ZEOTS-22 | QUESTIONS · focal length 10 cm in such a way that its end closer to the pole is 20 cm...

PAGE: 1 ZEOTS-22 NEET MODEL EXAMINATION

TRIVANDRUM KOLLAM ERNAKULAM ZEPHYR

ZEOTS-22 | QUESTIONS

1. The average energy density of an e.m . wave given by E = (50 N/C) sin (ωt-kx) will be nearly

(1) 10−8J/ m3 (2) 10−7J/ m3

(3) 10−6J/ m3 (4) 10−5J/ m3

2. A body executes SHM under the

influence of one force and has a period T1 seconds and the same body executes SHM with period T2 second when under the influence of another force. When both forces act simultaneously and in the same direction, then the time period of the same body is:

(1) 1 2( )secT T+ (2) 2 21 2 secT T+

(3) 2 2

1 2

1 2

secT TTT

+ (4) 2 2

1 22 2

1 2

sec( )

T TT T+

3. When the mass is doubled in figure (i)

the end of the spring descends an additional distance h. the frequency of oscillation for the arrangement in figure (ii), when all individual springs shown in figures (i) and (ii) are identical, is:

(1) 1/ 21 3

2 5gh

(2) 1/ 21 6

2 5ghπ

(3) 1/ 21 8

2 3ghπ

(4) 1/ 252

6gh

π

4. The bob of a simple pendulum executes

simple harmonic motion in water with a period t, while the period of oscillation of the bob is t0 in air. Neglect frictional force of water and given that the density of the bob is (4/3) × 1000 kg/m3. What relationship is true between t and t0?

(1) t = t0 (2) 0

2tt =

(3) t = 2t0 (4) t = 4t0 5. Two particles execute SHM of the same

amplitude and frequency along the same straight line. If they pass one another when going in opposite directions, each time their displacement is half their amplitude; the phase difference between them is:

(1) 3π (2)

4π

(3) 6π (4) 2

3π

6. A body executes simple harmonic

motion. The potential energy (PE), kinetic energy (KE) and total energy (TE) are measured as a function of displacement x. Which of the following statements is true?

(1) TE is zero when x = 0 (2) PE is maximum when x = 0 (3) KE is maximum when x = 0 (4) KE is maximum when x is maximum 7. One face of a rectangular glass plate 6cm

thick is silvered. An object held 8cm in front of the first face forms an image 12cm behind the silvered face. The refractive index of the glass is

(1) 0.4 (2) 0.8 (3) 1.2 (4) 1.6 8. An electromagnetic radian of frequency

n, wavelength λ, travelling with velocity

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v in air enters in a glass slab of refractive index (µ). The frequency, wavelength and velocity of light in the glass slab will be respectively

(1) ,n λµ

and vµ

(2) n, 2λ and vµ

(3) ,n λµ µ

and vµ

(4) 2 ,π λµ µ

and v

9. A boy is trying to start a fire by focusing

sunlight on a piece of paper using an equiconvex lens of focal length 10 cm. The diameter of the Sun is 1.39 × 109 m and its mean distance from the earth is 1.5 × 1011 m. What is the diameter of the Sun’s image on the paper

(1) 9.2 × 10–4 m (2) 6.5 × 10–4 m (3) 6.5 × 10–5 m (4) 12.4 × 10–4 m 10. A biconvex lens (µ = 1.5) has a radius of

curvature of magnitude 20 cm. Which one of the following options best describe the image formed of an object of height 2 cm placed 30 cm from the lens

(1) Virtual, upright, height = 1 cm (2) Virtual, upright, height = 0.5 cm (3) Real, inverted, height = 4 cm (4) Real, inverted, height = 1 cm 11. A thin prism of angle 15o made of glass

of refractive index µ1 = 1.5 is combined with another prism of glass of refractive index µ2 = 1.75. The combination of the prism produces dispersion without deviation. The angle of the second prism should be

(1) 7o (2) 10o (3) 12o (4) 5o

12. A converging beam of rays is incident on

a diverging lens. Having passed through the lens the rays intersect at a point 15 cm from the lens on the opposite side. If the lens is removed the point where the rays meet will move 5 cm closer to the lens. The focal length of the lens is

(1) –10 cm (2) 20 cm (3) –30 cm (4) 5 cm

13. Which of the following graphs is the magnification of a real image against the distance from the focus of a concave mirror

(1) (2) (3) (4) 14. A rod of length 10 cm lies along the

principal axis of a concave mirror of focal length 10 cm in such a way that its end closer to the pole is 20 cm away from the mirror. The length of the image is

(1) 10 cm (2) 15 cm (3) 2.5 cm (4) 5 cm 15. A plano convex lens fits exactly into a

plano concave lens. Their plane surfaces are parallel to each other. If lenses are made of different materials of refractive indices µ1 and µ2 and R is the radius of curvature of the curved surface of the lenses, then the focal length of the combination is

(1) 1 22( )R

µ µ− (2)

1 2( )R

µ µ−

(3) 2 1

2( )

Rµ µ−

(4) 1 22( )R

µ µ+

16. The head lights of a jeep are 1.2 m apart.

If the pupil of the eye of an observer has a diameter of 2 mm and light of wavelength 5896 Å is used, what should be the maximum distance of the jeep from the observer if the two head lights are just separated

(1) 3.9 km (2) 33.9 m (3) 3.39 km (4) 3.39 m

Distance

m

y

x Distance

m

y

x

Distance

m

y

x Distance

m

y

x



17. A simple telescope, consisting of an objective of focal length 60 cm and a single eye lens of focal length 5 cm is focused on a distant object is such a way that parallel rays come out from the eye lens. If the object subtends an angle 2o at the objective, the angular width of the image

(1) 10o (2) 24o (3) 50o (4) 1/6o 18. A bar is subjected to equal and opposite

forces as shown in the figure. PQ is plane making angle θ with the cross-section of the bar. If the area of cross-section be ‘a’, then what is the tensile stress on PQ?

(1) F/a (2) F cos θ/a

(3) 2cosF

aθ (4) F/a cosθ

19. A wire of initial length L and radius r is

stretched by a length l. Another wire of same material but with initial length 2L and radius 2r is stretched by a length 2l. The ratio of the stored elastic energy per unit volume in the first and second wire is:

(1) 1 : 4 (2) 1 : 2 (3) 2 : 1 (4) 1 : 1 20. Water rises in a capillary tube to a certain

height such that the upward force due to surface tension is balanced by 75 ×10–4 N force due to the weight of the liquid. If the surface tension of water is 6 × 10–2 N/m, the inner circumference of the capillary tube must be:

(1) 1.25 ×10–2 m (2) 0.50 ×10–2 m (3) 6.5 ×10–2 m (4) 12.5 ×10–2 m 21. A cylinder is filled with a liquid of

density d upto a height h. if the beaker is at rest, then the mean pressure on the wall is:

(1) zero (2) hdg

(3) 2h dg (4) 2hdg

22. A flat plate of area 10 cm2 is separated

from a large plate by a layer of glycerin 1 mm thick. If the coefficient of viscosity of glycerin is 20 poise, the force required to keep the plate moving with a velocity of 1 cm/sec is:

(1) 80 dyne (2) 200 dyne (3) 800 dyne (4) 2000 dyne 23. There are two identical small holes of

area of cross-section a on the opposite sides of a tank containing a liquid of density ρ. The difference in height between the holes is h. The tank is resting on a smooth horizontal surface. Horizontal force which will has to be applied on the tank to keep it in equilibrium is:

(1) ghρa (2) 2ghaρ

(3) 2ρagh (4) gha

ρ

24. The total area of cross-section is 0.25 m2.

If the blood is flowing at the rate if 100 cm3/sec, then the average velocity of flow of blood through the capillaries is:

(1) 0.4 mm/sec (2) 4 mm/sec (3) 25 mm/ sec (4) 400 mm/sec 25. If oxygen (O2) has root mean square

velocity of C ms–1, then root mean square velocity of hydrogen (H2) at the same temperature will be

(1) C ms–1 (2) 1C

ms–1

(3) 4C ms–1 (4) C/4 ms–1

F F

P

Q

θ



26. The tube shown in the figure is of uniform cross-section. Liquid flows through it at a constant speed in the direction shown by the arrows. The liquid exerts on the tube

(1) A net force to the right (2) A net force to the left (3) A clockwise torque (4) An anticlockwise torque 27. The equation of a wave is represented by

Y = 10–4 sin 10010xt −

m, then the

velocity of wave will be (1) 100 m/s (2) 4 m/s (3) 1000 m/s (4) zero 28. A 1000 Hz sound wave in air strikes the

surface of a lake and penetrates into water. If speed of sound in water is 1500 m/s, the frequency and wavelength of waves in water are

(1) 1500 Hz, 1 m (2) 1000 Hz, 1.5 m (3) 1000 Hz, 1 m (4) 1500 Hz, 1.5 m 29. If the temperature is raised by 1K from

300 K the percentage change in the speed of sound in the gaseous mixture is (R =8.31 J/mole-K)

(1) 0.167% (2) 0.334% (3) 1% (4) 2% 30. In an experiment with sonometer, a

tuning fork of frequency 256 Hz resonates with a length of 25 cm and another tuning fork resonates with a length of 16 cm. Tension of the string remaining constant, the frequency of the second turning fork is

(1) 204.8 Hz (2) 163.84 Hz (3) 400 Hz (4) 320 Hz

31. The general equation of a wave in a string is y = 0.1 sin π (0.10x – 8t + 1.3). The equation of the wave that would produce a stationary wave with the given wave is

(1) y = 0.1 cos π (0.10x – 8t + 1.3) (2) y = 0.1 sin π (0.10x + 8t + 1.3) (3) y = 0.1 sin π (8 x + 0.10 t + 1.3) (4) y = 0.1 cos π (8x + 0.10x + 1.3) 32. An organ pipe open at one end is

vibrating in first overtone and is in resonance with another pipe open at both ends and vibrating in third harmonic. The ratio of length of two pipes is

(1) 3 : 8 (2) 8 : 3 (3) 1 : 2 (4) 4 : 1 33. A person carrying a whistle emitting

continuously a note of 272 Hz is running towards a reflecting surface with a speed of 18 km/hour. The speed of sound in air is 345 m/s. The number of beats heard by him is

(1) 4 (2) 6 (3) 8 (4) 3 34. In brass, the velocity of longitudinal

wave is 100 times the velocity of the transverse wave. If Y = 1 × 1011 N/m2, then stress in the wire is

(1) 1 × 1013 N/m2 (2) 1 × 109 N/m2 (3) 1 × 1011 N/m2 (4) 1 × 107 N/m2 35. In Young’s double slit experiment, the

screen is 2 m from the double slits, the slits are 2 mm apart and are illuminated by photons of two wavelengths λ1 = 12000 Å and λ2 = 10000 Å. At what minimum distance from the common central bright fringe, a bright fringe from one interference pattern coincides with a bright fringe from the other

(1) 3 mm (2) 8 mm (3) 6 mm (4) 4 mm 36. A light source approaches the observer

with velocity 0.8 c. The doppler shift for the light of wavelength 5500 Å is

(1) 4400 Å (2) 1833 Å (3) 3167 Å (4) 7333 Å



37. A single slit of width 0.20 mm is illuminated with light of wavelength 500 nm. The observing screen is placed 80 cm from the slit. The width of the central bright fringe will be

(1) 1 mm (2) 2 mm (3) 4 mm (4) 5 mm 38. Diffraction and interference of light

suggest (1) Nature of light is electro-magnetic (2) Wave nature (3) Nature is quantum (4) Nature of light is transverse 39. Light of wavelength λ = 5000 Å falls

normally on a narrow slit. A screen placed at a distance of 1 m from the slit and perpendicular to the direction of light. The first minima of the diffraction pattern is situated at 5 mm from the centre of central maximum. The width of the slit is

(1) 0.1 mm (2) 1.0 mm (3) 0.5 mm (4) 0.2 mm 40. Two Nicols are oriented with their

principal planes making an angle of 60o. The percentage of incident unpolarized light which passes through the system is

(1) 50% (2) 100% (3) 12.5% (4) 37.5% 41. When a junction diode is reverse biased,

the flow of current across the junction is mainly due to

(1) Diffusion of charges (2) Depends upon the nature of material (3) Drift of charges (4) Both drift and diffusion of charges 42. The current gain β of a transistor is 50.

The input resistance of the transistor, when used in the common emitter configuration, is 1 kΩ. The peak value of the collector a.c. current for an alternating peak input voltage 0.01 V is

(1) 100 µA (2) 250 µA (3) 500 µA (4) 800 µA

43. The device that can act as a complete electronic circuit is

(1) Zener diode (2) Junctions diode (3) Integrated circuit (4) Junction transistor 44. For given n-p-n transistor RB = 100 kΩ.

The potential drop across the base emitter junction is 0.6 V then find VBB. If IC = 1 mA and β = 100

(1) 2.6 V (2) 1.6 V (3) 10.6 V (4) 2 V 45. The combination of the gates shown

represents

(1) AND gate (2) OR gate (3) NAND gate (4) NOR gate 46. 0.50 g of an organic compound was

Kjeldahlised and the NH3 evolved was absorbed in 50 ml of 0.5 M H2SO4. The residual acid required 70 ml of 0.5 M NaOH. The percentage of nitrogen in the organic compound is

(1) 14 (2) 28 (3) 56 (4) 42 47. The substance that would not be formed

during the reaction of methane and chlorine in the presence of sunlight is

(1) CH3Cl (2) CHCl3 (3) CH3CH3 (4) CH3CH2CH3



48. The major product obtained when isobutane is treated with chlorine in the presence of light is

(1) Isobutyl chloride (2) n-butyl chloride (3) tert-butyl chloride (4) sec-butyl chloride 49. Base strength of (i) H3CCH2

– (ii) H2C = CH– (iii) H – C ≡ C– is in the order (1) (iii) > (ii) > (i) (2) (i) > (iii) > (ii) (3) (i) > (ii) > (iii) (4) (ii) > (i) > (iii) 50. Complete combustion of a sample of a

hydrocarbon gives 0.66 g of CO2 and 0.36 g of H2O. The empirical formula of the compound is

(1) CH2 (2) C3H4 (3) C3H8 (4) C6H8 51. Impure glycerine is purified by (1) Steam distillation (2) Simple distillation (3) Vacuum distillation (4) Fractional distillation 52. In the following reaction, The major product is (1) (2)

(3) (4) 53. Ethylbenzene with bromine in presence

of FeBr3, predominantly gives (1)

(2)

(3) (4) 54. Which of the following carbocations is

least stable?

(1) 6 5 2C H C H+

(2) 2 6 4 2p NO C H CH +− − −

(3) 3 6 4 2p CH O C H CH +− − −

(4) 6 4 2p Cl C H CH +− − − 55. In which of the following species, all

types of hybrid carbons are present? (1) CH2 = C = CH2 (2) 3 2CH CH CH CH +− = −

(3) 3 2CH C C CH +− ≡ −

(4) 3 2CH CH CH CH −− = − 56. An alcohol on oxidation is found to give

CH3COOH and CH3CH2COOH. The structure of the alcohol is

(1) CH3CH2CH2OH (2) (CH3)2C(OH)CH2CH3 (3) CH3(CH2)3CH2OH (4) CH3CHOHCH2CH3

CH3 | CH3 – C – CH = CH2 2 /H O H +

→ A + B | (major (minor CH3 product) product)

CH3 | H3C – C – CH – CH3 | | HO CH3

CH3 | CH2 – C – CH2CH3 | | HO CH3

CH3 | H3C – C – CH – CH3 | | CH3 OH

CH3 | H3C – C – CH2 – CH2 | | CH3 OH

CH2CH3 Br

CH2CH2Br

CH2CH3 Br

CH(Br)CH3



57. Which of the following statements are correct for phenol?

I. Phenol is more acidic than alcohol II. Phenol is used in the production of

melamine plastic. III. Phenol gives violet colour with

neutral ferric chloride solution. IV. Phenol when heated with acetyl

chloride gives phenetole (1) I and II (2) I and III (3) III and IV (4) I and IV 58. CH3OC2H5 and (CH3)3COCH3 are

treated with hydriodic acid. The fragments after reaction obtained are

(1) CH3I + HOC2H5 ; CH3I + (CH3)3C – OH

(2) CH3OH + C2H5I ; (CH3)3C – I + HOCH3

(3) CH3OH + C2H5I ; (CH3)3C – OH + CH3I

(4) CH3I + HOC2H5 ; (CH3)3C – I + HOCH3

59. Choose the correct acidity order of the following phenols (1) p-nitrophenol > p-chlorophenol > p-

cresol > phenol > p-methoxyphenol (2) p-nitrophenol > p-methoxyphenol >

p-cresol > p-chlorophenol > phenol (3) p-nitrophenol > p-chlorophenol >

phenol > p-cresol > p-methoxyphenol (4) p-nitrophenol > phenol > p-

chlorophenol > p-methoxyphenol > p-cresol

60. Anisole 3 3 2 3

3

(CH ) CCl Cl /FeClAlCl

HBrHeat

→ → X

The product X in the above series of reactions is

(1) (2)

(3) (4)

61. Benzaldehyde and acetone can be best distinguished using

(1) Fehling’s solution (2) Hydroxylamine (3) 2, 4-DNP (4) Tollen’s reagent 62. Phenol

3

3 4

.

.

CH ClZn Alk

dust Anhyd AlCl KMnOX Y Z→ → → ,

the product Z is (1) benzaldehyde (2) benzoic acid (3) benzene (4) toluene 63. Match the following

Column I Column II (1) CH3CH2CH2

CH2OH PCC→ (p (CH3CH2)2C = O

(2) C6H5CH3 2 2CrO Cl→

(q CH3COCH3

(3) CH3COCl + (CH3)2 Cd →

(r) CH3CH2CH2CHO

(4) (CH3CH2COO)2

Ca ∆→ (s) C6H5CHO

(1) A – r; B – s; C – q; D – p (2) A – s; B – r; C – q; D – p

(3) A – r; B – s; C – p; D – q (4) A – p; B – q; C – r; D – s 64. Cyclohexene on reaction with ozone

followed by decomposition with zinc dust and water gives compound E. Compound E on further treatment with dilute NaOH yields compound F. Compound F is

(1) (2) (3) (4)

CHO

CHO

COOH

CO2H CO2H



65. Reagents required for the conversion respectively are (1) BH3/THF, H2O2/OH–,

C5H5NH+CrO3Cl– (2) BH3/THF, H2O2/OH–,

K2Cr2O7/H2SO4 (3) Hg(OAc)2THF – H2O, NaBH4/OH–,

C5H5NH+CrO3Cl– (4) H2O/H+, C5H5NH+CrO3Cl–

66. Among the isomers of C5H11Cl, the one which is chiral is

(1) 2-methyl-2-chlorobutane (2) 3-chloropentane

(3) 2-chloropentane (4) 1-chloro-2, 2-dimethylpropane

67. Which will undergo SN2 reaction fastest among the following halogen compounds?

(1) CH3CH2F (2) CH3CH2Cl (3) CH3CH2Br (4) CH3CH2I

68. Which of the following statement is incorrect?

(1) Carbon black is formed during incomplete combustion of alkanes

(2) Pyrolysis of alkanes is believed to be a free radical reaction

(3) Preparation of oil gas or petrol gas from kerosene oil or petrol involves the principle pyrolysis

(4) n-hexane is treated with anhydrous AlCl3 and HCl gas to give benzene.

69. Which one of the following is aromatic? (1) Cyclopropene (2) Cyclopropenyl cation

(3) Cyclopropenyl anion (4) Cyclopentadienyl cation 70. The polyhalogen compound used for the

production of the Freon refrigerant R-22 is (1) CH2Cl2 (2) CHCl3

(3) CH3Cl (4) DDT

71. The total number of benzene derivatives having the molecular formula C7H7Cl are

(1) 2 (2) 3 (3) 4 (4) 5

72. Which will exhibit geometrical isomerism?

(1) 3-phenylbut-1-ene (2) 1-phenylbut-2-ene (3) 2-phenylbut-1-ene (4) 1, 1-diphenylprop-1-ene 73. An alkyl halide by formation of its

Grignard reagent and heating with water gives propane. What is the original alkyl halide?

(1) Methyl iodide (2) Ethyl iodide (3) Ethyl bromide (4) Propyl bromide 74. Ethylene can be separated from acetylene

by passing the mixture through: (1) Fuming H2SO4 (2) Pyrogallol (3) Ammoniacal Cu2Cl2 (4) Charcoal power 75. Increasing order towards nucleophilic

substitution

(1) (i) < (ii) < (iii) (2) (i) < (iii) < (ii) (3) (iii) < (ii) < (i) (4) (ii) < (iii) < (i) 76. The product (4) of the following reaction

is 2

3

/3 ( )

( ) ( ) ( )

+

∆

→ →

→ →

KCN H O H

NH

CH Cl A

B C D

(1) 3 2 2CH CH NH (2) 3CH CN (3) 2HCONH (4) 3 2CH CONH

77. Which reaction sequence would he best to prepare 3-chloroaniline from benzene? (1) Chlorination, nitration, reduction (2) Nitration, chlorination, reduction (3) Nitration, reduction, chlorination (4) Nitration, reduction, acetylation, chlorination, hydrolysis.

CH2 CHO



78. The correct repeating structural unit of polystyrene is

(1) (2) (3) (4) 79. Which of the following is a condensation

homopolymer? (1) Nylon-6 (2) Nylon-6, 6 (3) Buna-N (4) Dacron

80. The order of basic strength among the

following amines in benzene solution is (1) CH3)2NH > CH3NH2 > (CH3)3N (2) (CH3)3N > (CH3)2NH > CH3NH2 (3) CH3NH2 > (CH3)2NH > (CH3)3N (4) (CH3)2NH > (CH3)3N > CH3NH2 81. The pentaacetate of glucose does not

react with hydroxylamine indicating the (1) presence of –OH group (2) presence of free – CHO group (3) absence of – OH group (4) absence of free – CHO group

82. In both DNA and RNA, hetrocyclic base

and phosphate ester linkages are at (1) C-5’ and C-2’ respectively of the

sugar molecule (2) C-2’ and C-5’ respectively of the



sugar molecule 83. A ketone upon reaction with ethyl

magnesium bromide followed by hydrolysis gave a product which on

dehydration gave an alkene. The alkene on ozonolysis gave diethyl ketone and acetaldehyde. The ketone is (1) dimethyl ketone (2) ethyl methyl ketone (3) diethyl ketone (4) ethyl propyl ketone

84. What is the product of the following

reaction? (1) (2) (3) (4)

85. Which of the following is the correct

decreasing order of boiling points of the given isomeric compounds? Ethyl dimethylamine (I), n-butylamine (II), Diethylamine(III) (1) II > III > I (2) I > II > III (3) III > II > I (4) III > I > II

86. The IUPAC name of the following

compound is

(1) 2-ethoxy-4-methoxypentan-2-one (2) 2-methoxy-4-ethoxy-pentan-3-one (3) 2-ethoxy-4-methoxy-pentan-3-one (4) 1-ethoxy-3-methoxy-1-methylbutan-2-one

O || CH3 – CH – C – CH – OCH2CH3 | | OCH3 CH3



87. Terfenadine is commonly used as a/ an (1) tranquilliser (2) antishistamine (3) antimicrobial (4) antifertility drug 88. Decreasing – I effect of given groups is

(i) CN (ii) NO2 (iii) NH2 (iv) F

(1) iii > ii >i> iv (2) ii > iii > iv > i (3) iii > ii > iv > i (4) ii > i > iv > iii

89. What will be the major product in the

following mononitration reaction (1) (2) (3) (4)

90. A student named a certain compound as 2, 3-diethylbutane its correct IUPAC name is

(1) 2, 3-dimethylhexane (2) 2-ethyl-3-methylpentane

(3) 3, 4-dimethylhexane (4) 2-ethylbutane

91. Which one is an organelle within organelle

(1) Nucleolus (2) Ribosome (3) Peroxisome (4) Mesosome 92. Leniar cytoskeletal elements of diameter

about 6 nm are (1) Microfilaments (2) Microtubules (3) Keratin (4) Intermediate filaments 93. Golgi complex play a major role in (1) Digesting protein and lipids (2) Packaging of enzymes (3) Glycosylation (4) Both (2) & (3) 94. Which of the following does not differ in

E. coli and Chlorella? (1) Cell wall (2) Ribosomes (3) Cell membrane (4) Both (2) & (3) 95. Polypeptide synthesis in cell takes place

in (1) Mitochondria (2) Ribosomes (3) Golgi (4) Cytoplasm 96. Membrane bound organelles are absent

in (1) Pseudomonas (2) Saccharomyces (3) Sporozoans (4) Both (1) & (3) 97. Plasma membrane consist mainly of (1) Proteins embedded within the

membrane (2) Phospholipid layers embedded in a

protein layer (3) Proteins embedded in a phospholipid

bilayer (4) Proteins embedded in a bilayer of

phospholipids and carbohydrates

O

N |

H

3

2 4

..

Conc HNOConc H SO→

O

N |

H NO2

O

N |

H

NO2

O

N |

H

O2N

O

N |

H

O2N



98. Which of the following have its own DNA?

(1) Dictyosomes (2) Mitochondria (3) Peroxisome (4) All of these 99. The main arena of metabolism in all cells

is (1) Karyon (2) Plasmalemma (3) Cytoplasm (4) Mitochondrion 100. Major site for synthesis of lipids is (1) RER (2) SER (3) Golgi (4) Cytoplasm 101. Cellular organelles with membranes are (1) Lysosomes, Microbodies, RER (2) Lysosomes, Nucleus, Microtubules (3) Golgi apparatus, Chromosomes,

Endoplasmic reticulum (4) Nuclei, Nucleolus, ER 102. Which of the following set contain a

structure which is not an inclusion body (1) Gas vacuole, cynophycean granule,

glycogen granule (2) Gas vacuole, polysome, glycogen

granule (3) Cyanophycean granule, phosphate

granule, Mesosome (4) Both (2) & (3) 103. RBC measures about (1) 5 n.m in diameter (2) 12 n.m in diameter (3) 7 n.m in diameter (4) 7 µm in diameter 104. Polysomes are (1) Ribosomes attached to RER

(2) Cluster of ribosomes in cytoplasm (3) Chain of ribosomes that translate the mRNA

(4) Chain of inactive ribosomes on the mrRNA

105. Mycoplasma has (1) A cell wall of lipids (2) No cell membrane and Cell wall (3) A cell wall made up of non-

peptidoglycan substance (4) No cell wall

106. Fluidity of the membrane is due to (1) Water content of the membrane (2) Coating of outer surface by sugars (3) Lateral movement of proteins within

the bilayer of lipids (4) Trilaminar structure

107. Secondary wall of plants are formed (1) Towards membrane side (2) Towards the outer side of the cell

wall (3) Just inner to middle lamella (4) Just on either side of middle lamella 108. Choose the correct description of

centrosome (1) A centrosome is bounded by

pericentriolar amorphous substances and has one centriole

(2) Centrosome is bouned by single membrane and contains 2 centrioles

(3) A centrosome contains 2 centrioles and surrounded by amorphous substances

(4) A centrosome contains 2 pairs of centrioles and not surrounded by any membrane

109. Microtubules are absent in (1) Prokaryotes (2) Plant cells (3) Animal cells (4) Both (1) & (2) 110. The dense staining organelles near to

nucleus is (1) Mitochondria (2) Golgi (3) Chromosomes (4) Both (2) & (3) 111. Choose the incorrect statement (1) The perinuclear space is about 50 n.m (2) Nucleoplasm contains chromatin and

nucleolus (3) The content of nucleolus is

continuous with rest of nucleoplasm (4) Passage of materials through nuclear

pores is unidirectional 112. Which of the following are correct? 1. The inner and outer membranes of

mitochondria are different in enzyme content

2. Mitochondria and chloroplast can divide by fission



3. The ‘biological energy currency’ is mitochondrion

4. Chloroplast has DNA and ribosomes but no RNA

(1) 1 & 4 only (2) 1, 2 & 4 only (3) 1 & 2 only (4) 1, 3 & 4 only 113. Gram – ve bacteria have an outer

membrane in the cell envelope which contain

(1) Porins (2) Chaperons (3) Lectins (4) Cyclins 114. Svedberg’s unit (s); indirectly is a

measure of (1) Molecular weight (2) Size (3) Density (4) Both (2) & (3) 115. The proportion of RNA and protein in

prokaryotic ribosome is about (1) 65% and 35% respectively (2) 50% and 50% respectively (3) 20% and 80% respectively (4) 80% and 20% respectively 116. Vacuole in plant is (1) Membrane bound and contain storage

proteins and metabolites (2) Membrane bound and contain water

and excretory substance (3) Membrane bound and contains air (4) Membraneless and contains stored

food substances of various kinds 117. the main organellel involved in

modification and routing of newly synthesised proteins to their destination is

(1) Mitochondria (2) Lysosome (3) E.R (4) Nucleus 118. Select the incorrect statement from the

following (1) Both chloroplast and mitochondria

contain an inner compartment and inter membrane space

(2) Both chloroplast and mitochondria have inter membrane space and thylakoid space in the inner compartment

(3) Both have chemiosmotic membranes (4) Chloroplasts are generally larger than

Mitochondria 119. Find the true aspect regarding plasmid (1) All bacteria have plasmid (2) They may contain some vital genes (3) Plasmid can replicate (4) They are single stranded DNA 120. Plasmodesmata are (1) Cell envelope of pro-karyotic cell (2) Layer of cell wall (3) Cytoplasmic connections between

neighbouring plant cells (4) Protenaceous covering of chromosome 121. Bundle of fibrils that constitute the

central core of cilia or flagella is called (1) Hub (2) Fimbriae

(3) Pili (4) Axoneme 122. Choose the wrong match? (1) Cisternae – Sacs of golgi (2) Middle lamella – Cell plate (3) Chromatin – Condensed chromosome

of interphase (4) Lysosome – Intracellular digestion 123. A chromosome with centromere towards

one end is called (1) Metacentric chromosome

(2) Acrocentric chromosome (3) Telocentric chromosome (4) Sub-metacentric chromosome 124. Which is a wrong statement? (1) Microtubules are made of tubulin (2) In bacterial cells, Pili are concerned

with adhesion (3) “Omnis cellula-e-cellula” was stated

by Rudolf Virchow (4) Basal body of cilia and flagella are

membraneless 125. Tonoplast refers to differentially

permeable membrane of (1) Lysosome

(2) Spherosome (3) Peroxisome (4) Vacuole



126. Which is not a function of integral proteins of plasma membrane?

(1) Enzyme synthesis (2) Active transport (3) Hormone reception (4) Passive transport

127. The main unit of golgi complex is

(1) Thylakoid (2) Cristae (3) Cisternae (4) Lomasomes

128. Which of the following pair is

mismatched? (1) Cytoskeleton – Microtubules (2) Lysosome – Lipid synthesis (3) Peroxisome – Microbody (4) Nucleolus – rRNA 129. Select the wrong statement (1) RER is seen in cells which are

actively involved in protein synthesis (2) Ribosomes of the chloroplast are

larger than the cytoplasmic ribosomes (3) Glucose is a primary metabolite (4) Fructose is a monosaccharide 130. Glycocalyx is associated with (1) Nucleus (2) Endoplasmic reticulum (3) Bacterial cell (4) Eukaryotic cell 131. RNA is not found in (1) Mitochondria (2) Chloroplast

(3) Ribosome (4) Plasma membrane

132. Which is not true about prokaryotes (1) DNA is single stranded and circular (2) Genome is single DNA

(3) Chloroplast absent (4) Non-flagellated forms exist

133. Which of the following is wrongly matched?

(1) Chromoplasts – Xanthophyll (2) Aleuroplasts – Protein (3) Sap vacuole – Prokaryotes (4) Cytoskeleton – Intermediate filaments

134. Which of the following is a part of endomembrane system of eukaryotic cell

(1) Peroxysomes (2) Glyoxysomes

(3) Golgi complex (4) Mitochondria

135. Which one of the cellular parts is

correctly described (1) Cristae – Flattened membraneous

sacs forming the matrix of mitochondria (2) Centrioles – Microtubule organising

centre (3) Lysosome – Optimaly acting at pH of

about 7.5 (4) Peroxisome – Microbody concerned

with glycosylation 136. Find out the wrong pairing of source and

product (1) Trichoderma polysporum –

Cyclosporin A (2) Monascus purpureus – Statins (3) Aspergillus niger – Citric acid (4) Clostridium butylicum –

Streptomycin 137. Azolla is used as a biofertilizer because it

(1) Has association of nitrogen fixing Cyanobacteria

(2) Has association of nitrogen fixing Rhizobium

(3) Multiplies very fast to produce massive biomass

(4) Has association of mycorrhiza 138. Study the following statements and select

the correct ones. i. Methanogens are archaebacteria which

produce methane in marshy areas. ii. Nostoc is a filamentous blue green

algae which fixes atmospheric nitrogen. iii. Biofertilisers have contributed

significantly to environmental pollution. iv. Baculoviruses are harmful to humans. (1) i and ii (2) i and iii (3) ii and iii (4) i, ii and iii

139. What advantage the fungal partner

derives from the symbiotic association with plants?



(1) Phosphorus from soil (2) Resistance to root-borne pathogens (3) Tolerance to salinity and drought (4) Carbohydrates from plants

140. Given below four statements. (i) The primary treatment of sewage is

not a microbial process. (ii) The primary effluent in sewage

treatment is initially passed into large anaerobic sludge digesters.

(iii)Secondary treatment of sewage involves aerobic bacteria only.

(iv) Digestion of cellulose in humans takes place by bacterial action.

How many of the above statements are true? (1) One (2) Two (3) Three (4) Four

141. Find the correct sequence of sewage

treatment procedure. 1. Primary effluent is passed in to large

aeration tanks. 2. Physical removal of particles from

sewage through filtration and sedimentation.

3. Effluent from secondary treatment plant is released into water bodies.

4. Activated sludge is passed into large anaerobic sludge digesters.

(1) 1-3-2-4 (2) 2-1-4-3 (3) 1-2-4-3 (4) 4-1-2-3 142. The use of biocontrol measures will: (1) help to eradicate agricultural pests

very effectively. (2) greatly reduce our dependence on

toxic chemicals and pesticides. (3) kill both useful and harmful life

forms indiscriminately. (4) employ the use of conventional

farming parctices 143. The purpose of biological treatment of

waste water is to: (1) increase BOD (2) reduce BOD (3) reduce sedimentation (4) increase sedimentation

144. Extinction of an ecologically unique assemblage of more than 200 species of cichlid fish in the lake Victoria in east Africa is attributed to: (1) Over-exploitation (2) Habitat fragmentation (3) Pesticide application (4) Alien species invasion

145. Which one is a method of in situ

conservation? (1) Arboratorium (2) Conservation of animals in

sanctuaries (3) Cryopreservation of gametes of

threatened species. (4) Propagation of plants using tissue

culture. 146. Mammals and birds requiring large

territories and certain animals with migratory habits are badly affected, leading to population declines due to: (1) Invasion by alien species (2) Habitat fragmentation (3) Over-exploitation (4) Environmental pollution

147. Which of the following is not an ex-situ

conservation method of biodiversity? (1) Breeding animals in zoo (2) Establishing a reserve forest (3) Setting up a biosphere reserve (4) Building a seed bank 148. Other than poaching, what are the

possible reasons for the decline in the population of Ganges River Dolphins?

i. Construction of dams on rivers ii. Increase in the population of

crocodiles in rivers iii. Getting trapped in fishing nets

accidentally iv. Use of synthetic fertilizers and other

agricultural chemicals in crop-fields in the vicinity (1) i and ii (2) ii and iii (3) i, iii and iv (4) i, ii, iii and iv

149. Which of the following criteria have

contributed to the recognition of Western



Ghats-Sri Lanka and Indo-Burma regions as hotspots of biodiversity?

(I) Species richness (II) Vegetation density (III) Endemism (IV) Ethno-botanical importance (V) Threat protection (VI) Adaptation of fauna and flora to

humid conditions Which three of the above are correct criteria in this context?

(1) I, II and VI only (2) II, IV and VI only (3) I, III and V only (4) III, IV and VI only 150. Which of the following is a pair of

endangered species in India? (1) Garden lizard and Blackbuck (2) Pigmy hog and Rhesus monkey (3) Bengal tiger and Green turtle (4) Indian peacock and Carrot grass

151. Tiger is not a resident in which of the

following National Park? (1) Gir (2) Ranthambore (3) Jim Corbett (4) Sunderbans

152. Biosphere reserves differ from National

Parks and Wildlife Sanctuaries because in the former: (1) No entry for human beings is allowed (2) People are an integral part of the

system (3) Plants are paid greater attention than

animals (4) Endangered species are brought from

all over the world and preserved 153. Which of the following is not a

component of “Evil Quartet”? (1) Habitat loss and fragmentation (2) Alien species invasions (3) Ecological successions (4) Overexploitation

154. What is incorrect about humus? (1) It is a dark coloured amorphous

substance.

(2) It is highly susceptible to microbial action. (3) It undergoes decomposition at an extremely slow rate. (4) It is colloidal in nature.

155. The gradual and fairly predictable change

in the species composition of a given area is called:

(1) adaptive radiation (2) ecological succession (3) species diversity (4) stratification 156. During the carbon cycle, which of the

following carbon compounds would be utilized as an energy source by heterotrophs? (1) calcium carbonate (2) carbonic acid (3) organic molecules (4) carbon dioxide

157. In the thermal stratification of a

freshwater lake, the transition layer between the warmer mixed water at the surface and the cooler deep water below is called:

(1) Hypolimnion (2) Dysphotic zone (3) Epilimnion (4) Thermocline 158. How is a food web model different from

a food chain? (1) In a web, energy moves from an

organism to only another. (2) In a web, energy may move to many

organisms from one. (3) In a web, an organism gets energy

from one source. (4) In a web, an organism receives less

energy than in a chain. 159. In which type of forest, humus is formed

by the dry and dead leaves? (1) Coniferous forest (2) Tundra forest (3) Alpine forest (4) Deciduous forest



160. The rate at which light energy is converted to chemical energy of organic molecules is the ecosystem’s: (1) Net primary productivity (2) Net secondary productivity (3) Gross primary productivity (4) Gross secondary productivity

161. The pyramid of number in a grassland

ecosystem is: (1) linear (2) upright (3) irregular (4) inverted 162. 10% law for flow of energy works

between: (1) sun and first trophic level (2) primary consumer and secondary

consumer (3) first trophic level and second trophic

level (4) all trophic levels 163. In the phosphorus cycle, weathering

makes phosphate available first to: (1) Decomposers (2) Consumers (3) Producers (4) All the above 164. Which of the following populations is

most likely to go extinct? (1) A very small population in an

unstable environment (2) A moderate-sized population of r

strategists (3) A large population with lots of

genetic variability (4) All would be equally likely to go

extinct 165. Which one of the following is the best

example for resource partitioning in nature? (1) Balanus and Chathamalus living in the rocky coasts of Scotland. (2) Abingdon tortoise and alien goats of Galapagos. (3) Visiting flamingoes and resident fishes of South American lake. (4) Five species of warblers living on the same tree.

166. The Mediterranean orchid Ophrys employs _______to get pollination done by a species of bee

(1) Pseudocopulation (2) Sexual deceit (3) Smell (4) Provides suitable place for egg laying

167. Find the incorrect statement (1) Average life expectancy of man

increased recently. (2) In logistic equation, ‘K’ represents

carrying capacity (3) The “r” value of population of India

remains constant during the last five decades.

(4) Logistic growth equation was derived by Verhulst and Pearl.

168. To obtain optimal yield, populations

should be harvested at what part of the sigmoid growth curve?

(1) Lag phase (2) Acceleration phase (3) Deceleration phase (4) Asymptote 169. Resource partitioning would be most

likely to occur between (1) sympatric populations of species with

similar ecological niches. (2) allopatric populations of species with

similar ecological niches. (3) allopatric populations of the same

animal species. (4) sympatric populations of a flowering

plant and its specialized insect pollinator.

170. Find the true statement

(1) Eels are marine fishes that migrate to rivers for spawning.

(2) Photoperiod is the only absolutely reliable environmental cue that triggers bird migration

(3) Many birds migrate from temperate to tropics to escape from the intolerable summer.

(4) Sweating is a mechanism of thermogenesis.



171. The plants that inhabit saline soil with high concentrations salt are:

(1) mesophytes (2) halophytes (3) hygrophytes (4) hydrophytes 172. Of the following types of interactions,

which one is least likely to limit population size? (1) Predation (2) Commensalism (3) Competition (4) Parasitism

173. Which of the following animal is an

endotherm? (1) Shark (2) Frog (3) Lizard (4) Rat 174. Which one of the following is most

appropriately defined? (1) Parasite is an organism that lives at

the expense of another organism and do harm to it.

(2) Amensalism is a relationship in which one species is benefited whereas the other is unaffected.

(3) Predator is an organism that competes with prey for similar resource.

(4) Commensal is an organism which always lives inside the body of other organism and gets some benefit.

175. Which of these is not a density-dependent

factor that could act to limit population growth as population size increases? (1) Waste accumulation (2) Forest fire (3) Inhibitory pheromones (4) Lowered immune function due to

stress 176. Kyoto Protocol (1997) was endorsed at (1) COP-6 (2) COP-4 (3) COP-3 (4) COP -5 177. Biochemical Oxygen Demand (BOD) in

a river water (1) Increases when sewage gets mixed

with river water (2) Remains unchanged when algal

bloom occurs (3) Has no relationship with

concentration of oxygen in the water

(4) Gives a measure of salmonella in the water

178. Which of the following is a major source

of thermal pollution in water bodies? (1) Sewage treatment plant (2) Solid waste disposal sites (3) Power plants using coal (4) Heat from solar radiation

179. Match the following 1. Mercury - (a) Catalytic converter 2. Rhodium - (b) ESP 3. Sulphur dioxide - (c) Heavy metal 4. Particulate matter- (d) Scrubber

(1) 1 – c, 2 – a, 3 – b, 4 – d (2) 1 – c, 2 – a, 3 – d, 4 – b (3) 1 – c, 2 – b, 3 – d, 4 – a (4) 1 – d, 2 – c, 3 – b, 4 – a 180. Cultural eutrophication can

(1) increase fish's reproductive rate (2) increase oxygen in water (3) never be reversed (4) harm commerical fisherman's income



ANSWER KEY

1 1 31 2 61 4 91 2 121 4 151 1

2 4 32 3 62 2 92 1 122 3 152 2

3 2 33 3 63 1 93 4 123 2 153 3

4 3 34 4 64 1 94 3 124 2 154 2

5 4 35 3 65 1 95 2 125 4 155 2

6 3 36 1 66 3 96 1 126 1 156 3

7 3 37 3 67 4 97 3 127 3 157 4

8 1 38 2 68 4 98 2 128 2 158 2

9 1 39 1 69 2 99 3 129 2 159 4

10 3 40 3 70 2 100 2 130 3 160 3

11 2 41 3 71 3 101 1 131 4 161 2

12 3 42 3 72 2 102 4 132 1 162 4

13 4 43 3 73 4 103 4 133 3 163 3

14 4 44 2 74 3 104 3 134 3 164 1

15 2 45 1 75 4 105 4 135 2 165 4

16 3 46 4 76 4 106 3 136 4 166 2

17 2 47 4 77 2 107 1 137 1 167 3

18 3 48 3 78 3 108 3 138 1 168 2

19 4 49 3 79 1 109 1 139 4 169 1

20 4 50 3 80 2 110 2 140 1 170 2

21 3 51 3 81 4 111 4 141 2 171 2

22 4 52 1 82 3 112 3 142 2 172 2

23 3 53 4 83 3 113 1 143 2 173 4

24 1 54 2 84 2 114 4 144 4 174 1

25 3 55 3 85 1 115 1 145 2 175 2

26 3 56 4 86 3 116 2 146 2 176 3

27 3 57 2 87 2 117 3 147 3 177 1

28 2 58 4 88 4 118 2 148 3 178 3

29 1 59 3 89 3 119 3 149 3 179 2

30 3 60 4 90 3 120 3 150 3 180 4



DETAILED ANSWERS

1. Hints : (1) Average energy density of em

wave 20 0

12avU E= ∈

12 21 (8.85 10 ) (50)2

−= × × ×

= 10–8 J/m3.

2. Hints: (4) Fext = mω2x ∝ 2

1T

; F = F1 +

F2

2 2 21 2

1 1 1T T T

= + ; T = 2 2

1 22 2

1 2

T TT T+

3. Hints: (2) From fig (1)

mg = 2k x

2mg = 2k (x + h); subtracting

Hence mg = 2kh ; k = 2mg

h

In fig (2)

f = 1 3 1 62 5 2 5

k gm h

=π π

after substituting

for k

4. Hints: (3) 0

0

b w

b

t gt g

ρ − ρ= =

ρ

ρb = 43

× 1000; ρw = ρl = 1000 kgm–3

0tt

= 12

; t = 2t0

5. Hints: (4) ∆t = 26T =

3T ; ∆φ = 2

3π

6. Hints: (3) T.E = constant PE is maximum and x = ±A KE is maximum at mean position (at x =

0) 7. Hints: (3) AB is silvered. Due to

refraction. Image of AB shifts to ' 'A B by distance x. Due to reflection at ' 'A B for object o and image 'I

x =1 cm For refraction at CD apparent depth = 6–

1 = 5 cm

Actual depth of AB = 6 cm 6 1.25

n = =

8. Hints: (10 Frequency does not depeud

on medium

λmed = airλµ

; airmed

VVµ

=

9. Hints: (1) Diameter of imageDiameter of sun

mean of distancef

Diameter of Sun = 1.39 ×109m mean distance = 1.5 × 1011m f = 0.1 m solve Diameter of image = 9.2 ×10 –4 m 10. Hints: (3) For n = 1.5 and R1 = R2 f = R = 20 cm u = –30 cm; f = + 20 cm

1 1 1v u f

− =

v = + 60 cm; m = 2 i

o

hvu h

= − =

hi = –4 cm ; Image is real, inverted 11. Hints: (2) No deviation 1 1 2 2( 1) ( 1)n A n A− = −

2(1.5 1)15 10

0.75

ooA−

= =

12. Hints: (3) u = + 10 ; v = + 15

1 1 1 ; 30f cmv u f

− = = −

13. Hints: (4) For concave mirror

( )

f fmf u u f

= =− −

Taking magnitudes

m = fx

x = distance from focus As x increases, m decreases



14. Hints: (4) One end at 2F for other end at 20 cm

U = –30 ; f = –10

1 1 1v u f

+ =

v = –15 cm Length of image = 20 – 15

= 5 cm

15. Hints: (2) 1 2

1 1 1f f f

= +

1

1

11f R

µ −= 2

2

( 1)1f R

µ− −=

1 2

Rfµ µ

=−

16. Hints: (3) 1.22a

λθ =

where a = 2 mm λ = 5896 Å

xD

θ = ; x = 1.2 m

Substitute D = 3.39 km

17. Hints: (2) fofe

βα

=

α = 2o ; fo = 60 cm fe = 5 cm β = 24o 18. Hints: (3) For tensile stress on PQ, component of force normal to PQ = F cos

θ If A = area of cross section of PQ, (direction is normal to PQ) its component A cos θ = a,

A = cos

aθ

Tensile stress = 2cos cosF F

A aθ θ

=

19. Hints: (4) E = 12

EnergyVolume

= Stress x

strain = y 2( )

2strain

∝ (strain)2 ∝ 2l

l∆

2 2

1 1 2

2 2 1

2 12

E l l l LE l l l L

∆ = = = ∆

20. Hints: (4) Upward force = S x 2πr

2πr = FS

= 4

2

75 106 10

−

−

××

= 12.5 x 10-2 m

21. Hints: (3) Mean pressure = pressure at

centre of gravity = h/2 dg

22. Hints: (4) F = η A dvdx

120 10 20000.1

F dyne= × × =

23. Hints: (3) F = aV2 ρ ; V2 = 2gh Fnet = a ρ2g (h1 – h2) = a ρ 2gh 24. Hints: (1) Volume per second = a V 100 = 0.25 x 104 V

V = 1125

cms− = 0.4 mm/sec

25. Hints: (3) 3 1rms

RTVM M

= ∝

2

2

32 42

H

O

VV

= =

2HV = 4 C

26. Hints: (3) Equal forces at the 2 junctions

are as shown. The force on the liquid is anticlockwise torque.

Hence liquid exerts clockwise torque.

F

P

Q

θ F cos θ

θ

F F



27. Hints: (3) V = kω = 100 × 10 =1000ms–1

28. Hints: (2) Frequency constant = 1000 Hz Wavelength in water

= 15001000

vf

= = 1.5m

29. Hints: (1) V ∝ T1/2; T = Temperature

1 1 12 2 300

VV T

∆ ∆= = ×

% = 1 100600

× = 16

% = 0.167%

30. Hints: (3) f ∝ 1l

2 1

1 2

f lf l

=

f2 = 256 2516

× = 400 Hz

31. Hints: (2) The two waves with same

speeds move in opposite directions. Hence y = 0.1 sin π (0.1x + 8t + 1.3)

32. Hints: (3) 1 2

3 34 2V Vl l

=

1

2

12

ll

=

33. Hints: (3) For reflected sound

1

1

( )'( )V Vf fV V

+=

−

272 345 5345 5

+ −

= 272 × 350340

= 280

Beat frequency = 280 – 272 = 8

34. Hints: (4) Strain = Stressy

Tw

Lw

V StressStrainV y

= =

21

100y

= stress = 107 Nm–2

35. Hints: (3) n1λ1 = n2λ2

1

2

10000 512000 6

nn

= =

5 th bright fringe of 12000Å coincides with 6th bright fringe of 10000Å

1 1Dy nd

λ=

73

2 5 12 102 10

−− × × ×

×

= 6 ×10–3m = 6 mm

36. Hints: (1) vCλ

λ∆ = = λ × 0.8 = 5500 ×

0.8 = 4400 Å 37. Hints: (3) width =

7

4

2 2 0.8 5 102 10

D ma

λ −

−

× × ×=

×

= 4 × 10–3 m = 4 mm 38. Hints: (2) Wave nature of light

39. Hints: (1) Dyaλ

=

7

3

1 5 105 10

Da maλ −

−

× ×= =

× = 0.1 mm

40. Hints: (3) Original intensity of

unpolarised light = I0

After I crystal, 0

2II =

After II crystal 0

2II = 2 0cos 60

8I

= .

Percentage of original intensity = 1 1008

× = 12.5%

41. Hints: (3) Due to drift current (Theory

bases)

42. Hints: (3) IB = 3

0.0110

i

i

vr

= = 10–5 A

IC = βIB = 50 × 10–5 A = 500 µA 43. Hints: (3) IC (Theory bases)



44. Hints: (2) β = C

B

II

IB = 310

100

−

A = 10–5 A

In Base – Emitter circuit, VBB = IB RB + VBE 10–5 × 100 × 103 + 0.6 = 1.6 V 45. Hints: (1)

A B A A + B

Y

0 0 1 1 0 1 1 0 1 1 1 0 0 0 0

Output is AND GATE 46. Hints: (4) Let the volume of 0.5 H2SO4

left unused = v Applying normality equation 2 × v × 0.5 = 1 × 70 × 0.5 or v = 35 ml ∴ Vol. of 0.5 H2SO4 used to neutralize NH3 = 50 – 35 = 15 ml

% N = 1.4tan

M V BMass of the subs ce taken

× × ×

= 1.4 0.5 15 20.5

× × × = 42

47. Hints: (4) Chlorination of methane gives

CH3Cl and CHCl3 while CH3CH3 is formed by combination of two CH3 radicals in the termination step. Therefore, CH3CH2CH3 cannot be formed.

48. Hints: (3) 3° hydrogen is abstracted

most readily and hence tert-butyl chloride is formed.

49. Hints: (3) Acid character increases in the

order:

CH3CH3 < CH2 = CH2 < HC ≡ CH Therefore, basic strength decreases in the

reverse order, ie., CH3CH2

– (i) > H2C = CH– (ii) > HC ≡ C– (iii)

50. Hints: (3) Let the wt. of the substance

taken = w g

∴ %C = 12 0.66 1810044 w w

× × =

and %H = 2 0.36 410018 w w

× × =

Now, C : H = 18 4: 3 : 812 1w w

=

Hence, E.F = C3H8 51. Hints: (3) Vacuum distillation 52. Hints: (1)

53. Hints: (4) In presence of Lewis acid (ie.,

FeBr3) nuclear halogenation occurs. Since CH3CH2 group is o, p-directing, therefore, halogenation occurs predominantly at the less hindered p-position to afford p-bromo ethylbenzene.

54. Hints : (2) 2 6 4 2p NO C H C H+

− − − is the least stable carbocation since electron-withdrawing – NO2 group destabilizes the carbocation.

55. Hints : (3) 3 2

3 2

sp sp sp spC H C C C H +− ≡ −

A

B

YA

AA

( A +B) ( A +B) . A

A



56. Hints : (4) CH3CHOHCH2CH3 [O]→ CH2 COCH2CH3 [O]→ CH3COOH + CH3CH2COOH

57. Hints: (2) Phenol is not used in the

production of melamine plastic and acetyl chloride is not used in the preparation of phenetole. Therefore, statements (I) and (III) are correct.

58. Hints : (4) In CH3OC2H5, attack of I–

occurs at CH3 group giving CH3I and C2H5OH. In contrast, in (CH3)3COCH3, reaction occurs by SN1 mechanism and I–

attacks the more stable (CH3)3C+ giving (CH3)3CI and CH3OH.

59. Hints: (3) Electron-withdrawing groups

increase while electron-donating groups decrease the acidity of phenols. Since – NO2 group is a much stronger electron-withdrawing group than – Cl and – OMe group is a much stronger electron donating group than Me, therefore, the overall acidity order is : p-nitrophenol > p-chlorophenol > phenol > p-cresol > p-methoxyphenol.

60. Hints : (4)

61. Hints: (4) Acetone neither reduces Fehling’s solution nor Tollens’ reagent. Benzaldehyde, however, reduces Tollens’ reagent but not Fehling’s solution.

62. Hints: (2)

63. Hints: (1) A CH3CH2CH2CH2OH PCC→

CH3CH2CH2CHO B C6H5CH3 2 2CrO Cl→ C6H5CHO C CH3COCl + (CH3)2 Cd →

CH3COCH3 D (CH3CH2COO)2Ca

∆→ (CH3CH2)2C = O 64. Hints: (1) 65. Hints: (1) 66. Hints: (3)

OH

3 3.. .

CH Cl Anhyd AlClF C alkylation

+→

Zn dustZnO−→

(X)

CH3

4

AlkalineKMnO→

(Y)

COOH

Benzoic acid (Z)

2

( )( ) /i Ozonolysisii Zn H O→

CH2CH2CHO | CH2CH2CHO Hexane-1, 6-dial (E)

.(

)

Aq KOHIntramolecular

aldol condensation→

CHO

Cyclopent-1-ene-1- carbaldehyde (F)

CH2 3BHTHF→

CH2

3

2 2H OOH −→ B

CH2OH

CHO

5 5 3

2 2

C H NH CrO ClCH Cl

+ −

→

CH3 – CH – CH – CH2 – CH3 | Cl



67. Hints: (4) Among halides iodides are more reactive in SN2 reaction.

68. Hints: (4) n-hexane is treated with

anhydrous AlCl3 and HCl gas to gives mainly 2-methyl pentane and 3-methyl pentane.

69. Hints: (2) Obeys Huckels rule. → Cyclopropenyl cation → 2 delocalised πe– are present 70. Hints: (2) CHCl3 is used for the

production of the Freon refrigerant R-22. 71. Hint : (3) , , ,o m p− − − chlorotoluene &

benzylchloride 72. Hints: (2) Geometrical isomers are the

compound which have at least one double bond and the groups attached with double bonded carbon atoms are different.

3 2 6 5CH CH CHCH C H= exhibits geometrical isomerism.

73. Hints: (4)

/3 2 2 3 2 2

Propylbromide

Mg ether

GrignardCH CH CH Br CH CH CH MgBr→

74. Hints: (3) Due to the presence of acidic

hydrogen ammoniacal Cu2Cl2 will give red copper acetylide ppt. with Cu2Cl2, whereas C2H4 will be neutral to ammoniacal Cu2Cl2.

75. Hints : (4) The rate of nucleophilic

substitution reaction decreases in presence of electron releasing group at ortho or para position. In compound (iii), electron releasing group is present at meta position w.r.t. chlorine, so the impact is less but in compound (ii) it will be present at ortho position.

76. Hints: (4)

3

3

3 3 3( ) ( )

3 4 3 2( ) ( )

H OKCN

A BNH heat

C D

CH Cl CH CN CH COOH

CH COONH CH CONH

+

→ →

→ →

77. Hints : (2)

78. Hints : (3) Styrene is C6H5-CH=CH2. So

the third one is correct. 79. Hints : (1) Nylon-6 is a condensation

homopolymer of caprolactam

80. Hints: (2) In non-aqueous solvents and in vapour phase basic strength of alkyl amines follows the order: Tertiary amines > secondary amines > primary amines

81. Hints : (4) Absence of free – CHO group

cyclic structure of glucose. 82. Hints : (3) 83. Hints : (3)

⊕



84. Hints: (2) LiAlH4 reduces group without affecting the double bond. 85. Hints: (1) 1° and 2° amines, unlike 3°

amines form intermolecular H-bonds. 1° amines with two H-atoms available for H-bonding have a higher boiling point than 2° amines which have only one H-atom.So, the order is : n-butylamine > diethylamine > ethyldimethylamine

86. Hints: (3) IUPAC name is 2-ethoxy-4-methoxy-

pentan-3-one 87. Hints: (2) Terfenadine is an antihistamine 88. Hints : (4) Order of I-power⇒ NO2 >CN

>F >NH2 89. Hints: (3) The reagent can attack only

the benzene ring contain activating group. Due to steric hindrance, the substitution takes place on para position and not on ortho position.

90. Hints: (3) The structure of given IUPAC

name is

Correct IUPAC name is 3, 4-dimethylhexane

91. Hints: Ribosomes present in organelle

like mitochondria and chloroplast 92. Hints: Microfilaments have diameter

about 6 nm. Microtubules are cytoskeleton of 24 nm diameter.

93. (4) Both (2) & (3) 94. Hints: Prokaryotes and Eukaryotes have

cell membrane basically similar. 95. Hints: Polypeptides (proteins) are

synthesised on ribosome translating the mRNA

96. Hints: Pseudomonas is prokaryotes

which lack membrane bound organelles. 97. Proteins embedded in a phospholipid

bilayer 98. Hints: In eukaryotes, only mitochondria

and chloroplast have own DNA. No other organelles have DNA.

99. Hints: Most of the cellular reactions

occur in cytosol. 100. (2) SER 101. Cellular organelles with membranes are

Lysosomes, Microbodies, RER 102. Hints: Polysome is chain of ribosomes

on mRNA mesosome is special membrane invaginatins of prokaryotic cell membrane.

103. RBC measures about 7 µm in diameter 104. (3) Chain of ribosomes that translate the

mRNA 105. Mycoplasma has No cell wall 106. Hints: Fluid mosaic model of membrane

satisfy the property like fluidity of the membrane.

107. Hints: First secreted wall layer is

primary wall. When cell become mature, secondary wall is secreted.

108. Hints: An animal cell has 2 centrooles at

one pole of the cytoplasm. 109. Microtubules are absent in Prokaryotes

C == O CHOH

O 5 4 3|| 2 CH3 – CH – C – CH – OCH2CH3 | | 1 OCH3 CH3

CH3 – CH – CH – CH3 | | C2H5 C2H5

3 4

1, 2 5, 6



110. Hints: In 1898, Camilo Golgi first observed these densely staining bodies near to the nucleus in the cytoplasm by special staining technique.

111. Hints: Different molecules like RNA,

proteins etc transport through nuclear pores in both directions

112. Hints: ‘Biological energy (power)

currency’ is ATP molecule and mitochondria which is the site of synthesis of ATP is ‘biological power house’

113. Hints: Porins are channel proteins in

membrane with huge pore for transport of even large sized proteins.

114. Hints: Sedimentation coefficient (s)

discovered by svedberg is a reflection of size and density of the particle.

115. Hints: Prokaryotic ribosome has an

RNA to protein ratio of 65 : 35. 116. Hints: Plant cell vacuole has a

hypertonic solution of various solutes including excretory substances.

117. Hints: RER is engaged in post

translational modification of polypeptide chains and transport of them to their destination like golgi complex in the form of vesicles.

118. Hints: Mitochondria have no thylakoids.

But both have suitable membrane for chemiosmosis. Mitochondria have cristae and chloroplast have thylakoids.

119. Hints: Some bacterial cells have extra

genomic DNA called palsmid. They can replicate and therefore one to may copies of plasmid may occur in a cell. They contain no vital genes but antibiotic resistance etc. may be present.

120. Hints: Channels between neighbouring

cells which facilitate cytoplasmic connections for the exchange of materials.

121. Hints: Axoneme contains 9 doublet fibrils in the centre and two singlet fibrils in the periphery

(9 + 2) 122. Hints: De- condensed chromosome is

called chromatin. 123. Hints: In aerocentric chromosome, one

arm is longer and the other arm is very short.

124. Hints: Pili are concerned with mating.

So they are called sex pili. 125. Hints: In plant cells the vacuolar

membrane is called tonoplast and the fluid inside is called cell sap.

126. Hints: Majority enzymes are proteins

plasma membrane is not engaged in protein synthesis.

127. Hints: The glogi complex contains

cisternae, vesicles and tubules. 128. Hints: Lysosomes are concerned with

intracellular digestion. Lipid synthesis is a function of SER.

129. Hints: In eukaryotic cells chloroplast

contain DNA, RNA and ribosomes. Ribosomes are smaller than (70s) that of cytoplasm (80s)

130. Hints: The outer most part of bacterial

cell envelope is called glycocalyx. 131. Hints: Plasma membrane is devoid of

RNA. 132. Hints: In prokaryotes, Genome is single,

circular and naked DNA 133. Hints: Sap vacuoles are seen in plant

cells. In certain prokaryotic cells gas vacuole is present.

134. Hints: The endomembrane system in

eukaryotes include ER, golgi, lysosomes and vacuole.



135. Hints: Centrioles act as organising centres for mitotic spindle, basal bodies etc.

136. Hint: - Clostridium butylicum is the

source of butyric acid. Streptomycin is an antibiotic drug, the first of a class of drugs called aminoglycosides to be discovered, and it was the first effective treatment for tuberculosis. It is derived from the actinobacterium Streptomyces griseus.

137. Hint: - Azolla (aquatic fern) form a

symbiotic relationship with the cyanobacterium Anabaena azollae, which fixes atmospheric nitrogen. The nitrogen-fixing capability of Azolla has led to Azolla being widely used as a biofertiliser.

138. Hint: - Use of biofertilisers which are

eco-friendly have significantly reduced environmental pollution. Baculoviruses never infect huamn cells.

139. Hint: - This mutualism gives the fungus

sugars, such as glucose and sucrose produced by the plant in photosynthesis. The carbohydrates move from their source (usually leaves) to the root and then to the fungal partner.

140. Hint: - Only the first statement is true.

The primary effluent in sewage treatment is initially passed into large aeration tanks. Secondary treatment of sewage involves both aerobic and anaerobic bacteria only. No cellulose digestion takes place in human gut.

141. Hint: - Sewage treatment is the process of

removing contaminants from wastewater, primarily from household sewage. It includes physical, chemical, and biological processes to remove these contaminants and produce environmentally safe treated wastewater (or treated effluent). A by-product of sewage treatment is usually a semi-solid waste or slurry, called sewage sludge that has to undergo further treatment before being suitable for disposal or land application.

142. Hint: - Organic farmer prefers to use biocontrol methods because it helps to keep pest species under manageable level.

143. Hint: - As aerobic oxidation of organic

waste progresses BOD decreases. 144. Hint: - Invasive alien species are plants,

animals, pathogens and other organisms that are non- native to an ecosystem, and which may cause economic or environmental harm or adversely affect human health.

145. Hint: - An animal sanctuary is a facility

where animals are brought to live and be protected for the rest of their lives. One of the most important missions of sanctuaries, beyond caring for the animals, is educating the public. The ultimate goal of many sanctuaries is to change the way that humans think of, and treat, non-human animals. An arboratorium is like a very large greenhouse, only it is fitted up with little waterfalls and ponds and has an artificially controlled temperature and humidity.

146. Hint; - Habitat fragmentation is the

process by which habitat loss results in the division of large, continuous habitats into smaller, more isolated remnants. Habitat fragmentation creates isolated patches of landscape, which can have harmful effects on biodiversity. One of the major ways that habitat fragmentation affects biodiversity is by reduction in the amount of available habitat (such as rainforests, boreal forests, oceans, marshlands, etc.) for all organisms in an ecological niche especially mammals and birds requiring large territories and certain animals with migratory habits.

147. Hint: - Ex-situ conservation is the

preservation of components of biological diversity outside their natural habitats. This involves conservation of genetic resources, as well as wild and cultivated



or species, and draws on a diverse body of techniques and facilities. In-situ conservation is the on-site conservation or the conservation of genetic resources in natural populations of plant or animal species, such as forest genetic resources in natural populations of tree species.

148. Hint: - The Ganges River dolphin or

Susu, lives in one of the most densely populated regions of the world. One of the main threats to the species is loss of habitat due in large part to the creation of dams and irrigation projects. River Dolphin is the National Aquatic Animal of India. This mammal is also said to represent the purity of the holy Ganga as it can only survive in pure and fresh water. The Ganges river dolphin can only live in freshwater and is essentially blind. They hunt by emitting ultrasonic sounds.

149. Hint: - A biodiversity hotspot is a

biogeographic region with significant levels of biodiversity that is under threat from humans. To qualify as a biodiversity hotspot, a region must meet two strict criteria: it must contain at least 0.5% or 1,500 species of vascular plants as endemics, and it has to have lost at least 70% of its primary vegetation. Around the world, 34 areas qualify under this definition, with nine other possible candidates. These sites support nearly 60% of the world's plant, bird, mammal, reptile, and amphibian species, with a very high share of those species as endemics.

150. Hint: - Endangered (EN) is a Red list category facing a high threat of extinction in the near future.

151. Hint: - Gir Forest National Park is a

wildlife sanctuary in Gujarat, western India. It was established to protect Asiatic lions.

152. Hint: - A biosphere reserve is an

ecosystem with plants and animals of unusual scientific and natural interest. It

is a label given by UNESCO to help protect the sites. The plan is to promote management, research and education in ecosystem conservation. This includes the 'sustainable use of natural resources'. If, for example, fish or trees are taken for human use, this is done in ways which least damage the ecosystem.

153. Hint; - Concept of Evil Quartet of species

extinction by Jared diamond does not include ecological succession.

154. Hint: - Humus is highly resistant to

microbial action and undergoes decomposition at an extremely slow rate. Being colloidal in nature it serves as a reservoir of nutrients.

155. Hint: - Ecological succession is the series

of changes in an ecosystem when one community is replaced by another community as a result of changes in biotic and abiotic factors. It can regenerate a damaged community. It can create a community in a previously unhabitated area.

Succession occurs in all types of ecosystem

156. Hint: - Heterotrophs cannot utilize

inorganic carbon. Only autotrophs can use it for photosynthesis. Heterotrophs depend on organic carbon synthesized by autotrophs.

157. Hint: - Lake stratification is the

separation of lakes into three layers: Epilimnion: the top of the lake with

warmer water. Thermocline or Metalimnion: the middle

layer, which may change depth throughout the day.

Hypolimnion: the bottom layer with cooler water

158. Hint: - A food web is the natural

interconnection of food chains and a graphical representation of what-eats-what in an ecological community. Another name for food web is a



consumer-resource system. Only food webs exist in ecosystems and not isolated food chains.

159. Hint: - Deciduous forest ground is rich in

fallen leaves. In deciduous forest, vegetation is composed primarily of broad-leaved trees that shed all their leaves during one season.

160. Hint: Gross primary productivity of an ecosystem is the rate of production of organic matter during photosynthesis.

161. Hint: - In upright pyramid, the base is

always broad and the number or biomass progressively decreases from lower to higher trophic level. of organisms that eat other organisms and may, in turn, be eaten themselves.

162. Hint: - The Ten percent law for the

transfer of energy from one trophic level to the next was introduced by Lindemann (1942). According to this law, during the transfer of energy from organic food from one trophic level to the next, only about ten percent of the of energy from organic matter is stored as flesh. The remaining is lost during transfer, broken down in respiration, or lost to incomplete digestion by higher trophic levels.

163. Hint: - When rocks are weathered,

minute amounts of phosphates get dissolved in soil solution and are absorbed by the roots of the plants.

164. Hint: - A population having a small size

and the one that lives in an unstable environment is more likely to become extinct. A small population size indicates low growth rate or “r” value.

(eg. Population of Giant Pandas) 165. Hint: - Resource partitioning is the

process whereby similar species exploit the limited resources in an ecological area without one species driving the others into extinction. Through natural selection, each species adopts a pattern of

resource usage that isn't competitive. Resource partitioning helps to reduce the intensity of interspecific competition. It permits co-existence rather than exclusion of species.

166. Hint: - The Mediterranean orchid Ophrys

is called “master of sexual deception”. These plants are remarkable in that they successfully reproduce through pseudocopulation, that is, their flowers mimic female insects to such a degree that males are fooled into mating with the flowers, thereby pollinating them.

167. Hint: - In 1981, the “r” value for human

population in India was 0.0205. In 2011 it has reduced to 0.0126. When “r” value decreases, doubling time increases.

168. Hint: - A population expresses high birth

rate or “r” value to the maximum during the acceleration phase of logistic growth. During this phase birth rate is high and death rate is minimum. When resources become limited, birth rate is opposed by death rate and deceleration phase starts.

169. Hint: - Resource partitioning is the

process whereby similar or sympatric species exploit the limited resources in an ecological area without one species driving the others into extinction. Through natural selection, each species adopts a pattern of resource usage that isn't competitive. Sympatric species are different species inhabiting the same area or niche.

170. Hint: - Eels are freshwater fishes that

migrate to sea for spawning. Many birds migrate from temperate to tropics to escape from the intolerable winter. Shivering is a mechanism of thermogenesis. Sweating is a mechanism of heat loss.

171. Hint: - Halophytes are plants that inhabit

saline soil with high concentrations salt. These plants have succulent leaves and sometimes the stem is also succulent.



Halophytes growing in marshy places near seashore form special vegetation known as the mangrove or tidal woodland. Eg. Mangrove plants

172. Hint: - Predation limits the population

size of prey population. Competition limits population density of both partners. Parasitism limits the population density of host species. Commensalism poses detrimental effect to neither partners and so it is the least likely interaction to limit population density.

173. Hint: - Endotherms are animals that raise

their body temperature through internal thermogenesis or metabolic heat (eg birds and mammals). Ectotherms raise body temperature by absorbing external heat (cold-blooded animals).

174. Hint: - Amensalism is any relationship

between organisms of different species in which one organism is inhibited or destroyed while the other organism remains unaffected. Predator actively hunts after the prey, kills and consumes it. Commensal gets benefit, while the other animal remains unaffected.

175. Hint: - Population size is regulated by

density-dependent factors (factors like diseases, competition, famine that increase when population size increases) and density-independent factors (factors such as natural calamities which have no relation with population size).

176. Hint: - Conference of Parties (COP) is a

body organized by the United Nations to assess progress in dealing with climate change. Conference of Parties III (COP3) was held in Kyoto, Japan in 1997. It is called the Kyoto Protocol. There, developed countries agreed to specific targets for cutting their emissions of greenhouse gases. COP1 was held in Berlin, Germany. COP8 was in New Delhi.

177. Hint: - Biochemical oxygen demand (BOD) is the amount of dissolved oxygen needed by aerobic biological organisms to break down organic material present in a given water sample at certain temperature over a specific time period. Higher the BOD, greater the organic pollution.

178. Hint: - Thermal pollution is the

degradation of water quality by any process that changes ambient water temperature. A common cause of thermal pollution is the use of water as a coolant by power plants and industrial manufacturers.

179. Hint: - Mercury poisoning causes

Minamata disease. Rhodium, palladium and platinum are used as catalysts in catalytic converters. Sulphur dioxide in industrial effluent can be removed by scrubber.

180. Hint: - Eutrophication causes algal bloom

which decreases the amount of dissolved oxygen in lakes causing increased mortality of fish population.

ZEOTS-22 | QUESTIONS · focal length 10 cm in such a way that its end closer to the pole is 20 cm...

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