for Medical-2019 All India Aakash Test Series for Medical - 2019 TEST - 1 (Code C) · 2019. 6....

Test - 1 (Code-C) (Answers & Hints) All India Aakash Test Series for Medical-2019

1/13

1. (3)

2. (3)

3. (2)

4. (2)

5. (4)

6. (3)

7. (2)

8. (1)

9. (3)

10. (4)

11. (1)

12. (4)

13. (1)

14. (1)

15. (3)

16. (2)

17. (4)

18. (3)

19. (4)

20. (1)

21. (3)

22. (2)

23. (3)

24. (1)

25. (2)

26. (4)

27. (3)

28. (4)

29. (2)

30. (4)

31. (2)

32. (1)

33. (3)

34. (2)

35. (1)

36. (1)

Test Date : 15/10/2017

ANSWERS

TEST - 1 (Code C)

All India Aakash Test Series for Medical - 2019

37. (4)

38. (2)

39. (4)

40. (3)

41. (1)

42. (3)

43. (1)

44. (4)

45. (1)

46. (1)

47. (1)

48. (3)

49. (4)

50. (4)

51. (1)

52. (1)

53. (2)

54. (1)

55. (3)

56. (3)

57. (2)

58. (2)

59. (3)

60. (1)

61. (3)

62. (3)

63. (4)

64. (1)

65. (4)

66. (1)

67. (2)

68. (3)

69. (4)

70. (1)

71. (3)

72. (3)

73. (4)

74. (2)

75. (2)

76. (4)

77. (1)

78. (3)

79. (2)

80. (3)

81. (2)

82. (4)

83. (2)

84. (3)

85. (2)

86. (4)

87. (3)

88. (3)

89. (1)

90. (3)

91. (3)

92. (2)

93. (3)

94. (1)

95. (4)

96. (1)

97. (3)

98. (2)

99. (4)

100. (3)

101. (2)

102. (4)

103. (3)

104. (4)

105. (1)

106. (1)

107. (3)

108. (4)

109. (4)

110. (3)

111. (2)

112. (2)

113. (2)

114. (1)

115. (4)

116. (3)

117. (4)

118. (4)

119. (2)

120. (3)

121. (4)

122. (2)

123. (2)

124. (1)

125. (2)

126. (3)

127. (4)

128. (3)

129. (3)

130. (3)

131. (2)

132. (3)

133. (4)

134. (2)

135. (4)

136. (4)

137. (4)

138. (3)

139. (4)

140. (1)

141. (3)

142. (2)

143. (3)

144. (2)

145. (2)

146. (4)

147. (2)

148. (1)

149. (2)

150. (2)

151. (3)

152. (3)

153. (3)

154. (4)

155. (3)

156. (3)

157. (3)

158. (3)

159. (3)

160. (4)

161. (4)

162. (4)

163. (4)

164. (3)

165. (2)

166. (1)

167. (3)

168. (4)

169. (4)

170. (4)

171. (4)

172. (4)

173. (1)

174. (3)

175. (3)

176. (3)

177. (4)

178. (2)

179. (4)

180. (4)

All India Aakash Test Series for Medical-2019 Test - 1 (Code-C) (Answers & Hints)

2/13

ANSWERS & HINTS

1. Answer (3)

v = 3 + 2t

dx = vdt

dx = (3 + 2t)dt

(3 2 )dx t dt ∫ ∫3

2

0

23

2f i

tx x t

⎡ ⎤ ⎢ ⎥

⎣ ⎦

s = (3 × 3 + 32) – 0

s = 18 m

2. Answer (3)

T = t1 + t

2 = 17 s

1st 2nd 3rd 4th 5th 6th 7th 8th

8.5

10th11th12th13th14th15th16th17th

t1 = 6 s

t2 = 11 s

Here acceleration is constant and body returns,

therefore initial velocity and acceleration must be in

opposite direction as shown above. Therefore

particle returns after 17 seconds.

3. Answer (2)

Acceleration

2( 2)0

dv d ta

dt dt

By solving we get t = 2

The velocity of particle = (t – 2)2 = (2 – 2)2 = 0

4. Answer (2)

When acceleration is zero, velocity may or may not

be zero similarly when velocity is zero, acceleration

may or may not be zero.

5. Answer (4)

During turn, direction of velocity changes, therefore

velocity changes.

6. Answer (3)

Here t1 and t

2 are two times when body is at same

height, then

2

2 1

2 2u ght t

g

[ PHYSICS]

22 2 10 160

410

u

220 3200 u

Squaring both sides, we get

400 = u2 – 3200

3600 = u2

u = ±60 ms–1

Body is thrown with speed 60 m/s

7. Answer (2)

1.5 1

2.5 2

tan 1

tan

2

h

v

hv

2

1

8. Answer (1)

Power of exponential is dimensionless

3t t is

0 0 0[M L T ]

3t⇒ is also

0 0 0[M L T ]

3T⇒ =

0 0 0[M L T ]

⇒ = 0 0 3[M L T ]

9. Answer (3)

Argument of trigonometric ratio is dimensionless

therefore t2 = M0L0T0

0 0 2[M L T ]⇒

and 0V

F

M1L1T–2 = 0

0 0 2M L T

V

1 1 2 0 0 2

0M LT M L T

V

= [M1L1T–4]


3/13

10. Answer (4)

Pressure = force

area=

1 1 2

2

MLT

L

= [M1L–1T–2]

In new system, pressure = (2 kg) (6 m)–1 (2 min)–2

= 1 2

2

2kgm min

6 (2)

= 1 21

kgm min12

11. Answer (1)

A

B

D

35 m

C

u = 30 (m/s)

For A to B, v2 – u2 = 2aS

02 – (30)2 = 2 (–10)S

–900 = –20S

S = 45 m

From B to C, S = BD + DC

= 45 + 35 = 80 m

Total distance covered is 80 + 45 = 125 m

For Time, s = –35 m

u = +30 m/s

a = –10 m/s2

s = 21

2ut at

–35 = 21

30 ( 10)2

t t

t2 – 6t – 7 = 0

6 8

2t

and 6 8

2t

t = 7 and t = –1

Neglect t = –1

t = 7 s

Average speed = Total distance

Total time

= 125 m

7 s = 17.86 m/s

12. Answer (4)

Slope of velocity-time graph is acceleration

OP slope negative

PQ slope positive

QR slope positive

RS slope negative

13. Answer (1)

Power of exponential is dimensionless,

2Ct = M0L0T0

CT1 = M0L0T0

C = M0L0T–1

and 3/2

dv

v = BC

1 1

1 1 3/2

[L T ]

[L T ]

= B[M0L0T–1]

B =

1 3

2 2[L T ]

14. Answer (1)

S b b c

S b b c

= 100 100 % ⎛ ⎞ ⎜ ⎟⎝ ⎠

b b c

b b c

0.1 (0.1 0.2)100 100 12%

5 5 2

15. Answer (3)

2 28 sinx t t

At t = 4, 2 2

8 4 sin4x

128 0 128 m x

Here 2

16 cosdx

v t tdt

= 16 2cost t

At t = 4, 16 4 2cos4v

= 64 + 2(1)

= 66 m/s

Here 16 2 sin4dv

adt

At t = 4, a = 16 – 2 sin4

a = 16 m/s2


4/13

16. Answer (2)

Till stone dropped from balloon,

v = u + at

= 0 + 2 × 4

= 8 m/s

During one second of drop, now consider initial

velocity after drop is 8 m/s upwards therefore stone

first stops then returns back

Time to stop (t0), v = u + at

0 = 8 – 10t0

t0 = 0.8

Distance covered =

2

2

0

1( )

2 2

ua t t

a

=

2

28 110 (1 0.8)

2 10 2

= 3.4 m

17. Answer (4)

a F m

a F m

100% 100 100 % ⎛ ⎞ ⎜ ⎟

⎝ ⎠

a F m

a F m

20 g2 100%

2 kg

0.020 kg2 100%

2 kg

20 12 100%

1000 2 = (2 + 1)% = 3%

18. Answer (3)

2s

GMg

R and

'

2

(1.8 )

(1.4 )s

G Mg

R

Percentage change in gs is =

'1 100

s

s

g

g

⎛ ⎞ ⎜ ⎟⎝ ⎠

2

1.81 100

(1.4)

⎡ ⎤ ⎢ ⎥⎣ ⎦

0.16100

1.96 = –8.16

19. Answer (4)

Angle has a unit but no dimensions.

20. Answer (1)

D5 + D

6 = 60

(2 5 1) (2 6 1) 602 2

a au u

(9) (11) 602 2

a au u

5 30u a

30 m/sv

21. Answer (3)

Distance between 1st and 11th drop is 80 m and let

t is time interval between two consecutive drops then

time difference between 1st and 11th drop is 10 t.

125 m

45 m

Roof

Ground

11

2

3

1

21

2s gt

2180 ( 10)(10 )

2t = 0.4 second

Distance covered by 10th drop in 0.4 s is

21

2s gt

21( 10)(0.4)

2s

s = 0.80 m

22. Answer (2)

From t = 0 to 2, change in velocity = 1

2 22

i.e., velocity of body decreases

2f i

v v

9 2f

v

7f

v m/s

After t = 2, change in velocity is positive i.e.,

velocity increases therefore velocity becomes greater

than 7 m/s.


5/13

23. Answer (3)

For A, slope decreases therefore velocity decreases

therefore negative acceleration.

For B, slope increases therefore velocity increases

therefore positive acceleration.

For C, slope decreases therefore velocity decreases


For D, slope increases, therefore velocity increases


24. Answer (1)

For 3

l distance,

2 2

1( ) 2

3

lv u a ...(1)

For A to B, 2 2

2v u al ...(2)

Taking ratios of (1) and (2) we get

2 2

2 2 2 21

12 2

( ) 13( ) 3

3

v u

v u v u

v u

⇒

2 2

2

1

2( )

3

v uv

2 2(60) 2(30)

3

2

1( ) 1800v

1

30 2 m/sv

25. Answer (2)

vav

= 1 2

1 2

2 2 40 6048 km/hr

40 60

v v

v v

Distance between A and B = vav

× t

= 48 × 2 = 96 km

26. Answer (4)

3va

x

3dv vvdx x

0

3

v x

u x

dxdv

x∫ ∫

0

[ ] 3(log )v x

u e xv x

v = u + 3loge|x/x

0|

27. Answer (3)

v(m/s)

t(s)

10

0 1 2

t

h

3

From 0 to 1 s, area 1

1 10 52

From 1 s to 3 s, area 1

2 10 102

From 0 to 3 s, total area = 5 + 10 = 15

For zero displacement, negative area should also be

15.

Therefore, 1

( 3) 152

t h

1 10

( 3) 10( 3) 152 3 1

⎡ ⎤ ⎢ ⎥⎣ ⎦

ht t

t

(t – 3)2 = 3

t – 3 = 3

t = 3 3 and 3 3t

Here t must be greater than 3

Answer is (3 3)s

28. Answer (4)

1 ly = 9.46 × 1015 m

1 AU = 1.496 × 1011 m

1 parsec = 3.084 × 1016 m

15

11

1Iy 9.46 10

1 AU 1.496 10

= 6.32 × 104

29. Answer (2)

m Gxc

ytz

M = 1 3 2 1 1 1[M L T ] [L T ] [T ]x y z

k

M1L0T0 = kM–xL3x+yT–2x–y+z

Compare dimensions of M, L and T are

1 = –x

x = –1

and 0 = 3x + y

y = 3

and 0 = –2x – y + z

1 = z

⇒m G–1c

3t1


6/13

30. Answer (4)

1

2

1

2

(2 )ht

g

1 1100 100 100

2 2

t h g

t h g

1 1( 4) ( 2)

2 2 = 3

31. Answer (2)

A must be most accurate, therefore option (2) and

(3) are preferable where B must be most precise

therefore from options, (2) is preferable.

Therefore answer is (2)

32. Answer (1)

2

1AU

r

lr

1 parsec = 1AU

1 second in radian...(1)

r = 1AU


1

1 parsec 2

r

1 parsec

2r

3.26 light year

2 = 1.63 light year

33. Answer (3)

1 fermi 1 angstrom

1 micron 1 giga

1510

6 9

10 10

10 10

= 10–28

34. Answer (2)

e2 and sin are dimensionless, therefore

x yz

x y

z = dimensionless

35. Answer (1)

Situation possible is

Time taken to reach top is

1

2

3

4

2 second

2 second

T = 4 second

v = u + at

0 = 40 – 10t

t = 4 s

and 2

3

1

2h gt = 80 m

2

2

1

2h ut gt

= 21

40 2 10 2 60 m2

36. Answer (1)

v = u + at

27 = 20 + a × 4

7

4a

a = 1.75 m/s2

Let the velocity at 3 s before the instant is u

After 3 s of u, velocity is 20 m/s

v = u + at

20 = u + 1.75 × 3

u = 14.75 m/s

37. Answer (4)

At the instant of overtake they will be at same point

SBC

= +100, uBC

= 20 2 , aBC

= –4

21

2S ut at


7/13

21100 20 2 ( 4)

2t t

By solving 5 2 st ,

At 5 2 st both have same velocity and after that

velocity of car increases so bus cannot overtake.

38. Answer (2)

urel

= 5 m/s

arel

= –(10 – 2) = –8 m/s2

srel

= 0

21

2s ut at

210 5 ( 8)

2t t

By solving t = 1.25 s

39. Answer (4)

Area = ad t∫

2( 2 )t t dt ∫

13 2

0

2

3 2

t t⎡ ⎤ ⎢ ⎥⎢ ⎥⎣ ⎦

31 2

13 3

⎡ ⎤ ⎢ ⎥⎢ ⎥⎣ ⎦

40. Answer (3)

First situation

v = u

2

uv

a = –a

s = 2

2 22v u as

2

22( ) 2

2

uu a

⎛ ⎞ ⎜ ⎟⎝ ⎠

23

44

ua ...(1)

Then , , , ?2 4

u uu v a a s

2 22v u as

2 2

2( )4 2

u ua s

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠...(2)

Solving (1) and (2) we get 1cm

2s

41. Answer (1)

Let AB = x

For BC, tBC

= tAC

– tAB

= 2 3 2x x

g g

For CD, tCD

= tAD

– tAC

2 6 2 3x x

g g

tAB

: tBC

: tCD

:: 1 : 3 1 : 6 3

42. Answer (3)

tan30 1

tan60 3

A

B

a

a

43. Answer (1)

At starting of collision, 2 2

2v u as

2 20 2( 10)(20)v

20i

v v

After collision 2 2

2v u as

2 20 2( 10)( 5)u

u = 10 = vf

Average acceleration time taken

f iv v

3

10 ( 20)

6 10

= 5000 m/s2


8/13

46. Answer (1)

24.8Moles = 0.3 for O

16

V.E. of O–2 = 8

Total number of valence e– = Mole × NA

× VE

= 0.3 × NA × 8

= 2.4 NA

47. Answer (1)

2 g butane moles = 2

58

Number of atoms = A A

2N 14 0.48N

58

2 g nitrogen moles = 2

28


2N 2 0.14 N

28

2 g silver moles =2

108


2N 0.0185 N

108

2 g water moles =2

18


2N 3 0.33 N

18

48. Answer (3)

H3BO

3 (Boric acid) is weak monobasic acid

49. Answer (4)

A + 2B 2C

5

1

8

2

B is L.R i.e B is completely consumed and 8

moles of C are obtained.

[ CHEMISTRY]

44. Answer (4)

2

2acceleration,d x

dt

accelerationdv

dt= rate of change in velocity

d v

dt= rate of change in speed,

2

dx

dt

⎛ ⎞⎜ ⎟⎝ ⎠

= square of speed

45. Answer (1)

2 22v u as

2 2(10 5) 2( )4

hu g

⎛ ⎞ ⎜ ⎟⎝ ⎠...(1)

and 2 2

0 2( )u g h ...(2)

By solving we get

h = 33.3 m

50. Answer (4)

Moles of nitrogen =2.8

0.128

Volume of nitrogen at STP = 0.1 × 22.4 L

= 2.24 L

= 2240 cc

51. Answer (1)

Weight of solute = 800 20

160 g100

Amount of solute remaining = 160 – 100 = 60 g

Mass of solution remaining = 800 – 100 = 700 g

% conc. of remaining solution = 60100 8.57%

700

52. Answer (1)

Moles of O2 =

16 1

32 2

Number of molecules = A

1N

2

(1) Mole = 22 1

44 2


1N

2

(2) Mole = 44

144

Number of molecules = 1 × NA

(3) Mole = 7

28


1N

4

(4) Mole = 28

128



9/13

53. Answer (2)

∵ 2.858 g O2 1 L

32 g O2 x L

3211.2 L

2.858x

54. Answer (1)

Urea (NH2 CONH

2)

12% C = 100 20%

60

55. Answer (3)

56. Answer (3)

1 M i.e. 1 mol of NaNO3 in 1000 mL of solution

Mass of solution =Volume × Density

=1000 × 1.20 = 1200 g

Mass of solvent =Mass of solution – Mass of solute

= 1200 – 85

= 1115 g

1 1000m = 0.89 m

1115

57. Answer (2)

58. Answer (2)

59. Answer (3)

∵ 1 amu is equal to =

A

1 g

N

20 amu is equal to = x g

x = 20/NA

60. Answer (1)

According to law of conservation of mass

Total mass of reactants = Total mass of products

61. Answer (3)

Normality = Molarity × n-factor

62. Answer (3)

Volume of one drop = 4

70

Number of moles in one drop

= d × v 2.4 4 1

Mol. wt 70 70

= 2

9.6

(70)

Number of molecules in one drop = A2

9.6N

(70)

63. Answer (4)

Let the number of moles of Ca3(PO

4)2 and H

3PO

3

are x and y respectively

x 1

y 2 (2x = y = Number of P atoms)

3 4 2

3 3

Moles of 'O' in Ca (PO ) 8x 8x 4

Moles of 'O' in H PO 3y 3 2x 3

64. Answer (1)

Moles of CO2 =

8.80.2

44

Moles of O = 0.4

Moles of NO2 =

4.60.1

46

Moles of 'O' = 0.1 × 2 = 0.2

Moles of H2O

2 =

13.60.4

34

Moles of 'O' = 0.4 × 2 = 0.8

Moles of SO2 =

6.40.1

64

Moles of 'O' = 0.1 × 2 = 0.2

65. Answer (4)

Nitrogen converted 80% of 14 g = 14 80

11.2 g100

Moles of nitrogen = 11.2

0.814

2 mol 'N' 3 mole 'O'

0.8 mol 'N' x mole 'O'

2.4x 1.2

2

Number of oxygen atom = 1.2 × 6 × 1023

= 7.2 × 1023

66. Answer (1)

1 Calory = 4.185 J

5 Calory = 5 × 4.185 J = 20.925 J

67. Answer (2)

68. Answer (3)

69. Answer (4)


10/13

70. Answer (1)

71. Answer (3)

C H

80% 20%

Moles 80

12

20

1

6.66 20

Moles ratio 6.66

6.66

20

6.66

1 3

Empirical Formula CH3

72. Answer (3)

CH4 + 2O

2 CO

2 + 2H

2O

∵ 1 Mol CH4 44 g CO

2

x Mol CH4 2.2 g CO

2

2.2 1x 0.05

44

73. Answer (4)

74. Answer (2)

According to Dulong and Petit's law

Atomic mass × Specific heat � 6.4

75. Answer (2)

Let the % abundance of Cu63 is x and Cu65 is

(100 – x) then

63x 65(100 x)63.546

100

x = 72.7% i.e Cu63 = 72.7%

76. Answer (4)

Mol. mass = Number of atoms per molecule × At. Mass ×100

Percentage of element

77. Answer (1)

S8(s)

+ 8O2(g)

8SO2(g)

...(i)

2SO2(g)

+ O2(g)

2SO3(g)

...(ii)

Equation number (ii) is multiplied by 4 and then add

equation (i) and (ii)

S8(s)

+ 8O2(g)

8SO2(g)

8SO2(g)

+ 4O2(g)

8SO3(g)

S8(s)

+ 12O2(g)

8SO3(g)

∵ 1 Mol. 8 Mol.SO3

2 Mole S8 x Mole of SO

3

x = 16 Mole of SO3

Mass of SO3 = 16 × 80 = 1280

78. Answer (3)

2H2(g)

+ O2(g)

2H2O

(g)

50 cc H2 will combine with 25 cc O

2 to form 50 cc

H2O

O2 left = 25 cc. At room temperature H

2O will

be in liquid state.

79. Answer (2)

80. Answer (3)

KClO4(s)

KCl(s)

+ 2O2(g)

loss is due to evolution of O2

% loss = 64

100 46138.5

81. Answer (2)

M.eq of HCl = M.eq of CaCO3

W

W0.1 500 1000

E

W = 2.5 g

Weight of sample = 2.5 × 2 = 5 g

82. Answer (4)

C2H

2 + 2H

2 C

2H

6

8 ml C2H

2 remains unreacted, 12 mL C

2H

2 reacts

with 24 mL H2 and forms 12 mL C

2H

6.

2 2 2 2 6T C H H C H

V V V V

VT = 8 + 16 + 12 = 36 mL

83. Answer (2)

Moles of H2 =

0.560.025

22.4

Mass of H2 = 0.025 × 2 = 0.05 g

2 2

M M

H H

E W

E W

ME 2.4

1 0.05

EM

= 48

84. Answer (3)

MO M = 53%, O = 47%

M MM

O O

E W 8 53 E = 9

E W 47

⇒

EM

= 9


11/13

85. Answer (2)

According to POAC

reagents

2 3 2 3 6 2K CO K Zn [Fe(CN) ]POAC at Carbon

2 3 2 3 6 2K CO K Zn [Fe(CN) ]1 × n = 12 × n

27.6 12 w

138 698

w = 11.6 g

86. Answer (4)

10 × d × x%M =

Mol. Mass

10 × d × 29

4 = 98

d = 1.35 g/mL

87. Answer (3)

N1V

1 = N

2V

2

N1

× 15 = 0.1 × 25

1

2.5 1N

15 6

N 1 1Molarity = 0.08

2 6 2 12

88. Answer (3)

A

A B

x 1000m

1 x M

A

A

x 10000.5

1 x 18

xA = 0.009

89. Answer (1)

Moles of AlCl3 =

500 0.20.1

1000

Moles of CI– = 3 × 0.1 = 0.3

Moles of MgCl2 =

500 0.20.1

1000

Moles of CI– = 0.1 × 2 = 0.2

Molarity of CI– =0.2 0.3 1

1 2

90. Answer (3)

CaO + 2HCl CaCl2 + H

2O

56 g 111g

∵ 56 g CaO 111 g CaCl2

2.46 g CaO x g CaCl2

111 2.46x = 4.87 g

56

Theoretical yield = 4.87 g

Actual yield = 3.7 g

% yield =3.7

100 75.9%4.87

= 76%

[ BIOLOGY]

91. Answer (3)

Binary fission in bacteria

92. Answer (2)

Mule, sterile worker bee not reproduce.

93. Answer (3)

New-systematics

94. Answer (1)

Genus

95. Answer (4)

Cortex of endospores

96. Answer (1)

Division-Angiospermae

97. Answer (3)

(b) Few similar characters

98. Answer (2)

99. Answer (4)

Insensitive of penicillin due to absence of cell wall.

100. Answer (3)

ICNB - For bacteria.

101. Answer (2)

Competent recipient living cell.

102. Answer (4)

Prokaryotic - two

Eukaryotic - four

103. Answer (3)

PPLO is Mycoplasma, a facultative anaerobe.

104. Answer (4)


12/13

105. Answer (1)

Three kingdom system.

106. Answer (1)

107. Answer (3)

(A) Vegetative cell

(B) Heterocyst

108. Answer (4)

109. Answer (4)

110. Answer (3)

Methanogens are obligate anaerobe.

111. Answer (2)

Corynebacterium

112. Answer (2)

Mycoplasma Mostly parasitic

Chloronema Photosynthetic

113. Answer (2)

114. Answer (1)

No cyclosis

115. Answer (4)

Heterotrophs are most abundant

116. Answer (3)

BGA : Non-motile

117. Answer (4)

Vibrio & Anabaena are prokaryotes

118. Answer (4)

119. Answer (2)

Botanical garden as ex situ conservation method.

120. Answer (3)

Herb, shrub, tree

121. Answer (4)

Nutrition, locomotion

122. Answer (2)

Circular DNA in bacteria and they commonly

reproduce by binary fission.

123. Answer (2)

Lactobacillus - Curd

Monerans - Motile / non-motile

124. Answer (1)

Bacteria as sole member of Monera.

125. Answer (2)

Nucleotide sequence in gene of 16S rRNA.

126. Answer (3)

Introns in archaebacteria.

127. Answer (4)

Acetobacter, Frankia = Eubacteria

Halobacterium = Archaebacteria

128. Answer (3)

Oxygenic photosynthesis

129. Answer (3)

130. Answer (3)

131. Answer (2)

Amoeba, Paramoecium, Halobacterium -

Heterotrophs

132. Answer (3)

133. Answer (4)

Animalia, fungi - Only heterotroph

134. Answer (2)

135. Answer (4)

� Discovered in Salmonella.

� Virulent phage can transport any gene.

136. Answer (4)

Fact

137. Answer (4)

Sycon, Leucosolenia show radial symmetry. They

can have purely calcareous or even siliceous

endoskeleton. Gastrovascular cavity is the feature of

cnidaria.

138. Answer (3)

Fact

139. Answer (4)

Ctenophores are also coelenterates that lack

nematocysts.

140. Answer (1)

Fact

141. Answer (3)

Centiped and millipedes show tracheal respiration.

142. Answer (2)

Insects show body divisible into head, thorax and

abdomen and two pairs of wings. Urochordates

show notochord only in tail of larva.

143. Answer (3)

Both are annelids hence excretion occurs through

nephridia.


13/13

144. Answer (2)

Pila shows respiration by ctenidia and pulmonary

sac. Branchiostoma shows respiration through

general body surface.

Cyclostomes have cartilaginous endoskeleton and

tornaria is the larva of hemichordates.

145. Answer (2)

146. Answer (4)

147. Answer (2)

148. Answer (1)

149. Answer (2)

150. Answer (2)

Fact

151. Answer (3)

Echinoderms show bilateral symmetry of larvae, like

chordates.

152. Answer (3)

153. Answer (3)

154. Answer (4)

155. Answer (3)

156. Answer (3)

Fact

157. Answer (3)

Parapodia are present in Nereis.

158. Answer (3)

159. Answer (3)

160. Answer (4)

161. Answer (4)

162. Answer (4)

Ctenophores like cnidaria excrete by diffusion

through general body surface and gastrovascular

cavity.

163. Answer (4)

164. Answer (3)

165. Answer (2)

166. Answer (1)

167. Answer (3)

Choanoflagellates are colonial flagellate protozoa.

168. Answer (4)

They are of equal length in both sexes.

169. Answer (4)

170. Answer (4)

171. Answer (4)

172. Answer (4)

Chitinous exoskeleton is the feature of arthropods.

173. Answer (1)

174. Answer (3)

175. Answer (3)

176. Answer (3)

177. Answer (4)

178. Answer (2)

179. Answer (4)

180. Answer (4)

Leech has open blood vascular system with

respiratory pigment in blood while centipede shows

direct respiration.

� � �

Test - 1 (Code-D) (Answers & Hints) All India Aakash Test Series for Medical-2019

1/13

1. (1)

2. (4)

3. (1)

4. (3)

5. (1)

6. (3)

7. (4)

8. (2)

9. (4)

10. (1)

11. (1)

12. (2)

13. (3)

14. (1)

15. (2)

16. (4)

17. (2)

18. (4)

19. (3)

20. (4)

21. (2)

22. (1)

23. (3)

24. (2)

25. (3)

26. (1)

27. (4)

28. (3)

29. (4)

30. (2)

31. (3)

32. (1)

33. (1)

34. (4)

35. (1)

36. (4)

Test Date : 15/10/2017

ANSWERS

TEST - 1 (Code D)

All India Aakash Test Series for Medical - 2019

37. (3)

38. (1)

39. (2)

40. (3)

41. (4)

42. (2)

43. (2)

44. (3)

45. (3)

46. (3)

47. (1)

48. (3)

49. (3)

50. (4)

51. (2)

52. (3)

53. (2)

54. (4)

55. (2)

56. (3)

57. (2)

58. (3)

59. (1)

60. (4)

61. (2)

62. (2)

63. (4)

64. (3)

65. (3)

66. (1)

67. (4)

68. (3)

69. (2)

70. (1)

71. (4)

72. (1)

73. (4)

74. (3)

75. (3)

76. (1)

77. (3)

78. (2)

79. (2)

80. (3)

81. (3)

82. (1)

83. (2)

84. (1)

85. (1)

86. (4)

87. (4)

88. (3)

89. (1)

90. (1)

91. (4)

92. (2)

93. (4)

94. (3)

95. (2)

96. (3)

97. (3)

98. (3)

99. (4)

100. (3)

101. (2)

102. (1)

103. (2)

104. (2)

105. (4)

106. (3)

107. (2)

108. (4)

109. (4)

110. (3)

111. (4)

112. (1)

113. (2)

114. (2)

115. (2)

116. (3)

117. (4)

118. (4)

119. (3)

120. (1)

121. (1)

122. (4)

123. (3)

124. (4)

125. (2)

126. (3)

127. (4)

128. (2)

129. (3)

130. (1)

131. (4)

132. (1)

133. (3)

134. (2)

135. (3)

136. (4)

137. (4)

138. (2)

139. (4)

140. (3)

141. (3)

142. (3)

143. (1)

144. (4)

145. (4)

146. (4)

147. (4)

148. (4)

149. (3)

150. (1)

151. (2)

152. (3)

153. (4)

154. (4)

155. (4)

156. (4)

157. (3)

158. (3)

159. (3)

160. (3)

161. (3)

162. (4)

163. (3)

164. (3)

165. (3)

166. (2)

167. (2)

168. (1)

169. (2)

170. (4)

171. (2)

172. (2)

173. (3)

174. (2)

175. (3)

176. (1)

177. (4)

178. (3)

179. (4)

180. (4)

All India Aakash Test Series for Medical-2019 Test - 1 (Code-D) (Answers & Hints)

2/13

ANSWERS & HINTS

1. Answer (1)

2 22v u as

2 2(10 5) 2( )4

hu g

⎛ ⎞ ⎜ ⎟⎝ ⎠...(1)

and 2 2

0 2( )u g h ...(2)

By solving we get

h = 33.3 m

2. Answer (4)

2

2acceleration,d x

dt

accelerationdv

dt= rate of change in velocity

d v

dt= rate of change in speed,

2

dx

dt

⎛ ⎞⎜ ⎟⎝ ⎠

= square of speed

3. Answer (1)

At starting of collision, 2 2

2v u as

2 20 2( 10)(20)v

20i

v v

After collision 2 2

2v u as

2 20 2( 10)( 5)u

u = 10 = vf

Average acceleration time taken

f iv v

3

10 ( 20)

6 10

= 5000 m/s2

4. Answer (3)

tan30 1

tan60 3

A

B

a

a

5. Answer (1)

Let AB = x

[ PHYSICS]

For BC, tBC

= tAC

– tAB

= 2 3 2x x

g g

For CD, tCD

= tAD

– tAC

2 6 2 3x x

g g

tAB

: tBC

: tCD

:: 1 : 3 1 : 6 3

6. Answer (3)

First situation

v = u

2

uv

a = –a

s = 2

2 22v u as

2

22( ) 2

2

uu a

⎛ ⎞ ⎜ ⎟⎝ ⎠

23

44

ua ...(1)

Then , , , ?2 4

u uu v a a s

2 22v u as

2 2

2( )4 2

u ua s

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠...(2)

Solving (1) and (2) we get 1cm

2s

7. Answer (4)

Area = ad t∫2( 2 )t t dt ∫

13 2

0

2

3 2

t t⎡ ⎤ ⎢ ⎥⎢ ⎥⎣ ⎦

31 2

13 3

⎡ ⎤ ⎢ ⎥⎢ ⎥⎣ ⎦


3/13

8. Answer (2)

urel

= 5 m/s

arel

= –(10 – 2) = –8 m/s2

srel

= 0

21

2s ut at

210 5 ( 8)

2t t

By solving t = 1.25 s

9. Answer (4)

At the instant of overtake they will be at same point

SBC

= +100, uBC

= 20 2 , aBC

= –4

21

2S ut at

21100 20 2 ( 4)

2 t t

By solving 5 2 st ,

At 5 2 st both have same velocity and after that

velocity of car increases so bus cannot overtake.

10. Answer (1)

v = u + at

27 = 20 + a × 4

7

4a

a = 1.75 m/s2

Let the velocity at 3 s before the instant is u

After 3 s of u, velocity is 20 m/s

v = u + at

20 = u + 1.75 × 3

u = 14.75 m/s

11. Answer (1)

Situation possible is

Time taken to reach top is

1

2

3

4

2 second

2 second

T = 4 second

v = u + at

0 = 40 – 10t

t = 4 s

and 2

3

1

2h gt = 80 m

2

2

1

2h ut gt

= 21

40 2 10 2 60 m2

12. Answer (2)

e2 and sin are dimensionless, therefore

x yz

x y

z = dimensionless

13. Answer (3)

1 fermi 1 angstrom

1 micron 1 giga

1510

6 9

10 10

10 10

= 10–28

14. Answer (1)

2

1AU

r

lr

1 parsec = 1AU


r = 1AU


1

1 parsec 2

r

1 parsec

2r

3.26 light year

2 = 1.63 light year


4/13

15. Answer (2)

A must be most accurate, therefore option (2) and

(3) are preferable where B must be most precise

therefore from options, (2) is preferable.

Therefore answer is (2)

16. Answer (4)

1

2

1

2

(2 )ht

g

1 1100 100 100

2 2

t h g

t h g

1 1( 4) ( 2)

2 2 = 3

17. Answer (2)

m Gxcytz

M = 1 3 2 1 1 1[M L T ] [L T ] [T ]x y z

k

M1L0T0 = kM–xL3x+yT–2x–y+z

Compare dimensions of M, L and T are

1 = –x

x = –1

and 0 = 3x + y

y = 3

and 0 = –2x – y + z

1 = z

⇒m G–1c3t1

18. Answer (4)

1 ly = 9.46 × 1015 m

1 AU = 1.496 × 1011 m

1 parsec = 3.084 × 1016 m

15

11

1Iy 9.46 10

1 AU 1.496 10

= 6.32 × 104

19. Answer (3)

v(m/s)

t(s)

10

0 1 2

t

h

3

From 0 to 1 s, area 1

1 10 52

From 1 s to 3 s, area 1

2 10 102

From 0 to 3 s, total area = 5 + 10 = 15

For zero displacement, negative area should also be

15.

Therefore, 1

( 3) 152

t h

1 10

( 3) 10( 3) 152 3 1

⎡ ⎤ ⎢ ⎥⎣ ⎦

ht t

t

(t – 3)2 = 3

t – 3 = 3

t = 3 3 and 3 3t

Here t must be greater than 3

Answer is (3 3)s

20. Answer (4)

3va

x

3dv vvdx x

0

3

v x

u x

dxdv

x∫ ∫

0

[ ] 3(log )v x

u e xv x

v = u + 3loge|x/x

0|

21. Answer (2)

vav

= 1 2

1 2

2 2 40 6048 km/hr

40 60

v v

v v

Distance between A and B = vav

× t

= 48 × 2 = 96 km

22. Answer (1)

For 3

l distance,

2 2

1( ) 2

3

lv u a ...(1)

For A to B, 2 2

2v u al ...(2)

Taking ratios of (1) and (2) we get


5/13

2 2

2 2 2 21

12 2

( ) 13( ) 3

3

v u

v u v u

v u

⇒

2 2

2

1

2( )

3

v uv

2 2(60) 2(30)

3

2

1( ) 1800v

1

30 2 m/sv

23. Answer (3)

For A, slope decreases therefore velocity decreases


For B, slope increases therefore velocity increases


For C, slope decreases therefore velocity decreases


For D, slope increases, therefore velocity increases


24. Answer (2)

From t = 0 to 2, change in velocity = 1

2 22

i.e., velocity of body decreases

2f i

v v

9 2f

v

7f

v m/s

After t = 2, change in velocity is positive i.e.,

velocity increases therefore velocity becomes greater

than 7 m/s.

25. Answer (3)

Distance between 1st and 11th drop is 80 m and let

t is time interval between two consecutive drops then

time difference between 1st and 11th drop is 10 t.

125 m

45 m

Roof

Ground

11

2

3

1

21

2s gt

2180 ( 10)(10 )

2t = 0.4 second

Distance covered by 10th drop in 0.4 s is

21

2s gt

21( 10)(0.4)

2s

s = 0.80 m

26. Answer (1)

D5 + D

6 = 60

(2 5 1) (2 6 1) 602 2

a au u

(9) (11) 602 2

a au u

5 30u a

30 m/sv

27. Answer (4)

Angle has a unit but no dimensions.

28. Answer (3)

2s

GMg

R and

'

2

(1.8 )

(1.4 )s

G Mg

R

Percentage change in gs is =

'1 100

s

s

g

g

⎛ ⎞ ⎜ ⎟⎝ ⎠

2

1.81 100

(1.4)

⎡ ⎤ ⎢ ⎥⎣ ⎦

0.16100

1.96 = –8.16

29. Answer (4)

a F m

a F m

100% 100 100 % ⎛ ⎞ ⎜ ⎟

⎝ ⎠

a F m

a F m


6/13

20 g2 100%

2 kg

0.020 kg2 100%

2 kg

20 12 100%

1000 2 = (2 + 1)% = 3%

30. Answer (2)

Till stone dropped from balloon,

v = u + at

= 0 + 2 × 4

= 8 m/s

During one second of drop, now consider initial

velocity after drop is 8 m/s upwards therefore stone

first stops then returns back

Time to stop (t0), v = u + at

0 = 8 – 10t0

t0 = 0.8

Distance covered =

2

2

0

1( )

2 2

ua t t

a

=

2

28 110 (1 0.8)

2 10 2

= 3.4 m

31. Answer (3)

2 28 sinx t t

At t = 4, 2 2

8 4 sin4x

128 0 128 m x

Here 2

16 cosdx

v t tdt

= 16 2cost t

At t = 4, 16 4 2cos4v

= 64 + 2(1)

= 66 m/s

Here 16 2 sin4dv

adt

At t = 4, a = 16 – 2 sin4a = 16 m/s2

32. Answer (1)

S b b c

S b b c

= 100 100 % ⎛ ⎞ ⎜ ⎟⎝ ⎠

b b c

b b c

0.1 (0.1 0.2)100 100 12%

5 5 2

33. Answer (1)

Power of exponential is dimensionless,

2Ct = M0L0T0

CT1 = M0L0T0

C = M0L0T–1

and 3/2

dv

v = BC

1 1

1 1 3/2

[L T ]

[L T ]

= B[M0L0T–1]

B =

1 3

2 2[L T ]

34. Answer (4)

Slope of velocity-time graph is acceleration

OP slope negative

PQ slope positive

QR slope positive

RS slope negative

35. Answer (1)

A

B

D

35 m

C

u = 30 (m/s)

For A to B, v2 – u2 = 2aS

02 – (30)2 = 2 (–10)S

–900 = –20S

S = 45 m

From B to C, S = BD + DC

= 45 + 35 = 80 m


7/13

Total distance covered is 80 + 45 = 125 m

For Time, s = –35 m

u = +30 m/s

a = –10 m/s2

s = 21

2ut at

–35 = 21

30 ( 10)2

t t

t2 – 6t – 7 = 0

6 8

2t

and 6 8

2t

t = 7 and t = –1

Neglect t = –1

t = 7 s

Average speed = Total distance

Total time

= 125 m

7 s = 17.86 m/s

36. Answer (4)

Pressure = force

area=

1 1 2

2

MLT

L

= [M1L–1T–2]

In new system, pressure = (2 kg) (6 m)–1 (2 min)–2

= 1 2

2

2kgm min

6 (2)

= 1 21

kgm min12

37. Answer (3)

Argument of trigonometric ratio is dimensionless

therefore t2 = M0L0T0 0 0 2[M L T ]

⇒

and 0V

F

M1L1T–2 = 0

0 0 2M L T

V

1 1 2 0 0 2

0M LT M L T

V

= [M1L1T–4]

38. Answer (1)

Power of exponential is dimensionless

3t t is

0 0 0[M L T ]

3t⇒ is also

0 0 0[M L T ]

3T⇒ =

0 0 0[M L T ]

⇒ = 0 0 3[M L T ]

39. Answer (2)

1.5 1

2.5 2

tan 1

tan

2

h

v

hv

2

1

40. Answer (3)

Here t1 and t

2 are two times when body is at same

height, then

2

2 1

2 2u ght t

g

22 2 10 160

410

u

220 3200u

Squaring both sides, we get

400 = u2 – 3200

3600 = u2

u = ±60 ms–1

Body is thrown with speed 60 m/s

41. Answer (4)

During turn, direction of velocity changes, therefore

velocity changes.

42. Answer (2)

When acceleration is zero, velocity may or may not

be zero similarly when velocity is zero, acceleration

may or may not be zero.

43. Answer (2)

Acceleration

2( 2)0

dv d ta

dt dt

By solving we get t = 2

The velocity of particle = (t – 2)2 = (2 – 2)2 = 0


8/13

46. Answer (3)

CaO + 2HCl CaCl2 + H

2O

56 g 111g

∵ 56 g CaO 111 g CaCl2

2.46 g CaO x g CaCl2

111 2.46x = 4.87 g

56

Theoretical yield = 4.87 g

Actual yield = 3.7 g

% yield =3.7

100 75.9%4.87

= 76%

47. Answer (1)

Moles of AlCl3 =

500 0.20.1

1000

Moles of CI– = 3 × 0.1 = 0.3

Moles of MgCl2 =

500 0.20.1

1000

Moles of CI– = 0.1 × 2 = 0.2

Molarity of CI– =0.2 0.3 1

1 2

48. Answer (3)

A

A B

x 1000m

1 x M

A

A

x 10000.5

1 x 18

xA = 0.009

[ CHEMISTRY]

44. Answer (3)

T = t1 + t

2 = 17 s

1st 2nd 3rd 4th 5th 6th 7th 8th

8.5

10th11th12th13th14th15th16th17th

t1 = 6 s

t2 = 11 s

Here acceleration is constant and body returns,

therefore initial velocity and acceleration must be in

opposite direction as shown above. Therefore

particle returns after 17 seconds.

45. Answer (3)

v = 3 + 2t

dx = vdt

dx = (3 + 2t)dt

(3 2 )dx t dt ∫ ∫3

2

0

23

2f i

tx x t

⎡ ⎤ ⎢ ⎥

⎣ ⎦

s = (3 × 3 + 32) – 0

s = 18 m

49. Answer (3)

N1V

1 = N

2V

2

N1

× 15 = 0.1 × 25

1

2.5 1N

15 6

N 1 1Molarity = 0.08

2 6 2 12

50. Answer (4)

10 × d × x%M =

Mol. Mass

10 × d × 29

4 = 98

d = 1.35 g/mL

51. Answer (2)

According to POAC

reagents

2 3 2 3 6 2K CO K Zn [Fe(CN) ]

POAC at Carbon

2 3 2 3 6 2K CO K Zn [Fe(CN) ]1 × n = 12 × n

27.6 12 w

138 698

w = 11.6 g

52. Answer (3)

MO M = 53%, O = 47%

M MM

O O

E W 8 53 E = 9

E W 47

⇒

EM

= 9


9/13

53. Answer (2)

Moles of H2 =

0.560.025

22.4

Mass of H2 = 0.025 × 2 = 0.05 g

2 2

M M

H H

E W

E W

ME 2.4

1 0.05

EM

= 48

54. Answer (4)

C2H

2 + 2H

2 C

2H

6

8 ml C2H

2 remains unreacted, 12 mL C

2H

2 reacts

with 24 mL H2 and forms 12 mL C

2H

6.

2 2 2 2 6T C H H C H

V V V V

VT = 8 + 16 + 12 = 36 mL

55. Answer (2)

M.eq of HCl = M.eq of CaCO3

W

W0.1 500 1000

E

W = 2.5 g

Weight of sample = 2.5 × 2 = 5 g

56. Answer (3)

KClO4(s)

KCl(s)

+ 2O2(g)

loss is due to evolution of O2

% loss = 64

100 46138.5

57. Answer (2)

58. Answer (3)

2H2(g)

+ O2(g)

2H2O

(g)

50 cc H2 will combine with 25 cc O

2 to form 50 cc

H2O

O2 left = 25 cc. At room temperature H

2O will

be in liquid state.

59. Answer (1)

S8(s)

+ 8O2(g)

8SO2(g)

2SO2(g)

+ O2(g)

2SO3(g)

Equation number (ii) is multiplied by 4 and then add

equation (i) and (ii)

S8(s)

+ 8O2(g)

8SO2(g)

...(i)

8SO2(g)

+ 4O2(g)

8SO3(g)

...(ii)

S8(s)

+ 12O2(g)

8SO3(g)

∵ 1 Mol. 8 Mol.SO3

2 Mole S8 x Mole of SO

3

x = 16 Mole of SO3

Mass of SO3 = 16 × 80 = 1280

60. Answer (4)

Mol. mass = Number of atoms per molecule × At. Mass ×100

Percentage of element

61. Answer (2)

Let the % abundance of Cu63 is x and Cu65 is

(100 – x) then

63x 65(100 x)63.546

100

x = 72.7% i.e Cu63 = 72.7%

62. Answer (2)

According to Dulong and Petit's law

Atomic mass × Specific heat � 6.4

63. Answer (4)

64. Answer (3)

CH4 + 2O

2 CO

2 + 2H

2O

∵ 1 Mol CH4 44 g CO

2

x Mol CH4 2.2 g CO

2

2.2 1x 0.05

44

65. Answer (3)

C H

80% 20%

Moles 80

12

20

1

6.66 20

Moles ratio 6.66

6.66

20

6.66

1 3

Empirical Formula CH3

66. Answer (1)

67. Answer (4)

68. Answer (3)


10/13

69. Answer (2)

70. Answer (1)

1 Calory = 4.185 J

5 Calory = 5 × 4.185 J = 20.925 J

71. Answer (4)

Nitrogen converted 80% of 14 g = 14 80

11.2 g100

Moles of nitrogen = 11.2

0.814

2 mol 'N' 3 mole 'O'

0.8 mol 'N' x mole 'O'

2.4x 1.2

2

Number of oxygen atom = 1.2 × 6 × 1023

= 7.2 × 1023

72. Answer (1)

Moles of CO2 =

8.80.2

44

Moles of O = 0.4

Moles of NO2 =

4.60.1

46

Moles of 'O' = 0.1 × 2 = 0.2

Moles of H2O

2 =

13.60.4

34

Moles of 'O' = 0.4 × 2 = 0.8

Moles of SO2 =

6.40.1

64

Moles of 'O' = 0.1 × 2 = 0.2

73. Answer (4)

Let the number of moles of Ca3(PO

4)2 and H

3PO

3

are x and y respectively

x 1

y 2 (2x = y = Number of P atoms)

3 4 2

3 3

Moles of 'O' in Ca (PO ) 8x 8x 4

Moles of 'O' in H PO 3y 3 2x 3

74. Answer (3)

Volume of one drop = 4

70

Number of moles in one drop

= d × v 2.4 4 1

Mol. wt 70 70

= 2

9.6

(70)

Number of molecules in one drop = A2

9.6N

(70)

75. Answer (3)

Normality = Molarity × n-factor

76. Answer (1)

According to law of conservation of mass

Total mass of reactants = Total mass of products

77. Answer (3)

∵ 1 amu is equal to = A

1 g

N

20 amu is equal to = x g

x = 20/NA

78. Answer (2)

79. Answer (2)

80. Answer (3)

1 M i.e. 1 mol of NaNO3 in 1000 mL of solution

Mass of solution =Volume × Density

=1000 × 1.20 = 1200 g

Mass of solvent =Mass of solution – Mass of solute

= 1200 – 85

= 1115 g

1 1000m = 0.89 m

1115

81. Answer (3)

82. Answer (1)

Urea (NH2 CONH

2)

12% C = 100 20%

60

83. Answer (2)

∵ 2.858 g O2 1 L

32 g O2 x L

3211.2 L

2.858x


11/13

[ BIOLOGY]

91. Answer (4)

� Discovered in Salmonella.

� Virulent phage can transport any gene.

92. Answer (2)

93. Answer (4)

Animalia, fungi - Only heterotroph

94. Answer (3)

84. Answer (1)

Moles of O2 =

16 1

32 2


1N

2

(1) Mole = 22 1

44 2


1N

2

(2) Mole = 44

144


(3) Mole = 7

28


1N

4

(4) Mole = 28

128


85. Answer (1)

Weight of solute = 800 20

160 g100

Amount of solute remaining = 160 – 100 = 60 g

Mass of solution remaining = 800 – 100 = 700 g

% conc. of remaining solution = 60100 8.57%

700

86. Answer (4)

Moles of nitrogen =2.8

0.128

Volume of nitrogen at STP = 0.1 × 22.4 L

= 2.24 L

= 2240 cc

87. Answer (4)

A + 2B 2C

5

1

8

2

B is L.R i.e B is completely consumed and 8

moles of C are obtained.

88. Answer (3)

H3BO

3 (Boric acid) is weak monobasic acid

89. Answer (1)

2 g butane moles = 2

58


2N 14 0.48N

58

2 g nitrogen moles = 2

28


2N 2 0.14 N

28

2 g silver moles =2

108


2N 0.0185 N

108

2 g water moles =2

18


2N 3 0.33 N

18

90. Answer (1)

24.8Moles = 0.3 for O

16

V.E. of O–2 = 8

Total number of valence e– = Mole × NA

× VE

= 0.3 × NA × 8

= 2.4 NA

95. Answer (2)

Amoeba, Paramoecium, Halobacterium -

Heterotrophs

96. Answer (3)

97. Answer (3)

98. Answer (3)

Oxygenic photosynthesis


12/13

99. Answer (4)

Acetobacter, Frankia = Eubacteria

Halobacterium = Archaebacteria

100. Answer (3)

Introns in archaebacteria.

101. Answer (2)

Nucleotide sequence in gene of 16S rRNA.

102. Answer (1)

Bacteria as sole member of Monera.

103. Answer (2)

Lactobacillus - Curd

Monerans - Motile / non-motile

104. Answer (2)

Circular DNA in bacteria and they commonly

reproduce by binary fission.

105. Answer (4)

Nutrition, locomotion

106. Answer (3)

Herb, shrub, tree

107. Answer (2)

Botanical garden as ex situ conservation method.

108. Answer (4)

109. Answer (4)

Vibrio & Anabaena are prokaryotes

110. Answer (3)

BGA : Non-motile

111. Answer (4)

Heterotrophs are most abundant

112. Answer (1)

No cyclosis

113. Answer (2)

114. Answer (2)

Mycoplasma Mostly parasitic

Chloronema Photosynthetic

115. Answer (2)

Corynebacterium

116. Answer (3)

Methanogens are obligate anaerobe.

117. Answer (4)

118. Answer (4)

119. Answer (3)

(A) Vegetative cell

(B) Heterocyst

120. Answer (1)

121. Answer (1)

Three kingdom system.

122. Answer (4)

123. Answer (3)

PPLO is Mycoplasma, a facultative anaerobe.

124. Answer (4)

Prokaryotic - two

Eukaryotic - four

125. Answer (2)

Competent recipient living cell.

126. Answer (3)

ICNB - For bacteria.

127. Answer (4)

Insensitive of penicillin due to absence of cell wall.

128. Answer (2)

129. Answer (3)

(b) Few similar characters

130. Answer (1)

Division-Angiospermae

131. Answer (4)

Cortex of endospores

132. Answer (1)

Genus

133. Answer (3)

New-systematics

134. Answer (2)

Mule, sterile worker bee not reproduce.

135. Answer (3)

Binary fission in bacteria

136. Answer (4)

Leech has open blood vascular system with

respiratory pigment in blood while centipede shows

direct respiration.

137. Answer (4)


13/13

138. Answer (2)

139. Answer (4)

140. Answer (3)

141. Answer (3)

142. Answer (3)

143. Answer (1)

144. Answer (4)

Chitinous exoskeleton is the feature of arthropods.

145. Answer (4)

146. Answer (4)

147. Answer (4)

148. Answer (4)

They are of equal length in both sexes.

149. Answer (3)

Choanoflagellates are colonial flagellate protozoa.

150. Answer (1)

151. Answer (2)

152. Answer (3)

153. Answer (4)

154. Answer (4)

Ctenophores like cnidaria excrete by diffusion

through general body surface and gastrovascular

cavity.

155. Answer (4)

156. Answer (4)

157. Answer (3)

158. Answer (3)

159. Answer (3)

Parapodia are present in Nereis.

160. Answer (3)

Fact

161. Answer (3)

162. Answer (4)

163. Answer (3)

164. Answer (3)

165. Answer (3)

Echinoderms show bilateral symmetry of larvae, like

chordates.

166. Answer (2)

Fact

167. Answer (2)

168. Answer (1)

169. Answer (2)

170. Answer (4)

171. Answer (2)

172. Answer (2)

Pila shows respiration by ctenidia and pulmonary

sac. Branchiostoma shows respiration through

general body surface.

Cyclostomes have cartilaginous endoskeleton and

tornaria is the larva of hemichordates.

173. Answer (3)

Both are annelids hence excretion occurs through

nephridia.

174. Answer (2)

Insects show body divisible into head, thorax and

abdomen and two pairs of wings. Urochordates

show notochord only in tail of larva.

175. Answer (3)

Centiped and millipedes show tracheal respiration.

176. Answer (1)

Fact

177. Answer (4)

Ctenophores are also coelenterates that lack

nematocysts.

178. Answer (3)

Fact

179. Answer (4)

Sycon, Leucosolenia show radial symmetry. They

can have purely calcareous or even siliceous

endoskeleton. Gastrovascular cavity is the feature of

cnidaria.

180. Answer (4)

Fact

� � �

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