home assignment – 7€¦ · home assignment-7 (answers and hints) mock test for neet - 2020...

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ANSWERS

1. (4)

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3. (4)

4. (2)

5. (2)

6. (3)

7. (1)

8. (4)

9. (4)

10. (4)

11. (2)

12. (4)

13. (3)

14. (3)

15. (3)

16. (3)

17. (3)

18. (3)

19. (1)

20. (2)

21. (2)

22. (2)

23. (2)

24. (1)

25. (2)

26. (4)

27. (1)

28. (2)

29. (3)

30. (1)

31. (3)

32. (1)

33. (4)

34. (3)

35. (4)

36. (3)

37. (1)

38. (2)

39. (4)

40. (4)

41. (2)

42. (2)

43. (2)

44. (1)

45. (4)

46. (4)

47. (3)

48. (2)

49. (3)

50. (2)

51. (4)

52. (2)

53. (2)

54. (2)

55. (1)

56. (4)

57. (2)

58. (2)

59. (1)

60. (1)

61. (2)

62. (1)

63. (2)

64. (1)

65. (3)

66. (4)

67. (4)

68. (3)

69. (1)

70. (1)

71. (1)

72. (2)

73. (2)

74. (4)

75. (1)

76. (1)

77. (4)

78. (1)

79. (4)

80. (1)

81. (2)

82. (2)

83. (1)

84. (3)

85. (3)

86. (3)

87. (4)

88. (4)

89. (4)

90. (2)

91. (2)

92. (2)

93. (3)

94. (2)

95. (1)

96. (3)

97. (3)

98. (4)

99. (4)

100. (3)

101. (3)

102. (1)

103. (3)

104. (1)

105. (3)

106. (1)

107. (3)

108. (3)

109. (2)

110. (1)

111. (3)

112. (1)

113. (2)

114. (4)

115. (3)

116. (3)

117. (3)

118. (4)

119. (4)

120. (1)

121. (3)

122. (4)

123. (2)

124. (1)

125. (4)

126. (4)

127. (2)

128. (2)

129. (2)

130. (2)

131. (3)

132. (2)

133. (3)

134. (1)

135. (2)

136. (4)

137. (2)

138. (3)

139. (3)

140. (4)

141. (3)

142. (2)

143. (4)

144. (1)

145. (3)

146. (3)

147. (2)

148. (2)

149. (1)

150. (3)

151. (4)

152. (2)

153. (1)

154. (3)

155. (3)

156. (2)

157. (3)

158. (3)

159. (2)

160. (2)

161. (2)

162. (4)

163. (3)

164. (4)

165. (4)

166. (2)

167. (2)

168. (2)

169. (2)

170. (3)

171. (4)

172. (1)

173 (3)

174. (3)

175. (2)

176. (1)

177. (2)

178. (4)

179. (3)

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 – 7

04/07/2020

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Home Assignment-7 (Answers and Hints) Mock Test for NEET - 2020

Answers and Hints

1. Answer (4)

F

x

L

P

T at P = Fx

L

L = 10m, x = 8m, f = 5N

2. Answer (2)

(constant)w

Pt

kp

t

(change in kinetic energy)

210

2

0

mv

Pt

2 2ptv

m

v t1/2

3. Answer (4)

t = ax2 + x

2dt

axdx

12ax

v

112

2v ax

ax

...(i)

dva v

dx

a = (2x + )–1 (–1) (2x + )–2 (2)

3

2

(2 )a

x

From (i)

a = – 2v3

4. Answer (2)

h

R cos

OR

P

Q

= tan

h = R – Rcos

h = R(1 – cos)

2

11

1

h R

1

1 + 2

5. Answer (2)

' 1d

g gR

1g d

gx R

1xd R

x

6. Answer (3)

l1 : l

2 : l

3 = 1 : 5 : 8

2v

HR

(for constant v)

r1 : r

2 : r

3 = 1 : 2 : 3

21 A r

HR l l

H1 : H

2 : H

3

32 2

31 2

1 2 3

: :

rr r

l l l

1 4 9: :

1 5 6

40 : 32 : 45

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 – 7

04/07/2020

(3)

Mock Test for NEET - 2020 Home Assignment-7 (Answers and Hints)

7. Answer (1)

12375eV

(Å)

Eg

= 12375

eV17000

= 0.73 eV

8. Answer (4)

3

0

tA A e

Power is dimensionless

t3 = (T°)

= [T–3]

9. Answer (4)

2

11 tan

n

2

11 tan45

3

8

9

10. Answer (4)

4

34

tx

34

04

dx t

dt

v = t3

2 21 1(2) (0)

2 2w k m v m v

= 3 2 31

(2) (2 ) 02

= 64J

11. Answer (2)

Tcos = mg

2

3T mg

3

cos2

Tcos

mg

T

= 30°

12. Answer (4)

a b c �

� �

a b c �

� �

a2 + b2 + 2abcos = c2

(a + b)2 – 2ab + 2abcos = c2 [a + b = c]

2ab(cos – 1)= 0

cos = 1

= 0

13. Answer (3)

Rcos60°

R

60°

P

v0

5gRv =

Under critical condition

Work energy theorem between A & P

22

0

1 15 ( cos60 )

2 2mv m gr mg R R

2

05 2

2

Rv gr g R

02v gR

14. Answer (3)

2

3

p

G M GMv

R R

= 7GM

R

15. Answer (3)

2 21 1

2 2mv Iw

I = MR2 (inertia of ring)

16. Answer (3)

In first case 2

3w

vg W

second case 2 l

vg W

4

3

l

w

17. Answer (3)

At x distance from end

Iv

xgv

x

v xg

v x18. Answer (3)

Total energy in constant

f = 0 , T

(4)


19. Answer (1)

0

qF

0

'2

qF

'2

FF

20. Answer (2)

L

die L

dt

= L(tcost + sint) (L = 2 Henry)

= 2tcost + 2sint

21. Answer (2)

D

d

g <

y

2 <

1

1

2

1

22. Answer (2)

Through D2, i

2 = 0 due to reverse biasing of diode.

through D1, i

1 =

50.1

50A

23. Answer (2)

In series

1 2

1 1 1

K K K

2

2

4 mK

T

2 2 2

1 2T T T 2

1K

T

24. Answer (1)

Neither the relative motion with S1 nor with S

2

25. Answer (2)

TVV–1 = constant

5 3 1

5 3 1

2

8(300)

27

vv T

2 3

2

8300

27T

2

4300

9T

T2 = 675 K

Rise in temperature = 675 – 300

= 375 K

26. Answer (4)

5

42

3

6A CB

v

v

i

120

9

20,

9AB BC

v i v i

Heat produced per sec = 2

v

R

Heat produced in 3 = 2

3

i


6

i


2

i

Heat produced in 5 =

220

9

5

i

Heat produced in 4 =

220

9

4

i

(highest)

27. Answer (1)

Current value depend on intensity

28. Answer (2)

min

1

R (Lyman)

1x

R

36

5R First member of Balmer series

36

5

x

(5)


29. Answer (3)

3 2

l lT mg

3

2

mgT

30. Answer (1)

Poissons ratio

r

r

l

l

(Longitudinal strain) l T

l Ay

2A r

A r

= 2l

l

= 2 T

Ay

31. Answer (3)

4 c

–1 c

4 c

4a

3a

2a

Net force on all the changes in equal to zero. Theentire charge system is in equilibrium.

32. Answer (1)

2 f 2 f

2 f

A

6V

Q = ceq

× v

= 2

63

= 4 C

0

4

2A

v v

VA = 2 volt

33. Answer (4)

2

0

2 2 3 22 ( )

NiaB

a x

At centre 0

0(max)

2

NiB

a

34. Answer (3)

cot21 + cot2

2 = cot2

1 and

2 are apparent dip in two mutally

perpendicular planes. is true dip.

1 = cot–12

2 = cot–1

2 3

22 2(2) 2 3 cot

cot = 4

= cot–1(4)

35. Answer (4)

c 2

2

gsina

k1

R

k = gyration radius

11

a(ring) 32

a(disc) 1 1 4

2

2

k 1(Disc)

2R

2

2

k1(Ring)

R

36. Answer (3)

PV–3 = constant n = –3

R RC

1 n 1

5

3

R RC

5 3 11

3

= 3R R

2 4

= 7R

4

37. Answer (1)

A

B

A + BA + B

Y = A + B

Y = 1 + 0 = 1

Y = 0 + 1 = 1

(6)


38. Answer (2)

v (man)m

v RM

–v RM

10 m/s NN

vR5 m/s

5tan

10

1 1tan

2

above north horizon.

39. Answer (4)

May increase decrease or remain same.

40. Answer (4)

Only rotational, only translational or rotational andtranslational both are possible.

41. Answer (2)

m

5M q 2 R

6 Initially

now it is bent

M = qm

R

3MM'

5

42. Answer (2)

B

A

P v

x

E.M.F. criss AB = Bv(2R)

Bv(2R)E Bv

2R

43. Answer (2)

At steady state

E is constant between the plates and B is zero.

44. Answer (1)

d

4E (Total energy)

A 4

(mass of parent = A + 4)

ETotal energy (A 4)

4

45. Answer (4)

On applying the high electric field collision of e–

results a breakdown called avalanche breakdown.

46. Answer (4)

In absence of intermolecular forces, water will existas gas

W

PV RTm

34.5 10

1 V 0.0821 27318

V = 5.603 L = 5.6 m3

47. Answer (3)

2x + 7 (–2) = – 4

2x – 14 = – 4

x = + 5

48. Answer (2)

Cu2+ + 2KI CuI2 + 2K+

2CuI2 Cu

2I2 + I

2

49. Answer (3)

Fe2+ + 2e– Fe –G1 = 2 × (–0.44) ×F

Fe3+ + 2e– Fe+ –G2 = 3 × (–0.036) ×F

Fe2+ Fe3+ + e– –G3 = –0.88F + 0.108 F

= – 0.772 F

For the reaction

Fe3+ + e– Fe2+ E° = + 0.772 V

50. Answer (2)

sP P n

P n N

n 760 750 1

n N 760 76

51. Answer (4)

Fact

52. Answer (2)

For the reaction

A B

10g 5g 2.5g30 min30 min

1.25 g30 min

90 min

Since t1/2

is not changing.

Order = 1

53. Answer (2)

5 3 2

01 0

1

PCl PCl +Cl

��⇀↽�� , (total pressure = P)

p

P1 1

K1

P1

neglecting ‘’

2pK P

p

K

P

(7)


54. Answer (2)

– –

21 h h h

X H O HX OH �

14

w

5a

K 10h

K C 10 0.1

= 10–4 or 10–2%

i.e. 0.01%

55. Answer (1)

CH – CH – CH – Cl3 2

AlCl3

CH3

CH – C – CH3 2

CH3

H

CH – C – CH3 3

CH3

CH – C – CH3 3

CH3

+ AlCl4

–

56. Answer (4)

CH = CH + Br2 2 2

CH – CH2 2

Br Br

NaCl

H O2

CH – Br2

CH2 – OH

+

+

CH – Br2

CH2 – Cl

57. Answer (2)

OH

+ 3Br2

OH

Br Br

Br

(2,4,6-tribromophenol)

1 mol of phenol (94 g) reacts with 3 × 160 g ofbromine

2 g of phenol requires 3 160

2 10.21g94

58. Answer (2)

Ph – CH – CH – OMgBr2 2

Ph – CH – CH – OH2 2

H O/H2

+

(B)

PhMgBr + CH – CH 2 2

O

– +

(A)

59. Answer (1)

C H NH + HONO 2 5 2

C H OH + N + H O2 5 2 2

1 mole 1 mole(22.4 L at STP)

60. Answer (1)

CHCl3

KOH

NH2

NCHCl/H O

2

300 K

NH2

HCOOH +

(Y)

(Z)

61. Answer (2)

HF > HCl > HBr > HI

62. Answer (1)

Co

O

O

OO

O

O

.

63. Answer (2)

Fe2+ has four unpaired electrons.

Cr3+ = d3, Fe2+ = d6, Cu2+ = d9, Zn2+ = d

64. Answer (1)

BP mass

65. Answer (3)

S

O

O OHO

S

O

O OH

(Oleum)

66. Answer (4)

Fact

67. Answer (4)

All are correct

68. Answer (3)

2

2

O

T

W

Po 132X

W WP 3

16 32

69. Answer (1)

Angular momentum = nh 5h 2.5h

2 2

70. Answer (1)

Smaller is the size of atom more will be the I.E.Moreover removal of 2 electrons requires moreenergy.

71. Answer (1)

AlCl3 will ionize to form more number of ions as

compared to when CaCl2 gets ionized.

(8)


72. Answer (2)

For two solutions to be isotonic

2 4Na SO glucose

(1 + 2) C1RT = C

2RT

(1 + 2) × 0.004 = 0.01

= 0.75 or 75%

73. Answer (2)

Fact.

74. Answer (4)

Fact.

75. Answer (1)

For an endothermic reaction H = positive andH = E

a(f)– E

a(b).

76. Answer (1)

Fact

77. Answer (4)

1,3-butadiene being most stable alkene has thelowest heat of hydrogenation.

78. Answer (1)

Chlorobenzene is less reactive than benzene towardselectrophilic substitution reaction due to strong - Ieffect of –Cl.

79. Answer (4)

Ph – O – H O = C = O

Ph – OPyridine

(X)4–7 atm. 410 K

C – O–

OO–

C – OH

OOH

H O3

+

(Y)(Z)

80. Answer (1)

OH

NO2

is a weak acid

81. Answer (2)

CH CHO + CH CH – NH3 3 2 2 CH CH – N = CH – CH

3 2 3

CH – CH – NH – CH – CH3 32 2

(Diethylamine)

H /Pt2

(Y)

82. Answer (2)

CH CH CH NHCH3 2 2 3

[CH – CH CH N(CH ) ] I3 2 2 3 3

– +CHI

excess

3

Salt

83. Answer (1)

Fact.

84. Answer (3)

CaCO CaO + CO3 2

Heat

(basic)

CaO + SiO CaSiO2 3

(slag)

85. Answer (3)

Li2CO

3 is highly decompossible

86. Answer (3)

4p4sd10

Zn : 2+

87. Answer (4)

3KClO3

+ 4H2SO

4 3KHSO

4 + ClO

2+ H

2O + O

2

88. Answer (4)

Fact.

89. Answer (4)

Fact

90. Answer (2)

CH – C – CH3 3

O

CH3

CH3

CH3

H SO

298K

2 4

mesitylene

91. Answer (2)

Division - Angiospermae

92. Answer (2)

(3) Metaphase

(4) Anaphase

93. Answer (3)

Glutamine is amide. Glutamic acid can form other17 amino acids by transamination.

94. Answer (2)

Stroma lamellae is site of cyclicphotophosphorylation.

95. Answer (1)

No ETC in fermentation.

96. Answer (3)

Maize, tomato, cotton are DNP, sugarbeet, oat isLDP, rice – SDP.

97. Answer (3)

Shift from grain to meat as food by man cannot beprevented.

(9)


98. Answer (4)

Cyclosporin A

99. Answer (4)

100. Answer (3)

Single tree can support many birds and the birds inturn can support less number of predators likehawks.

101. Answer (3)

Catalytic converters can reduce gaseous air pollutantreleased from automobiles exhaust.

102. Answer (1)

Archaebacteria have lipid monolayer in their cellmembrane. They have introns.

103. Answer (3)

Zygote does not divide by meiosis immediately.

104. Answer (1)

Male and female gametophytes are not of free livingexistence.

105. Answer (3)

All seed-bearing plants have diplontic life cycle.

106. Answer (1)

Thorns are modified axillary buds.

107. Answer (3)

Synaptonemal complex is dissolved in diplotene.

108. Answer (3)

NAD+ is not associated with ETS in chloroplast.

109. Answer (2)

AaBBCC × Aabbcc

aaBbCc = aa = 1

4,

4Bb

4 ,

4Cc

4

i.e.1 4 4 16 1

100 25%4 4 4 64 4

110. Answer (1)

VNTRs are inherited and have different lengths ineach individual.

111. Answer (3)

Population does not face a crash.

112. Answer (1)

It helps in ecotype formation and evolution of newspecies.

113. Answer (2)

114. Answer (4)

In Trifolium, gynoecium is G1

115. Answer (3)

‘C’ is secondary cortex (phelloderm) and developedfrom phellogen.

116. Answer (3)

Typhoid is bacterial disease.

117. Answer (3)

Ginkgo, Sequoia endosperm is haploid.

118. Answer (4)

The epidermis is covered with thin layer of cuticle.

119. Answer (4)

Gibberellins help in early seed production inconifers.

120. Answer (1)

A – Multicarpellary syncarpous pistil

B – Oogonium (nucule) of Chara

C – Archegoniophore of Marchantia

D – Offset of water hyacinth

121. Answer (3)

CcDd Xcy (Taylor)

Gametes = cdXc = 1 1 1

2 2 2

= 1

8

122. Answer (4)

123. Answer (2)

Predators can reduce the intensity of competitionamong competiting prey species.

124. Answer (1)

TtYy × Ttyy

Tall and green 3 1 3

(T yy)4 2 8

125. Answer (4)

Dodo from Mauritius.

126. Answer (4)

(1) Mesophyll companion cells sieve tube

(2) Hypertonic condition

(3) Difference of T.P. between source and sink isrequired

127. Answer (2)

In mitochondria, glycine (2 molecule) is convertedinto serine (one molecule) & CO

2.

128. Answer (2)

Pollen grains are light and non-sticky.

129. Answer (2)

(1) ETC

2

ETC

2

NADH 3ATP

FADH 2ATP

(3) Small protein

(4) Two copper centres

(10)


130. Answer (2)

Decomposition is fast in warm and moistenvironment.

131. Answer (3)

1974 - Chipko movement

1988 - National Forest Policy

1989 - Montreal Protocol (effective)

132. Answer (2)

Mesosome helps in secretion.

133. Answer (3)

The central part of centriole is proteinaceous andcalled the hub.

134. Answer (1)

Dianthus – free central placentation

Lily – Stamens are epitepalous

Mustard – hypogynous flower

135. Answer (2)

Deuteromycetes

136. Answer (4)

Echinoderms lack excretory organs. Hemichordateshave open blood vascular system. All annelids do nothave parapodia, only polychaetes.

137. Answer (2)

Diagram of multicellular exocrine gland. A hormonecannot be secreted from it.

138. Answer (3)

Rise per base pair in B-DNA is 3.4Å.

139. Answer (3)

140. Answer (4)

Amylases are found in saliva and pancreatic juice.

141. Answer (3)

Acetylcholine, released by parasympathetic nervoussystem decreases heart rate.

142. Answer (2)

143. Answer (4)

Blood group AB has both A and B antigens on RBCsand no antibodies in the plasma.

144. Answer (1)

Absence of P wave in the ECG is indicative of SANfailure.

145. Answer (3)

The glomerular filtrate is termed urine only after itreaches collecting duct.

146. Answer (3)

Scapula is labelled as clavicle and vice versa.

147. Answer (2)

Actin-myosin bridges dissociate when ATP binds tomyosin head and its affinity for actin decreases.

148. Answer (2)

Synaptic vesicles are part of chemical synapsis.

149. Answer (1)

Acromegaly is due to hypersecretion of GH in adults.

Myxedema – hyposecretion of thyroxine in adults.

Addison's disease – hyposecretion ofmineralocorticoids & glucocorticoids.

Diabetes mellitus – hyposecretion of insulin.

150. Answer (3)

With every cleavage division the size of resultantblastomeres decrease due to very short interphases.

151. Answer (4)

In GIFT, both gametes are transferred into thefallopian tubes of females.

152. Answer (2)

153. Answer (1)

Heroin, smack, brown sugar are various names ofdiacetyl morphine.

154. Answer (3)

Darwinian variations are small and directional.

155. Answer (3)

Competitive inhibitors bind to active site of enzymeby non-covalent linkage.

156. Answer (2)

Veins have thin walls and broader lumens.Pacemakers are used if there is defect in theconducting system of the heart.

157. Answer (3)

158. Answer (3)

ADH is secreted when osmolarity of the blood ishigh and thus there is hypovolemia.

159. Answer (2)

160. Answer (2)

Recticular tissue helps in filtering blood, removingantigens and old RBCs in spleen.

Cornea has no sensory epithelium.

161. Answer (2)

Cortisol induces gluconeogenesis.

162. Answer (4)

Melatonin suppresses gonadal function.

163. Answer (3)

At puberty, about 60,000 - 80,000 primary folliclesare remaining in each ovary.

(11)


��

164. Answer (4)

Lactational amenorrhea is due to high prolactin.

165. Answer (4)

Homologous organs.

166. Answer (2)

Psilophyton was pteridophyte ancestor from whichferns, conifers, progymnosperms arose.

167. Answer (2)

Isolated area without competitors and with vacantniches is prime for adaptive radiation.

168. Answer (2)

The Ti-plasmid of Agrobacterium has the ability tointegrate with DNA of dicot plants.

169. Answer (2)

170. Answer (3)

Emergency contraceptives if used with 72 hrs ofcoitus aim at preventing implantation.

171. Answer (4)

ANF is a vasodilators and hence lowers BP.

172. Answer (1)

Saddle joint between carpal and metacarpal ofthumb.

173 Answer (3)

174. Answer (3)

Suspensory ligaments are attached to lens and tociliary muscles.

175. Answer (2)

B-cells are stimulated by helper T-cells to secreteantibodies.

176. Answer (1)

177. Answer (2)

178. Answer (4)

MRI uses radio waves and thus can be used multipletimes.

179. Answer (3)

A population remains in Hardy-Weinberg equilibriumonly if it is not evolving.

180. Answer (2)

Edition: 2020-21

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