O

Initially 1 mole 0.5 mole 0 0

at equil. 0.5 mole 0 mole 0.5 mole

Solution containing weak acid (CH3COOH, 0.5 mole)

and its salt (CH3COONa, 0.5 mole) will act as a

buffer.

55. Answer (4)

CH –C–OH + C H OH3 2 5

O

H+

CH –C–OC H

3 2 5

O

Ethyl acetate

56. Answer (3)

Fact

57. Answer (2)

Rate = K [conc]1 = [for Ist order reaction]

K = 0.6932 × 10–2 min–1

t1/2

= 0.6932

K = 100 min

58. Answer (2)

CH3CN 2 5

Na/C H OH

Reduction

3 2 2

X

CH CH NH 2HNO

O

3 2 3

Y Z

CH CH OH CH COOH

59. Answer (3)

CH3–CH

2–CH = CH – CH

2 – CH

3

H2

KMnO /H4

+

Pt

C H6 14

2CH –CH –C–OH3 2

O

60. Answer (3)

Number of gram equivalent of KMnO4 = Number of

gram equivalent of O2

100 0.5 xL

1000 5.6L

x = 0.28 L

volume of O2 released = 0.28 L

61. Answer (1)

Fact

62. Answer (3)

2KOH + 5O3 3

Potassiumozonide(Orange coloured solid)

2KO + 5O2 + H

2O

(5)


63. Answer (1)

R – OH Na R – ONa + 1

2 H

2

nH2 =

760 100 1 1

760 1000 0.821 300 0.0041 mole

Number of moles of ehtanol = 0.0041 × 2

= 0.0082

mass of ethanol = 0.0082 × 46

0.38 g

% of ethanol = 0.38

1005

= 7.5%

64. Answer (1)

Fact

65. Answer (3)

Qgiven

= Qout

20 × 0.45 (100 – T) = 50 × 4.2 × (T – 27)

T = 30°C

66. Answer (3)

Area of the circle = work done

67. Answer (4)

Fact

68. Answer (1)

For salt of Trpe

AB : Ksp

= s2 s =

1

2sp

(K )

A2B : K

sp = 4s3 s =

1

3sp

(K /4)

AB3 : K

sp = 27s4 s =

1

4sp

(K /27)

69. Answer (3)

Amount of Ag deposited = 108 × 0.1 = 10.8

and amount of O2 released = 8 × 0.1 = 0.8

loss in mass of solution = 11.6

mass of resulting solution = 108 – 11.6 = 96.4 g

70. Answer (2)

Fact

71. Answer (1)

Fact

72. Answer (2)

Ratio of number of atoms =

1

Ratio of coordination number

73. Answer (2)

2 3 2 3 3 2

– 3 – 2 – – 2 – 3 –

3 3 3 3 3 3

sp sp sp sp sp sp

NO ; AsO ;CO ;ClO ;SO ;BO

74. Answer (1)

Cr

O

O

OCrO

(–)

O

O

O(–)

Si

O–

O–

O–

O–

Si

O–

O–

O–

75. Answer (4)

Atomic number 1

size for (isoelectronic species)

76. Answer (1)

B

N

B

N

B

N

H

H

H

H

H

H

Inorganic benzene

77. Answer (3)

O3

N2

Co Co(+)

Bond order 1.5 3 3 3.5

78. Answer (3)

Fact

79. Answer (2)

Fact

80. Answer (3)

Na2S

2O

3 + 2HCl 2 NaCl + S + SO

2 + H

2O

81. Answer (4)

Fact

82. Answer (4)

Na2B

4O

7

HCl

–2NaCl

H3BO

3 B

2O

3

Na

B

83. Answer (3)

CH –CH–CH –CH +Cl3 2 3 2

CH3

h H C–CH–CH –CH2 2 3

Cl

CH3

*

(optically active)

+ H C–C–CH –CH3 2 3

CH3

Cl

+ H C–CH–CH–CH3 3

CH3

(*)

Cl

(optically active)

+ H C–CH–CH –CH –Cl3 2 2

CH3

(6)


84. Answer (1)

Fact

85. Answer (1)

Stability of carbocation reactivity for SN1 reaction

86. Answer (3)

Fact

87. Answer (2)

Fact

88. Answer (4)

Solvolysis stability of carbocation

89. Answer (2)

Free radical allylic substitution with rearrangement

90. Answer (3)

Fact

91. Answer (3)

Higher the category lesser the similarities between

organisms

Order (less similar characters than family)

Family (less similar characters than )genus

Genus (less similar characters than )species

Species (maximum similar characters)

92. Answer (1)

Multicellular transducerswith tissue or organgrade of organisation

KingdomPlantae

93. Answer (4)

Chemoautotrophic bacteria play great role in

recycling in nutrients like N, P, Fe and S

94. Answer (2)

Chemical nature of cell wall is different from

eubacteria which helps in survival of archaebacteria

in extreme conditions

95. Answer (2)

Shorter dikaryophase, crozier Conidia Ascomycetes

96. Answer (3)

In Bentham and Hooker system of classification–

Gamopetalae was considered primitive to

Monochlamydae in dicots

Gymnosperms were considered advanced than

dicots

“Ordines anomali” is a series of dicots

97. Answer (4)

Red algae shows presence of sulphated

phycocolloids

98. Answer (3)

Bryophytes– [Archegoniates with gametophytic main

plant body]

Motile chemotactic male gametes

Internal fertilisation

Heteromorphic alternation of generation

99. Answer (1)

Dryopteris [Male shield fern] has frond, true indusium,

prothallus, multiflagellated sperms

Selaginella has rhizopore and shows heterosporous

condition

100. Answer (1)

Bipinnately compound leaf is present in Acacia and

Delonix

101. Answer (2)

Fabaceae – Pea family

–Non-endospermic seed

–Zygomorphic flowers

–Vexillary aestivation

–Racemose inflorescence

–Monocarpellary

–Marginal placentation

102. Answer (4)

Collenchyma–Intercellular spaces are absent

103. Answer (2)

Dicot stem differs from monocot stem in absence of

shizo-lysigenous water cavity

104. Answer (4)

Peripheral region of secondary xylem is sap wood.

It is functional and light coloured.

105. Answer (3)

Redifferentiated tissue–Bark and secondary phloem

Dedifferentiated tissue–Phellogen, Interfascicular

cambium

106. Answer (2)

Integral proteins of plasma membrane can be

partially or fully buried in the membrane

107. Answer (3)

Proteins synthesised by ribosomes are modified in

golgi bodies

108. Answer (4)

Spindle apparatus in plants is without centrioles

(Acentric) and astral rays (anastral)

(7)


109. Answer (1)

–Synaptonemal complex formation – Zygotene

–Attachment plate – Leptotene

–Terminalisation starts–Diplotene

–Recombinase activity–Pachytene

110. Answer (1)

DPD = 8 w = –15

w = –4

A

so direction of water movement is

A B

C

111. Answer (3)

Endodermis has casparian strips which are

impermeable to water thus changes apoplast into

symplast.

112. Answer (1)

PEPcase – Mg

Alcohol dehydrogenase – Zn

Nitrogenase – Mo

Ascorbic acid oxidase – Cu

113. Answer (3)

NADP+ reductase enzyme operates in non-cyclic

photophosphorylation

114. Answer (1)

CAM plants differs from C4 plant in absence of

wreath anatomy

115. Answer (2)

NADH is oxidised slowly to NAD+ during

fermentation

116. Answer (2)

Brewing industry–Gibberellins

Weed free lawns–Auxins

Richmond lang’s effect–Cytokinins

Flowering in mango–Ethylene

117. Answer (1)

Gootee is practised in tropical & sub-tropical trees.

118. Answer (3)

Most common pollen tetrads in flowering plants is

tetrahedral formed as the result of simultaneous

cytokinesis of meiosis.

119. Answer (3)

Castor seed–Proliferation of outer integument is

known as strophiole

120. Answer (2)

Ficus, Aristolochia, salvia, Ophrys–Insect pollination

Arum, Lilies, Arisaema, Lemma–Snails pollination

Potamogeton, Myriophyllum, Maize

Bamboo, Papaya–Wind pollination

Callistemon–Bird pollination

Kigelia–Bat pollination

Michelia–Snake pollination

121. Answer (4)

Haplospory produces inviable seeds/embryos

122. Answer (1)

Flower colour of Lathyrus odoratus

Complementry gene interaction

CcPp × ccpp

(Purple) (white)

CcPp Ccpp ccPp ccpp

CP Cp cP cp

(Purple) (white) (white) (white)

cp

Ratio 1 : 3

123. Answer (1)

The given pedigree is for autosomal recessive trait

124. Answer (2)

DNA polymerase I removes RNA primer in 53direction

125. Answer (4)

Translocase enzyme is involved in translation

126. Answer (1)

Ambiguous codon – GUG

Non-degenerate codon – UGG

Mis-sense mutation – 3CTC5 3CAC5

Silent mutation – 3GAA5 3GAT5

127. Answer (3)

Sequence anmotation involves sequencing both

exons & introns

128. Answer (3)

Himgiri is leaf and stripe rust as well as hill bunt

resistant variety of wheat

129. Answer (4)

Turbidity in waste water is removed by tertiary

treatment.

130. Answer (2)

Cyclosporin A–Trichoderma polysporum

Proteases–Mortierella renispora

(8)


131. Answer (3)

r–represents intrinsic rate of natural increase in

population

–K N

K represents environmental resistance

132. Answer (4)

Competitive exclusion principle. The superior species

exterminates inferior species during competition.

133. Answer (3)

NPP–Net primary productivity

GPP–Gross primary productivity

TLE–Trophic level efficiency

134. Answer (4)

Sixth extinction of species is purely anthropogenic

135. Answer (3)

Excess use of fertilisers increases soil salinity

Catalytic converters convert

NOX N

2 + O

2

CO CO2

Unburnt hydrocarbons CO2 + H

2O

Most effective GHG is CFCs

136. Answer (1)

Annelids have closed circulating system. Their body

is triploblastic, metamerically segmented.

137. Answer (2)

Radula is a file like rasping organ of mollusc

138. Answer (2)

Peripatus is a connecting link between annelida and

arthropoda.

139. Answer (4)

Internal ear and neuromast organs are present in all

fishes

140. Answer (4)

Reptiles are truly land vertebrate because of internal

fertilisation, presence of amnion, and cleidoic eggs.

141. Answer (3)

Mammary gland and diaphragm are found in all

mammals

142. Answer (1)

Lipids are not formed by polymerisation of

biomicromolecules. These are the ester of glycerol

and fatty acid.

143. Answer (2)

Deficiency of vit B1 (Thiamine) cause Beri-Beri

144. Answer (4)

In male cockroach, sperms are stored in seminal

vesicle and are glued together in the form of bundle

called spermatophore.

145. Answer (3)

Heparin is an anticoagulant which prevents blood

clotting

146. Answer (3)

In micturition reflex, as the urinary bladder become

half filled, stretch receptors sends sensory signals

to CNS. Conduction of motor impulse through

parasympathetic nerve leads to contraction of

smooth muscles of the bladder and relaxation of the

urethral sphincter

147. Answer (2)

In smooth muscles, calmodullin binds to Ca2+ in the

cytosol of smooth muscle.

148. Answer (3)

Patella/knee cap is a sesamoid bone formed by the

ossification of quadriceps extensor muscle.

149. Answer (4)

Rapid closure of AV valve produces a loud sound

“Lubb” at the end of atrial systole or at the

beginning of ventricular systole

150. Answer (2)

Heart failure is the inability of heart to pump enough

blood to the body.

151. Answer (2)

Liver eliminates bilirubin and biliverdin

152. Answer (4)

Human teeth are thecodont, diphyodont, heterodont

and bunodont

153. Answer (2)

Alveolar ventilation = Alveolar volume × Breathing

Pulmonary ventilation = Pulmonary volume ×

Breathing rate

7200 = Pulmonary volume × 12

Pulmonary volume = 600 ml

Alveolar volume = Pulmonary volume – Dead space

volume

= 600 – 200

= 400 ml

So, alveolar ventilation = 400 × 12 = 4800 ml

154. Answer (4)

High pH and low concentration of DPG favour the

association of oxygen and haemoglobin

(9)


155. Answer (2)

Limbic system is associated with the regulation of

sexual behaviour expression of emotional reaction

and motivation

156. Answer (3)

Ear pinna External auditory meatus Tympanum Ear ossicles Oval window Perilymph of scala vestibula Reissner’s

membrane

157. Answer (4)

Acetylcholine esterase in synaptic cleft is

associated with the chemical synapse while gap

junctions are associated with electrical synapse.

158. Answer (1)

Myasthenia gravis is an autoimmune disorder

associated with neuromuscular junction

159. Answer (4)

Thymus gland is associated with differentiation of T-

lymphocytes and production of antibodies

160. Answer (3)

Secretion of prostate gland form 20-25% of volume

of semen

161. Answer (4)

Cremaster and dartos muscles are present in

scrotum

162. Answer (2)

Hypothalamus

GnRH

Anterior pituitary

Gonadotropin (FSH & LH)

secretes

acts on

secretes

163. Answer (2)

Proliferative phase of menstrual cycle is

characterised by high concentration of FSH, LH and

estrogen not progesterone

164. Answer (3)

Foetal ejection reflex is induced by foetus and

placenta which trigger the release of oxytocin from

maternal pituitary.

165. Answer (2)

MTP during 2nd trimester is not safe

166. Answer (3)

Implant are the contraceptive method in which

progesterone is used.

167. Answer (4)

Forelimbs of different mammals are homologous

organs which show common ancestry and divergent

evolution

168. Answer (2)

When more than one adaptive radiations appeared

to have occured in an isolated geographical area,

one can call this convergent evolution.

169. Answer (4)

Genetic drift is change in gene frequency in a small

population occurring by chance.

170. Answer (4)

Homo sapiens fossils–Cromagnon man

171. Answer (4)

HIV is a retrovirus and is diploid. It enters into the

host cell through receptor mediated endocytosis

172. Answer (1)

Thymus is primary lymphoid organ while spleen,

lymph nodes and MALT are secondary lymphoid

organ

173. Answer (3)

Morphine is useful in patient who have undergone

surgery

174. Answer (2)

HLA genes present on 6th chromosome which is

responsible for MHC antigen

175. Answer (3)

Xenograft is the graft taken from the body of animal

of another species. It has maximum chances of

rejection.

176. Answer (4)

Brown swiss is an exotic breeds of cattle (cow)

177. Answer (3)

cry I Ab is used to control corn borer

178. Answer (3)

In PCR, DNA ligase is not used

179. Answer (4)

In RNA interference, ds RNA initiated RNA

interference and thus silenced the specific mRNA of

nematode.

180. Answer (2)

When an alien DNA is inserted to pBR-322 of E.coli

at selectable marker then it shows insertional

inactivation of one antibiotic resistance gene.

��

Edition: 2020-21

home assignment-6 sol · 2020-06-30 · (4) home assignment-6 (answers and hints) mock test for...

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