for medical-2018 all india aakash test series for medical ... · all india aakash test series for...

Test - 3 (Code-A) (Answers & Hints) All India Aakash Test Series for Medical-2018

1/11

1. (4)

2. (4)

3. (1)

4. (4)

5. (4)

6. (1)

7. (3)

8. (1)

9. (4)

10. (3)

11. (4)

12. (1)

13. (1)

14. (1)

15. (1)

16. (3)

17. (3)

18. (4)

19. (1)

20. (4)

21. (3)

22. (4)

23. (3)

24. (3)

25. (4)

26. (1)

27. (3)

28. (3)

29. (4)

30. (3)

31. (3)

32. (4)

33. (3)

34. (2)

35. (1)

36. (3)

Test Date : 12/11/2017

ANSWERS

TEST - 3 (Code A)

All India Aakash Test Series for Medical - 2018

37. (4)

38. (4)

39. (3)

40. (3)

41. (3)

42. (3)

43. (1)

44. (1)

45. (3)

46. (2)

47. (4)

48. (2)

49. (3)

50. (1)

51. (2)

52. (1)

53. (1)

54. (3)

55. (3)

56. (1)

57. (1)

58. (3)

59. (3)

60. (3)

61. (4)

62. (2)

63. (4)

64. (4)

65. (3)

66. (2)

67. (2)

68. (1)

69. (1)

70. (3)

71. (4)

72. (3)

73. (3)

74. (2)

75. (1)

76. (2)

77. (3)

78. (2)

79. (2)

80. (3)

81. (3)

82. (2)

83. (4)

84. (3)

85. (3)

86. (3)

87. (4)

88. (3)

89. (2)

90. (2)

91. (3)

92. (3)

93. (4)

94. (2)

95. (1)

96. (3)

97. (2)

98. (2)

99. (3)

100. (3)

101. (3)

102. (4)

103. (4)

104. (4)

105. (2)

106. (2)

107. (2)

108. (2)

109. (3)

110. (3)

111. (1)

112. (2)

113. (4)

114. (3)

115. (2)

116. (1)

117. (4)

118. (2)

119. (4)

120. (4)

121. (3)

122. (2)

123. (1)

124. (2)

125. (4)

126. (3)

127. (1)

128. (3)

129. (4)

130. (2)

131. (1)

132. (3)

133. (2)

134. (3)

135. (2)

136. (3)

137. (2)

138. (4)

139. (4)

140. (3)

141. (1)

142. (1)

143. (3)

144. (4)

145. (3)

146 (3)

147. (4)

148. (4)

149. (2)

150. (3)

151. (4)

152. (2)

153. (4)

154. (2)

155. (4)

156. (2)

157. (3)

158. (2)

159. (4)

160. (2)

161. (3)

162. (1)

163. (2)

164. (3)

165. (4)

166. (3)

167. (1)

168. (3)

169. (2)

170. (2)

171. (2)

172. (3)

173. (2)

174. (2)

175. (2)

176. (1)

177. (4)

178. (1)

179. (4)

180. (1)

All India Aakash Test Series for Medical-2018 Test - 3 (Code-A) (Answers & Hints)

2/11

ANSWERS & HINTS

1. Answer (4)

An air bubble behaves as a diverging lens inside

water. So it forms a virtual image.

2. Answer (4)

1 2( )i i A

2 230 (60 ) 30 0i i ⇒

Also: 1

1

sin( ) sin603

sin( ) sin30

i

r

3. Answer (1)

When tube length is decreased, the (real)

intermediate image formed by the objective will lie

between the eyepiece and its focus. This will cause

a virtual image to be formed.

4. Answer (4)

22

1

2min

m 1

2

131 142

1: 253 25

112 4

ax

I

II

I I

I

⎛ ⎞ ⎛ ⎞⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ ⎠ ⎜ ⎟⎜ ⎟ ⎛ ⎞⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠ ⎝ ⎠⎝ ⎠

5. Answer (4)

i

r1

r1

r2

r2

r3

For 1st refraction, sini × 1 = sinr1×

1

For 2nd refraction, sinr1 ×

1 = sinr

2 ×

2

For 3rd refraction, sinr2 ×

2 = sinr

3 ×

3

1

3

3

sinsin

⎛ ⎞ ⎜ ⎟⎝ ⎠

ir

6. Answer (1)

Position of image of O w.r.t fish will be AO

= 2

11

2

xx

⎛ ⎞

⎜ ⎟⎝ ⎠

x

y

A

oo

F

5 ms–1

2 ms–1

1 = 2

2

= 4/3

Now let relative separation is equal to r.

Then r = 2

1

x y

dr

dt =

2

1

dx dy

dt dt

= 2

1

(2) 5

= 6.33 ms–1

7. Answer (3)

In figure, CA = 20 cm = R

Using Sine law,

30°

MC

30°

120°

A

CA CM

sin120 sin30

CM = 20

3

[ PHYSICS]


3/11

8. Answer (1)

0

f

y

A

tan , 45 , 10 cmy

ff

y = 10 cm

9. Answer (4)

(i)

10 cm 5 cm

(ii)

10 cm

10. Answer (3)

Because image forms at infinity fo + f

e = 40 cm and

o

e

f

f= 8

Solving, 320

9of cm and

40

9ef cm

11. Answer (4)

Intensity of light from 1st polarising sheet =0

12

Ii

Intensity of light from 2nd sheet,

20 02 cos 45

2 4

I Ii

Intensity of light from 3rd sheet,

20 0

3 cos 454 8

I Ii

Intensity of final transmitted light from last sheet,

20 0

4cos (45)

8 16

I Ii

12. Answer (1)

Locus of a point having a given path difference of

light from two sources on the screen will be a circle

13. Answer (1)

y = ( 1)D n D

td d

n = 2

t

14. Answer (1)

Intensity of nth order secondary maxima

0

2

4

2 1

⎡ ⎤ ⎣ ⎦

n

Il

n

15. Answer (1)

Kinetic nergy of particle

Energy of photon

e

=

21

2mv

hc

=

21

2

hv

v

hc

⎛ ⎞⎜ ⎟⎝ ⎠

8

8

2 10 1

2 32 3 10

v

c

16. Answer (3)

Photoelectric emission rate intensity 1

r

(for

linear source), maximum kinetic energy does not

depend upon intensity of light. Hence, kinetic

energy = k, and emission become three times.

17. Answer (3)

1hf k and

25hf k

Solving for , = 15

3 10h .

18. Answer (4)

=

2

h

mqV

, P

P P

m q

m q

= 8

1= 2 2 :1

19. Answer (1)

( )ms T n hf

26

34 10

2 4200 16 10

7 10 2 10

ms Tn

hf


4/11

20. Answer (4)

Making the incident frequency one third, means

making the incident frequency less than Threshold

frequency.

21. Answer (3)

T1

T+ m g21

m g2

T w c2+2 /

For 1 1 2 1, m T T m g

For 2 2 2

2, w

m m g Tc

Solving, 1 1 2

2= m w

T g m gc

22. Answer (4)

V0 =

1hc

e e

⎛ ⎞ ⎜ ⎟ ⎝ ⎠ m =

hc

e

Nature must be straight line.

23. Answer (3)

24. Answer (3)

In Davisson and Germer experiment

12.27 V

New, 12.27 12.27 20

.215

100

V

V V

25. Answer (4)

1 12 2

n n

nh hmv r v r

m

⇒

26. Answer (1)

1 1n n nr r r

2 220 0

0

( 1) ( 1)

r n r nnr

Z Z Z

24 0 4 ⇒ n n n

22

13.6 0.8516

n

ZE Z

27. Answer (3)

We know that

2 2

2 2 2 2

1 1 ( 1)

( 1) ( 1)

E n nf

h n n n n

⎡ ⎤⎡ ⎤ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

2 2 3

(2 1) 1

( 1)

n

n n n

[ 1]n ∵

28. Answer (3)

2 19, 5n n

2 1 2 1( )( 1)

2

(9 5)(9 5 1)

2

10

n n n nN

29. Answer (4)

30. Answer (3)

1

0

tR R e

and 2

0

1 tR e

R

Dividing

2 1( )2 t t

R e

2 12ln( ) ( )R t t

2 1( )1

2ln( )avg

t tT

R

31. Answer (3)

1 1 3

2000 400 1000

Now, 0 0 0

5

6

tN N N N e

⎛ ⎞ ⎜ ⎟⎝ ⎠

ln6597t t⇒ ⇒

years.

32. Answer (4)

33. Answer (3)

N increases exponentially, N = N0(1 – e–t)


5/11

34. Answer (2)

K1

K2

1 2K K E ,

Also 1 2

P P

1 22 250 50

M MK M K

⎛ ⎞⇒ ⎜ ⎟

⎝ ⎠

1 2 1 149 49( ) 49 49K k E K E K⇒

1

49

50

EK⇒

35. Answer (1)

The angular momentum is

2

hL n

or 2 L

nh

Then, the energy of the electron is

2 4

2 2 208

mZ eE

n h

2 4 2 4

2 2 2 22 2 00

3228

⎛ ⎞ ⎜ ⎟

⎝ ⎠

mZ e mZ e

LLh

h

36. Answer (3)

3 3

6 3

1.5 10 4 1060

50 10 2 10

outgain ac

in

RV

R

37. Answer (4)

38. Answer (4)

39. Answer (3)

VCC

– 0 = IBR

B + V

BE

VCC

= IBR

B(∵ V

BE = 0)

So IB = Base current

= 3

8

500 10

CC

B

V

R

= 16A

Now, IC = I

B = 1.28 mA

40. Answer (3)

T/2 Tt

E

avg rms rmsP v i =

2

rms

4

v

R=

2

04E

R

02

[ ]2

rms

EE ∵

41. Answer (3)

42. Answer (3)

Output voltage across Zener diode will be constant.

So potential difference across RL will be 10 V.

Current, I = 3

10

10 = 10 mA

43. Answer (1)

2 resistor connected to diode can be treated as

shortcircuit. So Req

= 2

23 =

8

3

Ibattery

= 3

8

V

VAB

= 3 2

8 3 4

V V

44. Answer (1)

Truth table is identical to that of AND gate

Also A B = AB

45. Answer (3)

The circuit can be redrawn

I = 2 1

A12 6

I

2 V 1 10

1


6/11

46. Answer (2)

Number of chiral centre in aldoheptose will be 5 and

that in ketoheptose will be 4, so that

5

4

a (2) 2

b 1(2)

47. Answer (4)

CH3 CH

OH

CH2

CH2 OH

PCC

CH3 C

O

CH2

CHO

(A)

HOHO

CH3 C

O

CH2

CH

(B)

O

O

MeMgBr,H O

3

+

CH3 C CH

2 CHO

CH3

OH (C)

CH3 C

CH3

OH

NaBH4EtOH

CH2

CH2 OH

(D)

48. Answer (2)

3CH3OH + Al (CH

3O)

3Al + 2

3H (g)

2

CH3OH is limiting reagent.

3 mole of CH3OH gives 2

3moles of H (g)

2

1.5 mole of CH3OH gives

3 11.5

2 3 =

3

4mol

49. Answer (3)

Benzyl halides are more reactive than Aryl halides

due to more stability of benzyl carbocation /

intermediate. Moreover, reactivity of aryl halides is

increased by EWG like –NO2 so that order is

II > III > IV > I.

50. Answer (1)

Common antihistamines are brompheniramine

(Dimetapp) and terfenadine (seldane).

51. Answer (2)

CH3CH MgBr + CH

2 3 C

O

Cl

CH3 C

O

CH2

CH + Mg3

Br

Cl

CH2

C

OH

CH2

CH3 CH

3C

OH

CH CH3

enol forms

Tautomers

52. Answer (1)

C6H

12O

6

Zymase 2C2H

5OH + 2CO

2

Glucose/Fructose

Last step during commercial preparation of alcohol

from fermentation of sugar.

53. Answer (1)

54. Answer (3)

I

NBS

Allylic halogenation

I

CH2

Br

I

CH2

S CH3

+ NaBr

CH3

S 2N Na S CH

+

3

–

55. Answer (3)

LDP is obtained by polymerisation of ethene under

high pressure of 1000–2000 atm. and at high

temperature of 350 K to 570 K in presence of

peroxide initiator.

56. Answer (1)

O

HCN

HO C N

LiAlH4

HO CH2

NH2

57. Answer (1)

C C

F

Cl

Br

FC C

Cl

Br

F

FC C

Cl

F

Br

F

Geometrical isomers

[ CHEMISTRY]


7/11

58. Answer (3)

Biuret test : When alkaline solution of protein is

treated with a drop of aqueous copper sulphate, a

bluish violet colour is obtained.

59. Answer (3)

HOCH CH OH + n HO22

n

Ethylene glycol

C

O

C

O

OH

Phthalic acid

Polymerisation

O CH2

CH2

O C C

n

Glyptal

O O

60. Answer (3)

NPh2

HNO2

–H2O

NPh2

N O3° aromatic amine

p-Nitroso derivative

61. Answer (4)

(Fact)

Various bacteriostatic antibiotic are erythromycin,

tetracycline, chloramphenicol.

62. Answer (2)

A = CH3

C CH3

OKetone

B = CH3

CHOCH2

Aldehyde

Both can be differentiated by Tollen’s reagent.

63. Answer (4)

(A)

C

COOH

O

64. Answer (4)

No ortho effect is found in o-Methylphenol. It is less

acidic due to electron donating nature of methyl

group.

65. Answer (3)

A = C6H

5MgCl

B = C H6 5

C OMgCl

O

C = C6H

5COOH

66. Answer (2)

—COOH group present on bridge head C does not

take part in any reaction, -ketoacid is always very

difficult to undergo decarboxylation.

67. Answer (2)

Intermediate is cumene hydroperoxide

CH3

CH

3C

O O H

68. Answer (1)

2

4

4

Pd/H

2LiAlH ,1 alcoholNaBH

RCHO R—CH OH

69. Answer (1)

COOH

OH

70. Answer (3)

F

NO2

CH3

NaOH

(aq)

OH

NO2

CH3

(A)

—NO at p-position

favours attack of nucleophile

2

F

NO2

CH3

NaOH

(aq)—CH at p-position

unfavourable for attackof nucleophile.

3No reaction

71. Answer (4)

Higher-I effect, more will be acidic strength.

72. Answer (3)

Weak oxidising agents like PCC, PDC etc. oxidise

1° alcohol to aldehyde.


8/11

73. Answer (3)

A =

NO2

B =

NH2

C =

NHCOCH3

D =

NHCOCH3

Br

74. Answer (2)

75. Answer (1)

76. Answer (2)

A =

NH2

OH

B =

N2Cl

OH

–

C =

OH

77. Answer (3)

N

N

N

N 8

761

2

3

49

5

78. Answer (2)

79. Answer (2)

80. Answer (3)

CH3CHO + CH

3CH

2CHO

NaOH,

CH3

– CH = CH – CHO

+ CH CH = C – CHO3 2

CH

CH3

+ CH – = C – CHO3

CH

CH3

+ CH3

– CH2 – CH = CH – CHO

81. Answer (3)

A =

C

C

NK

– +

O

O

B =

C

C

N

O

O

CH2

Ph

C =

C

C

O

O

NR

NR

82. Answer (2)

83. Answer (4)

84. Answer (3)

85. Answer (3)

86. Answer (3)

87. Answer (4)

88. Answer (3)

X =

NH4

O

C – O–

NH4

C – O–

O

; Y =

NH2

O

O

NH2

Z = NH

O

O

89. Answer (2)

90. Answer (2)

No rearrangement of carbocation as stabilised by

adjacent methoxy group.

CH3

O C

CH3

CH3

C H52

Br–

CH3O C

CH3

CH3

C H52CH

Br

CH


9/11

91. Answer (3)

Mutualism is beneficial to both organisms and is

obligatory.

92. Answer (3)

Rate of population growth reduced if fewer pre-

reproductive and reproductive individuals present in a

community.

93. Answer (4)

dN K NrN

dt N

⎛ ⎞ ⎜ ⎟⎝ ⎠

94. Answer (2)

95. Answer (1)

Low percentage of pre- reproductive individuals.

96. Answer (3)

97. Answer (2)

Efficient predator is prudent which will not allow prey

to extinction.

98. Answer (2)

Birth rate = 8

0.180

99. Answer (3)

Phytoplanktons – Pioneer

Transitional communities

Hydrilla, Vallisneria submerged plants.

Typha – Rooted floating plant.

100. Answer (3)

Species diversity increases from high altitude to low

altitude and high latitude to low latitude.

101. Answer (3)

Detritus-raw material for decompostion.

102. Answer (4)

Effluents from industries rich in - nitrates and

phosphates.

103. Answer (4)

104. Answer (4)

105. Answer (2)

A – 12, B – 23, C –32, D – 31.

106. Answer (2)

a–Rabbit, b–Grasshopper, c–Frog, d–Hawks (2012).

107. Answer (2)

Aromatic compounds - 42%

108. Answer (2)

109. Answer (3)

Sea-inverted (pyramid of biomass)

110. Answer (3)

Phosphorus cycle.

111. Answer (1)

112. Answer (2)

Biomass increases from lower trophic level to higher

trophic level in pond ecosystem.

113. Answer (4)

Stratification and species composition.

114. Answer (3)

Alnus, Juncus, Typha, Hydrilla (2009).

115. Answer (2)

116. Answer (1)

117. Answer (4)

Richest and most threatened plant and animals

reserviors of life on the earth.

118. Answer (2)

Leads to decline in plant production, lowered

resistance to environmental perturbations, increased

variability in certain ecosystem processes.

119. Answer (4)

120. Answer (4)

Statement (b) only correct i.e. Western Ghats are

one of the Hotspots in India.

121. Answer (3)

122. Answer (2)

Aravalli hills are in Rajasthan

123. Answer (1)

124. Answer (2)

Seed banks and wildlife safari park are ex-situ

conservation strategy.

125. Answer (4)

UV-B– Snow blindness

Ozone- DU

[ BIOLOGY]


10/11

126. Answer (3)

(a) statement is only incorrect i.e. DFC may be

connected with GFC at some levels.

127. Answer (1)

Given diagram is representing effect of sewage

discharge on important characteristics of a river.

128. Answer (3)

129. Answer (4)

130. Answer (2)

A water logged soil has poor aeration and also draws

salts to the surface of the soil.

131. Answer (1)

COP : 2014 – Lima, Peru

132. Answer (3)

133. Answer (2)

Skin cancer, reduction in rate of photosynthesis,

more UV-rays reach to earth.

134. Answer (3)

135. Answer (2)

Industries including power plants and oil refineries,

use water as coolant for the machinery release hot

water in water body causes thermal pollution.

136. Answer (3)

137. Answer (2)

138. Answer (4)

139. Answer (4)

140. Answer (3)

141. Answer (1)

142. Answer (1)

143. Answer (3)

Inbreeding exposes harmful recessive traits that are

eliminated by selection process

144. Answer (4)

Hisardale was developed by crossing Bikaneri ewes

with Marino rams & it is the example of cross

breeding

145. Answer (3)

146 Answer (3)

147. Answer (4)

148. Answer (4)

149. Answer (2)

150. Answer (3)

Cocaine interferes with the transport of the neuro-

transmitter dopamine & its excess dosage causes

hallucinations.

151. Answer (4)

152. Answer (2)

MRI uses strong magnetic rays and non-ionising

radiations to accurately detect pathological changes

only in the living tissues

153. Answer (4)

154. Answer (2)

155. Answer (4)

156. Answer (2)

157. Answer (3)

158. Answer (2)

Disarmed retroviruses are used to deliver desirable

genes into animal cells.

159. Answer (4)

R.E. cleaves within the recognition site.

160. Answer (2)

161. Answer (3)

Sma I produces non-cohesive/blunt ends.

162. Answer (1)

Restriction enzymes are obtained from prokaryotic

cells only.

163. Answer (2)

164. Answer (3)

165. Answer (4)

166. Answer (3)

Zidovudine & nevirapine are given to HIV positive

pregnant women to ensure that their babies do not

carry the infection

167. Answer (1)

Cancer occurs when the normal cells of the body

lose the property of contact inhibition.

168. Answer (3)

Flavr-Savr tomato was produced by anti-sense mRNA

technique.

169. Answer (2)

170. Answer (2)

171. Answer (2)

172. Answer (3)


11/11

173. Answer (2)

174. Answer (2)

175. Answer (2)

176. Answer (1)

177. Answer (4)

178. Answer (1)

179. Answer (4)

180. Answer (1)

� � �

Test - 3 (Code-B) (Answers & Hints) All India Aakash Test Series for Medical-2018

1/11

1. (3)

2. (1)

3. (1)

4. (3)

5. (3)

6. (3)

7. (3)

8. (4)

9. (4)

10. (3)

11. (1)

12. (2)

13. (3)

14. (4)

15. (3)

16. (3)

17. (4)

18. (3)

19. (3)

20. (1)

21. (4)

22. (3)

23. (3)

24. (4)

25. (3)

26. (4)

27. (1)

28. (4)

29. (3)

30. (3)

31. (1)

32. (1)

33. (1)

34. (1)

35. (4)

36. (3)

Test Date : 12/11/2017

ANSWERS

TEST - 3 (Code B)

All India Aakash Test Series for Medical - 2018

37. (4)

38. (1)

39. (3)

40. (1)

41. (4)

42. (4)

43. (1)

44. (4)

45. (4)

46. (2)

47. (2)

48. (3)

49. (4)

50. (3)

51. (3)

52. (3)

53. (4)

54. (2)

55. (3)

56. (3)

57. (2)

58. (2)

59. (3)

60. (2)

61. (1)

62. (2)

63. (3)

64. (3)

65. (4)

66. (3)

67. (1)

68. (1)

69. (2)

70. (2)

71. (3)

72. (4)

73. (4)

74. (2)

75. (4)

76. (3)

77. (3)

78. (3)

79. (1)

80. (1)

81. (3)

82. (3)

83. (1)

84. (1)

85. (2)

86. (1)

87. (3)

88. (2)

89. (4)

90. (2)

91. (2)

92. (3)

93. (2)

94. (3)

95. (1)

96. (2)

97. (4)

98. (3)

99. (1)

100. (3)

101. (4)

102. (2)

103. (1)

104. (2)

105. (3)

106. (4)

107. (4)

108. (2)

109. (4)

110. (1)

111. (2)

112. (3)

113. (4)

114. (2)

115. (1)

116. (3)

117. (3)

118. (2)

119. (2)

120. (2)

121. (2)

122. (4)

123. (4)

124. (4)

125. (3)

126. (3)

127. (3)

128. (2)

129. (2)

130. (3)

131. (1)

132. (2)

133. (4)

134. (3)

135. (3)

136. (1)

137. (4)

138. (1)

139. (4)

140. (1)

141. (2)

142. (2)

143. (2)

144. (3)

145. (2)

146. (2)

147. (2)

148. (3)

149. (1)

150. (3)

151. (4)

152. (3)

153. (2)

154. (1)

155. (3)

156. (2)

157. (4)

158. (2)

159. (3)

160. (2)

161. (4)

162. (2)

163. (4)

164. (2)

165. (4)

166. (3)

167. (2)

168. (4)

169. (4)

170. (3)

171. (3)

172. (4)

173. (3)

174. (1)

175. (1)

176. (3)

177. (4)

178. (4)

179. (2)

180. (3)

All India Aakash Test Series for Medical-2018 Test - 3 (Code-B) (Answers & Hints)

2/11

ANSWERS & HINTS

1. Answer (3)

The circuit can be redrawn

I = 2 1

A12 6

I

2 V 1 10

1

2. Answer (1)

Truth table is identical to that of AND gate.

Also A B = AB

3. Answer (1)

2 resistor connected to diode can be treated as

shortcircuit. So Req

= 2

23 =

8

3

Ibattery

= 3

8

V

VAB

= 3 2

8 3 4

V V

4. Answer (3)

Output voltage across Zener diode will be constant.

So potential difference across RL will be 10 V.

Current, I = 3

10

10 = 10 mA

5. Answer (3)

6. Answer (3)

T/2 Tt

E

avg rms rmsP v i =

2

rms

4

v

R=

2

04E

R

02

[ ]2

rms

EE ∵

[ PHYSICS]

7. Answer (3)

VCC

– 0 = IBR

B + V

BE

VCC

= IBR

B(∵ V

BE = 0)

So IB = Base current

= 3

8

500 10

CC

B

V

R

= 16A

Now, IC = I

B = 1.28 mA

8. Answer (4)

9. Answer (4)

10. Answer (3)

3 3

6 3

1.5 10 4 1060

50 10 2 10

outgain ac

in

RV

R

11. Answer (1)

The angular momentum is

2

hL n

or 2 L

nh

Then, the energy of the electron is

2 4

2 2 208

mZ eE

n h

2 4 2 4

2 2 2 22 2 00

3228

⎛ ⎞ ⎜ ⎟

⎝ ⎠

mZ e mZ e

LLh

h

12. Answer (2)

K1

K2

1 2K K E ,

Also 1 2

P P

1 22 250 50

M MK M K

⎛ ⎞⇒ ⎜ ⎟

⎝ ⎠

1 2 1 149 49( ) 49 49K k E K E K⇒

1

49

50

EK⇒


3/11

13. Answer (3)

N increases exponentially, N = N0(1 – e–t)

14. Answer (4)

15. Answer (3)

1 1 3

2000 400 1000

Now, 0 0 0

5

6

tN N N N e

⎛ ⎞ ⎜ ⎟⎝ ⎠

ln6597t t⇒ ⇒

years.

16. Answer (3)

1

0

tR R e

and 2

0

1 tR e

R

Dividing

2 1( )2 t t

R e

2 12ln( ) ( )R t t

2 1( )1

2ln( )avg

t tT

R

17. Answer (4)

18. Answer (3)

2 19, 5n n

2 1 2 1( )( 1)

2

(9 5)(9 5 1)

2

10

n n n nN

19. Answer (3)

We know that

2 2

2 2 2 2

1 1 ( 1)

( 1) ( 1)

E n nf

h n n n n

⎡ ⎤⎡ ⎤ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

2 2 3

(2 1) 1

( 1)

n

n n n

[ 1]n ∵

20. Answer (1)

1 1n n nr r r

2 220 0

0

( 1) ( 1)

r n r nnr

Z Z Z

24 0 4 ⇒ n n n

22

13.6 0.8516

n

ZE Z

21. Answer (4)

1 12 2

n n

nh hmv r v r

m

⇒

22. Answer (3)

In Davisson and Germer experiment

12.27 V

New, 12.27 12.27 20

.215

100

V

V V

23. Answer (3)

24. Answer (4)

V0 =

1hc

e e

⎛ ⎞ ⎜ ⎟ ⎝ ⎠ m =

hc

e

Nature must be straight line.

25. Answer (3)

T1

T+ m g21

m g2

T w c2+2 /

For 1 1 2 1, m T T m g

For 2 2 2

2, w

m m g Tc

Solving, 1 1 2

2= m w

T g m gc

26. Answer (4)

Making the incident frequency one third, means

making the incident frequency less than Threshold

frequency.


4/11

27. Answer (1)

( )ms T n hf

26

34 10

2 4200 16 10

7 10 2 10

ms Tn

hf

28. Answer (4)

=

2

h

mqV

, P

P P

m q

m q

= 8

1= 2 2 :1

29. Answer (3)

1hf k and

25hf k

Solving for , = 15

3 10h .

30. Answer (3)

Photoelectric emission rate intensity 1

r

(for

linear source), maximum kinetic energy does not

depend upon intensity of light. Hence, kinetic

energy = k, and emission become three times.

31. Answer (1)

Kinetic nergy of particle

Energy of photon

e

=

21

2mv

hc

=

21

2

hv

v

hc

⎛ ⎞⎜ ⎟⎝ ⎠

8

8

2 10 1

2 32 3 10

v

c

32. Answer (1)

Intensity of nth order secondary maxima

0

2

4

2 1

⎡ ⎤ ⎣ ⎦

n

Il

n

33. Answer (1)

y = ( 1)D n D

td d

n = 2

t

34. Answer (1)

Locus of a point having a given path difference of

light from two sources on the screen will be a circle.

35. Answer (4)

Intensity of light from 1st polarising sheet =0

12

Ii

Intensity of light from 2nd sheet,

20 02 cos 45

2 4

I Ii

Intensity of light from 3rd sheet,

20 0

3 cos 454 8

I Ii

Intensity of final transmitted light from last sheet,

20 0

4cos (45)

8 16

I Ii

36. Answer (3)

Because image forms at infinity fo + f

e = 40 cm and

o

e

f

f= 8

Solving, 320

9of cm and

40

9ef cm

37. Answer (4)

(i)

10 cm 5 cm

(ii)

10 cm

38. Answer (1)

0

f

y

A

tan , 45 , 10 cmy

ff

y = 10 cm


5/11

39. Answer (3)

In figure, CA = 20 cm = R

Using Sine law,

30°

MC

30°

120°

A

CA CM

sin120 sin30

CM = 20

3

40. Answer (1)

Position of image of O w.r.t fish will be AO

= 2

11

2

xx

⎛ ⎞

⎜ ⎟⎝ ⎠

x

y

A

oo

F

5 ms–1

2 ms–1

1 = 2

2 = 4/3

Now let relative separation is equal to r.

Then r = 2

1

x y

dr

dt =

2

1

dx dy

dt dt

= 2

1

(2) 5

= 6.33 ms–1

41. Answer (4)

i

r1

r1

r2

r2

r3

For 1st refraction, sini × 1 = sinr1×

1

For 2nd refraction, sinr1 ×

1 = sinr

2 ×

2

For 3rd refraction, sinr2 ×

2 = sinr

3 ×

3

1

3

3

sinsin

⎛ ⎞ ⎜ ⎟⎝ ⎠

ir

42. Answer (4)

22

1

2min

m 1

2

131 142

1: 253 25

112 4

ax

I

II

I I

I

⎛ ⎞ ⎛ ⎞⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ ⎠ ⎜ ⎟⎜ ⎟ ⎛ ⎞⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠ ⎝ ⎠⎝ ⎠

43. Answer (1)

When tube length is decreased, the (real)

intermediate image formed by the objective will lie

between the eyepiece and its focus. This will cause

a virtual image to be formed.

44. Answer (4)

1 2( )i i A

2 230 (60 ) 30 0i i ⇒

Also: 1

1

sin( ) sin603

sin( ) sin30

i

r

45. Answer (4)

An air bubble behaves as a diverging lens inside

water. So it forms a virtual image.


6/11

46. Answer (2)

No rearrangement of carbocation as stabilised by

adjacent methoxy group.

CH3

O C

CH3

CH3

C H52

Br–

CH3O C

CH3

CH3

C H52CH

Br

CH

47. Answer (2)

48. Answer (3)

X =

NH4

O

C – O–

NH4

C – O–

O

; Y =

NH2

O

O

NH2

Z = NH

O

O

49. Answer (4)

50. Answer (3)

51. Answer (3)

52. Answer (3)

53. Answer (4)

54. Answer (2)

55. Answer (3)

A =

C

C

NK

– +

O

O

B =

C

C

N

O

O

CH2

Ph

[ CHEMISTRY]

C =

C

C

O

O

NR

NR

56. Answer (3)

CH3CHO + CH

3CH

2CHO

NaOH,

CH3

– CH = CH – CHO

+ CH CH = C – CHO3 2

CH

CH3

+ CH – = C – CHO3

CH

CH3

+ CH3

– CH2 – CH = CH – CHO

57. Answer (2)

58. Answer (2)

59. Answer (3)

N

N

N

N 8

761

2

3

49

5

60. Answer (2)

A =

NH2

OH

B =

N2Cl

OH

–

C =

OH

61. Answer (1)

62. Answer (2)

63. Answer (3)

A =

NO2

B =

NH2

C =

NHCOCH3

D =

NHCOCH3

Br


7/11

64. Answer (3)

Weak oxidising agents like PCC, PDC etc. oxidise

1° alcohol to aldehyde.

65. Answer (4)

Higher-I effect, more will be acidic strength.

66. Answer (3)

F

NO2

CH3

NaOH

(aq)

OH

NO2

CH3

(A)

—NO at p-position

favours attack of nucleophile

2

F

NO2

CH3

NaOH

(aq)—CH at p-position

unfavourable for attackof nucleophile.

3No reaction

67. Answer (1)

COOH

OH

68. Answer (1)

2

4

4

Pd/H

2LiAlH ,1 alcoholNaBH

RCHO R—CH OH

69. Answer (2)

Intermediate is cumene hydroperoxide

CH3

CH

3C

O O H

70. Answer (2)

—COOH group present on bridge head C does not

take part in any reaction, -ketoacid is always very

difficult to undergo decarboxylation.

71. Answer (3)

A = C6H

5MgCl

B = C H6 5

C OMgCl

O

C = C6H

5COOH

72. Answer (4)

No ortho effect is found in o-Methylphenol. It is less

acidic due to electron donating nature of methyl

group.

73. Answer (4)

(A)

C

COOH

O

74. Answer (2)

A = CH3

C CH3

OKetone

B = CH3

CHOCH2

Aldehyde

Both can be differentiated by Tollen’s reagent.

75. Answer (4)

(Fact)

Various bacteriostatic antibiotic are erythromycin,

tetracycline, chloramphenicol.

76. Answer (3)

NPh2

HNO2

–H2O

NPh2

N O3° aromatic amine

p-Nitroso derivative

77. Answer (3)

HOCH CH OH + n HO22

n

Ethylene glycol

C

O

C

O

OH

Phthalic acid

Polymerisation

O CH2

CH2

O C C

n

Glyptal

O O

78. Answer (3)

Biuret test : When alkaline solution of protein is

treated with a drop of aqueous copper sulphate, a

bluish violet colour is obtained.


8/11

79. Answer (1)

C C

F

Cl

Br

FC C

Cl

Br

F

FC C

Cl

F

Br

F

Geometrical isomers

80. Answer (1)

O

HCN

HO C N

LiAlH4

HO CH2

NH2

81. Answer (3)

LDP is obtained by polymerisation of ethene under

high pressure of 1000–2000 atm. and at high

temperature of 350 K to 570 K in presence of

peroxide initiator.

82. Answer (3)

I

NBS

Allylic halogenation

I

CH2

Br

I

CH2

S CH3

+ NaBr

CH3

S 2N Na S CH

+

3

–

83. Answer (1)

84. Answer (1)

C6H

12O

6

Zymase 2C2H

5OH + 2CO

2

Glucose/Fructose

Last step during commercial preparation of alcohol

from fermentation of sugar.

85. Answer (2)

CH3CH MgBr + CH

2 3C

O

Cl

CH3 C

O

CH2

CH + Mg3

Br

Cl

CH2

C

OH

CH2

CH3 CH

3C

OH

CH CH3

enol forms

Tautomers

86. Answer (1)

Common antihistamines are brompheniramine

(Dimetapp) and terfenadine (seldane).

87. Answer (3)

Benzyl halides are more reactive than Aryl halides

due to more stability of benzyl carbocation /

intermediate. Moreover, reactivity of aryl halides is

increased by EWG like –NO2 so that order is

II > III > IV > I.

88. Answer (2)

3CH3OH + Al (CH

3O)

3Al + 2

3H (g)

2

CH3OH is limiting reagent.

3 mole of CH3OH gives 2

3moles of H (g)

2

1.5 mole of CH3OH gives

3 11.5

2 3 =

3

4mol

89. Answer (4)

CH3 CH

OH

CH2

CH2 OH

PCC

CH3 C

O

CH2

CHO

(A)

HOHO

CH3 C

O

CH2

CH

(B)

O

O

MeMgBr,H O

3

+

CH3 C CH

2 CHO

CH3

OH (C)

CH3 C

CH3

OH

NaBH4EtOH

CH2

CH2 OH

(D)

90. Answer (2)

Number of chiral centre in aldoheptose will be 5 and

that in ketoheptose will be 4, so that

5

4

a (2) 2

b 1(2)


9/11

91. Answer (2)

Industries including power plants and oil refineries,

use water as coolant for the machinery release hot

water in water body causes thermal pollution.

92. Answer (3)

93. Answer (2)

Skin cancer, reduction in rate of photosynthesis,

more UV-rays reach to earth.

94. Answer (3)

95. Answer (1)

COP : 2014 – Lima, Peru

96. Answer (2)

A water logged soil has poor aeration and also draws

salts to the surface of the soil.

97. Answer (4)

98. Answer (3)

99. Answer (1)

Given diagram is representing effect of sewage

discharge on important characteristics of a river.

100. Answer (3)

(a) statement is only incorrect i.e. DFC may be

connected with GFC at some levels.

101. Answer (4)

UV-B– Snow blindness

Ozone- DU

102. Answer (2)

Seed banks and wildlife safari park are ex-situ

conservation strategy.

103. Answer (1)

104. Answer (2)

Aravalli hills are in Rajasthan

105. Answer (3)

106. Answer (4)

Statement (b) only correct i.e. Western Ghats are

one of the Hotspots in India.

107. Answer (4)

[ BIOLOGY]

108. Answer (2)

Leads to decline in plant production, lowered

resistance to environmental perturbations, increased

variability in certain ecosystem processes.

109. Answer (4)

Richest and most threatened plant and animals

reserviors of life on the earth.

110. Answer (1)

111. Answer (2)

112. Answer (3)

Alnus, Juncus, Typha, Hydrilla (2009).

113. Answer (4)

Stratification and species composition.

114. Answer (2)

Biomass increases from lower trophic level to higher

trophic level in pond ecosystem.

115. Answer (1)

116. Answer (3)

Phosphorus cycle.

117. Answer (3)

Sea-inverted (pyramid of biomass)

118. Answer (2)

119. Answer (2)

Aromatic compounds - 42%

120. Answer (2)

a–Rabbit, b–Grasshopper, c–Frog, d–Hawks (2012).

121. Answer (2)

A – 12, B – 23, C –32, D – 31.

122. Answer (4)

123. Answer (4)

124. Answer (4)

Effluents from industries rich in - nitrates and

phosphates.

125. Answer (3)

Detritus-raw material for decompostion.

126. Answer (3)

Species diversity increases from high altitude to low

altitude and high latitude to low latitude.


10/11

127. Answer (3)

Phytoplanktons – Pioneer

Transitional communities

Hydrilla, Vallisneria submerged plants.

Typha – Rooted floating plant.

128. Answer (2)

Birth rate = 8

0.180

129. Answer (2)

Efficient predator is prudent which will not allow prey

to extinction.

130. Answer (3)

131. Answer (1)

Low percentage of pre- reproductive individuals.

132. Answer (2)

133. Answer (4)

dN K NrN

dt N

⎛ ⎞ ⎜ ⎟⎝ ⎠

134. Answer (3)

Rate of population growth reduced if fewer pre-

reproductive and reproductive individuals present in a

community.

135. Answer (3)

Mutualism is beneficial to both organisms and is

obligatory.

136. Answer (1)

137. Answer (4)

138. Answer (1)

139. Answer (4)

140. Answer (1)

141. Answer (2)

142. Answer (2)

143. Answer (2)

144. Answer (3)

145. Answer (2)

146. Answer (2)

147. Answer (2)

148. Answer (3)

Flavr-Savr tomato was produced by anti-sense mRNA

technique.

149. Answer (1)

Cancer occurs when the normal cells of the body

lose the property of contact inhibition.

150. Answer (3)

Zidovudine and nevirapine are given to HIV positive

pregnant women to ensure that their babies do not

carry the infection

151. Answer (4)

152. Answer (3)

153. Answer (2)

154. Answer (1)

Restriction enzymes are obtained from prokaryotic

cells only.

155. Answer (3)

Sma I produces non-cohesive/blunt ends.

156. Answer (2)

157. Answer (4)

R.E. cleaves within the recognition site.

158. Answer (2)

Disarmed retroviruses are used to deliver desirable

genes into animal cells.

159. Answer (3)

160. Answer (2)

161. Answer (4)

162. Answer (2)

163. Answer (4)

164. Answer (2)

MRI uses strong magnetic rays and non-ionising

radiations to accurately detect pathological changes

only in the living tissues

165. Answer (4)

166. Answer (3)

Cocaine interferes with the transport of the neuro-

transmitter dopamine & its excess dosage causes

hallucinations.

167. Answer (2)

168. Answer (4)

169. Answer (4)

170 Answer (3)

171. Answer (3)

172. Answer (4)

Hisardale was developed by crossing Bikaneri ewes

with Marino rams & it is the example of cross

breeding


11/11

� � �

173. Answer (3)

Inbreeding exposes harmful recessive traits that are

eliminated by selection process

174. Answer (1)

175. Answer (1)

176. Answer (3)

177. Answer (4)

178. Answer (4)

179. Answer (2)

180. Answer (3)

for medical-2018 all india aakash test series for medical ... · all india aakash test series for...

Documents