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Test - 1 (Code-E) (Answers) All India Aakash Test Series for Medical-2018

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1. (2)

2. (4)

3. (2)

4. (2)

5. (2)

6. (1)

7. (1)

8. (2)

9. (3)

10. (4)

11. (1)

12. (1)

13. (4)

14. (2)

15. (4)

16. (4)

17. (4)

18. (1)

19. (4)

20. (2)

21. (1)

22. (2)

23. (2)

24. (4)

25. (3)

26. (4)

27. (1)

28. (2)

29. (2)

30. (2)

31. (2)

32. (3)

33. (4)

34. (2)

35. (1)

36. (1)

Test Date : 26/11/2017

ANSWERS

TEST - 1 (Code E)

All India Aakash Test Series for Medical - 2018

37. (3)

38. (2)

39. (3)

40. (4)

41. (3)

42. (2)

43. (1)

44. (4)

45. (1)

46. (3)

47. (4)

48. (3)

49. (4)

50. (1)

51. (2)

52. (3)

53. (2)

54. (3)

55. (1)

56. (4)

57. (3)

58. (3)

59. (2)

60. (4)

61. (1)

62. (4)

63. (2)

64. (4)

65. (2)

66. (4)

67. (1)

68. (1)

69. (4)

70. (1)

71. (3)

72. (2)

73. (3)

74. (2)

75. (1)

76. (2)

77. (1)

78. (3)

79. (3)

80. (4)

81. (1)

82. (2)

83. (3)

84. (2)

85. (4)

86. (2)

87. (3)

88. (3)

89. (3)

90. (2)

91. (3)

92. (2)

93. (3)

94. (3)

95. (3)

96. (4)

97. (3)

98. (1)

99. (2)

100. (4)

101. (2)

102. (2)

103. (4)

104. (3)

105. (2)

106. (3)

107. (3)

108. (1)

109. (3)

110. (4)

111. (1)

112. (2)

113. (1)

114. (3)

115. (3)

116. (4)

117. (2)

118. (1)

119. (3)

120. (2)

121. (4)

122. (4)

123. (1)

124. (2)

125. (2)

126. (4)

127. (2)

128. (1)

129. (1)

130. (1)

131. (4)

132. (1)

133. (2)

134. (2)

135. (2)

136. (1)

137. (2)

138. (2)

139. (2)

140. (1)

141. (3)

142. (3)

143. (2)

144. (2)

145. (2)

146 (2)

147. (2)

148. (4)

149. (2)

150. (4)

151. (3)

152. (2)

153. (4)

154. (2)

155. (1)

156. (1)

157. (2)

158. (3)

159. (3)

160. (2)

161. (3)

162. (3)

163. (4)

164. (1)

165. (1)

166. (4)

167. (1)

168. (3)

169. (4)

170. (3)

171. (4)

172. (3)

173. (3)

174. (1)

175. (1)

176. (4)

177. (1)

178. (4)

179. (1)

180. (2)

All India Aakash Test Series for Medical-2018 Test - 1 (Code-E) (Answers & Hints)

2/10

ANSWERS & HINTS

1. Answer (2)

1 1 2

1 1 12 1

2 2 2

M L Tn n

M L T

⎛ ⎞ ⎛ ⎞ ⎛ ⎞

⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

21 1 1

110 10 10

⎛ ⎞ ⎜ ⎟⎝ ⎠

= 1

2. Answer (4)

180° > > 90°

a ta

v

ac

3. Answer (2)

4. Answer (2)

5. Answer (2)

At t = 4.5 s, it is at highest point. At 4 s and 5 s,

it is 0.5 s below the highest point so

Distance = 21

2 0.5 10 0.25 2.5m2g

⎡ ⎤ ⎢ ⎥⎣ ⎦

6. Answer (1)

2 1[ ] [M L T ]E

E h h h ⇒ ⇒

7. Answer (1)

1/2P A

ET

8. Answer (2)

1 2

1 110 4 9 180 m

2 2h gt t

9. Answer (3)

1 2

2[M L T ]

aP

V

10. Answer (4)

13 2

2

X a b c d

X a b c d

13 0.1% 2 0.2% 0.4% 0.4%

2 = 1.3%

11. Answer (1)

100 100 100I V R

I V R

0.1 0.2100 100

20 80

= 0.5 + 0.25

= 0.75%

12. Answer (1)

x = 12t – t3

v = 2

12 3dx

tdt

0 = 12 – 3t2

t = 2 s

x (t = 2 s) = 12(2) – 23 = 24 – 8 = 16 m

x (t = 0 s) = 0

13. Answer (4)

Kads

m

∫

14. Answer (2)

6 5a t

2

26 5 d St

dt

23 5 dSt t

dt

4

2

0

(3 5 ) 104 mS t t dt ∫

15. Answer (4)

max

2 412

2 4v T

⎛ ⎞ ⎜ ⎟ ⎝ ⎠= 16 m/s

[ PHYSICS]

Test - 1 (Code-E) (Answers & Hints) All India Aakash Test Series for Medical-2018

3/10

16. Answer (4)

2 2 2 2 23

x yv v v t

17. Answer (4)

2u b

a c

18. Answer (1)

21

2S at

1 1

2 2

t ht S

t h ⇒

19. Answer (4)

Slope is rate of change of speed = tangential

acceleration

Upward motion

at = –(g + kv)

v at

Downward motion

at = g – kv

v at

20. Answer (2)

21. Answer (1)

h v2

2

1

2h

h

22. Answer (2)

v cos

v

v sinr

d = 50 m

B

2cos cos 20 1

50 2cos

v v v

dr d

= 0.2 rad/s

23. Answer (2)

602 sin 2 20 sin 20 m/s

2 2v v

24. Answer (4)

25. Answer (3)

A2 + B2 + 2ABcos = A2 + B2 – 2AB

= 90°

26. Answer (4)

1 1 1ˆ ˆ ˆA a i b j c k

�

2 2 2ˆ ˆ ˆB a i b j c k

�

If ||A B� �

then 1 1 1

2 2 2

ve

a b c

a b c

27. Answer (1)

28. Answer (2)

v u at ��

29. Answer (2)

2 2 2

1 2

sin;

2 2

u uh h H

g g

⎛ ⎞ ⎜ ⎟⎜ ⎟⎝ ⎠

30. Answer (2)

vav

(speed) = 0

vav

(velocity) = Total displacement

Total time taken

31. Answer (2)

2 21 1

2 22 2

cos cos 30

cos cos 60

H

H

= 3 : 1

32. Answer (3)

2sin2 u

Rg

33. Answer (4)

40 m/sH

Rv

T

1 2

2 1

sin 3

sin 2

u

u

34. Answer (2)


4/10

46. Answer (3)

A

B B

1000 0.25 1000m 7.246 molal

M 0.75 46

47. Answer (4)

Na2CO

3 does not decompose on heating.

CaCO3 CaO + CO

2

100 gm 22.4 L

50 gm 11.2 L

mass of CaCO3 = 50 g

mass of Na2CO

3 = 50 g

[ CHEMISTRY]

35. Answer (1)

If = constant

H u2

R u2

R% = H%

36. Answer (1)

45° 2 2

VRM

VM

2

2

VR

RM R Mv v v � � �

R RM Mv v v � � �

37. Answer (3)

38. Answer (2)

When R = xH, 2

2

16

2 16

uH

g x

Put x = 3 0 52

gu⇒

39. Answer (3)

BA B AV V V � � �

BA

100cos5 hr

| |t

v

A

100 kmvB = 10 km/h

B

10 2

Path of w.r.t B A

40. Answer (4)

|Displacement| Distance

41. Answer (3)

4tan

n

; when R = nH

42. Answer (2)

2 2

x ya a a

43. Answer (1)

uH

= constant

u cos60°= ucos30°

10'

3u⇒ m/s

44. Answer (4)

45. Answer (1)

22

2 2

1 1

c

a vva

R a v

⎛ ⎞ ⇒ ⎜ ⎟

⎝ ⎠

48. Answer (3)

1 L of H2O = 1000 gm =

1000

18 = 55.5 mole

= 55.5 × 6 × 1023 × 3 atoms

22.4 L of CO2

= 1 mole = 3 × 6 × 1023 atoms

44.8 L of O2

= 2 mole = 2 × 2 × 6 × 1023 atoms

11.2 L of SO3 = 0.5 mole = 0.5 × 4 × 6 × 1023 atoms

49. Answer (4)

∵ 2 g oxygen combines with 3 g of metal

8 g oxygen combines with 12 g of metal

EMO

= EM

+ 8 = 12 + 8 = 20 g


5/10

eq.MO

(ml)

n1000 1000M n

V n.f 500

0.030.01 2 n.f 6

n.f ⇒

Molar mass of metal oxide = 20 × 6 = 120 g

50. Answer (1)

(1 2

2O contains 18e– + 16p + 16n = 50)

32 g = 50 NA

16 g = 25 NA

51. Answer (2)

H2SO

4 + Na

2CO

3 Na

2SO

4 + H

2O + CO

2

98 g 106 g

% purity = cal. mass

×100expt. mass

expt. mass = 98 100

122.5 gm80

52. Answer (3)

C3H

8 + 5O

2 3CO

2 +

4H

2O(l)

1 : 5 : 3

20 ml 100 ml 60 ml of CO2 formed

80 ml left unreacted 80 + 60 = 140 ml

53. Answer (2)

Fact – sp3 hybridization

54. Answer (3)

O2, O

3, O

2

2–

B.O. = 2, 1.5, 1

55. Answer (1)

% of ionic character = obs

cal

cal

µ100; e d

µ

56. Answer (4)

Fact

57. Answer (3)

Fact

58. Answer (3)

XeF2 – 3 lone pair and 2 bond pair

–

3I – 3 lone pair and 2 bond pair

59. Answer (2)

4 3 2NH , NH and NH

○ have 0, 1 and 2 non-bonding

electron pairs respectively on nitrogen atom.

60. Answer (4)

Polarising power depends on charge

size of the cation.

61. Answer (1)

BrF5 contains one lone pair on ‘Br’ atom hence no

bond angle is of 90°.

O

SO O 6 lone pairs are present.

5 4PBr (s) [PBr ] [Br ]

⇒

○

Br

PBrBr

Br

sp3 hybrid central atom and all the

bond lengths are equal.

ClO4

–

:

O

Cl O–

O

Osp

3

∵ No p-orbitals are available on central atom hence

all the 3 bonds are formed by d & p orbitals.

62. Answer (4)

Fact

63. Answer (2)

2 2 2 2

1 2 1

2 2 2 2

2 21 2

E n Z 13.6 2 1

E En Z 1 3 ⇒

2

9E 13.6 30.6eV

4⇒

64. Answer (4)

h

mv

65. Answer (2)

–

–

– –

p pe

pep e e

m .v v 1835v

m .v

66. Answer (4)

67. Answer (1)

H2O has more number of hydrogen bonds than that

of HF.

68. Answer (1)

Esep.

= – EActual state

=

2

2

Z13.6

n


6/10

For H; EActual state

= 1 13.6

13.64 4

For He; EActual state

= 4

13.6 13.64

For Li2+; EActual state

= 9

13.6 13.69

For Be3+; EActual state

= 16

13.6 13.616

Removal of electron from excited H-atom (n = 2)

shall be easiest among them.

69. Answer (4)

h= given

mvv

2 h hv = v

m m ⇒

70. Answer (1)

Mg(g) Mg+(g) + e–; H1=737.76 kJ/mole

Mg+(g) Mg+2(g) + e–; H2=1450.73 kJ/mole

Mg(g) Mg+2(g) + 2e–; H=2188.49 kJ

71. Answer (3)

Fact

72. Answer (2)

Fact

73. Answer (3)

Cl has highest electron affinity value.

74. Answer (2)

75. Answer (1)

I.P. suggest 13th group.

76. Answer (2)

Fact

77. Answer (1)

BeF2 possess vacant p-orbitals on Be atom which

co-ordinates with OH○

ions resulting into formation

of ions which further remains in aqueous state

making BeF2 highly soluble in water.

F Be F + H O2

[BeF (OH) ] + 2H2 2

2–

78. Answer (3)

79. Answer (3)

‘S’ has more electron affinity than ‘O’.

80. Answer (4)

In B2 molecule, valence bonding electrons occupy

-BMOs only.

81. Answer (1)

82. Answer (2)

4th excited state = 5th level

n(n – 1) 5(5 – 1) = = 10

2 2

83. Answer (3)

Frequency

2v Z n

v and r2 n Z

r

2

3

Z

n

2

3He

2H

3

2

322

11

4

84. Answer (2)

2

2 2

1 2

1 1 1v

⎡ ⎤ ⎢ ⎥ ⎢ ⎥⎣ ⎦

R Z

n n for limiting series

n2 =

n1 = 1

85. Answer (4)

70 1.25A 1.25 1

56 1.25 ⇒

30 1.88B 1.875 1.5

16 1.25 ⇒

86. Answer (2)

For 2z

d orbital, m = 0

87. Answer (3)

Lower the ionisation potential smaller will be

threshold frequency of metal.

88. Answer (3)

Fact

89. Answer (3)

5 bonds and 19 bonds

90. Answer (2)

NO (7 + 8e– = 15e–) paramagnetic

NO+ (7 + 8 – 1e– = 14e–) diamagnetic


7/10

91. Answer (3)

92. Answer (2)

93. Answer (3)

94. Answer (3)

95. Answer (3)

Spores of slime mould are dispersed by air currents.

96. Answer (4)

97. Answer (3)

Chrysophytes are photosynthetic.

98. Answer (1)

Protein rich layer – Pellicle

All protozoans are heterotrophs

99. Answer (2)

Entamoeba is an amoeboid protozoan.

100. Answer (4)

Physarum – Heterotroph

101. Answer (2)

Marine forms have silica shell

102. Answer (2)

103. Answer (4)

104. Answer (3)

105. Answer (2)

106. Answer (3)

107. Answer (3)

Sexual reproduction is absent in deuteromycetes.

108. Answer (1)

Asexual spores are endogenously produced.

109. Answer (3)

Zoospores are asexual spores.

110. Answer (4)

Albugo, Mucor.

111. Answer (1)

112. Answer (2)

113. Answer (1)

114. Answer (3)

Lichen do not grow in SO2 polluted area.

Phycobiont provides food and mycobiont provides

shelter.

115. Answer (3)

116. Answer (4)

117. Answer (2)

118. Answer (1)

TMV genetic material – ssRNA

119. Answer (3)

120. Answer (2)

121. Answer (4)

122. Answer (4)

Kingdom Protista

123. Answer (1)

Fruiting body

124. Answer (2)

125. Answer (2)

Monera, Protista and Plantae.

126. Answer (4)

Mycoplasma can survive without oxygen and are

non-motile.

127. Answer (2)

128. Answer (1)

129. Answer (1)

130. Answer (1)

Bacterial structure is very simple but it has complex

behaviour

131. Answer (4)

Actinomycetes (Ray Fungi)

132. Answer (1)

Muscidae – Family

133. Answer (2)

134. Answer (2)

Endospores are formed during unfavourable condition.

135. Answer (2)

Soredia – Involve in asexual reproduction.

136. Answer (1)

– Three germinal layers ecto, meso, endoderm are

formed first in flatworms that belong to phylum

platyhelminthes. Mesoderm is solid or dense in

Platyhelminthes.

– All flatworms can be cut into two identical left

and right halves only in one plane passing

through the central axis.

[ BIOLOGY]


8/10

– In between ecto and endoderm, there is a solid

mesoderm or dense parenchyma which is

involved in transport of nutrients.

137. Answer (2)

Sponges have a special water transport system

through which water, nutrients, oxygen, enter through

ostia and CO2, nitrogenous wastes, water and

sperms leave through single opening called

osculum.

138. Answer (2)

Nerve cells are loosely aggregated but not compactly

arranged to form a distinct brain in coelenterates.

139. Answer (2)

– Taenia solium (Tape worm) has both hooks and

suckers meant for anchorage or attachment to

the surface of a host.

– Fasciola (Liver fluke) has only suckers but no

hooks for attachment.

– Ascaris (Round worm) belongs to aschelminthes

does not possess hooks and suckers.

140. Answer (1)

– Poriferans have minute openings called ostia and

wide openings for exit called oscula.

– Cnidarians have stinging cells in their epidermis

meant for locomotion, food capturing, anchorage,

offense and defense.

– Ctenophores are characterised by the presence

of eight longitudinal or vertical rows of comb

plates with cilia for locomotion.

– In platyhelminthes, since most of the organ

systems are yet to be formed and not found,

therefore, their body is compressed or flattened

dorsoventrally.

141. Answer (3)

Flame cells found in flat worms help in excretion and

osmoregulation.

142. Answer (3)

– In Aschelminthes, their coelom derived from

blastocoel is occupied by scattered and irregular

pouches of mesoderm, therefore called

pseudocoelom.

– Their pharynx is muscular to suck food from the

host but the intestine is non-muscular.

– Organ system level of organisation is present.

143. Answer (2)

– In polychaetes annelids, both parapodia and

setae are found. Parapodia help in swimming

and respiration whereas setae help in locomotion

e.g., Nereis (Clam worm), Chaetopterus (Paddle

worms). Aphrodite (Sea mouse)

– In Pheretima (Earthworm) parapodia are absent,

but setae help in locomotion.

– Aplysia, a mollusc has a muscular foot.

144. Answer (2)

145. Answer (2)

Locust & butterfly respire through trachea.

146. Answer (2)

Chitinous exoskeleton helps in conservation of water.

So it played a major role in the success of insects

on land.

147. Answer (2)

Blood vascular system with heart, dorsal and ventral

vessels is seen for the first time in phylum annelida,

148. Answer (4)

Metamerism is seen in annelids, arthropods and

chordates.

149. Answer (2)

Flat worms are first bilaterally symmetrical animals

with blindsac body plan in which digestive system

ends blindly without a separate opening as anus.

150. Answer (4)

Coelenterates have true tissues, most primitive

diffusible nervous system and sensory cells in body

wall to receive a stimulus.

151. Answer (3)

Digestion in sponges is intracellular.

152. Answer (2)

In most of the echinoderms, protective spines are

found throughout their body surface.

153. Answer (4)

Hemichordates have an open blood vascular system

with a dorsal heart.

154. Answer (2)

Notochord is otherwise known as chorda dorsalis,

which is derived from mesoderm. It is seen only in

Chordates.


9/10

155. Answer (1)

– In Schizocoelomates like annelids, arthropods

and molluscs, a mesoderm is formed between

ecto and endoderm.

– Mesoderm is splitted and confined to both

ectoderm and endoderm. A space found between

ectoderm and endoderm lined by mesoderm on

both sides of coelom.

156. Answer (1)

In parasitic nematodes, their body wall contains a

syncytial epidermis with several nuclei scattered in

a common cytoplasm.

157. Answer (2)

Mammals generally have heterodont dentition with

four types of teeth but in lagomorphs like rabbit,

canines are absent.

158. Answer (3)

There are no alveoli in air sacs of birds.

159. Answer (3)

A cartilaginous skull and vertebral column are found

in cyclostomes and chondrichthyes fishes. It is bony

in osteichthyes, amphibians, reptiles, aves and

mammalia.

160. Answer (2)

Cornified skin with epidermal scales prevents loss of

water from their body.

161. Answer (3)

– Mesonephric kidneys are found in pisces and

amphibians.

– Metanephric kidneys are found in reptiles, aves

and mammals.

162. Answer (3)

163. Answer (4)

– Pavo cristatus is the scientific name of our

national bird, peacock.

– Parrot is scientifically known as Psittacula.

– House sparrow is scientifically known as Passer

domesticus.

– Aptenodytes (Penguin) is a flightless bird in polar

regions.

164. Answer (1)

In amphibians, their body is divided into head and

trunk.

165. Answer (1)

Anadromous migration is seen in Petromyzon or

lamprey migrating from marine water to fresh water

for spawning. Myxine or slime eel or hagfish strictly

inhabits marine water.

166. Answer (4)

Notochord persists throughout life in

cephalochordates, cyclostomes and chondrichthyes.

167. Answer (1)

Retrogressive metamorphosis is seen in all

Urochordates or tunicates e.g., ascidian.

168. Answer (3)

– Urochordates (e.g., Salpa, Ascidia)

– Cephalochordates (e.g., Amphioxus are ciliary

feeders).

– Chondrichthyes fishes (e.g., Scoliodon) is a

vertebrate.

169. Answer (4)

Non-chordates possess double ventral solid nerve

cord, chordates possess dorsal single hollow nerve

cord.

170. Answer (3)

Viviparity is seen in reptiles like Chameleon and

Phrynosoma (Horned toad)

171. Answer (4)

– In flying birds, oil glands or preen glands or

uropygial glands are present at tail base which

lubricates feathers and prevent from being wet.

– In flightless birds, oil or preen or uropygial glands

are absent.

172. Answer (3)

In Torpedo (electric ray), electric organs are modified

dorsal muscles generating a voltage current to shock

or stun the prey or predator.

173. Answer (3)

– Ctenophores or combjellies or sea walnuts or

sea goose berries.

– Echinoderms or spiny skinned animals.

– Hemichordates or worm like marine animals or

tongue worms (or) acorn worms are exclusively

marine in their habitat.

174. Answer (1)

175. Answer (1)

176. Answer (4)

– Anadromous migration is seen in marine water

bony fishes like Salmon, Hilsa migrate for

spawning from marine water to fresh water.

– Catadromous migration is seen in Anguilla (eel)

which moves from fresh water or marine water.


10/10

� � �

– Edible or eatable carps include Labeo rohita

(Rohu), Labeo calabasu (Calabasu), Catla catla

(Thiela or catla).

– Cat fishes living in turbid water include Clarias

batrachus (Indian catfish or magur), Wallago attu

(Malli).

– Man eating shark is Carcharodon or great white

shark.

– Latimeria (coelacanth) is living fossil fish. It is an

oldest living fish surviving today without

undergoing any change during long evolutionary

time period.

177. Answer (1)

In Caecilians or limbless amphibians or blind worms,

have scales in dermis of the skin.

178. Answer (4)

– Urinary bladder is present in lizards but absent

in snakes.

– A hemipenis or paired penis is found in snakes

and lizards.

179. Answer (1)

Balaenoptera (Blue whale) being measured a weight

about 1,36,000 kgs (136 tonnes), it is considered as

the largest animal.

180. Answer (2)

Flying mammals include Pteropus (Flying fox) and

Vespertilio (Vampire bat).

Test - 1 (Code-F) (Answers) All India Aakash Test Series for Medical-2018

1/10

1. (1)

2. (4)

3. (1)

4. (2)

5. (3)

6. (4)

7. (3)

8. (2)

9. (3)

10. (1)

11. (1)

12. (2)

13. (4)

14. (3)

15. (2)

16. (2)

17. (2)

18. (2)

19. (1)

20. (4)

21. (3)

22. (4)

23. (2)

24. (2)

25. (1)

26. (2)

27. (4)

28. (1)

29. (4)

30. (4)

31. (4)

32. (2)

33. (4)

34. (1)

35. (1)

36. (4)

Test Date : 26/11/2017

ANSWERS

TEST - 1 (Code F)

All India Aakash Test Series for Medical - 2018

37. (3)

38. (2)

39. (1)

40. (1)

41. (2)

42. (2)

43. (2)

44. (4)

45. (2)

46. (2)

47. (3)

48. (3)

49. (3)

50. (2)

51. (4)

52. (2)

53. (3)

54. (2)

55. (1)

56. (4)

57. (3)

58. (3)

59. (1)

60. (2)

61. (1)

62. (2)

63. (3)

64. (2)

65. (3)

66. (1)

67. (4)

68. (1)

69. (1)

70. (4)

71. (2)

72. (4)

73. (2)

74. (4)

75. (1)

76. (4)

77. (2)

78. (3)

79. (3)

80. (4)

81. (1)

82. (3)

83. (2)

84. (3)

85. (2)

86. (1)

87. (4)

88. (3)

89. (4)

90. (3)

91. (2)

92. (2)

93. (2)

94. (1)

95. (4)

96. (1)

97. (1)

98. (1)

99. (2)

100. (4)

101. (2)

102. (2)

103. (1)

104. (4)

105. (4)

106. (2)

107. (3)

108. (1)

109. (2)

110. (4)

111. (3)

112. (3)

113. (1)

114. (2)

115. (1)

116. (4)

117. (3)

118. (1)

119. (3)

120. (3)

121. (2)

122. (3)

123. (4)

124. (2)

125. (2)

126. (4)

127. (2)

128. (1)

129. (3)

130. (4)

131. (3)

132. (3)

133. (3)

134. (2)

135. (3)

136. (2)

137. (1)

138. (4)

139. (1)

140. (4)

141. (1)

142. (1)

143. (3)

144. (3)

145. (4)

146. (3)

147. (4)

148. (3)

149. (1)

150. (4)

151. (1)

152. (1)

153. (4)

154. (3)

155. (3)

156. (2)

157. (3)

158. (3)

159. (2)

160. (1)

161. (1)

162. (2)

163. (4)

164. (2)

165. (3)

166. (4)

167. (2)

168. (4)

169. (2)

170. (2)

171. (2)

172. (2)

173. (2)

174. (3)

175. (3)

176. (1)

177. (2)

178. (2)

179. (2)

180. (1)

All India Aakash Test Series for Medical-2018 Test - 1 (Code-F) (Answers & Hints)

2/10

ANSWERS & HINTS

1. Answer (1)

22

2 2

1 1

c

a vva

R a v

⎛ ⎞ ⇒ ⎜ ⎟

⎝ ⎠

2. Answer (4)

3. Answer (1)

uH = constant

u cos60°= ucos30°

10'

3u⇒ m/s

4. Answer (2)

2 2

x ya a a

5. Answer (3)

4tan

n

; when R = nH

6. Answer (4)

|Displacement| Distance

7. Answer (3)

BA B AV V V � � �

BA

100cos5 hr

| |t

v

A

100 kmvB = 10 km/h

B

10 2

Path of w.r.t B A

8. Answer (2)

When R = xH, 2

2

16

2 16

uH

g x

Put x = 3 0 52

gu⇒

[ PHYSICS]

9. Answer (3)

10. Answer (1)

45° 2 2

VRM

VM

2

2

VR

RM R Mv v v � � �

R RM Mv v v � � �

11. Answer (1)

If = constant

H u2

R u2

R% = H%

12. Answer (2)

13. Answer (4)

40 m/sH

Rv

T

1 2

2 1

sin 3

sin 2

u

u

14. Answer (3)

2sin2 u

Rg

15. Answer (2)

2 21 1

2 22 2

cos cos 30

cos cos 60

H

H

= 3 : 1

16. Answer (2)

vav (speed) = 0

vav (velocity) = Total displacement

Total time taken

17. Answer (2)

2 2 2

1 2

sin;

2 2

u uh h H

g g

⎛ ⎞ ⎜ ⎟⎜ ⎟⎝ ⎠

Test - 1 (Code-F) (Answers & Hints) All India Aakash Test Series for Medical-2018

3/10

18. Answer (2)

v u at ��

19. Answer (1)

20. Answer (4)

1 1 1ˆ ˆ ˆA a i b j c k

�

2 2 2ˆ ˆ ˆB a i b j c k

�

If ||A B� �

then 1 1 1

2 2 2

ve

a b c

a b c

21. Answer (3)

A2 + B2 + 2ABcos = A2 + B2 – 2AB

= 90°

22. Answer (4)

23. Answer (2)

602 sin 2 20 sin 20 m/s

2 2v v

24. Answer (2)

v cos

v

v sinr

d = 50 m

B

2cos cos 20 1

50 2cos

v v v

dr d

= 0.2 rad/s

25. Answer (1)

h v2

2

1

2h

h

26. Answer (2)

27. Answer (4)

Slope is rate of change of speed = tangential

acceleration

Upward motion

at = –(g + kv)

v at

Downward motion

at = g – kv

v at

28. Answer (1)

21

2S at

1 1

2 2

t ht S

t h ⇒

29. Answer (4)

2u b

a c

30. Answer (4)

2 2 2 2 23

x yv v v t

31. Answer (4)

max

2 412

2 4v T

⎛ ⎞ ⎜ ⎟ ⎝ ⎠= 16 m/s

32. Answer (2)

6 5a t

2

26 5 d St

dt

23 5 dSt t

dt

4

2

0

(3 5 ) 104 mS t t dt ∫

33. Answer (4)

Kads

m

∫

34. Answer (1)

x = 12t – t3

v = 2

12 3dx

tdt

0 = 12 – 3t2

t = 2 s

x (t = 2 s) = 12(2) – 23 = 24 – 8 = 16 m

x (t = 0 s) = 0

35. Answer (1)

100 100 100I V R

I V R

0.1 0.2100 100

20 80 = 0.5 + 0.25= 0.75%


4/10

46. Answer (2)

NO (7 + 8e– = 15e–) paramagnetic

NO+ (7 + 8 – 1e– = 14e–) diamagnetic

47. Answer (3)

5 bonds and 19 bonds

48. Answer (3)

Fact

49. Answer (3)

Lower the ionisation potential smaller will be

threshold frequency of metal.

50. Answer (2)

For 2z

d orbital, m = 0

51. Answer (4)

70 1.25A 1.25 1

56 1.25 ⇒

30 1.88B 1.875 1.5

16 1.25 ⇒

[ CHEMISTRY]

36. Answer (4)

13 2

2

X a b c d

X a b c d

13 0.1% 2 0.2% 0.4% 0.4%

2 = 1.3%

37. Answer (3)

1 2

2[M L T ]

aP

V

38. Answer (2)

1 2

1 110 4 9 180 m

2 2h gt t

39. Answer (1)

1/2P A

ET

40. Answer (1)

2 1[ ] [M L T ]E

E h h h ⇒ ⇒

41. Answer (2)

At t = 4.5 s, it is at highest point. At 4 s and 5 s,

it is 0.5 s below the highest point so

Distance = 21

2 0.5 10 0.25 2.5m2g

⎡ ⎤ ⎢ ⎥⎣ ⎦

42. Answer (2)

43. Answer (2)

44. Answer (4)

180° > > 90°

a ta

v

ac

45. Answer (2)

1 1 2

1 1 12 1

2 2 2

M L Tn n

M L T

⎛ ⎞ ⎛ ⎞ ⎛ ⎞

⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

21 1 1

110 10 10

⎛ ⎞ ⎜ ⎟⎝ ⎠

= 1

52. Answer (2)

2

2 2

1 2

1 1 1v

⎡ ⎤ ⎢ ⎥ ⎢ ⎥⎣ ⎦

R Z

n n for limiting series

n2 = n1 = 1

53. Answer (3)

Frequency

2v Z n

v and r2 n Z

r

2

3

Z

n

2

3He

2H

3

2

322

11

4

54. Answer (2)

4th excited state = 5th level

n(n – 1) 5(5 – 1) = = 10

2 2

55. Answer (1)


5/10

56. Answer (4)

In B2 molecule, valence bonding electrons occupy

-BMOs only.

57. Answer (3)

‘S’ has more electron affinity than ‘O’.

58. Answer (3)

59. Answer (1)

BeF2 possess vacant p-orbitals on Be atom which

co-ordinates with OH○

ions resulting into formation

of ions which further remains in aqueous state

making BeF2 highly soluble in water.

F Be F + H O2

[BeF (OH) ] + 2H2 2

2–

60. Answer (2)

Fact

61. Answer (1)

I.P. suggest 13th group.

62. Answer (2)

63. Answer (3)

Cl has highest electron affinity value.

64. Answer (2)

Fact

65. Answer (3)

Fact

66. Answer (1)

Mg(g) Mg+(g) + e–; H1=737.76 kJ/mole

Mg+(g) Mg+2(g) + e–; H2=1450.73 kJ/mole

Mg(g) Mg+2(g) + 2e–; H=2188.49 kJ

67. Answer (4)

h= given

mvv

2 h hv = v

m m ⇒

68. Answer (1)

Esep. = – EActual state =

2

2

Z13.6

n

For H; EActual state = 1 13.6

13.64 4

For He; EActual state = 4

13.6 13.64

For Li2+; EActual state = 9

13.6 13.69

For Be3+; EActual state = 16

13.6 13.616

Removal of electron from excited H-atom (n = 2)

shall be easiest among them.

69. Answer (1)

H2O has more number of hydrogen bonds than that

of HF.

70. Answer (4)

71. Answer (2)

–

–

– –

p pe

pep e e

m .v v 1835v

m .v

72. Answer (4)

h

mv

73. Answer (2)

2 2 2 2

1 2 1

2 2 2 2

2 21 2

E n Z 13.6 2 1

E En Z 1 3 ⇒

2

9E 13.6 30.6eV

4⇒

74. Answer (4)

Fact

75. Answer (1)

BrF5 contains one lone pair on ‘Br’ atom hence no

bond angle is of 90°.

O

SO O 6 lone pairs are present.

5 4PBr (s) [PBr ] [Br ]

⇒

○

Br

PBrBr

Br

sp3 hybrid central atom and all the

bond lengths are equal.

ClO4

–

:

O

Cl O–

O

Osp

3

∵ No p-orbitals are available on central atom hence

all the 3 bonds are formed by d & p orbitals.


6/10

76. Answer (4)

Polarising power depends on charge

size of the cation.

77. Answer (2)

4 3 2NH , NH and NH

○ have 0, 1 and 2 non-bonding

electron pairs respectively on nitrogen atom.

78. Answer (3)

XeF2 – 3 lone pair and 2 bond pair

–

3I – 3 lone pair and 2 bond pair

79. Answer (3)

Fact

80. Answer (4)

Fact

81. Answer (1)

% of ionic character = obs

cal

cal

µ100; e d

µ

82. Answer (3)

O2, O3, O22–

B.O. = 2, 1.5, 1

83. Answer (2)

Fact – sp3 hybridization

84. Answer (3)

C3H8 + 5O2 3CO2 + 4H2O(l)

1 : 5 : 3

20 ml 100 ml 60 ml of CO2 formed

80 ml left unreacted 80 + 60 = 140 ml

85. Answer (2)

H2SO4 + Na2CO3 Na2SO4 + H2O + CO2

98 g 106 g

% purity = cal. mass

×100expt. mass

expt. mass = 98 100

122.5 gm80

86. Answer (1)

(1 2

2O contains 18e– + 16p + 16n = 50)

32 g = 50 NA

16 g = 25 NA

87. Answer (4)

∵ 2 g oxygen combines with 3 g of metal

8 g oxygen combines with 12 g of metal

EMO = EM + 8 = 12 + 8 = 20 g

eq.MO

(ml)

n1000 1000M n

V n.f 500

0.030.01 2 n.f 6

n.f ⇒

Molar mass of metal oxide = 20 × 6 = 120 g

88. Answer (3)

1 L of H2O = 1000 gm = 1000

18 = 55.5 mole

= 55.5 × 6 × 1023 × 3 atoms

22.4 L of CO2 = 1 mole = 3 × 6 × 1023 atoms

44.8 L of O2 = 2 mole = 2 × 2 × 6 × 1023 atoms

11.2 L of SO3 = 0.5 mole = 0.5 × 4 × 6 × 1023 atoms

89. Answer (4)

Na2CO3 does not decompose on heating.

CaCO3 CaO + CO2

100 gm 22.4 L

50 gm 11.2 L

mass of CaCO3 = 50 g

mass of Na2CO3 = 50 g

90. Answer (3)

A

B B

1000 0.25 1000m 7.246 molal

M 0.75 46

[ BIOLOGY]

91. Answer (2)

Soredia – Involve in asexual reproduction.

92. Answer (2)

Endospores are formed during unfavourable condition.

93. Answer (2)

94. Answer (1)

Muscidae – Family

95. Answer (4)

Actinomycetes (Ray Fungi)

96. Answer (1)

Bacterial structure is very simple but it has complex

behaviour


7/10

97. Answer (1)

98. Answer (1)

99. Answer (2)

100. Answer (4)

Mycoplasma can survive without oxygen and are

non-motile.

101. Answer (2)

Monera, Protista and Plantae.

102. Answer (2)

103. Answer (1)

Fruiting body

104. Answer (4)

Kingdom Protista

105. Answer (4)

106. Answer (2)

107. Answer (3)

108. Answer (1)

TMV genetic material – ssRNA

109. Answer (2)

110. Answer (4)

111. Answer (3)

112. Answer (3)

Lichen donot grow in SO2 polluted area.

Phycobiont provides food and mycobiont provides

shelter.

113. Answer (1)

114. Answer (2)

115. Answer (1)

116. Answer (4)

Albugo, Mucor.

117. Answer (3)

Zoospores are asexual spores.

118. Answer (1)

Asexual spores are endogenously produced.

119. Answer (3)

Sexual reproduction is absent in deuteromycetes.

120. Answer (3)

121. Answer (2)

122. Answer (3)

123. Answer (4)

124. Answer (2)

125. Answer (2)

Marine forms have silica shell

126. Answer (4)

Physarum – Heterotroph

127. Answer (2)

Entamoeba is an amoeboid protozoan.

128. Answer (1)

Protein rich layer – Pellicle

All protozoans are heterotrophs

129. Answer (3)

Chrysophytes are photosynthetic.

130. Answer (4)

131. Answer (3)

Spores of slime mould are dispersed by air currents.

132. Answer (3)

133. Answer (3)

134. Answer (2)

135. Answer (3)

136. Answer (2)

Flying mammals include Pteropus (Flying fox) and

Vespertilio (Vampire bat).

137. Answer (1)

Balaenoptera (Blue whale) being measured a weight

about 1,36,000 kgs (136 tonnes), it is considered as

the largest animal.

138. Answer (4)

– Urinary bladder is present in lizards but absent

in snakes.

– A hemipenis or paired penis is found in snakes

and lizards.

139. Answer (1)

In Caecilians or limbless amphibians or blind worms,

have scales in dermis of the skin.

140. Answer (4)

– Anadromous migration is seen in marine water

bony fishes like Salmon, Hilsa migrate for

spawning from marine water to fresh water.

– Catadromous migration is seen in Anguilla (eel)

which moves from fresh water or marine water.


8/10

– Edible or eatable carps include Labeo rohita

(Rohu), Labeo calabasu (Calabasu), Catla catla

(Thiela or catla).

– Cat fishes living in turbid water include Clarias

batrachus (Indian catfish or magur), Wallago attu

(Malli).

– Man eating shark is Carcharodon or great white

shark.

– Latimeria (coelacanth) is living fossil fish. It is an

oldest living fish surviving today without

undergoing any change during long evolutionary

time period.

141. Answer (1)

142. Answer (1)

143. Answer (3)

– Ctenophores or combjellies or sea walnuts or

sea goose berries.

– Echinoderms or spiny skinned animals.

– Hemichordates or worm like marine animals or

tongue worms (or) acorn worms are exclusively

marine in their habitat.

144. Answer (3)

In Torpedo (electric ray), electric organs are modified

dorsal muscles generating a voltage current to shock

or stun the prey or predator.

145. Answer (4)

– In flying birds, oil glands or preen glands or

uropygial glands are present at tail base which

lubricates feathers and prevent from being wet.

– In flightless birds, oil or preen or uropygial glands

are absent.

146. Answer (3)

Viviparity is seen in reptiles like Chameleon and

Phrynosoma (Horned toad)

147. Answer (4)

Non chordates possess double ventral solid nerve

cord, chordates possess dorsal single hollow nerve

cord.

148. Answer (3)

– Urochordates (e.g., Salpa, Ascidia)

– Cephalochordates (e.g., Amphioxus are ciliary

feeders).

– Chondrichthyes fishes (e.g., Scoliodon) is a

vertebrate.

149. Answer (1)

Retrogressive metamorphosis is seen in all

Urochordates or tunicates e.g., ascidian.

150. Answer (4)

Notochord persists throughout life in

cephalochordates, cyclostomes and chondrichthyes.

151. Answer (1)

Anadromous migration is seen in Petromyzon or

lamprey migrating from marine water to fresh water

for spawning. Myxine or slime eel or hagfish strictly

inhabits marine water.

152. Answer (1)

In amphibians, their body is divided into head and

trunk.

153. Answer (4)

– Pavo cristatus is the scientific name of our

national bird, peacock.

– Parrot is scientifically known as Psittacula.

– House sparrow is scientifically known as Passer

domesticus.

– Aptenodytes (Penguin) is a flightless bird in polar

regions.

154. Answer (3)

155. Answer (3)

– Mesonephric kidneys are found in pisces and

amphibians.

– Metanephric kidneys are found in reptiles, aves

and mammals.

156. Answer (2)

Cornified skin with epidermal scales prevents loss of

water from their body.

157. Answer (3)

A cartilaginous skull and vertebral column are found

in cyclostomes and chondrichthyes fishes. It is bony

in osteichthyes, amphibians, reptiles, aves and

mammalia.

158. Answer (3)

There are no alveoli in air sacs of birds.

159. Answer (2)

Mammals generally have heterodont dentition with

four types of teeth but in lagomorphs like rabbit,

canines are absent.

160. Answer (1)

In parasitic nematodes, their body wall contains a

syncytial epidermis with several nuclei scattered in

a common cytoplasm.


9/10

161. Answer (1)

– In Schizocoelomates like annelids, arthropods

and molluscs, a mesoderm is formed between

ecto and endoderm.

– Mesoderm is splitted and confined to both

ectoderm and endoderm. A space found between

ectoderm and endoderm lined by mesoderm on

both sides of coelom.

162. Answer (2)

Notochord is otherwise known as chorda dorsalis,

which is derived from mesoderm. It is seen only in

Chordates.

163. Answer (4)

Hemichordates have an open blood vascular system

with a dorsal heart.

164. Answer (2)

In most of the echinoderms, protective spines are

found throughout their body surface.

165. Answer (3)

Digestion in sponges is intracellular.

166. Answer (4)

Coelenterates have true tissues, most primitive

diffusible nervous system and sensory cells in body

wall to receive a stimulus.

167. Answer (2)

Flat worms are first bilaterally symmetrical animals

with blindsac body plan in which digestive system

ends blindly without a separate opening as anus.

168. Answer (4)

Metamerism is seen in annelids, arthropods and

chordates.

169. Answer (2)

Blood vascular system with heart, dorsal and ventral

vessels is seen for the first time in phylum annelida,

170. Answer (2)

Chitinous exoskeleton helps in conservation of water.

So it played a major role in the success of insects

on land.

171. Answer (2)

Locust & butterfly respire through trachea.

172. Answer (2)

173. Answer (2)

– In polychaetes annelids, both parapodia and

setae are found. Parapodia help in swimming

and respiration whereas setae help in locomotion

e.g., Nereis (Clam worm), Chaetopterus (Paddle

worms). Aphrodite (Sea mouse)

– In Pheretima (Earthworm) parapodia are absent,

but setae help in locomotion.

– Aplysia, a mollusc has a muscular foot.

174. Answer (3)

– In Aschelminthes, their coelom derived from

blastocoel is occupied by scattered and irregular

pouches of mesoderm, therefore called

pseudocoelom.

– Their pharynx is muscular to suck food from the

host but the intestine is non-muscular.

– Organ system level of organisation is present.

175. Answer (3)

Flame cells found in flat worms help in excretion and

osmoregulation.

176. Answer (1)

– Poriferans have minute openings called ostia and

wide openings for exit called oscula.

– Cnidarians have stinging cells in their epidermis

meant for locomotion, food capturing, anchorage,

offense and defense.

– Ctenophores are characterised by the presence

of eight longitudinal or vertical rows of comb

plates with cilia for locomotion.

– In platyhelminthes, since most of the organ

systems are yet to be formed and not found,

therefore, their body is compressed or flattened

dorsoventrally.

177. Answer (2)

– Taenia solium (Tape worm) has both hooks and

suckers meant for anchorage or attachment to

the surface of a host.

– Fasciola (Liver fluke) has only suckers but no

hooks for attachment.

– Ascaris (Round worm) belongs to aschelminthes

does not possess hooks and suckers.


10/10

� � �

178. Answer (2)

Nerve cells are loosely aggregated but not compactly

arranged to form a distinct brain in coelenterates.

179. Answer (2)

Sponges have a special water transport system

through which water, nutrients, oxygen, enter through

ostia and CO2, nitrogenous wastes, water and

sperms leave through single opening called

osculum.

180. Answer (1)

– Three germinal layers ecto, meso, endoderm are

formed first in flatworms that belong to phylum

platyhelminthes. Mesoderm is solid or dense in

Platyhelminthes.

– All flatworms can be cut into two identical left

and right halves only in one plane passing

through the central axis.

– In between ecto and endoderm, there is a solid

mesoderm or dense parenchyma which is

involved in transport of nutrients.

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