Solutions_AIATS Medical-2016 Test-09_(Code-A & B)

8/18/2019 Solutions_AIATS Medical-2016 Test-09_(Code-A & B)

1/19

Test - 9 (Code A) (Answers & Hints) All India Aakash Test Series for Medical-2016

1/9

1. (2)

2. (3)

3. (3)

4. (3)

5. (3)

6. (4)

7. (1)

8. (3)

9. (1)

10. (2)

11. (4)

12. (1)

13. (2)

14. (3)

15. (3)

16. (4)

17. (3)

18. (3)

19. (2)

20. (2)

21. (1)

22. (4)

23. (1)

24. (2)

25. (1)

26. (4)

27. (2)

28. (2)

29. (1)

30. (3)31. (3)

32. (4)

33. (2)

34. (2)

35. (1)

36. (4)

37. (2)38. (2)

39. (4)

40. (1)

41. (4)

42. (1)

43. (3)

44. (4)

45. (1)

46. (2)

47. (1)

48. (3)

49. (4)

50. (1)

51. (3)

52. (4)

53. (3)

54. (2)

55. (2)

56. (2)

57. (4)

58. (2)

59. (1)

60. (1)

61. (2)

62. (4)

63. (3)

64. (3)

65. (2)

66. (3)67. (1)

68. (2)

69. (3)

70. (1)

71. (2)

72. (4)

ANSWERS

TEST - 9 (Code-A)

All India Aakash Test Series for Medical-2016

Test Date : 03-04-2016

73. (2)74. (3)

75. (4)

76. (4)

77. (2)

78. (4)

79. (4)

80. (4)

81. (3)

82. (3)

83. (4)

84. (4)

85. (3)

86. (1)

87. (4)

88. (4)

89. (4)

90. (3)

91. (3)

92. (1)

93. (4)

94. (3)

95. (3)

96. (2)

97. (3)

98. (2)

99. (1)

100. (3)

101. (3)

102. (2)103. (3)

104. (3)

105. (3)

106. (2)

107. (2)

108. (3)

109. (3)110. (4)

111. (2)

112. (1)

113. (2)

114. (1)

115. (4)

116. (2)

117. (3)

118. (4)

119. (4)

120. (2)

121. (4)

122. (2)

123. (4)

124. (2)

125. (4)

126. (3)

127. (3)

128. (2)

129. (1)

130. (3)

131. (4)

132. (2)

133. (3)

134. (3)

135. (2)

136. (1)

137. (3)

138. (3)139. (1)

140. (3)

141. (4)

142. (2)

143. (2)

144. (4)

145. (2)146. (1)

147. (3)

148. (2)

149. (4)

150. (3)

151. (3)

152. (1)

153. (3)

154. (4)

155. (1)

156. (2)

157. (4)

158. (4)

159. (3)

160. (3)

161. (2)

162. (4)

163. (3)

164. (2)

165. (1)

166. (4)

167. (2)

168. (2)

169. (2)

170. (3)

171. (3)

172. (1)

173. (2)

174. (1)175. (1)

176. (3)

177. (3)

178. (2)

179. (1)

180. (1)

click here for code-B solutions


2/19

All India Aakash Test Series for Medical-2016 Test - 9 (Code A) (Answers & Hints)

2/9

Hints to Selected Questions

[ PHYSICS]

1. Answer (2)

2. Answer (3)

R =2

PQ

r

R% = l % + 2r %

=1 0.02

2 100100 1

3. Answer (3)

4. Answer (3)

v x

= 1 m/s v y = 2 m/s, tan =

y

x

v

v

y = tan –2

2

1

2x

x g

v

5. Answer (3)

A E

W

V pw

V pg

V wg

N

S

V wg

pg V

sin = wg V

40 sin = 20

= 30°

6. Answer (4)

P will be the maximum height.

7. Answer (1)

A

T 1

T 2

C

B90°–

mg

30°

1

sin120

T

=

sin(60 )

Mg

(T 1)min

where sin (60° + ) = 1

= 30°

8. Answer (3)

AN

B

mg sin

V

m g c o

s

Between A and B

mg (R – R cos) + 0 = 21

2mv

mg cos – N =2

mv

R , N = 0

9. Answer (1)

qE

qE cos

mg q E

c o s s i n

N N

mg s i n

N = mg cos + q E sin

mg sin = N + q E cos

10. Answer (2)

V 12

is uniform along – y axis.

12V

= 1 1

ˆ ˆcos 37 sin 37u i u j –

2 2

ˆ ˆcos 53 sin 53u i u j

= 1 2

ˆcos 37 cos 53u u i

1 2

ˆsin 37 sin 53u u j

1 2cos 37 cos 53 0u u

1 2

4 3

5 5u u

1 24 3u u

11. Answer (4)

12. Answer (1)


3/19


3/9

13. Answer (2)

14. Answer (3)

15. Answer (3)

C = C v –

1

R

u , PV n = constant

P =K

T constant

PV

T

P 2V = K

1

2PV = K

16. Answer (4)

mV 0 = 2m v

21 (2 )

2

m v = 21

2K x

17. Answer (3)

18. Answer (3)

A1V 1 = A

2V 2

19. Answer (2)

' w o

w s

V V V f f

V V V

20. Answer (2)

21. Answer (1)

22. Answer (4)

ˆi

P mv i

ˆ

3i

v P m j mv

i f P P

23. Answer (1)

5 – 10i = 2 + i x = 4

24. Answer (2)

25. Answer (1)

2 6 61

5 12 10 120 12 102

= 2 61

5 12 102

Heat

26. Answer (4)

0

0

640 693 2.3log

10

N

t N

27. Answer (2)

28. Answer (2)

3

h

kTm

29. Answer (1)

30. Answer (3)

E

S

X

W

Y Y

N

M = NiA

M = m x – my

31. Answer (3)

32. Answer (4)

33. Answer (2)

2 2 2

rms

200 100 50

2 2 2

V

34. Answer (2)

0 1.5 1

I f f

u f

1

2 .5

I

4I cm

35. Answer (1)

36. Answer (4)

– eV = 2 21

2m v u

37. Answer (2)

No. of fission =3

3

10

200 1.6 10


4/19


4/9

38. Answer (2)

39. Answer (4)

S 1 + S

2 = 0

( –1) A1 + ( –1) A

2 = 0

40. Answer (1)

41. Answer (4)

42. Answer (1)

43. Answer (3)

– ray is E.M. radiation

44. Answer (4)

f o + f

e= 60

f o + f

e+10 = V

o + f

e

V o

= 65, f o = 55

1 1

o oV u

=1

of

45. Answer (1)

E = B C

46. Answer (2)

Number of gram equivalents of oxygen = number of

gram equivalents of aluminium

⇒

w

8 =

1.35

9

⇒ w = 1.2 g

47. Answer (1)

Basicity of H3PO

2 is one

48. Answer (3)

2

2 2

1 2

1 1 1R ( )

n nZ

2

2 2

1 1 1

R (2)(2) (4)

Divide and multiply by (4)2

2

2 2

1 1 1R (4)

(4) (8)

49. Answer (4)

50. Answer (1)

Cr belongs to 6th group

51. Answer (3)

O sp

2

52. Answer (4)

F–Be–F

53. Answer (3)

Vc = 3 b

[ CHEMISTRY]

54. Answer (2)

1 mole NaOH + 1 mole HF give 68 kJ/mole

55. Answer (2)

A + B B

G =

°pG G R

= 500 [ 100 ( 200)]

= – 500 + 300

= – 200

G = – 2.303 RT Log Ke

– 200000 = – 2.303 8.31 T log 103

T = 3483.4 K

56. Answer (2)

Kp and total pressure remain constant

57. Answer (4)

MnO +5I I + Mn– – 2+

24

+7

–5

2

58. Answer (2)

Fact

59. Answer (1)

60. Answer (1)

Conc. H SO2 4

COOH

COOH CO + CO + H O2 2

61. Answer (2)

62. Answer (4)

Correct according to IUPAC rules


5/19


5/9

63. Answer (3)

O

O

will not show tautomerism

64. Answer (3)It is dissymmetric

65. Answer (2)

Acidic character – I

66. Answer (3)

Fact

67. Answer (1)

is aromatic

68. Answer (2)

FeCl3 + NaSCN 3

Blood red

Fe(SCN) +NaCl

69. Answer (3)

CH – CH – CH – CH3 3

CH3

Cl

, the total reactivity.

is 2 3.8 = 7.6 for 2° - carbon atom

70. Answer (1)

CH – C = CH CH – C = O + HCHO3 2 3

CH3

O3

Zn/H O2

CH3

71. Answer (2)

SO2 is reducing agent

72. Answer (4)

C

B

A

D

AB =a

2, AD = 2a ,

3AC a

2

73. Answer (2)

P = 100 m

A

P

P

= B

100 1X

700 7

bT =

bK m

= BbB A

X 1000K

1 X M

=

1

1000710

1 1671

1

7

= 10 K

74. Answer (3)

2Cl

red red 2

P0.059E E log

n Cl

.

75. Answer (4)

For zero order reaction

t1 + t

2 + t

3 = 10 + 5 + 2.5 = 17.5 s.

76. Answer (4)

Argyrol is Ag sol (–ve)

77. Answer (2)

2 2 2 Ag S 4NaCN 2Na[Ag(CN) ] Na S .

78. Answer (4)

Ionic PCl5 is [ PCl

4 ]+ [ PCl

6]–

79. Answer (4)

Fact

80. Answer (4)

All are correct

81. Answer (3)

Fe3+ has d 5 system

82. Answer (3)

Ionization and linkage both isomerism are possible.

83. Answer (4)

C6H5Cl has partial double bond character.


6/19


6/9

84. Answer (4)

Fact

85. Answer (3)

A is CH – C – CH3 3

NOH

Oxime

CH – CH – CH

3 3

NO

2- Nitrosopropane

86. Answer (1)

CH CH3 2

CH CH COOH CH CH CH CH NH2 2 3 2 2 2 2

HN3

CH CH CH3 2 2

CH NH CH – CH CH CH OH2 2 3 2 2 2

HNO

2

Conc.

H SO

443 K

2 4

HCl

87. Answer (4)

Br 2

(aq) forms saccharic acid from glucose but not

from fructose.

88. Answer (4)

CH = C – CH = CH and CH = C – CH = CH2 2 2 2

CH3

Isoprene

Cl

Chloroprene

89. Answer (4)

MgCO3 and Al(OH)

3 are antacid.

90. Answer (3)

OH

CHCl2

OO

OH

+ NaOH + CHCl3

O–H

C = O

O

NaOH

H

In this reaction electrophile is CCl2

91. Answer (3)

X X aaC C

XX aaC

X YaaC

X XYaa

XaYa

X aC

92. Answer (1)

n = 20

Caruncle (2n) = 40

tube nucleus (n) = 20

PEC (3n) = 60

Nucellus (2n) = 40

93. Answer (4)Zea mays – C

4 plant

94. Answer (3)

Collenchyma

95. Answer (3)

Insectivorous plants

[ BIOLOGY

]

96. Answer (2)

TTRR

TR

TtRR

(Tall-Red)

XttRr

tR tr

TtRr

(Tall-Pink)

97. Answer (3)

Oxidative phosphorylation in mitochondria.

98. Answer (2)

Diapause is a stage of suspended development.

99. Answer (1)

Dry to mesic conditions.

100. Answer (3)

23 S rRNA.

101. Answer (3)

Hypogynous and perigynous flowers.

102. Answer (2)

(b) Metaphase I

(d) Anaphase I

(a) Anaphase II


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7/9

103. Answer (3)

Largest tiger reserve.

104. Answer (3)

Parietal placentation

105. Answer (3)

106. Answer (2)80 dB

107. Answer (2)

Autosomal dominant trait

108. Answer (3)

Pyruvate act as oxidant in lactic acid fermentation.

109. Answer (3)

ER, golgi body, lysosomes and vacuoles in

endomembrane system.

110. Answer (4)

LDP.

111. Answer (2)

Secondary phloem is developed on outer side of

vascular cambium.

112. Answer (1)

Ethylene

113. Answer (2)

Fucus – Brown algae

Polysiphonia – Red algae

114. Answer (1)

CMP-11, 2015, Paris.

115. Answer (4)

Dorsiventral thalloid gametophyte in liverworts.

116. Answer (2)

Nucleosomes in eukaryotes.

117. Answer (3)

Diatoms.

118. Answer (4)

(a) Gases (b) Bacteria, fungi (c) Public transport

119. Answer (4)

Viroid is naked RNA.

120. Answer (2)

121. Answer (4)

Ulothrix , Spirogyra – cellulosic wall.

122. Answer (2)

Hydrophyte.

123. Answer (4)

Axillary buds.124. Answer (2)

Consciousness.

125. Answer (4)

Functionally similar.

126. Answer (3)

Night and early morning when absorption is high.

127. Answer (3)

Pericarp, placenta and thalamus are edible in guava.

128. Answer (2)

129. Answer (1)

130. Answer (3)

131. Answer (4)

Fungi, bacteria and cyanobacteria in organic farming

132. Answer (2)

PEG and NaNO3

133. Answer (3)

ETC and oxidative phosphorylation.

134. Answer (3)

Albugo candida is parasite on mustard.

135. Answer (2)

Haplo-diplontic.

136. Answer (1)

The first digit in E.C. indicates the class of theenzyme and hence the mode of its activity.

137. Answer (3)

Proton pump inhibition results in reduced HClrelease.

138. Answer (3)

Thymosine is a hormone from thymus gland.

139. Answer (1)

Regression of corpus luteum is triggered by reducedlevel of LH of secretory phase.


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8/9

140. Answer (3)

The first haploid cells in testes are secondary

spermatocytes while in ovaries are secondary

oocyte and first polar body.

141. Answer (4)

Annelids generally have closed circulatory system.

142. Answer (2)

Mitochondrial DNA from Africans populations has

maximum variations through mutations.

143. Answer (2)

Elytra are mesothoracic wings. Two pairs of thoracic

and eight pairs of abdominal spiracles are present in

cockroach. Unjointed anal styles are found only in

male cockroaches.

144. Answer (4)

Urethra has pseudostratified non-ciliated epithelium

while all other parts have transitional epithelium.

145. Answer (2)

Haem is the prosthetic group and it is a part of the

active site of enzyme.

146. Answer (1)

Lens and cornea, both are easily transplantable due

to avascular nature.

147. Answer (3)

Male amphibians have copulatory pads on first digit

of forelimbs.

148. Answer (2)

GIFT – Gametic Intrafallopian Transfer.

149. Answer (4)

Vector mediated gene transfer is indirect method of

gene transfer.

150. Answer (3)

Dystrophy is a genetic disorder, characterized by

lack of dystrophin protein.

151. Answer (3)

HIV reduces number of helper T cells by destroying

them.

152. Answer (1)

Frequency of recessive allele (q) = 0.3

So, frequency of dominant allele (p) = 0.7

Hence frequency of heterozygotes (2 pq) = 0.42

153. Answer (3)

AB positive blood group is a universal recipient due

to absence of agglutinins/antibodies.

154. Answer (4)

Interferons are secreted by virally infected cells and

other cells of the body to protect the non-infected

cells.155. Answer (1)

Autonomic neurons in submucosa (meissner’s

plexus) and muscularis externa (auerbach’s plexus)

control the muscles of gut wall.

156. Answer (2)

Melatonin hormone from pineal gland sets up the

biological clock of the body.

157. Answer (4)

Ca2+ ions trigger the exocytosis of synaptic vesicles

from synaptic knots.

158. Answer (4)

Deglutition is a voluntary process in which uvula

prevents the food from rising up in the nasopharynx

and epiglottis prevents the food entry in glottis

159. Answer (3)

Annelids and arthropods are metamerically

segmented schizocoelomates.

160. Answer (3)

Macroglia are ectodermal in origin.

161. Answer (2)

Malpighian tubules are located at the junction of

midgut and hindgut.

Pharyngeal glands of cockroach have four secretory

parts and two receptacles.

162. Answer (4)

Hemidesmosomes connect the cells to their

basement membranes.

163. Answer (3)164. Answer (2)

Charge and size both play an important role in

separation of DNA fragments.

165. Answer (1)

Synarthrose/fibrous joints are completely immovable

joints.


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9/9

166. Answer (4)

167. Answer (2)

Ca2+ ions are involved in blood clotting.

168. Answer (2)

Syphilis is curable in primary stage and its chancres

are painless.

169. Answer (2)

Insertionally inactivated lac z’ gene results in failure

to utilize ‘X–gal’ by recombinant bacteria, which fail

to produce blue color, hence appearing white.

170. Answer (3)

171. Answer (3)

Dissociation of oxygen from (Hb) causes a right

shift in the curve.

172. Answer (1)

Secondary/anamnestic response is always vigorous

due to memory cells.

173. Answer (2)

Earliest autotrophs were chemoautotrophs and were

anoxygenic.

174. Answer (1)

Dehydration triggers on ‘RAAS’ and enhanced tubular

reabsorption of Na+ ions.

175. Answer (1)

Outcrossing is mating of animals of same breed but

not having any common ancestors for at least 4–6

generations, introducing variations but not causing

drastic changes.

176. Answer (3)

Corpora allata produce juvenile hormone.

177. Answer (3)

Ringworms are fungal infections.

178. Answer (2)

179. Answer (1)

180. Answer (1)

Retroviruses are used as ideal cloning vectors for

eukaryotic host cells.


10/19


11/19

Test - 9 (Code B) (Answers & Hints) All India Aakash Test Series for Medical-2016

1/9

1. (3)

2. (2)

3. (1)

4. (3)

5. (2)

6. (3)

7. (2)

8. (4)

9. (4)

10. (2)

11. (3)

12. (4)

13. (4)

14. (4)

15. (1)

16. (1)

17. (3)

18. (4)

19. (4)

20. (2)21. (3)

22. (4)

23. (3)

24. (2)

25. (3)

26. (4)

27. (4)

28. (1)

29. (1)

30. (2)31. (1)

32. (1)

33. (4)

34. (3)

35. (2)

36. (4)

37. (3)38. (1)

39. (3)

40. (2)

41. (1)

42. (1)

43. (1)

44. (1)

45. (4)

46. (1)

47. (4)

48. (4)

49. (2)

50. (3)

51. (1)

52. (4)

53. (2)

54. (1)

55. (1)

56. (4)

57. (2)

58. (2)

59. (4)

60. (2)

61. (2)

62. (1)

63. (4)

64. (4)

65. (4)

66. (3)67. (1)

68. (4)

69. (3)

70. (1)

71. (4)

72. (1)

ANSWERS

TEST - 9 (Code-B)

All India Aakash Test Series for Medical-2016

Test Date : 03-04-2016

73. (1)74. (2)

75. (2)

76. (3)

77. (1)

78. (4)

79. (2)

80. (4)

81. (4)

82. (4)

83. (1)

84. (2)

85. (1)

86. (3)

87. (2)

88. (1)

89. (3)

90. (4)

91. (4)

92. (1)

93. (1)

94. (4)

95. (2)

96. (3)

97. (3)

98. (4)

99. (1)

100. (1)

101. (2)

102. (4)103. (2)

104. (4)

105. (2)

106. (4)

107. (2)

108. (2)

109. (1)110. (4)

111. (2)

112. (3)

113. (4)

114. (3)

115. (4)

116. (2)

117. (1)

118. (1)

119. (4)

120. (4)

121. (1)

122. (1)

123. (1)

124. (4)

125. (1)

126. (3)

127. (3)

128. (4)

129. (1)

130. (4)

131. (1)

132. (1)

133. (2)

134. (3)

135. (1)

136. (3)

137. (3)

138. (4)139. (1)

140. (1)

141. (3)

142. (1)

143. (4)

144. (3)

145. (1)146. (1)

147. (2)

148. (4)

149. (4)

150. (2)

151. (3)

152. (4)

153. (1)

154. (2)

155. (2)

156. (1)

157. (1)

158. (2)

159. (2)

160. (4)

161. (3)

162. (2)

163. (3)

164. (3)

165. (1)

166. (1)

167. (2)

168. (4)

169. (1)

170. (1)

171. (4)

172. (2)

173. (4)

174. (4)175. (2)

176. (1)

177. (3)

178. (1)

179. (1)

180. (3)

click here for code-A solutions


12/19

All India Aakash Test Series for Medical-2016 Test - 9 (Code B) (Answers & Hints)

2/9

Hints to Selected Questions

[ PHYSICS]

1. Answer (3)

E = B C

2. Answer (2)f o + f

e= 60

f o + f

e+10 = V

o + f

e

V o

= 65, f o = 55

1 1

o oV u

=1

of

3. Answer (1)

– ray is E.M. radiation

4. Answer (3)

5. Answer (2)

6. Answer (3)

7. Answer (2)

S 1 + S

2 = 0

( –1) A1 + ( –1) A

2 = 0

8. Answer (4)

9. Answer (4)

No. of fission =3

3

10

200 1.6 10

10. Answer (2)

– eV = 2 21

2m v u

11. Answer (3)

12. Answer (4)

0 1.5 1

I f f

u f

1

2 .5

I

4I cm

13. Answer (4)

2 2 2

rms

200 100 50

2 2 2V

14. Answer (4)

15. Answer (1)

16. Answer (1)

E

S

X

W

Y Y

N

M = NiA

M = m x – my

17. Answer (3)

18. Answer (4)

3

h

kTm

19. Answer (4)

20. Answer (2)

0

0

640 693 2.3log

10

N

t N

21. Answer (3)

2 6 61

5 12 10 120 12 102

= 2 61

5 12 102

Heat

22. Answer (4)

23. Answer (3)

5 – 10i = 2 + i x = 4

24. Answer (2)

ˆi

P mv i

ˆ

3i

v P m j mv

i f P P


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25. Answer (3)

26. Answer (4)

27. Answer (4)

' w o

w s

V V V f f

V V V

28. Answer (1)

A1V 1 = A

2V 2

29. Answer (1)

30. Answer (2)

mV 0 = 2m v

21 (2 )2

m v = 21

2K x

31. Answer (1)

C = C v –

1

R

u , PV n = constant

P =K

T constant

PV

T

P 2V = K

1

2PV = K

32. Answer (1)

33. Answer (4)

34. Answer (3)

35. Answer (2)

36. Answer (4)

V 12

is uniform along – y axis.

12V

= 1 1

ˆ ˆcos 37 sin 37u i u j –

2 2ˆ ˆcos 53 sin 53u i u j

= 1 2

ˆcos 37 cos 53u u i

1 2

ˆsin 37 sin 53u u j

1 2cos 37 cos53 0u u

1 2

4 3

5 5u u

1 24 3u u

37. Answer (3)

qE

qEcos

mg q E

c o s s i n

N N

mg s i n

N = mg cos + q E sin

mg sin = N + q E cos

38. Answer (1)

AN

B

mg sin

V

m g c o

s

Between A and B

mg (R – R cos) + 0 = 21

2mv

mg cos – N =2

mv

R , N = 0

39. Answer (3)

A

T 1

T 2

C

B

90°–

mg

30°

1

sin120

T

= sin(60 )

Mg

(T 1)min

where sin (60° + ) = 1

= 30°

40. Answer (2)

P will be the maximum height.


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41. Answer (1)

A E

W

V pw

V pg

V wg

N

S

V wg

pg V

sin = wg V

40 sin = 20

= 30°

42. Answer (1)

v x

= 1 m/s v y = 2 m/s, tan =

y

x

v

v

y = tan –2

2

1

2x

x g

v

43. Answer (1)

44. Answer (1)

R =2

PQr

R% = l % + 2r %

=1 0.02

2 100100 1

45. Answer (4)

46. Answer (1)

OH

CHCl2

OO

OH

+ NaOH + CHCl3

O–H

C = O

O

NaOH

H

In this reaction electrophile is CCl2

47. Answer (4)

MgCO3 and Al(OH)

3 are antacid.

48. Answer (4)

CH = C – CH = CH and CH = C – CH = CH2 2 2 2

CH3

Isoprene

Cl

Chloroprene

49. Answer (2)

Br 2 (aq) forms saccharic acid from glucose but not

from fructose.

50. Answer (3)

CH CH3 2

CH CH COOH CH CH CH CH NH2 2 3 2 2 2 2

HN3

CH CH CH3 2 2

CH NH CH – CH CH CH OH2 2 3 2 2 2

HNO2

Conc.

H SO

443 K

2 4

HCl

[ CHEMISTRY]

51. Answer (1)

A is CH – C – CH3 3

NOH

Oxime

CH – CH – CH

3 3

NO

2- Nitrosopropane

52. Answer (4)

Fact

53. Answer (2)

C6H5Cl has partial double bond character.

54. Answer (1)

Ionization and linkage both isomerism are possible.

55. Answer (1)

Fe3+ has d 5 system

56. Answer (4)

All are correct

57. Answer (2)

Fact

58. Answer (2)

Ionic PCl5 is [ PCl

4 ]+ [ PCl

6]–

59. Answer (4)

2 2 2 Ag S 4NaCN 2Na[Ag(CN) ] Na S .

60. Answer (2)

Argyrol is Ag sol (–ve)


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61. Answer (2)

For zero order reaction

t1 + t

2 + t

3 = 10 + 5 + 2.5 = 17.5 s.

62. Answer (1)

2Cl

red red 2

P0.059E E logn Cl

.

63. Answer (4)

P = 100 m

A

P

P

= B

100 1X

700 7

bT =

bK m

= BbB A

X 1000K

1 X M

=

1

1000710

1 1671

1

7

= 10 K

64. Answer (4)

C

B

A

D

AB =a

2, AD = 2a ,

3AC a

2

65. Answer (4)

SO2 is reducing agent

66. Answer (3)

CH – C = CH CH – C = O + HCHO3 2 3

CH3

O3

Zn/H O2

CH3

67. Answer (1)

CH – CH – CH – CH3 3

CH3

Cl

, the total reactivity.

is 2 3.8 = 7.6 for 2° - carbon atom

68. Answer (4)

FeCl3 + NaSCN 3

Blood red

Fe(SCN) +NaCl

69. Answer (3)

is aromatic

70. Answer (1)

Fact

71. Answer (4)

Acidic character – I

72. Answer (1)

It is dissymmetric

73. Answer (1)

O

O

will not show tautomerism

74. Answer (2)

Correct according to IUPAC rules

75. Answer (2)

76. Answer (3)

Conc. H SO2 4

COOH

COOH CO + CO + H O2 2

77. Answer (1)

78. Answer (4)

Fact

79. Answer (2)

MnO +5I I + Mn– – 2+

24

+7

–5

2


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80. Answer (4)

Kp and total pressure remain constant

81. Answer (4)

A + B B

G =

°pG G R

= 500 [ 100 ( 200)]

= – 500 + 300

= – 200

G = – 2.303 RT Log Ke

– 200000 = – 2.303 8.31 T log 103

T = 3483.4 K

82. Answer (4)

1 mole NaOH + 1 mole HF give 68 kJ/mole

83. Answer (1)

Vc = 3b84. Answer (2)

F–Be–F

85. Answer (1)

O sp

2

86. Answer (3)

Cr belongs to 6th group

87. Answer (2)

88. Answer (1)

2

2 2

1 2

1 1 1R ( )

n nZ

2

2 2

1 1 1R (2)

(2) (4)

Divide and multiply by (4)2

2

2 2

1 1 1R (4)

(4) (8)

89. Answer (3)

Basicity of H3PO

2 is one

90. Answer (4)

Number of gram equivalents of oxygen = number of

gram equivalents of aluminium

⇒ w

8 =

1.35

9

⇒ w = 1.2 g

91. Answer (4)

Haplo-diplontic.92. Answer (1)

Albugo candida is parasite on mustard.

93. Answer (1)

ETC and oxidative phosphorylation.

94. Answer (4)

PEG and NaNO3

95. Answer (2)

Fungi, bacteria and cyanobacteria in organic farming

96. Answer (3)

97. Answer (3)

98. Answer (4)

99. Answer (1)

Pericarp, placenta and thalamus are edible in guava.

[ BIOLOGY

]

100. Answer (1)

Night and early morning when absorption is high.

101. Answer (2)

Functionally similar.

102. Answer (4)

Consciousness.

103. Answer (2)

Axillary buds.

104. Answer (4)

Hydrophyte.

105. Answer (2)

Ulothrix , Spirogyra – cellulosic wall.

106. Answer (4)

107. Answer (2)

Viroid is naked RNA.


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108. Answer (2)

(a) Gases (b) Bacteria, fungi (c) Public transport

109. Answer (1)

Diatoms.

110. Answer (4)

Nucleosomes in eukaryotes.

111. Answer (2)

Dorsiventral thalloid gametophyte in liverworts.

112. Answer (3)

CMP-11, 2015, Paris.

113. Answer (4)

Fucus – Brown algae

Polysiphonia – Red algae

114. Answer (3)

Ethylene

115. Answer (4)

Secondary phloem is developed on outer side of

vascular cambium.

116. Answer (2)

LDP.

117. Answer (1)

ER, golgi body, lysosomes and vacuoles in

endomembrane system.

118. Answer (1)Pyruvate act as oxidant in lactic acid fermentation.

119. Answer (4)

Autosomal dominant trait

120. Answer (4)

80 dB

121. Answer (1)

122. Answer (1)

Parietal placentation

123. Answer (1)

Largest tiger reserve.

124. Answer (4)

(b) Metaphase I

(d) Anaphase I

(a) Anaphase II

125. Answer (1)

Hypogynous and perigynous flowers.

126. Answer (3)

23 S rRNA.

127. Answer (3)

Dry to mesic conditions.

128. Answer (4)

Diapause is a stage of suspended development.

129. Answer (1)

Oxidative phosphorylation in mitochondria.

130. Answer (4)

TTRR

TR

TtRR

(Tall-Red)

XttRr

tR tr

TtRr

(Tall-Pink)

131. Answer (1)

Insectivorous plants

132. Answer (1)

Collenchyma

133. Answer (2)

Zea mays – C4 plant

134. Answer (3)

n = 20

Caruncle (2n) = 40

tube nucleus (n) = 20

PEC (3n) = 60

Nucellus (2n) = 40

135. Answer (1)

X X aaC C

XX aaC

X YaaC

X XYaa

Xa Ya

X aC

136. Answer (3)

Retroviruses are used as ideal cloning vectors for

eukaryotic host cells.


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137. Answer (3)

138. Answer (4)

139. Answer (1)

Ringworms are fungal infections.

140. Answer (1)

Corpora allata produce juvenile hormone.

141. Answer (3)

Outcrossing is mating of animals of same breed but

not having any common ancestors for at least 4–6

generations, introducing variations but not causing

drastic changes.

142. Answer (1)

Dehydration triggers on ‘RAAS’ and enhanced tubular

reabsorption of Na+ ions.

143. Answer (4)

Earliest autotrophs were chemoautotrophs and were

anoxygenic.

144. Answer (3)

Secondary/anamnestic response is always vigorous

due to memory cells.

145. Answer (1)

Dissociation of oxygen from (Hb) causes a right

shift in the curve.

146. Answer (1)

147. Answer (2)

Insertionally inactivated lac z’ gene results in failure

to utilize ‘X–gal’ by recombinant bacteria, which fail

to produce blue color, hence appearing white.

148. Answer (4)

Syphilis is curable in primary stage and its chancres

are painless.

149. Answer (4)

Ca2+ ions are involved in blood clotting.

150. Answer (2)

151. Answer (3)

Synarthrose/fibrous joints are completely immovable

joints.

152. Answer (4)

Charge and size both play an important role in

separation of DNA fragments.

153. Answer (1)

154. Answer (2)

Hemidesmosomes connect the cells to their

basement membranes.

155. Answer (2)

Malpighian tubules are located at the junction of

midgut and hindgut.

Pharyngeal glands of cockroach have four secretory

parts and two receptacles.

156. Answer (1)

Macroglia are ectodermal in origin.

157. Answer (1)

Annelids and arthropods are metamerically

segmented schizocoelomates.

158. Answer (2)Deglutition is a voluntary process in which uvula

prevents the food from rising up in the nasopharynx

and epiglottis prevents the food entry in glottis

159. Answer (2)

Ca2+ ions trigger the exocytosis of synaptic vesicles

from synaptic knots.

160. Answer (4)

Melatonin hormone from pineal gland sets up the

biological clock of the body.

161. Answer (3)

Autonomic neurons in submucosa (meissner’s

plexus) and muscularis externa (auerbach’s plexus)

control the muscles of gut wall.

162. Answer (2)

Interferons are secreted by virally infected cells and

other cells of the body to protect the non-infected

cells.

163. Answer (3)

AB positive blood group is a universal recipient due

to absence of agglutinins/antibodies.

164. Answer (3)

Frequency of recessive allele (q) = 0.3

So, frequency of dominant allele (p) = 0.7

Hence frequency of heterozygotes (2 pq) = 0.42


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165. Answer (1)

HIV reduces number of helper T cells by destroying

them.

166. Answer (1)

Dystrophy is a genetic disorder, characterized by

lack of dystrophin protein.

167. Answer (2)

Vector mediated gene transfer is indirect method of

gene transfer.

168. Answer (4)

GIFT – Gametic Intrafallopian Transfer.

169. Answer (1)

Male amphibians have copulatory pads on first digit

of forelimbs.

170. Answer (1)

Lens and cornea, both are easily transplantable due

to avascular nature.

171. Answer (4)

Haem is the prosthetic group and it is a part of the

active site of enzyme.

172. Answer (2)

Urethra has pseudostratified non-ciliated epithelium

while all other parts have transitional epithelium.

173. Answer (4)

Elytra are mesothoracic wings. Two pairs of thoracic

and eight pairs of abdominal spiracles are present in

cockroach. Unjointed anal styles are found only in

male cockroaches.

174. Answer (4)

Mitochondrial DNA from Africans populations has

maximum variations through mutations.

175. Answer (2)

Annelids generally have closed circulatory system.

176. Answer (1)

The first haploid cells in testes are secondary

spermatocytes while in ovaries are secondary

oocyte and first polar body.

177. Answer (3)

Regression of corpus luteum is triggered by reduced

level of LH of secretory phase.178. Answer (1)

Thymosine is a hormone from thymus gland.

179. Answer (1)

Proton pump inhibition results in reduced HCl

release.

180. Answer (3)

The first digit in E.C. indicates the class of the

enzyme and hence the mode of its activity.

Solutions_AIATS Medical-2016 Test-09_(Code-A & B)

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