aiats medical 2014 test-9 solution

Test - 9 (Answers & Hints ) All India Aakash Test Series for Medical-2014

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TEST – 9

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

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

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

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

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

All India Aakash Test Series for Medical-2014 Test - 9 (Answers & Hints)

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Hints to Selected Questions[ PHYSICS]

1. Answer (2)

Gravitational potential Potential energy

Mass=

Latent heatHeat energy

Mass=

2. Answer (3)

2g T Lg T LΔ Δ Δ

= +

3. Answer (4)

Equations involving addition, subtraction anddimensionless quantities cannot be derived.

4. Answer (4)

1 22ut tg

+ =

5. Answer (1)

For equal changes in velocities, ratio of distancesfrom rest are 1 : 3 : 5 : ......

6. Answer (1)

Slope of the position versus time graph increasesgraphically.

7. Answer (3)

If R = nH, angle of projection, 1 4tann

− ⎛ ⎞θ = ⎜ ⎟⎝ ⎠

.

8. Answer (2)

Diagonals of rhombus are perpendicular to eachother.

9. Answer (4)

602 sin 2 20 sin 20 m/s2 2

v v θ °Δ = = × × =

10. Answer (2)

FBA = 3 × a = 163 6 N3 5

× =+

11. Answer (2)

fL = 0.5 × 4 g = 20 N,

Driving force = 1 g = 10 N

Body is at rest so static friction is acting on body.

12. Answer (1)

Normal reaction provides all necessary forces.

13. Answer (2)

14. Answer (3)

2 2duF rdr

= − = − +

For equilibrium, F = 0 ⇒ r = 1 m

15. Answer (4)

Exchange of velocities.

16. Answer (3)

1 1 2 2CM

1 2

m v m vv

m m+

=+

17. Answer (4)

2

2

sin

1

gakr

θ=

+

18. Answer (3)

No external torque is acting in the system.

19. Answer (3)

KE = – ME = PE2

20. Answer (2)

mgRhWR h

=+

21. Answer (1)

r < R 3inGMrER

=

r = R 2sGMER

=

r > R2

GMEr

=

22. Answer (4)

0 0.5≤ σ ≤

23. Answer (2)

W1 : W2 : W3 : W4 : ...... = 1 : 3 : 5 : .....

For consecutive equal elongations.


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24. Answer (4)

25. Answer (2)

Balloon gets thrust in vertically upward directionalong with the thrust in horizontally to the right.

26. Answer (1)

27. Answer (4)

28. Answer (3)

2 1

4 4 4T T Tpr r r

Δ = = −

29. Answer (3)

Melting of ice will be incomplete.

30. Answer (1)

1 2 3 50°C3

θ + θ + θθ = =

31. Answer (4)

Random motion of gas molecules.

32. Answer (2)

1 1 2 2 1 2max

1 2 2n F n F F F

Fn n+ +

= =+

and 22vFC R R= = .

33. Answer (4)

10dpB V CdV B

= − = ⇒ = = ∞

34. Answer (1)

2max

1

300 11 1 0.25 25%400 4

TT

η = − = − = = =

35. Answer (2)

36. Answer (4)

37. Answer (4)

V

P

O

T3T2

T1

T3 > T2 > T1

38. Answer (1)

39. Answer (2)

2 2 2 2 21 1 1( )2 2 2

m A y m A⎛ ⎞ω − = ω⎜ ⎟⎝ ⎠

40. Answer (3)

2 21 2A a a= +

41. Answer (3)

2 mTk

= π

Now, k′ = 4k

24 2m TTk

′ = π =

42. Answer (1)

314 100 m/s3.14

vkω

= = =

43. Answer (2)

44. Answer (3)

3000 320 3200 Hz320 20s

nvnv v

×′ = = =− −

45. Answer (1)

HCF is the fundamental.

[ CHEMISTRY ]46. Answer (3)47. Answer (1)

Fact.48. Answer (1)

Fact.49. Answer (1)

2 2 2 2 2 2 2A A A B B B A BP . M (n n ) P . M .(n n )P P

2 2

BA B

A

MP P

M

2 2 2A B B24P P 1.2 P20

50. Answer (3)

C + O2 ⎯→ CO2

32 g 22.4 L

8 g 5.6 L

2KClO3 ⎯→ 2KCl + 3O2

2 × 122.5 g 3 × 32 g

20.34 g 8 g


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51. Answer (2)In benzyne all C atoms are sp2 hybridised.

52. Answer (1)

Br

Br

being symmetrical its dipole moment is zero.

53. Answer (2)Fact.

54. Answer (2)

1Vn

55. Answer (1)

H PO3 3

+3H PO + PH3 4 3

+5 –3

6

2M M 2M

Eq. M = 6 2 3

n-factor = 32

56. Answer (1)Fact.

57. Answer (1)

To have the total charge on the silicate to be zero3 Ca2+ must be present.

58. Answer (4)

Change in concentration has no effect on the pH ofsolution of salt of weak acid and weak base.

59. Answer (2)

60. Answer (1)

61. Answer (1)

Fact.

62. Answer (2)

Fact.

63. Answer (1)

64. Answer (3)

65. Answer (4)

Water remain neutral at all temperatures though itspH changes.

66. Answer (3)

Fact

67. Answer (3)

AB2lp3 type of structure are linear.

68. Answer (3)

69. Answer (1)

KMnO /H4+

COOH

70. Answer (4)

CH CH CH CH3 2 2 2 OH,

CH CH CH OH3 2

CH3

,

CH CH CH CH3 2 3, CH C CH3 3

CH3

OHOH

,

CH CH O CH3 3,

CH3

CH CH CH O CH3 2 2 3,

CH CH O CH CH3 2 2 3

71. Answer (1)

C CH 1% HgSO4

20% H SO2 4

C CH2

C CH3

OH

O

Tautomerism

72. Answer (1)Fact

73. Answer (1)74. Answer (1)75. Answer (4)76. Answer (1)

Least charge separation.77. Answer (4)

Cis and trans.


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[ BIOLOGY ]

78. Answer (2)Only terminal alkynes with acidic hydrogen give thistest.

79. Answer (3)Fact

80. Answer (1)81. Answer (1)

IUPAC - rule82. Answer (3)

It forms sodium meta aluminate.83. Answer (3)

H C O + HI2 2 4 C H2 6

–3+3

12No. of gram equivalent of HI = 12

91. Answer (3)

92. Answer (1)

Category which contains related order is class.

93. Answer (3)

Spirogyra – Plantae

Nostoc, Anabaena – Monera

Mucor, Puccinia – Fungi

Euglena, Noctiluca – Protista

94. Answer (3)

Spore, unfavourable conditions.

95. Answer (4)

Linnaeus system.

96. Answer (2)

Sporangiospore, zoospore, conidia – Asexual spore.

97. Answer (3)

Fossil evidence – Phylogenetic system

98. Answer (1)

Ectocarpus – Filamentous branched

99. Answer (4)

Bryophytes – Liverworts + Mosses

Pteridopyte – Ferns + Horse tails

Gymnosperms – Cycads + Conifers

Angiosperms – Monocots + Dicots

100. Answer (3)

Sporic meiosis is present in bryophytes,pteridophytes, gymnosperms, angiosperms.

101. Answer (3)

Dryopteris – Ferns – Frond – Large size leaf.

102. Answer (3)

Turmeric – Underground storage stem.

103. Answer (2)

(a) Unipinnate leaf of neem.

(b) Palmate leaf of silk cotton tree.

104. Answer (3)

Plum, rose – Perigynous

Guava – Epigynous.

105. Answer (2)

Fabaceae family.

106. Answer (2)

Lignin thickening.

107. Answer (3)

Only green cells of epidermis are guard cells.

108. Answer (2)

Bundle cap – Dicot – Sclerenchymatous pericycle

Bundle sheath – Monocot – Sclerenchyma

84. Answer (2)Volume strength = 5.6 × N

N = 3.57

85. Answer (4)

All have vacant orbital.

86. Answer (1)

Fact.

87. Answer (2)

88. Answer (4)

89. Answer (2)

Fact.

90. Answer (1)

CaCO3(s) → CaO(s) + CO2(g) ↑


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80S ribosomes are formed by nucleolus.111. Answer (3)112. Answer (2)113. Answer (4)

Fluid nature of membrane is due to the mobility ofprotein and lipids.

114. Answer (1)More clearly visible tetrad, recombination nodule,crossing over.


2n = 46 chromosome.In S-phase each chromosome will have twochromatids, so total number of chromatids are2 × 46 = 92.

117. Answer (4)All are correct.

118. Answer (1)Specific, for hydrophilic molecule, no energy used,transport saturation present.

119. Answer (2)Guttation.

120. Answer (1)Phloem transport is bidirectional because source andsink relationship can be variable.


Leghaemoglobin activates the nitrogenase.123. Answer (4)

Mainly NO3– is absorbed.

124. Answer (3)Macroelements, microelements.


PS II - P680, PS I - P700

127. Answer (3)Ribulose 1, 5, bisphosphate.

128. Answer (1)Maize – C4 plant – Bundle sheath cell, mesophyllWheat – C3 plant – Mesophyll cell

129. Answer (3)

130. Answer (2)

Pyruvate Acetyl CoA

NAD+

NADH CO2

CoA

131. Answer (4)Cytochrome a–a3 complex.

132. Answer (3)0.7 – Fat> 1 – Organic acid1 – Carbohydrates

133. Answer (2)Cytokinin counteracts apical dominance.

134. Answer (4)Secondary meristem.


Enzymes generally function in a narrow range oftemperature and pH. Each enzyme shows itshighest activity at a particular temperature and pHcalled as the optimum temperature and optimum pH.With the increase in substrate concentration, thevelocity of the enzymatic reaction rises at first. Thereaction ultimately reaches maximum velocity (Vmax)which is not exceeded by any further rise inconcentration of the substrate.

137. Answer (2)The excess of secretion of growth hormone afteradolescence causes excessive growth of the bonesof limbs and lower jaw, and the person has gorillalike appearance.

138. Answer (2)Overproduction of hormones by adrenal cortexcauses Cushing’s syndrome and Conn’s disease.Cortisol stimulates RBC production causingpolycythemia.

139. Answer (3)Marasmus occurs in children less than one year ofage. Whereas, kwashiorkor occurs in children morethan one year of age.

140. Answer (2)Tunica media is comparatively thin in veins becausesmooth muscles are negligible.

141. Answer (4)

Coenzymes are nucleotides of vitamins.


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142. Answer (3)

Estrogen hormone is a lipophilic hormone, whichdirectly enters inside the cell and does not generatea second messenger. It binds with intracellularreceptors in nucleus.

143. Answer (2)Point A in the graph indicates depolarisation andpoint C in the graph indicates repolarisation.


Hypoxia is one of the factors which stimulates theJG cells to release erythropoietin hormone, tostimulate erythropoiesis.

146 Answer (4)147. Answer (3)

Knee joint, ankle joint and elbow joint all are theexamples of hinge joint.



152. Answer (4)Ciliated epithelium move particles, cells or mucus ina specific direction over the epithelium.

153. Answer (3)

154. Answer (4)Stratified compound epithelium covers the drysurface of the skin, buccal cavity, pharynx, ducts ofsalivary and pancreatic glands. A, B and C all aresimple epithelial tissues.

155. Answer (4)

The single letter Y depicts tyrosine amino acid.156. Answer (4)

A is nucleoside (adenosine).B is uracil.C is adenine (purine).

D is nucleotide (adenylic acid) i.e., AMP.157. Answer (2)158. Answer (3)

The diagram illustrated in the choice number three isof Ophiura (brittle star).

159. Answer (2)In Ascaris males are shorter than females and thetail is curved.In Hippocampus, males bear a brood pouch.

160. Answer (3)161. Answer (3)162. Answer (2)

Pivot joint is present between atlas and axis.Between the skull and atlas ellipsoid joint ispresent.


Chameleon is tree lizard, with prehensile tail.166. Answer (4)167. Answer (3)

In female cockroach ovaries are present in 2 to 6thabdominal segments.

168. Answer (1)169. Answer (3)170. Answer (1)171. Answer (4)172. Answer (1)173. Answer (3)

Collagen is the most abundant protein in animalworld. The most abundant protein in biosphere isRUBISCO.


Rods are evenly distributed all over the retina.Whereas, cones are more concentrated in the centreof retina.

177. Answer (2)Lymph can coagulate, it has fibrinogen and calcium.


Hippocampus is bony fish.180. Answer (3)

All the enzymes of intestinal juice are brush borderenzymes.

aiats medical 2014 test-9 solution

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