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Test - 7 (Code-C) (Answers & Hints) All India Aakash Test Series for Medical-2018

1/9

1. (1)

2. (3)

3. (4)

4. (3)

5. (2)

6. (1)

7. (2)

8. (1)

9. (1)

10. (3)

11. (2)

12. (4)

13. (4)

14. (3)

15. (2)

16. (4)

17. (4)

18. (2)

19. (1)

20. (2)

21. (3)

22. (1)

23. (1)

24. (3)

25. (3)

26. (1)

27. (1)

28. (3)

29. (3)

30. (2)

31. (3)

32. (1)

33. (2)

34. (3)

35. (1)

36. (3)

37. (1)

38. (4)

39. (2)

40. (1)

41. (3)

42. (1)

43. (3)

44. (3)

45. (1)

46. (3)

47. (3)

48. (1)

49. (3)

50. (1)

51. (2)

52. (4)

53. (4)

54. (3)

55. (1)

56. (2)

57. (2)

58. (3)

59. (3)

60. (2)

61. (1)

62. (1)

63. (Deleted)

64. (4)

65. (1)

66. (3)

67. (3)

68. (3)

69. (4)

70. (1)

71. (3)

72. (3)

73. (1)

74. (1)

75. (4)

76. (4)

77. (4)

78. (3)

79. (3)

80. (2)

81. (1)

82. (4)

83. (3)

84. (2)

85. (2)

86. (4)

87. (3)

88. (4)

89. (3)

90. (2)

91. (2)

92. (3)

93. (2)

94. (1)

95. (3)

96. (2)

97. (2)

98. (2)

99. (1)

100. (4)

101. (2)

102. (2)

103. (3)

104. (3)

105. (1)

106. (2)

107. (1)

108. (1)

109. (3)

110. (4)

111. (4)

112. (2)

113. (3)

114. (3)

115. (3)

116. (3)

117. (4)

118. (4)

119. (4)

120. (2)

121. (1)

122. (2)

123. (2)

124. (2)

125. (2)

126. (4)

127. (2)

128. (3)

129. (4)

130. (3)

131. (3)

132. (1)

133. (2)

134. (1)

135. (3)

136. (1)

137. (1)

138. (3)

139. (2)

140. (1)

141. (2)

142. (1)

143. (4)

144. (3)

145. (2)

146. (2)

147. (4)

148. (3)

149. (1)

150. (3)

151. (3)

152. (1)

153. (1)

154. (3)

155. (1)

156. (2)

157. (4)

158. (3)

159. (1)

160. (3)

161. (2)

162. (3)

163. (4)

164. (2)

165. (1)

166. (2)

167. (4)

168. (3)

169. (2)

170. (4)

171. (1)

172. (3)

173. (2)

174. (1)

175. (1)

176. (2)

177. (4)

178. (3)

179. (3)

180. (3)

ANSWERS

TEST - 7 (Code-C)

All India Aakash Test Series for Medical-2018

Test Date : 19-02-2017

All India Aakash Test Series for Medical-2018 Test - 7 (Code-C) (Answers & Hints)

2/9

Hints to Selected Questions

[ PHYSICS]

1. Answer (1)

Efficiency of Carnot engine and coefficient of

performance of refrigerator are related as

1

...(i)

Here, percentage = 10%

or = 0.1 ...(ii)

Putting value of in (i),

1 0.1 0.99

0.1 0.1

= 9

2. Answer (3)

1 2

v

700 Hz 700 Hz

0

1

⎧ ⎫ ⎨ ⎬⎩ ⎭

v v

n n

v

0

2

⎧ ⎫ ⎨ ⎬⎩ ⎭

v v

n n

v

Given 2 1

8 n n

0 0{ } 8 nv v v v

v

02

8nv

v

02 700

8350

v

v0 = 2 m/s

3. Answer (4)

By law of mixture,

Heat given = Heat taken

18(1)(50 – ) = (10)(0.5)(0 + 20) + 10(80)

900 – 18 = 100 + 800

18 = 0

= 0°C

Temperature of mixture will be 0°C and whole ice

will melt.

4. Answer (3)

2FP F PA P VT

A ⇒

5. Answer (2)

rel boat water v v v

ˆ ˆ ˆ ˆ3 5 2 3i j i j

ˆ ˆ ˆ ˆ3 5 2 3i j i j ˆ ˆ5 8i j

6. Answer (1)

F T MA F T

AM

∵T ma

F ma

M

7. Answer (2)

8. Answer (1)

3 22 2 4 8

3 3

e

e e e

e e

GM Gv R G R

R R

ve R

e

9. Answer (1)

230

dUF r

dr

|F | = 30r2 ...(i)

2

mvF

r

2 230 m

r v

r

mv2 = 30r

3 = 30000

2 4115000 1.5 10 J

2 mv

10. Answer (3)

For simple pendulum in liquid,

11

2

⎛ ⎞⎜ ⎟⎝ ⎠

gd

nl

For spring pendulum in liquid, frequency remain

unchanged.


3/9

11. Answer (2)

Given, l1 = 1.44 m

l2 = 1.00 m

When two pendulums are in out of phase, set number

of vibration completed by them are n1 and n

2

respectively. We can say if n1 = n, then n

2 = n +

1

2.

Now, by question set they are out of phase after time

t. So,

t = n1T1 = n

2T2

1 2

2 1

n T

n T

2

1

1 10

1 1.44 12

2

n

n

l

l

12n = 10n + 5

2n = 5

n = 2.5

Now, t = n1T1

= 1(2.5)(2 )g

l

= 1.44

(2.5)(2)( )10

6 s

12. Answer (4)

Here, H1 = H

2 = H

3

1 1 2 2(120 ) 1.5 ( ) 3 ( 0)

KA T KA T T KA T

l l l

(120 – ) = 1.5( – ) = 3T T T T1 1 2 2

120 – = 3T T1 2

1.5( – ) = 3T T1 2

T2

T T1 2 + 3 = 120 ...(i) T T

1 2 = 3 ...(ii)

Solving (i) and (ii),

T1 = 60°C

T2 = 20°C

13. Answer (4)

E T4

4

1 1

2 2

⎛ ⎞ ⎜ ⎟⎝ ⎠

E T

E T

Put T1 = T

T2 = 1.5T

1

2

1

5

E

E

E2 5E

1

E2 = 500% of E

1

14. Answer (3)

2

12 m

Il

I l2

2I

I

l

l

⎛ ⎞ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠

l

l

l

l 2 2I I I l

l

I = 2I

15. Answer (2)

2

' 1 9

hg g h R

R

⎛ ⎞ ⇒ ⎜ ⎟⎝ ⎠

16. Answer (4)

Final sphere

Final particle

90°

30° = 1

60° = 1

1 2+ = 90°

Initial particle

17. Answer (4)

18. Answer (2)

Temporary loss =

2

21 1(2 )

2 2 2

⎛ ⎞ ⎜ ⎟⎝ ⎠

vmv m

= 2 21 1 1

2 2 2

⎛ ⎞ ⎜ ⎟⎝ ⎠

mv mv

= 2

EE

= 2

E

19. Answer (1)

m1

m2

l1

l2

l

COM


4/9

I = m1l1

2 + m2l2

2 ...(i)

Here,2

1

1 2

m

m m

ll

1

2

1 2

m

m m

ll

Put value of l1 and l

2 in (i) and get answer.

20. Answer (2)

O

45°

Hmax

u cos 45°

L = mvr

= m(u cos)Hmax

=

2 2sin 45

22u u

mg

L =

3

4 2

mu

g

21. Answer (3)

f iE ME ME ME

22. Answer (1)

23. Answer (1)

A

Bmg

L/2

L cos

2

I 2

cos2 3

L MLmg

3cos

2

g

L

24. Answer (3)

nB – n

A = n

1...(A)

nC – n

A = n

2...(B)

From (A) and (B),

nC – n

B = n

2 – n

1

25. Answer (3)

2 kg block do not tendency to move. So there is no

force between them.

26. Answer (1)

27. Answer (1)

2 21 1

2 2 mv mgx Kx

28. Answer (3)

2

2 2 23 33

2 2 2 MR

MR MR MK

2 273

2MR MK

7

6K R

29. Answer (3)

Work = Change in rotational kinetic energy

W = 2 2

2 1

1( )

2 I

=

2

2 2 2

2 1

1(4 )( )

2 2

⎛ ⎞ ⎜ ⎟

⎝ ⎠

MRn n

= 1500 J

30. Answer (2)

3

5A

B...(i)

2 235 2 cos 60 A B AB ...(ii)

Solving (i) and (ii), we get

A = 15, B = 25

31. Answer (3)

32. Answer (1)

a

1x

x2 – 1

For block to remain stationary over wedge,

mg sin = ma cos

2

1tan

1

⎛ ⎞ ⎜ ⎟⎝ ⎠

a g g

x

21

ga

x


5/9

33. Answer (2)

For fourth harmonic, the string will vibrate in four

loops.

L/4

L/8Pressed

Plucked

34. Answer (3)

Given, 2

2

v

l = n

or n = v

l...(i)

For closed pipe, frequency of second overtone is

5 5

4 4 v n

n

l.

35. Answer (1)

2 23 38 2 8 m/sv n v

36. Answer (3)

U = (N1/3 – 1)U

37. Answer (1)

P h g

V V

V V

100 V

V= 100

h g

=

3

8

100 10 9.8 100

9.8 10

= 0.1

38. Answer (4)

39. Answer (2)

40. Answer (1)

L = 10

0

10 logI

I

= 10 log10

105

= 50 log10

10

= 50 dB

41. Answer (3)

22

1 2max 1 2

min 1 2 1 2

9

1

⎛ ⎞⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠

I II a a

I a a I I

42. Answer (1)

43. Answer (3)

44. Answer (3)

L = MvR + Ic

45. Answer (1)

The elongation produced by m1 at equilibrium is the

amplitude for m2

1m g Ka

1m g

aK

⇒

[ CHEMISTRY]

46. Answer (3)

N1V1 = N

2V2

2

1 1200 V

2 10

V2 = 1000

Water added = 1000 – 200 = 800 cc

47. Answer (3)

N = M × nfactor

48. Answer (1)

Atomic mass × specific heat = 6.4

Atomic mass = 6.4 6.4

Specific heat 0.16 = 40

49. Answer (3)

3 2 2 21mol2 mol100 g

CaCO 2HCl CaCl H O CO

Number of moles of HCl taken = 1 × 1

1 mol HCl will produce 1

2 mol CO

2.

50. Answer (1)

8

6

c 3 100.75 m

v 400 10


6/9

51. Answer (2)

2 1

4 3

(E E ) ( 3.4) ( 13.6) 10.2eV

(E E ) ( 0.85) ( 1.5) 0.66

52. Answer (4)

2

2

ZE

n

2n

rZ

Zv

n

53. Answer (4)

B

O O – H

H

O

H

54. Answer (3)

BrF5 (sp

3d2)

Br

F

F

F

F

F

Due to lp–bp repulsion none of the bond angle is at

90°.

55. Answer (1)

Fact.

56. Answer (2)

2

1Cl Cl (aq)

2

H = 1

2 × 240 + (–349) + (–381)

= –610 kJ mol–1

57. Answer (2)

Fact.

58. Answer (3)

2O

W 16n

M 32

W = 2

10

1

V2.303 nRT log

V

= 10

16 502.303 8.314 300 log joule

32 5

= –2.87 × 103 J

= –2.87 kJ

59. Answer (3)

Fact.

60. Answer (2)

rH° =

i f i fi i

a H (Product) b H (Reactant)

= 6 × (–94) + 3 × (–68.3) – 11.7

= –564 – 204.9 – 11.7

rH° = –780.6 kcal

61. Answer (1)

Fact.

62. Answer (1)

2CO

P

CO

pK

p

21.4 0.800

1.4 p 0.80 p

FeO CO(g) Fe CO (g)

��⇀↽��

0.80 p0.625

1.4 p

0.875 – 0.625p = 0.80 + p

0.875 – 0.80 = p + 0.625p

0.075 = 1.625p

p = 0.046

pCO

= 1.4 – 0.046 = 1.354

2CO

p 0.80 0.046 0.846

63. Deleted

64. Answer (4)

H+ = 14

9.62 10

= 3.10 × 10–7

pH = –log[3.10 × 10–7]

= 7 – 0.491

= 6.509

65. Answer (1)

Fact.

66. Answer (3)

67. Answer (3)

Fact.

68. Answer (3)

5 3

6 6 3 6 5 2C H HNO C H NO

69. Answer (4)

2 3

2 7Cr O 6e 2Cr


7/9

70. Answer (1)

Gram equivalent KMnO4

= Gram equivalent FeSO4


= N × V (L)

= 1 × 200

1000 = 0.2

Gram equivalent FeSO4 =

Weight

Equivalent weight

Weight0.2 30.4 g

152

71. Answer (3)

Fact.

72. Answer (3)

Na, Al, Si, P, S, Cl are paramagnetic.

73. Answer (1)

NaHCO3 + NaOH Na

2CO

3 + H

2O

74. Answer (1)

Fact.

75. Answer (4)

Fact.

76. Answer (4)

Both acidic oxide and amphoteric oxide react with

NaOH.

77. Answer (4)

N

H H

H

H+

78. Answer (3)

Zeolites are 3-D silicates.

79. Answer (3)

CH

H

CH3

CH3

C

, ,

CH – C – C – CH3 3

H CH3

H CH3

(Alkanes does not exhibitgeometrical isomerism)

,

80. Answer (2)

Homolytic fission produces free radical

3 3 3CH CH 2CH

81. Answer (1)

H w.r.t. carbonyl group should be present for

tautomerism.

82. Answer (4)

Fact.

83. Answer (3)

12 0.396Percentage of C 100 21.95%

44 0.492

2 0.2028Percentage of H 100 4.5%

18 0.492

Percentage of O = (100 – 21.95 – 4.5) = 73.47%

84. Answer (2)

Percentage of N = 1.4 m.eq. of acid used

Weight of compound

85. Answer (2)

SO3 is a neutral electrophile due to presence of

vacant d-orbital.

86. Answer (4)

Fact.

87. Answer (3)

Br

Cl

2Na/ether

Wurtz reaction

88. Answer (4)

CH = CH2

CH CH

OH

Hg /H SO2 2 4

+

CH CHO3

(X)

89. Answer (3)

CH –CH=CH2 2

CH–CH –CH2 2

CH –CH–CH2 3

HBr

Br–

Br

CH–CH –CH2 3

H shifting–

Benzyl carbocation

90. Answer (2)

Br

Br2

H O2

Br

BrOH

Br

H O2


8/9

[ BIOLOGY

]

91. Answer (2)

Thermoacidophiles

92. Answer (3)

Protista

93. Answer (2)

PS-II is present.

94. Answer (1)

Zygote

95. Answer (3)

Dinoflagellates

96. Answer (2)

97. Answer (2)

98. Answer (2)

99. Answer (1)

Heterospory – Pteridophytes

Seed habit – Gymnosperms

Fruit habit – Angiosperms

100. Answer (4)

Red algae

101. Answer (2)

102. Answer (2)

Fabaceae

103. Answer (3)

Analogous to tuber of sweet potato and possess

axillary buds.

104. Answer (3)

105. Answer (1)

All the tissue inside endodermis.

106. Answer (2)

Completely secondary in origin.

107. Answer (1)

108. Answer (1)

Oxidative phosphorylation

109. Answer (3)

Ovum (n), meiocyte (2n)

110. Answer (4)

Metaphase is followed by anaphase.

111. Answer (4)

112. Answer (2)

Consists of living part.

113. Answer (3)

ABA closes stomata.

114. Answer (3)

115. Answer (3)

116. Answer (3)

Legumes

117. Answer (4)

118. Answer (4)

C3

119. Answer (4)

C4 – 5, 2

120. Answer (2)

Glycolysis – 1 NADH

Link – 1 NADH

TCA cycle – 3 NADH and 1 FADH2

121. Answer (1)

122. Answer (2)

Fats are poor in oxygen.

123. Answer (2)

124. Answer (2)

Secondary meristem

125. Answer (2)

126. Answer (4)

Precursor to seed habit – Selaginella.

127. Answer (2)

128. Answer (3)

Tuber aestivum

129. Answer (4)

130. Answer (3)

Herbarium is quick referral system.

131. Answer (3)

Polymoniales

132. Answer (1)


9/9

133. Answer (2)

134. Answer (1)

135. Answer (3)

136. Answer (1)

Cnidaria and ctenophora are diploblastic.

137. Answer (1)

Echinoderms and hemichordates are also

deuterostomes.

138. Answer (3)

Sea-anemones are carnivores.

139. Answer (2)

140. Answer (1)

141. Answer (2)

142. Answer (1)

Sperms are formed in testes and stored in seminal

vesicles.

143. Answer (4)

144. Answer (3)

Pancreatic acini secrete pancreatic enzymes.

145. Answer (2)

146. Answer (2)

Carbohydrates are absorbed as monosaccharides

and amino acids are absorbed actively.

147. Answer (4)

Enterokinase activates trypsinogen.

148. Answer (3)

149. Answer (1)

150. Answer (3)

During inspiration, the intrapleural pressure becomes

negative.

151. Answer (3)

152. Answer (1)

153. Answer (1)

154. Answer (3)

Cardiac output to muscles increases during

exercise.

The resistance to flow in the pulmonary system is

much lower than in systemic circulation, hence

pressure in pulmonary artery is lower than in aorta.

155. Answer (1)

156. Answer (2)

157. Answer (4)

Descending limb of the loop of Henle is impermeable

to salts.

158. Answer (3)

These are in the cortex.

159. Answer (1)

160. Answer (3)

Alcohol suppresses ADH secretion.

161. Answer (2)

162. Answer (3)

163. Answer (4)

164. Answer (2)

165. Answer (1)

166. Answer (2)

167. Answer (4)

168. Answer (3)

Vitamin A itself is retinol and its aldehyde is retinal.

169. Answer (2)

170. Answer (4)

171. Answer (1)

172. Answer (3)

Hormones are required in traces.

173. Answer (2)

174. Answer (1)

175. Answer (1)

Pineal gland

176. Answer (2)

177. Answer (4)

178. Answer (3)

179. Answer (3)

180. Answer (3)

� � �

Test - 7 (Code-D) (Answers & Hints) All India Aakash Test Series for Medical-2018

1/9

1. (3)

2. (1)

3. (1)

4. (3)

5. (1)

6. (3)

7. (4)

8. (2)

9. (3)

10. (1)

11. (3)

12. (1)

13. (4)

14. (3)

15. (1)

16. (4)

17. (1)

18. (1)

19. (3)

20. (3)

21. (1)

22. (1)

23. (3)

24. (3)

25. (1)

26. (4)

27. (3)

28. (4)

29. (2)

30. (2)

31. (4)

32. (1)

33. (2)

34. (2)

35. (4)

36. (1)

37. (3)

38. (3)

39. (4)

40. (3)

41. (4)

42. (1)

43. (2)

44. (1)

45. (3)

46. (4)

47. (1)

48. (2)

49. (1)

50. (2)

51. (4)

52. (4)

53. (1)

54. (4)

55. (3)

56. (4)

57. (1)

58. (1)

59. (2)

60. (4)

61. (2)

62. (3)

63. (3)

64. (1)

65. (1)

66. (3)

67. (2)

68. (1)

69. (1)

70. (1)

71. (3)

72. (2)

73. (Deleted)

74. (3)

75. (3)

76. (4)

77. (1)

78. (1)

79. (4)

80. (4)

81. (3)

82. (1)

83. (2)

84. (2)

85. (4)

86. (3)

87. (1)

88. (3)

89. (1)

90. (1)

91. (1)

92. (3)

93. (4)

94. (3)

95. (1)

96. (1)

97. (2)

98. (1)

99. (4)

100. (2)

101. (4)

102. (4)

103. (4)

104. (4)

105. (3)

106. (4)

107. (2)

108. (2)

109. (4)

110. (1)

111. (1)

112. (1)

113. (1)

114. (4)

115. (4)

116. (2)

117. (1)

118. (3)

119. (3)

120. (4)

121. (3)

122. (1)

123. (1)

124. (4)

125. (4)

126. (2)

127. (3)

128. (2)

129. (4)

130. (4)

131. (1)

132. (3)

133. (4)

134. (1)

135. (4)

136. (1)

137. (1)

138. (1)

139. (4)

140. (4)

141. (3)

142. (3)

143. (4)

144. (1)

145. (3)

146. (2)

147. (4)

148. (1)

149. (2)

150. (4)

151. (3)

152. (4)

153. (4)

154. (1)

155. (4)

156. (1)

157. (3)

158. (1)

159. (2)

160. (4)

161. (3)

162. (1)

163. (1)

164. (3)

165. (1)

166. (3)

167. (3)

168. (1)

169. (4)

170. (4)

171. (2)

172. (1)

173. (4)

174. (3)

175. (4)

176. (3)

177. (4)

178. (1)

179. (3)

180. (3)

ANSWERS

TEST - 7 (Code-D)

All India Aakash Test Series for Medical-2018

Test Date : 19-02-2017

All India Aakash Test Series for Medical-2018 Test - 7 (Code-D) (Answers & Hints)

2/9

Hints to Selected Questions

[ PHYSICS]

1. Answer (3)

The elongation produced by m1 at equilibrium is the

amplitude for m2

1m g Ka

1m g

aK

⇒

2. Answer (1)

L = MvR + Ic

3. Answer (1)

4. Answer (3)

5. Answer (1)

22

1 2max 1 2

min 1 2 1 2

9

1

⎛ ⎞⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠

I II a a

I a a I I

6. Answer (3)

L = 10

0

10 logI

I

= 10 log10

105

= 50 log10

10

= 50 dB

7. Answer (4)

8. Answer (2)

9. Answer (3)

P h g

V V

V V

100 V

V= 100

h g

=

3

8

100 10 9.8 100

9.8 10

= 0.1

10. Answer (1)

U = (N1/3 – 1)U

11. Answer (3)

2 23 38 2 8 m/sv n v

12. Answer (1)

Given, 2

2

v

l = n

or n = v

l...(i)

For closed pipe, frequency of second overtone is

5 5

4 4 v n

n

l.

13. Answer (4)

For fourth harmonic, the string will vibrate in four

loops.

L/4

L/8Pressed

Plucked

14. Answer (3)

a

1x

x2 – 1

For block to remain stationary over wedge,

mg sin = ma cos

2

1tan

1

⎛ ⎞ ⎜ ⎟⎝ ⎠

a g g

x

21

ga

x

15. Answer (1)

16. Answer (4)

3

5A

B...(i)

2 235 2 cos 60 A B AB ...(ii)

Solving (i) and (ii), we get

A = 15, B = 25

17. Answer (1)

Work = Change in rotational kinetic energy

W = 2 2

2 1

1( )

2 I

=

2

2 2 2

2 1

1(4 )( )

2 2

⎛ ⎞ ⎜ ⎟

⎝ ⎠

MRn n

= 1500 J


3/9

18. Answer (1)

2

2 2 23 33

2 2 2 MR

MR MR MK

2 273

2MR MK

7

6K R

19. Answer (3)

2 21 1

2 2 mv mgx Kx

20. Answer (3)

21. Answer (1)

2 kg block do not tendency to move. So there is no

force between them.

22. Answer (1)

nB – n

A = n

1...(A)

nC – n

A = n

2...(B)

From (A) and (B),

nC – n

B = n

2 – n

1

23. Answer (3)

A

Bmg

L/2

L cos

2

I 2

cos2 3

L MLmg

3cos

2

g

L

24. Answer (3)

25. Answer (1)

f iE ME ME ME

26. Answer (4)

O

45°

Hmax

u cos 45°

L = mvr

= m(u cos)Hmax

=

2 2sin 45

22u u

mg

L =

3

4 2

mu

g

27. Answer (3)

m1

m2

l1

l2

l

COM

I = m1l1

2 + m2l2

2 ...(i)

Here,2

1

1 2

m

m m

ll

1

2

1 2

m

m m

ll

Put value of l1 and l

2 in (i) and get answer.

28. Answer (4)

Temporary loss =

2

21 1(2 )

2 2 2

⎛ ⎞ ⎜ ⎟⎝ ⎠

vmv m

= 2 21 1 1

2 2 2

⎛ ⎞ ⎜ ⎟⎝ ⎠

mv mv

= 2

EE

= 2

E

29. Answer (2)

30. Answer (2)

Final sphere

Final particle

90°

30° = 1

60° = 1

1 2+ = 90°

Initial particle

31. Answer (4)

2

' 1 9

hg g h R

R

⎛ ⎞ ⇒ ⎜ ⎟⎝ ⎠


4/9

32. Answer (1)

2

12 m

Il

I l2

2I

I

l

l

⎛ ⎞ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠

l

l

l

l 2 2I I I l

l

I = 2I

33. Answer (2)

E T4

4

1 1

2 2

⎛ ⎞ ⎜ ⎟⎝ ⎠

E T

E T

Put T1 = T

T2 = 1.5T

1

2

1

5

E

E

E2 5E

1

E2 = 500% of E

1

34. Answer (2)

Here, H1 = H

2 = H

3

1 1 2 2(120 ) 1.5 ( ) 3 ( 0)

KA T KA T T KA T

l l l

(120 – ) = 1.5( – ) = 3T T T T1 1 2 2

120 – = 3T T1 2

1.5( – ) = 3T T1 2

T2

T T1 2 + 3 = 120 ...(i) T T

1 2 = 3 ...(ii)

Solving (i) and (ii),

T1 = 60°C

T2 = 20°C

35. Answer (4)

Given, l1 = 1.44 m

l2 = 1.00 m

When two pendulums are in out of phase, set number

of vibration completed by them are n1 and n

2

respectively. We can say if n1 = n, then n

2 = n +

1

2.

Now, by question set they are out of phase after time

t. So,

t = n1T1 = n

2T2

1 2

2 1

n T

n T

2

1

1 10

1 1.44 12

2

n

n

l

l

12n = 10n + 5

2n = 5

n = 2.5

Now, t = n1T1

= 1(2.5)(2 )g

l

= 1.44

(2.5)(2)( )10

6 s

36. Answer (1)

For simple pendulum in liquid,

11

2

⎛ ⎞⎜ ⎟⎝ ⎠

gd

nl

For spring pendulum in liquid, frequency remain

unchanged.

37. Answer (3)

230

dUF r

dr

|F | = 30r2 ...(i)

2

mvF

r

2 230 m

r v

r

mv2 = 30r

3 = 30000

2 4115000 1.5 10 J

2 mv

38. Answer (3)

3 22 2 4 8

3 3

e

e e e

e e

GM Gv R G R

R R

ve R

e

39. Answer (4)

40. Answer (3)

F T MA F T

AM

∵T ma

F ma

M


5/9

[ CHEMISTRY]

46. Answer (4)

Br

Br2

H O2

Br

BrOH

Br

H O2

47. Answer (1)

CH –CH=CH2 2

CH–CH –CH2 2

CH –CH–CH2 3

HBr

Br–

Br

CH–CH –CH2 3

H shifting–

Benzyl carbocation

48. Answer (2)

CH = CH2

CH CH

OH

Hg /H SO2 2 4

+

CH CHO3

(X)

49. Answer (1)

Br

Cl

2Na/ether

Wurtz reaction

50. Answer (2)

Fact.

41. Answer (4)

rel boat water v v v

ˆ ˆ ˆ ˆ3 5 2 3i j i j

ˆ ˆ ˆ ˆ3 5 2 3i j i j ˆ ˆ5 8i j

42. Answer (1)

2FP F PA P VT

A ⇒

43. Answer (2)

By law of mixture,

Heat given = Heat taken

18(1)(50 – ) = (10)(0.5)(0 + 20) + 10(80)

900 – 18 = 100 + 800

18 = 0

= 0°C

Temperature of mixture will be 0°C and whole ice

will melt.

44. Answer (1)

1 2

v

700 Hz 700 Hz

0

1

⎧ ⎫ ⎨ ⎬⎩ ⎭

v v

n n

v

0

2

⎧ ⎫ ⎨ ⎬⎩ ⎭

v v

n n

v

Given 2 1

8 n n

0 0{ } 8 nv v v v

v

02

8nv

v

02 700

8350

v

v0 = 2 m/s

45. Answer (3)

Efficiency of Carnot engine and coefficient of

performance of refrigerator are related as

1

...(i)

Here, percentage = 10%

or = 0.1 ...(ii)

Putting value of in (i),

1 0.1 0.99

0.1 0.1

= 9


6/9

51. Answer (4)

SO3 is a neutral electrophile due to presence of

vacant d-orbital.

52. Answer (4)

Percentage of N = 1.4 m.eq. of acid used

Weight of compound

53. Answer (1)

12 0.396Percentage of C 100 21.95%

44 0.492

2 0.2028Percentage of H 100 4.5%

18 0.492

Percentage of O = (100 – 21.95 – 4.5) = 73.47%

54. Answer (4)

Fact.

55. Answer (3)

H w.r.t. carbonyl group should be present for

tautomerism.

56. Answer (4)

Homolytic fission produces free radical

3 3 3CH CH 2CH

57. Answer (1)

CH

H

CH3

CH3

C

, ,

CH – C – C – CH3 3

H CH3

H CH3

(Alkanes does not exhibitgeometrical isomerism)

,

58. Answer (1)

Zeolites are 3-D silicates.

59. Answer (2)

N

H H

H

H+

60. Answer (4)

Both acidic oxide and amphoteric oxide react with

NaOH.

61. Answer (2)

Fact.

62. Answer (3)

Fact.

63. Answer (3)

NaHCO3 + NaOH Na

2CO

3 + H

2O

64. Answer (1)

Na, Al, Si, P, S, Cl are paramagnetic.

65. Answer (1)

Fact.

66. Answer (3)


= Gram equivalent FeSO4


= N × V (L)

= 1 × 200

1000 = 0.2

Gram equivalent FeSO4 =

Weight

Equivalent weight

Weight0.2 30.4 g

152

67. Answer (2)

2 3

2 7Cr O 6e 2Cr

68. Answer (1)

5 3

6 6 3 6 5 2C H HNO C H NO

69. Answer (1)

Fact.

70. Answer (1)

71. Answer (3)

Fact.

72. Answer (2)

H+ = 14

9.62 10

= 3.10 × 10–7

pH = –log[3.10 × 10–7]

= 7 – 0.491

= 6.509

73. Deleted

74. Answer (3)

2CO

P

CO

pK

p

21.4 0.800

1.4 p 0.80 p

FeO CO(g) Fe CO (g)

��⇀↽��

0.80 p0.625

1.4 p

0.875 – 0.625p = 0.80 + p


7/9

0.875 – 0.80 = p + 0.625p

0.075 = 1.625p

p = 0.046

pCO

= 1.4 – 0.046 = 1.354

2CO

p 0.80 0.046 0.846

75. Answer (3)

Fact.

76. Answer (4)

rH° =

i f i fi i

a H (Product) b H (Reactant)

= 6 × (–94) + 3 × (–68.3) – 11.7

= –564 – 204.9 – 11.7

rH° = –780.6 kcal

77. Answer (1)

Fact.

78. Answer (1)

2O

W 16n

M 32

W = 2

10

1

V2.303 nRT log

V

= 10

16 502.303 8.314 300 log joule

32 5

= –2.87 × 103 J

= –2.87 kJ

79. Answer (4)

Fact.

80. Answer (4)

2

1Cl Cl (aq)

2

H = 1

2 × 240 + (–349) + (–381)

= –610 kJ mol–1

81. Answer (3)

Fact.

82. Answer (1)

BrF5 (sp

3d2)

Br

F

F

F

F

F

Due to lp–bp repulsion none of the bond angle is at

90°.

83. Answer (2)

B

O O – H

H

O

H

84. Answer (2)

2

2

ZE

n

2n

rZ

Zv

n

85. Answer (4)

2 1

4 3

(E E ) ( 3.4) ( 13.6) 10.2eV

(E E ) ( 0.85) ( 1.5) 0.66

86. Answer (3)

8

6

c 3 100.75 m

v 400 10

87. Answer (1)

3 2 2 21mol2 mol100 g

CaCO 2HCl CaCl H O CO

Number of moles of HCl taken = 1 × 1

1 mol HCl will produce 1

2 mol CO

2.

88. Answer (3)

Atomic mass × specific heat = 6.4

Atomic mass = 6.4 6.4

Specific heat 0.16 = 40

89. Answer (1)

N = M × nfactor

90. Answer (1)

N1V1 = N

2V2

2

1 1200 V

2 10

V2 = 1000

Water added = 1000 – 200 = 800 cc


8/9

[ BIOLOGY

]

91. Answer (1)

92. Answer (3)

93. Answer (4)

94. Answer (3)

95. Answer (1)

Polymoniales

96. Answer (1)

Herbarium is quick referral system.

97. Answer (2)

98. Answer (1)

Tuber aestivum

99. Answer (4)

100. Answer (2)

Precursor to seed habit – Selaginella.

101. Answer (4)

102. Answer (4)

Secondary meristem

103. Answer (4)

104. Answer (4)

Fats are poor in oxygen.

105. Answer (3)

106. Answer (4)

Glycolysis – 1 NADH

Link – 1 NADH

TCA cycle – 3 NADH and 1 FADH2

107. Answer (2)

C4 – 5, 2

108. Answer (2)

C3

109. Answer (4)

110. Answer (1)

Legumes

111. Answer (1)

112. Answer (1)

113. Answer (1)

ABA closes stomata.

114. Answer (4)

Consists of living part.

115. Answer (4)

116. Answer (2)

Metaphase is followed by anaphase.

117. Answer (1)

Ovum (n), meiocyte (2n)

118. Answer (3)

Oxidative phosphorylation

119. Answer (3)

120. Answer (4)

Completely secondary in origin.

121. Answer (3)

All the tissue inside endodermis.

122. Answer (1)

123. Answer (1)

Analogous to tuber of sweet potato and possess

axillary buds.

124. Answer (4)

Fabaceae

125. Answer (4)

126. Answer (2)

Red algae

127. Answer (3)

Heterospory – Pteridophytes

Seed habit – Gymnosperms

Fruit habit – Angiosperms

128. Answer (2)

129. Answer (4)

130. Answer (4)

131. Answer (1)

Dinoflagellates

132. Answer (3)

Zygote

133. Answer (4)

PS-II is present.


9/9

134. Answer (1)

Protista

135. Answer (4)

Thermoacidophiles

136. Answer (1)

137. Answer (1)

138. Answer (1)

139. Answer (4)

140. Answer (4)

141. Answer (3)

Pineal gland

142. Answer (3)

143. Answer (4)

144. Answer (1)

Hormones are required in traces.

145. Answer (3)

146. Answer (2)

147. Answer (4)

148. Answer (1)

Vitamin A itself is retinol and its aldehyde is retinal.

149. Answer (2)

150. Answer (4)

151. Answer (3)

152. Answer (4)

153. Answer (4)

154. Answer (1)

155. Answer (4)

156. Answer (1)

Alcohol suppresses ADH secretion.

157. Answer (3)

158. Answer (1)

These are in the cortex.

159. Answer (2)

Descending limb of the loop of Henle is impermeable

to salts.

160. Answer (4)

161. Answer (3)

162. Answer (1)

Cardiac output to muscles increases during

exercise.

The resistance to flow in the pulmonary system is

much lower than in systemic circulation, hence

pressure in pulmonary artery is lower than in aorta.

163. Answer (1)

164. Answer (3)

165. Answer (1)

166. Answer (3)

During inspiration, the intrapleural pressure becomes

negative.

167. Answer (3)

168. Answer (1)

169. Answer (4)

Enterokinase activates trypsinogen.

170. Answer (4)

Carbohydrates are absorbed as monosaccharides

and amino acids are absorbed actively.

171. Answer (2)

172. Answer (1)

Pancreatic acini secrete pancreatic enzymes.

173. Answer (4)

174. Answer (3)

Sperms are formed in testes and stored in seminal

vesicles.

175. Answer (4)

176. Answer (3)

177. Answer (4)

178. Answer (1)

Sea-anemones are carnivores.

179. Answer (3)

Echinoderms and hemichordates are also

deuterostomes.

180. Answer (3)

Cnidaria and ctenophora are diploblastic.

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