IATS Medical-2011 T1 Solution 10-4-50

8/8/2019 IATS Medical-2011 T1 Solution 10-4-50

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Test - 1 (Answers & Hints) All India Aakash Test Series for Medical-2011

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TEST 1

ANSWERSANSWERSANSWERSANSWERSANSWERS1. (2)

2. (1)

3. (3)

4. (1)

5. (3)

6. (4)

7. (1)

8. (1)

9. (3)10. (2)

11. (4)

12. (3)

13. (4)

14. (3)

15. (2)

16. (2)

17. (1)

18. (3)

19. (4)

20. (1)

21. (2)

22. (2)

23. (1)

24. (4)

25. (2)

26. (2)

27. (3)

28. (2)

29. (3)

30. (4)31. (4)

32. (1)

33. (3)

34. (2)

35. (3)

36. (2)

37. (4)

38. (1)

39. (1)

40. (3)

41. (4)

42. (1)

43. (1)

44. (2)

45. (3)

46. (2)

47. (1)

48. (4)

49. (1)50. (2)

51. (1)

52. (1)

53. (2)

54. (4)

55. (3)

56. (4)

57. (4)

58. (1)

59. (2)

60. (4)

61. (1)

62. (1)

63. (4)

64. (1)

65. (1)

66. (4)

67. (1)

68. (1)

69. (3)

70. (4)71. (1)

72. (2)

73. (4)

74. (2)

75. (1)

76. (1)

77. (4)

78. (2)

79. (3)

80. (1)

81. (2)

82. (4)

83. (1)

84. (3)

85. (1)

86. (2)

87. (2)

88. (4)

89. (3)90. (2)

91. (1)

92. (1)

93. (3)

94. (1)

95. (4)

96. (1)

97. (4)

98. (4)

99. (1)

100. (1)

101. (2)

102. (4)

103. (3)

104. (3)

105. (4)

106. (3)

107. (1)

108. (1)

109. (2)

110. (2)111. (2)

112. (4)

113. (3)

114. (4)

115. (1)

116. (3)

117. (2)

118. (2)

119. (4)

120. (4)

121. (3)

122. (4)

123. (2)

124. (2)

125. (3)

126. (4)

127. (4)

128. (3)

129. (3)130. (2)

131. (1)

132. (4)

133. (3)

134. (3)

135. (3)

136. (1)

137. (3)

138. (4)

139. (2)

140. (2)

141. (3)

142. (1)

143. (1)

144. (2)

145. (4)

146. (1)

147. (4)

148. (4)

149. (4)

150. (2)151. (2)

152. (4)

153. (3)

154. (2)

155. (1)

156. (1)

157. (3)

158. (2)

159. (3)

160. (3)

161. (4)

162. (3)

163. (4)

164. (3)

165. (4)

166. (3)

167. (3)

168. (3)

169. (4)170. (3)

171. (4)

172. (3)

173. (3)

174. (2)

175. (1)

176. (2)

177. (3)

178. (1)

179. (4)

180. (1)

181. (1)

182. (4)

183. (4)

184. (4)

185. (2)

186. (1)

187. (1)

188. (1)

189. (1)

190. (1)191. (2)

192. (3)

193. (1)

194. (1)

195. (4)

196. (1)

197. (1)

198. (2)

199. (2)

200. (1)


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All India Aakash Test Series for Medical-2011 Test - 1 (Answers & Hints)

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Hints to Selected Questions

[PHYSICS]

1. Answer (2)

2. Answer (1)

From principle of homogenity

[d] = [Pt] = [ML1T1]

3. Answer (3)

Thickness =2 1 4.23 3.89 0.17

2 2

d d

Error2 1 0.01 0.01 0.01

2 2

d d

(0.17 0.01) cm

4. Answer (1)

5. Answer (3)

A B

C

D

60

30

2R

AC= 2Rcos30 = 3 R

AD= 2Rcos60 = R

displacement

timeavv

Ratio = 3

6. Answer (4)

distance 7014 m/s

time 5

avv

7. Answer (1)

8. Answer (1)

If tis time of flight of first ball before meeting then(t 4) is the time of flight of 2nd body h1 = h2

2 21 198 98( 4) ( 4)2 2

t gt t g t

9. Answer (3)

R=

2 22 2

cos sinN f mg mg = mg

10. Answer (2)

Velocity when decreases from 2V to V, the distance

travelled is 3x. Velocity when decreases from V to

zero, the distance travelled is x.

11. Answer (4)

12. Answer (3)

13. Answer (4)

14. Answer (3)

Velocity vector is along tangent to the curve at point

(1, 2)

Angle = tan = slope =(1, 2)

dy

dx

tan = x= 1 = 45

15. Answer (2)

For movable pulley,

a=0 a1a21

20

2

aa

1 22a a

16. Answer (2)

v= u+ at

6 = 0 + a(0.2)

a= 30 m/s2

2

2

VS

a

=26

2 30

= 0.6 m

17. Answer (1)

4 kg block move only due to friction

f= N

f4 kg

54 10

10f

f= 20 N

Max. acceleration

220

5 m/s4

f

m


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18. Answer (3)

max2

N 3 10 6 N10

f

Hence for 0 3t acceleration will be zero after

3 s

2 6

3

ta

22

3a t

19. Answer (4)

2f 20

50 N50 N

ff

20 N

f 10

N 10

50 10

1

5

0.2

20. Answer (1)

2tan

H

R

H

R/2

y

xPutting values of H

and Rwill get,

1tan tan

2

21. Answer (2)

50 N

F =100

50 N

6 kg block will not move.

Hence a1

= 0

50

40

4 22

102.5 m/s

4a

22. Answer (2)

Net accelerating force on the system

F= (2gsin30 1g)

F= 1g 1g= 0

Hence friction will not generate

23. Answer (1)

24. Answer (4)

25. Answer (2)

tan =0v

T v

tT

v0

K =0v

T

v0

= KT

Distance = Area

=1

( )2

T KT

=2

2

KT

26. Answer (2)

Since length of the rod is constant

cos sinB Av v

3

vB

vA

vA sin

5

vB

cos

tanB Av v

34 3 m/s

4

27. Answer (3)

8Time taken = 1 h

8sR

d

v

28. Answer (2)

29. Answer (3)

30. Answer (4)

Change in velocity,

2 sin2

v v

60

2 6sin2

= 6 m/s

26 2 m/s3

va

t


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31. Answer (4)

32. Answer (1)

1 22h

t tg

2

3 15 m10

hh

33. Answer (3)

34. Answer (2)

35. Answer (3)

36. Answer (2)

37. Answer (4)

38. Answer (1)

39. Answer (1)

40. Answer (3)

A B

D

CO

Resultant of AB and BC will be along OB and of

1 N and OB and CD are in same direction.

Hence Resultant = 1 + 1 = 2 N

41. Answer (4)

If vector is rotated then its magnitude will notchange.

a 45

y

x

Hence, cos45 sin45 a i j

1 1 2 2

a i j

42. Answer (1)

22

0

Area 3x dx

23 2

0

38 m

3x

43. Answer (1)

mgmrw2

2rw

g

2

min

rw

g

44. Answer (2)

OA

/2

If angular displacement about O is then angular

displacement about A will be /2.

Hence A

=2

45. Answer (3)

Acceleration of system,am

T

mgsin30

2 4

mg mg ga a

m

Tension in string,3

4 4

mg mg mg T T

For pulley,

TT

603 33

4net

mgF T

46. Answer (2)

47. Answer (1)

48. Answer (4)

49. Answer (1)

50. Answer (2)


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54. Answer (4)

1 dm3 = 1000 mL = 1000 g

dH2O = 1g mL1

1 mole of N3contains 22 N

Ae

55. Answer (3)

56. Answer (4)

200 2 100 1 3 100N

400

N = 2

[H+] = 2 mole

57. Answer (4)

1 g eq. of O2 at STP occupies volume of 5.6 L

4 g eq. of O2 at STP occupies volume of 22.4 L

58. Answer (1)

3s3p

In third period there is filling in 3sand 3psubshellwhich posses 4 orbitals and in one orbital threeelectrons can be filled (three spin quantum number).Hence there can be total of 12 elements possible.

59. Answer (2)

Number of peaks = n = 2 1 = 1

60. Answer (4)

Fact

61. Answer (1)

Number of revolution per second = n2

n

V Z

2 r nn.

Z

2

3

Z

n

62. Answer (1)

Fact.

63. Answer (4)

6th excited state means eis present in 7th energy level

Number of spectral line for a particular series= (7 6) + (7 5) + (7 4) + (7 3)

= 1 + 2 + 3 + 4 = 10

51. Answer (1)

Suppose, wt. of Fe2O

3= x g

wt. of Fe3O

4= x g

Number of oxygen atoms in 2 3 1 0x

Fe O (n ) 3 N160

Number of oxygen atoms in 3 4 2 0x

Fe O (n ) 4 N232

1

2

n 174

n 160

52. Answer (1)

64

WO

2= x WO

3= 64 x

2O

xn

32 ;

3O

64 xn

48

2 3O O mix

V V V

x 64 x 522.4 22.4 22.4

32 48 3

x 64 x 5

32 48 3

3x 128 2x 5

96 3

x 128 5

96 3

x + 128 = 160

x 32

2O

32V 22.4

32 = 22.4 L

53. Answer (2)

1 mole of AB2

weighs = 64 g

1 mole of AB3

weighs = 80 g

WA + 2WB = 64 Where WA = atomic weight of A

WA

+ 3WB

= 80 WB= atomic weight of B

On solving, BW 16

AW 32

[CHEMISTRY]


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74. Answer (2)

B2

is the limiting reagent.

75. Answer (1)

The valency of A = 2

The valency of B = 2

2 2A B AB

76. Answer (1)

Fact

77. Answer (4)

30ob

19 12T

0.816 3.336 10100 100 17%

1.602 10 100 10

78. Answer (2)

Fact79. Answer (3)

Fact

80. Answer (1)

Z > 14, bonding e antibondinge

B.O.2

81. Answer (2)

* 2px antibonding molecular orbitals will formed

by subtraction of atomic orbitals by LCAO method

82. Answer (4)

1H (7 0 2 1) 5

2

sp3d 4 bond pair,1lone pair See saw

83. Answer (1)

NA

+ 2 2NA

+ 3 3NA

= 14NA

84. Answer (3)

% % / At. weight Simplest ratio

C =12

10013

10013

1

H =1

10013

100

131

(E.F.) = CH (E.F.M) = 13

(M.F.M) = 2 Vd

= 2 39 = 78

M.F.Mn 6

E.F.M.

(M.F.) = (E.F.)n = C6H6

64. Answer (1)

H HeI

H2 He

Z nV(H) 1 2 1

n . Z 2 1 1V (He )

65. Answer (1)

hx.m. v

4

343 2 6.626 10x 40 10 45

100 4 3.14

x = 1.46 1033m

66. Answer (4)

Radial nodes = n l 1

Angular nodes = l

67. Answer (1)

Fact

68. Answer (1)

11 0

hc hcK.E

22 0

hc hcK.E

1 0

1 0 2 2 0 1

2 0

1 1

3 3 ( ) ( )1 1

69. Answer (3)

A d orbital can accommodate only 2 electrons.

70. Answer (4)

Orbit angular momentum =nh

2

change = 5h h 4h 2h2 2 2

71. Answer (1)

Fact.

72. Answer (2)

Cr [Ar] 4s13d5

Cr +3 [Ar] 4s03d3

73. Answer (4)

d10 configuration has maximum number of exchanges

i.e., 20.


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85. Answer (1)

Total moles of 2 4196

H SO 298

Left moles of H2SO

4= 2 0.5

= 1.5Left atoms = 1.5 7 6.023 1023

= 6.32 1024

86. Answer (2)

Fact

87. Answer (2)

Fact

88. Answer (4)

56 [Xe] 6s2 s block

Period

Group 2

89. Answer (3)

Fact

90. Answer (2)

Fact

91. Answer (1)

3

Li

7 10n

7

3

Na

23 10n

23

ELi

= 103 x kJ = xJ

ENa

= 103 y kJ = yJ

92. Answer (1)

HF

= HS

+ HI.E.

+ HBD.E.

+ Hv+ 2H

EA+ H

L.E.

93. Answer (3)

Fact.

94. Answer (1)

Increasing the s-character, lesser will be bond length.

95. Answer (4)

Fact

96. Answer (1)

Fact

97. Answer (4)

Hybridisation is mixing of atomic orbitals having less

difference in energy.

98. Answer (4)

Bohrs model can be applied for single electronic

species.

99. Answer (1)

Fact.

100. Answer (1)

Fact.

101. Answer (2)

Ovules are always exposed, heterosporous nature.

102. Answer (4)

103. Answer (3)

Juniperus virginiana- Cedar wood oil.

104. Answer (3)

105. Answer (4)

Both (1) & (3)

All embryophytes do not show endoscopicembryogeny.

106. Answer (3)

[BIOLOGY

]

107. Answer (1)

108. Answer (1)

Dryopteris True fern, leptosporangiate.

109. Answer (2)

110. Answer (2)

It is protonema

111. Answer (2)

Sporophyte is a partial parasite on gametophyte inmosses.

112. Answer (4)

113. Answer (3)


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114. Answer (4)

Number of zygospore = number of meiosis = numberof new filaments for Spirogyra

115. Answer (1)

Oedogonium reticulate chloroplast

Pleurosigma discoid chloroplast

116. Answer (3)

117. Answer (2)

Viroids ss, naked RNA

118. Answer (2)

Isidia performs photosynthesis also.

119. Answer (4)

120. Answer (4)

121. Answer (3)

Each cell is dikaryotic in teleutospores.

Spermogonium/pycnium is found towards lower leafsurface of Berberis.

122. Answer (4)

123. Answer (2)

124. Answer (2)

Rami 1st branch, Metulae 2nd branch, Sterigmata= Phialides

125. Answer (3)

Clavicepsand Neurospora Perithecium ; PezizaApothecium

126. Answer (4)

127. Answer (4)

128. Answer (3)

Pseudomonas rubrilineans red stripe disease ofsugarcane; Trichodesmium erythraeum(BGA) red sea

129. Answer (3)

130. Answer (2)

131. Answer (1)

132. Answer (4)

133. Answer (3)

134. Answer (3)

135. Answer (3)

136. Answer (1)

Option (1) = photoorganotroph

137. Answer (3)

138. Answer (4)

139. Answer (2)

140. Answer (2)

141. Answer (3)

142. Answer (1)

143. Answer (1)

phyceae suffix for class, Arecaceae Palmae,Tigon Fertile hybrid

144. Answer (2)

145. Answer (4)

146. Answer (1)

147. Answer (4)

Flatworms are the only acoelomate bilateria.

148. Answer (4)

Notochord is a solid rod of mesodermal origin and ispresent dorsally between alimentary canal and nervecord.

149. Answer (4)

Echinoderms, Hemichordates, Chordates areenterocoelomate.

150. Answer (2)

Aschelminthes have complete alimentary canal,whereas, coelenterates have tissue grade level oforganisation.

151. Answer (2)

Mostly marine and belong to parazoa noteumetazoa.

152. Answer (4)

Aureliahas Medusa only, Obeliahas both polyp andmedusa stages.

153. Answer (3)

Ctenophores have indirect development and arehermaphrodite (Monoecious).

154. Answer (2)

Hooks and suckers are found in parasitic flatwormsonly.

155. Answer (1)Wuchereria, Dracunculusand Loa loa are digenetic.

156. Answer (1)

Entamoeba has no contractile vacuole due toparasitic life. It is endoparasite.

157. Answer (3)

Metacyst is formed after excystation ofquadrinucleated cyst.

158. Answer (2)

Plasmodium is haploid and undergoes zygotic

meiosis.


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159. Answer (3)

Haemozoin is released after bursting of RBCs.

160. Answer (3)

Stinging cells and germ cells are ectodermal inHydra.

161. Answer (4)

Blind sac plan in Hydra.

162. Answer (3)

Second juvenile is infective stage of Ascaris.

163. Answer (4)

In both male and female Ascarisexcretory pore ispresent.

164. Answer (3)

Cortical layer of cuticle is made up of keratin.

165. Answer (4)

As housefly and mosquito are holometabolous.

166. Answer (3)

167. Answer (3)

Johnstons organs are present in pedicel of antennaof insects. They have mechanosensory chordotonalneurons.

168. Answer (3)

Bottom feeder larva is of Culex.

169. Answer (4)

Drum stick shaped balancers are halteres.

170. Answer (3)

The larva of butterfly is caterpillar.

171. Answer (4)

All are insects (bed bug, sandfly and silkworm).

172. Answer (3)

Annelids, Arthropods, and Molluscs are protostomic.

173. Answer (3)

Hookworm belongs to the phylum Aschelminthes.

174. Answer (2)

Characteristic of sponges is canal system.

175. Answer (1)

The larva of echinoderms has bilateral symmetry butthe adult has radial symmetry.

176. Answer (2)

Blood fluke is unisexual.

177. Answer (3)

Nuclear dimorphism is present in Paramecium.

178. Answer (1)

Flat worms are acoelomate.

179. Answer (4)

Miracidium infects snail, Metacercaria is infective

stage for sheep.180. Answer (1)

Echinoderms, arthropods and molluscs have opencirculatory system.

181. Answer (1)

Syncytial epidermis is characteristic of roundworm.

182. Answer (4)

183. Answer (4)

Gastropods become asymmetrical due to torsion.

184. Answer (4)Botryoidal tissue is supposed to be excretory.

185. Answer (2)

Teredois shipworm.

186. Answer (1)

Masticatory structure of Echinoidea is Aristotleslantern.

187. Answer (1)

188. Answer (1)

The larva of Antedonis Doliolaria189. Answer (1)

Class insecta is also called Hexapoda.

190. Answer (1)

Trophozoite damages intestinal wall and activelyfeeds on the tissues.

191. Answer (2)

192. Answer (3)

193. Answer (1)

194. Answer (1)

195. Answer (4)

Only karyokinesis occurs during coenocyteformation, hence no septa formation.

196. Answer (1)

197. Answer (1)

198. Answer (2)

199. Answer (2)

200. Answer (1)

IATS Medical-2011 T1 Solution 10-4-50

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Transcript of IATS Medical-2011 T1 Solution 10-4-50