Aiats Medical Two Yr Test-2 (1)

Test - 2 (Code-C) (Answers & Hints) All India Aakash Test Series for Medical-2017

1/10

1. (1)

2. (2)

3. (3)

4. (4)

5. (2)

6. (3)

7. (4)

8. (1)

9. (3)

10. (2)

11. (4)

12. (1)

13. (4)

14. (4)

15. (3)

16. (4)

17. (1)

18. (2)

19. (2)

20. (1)

21. (1)

22. (3)

23. (2)

24. (1)

25. (3)

26. (2)

27. (2)

28. (2)

29. (3)

30. (4)

31. (2)

32. (3)

33. (2)

34. (3)

35. (1)

36. (1)

37. (2)

38. (1)

39. (2)

40. (3)

41. (4)

42. (4)

43. (2)

44. (4)

45. (4)

46. (2)

47. (1)

48. (1)

49. (2)

50. (4)

51. (3)

52. (4)

53. (1)

54. (2)

55. (3)

56. (2)

57. (1)

58. (2)

59. (2)

60. (4)

61. (2)

62. (2)

63. (1)

64. (2)

65. (4)

66. (1)

67. (1)

68. (4)

69. (1)

70. (1)

71. (4)

72. (1)

73. (1)

74. (1)

75. (3)

76. (1)

77. (2)

78. (1)

79. (4)

80. (2)

81. (1)

82. (4)

83. (1)

84. (1)

85. (3)

86. (1)

87. (3)

88. (1)

89. (1)

90. (4)

91. (4)

92. (3)

93. (2)

94. (4)

95. (3)

96. (4)

97. (1)

98. (3)

99. (2)

100. (3)

101. (3)

102. (4)

103. (3)

104. (4)

105. (3)

106. (4)

107. (4)

108. (4)

109. (3)

110. (4)

111. (1)

112. (1)

113. (3)

114. (1)

115. (4)

116. (2)

117. (2)

118. (3)

119. (2)

120. (4)

121. (3)

122. (1)

123. (2)

124. (3)

125. (2)

126. (4)

127. (3)

128. (4)

129. (2)

130. (3)

131. (2)

132. (2)

133. (4)

134. (1)

135. (4)

136. (1)

137. (1)

138. (4)

139. (3)

140. (2)

141. (1)

142. (3)

143. (2)

144. (1)

145. (3)

146 (4)

147. (4)

148. (3)

149. (4)

150. (2)

151. (2)

152. (4)

153. (3)

154. (1)

155. (1)

156. (4)

157. (4)

158. (3)

159. (3)

160. (2)

161. (1)

162. (3)

163. (3)

164. (4)

165. (2)

166. (4)

167. (4)

168. (1)

169. (3)

170. (3)

171. (2)

172. (2)

173. (4)

174. (1)

175. (4)

176. (1)

177. (4)

178. (2)

179. (2)

180. (4)

ANSWERS

TEST - 2 (Code-C)

All India Aakash Test Series for Medical-2017

Test Date : 15-11-2015

All India Aakash Test Series for Medical-2017 Test - 2 (Code-C) (Answers & Hints)

2/10

Hints to Selected Questions

[ PHYSICS]

1. Answer (1)

ˆ ˆ(6 )A B i j � �

ˆ ˆ( 4 7 )A B i j � �

than ˆ ˆ2 2 8A i j �

4 , | | 4.1A i j A � �

2. Answer (2)

P = 5

Q R

P + Q = 18, Q = 13

R2 + P2 = Q2

R2 = Q2 – P2

= (13)2 – (5)2

R = 12

3. Answer (3)

4. Answer (4)

v1

1v

1v

1v

2v

1 2| | | |v v v � �

2 1v v v � � �

2 2| | – 2 cos �

v v v v v

2| | 2 sin �

v v

5. Answer (2)

P

Q

R

A

B

C

From PQR

tanRQ

PQ

= 6 3

378 4

6. Answer (3)

120º

4N

x-axis4N

4N

y-axis

Resultant force

= 2 2(4) (4) 244cos120 = 4 N towards (–ve) x-axis

7. Answer (4)

x

1.96 km

600 km/h

tanx

h (i)

211960 9.8

2t (ii)

5600

18x t (iii)

From (i), (ii) & (iii)

= 60°


3/10

8. Answer (1)

9. Answer (3)

2 2sin

2

uH

g

2

2 2

1 1

2 2 22 2

1

sin sin 30 2

sin sin 60 3

2

H

H

⎛ ⎞⎜ ⎟ ⎝ ⎠

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

H1 : H

2 = 1 : 3

10. Answer (2)

(90 – )u v

P

ux = u cos

vx = v sin

ux

= vx

ucos = v sin

v = ucot

11. Answer (4)

12. Answer (1)

13. Answer (4)

14. Answer (4)

= 45º for maximum range

2 22 2 2

0 0( )sin

2 4 4

u vu u

Hg g g

15. Answer (3)

2

T

16. Answer (4)

ˆ ˆcos sinr a ti a tj �

ˆ ˆsin cosdr

v a ti a tjdt

�

�

2| | , | |v a a a ��

17. Answer (1)

R = 4H cot

3H = 4H cot

3cot

4

4 4 3tan , sin , cos

3 5 5

22sin cosu

Rg

224

25

uR

g

18. Answer (2)

| |av

va

t

�

0av

a

| | 0v �

for complete cycle

19. Answer (2)

2sin2 30

3u

g

(i)

2sin2 45u

Rg

(ii)

From (i) & (ii)

R = 3.5 m

20. Answer (1)

h h

t = 1 t = 3

u sin H

max

2 sin1 3 4

uT

g

usin = 2g

2

max

( sin )2

2

uH g

g

= 20 m


4/10

21. Answer (1)

30°

E

vN

W

S

Av 5i j

ˆ5A

v i�

ˆ ˆsin30 cos30B

v v i v j �

ˆ

BAv vj�

BA B Av v v � � �

then | | 10B

v �

km/h

22. Answer (3)

vw

s

x

vs

1

2d

1 55

6 6x

1

2d ,

25 1 34

36 4 6s km

23. Answer (2)

Initial speed = 10 m/s

Speed at heighest point = 10 cos 60 = 5 m/s

Change in speed = 10 – 5 = 5 m/s

24. Answer (1)

R = 4 H cot2

4cot

8

gTR

Comparing with

R = 5T2

5 cot = 5

= 45º

25. Answer (3)

H

R/2

tan

2

H

R

R = 4Hcot

2tan

4 cot

H

H

1tan tan

2

26. Answer (2)

H

R/2

60º

S

2

2

4

RS H

av

Sv

T

27. Answer (2)

200

50

st

v 4 s

2ut

g , u = 19.6 m/s

28. Answer (2)

| |sin30

| |

mg

Rg

v

v

�

�

30ºRgv

Rmv

mgv2Rg

v v�

29. Answer (3)

uA = 2 m/s 8 um/s =

B

Av S

v

At any time ‘t’

ˆ ˆ2A

v i gtj �

ˆ ˆ8B

v i gtj �

A Bv v� �

then

. 0

A Bv v � �

2 × 8 = g2t2

4t

g

4( ). 10

10A B

x u u t

x = 4 m


5/10

30. Answer (4)

1 2

1 2

tan 4m

3

X XH

X X

31. Answer (2)

60º

30º

v

v cos30

g cos30u sin60

u cos60

v sin30

g

u cos 60 = v cos 30

2

2 3

uv

3

uv

3cos30

2c

a g g

2 22

3 3c

v ur

a g

22

3 3

ur

g

32. Answer (3)

N

EW

S

45º

vA = 20 km/h

vAB

x

200 km

vB = 20 km/h

2 2

AB A Bv v v

= 2 2(20) (20)

= 20 2

= sin45º200

x

1

200 sin45º 2002

x

= 100 2 km

33. Answer (2)

m

xy

a

x2 = y2 + a2

2 2dx dy

x ydt dt

xu = yv

u = vcos34. Answer (3)

B Ar r r � � �


BAtv r

��

35. Answer (1)

1

2 sin uT

g (i)

2

2 cos uT

g (ii)

12 2 2

1 2

1

2 g t t u

36. Answer (1)

ˆ ˆ( )v k yi xj �

,x

dxv y y

dt

x = yt + c

vy = x, y = xt + c

then y2 = x2 + c

37. Answer (2)

2

c

va

R = v

38. Answer (1)

x

y

R

ˆ ˆcos sinR R i R j �

ˆ ˆsin cosv v i v j �


6/10

39. Answer (2)

ˆ3u i�

m/s

ˆˆ

2

ja i

⎛ ⎞ ⎜ ⎟⎜ ⎟⎝ ⎠

�

m/s2

vx = 0

0 = 3 – t

t = 3, 3

2

yV V m/s2

40. Answer (3)

2 2

60 60 1800T

rad/s

41. Answer (4)

ˆ10u i�

ˆ10a j�

.v u a t ��

ˆ ˆ10 10v i j �

| | 10 2v �

m/s

42. Answer (4)

43. Answer (2)

44. Answer (4)

/2

P

Q

| | | |P Q��

45. Answer (4)

[ CHEMISTRY]

46. Answer (2)

1

6 1

nE 1.312 10 J mol

6 25

n 2

1.312 10 (1)E 3.28 10

(2)

2

J mol–1

2 1

5

n nE E E 9.84 10 J mol–1

47. Answer (1)

m = 2 (highest)

So l = 2 (highest)

and l = 0 to n – 1

So n = 3

Angular momentum of orbit =nh

2

=3h

2

=1.5h

48. Answer (1)

In sodium hydride, anion is H–

So, EC of H– is 1s2

49. Answer (2)

Fact based.

50. Answer (4)

Isotones are the atoms having same number of

neutrons.

51. Answer (3)

Outer EC : 2s2 2p5 represents halogen, with highest

electronegativity in respective period.

52. Answer (4)

Ne shows +ve eg

H due to stable EC.

O– + e– O2– is endothermic process.

53. Answer (1)

n1 + n

2 = 4, So n

1 = 1

n2 – n

1 = 2 n

2 = 3

2

2 2

1 1 32RR(2)

91 3

⎡ ⎤ ⎢ ⎥⎣ ⎦

54. Answer (2)

It is d-block element, as last e– goes to d-subshell.

group No. is number of e– in (n – 1)d and n s

subshells.

55. Answer (3)

3

3hmvr

2

3

3h2 r 3

mv

56. Answer (2)

As option (2) EC follows Hund’s rule.


7/10

57. Answer (1)

Orbital 2 2

x y

3d

lies along X and Y axis

So shows maximum electron probable density.

58. Answer (2)

I.EI is energy required to remove an electron from

the valence shell of isolated and gaseous atom of

Ba.

59. Answer (2)

60. Answer (4)

61. Answer (2)

pZ has zero electron density in XY plane.

62. Answer (2)

2

n

0.529nr Å

Z

2

2

0.529 2r (H) 0.529 4Å

1

23

4

0.529 4r (Be ) 0.529 4Å

4

63. Answer (1)

Suppose velocity = x

4x

h

mv so

h4x

mx

2 hx

4m so

1 hx

2 m

1 hv

2 m

h4v 2

m

64. Answer (2)

4th IE is very high

So there are 3 valence e– and possible oxidation

state of metal is +3.

and oxide will be M2O

3.

65. Answer (4)

Al is forming six bonds with 1Cl and 5H2O

molecules, hence covalency is six.

66. Answer (1)

eg

H of gp16

S > Se > Te > Po > O

67. Answer (1)

E.C. (Z = 117)

[Rn] 5f14 6d10 7s2 7p5

68. Answer (4)

Fact.

69. Answer (1)

CO and NO are neutral oxides

70. Answer (1)

71. Answer (4)

Li & Mg

Be & Al

B & Si

Show similar properties due to diagonal relationship.

72. Answer (1)

Name of element with Z = 102 is Unnilbiium so

symbol is Unb.

73. Answer (1)

Ne 1s2 2s2 2p6 4 s electrons

O 1s2 2s2 2p4 4 p electrons

74. Answer (1)

Change in PE = 2E

2E4

= 3E

2

75. Answer (3)

h

2m KE

76. Answer (1)

77. Answer (2)

Na 1s2 2s2 2p6 3s1

7 e– possess m = 0

78. Answer (1)

Number of peaks in graph of 4r22 and r is n – l

79. Answer (4)

Number spectral line will be formed for 2px – 2p

z

transition.

80. Answer (2)

h

p


8/10

81. Answer (1)

When cation is formed

p

e ratio increases, so size decreases.

82. Answer (4)

He, Ne, Ar, Kr, Xe, Rn have all paired e– s.

83. Answer (1)

More electronegativity more acidic character and vice

versa.

84. Answer (1)

85. Answer (3)

H = IE of Li + eg

H of Cl

= + 5.4 – 3.61 = 1.79 eV atom–1

[ BIOLOGY

]

86. Answer (1)

87. Answer (3)

Particle nature of e– was demonstrated by J.J.

Thomson.

88. Answer (1)

There is 7 periods and 18 groups in long form of

periodic table.

89. Answer (1)

Mg is 3s2

Al is 3s2 3p1

So IEI Mg > Al

90. Answer (4)

Iodine has most electropositive character.

91. Answer (4)

In Trichoderma (Deuteromycetes) conidia are asexual

structure.

92. Answer (3)

Claviceps - Chitinous wall

93. Answer (2)

Agaricus is saprophyte

94. Answer (4)

Nitrogenase is absent in fungi.

95. Answer (3)

Rhinovirus – ssRNA

96. Answer (4)

Protozoans/primitive relatives of animals.

97. Answer (1)

98. Answer (3)

Plague, Diphtheria – bacterial diseases.

99. Answer (2)

T2 phage – dsDNA virus

100. Answer (3)

Yersinia is bacterium, TNV = Tobacco Necrosis Virus.

101. Answer (3)

Aseptate

102. Answer (4)

Spawn in Agaricus life cycle.

103. Answer (3)

Penicillium, Claviceps – Ascocarp.

Agaricus – Basidiocarp

104. Answer (4)

105. Answer (3)

Basidiospores

106. Answer (4)

In basidiomycetes, asexual spores are generally not

found.

107. Answer (4)

In basidiomycetes, mycelium is branched and septate.

108. Answer (4)

Viroids are smaller than bacteria

109. Answer (3)

1,3-glucan (Paramylon starch).

110. Answer (4)

111. Answer (1)

112. Answer (1)

Citrus canker is caused by Xanthomonas.

113. Answer (3)

Basidiomycetes

114. Answer (1)

115. Answer (4)

116. Answer (2)

It is smaller than bacteria.

117. Answer (2)

Plasmodium is a sporozoan.

118. Answer (3)

Late blight of potato.

119. Answer (2)


9/10

120. Answer (4)

121. Answer (3)

122. Answer (1)

Aplanospores

123. Answer (2)

124. Answer (3)

Facultative heterotrophs

125. Answer (2)

Citrus exocortis by viroids

126. Answer (4)

127. Answer (3)

Satellite RNA, discovered by Randle.

128. Answer (4)

Mucor – no dikaryophase

129. Answer (2)

They are endomycorrhiza.

130. Answer (3)

Float passively.

131. Answer (2)

Aspergillus, Neurospora, Penicillium

132. Answer (2)

133. Answer (4)

The mycobionts include members of ascomycetes or

basidiomycetes.

134. Answer (1)

135. Answer (4)

Parasites/predators

136. Answer (1)

137. Answer (1)

Fibroblasts secrete fibres and matrix.

138. Answer (4)

In rest all options, muscle fibres differ from nerve

fibres.

139. Answer (3)

Schwann cells and neurolemmocytes are name of

same cells meant for synthesis of myelin sheath in

PNS; Astrocytes act as blood brain barrier.

140. Answer (2)

Cancellous/spongy bone do not have Haversian

system

141. Answer (1)

Sudoriferous/sweat glands modify to make mammary

gland.

142. Answer (3)

PMNL – Neutrophils

143. Answer (2)

144. Answer (1)

Epithelial tissue protects internal tissue/organ and it

has fast repairing ability.

145. Answer (3)

Rest all options are of single unit smooth muscle.

146. Answer (4)

Oxalate withdraw Ca2+ ions from blood.

147. Answer (4)

Submandibular glands are compound alveolar glands.

148. Answer (3)

Smooth muscle gets scanty blood supply

149. Answer (4)

Osteon is other name of Haversian system whereas

ossein is protein in bones.

150. Answer (2)

151. Answer (2)

Skeletal muscle fibres are multinucleated.

152. Answer (4)

Aptenodytes is penguin whereas flying fox is

Pteropus.

153. Answer (3)

The main function of amnion is to prevent

dessication/drying up of embryo.

154. Answer (1)

155. Answer (1)

156. Answer (4)

By this age red bone marrow exists in the terminal

parts of long bones.

157. Answer (4)

Articular cartilage is modified hyaline cartilage but is

without blood vessel.

158. Answer (3)

In holocrine, sloughed off cell is replaced by a new

cell.


10/10

� � �

159. Answer (3)

Heart is absent in cyclostomates.

160. Answer (2)

Balanoglossus – Hemichordate;

Ascidia, Herdmania – Urochordate – Tail absent in

adult; Nerve cord turns into dorsal ganglia;

Cephalochordates are blue print of chordates.

161. Answer (1)

Pneumatic bones are character of Aves only.

162. Answer (3)

Whales are largest mammals.

163. Answer (3)

164. Answer (4)

Homeothermic – Birds, Mammals

Diaphragm – Mammals and Crocodiles

Seven cervical vertebrae –Mammals

165. Answer (2)

All these characters are found in most of reptiles

except second option.

166. Answer (4)

Lateral line sense organ check fluctuation in water

current.

167. Answer (4)

Aptenodytes is penguin, native of Antarctica.

168. Answer (1)

Trygon is sting ray.

169. Answer (3)

170. Answer (3)

In chondrichthyes, placoid scales modify to make

teeth.

171. Answer (2)

Adult stage of urochordate is tailess, without

notochord and nerve cord change to dorsal ganglia.

172. Answer (2)

Only amphibians and reptiles are dicondylic.

173. Answer (4)

Birds possess only right systemic arch.

174. Answer (1)

Cartilaginous fishes remove waste as urea.

175. Answer (4)

Bony fish and amphibians have bony vertebral

column.

176. Answer (1)

Cartilagenous fish do not have swim bladder.

177. Answer (4)

Chelone is reptile so poikilotherm.

178. Answer (2)

Second option is of prototherians.

179. Answer (2)

Pinna absent in prototherians; many land mammals

and all aquatic mammals lack body hair.

180. Answer (4)

They have paired appendages which may be fins or

limbs.

Test - 2 (Code-D) (Answers & Hints) All India Aakash Test Series for Medical-2017

1/10

1. (2)

2. (2)

3. (4)

4. (4)

5. (2)

6. (1)

7. (4)

8. (3)

9. (4)

10. (3)

11. (3)

12. (1)

13. (4)

14. (1)

15. (4)

16. (2)

17. (1)

18. (4)

19. (4)

20. (4)

21. (1)

22. (3)

23. (4)

24. (1)

25. (3)

26. (3)

27. (4)

28. (2)

29. (3)

30. (4)

31. (1)

32. (2)

33. (4)

34. (1)

35. (4)

36. (4)

37. (1)

38. (3)

39. (2)

40. (1)

41. (4)

42. (2)

43. (3)

44. (4)

45. (3)

46. (2)

47. (3)

48. (3)

49. (1)

50. (3)

51. (1)

52. (3)

53. (3)

54. (2)

55. (3)

56. (4)

57. (2)

58. (3)

59. (4)

60. (3)

61. (1)

62. (3)

63. (1)

64. (3)

65. (4)

66. (3)

67. (1)

68. (2)

69. (3)

70. (3)

71. (2)

72. (4)

73. (3)

74. (4)

75. (2)

76. (4)

77. (4)

78. (4)

79. (3)

80. (4)

81. (1)

82. (4)

83. (3)

84. (4)

85. (1)

86. (2)

87. (4)

88. (3)

89. (3)

90. (4)

91. (2)

92. (3)

93. (2)

94. (4)

95. (4)

96. (1)

97. (2)

98. (2)

99. (1)

100. (2)

101. (4)

102. (1)

103. (4)

104. (3)

105. (1)

106. (2)

107. (4)

108. (3)

109. (4)

110. (4)

111. (4)

112. (3)

113. (1)

114. (3)

115. (3)

116. (2)

117. (1)

118. (2)

119. (4)

120. (2)

121. (1)

122. (2)

123. (1)

124. (2)

125. (1)

126. (1)

127. (4)

128. (1)

129. (3)

130. (2)

131. (1)

132. (2)

133. (4)

134. (1)

135. (2)

136 (2)

137 (4)

138 (4)

139 (2)

140 (3)

141 (2)

142 (3)

143 (2)

144 (4)

145 (4)

146 (1)

147 (1)

148 (3)

149 (2)

150 (2)

151 (4)

152 (2)

153 (1)

154 (1)

155 (1)

156 (4)

157 (1)

158 (1)

159 (2)

160 (2)

161 (3)

162 (3)

163 (1)

164 (2)

165 (4)

166 (4)

167 (2)

168 (1)

169 (2)

170 (2)

171 (1)

172 (3)

173 (4)

174 (1)

175 (3)

176 (2)

177 (1)

178 (2)

179 (3)

180 (3)

ANSWERS

TEST - 2 (Code-D)

All India Aakash Test Series for Medical-2017

Test Date : 15-11-2015

All India Aakash Test Series for Medical-2017 Test - 2 (Code-D) (Answers & Hints)

2/10

Hints to Selected Questions

[ PHYSICS]

1. Answer (2)

2. Answer (2)

/2

P

Q

| | | |P Q��

3. Answer (4)

4. Answer (4)

5. Answer (2)

ˆ10u i�

ˆ10a j�

.v u a t ��

ˆ ˆ10 10v i j �

| | 10 2v �

m/s

6. Answer (1)

2 2

60 60 1800T

rad/s

7. Answer (4)

ˆ3u i�

m/s

ˆˆ

2

ja i

⎛ ⎞ ⎜ ⎟⎜ ⎟⎝ ⎠

�

m/s2

vx = 0

0 = 3 – t

t = 3, 3

2

yV V m/s2

8. Answer (3)

x

y

R

ˆ ˆcos sinR R i R j �

ˆ ˆsin cosv v i v j �

9. Answer (4)2

c

va

R = v

10. Answer (3)ˆ ˆ( )v k yi xj

�

,x

dxv y y

dt

x = yt + cvy = x, y = xt + c

then y2 = x2 + c11. Answer (3)

1

2 sin uT

g (i)

2

2 cos uT

g (ii)

12 2 2

1 2

1

2 g t t u

12. Answer (1)

B Ar r r � � �


BAtv r

��

13. Answer (4)

m

xy

a

x2 = y2 + a2

2 2dx dy

x ydt dt

xu = yv

u = vcos14. Answer (1)

N

EW

S

45º

vA = 20 km/h

vAB

x

200 km

vB = 20 km/h


3/10

2 2

AB A Bv v v

= 2 2(20) (20)

= 20 2

= sin45º200

x

1

200 sin45º 200

2

x

= 100 2 km

15. Answer (4)

60º

30º

v

v cos30

g cos30u sin60

u cos60

v sin30

g

u cos 60 = v cos 302

2 3

uv

3

uv

3cos30

2c

a g g

2 22

3 3c

v ur

a g

22

3 3

ur

g

16. Answer (2)

1 2

1 2

tan 4m

3

X XH

X X

17. Answer (1)u

A = 2 m/s 8 u m/s =

B

Av S

v

At any time ‘t’

ˆ ˆ2A

v i gtj �

ˆ ˆ8B

v i gtj �

A Bv v� �

then

. 0

A Bv v � �

2 × 8 = g2t2

4t

g

4( ). 10

10A B

x u u t

x = 4 m18. Answer (4)

| |sin30

| |

mg

Rg

v

v

�

�

30ºRgv

Rmv

mgv2Rg

v v�

19. Answer (4)

200

50

st

v 4 s

2ut

g , u = 19.6 m/s

20. Answer (4)

H

R/2

60º

S

22

4

RS H

av

Sv

T

21. Answer (1)

H

R/2

tan

2

H

R

R = 4Hcot

2tan

4 cot

H

H

1tan tan

2


4/10

22. Answer (3)R = 4 H cot

24

cot8

gTR

Comparing withR = 5T2

5 cot = 5 = 45º

23. Answer (4)Initial speed = 10 m/sSpeed at heighest point = 10 cos 60 = 5 m/sChange in speed = 10 – 5 = 5 m/s

24. Answer (1)

vw

s

x

vs

1

2d

1 55

6 6x

1

2d ,

25 1 34

36 4 6s km

25. Answer (3)

30º

E

vN

W

S

Av 5i j

ˆ5A

v i�

ˆ ˆsin30 cos30B

v v i v j �

ˆ

BAv vj�


then | | 10B

v �

km/h

26. Answer (3)

h h

t = 1 t = 3

u sin Hmax

2 sin1 3 4

uT

g

usin = 2g

2

max

( sin )2

2

uH g

g

= 20 m

27. Answer (4)

2sin2 30

3u

g

(i)

2sin2 45u

Rg

(ii)

From (i) & (ii)

R = 3.5 m

28. Answer (2)

| |av

va

t

�

0av

a

| | 0v �

for complete cycle

29. Answer (3)

R = 4H cot

3H = 4H cot

3cot

4

4 4 3tan , sin , cos

3 5 5

22sin cosu

Rg

224

25

uR

g

30. Answer (4)

ˆ ˆcos sinr a ti a tj �

ˆ ˆsin cosdr

v a ti a tjdt

�

�

2| | , | |v a a a ��

31. Answer (1)

2

T


5/10

32. Answer (2)

= 45º for maximum range

2 22 2 2

0 0( )sin

2 4 4

u vu u

Hg g g

33. Answer (4)

34. Answer (1)

35. Answer (4)

36. Answer (4)

(90 – )u v

P

ux = u cos

vx = v sin

ux = v

x

ucos = v sin

v = ucot

37. Answer (1)

2 2sin

2

uH

g

2

2 2

1 1

2 2 22 2

1

sin sin 30 2

sin sin 60 3

2

H

H

⎛ ⎞⎜ ⎟ ⎝ ⎠

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

H1 : H2 = 1 : 3

38. Answer (3)

39. Answer (2)

x

1.96 km

600 km/h

tanx

h (i)

211960 9.8

2t (ii)

5600

18x t (iii)

From (i), (ii) & (iii)

= 60º

40. Answer (1)

120º

4N

x-axis4N

4N

y-axis

Resultant force

= 2 2(4) (4) 244cos120

= 4 N towards (–ve) x-axis

41. Answer (4)

P

Q

R

A

B

C

From PQR

tanRQ

PQ

=

6 337º

8 4

42. Answer (2)

v1

1v

1v

1v

2v

1 2| | | |v v v � �

2 1v v v � � �

2 2| | – 2 cos �

v v v vv

2| | 2 sin �

v v

43. Answer (3)


6/10

44. Answer (4)

P = 5

Q R

P + Q = 18, Q = 13

R2 + P2 = Q2

R2 = Q2 – P2

= (13)2 – (5)2

R = 12

[ CHEMISTRY]

46. Answer (2)

Iodine has most electropositive character.

47. Answer (3)

Mg is 3s2

Al is 3s2 3p1

So IEI Mg > Al

48. Answer (3)

There is 7 periods and 18 groups in long form of

periodic table.

49. Answer (1)

Particle nature of e– was demonstrated by J.J.

Thomson.

50. Answer (3)

51. Answer (1)

H = IE of Li + eg

H of Cl

= + 5.4 – 3.61 = 1.79 eV atom–1

52. Answer (3)

53. Answer (3)

More electronegativity more acidic character and vice

versa.

54. Answer (2)

He, Ne, Ar, Kr, Xe, Rn have all paired e– s.

55. Answer (3)

When cation is formed

p

e ratio increases, so size decreases.

56. Answer (4)

h

p

57. Answer (2)

Number spectral line will be formed for 2px – 2p

z

transition.

58. Answer (3)

Number of peaks in graph of 4r22 and r is n – l

59. Answer (4)

Na 1s2 2s2 2p6 3s1

7 e– possess m = 0

60. Answer (3)

61. Answer (1)

h

2m KE

62. Answer (3)

Change in PE = 2E

2E4

= 3E

2

63. Answer (1)

Ne 1s2 2s2 2p6 4 s electrons

O 1s2 2s2 2p4 4 p electrons

64. Answer (3)

Name of element with Z = 102 is Unnilbiium so

symbol is Unb.

45. Answer (3)

ˆ ˆ(6 )A B i j � �

ˆ ˆ( 4 7 )A B i j � �

than ˆ ˆ2 2 8A i j �

4 , | | 4.1A i j A � �


7/10

65. Answer (4)

Li & Mg

Be & Al

B & Si

Show similar properties due to diagonal relationship.

66. Answer (3)

67. Answer (1)

CO and NO are neutral oxides.

68. Answer (2)

Fact.

69. Answer (3)

E.C. (Z = 117)

[Rn] 5f14 6d10 7s2 7p5

70. Answer (3)

eg

H of gp16

S > Se > Te > Po > O

71. Answer (2)

Al is forming six bonds with 1Cl and 5H2O

molecules, hence covalency is six.

72. Answer (4)

4th IE is very high

So there are 3 valence e– and possible oxidation

state of metal is +3.

and oxide will be M2O

3.

73. Answer (3)

Suppose velocity = x

4x

h

mv so

h4x

mx

2 hx

4m so

1 hx

2 m

1 hv

2 m

h4v 2

m

74. Answer (4)

2

n

0.529nr Å

Z

2

2

0.529 2r (H) 0.529 4Å

1

2

3

4

0.529 4r (Be ) 0.529 4Å

4

75. Answer (2)

pZ has zero electron density in XY plane.

76. Answer (4)

77. Answer (4)

78. Answer (4)

I.EI is energy required to remove an electron from

the valence shell of isolated and gaseous atom of

Ba.

79. Answer (3)

Orbital 2 2

x y

3d

lies along X and Y axis

So shows maximum electron probable density.

80. Answer (4)

As option (4) EC follows Hund’s rule.

81. Answer (1)

3

3hmvr

2

3

3h2 r 3

mv

82. Answer (4)

It is d-block element, as last e– goes to d-subshell.

group No. is number of e– in (n – 1)d and n s

subshells.

83. Answer (3)

n1 + n

2 = 4, So n

1 = 1

n2 – n

1 = 2 n

2 = 3

2

2 2

1 1 32RR(2)

91 3

⎡ ⎤ ⎢ ⎥⎣ ⎦

84. Answer (4)

Ne shows +ve eg

H due to stable EC.

O– + e– O2– is endothermic process.

85. Answer (1)

Outer EC : 2s2 2p5 represents halogen, with highest

electronegativity in respective period.

86. Answer (2)

Isotones are the atoms having same number of

neutrons.


8/10

87. Answer (4)

Fact based.

88. Answer (3)

In sodium hydride, anion is H–

So, EC of H– is 1s2

89. Answer (3)

m = 2 (highest)

So l = 2 (highest)

and l = 0 to n – 1

So n = 3

[ BIOLOGY

]

91. Answer (2)

Parasites/predators

92. Answer (3)

93. Answer (2)

The mycobionts include members of ascomycetes or

basidiomycetes.

94. Answer (4)

95. Answer (4)

Aspergillus, Neurospora, Penicillium

96. Answer (1)

Float passively.

97. Answer (2)

They are endomycorrhiza.

98. Answer (2)

Mucor – no dikaryophase

99. Answer (1)

Satellite RNA, discovered by Randle.

100. Answer (2)

101. Answer (4)

Citrus exocortis by viroids

102. Answer (1)

Facultative heterotrophs

103. Answer (4)

104. Answer (3)

Aplanospores

105. Answer (1)

106. Answer (2)

107. Answer (4)

108. Answer (3)

Late blight of potato.

109. Answer (4)

Plasmodium is a sporozoan.

110. Answer (4)

It is smaller than bacteria.

111. Answer (4)

112. Answer (3)

113. Answer (1)

Basidiomycetes

114. Answer (3)

Citrus canker is caused by Xanthomonas.

115. Answer (3)

116. Answer (2)

117. Answer (1)

1,3-glucan (Paramylon starch).

118. Answer (2)

Viroids are smaller than bacteria.

119. Answer (4)

In basidiomycetes, mycelium is branched and septate.

120. Answer (2)

In basidiomycetes, asexual spores are generally not

found.

121. Answer (1)

Basidiospores

122. Answer (2)

Angular momentum of orbit =nh

2

=3h

2

=1.5h

90. Answer (4)

1

6 1

nE 1.312 10 Jmol

6 2

5

n 2

1.312 10 (1)E 3.28 10

(2)

2

J mol–1

2 1

5

n nE E E 9.84 10 J mol–1


9/10

123. Answer (1)

Penicillium, Claviceps – Ascocarp.

Agaricus – Basidiocarp

124. Answer (2)

Spawn in Agaricus life cycle.

125. Answer (1)

Aseptate

126. Answer (1)

Yersinia is bacterium, TNV = Tobacco Necrosis Virus.

127. Answer (4)

T2 phage – dsDNA virus

128. Answer (1)

Plague, Diphtheria – bacterial diseases.

129. Answer (3)

130. Answer (2)

Protozoans/primitive relatives of animals.

131. Answer (1)

Rhinovirus – ssRNA

132. Answer (2)

Nitrogenase is absent in fungi.

133. Answer (4)

Agaricus is saprophyte

134. Answer (1)

Claviceps - Chitinous wall.

135. Answer (2)

In Trichoderma (Deuteromycetes) conidia are asexual

structure.

136. Answer (2)

They have paired appendages which may be fins or

limbs.

137. Answer (4)

Pinna absent in prototherians; many land mammals

and all aquatic mammals lack body hair.

138. Answer (4)

Fourth option is of prototherians.

139. Answer (2)

Chelone is reptile so poikilotherm.

140. Answer (3)

Cartilagenous fish do not have swim bladder.

141. Answer (2)

Bony fish and amphibians have bony vertebral

column.

142. Answer (3)

Cartilaginous fishes remove waste as urea.

143. Answer (2)

Birds possess only right systemic arch.

144. Answer (4)

Only amphibians and reptiles are dicondylic.

145. Answer (4)

Adult stage of urochordate is tailess, without

notochord and nerve cord change to dorsal ganglia.

146. Answer (1)

In chondrichthyes, placoid scales modify to make

teeth.

147. Answer (1)

148. Answer (3)

Trygon is sting ray.

149. Answer (2)

Aptenodytes is penguin, native of Antarctica.

150. Answer (2)

Lateral line sense organs check fluctuation in water

current.

151. Answer (4)

All these characters are found in most of reptiles

except fourth option.

152. Answer (2)

Homeothermic – Birds, Mammals

Diaphragm – Mammals and Crocodiles

Seven cervical vertebrae –Mammals

153. Answer (1)

154. Answer (1)

Whales are largest mammals.

155. Answer (1)

Pneumatic bones are character of Aves only.

156. Answer (4)

Balanoglossus – Hemichordate;

Ascidia, Herdmania – Urochordate – Tail absent in

adult; Nerve cord turns into dorsal ganglia;

Cephalochordates are blue print of chordates.

157. Answer (1)

Heart is absent in cyclostomates.

158. Answer (1)

In holocrine, sloughed off cell is replaced by a new

cell.


10/10

� � �

159. Answer (2)

Articular cartilage is modified hyaline cartilage but is

without blood vessel.

160. Answer (2)

By this age red bone marrow exists in the terminal

parts of long bones.

161. Answer (3)

162. Answer (3)

163. Answer (1)

The main function of amnion is to prevent

dessication/drying up of embryo.

164. Answer (2)

Aptenodytes is penguin whereas flying fox is

Pteropus.

165. Answer (4)

Skeletal muscle fibres are multinucleated.

166. Answer (4)

167. Answer (2)

Osteon is other name of Haversian system whereas

ossein is protein in bones.

168. Answer (1)

Smooth muscle gets scanty blood supply.

169. Answer (2)

Submandibular glands are compound alveolar glands.

170. Answer (2)

Oxalate withdraw Ca2+ ions from blood.

171. Answer (1)

Rest all options are of single unit smooth muscle.

172. Answer (3)

Epithelial tissue protects internal tissue/organ and it

has fast repairing ability.

173. Answer (4)

174. Answer (1)

PMNL – Neutrophils

175. Answer (3)

Sudoriferous/sweat glands modify to make mammary

gland.

176. Answer (2)

Cancellous/spongy bone do not have Haversian

system.

177. Answer (1)

Schwann cells and neurolemmocytes are name of

same cells meant for synthesis of myelin sheath in

PNS; Astrocytes act as blood brain barrier.

178. Answer (2)

In rest all options, muscle fibres differ from nerve

fibres.

179. Answer (3)

Fibroblasts secrete fibres and matrix.

180. Answer (3)

Aiats Medical Two Yr Test-2 (1)

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