Question PaPer PHYsiCs - Wiley IndiaO… · An observer looks at a distant tree of height 10 m with a telescope of magnifying power of 20. To the observer, the tree appears: (1) 20

2016 JEE MAIN (OFFLINE)

Question PaPer

PHYsiCs

1. A combination of capacitors is set up as shown in the figure. The magnitude of the electric field, due to a point charge Q (having a charge equal to the sum of the charges on the 4 μF and 9 μF capacitors), at a point 30 m from it, would equal

3 µF

9 µF4 µF

2 µF

8 V

+ –

(1) 480 N/C (2) 240 N/C

(3) 360 N/C (4) 420 N/C

2. An observer looks at a distant tree of height 10 m with a telescope of magnifying power of 20. To the observer, the tree appears:

(1) 20 times nearer (2) 10 times taller

(3) 10 times nearer (4) 20 times taller

3. Hysteresis loops for two magnetic materials A and B are given below:

B

H

(a) (b)

B

H

These materials are used to make magnets for electric generators, transformer core and electromagnet core. Then it is proper to use:(1) B for electromagnets and transformers.

(2) A for electric generators and transformers.

(3) A for electromagnets and B for electric transformers.

(4) A for transformers and B for electric generators.

4. Half-lives of two radioactive elements A and B are 20 min and 40 min, respectively. Initially, the samples have equal number of nuclei. After 80 minutes, the ratio of decayed numbers of A and B nuclei will be:

(1) 5 : 4 (2) 1 : 16

(3) 4 : 1 (4) 1 : 4

5. A particle of mass m is moving along the side of a square of side a, with a uniform speed v in the xy-plane as shown in the figure:

a

a

xO

y

R

45°

D C

A Bv

vv v aa

Which of the following statement is false for the angular momentum

L about the origin?

(1) L

mvRk=

2ˆ when the particle is moving from D

to A.

(2) L

mvRk= −

2ˆ when the particle is moving from

A to B.

(3) L mv

Ra k= −⎡

⎣⎢⎤⎦⎥2

ˆ when the particle is moving

from C to D.

(4) L mv

Ra k= +⎡

⎣⎢⎤⎦⎥2

ˆ when the particle is moving

from B to C.

JEE Main (OFFLINE) -2016.indd 643 5/18/2016 5:38:58 PM

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JEE Main 2016 (OFFLinE) 644

6. Choose the correct statement:

(1) In frequency modulation, the amplitude of the high frequency carrier wave is made to vary in propor-tion to the frequency of the audio signal.

(2) In amplitude modulation, the amplitude of the high frequency carrier wave is made to vary in propor-tion to the amplitude of the audio signal.

(3) In amplitude modulation, the frequency of the high frequency carrier wave is made to vary in propor-tion to the amplitude of the audio signal.

(4) In frequency modulation, the amplitude of the high frequency carrier wave is made to vary in propor-tion to the amplitude of the audio signal.

7. In an experiment for determination of refractive index of glass of a prism by i − δ , plot, it was found that a ray incident at angle 35°, suffers a deviation of 40° and that it emerges at angle 79°. In that case, which of the following is closest to the maximum possible value of the refractive index?

(1) 1.8 (2) 1.5

(3) 1.6 (4) 1.7

8. n moles of an ideal gas undergoes a process A→B as shown in the figure. The maximum temperature of the gas during the process will be

P

A

B

V

2Po

Po

Vo 2Vo

(1) 9 0 0PV

nR (2)

9

40 0PV

nR

(3) 3

20 0PV

nR (4)

9

20 0PV

nR

9. Two identical wires A and B, each of length l, carry the same current I. Wire A is bent into a circle of radius R and wire B is bent to form a square of side a. If B

A and B

B are

the values of the magnetic field at the centres of the circle and square, respectively, then the ratio B BA B/ is

(1) π 2

8 2 (2)

π 2

8

(3) π 2

16 2 (4)

π 2

16

10. A screw gauge with a pitch of 0.5 mm and a circular scale with 50 divisions is used to measure the thickness of a thin sheet of aluminium. Before starting the mea-surement, it is found that when the two jaws of the screw

gauge are brought in contact, the 45th division coincides with the main scale line and that the zero of the main scale is barely visible. What is the thickness of the sheet if the main scale reading is 0.5 mm and the 25th division coincides with the main scale line?

(1) 0.50 mm (2) 0.75 mm

(3) 0.80 mm (4) 0.70 mm

11. For a common emitter configuration, if α and β have their usual meanings, the incorrect relationship between α and β is

(1) α ββ

=+

2

21 (2)

1 11

α β= +

(3) α ββ

=−1

(4) α ββ

=+1

12. The box of a pin hole camera, of length L, has a hole of radius a. It is assumed that when the hole is illuminated by a parallel beam of light of wavelength λ the spread of the spot (obtained on the opposite wall of the camera) is the sum of its geometrical spread and the spread due to diffraction. The spot would then have its minimum size (say, b

min) when

(1) aL

=λ 2

and b Lmin = 4λ

(2) aL

=λ 2

and bLmin =

⎛⎝⎜

⎞⎠⎟

2 2λ

(3) a L= λ and bLmin =

⎛⎝⎜

⎞⎠⎟

2 2λ

(4) a L= λ and b Lmin = 4λ

13. A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up? Fat supplies 3.8 × 107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate. Take g = 9.8 ms−2.

(1) 12.89 × 10−3 kg (2) 2.45 × 10−3 kg

(3) 6.45 × 10−3 kg (4) 9.89 × 10−3 kg

14. Arrange the following electromagnetic radiations per quantum in the order of increasing energy:

A: Blue light B: Yellow light C: X-ray D: Radio waves

(1) B, A, D, C (2) D, B, A, C

(3) A, B, D, C (4) C, A, B, D



Question PaPer 645

15. An ideal gas undergoes a quasi-static, reversible process in which its molar heat capacity C remains constant. If during this process, the relation of pressure P and volume V is given by PV n = constant, then n is given by (Here, C

P

and CV are molar specific heat at constant pressure and

constant volume, respectively)

(1) nC C

C C=

−−

V

P

(2) nC

C= P

V

(3) nC C

C C=

−−

P

V

(4) nC C

C C=

−−

P

V

16. A satellite is revolving in a circular orbit at a height h from the Earth’s surface (radius of Earth R: h R). The maximum increase in its orbital velocity required, so that the satellite could escape from the Earth’s gravitational field, is close to: (Neglect the effect of atmosphere.)

(1) gR( )2 1− (2) 2gR

(3) gR (4) gR / 2

17. A galvanometer having a coil resistance of 100 Ω gives a full scale deflection, when a current of 1 mA is passed through it. The value of the resistance, which can convert this galvanometer into ammeter giving a full scale deflec-tion for a current of 10 A, is

(1) 3 Ω (2) 0.01 Ω(3) 2 Ω (4) 0.1 Ω

18. Radiation of wavelength λ, is incident on a photocell. The fastest emitted electron has speed v. If the wavelength of changed to 3 4λ / the speed of the faster emitted electron will be

(1) = ⎛⎝⎜

⎞⎠⎟

v3

5

1 2/

(2) > ⎛⎝⎜

⎞⎠⎟

v4

3

1 2/

(3) < ⎛⎝⎜

⎞⎠⎟

v4

3

1 2/

(4) = ⎛⎝⎜

⎞⎠⎟

v4

3

1 2/

19. If a, b, c, d are inputs to a gate and x is its output, then as per the following time graph, the gate is

d

c

b

a

x

(1) NAND (2) NOT

(3) AND (4) OR

20. The region between two concentric sphere of radii a and b, respectively (see figure), has volume charge density

ρ =A

r, where A is a constant and r is the distance from

the centre. At the centre of the sphere is a point charge Q. The value of A such that the electric field in the region between the sphere will be constant, is

a

b

Q

(1) 2

2

Q

aπ (2)

Q

a2 2π

(3) Q

b a2 2 2π( )− (4)

22 2

Q

a bπ( )−

21. A student measures the time period of 100 oscillations of a simple pendulum four times. The data set is 90 s, 91 s, 95 s and 92 s. If the minimum division in the measuring clock is 1 s, then the reported mean time should be

(1) 92 ± 3 s (2) 92 ± 2 s

(3) 92 ± 5.0 s (4) 92 ± 1.8 s

22. The temperature dependence of resistances of Cu and undoped Si in the temperature range 300–400 K, is best described by:

(1) Linear decrease for Cu, linear decrease for Si.

(2) Linear increase for Cu, linear increase for Si.

(3) Linear increase for Cu, exponential increase for Si

(4) Linear increase for Cu, exponential decrease for Si

23. Identify the semiconductor devices whose characteristics are given below, in the order (a), (b), (c), (d).

I

V

(a) (b)

I

V

(c)

Dark

Illuminated

I

V

(d)

Intensityof light

Resistance

(1) Zener diode, Solar cell, Simple diode, Light depen-dent resistance.

(2) Simple diode, Zener diode, Solar cell, Light depen-dent resistance.

(3) Zener diode, Simple diode, Light dependent resis-tance, Solar cell.

(4) Solar cell, Light dependent resistance, Zener diode, Simple diode.




24. A roller is made by joining together two cones at their vertices O. It is kept on two rails AB and CD which are placed asymmetrically (see figure), with its axis perpen-dicular to CD and its centre O at the centre of line joining AB and CD (see figure). It is given a light push so that it starts rolling with its centre O moving parallel to CD in the direction shown. As it moves, the roller will tend to

DB

C

O

A

(1) turn left and right alternately.

(2) turn left.

(3) turn right.

(4) go straight.

25. A pendulum clock loses 12 s a day if the temperature is 40°C and gains 4 s a day if the temperature is 20°C. The temperature at which the clock will show correct time, and the coefficient of linear expansion (α) of the metal of the pendulum shaft are, respectively

(1) 55°C; α = 1.85 × 10−2/°C

(2) 25°C; α = 1.85 × 10−5/°C

(3) 60°C; α = 1.85 × 10−4/°C

(4) 30°C; α = 1.85 × 10−3/°C

26. A uniform string of length 20 m is suspended from a rigid support. A short wave pulse is introduced at its lowest end. It starts moving up the string. The time taken to reach the support is (take g = 10 ms–2)

(1) 2 s (2) ( )2 2π s

(3) 2 s (4) 2 2 s

27. A point particle of mass m moves along the uniformly rough track PQR as shown in the figure. The coefficient of friction, between the particle and the rough track equals μ. The particle is released, from rest, from the point P and it comes to rest at a point R. The energies, lost by the ball, over the parts, PQ and PR, of the track, are equal to each other, and no energy is lost when particle changes direction from PQ to QR. The values of the coefficient of friction μ and the distance x (= QR) are, respectively, close to

h = 2 m

Horizontalsurface

Q30°

P

R

(1) 0.29 and 6.5 m (2) 0.2 and 6.5 m

(3) 0.2 and 3.5 m (4) 0.29 and 3.5 m

28. A pipe open at both ends has a fundamental frequency f in air. The pipe is dipped vertically in water so that half of it is in water. The fundamental frequency of the air column is now

(1) f (2) f

2

(3) 3

4

f (4) 2f

29. A particle performs simple harmonic motion with amplitude A. Its speed is trebled at the instant that it is at a distance 2 3A / from equilibrium position. The new amplitude of the motion is

(1) 7

3

A (2)

A

341

(3) 3A (4) A 3

30. An arc lamp requires a direct current of 10 A at 80 V to function. If it is connected to a 220 V (rms), 50 Hz AC supply, the series inductor needed for it to work is close to

(1) 0.065 H (2) 80 H

(3) 0.08 H (4) 0.044 H

CHeMistrY

31. Which one of the following statements about water is FALSE?

(1) Ice formed by heavy water sinks in normal water.

(2) Water is oxidized to oxygen during photosynthesis.

(3) Water can act both as an acid as a base.

(4) There is extensive intramolecular hydrogen bond-ing in the condensed phase.

32. The concentration of fluoride, lead, nitrate and iron in a water sample from an underground lake was found to be 1000 ppb, 40 ppb, 100 ppm and 0.2 ppm, respectively. This water is unsuitable for drinking due to high concentration of

(1) iron. (2) fluoride.

(3) lead. (4) nitrate.

33. Galvanization is applying a coating of

(1) Zn (2) Pb

(3) Cr (4) Cu

34. Which one of the following complexes shows optical isomerism

(1) [Co(NH3)

4Cl

2]Cl.

(2) [Co(NH3)

3Cl

3].

(3) cis-[Co(en)2Cl

2]Cl.

(4) trans-[Co(en)2Cl

2]Cl.

(en = ethylenediamine)



Question PaPer 647

35. Two closed bulbs of equal volume (V) containing an ideal gas initially at pressure p

i and temperature T

1 are

connected through a narrow tube of negligible volume as shown in the figure below. The temperature of one of the bulbs is then raised to T

2. The final pressure p

f is

p i, V

T1 T1 T1 T1

p i, V pf, V⇒ pf, V

(1) 2 1 2

1 2

pTT

T Ti +⎛⎝⎜

⎞⎠⎟

(2) pTT

T Ti1 2

1 2+⎛⎝⎜

⎞⎠⎟

(3) 2 1

1 2

pT

T Ti +⎛⎝⎜

⎞⎠⎟

(4) 2 2

1 2

pT

T Ti +⎛⎝⎜

⎞⎠⎟

36. The heats of combustion of carbon and carbon monoxide are –393.5 and –285.5 kJ mol–1, respectively. The heat of formation (in kJ) of carbon monoxide per mole is

(1) –110.5 (2) 110.5

(3) 676.5 (4) –676.5

37. At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon requires 375 mL air containing 20% O

2 by volume for

complete combustion. After combustion the gases occupy 330 mL. Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure, the formula of the hydrocarbon is

(1) C4H

10 (2) C

3H

6

(3) C3H

8 (4) C

4H

8

38. Decomposition of H2O

2 follows a first order reaction. In

fifty minutes the concentration of H2O

2 decreases from

0.5 to 0.125 M in one such decomposition. When the con-centration of H

2O

2 reaches 0.05 M, the rate of formation

of O2 will be

(1) 1.34 × 10–2 mol min–1

(2) 6.93 × 10–2 mol min–1

(3) 6.93 × 10–4 mol min–1

(4) 2.66 L min–1 at STP

39. The pair having the same magnetic moment is [At. No.: Cr = 24, Mn = 25, Fe = 26, Co = 27]

(1) [CoCl4]2– and [Fe(H

2O)

6]2+

(2) [Cr(H2O)

6]2+ and [CoCl

4]2–

(3) [Cr(H2O)

6]2+ and [Fe(H

2O)

6]2+

(4) [Mn(H2O)

6]2+ and [Cr(H

2O)

6]2+

40. The species in which the N-atom is in a state of sp-hybridization is

(1) NO2 (2) NO2

+

(3) NO2− (4) NO3

−

41. Thiol group is present in

(1) methionine. (2) cytosine.

(3) cystine. (4) cysteine.

42. The pair in which phosphorous atoms have a formal oxidation state of +3 is

(1) pyrophosphorous and pyrophosphoric acids.

(2) orthophosphorous and pyrophosphorous acids.

(3) pyrophosphorous and hypophosphoric acids.

(4) orthophosphorous and hypophosphoric acids.

43. The distillation technique most suited for separating glycerol from spent-lye in the soap industry is

(1) distillation under reduced pressure.

(2) simple distillation.

(3) fractional distillation.

(4) steam distillation.

44. Which one of the following ores is best concentrated by froth floatation method?

(1) Malachite (2) Magnetite

(3) Siderite (4) Galena

45. Which of the following atoms has the highest first ionization energy?

(1) Sc (2) Rb

(3) Na (4) K

46. In the Hofmann bromamide degradation reaction, the number of moles of NaOH and Br

2 used per mole of

amine produced are

(1) four moles of NaOH and one mole of Br2.

(2) one mole of NaOH and one mole of Br2.

(3) four moles of NaOH and two moles of Br2.

(4) two moles of NaOH and two moles of Br2.

47. Which of the following compounds is metallic and ferromagnetic?

(1) MnO2 (2) TiO

2

(3) CrO2 (4) VO

2

48. Which of the following statements about low density polythene is FALSE?

(1) It is used in the manufacture of buckets, dust-bins etc.

(2) Its synthesis requires high pressure.

(3) It is a poor conductor of electricity.

(4) Its synthesis requires dioxygen or a peroxide initiator as a catalyst.

49. 2-chloro-2-methylpentane on reaction with sodium methoxide in methanol yields:




(a)

C2H5CH2C — OCH3

CH3

CH3

—— (b) C2H5CH2C —— CH2

CH3

—

(c) C2H5CH2 —— C — CH3

CH3

—

(1) (a) and (b) (2) All of these

(3) (a) and (c) (4) (c) only

50. A stream of electrons from a heated filament was passed between two charged plates kept at a potential difference V esu. If e and m are charge and mass of an electron respectively, then the value of h/λ (where λ is wavelength associated with electron wave) is given by:

(1) 2meV (2) meV

(3) 2meV (4) meV

51. 18 g glucose (C6H

12O

6) is added to 178.2 g water. The

vapour pressure of water (in torr) for this aqueous solution is:

(1) 759.0 (2) 7.6

(3) 76.0 (4) 752.4

52. The product of the reaction given below is:

1. NBS/hν2. H2O/K2CO3

X

(1) CO2H (2)

(3) OH (4) O

53. The hottest region of Bunsen flame shown in the figure below is:

Region 4

Region 3

Region 2

Region 1

(1) region 4 (2) region 1

(3) region 2 (4) region 3

54. The reaction of zinc with dilute and concentrated nitric acid, respectively produces:

(1) NO2 and N

2O (2) N

2O and NO

2

(3) NO2 and NO (4) NO and N

2O

55. Which of the following is an anionic detergent?

(1) Glyceryl oleate.

(2) Sodium stearate.

(3) Sodium lauryl sulphate.

(4) Cetyltrimethyl ammonium bromide.

56. The reaction of propene with HOCl (Cl2 + H

2O) proceeds

through the intermediate:

(1) CH3–CHCl–CH

2+

(2) CH3–CH+–CH

2–OH

(3) CH3–CH+–CH

2–Cl

(4) CH3–CH(OH)–CH

2+

57. For a linear plot of log(x/m) versus log p in a Freundlich adsorption isotherm, which of the following statements is correct? (k and n are constants)

(1) log(1/n) appears as the intercept.

(2) Both k and 1/n appear in the slope term.

(3) 1/n appears as the intercept.

(4) Only 1/n appears as the slope.

58. The main oxides formed on combustion of Li, Na and K in excess of air is respectively

(1) Li2O, Na

2O

2 and KO

2.

(2) Li2O, Na

2O and KO

2.

(3) LiO2, Na

2O

2 and K

2O.

(4) Li2O

2, Na

2O

2 and KO

2.

59. The equilibrium constants at 298 K for a reaction A B C D+ + is 100. If the initial concentration of all the four species were 1 M each, then equilibrium concen-tration of D (in mol L−1) will be:

(1) 1.182 (2) 0.182

(3) 0.818 (4) 1.818

60. The absolute configuration of:

CO2H

CH3

OH

Cl

H

H

(1) (2R, 3R) (2) (2R, 3S)

(3) (2S, 3R) (4) (2S, 3S)



Question PaPer 649

MatHeMatiCs

61. A value of θ for which 2 3

1 2

+−

i

i

sin

sin

θθ

is purely imaginary is

(1) sin− ⎛⎝⎜

⎞⎠⎟

1 1

3 (2)

π3

(3) π6

(4) sin− ⎛

⎝⎜⎞

⎠⎟1 3

4

62. The system of linear equations

x + λy – z = 0 λx – y – z = 0 x + y – λz = 0 has a non-trivial solution for:

(1) exactly three values of λ.

(2) infinitely many values of λ.

(3) exactly one value of λ.

(4) exactly two values of λ.

63. A wire of length 2 units is cut into two parts which are bent respectively to form a square of side x units and a circle of radius r units. If the sum of the areas of the square and the circle so formed is minimum, then

(1) 2x = r (2) 2x = (π + 4)r

(3) (4 – π)x = πr (4) x = 2r

64. A man is walking towards a vertical pillar in a straight path, at a uniform speed. At a certain point A on the path, he observes that the angle of elevation of the top of the pillar is 30°. After walking for 10 min from A in the same direction, at a point B, he observes that the angle of ele-vation of the top of the pillar is 60°. Then, the time taken (in minutes) by him, from B to reach the pillar, is

(1) 5 (2) 6

(3) 10 (4) 20

65. Let two fair six-faced dice A and B be thrown simultane-ously. If E

1 is the event that die A shows up four, E

2 is the

event that die B shows up two and E3 is the event that the

sum of numbers on both dice is odd, then which of the following statements is NOT true?

(1) E1, E

2 and E

3 are independent.

(2) E1 and E

2 are independent.

(3) E2 and E

3 are independent.

(4) E1 and E

3 are independent.

66. If the standard deviation of the numbers 2, 3, a and 11 is 3.5, then which of the following is true?

(1) 3a2 – 23a + 44 = 0 (2) 3a2 – 26a + 55 = 0

(3) 3a2 – 32a + 84 = 0 (4) 3a2 – 34a + 91 = 0

67. For x R∈ , f (x) = |log2 – sinx| and g(x) = f (f (x)), then

(1) g is differentiable at x = 0 and ′ = −g ( ) sin(log ).0 2

(2) g is not differentiable at x = 0.

(3) ′ =g ( ) cos(log ).0 2

(4) ′ = −g ( ) cos(log ).0 2

68. The distance of the point (1, −5, 9) from the plane x – y + z = 5 measured along the line x = y = z is

(1) 20

3 (2) 3 10

(3) 10 3 (4) 10

3

69. The eccentricity of the hyperbola, whose length of the latus rectum is equal to 8 and the length of its conjugate axis is equal to half of the distance between its foci, is

(1) 3 (2) 4

3

(3) 4

3 (4)

2

3

70. Let P be the point on the parabola, y2 = 8x which is at a minimum distance from the centre C of the circle, x2 + (y + 6)2 = 1. Then, the equation of the circle, passing through C and having its centre at P is

(1) x2 + y2 – 4x + 9y + 18 = 0

(2) x2 + y2 – 4x + 8y + 12 = 0

(3) x2 + y2 – x + 4y – 12 = 0

(4) x yx

y2 2

42 24 0+ − + − =

71. If Aa b

=−⎡

⎣⎢

⎤

⎦⎥

5

3 2 and Aadj A = AAT, then 5a + b is equal

to

(1) 13 (2) −1

(3) 5 (4) 4

72. Consider f xx

xx( ) tan

sin

sin, , .=

+−

⎛

⎝⎜⎞

⎠⎟∈⎛

⎝⎜⎞⎠⎟

−1 1

10

2

π

A normal to y = f (x) at x =π6

also passes through the point

(1) π4

0,⎛⎝⎜

⎞⎠⎟

(2) (0, 0)

(3) 02

3,

π⎛⎝⎜

⎞⎠⎟

(4) π6

0,⎛⎝⎜

⎞⎠⎟

73. Two sides of a rhombus are along the lines, x – y + 1 = 0 and 7x – y – 5 = 0. If its diagonals intersect at (−1, −2), then which one of the following is a vertex of this rhombus?




(1) −⎛⎝⎜

⎞⎠⎟

10

3

7

3, (2) (−3, −9)

(3) (−3, −8) (4) 1

3

8

3,−⎛

⎝⎜⎞⎠⎟

74. If a curve y = f (x) passes through the point (1, −1) and satisfies the differential equation, y(1 + xy)dx = xdy, then

f −⎛⎝⎜

⎞⎠⎟

1

2 is equal to

(1) 4

5 (2) −

2

5

(3) −4

5 (4)

2

5

75. If all the words (with or without meaning) having five letters, formed using the letters of the word SMALL and arranged as in a dictionary; then the position of the word SMALL is

(1) 58th (2) 46th

(3) 59th (4) 52nd

76. If the 2nd, 5th and 9th terms of a non-constant A.P. are in G.P., then the common ratio of this G.P. is

(1) 7

4 (2)

8

5

(3) 4

3 (4) 1

77. If the number of terms in the expansion of 12 4

2− +⎛

⎝⎜⎞⎠⎟x x

n

,

x ≠ 0, is 28, then the sum of the coefficients of all terms in this expansion, is

(1) 729 (2) 64

(3) 2187 (4) 243

78. If the sum of the first ten terms of the series

135

225

315

4 445

165

2 2 22

2⎛⎝⎜

⎞⎠⎟

+ ⎛⎝⎜

⎞⎠⎟

+ ⎛⎝⎜

⎞⎠⎟

+ + ⎛⎝⎜

⎞⎠⎟

+, ,is m

then m is equal to

(1) 99 (2) 102(3) 101 (4) 100

79. If the line, x y z−

=+−

=+3

2

2

1

4

3 lies in the plane,

lx + my – z = 9, then l 2 + m2 is equal to

(1) 2 (2) 26

(3) 18 (4) 5

80. The Boolean expression ( ~ ) (~ )p q q p q∧ ∨ ∨ ∧ is equivalent to

(1) p q∨ ~ (2) ~ p q∧(3) p q∧ (4) p q∨

81. The integral 2 5

1

12 9

5 3 3

x x

x xdx

++ +∫ ( )

is equal to

(1) −+ +

+x

x xC

10

5 3 22 1( ) (2)

−+ +

+x

x xC

5

5 3 21( )

(3) x

x xC

10

5 3 22 1( )+ ++ (4)

x

x xC

5

5 3 22 1( )+ ++

82. If one of the diameters of the circle, given by the equation x2 + y2 – 4x + 6y – 12 = 0, is a chord of a circle S, whose centre is at (−3, 2), then the radius of S is

(1) 10 (2) 5 2

(3) 5 3 (4) 5

83. lim( )( ).....

/

n n

nn n n

n→∞

+ +⎛⎝⎜

⎞⎠⎟

1 2 32

1

is equal to:

(1) 3 log3 − 2 (2) 18

4e

(3) 27

2e (4)

92e

84. The centres of those circles which touch the circle, x2 + y2 – 8x – 8y – 4 = 0, externally and also touch the x-axis, lie on

(1) a parabola

(2) a circle

(3) an ellipse which is not a circle

(4) a hyperbola

85. Let a b, and

c be three unit vectors such that

a b c b c× × = +( ) ( ).

3

2 If b is not parallel to

c, then

the angle between a and

b is

(1) 5

6

π (2)

3

4

π

(3) π2

(4) 2

3

π

86. Let p xx

x= +→ +lim ( tan ) /

0

2 1 21 then log p is equal to

(1) 1

4 (2) 2

(3) 1 (4) 1

2

87. If 0 ≤ x < 2π, then the number of real values of x, which satisfy the equation cosx + cos2x + cos3x + cos4x = 0, is

(1) 9 (2) 3

(3) 5 (4) 7



Question PaPer 651

88. The sum of all real values of x satisfying the equation

( )x x x x2 4 605 5 12

− + =+ − is

(1) 5 (2) 3

(3) −4 (4) 6

89. The area (in sq. units) of the region (x, y):y2 ≥ 2x and x2 + y2 ≤ 4x, x ≥ 0, y ≥ 0 is

(1) π2

2 2

3− (2) π −

4

3

(3) π −8

3 (4) π −

4 2

3

90. If f x fx

x( ) ,+ ⎛⎝⎜

⎞⎠⎟

=21

3 x ≠0 and

S x f x f x= ∈ = − : ( ) ( ), then S

(1) contains more than two elements.

(2) is an empty set.

(3) contains exactly one element.

(4) contains exactly two elements.

ansWer KeY

1. (4) 2. (4) 3. (1) 4. (1) 5. (1) or (3) 6. (2) 7. (2) 8. (2) 9. (1) 10. (3) 11. (1) or (3) 12. (4) 13. (1) 14. (2) 15. (3) 16. (1) 17. (2) 18. (2) 19. (4) 20. (2) 21. (2) 22. (4) 23. (2) 24. (2) 25. (2) 26. (4) 27. (4) 28. (4) 29. (1) 30. (1) 31. (4) 32. (4) 33. (1) 34. (3) 35. (4) 36. (1) 37. (3) 38. (3) 39. (3) 40. (2) 41. (4) 42. (2) 43. (1) 44. (4) 45. (1) 46. (1) 47. (3) 48. (1) 49. (2) 50. (1) 51. (4) 52. (3) 53. (3) 54. (2) 55. (3) 56. (3) 57. (4) 58. (1) 59. (4) 60. (3) 61. (1) 62. (1) 63. (4) 64. (1) 65. (1) 66. (3) 67. (3) 68. (3) 69. (4) 70. (2) 71. (3) 72. (3) 73. (4) 74. (1) 75. (1) 76. (3) 77. (1) 78. (3) 79. (1) 80. (4) 81. (3) 82. (3) 83. (3) 84. (1) 85. (1) 86. (4) 87. (4) 88. (2) 89. (3) 90. (4)




PHYsiCs

1. The given circuit can be simplified as shown in the following figure:

4 µF 12 µF

2 µF

8 V

V4 812

166= × = V

V12

= V3 = V

9 = 2 V

Therefore, q

4 = 4V

4 = 24 μC

q9 = 9V

12 = 18 μC

Q = q4 + q

9 = 42 μC

Therefore, the magnitude of the electric filed is

EkQ

r= =

× × ××

=−

2

9 6

2

9 10 42 10

9 10420 N/C

Answer (4)

2. Telescope is an optical instrument which makes the distant objects to appear as if they are nearer to us by magnifying the object’s image. Hence, with a magnifying power of 20, a telescope creates the image of a distant tree, which appears 20 times taller.

Answer (4)

3. The area enclosed by a hysteresis loop gives loss per unit volume during one cycle of magnetization. Since in electro-magnets and transformers, the magnetic flux vary with time, it is necessary to keep the losses to the minimum. Hence, a hysteresis loop with the minimum area is preferred.

Answer (1)

4. We have the following half-lives:

t1 2/ A = 20 min

t1 2/ B = 40 min

It is given that at t = 0 min: N

0A = N

0B = N

0

At t = 80 min, we have

4 t1 2/ A = 2 t1 2/ B

Therefore, after 80 min, we get

NA(remaining) =

N0

16

NB(remaining) =

N0

4

ansWers WitH exPlanation



Answers with explAnAtion 653

Thus,

′NA (decayed) = − =NN N

00 0

16

15

16

′NB (decayed) = − =NN N

00 0

4

3

4 That is,

′′

= × =N

N

N

NA

B

15

16

4

35 40

0

:

Answer (1)

5. The angular momentum is

L m r v= ×( )

For the particle moving from D→A, we have

ˆ ˆ ˆ( )2 2

= + × −

R RL m i j v j

ˆ( )2

= −mvRk

A

45°

y

xR

R B

D Ca

a

a a

For the particle moving from A→B, we have

L m

Ri

Rj vi= +⎛

⎝⎜⎞⎠⎟

×⎡

⎣⎢

⎤

⎦⎥

2 2ˆ ˆ ˆ

= − ˆ( )2

mvRk

For the particle moving from C→D, we have

= × ⊥ × ˆ( distance) ( )L mv k

= −

ˆ( )2

Rmv a k

For the particle moving from B→C, we get

= × ⊥ × ˆ( distance) ( )L mv k

= +

ˆ( )mv a k

Answer (1) or (3)

6. If message (modulating) signal is m(t) = AM

sinωM

t and carrier signal is c(t) = Acsinω

ct, the modulated signal is

cM

(t) = Ac(1 + μsinω

Mt)sinω

ct

where μ = modulation index = A AM c/ . The amplitude of the carrier wave varies in accordance with the modulating signal where the amplitude of the modulated wave is given by

Ac(1 + μsinω

Mt)

Answer (2)




7. Given that i = 35°, δ = 40° and e = 79°.

i + e = A + δ Therefore,

δ = i + e – A40° = 35° + 79° − A

⇒ A = 74° For minimum deviation, we have

′ = ′ =+

i eA δM

2

40°

d

35° 79°i

i ′ = e ′

From (i – δ) plot as depicted in the above figure, we get

′ = ′ =+

= = °i e35 79

2

114

257

Therefore,

δM = − = ° − °2 2 57 74i A ( )

= 114° − 74°

= 40° Therefore,

μ

δ=

+

=° + °

sin[( ) / ]

sin( / )

sin[( ) / ]

sin( / )

A

A

A

M 2

2

74 40 2

2

=°°

sin

sin

57

37≈

°°

sin

sin

60

37

= × = ≈3

2

5

3

2 5

1 731 5

.

..

Answer (2)

8. We have the equation of the process AB as

P PP

VV V− =

−−2 0

00( )

P PP

VV= −3 0

0

0

Applying PV = nRT, we have

3 00

0

PP

VV V nRT−

⎛⎝⎜

⎞⎠⎟

=

TnR

PVPV

V= −

⎛⎝⎜

⎞⎠⎟

13 0

02

0




To maximize T, we have dT

dV= 0.

32

000

0

PPV

V− =

⇒ =VV3

20

Therefore,

TnR

PVPVmax = −⎡

⎣⎢⎤⎦⎥

1 9

2

9

40 0

0 0

=9

40 0PV

nR Answer (2)

9. When the wire is bent into a circle: l = 2R and R = l/2π.

R

I

When the wire is bent into a square: l = 4a and a = l/4.

a

I

a

BI

R

I

l

I

lA = = × =μ μ

πμ π0 0 0

2 22

BI

a

I

l

I

lB = = × =2 2 2 2 48 20 0 0μ

πμπ

μπ

Therefore,

B

B

I

I

l

IA

B

= × =μ π π

μπ0

0

2

8 2 8 2 Answer (1)

10. For a screw gauge, we have

Least countPitch

Total no. of divisions on the circular scale=

=0 5

50

.= 0.01 mm

Since 45th division of the screw is considered with the main scale line, it is a case of negative zero error of magnitude 5 × (L.C.) = 0.05 mm. Therefore,

Measured value = M.S.R. + [C.S.D × L.C. + Zero error] = 0.5 + [(25 × 0.01) + 0.05] = 0.5 + 0.30 = 0.80 mm Answer (3)




11. For a common-emitter configuration, we have

α =I

IC

E

, β =I

IC

B

, IE = I

C + I

B

From the above, we get

α ββ

=+1

β αα

=−1

So the incorrect relation are given in options (1) and (3). Answer (1) and (3)

12. The difference angle λ / a (due to the radius of the hole, a) causes a spreading of λL a/ in the size of the spot. When a is small, this becomes large; therefore, we get the following spot size:

aL

a+

λ

The spot would have its minimum size when

1 02

− =λL

a

⇒ =a Lλ

The geometric spread and the diffraction spread are equal; therefore, the minimum spot size is

b Lmin = 4λ Answer (4)

13. The work done by the person against gravity during lifting of 10 kg weight 1000 times is

W = 1000 × mgh = 1000 × 10 × 9.8 × 1 = 9.8 × 104 J

The energy needed is obtained by burning fat. Let the mass of fat that is burnt be m. Then

20

1003 8 10 9 8 107 4× × × = ×. .m

⇒ m = 12.89 × 10−3 kg Answer (1)

14. Energy is

In terms of increasing wavelength, we have

λX-rays

< λBlue light

< λYellow light

< λRadio waves

Ehc

=λ

Therefore, order of increasing energy is E

Radio waves < E

Yellow light < E

Blue light < E

X-Rays

Answer (2)

15. For a polytropic process, PVn = Constant. The molar heat capacity is given by

C CR

n= +

−V 1

C CC C

n= +

−−V

P V

1 (∵R = C

P – C

V)




1− =

−−

nC C

C CP V

V

nC C

C C= −

−−

1 P V

V

nC C

C C=

−−

P

V

Answer (3)

16. For low-orbit satellites, we have v gR0 Escape speed is v gRe = 2

Therefore, Δv = ve – v

0 = −gR( )2 1

Answer (1)

17. To convert a galvanometer into an ammeter, a shunt resistance is connected across it as shown in the following figure:

G

Rg = 100 Ω

Ig

I – Ig

I

S

Then, RgI

g = S(I – I

g)

Therefore, SR I

I I=

−=

×−

= =g g

g

100 0 001

10 0 001

0 10

9 990 01

.

.

.

.. Ω

Answer (2)

18. From Einstein’s photoelectric equation, we get

hc

mvλ

ϕ= + ⎛⎝⎜

⎞⎠⎟

1

22

max

(1)

where ϕ in the work function of the emitter. On changing the wavelength to 3λ/4, we get

hc

m v3 4

1

22

λϕ

/( )

max

= + ′⎡⎣⎢

⎤⎦⎥

(2)

Dividing Eq. (1) by Eq. (2), we get

3

4

1 2

1 2

2

2= +

+ ′ϕϕ

( / )

( / )

mv

mv

⇒ + ′⎡⎣⎢

⎤⎦⎥

= + ′⎡⎣⎢

⎤⎦⎥

4 41

23 3

1

22 2ϕ ϕm v m v( ) ( )

⇒ + = ′ϕ 23

22 2mv m v( )

Therefore,

( )′ = +vm

v2 22

3

4

3

ϕ

Since 2

30

ϕm

> , we get

′ > ⎛⎝⎜

⎞⎠⎟

v v4

3

1 2/

Answer (2)




19. According to the truth table of OR gate, the output is 1 when any of the inputs is 1 and output is 0 when all inputs are zero.

Answer (4)

20. Applying Gauss’s law, we have

E rQ q

( )4 2

0

× =+ ′ε

′ = ∫q x dxa

r

ρ π4 2 (where ρ =

A

x for a < x < N)

= ∫A x dxa

r

4π = 2πA(r2 – a2)

Therefore,

EQ

rA

a

r= + −

⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥

1

42 1

02

2

2πεπ

= − +⎡⎣⎢

⎤⎦⎥

1

4

12 2

02

2

πεπ π

rQ Aa A( )

Q a

b

r

r = Ar

If E is constant for a < r < b, then E should be independent of r. That is,

Q − 2πAa2 = 0

⇒ =AQ

a2 2π Answer (2)

21. We have T =

+ + +90 91 95 92

4 = 92 s

Δ = − =T T T1 1 2| | s

Δ = − =T T T2 2 2| | s

Δ = − =T T T3 3 3| | s

Δ = − =T T T4 4 0| | s

Therefore,

Δ =Δ + Δ + Δ + Δ

=TT T T T

avs1 2 3 4

41 5.

Since the minimum division in the measuring clock is 1 s, the reported mean time is 92 ± 2 s. Answer (2)

22. Copper (Cu) is a conductor. Its resistance varies with temperature as

R = k0 (1 + αΔT)

Silicon (Si) being a semiconductor, the resistance shows an exponential decrease with temperature. Answer (4)




23. Concept based

Answer (2)

24. As the roller rolls forward, ′v (left) becomes smaller than v (right). Therefore,

ω ω′ <r r ⇒ ′ <v v

DB

C

v

A

Answer (2)

Thus, the roller tends to turn towards left.

25. We have

Tl

g= 2π

and l = l0(1 + αΔθ)

Δ

=−

= Δl

l

l l

l0

0

0

α θ

Using error analysis, we get

Δ

=ΔT

T

l

l

1

2 Case 1: When the clock gains 12 s, we get

12 1

240 0T

= −α θ( ) (1)

where θ0 is the optimum temperature at which it shows the correct time.

Case 2: When the clock loses 4 s, we get

4 1

2200T

= −α θ( ) (2)


340

200

0

=−−

θθ

3θ0 – 60 = 40 – θ

0

4θ0 = 100

θ0 = 25°C

Therefore, from Eq. (2), we get

4

24 3600

1

225 20

×= −α( )

⇒ =× ×

= × °−α 8

24 3600 51 85 10 5. / C

Answer (2)

26. The velocity of the wave in a stretched string is

vT

=μ

where μ is mass/length of the string.




dy

y

l = 20 m

At any y, as shown in the figure, T = μyg

vyg

ygdy

dt= = =μ

μ⇒ =g dt

dy

y On integrating, we get

g dtdy

y

lt

= ∫∫00

g t l= 2

tl

g= = =2 2

20

102 2 s Answer (4)

27. The energy lost against friction for the path PQ is

W mgh

PQ = [ ( ) cos ]sin

μ θθ

Similarly,

WQR

= μ(mg)x

It is given that WPQ

= WQR

[ ( ) cos ]sin

( )μ θθ

μmgh

mg x=

⇒ x = hcotθ = hcot30°

⇒ = =x h3 3 5. m

30°

P

Q R

m

m

h = 2 m

Also, we have

WPQ

+ WQR

= mgh That is,

[ ( ) cos ]sin

( )μ θθ

μmgh

mg x mgh+ =

⇒ μh(cotθ) + μx = h

⇒ ( cot ) ( )μ μ× × ° + =2 30 3h h

⇒ 2 3 2 3 2μ μ+ = =h

⇒ 4 3 2μ =

⇒ μ = ≈1

2 30 29.

Answer (4)




28. For the open pipe as shown in the following figure, we have

ν0 2= =

v

lf

Airl

For the open pipe as shown in the following figure, we have

′ = = = =ν ν0 04 2 2

v

l

v

lf

( )/

Airl /2

l /2

Here, it acts as a closed pipe with vibrating length, l/2. Answer (1)

29. For simple harmonic motion, we have

v A x= −ω 2 2

⇒ = −v A x2 2 2 2ω ( )

where A is the amplitude of the simple harmonic motion. Initially, at x A= 2 3/ , we have

v AA

A2 2 22

2 24

9

5

9= −

⎛⎝⎜

⎞⎠⎟

=ω ω (1)

Now, at x A= 2 3/ , ′ =v v3 , let the new amplitude be ′A . Therefore,

( ) ( )34

92 2 2

2

v AA

= ′ −⎡

⎣⎢

⎤

⎦⎥ω (2)


1

9

5

9 4 9

5

9 4

2

2 2

2

2=

′ −=

′ −A

A A

A

A A( ) ( / ) ( )

Therefore, 9(A′)2 – 4A2 = 45A2

9(A′)2 = 49A2

⇒ ′ =A a7

3 Answer (1)

30. From the circuit, we have

XL = wL8 Ω

220 V, 50 Hz




R = =

80

108 Ω

102202 2

=+R X L

⇒ + =64 222X L

⇒ = − =X L2 484 64 420

⇒ = =X L 420 20 5. Ω⇒ ωL = 20.5

L =×

= =20 5

2 50

20 5

3140 065

. ..

πH

Answer (1)

CHeMistrY

31. In case of H2O molecule, there is extensive intermolecular hydrogen bonding in the condensed phase.

Answer (4)

32. The concentration of fluoride and lead are in ppb (parts per billion) while the concentration of nitrate and iron are in ppm (parts per million) which is more than ppb. So, the highest concentration is of nitrate, that is, 100 ppm.

Answer (4)

33. Galvanization is a process for cathodic protection of iron to prevent it from corrosion. It is a process in which coat of Zn is applied on the iron or steel article.

Answer (1)

34. The compound cis [Co(en)2Cl

2]Cl shows optical isomerism.

CO

en

en

Cl

Cl

CO

en

en

Cl

Cl

Answer (3)

35. Applying combined gas law (for both initial and final states)

(Initial moles) n1 = n

2 (Final moles)

pV

RT

p V

RT

p V

RT

p V

RTi i f f

1 1 2 1

+×

=×

+×

V

R

p

T

p

T

V

R

p

T

p

Ti i f f

1 1 2 1

+⎛⎝⎜

⎞⎠⎟

= +⎛⎝⎜

⎞⎠⎟

2 1 1

1 2 1

p

Tp

T Ti

f= +⎛⎝⎜

⎞⎠⎟

2

1

1 2

1 2

p

Tp

T T

T Ti

f=+⎛

⎝⎜⎞⎠⎟

p pT

T Tf i=+

⎛⎝⎜

⎞⎠⎟

2 2

1 2

Answer (4)




36. The given equations are

C(s) + O2 (g) → CO

2 (g) (1) ΔH = −393.5 kJ mol−1

CO g O g CO g( ) ( ) ( )+ →1

2 2 2 (2) ΔH = −283.5 kJ mol−1

The required equation is

C s O g CO g( ) ( ) ( )+ →

1

2 2 ΔH = ?

On reversing Eq. (2), we get

CO g CO(g) O (g)22

1

2( ) → + (3) ΔH = 283.5 kJ mol−1

On adding Eq. (1) and Eq. (3), we get the required equation:

C(s) O (g) CO (g)2 2+ →1

2

ΔH = −393.5 + 283.5 = −110.0 kJ mol−1

Answer (1)

37. General equation for the combustion reaction of hydrocarbons is:

C H (g) O (g CO H O2 2x y xy

x gy

+ +⎛⎝⎜

⎞⎠⎟

→ +4 22) ( ) (l)

Volume of O2 used = 20% of 375 mL

= × =20

100375 75 mL

Volume of air remaining = 375 – 75 = 300 mL Total volume of gas left after combustion = 330 mL Volume of CO

2 after combustion = 330 – 300 = 30 mL

C H g O g CO g H O lmL

mL

230 mL

x y xy

xy

1575

2 24 2( ) ( ) ( ) ( )+ +⎛

⎝⎜⎞⎠⎟

→ +

x

1

30

15= ⇒ x = 2

x

y+=4

1

75

15⇒ x

y+ =

45

⇒y = 12

C2H

12 is not in the given options, but out of the given options only C

3H

8 is able to consume 75 mL of O

2, so, it can be con-

sidered as the correct answer. Answer (3)

38. In 50 min, conc. of H2O

2 =

1

4 of initial conc.

Therefore, 2 × t1/2

= 50 min and hence t1/2

= 25 min.

As tk1 2

0 693/

.=

k =0 693

25

.

Rate = k[H2O

2] (given it is a first order reaction)

Rate (H2O

2) = ×

0 693

250 05

..

= 1.386 × 10−3 m min−1




2H2O

2 → 2H

2O + O

2

− = + =1

2

1

22 2d

dt

d

dt

d

dt

[ ] [ ] [ ]H O H O O2 2

− =1

22 2d

dt

d

dt

[ ] [ ]H O O2

d

dt

[ ].

O2 31

21 386 10= × × − = 6.93 × 10−4 mol min−1

Answer (3)

39. We have

[Cr(H2O)

6]2+ [Fe(H

2O)

6]2+

Cr (Z = 24) = 3d5 4s1 Fe (Z = 26) = 3d6 4s2

Cr2+ = 3d4 Fe2+ = 3d6

↑ ↑ ↑ ↑ ↑↓ ↑ ↑ ↑ ↑ Four unpaired electrons Four unpaired electrons

Magnetic moment depends on the number of unpaired electrons which is same in both the above cases; therefore, both of them will have same magnetic moment.

Answer (3)

40. The shape and hybridization of N atom in the given species are as follows: NO22− = sp (bent); NO2

+ = sp (linear); NO

2 = sp2 (bent); NO3

2− = sp (trigonal planar). Answer (2)

41. The structures of the given amino acids are as follows:

Methionine

CH3 — S — CH2 — CH2 — CH — C — OH

—

— —

NH2

O

Cytosine NH2

NO

N

H

Cystine NH2

HO

O NH2

COOHC

SS

Cysteine

NH2

COOH

HS

Thiolgroup is present Answer (4)

42. Orthophosphorous acid (H3PO

3)

3 6

3

+ == +

x

x

– 0

Pyrophosphorous acid (H4P

2O

5)

4 2 10 0

2 6 0

3

+ ==

= +

x

x

x

–

–

Answer (2)




43. Glycerol is a high boiling liquid. It is made to boil at lower temperature than normal temperature by lowering pressure on its surface due to which boiling point of liquid is lowered and hence glycerol is obtained without decomposition at high temperature.

Answer (1)

44. Froth floatation method is used to concentrate sulphide ore, so, the correct option is Galena PbS (sulphide ore of lead).

Answer (4)

45. The effective nuclear change increases in a period, hence ionization energy increases while it generally decreases down the group. Thus, the increasing order of first ionization enthalpy is as follows:

Rb < K < Na < Sc Answer (1)

46. According to the reaction, for the production of one mole of amine, four moles of NaOH and one mole of Br2 are consumed.

RCONH Br NaOH R Na CO NaBr + 2H O22 2 2 34 2+ + → + +− ΝΗ2 Answer (1)

47. CrO2 is strongly attracted by magnetic field and can be permanently magnetized. The alignment of CrO

2 domains in mag-

netic field is as follows:

Answer (3)

48. Low density polyethene (LDP) is obtained when ethene is polymerized under high pressure and temperature in the pres-ence of peroxide or dioxygen initiator. It is chemically inert and flexible and also a poor conductor of electricity. It is not used in the manufacturing of buckets, dustbin etc., instead used in manufacturing toys, carrier bags, etc.

Answer (1)

49.

CH3 — C — CH2 — CH2 — CH3

CH3ONa

CH3OHCH3

Cl

— CH3 — C —— CH — CH2 — CH3

CH3

—

—

(I)

+ CH2 —— C — CH2 — CH2 — CH3

CH3

—

(II)

+ CH3 — C — CH2 — CH2 — CH3

CH3

OCH3

——

(III)

In case of products I and II, β-elimination takes place as methoxide acts as a strong base in presence of methanol. Product (III) is also possible because the reaction is carried out between haloalkane and sodium methoxide (Williamson’s synthesis).

Answer (2)

50. K.E. of an electron = eV mv=1

22

According to de-Broglie relationship, λ =h

mv

1

22mv eV=

⇒

m v

meV

2 2

2=

⇒ mv meV= 2

λ =h

meV2 ⇒

hmeV

λ= 2

Answer (1)




51. No. of moles of H2O (n

1) = =

178 2

189 9

..

No. of moles of glucose (n2) = =

18

1800 1.

Total number of moles is n n ntotal 1 1= + =2 0

According to relative lowering in vapor pressure

p p

p

n

n n

°°

−=

+S 2

1 2

760

760

0 1

10

−=

pS .

⇒p

S = 760 – 7.6 ⇒ p

S = 752.4 torr

Answer (4)

52.

1. NBS

hn

• •

Br Br

2. H2O K2CO3

+

OH OH

+

Answer (3)

53.

Reactive gasesincomplete combustion

Hottest part of flame

Unburnt gas and air

Answer (3)

54. 4Zn + 10HNO3 (dil.) → 4Zn (NO

3)

2 (aq.) + N

2O + 5H

2O

Zn + 4HNO3 (conc.) → Zn (NO

3)

2(eq) + 2NO

2 + 2H

2O

Answer (2)

55. Sodium lauryl sulphate is an anionic detergent. The structure is CH3 — (CH2)16 — CH2 — O — S — ONa+

— —— —

–O

O Glyceryl oleateis a non-ionic detergent. Sodium stereate is a soap (anionic). Cetyltrimethyl ammonium bromide is a cationic detergent




Answer (3)

56. Reaction: + HO — Cl

+— OH– COOHCl OH–

OH

Mechanism: + Cl — Clδ+ δ−

+ Cl

Intermediate Answer (3)

57. According to Frendlich adsorption isotherm

x

mkp n= 1/

log log logx

m np k= +

1

Comparing with straight line equation y = mx + c, m is the slope which is equal to 1/n. Answer (4)

58. We have

Li O g Li O(excess

2+ →2 ( ))

(Monoxide)

Na O (g) Na O2(excess)

2 2+ → (Peroxide)

K O (g) KO2(excess)

2+ → (Superoxide)

Answer (1)

59. For the given equation, let x be the amount of substance dissociated.

A + B C + D

Initial mole 1 1 1 1

At teq

1 – x 1 – x 1 + x 1 + x

keq

C D

A B= [ ][ ]

[ ][ ]

1001

1

2

2=

+−

( )

( )

x

x⇒1 + x = 10 – 10x ⇒ x =

9

11

[D] = 1 + x = +19

11= 1.818

Answer (4)

60.

COOH3

CH3

OH ≡

≡Cl1

2

3

2

1

4

4H

H

3

2

2S4 1

2

3

3R4 1

Answer (3)




MatHeMatiCs

61. We have

Resin

sin

2 3

1 20

+−

⎛⎝⎜

⎞⎠⎟

=i

i

θθ

That is,

Re( sin )( sin )

( sin )( sin )

2 3 1 2

1 2 1 20

+ +− +

⎛⎝⎜

⎞⎠⎟

=i i

i i

θ θθ θ

⇒ Re( sin ) ( sin )

sin

2 6 2 5

1 40

2

− ++

⎛⎝⎜

⎞⎠⎟

=θ θ

θi

⇒2 6

1 40

2

2

−+

=sin

sin

θθ

⇒ 6sin2θ = 2

⇒ sin2 1

3θ =

⇒ θ =⎛⎝⎜

⎞⎠⎟

−sin 1 1

3

Answer (1) 62. For non-trivial solution, we have

1 1

1 1

1 1

0

λλ

λ

−− −

−=

⇒ 1(λ + 1) – λ(−λ2 + 1) – 1(λ + 1) = 0

⇒ λ(λ2 – 1) = 0

⇒ λ = −1, 0, 1 Answer (1)

63. Side length of the square = x. Radius of circle = r. Perimeter of square + Perimeter of circle = 2 4x + 2πr = 2 ⇒ 2x + πr = 1 ⇒ πr = 1 – 2x

⇒ rx

=−⎛

⎝⎜⎞⎠⎟

1 2

π The sum of the area is x2 + πr2 = A

Substituting the value of r we get

A xx

= +−2

2

2

1 2ππ

( ) = +

−x

x221 2( )

π Differentiating w. r. t. x we get

dA

dxx x= + − −2

12 1 2 2

π( )( ) = − − =2

41 2 0x x

π( )

Again differentiating we get

d A

dxx

2

22

42 0= + >

π( )




Now, from equation calculation dA/dx we have

x x− − =2

1 2 0π

( )

⇒ πx – 2 + 4x = 0

⇒ =+

⎛⎝⎜

⎞⎠⎟

x2

4π

The minimum value occurs at

x =+2

4π The minimum value is

ππ

ππ

r = −+

=+

14

4 4

⇒ =+

r1

4( )π ⇒ x = 2r Answer (4)

64. The given situation is depicted in the following figure. We have

AB = x; BP = y; PQ = h

Q

P AB60° 30°

xy

h

tan301

33° =

+⇒ =

+⇒ + =

h

x y

h

x yx y h

tan60 3° = ⇒ =h

yh y

Now, x y y y+ = × =3 3 3 ⇒ x = 2y

Let the speed of man be u. Therefore,

u

x

t

x

uy

t

x y

t

y y

t tt

= =

=

⎤

⎦

⎥⎥⎥⎥

⇒ = ⇒ = ⇒ = ⇒ =1010

2

10

1

5

15 min

Answer (1)

65. We have

E1 → Dice A shows 4

E2 → Dice B shows 2

E3 → Sum odd (No. on A to be odd + No. on B to be even or No. on B to be odd + No. on A to be even)




Therefore,

P(E1) =

1

6

P E( )2

1

6=

P EC C C

( )3

21

31

31

6 6

1

2=

⋅ ⋅⋅

=

Now, P E E P E P E( ) ( ) ( )1 2 1 2∩ = ⋅

P E E P E P E( ) ( ) ( )1 3 1 3∩ = ⋅

P E E P E P E( ) ( ) ( )2 3 2 3∩ = ⋅

This shows that options (2) to (4) are correct and hence option (1) that the events are independent is not correct. Answer (1)

66. We have

Standard deviation = Σ Σx

n

x

ni i2 2

− ⎛⎝⎜

⎞⎠⎟

⇒ 3 54 9 121

4

16

4

2 2

2.

( )=

+ + +−

+a a

⇒49

4

4 134 16

4

2 2

2=

+ − +( ) ( )a a

⇒ 3a2 – 32a + 84 = 0 Answer (3)

67. We have

f (x) = |log2 – sinx| and g(x) = f (f (x))

Therefore, g(x) = |log2 – sin f (x)|

But f (x) = log2 – sinx(log2 – sin x > 0 in neighbourhood of x = 0). So g(x) = |log2 – sin(log2 – sinx)

= log 2 – sin(log 2 – sin x) [g(x) is constant function at x = 0]

Differentiating we get ′ = − − −g x x x( ) cos(log sin )( cos )2

= ⋅ −cos cos(log sin )x x2

Substituting x = 0, we get ′ =g ( ) cos(log )0 2 Answer (3)

68. The line passing through the point (1, 5, 9) and parallel to the line x = y = z is given by

x y z−

=+

=−1

1

5

1

9

1 Any point on this line is x – 1 = y + 5 = z – 9 = λ

Now, (1 + λ, λ – 5, 9 + λ) lies on the given plane. Therefore,

x – y + z = 5 λ λ λ+ − + + + =1 5 9 5

⇒ λ + 15 = 5 ⇒ λ = −10




The point is (−9, −15, −1). The distance between the points is

( ) ( ) ( )1 9 5 15 9 12 2 2+ + − + + + = 10 3

Answer (3)

69. Length of the latus rectum of hyperbola is 8. So

2

82b

a= ⇒ b2 = 4a

Also given Length of conjugate axis =1

2× (Distance between the foci)

⇒ 2b = 1

22 2× ⇒ =ae b ae

Eccentricity is given by eb

a= +1

2

2

Squaring both sides we get

e

b

a

b

ae2

2

2

2

221 1= + ⇒ = −

But 22 4

2

2

2

b aeb

a

e b

a

e= ⇒ = ⇒ = . So we have

e

e2

2

41= − ⇒ 1

3

4

2

=e

⇒ e =2

3 Answer (4) 70. The parabola is y2 = 8x. Therefore, 4a = 8 ⇒ a = 2.

P(2, – 4)

C(0, –6)

The normal to the parabola that meets at points P( , )am am2 2− = P( , )2 42m m− given by

y = mx – 4m – 2m3

passes through the centre of the circle (0, −6). So −6 = −4m – 2m3

⇒ 2m3 + 4m – 6 = 0

⇒ m3 + 2m – 3 = 0 Therefore, ⇒ m2(m – 1) + m(m – 1) + 3(m – 1) = 0

⇒ m = 1 and m2 + m + 3 ≠ 0

Hence, the point on the parabola is P(2, –4). Therefore,

CP = Radius = − + − + = + =( ) ( )0 2 6 4 4 4 2 22 2

The equation of circle is (x – 2)2 + (y + 4)2 = 8

⇒ x2 + y2 – 4x + 8y + 12 = 0

Answer (2)




71. Given

Aa b

=−⎡

⎣⎢

⎤

⎦⎥

5

3 2 and Aadj A = AAT

Now

adj Ab

a=

−⎡

⎣⎢

⎤

⎦⎥

2

3 5 and A

a

bT =

−⎡

⎣⎢

⎤

⎦⎥

5 3

2

Substituting these values in Aadj A = AAT, we get

5

3 2

2

3 5

5

3 2

5 3

2

a b b

a

a b a

b

−⎡

⎣⎢

⎤

⎦⎥ −

⎡

⎣⎢

⎤

⎦⎥ =

−⎡

⎣⎢

⎤

⎦⎥ −

⎡

⎣⎢

⎤

⎦⎥

⇒10 3 5 5

0 3 10

25 15 2

15 2 13

2 2a b ab ab

b a

a b a b

a b

+ −+

⎡

⎣⎢

⎤

⎦⎥ =

+ −−

⎡

⎣⎢

⎤

⎦⎥

⇒10 3 0

0 10 3

25 15 2

15 2 13

2 2a b

a b

a b a b

a b

++

⎡

⎣⎢

⎤

⎦⎥ =

+ −−

⎡

⎣⎢

⎤

⎦⎥

Comparing respective entries of equal matrices we get 10a + 3b = 25a2 + b2 (1) 10a + 3b = 13 (2)

and 15a −2b = 0 ⇒ =ba15

2 (3) & (4)

Substituting the value of b in Eq. (2), we get

10 315

213a

a+ ⎛

⎝⎜⎞⎠⎟

=

⇒+

=20 45

213

a a

⇒ 65a = 13 × 2 ⇒ a = 2

5 Substituting the value of a in Eq. (4), we get

b = × =15

2

2

53

Hence 5 52

53 5a b+ = ⎛

⎝⎜⎞⎠⎟

+ =

Answer (3)

72. We have

f xx

x( ) tan

sin

sin,=

+−

⎛

⎝⎜⎞

⎠⎟−1 1

1 x ∈⎛

⎝⎜⎞⎠⎟

02

,π

⇒ f xx x

x x( ) tan

[cos( ) sin( )]

[cos( ) sin( )]=

+−

⎛

⎝⎜

⎞

⎠⎟−1

2

2

2 2

2 2

/ /

/ /

⇒ f xx x

x x( ) tan

cos( ) sin( )

cos( ) sin( ),=

+−

⎛

⎝⎜⎞

⎠⎟−1 2 2

2 2

/ /

/ / x ∈⎛

⎝⎜⎞⎠⎟

02

,π

⇒ f xx x

x x( ) tan

cos( ) sin( )

cos( ) sin( )=

+−

⎡

⎣⎢

⎤

⎦⎥

−1 2 2

2 2

/ /

/ /

⇒ f xx

x( ) tan

tan( )

tan( )=

+−

⎡

⎣⎢

⎤

⎦⎥

−1 1 2

1 2

/

/




⇒ f xx

( ) tan tan= +⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥

−1

4 2

π

⇒ f xx

( ) = +π4 2

Hence fπ π π π π π6 4 12

3

12 3⎛⎝⎜

⎞⎠⎟

= + =+

=

The point is π π6 3

, .⎛⎝⎜

⎞⎠⎟

Also

′ =f x( )1

2 ⇒ ′ ⎛

⎝⎜⎞⎠⎟

=fπ6

1

2

The slope of normal is −2 and the equation of normal is

y x− = − −⎛⎝⎜

⎞⎠⎟

π π3

26

= − +23

xπ

Therefore,

y x= − +2

2

3

π

⇒ 22

3x y+ =

π

⇒x y

π π/ /3 2 31+ =

Thus, the normal passes through the point 02

3, .

π⎛⎝⎜

⎞⎠⎟ Answer (3)

73. We have

7 5 0

1 0

6 6 0

x y

x y

x

– –

–

–

=+ =

− + −

=

⇒ x = 1

D C

A

7x – y – 5 = 0

x – y + 1 = 0 B

O(–1, –2)

x – y + 1 = 0

1 – y + 1 = 0 ⇒ y = 2

Therefore, point A is (1, 2). C(α, β); O is mid-point of points A and C.

α +

= −1

21 ⇒ α + 1 = −2 ⇒ α = −3

β +

= −2

22 ⇒ β + 2 = −4 ⇒ β = −6




Therefore, we get point C(−3, −6). Point B is (x

1, y

1) and point D is (x

2, y

2), that is,

x1 – y

1 + 1 = 0 and 7x

2 – y

2 – 5 = 0

x x1 2

21

+= − and

y y1 2

22

+= −

x2 = −2 – x

1 and y

2 = −4 – 2y

1

Therefore, 7(−2 – x

1) + 4 + y

1 – 5 = 0

⇒ −14 – 7x1 + y

1 – 1 = 0

⇒ −7x1 + y

1 – 15 = 0

⇒ −7x1 + x

1 + 1 – 15 = 0

⇒ −6x1 – 14 = 0

⇒ x1

7

3= − and y1

7

31

4

3=

−+ =

−

Therefore, we get point B − −⎛⎝⎜

⎞⎠⎟

7

3

4

3, .

Now,

x2 27

3

1

3= − + = +

and

y2 4

4

3

8

3= − + = −

Therefore, we get point D1

3

8

3, .−⎛

⎝⎜⎞⎠⎟ Answer (4)

74. We have

y (1 + xy)dx = xdy

⇒ ydx – xy2dx = xy2dx

⇒ ydy – ydx = xy2dx

⇒ ydx – xdy = −xy2dx

⇒y dx x dy

yx dx

−⎛⎝⎜

⎞⎠⎟

= −2

Therefore,

dx

yx dx

⎛⎝⎜

⎞⎠⎟

= −∫∫

⇒x

y

xC= − +

2

2 (1)

which passes through (1, −1).

− = − +11

2C

⇒ C = − + = −11

2

1

2

Substituting the value of C in Eq. (1), we get

x

y

x= − −

2

2

1

2

⇒x

y

x= −

+( )2 1

2

⇒ yx

x=

−+

2

12 and x = −

1

2




Therefore,

f −⎛⎝⎜

⎞⎠⎟

= −− −

+=

1

2

2 1 2

1 1 4

4

5

( )

( / )

/

Answer (1)

75. For the word SMALL:

• The number of word starting from A is

4

212

!

!=

• The number of words starting from L is 4! = 24 • The number of words starting from M is

4

212

!

!=

• The number of words starting with SA is

3

23

!

!=

• The number of words starting with SL is 3! = 6 • The number of words starting with SMALL is 1.

Therefore, the position of the word SMALL to occur is 12 + 24 + 12 + 3 + 6 + 1 = 58th position. Answer (1)

76. Let a and b be the 1st term and the common difference of A.P., namely, t2 = a + d, t

5 = a + 4d and t

9 = a + 8d are in G.P.

That is, t2, t

5 and t

9 in G.P.

t

t

t

tt t t5

2

9

552

2 9= ⇒ =

⇒ (a + 4d)2 = (a + d) (a + 8d)

⇒ a d ad a ad ad d2 2 2 216 8 8 8+ + = + + +

⇒ 8d2 – ad = 0

⇒ d(8d – a) = 0 ⇒ d = 0 which is not possible and a = 8d. Now, the common ratio of G.P. is

t

t

a d

a d

d

d5

2

4 12

9

4

3=

++

= = Answer (3)

77. We have

12 4

2− +⎛

⎝⎜⎞⎠⎟x x

n

The number of terms in the expansion is

n+2C2 = 28

⇒+ +

=( )( )n n2 1

228

⇒ (n + 2) (n + 1) = 56

⇒ n2 + 3n – 54 = 0

⇒ n = 6

The sum of coefficient is

(1 – 2 + 4)6 = 36 = 729




This is possible only when we are not considering the number of dissimilar terms.

Note: If we consider the dissimilar term, then number of terms is 2n + 1 and hence

2n + 1 = 28 ⇒ =n27

2 which is not possible (and hence the question may be considered wrong). Answer (1)

78. We have

8

5

12

5

16

5

20

5

2 2 2 2⎛⎝⎜

⎞⎠⎟

+ ⎛⎝⎜

⎞⎠⎟

+ ⎛⎝⎜

⎞⎠⎟

+ ⎛⎝⎜

⎞⎠⎟

+ upto 10 terms = 16

5m

That is,

4

52 3 4 5 11

16

5

2

2 2 2 2 2⎛⎝⎜

⎞⎠⎟

+ + + + + =( ) m

⇒ 12 + 22 + 32 + 42 + 52 + … + 12 = 5m + 12

⇒+ +

= +11 11 1 2 11 1

65 1

( )[ ( ) ]m

⇒ 22 × 23 = 5m +1

⇒ 506 = 5m + 1

⇒ 5m = 505

⇒ m = 101 Answer (3)

79. The line x y z−

=+−

=+3

2

2

1

4

3 lies in the plane lx + my – z = 9.

Point (3, −2, −4) lies in the plane and direction ratios of the line of plane are at 90°. Therefore, l (3) + m(−2) + 4 = 0 ⇒ 3l – 2m = 5

⇒ l (2) + m(−1) + 3(−1) = 0

⇒ 2l – m = 3

⇒ m = 2l – 3

Now, 3l – 2 (2l – 3) = 5

⇒ 3l – 4l + 6 = 5

⇒ −l = −1 ⇒ l = 1

⇒ m = 2 – 3 = −1

⇒ l = 1 and m = −1 Therefore, l2 + m2 = 2. Answer (1)

80. The given Boolean expression is

( ~ ) ( )p q q p q∧ ∨ ∨ − ∧ Therefore, ( ~ ) ( ) (~ ) ( ) ( )p q q p q q q p q t p q∧ ∨ = ∨ ∧ ∨ = ∨ ∧ = ∨

Now, ( ) (~ ) ( )p q p q p q∨ ∨ ∧ = ∨

Answer (4)

81. We have

2 5

1

12 9

5 3 3

x x

x xdx

++ +∫ ( )




⇒+

+ +⎛⎝⎜

⎞⎠⎟

∫2 5

11 1

12 9

152 5

3

x x

xx x

dx

⇒

2 5

11 1

3 6

2 5

3

x x

x x

dx

+⎛⎝⎜

⎞⎠⎟

+ +⎛⎝⎜

⎞⎠⎟

∫

Let 1 + x−2 + x−5 = t. Then

− −⎛⎝⎜

⎞⎠⎟

=2 5

3 6x xdx dt

⇒2 5

3 6x xdx dt+⎛

⎝⎜⎞⎠⎟

= −

⇒ 1 1

23 2tdt

tC= − +∫ =

+ +⎛⎝⎜

⎞⎠⎟

+ ≡+ +

+1

2 11 1 2 1

2 5

2

10

5 3

x x

Cx

x xC

( )

Answer (3)

82. We have

x2 + y2 – 4x + 6y – 12 = 0

⇒ (x2 – 4x + 4) + (y2 + 6y + 9) – 25 = 0

⇒ (x – 2)2 + (y + 3)2 = 25

P

5

(–3, 2)

(2, –3)C1

C2

Therefore,

C C1 2 = + + − −( ) ( )2 3 3 22 2 = 5 2

(C P) = C C ) +(C P)22

1 22

12( = 50 + 25 = 75

Therefore, the required radius of the circle is

C P2 = 5 3

Answer (3)

83. We have

yn n n n

n n

n

=+ + +⎛

⎝⎜⎞⎠⎟→∞

lim( )( ) ( )

/1 2 2

22

1

⇒ =+⎛

⎝⎜⎞⎠⎟

++⎛

⎝⎜⎞⎠⎟

+ ++⎛

⎝⎜⎞⎠⎟

⎡

⎣⎢

⎤→∞

ln ln ln lnyn

n

n

n

n

n n

nnlim

1 1 2 2

⎦⎦⎥

= +⎛⎝⎜

⎞⎠⎟

+ +⎛⎝⎜

⎞⎠⎟

+ + +⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥→∞

lim ln ln lnn n n n

n

n

11

11

21

2




= +⎛⎝⎜

⎞⎠⎟→∞

=∑lim

nr

n

n

r

n

11

1

2

ln

= + = + + − +∫ ln( ) ( ) ln( ) ( )1 1 1 10

2

0

2

x dx x x x

= 3(ln3) – 3 + 1

= 3(ln3) – 2 = ln 27 – ln e2

ln lnye

= ⎛⎝⎜

⎞⎠⎟

272

⇒ ye

=27

2

Answer (3) 84. We have

(x2 – 3x + 16) + (y2 – 8y + 16) = 36

⇒ (x − 4)2 + (y – 4)2 = 36

The centre of the circle is (α, β) and the radius of the circle is β.

y

x

C1

C2

(a, b)(4, 4)

Now, C1C

2 = 4 + β

( ) ( ) ( )4 4 42 2− + − = +α β β

Squaring both sides

16 + α2 − 3α + 16 + β2 − 8β = β2 + 16 + 8β ⇒ (α – 4)2 = 8β – 16

⇒ (α – 4)2 = 8(β – 2)

⇒ (x – 4)2 = 8(y – 2)

which is a parabola. Answer (1)

85. It is given that a b c, and are three unit vectors such that

a b c b c× × = +( ) ( )

3

2

Therefore,

( ) ( ) ( ) a c b a b c b c⋅ − ⋅ = +

3

2

That is,

a c a b⋅ = ⋅ = −

3

2

3

2and




Here, cos cosα β= = −3

2

3

2and , where α is angle between

a cand ; β is the angle between

a band . Therefore,

β π=

5

6 Answer (1)

86. We have

p xx

x= +→ +lim

0

2 1 21( tan ) /

Therefore,

p exx

x= → + + −lim0

21 11

2( tan ).

= =→ +e ex

x

xlim

tan

( ) /0

2

22 1 2

Hence, ln / .p = 1 2

Answer (4)

87. We have

(cosx + cos3x) + (cos2x + cos4x) = 0, x ∈[ , ]0 2π ⇒ 2cos2xcos x + 2cos3x cos x = 0 ⇒ 2cosx(cos2x + cos3x) = 0 Therefore,

45

2 20cos cos cosx

x x=

cosx = 0 or cos5

20

x= or cos

x

20=

x = 90°, 270° 5

22 1

2

xn= + +( )

π

xm

22 1

2= +( )

π

x n= +( )2 16

π x m= +( )2 1 π

x =π π π π5

3

5

7

5

9

5, , , Required value = π

Therefore, the total number of solutions is 7. Answer (4)

88. We have

( )x x x x2 4 605 5 12

− + =+ −

Case 1: x2 + 4x – 60 = 0 (x + 10) (x – 6) = 0 ⇒ x = −10, 6 Case 2: x2 – 5x + 5 = 1 or x2 – 5x + 5 = −1 x2 – 5x + 4 = 0 x2 – 5x + 6 = 0 x = 1, 4 x = 2 and 3 x2 + 4x – 60 must be even number for these two values. 4 + 8 – 60 = −48 9 + 12 – 60x Therefore, let us reject x = 3. Thus, the sum of values is −10 + 6 + 1 + 4 + 2 = 3 Answer (2)




89. We have y2 − 2x ≥ 0 and x2 + y2 – 4x ≤ 0, x ≥ 0, y ≥ 0.

(x – 2)2 + y2 ≤ 4 Point of intersection of both curves y2 = 2x and (x – 2)2 + y2 = 4 is (0, 0) and (2, 2).

Q(2, 2)y

P(0, 0) (2, 0)x

The required area is

( )y y dx1 2

0

2

−∫ = − −∫ ( )4 22

0

2

x x x dx

= − ∫π( )2

42

2

0

2

x dx

= −π 8

3 Answer (3)

90. We have

f x fx

x( ) ,+ ⎛⎝⎜

⎞⎠⎟

=21

3 x ≠ 0

Replacing x by 1/2, we get

fx

f x x1

2 3⎛⎝⎜

⎞⎠⎟

+ =( ) /

⇒ fx x

f x1 3

2⎛⎝⎜

⎞⎠⎟

= − ( )

⇒ f xx

f x x( ) ( )+ − =6

4 3

⇒6

3 3x

x f x− = ( )

⇒ f xx

x( ) = −2

Now, f x f x( ) ( ).= − So

2 2

xx

xx− = − + ⇒

42

xx= ⇒

222

xx x= ⇒ = ⇒ x = − 2 2,

Therefore, S contains only two elements. Answer (4)



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