Test-7 (Code-E) (Answers) All India Aakash Test …...Test-7 (Code-E)_(Answers) All India Aakash...

Test-7 (Code-E)_(Answers) All India Aakash Test Series for NEET-2020

Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456 1/16

All India Aakash Test Series for NEET-2020

Test Date : 15/03/2020

ANSWERS

1. (1) 2. (3) 3. (1) 4. (2) 5. (4) 6. (2) 7. (4) 8. (3) 9. (2) 10. (3) 11. (3) 12. (3) 13. (2) 14. (1) 15. (4) 16. (4) 17. (1) 18. (3) 19. (2) 20. (3) 21. (3) 22. (2) 23. (3) 24. (4) 25. (3) 26. (2) 27. (4) 28. (2) 29. (2) 30. (1) 31. (1) 32. (3) 33. (2) 34. (2) 35. (1) 36. (3)

37. (2) 38. (1) 39. (3) 40. (1) 41. (1) 42. (4) 43. (2) 44. (2) 45. (2) 46. (3) 47. (1) 48. (4) 49. (3) 50. (4) 51. (4) 52. (2) 53. (2) 54. (2) 55. (4) 56. (4) 57. (3) 58. (4) 59. (1) 60. (4) 61. (2) 62. (3) 63. (1) 64. (4) 65. (1) 66. (4) 67. (4) 68. (3) 69. (2) 70. (1) 71. (1) 72. (2)

73. (1) 74. (4) 75. (3) 76. (1) 77. (4) 78. (4) 79. (3) 80. (2) 81. (2) 82. (4) 83. (4) 84. (4) 85. (1) 86. (1) 87. (2) 88. (2) 89. (4) 90. (3) 91. (3) 92. (2) 93. (4) 94. (2) 95. (1) 96. (3) 97. (2) 98. (4) 99. (3) 100. (3) 101. (2) 102. (1) 103. (2) 104. (3) 105. (2) 106. (4) 107. (4) 108. (3)

109. (3) 110. (2) 111. (4) 112. (1) 113. (4) 114. (1) 115. (2) 116. (4) 117. (4) 118. (3) 119. (2) 120. (3) 121. (4) 122. (2) 123. (2) 124. (1) 125. (3) 126. (3) 127. (2) 128. (3) 129. (4) 130. (3) 131. (3) 132. (4) 133. (1) 134. (2) 135. (2) 136. (3) 137. (4) 138. (2) 139. (3) 140. (4) 141. (1) 142. (2) 143. (3) 144. (2)

145. (4) 146. (2) 147. (2) 148. (3) 149. (1) 150. (2) 151. (4) 152. (2) 153. (1) 154. (2) 155. (4) 156. (1) 157. (3) 158. (1) 159. (3) 160. (4) 161. (4) 162. (1) 163. (1) 164. (4) 165. (4) 166. (2) 167. (3) 168. (2) 169. (1) 170. (4) 171. (2) 172. (2) 173. (3) 174. (1) 175. (2) 176. (4) 177. (2) 178. (1) 179. (1) 180. (3)

TEST - 7 (Code-E)

All India Aakash Test Series for NEET-2020 Test-7 (Code-E)_(Hints & Solutions)


HINTS & SOLUTIONS

[PHYSICS] 1. Answer (1)

Hint & Sol. : When a real object is placed at centre of curvature, image is also formed at centre of curvature. So least distance is zero.

2. Answer (3)

Hint : Length of image = 1 2v v− .

Sol. : For near end of object, u1 = –30 cm, f = –15 cm From mirror equation,

1

1 1 130 15v

− = −

1

1 1 1 1 230 15 30v

−⇒ = − =

⇒ v1 = –30 cm For far end of object u2 = –45 cm,

So, 2

1 1 1 3 115 45 45v

− += − + =

245 22.5 cm2

v⇒ = − = −

So, length of image 1 2 7.5 cmv v− =

3. Answer (1)

Hint : 1 2

1 1 1( 1)f R R

= µ − −

Sol. :

⇒ 1 10.460 2R

= ×

⇒ R = 12 cm and 2R = 24 cm

4. Answer (2) Hint & Sol. : In compound microscope both the lenses are of short focal length and difference of focal should be small and f0 < fe.

5. Answer (4)

Hint : µ =cv

2 1

1 2sin

vC

vµ

= =µ

Sol. : 8

82 10 8sin

92.25 10C ×

= =×

1 8sin9

C − =

6. Answer (2)

Hint : For concave lens, virtual image will form at focus, which acts as an object for convex lens.

Sol. :

For convex lens

u = – 30 cm, f = 15 cm, v = ?

1 1 1f v u

= −

1 1 115 30v

= +

1 1 1 115 30 30v

= − =

v = 30 cm

Distance of final image from concave lens = 50 cm

7. Answer (4)

Hint : sin

2

sin2

+ δ µ =

mA

A

Test-7 (Code-E)_(Hints & Solutions) All India Aakash Test Series for NEET-2020


Sol. : Given A = 60°

60sin22

sin30

m° + δ =

°

1 60sin2 2

m° + δ =

6045

2m° + δ

= °

δm = 30°

8. Answer (3) Hint : Length of instrument L = f0 + ue

Sol. : Image formed by objective is at focus of objective lens For eyepiece ve = – 25 cm. fe = 5 cm ue = ?

1 1 1

e e ef v u= −

1 1 15 25 eu

+ = −

5 1 125 eu+

= −

25 4.17 cm6eu = − = −

L = 60 + 4.17 = 64.17

9. Answer (2)

Hint & Sol. : Real depth (R.D)Apparent depth (A.D)

µ =

From 1st surface (R.D)1 = µ × (A.D)1 = 1.5 × 5 = 7. 5 cm

From 2nd surface (R.D)2 = µ × (A.D)2 = 1.5 × 7 = 10.5 cm Thickness of slab = 18 cm.

10. Answer (3) Hint & Sol. : If wave front is cylindrical.

2= =

πE ElAt r l t

or 1lr

α

11. Answer (3) Hint : For bright fringe, path difference is nλ.

Sol. : Path difference ∆y = 4 × 10–6 m

Wavelength λ = 5 × 10–7 m

nλ = ∆x 6

74 105 10

xn−

−

∆ ×= =

λ ×

= 8

Path difference is integer multiple of λ, so bright fringe will obtain and point will be bright.

12. Answer (3)

Hint : When light enters in liquid its wavelength will change.

Fringe width = Dd

λ .

Sol. : Wavelength in water aw

λλ =

µ

λ βλβ = = =

µ µa aw

wDD

d d

0.6 3 0.45 mm4w×

β = =

13. Answer (2)

Hint : Light transmitted through a polaroid

I = l0 cos2θ

Sol. : Light transmitted through two polaroid at 30°

I2 = I1cos230°

0 0332 4 8l l

= × =

Light not transmitted 0 00

3 58 8l l

l= − =

% of light not transmitted 5 100 62.5%8

= × =

14. Answer (1) Hint : In diffraction, for subsidiary maxima, path

difference = (2n + 1) 2λ

For minima path difference = nλ n = 1, 2, 3 ……….



Sol. : For 1st subsidiary maxima

sin30 32

d λ° = …(i)

For second minima

dsinθ = 2λ …(ii)

Divide (ii) by (i)

sin 23

2 2

dd

θ λ=

λ

42sin3

θ =

1 2sin3

− θ =

15. Answer (4)

Hint : 1 2m D n Dd dλ λ

=

Sol. : d = 1 mm, D = 2 m

λ1 = 6300 Å λ2 = 5400 Å

Two bright fringes coincide if m × 6300 = n × 5400

⇒ m = 6, n = 7

Distance on screen, 7

13

6 6 6.3 10 21 10

−

−

λ × × ×= =

×

Dy

d

= 7.56 mm

16. Answer (4)

Hint & Sol. : Resolving power of compound

microscope 2 sinµ θ=

λ

So, resolving power ∝ µ.

17. Answer (1)

Hint : Path difference at any point at distance y.

ydyD

∆ =

For dark fringe ∆y = (2n – 1) 2λ n = 1, 2, 3……..

Sol. : 2dy = n = 2

For dark fringe ( )2 1 2yd

nDλ

= −

2 32 2dD

λ=

2

3λ =

dD

18. Answer (3)

Hint : Shift of fringe on introducing a film

( )1x tβ= µ −

λ

Sol. : µ = 1.6

Shift 2

x β=

( 1)2

tβ β= µ −

λ

6000 5000 Å2( 1) 2(1.6 1)

t λ= = =

µ − −

t = 5 × 10–7 m

19. Answer (2)

Hint : Energy of photon hν = φ + max (K.E)

Sol. : ν1 = ν, ν2 = 1.25 ν, K1 = 0.5 eV, K2 = 0.75 eV

Case 1 hν = φ + 0.5 eV …(1)

Case 2 1.25 hν = φ + 0.75 eV …(2)

Multiply equation (1) by 5 and (2) by 4 and on subtracting

φ = 0.5 eV

20. Answer (3)

Hint & Sol. : The wavelength of red light is greater than threshold wavelength, so electrons will not emit for red light.

21. Answer (3)

Hint : de-Broglie wavelength λ =h

mv

Sol. : Force on electron ˆ=

F eE i

Acceleration of electron ˆeEa im

=

Speed of electron 2 20 2 eEv v d

m= +

0 20

21 eEv v dmv

= +



de-Broglie wavelength

0 20

21

heEdmv

mv

λ =+

0

20

21 eEdmv

λλ =

+

22. Answer (2)

Hint : 12400 eV(Å)

E =λ

eVchc

= − φλ

Sol. : λ1 = 5000 Å, λ2 = 3000 Å

Case – 1

eV1c = 2.48 – φ

Case – 2

eV2c = 4.13 – φ

V2c – V1c = 1.65 V

23. Answer (3)

Hint : 3

hkTm

λ =

1T

λ ∝

Sol. : T1 = 27 + 273 = 300 K

T2 = 927 + 273 = 1200 K

2 1

1 2

300 11200 2

TT

λ= = =

λ

12 2

λλ =

24. Answer (4)

Hint & Sol. : Photoelectrons will be emitted only when wavelength of incident light is less than certain limiting value called threshold wavelength. Then photocurrent is proportional to intensity of light.

25. Answer (3) Sol. : Intensity of photons.

nlAt

=

From point source wavefront will be spherical

24nld t

=π

21

αld

so 21

αid

26. Answer (2)

Hint : 1v

λ ∝

Sol. : λ =h

mv

137

λ =hc zm

n

137λ =

nmcz

λ α n 27. Answer (4)

Hint : 12400 eV(Å)

E =λ

Sol. : λ = 1215 Å

12400 10.2 eV1215

= =E

So the wavelength of 1215 Å may excite the hydrogen atom form n = 1 to n = 2. So the number of spectral lines in emission spectrum is 1.

28. Answer (2) Hint & Sol. : Energy gap is maximum between energy level 1 and 2. So radiation of maximum frequency will be emitted, when electron makes transition from n = 2 to n = 1.

29. Answer (2) Hint : Time period in any orbit T ∝ n3

Sol. : T1 = 27 T2 3

11

22

nTnT

=

⇒ 3

12

22

27 nTnT

=

⇒ 1

23

nn

=



30. Answer (1) Hint : Energy of electron in any orbit

213.6 eV,= −nEn

angular momentum of electron =

2nh

π

Sol. : Energy of electron in n = 1, E1 = –13.6 eV Energy given to atom E = 12.75 eV Net final energy of electron, En = – 13.6 + (12.75) = – 0.85 eV

213.6– 0.85n

= −

2 13.6 160.85

n = =

n = 4

So 4 22 2nh h hL = = =

π π π

31. Answer (1)

Hint : 2·e h in n n=

Sol. : In intrinsic semi conductor e h in n n= =

ni = 1.5 × 1016 per m3

In extrinsic semiconductor hole density nh = 4.5 × 1022 per m3

ne × 4.5 × 1022 = (1.5 × 1016)2

32

221.5 1.5 10

4.5 10en × ×=

×

= 5 × 109 per m3

32. Answer (3)

Hint : Current gain β = c

b

ii

Sol. : IE = IB + IC

Current in base = 2% of 2100e ei i=

Current in collector = 98% of 98100e ei i=

49c

b

ii

β = =

33. Answer (2) Hint : Voltage across Zener diode is equal to voltage across 1 kΩ resistance.

Sol.: Breakdown voltage of Zener diode = 10 V Voltage drop across 250 Ω = 10 V

Current through dropping resistance 10250

VR

= =

= 40 mA

Current through 1 kΩ resistance 10 10 mA1000

= =

Current through Zener diode = 40 – 10 = 30 mA

34. Answer (2)

Hint & Sol. : Energy gap of GaAs0.6P0.4 is 1.9 eV and emits red colour.

35. Answer (1)

Hint & Sol. : The potential barrier of junction opposes the movement of majority carriers towards junction but supports the minority carriers to move towards junction.

36. Answer (3) Hint & Sol. : Truth table

A B Y′ C Y

0 0 1 1 1

1 0 0 0 0

0 1 0 1 0

1 1 0 0 0

37. Answer (2) Hint : Diode is reversed biased. Sol. :

If VA < VB then junction diode will be reverse biased and no current will pass through junction diode. So effective resistance is 12 Ω.



38. Answer (1)

Hint : Radius of nuclei 13

0R R A=

Sol. : Radius of nucleus of 4Be9 = R Radius of 32X nucleus = 2R

139

2RR A

=

3 912 A

=

A = 72 Number of neutrons in nucleus X = 72 – 32 = 40

39. Answer (3)

Hint : Momentum conservation law.

Sol. : 0 = +E mvc

= −Ev

mc

Kinetic energy of nucleus 212

mv=

21

2Em

mc =

2

22Emc

=

40. Answer (1)

Hint : Remaining value after n half lives.

012

nN N =

Sol. : Half life period T = 40 minute

Number of half lives 120 340

tnT

= = =

30

01

82N

N N = =

Amount decayed = N0 – N

0 018

= −N N

078

N=

41. Answer (1)

Hint & Sol. :

4 4 42 2He 2− − −

−→ + → +A A AZ Z ZP Q R e

So P and R are isotopes

42. Answer (4)

Hint : Photo diode will absorb those frequencies, by absorbing which, electron excite to higher energy state.

Sol. : Eg = hνmin

1.65 × 1.6 × 10–19 = 6.6 × 10–34 ν

19

min 341.65 1.6 10

6.6 10

−

−

× ×ν =

×

= 4 × 1014 Hz

43. Answer (2)

Hint : Voltage gain = –gmR0, where gm is transconductance and R0 is output resistance.

Sol. : AV = –gmR0

⇒ AV ∝ gm

1 1

2 2

=V m

V m

A g

A g

50 0.040.02

=VA

AV = 25

44. Answer (2)

Hint & Sol : Reverse bias applied to a junction diode increases the potential barrier.

45. Answer (2)

Hint & Sol. : 0

112 1 13 136 0 6 7C n C N− β+ → →



[CHEMISTRY]46. Answer (3)

Hint : Melting point of B is exceptionally high. Sol. : Element Melting Point B 2453 K Al 933 K Ga 303 K In 430 K

47. Answer (1) Hint : In heavier members the tendency to show +2 oxidation state increases among 14 group elements.

48. Answer (4) Hint : In cyclic and chain silicates, two corners of each tetrahedron are shared.

49. Answer (3)

Hint : 2H2O2 → 2H2O + O2 Sol. : 22.4 L of O2 at STP is obtained from 68 g of H2O2

50 L of O2 at STP is obtained from = 68 5022.4

×

= 151.78g L–1

15%

50. Answer (4) Hint : Isotopes have similar chemical properties. Sol. : Dielectric constant of ordinary water is greater than heavy water.

51. Answer (4) Hint : Electron rich hydrides contain lone pairs of electrons.

Sol. : HF⋅⋅

, 2H O⋅⋅ , 3NH

52. Answer (2) Hint : Copper glance : Cu2S.

53. Answer (2)

Hint : van Arkel process is used for obtaining ultra pure metals.

Sol. : 2 4Zr 2I ZrI+ → ↑

1800 K4 2ZrI Zr 2I→ +

54. Answer (2) Hint : 2[Au(CN)2]–(aq) + Zn(s) →

2Au + [Zn(CN)4]2–(aq)

55. Answer (4)

Hint : Cryolite : Na3AlF6, Fluorspar : CaF2.

Sol. : The fusion temperature of mixture [Al2O3, Na3AIF6, CaF2] lowers to 900°C.

56. Answer (4)

Hint : Copper matte contains mostly Cu2S and little FeS.

57. Answer (3)

Hint : 2Cu2O + Cu2S → 6Cu + SO2↑.

58. Answer (4)

Hint : Monel metal is an alloy of Ni, Cu and Fe.

59. Answer (1)

Hint : Cupellation is used for refining Ag.

60. Answer (4)

Hint : Impure Zn contains Pb, Cd and Fe as major impurities.

61. Answer (2)

Hint : H-X bond dissociation enthalpy decreases down the group.

Sol. : HF HCl HBr HI

∆dissH°(kJ/mol) 574 432 363 295

62. Answer (3) Hint : 2NaOH + Cl2 → NaCl + NaOCl + H2O. (cold and dilute)

Sol. : 6NaOH + 3Cl2 → 5NaCl + NaClO3 + 3H2O. (hot and conc)

63. Answer (1) Hint : (HPO3)3 is cyclotrimetaphosphoric acid. Sol. :

It contain three P — O — P bonds.



64. Answer (4) Hint : P4 + 8SOCl2 → 4PCl3 + 4SO2 + 2S2Cl2.

65. Answer (1)

Hint : XeF6 + 3H2O → XeO3 + 6HF

Sol. : XeF6 + H2O → XeOF4 + 2HF

XeF6 + 2H2O → XeO2F2 + 4HF

66. Answer (4)

Hint : Order of oxidising power:

HCIO > HCIO2 > HCIO3 > HCIO4.

67. Answer (4) Hint : N2O : N ≡ N → O.

Sol. : N2O3 :

N2O4 :

N2O5 :

68. Answer (3)

Hint : Zn + 4HNO3(conc) → Zn(NO3)2 + 2H2O + 2NO2

Sol. : • 4Zn + 10HNO3(dilute) → 4Zn(NO3)2 + 5H2O

+ N2O

• Cu + 4HNO3(conc) → Cu(NO3)2 + 2NO2

+ 2H2O

• 3Cu + 8HNO3(dilute) → 3Cu(NO3)2 + 2NO

+ 4H2O

69. Answer (2) Hint : Phosphine burns spontaneously in contact with air.

70. Answer (1) Hint : Oxidation state of Cl in Cl2O7 is +7. Sol. : Acidic nature : Cl2O7 > SO3 > N2O5 > N2O3.

71. Answer (1)

Hint : NH4NO3 ∆→ N2O + 2H2O Sol. :

• (NH4)2Cr2O7∆→ N2 + 4H2O + Cr2O3

• 2H2O + NH2CONH2 → (NH4)2CO3 • (NH4)2SO4 + 2NaOH → 2NH3 + 2H2O + Na2SO4

72. Answer (2) Hint : Monosaccharides can not be hydrolysed further. Sol. : Glucose is a monosaccharide.

73. Answer (1) Hint : Glucose get oxidised by Br2 water.

Sol. : 4

2

CHO|

(CHOH)|

CH OH

2Br water→4

2

COOH|

(CHOH)|

CH OH

Glucose Gluconic acid

74. Answer (4) Hint : Lactose formed by the linkage between C1 of galactose and C4 of glucose. Sol. : Lactose is a reducing sugar.

75. Answer (3)

Hint : Glycine 2

COOH

H N H

H

is non-essential

amino acid. 76. Answer (1)

Hint : Night blindness is caused by vitamin A deficiency.

77. Answer (4) Hint : Adenine, Guanine and Uracil bases are present in RNA. Sol. : Thymine is present in DNA.

78. Answer (4) Hint : Glucose can form six membered ring in which –OH at C-5 can add to the –CHO group.

79. Answer (3) Hint : In globular proteins, polypeptide chains coil around to give a spherical shape and these are usually soluble in water.



80. Answer (2)

Hint : Glucose Zymase→ C2H5OH

Sol. : Sucrose Invertase→ glucose and fructose

Urea Urease→ CO2 + NH3

Cellulose Cellulase→ Glucose

81. Answer (2)

Hint : Buna-S is addition polymer of butadiene and styrene.

82. Answer (4)

Hint : Cross-linked polymers have 3D-network structure.

83. Answer (4)

Hint : is Buna-N.

84. Answer (4) Hint : In vulcanisation, sulphur forms cross links at the reactive sites of double bonds.

85. Answer (1) Hint : Bakelite is a cross-linked polymer.

86. Answer (1) Hint : Phenelzine is an antidepressant drug.

87. Answer (2) Hint : Bithional is an antiseptic.

88. Answer (2) Hint : Maximum limit for nitrate in drinking water is 50 ppm.

89. Answer (4) Hint : Green fuel is obtained by recycling plastic waste.

90. Answer (3) Hint : Zn is refined by distillation and Sn by liquation.

[BIOLOGY]91. Answer (3)

Sol. : Grasslands and tundra biomes do not have trees.

92. Answer (2) Hint : Kangaroo rat shows physiological adaptations to survive in North American desert. Sol. : Kangaroo Rat of North American desert is capable of meeting all its water requirements through its internal fat oxidation.

93. Answer (4) Hint : A population showing declined growth has less number of pre-reproductive individuals as compare to reproductive and post-reproductive individuals. Sol. : Age pyramid of population showing declined growth is urn-shaped.

94. Answer (2) Hint : Logistic growth of a population is represented by sigmoid (S-shaped) curve. Sol. : Logistic growth occurs when the resource in a habitat are limited.

95. Answer (1) Hint : Darwin’s finches present in Galapagos island are the example of competitive co-existence. Sol. : Parasitism – Cuscuta on hedge plants.

Commensalism – Epiphytes (orchids) growing on other plants.

Mutualism – Fig and fig wasp.

96. Answer (3) Hint : Most ecologically relevant environmental factor is temperature. Sol. : Temperature gradually decreases latitude wise from equator towards the poles and altitude wise from plains to mountain tops.

97. Answer (2) Hint : Reduction of leaves into spines is an adaptation to prevent water loss in xerophytes. Sol. : Spines and flattened phylloclade (photosynthetic modified stem) are the adaptations seen in Opuntia.



98. Answer (4) Hint : All mammals and birds are regulators. Sol. : All mammals and birds are capable of thermoregulation and osmoregulation to maintain constant internal environment.

99. Answer (3) Hint : Anthropogenic ecosystems are man-made ecosystems such as agricultural farms. Sol. : Anthropogenic ecosystems have high productivity.

100. Answer (3) Hint : For decomposition, availability of air and saprophytic microorganisms is important. Sol. : Decomposition is better/ faster if detritus lack lignin, tannin, cellulose etc and if soil has optimum temperature and moisture.

101. Answer (2) Sol. : The rate of formation of new organic matter by consumers is called secondary productivity.

102. Answer (1) Hint : For aquatic ecosystems, GFC is the major conduit of energy flow. Sol. : In terrestrial ecosystems, much larger fraction of energy flows through DFC than through GFC.

103. Answer (2) Hint : Pyramid of number is not upright for the ecosystem having parasitic food chain. Sol. : In pond ecosystems and grassland ecosystems, pyramid of number is upright.

104. Answer (3) Hint : Humus has slightly acidic pH. Sol. : Humus is dark coloured, amorphous and more or less decomposed organic matter.

105. Answer (2) Hint : Secondary succession starts in areas that somehow lost all the living organisms that existed there. Sol. : Secondary succession can occur at abandoned farm lands, burned and cut forests and flooded lands.

106. Answer (4) Hint : Plants capture 2-10% of photosynthetically active radiation (PAR) for photosynthesis.

Sol. : Energy flow in an ecosystem is always unidirectional (from producers to consumers) and energy does not remain trapped in any organism permanently as it either flows through trophic levels or becomes available to decomposers for decomposition. Energy flow follows the laws of thermodynamics.

107. Answer (4) Hint : Hydrarch succession starts in aquatic habitat. Sol. : All kinds of successions including hydrarch succession, increase total biomass and humus content of the soil. In hydrarch succession, phytoplanktons are pioneer community and it progresses from hydric to mesic conditions.

108. Answer (3) Sol. : Amount of all the inorganic substances present in an ecosystem per unit area at a given time is called standing state.

109. Answer (3) Hint : In a food web, one animal may feed upon organism of different trophic levels. Sol. : In a food web, sparrow is a primary consumer when it eats seeds, fruits etc. and a secondary consumer when it eats insects and worms.

110. Answer (2) Hint : Phosphorus is a mineral absorbed by plants directly form the soil. Sol. : Herbivores and other animals obtain phosphorus from plants.

111. Answer (4) Sol. : Sulphur cycle is a sedimentary cycle

112. Answer (1) Hint : Ecological pyramids do not take into account the same species belonging to two or more trophic levels. Sol. : Ecological pyramids accommodate a simple food chain and do not accommodate a food web.

113. Answer (4) Hint : Dodo, an extinct species of bird was native to Mauritius. Sol. : Quagga – Africa Steller’s sea cow – Russia Thylacine – Australia



114. Answer (1) Sol. : In vertebrates, highest number of species are of fishes.

115. Answer (2) Hint : A stable community shows low variability of ecosystem processes like productivity, water use, pest and disease cycle. Sol. : A stable community does not show too much variation in productivity from year to year.

116. Answer (4) Hint : Warm temperature and high humidity in tropical areas provide favourable conditions throughout the year. Sol. : Tropics receive more solar radiations compare to poles. Its environment is less seasonal, relatively more constant and predictable.

117. Answer (4) Hint : New species entering a geographical region is called exotic species. Sol. : Cichlid fish is the native species of Lake Victoria of East Africa.

118. Answer (3) Sol. : Kanha National park (Madhya Pradesh) is for the protection of tigers.

119. Answer (2) Hint : Directly derived economical benefits for human are called narrowly utilitarian benefits of ecosystem. Sol. : Oxygen, pollination and erosion control are broadly utilitarian benefits of ecosystem.

120. Answer (3) Sol. : National parks are in-situ conservation strategies.

121. Answer (4) Hint : Transition zone which is the outermost part of biosphere reserve is the area of active cooperation between reserve management and local tribals. Sol. : Core zone of biosphere reserve is the undisturbed and legally protected ecosystem.

122. Answer (2) Sol. : High species richness, high degree of endemism and degree of threat, measured in terms of habitat loss are the criteria used to determine a hot spot.

123. Answer (2) Sol. : Ozone (O3) is a secondary pollutant.

124. Answer (1) Hint : Eutrophication is natural ageing of lakes due to nutrient enrichment particularly with nitrogen and phosphorus. Sol. : Electrostatic precipitator – Removal of particulate matter Scrubber – Removal of SO2 Green muffler – Reduction of noise pollution

125. Answer (3) Sol. : Leakage of poisonous gas MIC (Methyl Isocyanate) was cause of Bhopal gas tragedy.

126. Answer (3) Hint : Large amount of organic material in water bodies increases the BOD of water. Sol. : In water bodies, presence of high amount of organic materials in water causes algal bloom, promotes the growth of water hyacinth and causes death of fishes and other aquatic animals.

127. Answer (2) Sol. : Incineration is the process used for disposal of hospital wastes.

128. Answer (3) Sol. : CNG unlike petrol or diesel cannot be siphoned off by thieves.

129. Answer (4) Hint : Biomagnification of DDT in birds disturb calcium metabolism and causes thinning of egg shells. Sol. : Addition of pesticides and fertilizers in a soil is a positive soil pollution.

130. Answer (3) Sol. : Painful skeletal deformities (itai-itai) is an effect caused by cadmium.

131. Answer (3) Sol. : Montreal protocol was an international treaty to control the emission of ozone depleting substances.

132. Answer (4) Sol. : Skin cancer, damage of cornea, snow-blindness are the effect caused by UV-B. Methemoglobinemia is because of binding of nitrite with haemoglobin.



133. Answer (1) Sol. : Sunder Lal Bahuguna is related to Chipko movement for the conservation of forests.

134. Answer (2)

Hint : Jhum cultivation is slash and burn agriculture practiced in north eastern states of India. Sol. : Jhum cultivation leads to deforestation.

135. Answer (2)

Sol. : Sound of intensity more than 80 dB causes noise pollution.

136. Answer (3)

Hint : Identify transgenic sheep.

Sol. : ANDi is transgenic monkey.

Dolly is first cloned sheep.

Rosie is transgenic cow.

137. Answer (4)

Hint : Agrobacterium tumefaciens infects dicots.

Sol. : Cereals are monocots, therefore can’t usually be infected by Ti plasmid. Tomato, soyabean and sunflower are dicots.

138. Answer (2)

Hint : Endo mean inside/between.

Sol. : Exonucleases remove nucleotides from ends by breaking phosphodiester bonds.

139. Answer (3)

Hint : Sticky ends are produced by this restriction enzyme.

Sol. : Eco RI produces sticky ends.

Eco RV, Hind II and Sma I produce blunt ends upon cleaving DNA.

140. Answer (4)

Hint : Proinsulin has an additional polypeptide chain.

Sol. : Insulin consists of two short polypeptide chains: Chain A and Chain B. Insulin is synthesised as a prohormone which has an additional ‘C’ peptide chain.

141. Answer (1) Hint : Identify an algae.

Sol. : Bacteriophage is a virus that infects bacteria. Taq polymerase is extracted from thermostable Thermus aquaticus bacterium. Agrobacterium tumefaciens is a natural genetic engineer of plants.

142. Answer (2) Hint : Bt produces proteins that aggregate to form crystals. Sol. : The crystal proteins are toxic to specific species of insects yet harmless to host. They are pore forming proteins that bind to receptors on insect’s midgut & cause death.

143. Answer (3) Hint : Amplification of DNA. Sol. : Very low count of viruses can be detected by PCR by multiplication of their nucleic acid. PCR and ELISA are usually used to detect HIV in suspected patients.

144. Answer (2) Hint : Meloidogyne incognita infects tobacco plants and not bacteria. Sol. : RNAi takes place only in eukaryotes and not in prokaryotes, as a method of cellular defence.

145. Answer (4) Hint : RNA to DNA. Sol. : Alkaline phosphatase cleaves ‘OH group at 5’ end of DNA, hence prevents self ligation of vector.

146. Answer (2) Hint : It causes anthrax. Sol. : Bioweapons are highly infectious pathogens, their spores & toxins are used against humans, crop & animals of enemy country. E.coli are not harmful. B.thuringiensis is a biological pesticide.

147. Answer (2) Hint : DNA fragments move towards anode. Sol. : DNA fragments can be separated by forcing them to move towards anode owing to their negative charge.

148. Answer (3) Hint : This step follows staining. Sol. : DNA fingerprinting is used to identify living beings based on samples of their DNA. Method of obtaining DNA in form of a spool over a glass rod is called spooling.



149. Answer (1) Hint : REs cleave foreign DNA of virus in bacteria. Sol. : RE is called so because it restricts bacteriophages to grow in bacteria.

150. Answer (2) Hint : It is present in A. tumefaciens. Sol. : Ti plasmid of Agrobacterium tumefaciens which is a pathogen of several dicot plants, it invades plants at the site of wound and transforms them and nearby cells to form a tumour called crown gall tumour. BAC is bacterial artificial chromosome. Lambda phage is a bacteriophage. Phagemid’s structure is made up of bacteriophage and plasmid.

151. Answer (4) Hint : Amplication of DNA. Sol. : A radioactively labelled DNA probe containing the sequence of interest is then hybridised with the DNA on the membrane and detected by autoradiography. ELISA is based on antigen-antibody interaction.

152. Answer (2) Hint : Analyses of mammalian genomes. Sol. : Bacterial artificial chromosomes are used for mapping of large eukaryotic genome and can accomodate large DNA fragments. YAC : Yeast artificial chromosome. Phagemid’s structure is made up of bacteriophage and plasmid.

153. Answer (1) Hint : Microneedle is used. Sol. : The alien gene is introduced into a cell by injecting it in the nucleus of the target cell under a microscope. Biolistics/gene gun method is suitable for plants. Short impulses of high field strength are given in electroporation.

154. Answer (2) Hint : Identify a nematode. Sol. : Hind III – Haemophilus influenzae Sal I – Streptococcus albus Sma I – Serratia marcescens

155. Answer (4) Hint : It contains agitator blades. Sol. : The design of agitator blades is crucial for proper mixing of the reactor contents.

156. Answer (1) Hint : Plasmid is double stranded. Sol. : Baculoviruses are biological pesticides. They inhibit larval development in insects. Transposons are jumping genes.

157. Answer (3)

Hint : β-carotene gene was taken from Daffodil. Sol. : Transgenic rice express high level of β-carotene which is a precursor of Vit. A.

158. Answer (1) Hint : These genes code for Cry protein. Sol. : cry I Ac and cry II Ab are known to control cotton bollworm.

159. Answer (3) Hint : Antigen – antibody reaction. Sol. : Enzymes used in ELISA catalyse formation of coloured product from colourless substrate indicating the presence of Ag and Ab.

160. Answer (4) Hint : Restriction site for Pvu I is closer to Pst I. Sol. : Insertion at Pvu I recognition site leads to insertional inactivation of ampicillin resistance gene but recombinants will be resistant to tetracycline.

161. Answer (4) Hint : Gene therapy. Sol. : Functional ADA gene is introduced into the lymphocytes with the help of retroviruses. These lymphocytes are introduced into patient. If the gene isolated from marrow cells producing ADA is introduced into cells at early embryonic stages, it could be a permanent cure.

162. Answer (1) Hint : Penicillium. Sol. : Fungal cell wall is rich in chitin.

163. Answer (1) Hint : Human insulin. Sol. : They prepared two DNA sequences by reverse transcription and linked them separately with plasmid of E.coli. Dolly was cloned at the Roslin institute, Scotland. Saheli is a research product of CDRI, Lucknow. RCGM takes decisions on biosafety research.



164. Answer (4) Hint : It digests hydrocarbons of crude oil. Sol. : It is an organism having ability to digest hydrocarbons and has been used for bioremediation.

165. Answer (4)

Hint : Gene subtraction.

Sol. : mRNA silencing is also called RNAi. In this method silencing of a specific mRNA due to formation of dsRNA molecule formed by binding of complementary RNA (anti-sense RNA) molecule to original mRNA occurs, thereby preventing translation of the original mRNA (silencing).

166. Answer (2)

Hint : Abiotic conditions in midgut of bacterium.

Sol. : The cry I Ab and cry II Ab genes code for Cry proteins which control corn borer and cotton bollworm respectively.

Bt protoxin gets activated in alkaline pH in the midgut of B. thuringiensis.

167. Answer (3)

Hint : Antigen specificity.

Sol. : Transgenic animals have not been used for early detection of diseases. Various tests which detect antigens have been employed for this purpose.

168. Answer (2)

Hint : Identify an oilseed crop.

Sol. : Roundup Ready canola is genetically modified oilseed crop.

169. Answer (1)

Hint : Bt for Bacillus thuringiensis.

Sol. : GM crops were insect resistant e.g. Bt cotton, Bt brinjal. The toxin expression is contained within the plant system and insects feeding on the crop perish.

170. Answer (4)

Hint : Pseudomonas feeds on hydrocarbons.

Sol. : Crops engineered for glyphosate are tolerant to herbicides. Pseudomonas putida has been used for bioremediation.

171. Answer (2) Hint : These enzymes are considered as molecular scissors.

Sol. : Restriction endonucleases were used for cutting DNA at specific sequences.

172. Answer (2) Hint : Circle will open. Sol. :

173. Answer (3)

Hint : Identify a closed system.

Sol. : A selectable marker helps in identifying non-transformants. Continuous culture allow cells to be maintained in exponential phase continuously.

174. Answer (1)

Hint : It doesn’t affect human respiratory system.

Sol. : Rhinovirus causes common cold

Haemophilus influenzae causes severe pneumonia

H1N1 causes swine flu

175. Answer (2)

Hint : Exotic breeds brought to India.

Sol. : Shahtoosh – Chiru

Angoora – Rabbit

176. Answer (4)

Hint : Method to overcome inbreeding depression.

Sol. : Cross breeding allows the desirable qualities of two different breeds to be combined. Interspecific hybridization is mating of two different related species. A single outcross often helps to overcome inbreeding depression.

Outbreeding is the breeding of unrelated animals within same breed, but having no common ancestors upto 4-6 generations.



177. Answer (2) Hint : To avoid competition for food. Sol. : Mode of nutrition of fish is heterotrophic. Different species of fish are cultured together independent of their size and type of excretory products.

178. Answer (1) Hint : White diarrhoea. Sol. : Pullorum in fowls is caused by Salmonella pullorum. It spreads through contaminated food and water and is characterised by loss of appetite and difficulty in breathing. Clostridium tetani causes tetany, H. influenzae causes pneumonia while M. leprae causes leprosy.

179. Answer (1) Hint : Superior parents are chosen.

Sol. : Multiple ovulation embryo transfer is a program in which superior cows and bulls are chosen and are hybridised to obtain superior progeny.

180. Answer (3)

Hint : Aquaculture.

Sol. : Aquaculture has led to increase in the production of aquatic plants and animals, both fresh water and marine.

Yellow revolution in India was launched to increase the production of edible oil seeds in the country to meet the domestic demand.

White revolution is associated with a sharp increase in milk production. Green revolution is associated with agricultural productivity improvement.

Test-7 (Code-F)_(Answers) All India Aakash Test Series for NEET-2020


All India Aakash Test Series for NEET-2020

Test Date : 15/03/2020

ANSWERS

1. (2) 2. (2) 3. (2) 4. (4) 5. (1) 6. (1) 7. (3) 8. (1) 9. (2) 10. (3) 11. (1) 12. (2) 13. (2) 14. (3) 15. (1) 16. (1) 17. (2) 18. (2) 19. (4) 20. (2) 21. (3) 22. (4) 23. (3) 24. (2) 25. (3) 26. (3) 27. (2) 28. (3) 29. (1) 30. (4) 31. (4) 32. (1) 33. (2) 34. (3) 35. (3) 36. (3)

37. (2) 38. (3) 39. (4) 40. (2) 41. (4) 42. (2) 43. (1) 44. (3) 45. (1) 46. (3) 47. (4) 48. (2) 49. (2) 50. (1) 51. (1) 52. (4) 53. (4) 54. (4) 55. (2) 56. (2) 57. (3) 58. (4) 59. (4) 60. (1) 61. (3) 62. (4) 63. (1) 64. (2) 65. (1) 66. (1) 67. (2) 68. (3) 69. (4) 70. (4) 71. (1) 72. (4)

73. (1) 74. (3) 75. (2) 76. (4) 77. (1) 78. (4) 79. (3) 80. (4) 81. (4) 82. (2) 83. (2) 84. (2) 85. (4) 86. (4) 87. (3) 88. (4) 89. (1) 90. (3) 91. (2) 92. (2) 93. (1) 94. (4) 95. (3) 96. (3) 97. (4) 98. (3) 99. (2) 100. (3) 101. (3) 102. (1) 103. (2) 104. (2) 105. (4) 106. (3) 107. (2) 108. (3)

109. (4) 110. (4) 111. (2) 112. (1) 113. (4) 114. (1) 115. (4) 116. (2) 117. (3) 118. (3) 119. (4) 120. (4) 121. (2) 122. (3) 123. (2) 124. (1) 125. (2) 126. (3) 127. (3) 128. (4) 129. (2) 130. (3) 131. (1) 132. (2) 133. (4) 134. (2) 135. (3) 136. (3) 137. (1) 138. (1) 139. (2) 140. (4) 141. (2) 142. (1) 143. (3) 144. (2)

145. (2) 146. (4) 147. (1) 148. (2) 149. (3) 150. (2) 151. (4) 152. (4) 153. (1) 154. (1) 155. (4) 156. (4) 157. (3) 158. (1) 159. (3) 160. (1) 161. (4) 162. (2) 163. (1) 164. (2) 165. (4) 166. (2) 167. (1) 168. (3) 169. (2) 170. (2) 171. (4) 172. (2) 173. (3) 174. (2) 175. (1) 176. (4) 177. (3) 178. (2) 179. (4) 180. (3)

TEST - 7 (Code-F)

All India Aakash Test Series for NEET-2020 Test-7 (Code-F)_(Hints & Solutions)


HINTS & SOLUTIONS

[PHYSICS] 1. Answer (2)

Hint & Sol. : 0

112 1 13 136 0 6 7C n C N− β+ → →

2. Answer (2)

Hint & Sol : Reverse bias applied to a junction diode increases the potential barrier.

3. Answer (2)

Hint : Voltage gain = –gmR0, where gm is transconductance and R0 is output resistance.

Sol. : AV = –gmR0

⇒ AV ∝ gm

1 1

2 2

=V m

V m

A g

A g

50 0.040.02

=VA

AV = 25

4. Answer (4) Hint : Photo diode will absorb those frequencies, by absorbing which, electron excite to higher energy state.

Sol. : Eg = hνmin

1.65 × 1.6 × 10–19 = 6.6 × 10–34 ν 19

min 341.65 1.6 10

6.6 10

−

−

× ×ν =

×

= 4 × 1014 Hz

5. Answer (1) Hint & Sol. :

4 4 42 2He 2− − −

−→ + → +A A AZ Z ZP Q R e

So P and R are isotopes

6. Answer (1) Hint : Remaining value after n half lives.

012

nN N =

Sol. : Half life period T = 40 minute

Number of half lives 120 340

tnT

= = =

30

01

82N

N N = =

Amount decayed = N0 – N

0 018

= −N N

078

N=

7. Answer (3)

Hint : Momentum conservation law.

Sol. : 0 = +E mvc

= −Ev

mc

Kinetic energy of nucleus 212

mv=

21

2Em

mc =

2

22Emc

=

8. Answer (1)

Hint : Radius of nuclei 13

0R R A=

Sol. : Radius of nucleus of 4Be9 = R

Radius of 32X nucleus = 2R

139

2RR A

=

3 912 A

=

A = 72

Number of neutrons in nucleus X = 72 – 32 = 40

Test-7 (Code-F)_(Hints & Solutions) All India Aakash Test Series for NEET-2020


9. Answer (2) Hint : Diode is reversed biased. Sol. :

If VA < VB then junction diode will be reverse biased and no current will pass through junction diode. So effective resistance is 12 Ω.

10. Answer (3) Hint & Sol. : Truth table

A B Y′ C Y

0 0 1 1 1

1 0 0 0 0

0 1 0 1 0

1 1 0 0 0

11. Answer (1)

Hint & Sol. : The potential barrier of junction opposes the movement of majority carriers towards junction but supports the minority carriers to move towards junction.

12. Answer (2) Hint & Sol. : Energy gap of GaAs0.6P0.4 is 1.9 eV and emits red colour.

13. Answer (2) Hint : Voltage across Zener diode is equal to voltage across 1 kΩ resistance. Sol.: Breakdown voltage of Zener diode = 10 V Voltage drop across 250 Ω = 10 V

Current through dropping resistance 10250

VR

= =

= 40 mA

Current through 1 kΩ resistance 10 10 mA1000

= =

Current through Zener diode = 40 – 10 = 30 mA

14. Answer (3)

Hint : Current gain β = c

b

ii

Sol. : IE = IB + IC

Current in base = 2% of 2100e ei i=

Current in collector = 98% of 98100e ei i=

49c

b

ii

β = =

15. Answer (1)

Hint : 2·e h in n n=

Sol. : In intrinsic semi conductor e h in n n= =

ni = 1.5 × 1016 per m3

In extrinsic semiconductor hole density

nh = 4.5 × 1022 per m3

ne × 4.5 × 1022 = (1.5 × 1016)2

32

221.5 1.5 10

4.5 10en × ×=

×

= 5 × 109 per m3

16. Answer (1)

Hint : Energy of electron in any orbit

213.6 eV,= −nEn

angular momentum of electron =

2nh

π

Sol. : Energy of electron in n = 1, E1 = –13.6 eV

Energy given to atom E = 12.75 eV

Net final energy of electron, En = – 13.6 + (12.75)

= – 0.85 eV

213.6– 0.85n

= −

2 13.6 160.85

n = =

n = 4

So 4 22 2nh h hL = = =

π π π



17. Answer (2)

Hint : Time period in any orbit T ∝ n3

Sol. : T1 = 27 T2 3

11

22

nTnT

=

⇒ 3

12

22

27 nTnT

=

⇒ 1

23

nn

=

18. Answer (2) Hint & Sol. : Energy gap is maximum between energy level 1 and 2. So radiation of maximum frequency will be emitted, when electron makes transition from n = 2 to n = 1.

19. Answer (4)

Hint : 12400 eV(Å)

E =λ

Sol. : λ = 1215 Å

12400 10.2 eV1215

= =E

So the wavelength of 1215 Å may excite the hydrogen atom form n = 1 to n = 2. So the number of spectral lines in emission spectrum is 1.

20. Answer (2)

Hint : 1v

λ ∝

Sol. : λ =h

mv

137

λ =hc zm

n

137λ =

nmcz

λ α n

21. Answer (3) Sol. : Intensity of photons.

nlAt

=

From point source wavefront will be spherical

24nld t

=π

21

αld

so 21

αid

22. Answer (4)

Hint & Sol. : Photoelectrons will be emitted only when wavelength of incident light is less than certain limiting value called threshold wavelength. Then photocurrent is proportional to intensity of light.

23. Answer (3)

Hint : 3

hkTm

λ =

1T

λ ∝

Sol. : T1 = 27 + 273 = 300 K

T2 = 927 + 273 = 1200 K

2 1

1 2

300 11200 2

TT

λ= = =

λ

12 2

λλ =

24. Answer (2)

Hint : 12400 eV(Å)

E =λ

eVchc

= − φλ

Sol. : λ1 = 5000 Å, λ2 = 3000 Å

Case – 1

eV1c = 2.48 – φ

Case – 2

eV2c = 4.13 – φ

V2c – V1c = 1.65 V

25. Answer (3)

Hint : de-Broglie wavelength λ =h

mv

Sol. : Force on electron ˆ=

F eE i

Acceleration of electron ˆeEa im

=



Speed of electron 2 20 2 eEv v d

m= +

0 20

21 eEv v dmv

= +

de-Broglie wavelength

0 20

21

heEdmv

mv

λ =+

0

20

21 eEdmv

λλ =

+

26. Answer (3) Hint & Sol. : The wavelength of red light is greater than threshold wavelength, so electrons will not emit for red light.

27. Answer (2) Hint : Energy of photon hν = φ + max (K.E) Sol. : ν1 = ν, ν2 = 1.25 ν, K1 = 0.5 eV, K2 = 0.75 eV Case 1 hν = φ + 0.5 eV …(1) Case 2 1.25 hν = φ + 0.75 eV …(2) Multiply equation (1) by 5 and (2) by 4 and on subtracting φ = 0.5 eV

28. Answer (3) Hint : Shift of fringe on introducing a film

( )1x tβ= µ −

λ

Sol. : µ = 1.6

Shift 2

x β=

( 1)2

tβ β= µ −

λ

6000 5000 Å2( 1) 2(1.6 1)

t λ= = =

µ − −

t = 5 × 10–7 m

29. Answer (1) Hint : Path difference at any point at distance y.

ydyD

∆ =

For dark fringe ∆y = (2n – 1) 2λ n = 1, 2, 3……..

Sol. : 2dy = n = 2

For dark fringe ( )2 1 2yd

nDλ

= −

2 32 2dD

λ=

2

3λ =

dD

30. Answer (4) Hint & Sol. : Resolving power of compound

microscope 2 sinµ θ=

λ

So, resolving power ∝ µ.

31. Answer (4)

Hint : 1 2m D n Dd dλ λ

=

Sol. : d = 1 mm, D = 2 m λ1 = 6300 Å λ2 = 5400 Å Two bright fringes coincide if m × 6300 = n × 5400 ⇒ m = 6, n = 7

Distance on screen, 7

13

6 6 6.3 10 21 10

−

−

λ × × ×= =

×

Dy

d

= 7.56 mm

32. Answer (1) Hint : In diffraction, for subsidiary maxima, path

difference = (2n + 1) 2λ

For minima path difference = nλ n = 1, 2, 3 ………. Sol. : For 1st subsidiary maxima

sin30 32

d λ° = …(i)

For second minima dsinθ = 2λ …(ii) Divide (ii) by (i)

sin 23

2 2

dd

θ λ=

λ

42sin3

θ =

1 2sin3

− θ =



33. Answer (2)

Hint : Light transmitted through a polaroid

I = l0 cos2θ Sol. : Light transmitted through two polaroid at 30°

I2 = I1cos230°

0 0332 4 8l l

= × =

Light not transmitted 0 00

3 58 8l l

l= − =

% of light not transmitted 5 100 62.5%8

= × =

34. Answer (3)

Hint : When light enters in liquid its wavelength will change.

Fringe width = Dd

λ .

Sol. : Wavelength in water aw

λλ =

µ

λ βλβ = = =

µ µa aw

wDD

d d

0.6 3 0.45 mm4w×

β = =

35. Answer (3)

Hint : For bright fringe, path difference is nλ.

Sol. : Path difference ∆y = 4 × 10–6 m

Wavelength λ = 5 × 10–7 m

nλ = ∆x 6

74 105 10

xn−

−

∆ ×= =

λ ×

= 8

Path difference is integer multiple of λ, so bright fringe will obtain and point will be bright.

36. Answer (3)

Hint & Sol. : If wave front is cylindrical.

2= =

πE ElAt r l t

or 1lr

α

37. Answer (2)

Hint & Sol. : Real depth (R.D)Apparent depth (A.D)

µ =

From 1st surface (R.D)1 = µ × (A.D)1 = 1.5 × 5

= 7. 5 cm

From 2nd surface (R.D)2 = µ × (A.D)2 = 1.5 × 7

= 10.5 cm

Thickness of slab = 18 cm.

38. Answer (3)

Hint : Length of instrument L = f0 + ue

Sol. : Image formed by objective is at focus of objective lens

For eyepiece

ve = – 25 cm. fe = 5 cm ue = ?

1 1 1

e e ef v u= −

1 1 15 25 eu

+ = −

5 1 125 eu+

= −

25 4.17 cm6eu = − = −

L = 60 + 4.17

= 64.17

39. Answer (4)

Hint : sin

2

sin2

+ δ µ =

mA

A

Sol. : Given A = 60°

60sin22

sin30

m° + δ =

°

1 60sin2 2

m° + δ =

6045

2m° + δ

= °

δm = 30°



40. Answer (2)

Hint : For concave lens, virtual image will form at focus, which acts as an object for convex lens.

Sol. :

For convex lens

u = – 30 cm, f = 15 cm, v = ?

1 1 1f v u

= −

1 1 115 30v

= +

1 1 1 115 30 30v

= − =

v = 30 cm

Distance of final image from concave lens = 50 cm

41. Answer (4)

Hint : µ =cv

2 1

1 2sin

vC

vµ

= =µ

Sol. : 8

82 10 8sin

92.25 10C ×

= =×

1 8sin9

C − =

42. Answer (2)

Hint & Sol. : In compound microscope both the lenses are of short focal length and difference of focal should be small and f0 < fe.

43. Answer (1)

Hint : 1 2

1 1 1( 1)f R R

= µ − −

Sol. :

⇒ 1 10.460 2R

= ×

⇒ R = 12 cm

and 2R = 24 cm

44. Answer (3)

Hint : Length of image = 1 2v v− .

Sol. : For near end of object,

u1 = –30 cm, f = –15 cm

From mirror equation,

1

1 1 130 15v

− = −

1

1 1 1 1 230 15 30v

−⇒ = − =

⇒ v1 = –30 cm

For far end of object

u2 = –45 cm,

So, 2

1 1 1 3 115 45 45v

− += − + =

245 22.5 cm2

v⇒ = − = −

So, length of image 1 2 7.5 cmv v− =

45. Answer (1)

Hint & Sol. : When a real object is placed at centre of curvature, image is also formed at centre of curvature. So least distance is zero.



[CHEMISTRY]46. Answer (3)

Hint : Zn is refined by distillation and Sn by liquation.

47. Answer (4) Hint : Green fuel is obtained by recycling plastic waste.

48. Answer (2) Hint : Maximum limit for nitrate in drinking water is 50 ppm.

49. Answer (2) Hint : Bithional is an antiseptic.

50. Answer (1) Hint : Phenelzine is an antidepressant drug.

51. Answer (1) Hint : Bakelite is a cross-linked polymer.

52. Answer (4) Hint : In vulcanisation, sulphur forms cross links at the reactive sites of double bonds.

53. Answer (4)

Hint : is Buna-N.

54. Answer (4) Hint : Cross-linked polymers have 3D-network structure.

55. Answer (2) Hint : Buna-S is addition polymer of butadiene and styrene.

56. Answer (2)

Hint : Glucose Zymase→ C2H5OH

Sol. : Sucrose Invertase→ glucose and fructose

Urea Urease→ CO2 + NH3

Cellulose Cellulase→ Glucose

57. Answer (3) Hint : In globular proteins, polypeptide chains coil around to give a spherical shape and these are usually soluble in water.

58. Answer (4)

Hint : Glucose can form six membered ring in which –OH at C-5 can add to the –CHO group.

59. Answer (4)

Hint : Adenine, Guanine and Uracil bases are present in RNA.

Sol. : Thymine is present in DNA.

60. Answer (1)

Hint : Night blindness is caused by vitamin A deficiency.

61. Answer (3)

Hint : Glycine 2

COOH

H N H

H

is non-essential

amino acid.

62. Answer (4)

Hint : Lactose formed by the linkage between C1 of galactose and C4 of glucose.

Sol. : Lactose is a reducing sugar.

63. Answer (1)

Hint : Glucose get oxidised by Br2 water.

Sol. : 4

2

CHO|

(CHOH)|

CH OH

2Br water→4

2

COOH|

(CHOH)|

CH OH

Glucose Gluconic acid

64. Answer (2)

Hint : Monosaccharides can not be hydrolysed further.

Sol. : Glucose is a monosaccharide.

65. Answer (1)

Hint : NH4NO3 ∆→ N2O + 2H2O

Sol. :

• (NH4)2Cr2O7∆→ N2 + 4H2O + Cr2O3

• 2H2O + NH2CONH2 → (NH4)2CO3

• (NH4)2SO4 + 2NaOH → 2NH3 + 2H2O + Na2SO4



66. Answer (1)

Hint : Oxidation state of Cl in Cl2O7 is +7.

Sol. : Acidic nature : Cl2O7 > SO3 > N2O5 > N2O3.

67. Answer (2)

Hint : Phosphine burns spontaneously in contact with air.

68. Answer (3)

Hint : Zn + 4HNO3(conc) → Zn(NO3)2 + 2H2O +

2NO2

Sol. :

• 4Zn + 10HNO3(dilute) → 4Zn(NO3)2 + 5H2O

+ N2O

• Cu + 4HNO3(conc) → Cu(NO3)2 + 2NO2

+ 2H2O

• 3Cu + 8HNO3(dilute) → 3Cu(NO3)2 + 2NO

+ 4H2O

69. Answer (4)

Hint : N2O : N ≡ N → O.

Sol. : N2O3 :

N2O4 :

N2O5 :

70. Answer (4)

Hint : Order of oxidising power:

HCIO > HCIO2 > HCIO3 > HCIO4.

71. Answer (1)

Hint : XeF6 + 3H2O → XeO3 + 6HF

Sol. : XeF6 + H2O → XeOF4 + 2HF

XeF6 + 2H2O → XeO2F2 + 4HF

72. Answer (4)

Hint : P4 + 8SOCl2 → 4PCl3 + 4SO2 + 2S2Cl2.

73. Answer (1)

Hint : (HPO3)3 is cyclotrimetaphosphoric acid.

Sol. :

It contain three P — O — P bonds.

74. Answer (3)

Hint : 2NaOH + Cl2 → NaCl + NaOCl + H2O. (cold and dilute)

Sol. : 6NaOH + 3Cl2 → 5NaCl + NaClO3 + 3H2O. (hot and conc)

75. Answer (2)

Hint : H-X bond dissociation enthalpy decreases down the group.

Sol. : HF HCl HBr HI

∆dissH°(kJ/mol) 574 432 363 295

76. Answer (4)

Hint : Impure Zn contains Pb, Cd and Fe as major impurities.

77. Answer (1)

Hint : Cupellation is used for refining Ag.

78. Answer (4)

Hint : Monel metal is an alloy of Ni, Cu and Fe.

79. Answer (3)

Hint : 2Cu2O + Cu2S → 6Cu + SO2↑.

80. Answer (4)

Hint : Copper matte contains mostly Cu2S and little FeS.

81. Answer (4)

Hint : Cryolite : Na3AlF6, Fluorspar : CaF2.

Sol. : The fusion temperature of mixture [Al2O3, Na3AIF6, CaF2] lowers to 900°C.

82. Answer (2)

Hint : 2[Au(CN)2]–(aq) + Zn(s) →

2Au + [Zn(CN)4]2–(aq)



83. Answer (2)

Hint : van Arkel process is used for obtaining ultra pure metals.

Sol. : 2 4Zr 2I ZrI+ → ↑

1800 K4 2ZrI Zr 2I→ +

84. Answer (2)

Hint : Copper glance : Cu2S.

85. Answer (4)

Hint : Electron rich hydrides contain lone pairs of electrons.

Sol. : HF⋅⋅

, 2H O⋅⋅ , 3NH

86. Answer (4)

Hint : Isotopes have similar chemical properties.

Sol. : Dielectric constant of ordinary water is greater than heavy water.

87. Answer (3) Hint : 2H2O2 → 2H2O + O2

Sol. : 22.4 L of O2 at STP is obtained from 68 g of H2O2

50 L of O2 at STP is obtained from = 68 5022.4

×

= 151.78g L–1

15%

88. Answer (4) Hint : In cyclic and chain silicates, two corners of each tetrahedron are shared.

89. Answer (1) Hint : In heavier members the tendency to show +2 oxidation state increases among 14 group elements.

90. Answer (3) Hint : Melting point of B is exceptionally high. Sol. : Element Melting Point B 2453 K Al 933 K

Ga 303 K In 430 K

[BIOLOGY]91. Answer (2)

Sol. : Sound of intensity more than 80 dB causes noise pollution.

92. Answer (2) Hint : Jhum cultivation is slash and burn agriculture practiced in north eastern states of India. Sol. : Jhum cultivation leads to deforestation.

93. Answer (1) Sol. : Sunder Lal Bahuguna is related to Chipko movement for the conservation of forests.

94. Answer (4) Sol. : Skin cancer, damage of cornea, snow-blindness are the effect caused by UV-B. Methemoglobinemia is because of binding of nitrite with haemoglobin.

95. Answer (3) Sol. : Montreal protocol was an international treaty to control the emission of ozone depleting substances.

96. Answer (3) Sol. : Painful skeletal deformities (itai-itai) is an effect caused by cadmium.

97. Answer (4) Hint : Biomagnification of DDT in birds disturb calcium metabolism and causes thinning of egg shells. Sol. : Addition of pesticides and fertilizers in a soil is a positive soil pollution.

98. Answer (3) Sol. : CNG unlike petrol or diesel cannot be siphoned off by thieves.

99. Answer (2) Sol. : Incineration is the process used for disposal of hospital wastes.

100. Answer (3) Hint : Large amount of organic material in water bodies increases the BOD of water. Sol. : In water bodies, presence of high amount of organic materials in water causes algal bloom, promotes the growth of water hyacinth and causes death of fishes and other aquatic animals.



101. Answer (3) Sol. : Leakage of poisonous gas MIC (Methyl Isocyanate) was cause of Bhopal gas tragedy.

102. Answer (1) Hint : Eutrophication is natural ageing of lakes due to nutrient enrichment particularly with nitrogen and phosphorus. Sol. : Electrostatic precipitator – Removal of particulate matter Scrubber – Removal of SO2 Green muffler – Reduction of noise pollution

103. Answer (2) Sol. : Ozone (O3) is a secondary pollutant.

104. Answer (2) Sol. : High species richness, high degree of endemism and degree of threat, measured in terms of habitat loss are the criteria used to determine a hot spot.

105. Answer (4) Hint : Transition zone which is the outermost part of biosphere reserve is the area of active cooperation between reserve management and local tribals. Sol. : Core zone of biosphere reserve is the undisturbed and legally protected ecosystem.

106. Answer (3) Sol. : National parks are in-situ conservation strategies.

107. Answer (2) Hint : Directly derived economical benefits for human are called narrowly utilitarian benefits of ecosystem. Sol. : Oxygen, pollination and erosion control are broadly utilitarian benefits of ecosystem.

108. Answer (3) Sol. : Kanha National park (Madhya Pradesh) is for the protection of tigers.

109. Answer (4) Hint : New species entering a geographical region is called exotic species. Sol. : Cichlid fish is the native species of Lake Victoria of East Africa.

110. Answer (4)

Hint : Warm temperature and high humidity in tropical areas provide favourable conditions throughout the year. Sol. : Tropics receive more solar radiations compare to poles. Its environment is less seasonal, relatively more constant and predictable.

111. Answer (2)

Hint : A stable community shows low variability of ecosystem processes like productivity, water use, pest and disease cycle. Sol. : A stable community does not show too much variation in productivity from year to year.

112. Answer (1)

Sol. : In vertebrates, highest number of species are of fishes.

113. Answer (4)

Hint : Dodo, an extinct species of bird was native to Mauritius. Sol. : Quagga – Africa

Steller’s sea cow – Russia

Thylacine – Australia

114. Answer (1)

Hint : Ecological pyramids do not take into account the same species belonging to two or more trophic levels. Sol. : Ecological pyramids accommodate a simple food chain and do not accommodate a food web.

115. Answer (4)

Sol. : Sulphur cycle is a sedimentary cycle

116. Answer (2)

Hint : Phosphorus is a mineral absorbed by plants directly form the soil. Sol. : Herbivores and other animals obtain phosphorus from plants.

117. Answer (3)

Hint : In a food web, one animal may feed upon organism of different trophic levels. Sol. : In a food web, sparrow is a primary consumer when it eats seeds, fruits etc. and a secondary consumer when it eats insects and worms.



118. Answer (3) Sol. : Amount of all the inorganic substances present in an ecosystem per unit area at a given time is called standing state.

119. Answer (4) Hint : Hydrarch succession starts in aquatic habitat. Sol. : All kinds of successions including hydrarch succession, increase total biomass and humus content of the soil. In hydrarch succession, phytoplanktons are pioneer community and it progresses from hydric to mesic conditions.

120. Answer (4) Hint : Plants capture 2-10% of photosynthetically active radiation (PAR) for photosynthesis. Sol. : Energy flow in an ecosystem is always unidirectional (from producers to consumers) and energy does not remain trapped in any organism permanently as it either flows through trophic levels or becomes available to decomposers for decomposition. Energy flow follows the laws of thermodynamics.

121. Answer (2) Hint : Secondary succession starts in areas that somehow lost all the living organisms that existed there. Sol. : Secondary succession can occur at abandoned farm lands, burned and cut forests and flooded lands.

122. Answer (3) Hint : Humus has slightly acidic pH. Sol. : Humus is dark coloured, amorphous and more or less decomposed organic matter.

123. Answer (2) Hint : Pyramid of number is not upright for the ecosystem having parasitic food chain. Sol. : In pond ecosystems and grassland ecosystems, pyramid of number is upright.

124. Answer (1) Hint : For aquatic ecosystems, GFC is the major conduit of energy flow. Sol. : In terrestrial ecosystems, much larger fraction of energy flows through DFC than through GFC.

125. Answer (2)

Sol. : The rate of formation of new organic matter by consumers is called secondary productivity.

126. Answer (3)

Hint : For decomposition, availability of air and saprophytic microorganisms is important. Sol. : Decomposition is better/ faster if detritus lack lignin, tannin, cellulose etc and if soil has optimum temperature and moisture.

127. Answer (3)

Hint : Anthropogenic ecosystems are man-made ecosystems such as agricultural farms. Sol. : Anthropogenic ecosystems have high productivity.

128. Answer (4)

Hint : All mammals and birds are regulators. Sol. : All mammals and birds are capable of thermoregulation and osmoregulation to maintain constant internal environment.

129. Answer (2)

Hint : Reduction of leaves into spines is an adaptation to prevent water loss in xerophytes.

Sol. : Spines and flattened phylloclade (photosynthetic modified stem) are the adaptations seen in Opuntia.

130. Answer (3)

Hint : Most ecologically relevant environmental factor is temperature. Sol. : Temperature gradually decreases latitude wise from equator towards the poles and altitude wise from plains to mountain tops.

131. Answer (1)

Hint : Darwin’s finches present in Galapagos island are the example of competitive co-existence. Sol. : Parasitism – Cuscuta on hedge plants.

Commensalism – Epiphytes (orchids) growing on other plants.

Mutualism – Fig and fig wasp.



132. Answer (2) Hint : Logistic growth of a population is represented by sigmoid (S-shaped) curve. Sol. : Logistic growth occurs when the resource in a habitat are limited.

133. Answer (4) Hint : A population showing declined growth has less number of pre-reproductive individuals as compare to reproductive and post-reproductive individuals. Sol. : Age pyramid of population showing declined growth is urn-shaped.

134. Answer (2) Hint : Kangaroo rat shows physiological adaptations to survive in North American desert. Sol. : Kangaroo Rat of North American desert is capable of meeting all its water requirements through its internal fat oxidation.

135. Answer (3) Sol. : Grasslands and tundra biomes do not have trees.

136. Answer (3) Hint : Aquaculture. Sol. : Aquaculture has led to increase in the production of aquatic plants and animals, both fresh water and marine. Yellow revolution in India was launched to increase the production of edible oil seeds in the country to meet the domestic demand. White revolution is associated with a sharp increase in milk production. Green revolution is associated with agricultural productivity improvement.

137. Answer (1) Hint : Superior parents are chosen. Sol. : Multiple ovulation embryo transfer is a program in which superior cows and bulls are chosen and are hybridised to obtain superior progeny.

138. Answer (1) Hint : White diarrhoea. Sol. : Pullorum in fowls is caused by Salmonella pullorum. It spreads through contaminated food and water and is characterised by loss of appetite and difficulty in breathing. Clostridium tetani causes tetany, H. influenzae causes pneumonia while M. leprae causes leprosy.

139. Answer (2) Hint : To avoid competition for food. Sol. : Mode of nutrition of fish is heterotrophic. Different species of fish are cultured together independent of their size and type of excretory products.

140. Answer (4) Hint : Method to overcome inbreeding depression. Sol. : Cross breeding allows the desirable qualities of two different breeds to be combined. Interspecific hybridization is mating of two different related species. A single outcross often helps to overcome inbreeding depression. Outbreeding is the breeding of unrelated animals within same breed, but having no common ancestors upto 4-6 generations.

141. Answer (2) Hint : Exotic breeds brought to India. Sol. : Shahtoosh – Chiru Angoora – Rabbit

142. Answer (1) Hint : It doesn’t affect human respiratory system. Sol. : Rhinovirus causes common cold Haemophilus influenzae causes severe pneumonia H1N1 causes swine flu

143. Answer (3) Hint : Identify a closed system. Sol. : A selectable marker helps in identifying non-transformants. Continuous culture allow cells to be maintained in exponential phase continuously.

144. Answer (2) Hint : Circle will open. Sol. :



145. Answer (2)

Hint : These enzymes are considered as molecular scissors.

Sol. : Restriction endonucleases were used for cutting DNA at specific sequences.

146. Answer (4)

Hint : Pseudomonas feeds on hydrocarbons.

Sol. : Crops engineered for glyphosate are tolerant to herbicides. Pseudomonas putida has been used for bioremediation.

147. Answer (1)

Hint : Bt for Bacillus thuringiensis.

Sol. : GM crops were insect resistant e.g. Bt cotton, Bt brinjal. The toxin expression is contained within the plant system and insects feeding on the crop perish.

148. Answer (2)

Hint : Identify an oilseed crop.

Sol. : Roundup Ready canola is genetically modified oilseed crop.

149. Answer (3)

Hint : Antigen specificity.

Sol. : Transgenic animals have not been used for early detection of diseases. Various tests which detect antigens have been employed for this purpose.

150. Answer (2)

Hint : Abiotic conditions in midgut of bacterium.

Sol. : The cry I Ab and cry II Ab genes code for Cry proteins which control corn borer and cotton bollworm respectively.

Bt protoxin gets activated in alkaline pH in the midgut of B. thuringiensis.

151. Answer (4)

Hint : Gene subtraction.

Sol. : mRNA silencing is also called RNAi. In this method silencing of a specific mRNA due to formation of dsRNA molecule formed by binding of complementary RNA (anti-sense RNA) molecule to original mRNA occurs, thereby preventing translation of the original mRNA (silencing).

152. Answer (4)

Hint : It digests hydrocarbons of crude oil.

Sol. : It is an organism having ability to digest hydrocarbons and has been used for bioremediation.

153. Answer (1)

Hint : Human insulin.

Sol. : They prepared two DNA sequences by reverse transcription and linked them separately with plasmid of E.coli. Dolly was cloned at the Roslin institute, Scotland. Saheli is a research product of CDRI, Lucknow. RCGM takes decisions on biosafety research.

154. Answer (1)

Hint : Penicillium.

Sol. : Fungal cell wall is rich in chitin.

155. Answer (4)

Hint : Gene therapy.

Sol. : Functional ADA gene is introduced into the lymphocytes with the help of retroviruses. These lymphocytes are introduced into patient. If the gene isolated from marrow cells producing ADA is introduced into cells at early embryonic stages, it could be a permanent cure.

156. Answer (4)

Hint : Restriction site for Pvu I is closer to Pst I.

Sol. : Insertion at Pvu I recognition site leads to insertional inactivation of ampicillin resistance gene but recombinants will be resistant to tetracycline.

157. Answer (3)

Hint : Antigen – antibody reaction.

Sol. : Enzymes used in ELISA catalyse formation of coloured product from colourless substrate indicating the presence of Ag and Ab.

158. Answer (1)

Hint : These genes code for Cry protein.

Sol. : cry I Ac and cry II Ab are known to control cotton bollworm.



159. Answer (3)

Hint : β-carotene gene was taken from Daffodil.

Sol. : Transgenic rice express high level of β-carotene which is a precursor of Vit. A.

160. Answer (1)

Hint : Plasmid is double stranded.

Sol. : Baculoviruses are biological pesticides. They inhibit larval development in insects. Transposons are jumping genes.

161. Answer (4)

Hint : It contains agitator blades.

Sol. : The design of agitator blades is crucial for proper mixing of the reactor contents.

162. Answer (2)

Hint : Identify a nematode.

Sol. : Hind III – Haemophilus influenzae

Sal I – Streptococcus albus

Sma I – Serratia marcescens

163. Answer (1)

Hint : Microneedle is used.

Sol. : The alien gene is introduced into a cell by injecting it in the nucleus of the target cell under a microscope. Biolistics/gene gun method is suitable for plants. Short impulses of high field strength are given in electroporation.

164. Answer (2)

Hint : Analyses of mammalian genomes.

Sol. : Bacterial artificial chromosomes are used for mapping of large eukaryotic genome and can accomodate large DNA fragments. YAC : Yeast artificial chromosome. Phagemid’s structure is made up of bacteriophage and plasmid.

165. Answer (4)

Hint : Amplication of DNA.

Sol. : A radioactively labelled DNA probe containing the sequence of interest is then hybridised with the DNA on the membrane and detected by autoradiography. ELISA is based on antigen-antibody interaction.

166. Answer (2) Hint : It is present in A. tumefaciens. Sol. : Ti plasmid of Agrobacterium tumefaciens which is a pathogen of several dicot plants, it invades plants at the site of wound and transforms them and nearby cells to form a tumour called crown gall tumour. BAC is bacterial artificial chromosome. Lambda phage is a bacteriophage. Phagemid’s structure is made up of bacteriophage and plasmid.

167. Answer (1) Hint : REs cleave foreign DNA of virus in bacteria. Sol. : RE is called so because it restricts bacteriophages to grow in bacteria.

168. Answer (3) Hint : This step follows staining. Sol. : DNA fingerprinting is used to identify living beings based on samples of their DNA. Method of obtaining DNA in form of a spool over a glass rod is called spooling.

169. Answer (2) Hint : DNA fragments move towards anode. Sol. : DNA fragments can be separated by forcing them to move towards anode owing to their negative charge.

170. Answer (2) Hint : It causes anthrax. Sol. : Bioweapons are highly infectious pathogens, their spores & toxins are used against humans, crop & animals of enemy country. E.coli are not harmful. B.thuringiensis is a biological pesticide.

171. Answer (4) Hint : RNA to DNA. Sol. : Alkaline phosphatase cleaves ‘OH group at 5’ end of DNA, hence prevents self ligation of vector.

172. Answer (2) Hint : Meloidogyne incognita infects tobacco plants and not bacteria. Sol. : RNAi takes place only in eukaryotes and not in prokaryotes, as a method of cellular defence.



173. Answer (3)

Hint : Amplification of DNA.

Sol. : Very low count of viruses can be detected by PCR by multiplication of their nucleic acid. PCR and ELISA are usually used to detect HIV in suspected patients.

174. Answer (2)

Hint : Bt produces proteins that aggregate to form crystals.

Sol. : The crystal proteins are toxic to specific species of insects yet harmless to host. They are pore forming proteins that bind to receptors on insect’s midgut & cause death.

175. Answer (1)

Hint : Identify an algae.

Sol. : Bacteriophage is a virus that infects bacteria. Taq polymerase is extracted from thermostable Thermus aquaticus bacterium. Agrobacterium tumefaciens is a natural genetic engineer of plants.

176. Answer (4) Hint : Proinsulin has an additional polypeptide chain.

Sol. : Insulin consists of two short polypeptide chains: Chain A and Chain B. Insulin is synthesised as a prohormone which has an additional ‘C’ peptide chain.

177. Answer (3) Hint : Sticky ends are produced by this restriction enzyme. Sol. : Eco RI produces sticky ends. Eco RV, Hind II and Sma I produce blunt ends upon cleaving DNA.

178. Answer (2) Hint : Endo mean inside/between. Sol. : Exonucleases remove nucleotides from ends by breaking phosphodiester bonds.

179. Answer (4) Hint : Agrobacterium tumefaciens infects dicots. Sol. : Cereals are monocots, therefore can’t usually be infected by Ti plasmid. Tomato, soyabean and sunflower are dicots.

180. Answer (3) Hint : Identify transgenic sheep. Sol. : ANDi is transgenic monkey. Dolly is first cloned sheep. Rosie is transgenic cow.

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