ANSWERS - dcx0p3on5z8dw.cloudfront.net€¦ · All India Aakash Test Series for Medical-2017 Test -...

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

1/10

1. (3)

2. (4)

3. (1)

4. (3)

5. (2)

6. (1)

7. (2)

8. (3)

9. (3)

10. (2)

11 (1)

12. (4)

13. (2)

14. (3)

15. (3)

16. (2)

17. (4)

18. (3)

19. (2)

20. (2)

21. (4)

22. (1)

23. (3)

24. (3)

25. (2)

26. (2)

27. (3)

28. (1)

29. (3)

30. (3)

31. (4)

32. (1)

33. (2)

34. (4)

35. (2)

36. (2)

ANSWERS

TEST - 5 (Code-A)

All India Aakash Test Series for Medical-2017

Test Date : 29-01-2017

37. (3)

38. (1)

39. (1)

40. (2)

41. (4)

42. (1)

43. (1)

44. (4)

45. (1)

46. (2)

47. (2)

48. (3)

49. (3)

50. (1)

51. (2)

52. (3)

53. (4)

54. (2)

55. (4)

56. (4)

57. (4)

58. (4)

59. (4)

60. (2)

61. (4)

62. (3)

63. (1)

64. (3)

65. (3)

66. (2)

67. (2)

68. (2)

69. (3)

70. (4)

71. (3)

72. (1)

73. (4)

74. (3)

75. (2)

76. (2)

77. (1)

78. (2)

79. (4)

80. (3)

81. (3)

82. (1)

83. (4)

84. (2)

85. (1)

86. (1)

87. (1)

88. (4)

89. (4)

90. (2)

91. (4)

92. (2)

93. (1)

94. (1)

95. (3)

96. (3)

97. (3)

98. (2)

99. (1)

100. (3)

101. (2)

102. (4)

103. (4)

104. (4)

105. (4)

106. (1)

107. (4)

108. (2)

109. (2)

110. (3)

111. (2)

112. (3)

113. (1)

114. (1)

115. (2)

116. (1)

117. (1)

118. (3)

119. (2)

120. (4)

121. (2)

122. (1)

123. (4)

124. (3)

125. (1)

126. (4)

127. (3)

128. (3)

129. (4)

130. (2)

131. (4)

132. (3)

133. (1)

134. (1)

135. (4)

136. (1)

137. (4)

138. (4)

139. (2)

140. (3)

141. (4)

142. (1)

143. (4)

144. (3)

145. (2)

146 (3)

147. (3)

148. (3)

149. (4)

150. (2)

151. (2)

152. (4)

153. (2)

154. (2)

155. (3)

156. (1)

157. (1)

158. (3)

159. (4)

160. (3)

161. (2)

162. (2)

163. (4)

164. (2)

165. (4)

166. (3)

167. (4)

168. (4)

169. (1)

170. (2)

171. (2)

172. (3)

173. (4)

174. (3)

175. (2)

176. (1)

177. (4)

178. (1)

179. (1)

180. (2)

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

2/10

Hints to Selected Questions

[ PHYSICS]

1. Answer (3)

qv0

q

r

v

By angular momentum conservations

024mv mvr ...(i)

By energy conservations

22 2

0

1 1

2 2

kqmv mv

r

2

2 2

0 0 20

1 1 24. .

2 2 4

qmv m v

rr

2 2

2 0 0 0

0 20

12 41

2 4

mv mv

mv

rr

2

1 12 1

2 rr

r2 – 2r – 24 = 0

r = 6 m

2. Answer (4)

++

+

+

+

+

–

–

–

–

c q2R

Vc = V

q + V

shell

Vshell

= 0

As total charge on shell = 0

3. Answer (1)

21

2U CV

10 20 10dU dV

CVdt dt

= 2000 J/s = 2 mJ/s

4. Answer (3)

High resistance would ensure more heat production

in short circuit condition, low melting point would

melt the wire fairly easy and thus breaking circuit.

5. Answer (2)

Let charge/volume =

3

0

a

3

0

4

3a

Dividing, 3

4

4

3

6. Answer (1)

2

1 1

1

EH R t

R r

⎡ ⎤ ⎢ ⎥⎣ ⎦

2

2 2

2

EH R t

R r

⎡ ⎤ ⎢ ⎥⎣ ⎦

Equating H1 = H

2

1 2r R R

7. Answer (2)

2

1– –

x

VE

x x

2

1– –

y

VE

y y

2

1– –

z

VE

z z

2 2 2

1 1 1ˆ ˆ ˆ– – –E i j k

x y z

�

at (1, 1, 1)

ˆ ˆ ˆ– – –E i j k

�

| | 3 N/CE �


3/10

8. Answer (3)

dqi

dt = Slope of charge-time graph

i = –5 A

H = i2 Rt

H t

9. Answer (3)

I = neAVd

1 2

2 (2 )d d

neAV ne A V

1

2

1:1d

d

V

V

10. Answer (2)

50 V

R30 V

10

20

I

I

1 2

Loop-1

30 – 50 + 20 I = 0

I = 1 ampere

Loop-2

50 – IR – 20I = 0

50 – R – 20 = 0

R = 30 11. Answer (1)

C1

C2

A

B

120°

Resistance per unit central arc angle

36 18

/ rad2

Resistance of � �

1 2

18 212

3

AC B AC B

eq

1 1 1 1 1

24 24 12 12R

Req

= 4

12. Answer (4)

A B

2 F

1 F

1 F

VC

VA

VC

VA

1 F

VCV

B

VAV

B

AB

2 F

1 F2 FC

⇒

2 1 82 F

3 3

AB

C

13. Answer (2)

t

q

dq

dt decreases with time

14. Answer (3)

2 2H i Rdt R i dt ∫ ∫H = R × Area under i2-t graph

Area = 1

(6 2) 4 162

H = 160 J

15. Answer (3)

Uniformly accelerated motion can have parabolic or

straight line trajectory .

16. Answer (2)

2QQ

q

Final potential difference between spheres will be E

(2 – ) ( )–

K Q q K Q qE

R R

0– 2

2

Qq ER


4/10

17. Answer (4)

qE

T

mg

T cos = mg

T sin = qE

18. Answer (3)

Final charge on capacitor would be CE.

19. Answer (2)

( 3 3)3 3

2 3 3AB

R RR R

R

Solving R = 3

20. Answer (2)

10 V

2 k 3 k

6 k

Req

of 3 k and 6 k= 2

Potential drop across 2 kwill be 5 V and

across combinations of 3 kand 6 kwould also

be 5 V. Hence, reading = 5 V

21. Answer (4)

Value of E will determine direction of flow of current.

22. Answer (1)

–

g

VR G

I

–3

10– 20 9980

10R

23. Answer (3)

enc

0

q

, initially qenc

increases till l

tu

then

becomes constant and then decreases when rod

comes completely out of cube.

24. Answer (3)

All three capacitors are in parallel hence Ceq

= 9 C

25. Answer (2)

Potential first decreases to R, then remains constant

inside shell and then decreases to zero at infinity

26. Answer (2)

E

curved

= base

box

= ER2

27. Answer (3)

Charge distribution before closing the key

4 V

2 F 6 F

8 C 8 C+6 C +6 C

4 F 4 F+ +

Charge distribution after closing the switch

4 V

2 F 6 F

+10 C

+5 C +9 C

4 F 4 F

–5 C –9 C

–10 C

4 C

+6 C –6 C

Charge flown through the switch = 4 C

28. Answer (1)

No current will pass through the grounded resistor

Hence reading of ammeter will be zero.

29. Answer (3)

+10 0 0 2 –2 10


5/10

Final charge distribution on the plates will be

x = 10

y = 0

z = –2

x + y + z = 8 C

30. Answer (3)

r

q

kqV

r = 5 volt

when drops are merged, let radius be R

3 34 48

3 3r R

R = 2r

Hence, its potential will be

(8 ) (8 ) 4

2

k q k q kq

R r r = 20 V

Hence, difference = 15 V

31. Answer (4)

If charges are both positive or both negative PE

decreases if they are of opposite sign PE increases

32. Answer (1)

P

E

When PE is chosen to be zero when P E� �

PE = – PEcos

So at 3

–2

PEPE

and PE when dipole is parallel to the field is –PE.

Now when reference potential is chosen to be of

3

its PE becomes 0.

ie PE increases by 2

PE

hence PE in parallel position will be

– –2 2

PE PEPE

33. Answer (2)

10

20

i

1 2 4 t

Charge flown = area under i-t graph

= 35

34. Answer (4)

Terminal voltage = E – iR

Given E – 10 × R = 50

E – 1 × R = 60

Solving,10

9R

550V

9E

35. Answer (2)

Potential difference across 100 resistance should

be 5 V as voltmeter and 100 resistance are in

parallel

req

of voltmeter and 100 should be 50

r of voltmeter = 100

36. Answer (2)

R1

R2

3 V 2 V

In series current are same through both

3 = IR1

2 = IR2


6/10

1

1 1 1 2

22 2 1

2

3

2

l

R A l A

lR l A

A

2

2

3 6.

2 1

B

A

r

r

2

2

12 :1

4 ⇒ B A

BA

r r

rr

37. Answer (3)

Let charge induced on sphere at any moment = q

Potential of sphere at any time = 0

0kq kq

x R

–

qRq

x

2

dq qR dxi

dt dtx

⎛ ⎞ ⎜ ⎟⎝ ⎠

2

qRVi

x

38. Answer (1)

Just after closing the switch, branch containing only

capacitor will make the circuit short, hence E

iR

39. Answer (1)

Time period of SHM is not affected by constant force

hence will remain unchanged.

40. Answer (2)

l

Work done by electric force = PE

qEl sin = mgl(1 – cos)

2

2sin cos 2sin.

2 2 2qEl mgl

tan2

qE

mg

⎛ ⎞ ⎜ ⎟⎝ ⎠

1

2tanqE

mg

⎡ ⎤ ⎢ ⎥⎣ ⎦

41. Answer (4)

1

CR

42. Answer (1)

Only charge and electric field remain unchanged

43. Answer (1)

r

dr

di = J. 2r dr

0

0

. 2

R

di J r r dr ∫ ∫

3

02

3

J Ri

44. Answer (4)

45. Answer (1)

qi

t

i = 2, t = 1

q = 2 C

Charge of 1e– = +1.602 × 10–19C in magnitude

In 1 C, number of electron

– 18

–19

16.25 10

1.602 10e

2 C charge will have 12.5 × 1018 = 1.25 × 1019


7/10

[ CHEMISTRY]

46. Answer (2)

Siderite is FeCO3

47. Answer (2)

Fact

48. Answer (3)

ZnS + 4NaCN Na2S + Na

2[Zn(CN)

4]

49. Answer (3)

When Fe2O

3 is in excess, Bayer's process is

followed.

50. Answer (1)

Complex formed is Na[Ag(CN)2]

51. Answer (2)

In cynide process NaCN is used.

52. Answer (3)

Dolomite is CaCO3. MgCO

3

53. Answer (4)

All three are correct.

54. Answer (2)

Aluminium is highly reactive.

55. Answer (4)

Fact

56. Answer (4)

Fe2O

3, MnO

2 and P

4O

10 are reduced to Fe, Mn and

P respectively.

57. Answer (4)

In pig iron carbon is 4 g, in cast iron it is 3 g

(%) decrease = 4 – 3

100 25%4

58. Answer (4)

Fact

59. Answer (4)

60. Answer (2)

It is semiconductor

61. Answer (4)

Ag2+ has d9 and Ag3+ has d8 system.

62. Answer (3)

63. Answer (1)

Ni according to the trend

64. Answer (3)

Sc2+, Zn3+ are not possible.

65. Answer (3)

+7 is the highest oxidation number of Mn (Group

VII)

66. Answer (2)

2Cu /Cu

E 0.34 V

67. Answer (2)

SO2 + V

2O

5 SO

3 + V

2O

4

2 4 2 2 5

1V O O V O

2

68. Answer (2)

Ni2+ and Pt4+ are stable

69. Answer (3)

Fact

70. Answer (4)

pH 42– 2–2 7 4Cr O CrO

��⇀↽��

71. Answer (3)

Fact

72. Answer (1)

In CeO2, cerium has noble gas configuration

73. Answer (4)

They do not react with CaCO3

74. Answer (3)

75. Answer (2)

Fact

76. Answer (2)

ICH NH

2 2

CH2

ICH

2

CH NH2 2

NH (dien)

77. Answer (1)

According to the rules


8/10

78. Answer (2)

EAN = 28 – 2 + 2 × 4 = 26 + 8 = 34

79. Answer (4)

Fact

80. Answer (3)

All surrounding atoms are same

81. Answer (3)

Fe3+ has an unpair electron

82. Answer (1)

Cr3+ : xx xx

3d

xx

4s

xx xx xx

4p

83. Answer (4)

[PtCl4]2– is

Cl

Cl

PtCl

Cl

2–

dsp2

84. Answer (2)

Only (a) and (b) are correct

85. Answer (1)

Fact

86. Answer (1)

Fact

87. Answer (1)

Cl– is weak ligand

88. Answer (4)

All behave as acid ligand

89. Answer (4)

Mn+ is electron deficient hence strength of Mn–C

bond will be less

90. Answer (2)

It has 6 42g gt e configuration.

[ BIOLOGY

]

91. Answer (4)

Conidia is a feature of ascomycetes and

deutomycetes.

92. Answer (2)

93. Answer (1)

Orange and mango which are polycarpic.

94. Answer (1)

95. Answer (3)

Embryo stage is absent in algae.

96. Answer (3)

97. Answer (3)

98. Answer (2)

Foliar buds are present on leaves.

99. Answer (1)

100. Answer (3)

Zoidogamous in Funaria, Chara, Marchantia and

Adiantum.

101. Answer (2)

Embryo sac represents female gametophyte.

102. Answer (4)

A – It is the epidermis which is protective in function.

103. Answer (4)

All are correct.

104. Answer (4)

105. Answer (4)

106. Answer (1)

107. Answer (4)

Has an unwettable stigma.

108. Answer (2)

109. Answer (2)

110. Answer (3)

Chemotropic growth of pollen tube.

111. Answer (2)

112. Answer (3)

113. Answer (1)

114. Answer (1)

It is also called perisperm.

115. Answer (2)

A – Cock (ZZ)

B – Hen (ZW)

116. Answer (1)

117. Answer (1)


9/10

118. Answer (3)

119. Answer (2)

120. Answer (4)

Ratio is (1 : 1)3

121. Answer (2)

122. Answer (1)

123. Answer (4)

124. Answer (3)

It is a physical mutagen, others are chemical

mutagen.

125. Answer (1)

126. Answer (4)

127. Answer (3)

0.6%

128. Answer (3)

129. Answer (4)

The ratio is 9 : 7

130. Answer (2)

131. Answer (4)

Both (1) and (2)

132. Answer (3)

Thalassemia, Cystic fibrosis and sickle cell anaemia.

133. Answer (1)

134. Answer (1)

or Dizygotic twins

135. Answer (4)

136. Answer (1)

Genetic similarity between parent and young one

occurs in asexual reproduction.

137. Answer (4)

A(Horse) = 60 years

B(Cow) = 25 years

C(Crow) = 15 years

138. Answer (4)

Longitudinal binary fission - Euglena & Vorticella

Plasmotomy – Pelomyxa

139. Answer (2)

Sharks have internal fertilisation.

140. Answer (3)

Haploid males are produced through parthenogensis

in honey bee.

141. Answer (4)

Sertoli cells secrete ABP which concentrates

testosterone in seminiferous tubules.

142. Answer (1)

Fraternal twins are dizygotic twins.

143. Answer (4)

144. Answer (3)

145. Answer (2)

Ectodermal – Spinal cord, Pituitary, Iris muscles

Mesodermal – Notochord, spleen, adrenal cortex,

blood vessels

Endodermal – Pancreas, thymus, liver, lining of

urinary bladder and thyroid

146 Answer (3)

Expulsion of baby occurs in expulsion stage.

147. Answer (3)

FSH stimulates sertoli cells to convert spermatids

into sperms.

148. Answer (3)

Castration – Removal of primary sex organs

Oopherectomy – Removal of ovaries

Orchidectomy – Removal of testes

149. Answer (4)

First movement of foetus and appearance of hair on

head is observed in fifth month.

150. Answer (2)

151. Answer (2)

152. Answer (4)

During cleavage nuclear cytoplasmic ratio increases.

153. Answer (2)

Fast block to check polyspermy involves influx of

Na+ causing depolarisation of membrane of ovum.

154. Answer (2)

A preovulatory surge of LH is necessary for ovulation

to take place.

155. Answer (3)

156. Answer (1)

157. Answer (1)

158. Answer (3)


10/10

159. Answer (4)

FSH and LH increase to high levels.

160. Answer (3)

Spermiogenesis involves formation of sperm from

spermatid.

161. Answer (2)

Endoderm is first germinal layer to be developed

during embryonic development.

162. Answer (2)

Gonorrhoea – Neisseria gonorrhoeae

Chancroid – Haemophilus ducreyi

Genital Herpes – Herpes simplex virus

163. Answer (4)

164. Answer (2)

IUI-Intra uterine insemination

165. Answer (4)

166. Answer (3)

Contraceptive method Average failure rate

Coitus interruptus 23%

Barrier methods 10-15%

Oral contraceptives 2-2.5%

Rhythm method 20-30%

167. Answer (4)

168. Answer (4)

Norplant is a subcutaneous implant.

169. Answer (1)

170. Answer (2)

171. Answer (2)

172. Answer (3)

Circumcision is surgical removal of prepuce.

173. Answer (4)

174. Answer (3)

Demographic transition occurs when birth rate

equals death rate.

175. Answer (2)

176. Answer (1)

177. Answer (4)

Norgynon, femsheild and DMPA (Depot medroxy

progesterone acetate) are female contraceptives.

178. Answer (1)

ICSI – Intra cytoplasmic sperm injection

179. Answer (1)

180. Answer (2)

AIDS is a viral STD.

� � �

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

1/10

1. (3)

2. (2)

3. (3)

4. (3)

5. (2)

6. (4)

7. (3)

8. (3)

9. (1)

10. (4)

11. (4)

12. (2)

13. (4)

14. (3)

15. (2)

16. (1)

17. (1)

18. (3)

19. (1)

20. (4)

21. (4)

22. (1)

23. (1)

24. (3)

25. (2)

26. (4)

27. (4)

28. (1)

29. (2)

30. (4)

31. (3)

32. (1)

33. (4)

34. (2)

35. (3)

36. (4)

ANSWERS

TEST - 5 (Code-B)

All India Aakash Test Series for Medical-2017

Test Date : 29-01-2017

37. (1)

38. (1)

39. (4)

40. (3)

41. (4)

42. (1)

43. (3)

44. (2)

45. (1)

46. (4)

47. (2)

48. (4)

49. (3)

50. (3)

51. (3)

52. (4)

53. (2)

54. (3)

55. (1)

56. (1)

57. (2)

58. (4)

59. (3)

60. (4)

61. (4)

62. (1)

63. (2)

64. (3)

65. (1)

66. (2)

67. (3)

68. (4)

69. (2)

70. (4)

71. (1)

72. (1)

73. (3)

74. (1)

75. (4)

76. (4)

77. (4)

78. (4)

79. (2)

80. (2)

81. (4)

82. (4)

83. (2)

84. (1)

85. (4)

86. (3)

87. (1)

88. (1)

89. (4)

90. (4)

91. (4)

92. (3)

93. (1)

94. (1)

95. (4)

96. (4)

97. (2)

98. (1)

99. (1)

100. (2)

101. (3)

102. (1)

103. (4)

104. (3)

105. (4)

106. (2)

107. (4)

108. (1)

109. (3)

110. (3)

111. (4)

112. (1)

113. (3)

114. (1)

115. (2)

116. (1)

117. (4)

118. (2)

119. (2)

120. (1)

121. (4)

122. (2)

123. (4)

124. (2)

125. (4)

126. (1)

127. (3)

128. (4)

129. (1)

130. (1)

131. (3)

132. (1)

133. (3)

134. (4)

135. (2)

136. (2)

137. (3)

138. (3)

139. (2)

140. (3)

141. (2)

142. (1)

143. (2)

144. (1)

145. (4)

146. (4)

147. (1)

148. (2)

149. (2)

150. (1)

151. (2)

152. (4)

153. (2)

154. (4)

155. (4)

156. (1)

157. (2)

158. (1)

159. (3)

160. (3)

161. (1)

162. (4)

163. (4)

164. (2)

165. (4)

166. (4)

167. (2)

168. (1)

169. (1)

170. (1)

171. (4)

172. (1)

173. (2)

174. (3)

175. (2)

176. (1)

177. (4)

178. (2)

179. (2)

180. (3)

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

2/10

Hints to Selected Questions

[ PHYSICS]

1. Answer (3)

qi

t

i = 2, t = 1

q = 2 C

Charge of 1e– = +1.602 × 10–19C in magnitude

In 1 C, number of electron

– 18

–19

16.25 10

1.602 10e

2 C charge will have 12.5 × 1018 = 1.25 × 1019

2. Answer (2)

3. Answer (3)

r

dr

di = J. 2r dr

0

0

. 2

R

di J r r dr ∫ ∫

3

02

3

J Ri

4. Answer (3)

Only charge and electric field remain unchanged.

5. Answer (2)

1

CR

6. Answer (4)

l

Work done by electric force = PE

qEl sin = mgl(1 – cos)

2

2sin cos 2sin.

2 2 2qEl mgl

tan2

qE

mg

⎛ ⎞ ⎜ ⎟⎝ ⎠

12tan

qE

mg

⎡ ⎤ ⎢ ⎥⎣ ⎦

7. Answer (3)

Time period of SHM is not affected by constant force

hence will remain uncharged.

8. Answer (3)

Just after closing the switch, branch containing only

capacitor will make the circuit short, hence E

iR

9. Answer (1)

Let charge induced on sphere at any moment = q

Potential of sphere at any time = 0

0kq kq

x R

–

qRq

x

2

dq qR dxi

dt dtx

⎛ ⎞ ⎜ ⎟⎝ ⎠

2

qRVi

x

10. Answer (4)

R1

R2

3 V 2 V

In series current are same through both

3 = IR1

2 = IR2


3/10

1

1 1 1 2

22 2 1

2

3

2

l

R A l A

lR l A

A

2

2

3 6.

2 1

B

A

r

r

2

2

12 :1

4 ⇒ B A

BA

r r

rr

11. Answer (4)

Potential difference across 100 resistance should

be 5 V as voltmeter and 100 resistance are in

parallel

req

of voltmeter and 100 should be 50

r of voltmeter = 100

12. Answer (2)

Terminal voltage = E – iR

Given E – 10 × R = 50

E – 1 × R = 60

Solving,10

9R

550V

9E

13. Answer (4)

10

20

i

1 2 4 t

Charge flown = area under i-t graph

= 35

14. Answer (3)

P

E

When PE is chosen to be zero when P E� �

PE = – PEcos

So at 3

–2

PEPE

and PE when dipole is parallel to the field is –PE.

Now when reference potential is chosen to be of

3

its PE becomes 0.

i.e., PE increases by 2

PE

hence PE in parallel position will be

– –2 2

PE PEPE

15. Answer (2)

If charges are both positive or both negative PE

decreases if they are of opposite sign PE increases.

16. Answer (1)

r

q

kqV

r = 5 volt

when drops are merged, let radius be R

3 34 48

3 3r R

R = 2r

Hence, its potential will be

(8 ) (8 ) 4

2

k q k q kq

R r r = 20 V

Hence, difference = 15 V

17. Answer (1)

+10 0 0 2 –2 10


4/10

Final charge distribution on the plates will be

x = 10

y = 0

z = –2

x + y + z = 8 C

18. Answer (3)

No current will pass through the grounded resistor

Hence reading of ammeter will be zero.

19. Answer (1)

Charge distribution before closing the key

4 V

2 F 6 F

8 C 8 C+6 C +6 C

4 F 4 F+ +

Charge distribution after closing the switch

4 V

2 F 6 F

+10 C

+5 C +9 C

4 F 4 F

–5 C –9 C

–10 C

4 C

+6 C –6 C

Charge flown through the switch = 4 C

20. Answer (4)

E

curved

= base

box

= ER2

21. Answer (4)

Potential first decreases to R, then remains constant

inside shell and then decreases to zero at infinity.

22. Answer (1)

All three capacitors are in parallel hence Ceq

= 9 C

23. Answer (1)

enc

0

q

, initially qenc

increases till l

tu

then

becomes constant and then decreases when rod

comes completely out of cube.

24. Answer (3)

–

g

VR G

I

–3

10– 20 9980

10R

25. Answer (2)

Value of E will determine direction of flow of current.

26. Answer (4)

10 V

2 k 3 k

6 k

Req

of 3 k and 6 k= 2

Potential drop across 2 kwill be 5V and across

combinations of 3 kand 6 kwould also be 5 V.

Hence, reading = 5 V

27. Answer (4)

( 3 3)3 3

2 3 3AB

R RR R

R

Solving R = 3

28. Answer (1)

Final charge on capacitor would be CE.

29. Answer (2)

qE

T

mg

T cos = mg

T sin = qE


5/10

30. Answer (4)

2QQ

q

Final potential difference between spheres will be E

(2 – ) ( )–

K Q q K Q qE

R R

0– 2

2

Qq ER

31. Answer (3)

Uniformly accelerated motion can have parabolic or

straight line trajectory .

32. Answer (1)

2 2H i Rdt R i dt ∫ ∫H = R × Area under i2-t graph

Area = 1

(6 2) 4 162

H = 160 J

33. Answer (4)

t

q

dq

dt decreases with time

34. Answer (2)

A B

2 F

1 F

1 F

VC

VA

VC

VA

1 F

VCV

B

VAV

B

AB

2 F

1 F2 FC

⇒

2 1 82 F

3 3

AB

C

35. Answer (3)

C1

C2

A

B

120°

Resistance per unit central arc angle

36 18

/ rad2

Resistance of � �

1 2

18 212

3

AC B AC B

eq

1 1 1 1 1

24 24 12 12R

Req

= 4

36. Answer (4)

50 V

R30 V

10

20

I

I

1 2

Loop-1

30 – 50 + 20 I = 0

I = 1 ampere

Loop-2

50 – IR – 20I = 0

50 – R – 20 = 0

R = 30

37. Answer (1)

I = neAVd

1 2

2 (2 )d d

neAV ne A V

1

2

1:1d

d

V

V

38. Answer (1)

dqi

dt = Slope of charge-time graph

i = –5 A

H = i2 Rt

H t


6/10

39. Answer (4)

2

1– –

x

VE

x x

2

1– –

y

VE

y y

2

1– –

z

VE

z z

2 2 2

1 1 1ˆ ˆ ˆ– – –E i j k

x y z

�

at (1, 1, 1)

ˆ ˆ ˆ– – –E i j k

�

| | 3 N/CE �

40. Answer (3)

2

1 1

1

EH R t

R r

⎡ ⎤ ⎢ ⎥⎣ ⎦

2

2 2

2

EH R t

R r

⎡ ⎤ ⎢ ⎥⎣ ⎦

Equating H1

= H2

1 2r R R

41. Answer (4)

Let charge/volume =

3

0

a

3

0

4

3a

Dividing, 3

4

4

3

42. Answer (1)

High resistance would ensure more heat production

in short circuit condition, low melting point would

melt the wire fairly easy and thus breaking circuit.

43. Answer (3)

21

2U CV

10 20 10dU dV

CVdt dt

= 2000 J/s = 2 mJ/s

44. Answer (2)

++

+

+

+

+

–

–

–

–

c q2R

Vc = V

q + V

shell

Vshell

= 0

As total charge on shell = 0

45. Answer (1)

qv0

q

r

v

By angular momentum conservations

024mv mvr ...(i)

By energy conservations

22 2

0

1 1

2 2

kqmv mv

r

2

2 2

0 0 20

1 1 24. .

2 2 4

qmv m v

rr

2 2

2 0 0 0

0 20

12 41

2 4

mv mv

mv

rr

2

1 12 1

2 rr

r2 – 2r – 24 = 0

r = 6 m


7/10

[ CHEMISTRY]

46. Answer (4)

It has 6 42g gt e configuration.

47. Answer (2)

Mn+ is electron deficient hence strength of Mn–C

bond will be less

48. Answer (4)

All behave as acid ligand

49. Answer (3)

Cl– is weak ligand

50. Answer (3)

Fact

51. Answer (3)

Fact

52. Answer (4)

Only (a) and (b) are correct

53. Answer (2)

[PtCl4

]2– is

Cl

Cl

PtCl

Cl

2–

dsp2

54. Answer (3)

Cr3+ : xx xx

3d

xx

4s

xx xx xx

4p

55. Answer (1)

Fe3+ has an unpair electron

56. Answer (1)

All surrounding atoms are same

57. Answer (2)

Fact

58. Answer (4)

EAN = 28 – 2 + 2 × 4 = 26 + 8 = 34

59. Answer (3)

According to the rules

60. Answer (4)

ICH NH

2 2

CH2

ICH

2

CH NH2 2

NH (dien)

61. Answer (4)

Fact

62. Answer (1)

63. Answer (2)

They do not react with CaCO3

64. Answer (3)

In CeO2

, cerium has noble gas configuration

65. Answer (1)

Fact

66. Answer (2)

pH 42– 2–2 7 4Cr O CrO

��⇀↽��

67. Answer (3)

Fact

68. Answer (4)

Ni2+ and Pt4+ are stable

69. Answer (2)

SO2

+ V2

O5

SO3

+ V2

O4

2 4 2 2 5

1V O O V O

2

70. Answer (4)

2Cu /Cu

E 0.34 V

71. Answer (1)

+7 is the highest oxidation number of Mn (Group

VII)

72. Answer (1)

Sc2+, Zn3+ are not possible.

73. Answer (3)

Ni according to the trend

74. Answer (1)

75. Answer (4)

Ag2+ has d9 and Ag3+ has d8 system.

76. Answer (4)

It is semiconductor

77. Answer (4)


8/10

[ BIOLOGY

]

91. Answer (4)

92. Answer (3)

or Dizygotic twins

93. Answer (1)

94. Answer (1)

Thalassemia, Cystic fibrosis and sickle cell anaemia.

95. Answer (4)

Both (1) and (2)

96. Answer (4)

97. Answer (2)

The ratio is 9 : 7

98. Answer (1)

99. Answer (1)

0.6%

100. Answer (2)

101. Answer (3)

102. Answer (1)

It is a physical mutagen, others are chemical

mutagen.

103. Answer (4)

104. Answer (3)

105. Answer (4)

106. Answer (2)

Ratio is (1 : 1)3

107. Answer (4)

108. Answer (1)

109. Answer (3)

110. Answer (3)

111. Answer (4)

A – Cock (ZZ)

B – Hen (ZW)

112. Answer (1)

It is also called perisperm.

113. Answer (3)

114. Answer (1)

115. Answer (2)

116. Answer (1)

Chemotropic growth of pollen tube.

78. Answer (4)

Fact

79. Answer (2)

In pig iron carbon is 4 g, in cast iron it is 3 g

(%) decrease = 4 – 3

100 25%4

80. Answer (2)

Fe2

O3

, MnO2

and P4

O10

are reduced to Fe, Mn and

P respectively.

81. Answer (4)

Fact

82. Answer (4)

Aluminium is highly reactive.

83. Answer (2)

All three are correct.

84. Answer (1)

Dolomite is CaCO3

. MgCO3

85. Answer (4)

In cynide process NaCN is used.

86. Answer (3)

Complex formed is Na[Ag(CN)2

]

87. Answer (1)

When Fe2

O3

is in excess, Bayer's process is

followed.

88. Answer (1)

ZnS + 4NaCN Na2

S + Na2

[Zn(CN)4

]

89. Answer (4)

Fact

90. Answer (4)

Siderite is FeCO3


9/10

117. Answer (4)

118. Answer (2)

119. Answer (2)

Has an unwettable stigma.

120. Answer (1)

121. Answer (4)

122. Answer (2)

123. Answer (4)

All are correct.

124. Answer (2)

A – It is the epidermis which is protective in function.

125. Answer (4)

Embryo sac represents female gametophyte.

126. Answer (1)

Zoidogamous in Funaria, Chara, Marchantia and

Adiantum.

127. Answer (3)

128. Answer (4)

Foliar buds are present on leaves.

129. Answer (1)

130. Answer (1)

131. Answer (3)

Embryo stage is absent in algae.

132. Answer (1)

133. Answer (3)

Orange and mango which are polycarpic.

134. Answer (4)

135. Answer (2)

Conidia is a feature of ascomycetes and

deutomycetes.

136. Answer (2)

AIDS is a viral STD.

137. Answer (3)

138. Answer (3)

ICSI – Intra cytoplasmic sperm injection

139. Answer (2)

Norgynon, femsheild and DMPA (Depot medroxy

progesterone acetate) are female contraceptives.

140. Answer (3)

141. Answer (2)

142. Answer (1)

Demographic transition occurs when birth rate

equals death rate.

143. Answer (2)

144. Answer (1)

Circumcision is surgical removal of prepuce.

145. Answer (4)

146. Answer (4)

147. Answer (1)

148. Answer (2)

Norplant is a subcutaneous implant.

149. Answer (2)

150. Answer (1)

Contraceptive method Average failure rate

Coitus interruptus 23%

Barrier methods 10-15%

Oral contraceptives 2-2.5%

Rhythm method 20-30%

151. Answer (2)

152. Answer (4)

IUI-Intra uterine insemination

153. Answer (2)

154. Answer (4)

Gonorrhoea – Neisseria gonorrhoeae

Chancroid – Haemophilus ducreyi

Genital Herpes – Herpes simplex virus

155. Answer (4)

Endoderm is first germinal layer to be developed

during embryonic development.

156. Answer (1)

Spermiogenesis involves formation of sperm from

spermatid.


10/10

157. Answer (2)

FSH and LH increase to high levels.

158. Answer (1)

159. Answer (3)

160. Answer (3)

161. Answer (1)

162. Answer (4)

A preovulatory surge of LH is necessary for ovulation

to take place.

163. Answer (4)

Fast block to check polyspermy involves influx of

Na+ causing depolarisation of membrane of ovum.

164. Answer (2)

During cleavage nuclear cytoplasmic ratio increases.

165. Answer (4)

166. Answer (4)

167. Answer (2)

First movement of foetus and appearance of hair on

head is observed in fifth month.

168. Answer (1)

Castration – Removal of primary sex organs

Oopherectomy – Removal of ovaries

Orchidectomy – Removal of testes

169. Answer (1)

FSH stimulates sertoli cells to convert spermatids

into sperms.

170 Answer (1)

Expulsion of baby occurs in expulsion stage.

� � �

171. Answer (4)

Ectodermal – Spinal cord, Pituitary, Iris muscles

Mesodermal – Notochord, spleen, adrenal cortex,

blood vessels

Endodermal – Pancreas, thymus, liver, lining of

urinary bladder and thyroid

172. Answer (1)

173. Answer (2)

174. Answer (3)

Fraternal twins are dizygotic twins.

175. Answer (2)

Sertoli cells secrete ABP which concentrates

testosterone in seminiferous tubules.

176. Answer (1)

Haploid males are produced through parthenogensis

in honey bee.

177. Answer (4)

Sharks have internal fertilisation.

178. Answer (2)

Longitudinal binary fission - Euglena & Vorticella

Plasmotomy – Pelomyxa

179. Answer (2)

A(Horse) = 60 years

B(Cow) = 25 years

C(Crow) = 15 years

180. Answer (3)

Genetic similarity between parent and young one

occurs in asexual reproduction.

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