aiats_medical_aipmt2015_test1.pdfghjghjghjghjghjghjghjghjghjghjghjghjghjhj

Test - 1_Code B (Answers & Hints) All India Aakash Test Series for Medical-2015

1/8

TEST – 1

ANSWERS1. (3)2. (2)3. (4)4. (4)5. (1)6. (4)7. (3)8. (2)9. (3)10. (3)11. (4)12. (1)13. (1)14. (2)15. (2)16. (4)17. (2)18. (4)19. (3)20. (3)21. (4)22. (2)23. (3)24. (2)25. (4)26. (1)27. (2)28. (4)29. (1)30. (2)31. (1)32. (4)33. (2)34. (2)35. (2)36. (4)

37. (2)38. (2)39. (4)40. (3)41. (3)42. (4)43. (2)44. (2)45. (4)46. (2)47. (1)48. (2)49. (4)50. (2)51. (3)52. (2)53. (3)54. (3)55. (1)56. (2)57. (4)58. (1)59. (2)60. (4)61. (4)62. (4)63. (4)64. (2)65. (3)66. (4)67. (3)68. (2)69. (3)70. (1)71. (2)72. (2)

73. (2)74. (1)75. (4)76. (1)77. (4)78. (3)79. (4)80. (3)81. (3)82. (2)83. (4)84. (2)85. (4)86. (3)87. (4)88. (1)89. (4)90. (1)91. (4)92. (3)93. (3)94. (3)95. (2)96. (2)97. (3)98. (2)99. (4)100. (4)101. (3)102. (3)103. (3)104. (2)105. (4)106. (3)107. (3)108. (3)

109. (3)110. (4)111. (1)112. (3)113. (3)114. (3)115. (3)116. (3)117. (2)118. (4)119. (4)120. (3)121. (4)122. (4)123. (3)124. (4)125. (3)126. (3)127. (3)128. (1)129. (2)130. (1)131. (1)132. (1)133. (3)134. (3)135. (3)136. (1)137. (3)138. (4)139. (2)140. (3)141. (4)142. (3)143. (3)144. (4)

145. (3)146. (2)147. (3)148. (2)149. (1)150. (2)151. (2)152. (3)153. (2)154. (1)155. (1)156. (3)157. (1)158. (4)159. (1)160. (4)161. (3)162. (2)163. (1)164. (2)165. (2)166. (3)167. (2)168. (1)169. (1)170. (4)171. (3)172. (4)173. (3)174. (3)175. (2)176. (4)177. (3)178. (3)179. (2)180. (4)

All India Aakash Test Series for Medical-2015 Test - 1_Code B (Answers & Hints)

2/8

Hints to Selected Questions[ PHYSICS]

1. Answer (3)

Energy density = 2 2

1 23

[ML T ] [ML T ][L ]

−− −=

Pressure = [ML–1T–2]

2. Answer (2)

3. Answer (4)

4. Answer (4)

5. Answer (1)

n1u1 = n2u2

1 kg m2 s–2 = n2 (10 kg) (10 m)2 (1 minute)–2

n2 = 3.6 = 36 × 10–1

6. Answer (4)

7. Answer (3)

F = px + qt2 + r

[px] = [qt2] = [r] = [F]

22 [MLT ]pxr F

qt−= =

8. Answer (2)

x = Asin(ABt)

A = [L]

Dimension of ABt = [L0T0]

1 11 [L T ][L] [T]

− −= =B

9. Answer (3)

10. Answer (3)

2100

dLL

= +

1 12 2

dT dL dgT L g

= −

4100

dgg

= −

1 1% (2) ( 4)2 2

T = − −

= 3% (Increases)

11. Answer (4)

Least % error in 500 × 10–4 m.

12. Answer (1)

13. Answer (1)

S% = 4 = 2 % (s = 6 2)

% = 2%

v = 3 % (v = 3)

v = 6%

14. Answer (2)

15. Answer (2)

0.003% 100 1%0.3

m = × =

0.005% 100 1 %0.5

= × =r

0.06% 100 1 %6

= × =

2mdr

=π

d% = m% + 2r% + %

d% = 4%

16. Answer (4)

Maximum % error = 3L% + 2T%

= 3 × 1 + 2 × 0.5

= 4%

17. Answer (2)

18. Answer (4)

19. Answer (3)

20. Answer (3)

21. Answer (4)

22. Answer (2)

21

1 (2)2

h g=

2 22

1 1( ) (2)2 2

= −h g u g

1

2

13

hh

=


3/8

23. Answer (3)

h2 = 3h1

212

h gT=

221 1

2 2 4Th gT g ⎛ ⎞′ = − ⎜ ⎟⎝ ⎠

21 152 16

h gT ⎛ ⎞′ = ⎜ ⎟⎝ ⎠

1516

h h⎛ ⎞′ = ⎜ ⎟⎝ ⎠

24. Answer (2)

1 2 2

21 1

x xv x xv v v v

+= =

+ +

⇒1 2

2 1 1v v v= +

25. Answer (4)

26. Answer (1)

=v B x

v2 = B2x

uniformly accelerated

Acceleration. = 2

2β

221

2 2⎛ ⎞β

= ⎜ ⎟⎜ ⎟⎝ ⎠

s t

27. Answer (2)

Area = 1 13 30 1 ( 10)2 2× × + × × −

= 45 – 5

= 40 m

28. Answer (4)

29. Answer (1)

2u vs t+⎛ ⎞= ⎜ ⎟⎝ ⎠

30. Answer (2)

th1 (2 1)2

= + −t

s u a t

31. Answer (1)

α = 2 m/s2 T = 10 s

(Retardation)β = 3 m/s2 maxv T⎛ ⎞αβ= ⎜ ⎟α + β⎝ ⎠

32. Answer (4)

A B C40 m

64 m

tu = 0 = 12 m/s

For AB, 212

s ut at= +

a = –1 m/s2

For AC, 2164 12 (1)2

= −t t

t = 8 s and 16 s

33. Answer (2)

22 2 (30)

2⎛ ⎞ = −⎜ ⎟⎝ ⎠

v v g

800v =

20 2 m/sv =

28 m/s

34. Answer (2)

35. Answer (2)

36. Answer (4)

37. Answer (2)

x = t2 – t3

u = 2t – 3t2

For max. velocity

0dvdt

= ,

2

2 0d vdt

<

2 – 6t = 0

1 s3

=t


4/8

[ CHEMISTRY]

38. Answer (2)

(acc.) av

α =

dv adt v

=

0 0

=∫ ∫v t

v

vdv adt

0

20

2⎡ ⎤

= −⎢ ⎥⎢ ⎥⎣ ⎦

v

v

v at

220

2 2vv at− =

220

2 2= +

vv at

39. Answer (4)

40. Answer (3)

41. Answer (3)

42. Answer (4)2 3x t t= α − β

22 3dx t tdt

= α − β

v = 0

23α

=β

t

x = 0

t α=β

43. Answer (2)x = 10t – 2t2

10 4dxv tdt

= = −

v = 0, 5 s2

t =

Average speed = | (2.5) (2) | | (3) (2.5) |

1x x x x− + −

= 1 m/s44. Answer (2)

Area = v5 – v0

51 5 5 02

v× × = −

v5 = 12.5 m/s45. Answer (4)

46. Answer (2)

Relative mass of Mg

= 78.7 10.1 11.224 25 26100 100 100

⎛ ⎞ ⎛ ⎞ ⎛ ⎞× + × + ×⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

= 24.3

47. Answer (1)

Number of Cu atoms in 0.200 mol

= 230.2 6.022 10× ×

= 231.20 10×

48. Answer (2)

N (g)22.7

1

2 3H (g)68.1

3

2+::

⎯::

→ 2NH (g)45.4

2

3

The ratio of gases is consistent with thestoichiometry of the chemical equation.

49. Answer (4)Fact

50. Answer (2)M1V1 = M2V2

⇒ 0.5 × x = 0.02 × (250 + x)∴ x = 10.41 mL

51. Answer (3)

No. of moles of H2SO4 = mol50 mL 1L

×

= 50 × 10–3 mol∴ Mole of Zn required = 50 × 10–3 mol∴ Mass of Zn = 50 × 65 × 10–3 g

= 3.25 g


5/8

52. Answer (2)

Millimole of AgNO3 = 50 × 0.2 = 10

Millimole of NaI = 25 × 0.4 = 10

∴ Mole of AgI formed = 10 m mol

= 10 × 10–3 mol

∴ Amount of AgI (in g) = 10–2 × 235 g

= 2.35 g

53. Answer (3)

The empirical formula of the compound would beC3H7 and the molecular formula of alkane will beC6H14.

54. Answer (3)

No. of H-atom = 2339.6 8 6.02 10132.1

× × ×

= 1.44 × 1024 H-atom

55. Answer (1)

2CO2.2 1n mole4.4 20

= =

∴ No. of O-atoms = A1 N

10

2NO2.3 1n mole46 20

= =

∴ No. of O-atoms = A1 N

10

∴ Ratio would be 1 : 1

56. Answer (2)

Fact

57. Answer (4)

58. Answer (1)

200 drops ≡ 10 g

∴ 1 drop ≡ 10 g 0.05 g200

=

∴ No. of molecules= 230.05 6.023 1018

× ×

= 1.673 × 1021

59. Answer (2)

180 g H2O has maximum number of moles.

60. Answer (4)

Fact

61. Answer (4)

Fact62. Answer (4)

Fact63. Answer (4)

Fact64. Answer (2)

m = B

1000 M1000d M.M

×−

= 1000 3

1000 1.25 3 40×

× − ×

= 2.79 m65. Answer (3)

N = 7 100049 250

×

= 4N7

66. Answer (4)Fact

67. Answer (3)

m = 20 100040 80××

= 6.25 m68. Answer (2)

Relative moles of Fe = 69.9 1.2556

=

O = 30.1 1.8816

=

∴ Simplest ratio = 1 : 1.5= 2 : 3

∴ Empirical formula= Fe2O3

69. Answer (3)Fact

70. Answer (1)Fact

71. Answer (2)

30 g ethane 65 g Zn pure≡∵

3 g ethane 6.5 g Zn pure∴ ≡

Impure Zn = 10 g∴


6/8

72. Answer (2)

2H O60 / 18x 0.83340 6060 18

= =+

73. Answer (2)

1000M 55.55 M18

= =

74. Answer (1)1 g equivalent O2 ≡ 5.6 L

75. Answer (4)All has NA-number of electrons

76. Answer (1)Fact

77. Answer (4)28 9.11 10 g electron 1 electron−× ≡∵

281 1 g electron electron

9.11 10−∴ ≡×

∴ Mole of electron in 1 g

≡ 28 231

9.11 10 6.023 10−× × ×

≡ 51 10 mole9.11 6.023

××

78. Answer (3)

2 3 276 g Ag CO 1.5 22.4 L≡ ×∵

2 3 2.76 g Ag CO 0.336 L∴ ≡

79. Answer (4)Fact

[ BIOLOGY ]

91. Answer (4)Paramoecium, Chlamydomonas, Chlorella andAmoeba were placed in Protista.

92. Answer (3)Monera

93. Answer (3)Cell wall, locomotion, contractile system, nutrition,response to external stimulus.

94. Answer (3)Bacteria

95. Answer (2)96. Answer (2)

97. Answer (3)Cyanobacteria show oxygenic photosynthesis.


Heterocyst is non-green cell.100. Answer (4)

Cyanobacteria, Methanogens, Thermoacidophiles areautotrophic.

101. Answer (3)Salmonella - TyphoidClostridium - TetanusVibrio - Cholera

80. Answer (3)

Fact

81. Answer (3)

82. Answer (2)

Atomic mass =54 5 56 90 57 5

100× + × + ×

= 55.95

83. Answer (4)

84. Answer (2)

Equivalent mass of carbonate = 12 + 30 = 42

85. Answer (4)

Fact

86. Answer (3)

Fact

87. Answer (4)

Fact

88. Answer (1)

0.34% 56 g≡∵

56 100% 100 16470.58 g0.34

∴ ≡ × =

89. Answer (4)

Fact

90. Answer (1)

Fact


7/8

102. Answer (3)

Flagella help in locomotion.103. Answer (3)

Methanogens - Obligate anaerobe

Thermoacidophiles - Autotrophs104. Answer (2)

Bacteria105. Answer (4)

Chemoautotrophs oxidise inorganic substances.

106. Answer (3)Parasexuality

107. Answer (3)

Three domains are Bacteria, Archae and Eukarya108. Answer (3)


Influenza is viral disease.

111. Answer (1)Sap vacuoles are absent.

112. Answer (3)

Cell wall is made up of pseudomurein.113. Answer (3)


Erwinia causes fire blight of apple.

116. Answer (3)Specific for each category, used for identification forplants and animals.

117. Answer (2)


Herbarium – Specimens on sheetsEubacteria – Flagella present or absentChlorella – Cannot produce nitrogenase

Properties of cellular organelles arise as a result ofinteractions among the molecular componentscomprising the organelle.

120. Answer (3)Potato, Brinjal and Makoi, belong to Solanum.

121. Answer (4)

Diptera is an order.

122. Answer (4)

Lowest category is species.

123. Answer (3)

Problem of classification becomes difficult fromspecies to kingdom.

124. Answer (4)

125. Answer (3)

126. Answer (3)

International code for Botanical nomenclature.

127. Answer (3)

128. Answer (1)

- Derived from Latin word systema

- Based on evolutionary study

129. Answer (2)

Biological concept is based on reproductive isolation.

130. Answer (1)

131. Answer (1)

Unicellular organisms - Protists, Monera

132. Answer (1)

Protonema - Fragmentation, Planaria - Regeneration

133. Answer (3)

Reproduction is absent in mule, worker bees, somehuman couple.

134. Answer (3)

Growth of sand-dunes is due to the addition ofmaterial from outside.

135. Answer (3)

136. Answer (1)

Porifera : Lack neurons. All members of kingdomAnimalia are multicellular without exception.

137. Answer (3)

Nereis has closed circulatory system.

138. Answer (4)

Ascaris is Pseudocoelomate.

139. Answer (2)

Open circulatory system is present inBalanoglossus.

140. Answer (3)

In echinoderms there is external fertilization.

141. Answer (4)

Planaria is acoelomate.


8/8

142. Answer (3)


Molluscs are unsegmented.145. Answer (3)

Sea hare, tusk shell, squid, chiton all are molluscs.146. Answer (2)

Spongilla and Hydra are fresh water forms.147. Answer (3)

Corpora allata produces juvenile hormone.148. Answer (2)

Malpighian tubules are the excretory organs ofinsects e.g. Culex, Locusta.

149. Answer (1)Blood fluke is dioecious.

150. Answer (2)(i) Physalia is an hydrozoan, both polyp and

medusoid form present.(ii) Ancylostoma, is hookworm, it has filariform

larvae.(iii) Pleurobrachia, is ctenophore and fertilisation is

external.151. Answer (2)152. Answer (3)153. Answer (2)

Cephalopods are predators with ink gland.154. Answer (1)155. Answer (1)156. Answer (3)157. Answer (1)

In silver fish, there is direct development.158. Answer (4)159. Answer (1)

160. Answer (4)

161. Answer (3)

Metacercaria is infective stage for primary host i.e.sheep.

162. Answer (2)

Arthropods have white blood, they lack therespiratory pigment.

163. Answer (1)

Choanocytes are collar cells.

164. Answer (2)

Metamerism is present in arthropods too.

165. Answer (2)

166. Answer (3)

167. Answer (2)

168. Answer (1)

169. Answer (1)

Leech has no setae or parapodia.

170. Answer (4)

171. Answer (3)

172. Answer (4)

173. Answer (3)

Wuchereria is present in the lymph vessels.

174. Answer (3)

In ctenophores, bioluminescence is well marked.

175. Answer (2)

176. Answer (4)

Ringworm is fungus.

177. Answer (3)

Tapeworm exhibits pseudometamerism.

178. Answer (3)

179. Answer (2)

180. Answer (4)

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