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Test - 10 (Code-A) (Answers) All India Aakash Test Series for Medical-2018

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Test Date : 01/04/2018

TEST - 10 (Code A)

All India Aakash Test Series for Medical - 2018

ANSWERS

37. (1)

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180. (1)

All India Aakash Test Series for Medical-2018 Test - 10 (Code-A) (Hints)

2/8

PHYSICS

1.2 x m L T

x m L T

2.

0

=µ

BH

–1

⎡ ⎤⎡ ⎤⎢ ⎥= = ⎢ ⎥⎣ ⎦⎢ ⎥⎣ ⎦

IH AL

r

3. Relative velocity of the train w.r.t. other train is

V – v.

Hence 0 – (V – v)2 = 2ax a = ( )

2

–

2

V v

x

Minimum retardation = ( )

2

–

2

V v

x

5.

2 2 5 1010 m

10

⇒ x yu u

Rg

6. x2 = y

Differentiating w.r.t time

y

xv

ay

2 2x y

dx dyx xv vdt dt

⇒

Again differentiating w.r.t time

ay = 2v

x

2 + 2x ax

ax = 0, tangential acceleration of particle is zero

ay = 2v2 ( v

x = v)

Radius of curvature of origin is

21

0.5 unit2

y

vR

a ⇒

7. Acceleration, a = 24

2 m/s2

= =

f

m

Displacement along OE, S1 = vt = 3 × 4 = 12 m

Displacement pendendicular to OE

2 2

2

1 12 4 16 m

2 2= = × × =S at

Resultant displacement

HINTS

2 2

1 2144 256 400 20 mS S S= + = + = =

8. If a liquid of density is flowing through a pipe of

cross-section A at speed v, the mass coming out

per second will = ρ

dmAv

dt

In order to get n time water in the same time, we

get

⎛ ⎞⎟⎜= ⎟⎜ ⎟⎟⎜⎝ ⎠

dm dmn

dt dt i.e. AV = nAV

But as pipe and liquid are same

= , A = A

V = nV

2

/

dm dmV nV n

F dt dtn

dmF V dm dtV

dt

F = n2F

23

P F v n f nvn

P FV fv

P = n3P

9. Acceleration of centre of mass

1 1 2 2

cm

1 2

m a m aa

m m

22

3 3

ma maF

m m

2

2 F maa

m

10.Net external force sin

Total mass 2

mg mga

m

(1 sin )

2

g

11. Mass of each of the four parts = 3

m

Mass of the plate including the cut piece = 4

3

m

Test - 10 (Code-A) (Hints) All India Aakash Test Series for Medical-2018

3/8

MI of the whole plate (including the cut piece) about

the said axis =

24

3 6

⎛ ⎞⎟⎜ ⎟⎜ ⎟⎟⎜⎝ ⎠

m l

Now, M.I. of the remaining portion should be

2 23 4

4 3 6 6

⎛ ⎞⎟⎜ =⎟⎜ ⎟⎟⎜⎝ ⎠

m l ml

12.

A

vc

vP

v rc +

v r c –

B

For pure rolling, velocity of point of contact has to be

equal to the velocity of the surface.

Let cylinder rolls with angular velocity .

At point B,

vc – r = v

p

r = vc – v

p

At point A,

vA = v

c + r

= vc + v

c – v

p

= 2vc – v

p

13. Maximum height after nth collision

hn = h

0e2n

n = 2; h2 = h

0e4

100

6.25 m16

14.

2

2h

h

.1

.2

Applying Bernoulli’s theorem between points (1) & (2)

(Patm

+ 2gh) + g(2h) = Patm

+ 1

22v2

v = 2 gh

15.1KEr

(r = radius of orbit)

3/2T r

3/2( )T KE

3/2T x

16.Fl

YA l

Y, l and F are constant

2

1l

D

2

2 1

21 2

1

16

l D

l D

2

1

16Δ =l mm.

17.1

1

2m

tK

and 2

2

2m

tK

In series,

2

1 2

1

4 mK

t

and

2

2 2

2

4 mK

t

1 2

1 1 1

eqK K K

= +

2 22

1 2

2 2 24 4 4

t tT

m m m

,

2 2 2

1 2= +T t t

18. 1 =

l – 3

2 =

l – 3

1 2–

3

⎛ ⎞ ⎜ ⎟⎝ ⎠

19. TA = T

B U = 0

Q =W

W = +ve

20.5 3

2 42 2

= × + ×

RTU RT =11 RT

21. 1 0 2 0;

C V C Vf f f f

C C

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

0

0

2 6 3301 m/s

2 2 990

⇒ ⇒

f V fCf V

C f

22.1

2

4 2

1 1 A

A

2 2

1 2max

2 2

min 1 2

2 19

– 2 –1

A AI

I A A

= 10 log9 = 20 log3

23. F kx 0 0

0

3

4 4

x x

x x


4/8

0

4

kxF

21

2W kx

2

0 01

2 16 8 ⇒

x FxW k

24.

1 2

1 2

2 2

0

3

f f

kq kq

xx

��

q1 = – 9q

2

25. x

y

z

r 0

r

A

B

q 0

0

30

0

1

4

qE r r

r r

��

� �

08 5 2 3AB r r i j i j

��

= 6 8i j� �

0| | 36 64 10r r � ��

unit

9 6

3

9 10 50 106 8

10

E i j

��

� �2.7 3.6 � �

i j kNC–1

26. Let q1 and q

2 be the charges on the inner and outer

sphere.

1 2

2 24 4

q q

a b...(i)

q1 + q

2 = q ...(ii)

From (i) & (ii),

2 2

1 22 2 2 2;

qa qb

q qa b a b

1 2

2 20 0 0

( )

4 4 4 ( )

⇒

c

q q q a bV

q q a b

27. Heat = 21

2Cv =

1

2× 2 × 10–6 × (100)2 = 0.01 J

28. Resistance across AC is more than CB.

29. The electric current through ideal voltmeter is zero

According to loop rule

E – 1 × I – 1 × I = 0

I = 2

2 2E

= 1 A

Reading of voltmeter = VA – V

B

= [1 × I] = 1 × 1 = 1 V

30. mvR

qB

2 mT

qB

31. BAB

= 0 [2sin ]

4

I

OC

But OC = r cos or BAB

= 0

tan2

I

r

Magnetic field due to circular portion,

BAB

= 0 02 2

2 2 2

⎛ ⎞ ⎜ ⎟ ⎝ ⎠

I I

r r

Total magnetic field

0 0 0tan tan

2 2 2

I I I

r r r

32. Inside bar magnet, lines of force are from south to

north.

33. Emf = 210 50 250

d dt t

dt dt

= – (20t – 50)

= – (20 × 3 – 50)

= – 10 V

34. cos = 10

0.52 20

R

37. u = v = –40 cm

1 1

40

f

f = – 40 cm = –40

m100

Power = 1 100

40f = – 2.5 D

38. fo + f

e = 44 cm,

o

e

f

f = 10 , f

o = 10f

e

10fe + f

e = 44

fe = 4 cm

fo = 40 cm

39. = 5000 Å, d = 0.2 mm, D = 200 cm

xn = 2n

2

D

d

⎛ ⎞⎜ ⎟⎝ ⎠

= 2 × 32

D

d

⎛ ⎞⎜ ⎟⎝ ⎠

=

8

1

3 200 5000 101.5 cm

0.2 10


5/8

40. A = 2 2

4 3 2 4 3cos3

= 37 6

41. p =

2 2

p p

h h

m Q v m Q v

mpQpv = mQv

v = p p

m Q

v

m Q

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

= 1 1

4 2 8

vv

42.2

n

nr

43. 90% of the sample is decayed and 10% is

undecayed. Fraction of 1

10 lies between

31

2

⎛ ⎞⎜ ⎟⎝ ⎠

to

41

2

⎛ ⎞⎜ ⎟⎝ ⎠

; so the time is less than 4 half-life periods but

more than 3 half-life periods, i.e., between 30 to 40

days.

45.

A

B

A

B

Y

Y = AB A B

Y = A + B

CHEMISTRY

46. Mole fraction of pure component = 1

47. gram equivalent of oxide = gram equivalent of chloride

1 2

1 2

m m =

Eq. wt Eq. wt

3 5

x 8 x 35.5

x = 33.25

48. p orbital can accommodate maximum two electrons

with opposite spin.

51. Mg2C

3 + 4H

2O 2Mg(OH)

2 + CH

3 – C CH

Propyne

52. In BrF3, hybridisation is sp3d [two lone pair and

3 bond]

Br

F

F

F

Due to lone pair-bond pair repulsion, no

F–Br–F bond angle in it is equal to 90°.

54. Z > 1

real

ideal

V1

V

realV

122.4

Vreal

> 22.4 L.

55. Rate of diffusion or effusion area

mol. wt.

2

2

H

O

r 1 32 1 4 2

r 2 2 2 1 1

56. q = 0 U = W

For expansion, W = –ve

U = –ve

57. Due to high hydration energy of F– ion

58. Initial pH = 1 2a a

pK pK 7 119

2 2

Now, H+ + HCO3

– H2CO

3

t = 0, meq 0.9 0.1 0

Finally, meq 0.8 0 0.1

[H+] = 30.8

8 10 M100

pH = –log (8 × 10–3) = 3 – log8 = 3 – 3log2

pH = 9 – [3 – 3log2] = 6 + 3log2

59. If Qc < K

c then net reaction is take place in forward

direction.

60. In neutral medium Mn+7 is convert into Mn+4,

n-factor = 3

61. In group 6th only Cr form hydride.

63. Orthoboric acid is mono basic lewis acid.

65.

Br +

AgNO3 + AgBr

6e–

Aromatic

66.

H

C CHCH

2sp sp sp

2

sp2

C

67.C

CHCH2

CH2

CH3

H+

CCHCH

2

CH2

CH3

+

CCHCH

3

CH2

CH3

+

Br–

H

CCHCH

3

CH2

CH3

(Major)

++

Br

(Stable due to hyperconjugation)


6/8

68. 8 in CsCl type structure.

69.bcc

Fcc

d 2 3 3 3

d 4 4 4 4

27

128 1

5

70. Tb i c

For glucose 1 0.05 = 0.05

For KCl 2 0.02 = 0.04

For CaCl2

3 0.01 = 0.03

For NaCl 2 0.01 = 0.02

71.

2H /HE 0.0591 pH

= –0.0591 10 = –0.591 V

72. t = nt1/2

800 = n 160 n = 5

0 0

t n 5

a 1a

322 2 a

73. (I) For 1st order reaction, = 1 – e–kt

at t = 0, = 0 and at t 1

(II) x = kt for zero order reaction

0

0

ka kt t

a

⎛ ⎞ ⇒ ⎜ ⎟

⎝ ⎠

Straight line passing through origin.

74. Sb2S

3 is negatively charged colloids.

77. Racemic mixture + optically active substance

mixture of diastereomers

Separation by

fractional

distillation

Separated diastereomers

79.

+

CH –OH2

H+

–H O2

CH2

+

Ring expansion –H+

..

80.O

H+

O

H

+

R–CH O2

H

O

H

OCH R2

Acetal

–H+

81. Tollen’s reagent react with terminal alkyne, aldehyde

and formic acid.

82. Hydride shifting is slow step in Cannizzaro reaction.

83. R–NH–H + C H SO Cl6 5 2

R–NHSO C H2 6 5

Alkali

Soluble

1° Amine

84. and -glucose are anomers.

88. Fe2+ have 4 unpaired electrons.

90.

O

CH –CH –C–OH2 2


O

O

MnOO

dil OH , –

Aldol

Condensation

BIOLOGY

91. Angiospermae.

93. Basidiomycetes – Agaricus, Puffball, Bracket fungi,

Ustilago.

94. b. BSE-caused by prions.

c. Phagic reproduction – Lytic cycle – T4

bacteriophages

95. A – All bryophytes.

B – Many algae.

96. Except Chara, Volvox and Equisetum, Dryopteris.

97. Hypogynous.

98. Fabaceae

99. Sap wood changed to heart wood.

101. X and W - Telomere formed by reverse transcriptase.

Telomeres are absent in prokaryotes as they have

circular DNA.

102. Intermediate Filaments.

104. Crossing over.

106. B - Nitrite reductase.

A - Nitrate reductase.

108. OAA is formed in the mesophyll cells.

109. Oxygenic photosynthesis is shown by all eukaryotic

algae/plants.

110. High ADP, low ATP, low acetyl CoA promotes

respiratory breakdown.

112. ABA - Closure of stomata.

Ethylene - Break seed dormancy, Fruit ripening

Leaf senescence

Synchronising fruit set in pineapple.

GA - Break seed dormancy, Bolting, Mobilisation of

reserve food.

114. Callose release microspores.

115. If it is from egg, then it is parthenogenesis.


7/8

117. SBE deficient and yellow.

1 3 3

4 4 16

118. Non-disjunction - Aneuploidy

Endomitosis - Polyploidy

119. Shine dalgarno sequence helps in initiation of

translation.

120. � Are transcribed.

� S-phase synthesis.

121. cAMP activates CAP

124. Meristem

125. High organic matter pollution.

126. Yeast - Monascus purpureus.

128. Birth rate = N 5

0.1N t 50

131. Sedge meadow stage.

132. National parks - in-situ conservation.

133. Due to introduction of Nile perch.

136. Planaria is a platyhelminth and does not show

bioluminescence.

137. Wuchereria is ovoviviparous (viviparous)

138. True fish includes dog fish, cat fish, puffer fish, shark

and eel as they belong to pisces.

139. Airs sacs (Birds)

Diaphragm (Mammal)

Placoid scales (Cartilaginous fish)

140. Epithelial tissue has highest power of regeneration

and nervous tissue has least power of regeneration.

142. Muscles fibres in stomach and cardiac region are

uninucleated.

143. Hypopharynx, labium, labrum, frons are unpaired

structures.

144. Compound eye of cockroach has good sensitivity but

poor resolution.

145. Fructose is laevorotatory.

Maltose and sucrose are disaccharides.

146. PUFA are essential fatty acid.

147. Z-DNA has 12 bp/turn.

148. Activation energy is not a factor affecting rate of

reaction.

149. Fatty acid are absorbed passively.

150. Indigestion can lead to vomiting (emesis) and

abdominal pain.

151. RV, TLC and FRC cannot be measured directly by

spirometer.

152. External intercostals and phrenic muscles relax

during expiration.

153. Basophil increase in inflammation.

Neutrophils are most abundant WBCs with

polymorphic nucleus. Lymphocytes are not

macrophages.

154. Semilunar valves open at the end of ventricular

diastole.

155. End of T-wave marks end of ventricular systole.

157. 75% H2O is reabsorbed in PCT and conditional

reabsorption occurs in DCT.

158. Kidney stones are mostly crystallised form of Ca

oxalate.

159. Smooth and cardiac muscles utilise both intracellular

and extracellular Ca2+ for contraction.

160. Gliding-between carpals, Pivot-Axis-Atlas

Hinge-Knee joint, Saddle-1st carpal-metacarpal of

thumb.

161. Duct of Sylvius / Iter / cerebral aqueduct is in

midbrain.

162. Proprioreceptors provide sense of position.

163. Nasal epithelium has Bowman’s gland to secrete

mucous for chemoreception.

164. Growth hormone, glucocorticoids and glucagon

increase blood glucose (Diabetogenic).

165. Grave’s disease is an autoimmune disease leading

to excessive production of thyroxine.

Cretinism – Deficiency of thyroxine.

Diabetes insipidus – Deficiency of ADH.

Addison’s disease – Deficiency of Aldosterone etc.

166. In honey bee only drones (males) develop through

parthenogenesis.

167. Impotency is erectile dysfunction.

168. Prolactin causes secretion and synthesis of milk.

169. Oophorectomy — Ovary removal

Mastectomy — Mammary gland removal

Hysterectomy — Uterus removal

171. Coitus interruptus is a natural method of

contraception.

172. Flippers of whale and penguin are analogous, rest

are homologous structures.


8/8

174. Natural selection is directional.

Mutation is directionless.

175. Aedes ageypti is vector for dengue, yellow fever,

chikungunya and zika virus. Ebola spreads through

blood, body fluids.

176. Complement protein –30 proteins in blood

Mast cell – Basophil

PMNL (polymorphonuclear leucocyte) is neutrophil

177. Morphine, benzodiazepine and barbiturates are

depressant, cocaine is a stimulant.

178. When Sal-I is used for insertion of gene of interest

in pBR322 then insertional inactivation of tetracycline

resistant gene will occur.

179. Flavr-savr variety is modified tomato in which there

is delayed ripening.

180. Glyphosate is glycine derivative herbicide which

prevent synthesis of aromatic amino acid.

� � �

Test - 10 (Code-B) (Answers) All India Aakash Test Series for Medical-2018

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1. (2)

2. (1)

3. (3)

4. (2)

5. (4)

6. (3)

7. (2)

8. (2)

9. (1)

10. (4)

11. (4)

12. (3)

13. (2)

14. (4)

15. (3)

16. (2)

17. (1)

18. (2)

19. (2)

20. (3)

21. (3)

22. (4)

23. (2)

24. (2)

25. (3)

26. (4)

27. (4)

28. (4)

29. (2)

30. (4)

31. (3)

32. (2)

33. (2)

34. (3)

35. (1)

36. (2)

Test Date : 01/04/2018

TEST - 10 (Code B)

All India Aakash Test Series for Medical - 2018

ANSWERS

37. (4)

38. (2)

39. (2)

40. (4)

41. (4)

42. (1)

43. (4)

44. (1)

45. (2)

46. (1)

47. (2)

48. (2)

49. (2)

50. (3)

51. (2)

52. (2)

53. (1)

54. (2)

55. (4)

56. (2)

57. (2)

58. (2)

59. (3)

60. (1)

61. (2)

62. (3)

63. (3)

64. (4)

65. (3)

66. (2)

67. (1)

68. (2)

69. (4)

70. (2)

71. (2)

72. (2)

73. (4)

74. (3)

75. (4)

76. (4)

77. (4)

78. (4)

79. (1)

80. (4)

81. (3)

82. (2)

83. (2)

84. (4)

85. (3)

86. (1)

87. (2)

88. (4)

89. (4)

90. (2)

91. (3)

92. (3)

93. (2)

94. (4)

95. (2)

96. (4)

97. (4)

98. (3)

99. (3)

100. (2)

101. (3)

102. (1)

103. (4)

104. (3)

105. (1)

106. (3)

107. (2)

108. (3)

109. (2)

110. (2)

111. (4)

112. (2)

113. (4)

114. (1)

115. (2)

116. (3)

117. (2)

118. (4)

119. (3)

120. (3)

121. (2)

122. (1)

123. (3)

124. (3)

125. (1)

126. (1)

127. (2)

128. (3)

129. (2)

130. (4)

131. (3)

132. (4)

133. (2)

134. (3)

135. (4)

136. (1)

137. (3)

138. (3)

139. (4)

140. (2)

141. (4)

142. (4)

143. (3)

144. (4)

145. (3)

146. (2)

147. (3)

148. (3)

149. (2)

150. (1)

151. (3)

152. (3)

153. (3)

154. (1)

155. (3)

156. (3)

157. (4)

158. (2)

159. (4)

160. (2)

161. (3)

162. (1)

163. (4)

164. (1)

165. (2)

166. (4)

167. (4)

168. (2)

169. (4)

170. (1)

171. (1)

172. (4)

173. (2)

174. (1)

175. (1)

176. (3)

177. (4)

178. (1)

179. (4)

180. (4)

All India Aakash Test Series for Medical-2018 Test - 10 (Code-B) (Hints)

2/8

PHYSICS

1.

A

B

A

B

Y

Y = AB A B

Y = A + B

3. 90% of the sample is decayed and 10% is

undecayed. Fraction of 1

10 lies between

31

2

⎛ ⎞⎜ ⎟⎝ ⎠

to

41

2

⎛ ⎞⎜ ⎟⎝ ⎠

; so the time is less than 4 half-life periods but

more than 3 half-life periods, i.e., between 30 to 40

days.

4.2

n

nr

5. p =

2 2

p p

h h

m Q v m Q v

mpQpv = mQv

v = p p

m Q

v

m Q

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

= 1 1

4 2 8

vv

6. A = 2 2

4 3 2 4 3cos3

= 37 6

7. = 5000 Å, d = 0.2 mm, D = 200 cm

xn = 2n

2

D

d

⎛ ⎞⎜ ⎟⎝ ⎠

= 2 × 32

D

d

⎛ ⎞⎜ ⎟⎝ ⎠

=

8

1

3 200 5000 101.5 cm

0.2 10

8. fo + f

e = 44 cm,

o

e

f

f = 10 , f

o = 10f

e

10fe + f

e = 44

fe = 4 cm

fo = 40 cm

9. u = v = –40 cm

1 1

40

f

HINTS

f = – 40 cm = –40

m100

Power = 1 100

40f = – 2.5 D

12. cos = 10

0.52 20

R

13. Emf = 210 50 250

d dt t

dt dt

= – (20t – 50)

= – (20 × 3 – 50) = – 10 V

14. Inside bar magnet, lines of force are from south to

north.

15. BAB

= 0 [2sin ]

4

I

OC

But OC = r cos or BAB

= 0

tan2

I

r

Magnetic field due to circular portion,

BAB

= 0 02 2

2 2 2

⎛ ⎞ ⎜ ⎟ ⎝ ⎠

I I

r r

Total magnetic field

0 0 0tan tan

2 2 2

I I I

r r r

16. mvR

qB

2 mT

qB

17. The electric current through ideal voltmeter is zero

According to loop rule

E – 1 × I – 1 × I = 0

I = 2

2 2E

= 1 A

Reading of voltmeter = VA – V

B

= [1 × I] = 1 × 1 = 1 V

18. Resistance across AC is more than CB.

19. Heat = 21

2Cv =

1

2× 2 × 10–6 × (100)2 = 0.01 J

20. Let q1 and q

2 be the charges on the inner and outer

sphere.

1 2

2 24 4

q q

a b...(i)

q1 + q

2 = q ...(ii)

Test - 10 (Code-B) (Hints) All India Aakash Test Series for Medical-2018

3/8

From (i) & (ii),

2 2

1 22 2 2 2;

qa qb

q qa b a b

1 2

2 20 0 0

( )

4 4 4 ( )

⇒

c

q q q a bV

q q a b

21. x

y

z

r 0

r

A

B

q 0

0

30

0

1

4

qE r r

r r

��

� �

08 5 2 3AB r r i j i j

��

= 6 8i j� �

0| | 36 64 10r r � ��

unit

9 6

3

9 10 50 106 8

10

E i j

��

� �2.7 3.6 � �

i j kNC–1

22.

1 2

1 2

2 2

0

3

f f

kq kq

xx

��

q1 = – 9q

2

23. F kx 0 0

0

3

4 4

x x

x x

0

4

kxF

21

2W kx

2

0 01

2 16 8 ⇒

x FxW k

24.1

2

4 2

1 1 A

A

2 2

1 2max

2 2

min 1 2

2 19

– 2 –1

A AI

I A A

= 10 log9 = 20 log3

25. 1 0 2 0;

C V C Vf f f f

C C

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

0

0

2 6 3301 m/s

2 2 990

⇒ ⇒

f V fCf V

C f

26.5 3

2 42 2

= × + ×

RTU RT =11 RT

27. TA = T

B U = 0

Q =W

W = +ve

28. 1 =

l – 3

2 =

l – 3

1 2–

3

⎛ ⎞ ⎜ ⎟⎝ ⎠

29.1

1

2m

tK

and 2

2

2m

tK

In series,

2

1 2

1

4 mK

t

and

2

2 2

2

4 mK

t

1 2

1 1 1

eqK K K

= +

2 22

1 2

2 2 24 4 4

t tT

m m m

,

2 2 2

1 2= +T t t

30.Fl

YA l

Y, l and F are constant

2

1l

D

2

2 1

21 2

1

16

l D

l D

2

1

16Δ =l mm.

31.1KEr

(r = radius of orbit)

3/2T r

3/2( )T KE

3/2T x

32.

2

2h

h

.1

.2

Applying Bernoulli’s theorem between points (1) & (2)

(Patm

+ 2gh) + g(2h) = Patm

+ 1

22v2

v = 2 gh


4/8

33. Maximum height after nth collision

hn = h

0e2n

n = 2; h2 = h

0e4

100

6.25 m16

34.

A

vc

vP

v rc +

v r c –

B

For pure rolling, velocity of point of contact has to be

equal to the velocity of the surface.

Let cylinder rolls with angular velocity .

At point B,

vc – r = v

p

r = vc – v

p

At point A,

vA = v

c + r

= vc + v

c – v

p

= 2vc – v

p

35. Mass of each of the four parts = 3

m

Mass of the plate including the cut piece = 4

3

m

MI of the whole plate (including the cut piece) about

the said axis =

24

3 6

⎛ ⎞⎟⎜ ⎟⎜ ⎟⎟⎜⎝ ⎠

m l

Now, M.I. of the remaining portion should be

2 23 4

4 3 6 6

⎛ ⎞⎟⎜ =⎟⎜ ⎟⎟⎜⎝ ⎠

m l ml

36.Net external force sin

Total mass 2

mg mga

m

(1 sin )

2

g

37. Acceleration of centre of mass

1 1 2 2

cm

1 2

m a m aa

m m

22

3 3

ma maF

m m

2

2 F maa

m

38. If a liquid of density is flowing through a pipe of

cross-section A at speed v, the mass coming out

per second will = ρ

dmAv

dt

In order to get n time water in the same time, we

get

⎛ ⎞⎟⎜= ⎟⎜ ⎟⎟⎜⎝ ⎠

dm dmn

dt dt i.e. AV = nAV

But as pipe and liquid are same

= , A = A

V = nV

2

/

dm dmV nV n

F dt dtn

dmF V dm dtV

dt

F = n2F

23

P F v n f nvn

P FV fv

P = n3P

39. Acceleration, a = 24

2 m/s2

= =

f

m

Displacement along OE, S1 = vt = 3 × 4 = 12 m

Displacement pendendicular to OE

2 2

2

1 12 4 16 m

2 2= = × × =S at

Resultant displacement

2 2

1 2144 256 400 20 mS S S= + = + = =

40. x2 = y

Differentiating w.r.t time

y

xv

ay

2 2x y

dx dyx xv vdt dt

⇒


5/8

Again differentiating w.r.t time

ay = 2v

x

2 + 2x ax

ax = 0, tangential acceleration of particle is zero

ay = 2v2 ( v

x = v)

Radius of curvature of origin is

21

0.5 unit2

y

vR

a ⇒

41.

2 2 5 1010 m

10

⇒ x yu u

Rg

43. Relative velocity of the train w.r.t. other train is

V – v.

Hence 0 – (V – v)2 = 2ax a = ( )

2

–

2

V v

x

Minimum retardation = ( )

2

–

2

V v

x

44.

0

=µ

BH

–1

⎡ ⎤⎡ ⎤⎢ ⎥= = ⎢ ⎥⎣ ⎦⎢ ⎥⎣ ⎦

IH AL

r

45.2 x m L T

x m L T

CHEMISTRY

46.

O



O

O

MnOO

dil OH , –

Aldol

Condensation

48. Fe2+ have 4 unpaired electrons.

52. and -glucose are anomers.

53. R–NH–H + C H SO Cl6 5 2

R–NHSO C H2 6 5

Alkali

Soluble

1° Amine

54. Hydride shifting is slow step in Cannizzaro reaction.

55. Tollen’s reagent react with terminal alkyne, aldehyde

and formic acid.

56.O

H+

O

H

+

R–CH O2

H

O

H

OCH R2

Acetal

–H+

57.

+

CH –OH2

H+

–H O2

CH2

+

Ring expansion –H+

..

59. Racemic mixture + optically active substance

mixture of diastereomers

Separation by

fractional

distillation

Separated diastereomers

62. Sb2S

3 is negatively charged colloids.

63. (I) For 1st order reaction, = 1 – e–kt

at t = 0, = 0 and at t 1

(II) x = kt for zero order reaction

0

0

ka kt t

a

⎛ ⎞ ⇒ ⎜ ⎟

⎝ ⎠

Straight line passing through origin.

64. t = nt1/2

800 = n 160 n = 5

0 0

t n 5

a 1a

322 2 a

65.

2H /HE 0.0591 pH

= –0.0591 10 = –0.591 V

66. Tb i c

For glucose 1 0.05 = 0.05

For KCl 2 0.02 = 0.04

For CaCl2

3 0.01 = 0.03

For NaCl 2 0.01 = 0.02

67.bcc

Fcc

d 2 3 3 3

d 4 4 4 4

27

128 1

5

68. 8 in CsCl type structure.

69.C

CHCH2

CH2

CH3

H+

CCHCH

2

CH2

CH3

+

CCHCH

3

CH2

CH3

+

Br–

H

CCHCH

3

CH2

CH3

(Major)

++

Br

(Stable due to hyperconjugation)

70.

H

C CHCH

2sp sp sp

2

sp2

C


6/8

71.

Br +

AgNO3 + AgBr

6e–

Aromatic

73. Orthoboric acid is mono basic lewis acid.

75. In group 6th only Cr form hydride.

76. In neutral medium Mn+7 is convert into Mn+4,

n-factor = 3

77. If Qc < K

c then net reaction is take place in forward

direction.

78. Initial pH = 1 2a a

pK pK 7 119

2 2

Now, H+ + HCO3

– H2CO

3

t = 0, meq 0.9 0.1 0

Finally, meq 0.8 0 0.1

[H+] = 30.8

8 10 M100

pH = –log (8 × 10–3) = 3 – log8 = 3 – 3log2

pH = 9 – [3 – 3log2] = 6 + 3log2

79. Due to high hydration energy of F– ion

80. q = 0 U = W

For expansion, W = –ve

U = –ve

81. Rate of diffusion or effusion area

mol. wt.

2

2

H

O

r 1 32 1 4 2

r 2 2 2 1 1

82. Z > 1

real

ideal

V1

V

realV

122.4

Vreal

> 22.4 L.

84. In BrF3, hybridisation is sp3d [two lone pair and

3 bond]

Br

F

F

F

Due to lone pair-bond pair repulsion, no

F–Br–F bond angle in it is equal to 90°.

85. Mg2C

3 + 4H

2O 2Mg(OH)

2 + CH

3 – C CH

Propyne

88. p orbital can accommodate maximum two electrons

with opposite spin.

89. gram equivalent of oxide = gram equivalent of chloride

1 2

1 2

m m =

Eq. wt Eq. wt

3 5

x 8 x 35.5

x = 33.25

90. Mole fraction of pure component = 1

BIOLOGY

93. Due to introduction of Nile perch.

94. National parks - in-situ conservation.

95. Sedge meadow stage.

98. Birth rate = N 5

0.1N t 50

100. Yeast - Monascus purpureus.

101. High organic matter pollution.

102. Meristem

105. cAMP activates CAP

106. � Are transcribed.

� S-phase synthesis.

107. Shine dalgarno sequence helps in initiation of

translation.

108. Non-disjunction - Aneuploidy

Endomitosis - Polyploidy

109. SBE deficient and yellow.

1 3 3

4 4 16

111. If it is from egg, then it is parthenogenesis.

112. Callose release microspores.

114. ABA - Closure of stomata.

Ethylene - Break seed dormancy, Fruit ripening

Leaf senescence

Synchronising fruit set in pineapple.

GA - Break seed dormancy, Bolting, Mobilisation of

reserve food.

116. High ADP, low ATP, low acetyl CoA promotes

respiratory breakdown.

117. Oxygenic photosynthesis is shown by all eukaryotic

algae/plants.

118. OAA is formed in the mesophyll cells.

120. B - Nitrite reductase.

A - Nitrate reductase.


7/8

122. Crossing over.

124. Intermediate Filaments.

125. X and W - Telomere formed by reverse transcriptase.

Telomeres are absent in prokaryotes as they have

circular DNA.

127. Sap wood changed to heart wood.

128. Fabaceae

129. Hypogynous.

130. Except Chara, Volvox and Equisetum, Dryopteris.

131. A – All bryophytes.

B – Many algae.

132. b. BSE-caused by prions.

c. Phagic reproduction – Lytic cycle – T4

bacteriophages

133. Basidiomycetes – Agaricus, Puffball, Bracket fungi,

Ustilago.

135. Angiospermae.

136. Glyphosate is glycine derivative herbicide which

prevent synthesis of aromatic amino acid.

137. Flavr-savr variety is modified tomato in which there

is delayed ripening.

138. When Sal-I is used for insertion of gene of interest

in pBR322 then insertional inactivation of tetracycline

resistant gene will occur.

139. Morphine, benzodiazepine and barbiturates are

depressant, cocaine is a stimulant.

140. Complement protein –30 proteins in blood

Mast cell – Basophil

PMNL (polymorphonuclear leucocyte) is neutrophil

141. Aedes ageypti is vector for dengue, yellow fever,

chikungunya and zika virus. Ebola spreads through

blood, body fluids.

142. Natural selection is directional.

Mutation is directionless.

144. Flippers of whale and penguin are analogous, rest

are homologous structures.

145. Coitus interruptus is a natural method of

contraception.

147. Oophorectomy — Ovary removal

Mastectomy — Mammary gland removal

Hysterectomy — Uterus removal

148. Prolactin causes secretion and synthesis of milk.

149. Impotency is erectile dysfunction.

150. In honey bee only drones (males) develop through

parthenogenesis.

151. Grave’s disease is an autoimmune disease leading

to excessive production of thyroxine.

Cretinism – Deficiency of thyroxine.

Diabetes insipidus – Deficiency of ADH.

Addison’s disease – Deficiency of Aldosterone etc.

152. Growth hormone, glucocorticoids and glucagon

increase blood glucose (Diabetogenic).

153. Nasal epithelium has Bowman’s gland to secrete

mucous for chemoreception.

154. Proprioreceptors provide sense of position.

155. Duct of Sylvius / Iter / cerebral aqueduct is in

midbrain.

156. Gliding-between carpals, Pivot-Axis-Atlas

Hinge-Knee joint, Saddle-1st carpal-metacarpal of

thumb.

157. Smooth and cardiac muscles utilise both intracellular

and extracellular Ca2+ for contraction.

158. Kidney stones are mostly crystallised form of Ca

oxalate.

159. 75% H2O is reabsorbed in PCT and conditional

reabsorption occurs in DCT.

161. End of T-wave marks end of ventricular systole.

162. Semilunar valves open at the end of ventricular

diastole.

163. Basophil increase in inflammation.

Neutrophils are most abundant WBCs with

polymorphic nucleus. Lymphocytes are not

macrophages.

164. External intercostals and phrenic muscles relax

during expiration.

165. RV, TLC and FRC cannot be measured directly by

spirometer.

166. Indigestion can lead to vomiting (emesis) and

abdominal pain.

167. Fatty acid are absorbed passively.

168. Activation energy is not a factor affecting rate of

reaction.

169. Z-DNA has 12 bp/turn.

170. PUFA are essential fatty acid.


8/8

171. Fructose is laevorotatory.

Maltose and sucrose are disaccharides.

172. Compound eye of cockroach has good sensitivity but

poor resolution.

173. Hypopharynx, labium, labrum, frons are unpaired

structures.

174. Muscles fibres in stomach and cardiac region are

uninucleated.

176. Epithelial tissue has highest power of regeneration

and nervous tissue has least power of regeneration.

177. Airs sacs (Birds)

Diaphragm (Mammal)

Placoid scales (Cartilaginous fish)

178. True fish includes dog fish, cat fish, puffer fish, shark

and eel as they belong to pisces.

179. Wuchereria is ovoviviparous (viviparous)

180. Planaria is a platyhelminth and does not show

bioluminescence.

� � �

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