FULL TEST-3 Answers and Hint s · RISHI TYAGI (2) 16. Answer (1) P 2 P 0 P 1 R 2R P 1 – P 0 4 2 S R P 2 – P 1 4S R 2 6S P R 17. Answer (4) 3P – P 2 1 2 v 5 3 4410 20m/s 10 P

RISHI TYAGI

(1)

MM : 720 Time : 3 Hrs.

1. Answer (3)

2. Answer (3)

3. Answer (1)

t sin

u

g

H

2 2

2 2

1sin .

sin 2 sin 3

u u uu g

g g g

4. Answer (2)

5. Answer (1)

Bodies do not move

T = 0

f = 2g

6. Answer (2)

60º

N

mg

2cos60º

2

aN m tan 60º

2

2g

a

N sin 60º = mg 22

3

g

a

7. Answer (2)

8. Answer (3)

Loss of KE 2 21 2

rel

1 2

1(1 )

2

mmu e

m m

⎛ ⎞ ⎜ ⎟⎝ ⎠

2

21 3 27(9 )

2 2 4 16

m mvv

9. Answer (1)

2 dvmv P

dx

1

3v x

PHYSICS

2

2 31

KE2mv x

10. Answer (3)

2

ga

4x

ga

4y

ga

2

4 2 2

g ga

11. Answer (1)

2 21 1

2 2mv I I = mR2 Hence ring

12. Answer (2)

18 813 kg

2

13. Answer (3)

I1

1 = I

2

2

2 2

1 2

1 2

R R

T T T

2 = 6 hour

14. Answer (4)

( )

2 2or

a

V R HLV

m

22

a

or

V GM gRV

R H R H R H

2

2

4a

VH R

gR

15. Answer (2)

Average force

3

2 2

2

F F

F

Extension Fl

AY

Test Series : Test – 13

(for NEET - 2019-2020 )

Answers and Hints

DOT INSTITUTE OF PHYSICS

TEST-SERIES FOR NEET 2019-2020

DOT INSTITUTE OF PHYSICS RUN BY RISHI TYAGI 582,SUNDRAM PLAZA P.L. SHARMA ROAD, MEERUT(UP)

FULL TEST-3

RISHI TYAGI

(2)

16. Answer (1)

P2

P0

P1

R

2R

P1 – P

0

4

2

S

R

P2 – P

1

4S

R

2

6SP

R

17. Answer (4)

3P – P 21

2v

5

3

4 4 1020 m/s

10

Pv

18. Answer (3)

1 1 2 2 3 3

1 2 3

45ºC

j

K T K T K TT

K K K

19. Answer (3)

w pdv ∫0

0

0

3T VpV

TR V

0 0

0

3RT RTp

V V

w = RT0 (3 ln 2 –1)

20. Answer (4)

2

1

1T

T

2

1

0.8 21.2 1 1 (1 ) 62.5%

1.2 3

T

T ⇒

21. Answer (1)

2 1

2

m gA

m

2

1

2

24

m gTA

m

22. Answer (1)

Average speed Distance (2 3) 6

Time 2

A

23. Answer (4)

VP

dy

dt

Vpmax

= 2 A

24. Answer (4)

1

2

f T

f T

1 2

400 2 100

f ⎛ ⎞ ⎜ ⎟⎝ ⎠

4 Hzf

25. Answer (1)

0

c atf f

c

⎡ ⎤ ⎢ ⎥⎣ ⎦

26. Answer (1)

p sin60º experiences no force.

Force on 2

0

cos60º cos60º4

dE pp p

dr R

27. Answer (2)

28. Answer (4)

29. Answer (2)

2 21 2 2

2 3 9

E CEU C

⎛ ⎞ ⎜ ⎟⎝ ⎠

30. Answer (4)

31. Answer (3)

32. Answer (3)

2

0

22

IrB r

R

0 0

22 8

Ir IB

R

33. Answer (2)

34. Answer (2)

2

2 2

RMS RMS

42 12 2 3

2I I

⎛ ⎞ ⇒ ⎜ ⎟⎝ ⎠

35. Answer (1)

1 s2

TT

36. Answer (4)

0

0 0 0

1 1 16 cmv

v u f

1 1 15 cm

e

e e

uv u fe

L = 6 + 5 = 11 cm

37. Answer (1)

cos

cos

d rd

i and

sin

sin

i

r 2

cos3

r

4 cmd



RISHI TYAGI

(3)

38. Answer (3)

2

2 22 2

dx D d D

D

Now,

2

2

d

D

2

Dd

39. Answer (1)

1.22

D

10

6

1.22 1.22 5460 100.145 m

4.6 10D

40. Answer (2)

54º + 54º + 90º = 198º

41. Answer (4)

6000

hck

24000

hck

1.03 eV

42. Answer (3)

36

5

4

3

B

L

R

R

1081215 6561 Å

20B

43. Answer (3)

44. Answer (4)

45. Answer (2)

502.5 mA

20

200 5030 mA

5

I0 = 30 – 2.5

= 27.5 mA

CHEMISTRY

46. Answer (1)

No. of formula units = 14.35

0.1143.5

No. of g ions = –

Ag Cl

0.1 0.1 0.2

47. Answer (2)

2KMnO4 K

2MnO

4 + MnO

2 + O

2...(i)

4Al + 3O2 2Al

2O

3...(ii)

From equation (i) and (ii)

3 mole of O2 6 mole of KMnO

4 4 mole of Al

0.1 mole of Al 0.15 mole of KMnO4

= 0.15 × 158 = 23.7 g

48. Answer (4)

r n2

2

1

2

2

n 25

64n ,

1

2

n 5

n 8 =

2

1

v

v

49. Answer (3)

Fact

50. Answer (4)

Fact

51. Answer (2)

T-shape See saw

52. Answer (1)

53. Answer (4)

Cp for diatomic gas N

2 = O

2

= 7R

2

54. Answer (4)

For A2B

3, K

sp = 108 S5

55. Answer (1)

n-factor = 1

56. Answer (2)

x x

A(s) B(g) C(g)�Kp

1 = 180 atm2

y

B(s) B(g)� Kp2 = 10 atm

Now, y = 10 atm and Kp1 = (x + y)y = 80

(x + 10) × 10 = 180

x = 18 – 10 = 8 atm

PB = x + y = 10 + 8 = 18

57. Answer (4)

Ca(HCO3)2

Temporary of hardness

CaCO3

Degree of hardness

CaSO4, CaCl

2 Permanent hardness

58. Answer (4)

Fact

59. Answer (3)

K2CO

3 is carbonate of alkali metal

60. Answer (2)

B2O

3 form metaborate of metal



RISHI TYAGI

(4)

61. Answer (3)

Two sp2 and sp3 boron are present

62. Answer (1)

Fact

63. Answer (3)

CH C CH + 4HC3 3

O

P

A

4CH CH + 2I

3 2 2CH

2

64. Answer (1)

is aromatic

65. Answer (4)

66. Answer (2)

A is OH

OH

O

O

, B is O, C is O

||O

67. Answer (4)

Fact

68. Answer (3)

Fact

69. Answer (2)

Fact

70. Answer (2)

r+ + r – = 3 a

4

S2– ion is present in tetrahedral voids

71. Answer (1)

PA = P°

A

A

A

2P 100

5 = 40 mm Hg

B

3P 300

5 = 180 mm Hg

Mole fraction in vapour phase

A

40 40 2Y

40 180 220 11

72. Answer (2)

Ered

= –0.059 × pH

= –0.059 × 4

= –0.0236 V

73. Answer (4)

Absolute ionic mobility

100

F 96500

= 1.03 × 10–4 cm2s–1 volt–1

74. Answer (1)

It is a first order reaction as

2.303 5K log 0.023

10 4

which is constant throughout.

75. Answer (4)

Fact

76. Answer (4)

Fact

77. Answer (2)

Highest oxidation state

78. Answer (4)

Fact

79. Answer (1)

All gives NO2

80. Answer (1)

Unpaired electron in eg orbital

81. Answer (3)

EAN = z – (O.N.) + 2 ×(C.N.)

82. Answer (3)

Fact

83. Answer (2)

By ring expansion

84. Answer (3)

OH

+ CO + NaOH2

4-7atm

120°C

OHNA

COONa

H+

OH

COOH

85. Answer (4)

Fact

86. Answer (2)

OH

CH CH CH + R C R3 3

O

(CH CHO) Al3 3

R CH R + CH C CH3 3

OH O

87. Answer (3)

CH NH + HNO CH N NO3 2 3

CH3

(Yellow oily liquid)

CH3

88. Answer (1)

Both are dextrorotatory

89. Answer (4)

Fact

90. Answer (2)

Fact



RISHI TYAGI

(5)

BOTANY

91. Answer (3)

B-Bacteria and Protista

92. Answer (4)

Most general characters

93. Answer (3)

Unicellular eukaryotes in protista in 5K-system.

94. Answer (4)

Slime moulds

95. Answer (1)

Ascomycetes

96. Answer (3)

Chemotaxonomy

97. Answer (1)

98. Answer (2)

Bryophytes

99. Answer (3)

Gymnosperms and Angiosperms

100. Answer (4)

Rhizophora

101. Answer (2)

Pea family-vexillary

102. Answer (4)

Basal

103. Answer (1)

104. Answer (2)

In stem the condition is Endarch.

105. Answer (3)

Few layered, thick-walled pericycle.

106. Answer (1)

107. Answer (4)

108. Answer (2)

Semi-autonomous organelles-Mitochondria and

Chloroplast

109. Answer (2)

Spireme stage-Prophase

110. Answer (2)

G1 S M I M II

80 pg 160 pg 80 pg 40 pg

(2C) (4C) (2C) (1C)

111. Answer (1)

Hypertonic solution

112. Answer (4)

113. Answer (3)

Malic acid (4C) Pyruvate (3C) (in Bundle sheath

cell)

PEP (3C) Malic acid (Mesophyll)

114. Answer (2)

During reduction of one NAD+, 2 redox equivalents

are removed from respiratory substrate.

115. Answer (2)

Growth is determinate for some plant organs i.e. leaf,

flower.

116. Answer (3)

A a

a B

Trans-arrangement of genes (AAbb × aaBB)

P-type Ab, aB

NP-type AB, ab

117. Answer (3)

AaBbCc × aabbcc

118. Answer (1)

119. Answer (1)

120. Answer (4)

121. Answer (2)

122. Answer (3)

123. Answer (1)

124. Answer (4)

VNTR/Mini-satellites.

125. Answer (3)

Lysine and Tryptophan.

126. Answer (2)

127. Answer (3)

128. Answer (1)

129. Answer (2)

Charging of tRNA.

130. Answer (4)

Exponential growth curve is J-shaped.

131. Answer (2)

132. Answer (4)

Key industry animal – Herbivore (T2)

133. Answer (3)

b and c correct.

134. Answer (3)

Broadly utilitarian aspect pollination.

135. Answer (3)

Bad O3 – Secondary pollutant in troposphere.



RISHI TYAGI

(6)

ZOOLOGY

136. Answer (3)

Peripheral nervous system includes somatic nervous

system which will carry the impulse from CNS to

skeletal muscles and autonomic nervous system or

visceral nervous system which carry the impulses

from CNS to smooth muscle or visceral organs not

under conscious control.

137. Answer (4)

The person with blood group O has I°I° genotype, so

there are neither antigen A nor antigen B, but both

antibodies a and b are present.

138. Answer (2)

Level of organisation is

Cellular level Tissue level Organ level Organ

system

139. Answer (4)

140. Answer (4)

Physician cannot test the foetus for eye colour by

amniocentesis; as we cannot determine whether the

gene is dominant or recessive. Cleft palate can be

both due to deficiency of folic acid or due to trisomy

of chromosome number 13.

141. Answer (2)

142. Answer (4)

During the follicular phase, the secretion of

gonadotropins will decrease gradually due to negative

feedback control of estrogen on the anterior lobe of

pituitary.

143. Answer (4)

r DNA = Plasmid + Insert

is vector is

or carrier foreign DNA

144. Answer (2)

We amplify the desired gene to be detected with

PCR. Now we take single stranded copy of the DNA

of the gene, combine it with radioactive isotopes this

is called probe, used for screening.

145. Answer (2)

Glyphosate is a powerful biodegradable herbicide; its

trade name is round up. We are making plants

resistant to glyphosate to use the herbicides.

146. Answer (4)

Affinity of haemoglobin for oxgyen

50

1

P value. If

P50

value decreases, the affinity of hemoglobin for

oxgyen increases and the curve will shift towards

left.

147. Answer (4)

DCT is concerned with conditional reabsorption of

water and Na+.

148. Answer (2)

Viral diseases e.g., genital herpes, hepatitis-B, AIDS

cannot be treated by antibiotics.

149. Answer (3)

Cortisol is anti-allergic, anti-inflammatory immuno

suppressive hormone.

150. Answer (2)

HIV infects the cells with CD4 receptors, they are

macrophages and Helper T cells. It depletes helper

T cells but not macrophages.

151. Answer (2)

Artificial pacemaker is required when heart beat

becomes irregular. It can be due to damage to SA

node or due to heart block.

152. Answer (4)

Endothermal animals have high rate of metabolism

so they eat more than ectothermal animals.

153. Answer (3)

In our body there are genes called c-onc/cellular

oncogenes they code for normal body growth and

development, but if their expression becomes

uncontrolled, they will be called oncogenes.

Mutations in tumor-suppressor genes can also lead

to cancer.

154. Answer (2)

Hair provide thermal insulation to the organism, they

trap a layer of air which is poor conductor of heat.

155. Answer (2)

156. Answer (3)

Bats are mammals and their most unique character

is presence of mammary glands. Birds are

oviparous.

157. Answer (1)

Woodpecker finch and warbler finch have evolved

from a common ancestor and show adaptive

radiation, but do not interbreed.

158. Answer (4)

The second line of non-specific defense mechanism

is chemical defense called inflammatory response.

159. Answer (3)

GIFT : Transfer of an ovum from the donor female into

the fallopian tube of another female, who cannot

produce ova but can provide a suitable environment

for fertilisation and further development.

160. Answer (2)



RISHI TYAGI

(7)

161. Answer (2)

Heparin, citrate and oxaloacetates are anticoagulants.

162. Answer (2)

163. Answer (2)

164. Answer (2)

Cockroach is protostomic with schizocoely type of

the development of the coelom.

165. Answer (1)

Spider respires by book lungs and/or trachea.

166. Answer (4)

Amphioxus belongs to the subphylum

Cephalochordata / Acraniata, no brain box.

167. Answer (4)

168. Answer (4)

Random mating does not bring about change in gene

frequency.

169. Answer (3)

170. Answer (1)

171. Answer (3)

Genetic drift is a non-directional factor which can

bring about change in gene frequency.

172. Answer (2)

Marine water bony fishes remove the nitrogenous

waste mainly in the form of NH3, only one third is

removed in the form of trimethyl amine oxide.

173. Answer (3)

Mast cells and basophils have similar functions.

174. Answer (4)

175. Answer (3)

Lyases catalyse breakdown of covalent bond in

absence of water with the removal of group, leaving

covalent bond behind.

176. Answer (2)

Nicotine receptors are cholinergic.

177. Answer (2)

178. Answer (4)

179. Answer (2)

Iodothyronines are lipid soluble hormones.

180. Answer (2)

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Documents

FULL TEST-3 Answers and Hint s · RISHI TYAGI (2) 16. Answer (1) P 2 P 0 P 1 R 2R P 1 – P 0 4 2 S R P 2 – P 1 4S R 2 6S P R 17. Answer (4) 3P – P 2 1 2 v 5 3 4410 20m/s 10 P