ANSWERS · All India Aakash Test Series for Medical-2017 Test - 5 (Code E) (Answers & Hints) 2/9...

Test - 5 (Code E) (Answers & Hints) All India Aakash Test Series for Medical-2017

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1. (2)

2. (4)

3. (2)

4. (4)

5. (1)

6. (4)

7. (2)

8. (2)

9. (1)

10. (1)

11 (4)

12. (1)

13. (2)

14. (3)

15. (3)

16. (4)

17. (4)

18. (2)

19. (2)

20. (2)

21. (3)

22. (1)

23. (4)

24. (1)

25. (4)

26. (3)

27. (2)

28. (1)

29. (2)

30. (1)

31. (3)

32. (3)

33. (1)

34. (3)

35. (3)

36. (3)

ANSWERS

TEST - 5 (Code-E)

All India Aakash Test Series for Medical-2017

Test Date : 19-02-2017

37. (3)

38. (1)

39. (2)

40. (2)

41. (3)

42. (1)

43. (2)

44. (3)

45. (1)

46. (1)

47. (3)

48. (2)

49. (4)

50. (1)

51. (4)

52. (2)

53. (1)

54. (2)

55. (4)

56. (4)

57. (2)

58. (2)

59. (2)

60. (4)

61. (3)

62. (3)

63. (1)

64. (3)

65. (3)

66. (1)

67. (2)

68. (4)

69. (2)

70. (1)

71. (1)

72. (3)

73. (3)

74. (3)

75. (1)

76. (4)

77. (2)

78. (1)

79. (3)

80. (3)

81. (4)

82. (4)

83. (2)

84. (1)

85. (2)

86. (1)

87. (2)

88. (3)

89. (3)

90. (3)

91. (2)

92. (2)

93. (3)

94. (1)

95. (4)

96. (2)

97. (3)

98. (2)

99. (4)

100. (3)

101. (1)

102. (2)

103. (3)

104. (2)

105. (3)

106. (1)

107. (2)

108. (3)

109. (1)

110. (4)

111. (4)

112. (3)

113. (2)

114. (2)

115. (2)

116. (3)

117. (1)

118. (2)

119. (2)

120. (2)

121. (1)

122. (3)

123. (2)

124. (1)

125. (2)

126. (3)

127. (1)

128. (3)

129. (1)

130. (1)

131. (4)

132. (4)

133. (2)

134. (2)

135. (1)

136. (2)

137. (2)

138. (1)

139. (3)

140. (3)

141. (4)

142. (2)

143. (2)

144. (4)

145. (4)

146 (2)

147. (3)

148. (2)

149. (3)

150. (2)

151. (4)

152. (2)

153. (2)

154. (3)

155. (4)

156. (3)

157. (3)

158. (4)

159. (2)

160. (3)

161. (4)

162. (1)

163. (3)

164. (3)

165. (3)

166. (1)

167. (4)

168. (4)

169. (3)

170. (3)

171. (2)

172. (1)

173. (3)

174. (1)

175. (2)

176. (1)

177. (3)

178. (2)

179. (4)

180. (3)

All India Aakash Test Series for Medical-2017 Test - 5 (Code E) (Answers & Hints)

2/9

Hints to Selected Questions

[ PHYSICS]

1. Answer (2)

Charge distribution would be unsymmetrical inside

cavity but symmetry on outer surface of cavity.

Potential will be

0

kq

r.

O

++

+

+

++

+

+

+

+

q

2. Answer (4)

kqV

r

(27 )

k qV

r,

3 3( ) 27r r

3. Answer (2)

enc

0

q

4. Answer (4)

Fq

⇓

E

P

Fdipole

5. Answer (1)

E

Equipotential surface

6. Answer (4)

Work done = PE = qV

Now, VA = V

B = V

C = V

D = V

E

Hence V is zero in all cases

Work done = 0

7. Answer (2)

E�

will change but | |E�

doesn't change also

potential doesn't change.

8. Answer (2)

When key is closed, to make the potential of both

the sphere same, all charge of inner sphere flows to

the outer sphere.

9. Answer (1)

10. Answer (1)

The equivalent circuit will be

CC

C C

C

C CA B

eq

77 F

8

CC

11 Answer (4)

We require 16 such branches in parallel

Hence total number of capacitors required = 32

8 F 8 F 8 F 8 F1000 V

eq

82 F

4C

12. Answer (1)

eq

1 2 3 4–

1 1 1 1

1 1 1 1

1 1 1 1

E

eq

41V

4E

eq

1 1 1 1 1

1 1 1 1r

eq

1

4r

13. Answer (2)

R l

Two parts would be parallel to each other.

1 2eq

1 2

R RR

R R

, eq

8

9R

1

2.3

R

RA

,

2

4.3

R

RA


3/9

14. Answer (3)

Under no deflection condition it will be balanced

Wheatstone bridge

80

20 80

R

R = 20

15. Answer (3)

No current will pass through resistance 3 and 6

Hence i1 = i

2

1

2

1i

i

16. Answer (4)

Rt = R

0(1 + t)

1 = R0(1 + 0.00125 × 27)

2 = R0(1 + 0.00125 × t)

1 1 0.00125 27

2 1 0.00125 t

1 + 0.00125 t = 2(1 + 0.00125 × 27)

t = 854°C

1127 K

17. Answer (4)

200 k

100 k

a

b

30 k

Equivalent resistance between a and ground

200 30100 k 126 k

230

⎛ ⎞ ⎜ ⎟⎝ ⎠

Current = 300

mA = 2.38 mA126

Now reading of voltmeter = 200 30 300

230 126

= 62 volt

18. Answer (2)

Average speed is nonzero.

19. Answer (2)

Equilibrium is unstable if Q and q are of opposite

polarity and Q is displaced along the line joining two

q charges.

20. Answer (2)

+Q

Flux of field by symmetry = 0

4

Q

21. Answer (3)

a

a

a

Electric field = 0 by symmetry, all fields will cancel

each other

Distance of charge from centre of cube = 3

2

a

Potential due to each

0

34

2

qV

a

Net potential = 8 V

0

4

3

q

a

22. Answer (1)

The given figure is equivalent to two identical

capacitors in parallel.

Hence, 02 A

Cd


4/9

23. Answer (4)

2 x

VE

x

3 y

VE

y

4 z

VE

z

ˆ ˆ ˆ–2 – 3 4E i j k �

24. Answer (1)

old2

LT

g

new

22 2 2

– –2

L L LT

qE g gg g

m

Ratio, old

new

1

2

T

T ,

new

old

2 :1T

T

25. Answer (4)

.E A � �

Here E A� �

= 0

26. Answer (3)

2.5 101

10 2

⎛ ⎞ ⎜ ⎟⎝ ⎠

L

R L

2.5 101

10 2

⎛ ⎞ ⎜ ⎟⎝ ⎠x

R L

27. Answer (2)

Energy of system =

2

0

64

q

a

Since there will be 6 pair interactions.

28. Answer (1)

2 sin2

k

Er

2

q

r

29. Answer (2)

Consider potential of point P due to complete sphere

be V1 then

2

1

2

3V k R

Now consider potential that would have been created

by removed position

2

29

k RV

Hence potential of point P = V1 – V

2

2 2

22 5–

3 9 9

k R k Rk R

2

0

5

36

R

30. Answer (1)

Let q be the charge on inner shell upon earthing

Vinner

= 0

0

10

4 3

q q

r r

⎡ ⎤ ⎢ ⎥ ⎣ ⎦

–3

qq

i.e., 3

q will flow from inner shell to the earth.

31. Answer (3)

i

3

10 V 4 V

10 – 42 A

3i

Power dissipated across resistor = i2 R = 12 W

32. Answer (3)

From graph,

i = 2t

Heat =

3 3

2 2

0 0

4 . 5 180 Ji Rdt t dt ∫ ∫


5/9

33. Answer (1)

q2

q1

q4q

3

rWhen broughtto one point

r

( + + + )q q q q1 2 3 4

1 2 3 4

centre

( )k q q q qV

r

1 2 3 4

centre

( )k q q q qV

r

Hence change in potential = zero

34. Answer (3)

Only option (3) shows correct variation.

35. Answer (3)

The figure can be reduced as

A

B

C3

C

Both capacitors are in parallel.

Hence, eq

4

3 3

C CC C

36. Answer (3)

2

1

(120)P

R

2

2

(110)P

R

% drop =

2 2

1 2

21

– 120 –110100 16%

120

P P

P

37. Answer (3)

Current through both the sections will be same.

Heat energy per unit volume

2i Rt

V

Now, 2

dx

R A

V A dx A

2

2

. .i tH

A

2

1H

A

Now,

2 21 1

2 4

A B

B A

H A

H A

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

38. Answer (1)

Q

q

q q

2

3kqQ

a2

2

kq

a2

2

kq

a

For equilibrium,

2

2 2

33

kqQ kq

a a

3

qQ

But Q should be of opposite polarity from q

Hence, –3

qQ

39. Answer (2)

qE

6rVT

6rVT = qE

6T

qEV

r

40. Answer (2)

Across PQ

R/3

R/2

R

P

RQ

5

11PQ

RR

41. Answer (3)

lR

A

l

AR


6/9

[ CHEMISTRY]

42. Answer (1)

.E A � �

ˆ ˆ ˆ(5 2 ). (1 ) 5i j i

43. Answer (2)

18– 2 2 – 6 – 3 5

3A B

V V

VA – V

B = 19 volt

44. Answer (3)

When switch 1 is connected

+q –q

q CE =

When switch 2 is connected and after reaching

steady state

–q +q

Work done by battery = 2qE = 2CE2

Heat generated = 2CE2

45. Answer (1)

In steady state condition no current will pass through

3 ohm resistor and current in the circuit will be 1 A

through 2 .

Potential difference across 10 F is 2 V

Hence charge = 20 C

46. Answer (1)

Cresol is used as froath stabilizer.

47. Answer (3)

Sphalerite is ZnS.

48. Answer (2)

SnO2 has magnetic impurities.

49. Answer (4)

Pb is more reactive than the other three.

50. Answer (1)

When Fe2O

3 impurity is in excess, Bayer's process

is used.

51. Answer (4)

All will decompose because these are carbonates

and hydroxides ore.

52. Answer (2)

2O

2A

1HgS HgO Hg O

2

53. Answer (1)

2Na[Ag(CN)2] + Zn 2Ag + Na

2[Zn(CN)

4]

54. Answer (2)

According to definition.

55. Answer (4)

Al powder can reduce all these.

56. Answer (4)

Fact.

57. Answer (2)

2 2

1Ag O 2Ag O

2

58. Answer (2)

Spongy iron forms in fusion zone and then converted

to pig iron.

59. Answer (2)

Carbon (graphite rods) are used as anode.

60. Answer (4)

Mond’s process.

61. Answer (3)

Hg is not transition elements.

62. Answer (3)

Fact.

63. Answer (1)

64. Answer (3)

65. Answer (3)

Ru shows +8.

66. Answer (1)

Mn2+ has d5 system (half filled d-orbital).

67. Answer (2)

Sc3+ has 18e– [Ne] 3s2 3p6


7/9

68. Answer (4)

Mn2O

7 has highest oxidation state.

69. Answer (2)

Zn2+ has d10 system.

70. Answer (1)

2

4 4 2 23MnO 4H 2MnO MnO 2H O

1 mole MnO4

2– = 2

3 mole MnO

4

–, 1

3 mole MnO

2

71. Answer (1)

4 2 4 2 22KMnO K MnO MnO O (g)

72. Answer (3)

HNO3 is good oxidizing and HCl is reducing agent.

73. Answer (3)

o

redE (KMnO

4) is maximum in neutral medium.

o 7 4

redE (Mn , Mn ) 1.69 V

74. Answer (3)

f 0 , f 7 and f 14 are colourless.

75. Answer (1)

76. Answer (4)

COO–

COO–

(ox), CH2

COO–

NH2

(gly)

77. Answer (2)

It has different ligands.

78. Answer (1)

EAN = 28 + 0 + 2 × 4 = 36

79. Answer (3)

(en) has, di, in its name.

80. Answer (3)

In K4[Fe(CN)

6], Fe does not exist as free Fe2+ ion in

anion sphere of [Fe(CN)6]4–.

81. Answer (4)

Fact.

82. Answer (4)

All does not show optical isomerism.

83. Answer (2)

It has d3 system.

84. Answer (1)

For low spin complex, splitting energy is large

(0 > P) compare to pairing energy.

Orbital splitting energy in octahedral coordination

entities is more than tetrahedral entities for same

metal ion and ligand.

85. Answer (2)

According to CFT.

86. Answer (1)

0 t

9

4 is correct

87. Answer (2)

4d has 50% more CFSE than 3d.

88. Answer (3)

Pb

C H2 5

C H2 5

C H2 5 C H

2 5

89. Answer (3)

It has 6 32g gt e configuration and unpaired electron

in eg.

90. Answer (3)

[ BIOLOGY

]

91. Answer (2)

Dioscorea, XO

92. Answer (2)

93. Answer (3)

It is an intersex.

94. Answer (1)

95. Answer (4)

(a) 0%

(b) Two

96. Answer (2)

X-body was discovered by Henking.

97. Answer (3)


8/9

98. Answer (2)

2 2 425%

4 4 16

99. Answer (4)

100. Answer (3)

101. Answer (1)

It can be expressed in females in homozygous

condition.

102. Answer (2)

103. Answer (3)

104. Answer (2)

GAG is replaced by GTG.

105. Answer (3)

106. Answer (1)

107. Answer (2)

108. Answer (3)

Tall plants = 12/16

Plants with long sized starch grains = 4/16

109. Answer (1)

Phenotypes = 4, Genotypes = 8.

110. Answer (4)

111. Answer (4)

112. Answer (3)

113. Answer (2)

Algae show external fertilization.

114. Answer (2)

115. Answer (2)

116. Answer (3)

Agave produces fleshy buds called bulbils.

117. Answer (1)

118. Answer (2)

119. Answer (2)

Rice, bamboo, radish.

120. Answer (2)

121. Answer (1)

122. Answer (3)

Monocliny involves formation of bisexual flowers.

123. Answer (2)

-cellulose bands are present in endothecium.

124. Answer (1)

125. Answer (2)

126. Answer (3)

127. Answer (1)

Castor, maize, rice.

128. Answer (3)

129. Answer (1)

130. Answer (1)

131. Answer (4)

132. Answer (4)

133. Answer (2)

Abiotic pollinator in Zea mays.

134. Answer (2)

135. Answer (1)

136. Answer (2)

137. Answer (2)

Tunica albuginea is situated under tunica vaginalis.

138. Answer (1)

139. Answer (3)

140. Answer (3)

Number of chromosomes in gamete of dog = 39

141. Answer (4)

In haploid organism gametes are formed through

mitosis.

142. Answer (2)

143. Answer (2)

Fraternal twins or dizygotic twins are developed from

two different ova.

144. Answer (4)


9/9

145. Answer (4)

Fructose is present in secretion of seminal vesicles.

146 Answer (2)

147. Answer (3)

Ovulation will take place on 40 – 14 = 26th day.

148. Answer (2)

Unejaculated sperms are reabsorbed in epididymis

and vas deferens.

149. Answer (3)

150. Answer (2)

hCG keeps corpus luteum active. Relaxin softens

pubic symphysis. Progesterone maintains cervical

mucus plug.

151. Answer (4)

152. Answer (2)

153. Answer (2)

Human placenta is deciduous, haemochorial,

chorionic and metadiscoidal.

154. Answer (3)

155. Answer (4)

156. Answer (3)

157. Answer (3)

Adrenal medulla and pineal gland are ectodermal.

Lining of urinary bladder is endodermal.

158. Answer (4)

159. Answer (2)

Human Immunodeficiency Virus causes AIDS.

160. Answer (3)

The incidence of STDs are reported to be very high

among persons in the age group of 15-24 years.

Genital warts is caused by Human Papilloma Virus

(HPV).

161. Answer (4)

162. Answer (1)

163. Answer (3)

164. Answer (3)

165. Answer (3)

166. Answer (1)

167. Answer (4)

Lactational amenorrhoea = 6 months.

168. Answer (4)

169. Answer (3)

170. Answer (3)

In vasectomy the two vasa deferentia are interrupted

by giving cuts and the ends tied.

171. Answer (2)

ICSI (Intra cytoplasmic sperm injection) involves in

vitro fertilisation.

172. Answer (1)

Trichomoniasis is caused by Trichomonas vaginalis

a flagellate protozoan.

173. Answer (3)

AIH is artificial insemination husband.

174. Answer (1)

Gossypol inhibits spermatogenesis.

175. Answer (2)

Teratozoospermia – Defective sperm morphology.

176. Answer (1)

VDRL is antibody detection.

177. Answer (3)

178. Answer (2)

179. Answer (4)

180. Answer (3)

Genital herps is caused by a virus.

� � �

Test - 5 (Code F) (Answers & Hints) All India Aakash Test Series for Medical-2017

1/9

1. (3)

2. (1)

3. (4)

4. (3)

5. (1)

6. (4)

7. (4)

8. (3)

9. (1)

10. (1)

11. (1)

12. (1)

13. (3)

14. (1)

15. (1)

16. (3)

17. (3)

18. (4)

19. (4)

20. (1)

21. (2)

22. (3)

23. (2)

24. (3)

25. (1)

26. (4)

27. (4)

28. (4)

29. (2)

30. (2)

31. (1)

32. (1)

33. (4)

34. (3)

35. (2)

36. (3)

ANSWERS

TEST - 5 (Code-F)

All India Aakash Test Series for Medical-2017

Test Date : 19-02-2017

37. (1)

38. (4)

39. (4)

40. (2)

41. (3)

42. (2)

43. (4)

44. (2)

45. (4)

46. (1)

47. (1)

48. (3)

49. (4)

50. (3)

51. (4)

52. (3)

53. (4)

54. (4)

55. (2)

56. (1)

57. (1)

58. (3)

59. (4)

60. (4)

61. (3)

62. (1)

63. (3)

64. (3)

65. (3)

66. (3)

67. (4)

68. (2)

69. (4)

70. (3)

71. (1)

72. (1)

73. (3)

74. (1)

75. (1)

76. (2)

77. (4)

78. (4)

79. (4)

80. (2)

81. (4)

82. (2)

83. (3)

84. (4)

85. (2)

86. (3)

87. (2)

88. (4)

89. (1)

90. (3)

91. (3)

92. (4)

93. (4)

94. (2)

95. (2)

96. (3)

97. (3)

98. (1)

99. (3)

100. (3)

101. (4)

102. (3)

103. (4)

104. (1)

105. (3)

106. (4)

107. (4)

108. (4)

109. (3)

110. (1)

111. (4)

112. (4)

113. (4)

114. (1)

115. (2)

116. (4)

117. (3)

118. (1)

119. (4)

120. (1)

121. (1)

122. (4)

123. (1)

124. (4)

125. (3)

126. (1)

127. (2)

128. (4)

129. (1)

130. (4)

131. (2)

132. (3)

133. (1)

134. (4)

135. (4)

136. (1)

137. (2)

138. (4)

139. (1)

140. (3)

141. (4)

142. (3)

143. (1)

144. (3)

145. (2)

146. (3)

147. (1)

148. (2)

149. (2)

150. (3)

151. (1)

152. (1)

153. (1)

154. (3)

155. (4)

156. (1)

157. (4)

158. (2)

159. (1)

160. (1)

161. (2)

162. (1)

163. (4)

164. (4)

165. (2)

166. (4)

167. (3)

168. (4)

169. (1)

170. (4)

171. (2)

172. (2)

173. (4)

174. (4)

175. (2)

176. (1)

177. (1)

178. (3)

179. (4)

180. (4)

All India Aakash Test Series for Medical-2017 Test - 5 (Code F) (Answers & Hints)

2/9

Hints to Selected Questions

[ PHYSICS]

1. Answer (3)

In steady state condition no current will pass through

3 ohm resistor and current in the circuit will be 1 A

through 2 .

Potential difference across 10 F is 2 V

Hence charge = 20 C

2. Answer (1)

When switch 1 is connected

+q –q

q CE =

When switch 2 is connected and after reaching

steady state

–q +q

Work done by battery = 2qE = 2CE2

Heat generated = 2CE2

3. Answer (4)

18– 2 2 – 6 – 3 5

3A B

V V

VA – V

B = 19 volt

4. Answer (3)

.E A � �

ˆ ˆ ˆ(5 2 ). (1 ) 5i j i

5. Answer (1)

lR

A

l

AR

6. Answer (4)

Across PQ

R/3

R/2

R

P

RQ

5

11PQ

RR

7. Answer (4)

qE

6rVT

6rVT = qE

6T

qEV

r

8. Answer (3)

Q

q

q q

2

3kqQ

a2

2

kq

a2

2

kq

a

For equilibrium

2

2 2

33

kqQ kq

a a

3

qQ

But Q should be of opposite polarity from q

Hence, –3

qQ


3/9

9. Answer (1)

Current through both the sections will be same.

Heat energy per unit volume

2i Rt

V

Now, 2

dx

R A

V A dx A

2

2

. .i tH

A

2

1H

A

Now,

2 21 1

2 4

A B

B A

H A

H A

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

10. Answer (1)

2

1

(120)P

R

2

2

(110)P

R

% drop =

2 2

1 2

21

– 120 –110100 16%

120

P P

P

11. Answer (1)

The figure can be reduced as

A

B

C3

C

Both capacitors are in parallel.

Hence, eq

4

3 3

C CC C

12. Answer (1)

Only option (1) shows correct variation.

13. Answer (3)

q2

q1

q4q

3

rWhen broughtto one point

r

( + + + )q q q q1 2 3 4

1 2 3 4

centre

( )k q q q qV

r

1 2 3 4

centre

( )k q q q qV

r

Hence change in potential = zero

14. Answer (1)

From graph,

i = 2t

Heat =

3 3

2 2

0 0

4 . 5 180 Ji Rdt t dt ∫ ∫

15. Answer (1)

i

3

10 V 4 V

10 – 42 A

3i

Power dissipated across resistor = i2 R = 12 W

16. Answer (3)

Let q be the charge on inner shell upon earthing

Vinner

= 0

0

10

4 3

q q

r r

⎡ ⎤ ⎢ ⎥ ⎣ ⎦

–3

qq

i.e., 3

q will flow from inner shell to the earth.

17. Answer (3)

2 sin2

k

Er

2

q

r

18. Answer (4)

Consider potential of point P due to complete sphere

be V1 then

2

1

2

3V k R

Now consider potential that would have been created

by removed position

2

29

k RV


4/9

Hence potential of point

P = V1 – V

2

2 2

22 5–

3 9 9

k R k Rk R

2

0

5

36

R

19. Answer (4)

Energy of system =

2

0

64

q

a

Since there will be 6 pair interactions.

20. Answer (1)

2.5 101

10 2

⎛ ⎞ ⎜ ⎟⎝ ⎠

L

R L

2.5 101

10 2

⎛ ⎞ ⎜ ⎟⎝ ⎠x

R L

21. Answer (2)

.E A � �

Here, E A� �

= 0

22. Answer (3)

old2

LT

g

new

22 2 2

– –2

L L LT

qE g gg g

m

Ratio, old

new

1

2

T

T ,

new

old

2 :1T

T

23. Answer (2)

2 x

VE

x

3 y

VE

y

4 z

VE

z

ˆ ˆ ˆ–2 – 3 4E i j k �

24. Answer (3)

The given figure is equivalent to two identical

capacitors in parallel.

Hence, 02 A

Cd

25. Answer (1)

a

a

a

Electric field = 0 by symmetry, all fields will cancel

each other

Distance of charge from centre of cube = 3

2

a

Potential due to each

0

34

2

qV

a

Net potential = 8 V

0

4

3

q

a

26. Answer (4)

+Q

Flux of field by symmetry = 0

4

Q

27. Answer (4)

Equilibrium is unstable if Q and q are of opposite

polarity and Q is displaced along the line joining two

q charges.

28. Answer (4)

Average speed is nonzero.


5/9

29. Answer (2)

200 k

100 k

a

b

30 k

Equivalent resistance between a and ground

200 30100 k 126 k

230

⎛ ⎞ ⎜ ⎟⎝ ⎠

Current = 300

mA = 2.38 mA126

Now reading of voltmeter = 200 30 300

230 126

= 62 volt

30. Answer (2)

Rt = R

0(1 + t)

1 = R0(1 + 0.00125 × 27)

2 = R0(1 + 0.00125 × t)

1 1 0.00125 27

2 1 0.00125 t

1 + 0.00125 t = 2(1 + 0.00125 × 27)

t = 854°C

1127 K

31. Answer (1)

No current will pass through resistance 3 and 6

Hence i1 = i

2

1

2

1i

i

32. Answer (1)

Under no deflection condition it will be balanced

Wheatstone bridge

80

20 80

R

R = 20

33. Answer (4)

R l

Two parts would be parallel to each other.

1 2eq

1 2

R RR

R R

, eq

8

9R

1

2.3

R

RA

,

2

4.3

R

RA

34. Answer (3)

eq

1 2 3 4–

1 1 1 1

1 1 1 1

1 1 1 1

E

eq

41V

4E

eq

1 1 1 1 1

1 1 1 1r

eq

1

4r

35. Answer (2)

We require 16 such branches in parallel

Hence total number of capacitors required = 32

8 F 8 F 8 F 8 F1000 V

eq

82 F

4C

36. Answer (3)

The equivalent circuit will be

CC

C C

C

C CA B

eq

77 F

8

CC

37. Answer (1)

38. Answer (4)

When key is closed, to make the potential of both

the sphere same, all charge of inner sphere flows to

the outer sphere.

39. Answer (4)

E�

will change but | |E�

doesn't change also

potential doesn't change.


6/9

[ CHEMISTRY]

46. Answer (1)

47. Answer (1)

It has 6 32g gt e configuration and unpaired electron

in eg.

48. Answer (3)

Pb

C H2 5

C H2 5

C H2 5 C H

2 5

49. Answer (4)

4d has 50% more CFSE than 3d.

50. Answer (3)

0 t

9

4 is correct.

51. Answer (4)

According to CFT.

52. Answer (3)

For low spin complex, splitting energy is large

(0 > P) compare to pairing energy.

Orbital splitting energy in octahedral coordination

entities is more than tetrahedral entities for same

metal ion and ligand.

40. Answer (2)

Work done = PE = qV

Now, VA = V

B = V

C = V

D = V

E

Hence V is zero in all cases

Work done = 0

41. Answer (3)

E

Equipotential surface

42. Answer (2)

Fq

⇓

E

P

Fdipole

43. Answer (4)

enc

0

q

44. Answer (2)

kqV

r

(27 )

k qV

r,

3 3( ) 27r r

45. Answer (4)

Charge distribution would be unsymmetrical inside

cavity but symmetry on outer surface of cavity.

Potential will be 0

kq

r.

O

++

+

+

++

+

+

+

+

q

53. Answer (4)

It has d3 system.

54. Answer (4)

All does not show optical isomerism.

55. Answer (2)

Fact.

56. Answer (1)

In K4[Fe(CN)

6], Fe does not exist as free Fe2+ ion in

anion sphere of [Fe(CN)6]4–.

57. Answer (1)

(en) has, di, in its name.

58. Answer (3)

EAN = 28 + 0 + 2 × 4 = 36

59. Answer (4)

It has different ligands.

60. Answer (4)

COO–

COO–

(ox), CH2

COO–

NH2

(gly)

61. Answer (3)

62. Answer (1)

f 0 , f 7 and f 14 are colourless.


7/9

[ BIOLOGY

]

91. Answer (3)

92. Answer (4)

93. Answer (4)

Abiotic pollinator in Zea mays.

94. Answer (2)

95. Answer (2)

96. Answer (3)

97. Answer (3)

98. Answer (1)

99. Answer (3)

Castor, maize, rice.

100. Answer (3)

63. Answer (3)

o

redE (KMnO

4) is maximum in neutral medium.

o 7 4

redE (Mn , Mn ) 1.69 V

64. Answer (3)

HNO3 is good oxidizing and HCl is reducing agent.

65. Answer (3)

4 2 4 2 22KMnO K MnO MnO O (g)

66. Answer (3)

2

4 4 2 23MnO 4H 2MnO MnO 2H O

1 mole MnO4

2– = 2

3 mole MnO

4

–, 1

3 mole MnO

2

67. Answer (4)

Zn2+ has d10 system.

68. Answer (2)

Mn2O

7 has highest oxidation state.

69. Answer (4)

Sc3+ has 18e– [Ne] 3s2 3p6

70. Answer (3)

Mn2+ has d5 system (half filled d-orbital).

71. Answer (1)

Ru shows +8.

72. Answer (1)

73. Answer (3)

74. Answer (1)

Fact.

75. Answer (1)

Hg is not transition elements.

76. Answer (2)

Mond’s process.

77. Answer (4)

Carbon (graphite rods) are used as anode.

78. Answer (4)

Spongy iron forms in fusion zone and then converted

to pig iron.

79. Answer (4)

2 2

1Ag O 2Ag O

2

80. Answer (2)

Fact.

81. Answer (4)

Al powder can reduce all these.

82. Answer (2)

According to definition.

83. Answer (3)

2Na[Ag(CN)2] + Zn 2Ag + Na

2[Zn(CN)

4]

84. Answer (4)

2O

2A

1HgS HgO Hg O

2

85. Answer (2)

All will decompose because these are carbonates

and hydroxides ore.

86. Answer (3)

When Fe2O

3 impurity is in excess, Bayer's process

is used.

87. Answer (2)

Pb is more reactive than the other three.

88. Answer (4)

SnO2 has magnetic impurities.

89. Answer (1)

Sphalerite is ZnS.

90. Answer (3)

Cresol is used as froath stabilizer.


8/9

101. Answer (4)

102. Answer (3)

103. Answer (4)

-cellulose bands are present in endothecium.

104. Answer (1)

Monocliny involves formation of bisexual flowers.

105. Answer (3)

106. Answer (4)

107. Answer (4)

Rice, bamboo, radish.

108. Answer (4)

109. Answer (3)

110. Answer (1)

Agave produces fleshy buds called bulbils.

111. Answer (4)

112. Answer (4)

113. Answer (4)

Algae show external fertilization.

114. Answer (1)

115. Answer (2)

116. Answer (4)

117. Answer (3)

Phenotypes = 4, Genotypes = 8.

118. Answer (1)

Tall plants = 12/16

Plants with long sized starch grains = 4/16

119. Answer (4)

120. Answer (1)

121. Answer (1)

122. Answer (4)

GAG is replaced by GTG.

123. Answer (1)

124. Answer (4)

125. Answer (3)

It can be expressed in females in homozygous

condition.

126. Answer (1)

127. Answer (2)

128. Answer (4)

2 2 425%

4 4 16

129. Answer (1)

130. Answer (4)

X-body was discovered by Henking.

131. Answer (2)

(a) 0%

(b) Two

132. Answer (3)

133. Answer (1)

It is an intersex.

134. Answer (4)

135. Answer (4)

Dioscorea, XO

136. Answer (1)

Genital herps is caused by a virus.

137. Answer (2)

138. Answer (4)

139. Answer (1)

140. Answer (3)

VDRL is antibody detection.

141. Answer (4)

Teratozoospermia – Defective sperm morphology.

142. Answer (3)

Gossypol inhibits spermatogenesis.

143. Answer (1)

AIH is artificial insemination husband.


9/9

144. Answer (3)

Trichomoniasis is caused by Trichomonas vaginalis

a flagellate protozoan.

145. Answer (2)

ICSI (Intra cytoplasmic sperm injection) involves in

vitro fertilisation.

146. Answer (3)

In vasectomy the two vasa deferentia are interrupted

by giving cuts and the ends tied.

147. Answer (1)

148. Answer (2)

149. Answer (2)

Lactational amenorrhoea = 6 months.

150. Answer (3)

151. Answer (1)

152. Answer (1)

153. Answer (1)

154. Answer (3)

155. Answer (4)

156. Answer (1)

The incidence of STDs are reported to be very high

among persons in the age group of 15-24 years.

Genital warts is caused by Human Papilloma Virus

(HPV).

157. Answer (4)

Human Immunodeficiency Virus causes AIDS.

158. Answer (2)

159. Answer (1)

Adrenal medulla and pineal gland are ectodermal.

Lining of urinary bladder is endodermal.

160. Answer (1)

161. Answer (2)

162. Answer (1)

� � �

163. Answer (4)

Human placenta is deciduous, haemochorial,

chorionic and metadiscoidal.

164. Answer (4)

165. Answer (2)

166. Answer (4)

hCG keeps corpus luteum active. Relaxin softens

pubic symphysis. Progesterone maintains cervical

mucus plug.

167. Answer (3)

168. Answer (4)

Unejaculated sperms are reabsorbed in epididymis

and vas deferens.

169. Answer (1)

Ovulation will take place on 40 – 14 = 26th day

170 Answer (4)

171. Answer (2)

Fructose is present in secretion of seminal vesicles.

172. Answer (2)

173. Answer (4)

Fraternal twins or dizygotic twins are developed from

two different ova.

174. Answer (4)

175. Answer (2)

In haploid organism gametes are formed through

mitosis.

176. Answer (1)

Number of chromosomes in gamete of dog = 39.

177. Answer (1)

178. Answer (3)

179. Answer (4)

Tunica albuginea is situated under tunica vaginalis.

180. Answer (4)

ANSWERS · All India Aakash Test Series for Medical-2017 Test - 5 (Code E) (Answers & Hints) 2/9...

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Transcript of ANSWERS · All India Aakash Test Series for Medical-2017 Test - 5 (Code E) (Answers & Hints) 2/9...