Physics Challenger by Editora VestSeller

A physical quantity Q is related to four observables x, y, z and t by the relation Q=

x2 / 5 z3 y t

The percentage errors of measurement in x, y, z and t are 2.5%, 2%, 0.5% and 1% respectively. The percentage error in Q will be (a) 5% (b) 4.5% (c) 8% (d) 7.75% If area (A), velocity (v) and density ( ) are base units, then the dimensional formula of force can be represented as (a) Av (b) Av² (c) Av ² (d) A²v

1 ÷ 2

(

The dimension of

E electric field) is (a) MLT–1 (b) ML2T–2 –1 –2 (c) ML T (d) ML2T–1 The resistance R of a wire is given by the relation . Percentage error in the measurement of , and r2 r is 1%, 2% and 3% respectively. Then the percentage error in the measurement of R is (a) 6 (b) 9 (c) 8 (d) 10 If force, acceleration and time are taken as fundamental quantities, then the dimensions of length will be (a) FT2 (b) F–1 A2 T–1 (c) FA2T (d) AT2

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(a) (19.0 ± 1.5) sq.cm.

(b) (19.0 ± 2.5) sq.cm.

(d) (19.0 ± 4.5) sq.cm.

A quantity X is given by

V where t

is the

permittivity of the free space, L is a length, V is a potential difference and t is a time interval. The dimensional formula for X is the same as that of

0 : permittivity of free space,

The length and breadth of a rectangle are (5.7 ± 0.1) cm and (3.4 ± 0.2) cm. The area of the rectangle with error limits is

10.

(a) resistance

(b) charge

(d) current

When a small sphere moves at low speed through a fluid, the viscous force F, opposing the motion is experimentally found to depend upon the radius r, the velocity v of the sphere and the viscosity of the fluid. Expression for force is (a) 4

rv2

(b) 5

r2v

(d) 6

A gas bubble from an explosion under water, oscillates with a period T proportional to pa db Ec where p is the static pressure, d is the density of water and E is the total energy of explosion. Find the values of a, b and c. (a) –5/6, 1/2, 1/3

(b)

–5/6, 1/3, 1/2

(d) –1/3, 1/3, 1/4

= 2.0 ± 0.1, z = 1.0 ± 0.1 then where z the value of y is given by

Let y

(a) + 2 ± 0.8

(b) – 4 ± 1.6

(d) None of these

10.

IIT-JEE PHYSICS Challenger

11.

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A cube has a side of length 1.2 × 10–2m. Calculate its volume. (a) 1.7 × 10–6 m3 (b) 1.73 × 10–6 m3 –6 3 (c) 1.70 × 10 m (d) 1.732 × 10–6 m3 Given that ( /p ) = az/KB where p is pressure, z is distance, KB is Boltzmann constant and is temperature, the dimensions of are (a) [L0M 0 T 0] (b) [L1M –1T 2] 2 0 0 (c) [L M T ] (d) [L–1 M 1T –2] If E is the electric field intensity and 0 is the permeability of free space, then the quantity E2/ 0 has the dimensions of (a) [M 0L1T –2] (b) [M 1L1T –4] (c) [M1L0T –4] (d) [M 2L2T 0] The speed of light in vacuum, c, depends on two fundamental constants, the permeability of free space, 0 and the permittivity of free space, 0. The speed of light is given by c = units for 0 are (a) kg–1m–1C2 (c) kg ms–4C–2

1 0

. The units of

18.

To find the value of g using simple pendulum, T = 2.00 second and = 50cm was measured. The maximum permissible error in g is : ( T = ± 0.01,

19.

20.

= ± 0.1 )

(a) 1.4%

(b) 1.1%

(d) 1.2%

A wire has a length = 6 ± 0.06 cm, radius r = 0.5 ± 0.005 cm and mass m = 0.3 ± 0.003 gm. Maximum percentage error in density is (a) 4

(b) 2

(d) 6.8

The frequency ( f ) of a wire oscillating with a length , in p loops, under a tension T is given by f =

are N–1C2m–2. The

(b) kg mC (d) kg–1s–3C–2

15.

Pressure depends on distance as, P

÷ ,

(a) – 4%

(b) – 2%

(d) –5%

16.

where , are constants, z is distance, k is Boltzmann’s constant and is temperature. The dimensions of are (a) M0L0T0 (b) M–1L–1T –1 0 2 0 (c) M L T (d) M–1L1T 2 The velocity v of surface waves on a liquid may be related to their wavelength , the surface tension of liquid and its density by the following equation : v = k where k is a dimensionless constant. The values of , and should respectively be given as (a) (c)

17.

1 1 1 , , 2 2 2

21.

22.

where

The density of a sphere is measured by measuring its mass and diameter. If, it is known that the maximum percentage errors in the measurement are 2% and 3%, then find the maximum percentage error in the measurement of density? (a) 15%

(b) 18%

(d) 11%

Which of the following set have different dimensions? (a) Pressure, Young’s modulus , Stress

(b) – 1, +1 – 1

(b) EMF, Potential difference, Electric potential

1 1 1 + , ,+ 2 2 2

(d) + 1 –1 + 1

(d) Dipole moment, Electric flux, Electric field

A physical quantity

is calulated using the formula

23.

1 2 1/ 3 xy / z , where x, y and z are experimentally 10 measured quantities. If the fractional error in the measurement of x, y and z are 2 %, 1% and 3% respectively, =

then the fractional error in (a) 0.5% (c) 6%

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µ = linear density of the wire. If the error made in determining length, tension and linear density be 1%, –2% and 4%, then find the percentage error in the calculated frequency is

–2

exp

p 2

While measuring length of an object it was observed that the zero of the vernier lies between 1.4 and 1.5 of the main scale and the fifth vernier division coincides with a main scale division. If the length of the object measured is , then the value of ( the instrument is

will be (b) 5% (d) 7%

– 1.4) in terms of the least count C of

(a) C

(b) 1.45 C

(d) 5 C

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GENERAL PHYSICS 24.

In a screw gauge, the zero of mainscale coincides with fifth division of circular scale in figure (i). The circular division of screw gauge are 50. It moves 0.5 mm on main scale in one rotation. The diameter of the ball in figure (ii) is 0

29.

(b) Spherometer and screw gauge (c) Slide callipers and spherometer (d) Slide callipers, screw gauge and spherometer Intensity observed in an interference pattern is

10 5 0

I 0 sin 2 . At

= 30° intensity I = 5 ± 0.0020 W/m2.

Find percentage error in angle if I0 = 20 W/m2. (a) (c)

3 10 2 %

(b)

3 10 2 %

(d)

3 10 2 %

Figure (i)

30. 0

30 25 20

The momentum of an electron in an orbit is h/ where h is a constant and is wavelength associated with it. The nuclear magneton of electron of charge e and mass me is given as µn =

Figure (ii)

25.

(a) 2.25 mm (b) 2.20 mm (c) 1.20 mm (d) 1.25 mm The force experienced by a particle is F0 when its speed is v = v0. If the speed varies with displacement as v = kx, then the force acting on the particle when its speed becomes 2v0 is (a) 4F0 (b) F0/2

(a)

(d)

1 mm 25

2 mm 49 A formula is given as (c)

27.

(b)

1 mm 50

(d)

1 mm 49

b k . .t 3 1+ a m.a where P = pressure; k = Boltzmann’s constant; = temperature; t = time; ‘a’ and ‘b’ are constants. Dimensional formula of ‘b’ is same as (a) Formula (b) Linear momentum (c) Angular momentum (d) Torque Backlash error may occur in which of the following instrument? (a) Slide callipers and screw gauge P=

28.

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(A current) (a) [ML2A] (b) [ML3A] 2 (c) [L A] (d) [ML2] In an experiment to determine the inertial mass of an object using Newton’s second law, following graph is obtained between net force on the object and the acceleration produced in it. The mass of the object within error limits is

a(m/s2)

F02 49 divisions on the vernier scale coincides with 50 divisions on the main scale of a vernier calliper. The least count of the instrument is, if graduation on the main scale is 2 mm

31.

eh . The dimensions of µn is 3672 me

3 2 1 F(N) 3 (a) 1.0 kg (b) 1 kg (c) (1.0 ± 0.1) kg (d) (1.0 ± 0.2) kg A physical quantity A is dependent on other four physical 1

32.

quantities p, q, r and s as given below A =

pq 2 3

. The

r s

percentage error of measurement in p, q, r and s are 1%, 3%, 0.5% and 0.33% respectively, then the maximum percentage error in A is (a) 2% (c) 4%

(b) 0% (d) 3%

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IIT-JEE PHYSICS Challenger

33.

(a) ML3T–3I–1 (c) ML3T3I–1 34.

35.

of the spring k is measured in kg/s2 and the chair’s frequency f is 0.50 s–1 for a 62 kg astronaut, what is the chair’s frequency for a 75kg astronaut ? The chair itself has a mass of 10.0kg.

The dimensional equation for electric flux is (symbols have usual meaning, I current) (b) ML–3 T3I–1 (d) ML–3T3I

Dimensionally wavelength is equivalent to (a)

E LC B

(b)

(c)

B LC E

(d)

B E LC

A student performs an experiment for determination of

g =

39.

B LC

40.

÷ . The error in length is

and in time T is T

and n is number of times the reading is taken. The measurement of g is most accurate for

36.

38.

(a) 5 mm

0.2 sec

(b) 5 mm

0.2 sec

0.1 sec.

(d) 1 mm

0.1 sec

41.

The length of the string of a simple pendulum is measured with a metre scale to be 90.0 cm. The radius of the bob plus the length of the hook is calculated to be 2.13 cm using measurements with a slide callipers. What is the effective length of the pendulum ? (This effective length is defined as the distance between the point of suspension and the centre of the bob) (a) 87.87cm (c) 91.2 cm

37.

(a) 3.13 cm

(b) 3.33 cm

(d) 3.93 cm

42.

43.

The space shuttle astronauts use a massing chair to measure their mass. The chair is attached to a spring and is free to oscillate back and forth. The frequency of the oscillation is measured and that is used to calculate the total mass m attached to the spring. If the spring constant

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(b) 0.12 s–1

(d) 0.92 s–1

Vernier callipers has 20 divisions on its vernier scale which coincide with 19 divisions on the main scale. Least count of the instrument is 0.1mm. The main scale division is (a) 1 mm

(b) 4 mm

(d) – 2 mm

A student performs an experiment to determine the Young's modulus of a wire, exactly 2 m long, by Searle's method. In a particular reading, the student measures the extension in the length of the wire to be 0.8 mm with an uncertainty of ± 0.05 mm at a load of exactly 1.0 kg. The student also measures the diameter of the wire to be 0.4 mm with an uncertainty of ± 0.01 mm. Take g = 9.8 m/s2 (exact). The Young's modulus obtained from the reading is (a) (2.0 ± 0.3) × 1011 N/m2

(b) (2.0 ± 0.2) × 1011 N/m2

(d) (2.0 ± 0.05) × 1011 N/m2

In a new system of units, the fundamental quantities mass, length and time are replaced by acceleration ‘a’, density ‘ ’ and frequency ‘f’. The dimensional formula for force in this system is (a) [ a4 f ] (c) [ –1a–4f 6]

(b) 92.1 cm (d) 90.2 cm

The length of a cylinder is measured with the help of a vernier callipers whose smallest division on the main scale is 0.5mm and nine divisions of the main scale are equal to ten divisions of the vernier scale. It is observed that 78th divisions of the main scale coincides with the sixth division of the vernier scale. Find the length of the cylinder.

(a) 0.46 s–1

(b) [ a4 f –6] (d) [ –1a–4 f –1]

The pitch of a screw gauge is 1mm and there are 100 division on its circular scale. When nothing is put in between its jaws, the zero of the circular scale lies 4 divisions below the reference line. When a steel wire is placed between the jaws, two main scale divisions are clearly visible and 67 divisions on the circular scale are observed. The diameter of the wire is (a) 2.71 mm (b) 2.67 mm (c) 2.63 mm (d) 2.65 mm When the callipers are completely closed, zero of vernier scale lies to the right of zero of main scale and coinciding vernier division is 7. If L.C. is (1/20) cm., the zero error is (a)

7 cm 20

(b)

7 cm 20

(c)

7 cm 10

(d)

7 cm 10

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GENERAL PHYSICS 44.

The formula for the period of a simple pendulum is T= 2

45.

/g .

The fractional error in the measurement of the time period T is ± x and that in the measurement of the length is ± y. The maximum fractional error in the calculated value of g is (a) ± (x + y) (b) ± (x – y) (c) ± (2x + y) (d) ± (2x – y) In vernier callipers, m divisions of main scale coincide with (m + 1) divisions of vernier scale. If each division of main scale is d units, the least count of the instrument is

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45.

44.

Parallax refers to the different views that you see from two different positions. Try this experiment. Hold the index finger of your left hand vertical, 20 cm in front of you. Hold the index finger of your right hand vertical, 40 cm in front of you. Now close your left eye and, using just your right eye, move the two fingers sideways until they line up. Now close your right eye and open the left. The closer finger has ‘jumped’ to the right of the further finger. Repeat a few times. Compared to a distant background, both fingers have jumped to the right, but the closer one jumps farther. If you measure the angles through which they jump and the distance between your eyes, you can work out how far away the fingers are. For distant objects, the distance between our viewing positions must be greater than the distance between your eyes. Fortunately for astronomers, the Earth shifts our telescopes round the sun, so we can get a separation equal to the diameter of the orbit of the Earth (16 light minutes) if we wait six months, as shown in this diagram.

Sun

Earth six months ago

d (m 1)

(b)

d ( m + 1)

(c)

d m

(d)

d ( m + 1)

The circular scale of screw gauge has 100 equal divisions. When it is given 4 complete rotations, it moves through 2mm. The L.C. of screw gauge is (a) 0.005 cm (b) 0.0005 cm (c) 0.001 cm (d) 0.0001 cm

46.

PASSAGE-1

Earth now Sun

46.

(a)

In this sketch, which is not to scale, imagine an observer looking at objects A and B, standing at the pole of the Earth with his head towards us. Now he sees object A to be to the right of B. Six months ago, he saw it to be to the left of B. Now most stars are so far away from us that we cannot observe any relative motion in this way. However, for close stars it is possible. The next sketch shows the path of light from a close object and from a very distant star.

Earth now Sun

light from a very distant star D

R Earth six months ago

From trigonometry, D = A

light from same star, six months ago

R R = tan

where we have used the small angle approximation for measured in radians. A parsec is defined as the distance to an object that ‘moves’ (w.r.t. to the distant stars) by an angle of 1 second (1/3600 of a degree) when the Earth moves by the mean radius of its orbit. In terms of this sketch, if = one second, D = 1 parsec. Now all stars except the sun are more than one parsec distant, so to measure their distance by parallax, we need to be able to resolve angles of about 1 second or better.

IIT-JEE PHYSICS Challenger

If the first experiment if we have two point objects A and B in place of your left and right fingers respectively then (a) your two eyes and A and B will lie on a circle (b) your two eyes and A and B will always lie in a plane (c) your two eye and A will form a triangle area of which is equal to the area of the triangle formed by A, B and your left eye (d) Both (b) and (c) In above if we have three objects A, B and C such that you see only A & B from your left eye and A & C from your right eye. Angle formed by AB on your left eye equals that formed by AC on your right eye. Then AB will (a) always be equal to AC (b) be equal to AC if your two eye and A form an isosceles triangle (c) be equal to AC if your two eye and A form a right angled triangle (d) none of these Parallax method is useful only if the star whose distance is to be measured should be closer to the earth and the reference star should be (a) very close (b) very far away (c) any other star (d) only pole star

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Statement-1

Statement-2

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: In Searle’s experiment after every step of loading we wait for some time (generally two to three minutes) before taking the readings. : In this time the wire gets free from kinks.

PASSAGE-2 Three of the fundamental constants of physics are the universal gravitational constant, G = 6.7 × 10–11m3kg–1s–2, the speed of light, c = 3.0 × 108 m/s, and Planck’s constant, h = 6.6 × 10–34 kg m2s–1. 4. Find a combination of these three constants that has the dimensions of time. This time is called the Planck time and represents the age of the universe before which the laws of physics as presently understood cannot be applied. (a)

hG c

(d)

hG c3

hG c5

Find the value of Planck time in seconds (b) 1.3 × 10– 43 s (a) 1.3 × 10– 33 s (c) 2.3 × 10– 13 s (d) 0.3 × 10– 23 s

(b)

Statement-1

: A screw gauge having a smaller value of pitch has greater accuracy.

Statement-2

: The least count of screw gauge is directly proportional to the number of divisions on circular scale.

A particle P is projected from a point on the surface of smooth inclined plane (see figure). Simultaneously another particle Q is released on the smooth inclined plane from the same position. P and Q collide on the inclined plane after t = 4 second. The speed of projection of P is

Two particles A and B separated by a distance 2R are moving counter clockwise along the same circular path of radius R each with uniform speed v. At time t = 0, A is given a tangential 72v 2 acceleration of magnitude a 25 R 6 R (a) the time lapse for the two bodies to collide is 5v (b) the angle covered by A is 11 /6 11v 5R (d) radial acceleration of A is 289v2/5R

60°

(a) 5 m/s (b) 10 m/s (c) 15 m/s (d) 20 m/s The figure shows the velocity and acceleration of a point like body at the initial moment of its motion. The acceleration vector of the body remains constant. The minimum radius of curvature of trajectory of the body is =150°

v=8m/s 0

a=2m/s²

(a) 2 m (b) 4 m (c) 8 m (d) 16 m A stone projected at an angle with horizontal from the roof of a tall building falls on the ground after three second. Two second after the projection it was again at the level of projection. Then the height of the building is (a) 15 m (b) 5 m (c) 25 m (d) 20 m

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1. 6.

A cannon ball has the same range R on a horizontal plane for two angles of projection. If h1 and h2 are the greatest heights in the two paths for which this is possible, then

4 h1 h2

(b)

(d) R = h1 h2

1 4

A passenger in an open car travelling at 30 m/s throws a ball out over the bonnet. Relative to the car the initial velocity of the ball is 20 m/s at 600 to the horizontal. The angle of projection of the ball with respect to the horizontal road will be (a) tan–1

(a) R = h1 h2

2 3

(b) tan–1

(d) tan–1

3 4

IIT-JEE P HYSICS Challenger

The height y and the distance x along the horizontal plane of a projectile on a certain planet (with no surrounding atmosphere) are given by y = (8t – 5t2) meter and x = 6t meter, where t is in second. The velocity with which the projectile is projected is (a) 8 m/sec (b) 6 m/sec (c) 10 m/sec (d) Not obtainable from the data A particle is projected at an angle of elevation and after t seconds it appears to have an angle of elevation as seen from point of projection. The initial velocity will be (a)

gt 2 sin (

(c)

sin ( 2gt

)

(b)

gt cos 2sin (

(d)

2 sin ( gt cos

)

13.

5 rad 36

(b)

least possible length plank?

11 rad 36

7 13 rad rad (d) 36 36 10. A cricket ball is hit at 30º with the horizontal with kinetic energy K. The kinetic energy at the highest point is (a) 0 (b) K/4 (c) K/2 (d) 3K/4 11. A body of mass m is projected at an angle of 45º with the horizontal with velocity u. If air resistance is negligible, then total change in momentum when it strikes the ground is (a) 2 mu (b) 2 mu (c) mu (d) mu/ 2 12. A ball is projected horizontally with a speed v from the top of a plane inclined at an angle 45° with the horizontal. How far from the point of projection will the ball strike the plane (a)

v2 g

(c)

2v 2 g

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(d)

2v 2 2 g

0.75 m

l (a) 4.84 m (b) 3.84 m (c) 5.62 m (d) 4.62 m 16. For a stone thrown from a lower of unknown height, the maximum range for a projection speed of 10 m/s is obtained for a projection angle of 30°. The corresponding distance between the foot of the lower and the point of landing of the stone is (a) 10 m (b) 20 m (c)

20 / 3 m

(d)

10 / 3 m

17. A particle moves with a constant speed u along the curve y = sin x. The magnitude of its acceleration at the point corresponding to x = /2 is

v2 g

(b)

, so that the rod doesn’t fall on the

37°

(c)

u2 2

(b)

(d)

(a)

u2 2 2 u2

10.

11.

12.

13.

14.

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17.

2gh from

the top of a tower of height h. It strikes the level ground through the foot of the tower at a distance x from the tower. The value of x is (a) h (b) h/2 (c) 2h (d) 2h/3 14. A particle is projected at angle 37° with the incline plane in upward direction with speed 10 m/s. The angle of incline plane is given 53°. Then the maximum height attained by the particle from the incline plane will be (a) 3 m (b) 4 m (c) 5 m (d) zero 15. From the given position, as shown in the figure, the plank starts moving towards left with initial velocity zero and acceleration 8 m/s2. The rod flies in the air and falls back on the plank. With all surfaces smooth, what should be the

It was calculated that a shell when fired from a gun with a certain velocity and at an angle of elevation 5 /36 rad should strike a given target. In actual practice, it was found that a hill just prevented the trajectory. At what angle of elevation should the gun be fired to hit the target (a)

A body is thrown horizontally with a velocity

MOTION IN ONE AND TWO DIMENSIONS 18. A skier travels with a constant speed of 6 m/s along a

x2 . Find the acceleration of the skier 20 when he is at (10, 5). Neglect the size of skier.

parabolic path y

(a)

9 2 units 10 (d) none of these

units

(b)

10 2 (c) 1 unit

19. A swimmer can swim in still water with a speed of 5 m/s. While crossing a river his average speed is 3 m/s. If he crosses the river in the shortest possible time, what is the speed of flow of water? (a) 2m/s (b) 4 m/s (c) 6 m/s (d) 8 m/s 20. A particle starts from rest and moves with an acceleration of a = {2 + |t – 2|} m/s2, the velocity of the particle at t = 4 sec is (a) 2 m/s (b) 4 m/s (c) zero (d) 12 m/s 21. A car accelerates from rest with a constant acceleration on a straight road. After gaining a velocity v1 the car moves with the same velocity for some-time. Then the car decelerated to rest with a retardation . If the total distance covered by the car is equal to S, the total time taken for its motion is (a)

v 1 2

S v v

23. A boat B is moving upstream with velocity 3 m/s with respect to ground. An observer standing on boat observes that a swimmer S is crossing the river perpendicular to the direction of motion of boat. If river flow velocity is 4 m/s and swimmer crosses the river of width 100 m in 50 sec, then

(b)

S v

(d)

S v v – v 2

B j oS

(a) velocity of swimmer w.r.t ground is 13 m/s (b) drift of swimmer along river is zero (c) drift of swimmer along river will be 50 m (d) velocity of swimmer w.r.t ground is 2 m/s 24. For an observer on trolley direction of projection of particle is shown in the figure, while for observer on ground ball rise vertically. The maximum height reached by ball from trolley is

v (w. r. t. trolley)

10 m/s

A disc having plane parallel to the horizontal is moving such that velocity of point P with respect to ground on its periphery is 2 m/s j as shown in the figure. If radius of disc is R = 1 m and angular speed of disc about vertical axis passing through disc is = 2 rad/s, the velocity of centre of disc in m/s is 2 ˆj

(a) 10 m

(b) 15 m

(d) 5 m

25. A particle is projected with a velocity u at an angle with the horizontal. After some time velocity of particle becomes perpendicular to initial velocity when the particle is still above the horizontal. The angle of projection may be (choose the most appropriate option)

j C

(a) 30° (b) 60°

(a) 2 j

(b) 2 i + 2 j

(d) none of these

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18.

19.

20.

23.

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25.

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IIT-JEE P HYSICS Challenger

26. AB is an inclined plane of inclination 30° with horizontal. Point O is 20 m above point A. A particle is projected horizontally and it collides with the plane AB, perpendicularly. Speed of the particle must be (g = 10 m/s2)

20m X

20m 25m

30°

Horizontal

A (a) 13 m/s

(b)

75m

8 3m / s

(d) 2 5m / s 4 5m / s 27. A boy is standing on a cart moving along x-axis with the speed of 10 m/s. When the cart reaches the origin he throws a stone in the horizontal x-y plane with the speed of 5 m/s with respect to himself at an angle with the x-axis. It is found that the stone hits a ball lying at rest at a point whose

(a)

(c)

co-ordinates are ( 3m,1m ). The value of

is (gravitational

effect is to be ignored) (a) 30° (b) 60° (c) 90° (d) 120° 28. A particle is moving with a constant speed of m/s on a circular track of radius 12 m. The magnitude of its average acceleration for the time interval of 6s is 2

(a)

m / s2

(b)

3 2

tan

4 3

(b)

tan

3 4

3 12 (d) tan 3 2 31. A person throws vertically n balls per second with the same velocity. He throws a ball whenever the previous one is at its highest point. The height to which the balls rise is (a) g/n 2 (b) 2gn (c) g/2n2 (d) 2gn2 32. The velocity-displacement graph of a particle moving along a straight line is shown (c)

tan

m / s2

m/s (d) 0 m/s2 6 29. A particle, of mass m, is falling under the influence of gravity through a medium whose resistance equals µ times the velocity. If the particle were released from rest, the distance fallen through in time t is (c)

(a)

(c)

m2 2

t/m

t (b) 1 m

2 t (d) m

m2 2

t/m

The most suitable acceleration-displacement graph will be

t 1 m 1

26.

27.

31.

32.

(b)

(c)

(d)

t m

30. A ball thrown down the incline strikes at a point on the incline 25m below the horizontal as shown in the figure. If the ball rises to a maximum height of 20m above the point of projection, the angle of projection (with horizontal x-axis) is

MARK YOUR RESPONSE

(a)

28.

29.

30.

33.

A point moves in x-y plane according to the law x = 3 cos 4t and y = 3 (1 – sin 4t). The distance travelled by the particle in 2 sec is (where x and y are in meters) (a) 48 m (b) 24 m (c)

34.

(d)

48 2 m

24 2 m

Two particles 1 and 2 are projected with same speed v as shown in figure. Particle 2 is on the ground and particle 1 is at a height h from the ground and at a horizontal distance s from particle 2. If a graph is plotted between v and s for the condition of collision of the two then (v on y-axis and s on x-axis)

Velocity (m/s) [west]

MOTION IN ONE AND TWO DIMENSIONS 14.0 12.0 10.0 8.0 6.0 4.0 2.0 0.0 0.0

Car A

Car B 2.0

4.0

6.0

8.0 Time (s)

10.0

12.0

14.0

Which of the following is correct? (a) Only statement 1 is true

(b) Only statement 2 is true (c) Only statement 3 is true

(d) Only statements 1 and 2 are true

s (a) It will be a parabola passing through the origin (b) It will be a straight line passing through the origin and

g having a slope of 8h (c) It will be a straight line passing through the origin and having a slope of

36.

(a) 2m behind car A

(b) 2m behind car A

(d) 4 m ahead of car A

38. A particle starts from rest at time t = 0 and moves on a straight line with acceleration as plotted in figure. The speed of the particle will be maximum at time

(d) None of these A particle of unit mass is projected with velocity u at an inclination above the horizon in a medium whose resistance is k times the velocity. Its direction will again make an angle with the horizon after a time (a)

1 2ku log 1 sin k g

(c)

1 log 1 k

ku sin g

(b)

1 2ku log 1 sin k g

(d)

1 2ku log 1 sin k 3g

The graph shown illustrates velocity versus time for two cars A and B constrained to move in a straight line. Both cars were at the same position at t = 0s. Consider the following statements. (1) Car A is travelling west and Car B is travelling east. (2) Car A overtakes Car B at t = 5 s. (3) Car A overtakes Car B at t = 10 s.

MARK YOUR RESPONSE

33.

34.

38.

39.

a (in m/s2 )

35.

8h g

37. Two drag racers accelerate from rest down a drag strip. The engine of each car produces a constant forward force of 1200 N on the car. Car A has a mass of 1.25× 103 kg, while car B has a mass of 1.20× 103 kg. When car A has gone 1.00 × 102 m, car B will be

10 0 –10

(a) 1s (c) 3s

4 (in seconds)

(b) 2s (d) 4s

39. A car moves with a speed of 60 km/hr from point A to point B and then with the speed of 40 km/hr from point B to point C. Further it moves to a point D with a speed equal to its average speed between A and C. Points A, B, C and D are collinear and equidistant. The average speed of the car between A and D is (a) 30 km/hr (c) 48 km/hr

35.

(b) 50 km/hr (d) 60 km/hr

36.

37.

IIT-JEE P HYSICS Challenger

40. Two particles start moving from rest from the same point along the same straight line. The first moves with constant velocity v and the second with constant acceleration a. During the time that elapse before the second catches the first, the greatest distance between the particles is

v2 v2 2v 2 v2 (c) (b) (d) a 2a a 4a 41. If a particle is projected with speed u from ground at an angle with horizontal ,then radius of curvature of a point where velocity vector is perpendicular to initial velocity vector is given by (a)

(a)

u 2 cos 2 g

(b)

u 2 cot 2 g sin

u2 u 2 tan 2 (d) g cos g 42. Rishabh of Raxaul skated the 10,000m race in Salt Lake City in 12min, 58.92 seconds. The oval track is made up of two straight 112.00m sections and two essentially identical semicircular curves. There are two lanes, each 5.00m wide. The 400m lap starts at A on the inner straightway, rounds the inner curve, crosses over in the next straight section in the shortest diagonal path to the outside lane (the other skater crosses over the other way), and rounds the outer curve, ending up on the adjacent lane at B (see dotted line). The measurement is made 5cm out from the inner edge of the lane, and is exactly 400m for one lap. What is the radius of the inner curve, R, in m?

(c)

2ab ba (b) c c 3ba (c) (d) None of these c 45. The greatest range of a particle, projected with a given velocity on an inclined plane, is x times the greatest vertical altitude above the inclined plane. Find the value of x. (a) 2 (b) 4 (c) 3 (d) 1/2 46. An object has velocity v1 relative to the ground. An observer moving with a constant velocity v0 relative to the ground measures the velocity of the object to v2 be (relative to the observer). The magnitudes of these velocities are related by (b) v1 v2 v0 (a) v0 v1 v 2 (c) v2 v0 v1 (d) All the above are true from 47. A particle is projected with speed u at an angle horizontal at t = 0. Its horizontal component of velocity (vx) varies with time as following graph : (a)

(a)

112m

T/2

(b)

5m A

5cm

(a) 24.23 (b) 25.47 (c) 25.31 (d) 25.44 43. A particle starts sliding down a frictionless inclined plane. If Sn is the distance travelled by it from time t = n â&#x20AC;&#x201C; 1 sec to t = n sec, the ratio Sn/Sn+1 is (a)

5cm

T/2

2n 1 2n 1

(b)

(c)

2n 1 2n

2n 2n 1 (d) 2n 1 2n 1 44. A particle is projected from a horizontal plane (x-z plane) such that its velocity vector at time t is given by v aiË&#x2020; (b ct ) Ë&#x2020;j . Its range on the horizontal plane is given

(c)

(d)

T/2

MARK YOUR RESPONSE

40.

41.

42.

45.

46.

47.

43.

44.

MOTION IN ONE AND TWO DIMENSIONS

49.

50.

An airplane flies from a town A to a town B when there is no wind and takes a total time T0 for a return trip. When there is a wind blowing in a direction from town A to town B, the plane’s time for a similar return trip, Tw, would satisfy (a) T0 < Tw (b) T0 > Tw (c) T0 = Tw (d) the result depends on the wind velocity between the towns A bird flies with a speed of 10 km/h and a car moves with uniform speed of 8 km/h. Both start from B towards A (BA = 40km) at the same instant. The bird having reached A, flies back immediately to meet the approaching car. As soon as it reaches the car, it flies back to A. The bird repeats this till both the car and the bird reach A simultaneously. The total distance flown by the bird is (a) 80 km (b) 40 km (c) 50 km (d) cannot be determined. A block B moves with a velocity u relative to the wedge A. If the velocity of the wedge is v as shown in figure, what is the value of so that the block B moves vertically as seen from ground ?

u A

(a)

cos

B v

u v

(b)

cos

v u

u 1 v (d) sin u v 51. A hunter tries to hunt a monkey with a small, very poisonous arrow, blown from a pipe with initial speed v0. The monkey is hanging on a branch of a tree at height H above the ground. The hunter is at a distance L from the bottom of the tree. The monkey sees the arrow leaving the blow pipe and immediately loses the grip on the tree, falling freely down with zero initial velocity. The minimum initial speed v0 of the arrow for hunter to succeed while monkey is in air is (c)

52.

sin

(a)

g ( H 2 L2 ) 2H

(c)

g H 2 L2 H

gH 2 (b)

2gH 2 (d)

Ship A is moving with velocity 30m/s due east and ship B with velocity 40m/s due north. Initial separation between the ships is 10km as shown in figure. After what time ships are closest to each other ?

MARK YOUR RESPONSE

40m/s

30m/s A 10km

(a) 80 sec (b) 120 sec (c) 160 sec (d) None of these 53. A particle is projected from a tower as shown in figure, then the distance from the foot of the tower where it will strike the ground will be

37° 500 m/s 3

1500m

48.

/////////////////////////////////////////////////// (a) 4000/3 m (b) 2000/m (c) 1000/3 m (d) 2500/3 m 54. If a boat can travel with a speed of v in still water, which of the following trips will take the least amount of time ? (a) travelling a distance of 2d in still water (b) travelling a distance of 2d across (perpendicular to) the current in a stream (c) travelling a distance d downstream and returning a distance d upstream (d) travelling a distance d upstream and returning a distance d downstream 55. A body A is thrown vertically upward with the initial velocity v1. Another body B is dropped from a height h. Find how the distance x between the bodies depends on the time t if the bodies begin to move simultaneously. (a) x = h – v1t (b) x = (h – v1) t

v1 h (d) x = – v1 t t 56. Two swimmers start from point A on one bank of a river to reach a point B on the other bank lying directly opposite to point A. One of them crosses the river along the straight line AB while the other swims at right angles to the stream and then walks the distance which he has been carried away by the stream to get to point B. What was the velocity (assumed uniform) of his walking if both the swimmers reached point B simultaneously ? Velocity of each swimmer in still water is 2.5 km/hr and the stream velocity is 2 km/hr. (a) 4 km/hr (b) 3 km/hr. (c) 5 km/hr (d) 8 km/hr (c) x = h –

48.

49.

50.

51.

53.

54.

55.

56.

52.

MOTION IN ONE AND TWO DIMENSIONS 67.

A man in a row boat must get from point A to point B on the opposite bank of the river (figure). The distance BC = a. The width of the river AC = b. At what minimum speed u relative to still water should the boat travel to reach the point B ? The velocity of flow of the river is v0. C B

A v0b

(a)

a 2 b2

v0b

(c)

69.

v0b

(d)

a2 2b2

a 2 b2 68.

2v0b

(b) 2

A particle has initial velocity 10m/s. It moves due to a constant force along the line of velocity which initially produces retardation of 5m/s2. Then (a) the distance travelled in first 3 seconds is 10.0m (b) the distance travelled in first 3 seconds is 7.5m (c) the distance travelled in first 3 seconds is 12.5m (d) the distance travelled in first 3 seconds is 17.5m A particle is moving in a plane with velocity given by

u u0iˆ ( a cos t ) ˆj , where iˆ and ˆj are unit vectors along x and y axis respectively. If the particle is at the origin at t = 0, find its distance from the origin at time

(a)

3 2u02

2 2 u0 2

71. When a shot is projected from a gun at any angle of elevation, the shot as seen from the point of projection will appear to descend past a vertical target with (a) uniform velocity (b) uniform acceleration (c) non-uniform velocity (d) None of these 72. The drawing shows velocity (v) versus time (t) graphs for two cyclists moving along the same straight segment of a highway from the same point. The first cyclist starts at t = 0 min and the second cyclist starts moving at t = 3.0 min. The time at which the two cyclists meet is (Both velocity-time curves intersect at t = 4 min)

(b)

2 7

(d)

A shot is fired with velocity

2 2 u0 2

3 . 2

4.0

(a) 4.0 min (b) 6.0 min (c) 8 min (d) 12 min 73. A sandbag ballast is dropped from a balloon that is ascending with a velocity of 40 ft/s. If the sandbag reaches the ground in 20s, how high was the balloon when the bag was dropped? Neglect air resistance. (a) 5200 ft (b) 6000 ft (c) 5000 ft (d) 5600 ft 74. A particle is projected with speed 10m/s at an angle 60° with the horizontal. Then the time after which its speed becomes half of initial is

1 sec 2

(b)

3 sec 2

(d)

3 sec 2

(a)

2gh from the top of a

mountain which is in the form of hemisphere of radius r. The farthest points which can be reached by the shot are at a distance (measured in a straight line) from the point of projection is

3.0

(a)

r 2 4rh )

(b)

r 2 4rh )

(c)

r 2 4rh )

(d)

r 2 2rh )

75. An arrow is shot into the air on a parabolic path to a target. Neglecting air resistance, at its highest point (a) both velocity and acceleration vectors are horizontal (b) the acceleration vector is zero but not the velocity (c) the velocity and acceleration vectors are both zero (d) the upward component of velocity is zero but not the acceleration

67.

68.

69.

70.

72.

73.

74.

75.

MARK YOUR RESPONSE

71.

MOTION IN ONE AND TWO DIMENSIONS 7.

(a) > (b) t cos = (t – T) cos (c) (t – T) cos = t cos

gh at an angle with the horizontal, so that it strikes a vertical wall distant c from the point of projection, and returns to the point of projection. If e is the coefficient of restitution, then An imperfectly elastic ball is projected with velocity

(d) (u sin ) t –

1 2 1 gt = (u sin ) (t – T) – g (t T )2 2 2

10. A stunt man is to drive an auto across the water-filled gap shown in figure. Choose the correct option(s). (In FPS g = 32.2 ft/s2)

v0 28.56

(a) Time from A to O =

(b) Time from A to O =

1 2

c gh cos c

B /// /// /// /// /// ////////////////

/////////////////////////

///////////////////////////////

e gh cos

40ft (c) Time from O to A =

c gh cos

c (h sin 2 c) A particle is projected with velocity V along a smooth horizontal plane in a medium whose resistance per unit of mass is µ times the cube of the velocity. Choose the correct options(s). (a) The distance it has described in time t is (d) The coefficient of restitution =

11.

(a) The auto’s minimum take-off velocity is 16.76 ft/sec (b) The angle of the landing ramp is 45° (c) The angle of the landing ramp is 30° (d) The auto’s minimum take-off velocity is 32.76 ft/sec A ball is projected from a building of height 20m at a speed of 30m/sec making an angle of 30° with the horizontal. Then 30m/s 30°

1 [ (1 2 V 2t ) 1] V (b) Velocity at time t is

V (1 2 V 2 t )

. 20m

1 [ (1 2 V 2t ) 1] V (d) Velocity at time t is 9.

(a) time after which ball strikes the ground is 4sec (b) ball comes to a height of 20m again after 3 sec

(1 2 V 2 t )

Two shells are fired from a cannon with speed u each, at angles of and respectively with the horizontal. The time interval between the shots is T. They collide in mid air after time t from the first shot. Which of the following condition(s) must be satisfied ?

MARK YOUR RESPONSE

is tan–1

5 3 9

(d) Value of D is 60 3m

10.

11.

IIT-JEE P HYSICS Challenger

(b) P Q

u u=0

a=gcos

A h

uII

a=gsin a=gsin

It can be observed from figure that P and Q shall collide if the initial component of velocity of P on inclined plane i.e., along incline uII = 0 that is particle is projected perpendicular to incline. Time of flight on an inclined plane of inclination is given by u 2u sin T , g cos

Now y

1 2 ayt 2

uyt

u sin

1 10 9 2 h = 45 – 30 = 15m h 10 3

(b)

5 R 6v

v2 an

(b)

Radius of curvature rmin means v is minimum and an is maximum. Thus at point P when component of velocity parallel to acceleration vector becomes zero, that is u|| = 0

u 2 42 a 2 TAC = 2 sec.

(a)

So,

2u sin g

Put t ,

11 6

u = 8sin30° P = 4m/s a=2m/s² u =4m/s

(b)

u 2 sin 2 2g

u 2 sin 2 (90 2g

10m/s

( vbc ) x

( vb ) x

(vb ) x

( vc ) x

(vb ) x

20 cos 60

(vbc ) y

tan

u 2 cos 2 2g

4 h1h2

(vb ) y u sin

)

1 u 2 sin cos 4 g

(vb ) y

20 sin 60

8 metres.

angle covered by A

30°

vt R

h1h2

u|| = 8 cos 30°

a=2m/s²

Now, angle covered by A

(vc ) y

(vb ) y

10 3 (vb ) y (vb ) x

10 3 40

3 4

1 R 4 2

1 g 32 2

30 45

1 72v 2 2 t 2 25 R

As when they collide vt

t 2u sin 90 u 10 m/s 1 10 2 (c) The acceleration vector shall change the component of velocity u|| along the acceleration vector.

MOTION IN ONE AND TWO DIMENSIONS 7.

(c)

dx dt

x = 6t, v x

y = 8t – 5t2 , dy vy 8 10t dt Initial (t = 0) v y 8 v x2

(b)

(d)

10.

11.

(d)

(b)

13.

(c)

u2 Rg

10 m/s

1 2 mu cos 2 2

u2 g tan

u (cos i sin j )

dv dt

1 3 mu 2 2 4

d dt

sec2

d dt

mu ˆ mu ˆ i j 2 2

d d i cos j dt dt

u – sin dy dx

Now, tan

mu ˆ mu ˆ i j 2 2

...(1)

cos x

– sin x

dx dt

– cos 2 sin x

dx dt

Now, at x = /2, = 0°,

2mu ˆj

100

m 10 3 3 17. (c) Say at any instant, the velocity makes an angle with the x-axis. (d) tan

3 K 4 mu cos 45 iˆ mu sin 45 ˆj

0.75 m

1m 16.

KEhighestpoint =

(d)

37°

gt cos 2u sin ( ) t .; u 2 sin ( ) g cos Ranges for complementary angles are same 5 13 Required angle = 2 36 36

12.

v 2y

1 8 (0.6)2 = 4.84 m 2

= 1 + 4 × (0.6)

dx dt

d –u dt Putting this in (1), we get

45 in the formula for range down the Use incline plane.

(a)

0, aR

u= 2gh 18.

h (–h,x)

(b)

v2 R

36 R

d 2 y / dx 2

1 R

Using equation to trajectory

3 (dy / dx) 2 ] 2

3/ 2

14.

(a)

gx 2(2gh)(cos 2 0 )

x tan(0 ) u2 2g

10sin 37 10sin 37 2 g cos 53

3m 3 5 (a) Speed of plank when the rod leaves the plank v2 = 2 × 8 × 1 = 16 v = 4 m/s By constraint relation speed of the rod. u/4 = tan 37° u = 3 m/s (vertically upwards)

Time period of its flight =

2 3 10

0.6s

19.

x d2y and 2 10 dx

9 2 10

2 10

15.

dy dx

x = 2h

(vr t )2

(a) Avg. speed = 3

vr2 5 9 20. (d) a = 2 + |t – 2| For t 2 a=2–t+2 a=4–t

(vmr t )2 t

2m / s

1 10

MOTION IN ONE AND TWO DIMENSIONS 43.

(a) sn =

47.

a (2n – 1); 2

48.

a a sn + 1 = [2(n + 1) – 1] = (2n + 1) 2 2 sn 2n 1 sn 1 2n 1 44.

b (a) vy = 0 at t = , T c

2b , c

2ab c (b) P be the point where the tangent is parallel to the inclined plane. If PN = z be perpendicular from P on the inclined plane and PM the vertical altitude of P then evidently for all points on the path, P is the point where z is the greatest and consequently PM is greatest. Range = aT =

45.

(b) vx = const. No acceleration. 2s (a) T0 v

49.

s v vw

v w2

2s v2 v v 2 v w2

1 1 (v w / v ) 2 40km

(c)

vL = 8 km/h, s = v0 × t

40 5h , 8 Total distance flown by the bird = 10 × 5 = 50 km. t

50.

(a)

v BA

usin

u z A M N

vA = v

ucos Now for the point P, velocity perpendicular to the inclined plane is zero. Now the velocity and acceleration perp. to the plane at O is u sin ( – ) and g cos and this velocity becomes zero at P. 0 = u2 sin2 ( – ) – 2g cos .z z

u 2 sin 2 ( 2 g cos

)

For max. range

u cos

51.

(a)

52.

(b)

cos

srel vrel

u sin 2 2 g cos 4

2H g y

Hence, z

v u

90°

v BA

40m/s

= =

46.

u 4 g cos

1 cos

u2 (1 sin ) 4 g cos

(1 sin )

30m/s

or PM = z sec

u2 4 g (1 sin )

4 g cos 2 Maximum range = 4 × PM (d) By definition of relative velocity v1 v0 v2

1 (max. range) 4

v0 v2 ( v1 ) 0 v0, v1 and v2 will be sides of a triangle and we know that the sum of any two sides is greater than third side of the triangle.

vBA | v BA |

40 ˆj 30iˆ ; tan

vA 50m/s

Required time = 104 m 3 50ms 1 5

BD vBA

120sec.

AB cos vBA

40 30

A bob is hanging over a pulley inside a car through a string. The second end of the string is in the hand of a person standing in the car. The car is moving with constant acceleration a directed horizontally as shown in figure. Other end of the string is pulled with constant acceleration a vertically. The tension in the string is equal to car

(a) 20N (b) 10N (c) 12N (d) 15N Three blocks A, B and C of equal mass m are placed one over the other on a smooth horizontal ground as shown in figure. Coefficient of friction between any two blocks of A, B and C is 1/2.

a m

(a)

m g2

(c)

m g2

(b) ma

m g2

(d) m (g + a)

The two blocks, m = 10 kg and M = 50 kg are free to move as shown. The coefficient of static friction between the blocks is 0.5 and there is no friction between M and the ground. A minimum horizontal force F is applied to hold m against M that is equal to

Smooth

The maximum value of mass of block D so that the blocks A, B and C move without slipping over each other is (a) 6 m (b) 5 m (c) 3 m (d) 4 m Two blocks of masses m1= 4kg and m2 = 6 kg are connected by a string of negligible mass passing over a frictionless pulley as shown in the figure below. The coefficient of friction between the block m1 and the horizontal surface is 0.4. When the system is released, the masses m1 and m2 start accelerating. What additional mass m should be placed over mass m1 so that the masses (m1 + m) slide with a uniform speed? m

(a) 100 N (b) 50 N (c) 240 N (d) 180 N What is the maximum value of the force F such that the block shown in the arrangement, does not move?

F m2

1 60ยบ m

MARK YOUR RESPONSE

2 3

3kg

(a) 12 kg (c) 10 kg

(b) 11kg (d) 2 kg

IIT-JEE P HYSICS Challenger

A ship of mass 3 × 107 kg initially at rest, is pulled by a force of 5× 104 N through a distance of 3m. Assuming that the resistance due to water is negligible, the speed of the ship is (a) 1.5 m/sec. (b) 60 m/sec. (c) 0.1 m/sec. (d) 5 m/sec. is the angle of the incline when a block of mass m just starts slipping down. The distance covered by the block if thrown up the incline with an initial speed u0 is (a)

u02 / 4g sin

(b)

4u02 / g sin

(c)

u02 / sin / 4g

(d)

4u02 sin / g

A given object takes n times as much time to slide down a 45° rough incline as it takes to slide down a perfectly smooth 45° incline. The coefficient of friction between the object and the incline is (a) (1 – 1/n2) (b) 1/(1 – n2) (c)

(d) 1/ (1 n 2 )

(1 1/ n 2 )

g/4

The acceleration of the block B in the above figure, assuming the surfaces and the pulleys P1 and P2 are all smooth.

2m A

4m B

Q P

m2 F

A horizontal force F = 30 N is applied to the plank. Then the acceleration of bar and the plank in the reference frame of elevator are (a) 3.5 m/s2, 5 m/s2 (b) 5 m/s2, 50/8 m/s2 (c) 2.5 m/s2, 25/8 m/s2 (d) 4.5 m/s2, 4.5 m/s2 13. All the surfaces are frictionless then acceleration of the block B is (a) 2g/13 (b) 3g/13 (c) 4g/13 (d) g/13

(a) kA/2 (b) kA (c) s mg (d) zero A system is shown in the figure. A man standing on the block is pulling the rope. Velocity of the point of string in contact with the hand of the man is 2 m/s downwards.

MARK YOUR RESPONSE

(a) F/4m (b) F/6m (c) F/2m (d) 3F/17m 10. A block P of mass m is placed on a horizontal frictionless plane. A second block of same mass m is placed on it and is connected to a spring of spring constant k, the two blocks are pulled by distance A. Block Q oscillates without slipping. What is the maximum value of frictional force between the two blocks.

11.

The velocity of the block will be [assume that the block does not rotate] (a) 3 m/s (b) 2 m/s (c) 1/2 m/s (d) 1 m/s 12. A plank of mass M 1 = 8 kg with a bar of mass M2 = 2 kg placed on its rough surface, lie on a smooth floor of elevator ascending with an acceleration g/4. The coefficient of friction is µ = 1/5 between m1 and m2.

11.

12.

13.

10.

LAWS OF MOTION 14. A smooth ring P of mass m can slide on a fixed horizontal rod. A string tied to the ring passes over a fixed pulley and carries a block Q of mass (m/2) as shown in the figure. At an instant, the string between the ring and the pulley makes an angle 60° with the rod. The initial acceleration of the ring is

m 60°

Q m/2 2g g 2g 2g (b) (c) (d) 6 3 3 9 15. A massive wooden plate of unknown mass M remains in equilibrium in vacuum when n bullets are fired per second on it. The mass of each bullet is m (M >> m) and it strikes the plate at the centre with speed v. If the coefficient of restitution is e, then M is equal to (a)

h/4 h

(a) 30 N (b) 4 0 N (c) 120 N (d) none of the above 18. There are two blocks A and B in contact with vertical and horizontal smooth surfaces respectively, as shown in the figure. Acceleration of A and B are aA and aB respectively along their constrained direction of motions. Relation between aA and aB is (Assume sin 23° = 2/5).

16.

(a)

mev n g

mv n g

(b)

(c)

m v (1 e) n g

(d) n m

A block is resting on a horizontal plate in the xy plane and the coefficient of friction between block and plate is . The plate begins to move with velocity u = bt2 in x direction. At what time will the block start sliding on the plate. (a)

b g

(b)

m (a)

2a A

21 aB

(b) 2aA = 5aB

(c) 5aA = 2aB (d) 21 a A 2aB 19. A bead of mass m is located on a parabolic wire with its axis vertical and vertex at the origin as shown in the figure and whose equation is x2 = 4ay. The wire frame is fixed and the bead can slide on it without friction. The bead is released from point y = 4a on the frame from rest. The tangential acceleration of the bead when it reaches the position given by y = a is

bg 2

g g (d) b 2b 17. A block of mass m = 2 kg is placed on a plank of mass M = 10 kg which is placed on a smooth horizontal plane. The coefficient

(c)

1 . If a 3 horizontal force F is applied on the plank, then find the maximum value of F for which the block and the plank move together. (Take g = 10 m/s2)

of friction between the block and the plank is

MARK YOUR RESPONSE

14. 19.

15.

(a) (c)

16.

3g 2

(b)

g 2

(d)

17.

g 2

18.

IIT-JEE P HYSICS Challenger

20. In the figure acceleration of bodies A, B and C are shown with directions. Values b and c are w.r.t. ground whereas a is acceleration of block A w.r.t. wedge C. Acceleration of block A w.r.t. ground is

(a) u = 5m/s (b) u = 10 m/s (c) u = 20 m/s (d) u = 15 m/s 23. If 1 = 2 in figure, (pulley is frictionless). Choose the correct option

(b c) 2 a 2 (b) c – (a + b) cos

(a)

(c)

(b c )

(a) T1 = T2 = T3 = W2 =

W1 2 cos

(b) T1 T2 = T3 = W2 =

W1 2 cos

2(b c).c.cos

(b c)2 c 2 2(b c).c.cos 21. The system shown in the adjacent figure is in equilibrium. All the blocks are of the equal mass m and springs are ideal. When the string between A and B is cut then (d)

W1 k

C (a) the acceleration of block A is 2g while the acceleration of block C is g (b) the acceleration of block B is 2g while the acceleration of block C is zero (c) the acceleration of block A is 2g while the acceleration of block B is g (d) the acceleration of block B is g while the acceleration of block C is zero 22. In the arrangement shown in the adjacent figure, the two pulleys are fixed and the two blocks A and B are made to move downwards so that they decelerate at 10 m/s2. The arrangement achieving the above is not shown in the diagram. The block C which is fixed to the middle of the string, moves upward with a constant velocity u. At a certain

d instant, (shown in the figure) = 30 ° and dt It can be concluded that

MARK YOUR RESPONSE

20.

21.

W1 2 cos

(d) T1 = T2 = T3 W2

W1 2 cos

24. The inclined plane shown in figure has an acceleration ‘a’ to the right. The block will side on the plane if (µs tan is the coefficient of static friction for the contacting surfaces) Y

N f 90° – a mg

1 radian/s.

X (a) a < g tan ( – ) (c) a > g tan ( – )

22.

23.

(b) a > g tan ( + ) (d) a > g cot ( – )

24.

109

LAWS OF MOTION

F cos 60° = µ [F sin 60° + mg]

(c)

mg cos 60 sin 60

T 1

ma m

2 3 = 1 1 2 2 3

a mg (Force diagram in the frame of the car) Applying Newton’s law perpendicular to string

mg sin

ma cos

a g

tan

(c)

3 10 3 2

m g2

f max

100 0.5

mD g 3m mD

g 2

(b) For uniform speed of the blocks T = 6g ...(i) T = f = µN ...(ii) µN = 6g or 0.4 (4 + m)g = 6g m = 15 – 4 = 11 kg

N m T

f F cos 60o

4 kg

g 2

Further, amax

Same normal force would accelerated M,

f max m

amax

200 Newton

200 4 m / s2 thus aM 50 Taking m + M as system F = (m + M) 4 = 240 N (a) The forces acting on the block are shown. Since the block is not moving forward for the maximum force F applied, therefore F cos 60° = f = µN ... (i) (Horizontal Direction) For maximum force F, the frictional force is the limiting friction = µN] and F sin 60° + mg = N... (ii) From (i) and (ii)

1 mg 2

m Ag

Maximum acceleration of A can be

or T m g 2 a 2 ma (c) As m would slip in vertically downward direction, then mg = µN

20 N

Blocks A and C both move due to friction. But less friction is available to A as compared to C because normal reaction between A and B is less. Maximum friction between A and B can be :

Applying Newton’s law along string

5 1 4

(4 + m) g

T 6 kg

Fsin60o

A particle A of mass 10/7 kg is moving in the positive direction of x. Its initial position is x = 0 & initial velocity is 1 m/s. The velocity at x = 10 is : (use the graph given)

Power (in watts)

A uniform rope of linear mass density and length is coiled on a smooth horizontal surface. One end is pulled up with constant velocity v. Then the average power applied by the external agent in pulling the entire rope just off the horizontal surface is v

x (in m)

(a) 4 m/s

(b) 2 m/s

(c)

(d) 100/3 m/s

3 2 m/s

1 kg M

1 2

v2 +

(b)

1 3 vg 1 3 v + vg v (d) 2 2 2 In the above question the maximum power delivered by the agent in pulling up the rope is (c)

As shown in the figure, there is no friction between the horizontal surface and the lower block (M = 3 kg) but friction coefficient between both the blocks is 0.2. Both the blocks move together with initial speed v towards the spring, compresses it and due to the force exerted by the spring, moves in the reverse direction of the initial motion. What can be the maximum value of v (in cm/s) so that during the motion, there is no slipping between the blocks. (use g = 10m/sÂ˛) K=400N/m

(a)

Âľ=0.2

(b)

v3 2

gv v3 2 2 A particle falls from a height h on a fixed horizontal plane and rebounds. If e is the coefficient of restitution, the total distance travelled by the particle before it stops rebounding is (c)

gv v3

(d)

(a)

h [1 e 2 ] 2 [1 e 2 ]

(b)

(c)

h [1 e 2 ] 2 [1 e 2 ]

3 kg

h[1 e 2 ] [1 e 2 ]

smooth

(a) 20 cm/s (c) 30 cm/s

MARK YOUR RESPONSE

(b) 10 cm/s (d) 40 cm/s

h[1 e 2 ]

(d)

[1 e 2 ]

161

WORK, POWER, ENERGY & CONSERVATION LAWS

1 2 3 4 5 6 7 8 9 10 11 12 13

(a) (a) (c) (b) (d) (c) (d) (a) (a) (c). (b) (b) (c)

14 15 16 17 18 19 20 21 22 23 24 25 26

(b) (c) (d) (b) (c) (d) (c) (c) (d)

1 2 3 4 5 6 7 8 9

10 11 12 13 14 15 16 17 18

27 28 29 30 31 32 33 34 35 36 37 38 39

40 41 42 43 44 45 46 47 48 49 50 51 52

19 20 21 22 23 24 25 26 27

(a) (a) (b) (a) (b) (a) (a) (c) (a)

28 29 30 31 32 33 34 35 36

(a) (b) (d) (d) (a) (a) (c) (c) (b)

(a) (c) (c) (a) (c) (a) (d) (b) (b) (d) (c) (a) (a)

53 54 55 56 57 58 59 60 61 62 63 64 65

(d) (d) (b) (b) (c) (c) (a) (d) (a) (b) (b) (c). (b)

(a) (a) (d) (a) (b) (d) (a) (b) (b)

37 38 39 40 41 42 43 44 45

1 2 3

(d) (d) (c)

4 5 6

(a) (d) (a)

7 8 9

(a) (b) (b)

10 11 12

(d) (d) (c)

13 14

1 2 3 4

(a, b, c) (c) (a, d) (a)

5 6 7 8

(a, b) (b, d) (a, b) (a, c)

9 10 11 12

(a, b, c) (c, d) (a, c) (c, d)

13 14 15 16

17 18 19 20

(d) (a, b, c) (c, d) (c, d)

1. 4. 7.

A-s; B-s; C-p, q, r; D-p, q, r A-q, s; B-q; C-r; D-p, r, s A-p; B-q; C-s

6.53

44.2

2. 5.

A-q; B-s; C-r; D-p A-r; B-p; C-q; D-s

0.318

0.25

3. 6.

66 67 68 69 70 71 72 73 74 75

46 47 48 49 50 51

21 22 23

(b) (d) (b) (b) (c) (b) (b) (c) (a) (b)

(a, c) (b, c, d) (a, b)

A-p, s; B-p, s; C-q, r; D-p, s A-q; B-q; C-p, r; D-q, s

28.3

7.7

IIT-JEE PHYSICS Challenger

162

(a)

Area under P-x graphs

pdx

= v

mv 3 3

2 = mv dv 1

1 (2 4) 10 2

section of the rope and

vdm d v. ( ) v 2 , a constant. dt dt The force mg is maximum when the whole rope has to be pulled up

10 (v3 1) 30 7 3 v = 4 m/s (a) Maximum chance of slipping occurs when spring is maximum compressed. At this moment, as force exerted by the spring is maximum, acceleration of the system is maximum. Hence maximum friction force is required at this moment. By W/E theorem

been pulled up, (mg)max

Pmax

kxm M +m Force on upper block is provided by the friction force. kxm .m M +m

For limiting values v

(c)

M +m k

Using values vmax imum = 20 cm / s Initial KE. = 0, Initial P.E. = 0 When the rope is just pulled off the table, final K.E. =

1 ( 2

)v 2 , final P.E. = ( ) g / 2

time taken = t = / v Average power =

1 2

g /2 /v

1 3 v + 2

vg 2

dm v dt

v 2 dm 2 dt

v 2 and

1 3 v 2

(d) The velocity of particle after falling through height h

...(i)

2 gh

After first rebounding, the velocity of ball is eu and after attaining maximum height it will come to the ground with same velocity eu. So, after second rebounding its velocity will be e2u. Similarly, after third fourth ... etc reboundings its velocities will be e2u, e4u, ... etc. Since, it first rebounds with velocity eu so if it attains height h then from v2 = u2 – 2gh 0 = e2u2 – 2gh1 or

net change in energy time

where v = average velocity of mass dm which is pulled off the ground in the time interval at dt = v/2.

( M + m )v 2 k

Therefore ,

dm dt

Hence Fmax

Now for upper block am =

(

Now power due to the force mg = mg .v While power due to force

1 1 ( M + m )v 2 = kx m2 2 2 xm =

vdm where mg is the force dt needed for supporting the weight of already hanging

(b) Instantaneous force = mg

vdm is the force needed to dt supply momentum to the portion of the rope which is

10 (v 3 1) 7 3

From graph : area =

dv m vdx dt

e2u 2 2g

e 2 2gh 2g

e 2 h [from Eq. (i)]

The same height the ball travels while approaching ground. Now, it rebounds with velocity e2u so if it attains a height h2 then 0 = e4u2 – 2gh2 or h2 = e4h

IIT-JEE PHYSICS Challenger

172

The path OAC consists of the straight lines OA and AC. Along OA, y = 0 and dy = 0, so

Mv and v2 = 2gy, ( M + m) M 2 v2 f = (M + m) g + (M + m) 2 (M + m)2 d

Substituting v =

F2 .ds = 0 OA

Along AC, x = 1m and dx = 0, so

M 2 gy M +m d 23. (b) F = p/ t, where t, is the time interval from just after (or just before, since momentum is conserved) the impact to cessation of motion of the stake plus mallet. Then, = ( M + m) g +

F2 .ds = y dy Therefore, 1

y2 F2 .ds = y dy = 2 0 CA

= 0

1 J 2

WOAC =

F2 .ds + OA

1 1 F2 .ds = 0 + J = J 2 2 AC

F2 .ds = 0 .

24.

M gy ÷ m+m d

was

F2 .ds = x dx . 1

x2 Therefore, F2 .ds = x dx = 2 0 BC

= 0

F2 .ds +

1 J 2

1 1 F2 .ds = 0 + J = J 2 2 AC

Along the straight line OC, y = x, so dy = dx and

F2 .ds = xy dx + xy dy = 2 x 2 dx . Then 1

WOC =

F2 .ds = 2 x 2 dx = 0

22.

2x 3

= 0

1 Mv 2 2

m M +m 25. (a) Conserving linear momentum : m1v1 = m2v2 as m1 = m2 = m v1 = v2 Work energy theorem

2 J 3

26.

1 2 mv ÷ 2

v = gh

v = 2 gy . Momentum is conserved at the instant of collision, so that Mv = (M + m) v', where v' is the speed of the stake plus mallet just after impact. The resultant upward force on the stake plus mallet is F = f – (M + m) g, where f is the resistive force of the ground. Then the work-energy principle gives

m 0

1 m 2

mgh

mgh 4

gh ÷ 2

=[f

f = (M + m) g + ( M + m)

v2 2d

m) g ] ( d )

27.

mgh 2

h mgh h = 4 4 (a) Momentum conservation mgh

v =

2m v

gh 2

Work energy theorem

( F) ( d)

1 0 ( M + m)v 2

1 M2 2 m 1 v = Mv 2 ÷ M +m 2 2 M +m

or the fraction lost was

(a) The speed of the mallet on just striking the stake is

1 Mv 2 , just afterward it was 2

mgh = 2

2 gy

1 1 M2 2 ( M + m) v 2 = v 2 2 M +m So the amount lost by the mallet was

The work done along the entire path OBC is given by

WOBC =

M +m d M

(a) Just before impact the kinetic energy of the system

Along BC, y = 1m and dy = 0, so

The path OBC consists of the straight line OB and BC. Along OB, x = 0 and dx = 0, so

p 0 [ ( M + m) v ] = F f ( M + m) g

The work done along the entire path OAC is given by

m1v1 + m2 v2 = 2 m

gh 2

1 m ( gh )2 2

173

WORK, POWER, ENERGY & CONSERVATION LAWS Work energy theorem

mgh 1 2 1 2 mv1 + mv2 ÷ = 4 2 2 v1 = 0 , v2 =

1 2m 2

gh ÷ 2

Sol. 37-39: We need to know the velocities of particle and pile immediately after the inelastic collision. Let the velocity of the particle just before strike be v and velocity of the particle after the strike be v1 and velocity of pile after strike be v2.

v = 2 gh From momentum conservation

block comes to rest & wedge moves with velocity

m1i v1i + m 2 i v 2i = m1 f v1 f + m 2 f v 2 f

gh . 28.

29.

30. 31.

(a) Physically possible explosions are those in which both particles move in opposite directions. i.e. signs of velocities are opposite. II, IV & V (b) If mA = mB, magnitude of velocities are same II If mA > mB vA < vB V If mA < mB vA > vB IV (d) As in VI momentum is not conserved. Force must be acting on it. (d) v1 + v2 = v (remains same straight line)

C2 )2

R 32. 33.

v02

(1 3e) v , 4 Note that (1 – 3e) < 0 as e > 1/3, i.e.the velocity of particle is opposite to the assumed positive reference direction. This further means that it rebounds back in the upward direction.

v2 h = 1 (height attained after the collision) = 2g (3e 1) 2 h . 16 Now consider the motion of pile into the material. Two forces are acting on pile : (i) Constant resisting force F (ii) Force due to gravity,

therefore yield circle of smaller radius. (a) In inelastic collision radius will be less but straight line of momentum conservation will be same. (a) Velocity of incident I object will always be less than II object, therefore from above two solutions (1.5, 0.5) is correct.

F 3mg 3m Pile penetrates the distance, a=

(0.5, 1.5) A B

(1.5, 0.5) v1 2

34. 35.

36.

(b) At maximum extension Fnet (a)

38.

(a)

39. (d)

(1 + e)2 3mgh 16 ( F 3mg )

40.

dx =

(a) 0

41.

37.

(1 + e) v 4

v1 =

C 2 ( < v0 )

........... (2)

From (1), v1 = v – 3v2

On solving, v02 = v12 + v 22 + C 2

v12 + v22 = (v02

v2 v1 v 0 ev = v2 – v1 From (1) and (2), (1 + e) v = 4v2 v2 =

C1 is loss in kinetic energy

........... (1)

1 2 1 2 1 mv1 + mv2 + C1 = mv02 2 2 2

v12 + v 22 = v02

mv = mv1 + 3mv2 v = v1 + 3v2

t 8 8 5 dt ; x = ln (1 + 5t ) 5 5 0 1 + 5t

(b) Just after collision, v = 8 m/s (t = 0) (M + 1000) 8 = 1000 × 12 (Using momentum conservation) 2M = 1000 M = 500 kg 1

Fr = 1500 × a = 1500 × 8 ×

(1 + 5t )

60000 (1 + 5t ) 2

IIT-JEE PHYSICS Challenger

174

42.

(d) Impulse = p = 1500 8

8 ÷ 11

47.

(a) a µ x with jerk when string slacks.

48.

(c)

1500 80 120000 = N-s 11 11 43. (a) From conservation of momentum Mav0 = (Ma + Ms) vf

1 k (3 2

)2 =

1 2 1 mv + k ( 2 2

k.7 (

0.50 vf = ÷ 4.0 = 0.57 m / s 0.50 + 3.0 44.

(b) Only friction is responsible to move the slab f = Msas

MBg Ms

as = v 2f

45.

(b)

50.

(b)

........ (2)

7 m/s 10

v m

m 2

By COE,

1 M s v 2f 2

(c)

1 kg

vcm = 51.

m1a1 + m2 a2 m 0 m 2 g 3 g = = m1 + m2 m1 + m2 2

0 m/s

+ 2 a s x , where v0 = 0

acm =

........ (1)

From eq. (1) and (2), x = 0.49m. (b) W = Fd = µMBgx W = 0.20 (0.5) (10) (0.49) = 0.49 Joule or W

46.

v02

49.

)2 = mv2

1 3mg 1 2 k ÷ = mv 2 2 k

9mg 2 m = 3g k k m 0 mv v 3 g = = 2 2 m+m

(a) By COE in CM-frame,

m k

1 2 1 vref = kx 2 2 2

1m m 1 3g = kx 2 2 2 2 k÷

Total ext. =3

Slack F = ma = 100 (3

2. 3. 4.

) ; a = 3 m/s2

(d) When e = 0, velocity of separation along common normal zero, but there may be relative velocity along common tangent. (d) Statement – 1 is False but statement – 2 is true. (c) Statement – 1 is True but statement – 2 is False. (a) a =

9 2 m2 g = kx 2 ; 2 k

3mg 2k

When object slips k = 0 5.

(d)

(a)

7. 8. 9.

(a) (b) (b)

g sin 1+ k2 / R2

A = g sin ; which is independent on shape. The gain in KE is due to work done by muscles & work done by friction is zero as it is static friction. Work done by action reaction force may be zero only if displacement of both bodies are same. Statement-2 clearly explains statement-1. Both statements are true and independent. Both the statements are true but statement-2 is not correct explanation of statement-1. In fact the momentum is conserved both in elastic as well as in inelastic collision. But in elastic collision the total kinetic energy of the system is also conserved.

When a bucket containing water is rotated fast in a vertical circle of radius R, the water in the bucket doesnâ&#x20AC;&#x2122;t spill provided the bucket is (a) whirled with a maximum speed of

(a)

b ,0 2

(b)

b ,b 2

(c)

b ,b 3

(d)

2 b, b 5

2gR

(b) whirled around with a minimum speed of

(1/ 2)gR

900g /( 2 R)

Here m and 2m represent the masses of the particles.

3600g /( 2 R) A disc of uniform thickness and radius 50.0 cm is made of two zones. The central zone of radius 20.0 cm is made of metal and has a mass of 4.00 kg. The outer zone is of wood and has a mass of 3.00 kg. The moment of inertia of the disc about a transverse axis through its centre is (a) 0.510 kg-m2 (b) 0.515 kg-m2 (c) 0.500 kg-m2 (d) 0.525 kg-m2 A particle is confined to rotate in a circular path decreasing linear speed, then which of the following is correct? (d) having an r.p.m of n =

(a)

A uniform rod of mass m and length L lies radially on a disc rotating with angular speed

axis. The rod does not slip on the disc and the centre of the rod is at a distance R from the centre of the disc. Then the kinetic energy of the rod is

L (angular momentum) is conserved about the centre

L R

(b) only direction of angular momentum L is conserved (c) It spirals towards the centre (d) its acceleration is towards the centre. With O as the origin of the coordinate axis, the X and Ycoordinates of the centre of mass of the system of particles shown in the figure may be given as : Y

in a horizontal plane about its

m 2m

(a)

1 m 2

(c)

1 m 24

L2 12

(b)

1 m 2

(d)

1 m 12

h X b

O b

2 2

MARK YOUR RESPONSE

IIT-JEE PHYSICS Challenger

184

A disc of mass M and radius R is rolling with angular speed on a horizontal plane as shown in Figure. The magnitude of angular momentum of the disc about the origin O is

M X

(a) (1/2) MR2 (b) MR2 2 (c) (3/2) MR (d) 2 MR2 A sphere of mass ‘m’ is given some angular velocity about a horizontal axis through its centre and gently placed on a plank of mass ‘m’. The coefficient of friction between the two is µ. The plank rests on a smooth horizontal surface. The initial acceleration of the plank is m

A cylinder rolls up an inclined plane, reaches some height, and then rolls down (without slipping throughout these motions). The directions of the frictional force acting on the cylinder are (a) up the incline while ascending and down the incline descending (b) up the incline while ascending as well as descending (c) down the incline while ascending and up the incline while descending (d) down the incline while ascending as well as descending. 10. A train of mass M is moving on a circular track of radius R with constant speed V. The length of the train is half of the perimeter of the track. The linear momentum of the train will be (a) 0 (b) 2MV/ (c) MVR (d) MV 11. A small block of mass 'm' is rigidly attached at 'P' to a ring of mass '3m' and radius 'r'. The system is released from rest at = 90° and rolls without sliding.

(a) zero (b) (7/5) µg (c) µg (d) 2 µg A uniform triangular plate ABC of moment of inertia I (about an axis passing through A and perpendicular to plane of the plate) can rotate freely in the vertical plane about point 'A' as shown in figure. The plate is released from the position shown in the figure. Line AB is horizontal. The acceleration of centre of mass just after the release of plate is

A× a

B a

(a)

mga 2

(b)

mga 2 4I

(d)

mga 2 3I

(c)

mga 2 2 3I

MARK YOUR RESPONSE

The angular acceleration of hoop just after release is (a) g/4r (b) g/8r (c) g/3r (d) g/2r 12. A running man has half the K.E. that a body of half his mass. whenthe man speeds up by 1 ms–1 then he has the same K.E. as that of the body. The original speeds of the man and the boy in ms–1 are (a) 1.41 each (b) 2.42, 4.84 (c) 4.84, 0.8 (d) 2.41, 0.41 13. A horizontal circular plate is rotating about a vertical axis passing through its centre with an angular velocity o. A man sitting at the centre having two blocks in his hands stretches out his hands so that the moment of inertia of the system doubles. If the kinetic energy of the system is K initially, its final kinetic energy will be (a) 2 K (b) K/ 2 (c) K (d) K/ 4

11.

12.

13.

10.

185

CIRCULAR AND ROTATIONAL DYNAMICS 14.

A particle of mass m is attached to a thin uniform rod of length a and mass 4 m. The distance of the particle from the centre of mass of the rod is a/4. The moment of inertia of the combination about an axis passing through O normal to the rod is

17. A small object of uniform density rolls up a curved surface with an initial velocity v. It reaches up to a maximum height of

3v 2 with respect to the initial position. The object is 4g

a/2 a/4

(a)

64 2 ma 48

(b)

v (a) ring (b) solid sphere (c) hollow sphere (d) disc 18. Two thin rods of mass m and length each are joined to form L shape as shown. The moment of inertia of rods about an axis passing through free end (O) of a rod and perpendicular to both the rods is

91 2 ma 48

27 2 51 2 ma ma (d) 48 48 15. The spool shown in figure is placed on a rough horizontal surface has inner radius r and outer radius R. (c)

O R

2 m 7

(a)

cos

r R

(b)

cos

19.

2r R

r r (d) sin 1 R R 16. The free end of a thread wound on a bobbin is passed round a nail A hammered into the wall. The thread is pulled at a constant velocity. Assuming pure rolling of bobbin, find the velocity v0 of the centre of the bobbin at the instant when the thread forms an angle with the vertical. (c)

cos

vR R sin

(c)

2vR R sin r

MARK YOUR RESPONSE

(a)

14. 19.

(b)

vR R sin

(d)

v R sin

(b)

m 2 6

5m 2 3 The figure shows a hollow cube of side 'a' of volume V. There is a small chamber of volume V/4 in the cube as shown. This chamber is completely filled by m kg of water. Water leaks through a hole H and spreads in the whole cube. Then the work done by gravity in this process assuming that the complete water finally lies at the bottom of the cube is

(c)

The angle between the applied force and the horizontal can be varied. The critical angle ( ) for which the spool does not roll and remains stationary is given by

(d)

a/2 a

Hole

r (a) (1/2) mga (c) (5/8) mga

15.

16.

(b) (3/8) mga (d) (1/8) mga

17.

18.

IIT-JEE P HYSICS Challenger

186

20. A solid sphere of mass M and radius R having moment of inertia I about its diameter is recast into a solid disc of radius r and thickness t. The moment of inertia of the disc about an axis passing the edge and perpendicular to the plane remains I. Then R and r are related as (a)

2 R 15

(b)

24. A particle of mass m is attached to a rod of length L and it rotates in a circle with a constant angular velocity . An observer P is rigidly fixed on the rod at a distance L/2 from the centre. The acceleration of m and the pseudo force on m from the frame of reference of P must be respectively.

P 2 R 15

2 R (c) r (d) r 15 21. A heavy particle of weight W, attached to a fixed point by a light inextensible string describes a circle in a vertical plane. The tension in the string has the values mW and nW respectively, when the particle is at the highest and the lowest points in the path. The value of (n – m) is (a) 5 (b) 4 (c) 6 (d) 7 22. A sphere of mass M and radius R is moving on a rough fixed surface, having coefficient of friction µ as shown in figure. It will attain a minimum linear velocity after a time

2 L (b) zero, m 2

(a) zero, zero

(c)

L ,m 2

L 2

(d) zero, m

25. A solid cylinder is wrapped with a string and placed on an inclined plane as shown in the figure. Then the frictional force acting between cylinder and plane is

= 0.4 60°

(a) v0/µg (b) 0R/µg (c) (v0 – 0R)/µg (d) 2 (v0 – 0R)/7µg 23. A block of mass m is at rest under the action of force F against a wall as shown in figure. Which of the following statement is incorrect?

(a) zero (c)

(b) 5 mg

7m g 2

(d)

mg 5

26. A disc is rolling without slipping with angular velocity . P and Q are two points equidistant from the centre C. The order of magnitude of velocity is

a F

C P Q (a) (b) (c) (d)

f = mg [where f is the friction force] F = N [where N is the normal force] F will not produce torque N will not produce torque

MARK YOUR RESPONSE

20.

21.

25.

26.

22.

(a) vQ > vC > vP

(b) vP > vC > vQ

(d) vP < vC > vQ

23.

24.

187

CIRCULAR AND ROTATIONAL DYNAMICS 27.

A solid sphere of mass M and radius R is kept on a rough surface. The velocities of air (density ) around the sphere are as shown in the figure. Assuming R to be small and

R2 kg, what is the minimum value of coefficient g

30. A circular disc of mass m and radius R is rotating on a rough surface having a coefficient of friction µ with an initial angular velocity . Assuming a uniform normal reaction on the entire contact surface, the time after which the disc comes to rest is

of friction so that the sphere starts pure rolling? (Assume force due to pressure difference is acting on centre of mass of the sphere)

(a)

R µg

(b)

(c)

1 R 2 µg

(d)

14 m/s

7 m/s M

Horizontal (a) 0.25 (b) 0.50 (c) 0.75 (d) 1.0 28. A racing car driver drives his car on a flat circular track of radius 25/3 m and a coefficient of friction 0.5. He drives the car in such a manner that he may attain the maximum possible velocity on the track in a minimum possible time. The magnitude of his tangential acceleration at an instant when his speed is 5m/s is (a) 2 m/s2 (b) 3 m/s2 2 (c) 4 m/s (d) 1 m/s2 29. From a circular disc of radius R and mass 9M, a small disc of radius R/3 is removed from the disc. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through O is

3 R 4µg

3 R 2 µg

31. Consider the two bobs are shown in the figure. The bobs are pivoted to the hinges through massless rods. If tA be the time taken by the bob A to reach the lowest position and tB be the time taken by the bob B to reach the lowest position. (Both bobs are released from rest from a horizontal position) then ratio tA / tB is m

m A

B /2

(a)

(b)

(c)

(d)

1 2

32. A uniform solid cube of mass M has edge length a. The moment of inertia of the cube about its face diagonal will be (a)

R/3

2R/3 (c)

3 Ma 2 2 5 Ma 2 12

(b)

1 Ma 2 2

(d)

7 Ma 2 12

33. A particle undergoes uniform circular motion. About which point on the plane of the circle, will the angular momentum of the particle remain conserved? (a) centre of the circle (a)

4MR2

MARK YOUR RESPONSE

40 MR 2 (b) 9

(b) on the circumference of the circle.

37 MR 2 (d) 9

(d) outside the circle.

27.

28.

32.

33.

29.

30.

31.

IIT-JEE P HYSICS Challenger

188

34. A system consists a ball of mass M2 and a uniform thin rod of mass M1 and length d. The rod is attached to a frictionless horizontal table by a pivot at point P and initially rotates at an angular speed as shown in figure. The rod strikes the ball, which is initially at rest. As a result just after collision, the rod stops and ball moves in the direction shown. If collision is elastic, the ratio M1/M2 is

A O R v B

P d

After collision

(a) 3 (b) 2 (c) 1 : 2 (d) 1 : 3 35. A hemispherical shell of mass m and radius R is hinged at point O and placed on a horizontal surface. A ball of mass m moving with a velocity u inclined at an angle = tanâ&#x20AC;&#x201C;1(1/2) strikes the shell at point A (as shown in the figure) and stops. What is the minimum speed u if the given shell is to reach the horizontal surface OP ?

(c)

2 3v R

(d)

2v l 3

then the time after which the bead starts slipping is

L (a)

(b)

(c)

(a) Zero

(d) infinitesimal

38. A hollow sphere of mass 2 kg is kept on a rough horizontal surface. A force of 10 N is applied at the centre of the sphere as shown in the figure. Find the minimum value of so that the sphere starts pure rolling. (Take g = 10m/s2)

2gR 3

F = 10 N

(d) it cannot come on the surface for any value of u. 36. A disc is fixed at its centre O and rotating with constant angular velocity . There is a rod whose one end is connected at A on the disc and the other end is connected with a ring which can freely move along the fixed vertical smooth rod. At an instant when the rod is making an angle 30Â° with the vertical the ring is found to have a velocity v in the upward direction. Find of the disc. Given that the point A is R/2 distance above point O and length of the rod AB is l

MARK YOUR RESPONSE

v 3 2R

37. A long horizontal rod has a bead which can slide along its length and initially placed at a distance L from one end A of the rod. The rod is set in angular motion about A with constant angular acceleration . If the coefficient of friction between the rod and the bead is , and gravity is neglected,

(c)

(b)

v 3 2l

Before collision

(a)

34.

35.

30Â°

36.

3 0.08

(a)

3 0.16

(b)

(c)

3 0.1

(d) Data insufficient

37.

38.

IIT-JEE P HYSICS Challenger

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The moment of inertia of a hollow cylinder of mass M and inner radius R1 and outer radius R2 about its central axis is (a)

(b)

1 M ( R12 2

R22 )

1 M ( R2 – R1 )2 2 A uniform rod moves in a vertical circle. Its ends are constrained to move on the track without friction. The angular frequency of small oscillation is given by (c)

1 M ( R22 – R12 ) 2

M ( R12 – R22 )

(d)

(a) its velocity v on arriving at B is proportional to cos . (b) its velocity v on arriving at B is proportional to tan .

(d) time to arrive at B is independent of .

d 7.

R (a)

L2 4

L2 R – 4

1/ 2

(b)

A small ball is connected to a block by a light string of length . Both are initially on the ground. There is sufficeint friction om the ground to prevent the block from slipping. The ball is projected vertically up with a velocity u, where 2g < u2 < 3g . The centre of mass of the block + ball system is C.

L2 R – 6 2

(a) C will move along a circle.

(c)

(4 R

L ) 2

(R – L )

A torque

(d)

R2 –

(b) C will move along a parabola

L 6

(d) The horizontal component of the velocity of the ball will be maximum when the string makes an angle q = sin–1 (u2/3g ) with the horizontal

L 6

on a body about a given point is found to be

equal to C L , where C is a constant vector and L is the angular momentum of the body about that point. From this, it follows that

(a)

L does not change with time.

(b)

dL is perpendicular to L at all instants of time. dt

C a B v A

(c) the magnitude of L does not change with time. (d) all the above A bead is free to slide down a smooth wire tightly stretched between the points A and B on a fixed vertical circle. If the bead starts from rest at A, the highest point on the circle.

MARK YOUR RESPONSE

3. 8.

A wheel is rolling on a horizontal plane without slipping. At a certain instant, it has velocity ‘v’ and acceleration ‘a’ of c.m. as shown in the figure. Acceleration of

(a) A is vertically upwards (b) B may be vertically downwards (c) C cannot be horizontal (d) Some point on the rim may be horizontal leftwards.

209

CIRCULAR AND ROTATIONAL DYNAMICS 9.

A ball of mass 1 kg is thrown up with an initial speed of 4 m/s. A second ball of mass 2 kg is released from rest from some height as shown in the figure.

u=0

2 kg

4 m/s 1 kg

10.

11.

(a) The centre of mass of the two balls comes down with acceleration g/3. (b) The centre of mass first moves up and then comes down (c) The acceleration of the centre of mass is g downwards (d) The centre of mass of the two balls remains stationary. A disc is given an initial angular velocity 0 and placed on rough horizontal surface as shown. The quantities which will not depend on the coefficient of friction is/are

(a) The time until rolling begins. (b) The displacement of the disc until rolling begins. (c) The velocity when rolling begins. (d) The work done by the force of friction. The angular acceleration of the toppling pole shown in figure is given by = k sin , where is the angle between the axis of the pole and the vertical, and k is a constant. The pole starts from rest at = 0. Choose the correct options

(c) The tangential acceleration of the upper end of the pole is 2k (1 â&#x20AC;&#x201C; cos ) (d) The centripetal acceleration of the upper end of the pole is k sin 12. A thin rod AB of mass M and length L is rotating with angular speed 0 about vertical axis passing through its end B on a horizontal smooth table as shown. If at some instant the hinge at end B of rod is opened then which of the following statement is/are correct about motion of rod ? A M

L B 0

(a) The angular speed of rod after opening the hinge will remain 0. (b) The angular speed of rod after opening the hinge will be less than 0. (c) In the process of opening the hinge the kinetic energy of rod will remain conserved. (d) Angular momentum of rod will remain conserved about centre of mass of rod in the process of opening the hinge. 13. A rod leans against a stationary cylindrical body as shown in figure, and its right end slides to the right on the floor with a constant speed v. Choose the correct option(s).

v x

as an

(a) the angular speed

Rv 2 (2 x 2

x2 ( x2

(b) the angular acceleration

R 2 )3 / 2

x x2 O (a) The tangential acceleration of the upper end of the pole is k sin (b) The centripetal acceleration of the upper end of the pole is 2k (1 â&#x20AC;&#x201C; cos )

MARK YOUR RESPONSE

10.

12.

Rv x x2

(d) the angular acceleration

11.

R2 )

R2 Rv 2 (2 x 2 x2 ( x2

13.

R2 )

R 2 )3 / 2

IIT-JEE P HYSICS Challenger

244

1 2 1 mv mu 2 – mg sin 2 2 Horizontal component of v = v' = v sin

u 2 – 2 g sin For v to be maximum, dv/d = 0, which gives sin = u2/3g (a, b, c, d) At A : v R A R for a rolling wheel, a = R (a) is correct. At B :

k sin d

0 2

= sin

Then

12.

(a, c, d)

13.

(c, d)

2k (1 cos ) and

2 an 2k (1 cos ) As no external torque is present, using angular momentum conservation about any point we can say that f = i.

From the geometry, x

R sin

v x

v R

dx dt

v If a then aB may be vertically downwards R (b) is correct. At C : R C a

sin

d dt

v2 R

= a

(b, c)

10.

(c, d)

11.

(a, b)

dv d ( dt dt d (b) From , dt (a) as

k sin dt

)

k sin

dt d d

sin d

14.

R cos

sin 2

v sin 2 R sin

R (d) is correct. The initial velocity of c.m. is upward. The acceleration of the c.m. is 'g' downward. The velocity of the disc when rolling begins can be obtained using the conservation of angular momentum principle about the point through which the friction force acts. So, the coefficient of friction has no bearing on final velocity. The work done by the force of friction will simply be change in kinetic energy.

d R dt sin

R (d / dt ) cos

v R

d . Therefore, dt

Also,

Rv x x2

d Rv dt x x 2 R 2

Rv 2 (2 x 2 x2 ( x2

R2 )

R 2 )3 / 2

(a, c)

B acute

a C

obtuse

a D As the object moves from A to C via B the angle between acceleration vector and velocity vector decreases from 90° and then increases back to 90°. Since the angle between velocity and acceleration is acute, the object speeds up. As the object moves from C to A via D the angle between acceleration vector and velocity vector

A certain pendulum clock keeps good time on the earth.If the same clock were placed on the moon, where objects weigh only one sixth as much as on the earth,how many seconds will the clock tick out in an actual time of 1 minute? (a) 24 sec (b) 24.5 sec (c) 25 sec (d) 25.5 sec A satellite of mass ms revolving in a circular orbit of radius rs round the earth of mass M has a total energy E. Then its angular momentum will be (a)

(2 Ems rs 2 )1/ 2

(b)

(2 Ems rs 2 )

(c)

(2 Ems rs )1/ 2

(d)

(2 Ems rs ) .

Vesc , N where N is some number greater than one and Vesc is escape speed for the earth. Neglecting the rotation of the earth and air resistance, the maximum altitude attained by the shell will be (RE is radius of the earth) A shell is fired vertically from the earth with speed

(a)

N 2 RE N

(c)

(b)

NRE N2 1

(a)

(c)

Mm r 2 1÷ r2 Ms

(b)

100

100 (d)

100

Ms r1 r2 ÷ M m

(c)

Gm BL +

A a+L

1 a

(b)

Gm A

(d)

G m BL

1 ÷ + BL a+L

A a

A geo-stationary satellite orbits around the earth in a circular orbit of radius 36,000km. Then, the time period of a spy satellite orbiting a few hundred km above the earth's surface (Rearth = 6,400km) will approximately be (a) 1/2 hr (b) 1 hr (c) 2 hr (d) 4 hr In a certain region of space, gravitational field is given by I= –(K/r). Taking the reference point to be at r = r0 with V=V0, find the potential. r r (a) K log + V0 (b) K log 0 + V0 r r0 r0 r V0 r V0 (c) K log (d) log r0 r A projectile leaves the earth’s surface with a speed equal to

(a) 9.

from the earth would be

MARK YOUR RESPONSE

M s r2 M m r1 ÷

A + BL a

(a)

2 gRe , where Re is the radius of earth. Its speed far away

(d) N2 1 N2 The percentage change in the acceleration of the earth towards the sun from a total eclipse of the sun to the point where the moon is on a side of earth directly opposite to the sun is

M s r2 M m r1

A straight rod of length L extends from x = a to x = L +a.Find the gravitational force it, exerts on a point mass m at x = 0 if the linear density of rod µ =A + Bx2

2 gRe

(b)

(c) (d) g Re 2gRe An artificial satellite is moving in a circular orbit around the earth with a speed equal to half the magnitude of the escape velocity from the earth. Find the height (h) of the satellite above the earth’s surface : (Take radius of earth as Re)

h Re2 (c) h = 2Re

(a)

100

g Re

(b) h = Re (d) h = 4Re

IIT-JEE PHYSICS Challenger

254

10. A planet is revolving around a star in an elliptic orbit. The ratio of the farthest distance to the closest distance of the planet from the star is 4. The ratio of kinetic energies of the planet at the farthest to the closest position is (a) 1 : 16 (b) 16 : 1 (c) 1 : 4 (d) 4 : 1 11. A spherical uniform planet is rotating about its axis. The velocity of a point on its equator is v. Due to the rotation of planet about its axis the acceleration due to gravity g at equator is 1/2 of g at poles. The escape velocity of a particle on the pole of planet in terms of v is (a) ve = 2v (b) ve = v (c) ve = v/2 (d) ve = 3v 12. A point P lies on the axis of a fixed ring of mass M and radius R, at a distance 2R from its centre O. A small particle starts from P and reaches O under gravitational attraction only. Its speed at O will be (a) zero

(b)

2GM R

2GM 1 2GM (1 ) ( 5 1) (d) R 5 R 13. A planet of mass m is in an elliptical orbit about the sun (m << Msun) with an orbit period T. If A be the area of orbit, then its angular momentum would be (a) 2mA/T (b) mAT (c) mA/2T (d) 2mAT 14. If the radius of earth shrinks by one percent and its mass remains the same, then acceleration due to gravity on the earth surface will (a) increase by 1% (b) decrease by 1% (c) decrease by 2% (d) increase by 2% 15. The work done required to increase the separation distance from x1 to x1+ d between two masses m1 and m2 is

(a)

R 6

(b)

R 3

2R (d) R 3 18. A small satellite of mass m is revolving around earth in a circular orbit of radius r0 with speed v0. At certain point of its orbit, the direction of motion of satellite is suddenly changed by angle = cos–1 (3/5) by turning its velocity vector, such that speed remains constant. The satellite consequently goes to elliptical orbit around earth. The ratio of speed at perigee to speed at apogee is (c)

(a) 3

(b) 9

(d) 1/9

19. With what minimum speed should m be projected from point C in presence of two fixed masses M each at A and B as shown in the figure such that mass m should escape the gravitational attraction of A and B

(c)

(a)

(c)

G m1m2 d x1 ( x1 + d )

(b)

G m1m 2 x1 ( x1 + d )

30°

R M

G m1 m 2 x1 ( x1 + d ) d

(a)

(d) none

(c)

16. Two satellites of same mass are orbiting round the earth at the heights of R and 4R above the earth's surface; R being the radius of the earth. The kinetic energies are in the ratio (a) 4 : 1 (b) 3 : 2 (c) 4 : 3 (d) 5 : 2 17. A body is projected vertically upward from the surface of the earth with a velocity equal to half the escape velocity. If R is the radius of the earth, the maximum height attained by the body is

MARK YOUR RESPONSE

vmin

2GM R

(b)

GM R

(d)

B 2 2GM R

2 2

20. The radius of a planet is n times the radius of earth (R). A satellite revolves around it in a circle of radius 4nR with angular velocity . The acceleration due to gravity on planet’s surface is (a) R

(b) 16 R 2

(d) 64 nR 2.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

GM R

255

GRAVITATION 21.

Two concentric uniform shells of mass M1 and M2 are as shown in the figure. A particle of mass m is located just within the shell M2 on its inner surface. Gravitational force on ‘m’ due to M1 and M2 will be

where x is horizontal, y is vertically upwards; both being measured in meter. The x–component of the velocity of the plane is constant throughout, and has the value of 360 km/h. The effective g (“g – force”) experienced by an astronaut on the plane equals

M2 a

(a) zero

22.

(b)

G ( M1 + M 2 ) m

G M 1m b2

(c)

2GM , 3R

(d) 5g

R from the centre of the planet. Assume that the planet 2 has uniform density. The kinetic energy required to be given to a small body of mass m, projected radially outward from P, so that it gains a maximum altitude equal to the thrice the radius of the planet from its surface, is equal to

The minimum and maximum distances of a satellite from centre of earth are 2R and 4R respectively, where R is the radius of earth. The minimum and maximum speeds of the satellite will be

GM , R

g 5

25. A planet of radius R has an acceleration due to gravity of gs on its surface. A deep smooth tunnel is dug on this planet, radially inward, to reach a point P located at a distance of

(d) None

(a)

(b) 3g

(c)

b m

(c)

(a) 4g

2GM R

(b)

GM , 6R

2GM 3R

A OP

23.

4GM 3R

(d) None

Note that M represents the mass of the earth. An artificial satellite is first taken to a height equal to half the radius of earth. Assume that it is at rest on the earth’s surface initially and that it is at rest at this height. Let E1 be the energy required. It is then given the appropriate orbital speed such that it goes in a circular orbit at that height. Let

E1 E2 be the energy required. The ratio is E2

24.

(a) 4 : 1 (b) 3 : 1 (c) 1 : 1 (d) 1 : 2 In order to simulate different values of g, aspiring astronauts are put on a plane which dives in a parabola given by the equation : x2 = 500 y

(a)

63 mg s R 16

(b)

3 mg s R 8

(c)

9 mg s R 8

(d)

21 mg s R 16

26. Six stars of equal mass are moving about the centre of mass of the system such that they are always on the vertices of a regular hexagon of side length a. Their common time period will be

4 3a3

(a)

a3 Gm

(b)

(c)

3a3 Gm

(d) None of these

(

Gm 5 3+4

MARK YOUR RESPONSE

21. 26.

22.

23.

24.

25.

)

IIT-JEE PHYSICS Challenger

256

27. An earth satellite of mass m orbits along a circular orbit C1 at a height 2R from earth’s surface. It is to be transferred to a circular orbit C2, of bigger radius, at a height 5R from earth’s surface. The transfer is affected by following an elliptical path as shown in figure. Calculate the change in the energies required at the transfer points A and B. [R = radius of earth]

31. A spaceship is sent to investigate a planet of mass M and radius R. While hanging motionless in space at a distance 5R from the centre of the planet, the spaceship fires an instrument package with speed v0 as shown in the figure. The package has mass m, which is much smaller than the mass of the spaceship. For what angle will the package just graze the surface of the planet ? v0 m

v'1

v'2 B

A 5R

mgR 36

(b)

mgR 18

mgR mgR (d) 2 9 28. If the radius of the earth were to shrink by one percent, its mass remaining the same, the acceleration due to gravity on the earth’s surface would (a) decrease (b) remain unchanged (c) increase (d) be zero 29. The acceleration due to gravity on the surface of the moon is 1/6 that on the surface of earth and the diameter of the moon is one-fourth that of earth. The ratio of escape velocities on earth and moon will be (c)

6 3 (b) (c) 3 (d) 24 2 2 30. An artificial satellite of mass m of a planet of mass M, revolves in a circular orbit whose radius is n times the radius R of the planet. In the process of motion, the satellite experiences a slight resistance due to cosmic dust. Assuming resistance force on satellite depends on velocity as F = av2 where ‘a’ is constant. Calculate how long the satellite will stay in orbit before it falls onto the planet’s surface. (a)

(a)

(c)

m R ( n 1) a GM 2m R ( n 1) a GM

MARK YOUR RESPONSE

(a)

sin

1 8GM 1 + 2 ÷ (b) 2 5v0 R

sin

1 8GM 1+ 2 ÷ 5 5v0 R

(c)

sin

1 2GM 1+ 2 ÷ 5 5v0 R

sin

1 8GM 1+ 2 ÷ 5 3v0 R

C1 C2

(a)

(b)

(d)

m R ( n + 1) a GM

m R ( n 1) a 2GM

(d)

32. In older times, people used to think that the earth was flat. Imagine that the earth is indeed not a sphere of radius R, but an infinite plate of thickness H. What value of H is needed to allow the same gravitational acceleration to be experienced as on the surface of the actual earth ? (Assume that the earth’s density is uniform and equal in the two models (a) 2R/3 (b) 4R/3 (c) 8R/3 (d) R/3 33 A planet revolves about the sun is elliptical orbit. The areal velocity

dA 16 2 ÷ of the planet is 4.0 × 10 m /s. The least dt

distance between planet and the sun is 2 × 1012 m. Then the maximum speed of the planet in km/s is (a) 10 (b) 20 (c) 30 (d) 40 34. The mass M of a planet-earth is uniformly distributed over a spherical volume of radius R. Calculate the energy needed to disassemble the planet against the gravitational pull amongst its constituent particles. Given : MR = 2.5 × 1031 kg-m and g = 10 m/s2 (a) 3.0 × 1032 J (b) 2.5 × 1032 J 28 (c) 0.5 × 10 J (d) 1.5 × 1032 J 35. If gravitational forces alone prevent a spherical, rotating neutron star from disintegrating , the minimum mean density of a star that has a rotation period of one second will be (a) 2.4 × 1010 kg/m3 (b) 1.4 × 1010 kg/m3 11 3 (c) 1.4 × 10 kg/m (d) 3.4 × 1011 kg/m3

27.

28.

29.

30.

32.

33.

34.

35.

31.

257

GRAVITATION 36.

Inside a uniform sphere of density there is a spherical cavity whose centre is at a distance from the centre of the sphere. Find the strength F of the gravitational field inside the cavity at the point P.

(a)

4 G 3

(b)

1 G 3

40. The orbital speed of Jupiter is (a) greater than the orbital speed of earth (b) less than the orbital speed of earth (c) equal to the orbital speed of earth (d) zero 41. The eccentricity of the earth’s orbit is 0.0167. The ratio of its maximum speed in its orbit to its minimum speed is (a) 1.67 (b) 1.034 (c) 1 (d) 0.167 42. Two bodies of masses M1 and M2 are placed at a distance d apart. What is the potential at the position where the gravitational field due to them is zero ?

1 2 (d) G G 2 3 37. If g is the acceleration due to gravity on the earth’s surface, the change in the potential energy of an object of mass m raised from the surface of the earth to a height equal to the radius R of the earth is (a) mgR/2 (b) 2mgR (c) mgR (d) – mgR 38. If the distance between the earth and the sun were half its present value, the number of days in a year would have been (a) 64.5 (b) 129 (c) 182.5 (d) 730 39. A spherical cavity is made in a lead sphere of radius R such that its surface touches the outside surface of the lead sphere and passes through its centre . The mass of the lead sphere before hollowing was M. What is the force of attraction that this sphere would exert on a particle of mass m, which lies at a distance d from the centre of the lead sphere on the straight line joining the centres of the sphere and the centre of cavity as shown in the figure. (c)

(c)

(b)

2R2 GMm

(d)

MARK YOUR RESPONSE

(

)

P3 S

P2 P1

(a) P 1 (b) P 2 (c) P 3 (d) P 4 44. A missile, which missed its target when into orbit around the earth at a mean radius 3 times as great as the parking orbit of the satellite. The period of the missile is

2 day

(b) 3 day

(c) (d) 3 3 day 3 day 45. A satellite of mass m is orbiting the earth in a circular orbit of radius R. It starts losing energy due to small air resistance at the rate of C J/s. Find the time taken for the satellite to reach the earth.

3GMm

) )

(

(a)

( (

G M1 + M 2 + 2 M1 M 2 d G M1 M 2 2 M1 M 2 (b) d G 2M1 + M 2 + 2 M1 M 2 (c) d G M1 + M 2 + 2 M1 M 2 (d) 2d 43. The figure shows a planet in elliptical orbit around the sun S. Where is the kinetic energy of the planet maximum ? (a)

2GMm

(a)

GMm 1 C R

1 r

(b)

GMm 1 1 + 2C R r

(c)

GMm 1 2C R

1 r

(d)

2GMm 1 1 + C R r

R2 GMm 2R 2

36.

37.

38.

39.

40.

41.

42.

43.

44.

45.

267

GRAVITATION

1. (b) On earth, T

4. (c) During total eclipse : Total attraction due to sun and moon,

l g

F1 =

On moon, T

l g¢

g 6

or g

GM m M e r22

GM s M e

GM m M e

r12

r22

Change in force,

6l g

6 min

( T

F1 F2 =

sec = 24.5 sec 6 2. (a) The velocity vs of the satellite is given by G Mms

K.E. =

ms vs rs

Average force on earth, Fav =

1 1 GM ms vs 2 = ms P.E. 2 2 rs ÷ ;

Total energy E = K.E. + P.E.

GMms 2rs

L = ms vs rs = ms

100

r1 M m r2 ÷ M s

…(ii) dF =

5. (b)

x2 x

a + L

F = Gm

GMm H

Altitude = H

RE N2 1

dx x 1 ÷ + BL a+L

( A + Bx ) 2

3. (c) By conservation of energy

100

rs = (GMms 2 rs )1/ 2 …(iii)

L = (2 E ms rs 2 )1/ 2

100

Gm ( dx )

From eqs. (ii) and (iii), we get

N 2 RE

r22

r12 GM s

1/ 2

GMm 1 m GM + = 2 N 2 2 RE RE

2GM m

a = a av GMms rs

The angular momentum L is given by

GM rs ÷

Fav GM s = 2 % age change Me r1

in acceleration

GM rs ÷ …(i)

vs =

r22

F 2GM m a= M = r22 e

out

2GM m M e

Change in acceleration of earth

1min )

So, in actual time of 1 minute on earth, on moon its ticks

r12

When moon goes on the opposite side of earth. Effective force of attraction,

mg 6

but mg

GM s M e

Gm A

1 a

(c) A satellite revolving near the earth's surface has a time period of 84.6 min. We know that as the height increases, the time period increases. Thus the time period of the spy satellite should be slightly greater than 84.6 minutes. Ts = 2 hr

IIT-JEE PHYSICS Challenger

268

7. (a) We know that intensity is negative gradient of potential, i.e., I = – (dV/dr) and as here I = – (K/r), so

dV dr

K , r

or V V0 so V

i.e.,

K log

r r0

K log

E2 r12 1 = = E1 r22 16 11. (a) v = R g

8. (c) If a body is projected with ( 2 gRe ) greater than escape

2gRe then by conservation of energy

velocity

GMm Re

1 2 mv 2 i.e., (v )2

v2 v2

i.e., v

1 m (v ) 2 2 2GM Re ve2

Since, L is constant, thererfore 1 1 K .E. µ 2 K.E. µ I r

4gRe

g0 2

R [g = at equator, g0 = at poles] 2

R ;

g0 g0 R 2 ; v = 2 2

ve = 2g 0 R = 4v 2 = 2v 12. (d) Gravitational potential at P, V p =

ve2

potential at O, VO =

2gRe = 2gRe

M,R

2g Re

...(i)

Gravitational

GM R

9. (b) The escape velocity from earth is given by ve =

m P

The orbital velocity of a satellite revolving around earth is given by v0 =

GM e ( Re

By work energy theorem,

where, Me = mass of earth, Re = radius of earth, h = height of satellite from surface of earth. By the relation GMe = g Re2 So,

g Re2

( Re h ) Dividing equation (i) by (ii), we get

ve v0

2( Re

...(ii)

( Re ) v Given, v0 = e 2 2( Re h ) 2ve ve Re Squaring on both side, we get

2( Re h ) 4= or Re + h = 2Re i.e., h = Re Re 10. (a) Angular momentum remains conserved during the revolution of planet. Because gravitational force is a central force. Now

1 2 m2 v 2 r 2 L2 = K.E. = mv = 2 2I 2mr 2

GM R

1 2 mv 2

VO ] =

m[VP =

1 2 mv or 2

5R 13. (a) Using Kepler’s 2nd law

dA J = dt 2 m

2GM (1 R

[J = angular momentum]

A =J T

Integrating, 2m

14. (d) Acceleration due to gravity g =

R2 G = gravitational constant, M = mass and R = radius of earth

New radius R' = R –

g' =

GM 99 R 100

g' = g

R 100

100 99

99 R 100

GM 2

100 99

1 5

)

269

GRAVITATION % change in g =

100 99

(100)2

100 =

Using conservation of energy

g' g × 100 g (99)

(99)

1 2 – GM e m – GM e m 1 2 mv p + = = mv0 2 2 rA r0

100

9v02 r02 50r A2

199

100 2% (99) 2 15. (a) Work done, W = Uf – Ui =

G m1 m 2 rf

= Gm1m2

16. (a)

v1 = v2

r2 r1

1 rf

= Gm1m2

1 x1

1 x1 + d

17. (b) v

ve 2

20.

18.

(b)

1 2 GMm 2=0 mvmin – 2 R 2

(d) mr

GMm r2

3 1 = 4R R + h

R 3

(GM = gR2)

GM 2R

GM GMm = +0 ÷ R+h 2R

2GM 2 R

r 3. 2

GMm 1 + m R 2

(b)

GM = r3

If h is the height reached by the body, then by conservation of mechanical energy

19.

vmin =

1 2 K1 2 mv1 v2 = = 1 =4 K2 1 2 v2 mv 2 2 2 2GM R 2

r0 =x rA

vP rA = =9 v A rp

4R =2 R

Let

9x2 – 50x + 25 = 0 x = 5 or (5/9)

G m1m 2 ri

1 ri

v02 r0 v2 =+ 0 2 rA

g' =

(4nR)3 . 2

n 2 .R 2 g' = 64 nR 2. 21. (b) Since gravitational field intensity inside hollow sphere M2 will be zero. So, gravitational force will be due to M1 only. 22. (b) By conservation of angular momentum mv1 (2R) = mv2 (4R) v1 = 2v2 ... (1) By conservation of energy, 1 2 GMm 1 2 mv1 mv2 2 2R 2 From (1) and (2), we get

–GM e m TE = 2r0

vmin =

GM ; vmax = 6R

GMm 4R

2GM 3R

rA 23.

O rp

P Using conservation of angular momentum about O. mvp rp = mvA rA = mv0 r0 cos

3v0 r0 vA rA = vp rp = 5

E2 =

GM e m R Re + e 2

GM e m GM e m = Re ÷ 3R e

GM e GM e m 1 2 1 mv0 = m. = R 2 2 3R e R+ e 2

E1 = 1:1 E2

... (2)

A particle of mass m is executing oscillations about the origin on the X-axis with amplitude A. Its potential energy is given as U(x) = x4 where is a positive constant. The x coordinate of the particle where the potential energy is one third of the kinetic energy is (a)

(b)

A/2

(c)

( m1

(b)

m2 )

k m1

A2 +

B2 4

(b) the motion of the particle is not SHM, but oscillatory with a time period of T = / (c) the motion of the particle is oscillatory with a time period of T = /2 (d) the motion of the particle is aperiodic. The displacements y of a particle executing a certain periodic

k m2

(d)

( m1

2 motion is given by y = 4 cos

m2 )

Two bodies M and N of equal masses are suspended from

1 t รท sin (1000t ) . This 2

(a)

k1 k2

(b)

k1 / k2

expression may be considered to be the superposition of n independent harmonic motions. Then, n is equal to (a) 2 (b) 3 (c) 4 (d) 5 A particle is describing simple harmonic motion. If its velocities are v1 and v2 when the displacements from the mean position are y1 and y2 respectively, then its time period is

(c)

k2 k1

(d)

k2 / k1

(a)

y12 + y22

(c)

v12 + v22

two separate massless springs of spring constants k1 and

k2 respectively. If the two bodies oscillate vertically such that their maximum velocities are equal, the ratio of the amplitude of vibration of M to that of N is

A sin( 2 t ) B sin 2 t Then,

(a) the motion of the particle is SHM with an amplitude of

A/ 2

A/ 3 (c) (d) A/3 Two masses m 1 and m 2 are suspended together by a massless spring of spring constant k. When the masses are in equilibrium, m1 is removed without disturbing the system. Find the angular frequency of m2 (a)

(a) 0.283 sec (b) 0.0283 sec (c) 2.83 sec (d) 28.3 sec The displacement of a particle is given at time t, by:

A block of 4 kg produces an extension of 0.16 metre in a spring. The block is replaced by a body of mass 0.50 kg. If the spring is stretched and then released the time period of motion will be

MARK YOUR RESPONSE

v12 + v22

y12 + y22

(b)

(d)

v22

v12

y12

y22

y12

y22

v22

v12

IIT-JEE PHYSICS Challenger

326

A body of mass 0.01 kg executes simple harmonic motion (SHM) about x = 0 under the influence of a force shown below

x (t )

F(N) t

80 0.2

x(m)

– 0.2 – 80

The period of the SHM is (a) 1.05 s (b) 0.52 s (c) 0.25 (d) 0.03 s 9. A particle free to move along the x-axis has potential energy given by U(x) = k [1–exp(–x2)] for x , where k is a positive constant of appropriate dimensions. Then (a) at points away from the origin, the particle is in unstable equilibrium (b) for any finite nonzero value of x, there is a force directed away from the origin (c) if its total mechanical energy is k/2, it has its minimum kinetic energy at the origin. (d) for small displacements from x = 0, the motion is simple harmonic 10. A particle describes SHM in a straight line about O.

If T is the time period of the motion, then its kinetic energy at P be half its peak at O, if the time taken by the particle to travel from O to P is

(c)

1 T 2

1 2 2

11.

t 3 (a) + 180°, + 90°, – 90° (b) – 180°, – 90°, + 90° (c) + 360°, + 180°, – 180° (d) + 180°, + 180°, – 90° 12. The period of oscillation of a simple pendulum of length L suspended from the roof of a vehicle which moves without friction down an inclined plane of inclination , is given by (a)

L g cos

(b)

L g sin

(c)

L g

(d)

L g tan

13. The velocity of the bob on a pendulum of length 10 m (figure)

2 t where v0 = 1.00 m/s and T = T

(b)

1 T 4

is given by v = v0 cos

(d)

1 T 8

2 s.The radial acceleration at t =

The figure shows three situations of displacements x(t) of a pair of two SHM (A and B) that are identical except for the phase difference. For each pair, find out the phase shift (in degrees) needed to shift the curve for A to coincide with the curve for B.

x (t )

(a)

s is

(a) 1 m / s2 (b) 1/10 m / s2 (c) 1/20 m / s2 (d) 1/30 m / s2 14. A particle performs S.H.M. on x-axis with amplitude A and time period T. The time taken by the particle to travel a distance A/5 starting from rest is :

(a)

T 20

(c)

T 2

cos–1

FG 1IJ H 5K

(b)

T 2

cos–1

(d)

T 2

sin–1

MARK YOUR RESPONSE

13.

14.

10.

11.

12.

FG 4 IJ H 5K FG 1IJ H 5K

327

SIMPLE HARMONIC MOTION 15.

A block of mass ‘m’ is suspended from a spring and executes vertical SHM of time period T as shown in figure. The amplitude of the SHM is A

19. Four massless springs whose force constants are 2k, 2k, k and 2k respectively are attached to a mass M kept on a frictionless plane (as shown in figure). If the mass M is displaced in the horizontal direction, then the frequency of the system.

k M 2k

(a)

Spring is never in compressed state during the oscillation. The minimum force exerted by spring is never in compressed state during the oscillation. The minimum force exerted by spring on the block is (a)

4 2 mA T2

(b) mg +

(d) m g +

(a)

2 (M / 2k)1/ 2

(b) 2

1 K Adg ÷ 2 M

(c)

1 2

k m1

mm2 (a) system performs SHM with angular frequency given

1/ 2

(b)

K + Adg ÷ M

MARK YOUR RESPONSE

4k M

1 k 1 7k (d) 2 7M 2 M 20. m1 and m2 are connected with a light inextensible string with m1 lying on smooth table and m2 hanging as shown in figure. m1 is also connected to a light spring which is initially unstretched and the system is released from rest

(a)

1 2

2 Mg 2M (d) 2 ÷ ÷ k k A particle executes simple harmonic motion between x = A and x = +A. The time taken for it to go from 0 to A/2 is T1 and to go from A/2 to A is T2. Then (a) T1 < T2 (b) T1 > T2 (c) T1 = T2 (d) T1 = 2T2 A uniform cylinder of length L and mass M having crosssectional area A is suspended with its vertical length, from a fixed point by a massless spring, such that it is halfsubmerged in a liquid of density d at equilibrium position. When the cylinder is given a small downward push and released, it starts oscillating vertically with a small amplitude. If the force constant of the spring is K, the frequency of oscillation of the cylinder is :

(c)

18.

(b)

(c)

Mg sin ÷ 2k

1/ 2

17.

k 4M

mA T T2 On a smooth inclined surface a body of mass M is attached between two spring. The other ends of the springs are fixed to firm supports. If each spring has force constant k, the period of oscillation of the body is

(c)

16.

4 2 mA T2

1 2

1/ 2

1 2

1 (d) 2

K + dgL ÷ M K

k( m1 m2 ) m1m2

(b) system performs SHM with angular frequency given

k by

21. A simple pendulum with length L and mass M of the bob is vibrating with amplitude a. Then the maximum tension in the string is

1/ 2

Adg Adg ÷

(b) Mg 1 +

(a) Mg 1/ 2

15.

16.

20.

21.

m2 g k

(c)

17.

Mg 1 +

a L

(d) Mg 1 +

18.

a ÷ L a 2L

19.

IIT-JEE PHYSICS Challenger

328

22. A man of mass 60 kg is standing on a plateform which is oscillating up and down with frequency 2 oscillations/sec and amplitude 50 cm. A machine on the platform indicates weight of the man with respect to time, then the maximum reading of the machine will be (g = 10 m/s2) (a) 10 kg (b) 532.8 kg 3 (c) 10 kg (d) 104 kg 23. A uniform rod of mass m and length is hinged at its mid point in such a way that it can rotate in the vertical plane about a horizontal axis passing through the hinge. One of its ends is attached to a spring of spring constant k which is unstretched when the rod is horizontal. If this end is now given a small displacement and released angular frequency of the resulting motion is

25. A tube of given shape has total length 2 and area of cross section is S. Its bottom and upper halves are filled with two non-viscous, non-compressible liquids of densities 3 and respectively. If its motion is simple harmonic then find its angular frequency.

(a) It is not an S.H.M.

(a)

k/m

(b)

2k / m

(c)

3k / m

(d)

O A m

(c)

(d)

(a) I, III (b) II, IV (c) II, III (d) I, IV 27. Three charges q, q and – 2q are fixed on the vertices of an equilateral triangular plate of edge length a. This plate is in equilibrium between two very large plates having surface charge density 1 and 2 respectively. Find time period of small angular oscillation about an axis passing through its centroid and perpendicular to plane. Moment of inertia of the system about this axis is I.

rough

A sin t (a)

MARK YOUR RESPONSE

A sin t

( M + m)g +

3 m 2

22. 27.

IV x

( M + m)g 1 m 2

PE III t

60º

(b)

smooth

(d)

26. For a particle executing SHM the displacement x is given by x = A cos t. Identify the graph which represents the variation of potential energy (PE) as a function of time t and displacement x

24. A block of mass m, attached to a fixed position O on a smooth inclined wedge of mass M, oscillates with amplitude A and linear frequency f. The wedge is located on a rough horizontal surface. If the angle of the wedge is 60º, then the force of friction acting on the wedge is given by (coefficient of static friction = µ).

(a)

(c)

3g 2

(b)

(c)

A sin t

23.

24.

qa |

–

(b)

2 0I 3qa |

25.

–

(d)

2qa |

–

2 0I qa | 1 –

26.

329

SIMPLE HARMONIC MOTION 28.

A block of mass m is suspended by means of an ideal spring of force constant k from ceiling of a car which is moving along a circular path of radius r with acceleration a. The time period of oscillation of the block when it is displaced along the spring, will be

31. A particle of mass 1 kg is placed in a potential field. Its potential energy is given by U = 10x2 + 5. The frequency of oscillations of the particle is given by (a)

( 10 ) 10

(c)

mg + ma k

m k

(d)

32. ABC is an equilateral triangle structure made up of a light rigid material. Find the frequency of small vertical oscillations of mass m along AG. Consider k1 = k2 = k3 = k4 = k.

(a)

( 5)

(b)

k g +a 2

k1 (c) 29.

(d)

k 2 + g2 + a2

30° k2

A block of mass M is kept in gravity free space and touches the two springs as shown in the figure. Initially springs are in its natural length. Now, the block is shifted (l0/2) from the given position in such a way it compresses a spring and released. The time-period of oscillation of mass will be

k4 G

B k

(a)

2 (c) 30.

3 2

M k

(b)

M k

M 2k

(d)

(c)

7 6

MARK YOUR RESPONSE

28. 33.

(b)

4 3

(d)

2 3

(a)

5k 2m

(b)

4k m

(c)

3k 5m

(d)

10k 9m

M 5k

The particles execute simple harmonic motion of the same amplitude and frequency along close parallel lines. They pass each other moving in opposite directions. Each time their displacement is half of their amplitude. Their phase difference is (a)

29.

33. A point particle of mass 0.1 kg is executing SHM of amplitude of 0.1m. When the particle passes through the mean position, its kinetic energy is 18 × 10–3 J. The equation of motion of this particle when the initial phase of oscillation is 45° can be given by

30.

(a) 0.1 cos 6t + ÷ 4

(b) 0.1 sin 6t + ÷ 4

(d) 0.2 sin

31.

32.

+ 2t ÷

Two sound sources emitting sound each of wavelength are fixed at points A and B. A listener moves with velocity u from A to B. The number of beats heard by him per second is (a) 2u/ (b) u/ (c) u/3 (d) 2 /u If the fundamental frequency of a vibrating organ pipe is 200 Hz, then (a) the first overtone is 200 Hz (b) the first overtone may be 400 Hz (c) the first overtone is 300 Hz (d) none of the above Calculate the velocity of sound in a mixture of oxygen, nitrogen and argon at 0°C. The mixture consists of the gases oxygen, nitrogen and argon in the mass ratio 2 : 7 : 1. (Given R = 8.3 J mol–1 K–1. Ratio of specific heats of the gases are argon 1.67, oxygen 1.4, nitrogen 1.4. The molecular weights of the respective gases are 40, 32 and 28.) (a) 329.5 m/s (b) 219.0 m/s (c) 422.0 m/s (d) 380.2 m/s A listener moves towards a source with a speed of 10 ms–1. If the source emits a frequency 200 Hz and velocity of sound in air is 332 ms–1, the wavelength of the note received by the listener is (a) 1.685 m (b) 1.71 m (c) 1.66 m (d) 2 m A cylindrical hose open at both ends has a fundamental frequency f in air. The hose is dipped vertically in water so that half of it is in water. The fundamental frequency of the air column in now (a) f / 2 (b) 3f / 4 (c) f (d) 2f A wave represented by the equation y = a cos (kx t ) is superposed with another wave to form a stationary wave such that point x = 0 is a node. The equation for the other wave is (a) a sin (kx (c) – a cos (kx

MARK YOUR RESPONSE

(b) – a cos (kx

(d) – a sin ( kx

A train approaching a hill at a speed of 40 km/hour sounds a whistle of frequency 580 Hz when it is at a distance of 1 km from the hill. Wind is blowing in the direction of the train with a speed of 40 km/h. Find the frequency of the whistle heard by an observer on the hill: (Velocity of sound in air = 1200 km/h) (a) 585 Hz (b) 575 Hz (c) 599 Hz (d) 589 Hz 8. A siren placed at a railway platform is emitting sound of frequency 5kHz. A passenger sitting in a moving train A records a frequency of 5.5 kHz while the train approaches the siren. During his return journey in a different train B be records a frequency of 6.0 kHz while approaching the same siren. The ratio of the velocity of train B to that of train A is (a) 242/252 (b) 2 (c) 5/6 (d) 11/6 9. A note is produced when you blow air across the top of a test tube. Two students were asked about the effect of blowing harder. Student A : The pitch of sound would increase. Student B : The intensity of sound would increase. (a) A is correct, B is wrong (b) B is correct, A is wrong (c) Both are correct (d) Both are wrong 10. An object of specific gravity is hung from a thin steel wire. The fundamental frequency for transverse standing waves in the wire is 300 Hz. The object is immersed in water so that one half of its volume is submerged. The new fundamental frequency in Hz is (a)

300

(c)

300

t) t)

1 2 2

1/ 2

÷÷

(b)

300

÷ 1÷

(d)

300

10.

2 2 2

1/ 2

÷ 1÷ 1

÷÷

IIT-JEE PHYSICS Challenger

358

11.

Two vibrating strings of the same material but length L and 2L have radii 2r and r respectively. They are stretched under the same tension. Both the strings vibrate in their fundamental modes, the one of length L with frequency f1

17. A source emitting sound of frequency f0 is moving in a circle of radius R, having centre at the origin, with a uniform speed = c/3, where c = speed of sound. Find the maximum and minimum frequencies heard by a stationary listener at the point (R/2, 0).

and the other with frequency f 2 . The ratio f1 / f 2 is given by (a) 2 (b) 4 (c) 8 (d) 1 12. A wave disturbance in a medium is described by

y ( x, t )

0.02 cos 50 t

÷ cos(10 x) where x and y are

in metre and t is in second (a) A node occurs at x = 0.15 m (b) An antinode occurs at x = 0.3 m (c) The speed of wave is 5 ms–1 (d) The wavelength of wave is 0.3 m 13. String # 1 is connected with string # 2. The mass per unit length in string # 1 is 1 and the mass per unit length in string # 2 is 4 1. The tension in the strings is T. A travelling wave is coming from the left. What fraction of the energy in the incident wave goes into string # 2 ? incident 2

(a)

(b)

2 3 f0

2 3 1 2 3 +1

3 f 0 3 f0 , (d) None of these 2 5 18. A transverse wave in a medium is described by the equation y = A sin2( t – kx). The magnitude of the maximum velocity of particles in the medium is equal to that of the wave velocity, if the value of A is (a) /2 (b) /4 (c) / (d) 2 / 19. A complex wave is represented by an expression of the form: (c)

2 1 .sin 3 x , where = is the angular T 2 frequency and T is the period of the wave. Which of the following sketches best represents the wave?

y(t) = 1. sin x

y(t)

(a)

(a) 1/8 (b) 4/9 (c) 2/3 (d) 8/9 14. An open pipe is suddenly closed at one end with the result that the frequency of third harmonic of the closed pipe is found to be higher by 100 Hz than the fundamental frequency of the open pipe. The fundamental frequency of the open pipe is (a) 200 Hz (b) 300 Hz (c) 240 Hz (d) 480 Hz 15. The extension in a string, obeying Hooke’s law, is x. The speed of sound in the stretched string is v. If the extension in the string is increased to 1.5x, the speed of sound will be (a) 1.22v (b) 0.61v (c) 1.50v (d) 0.75v 2 16. The equation y = a cos (2 nt – 2 x/ ) represents a wave with : (a) amplitude a, frequency n and wavelength (b) amplitude a, frequency 2n and wavelength 2 (c) amplitude a/2, frequency 2n and wavelength (d) amplitude a/2, frequency 2n and wavelength /2

MARK YOUR RESPONSE

6 f0 6 f0 , 5 7

y(t) (b)

y(t)

(c)

y(t)

(d)

11.

12.

13.

14.

16.

17.

18.

19.

15.

359

WAVES 20.

21.

A whistle giving out 450 Hz approaches a stationary observer at a speed of 33 m/s. The frequency heard by the observer in Hz is (a) 409 (b) 429 (c) 517 (d) 500 Equations of a stationary and a travelling waves are as follows y1 = a sin kx cos t and y2 = a sin ( t – kx). The phase difference between two points x1 = is

in the standing wave (y1) and is

(y2), then ratio 22.

23.

and x2 =

3 2k

in travelling wave

(a) 1 (b) 5/6 (c) 3/4 (d) 6/7 A travelling wave in a stretched string is described by the equation y = A sin (kx – t) The maximum particle velocity is (a) A (b) /k (c) d / dk (d) x / t The equation of plane progressive wave motion is y = a sin2 / (vt – x). Velocity of particle is

dv dy dv dy (b) v (c) – y (d) – v dx dx dx dx 24. A train moves towards a stationary observer with speed 34 m/s. The train sounds a whistle and its frequency registered by the observer is f1. If the train's speed is reduced to 17 m/s, the frequency registered is f2. If the speed of sound is 340 m/s, then the ratio f1/f2 is (a) 18/19 (b) 1/2 (c) 2 (d) 19/18 25. Three coherent sonic sources emitting sound of single wavelength ' ' are placed on the x-axis at points (a) y

(–

11 6 , 0) , (0 , 0) , (

reaching a point ( 0 , 5

26.

11 6 , 0) . The intensity

6) from each source has the same

value I0 . Then the resultant intensity at this point due to the interference of the three waves will be : (a) 6 I0 (b) 7 I0 (c) 4 I0 (d) 5 I0 Two pulses on the same string are described by the following wave equations y1 =

5 (3 x 4t ) + 2 2

and

y2 =

5 (3 x 4t 6) 2 + 2

27. The amplitude of a wave disturbance propagating in the positive x–direction is given by y =

1 + x2

(a)

m1 m2

m2 m1

(b)

(c)

m1 m2

(d)

m2 m1

30. A sound source S, emitting a particular sound frequency can slide along a horizontal rough rod. It is tied with a string (of total length ) that passes over a pulley. An observer O is in a lift which is tied to the other end of the string as shown in the figure The system is made to move with a constant speed. Considering that is not very large, as distance between the pulley and the source S varies from to zero, the apparaent frequency perceived by the observer

O (a) (b) (c) (d)

first increases, then decreases first decreases, then increases continously increases continuously decreases

20.

21.

22.

23.

24.

25.

26.

27.

28.

29.

30.

at t = 0 and

at t = 2s, where x and y are in meter.. 2 x2 2x Assuming that the shape of the wave disturbance does not change during the propagation, the speed of the wave is (a) 0.5 m/s (b) 1 m/s (c) 1.5 m/s (d) 2 m/s 28. A composite string is made up by joining two strings of different masses per unit length µ and 4µ. The composite string is under the same tension. A transverse wave pulse Y = (6 mm) sin (5 t + 40 x) , where 't ' is in seconds and 'x' is in metres, is sent along the lighter string towards the joint. The joint is at x = 0. The equation of the wave pulse reflected from the joint is (a) Y = (2 mm) sin (5 t – 40 x) (b) Y = (4 mm) sin (40 x – 5 t) (c) Y = – (2 mm) sin (5 t – 40 x) (d) Y = (2 mm) sin (5 t – 10 x) 29. Two monotomic ideal gases 1 and 2 of molecular masses m1 and m2 respectively are enclosed in separate containers kept at the same temperature. The ratio of the speed of sound in gas 1 to that in gas 2 is given by

where symbols have usual meaning. Choose the incorrect statement (a) Pulse y1 and pulse y2 travel along +ve and –ve x-axis respectively (b) At t = 0.75s, y displacement at all points on the string is zero (c) At x = 1m, y displacement is zero for all time (d) Energy of string is zero at t = 0.75s

MARK YOUR RESPONSE

IIT-JEE PHYSICS Challenger

360

31. Two pulses in a stretched string whose centers are initially 8 cm apart are moving towards each other as shown in the figure. The speed of each pulse is 2 cm/s. After 2 seconds, the total energy of the pulses will be

8 cm

(a) zero (b) purely kinetic (c) purely potential (d) partly kinetic and partly potential 32. A stationary observer receives sound wave from two tuning forks, one of which approaches and the other recedes with the same velocity. As this takes place, the observer hears beats with frequency 2 Hz. What is the velocity of each tuning fork if their oscillation frequency is 0 = 680 Hz and the velocity of sound in air is vs = 340 m/s (a) 5 m/s (b) 0.5 m/s (c) 15 m /s (d) 50 m/s 33. The temperature of a mono-atomic gas in an uniform container of length L varies linearly from T0 to TL as shown in the figure. If the molecular weight of the gas is M0, then the time taken by a wave pulse in travelling from end A to end B is

TL L

(a)

(c)

( (

2L TL + T0 L TL – T0

) )

3M 5R

(b)

3M 5R

(d)

3(TL – T0 ) 5 RM 0 L

(a)

300

(c)

300

1–

1 2 ÷

2 –1÷

(b)

300

(d)

300

2 2 –1÷

2 –1 2 ÷

36. A sonometer wire resonates with a given tuning fork forming standing waves with five antinodes between the two bridges when a mass of 9 kg is suspended from the wire. When this mass is replaced by a mass M, the wire resonates with the same tuning fork forming three antinodes for the same positions of the bridges. The value of M is (a) 25 kg (b) 5 kg (c) 12.5 kg (d) 1/25 kg 37. How many frequencies below 1 kHz of natural oscillations of air column will be produced if a pipe of length 1 m is closed at one end ? [velocity of sound in air is 340 m/s] (a) 5 (b) 6 (c) 4 (d) 8 38. A sound source emits frequency of 180 Hz when moving towards a rigid wall with speed 5 m/s and an observer is moving away from wall with speed 5 m/s. Both source and observer moves on a straight line which is perpendicular to the wall. The number of beats per second heard by the observer will be [Speed of sound = 355 m/s] (a) 5 beats/s (b) 10 beats/s (c) 6 beats/s (d) 8 beats/s 39. A police car moving at 22 m/s, chases a motorcyclist. The police man sounds his horn at 176 Hz, while both of them move towards a stationary siren of frequency 165 Hz. Calculate the speed of the motorcycle, if it is given that he does not observes any beats. Police Car

M0 2 R (TL – T0 )

22 m/s

34. The ends of a stretched wire of length L are fixed at x = 0 and x = L. In one experiment, the displacement of the wire is y1 = A sin ( x/L) sin t and energy is E1 and in another experiment its displacement is y2 = A sin (2 x/L) sin 2 t and energy is E2. Then (a) E2 = E1 (b) E2 = 2E1 (c) E2 = 4E1 (d) E2 = 16 E1 35. An object of specific gravity is hung from a thin steel wire. The fundamental frequency of transverse standing waves in the wire is 300 Hz. The object is immersed in water,

MARK YOUR RESPONSE

so that one half of its volume is submerged. The new fundamental frequency (in Hz) is

176 Hz

Motorcycle

Stationary Siren (165 Hz)

(a) 33m/s (b) 22m/s (c) zero (d) 11m/s 40. A string of length 0.3 m and mass 10 –2 kg is clamped at both of its ends. The tension in the string is 1.2 N. When a pulse travels along the string, the shape of the string is found to be the same at times t and t + t. The minimum value of t is (a) 0.1 sec (b) 0.2 sec (c) 0.3 sec (d) 0.4 sec

31.

32.

33.

34.

35.

36.

37.

38.

39.

40.

IIT-JEE PHYSICS Challenger

376

observer u x.......................>

1. (a)

4. (b) The apparent frequency heard by the listener

n =n

v v

The observed frequency from A, A

( vs u ) vs

but n =

= 2.

.u =

2. (b) A vibrating organ pipe can be closed or open. For a closed organ pipe fundamental frequency

v 4l

v 332 1.71m n 194 5. (c) Frequency of the open organ pipe n = f. Initial length of the pipe = l l Final length = 2 We know that frequency of open organ pipe is v 2l and when it is dipped in water, it behaves as a closed end pipe f1 =

200Hz

3v 3 fc 600Hz 4l For an open organ pipe, fundamental frequency v 200Hz f0 = 2l Its first overtone =

2v 2 f 0 400Hz 2l 3. (a) The relation for the velocity of sound in a gas Its first overtone =

RT M Considering the mixture of gas while all the constituents of the mixture occupy the same volume their masses vary. Let mO, m N, m A be the fractions of masses of the respective gases and MO, MN, MA be their respective molecular weights. Now the velocity of sound in the mixture can be given by the relation, v=

v = RT

332 10 = 194 Hz 332

(vs + u ) = vs

Number of beats =

fc =

= 200

So, wavelength of the note received by the listener

The observed frequency from B, B

O mO

N mN

A mA

Its frequency f2 =

MA 8. (b)

2 7 1 1.4 1.67 ÷ 10 10 10 + + = 8.3 273 3 3 3÷ 32 10 28 10 40 10 ÷

= [8.3×273×1000 (8.75× 10–3 + 35 × 10–3 + 4.175 × 10–3)]1/2 = [8.3 × 273 × 47.925]1/2 = 329.5 m/s

v 2l

(v (v

w ) v0 w) vs

Here, = 580 Hz v + w = 1200 + 40 = 1240 km/h vs = 40 km/h v0 = 0 So,

1.4

v 4l/ 2

6. (c) Stationary wave is produced when two waves travel in opposite direction. Now, y = a cos (k x – t) – a cos (k x + t) y = 2a sin kx sin t is equation of stationary wave which gives a node at x = 0. 7. (c) The air is blowing in the direction of train i.e., in the direction of motion of sound. The frequency of sound heard by observer

1/ 2

1/2

v 4l2

= 580

1240 0 1240 40

599 Hz

v + vA v f , or v A ( fA v f v + vB v fB = f , or vB = ( f B v f fA

vB f = B vA f A

f) f)

f 6.0 5 1 = = =2 f 5.5 5 0.5

9. (b) Pitch is related with frequency (f) = v/ = const. But intensity is related with particle energy which increases.

377

WAVES

1 T ; 2 m In air : T = mg = Vg

f0 =

14. (a) Fundamental frequency of open pipe is

10. (a) f =

1 Vg ... (i) 2 m In water : T = mg – upthrust

Third harmonic of the closed pipe fc = 3

v 2

v ÷ 4

V 2

=V g–

Vg (2 2

1 2

f '=

Vg (2 2

)

Velocity of sound by a stretched string v =

)

Vg m

vµ T

)

Hence v µ T

or,

f' = f

v = v'

T T'

2 or v' = v

1/ 2

f '= f

T' =v T

1.5x = 1.22 v x

2 x

16. (d) y = a cos2 (2 nt = 300

f1 = f2

1 cos 2 2 nt

1 2

a 2

a cos 4 nt 2

4 x

Comparing with standard equation of harmonic wave y = A cos ( t – x), we get

4r 2

(since tension is same)

a , 2

Also k =

= 4 n = 2 f, where f = 2n

2 2 = ; ( / 2)

sin (90 R/2

50 = 5 m/s and k 10 = 2 /k = 2 /10 = 0.2 m 2

)

sin R

cos =

ABC, using

sin 2

R/2 O

R 2

V V 13. (d) Er = Ar = 2 1 ÷ = 1/ 9 Ei Ai ÷ V1 + V2 Et = 8/9 Ei

17. (a) For f to be greatest, should be minimum the rule

Therefore,

2 x

(since the wires are of same material) =1 12. (c) Comparing it with y (x, t) = A cos ( t + /2) cos k x If k x = /2, a node occurs; 10 x = /2 x = 0.05 m If k x = , an antinode occurs 10 x = x = 0.1 m Also speed of wave =

2L L

) can be written as

1/ 2

11. (d) Fundamental frequency f =

T m

where m is the mass per unit length

1 = 2

v v v = + 100 or = 200 Hz 4 2 2 15. (a) According to Hooke's law FR µ x [Restoring Force FR = T, tension of spring] Given : 3

For minimum sin should be max. = 60°

= 90°

Consider a pair of insulating blocks with thermal resistances R1 and R2 as shown in the figure. The temperature at the boundary between the two blocks is 1 2

(a) ( 1 2 R1R2) / ( 1 + 2) (R1 + R2) (b) ( 1R1 + 2R2) / (R1 + R2) (c) [( 1 + 2) R1R2] / (R12 + R22) (d) ( 1 R2+ 2R1) / (R1 + R2) One end of a uniform rod of length 1m is placed in boiling water while its other end is placed in melting ice. A point P on the rod is maintained at a constant temperature of 800Â°C. The mass of steam produced per second is equal to the mass of ice melted per second. If specific latent heat of steam is 7 times the specific latent heat of ice, the distance of P from the steam chamber must be (a) (1/7) m (b) (1/8) m (c) (1/9) m (d) (1/10) m Two thin walled spheres of different materials, one with double the radius and one-fourth wall thickness of the other, are filled with ice. If the time taken for complete melting of ice in the sphere of larger radius is 25 minutes and that for smaller one is 16 minutes, the ratio of thermal conductivities of the materials of larger sphere to the smaller sphere is (a) 4 : 5 (b) 25 : 1 (c) 1 : 25 (d) 8 : 25 In a vertical U-tube containing a liquid, the two arms are maintained at different temperatures, t1 and t2. The liquid columns in the two arms have heights 1 and 2 respectively. The coefficient of volume expansion of the liquid is equal to

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(a)

2 t1

(b)

1t 2

1+ 2 2 t1

(d)

1t 2

2t2

1+ 2 1t1 + 2 t 2

A body cools in a surrounding which is at a constant temperature of 0. Assume that it obeys Newtonâ&#x20AC;&#x2122;s law of cooling. Its temperature is plotted against time t. Tangents are drawn to the curve at the points P( = 2 ) and Q( = 1). These tangents meet the time axis at angle of 2 and 1, as shown.

1 2

(a)

(c)

1t1

tan tan tan tan

1 2

tan (b) tan

(d)

tan tan

2 1

IIT-JEE PHYSICS Challenger

396

The temperature of an air bubble while rising from bottom to surface of a lake remains constant but its diameter is doubled if the pressure on the surface is equal to h meter of mercury column and relative density of mercury is then the depth of lake in metre is (a) 2 h (b) 4 h (c) 8 h (d) 7 h Two ideal gases at temperature T1 and T2 are mixed. There is no loss of energy. If the masses of molecules of the two gases are m1 and m2 and number of their molecules are n1 and n2 respectively, the temperature of the mixture will be (a)

T1 + T2 n1 + n2

(b)

T1 T2 + n1 n2

(c)

n2T1 + n1T2 n1 + n2

(d)

n1T1 + n2T2 n1 + n2

A mixture of n1 moles of monoatomic gas and n2 moles of diatomic gas has

Cp Cv

= 1.5 .

(a) n1 = n2 (b) 2n1 = n2 (c) n1 = 2n2 (d) 2n1 = 3n2 9. Two identical containers A and B have frictionless pistons. They contain the same volume of an ideal gas at the same temperature. The mass of the gas in A is mA and that in B is mB. The gas in each cylinder is now allowed to expand isothermally to double the initial volume. The changes in the pressure in A and B are found to be p and 1.5 p respectively. (a) 4mA = 9mB (b) 2mA = 3mB (c) 3mA = 2mB (d) 9mA = 4mB 10. A monatomic ideal gas, initially at temperature T1, is enclosed in a cylinder fitted with a frictionless piston. The gas is allowed to expand adiabatically to a temperature T2 by releasing the piston suddenly. If L1 and L2 are the length of the gas column before and after expansion respectively, then

The density of water at 4°C is 1000.0 kg/m3 and at 100°C it is 958.4 kg/m3. The cubic expansivity of water between these temperatures is (a) 4.5 × 10–3 / K (b) 5.4 × 10–5 / K (c) 4.5 × 10–4 / K (d) 5.4 × 10–6 / K 12. The power radiated by a black body is P, and it radiates maximum energy around the wavelength 0 . If the temperature of the black body is now changed so that it radiates maximum energy around a wavelength 3 0/4, the power radiated by it will increase by a factor of (a) 4/3 (b) 16/9 (c) 64/27 (d) 256/81 13. Two gases occupy two containers A and B the gas in A, of volume 0.10m3, exerts a pressure of 1.40 MPa and that in B of volume 0.15m3 exerts a pressure 0.7 MPa. The two containers are united by a tube of negligible volume and the gases are allowed to intermingle. Then if the temperature remains constant, the final pressure in the container will be (in MPa) (a) 0.70 (b) 0.98 (c) 1.40 (d) 210 14. A diatomic ideal gas undergoes a thermodynamic change according to the P-V diagram shown in the figure. The total heat given to the gas is nearly (use ln 2 = 0.7) 11.

P B

2P0

isothermal A

P B C

(a)

(c)

2/3

(b)

L2 L1

MARK YOUR RESPONSE

2V0

(d)

L1 L2 L2 L1 ÷

(a) 2.5 P0V0 (b) 1.4 P0V0 (c) 1.1 P0V0 (d) 3.9 P0V0 15. One mole of an ideal gas is taken from state A to state B by three different processes, (i) ACB (ii) ADB (iii) AEB as shown in the P-V diagram. The heat absorbed by the gas is–

T1 T2 is given by L1 L2 ÷

D A

(a) (b) (c) (d)

2/3

E V

greater in process (ii) than in (i) the least in process (ii) the same in (i) and (iii) less in (iii) than in (ii)

10.

11.

12.

13.

14.

15.

397

HEAT AND THERMODYNAMICS 16.

Three rods of the same cross section and made of the same material from the sides of a triangle ABC as shown. The points A and B are maintained at temperatures T and 2T respectively in the steady state. Assuming that only heat conduction takes place, the temperature at point C is

19. When a system is taken from a state i to a state f along the path i a f (as shown in the figure), Q = 50 cal and W = 20 cal. Along path i b f Q = 36 cal. (i) What is W along path i b f ? (ii) If W = – 13 cal for the return path f i, what is Q for this path? (iii) Take Eint i= 10 cal. What is Eint f ?

60°

(a)

2 2+ 3 2+ 3 2

(b)

(d)

3 2 2+ 3

5 T 2

5 moles of nitrogen gas are enclosed in an adiabatic cylindrical vessel. The piston itself is a rigid light cylindrical container containing 3 moles of Helium gas. There is a heater which gives out a power 100 cal to the nitrogen gas. A power of 30 cal is transferred to Helium through the bottom surface of the piston. The rate of increment of temperature of the nitrogen gas assuming that the piston moves slowly :

He N2

b Volume

(a) + 6, – 43, + 40 (b) + 20, + 40, – 20 (c) + 13, + 30, – 20 (d) + 15, – 50, + 30 20. The ratio C P /C V for a gas mixture consisting of 8g of helium and 16 g of oxygen is (a) 24.2/15 (b) 15/23 (c) 27/17 (d) 17/27 21. An ideal gas heat engine operates in a cycle between 227°C and 127°C. It absorbs 6.0 × 104 cal at the higher temperature. How much work per cycle is this engine capable of performing? (a) 1.2 × 103 cal (b) 1.2 × 104 J (c) 12 × 104 J (d) 1.2 × 104 cal

CP for a mixture consisting of n1 moles CV

22. Find the value of

18.

(a) 2K/sec (b) 4K/sec (c) 6K/sec (d) 8K/sec Pressure p, volume V and temperature T for a certain gas

of a monoatomic gas and n2 moles of a gas of diatomic molecules:

AT BT 2 , where A and B are constants. V The work done by the gas as its temperature changes from T1 to T2 while pressure remains constant is are related by p =

B (T2 2

(a)

(b)

A(T2

(c)

A 2 (T2 T12 ) T

(d)

A(T2

MARK YOUR RESPONSE

T1 )

T1 ) B (T22

T1 ) 2

n1 n2

(b)

5n1 7 n2 3n1 5n2

(c)

3n1 5n2 5n1 7 n2

(d)

7 n2 3n1 5n1 3n2

23. Two identical containers A and B with frictionless pistons contain the ideal gas at the same temperature and the same volume V. The mass of the gas in A is mA, and in B is mB. The gas in each cylinder is now allowed to expand isothermally to the same final volume 2V. The changes in pressure in A and B are found to be P and 1.5 P respectively. Then: (a) 4mA = 9mB (b) 2mA = 3mB (c) 3mA = 2mB (d) 9mA = 4mB

T12 )

B 3 (T2 T13 ) 3 B (T2 3

(a)

T1 )3

16.

17.

18.

21.

22.

23.

19.

20.

IIT-JEE PHYSICS Challenger

398

24. In a given process on an ideal gas, dW = 0 and dQ < 0. Then for the gas (a) the temperature will decrease (b) the volume will increase (c) the pressure will remain constant (d) the temperature will increase 25. An ideal gas can be expanded from an initial state to a certain volume through two different processes, (A) PV2 = K and (B) P = KV2 , where K is a positive constant. Then, choose the correct option from the following. (a) Final temperature in (A) will be greater than in (B) (b) Final temperature in (B) will be greater than in (A) (c) Work done by the gas in both the processes would be equal (d) Total heat given to the gas in (A) is greater than in (B) 26. 0.5 mole of an ideal gas at constant temperature 27°C kept inside a cylinder of length L and cross-section area A closed by a massless piston.

300 K 1m

(a)

(

(c)

k ÷ m / sec 100R

(d)

24. 29.

T 100)%

T 100) 100% (d) (4

T 100)%

where P0 and

are +ve constants. 1/ 2

(a)

2 P0 P0 3nR 3 ÷

(c)

2nR 2 P0 ÷ P0 3 ÷

1/ 2

(b)

P0 2 P0 ÷ 2nR 3 ÷

(d)

2 P0 P0 ÷ nR 2 ÷

25. 30.

of air are related

as P/ = constant regardless of height (n is a constant here). The corresponding temperature gradient is (M is the molecular weight of air) (a)

Mg (n + 1) R

(b)

Mg (1 n) nR

(c)

Mg (1 n2 ) nR

(d)

Mg (n 2 + 1) R

k ÷ m / sec 10R

k ÷ m / sec 1000R

1/ 2

30. Suppose the pressure P and the density

31. Three moles of an ideal gas being initially at a temperature

T0 = 273 K were isothermally expanded = 5.0 time its initial volume and then osochorically heated so that the volume and pressure in the final state become equal to that in the initial state. The total amount of heat transferred to the gas during the process equals Q = 80 kJ. The ratio

27. Consider the shown diagram where the two chambers separated by piston-spring arrangement contain equal amounts of certain ideal gas. Initially when the temperatures of the gas in both the chambers are kept at 300 K. The compression in the spring is 1m. The temperature of the left and the right chambers are now raised to 400 K and 500 K respectively. If the pistons are free to slide,the compression in the spring is about.

MARK YOUR RESPONSE

(b) (2

1/ 2

The cylinder is attached with a conducting rod of length L , cross-section area (1/9) m2 and thermal conductivity k, whose other end is maintained at 0°C. If piston is moved such that rate of heat flow through the conducing rod is constant then velocity of piston when it is at height L/2 from the bottom of cylinder is : [ neglect any kind of heat loss from system ] (b)

T 100)%

29. The maximum attainable temperature of ideal gas in the process

(a)

(a) 1.3 m (b) 1.5 m (c) 1.1 m (d) 1.0 m 28. A rod PQ of length l is pivoted at an end P and freely rotated in a horizontal plane at an angular speed about a vertical axis passing through P. If coefficient of linear expansion of material of rod is , find the percentage change in its angular velocity if temperature of system is increased by T . is

k ÷ m / sec R

300 K vacuum

= C p / Cv for this gas is (a) 1.4 (c) 1.5

26. 31.

(b) 1.33 (d) 1.6

27.

28.

399

HEAT AND THERMODYNAMICS 32.

A certain amount of ideal monoatomic gas undergoes, process given by UV1/2 = C where U is the internal energy of the gas. The molar specific heat of the gas for the process will be (a) R/2 (b) 3R (c) 5R/2 (d) –R/2 33. Three pieces of metallic heat conductors X, Y, Z of equal lengths and cross-sectional area are connected in series as shown in figure. Ends A and D are maintained at 0°C and 100°C respectively. If Y is made up of same material as that of Z, point B is at 20°C. What will be the temp of B, if Y is made up of a material same as that of X. Consider only conduction and laterally insulated surfaces of conductors.

0°C

36. Three discs A, B and C having radii 2 m, 4m and 6 m respectively are coated with carbon black on their outer surfaces. The wavelengths corresponding to the maximum intensity are 300 nm, 400 nm, 500 nm respectively. The power radiated by the are QA, QB and QC respectively. Then (a) QA is maximum (b) QB is maximum (c) QC is maximum (d) QA = QB = QC 37. An ideal gas whose adiabatic exponent equals is expanded so that the amount of heat transferred to the gas is equal to the decrease of its internal energy. Then the equation of the process in the variables T and V ( –1)

100°C Y

TV 2

(b)

( –2) 2

(c)

( –1) TV 4

(d)

( –2) TV 4

D C (a) 25°C (b) 50°C (c) 40°C (d) 20°C 34. 12 identical rods made of same material are arranged in the form of a cube. The temperature of P and R are maintained at 90°C and 30°C respectively. Then the temperature of point V, when steady state is reached. T U A

(a)

38. One mole of a diatomic gas is taken through the process PV n = k, where n and k are constant. If the heat capacity of gas is negative, then the value of n may be

Q a

P 90°C W

(a)

5 7

(b)

–

5 7

(c)

9 7

(d)

–

9 7

39. An ideal monoatomic gas is taken round the cycle ABCDA as shown in the P – V diagram (see Fig.). The work done during the cycle is

30°C R (a) 65°C (b) 60°C (c) 20°C (d) 50°C 35. ABCDE is a regular pentagon of uniform wire. The rate of heat entering at A and leaving at C is equal. TB and TD are temperature of B and D. Find the temperature TC. TB B (d /dt )=1 (d /dt )=1 A C

D TD

(a)

3TB + 2TD 5

(b) 3TD – 2TB

MARK YOUR RESPONSE

(d) can have any value

2P, 2V C

A P, V

(a) PV (c)

2P, V B

P,2V V

(b) 2 PV

1 PV 2

(d) zero

40. A thermodynamical process is shown in the figure with PA = 3 × Patm, VA = 2 × 10–4 m3, PB = 8 × Patm, VC = 5 × 10–4 m3, in the process AB and BC, 600 J and 200 J heat are added to the system, respectively. Find the change in internal energy of the system in the process CA. (1 Patm = 105 N/m2)

32.

33.

34.

37.

38.

39.

35.

36.

IIT-JEE PHYSICS Challenger

400

44. Two adiabatic vessels, each containing the same mass m of water but at different temperatures, are connected by a rod of length L, cross – section A, and thermal conductivity K. The ends of the rod are inserted into the vessels, while the rest of the rod is insulated so that there is negligible loss of heat into the atmosphere. The specific heat capacity of water is s, while that of the rod is negligible. The temperature

P B

1 of its e original value after a time t. The thermal conductivity (K) of the rod may be expressed by

difference between the two vessels reduces to

(a) 560 J (b) – 560 J (c) – 40 J (d) + 40 J 41. If a given mass of an ideal gas followed a relation VT = constant during a process, then which of the following graphs correctly represents the process ?

(a)

(b)

(a)

msL A t

(b)

msL msL (d) 2eA t 2A t 45. What is the equivalent thermal conductivity of the rods in figure given below , if the length of each cylinder be and area of cylinder having thermal conductivities K1 and K3 be A while that of the middle cylinder having thermal conductivity K2 be 2A? (c)

K2 Area 2A

K1 P

(d)

(c)

a +b 3 5

1 1 1 + + K1 2 K 2 K 3

(b)

l 1 1 1 1 + + K1 2 K 2 K 3

(d) None

46. Heat is supplied to a certain homogeneous sample of matter at a uniform rate. Its temperature is plotted against time as shown in the figure.

Temp.

(a)

1 1 1 + + K1 2 K 2 K 3 2

(c)

where p is the pressure and V is the volume of the gas. The molar heat capacity of the gas in the above process, is (a) CP + 2R (b) CV – 2R (c) CV + 2R (d) CP + 2R 43. The heat (Q) supplied to a solid, which is otherwise thermally isolated from its surroundings, is plotted as a function of its absolute temperature, . It is found that they are related by the equation. Q = a 2 + b 4. (a, b are constants). The heat capacity of the solid is given by 3

1 2

(a)

42. A monoatomic ideal gas is taken through a reversible process whose equation is given by :

p = kV

(b) a + b

MARK YOUR RESPONSE

(d) 2a + 4b

40. 45.

Area A l

(c)

emsL A t

41.

Time

42.

43.

44.

401

HEAT AND THERMODYNAMICS

48.

dP ÷ for different processes, where symbols have dV / V usual meaning. Process Bulk modulus (A) isothermal process (i) 0 (B) adiabatic process (ii) (C) isobaric process (iii) P (D) isochoric process (iv) P (a) (A)-iv, (B)-iii, (C)-ii, (D)-i (b) (A)-iii, (B)-iv, (C)-ii, (D)-i (c) (A)-iii, (B)-iv, (C)-i, (D)-ii (d) (A)-iii, (B)-ii, (C)-iv, (D)-i 51. The relation between internal energy U, pressure P and volume V of an ideal gas in an adiabatic process is U = 2 + 3PV. What is the value of the ratio of the molar

P a

specific heats

The pressure-volume diagram shows six curved paths that can be followed by the gas (connected by vertical paths). Which two of them should be part of a closed cycle if the net work done by the gas is to be its maximum positive value? (a) af (b) ae (c) ac (d) ce An ideal gas is initially at temperature T and volume V. Its volume is increased by V due to an increase in temperature T, pressure remaining constant. The quantity

V V T

varies with temperature as

Y (a)

(b)

T T+ T

Temperature K

(c)

T T+ T

(d)

T T+ T

Temperature K

MARK YOUR RESPONSE

CP ? CV ÷

(a) 2/3 (b) 4/3 (c) 3/2 (d) 1 52. The graph below shows the variation with temperature of the volume of a fixed mass of water.

e f

49.

50. Match the column according to the bulk modulus

Volume/cm3

47.

Which of the following conclusions can be drawn? (a) Its specific heat capacity is greater in the solid state than in the liquid state (b) Its specific heat capacity is greater in the liquid state than in the solid state (c) Its latent heat of vaporization is smaller than its latent heat of fusion (d) None of these The value of Cp – Cv is 1.00R for a gas sample in state A and is 1.06R in state B. Let pA, pB denote the pressure and TA , TB denote the temperature of the states A and B respectively. Most likely – (a) pA < pB and TA > TB (b) pA > pB and TA < TB (c) pA = pB and TA < TB (d) pA > pB and TA = TB

2.0004 2.0003 2.0002 2.0001 2.0000 274

275

276

277

278

279

280

281

Temperature/K

A thermometer is to be designed that measures temperatures in the range between 274K and 280K. Which one of the following is the main reason why the variation with temperature of the volume of the water is not suitable to use in the thermometer ? (a) Water is a colourless liquid (b) Water freezes at 273 K (c) The change in volume is too small over the temperature range (d) The volume has the same value at more than one temperature 53. For a monoatomic ideal gas undergoing an adiabatic change, the relation between temperature and volume is TVx = constant where x is (a) 7/5 (b) 2/5 (c) 2/3 (d) 1/3

46.

47.

48.

51.

52.

53.

49.

50.

IIT-JEE PHYSICS Challenger

402

54. In the following P-V diagram of an ideal gas, two adiabates cut two isotherms at T1 = 300K and T2 = 200K. The value of VB/VC is

55.

56.

57.

58.

(a) = VA/VD (b) < VA/VD (c) > VA/VD (d) 3/2 A 2 kW continuous flow geyser (instant geyser) has inlet water temperature of 10°C, while water is coming out from the tap at the rate of 20 g/s. The temperature of the water coming out from the tap must be about [Assume no loss of energy in other forms. Take sw = 4.2 J/g°-C] (a) 45°C (b) 33.8°C (c) 23.8°C (d) None of these The quantity of gas in a closed vessel is halved and the velocities of its molecules are doubled. The final pressure of the gas will be (a) P (b) 2P (c) P/2 (d) 4P Soap bubbles filled with helium float in air. Which has the greater mass : the wall of a bubble or the gas enclosed within it (Assume pressure inside and outside the bubble to be same) ? (a) the wall of bubble (b) the enclosed gas (c) mass of both are equal (d) any one can be greater depending on size of bubble Three moles of an ideal monoatomic gas perform a cycle shown in figure. The gas temperatures in different states are T1 = 200K, T2 = 400K, T3 = 1600 K, and T4 = 800K. The work done by the gas during the cycle is (Take R = 25/3 J/mol-K)

P A

P0 P0 2

C T0

V (a) P0/2 (b) P0/4 (c) P0/6 (d) P0/8 61. A system is taken from a given initial state to a given final state along various paths represented on a P–V diagram. The quantity that is independent of the path is (a) amount of heat transferred Q (b) amount of work done W (c) (Q – W) (d) none of these 62. P-V plots for two gases during adiabatic processes are shown in the figure. Plots 1 and 2 should correspond respectively to P

1 2 V

4 T

(a) 5 kJ (c) 15 kJ

MARK YOUR RESPONSE

59. If the molecules in a tank of hydrogen have the same RMS speed as the molecules in another tank of oxygen, we may be sure that (a) the pressure are the same (b) the hydrogen is at the higher temperature (c) the temperature are the same (d) the oxygen is at the higher temperature 60. The state of an ideal gas is changed through an isothermal process at temperature T0 as shown in figure. The work done by gas in going from state B to C is double the work done by gas in going from state A to B. If the pressure in the state B is P0/2 then the pressure of the gas in state C is

(b) 25 kJ (d) 20 kJ

(a) He and O2 (b) O2 and He (c) He and Ar (d) O2 and N2 63. Why does the pressure of an ideal gas increase when it is heated at constant volume ? (a) The gas molecules expand (b) The molecules move at the same speed, but hit the walls more often (c) The molecules move faster and hit the walls more often (d) The number of molecules of gas increases

54.

55.

56.

57.

58.

59.

60.

61.

62.

63.

IIT-JEE PHYSICS Challenger

418

1 2 3 4 5 6 7 8 9 10 11 12 13

(d) (c) (d) (a) (b) (d) (d) (a) (c) (d) (c). (d) (b)

14 15 16 17 18 19 20 21 22 23 24 25 26

(d) (d) (a) (a) (b) (a) (a) (b) (b) (c) (a) (b) (c)

27 28 29 30 31 32 33 34 35 36 37 38 39

(a) (b) (a) (b) (a) (d) (a) (d) (b) (b) (a) (c) (a)

40 41 42 43 44 45 46 47 48 49 50 51 52

(b) (a) (c). (d) (d) (a) (a) (a) (b). (c) (c) (b) (d).

53 54 55 56 57 58 59 60 61 62 63 64 65

66 67 68 69 70 71 72 73 74

(a) (d) (c) (d) (b) (a) (c) (d) (a)

1 2 3 4 5 6

7 8 9 10 11 12

(b) (c) (c) (a) (a) (b)

13 14 15 16 17 18

(b) (a) (b) (a) (c) (c)

19 20 21 22 23 24

25 26 27 28 29 30

(a) (a) (d) (a) (d) (b)

31 32

1 2

(a) (d )

3 4

(d ) (d )

5 6

7 8

(a) (a)

9 10

(d ) (d )

11 12

(b ) (b )

1 2 3

(a, b) (a, c) (a, b)

4 5 6

1. 4. 7.

(a, c) (b, c) (a, b, c, d)

(a, b , c) (a, b) (c, d)

7 8 9

A-q; B-r; C-s; D-p A-p, s; B-s; C-p, r; D-q, r A-s; B-p, r; C-r; D-q, s

2. 5. 8.

10 11 12

(b, d). (a, c, d) (a, c, d)

(a, c, d) (a, b, d) (a, c, d)

13 14 15

A-p, r; B-p; C-q, r; D-q A-r, s; B-q, r, s; C-p, r; D-q, s A-p; B-s; C-r; D-q

3. 6. 9.

16 17

(a, c) (b, c)

A-p, s; B-q, r; C-p, r, s; D-p, s A-p, q; B-r, s; C-p, q A-p, s; B-q; C-q, r; D-p, s

1714

20.5

0.014

238

346

103

15.2

9000

3.6

800

10.8

0.002

595

273

419

HEAT AND THERMODYNAMICS

(d) Rate of transmission of heat

mL 4 K = time (t / r 2 )

Temperature difference = Thermal Resistance

L time

dQ d = dt R Here,

(

dQ dt

R1 R1

( R1 R2 )

R2 1 R1 R1 R2 2.

t/4 2R

R2 R1

constant

R2 2

K ÷ tr

4 3 r 3

time µ

tR K

t 2 RK s KL 8 25 4 1 KS ; = = = K S 25 16 tRK L 2 KL

(c) Let Q1 and Q2 amounts of heat flow from P in any time t. Let m be the masses of steam formed and ice melted in time t. Let k and A be the thermal conductivity and the area of cross-section respectively of the rod.

(a) Let 0, 1 and 2 be the densities of the liquid at temperature 0, t1 and t2 respectively To balance pressure, 1 1g = 2 2g 0

800 100 Q1 = kA ÷ t = mLsteam x

t1 ÷

t2 ÷

Solving we get

800 0 Q2 = kA ÷ t = mLice 1 x

1 x ÷ =7 800

700 Dividing, x or

1 – x = 8x or x =

(d) R = dR =

dr 2

4 r K

1 m 9

4 K r2 r1 r1r2 ÷

1 r2 r1 4 K r2 r1

4 K t r

2÷

(b) For -t plot, rate of cooling =

(

dQ = slope of the curve. dt

dQ = |tan(180°– 2)| = tan 2 = k( 2 – 1) dt

where k = constant. At Q,

dQ =| tan(180° dt tan tan

r1 = r2) 6.

2t1 - 1t2

AT P,

[R = thermal resistance] Q=

1- 2

2 1

1) |

tan

(d) From Boyle's law (T = constant) P1 V1 = P2 V2

IIT-JEE PHYSICS Challenger

420

(Hdwaer + h dmercury) g ( = hdmercury g(

4 3 r ) 3

4 (2r)3) 3

Hdwater = 8hdmercury – hdmercury H = 7h

for A, p A

pA 2

for B, pB

pB 2

1.5 p

d mercury

pA 2 = pB 3

d water

Also, p AV =

3 (d) Total energy of molecules of first gas = n1kT1 2

mA 2 = mB 3

3 Total energy of molecules of mixture = K(n1T1 + 2 n 2 T2 ) 3 3 (n1 + n2) KT = K(n1 T1 + n2T2) 2 2

or 10.

(d) Here TV

11.

n1 N A f1 + n2 N A f 2 n1 N A + n2 N A

i.e.,

pB 2

(d1 d 2 ) 1 . d2

1000 958.4 958.4 96

n1 = n2

constant temperature, their pressures fall to

V0 1

pA and 2

12.

V µ L);

2/3

m m d 2 d1 1 = . m/d1

2 = 1.5 or f = 4 f

3n1 + 5n2 =4 n1 + n2

n1 f1 + n2 f 2 3n1 + 5n2 = n1 + n1 n1 + n2

= 1+

5 , hence TV2/3 = constant 3

where NA = Avogadro constant. Here, f1 = 3, f2 = 5

Also,

constant

T L Hence, 1 = 2 ÷ T2 L1

total number of degree of freedom total number of molecules

f =

–1 =

Now T1 L12 / 3 = T2 L22 / 3 (

(a) The average number of degrees of freedom per molecule,

3mA = 2mB.

As =

n T + n2T2 T= 1 1 ( n1 + n2 )

f =

mA m RT and pBV = B RT M M

pA mA = pB mB

3 Total energy of molecules of second gas = n2kT2 2

2 p

3 p

H = 7h 7.

or p A

(d1 d 2 ) (d1d 2 )

4.5 10

per degree K.

(d) Let T0 = initial temperature of the black body. 0T0 = b (constant) Power radiated = P0 = C.T04. (C = constant) Let T = new temperature of black body.

T =b=

0T0

or T =

4T0 3

439

ELECTROSTATICS

A solid sphere of radius R carries a uniform volume charge density r. The magnitude of electric field inside the sphere at a distance r from the centre is (a)

rr 3e 0

Rr 3e 0

(b)

(c)

R3r

R 2r r e0

(d)

r 2 e0

Two identical metallic blocks resting on a frictionless horizontal surface are connected by a light metallic spring having the spring constant 100 N/m and an unstretched length of 0.2m, as shown in figure 1. A total charge Q is slowly placed on the system, causing the spring to stretch to an equilibrium length of 0.3m, as shown in figure 2. The value of charge Q, assuming that all the charge resides on the blocks and that the blocks are like point charges, is

ur ur q + q + q3 ur ur ( E1 + E 2 + E 3 ).d A = 1 2 s Î0

(a)

Ñò

(b)

Ñò s (E1 + E 2 + E3 + E 4 ).d A

(c)

Ñò s (E1 + E 2 + E3 + E 4 ).d A

q1 + q2 + q3 + Î0

q1 + q2 + q3 + q4 Î0

(d) none of the above Two thin flat metal plates having large surface area are charged separately to acquire charge densities + s and – s. The plates are then brought near to each other and held parallel to each other (Fig.) :

Figure 1

–vely charged

m +vely charged

Figure 2 3.

If EA, EB and EC denote the electric fields at the points A, B and C respectively, then which of the following will be true :

(a) 10 µC (b) 15 µC (c) 20 µC (d) 30 µC q1, q2, q3 and q4 are point charges located at point as shown in the figure and S is a spherical Gaussian surface of radius R. Which of the following in true according to the Gauss’s law : S

q1 q2

MARK YOUR RESPONSE

s e0

(a)

E A = EC =

(b)

E A = E B = EC =

(c)

E A = EC = 0, EB =

s e0

(d)

E A = EC = 0, EB =

2s e0

s e0

q4 q3

IIT-JEE PHYSICS Challenger

440

7p ˆ k (d) None of these 32p Î0 Four charges are rigidly fixed along the Y-axis as shown. A positive charge approaches the system along the X-axis with initial speed just enough to cross the origin. Then its total energy at the origin is – Y (c)

particle before coming to rest and acceleration of particle at that moment are respectively :

p Two point dipoles pkˆ and kˆ are located at (0, 0, 0) and 2 (1m, 0, 2m) respectively. The resultant electric field due to the two dipoles at the point (1m, 0, 0) is 9p ˆ -7 p ˆ k (a) 32p Î k (b) 32p Î0 0

(a)

2A ,0 B

(b)

2 A qA -2 A qA ,,(d) B m B m In the figure shown, A is a fixed charged B (of mass m) is given a velocity v perpendicular to line AB. At this moment the radius of curvature of the resultant path of B is

2 2q

(a)

–q

1 ] 4p Î0

T =p 2 r 1 T= 2pr

4p Î0 r 2 mv 2 q2

(b)

infinity

(d)

m 2k lq

(b)

2klq m

(d)

4p 2 m 3 T2 = r 2klq

a , where a is a positive constant and r is the r distance from the centre of the charge. The charge of the sphere for which the electric field intensity E outside the sphere is independent of r is – (a) pR2a (b) 4pR2a 2 (c) 2pR a (d) 3pR2a/4 11. Three identical positive charges Q are arranged at the vertices of an equilateral triangle. The side of the triangle is a. Find the intensity of the field at the vertex of a regular tetrahedron of which the triangle is the base. KQ KQ 6 2 2 2 (a) (b) a a KQ 3 2 (c) (d) None of these a 12. A network of six identical capacitors, each of value C is made as shown in the figure. Equivalent capacitance between points A and B is density r =

1 m T= 2pr 2klq

A particle of charge q and mass m moves rectilinearly under the action of electric field E = A – Bx, where A and B are positive constants and x is distance from the point where particle was initially at rest then the distance traveled by the

MARK YOUR RESPONSE

B +q

10. A system consists of a uniform charged sphere of radius R and a surrounding medium filled by a charge with the volume

(a)

r +q

(b) positive (a) zero (c) negative (d) data insufficient A particle of charge – q and mass m moves in a circle of radius r around an infinitely long line charge of linear charge density +l. Then time period will be –

[where k =

qA m

(c)

v Q

0, -

B (b) 3C/4 (d) 3C

(a) C/4 (c) 4C/3

10.

11.

12.

441

ELECTROSTATICS 13.

Three concentric charged metallic spherical shells A, B and C have radii a, b and c; charge densities s , – s and s and potentials VA, VB and VC respectively. Then which of the following relations is correct? (a)

(a) 0

(b) – 6 f0 e0

(d)

–X

1/ 2

(a)

(c)

é Kq 2 ù (2.7)ú ê êë mL úû

–

6f0 e0

me w 2 R 2 2e

1/ 2

(b)

é Kq 2 ù (1.35) ú ê úû ëê mL

1/ 2

(d) Zero

20. Three charges –q, +q and +q are situated in X-Y plane at points (0, –a), (0, 0) and (0, a) respectively. The potential at a point distantr (r > a) in a direction making an angle q from Y-axis will be

(b)

(d) +X

–X

A charge of 3 coulombs moving in a uniform electric field experiences a force of 3000 newtons. The potential difference between the two points situated in a field at a distance of 1 cm from each other will be : (a) 100 volt (b) 5000 volt (c) 10 volt (d) 50 volt

MARK YOUR RESPONSE

Kq æ 2a cos q ö ç1 ÷ r ø r è

2kq cos q

(c)

16.

(b)

é Kq 2 ù (5.4) ú ê úû ëê mL

(c)

–X

given

me w R 3 eme w R 2 (d) 3e 2 19. Four identical particles each of mass m and charge q are kept at the four corners of a square of length L. The final velocity of these particles after setting them free will be

(a)

were

z2),

(c)

(b) –X

(a) zero

A capacitor is charged to a potential difference of 100 V and is then connected across a resistor. The potential difference across the capacitor decays exponentially with respect to time. After 1s, the potential difference between the plates of the capacitor is 80 V. After 2 s, the potential difference between the plates will be (a) 40 V (b) 56 V (c) 60 V (d) 64 V Four identical charges are placed at the four vertices of a square lying in YZ plane. A fifth charge is moved along X axis. The variation of potential energy (U) along X axis is correctly represented by

(a)

potential

+ + where f0 is constant, then the charge density giving rise to the above potential would be.

æ a 2 + b2 ös VC = ç + c÷ è b ø e0

15.

electrostatic

(x2

18. A conducting disc of radius R about its axis with an angular velocity w. Then the potential difference between the centre of the disc and its edge is (no magnetic field is present)

s (d) V A = VB = VC = ( a + b + c ) e 14.

the

f = f0

s V A = (a + b + c ) e0

æ a2 ös - b + c÷ (b) VB = ç è b ø e0

(c)

17. If

Kq r

Kq æ 2a cos q ö ç1 + ÷ r ø r è 21. A large hollow metallic sphere A (of radius R) is positively charged to a potential of 100 V and a small sphere B (of radius R/5) is also positively charge to a potential of 100 V. Now B is placed inside A and they are connected by a wire. The final potential of A will be : (a) 200 V (b) 150 V (c) 120 V (d) none of above (d)

13.

14.

15.

16.

18.

19.

20.

21.

17.

445

ELECTROSTATICS 44.

A uniformly charged and infinitely long line having a linear charge density l is placed at a normal distance y from a point O. Consider an imaginary sphere of radius R with O as centre and R > y. Electric flux through the surface of the sphere is

(a)

(b)

h=R

+ B ++ + + + + + y + + O + + A+ R + + +

(a) zero

(d)

45.

h=R

(d)

(c)

(b)

2l R 2 – y 2 e0

h=R

47. The figures represent charged metallic plates. Find the incorrect distribution of charge.

2lR e0

–

+ (c)

l R2 + y 2 e0

(a)

(b)

Find equivalent capacitance between A and B. [Assume each conducting plate is having same dimensions and

–

e0 A = 7 µF where A is neglect the thickness of the plate, d area of plates, A > > d]

– s2 + s1 – s1 – s2 – – + –

–s

–

+ s 2 – s1 + s 1 – s2 – + – +

(c)

–

+ s1 – s2 + s2 + s1 – + + + (d)

d +

–

2d 48. Four identical square plates of side a are arranged as shown. The equivalent capacity between A and C

46.

(b) 11µF (a) 7 µF (c) 12 µF (d) 14 µF A positively fixed charged ring is placed on a horizontal surface on the earth. A positively charged particle q is released from a height R on its axis. The variation of total potential energy (gravitational + electrostatic) as a function of separation between the particle and the ring is correctly depicted by, if charge particle just reaches at the centre of ring

MARK YOUR RESPONSE

44.

45.

D C

46.

(a)

3e0 a 2 2d

(b)

3e0 a 2 5d

(c)

3e0 a 2 3d

(d)

5e 0 a 2 . 3d

47.

48.

IIT-JEE PHYSICS Challenger

446

49. An electric dipole is placed in an electric field whose direction is fixed but whose magnitude varies with distance. The incorrect option is

52. Two identical blocks are kept on a frictionless horizontal table connected by a spring of stiffness k and of original length l 0 . A total charge Q is distributed on the block such that maximum elongation of spring at equilibrium is equal to x. Value of Q is

(a) it experiences no net force and no torque (b) it experiences a net force but no torque (c) it experiences a net force and a torque (d) it experiences no net force but a torque 50. In the circuit shown, A and B are equal resistance. When S is closed, the capacitor C charge from the cell of emf e reaches a steady state. The correct option is

2l 0 4pe 0 k (l 0 + x )

(b)

2 x 4pe 0 k (l 0 + x )

(c)

2(l 0 + x ) 4 pe 0 kx

(d)

(l 0 + x ) 4 pe 0 kx .

53. Electrical field intensity is given as ur = E (2x + 1)yi$ + x(x + 1)$j. The potential of a point (1, 2) if potential at origin is 2 volts is, (a) 2 volts (b) 4 volts (c) – 2 volts (d) 0 volts 54. A point charge 50 µC is located in the x-y plane at a point r whose position vector is r = (2iˆ + 3 ˆj ) m . Then electric field

B +

(a)

r at the point whose position vector is r = (8iˆ - 5 ˆj ) m

(a) During charging, more heat is produced in A than in B

(in vector form) will be

(b) In the steady, heat is produced at the different rate in A and B (c) In the steady state, energy strored in C is

(d) In the steady state, energy stored in C is

1 2 Ce 4 1 2 Ce 2

51. Three identical capacitors A, B and C are charged to the same potential and then made to discharge through three resistanceRA, RB and RC, where RA > RB > RC. Their potential differences (V) are plotted against the time t, giving the curves 1, 2 and 3. The correlation between A, B, C and 1, 2, 3 is

(a)

90 ( -3iˆ + 4 ˆj ) V/m

(b)

900 (3iˆ - 4 ˆj ) V/m

(c)

90 (3iˆ - 4 ˆj ) V/m

(d)

900 ( -3iˆ + 4 ˆj ) V/m

55. Two capacitors of capacitances C1 and C2 are connected in series, assume that C1 < C2. The equivalent capacitance of this arrangement is C, where (a) C < C1/2 (b) C1/2 < C < C2/2 (c) C1 < C < C2 (d) C2 < C < 2C2 56. Figure shows a system of three concentric metal shells A, B and C with radii a, 2a and 3a respectively. Shell B is earthed and shell C is given a charge Q. Now if shell C is connected to shell A, then the final charge on the shell B, is equal to

t (a) 1 ® B, 2 ® C,3 ® A

(b) 1 ® C, 2 ® B,3 ® A

(d) 1 ® B, 2 ® A,3 ® C

MARK YOUR RESPONSE

(a) – 4Q/13 (c) – 5Q/3

49.

50.

51.

54.

55.

56.

A B C

(b) – 8Q/11 (d) – 3Q/7

52.

53.

447

ELECTROSTATICS 57.

The electric field intensity at all points in space is given by r E = 3iˆ - ˆj volt/metre. The nature of equipotential lines in xy-plane is given by Low potential

High potential y

30°

(a)

60. A battery, or batteries, connected to two parallel plates produce the equipotential lines between the plates as shown. Which of the following configurations is most likely to produce these equipotential lines ?

30°

(b)

–2V –1V

(a)

High potential

Low potential

60°

(c)

60°

(d)

59.

3µF

High potential

15V

MARK YOUR RESPONSE

62.

+ – 2V

–

+ 2V

r E must be the electric field due to the enclosed charge r (b) If net charge inside the Gaussian surface = 0, then E must be zero everywhere over the Gaussian surface. (c) If the only charge inside the Gaussian surface is an electric dipole, then the integral is zero. r uur (d) E is parallel to dA everywhere over the Gaussian surface. (a)

+4q

58.

+ 2V

+ –

(b) 1 : 3 (d) 3 : 1

57.

–

is true ?

+2q

+ 2V

61. The force on an electron located on the 0-volt potential line (in previous problem) is (a) 2N, directed to the right (b) zero (c) directed to the right, but its magnitude cannot be determined without knowing the distance between the lines (d) directed to the left, but its magnitude cannot be determined without knowing the distance between the lines r uur q 62. Consider Gauss’s law Ñ ò E.dA = Î0 . Which of the following

(a) 1 : 2 (c) 2 : 1

(d) –

– +

(c)

(b) 10V (a) 30V (c) 25V (d) None of these Three charges +q, +2q and +4q are connected by strings as shown in the figure. What is ratio of tensions in the strings AB and BC ? A

+ –

19V

Both capacitors are initially uncharged and then connected as shown and switch is closed. What is the potential difference across the 3µF capacitor ?

2µF

–

Low potential

58.

(b) +

High potential y

59.

60.

61.

IIT-JEE PHYSICS Challenger

448

63. An electrostatic field line leaves at angle a from point charge q1, and terminates at point charge –q2 at an angle b (shown in figure). Then the relationship between a and b is

–50µC C=20µF +50µC

+100µC C=5µF –100µC +q1

–q 2 a b = q2 sin 2 2 2

(a)

q1 sin 2 a = q2 sin 2 b

(b)

q1 sin 2

(c)

q1 tan a = q2 tan b

(d)

q1 cos a = q2 cos b

64. Two very long line charges of uniform charge density +l and –l are placed along same line with the separation between the nearest ends being 2a, as shown in figure. The electric field intensity at point O is a

–

(a)

(c)

l 2pe 0 a

(b) 90µC (a) 100 µC (c) 10µC (d) None of these 67. Three identical large metal plates of area A are arranged at distances d and 2d from other. Top metal plate is uncharged, while other metal plates have charges +Q and –Q. Top and bottom metal plates are connected by switch S through a resistor of unknown resistance. What energy (in mJ) is dissipated in the resistor when switch is closed ?

Î0 A = 6µF, Q = 60mC ) d

(Given :

+ ++++++

A (b) 0

l pe0 a

(d)

l 4pe 0 a

65. Three capacitors C1, C2 and C3 are connected to a battery as shown in the figure. The three capacitors have equal capacitances. Which capacitor stores the most energy ?

+ V –

d A –Q

(b) 0.5 mJ (a) 0.1 mJ (c) 1.0 mJ (d) 2.0 mJ 68. The figure shows the electric field lines in the vicinity of two point charges. Which one of the following statements concerning this situation is not true ?

(a) C2 or C3 as they store the same amount of energy (b) C2 (c) C1 (d) All three capacitors store the same amount of energy

(a) (b) (c) (d)

66. For the configuration of capacitors shown, both switches are closed simultaneously. After equilibrium is established, what is the charge on the top plate of the 5µF capacitor ?

MARK YOUR RESPONSE

63. 68.

64.

65.

q1 is negative and q2 is positive The magnitude of the ratio (q2/q1) is less than one Both q1 and q2 have the same sign of charge The electric field is strongest midway between the charges.

66.

67.

449

ELECTROSTATICS 69.

A point charge +Q is positioned at the center of the base of a square pyramid as shown. The flux through one of the four identical upper faces of the pyramid is

(a) t1 = t2 (b) t1 > t2 (c) t1 < t2 (d) No definite relation between t1 and t2 may be predicted by the given data 72. Two identical electric dipoles are arranged on x-axis as shown in figure. Electric field at the origin will be

y +Q

(a)

Q 16e 0

(b)

Q 8e 0

(d) None of these

Q 4e 0

A dielectric slab is attached to a string of mass per unit length µ, whose other end is fixed to a wall. Capacitor has square plates of side b and separation between the plates is d. Find the fundamental frequency of vibration of the string. (Dielectric slab remains in equilibrium)

////////////

V k

71.

1 Î0 bV 2 (k - 1)m (b) 2L 2d

1 Î0 bV 2 ( k + 1)m 2L 2d

(c)

1 Î0 bV (k - 1) 2L 2d m

1 Î0 bV 2 (k - 1) 2L 2d m

A dipole is kept in front of a conducting sphere containing a total charge Q. If the dipole is released from rest, it reaches the sphere in time t1. If the same experiment is repeated with an insulating sphere (Îr = 1) with same charge Q distributed uniformly over its surface, the dipole reaches in time t2. Then

kp 2 ˆ j r3

(d)

2kp ˆ kp ˆ i- 3 j r3 r

– q2

(a) The total force acting on the charge +q1 will increase. (b) The total force acting on the charge +q1 will decrease. (c) The total force acting on the charge +q 1 remain unchanged (d) None of these 74. Two small balls of mass m, bearing a charge q each, are connected by a nonconducting thread of length 2l. At a certain instant, the middle of the thread starts moving at a constant velocity v perpendicular to the direction of the thread at the initial instant. Determine the minimum distance d between the balls. (a)

(c)

74.

(b)

73. A thin insulator rod is placed between two unlike point charges +q1 and –q2 (figure). How will the forces acting on the charges change ?

69.

+ q1

(a)

MARK YOUR RESPONSE

- kp 2 ˆ j r3

//////////////////////

(d)

45°

(a) Zero

(c)

45°

70.

71.

2 lq 2

q 2 + 8pe 0 mv 2 l

2 lq 2

q 2 + 3pe 0 mv 2 l

72.

(b)

(d)

2lq 2

q 2 + 4pe 0 mv 2 l

3lq 2

q 2 + 8pe 0 mv 2 l

73.

457

ELECTROSTATICS

A dielectric cylinder of radius a is infinitely long. Its volume charge density r varies directly as the distance from the

(b) The electric field is zero at all points inside S (c) At a point just outside S, the electric field is double the field at a point just inside S.

cylinder. If r is zero at the axis and is rs on the surface, the electric intensity due to it is (a)

rs a 3

at a point distant ‘r’ outside it

(b)

rs r 2

at a point distant ‘r’ inside it

3e 0 r

3e0 a

(d) 4.

At any point inside S, the electric field is inversely proportional to the square of its distance from C

A small electric dipole is placed in a uniform electric field as shown in the diagram.

+q E

rs a at a point on its surface 3e 0

–q

Considering the situation above, choose the correct statement(s).

at a point on its surface 3e0 a 2 Three concentric conducting spherical shells have radii r, 2r & 3r and charge q1, q2 & q3 respectively as shown in the figure. Select the correct alternatives

(a) The torque on the dipole points into the plane of the paper (b) If allowed to rotate freely about its center, the dipole would initially swing counter-clockwise (c) Work done by the electric field on the dipole, in rotating it from q = 90° to q = 30° is positive

3r q1 2r

(d) The potential energy of the dipole is maximum when the electric field is perpendicular to the dipole moment.

q2 q3 5.

For the situation shown in the figure (assume r >> length of dipole) mark out the correct statement(s).

p (Small dipole)

Q (a) q1 + q3 = – q2

q3 =3 q1

(b)

(d)

q q1 = – 2 4

(a) Force acting on the dipole is zero

q3 1 =– 3 q2

(b) Force acting on the dipole is approximately and is acting upward

A thin-walled, spherical conducting shell S of radius R is given charge Q. The same amount of charge is also placed at its centre C. Which of the following statements are correct? (a) On the outer surface of S, the charge density is

2 pR 2

(d) Torque acting on dipole is direction.

MARK YOUR RESPONSE

pQ 4p Î0 r 2

pQ 4p Î0 r 3

in clockwise

in anti-clockwise

IIT-JEE P HYSICS Challenger

458

How does the total energy stored in the capacitors in the circuit shown in the figure change when first switch K1 is closed (process-1) and then switch K 2 is also closed (process-2). Assume that all capacitors were initially uncharged. C

-1 éæ mö æ ql æ Röö ù ln ç1 - ÷ ÷ ú êç1 - ÷ ç 2 gR + pe 0 m è r ø ø úû êëè M ø è

(d) The tension in the string is mg -

A very small earthed conducting sphere is at a distance ‘a’ from a point charge q1 and at a distance ‘b’ from a point charge q2 (a < b). At a certain instant, the sphere starts expanding so that its radius grows according to the law, R = vt. Choose the correct options for the time dependence I (t) of the current in the earthing conductor, assuming that the point charges and the centre of the sphere are at rest, and in due time the initial point charges get into the expanding sphere without touching it (through small holes).

(d) Magnitude of change in process-2 is less than that in process-1

(a)

a æq q ö -v ç 1 + 2 ÷ , t < (b) èa bø v

(c)

0, t ³

K2 K1

(a) Increases in process-1 (b) Increases in process-2

mv 2 R+r

Figure shows a smooth thin nonconducting rod of radius r charges with uniform linear charge density l, fixed horizontally. A neutral and smooth ring Q of mass M can slide freely on the rod which happens to just fit in it, and P is non-conducting particle having a mass m and charge q, attached to the ring Q by means of a non-conducting and inextensible string of length R. Choose the correct options when string becomes vertical, if P is released from the position shown in figure. +

Q + + + + + + + +

+ + + + +P + + + R

-1 éæ mö æ ql æ Röö ù ln ç 1 + ÷ ÷ ú êç1 + ÷ ç 2 gR + r ø ø úû pe0 m è êëè M ø è

(b) The tension in the string is mg +

(d)

(a)

v2 =

(c)

v2 =

mv R+r

where v is the speed of particle.

MARK YOUR RESPONSE

q2 a b , £t< b v v

a æq q ö -v ç 1 + 2 ÷ , t > è a bø v

Two balls of charge q1 and q2 initially have a velocity (v) of the same magnitude and direction. After a uniform electric field has been applied during a certain time, the direction of the velocity of the first ball changes by 60°, and the velocity magnitude is reduced by half. The direction of the velocity of the second ball changes thereby by 90°. The velocity of the second ball changes to v2. The magnitude of the chargeto-mass ratio for the second ball is x if it is equal to k for the first ball. The electrostatic interaction between the balls should be neglected. Choose the correct options

(a) The speed of particle P is 1/ 2

b v

-v

(b)

4 k1 3

(d)

2 k1 3

IIT-JEE PHYSICS Challenger

468

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

(a) (c) (b) (c) (b) (b) (a) (c) (c) (c) (a) (c) (b) (d) (b)

16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

31 32 33 34 35 36 37 38 39 40 41 42 43 44 45

46 47 48 49 50 51 52 53 54 55 56 57 58 59 60

(d) (c) (c) (a). (a) (b) (c) (c) (b) (b) (b) (c) (b) (b) (c)

61 62 63 64 65 66 67 68 69 70 71 72 73 74 75

76 77 78 79 80 81 82 83 84 85 86 87 88

(a) (b) (c) (b) (d) (d) (a) (c) (a) (d) (d) (a) (d)

1 2 3 4 5

(b) (a) (b) (a) (b)

6 7 8 9 10

(a) (b) (b) (c) (b)

11 12 13 14 15

(b) (c) (a) (a) (d)

16 17 18 19 20

(b) (b) (a) (c) (d)

21 22 23 24 25

(a) (c) (a) (b) (a)

26 27

(b) (c)

1 2

(a) (c)

3 4

5 6

(a) (d)

7 8

(d) (c)

9 10

(d) (a)

1 2

(a, b, c) (a, b, c)

3 4

(a, c, d) (a, c)

1. 4. 7. 10.

(b, c) (a, b, d)

5 6

A-p; B-q; C-r; D-q 2. A-s; B-q; C-p; D-r 5. A-s, t; B-q, r, t; C-p, r; D-q 8. A-q, r; B-q, s, t; C-q, s, t; D-q, s, t

7 8

(a, b) (a, b, c)

A-p, s; B-p, s; C-q, s; D-p, s A-s; B-p; C-r; D-s A-p, s; B-p, r; C-p, s; D-q, s

(a, b)

3. 6. 9.

A-q; B-r, s; C-p, r; D-q A-r; B-q; C-s; D-p A-p, s; B-q, s; C-q, s; D-s, t

100

0.63

0.6

8.48

3.19

4.42

0.9

469

ELECTROSTATICS

(a)

(c)

q r 1 qr r´ r ´ = 4 = 3 e 4pe 0 R 3 3e 0 0 pR 3 3 1 (Q / 2) (Q / 2) . = (100 N / m) (0.1m) 4p Î0 (0.3m 2 )

Q2 =

4 ´ 100 ´ (0.1) ´ (0.3) 2

9 ´ 109 Þ Q = 20 × 10–6 C = 20 µC r uur q E.ds = in (b) e0

Ñò

(b) There exists a point P on the x-axis (other than the origin), where net electric field is zero. Once the charge Q reaches point P, attractive forces of the two -ve charge will dominate and automatically cause the charge Q to cross the origin. Now if Q is projected with just enough velocity to reach P, its K.E. at P is zero, but while being attracted towards origin it acquires KE and hence its net energy at the origin is positive. (P.E. at origin = zero) (a) We have centripetal force equation 2 æ 2kl ö mv qç = è r ÷ø r

Where E ® field due to all the charges in space.

s 4.

–s

(c)

s s =0 2e0 2e0 EC = 0,

EA =

EB = 5.

2kql m

2pr m = 2pr 2kql v

Now, T = where k = 8.

1 4pe 0

s s s + = . 2e0 2e0 e0

r p dipole and at equatorial 2 r r r line of p dipole so that field at given point is E1 + E2

q (A – Bx) m

...(1)

vdv q q = ( A - Bx ) ; vdv = ( A - Bx) dx m dx m

(b). The given point is at axis of

ò vdv = 0

2 q ( A - Bx ) dx ; Ax - Bx = 0 ò m0 2

2A B From eq. (1) and (2)

x = 0, x =

...(2)

qæ q 2 Aö ( A - Bx) = ç A - B ´ ÷ m mè Bø r 2 K ( p / 2) Kp E1 = = (+ kˆ) 8 23 r Kp ˆ E2 = (- k ) 1 r r 7 E1 + E2 = - Kp( - kˆ) = - 7 p kˆ 8 32p Î0

(c)

q - qA ( A - 2 A) = . m m

Kq 2 r2

Þ RC =

Kq 2

mv 2 r 2

Kq 2

mv 2 RC

Þ RC =

4 pe 0 v 2 r 2 m q2

IIT-JEE PHYSICS Challenger

470

10.

(c)

Ñò E p dS =

Therefore, the sought intensity of the field is

Q + Q¢ where Q' is the charge outside the e0

sphere 12.

C C

Q1¢ =

a ´ 4pr 2 dr r R

æ 2C ´ C ö 4C = 2 çè ÷= 2C + C ø 3

E p ´ 4pr 2 = Q

4pr 2 e0

13.

Q + 2pa (r - R ) e0 2

Q 1 = 4 pe ´ r 0 while that inside the shell =

Q 4pr e0 2

aR 2 2r e 0 2

Q = 2pR2a

14.

(a) Each charge creates at point D a field intensity of KQ

. The total intensity will be the sum of three a2 vectors (fig.). The sum of the horizontal components of these vectors will be zero, since they are equal in magnitude and form angles of 120° with each other. The vectors from angles of 90° – a with the vertical, where a is the angle between the edge of the tetrahedron and the altitude h of triangle ABC.

15.

16.

But E =

V d

qV d

h E

a C

The vertical components are identical each being equal to

KQ a2

sin a =

sin a . It follows from triangle ADE that

2 . 3

17.

sR , where R is the radius e0

of the shell. Add potential due to all the shells at a given point. (d) During discharging of a capacitor, q, the charge after a time t = q0e–lt But q = CV. i.e. q µ V, C × 100ve–l1s = 80V. C \ e–l = 0.8 After 2 s, V2 = 100 e–2l = 100 (e–l)2 = 100 × 0.64 \ V after 2 s = 64 V. (b) At the centre of the square net force on the charge is zero. However, in this position the charge is in unstable equilibrium and hence its potential energy is maximum and finite. The four charges are assumed to be placed symmetrically about the origin in YZ plane. (c) Force on a charge in an electric field is F=qE

(b) Potential due to a charged shell at an outside point

a aR 2 - 2 2e0 2r e0

E is independent of r if

E1 =

Therefore equivalent capacitance = [2C series C] // [C series 2C]

æ r 2 R2 ö æ r2 ö 2 2 = 4pa ç 2 ÷ = 4pa ç 2 - 2 ÷ = 2pa ( r - R ) è øR è ø

11.

ò dV = ò R

Ep =

(b)

Fd 300N ´ 1 ´ 10-2 m = = 10 volt q 3C

E = -Ñf = -f0 2[ xiˆ + yiˆ + zxˆ] = e0Ñ.E = –2e0f0Ñ.( xiˆ + yiˆ + zxˆ) n = –6f0 e 0

471

ELECTROSTATICS 18.

(b)

e E = me Þ

w2r

ò E dr =

me w e

22.

2 R

ò rdr

1 (d) V = 4 pe 0

e r

me w 2 R 2 2e (c) Potential energy of the system

23.

Þ V =

19.

q log e 2 1 q log e (1 + 1) = 4pe0 x0 4 pe0 x0

21.

Kq é 2a cos q ù + 1ú ê û r ë r (c) The charge on sphere A and B are:

C= 25.

(a)

+ 0.4) dx = [ x 3 + 0.4 x]0.2 0

Q 0.88 = = 10mF V 0.088 ......... (1)

qr r2 = Þ qR R 2 From (1) and (2)

qA = (4pe0R)V

qr =

R V 5 When they are connected by conducting were, the entire charge will flow to outer sphere, so that their potentials become equal. Thus

6V 600 = = 120V. 5 5

qr + qR = Q

and qB = (4 pe 0 )

ò (3x

qr 4 pr 2 s r 2 = = q R 4 pR 2 s R 2

R/5

æ 6R ö (4pe 0 ) ç V ÷ è 5 ø 4pe 0 R

0.2

V = (0.2)³ + 0.4 × 0.2 = 0.088 volt

(d) V =

1 q = V' = (4pe 0 ) R

mv02 2e

20.

6R .V 5 The final potential of A now,

V =

V = ò Ex dx =

Kq + r

q = qA + qB = (4pe0 )

1 2 mv0 2

( )

1/ 2

1 q é 1 1 1 1 1 ù 1 - + - + - + ...ú a ê 4pe0 x0 ë 2 3 4 5 6 û

2pe 0 R mv02 mv02 = – \ Q = CV = 4pe 0 R ´ 2e e negative sign has been included because of electron. 24. (a) Ex = 3x² + 0.4 N/C

é Kq 2 ù (2.7)ú v= ê úû ëê mL

1 q V = 4pe r 0

1 = 4 × mv2 = 2 mv2 2

ù 1 é ( - q ) ( - q) ( - q) + + + ...ú ê 4pe0 ë 2 x0 4 x0 6 x0 û

eV =

= Final kinetic energy of the system

Kq.2a cos q

(a) Let V be the steady state potential of sphere, then

2 2 2 K æq + q + q ö Kq 2 ç ÷ = 5.4 =4× L L 2ø 2 è L L

éq ù q q + + ...ú ê + ë x0 3 x0 5 x0 û

Qr 2 R +r 2

So, V =

........ (2)

and q R =

QR 2 R2 + r 2

qr qR + 4pe 0 r 4pe 0 R

Q æ r+R ö 4pe 0 çè r 2 + R 2 ÷ø (a) Electric field intensity between plates Þ V =

26.

15 ´ 103 V E= = 3 × 106 V/m = 5 ´ 10 –3 d Weight of the oil drop = =

4 3 pa rg 3

4 p (5 × 10–6)3 × (0.92 × 103) (9.8) 3

483

ELECTROSTATICS

Þ q2 = Q2 =

VB 2

–3CE 4

5CE 5CE 3CE 7CE – + q2 = = 2 4 4 2

(2 =

n +1

) E = çæ 2

–1

n +1

è 2

1ö 1ö æ E = ç2- ÷ E n÷ è 2 ø 2n ø

(d)

2 q2 , U = q Ui = f 2Ceq 2Cair

Ceq d 3 Ui = = = Uf Cair t 2

16. (b), 17. (b)

®0

VA =

1 æ Q Q 3Q ö + ç ÷ 4pe 0 è R 2 R 3R ø

Work done by battery Q in n-cycle

VB =

Q 3Q ö 1 æ Q + ç ÷ 4pe 0 è 2 R 2 R 3R ø

VC =

Q 3Q ö 1 æ Q + ç ÷ 4pe 0 è 3R 3R 3R ø

For n ® ¥,

1 2

VB2 = 2 E.

= CE 2 + 10.

15.

CE 2 CE 2 æ 2n –1ö + ¼+ =ç CE 2 n –1 n –1 ÷ 2 2 è 2 ø

(b) q1 + q2 = 0

q2 Q

18.

\ 2VA = 3VB VBC = VB – VA = 0. (a) Q' ® final charge on B

B \

VB =

1 æ Q Q ' 3Q ö + + ç ÷ =0 4pe 0 è 2R 2R 3R ø

Q Q' +Q=0 + 2 2 VA =

kq1 kQ kq2 + + R 2R 4R

kq1 kQ kq2 + + 4R 4 R 4R VA = VC Þ q1 = – Q/3 and q2 = Q/3

VC =

11.

(b)

Q ù Q é –Q Q VA = k ê + + = ú ë 3R 2 R 12 R û 16pe0 R

12.

(c)

Q ù 5Q é –Q Q + + = VB = k ê ú ë 6 R 2 R 12 R û 48pe0 R

13.

(a)

e 0 kA e A , C2 = 0 t d –t Two capacitors are in series

C1 =

(a)

20.

1 1 d –t 1 t = + = + C C1 C2 e 0 kA e0 A

14.

19.

Ceq

3 Cair 2

e0 A 3 e0 A t 2 = Þ = d – (t / 2) 2 d d 3

(d) The potential is positive and decreases as the point of observation is moved away from origin along the yaxis. Hence the correct option is V

e0 A d – (t / 2)

Ceq =

Q' = – 3Q \ charge given by earth DQ = – 2Q. (c) The electric field at points on positive y-axis will be positive and negative at all points on negative y-axis. Hence the correct options is (c).

21.

(a) Since the distance of released particle from mean position is comparable to distance between the fixed charges, motion of negatively charged particle will be oscillatory but not SHM.

IIT-JEE PHYSICS Challenger

484

a 22.

(c)

mg t = mg a= 23.

25.

(a)

C2+C3

æ ml 2 ö l 9g + mg l = ç + ml 2 ÷ a ; a = 2 3 8l è ø

9g 9g ´l = 8l 8

(a)

VC1 =

Q m

V (C 2 + C 3 ) C1 + (C2 + C3 )

Initially C3 = 0 So VC1 =

VC2 =6 C1 + C2

........... (1)

Now, at VC1 = 10, C3 = ¥

ö l 1 æ ml 2 EQl + mg + mg l = ç + ml 2 ÷ w 2 2 2è 3 ø

Þ 10 =

mg 3mg l 2ml 2 2 = w l+ 2 3 2q

Þ 10 =

mg l = v=

24.

(b)

2 ml 2 2 w 3

Þ 10 = V .......... (2)

æC ö = 6 Þ 5 = 3 ç 1 + 1÷ C æ C1 ö è 2 ø çè C + 1÷ø 2

4ml 2 a ; 3

26.

4M l 2 mg l 4ml 2 3g EQl = µ = µ =a 2 3 8l 3

Þ 3g 2l 27.

æ 3g ö æ 3g ö a = ç ÷ +ç ÷ è 8ø è 2ø

3g 1 3g 17 +1 = 2 16 8

C1 5 2 = -1 = C2 3 3

(b) Now, VC1 =

3g 8

ar = w2 r =

æ C1 ö çè C + C + 1÷ø 2 3

3g mg l 4ml 2 =a = a ; 8l 2 3

at =

Eq. (1) and (2),

3g l 2

EQl =

V (C2 + C3 ) C1 + (C2 + C3 )

(c)

10 (C2 + C3 ) =8 (C1 + C2 + C3 )

æC C ö 10 ç 2 + 3 ÷ è C1 C1 ø æ C2 C3 ö çè 1 + C + C ÷ø 1 1

= 8 Þ C = 2.5C 3 1

1 1 1 + » C3 + C2 C1 C1 \ \

(where C3 ® ¥ )

Total energy = energy in C1 Required ratio = 1

485

ELECTROSTATICS

(a)

A B

Q1 Q2 5.

R2 VA =

1 Q1 + Q2 4pe 0 R2

VB =

1 4pe 0

VB – VA = 2. 3.

æ Q1 Q2 ö çè R + R ÷ø 1 2

V1 V2

æ 1 1 1 ö Q1 ç 4 pe 0 è R1 R2 ÷ø

r (c) E at any point on the Gaussian surface is due to all charges. (c) Net potential at centre

–q

10.

q (a) f = net Î0

If electric potential at a point is zero then the magnitude of electric field at that point is not necessarily to be

(a, b, c) Let us divide the charged cylinder into a large number of co-axial thin cylindrical shells. Consider one such co-axial shell of radii x and x + dx and of length L. If the thickness dx is very small, so that the charge density can be assumed to be constant at each point on it, then the charge on the shell is given by :

dq = 2pxLdxr As r varies directly as the distance x, hence, r= cx where c = constant which can be given as : c a = rs or c = rs /a The field E at a point P distant r outside the cylinder is given by :

s , hence E decreases. K Î0

\ Statement -1 is false and Statement - 2 is true.

–q a

Statement-1 Potential inside shell 2 is V2 alone and not V1 + V2. Statement-2 Potential is work done per unit charge hence it is scalar. (d) For isolated capacitor Q = constant, F = constant. But

+ =0 a/ 2 a/ 2 a/ 2 a/ 2 and field is zero due to symmetry.

zero. (c) Statement 1 : As net charge of dipole is zero. Statement 2 : Gauss law is valid for all charge distributions. (a) The electric field due to disc is superposition of electric fields due to its constituent rings as given in statement-2. statement-1 is true, statement-2 is true. Statement-2 is a correct explanation for statement-1. (d)

In uniform electric field fclosed surface = 0

E 2prL =

q 1 = r dq e0 e0

Since charge inside the Gaussian surface is the charge on the cylinder, hence, we get : a

ca3 rs a3 1 = 2pLcx 2 dx = 3r e 0 3r e 0 2prLe0

ò 0

At the points inside the cylinder, E is given by :

E 2prL =

1 q¢ 1 = dq = 2pLcx 2 dx e0 e0 e0

ò 0

Þ E=

cr 2 rs r 2 = 3e0 3ae0

491

ELECTROSTATICS =

x 2 + 32

12. 4.42 The adjacent figure is a case of parallel plate capacitor. The combined capacitance will be

1 4 9-4 5 1 = = = 45 45 9 5 45

\ x2 + 9 = 81 \ x = 8.48 m 3.19 Each mass will be in equilibrium under the act of three forces namely tension of string, weight, resultant electrostatic force of the two other charges. Out of these three forces, F and mg are perpendicular.

+ 1-x x

A m

q O

B m

T C

Tcosq

0.03

q T

Tsi

FCA

Let T make an angle q with the vertical OC =

2 (0.03) 2 - (0.015)2 = 0.0173 m 3

\ OM = 0.9997 Resolving T in the direction of mg and F and applying the condition of equilibrium, we get T cos q = mg; \

tan q = F=

e0 [ kx + kdx + 1 - x - dx ] d Charge of capacitance in time dt

3FCA = 3 ´

1 2

kq 2 (CA) 2

OC 0.0173 = OM 0.9997

From (i), (ii) and (iii) 3 ´ 9 ´ 109 ´ q 2 0.0173 = 0.9997 (0.03) 2 ´ 10-3 ´ 9.8

On solving, we get q = 3.19 × 10–9 C = 3.19 nC.

... (iii)

dx dt We know that q = CV where

dq dC =V dt dt

... (ii)

[where FCB = Force on C due to B FCA = Force on C due to A ur ur | F CB | = | F CA | and a = 60° ] Also, tan q =

e0 (k – 1) dx d

dC e 0 dx e0 = (k - 1)v = (k - 1) d dt d dt

2 2 + FCB + 2 FCA FCB cos a FCA

... (ii)

e0 [kx + kdx + 1 – x – dx – kx – 1 + x] d

dC = ... (i)

2 2 2 FCA + FCA + 2 FCA ´

e0 e [(1 - x - dx) ´ 1] [( x + dx ) ´ 1] + 0 d d

T sin q = F

F mg

ke 0 ( x ´ 1) e 0 [(1 - x) ´ 1) + d d

e0 [ dx + 1 - x] ... (i) d After time dt, the dielectric rises by dx. The new equivalent capacitance will be C + dC = C1' + C2'

FCB

1m d

C = C1 + C2

M q

M 1m

11.

... (iv)

e0 (k - 1)v d From (i) and (ii) Þ I= V

.... (iii)

500 ´ 8.85 ´ 1012 (11 – 1) × 0.001= 4.425 × 10–9 A 0.01 = 4.42 nA Alternatively : We can differentiate eq. (1) w.r.t 't ', we get I=

dx dC e 0 = (K - 1) and then proceed further.. d dt dt

IIT-JEE PHYSICS Challenger

492

13.

3 Let the particle be, at some instant, at a point P distant x from the origin. As shown in the figure, there are two forces of repulsion acting due to two charges of + 8 mC. The net force is 2F cos a towards right. Similarly, there are two forces of attraction due to two charges of – 1 mC. The net force due to these force is 2F' cos b towards left.

27 3ö æ 4 ç x2 + ÷ = x2 + è 2ø 2

Þ 3x2 =

27 12 2 2

Þ x=±

5 2

This means that we need to move the charge from – ¥ to 27 + 2 +

3 2

+8µC

5 . Thereafter the attractive forces will make the charge 2

–1µC F'

F +0.1µC Fcosa P a Fcosa F

F'cosb b O F'cosb x 3 F' 2 –1µC 27 2

move to origin. Thereafter the attractive forces will make the charge move to origin. The electric potential of the four charges at x =

+8µC V=

2 ´ 9 ´ 109 ´ 8 ´ 10 -6

The net force on charge 0.1 µC is zero when 2F cos a = 2F' cos b Þ

K ´ 8 ´ 10 -6 ´ 0.1 ´ 10-6 27 x + 2

x2 +

K ´ 1 ´ 10

-6

´ 0.1 ´ 10

x2 +

8 é 2 27 ù êë x + 2 úû

3/ 2

3 2

27 2

from ¥ to x =

é 2 3ù êë x + 2 úû

ìï 3 é 2 3 ù 3 / 2 üï í2 ê x + ú ý 2û ï ïî ë þ

5 3 + 2 2

Kinetic energy is required to overcome the force of repulsion

x 3 2

x2 +

3/ 2

-6

2 ´ 9 ´ 109 ´ 10 -6

é8 1ù 4 = 2 × 9 × 109 × 10–6 ê - ú = 2.7 ´ 10 V 4 2 ë û

5 27 + 2 2

5 is 2

The work done in this process is W = q (V) where V = p.d between ¥ and x = \

3/ 2

5 . 2

W = 0.1 × 10–6 × 2.7 × 104 = 2.7 × 10–3 J

By energy conservation 3/ 2

3/ 2 ìï é 27 ù üï = íê x2 + ú ý 2û ï ïî ë þ

5 . 2

1 mv02 = 2.7 × 10–3 2

1 ´ 6 ´ 10 -4 v02 = 2.7 ´ 10 -3 Þ v0 = 3 m/s 2

In the circuit shown in the given figure the resistances R1 and R2 are respectively R1

b 220 V.

0.5 A

69 V (a) 14 and 40 (b) 40 and 14 (c) 40 and 30 (d) 14 and 30 . A copper wire and an iron wire, each having an area of cross-section A and lengths L1 and L2 are joined end to end. The copper end is maintained at a potential V1 and the iron end at a lower potential V2. If 1 and 2 are the conductivities of copper and iron respectively, then the potential of the junction will be (a)

1V1

(

1 / L1 )

2V2

(

1V1

(b)

(

(d) 3.

/ L2 )

2V2

L1 L2 ( 1 / L1 ) ( 2 / L2 )

(c)

(

1 / L1 )

(

2V2

1V1

2V2

(

/ L2 )

1V1

1 / L1 )

/ L2 )

A potential difference of 220 V is maintained across a 12000 ohm rheostat, as shown in the figure. The voltmeter has a resistance of 6000 ohm and point c is at one-fourth of the distance from a to b. Therefore, the reading of the voltmeter will be

MARK YOUR RESPONSE

(a) 32 V (b) 36 V (c) 40 V (d) 42 V What is the voltage across resistor A in the following circuit? Each resistor has a resistance of 2 M and the capacitors have capacitances of 1 F. The battery voltage is 3V.

1. 6.

(a) 0 V (b) 0.5 V (c) 0.75 V (d) 1.5 V Calculate the drift velocity of electrons in silver wire with cross-sectional area 3.14 × 10–6 m2 carrying a current of 20 A. Given atomic weight of Ag = 108, density of silver = 10.5 × 103 kg/m3. (a) 2.798 × 10–4 m/sec. (b) 67.98 × 10–4 m/sec. (c) 0.67 × 10–4 m/sec. (d) 6.798 × 10–4 m/sec. All the edges of a block with parallel faces are unequal. Its longest edge is twice its shortest edge. The ratio of the maximum to minimum resistance between parallel faces is (a) 2 (b) 4 (c) 8 (d) indeterminate unless the length of the third edge is specified

IIT-JEE PHYSICS Challenger

494

A wire of length L and 3 identical cells of negligible internal resistances are connected in series. Due to the current, the temperature of the wire is raised by T in a time t. A number N of similar cells is now connected in series with a wire of the same material and cross-section but of length 2L. the temperature of the wire is raised by the same amount T in the same time t. The value of N is (a) 4 (b) 3 (c) 8 (d) 9 Two resistances equal at 0° C with temperature coefficient of resistance 1 and 2 joined in series act as a single resistance in a circuit. The temperature coefficient of their single resistance will be : (a)

1 2

(b) 1

2 (d) 2 2 The current density varies with radial distance r as J = a r2, in a cylindrical wire of radius R. The current passing through the wire between radial distance R/3 and R/2 is :

(c)

12 7

4 R

24 V (a) 12.5

(b) 15.5

(d) 19.5

12. A moving coil galvanometer of resistance 100 is used as an ammeter using a resistance 0.1 . The maximum deflection current in the galvanometer is 100 A. Find the minimum current in the circuit so that the ammeter shows maximum deflection (a) 100.1 mA

(b) 1000.1 mA

(d) 1.01 mA

13. Calculate the relaxation time and mean free path in Cu at room temperature 300 K, if number density of free electrons is 8.5 × 1028 /m³ and resistivity = 1.7 × 10–8 mho-m.

(a)

65 a R 4 2592

(b)

25 a R 4 72

(c)

65 a 2 R 3 2938

(d)

81 a 2 R 4 144

Given k = 1.38 × 10–23 J/K.

10. A simple ohm meter is made by connecting a 1.5 V battery in series with a resistance R and a ammeter that reads from 0– 1.0mA, as shown below. The resistance R is adjusted so that when the leads A and B shortened together, the meter reads 1 mA. What external resistance across A, B results in (1) 10% (2) 50% (3) 90% of the full scale deflection ?

(a) 25 Å

(b) 20 Å

(d) 30 Å

14. In the diagram shown, all the wires have resistance R. The equivalent resistance between the upper and lower dots shown in the diagram is :

mA A B

(a)

11.

(a) (1) 17,500 (2) 9500 (3) 365 (b) (1) 15000 (2) 7500 (3) 1500 (c) (1) 13500 (2) 1500 (3) 167 (d) (1) 11, 500 (2) 1300 W(3) 65 In the circuit shown, for what value of R, will the ideal battery transfer energy at the rate of 60 W ?

MARK YOUR RESPONSE

R/8

(b)

(d)

3R/8

15. The charge flowing through a resistance R varies with time t as Q = at – bt2. The total heat produced in R by the time current ceases is (a) a3R/6b (c)

(b) a3R/3b

a3 R 2b

(d)

10.

12.

13.

14.

15.

a3 R b

11.

495

CURRENT ELECTRICITY 16.

A wire has linear resistance (in Ohm/m). Find the resistance R between points A and B if the side of the larger square is 'd'.

20. Ends of two wires A and B having resistivity m and

6 10 –5

3 10 –5

m of same cross section area are

joined together to form a single wire. If the resistance of the joined wire does not change with temperature, then find the ratio of their lengths, given that temperature coefficient of resistivity of wire A and B is

(b) d/ 2 2 d (c) 2 d (d) None of these 17. A potentiometer wire AB as shown is 40 cm long of resistance 50 /m, free end of an ideal voltmeter is touching the potentiometer wire. What should be the velocity of the jockey as a function of time so that reading in voltmeter is varying with time as (2 sin t). Q 10 10 (a)

V B

A 4V

(a) 10 sin t cm/s (b) 10 cos t cm/s (c) 20 sin t cm/s (d) 20 cos t cm/s 18. For the shown circuit the effective resistance between the points A and B will be 2R

4 10 –5 / °C and

–6 10–6 / °C. Assume that mechanical dimensions do not change with temperature. 3 10 (a) (b) 7 3 3 1 (c) (d) 10 2 21. A resistance coil, wired to an external battery is placed inside a thermally insulated cylinder fitted with a frictionless piston and containing an ideal gas. A current I = 240 mA flows through the coil which has resistance R= 490 . At what constant speed v must the piston of mass m = 12 kg move upward in order that the temperature of the gas remains unchanged? [Assume that 40% energy is utilised to do work by the gas, and neglect the work done by atmospheric pressure]. v B

m Gas

2R R

A 4R

B R

2R (a) 2 R (b) 4 R (c) R (d) R/2 19. An n sided regular polygon is inscribed in a circular region of radius R using a wire of cross sectional radius r and resistivity . If the magnetic field in the circular region has a time rate of change given by B then the current in the wire loop at any time is given by

(c)

RB r cos( / n) 2 RB r 2 sin( / n) 2

MARK YOUR RESPONSE

(a) 0.141 m/s (b) 1.41 m/s (c) 0.24 m/s (d) 2.4 m/s 22. In the circuit shown in the figure, find the current in 45

180V

90 90 100

100

(a)

R/2

(b)

(d)

RB r cos( / n) 2 (a) 4 A (c) 2 A

RB r 2 cos( / n) 3

16.

17.

21.

22.

18.

100

19.

50 (b) 2.5A (d) none

20.

IIT-JEE PHYSICS Challenger

496

23. Figure shows a 2.0 V potentiometer used for the determination of internal resistance of a 1.5 V cell. The balance point of the cell in open circuit is 76.3 cm. When a resistor of 9.5 is used in the external circuit of the cell, the balance point shifts to 64.8 cm length of the potentiometer wire. The internal resistance of the cell is approximately

(a) decreasing the resistance R and decreasing V (b) decreasing the resistance R and increasing V (c) increasing the resistance R and increasing V (d) increasing the resistance R and decreasing V. 26. A meter bridge with two resistances connected in the two gaps has a balance point at 40 cm. If a resistance 10 is connected in series with the smaller resistor, the balance point shifts by 20 cm. Neglecting any corrections, the larger resistance is

2.0 V A

B 1.5 V

(a)

(b)

(c)

(d) 12

h h

27. For maximum power from battery the internal resistance of battery r is

9.5 (a) 1.7 (b) 1.5 (c) 1.2 (d) None of these 24. The incorrect option in the given circuit if the reading of ammeter is zero is

R R R

2R A (a) 10 R (a) The value of

(R + r) R

1 will be

(b) Current in R is

(c)

r+R (c) Value of 1 will be (d) Potential across 2R is zero. 25. A potentiometer circuit shown in the figure is set up to measure emf of cell E. As the point P moves from X to Y the galvanometer G shows deflection always in one direction, but the deflection decrease continuously until Y is reached. The balance point between X and Y may be obtained by

R 8

(b)

4R 9

(d)

10 R 9

28. As shown in circuit diagram an ideal voltmeter and an ideal ammeter reading are zero. Sum of potential drop across box and 5 resistances will be (given that there are no fault in voltmeter and ammeter)

Box

12 Volt 5 A

(a) (b) (c) (d)

MARK YOUR RESPONSE

23. 28.

24.

25.

7 volt 5 volt 12 volt Data insufficient

26.

27.

497

CURRENT ELECTRICITY 29.

In the diagrams, all light bulbs are identical, all cells are ideal and identical. In which circuit (a, b, c, d) will the bulbs be dimmest ?

C (a)

A D

(a)

(b)

R1 A D

C (c) 30.

31.

Rx (b)

(d)

A hank of uninsulated wire consisting of seven and a half turns is stretched between two nails hammered into a board to which the ends of the wire are fixed. The resistance of the circuit between the nails is determined with the help of electrical measuring instruments. Determine the proportion in which the resistance will change if the wire is unwound so that the ends remain to be fixed to the nails. (a) 225 (b) 15 (c) 240 (d) 250 The diagram below shows a junction with currents labeled I1 to I6. Which of the following statements is correct?

(a)

R1 n2 n1 n1 n2

(b)

2R1n2 n1 n1 n2

(c)

R1 n2 n1 2 n1 n2

(d)

3 R1 n2 n1 2 n1n2

33. Determine the resistance RAB between points A and B of the frame formed by nine identical wires of resistance R each (figure).

32.

(a) I1 + I3 = I6 + I4 (b) I1 + I2 = I6 + I4 (c) I4 + I3 = I6 (d) I2 = I6 + I4 Given two different ammeters in which the deflections of the pointers are proportional to current, and the scales are uniform. The first ammeter is connected to a resistor of resistance R1 and the second to a resistor of unknown resistance Rx. At first the ammeters are connected in series between points C and D (as shown in figure a ). In this case, the readings of the ammeters are n1 and n2 . Then the ammeters are connected in parallel between C and D (as shown in figure b) and indicates n1' and n2'. Determine the unknown resistance Rx of the second resistor.

MARK YOUR RESPONSE

29. 34.

A R

30.

(a)

2 R 11

(b)

9 R 11

(c)

15 R 11

(d)

1 R 11

R B

34. For the circuit shown, a shorting wire of negligible resistance is added to the circuit between points A and B. When this shorting wire is added, bulb-3 goes out. Which bulbs (all identical) in the circuit brighten ?

A 3 2

(a) only bulb 2 (c) only bulbs 1 and 4

31.

32.

B 4

(b) only bulb 4 (d) only bulbs 2 and 4

33.

IIT-JEE PHYSICS Challenger

498

35. A perfect voltmeter and a perfect ammeter are connected in turn between points E and F of a circuit whose diagram is shown in figure. The readings of the instruments are V0 and I0. Determine the current I through the resistor of resistance R connected between points E and F. A

F E

(a)

V0 I 0 RI 0 + V0

(b)

V0 I 0 RI 0 V0

(c)

2V0 I 0 RI 0 + V0

(d)

V0 I0 2RI0 + V0

(a) 1/9 (b) 2/9 (c) 1/4 (d) 10/9 38. Which one of the following modifications may increase the sensitivity of moving coil galvanometer ? 1st Way : By using spring of smaller torsion constant. 2nd Way : By using a smaller coil. 3rd Way : By using a stronger magnet. 4th Way : By using a coil having fewer number of turns (a) 1st and 4th ways only (b) 1st and 3rd ways only (c) 1st, 2nd and 3rd ways only (d) 2nd and 4th ways only 39. A circuit is connected as shown in the figure with the switch S open. When the switch is closed, the total amount of charge that flows from Y to X is 3 F

6 F

S 3

36. Five identical resistors (coils for hot plates) are connected as shown in the diagram. What will be the change in the voltage across the right upper spiral upon closing the key K?

6 Y 9V

(a) 0 (b) 54 C (c) 27 C (d) 81 C 40. The resistance RAB between points A and B of the frame made of thin homogeneous wire (figure), assuming that the number of successively embedded equilateral triangles (with sides decreasing by half) tends to infinity, if side AB is equal to a, and the resistance of unit length of the wire is

~ (a) Increases (b) Decreases (c) no change (d) Nothing can be said 37. The voltage across a load is controlled by using the circuit diagram shown in figure. The resistance of the load and of the potentiometer is R. The load is connected to the middle of the potentiometer. The input voltage is constant and equal to V. The change in the voltage across the load if its resistance is doubled is

A a

MARK YOUR RESPONSE

(a)

a ( 7 1) 3

(b)

a ( 7 + 1) 3

(c)

2a ( 7 + 1) 3

(d)

a ( 7 + 1)

35. 40.

36.

37.

38.

39.

IIT-JEE PHYSICS Challenger

500

46. In the figure ammeter A1 reads a current of 10mA, while the voltmeter reads a potential difference of 3V. What does ammeter A2 in mA read ? The ammeters are identical, the internal resistance of the battery is negligible. (Consider all ammeters and voltmeters as non-ideal.)

C 100

1 2

4V A2

100

(a) 6.67 mA (b) 3.12 mA (c) 1.12 mA (d) 5.14 mA 47. All the wires on the front and the back face have resistance R. All the wires along the length have resistance 2R. Find the equivalent resistance between P and Q.

(a)

C ( E2 R1 E1 R2 ) R1 + R2

(b)

C ( E2 R1 + E1 R2 ) R1 + R2

(c)

2C ( E2 R1 + E1 R2 ) R1 + R2

(d)

C ( E2 R1 + E1 R2 ) 2 ( R1 + R2 )

49. Six identical resistors are connected as shown in the figure. The equivalent resistance will be

P R

R R

(a)

23R 20

(b)

15R 20

(c)

13R 20

(d)

3R 20

(a) Maximum between P and R (b) Maximum between Q and R (c) Maximum between P and Q (d) all are equal 50. What will be the change in the resistance of a circuit consisting of five identical conductors if two similar conductors are added as shown by the dashed line in figure.

48. The key K (figure) is connected in turn to each of the contacts over short identical time intervals so that the change in the charge on the capacitor over each connection is small. What will be the final charge qf on the capacitor ?

MARK YOUR RESPONSE

46.

47.

(a) becomes 1/5 times (c) becomes 2/5 times

48.

49.

(b) becomes 3/5 times (d) becomes 1/2 times

50.

513

CURRENT ELECTRICITY

(a) Potential difference across 20 = 20 × 1 = 20 volt = potential difference across R2. Current in R2 = 0.5 A. R2 = 20/0.5 = 40 . Potential difference across R1 = 69 – 20 = 49 Volts. Current in R1 = 0.5 A +

(b)

49 3.5

vd =

20 1A = 3.5A 10

L2 . 2A

= 6.798 × 10–4 m/sec. (b) Let the edges be 2 , a and

Rmin

I ×12000 /6000=2000, 4

(c)

220 A 9000+2000 Potential difference across voltmeter = 40 V. (d) 1.5 V

(b)

2 a

[( N ) /(

9 8.

, in decreasing order..

2 L / A)]

( LA ) s T (2 L A ) s T

1 2

(d) R1 = R0(1 +

N2 = 9

N = 3.

1t) + R0(1 +

2t)

= 2 R0 1 + 1

2 t÷ 2 Comparing with R = R0 (1+ t)

A 9.

(d)

2 (a) Given ; J = ar2. 2

R/ 2

J 2 r dr

R/2

=2 a R/3

a 2

resistor A.

6.023 1026 108 Number of electrons per unit volume of silver

aR 4 = 2

Number of electrons per kg of silver =

108

Rmax =4 ; R 2a min

[(3 ) 2 /( L / A)]

(using V2/R = ms T) = resistivity; s = specific heat capacity of material of the wire. A = area of cross section

This is a DC circuit because the battery is the only source of voltage. Hence, the capacitors behave like open circuits. An equivalent circuit is then two parallel sets of two identical series resistors, see figure. The voltage drop across each parallel branch must be the battery voltage of 3V. Since the resistors are identical there is an equal voltage drop of 1.5 V across each resistor. In particular there is a drop of 1.5 V across

3.14 10

2 2 = a a

2Rmax

V = potential at junction.

V1 V R1

neA

6.023 10 26 10.5 103 1.6 10

L1 , R2 1A

6.023 1026 10.5 103 108

R 2

r 3dr

R/3

ar 2 2 r dr

2 a

R ÷ 3

r4 4

R/2

R/3

65 65 aR 4 = 81 16 2592

Two electron beams having their velocities in the ratio 1: 2 are subjected to identical magnetic fields acting at right angles to the direction of motion of electron beams. The ratio of deflection produced is: (a) 2 : 1 (b) 1: 2 (c) 4 : 1 (d) 1: 4 A current I flows in the anticlockwise direction through a square loop of side a lying in the xoy plane with its center at the origin. The magnetic induction at the center of the square loop is

2 2 0I 2 2 0I eˆx eˆz (b) a a 2 2 0I 2 2 0I eˆz eˆx (c) (d) 2 a a2 A particle of charge q and mass m moves in a circular orbit of radius r with angular speed . The ratio of the magnitude of its magnetic moment to that of its angular momentum depends on (a) and q (b) , q and m (c) q and m (d) and m A uniform magnetic field of magnitude 1T exists in region

(a)

(b)

(1, 3)

(1,

1 3 3 1 , , ÷ (d) 2 2 ÷ 2 2 A long straight wire along the z-axis carries a current I in the

negative z direction. The magnetic vector field B at a point having coordinates (x, y) in the z = 0 plane is – (a)

I ( yiˆ xjˆ) 2 ( x2 + y 2 )

(b)

I ( xiˆ + yjˆ) 2 ( x2 + y 2 )

(c)

I ( xjˆ yiˆ) 2 ( x2 + y 2 )

(d)

I ( xiˆ yjˆ) 2 ( x2 + y 2 )

(a)

An infinitely long wire carrying current I is along Y axis such that its one end is at point A (0, b) while the wire extends upto + . The magnitude of magnetic field strength at point (a, 0).

0 is along kˆ direction as shown.

I A

y B=1T

(0,0)

(a)

(– 3–1) A particle of charge 1C is projected from point ( 3, 1) towards origin with speed 1 m/sec. If mass of particle is 1 kg, then co-ordinates of centre of circle in which particle moves are –

MARK YOUR RESPONSE

1. 6.

(c)

(a,0)

b 0I 1+ ÷ 2 4 a a + b2 0I

4 a

b a + b2 2

(b)

0I 1 4 a

(d)

None of these

b a + b2 2

IIT-JEE PHYSICS Challenger

530

A battery is connected between two points A and B on the circumference of a uniform conducting ring of radius r and resistance R. One of the arcs AB of the ring subtends an angle at the centre.The value of the magnetic induction at the centre due to the current in the ring is (a) proportional to 2 (180° – ) (b) inversely proportional to r (c) zero, only if = 180° (d) zero for all values of A small current element of length d and carrying current is placed at (1, 1, 0) and is carrying current in ‘+z’ direction. If magnetic field at origin be B1 and at point (2, 2, 0) be B2 then: (a)

B1 = B2

(b) | B1 | = | 2 B2 |

12. A charged particle is released from rest in a region of steady and uniform electric and magnetic fields which are parallel to each other. The particle will move in a (a) straight line (b) circle (c) helix (d) cycloid 13. A cyclotron is operated at an oscillator frequency of 24 MHz and has a dee radius R= 60cm. What is magnitude of the magnetic field B (in Tesla) to accelerate deuterons (mass = 3.34 × 10–27) kg? (a) 9.5 (b) 7.2 (c) 5.0 (d) 3.2 14. A thin rod is bent in the shape of a small circle of radius r. If the charge per unit length of the rod is , and if the circle is rotated about its axis at a rate of n rotations per second, the magnetic induction at a point on the axis at a large distance y from the centre

(c)

(d) B1 B1 B2 2 B2 Two long parallel wires carry equal current i flowing in the same direction are at a distance 2d apart. The magnetic field B at a point lying on the perpendicular line joining the wires and at a distance x from the midpoint is – (a)

(c)

0id 2 2

0ix 2

)

(b)

(

(d)

0ix 2 2

0id 2

+ x2

)

(a)

µ0 i R

2÷

µ0 i

(b)

2÷

(d)

2µ0 i R2

the velocity of the electron after a time interval of t second?

vxiˆ v y ˆj

e v Bt kˆ m y

(c)

vxiˆ + v y +

e v y B t ÷ ˆj m

(d)

e vxiˆ + v y + v y Bt ÷ iˆ + v y ˆj m

MARK YOUR RESPONSE

r 3n 3 y

(d)

r 3n

r 3n 3 y

15. An infinitely long conductor is bent at a point O at an angle as shown. The magnetic field at a point P distant r along the angle bisector is

÷ .r

field B = Bi with an initial velocity v = v x i + v y j . What is

(b)

2 R An electron, charge –e, mass m, enters a uniform magnetic

e vxiˆ + v y ˆj + v y Bt kˆ m

)

µ0 i .r 2 R÷

(a)

(c)

10. A long straight wire of radius R carries current i. The magnetic field inside the wire at distance r from its centre is expressed as : (a)

r 3n

(a)

(

i / 4 ).(2 i / r )[1 cos( / 2)] / sin( / 2) 0

(b)

(

/ 4 ).(i / r ) cot( / 2)

(c)

(

/ 2 ).(i / r ) cot( / 2)

(d) ( 0 / 4 ).(i / r )sin( / 2) 16. Two particles A and B of masses mA and mB respectively and having the same charge are moving in a plane. A uniform magnetic field exists perpendicular to this plane. The speeds of the particles are vA and vB respectively and the trajectories are as shown in the figure. Then A B

(a) mAvA < mBvB (c) mA < mB and vA < vB

(b) mAvA > mBvB (d) mA = mB and vA = vB

10.

11.

12.

13.

14.

15.

16.

531

MAGNETIC EFFECT OF CURRENT 17.

Three infinitely long wires are placed equally apart on the circumference of a circle of radius a, perpendicular to its plane. Two of the wires carry current I each, in the same direction, while the third carries current 2I along the direction opposite to the other two. The magnitude of the magnetic indution B at a distance r from the centre of the circle, for r > a, is (a) 0

(b)

I r

2 0 Ia I (d) r r2 18. A long cylindrical wire kept along z-axis carries a current of (c)

–

(a) angle subtended by charged particle at the centre of circular path is 2 . (b) the charge will move on a circular path and will come out from magnetic field at a distance 4d from the point of insertion. (c) the time for which particle will be in the magnetic field is 2 / b. (d) the charged particle will subtend an angle of 90° at the centre of circular path 21. An equilateral triangular loop is made up of uniform wire as shown in figure.

density j = J 0 rkˆ , where J0 is a constant and r is the radial distance from the axis of the cylinder. The magnetic induction B̂ inside the conductor at a distance d from the axis of the cylinder is

(a)

0 J0

( c)

0 J0 d

0 J0d

(b) 2

(d)

0 J0d

–

A circular arc QTS is kept in an external magnetic field B0 as shown in figure. The arc carries a current I. The magnetic field is directed normal and into the page. The force acting on the arc is × B× × 0 × × × × × × Q × × ×

20.

3 19.

× × × ×

T × × × I × × × × × × 60° × × × R

× × × ×

× × × R ×

× × × × × S× × ×

(a)

2IB0 Rkˆ

(b)

IB0 Rkˆ

(c)

–2 IB0 Rkˆ

(d)

– IB0 Rkˆ

× × × ×

q m entering in a magnetic field of strength B at a speed v = (2 d) (B), then which of the following is correct If a charged particle of charge to mass ratio

4d × × × × ×× × × × × × × ×× × × × × × × ×× × × × × × × ×× × × × × × × ×× × × × × × × ×× × × × × × × ×× × × × × × × ×× × × × × × × ×× × × × × × × ×× × × × × × × ×× × × × × × × ×× × × × × × × ×× × ×

q/ m=

MARK YOUR RESPONSE

17.

18.

22.

23.

i i A current I enters through one of the vertices of triangle and exits from other vertices of the triangle of side ‘a’ as shown. The magnitude of magnetic field at the centre of the equilateral triangle loop due to itself is 3 0I 9 0I (a) (b) 2 a 2 a 3 3 0I (c) (d) none of these 2 a 22. An ionized gas contains both positive and negative ions. If it is subjected simultaneously to an electric field along the +x-direction and a magnetic field along the +z-direction, then (a) positive ions deflect towards +y-direction and negative ions towards -y direction (b) all ions deflect towards +y-direction (c) all ions deflect towards –y-direction (d) positive ions deflect towards –y-direction and negative ions towards + y-direction. 23. An particle is moving along a circle of radius R with a constant angular velocity . Point A lies in the same plane at a distance 2R from the centre. Point A records magnetic field produced by particle. If the minimum time interval between two successive times at which A records zero magnetic field is t, the angular speed , in terms of t is – (a) 2 /t (b) 2 /3t (c) /3t (d) /t

19.

20.

21.

551

MAGNETIC EFFECT OF CURRENT

1 2 3 4 5 6

(b) (b) (c) (c) (a) (b)

7 8 9 10 11 12

1 2

(b) (c)

3 4

(d)

1. 3. 5. 7. 9.

(c,d)

(d) (c) (b) (c) (a) (a)

(a) (d)

(d) (a) (a) (b) (a) (c)

13 14 15 16 17 18

(d)

(a,b,c,d)

(d) (c)

5 6

(b) (b) (d) (c) (b) (c)

19 20 21 22 23 24

25 26 27 28 29 30

(b) (d)

7 8

(b) (d) (d) (b) (d) (b)

(b) (b)

9 10

(d)

(c)

(a)

(b,d)

(a,b,c)

0.012 2

2 6

2. 4. 6. 8. 10.

4 0.7

37 38 39 40 41 42

(a) (c)

11 12

A-r, s; B-p; C-p, q; D-r, s A-r, s; B-r, s; C-p, s; D-p, s A-q; B-r; C-p; D-s A-s; B-q; C-p; D-r A-p, q; B-p, q, r; C-p, s; D-p, q, r, s

1 5

(d) (a) (b) (b) (b) (a)

31 32 33 34 35 36

(a) (b)

13 14

(a) (b) (d) (b) (b) (d)

43 44 45 46 47 48

(b) (d) (d) (c) (a) (a)

49 50 51 52 53 54

(d) (d)

15 16

55 56 57

(d) (c)

17 18

(a)

(b) (b) (a) (a) (a) (c)

(a,b)

(a,b,d)

(b,d)

A-p, q, r, s; B-p, q, s; C-s; D-r A-p, s; B-p; C-s; D-q A-q; B-s; C-r; D-p A-p, q, r; B-p, q, r, s; C-r; D-p, q, r, s A-r, s, t; B-r, s, t; C-q, r; D-p, r

4.74

0.2

(a,d)

(d) (c) (a)

IIT-JEE PHYSICS Challenger

552

(b) Since, magnetic field is acting at right angles to the direction of motion of electron beams, so their paths will be circular.

r1 v1 = = 1/2 r2 v2

mv or r µ v qB or r1 : r2 = 1: 2 r=

Coordinates of the centre are (r cos 60°, – r sin 60°) where r = radius of circle

5. 0I

(b) Field due to one side of loop at O =

a 2÷

1 mv 1 1 = = 1 i.e., 2 , Bq 1 1

(a) Magnetic field

(2sin 45°)

Field at O due to all four sides is along unit vector eˆz I a

(b)

(x + y )

0 I [sin 90° + sin( 4 a

a ÷ 2

(2 sin 45°) =

2 2 0I a

; Further i = q × n =

Magnetic moment, M = iA =

q 2

r2;

q 2

b a 2 + b2

I ( yiˆ xjˆ) 2 ( x2 + y 2 ) 0

)] =

0 I (1 sin ) 4 a

(d) I2

2 B

qr 2 2

M qr 2 q = = 2 L 2mr 2m (c) The centre will be at C as shown :

y B=1T x (– 3–1)

(a,0)

0 I = 4 a 1

So,

(0,0)

x +y

Total field = 4.

yiˆ xjˆ

45° 45° I

3 2 ÷

60° 30°

Magnetic field at the centre due to current in arc ABC is 0 I1 (Directed upwards) 4 r Magnetic field at the centre due to current in arc ADB is

B1 =

B2 =

µ0 I 2 (2 4 r

)

(Directed downwards)

553

MAGNETIC EFFECT OF CURRENT Therefore net magnetic field at the centre B=

0 I1

I2 (2 4 r 0

E = Also, I1 = R1

(c)

r r

Thus acceleration of the electron

EA r (2

4 2 2

az =

)

(2 )

) r

ˆj )

k ev y B m

v = vx i + v y j + 12.

kˆ ( iˆ

ev y B

Velocity vz = azt =

( iˆ ˆj )

for B1 , r

= – e [(vx i v y j) Bi] = evyB k

E EA = r (2 2/A

EA r

(a) Force on the electron F = e (v B)

)

E EA = A r / 1

E = R2

and I2 =

11.

............... (1)

(a) FE = qE FB = evB sin

ev y Bt m

kˆ

(Force due to electric field) = qvB sin 0 = 0 (Force due to magnetic field)

for B2, r = iˆ + ˆj B2 =

ikˆ (iˆ + ˆj )

............... (2)

2 2

+q B

From (1) and (2)

B2 and | B1 | = | B2 |

B1 9.

(b) The magnetic field due to two wires at P

d M

13.

Force due to electric field will make the charged particle released from rest to move in the straight line (that of electric field). Since the force due to magnetic field is zero, therefore, the charged particle will move in a straight line. (d) Given: f = 24 × 106 Hz, R = 0.60 m We know that, R =

B1 =

2 ( d + x)

; B2 =

2 (d

= 10.

x d+x d

1 d

0ix 2 2

x )

(3.34 10

14.

(a) Current, I =

r R B 2( y

B × 2 r = µ0

µ0 i r . 2 2 R

i R

1.6 10

B . d = µ0iin

...(i)

where v = R = 2 f R = 2 × 24 × 106 × 0.60 = 9.04 × 107 m/s From (i),

1 d+x

mv qR

Both the magnetic fields act in opposite direction.

mv qB

)(9.04 107 )

2 r /n

2 0 Ir 2 2 3/ 2

r )

0.60

2 r n. 2 0 Ir 3

If y > > r. B

2 0r 2

(2 r n)

r3 n y3

= 3.2 T..

An ac source of angular frequency is fed across a resistor R and a capacitor C in series. The current registered is I. If now the frequency of source is changed to /3 (but maintaining the same voltage), the current in the circuit is found to be halved. Then the ratio of reactance to resistance at the original frequency is (a)

(b)

3/ 5

(c)

5/3

the A.C. current is I = 10 sin

(d)

2/3 3/ 2 If the readings of V1 and V3 are 100 volt each then reading of V2 is L

7. V3

~ 200V,50Hz

(a) 0 volt (b) 100 volt (c) 200 volt (d) cannot be determined by given information. A beam of light of intensity 12 watt/cm² incident on a totally reflecting plane mirror of area 1.5 cm². The force in newton acting on the mirror will be(a) 2.4 × 10–6 (b) 1.2 × 10–7 –8 (c) 3.6 × 10 (d) 5.6 × 10–5 If i1 = 3 sin t and i2 = 4 cos t, i3 = i1 + i2 then i3 is (a) 5 sin ( t + 53°) (b) 5 sin ( t + 37°) (c) 5 sin ( t + 45°) (d) 5 cos ( t + 53°)

MARK YOUR RESPONSE

The equation of AC voltage is E = 220 sin

(

/ 6) and

/ 6) . The average power

dissipated is (a) 150 W (b) 550 W (c) 250 W (d) 50 W A uniform coil of self-inductance 1.8 × 10–4 H and resistance 6W is broken up into two identical coils. These two coils are then connected in parallel across a 12 V battery. The circuit time constant and steady state current through the battery respectively are : (a) 30 s, 8 A (b) 30 ms, 8 mA (c) 30 s, 8 A (d) 300 s, 800 A Current in an ac circuit is given by i = 3 sin t + 4 cos t then (a) rms value of current is 5 A (b) mean value of this current in one half period will be 6/ (c) if voltage applied is V = Vm sin t then the circuit must be containing resistance and capacitance. (d) if voltage applied is V = Vm sin t, the circuit may contain resistance and inductance. An LCR series circuit with 100 resistance is connected to an AC source of 200 V and angular frequency 300 radians per second. When only the capacitance is removed, the current lags behind the voltage by 60°. When only the inductance is removed, the current leads the voltage by 60°. Then the current and power dissipated in LCR circuit are respectively (a) 1A, 200 watt (b) 1A, 400 watt (c) 2A, 200 watt (d) 2A, 400 watt

IIT-JEE PHYSICS Challenger

566

In an L-R circuit, the value of L is (0.4/ ) henry and the value of R is 30 ohm. If in the circuit, an alternating emf of 200 volt at 50 cycles per second is connected, the impedance of the circuit and current will be : (a) 11.4 ohm, 17.5 ampere (b) 30.7 ohm, 6.5 ampere (c) 40.4 ohm, 5 ampere (d) 50 ohm, 4 ampere. 10. What is the amount of power delivered by the ac source in the circuit shown (in watts).

XC=12

15. The charge on a capacitor decreases times in time t, when it discharges through a circuit with a time constant (a) t = (b) t = ln 1

ln 1

16. A uniform magnetic field of induction B is confined to a cylindrical region of radius R.

R1=5

XL=8

(d) t

R2=6 P

dB dt (tesla/second). An electron of charge q, placed at the point P on the periphery of the field experiences an acceleration.

Rrms=130V

The magnetic field is increasing at a constant rate of

(a) 500 watt (b) 1014 watt (c) 1514 watt (d) 2013 watt 11. Calculate the power factor of L–C–R circuit at resonance ? (a) 0.1 (b) 1/4 (c) 1/2 (d) 1 12. For a LCR series circuit with an A.C. source of angular frequency . (a) circuit will be capacitive if

(b) circuit will be inductive if

> =

(b)

( 2 + 1) m

towards left

1 eR dB towards left 2 m dt

eR dB towards left m dt (d) zero 17. A wooden stick of length 3 is rotated about an end with constant angular velocity in a uniform magnetic field B perpendicular to the plane of motion. If the upper one third of its length is coated with copper, the potential difference across the whole length of the stick is

1 LC

× × × × × × × × × ×

13. An observer is at a distance of 18 meter from a point source of light. The output power of the source is 250 watt. The R.M.S. value of electric field at the position of the observer will be in volt/meter. (a) 0.48 (b) 4 8 (c) 4.8 (d) 4 80 14. A coil has an inductance of 0.7 henry and is joined in series with a resistance of 220 . When the alternating emf of 220 V at 50 Hz is applied to it then the phase through which current lags behind the applied emf and the wattless component of current in the circuit will be respectively (a) 30°, 1 A (b) 45°, 0.5 A (c) 60°, 1.5 A (d) None of these

MARK YOUR RESPONSE

(c)

LC (c) power factor of circuit will by unity if capacitive reactance equals inductive reactance (d) current will be leading voltage if

(a)

(c)

× × × × × × × × × ×

× × × × ×3 × × × × ×

× × × × × × × × × ×

(b)

2 2

(d)

10.

11.

12.

14.

15.

16.

17.

× × × × × × × × × ×

× × × ×copper × × ×coat× × × × × × × × × × × × × × × × × × × × × × ×

× × × × × × × × × ×

4B 2 2

B 2

13.

567

ELECTROMAGNETIC INDUCTION & AC CURRENT 18.

A superconducting loop of radius R has self inductance L. A uniform and constant magnetic field B is applied perpendicular to the plane of the loop. Initially current in this loop is zero. The loop is rotated by 180°. The current in the loop after rotation is equal to (a) zero

B R2 L

(b)

2B R 2 B R2 (d) L 2L 19. PQ is an infinite current carrying conductor. AB and CD are smooth conducting rods on which a conductor EF moves with constant velocity v as shown. The force needed to maintain constant speed of EF is (c)

R A

C F

cm2

22. A coil of area 7 and of 50 turns is kept with its plane normal to a magnetic field B. A resistance of 30 ohm is connected to the resistance-less coil. B is 75 exp (– 200t) gauss. The current passing through the resistance at t = 5 ms will be(a) 0.64 mA (b) 1.05 mA (c) 1.75 mA (d) 14 2.60 mA 23. A short-circuited coil is placed in a time-varying magnetic field. Electrical power is dissipated due to the current induced in the coil. If the number of turns were to be quadrupled and the wire radius halved, the electrical power dissipated would be (a) halved (b) the same (c) doubled (d) quadrupled 24. The figure shows a conducting loop consisting of half circle of area A = 0.06 m2 and three straight segments. The half circle lies in a uniform changing magnetic field B = 4r2 + 2t + 5 (SI unit), where t is the time in second. An ideal battery E = 2 V is connected as shown and the total resistance of the wire is 2 . The net current in the loop is at t = 5 second is:

v V

a B

Q (a)

21.

1 vR 0 Iv

0 Iv

b ÷ a

B b ÷ a

0 Iv

(b)

v (d) R

2 v R

0 Iv

a ÷ b

1 vR 2

A copper rod of length 0.19 m is moving parallel to a long wire with a uniform velocity of 10 m/s. The long wire carries 5 ampere current and is perpendicular to the rod. The ends of the rod are at distances 0.01 m and 0.2 m from the wire. The emf induced in the rod will be(a) 10 µV (b) 20 µV (c) 30 µV (d) 40 µV The current in an L–R circuit builds up to (3/4)th of its steady state value in 4 seconds. The time constant of this circuit is (a)

1 sec ln 2

(c)

3 sec ln 2

MARK YOUR RESPONSE

(b)

2 sec ln 2

(d)

4 sec ln 2

– + (a) 1 A (b) 1.5 A (c) 0.26 A (d) 0.10 A 25. A current of 1.5 A flows through a solenoid of length 20.0 cm, cross-section 20.0 cm2 and 400 turns. The current is suddenly switched off in a short time of 1.0 millisecond. Ignoring the variation in the magnetic field the ends, the average back emf induced in the solenoid is: (a) 0.3 V (b) 9.6 V (c) 30.0 V (d) 3.0V 26. A rod PQ of length L moves with a uniform velocity v parallel to a long straight wire carrying a current i, the end P remaining at a distance r from the wire. The emf induced across the rod is (a)

(c)

0 iv

2 0 iv

r+L ÷ (b) r

r+L ÷ r

18.

19.

20.

21.

23.

24.

25.

26.

(d)

2 2

2 0iv

r2 + L ÷ r ÷

r 2 + L2

22.

÷ ÷

IIT-JEE PHYSICS Challenger

568

27. In the given figure MNPQ which falls through the magnetic field has conductivity and mass density . The frame’ss terminal velocity assuming it to be small enough so that it reaches its final velocity before leaving the region occupied by the magnetic field is

B × × × × M

× N

1m (a) 1.5 × 10–6 A anticlockwise (b) 1.5 × 10–6 A clockwise (c) 0.75 × 10–6 A anticlockwise (d) 0.75 × 10–6 A clockwise 30. An equilateral triangular loop having a resistance R and length of each side is placed in a magnetic field which is

a v

12c 2 g

16c 2 g

(b)

eB 2

45°

(a)

B = 1 T, perpendicular into the paper. Find the current in the loop at the moment, when the heating starts. Resistance of the loop is 10 at any temperature. Coefficient of linear expansion = 10–6/°C.

eB 2

varying at

16c g 16c 2 2 g (c) (d) eB e2 B 28. The given assembly made of a conducting wire is rotated with a constant angular velocity about a vertical axis MO

dB = 1 T/s. The induced current in the loop will dt

as shown in the figure. The magnetic field B exists vertically upwards as shown in the figure. Find the potential difference between points M and N, |Vm – VN| (only the magnitude)

(a)

B R2 2

(c)

2R2 –

(b)

R2 4

3 R

power source of voltage 0 sin t . The maximum current through the resistance is half of the maximum current through the power source. Then value of R is

R2 ÷ (d) zero 4

27.

4 R 3 R (d) 4 2 3 2 31. A resistor of resistance R, capacitor of capacitance C and inductor of inductance L are connected in parallel to AC

(a)

29. A straight conducting metal wire is bent in the given shape and the loop is closed. Dimensions are as shown in the figure. Now the assembly is heated at a constant rate dT/dt = 1°C/s. The assembly is kept in a uniform magnetic field

MARK YOUR RESPONSE

(b)

(c)

O B

(a)

3 2 4 R

28.

(c)

29.

(b)

1 C– L

1 – L C

30.

1 – L C

(d) none of these

31.

569

ELECTROMAGNETIC INDUCTION & AC CURRENT 32.

In the circuit shown switch k2 is open and switch k1 is closed at t = 0. At time t = t0 switch k1 is opened and switch k2 is simultaneously closed. The variation of inductor current with time is

k2 L

(a) maximum charge in the capacitor can be 6C (b) maximum charge in the capacitor can be 8C (c) charge in the capacitor will be maximum after time 2 sin–1 (2/3) sec (d) None of these 34. In the circuit shown, the key (K) is closed at t = 0, the current through the key at the instant t = 10–3 ln 2, is

20V

i (a)

L = 5mH

Et0 L

K 6 t0

i (b)

C = 0.1 mF

t (a) 2A (b) 3.5A (c) 2.5A (d) zero 35. Find the current passing through battery immediately after key (K) is closed. It is given that initially all the capacitors are uncharged. (given that R = 6 and C = 4µF)

Et0 L t0

i (c)

Et0 L

E = 5V (d)

Et0 L

R R

t 33. In an L-C circuit shown in the figure, C = 1F, L = 4H. At time t = 0, charge in the capacitor is 4C and it is decreasing at a rate of

5 C/s. Choose the correct statements. q + – C

MARK YOUR RESPONSE

32.

33.

(a) 1 A (c) 3 A

(b) 5 A (d) 2 A

36. The magnetic flux in a closed circuit of resistance 10 varies with time as = (2t –4t2 +1). The current in the loop will change its direction after a time of (a) 0.25 sec (b) 0.5 sec (c) 1 sec (d) None

34.

35.

36.

IIT-JEE PHYSICS Challenger

578

PASSAGE-3

10. Ratio of

Two capacitors of capacitance C and 3C are charged to potential difference V0 and 2V0 respectively and connected to an inductor of inductance L as shown in the figure. Initially the current in the inductor is zero. Now the switch S is closed.

V0 +

2V0 + –

–

3C L

(b) V0

3C L

2V0

(d) V0

CV0 4

(c)

5CV0 4

(b)

R then the time varying expression of the current in 2 the capacitor is

3C L

C L

(a)

E RC e R

(c)

2 E 5RC e 3R

(b)

2E 5RC e 5R

(d)

3E 5 RC e 5R

3CV0 4

PASSAGE-5

(d) None of these

Potential difference across capacitor of capacitance 3C when the current in the circuit is maximum (a)

CV0 4

(b)

(c)

5CV0 4

(d) None of these

3CV0 4

Faraday’s Law says that the change of magnetic flux with any closed loop produces induced emf and current in the loop. Magnitude of current produced i =

e = R

d 1 . The cause ÷ dt R

of induced current is the rate of change of flux. Then using this concept in the situation where a wire frame of area 4 ×10–4 m2 and resistance 10 is suspended from a thread 2 m long in a vertical plane. Now a horizontal magnetic field of 1 tesla is introduced and the arrangement is made to oscillate from its equilibrium position as shown in the figure.

PASSAGE-4

The capacitor is charged by closing the switch S. The switch is then opened and the capacitor is allowed to discharge. t1 and t2 be the time constant of the circuit during the charging and discharging of the capacitor. Take R1 = R2 = R3 = R

l B

S E

MARK YOUR RESPONSE

1 3

Potential difference across capacitor of capacitance C when the current in the circuit is maximum. (a)

1 6

1 2 (d) 2 3 12. If the battery as shown in the figure has an internal resistance

The maximum current in the inductor is

3V0 2

(c)

(a)

(a) 2/3 (b) 1/1 (c) 1/3 (d) 3/1 What fraction of the heat generated during discharging is lost in R1? (a)

11.

12.

13.

13. The magnetic flux with loop will be maximum in the position (a) P (b) Q (c) between P and Q (d) = 45°

10.

11.

579

ELECTROMAGNETIC INDUCTION & AC CURRENT 14.

For small angle , such that maximum displacement of x0 = 1 cm and

(c)

dP/dt

(d) dP/dt

÷ ; find the induced e.m.f as a

function of time ‘t’ when t = 0 is equilibrium position (a) 5 ×10 –9 sin (10t)

(b)

5 5 10

sin (2 5t)

(d) 10 –6 sin (5t) 5sin ( 10t) 15. The maximum induced current in the loop is (c)

(a) 5 × 10 – 8A (c)

(b)

5 5 10 10 A

PASSAGE-7

2 5 10 9 A

(d) 10 –7A

PASSAGE-6 A resistor and inductor are connected in series through a battery. The switch S is closed at time t = 0.

The conducting connector of mass m and length L can freely slide on horizontal long conducting parallel rails connected by capacitor C at one end as shown in figure. A non conducting light spring (spring constant K) is connected to the connector ab and is in a relaxed state. The whole system is placed in uniform magnetic field of strength B directed into the plane of rails (figure). Now at time t = 0, connector is suddenly given velocity V0 in rightward direction. If resistance and self inductance of circuit is negligible then

a B V

spring L V0

16.

The rate of Joule heating (P) in resistor varies with the time ‘t’ is best represented by the graph. P P

(a)

b x=0 19. The magnitude of acceleration of connector as a function of x is

(b)

(a)

P (c)

18.

(d)

(b) dP/dt

dP/dt

Kx m + B 2 L2C

Kx m B 2 L2C

(d) None of these

m + B 2 L2 C K

(a)

m B 2 L2 C K

(b) V0

(c)

m K

(d) None of these

21. The rod will execute (after projecting it at t = 0) (a) SHM (b) harmonic motion but not SHM (c) the rod will come at rest at certain position then afterward it will not move (d) None of these

MARK YOUR RESPONSE

(b)

20. The maximum compression in the spring is

t t What is the magnitude of current flowing when the rate of increase of magnetic energy in the inductor is maximum (a) I = V/R (b) I = V/2R (c) I = V/4R (d) I = (V/R) ln 2 Which of the following graph best represent rate of change of power dissipated in resistor as a function of time. (a)

Kx m

14.

15.

16.

19.

20.

21.

17.

18.

IIT-JEE PHYSICS Challenger

580

25. Which plot does correctly represents rms current against frequency ?

PASSAGE-8 In a series L-R circuit, connected with a sinusoidal as source, the maximum potential difference across L and R are respectively 3 volts and 4 volts.

22. At an instant the potential difference across resistor is 2 volts. The potential difference in volt, across the inductor at the same instant will be (a) 3 cos 30° (c) 3 cos 45°

Tweeter (a)

(b) 3 cos 60° (d) None of these

Tweeter I

23. At the same instant, the magnitude of the potential difference in volt, across the ac source may be (a)

4+3 3

Woofer

3 2

(b)

4+3 3 2

(d)

Woofer (b)

f Woofer

3 2

I Tweeter

24. If the current at this instant is decreasing the magnitude of potential difference at that instant across the ac source is (a) increasing (c) constant

(c)

(b) decreasing (d) cannot be said PASSAGE-9

Tweeter

A loudspeaker system uses alternating current to amplify sound of certain frequencies. It consists of 2 speakers. Tweeter-which has smaller diameter produces high frequency sounds. Woffer-which has larger diameter produces low frequency sound. For purpose of circuit analysis, we can take both speakers to be of equal resistance R. The equivalent circuit is shown in the figure. The 2 speakers are connected to the amplifier via capacitance and inductance respectively. The capacitor in tweeter branch blocks the low frequency sound but passes the high frequency. The inductor in woffer branch does the opposite.

(d)

f 26. What is the frequency which is sounded loudly in equal amounts by both speakers ?

Tweeter R L

~ V0sin( t)

MARK YOUR RESPONSE

22. 27.

(a)

1 2

(c)

1 2

1 LC

1 LC R2 2

(b)

4R2

1 2

1 LC

1 (d)

27. For a combination of L, R and C the current in woofer and tweeter are always found to have a phase difference of /2. What is the relation between L, R and C ?

Woofer

23.

24.

(a) L = 2R2C

(b)

L = 2 R 2C

(d)

25.

R 2C

26.

IIT-JEE PHYSICS Challenger

592

1 2 3 4 5 6 7 8 9 10

(a) (c) (b) (a) (b) (a) (c) (d) (d) (c)

11 12 13 14 15 16 17 18 19 20

(d) (c) (c) (b) (b) (b) (c) (c) (a) (c)

21 22 23 24 25 26 27 28 29 30

(b) (a) (b) (c) (a) (c) (b) (a) (a) (a)

31 32 33 34 35 36 37 38 39 40

(a) (a) (a) (c) (a) (a) (c) (a) (c) (b)

41 42 43 44 45 46 47 48 49 50

(b) (a) (c) (b) (b) (c) (c) (c) (d) (c)

51 52 53 54 55 56 57 58 59 60

(b) (b) (a) (a) (c) (d) (c) (d) (d) (d)

61 62 63 64 65 66 67 68 69 70

(d) (a) (a) (d) (d) (d) (b) (d) (a) (a)

1 2 3 4

(a) (a) (a) (b)

5 6 7 8

9 10 11 12

13 14 15 16

(a) (b) (c) (c)

17 18 19 20

(b) (d) (c) (b)

21 22 23 24

(a) (a) (b) (d)

25 26 27

(b) (d) (c)

1 2

(a) (a)

(b, c) (b, c, d) (a, b, c)

1 2 3

(d) (b)

3 4

4 5 6

(a) (c)

5 6

(b, c) (b, d) (a, c)

7 8 9

7 8

(a, b, c, d) (a, b, c, d) (a, d)

(d) (d)

9 10

10 11 12

(a, b) (a, c, d) (a, b, c)

(d)

13 (a, b, d) 16 (a, c) 14 (a, b, c) 15 (a, c)

A-r; B-p; C-s; D-p, q

A-q; B-r, p; C-r, s; D-q, r

3. 5. 7.

A-p, q; B-r, s; C-r, s; D-p, q A-s; B-q; C-p; D-r A-p, r, s, t; B-q, r, s; C-p, r, s, t; D-p, r, s, t

4. 6. 8.

A-q; B-p; C-s; D-r A-p, r, s; B-q; C-t; D-p, r, s A-q; B-r, s; C-s; D-p, q, r

1 5

1 25

2 6

5 4

0.01

71 72 73 74 75

(a) (a) (b) (c) (b)

593

ELECTROMAGNETIC INDUCTION & AC CURRENT

(a)

According to given problem, V I= = 2 Z [R

and

I = 2 [R2

....... (1)

(1/ C ) 2 ]1/ 2

E0 (b) We know that, Z = I 0 Given, E0 = 220 and I0 = 10

....... (2)

(3 / C ) 2 ]1/ 2

LM FG IJ OP = N6 H 6KQ 3

Substituting the value of I from eq. (1) in (2),

4 R2 +

i.e.,

1 C2

1 C

2÷

= R2 +

9 C

(c)

1/ 2

3 5

L' =

L 1.8 = 10 –4 2 2

p 2 A = t c 4

L' R'

L 1.8 10 –4 = = 30 s R 6

R' R' R '+ R ' So, maximum current drawn from battery R" =

3 10

= 1.2 x 10–7 newton (a) From Kirchoff’s current law,

i3 = i1 + i2 = 3sin t 4sin ( t + 90°)

32 + 42 + 2(3)(4) cos 90° sin ( t

)

4sin 90° 4 = 3 + 4cos 90° 3

E 12 = = 8A R" 6 / 9

5 [sin ( t

)]

3 sin t 5

4 cos t 5

...... (1)

i3 = 5 sin ( t + 53°)

R 2

0.9 10 –4 H

Now, effective resistance when both coils are connected in parallel

2 12 104 1.5 10

where tan =

2 U 2 = p = .A. t c c

Now, time constant

perfectly reflecting surface

R 6 = =3 2 2 Inductance of each part

p for t

(b) According to Newton’s Second law, F =

cos =

R' =

Resultant voltage = 200 volt Since V1 and V3 are 180° out of phase, the resultant voltage is equal to V2. V2 = 200 volt

220

cos 3 2 2 2 2 = 550 W (a) When coil is broken into two identical parts, then resistance of each part

3 = R2 5

3 2 R ÷ X (1/ C ) 5 = = So that R R R

220 = 22 ohm 10

i dt

rms value =

5 2

mean value =

T1 T2

dt T1

IIT-JEE PHYSICS Challenger

594

initial value is not given hence the mean value will be difference for various time intervals. If voltage applied is V = Vm sin t then i given by eq. (1) indicates that it is ahead of V by where 0 < < 90° which indicates that the circuit contains R and C. (d) When capacitance is removed

L or L = 100 tan 60° R when inductance is removed tan =

(c)

Here, X L

(c)

P 0.4

4 r2 From electromagnetic theory

Vrms 200 = = 4A Z 50

X C2

R12

Erms

i2rms =

1 . C

= cu = c 0 E 2 RMS

Erms

i1rms =

1 ÷ C

= 1 if

given by =

Z = R 2 + X L2 = 302 + 402 = 50

10.

The intensity of light at a distance r from source is

R = 30

I rms =

R2 13.

1 ÷ the circuit will have C LC resistance nature. Hence (b) is false. Power factor

....... (2)

2 fL

cos =

1 = 100 tan 60° C z = R = 100

Power P = I 2 R = 4 100 = 400W (d)

Hence (a) is false. Also if circuit has inductive nature the current will log behind voltage. Hence (d) is also false.

....... (1)

I = v / R = 200 /100 = 2A

The circuit will have inductive nature if

1 ( C ) ( R)

tan = or

12.

= 13A

X L2 + R22

4 r2

ERMS =

tan

= 10

2 ERMS 0c 7

0 cP 2

ERMS =

3 108

4 r

250

= 4.8 V/m 18 14. (b) L = 0.7H, R = 220 , E0 = 220V, v = 50 Hz. This is an L-R circuit Phase difference, or

130 = 10A 13

XL R

L R

[XL= 2 vL = 2 ×

2 vL R

22 × 50 × 0.7 = 220 ] 7

220 = 1 or, = 45° 220 Wattless component of current =

11.

Power dissipated = i21rms R1 + i22rms R2 = 102 × 5 + 132 + 6 = power delivered by battery = 500 + 169 × 6 = 1514 watt (d) We know that,

at resonance L cos =

R =1 R

L 1 . C

1 ÷ C

cos =

= I0 sin =

15.

(b)

1 E0 . 2 Z

220 X L2

1 R2

1 = 0.5 A 2

Q = Q0 e t/

et /

= Q0 /

220 220 2

2202

A fish looking up through the water sees the outside world contained in a circular horizon. If the refractive index of water is 4/3 and the fish is 12 cm below the surface, the radius of this circle (in cm) is – (a)

(b)

36 5

r and r' denote the angles inside an equilateral prism, as usual, in degrees.

4 5 r

(c) 36 7 (d) 36 / 7 A mango tree is at the bank of river and one of the branch of tree extends over the river. A tortoise lives in river. A mango falls just above the tortoise. The acceleration of the mango falling from tree appearing to the tortoise is (Refractive index of water is 4/3 and the tortoise is stationary) (a) g (b) 3g/4 (c) 4g/3 (d) None of these The given lens is broken into four parts and rearranged as shown. If the initial focal length is f then after rearrangement the equivalent focal length is –

Consider that during some time interval from t = 0 to t = t, r' varies with time as r' = 10 + t2. During this time r will vary as (assume that r and r' are in degree)

(a) 50 – t2

(b) 50 + t2

(d) 60 + t2

A triangular prism of glass is inside water. A ray, incident normally, on one of the faces, is totally reflected from face BC. Then the minimum refractive index of glass is –

1 2

A 3 4

in air (a) f (c) f / 4

MARK YOUR RESPONSE

(b) f / 2 (d) 4 f

(a)

3 2

(b)

5 3

(c)

2 2 5

(d)

4 2 3

IIT-JEE PHYSICS Challenger

612

A ray of light is incident on a thick slab of glass (thickness t) as shown below. The emergent ray is parallel to the incident ray but displaced sideways by a distance d. If the angles are small then d is: t

(a)

(c)

(cos i + cos B )2 sin B

(b)

cos i sin B cos B

(d)

sin i cos B sin B

10. On the earth a child focuses the sun light on a screen, using a convex lens of focal length f and radius of aperture r. What is the intensity of light at the focus point (S = solar constant)

d (a) t (1–i/r)

sin i + cos B sin B

1 2 2

(b) r t (1–i/r)

r r (d) t 1 ÷ ÷ i i An object is moving with speed v0 towards a spherical mirror with radius of curvature R, along the central axis of the mirror. The speed of the image with respect to the mirror is (U is the distance of the object from mirror at any given time t):

(a)

R ÷ v0 R 2U

(b)

focus point

R 2 ÷ v0 U 2R R

÷ v0 (d)

(a)

Sr 2 d 2 2R2 f 2

(b)

Sr 2 d 2 4R2 f 2

(c)

Sr 2 d 2 R2 f 2

(d)

2Sr 2 d 2 R2 f 2

R 2 ÷ v0 2U 2

A point source of light is placed at a depth of h below the surface of water of refractive index . A floating opaque disc is placed on the surface of water so that light from the sources is not visible from the surface. The minimum diameter of the disc is (a) 2h/ ( 2 – 1)1/2 (b) 2h ( 2 – 1)1/2 2 1/2 (c) h/ [2( – 1) ] (d) h( 2 – 1)1/2 A ray of light PQ is incident at angle i on a prism face. (See figure) After 2 refractions it leaves the prism at a grazing angle.

11.

In a thick glass slab of thickness and refractive index n1 a cuboidal cavity of thickness m is carved as shown in the figure and is filled with liquid of R.I. n2 (n1 > n2). The ratio of / m , so that shift produced by this slab is zero when an observer A observes an object B with paraxial rays is A air n1 (1)

air

P If is the refractive index and B is the refractive index and B is the refracting angle of the prism, the refracting index ( ) is:

MARK YOUR RESPONSE

6. 11.

(2)

(a)

n1 n2 n2 1

(b)

n1 n2 n2 ( n1 1)

(c)

n1 n2 n1 1

(d)

n1 n2 n1 ( n2 1)

10.

629

RAY OPTICS

A ray of light is incident at an angle of 60° on one face of prism which has an angle of 30°. The ray emerging out of the prism makes an angle of 30° with the incident ray. Calculate the refractive index of the material of the prism. An object is placed 21 cm in front of a concave mirror of radius of curvature 10 cm. A glass slab of thickness 3 cm and refractive index 1.5 is then placed close to the mirror in the space between the object and the mirror. Find the image distance (in cm) of the final image formed. (You may take the distance of the near surface of the slab from the mirror to be 1 cm). A telescope has an objective of focal length 50 cm and an eye piece of focal length 5 cm. The least distance of distinct vision is 25 cm. The telescope is focussed for distinct vision on a scale 200 cm away from the objective. Calculate the separation (in cm) between the objective and the eye-piece. The convex surface of a thin concavo-convex lens of glass of refractive index 1.5 has a radius of curvature 20 cm. The concave surface has a radius of curvature 60 cm. The convex side is silvered and placed on a horizontal surface.

At what distance (in metre) should a pin be placed on the optic axis such that its image is formed at the same place? Calculate your answer in cm. A plano convex lens has a thickness of 4 cm . When placed on a horizontal table, with the curved surface in contact with it, the apparent depth of the bottom most point of the lens is found to be 3 cm. If the lens is inverted such that the plane face is in contact with the table, the apparent depth of the centre of the plane face is found to be 25/8 cm. Find the focal length (in cm) of the lens.

Light is incident at an angle on one planar end of a transparent cylindrical rod of refractive index µ. Determine the least value of µ so that the light entering the rod does not emerge from the curved surface of rod irrespective of the value of . µ

A thin plano-convex lens of focal length f is split into two halves: one of the halves is shifted along the optical axis . The separation between object and image planes is 1.8 m. The magnification of the image formed by one of the halflenses is 2. Find the focal-length of the lens (in m).

Two plane mirrors A and B are aligned parallel to each other, as shown in the figure. A light ray is incident at an angle of 30° at a point just inside one end of A. The plane of incidence coincides with plane of the figure. Find the maximum number of times the ray undergoes reflections (including the first one) before it emerges out.

2 3m B 0.2m

30 °

MARK YOUR RESPONSE

IIT-JEE PHYSICS Challenger

630

1 2 3 4 5 6

(d) (c) (b) (a) (d) (c)

(b) (a) (a) (c) (b) (d)

7 8 9 10 11 12

13 (a) 14 (b) 15 (c) 16 (d) 17 (c) 18 (a)

19 (a) 20 (a) 21 (b) 22 (c) 23 (a) 24 (d)

(d) (b) (d) (c) (b) (c)

25 26 27 28 29 30

31 (c) 32 (a) 33 (d) 34 (c) 35 (d) 36 (a)

37 (a) 38 (d) 39 (c) 40 (a) 41 (a) 42 (a)

43 44 45 46 47 48

(c)

(d)

(a)

(d)

(b)

(d)

(a)

(b)

(a)

(d)

(b)

(a)

1 2

(a, b, c, d) (a, b, c, d)

(d)

3 4

(d)

(c, d) (a, b, c, d)

5 6

(d)

(c)

(a, c) (a, b, c, d)

7 8

49 50 51 52 53 54

(a) (b) (a) (a) (a) (b)

(d)

(a, b, c) 9 (a, c) (b, d) 10 (a, b, c)

A-s; B-q; C-r; D-p

A-q; B-p; C-s; D-r

A-q; B-r; C-q; D-p

A-s; B-p; C-q; D-r

A-p, q, s; B-p, s; C-p; D-r

A-q, r; B-s; C-s; D-q, r

A-r, B-s; C-p; D-q

A-p, q; B-p, r, s; C-p, r; D-p, r

A-q, r, t; B-q, r; C-q, s; D-q, s, t

10.

A-s; B-q; C-p; D-r

11.

A-p, q, s, t; B-r, t; C-p, q, s, t; D-p, q, t

12.

A-p, q, r, s, t; B-q, s; C-q, s; D-p, q, r, s, t

13.

A-q, r; B-p, s; C-q, r; D-p, q, r

1.73

7.67

70.8

1.41

0.4

55 56 57 58 59 60

(a)

(a) (b) (d) (d) (c) (b)

61 62 63 64

(a) (c) (a) (d)

631

RAY OPTICS

(d)

sin ic =

r 2 + h2 Using h = 12 cm, µ = 4/3

For small angle i, angle r is also small, and so d = t (i – r) = t i (1 – r/i) (b) For concave mirror

2 1 1 = + R v u

ic ic

1 v

Fish

We get

(c)

x 1

xrel

RU R 2U In spherical mirror, image velocity

x vi =

d 2 xrel

d 2x

2 g arel dt 2 dt 2 (b) Cutting a lens in transverse direction doubles their focal length i.e. 2f. Using the formula of equivalent focal length,

v2 u

RU R 2U

v0 U2

1 1 1 1 1 = + + + f f1 f2 f3 f4 4.

2 R 2U = R UR

or v =

cm.

xrel

1 U

R v0 R 2U (a) The figure shows incidence from water at critical angle c for the limiting case. disc

air water

We get equivalent focal length as f/2. (a) In a prism : r + r' = A r = A – r' r = 60° – (10 + t2) = 50 – t2

light source 5.

45°

(d)

45° 9. For T.I.R. 45° > C

sin 45° > sin C

1 2

4/3 n

Now, sin c = 1/ so that tan c =1/ ( 2 –1)1/2 which is also equal to r/h where r is the radius of the disc. Therefore, diameter of the disc is 2r = 2h tan c. (a) A ray of light incident on face AB will just suffer internal reflection at the other face BC. If it gets incident on face BC at critical C angle for the material of the prism. If critical angle for material is C, then

sin C =

1 …(i)

4 2 3 (c) The lateral displacement is given by n>

t sin (i r ) d= cos r

r1r2

Q C

647

WAVE OPTICS

Two sources S1 and S2 emitting coherent light waves of wavelength in the same phase are situated as shown. The distance OP, so that the light intensity detected at P is equal to that at O is

(b)

(a)

P 2 S1

O I0

(d)

(c)

Screen

(a)

D 2

(b)

D/2

(c)

D 3

(d)

D/ 3

In Young’s double-slit experiment, the separation between the slits is halved and the distance between the slits and the screen is doubled. The fringe width is (a) unchanged.

(b) halved.

(d) quadrupled

Two coherent light sources each of wavelength are separated by a distance 3 . The maximum number of minima formed on line AB which runs from – to + is (a) 2

(b) 4

(d) 8

Q A

In a YDSE experiment if a slab whose refraction index can be varied is placed in front of one of the slits then the variation of resultant intensity at mid-point of screen with ‘ ’ will be best represented by ( 1). [Assume slits of equal width and there is no absorption by slab; mid point of screen is the point where waves interfere with zero phase difference in absence of slab]

MARK YOUR RESPONSE

Two coherent monochromatic light beams of intensities I and 4 I are superposed. The maximum and minimum possible intensities in the resulting beam are (a) 5I and I (b) 5I and 3I (c) 9I and I (d) 9I and 3I In Fresnel's biprism experiment the width of 10 fringes is 2cm which are formed at a distance of two 2 meter from the slit. If the wavelength of light is 5100 Å then the distance between two coherent sources will be (a) 5.1 × 10–4 m (b) 5.1 × 104 cm. –4 (c) 5.1 × 10 mm (d) 10.1 × 10–4 cm A beam of light of wave length 600 nm from a distance source falls on a single slit 1 mm wide and a resulting diffraction pattern is observed on a screen 2m away. The distance between the first dark fringes on either side of central bright fringe is (a) 1.2 cm (b) 1.2 mm (c) 2.4 cm (d) 2.4 mm When a plastic thin film of refractive index 1.45 is placed in the path of one of the interfering waves then the central fringe is displaced through width of five fringes. The thickness of the film, if the wavelength of light is 05890Å, will be (a) 6.544 × 10–4 cm (b) 6.544 × 10–4 m –4 (c) 6.54 × 10 cm (d) 6.5 × 10–4 cm

IIT-JEE PHYSICS Challenger

648

The equation of two light waves are y1 = 6 cos t, y2 = 8 cos ( t + f ).

(a)

The ratio of maximum to minimum intensities produced by the superposition of these waves will be (a) 49 : 1 (b) 1 : 49 (c) 1 : 7 (d) 7 : 1 10. In an interference arrangement similar to Young’s doubleslit experiment, the slits S1 and S2 are illuminated with coherent microwave sources, each of frequency 106 Hz. The sources are synchronized to have zero phase difference. The slits are separated by a distance d = 150.0 m. The intensity I ( ) is measured as a function of , where is defined as shown. If I0 is the maximum intensity, then I ( ) for 0 90° is given by S1 d/2

(b)

(c) 2 (d) 14. Figure shows two coherent sources S1 – S2 vibrating in same phase. AB is an irregular wire lying at a far distance

= 10 3 . d BOA = 0.12 ° . How many bright spots will be seen on the wire, including points A and B. from the sources S1 and S2. Let

S1 d

O S2

(a) 2 (b) 3 (c) 4 (d) more than 4 15. Yellow light is used in a single slit diffraction experiment with slit width of 0.6 mm. If yellow light is replaced by X– rays, then the observed pattern will reveal, (a) that the central maximum is narrower (b) more number of fringes

d/2

(d) no diffraction pattern (a)

I( )

I0 / 2 for

= 30°

(b)

I( )

I0 / 4 for

= 90°

(c)

I( )

I0 for

(d)

I ( ) is constant for all values of .

= 0°

11.

Interference fringes were produced using white light in a double slit arrangement. When a mica sheet of uniform thickness of refractive index 1.6 (relative to air) is placed in the path of light from one of the slits, the central fringe moves through some a distance. This distance is equal to the width of 30 interfernece bands if light of wavelength 4800 Å is used. The thickness (in 1 µm) of mica is (a) 9 0 (b) 12 (c) 14 (d) 24 12. In Young's experiment the wavelength of red light is 7.5 × 10–5 cm and that of blue light 5.0 × 10–5 cm. The value of n for which (n +1)th the blue bright band coincides with nth red band is (a) 8 (b) 4 (c) 2 (d) 1 13. Consider Fraunhoffer diffraction pattern obtained with a single slit illuminated at normal incidence. At the angular position of the first diffraction minimum the phase difference (in radians) between the wavelets from the opposite edges of the slit is

MARK YOUR RESPONSE

16. Two slits separated by a distance of 1 mm are illuminated with red light of wavelength 6.5 × 10–7 m. The interference fringes are observed on a screen placed 1 m from the silts. The distance of the third dark fringe from the central fringe will be equal to : (a) 0.65 mm (b) 1.30 mm (c) 1.62 mm (d) 1.95 mm 17. A diffraction grating 1 cm wide has 1000 lines and is used to third order. Find the difference between the diffraction angles for 400 nm and 600 nm light. (a) 2° (b) 3.4° (c) 4° (d) 6° 18. A thin slice is cut out of a glass cylinder along a plane parallel to its axis. The slice is placed on a flat glass plate as shown in figure. The observed interference fringes from this combination shall be

(a) straight (b) circular (c) equally spaced (d) having fringe spacing which increases as we go outwards

10.

11.

12.

13.

14.

15.

16.

17.

18.

649

WAVE OPTICS 19.

A broad source of light (I = 680 nm) illuminates normally two glass plates 120 mm long that touch at one end and are separated by a wire 0.034 mm in diameter at the other end. The total number of bright fringes that appear over the 120 mm distance is –

incident light 0.034mm

t 120mm

20.

(a) 50 (b) 100 (c) 200 (d) 400 In the figure shown in a YDSE, a parallel beam of light is incident on the slits from a medium of refractive index n1. The wavelength of light in this medium is 1. A transparent slab of thickness t and refractive index is put infront of one slit. The medium between the screen and the plane of the slits is n2. The phase difference between the light waves reaching point O (symmetrical, relative to the slit) is

22. A beam of light consisting of two wavelength 6500Aº & 5200Aº is used to obtain interference fringes in a young's double slit experiment. The distance between the slits is 2.0 mm and the distance between the plane of the slits and the screen is 120 cm. What is the least distance from the central maximum where the bright fringes due to both the wave length coincide ?

2 n1

( n3

n2 ) t

(b)

( n3

1÷ t

(d)

2 n1

( n3

n2 ) t

(a) 1.0

(b) 1.5

(d) 1.25

(a) 2I

(b) 4I

(d) 7I

when maximum light is transmitted if ( n1 < n2 < n3 ) is (a)

n2 ) t

In a double slit experiment instead of taking slits of equal widths, one slit is made twice as wide as the other. Then, in the interference pattern (a) the intensities of both the maxima and the minima increase (b) the intensity of the maxima increases and the minima has zero intensity (c) the intensity of the maxima decreases and that of the minima increases (d) the intensity of the maxima decreases and the minima has zero intensity

MARK YOUR RESPONSE

(d) 0.16 cm

25. From a medium of index of refraction n1, monochromatic light of wavelength is incident normally on a thin film of uniform thickness L (where L > 0.1 ) and index of refraction n2. The light transmitted by the film travels into a medium with refractive index n3. The value of minimum film thickness

2 n1 n3 n2 1 n2

24. Two beams of light having intensities I and 4I interfere to produce a fringe pattern on a screen. The phase difference between the beams is /2 at point A and at point B. Then the difference between the resultant intensities at A and B is

(a)

(b) 0.152 cm

23. A thin sheet of mica ( 12 × 10–7 m thick) is placed in the path of one of the interfering beams in a biprism arrangement. It is found that the central bright band shifts a distance equal to the width of a bright fringe. The refractive index of mica (Given = 6 × 10–7 m) is

n3 O

(a) 0.156 cm

(c)

n1 2 n2

(b)

(d)

4n2

n1 4 n2

2n2

26. Interference fringes were produced in Young's double slit experiment using light of wave length 5000 Å. When a film of material 2.5 × 10–3 cm thick was placed over one of the slits, the fringe pattern shifted by a distance equal to 20 fringe width. The refractive index of the material of the film is -

19.

20.

21.

24.

25.

26.

(a) 1.25

(b) 1.33

(d) 1.5

22.

23.

IIT-JEE PHYSICS Challenger

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27. In a Young’s double slit experiment, 12 fringes are observed to be formed in a certain segment of the screen when light of wavelength 600 nm is used. If the wavelength of light is changed to 400 nm, number of fringes observed in the same segment of the screen is given by (a) 12 (b) 18 (c) 24 (d) 30 28. A plastic sheet (refractive index = 1.6) covers one slit of a double slit arrangement meant for the Young's experiment. When the double slit is illuminated by monochromatic light (wavelength in air = 6600 Å), the centre of the screen appears dark rather than bright. The minimum thickness of the plastic sheet to be used for this to happen is: (a) 3300 Å (b) 6600 Å (c) 2062 Å (d) 5500 Å 29. The intensity of a point source of light, S, placed at a distance d in front of a screen A, is I0 at the center of the screen. Find the light intensity at the center of the screen if a completely reflecting plane mirror M is placed at a distance d behind the source, as shown in figure.

A d

s ×

(a)

27 I0 9

(b)

25 I0 9

(c)

17 I0 9

(d)

10 I0 9

O S2 D

(a) Larger the wavelength, larger will be the fringe width. (b) If white light is used, violet colour forms its first maxima closest to the central maxima. (c) The central maxima of all wavelength coincide. (d) All of the above 32. Monochromatic light of wavelength 400 nm and 560 nm are incident simultaneously and normally on double slits apparatus whose slits separation is 0.1 mm and screen distance is 1m. Distance between areas of total darkness will be (a) 4 mm (b) 5.6 mm (c) 14 mm (d) 28 mm 33. Consider the YDSE arrangement shown in figure. If d = 10 then position of 8th maxima is

D >>d

(b) 2 /3

(c) /3 (d) 31. The figure shows a schematic diagram for Young’s double slit experiment. Given d << , d << D, / D << 1. Which of the following is a right statement about the wavelength of light used?

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30. In the ideal double-slit experiment, when a glass-plate (refractive index 1.5) of thickness t is introduced in the path of one of the interfering beams (wave-lenght ), the intensity at the position where the central maximum occurred previously remains unchanged. The minimum thickness of the glass-plate is (a) 2

(a) y =

D 10

(b) y =

D 3

4 D 5

(d) y =

4D 3

34. In a Young’s double slit experiment, the fringes are displaced by a distance x when a glass plate of refractive index 1.5 is introduced in the path of one of the beams. When this plate is replaced by another plate of same thickness, the shift of fringes is (3/2) x. The refractive index of second plate is (a) 1.75 (b) 1.50 (c) 1.25 (d) 1.00

27.

28.

29.

32.

33.

34.

30.

31.

651

WAVE OPTICS 35. In Young’s double slit experiment intensity at a point is (1/4) of the maximum intensity. Angular position of this point is (a) sin–1( /d) (b) sin–1( /2d) –1 (c) sin ( /3d) (d) sin–1( /4d) 36. A Young’s double slit experiment is conducted in a liquid of refractive index 1 and a glass plate of thickness t and refractive index 2 is placed in path of one slit. The magnitude of the optical path difference at centre of screen will be 2

(a)

1÷ t

37.

(b)

(a) most polluting gases and dust particles in the air are bluish in colour and lend their colour to that of the sky. (b) air molecules absorb red light more efficiently than they do blue light because of their electron orbitals. (c) tiny particles in the air are more efficient at scattering short wavelength light than they are at scattering long wavelength light. (d) air molecules absorb blue light more efficiently than they do red light because of their electron orbitals

1÷ t

(c) | ( 2 – 1) t | (d) | ( 2 – 1) t | A thin convex lens of focal length f = 0.6m is a cut into two unequal parts L1 and L2. One part is shifted along the cutting plane axis as shown in figure. A monochromatic line source S, perpendicular to the plane of paper, emitting light of wavelength = 600nm, is placed on the cutting plane axis. A screen with slits where the images of S is formed by these two pieces of the lens separately is placed perpendicular to the optical axis from the source at 4.9m. There is an another screen placed at distance 0.6m normal to optical axis where fringes are observed due to interference of the light passes through the holes. Find the position of central maximum from P. [Dotted line represent the principal axis of lens L1] Screen

40. At night approximately 500 photons per second must enter an unaided human eye for an object to be seen. A light bulb emits about 5.00 × 1018 photons per second uniformly in all directions. The radius of the pupil of the eye is about 4×10–3 meters. What is the maximum distance from which the bulb could be seen ? (a) 2.0 × 104 m (c) 2.0 × 102 m 41.

6mm P

0.6m

(a) 24.5mm. (b) 12.5mm. (c) 18.5mm. (d) 6.5mm. A physics professor wants to find the diameter of a human hair by placing it between two flat glass plates, illuminating the plates with light of vaccum wavelength = 552nm and counting the number of bright fringes produced along the plates. The Professor find 125 bright fringes between the edge of the plates and the hair. What is the diameter of the hair? (a) 525

10–9m

(b) 344

10–5m

(d) none of the above

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(a) 1.2 10–7m

(b) 650 10–9m

(d) 650 105m

(a) 0.3° (c) 0.8° 43.

38.

A certain region of a soap bubble reflects red light of

42. A double slit arrangement produces fringes for = 5890 Å that are 0.4° apart. What is the angular width if the entire arrangement is immersed in water ? (µ w = 4/3)

4.9m

(b) 2.0 × 105 m (d) 5.0 × 103 m

vaccum wavelength = 650nm. What is the minimum thickness that this region of the soap bubble could have? Take the index of reflection of the soap film to be 1.41.

Screen

39. The sky is blue because

Screen S is illuminated by two points source A and B. Another source C sends a parallel beam of light towards the point P on the screen. Line AP is normal to the screen and line AP, BP and CP are in one plane. The distance AP, BP and CP are 3m, 1.5m and 1.5m respectively. The radiant powers of source A and B are 90 and 180W respectively and the beam from C is of the intensity 20W/m2. Calculate the intensity at P on the screen. (a) 14

10–3m

(b) 2.3° (d) 1.3°

W m

(b) 10

m2 W

(d) 14

35.

36.

37.

38.

40.

41.

42.

43.

W m2 m2 W

39.

IIT-JEE PHYSICS Challenger

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////////////////////////////////////////////////

44. A ray of light travels through a slab as shown. x = 1m

x=0

1.2 +

The refractive index of the material of the slab varies as µ =

x2 , where 0 x 1m. What is the equivalent optical 2

path of the glass slab ? (a) 1.212 m (c) 0.123 m 45.

(b) 1.367 m (d) 2.124 m

White light used to illuminate the two slits in Young’s double slit experiment. The separation between the slits is d and the distance between the screen and the slit is D (>>d). At a point on the screen in front of one of the slits, certain wavelengths are missing. The missing wavelengths are

(a)

d2 ( 2n + 1)D

(b)

(c)

d2 (n + 1)D

(d)

(2n + 1)d 2 D

(n + 1)D d2

46. Young’s double slit experiment is made in a liquid. The 10th bright fringe lies in liquid where 6th dark fringe lies in vacuum. The refractive index of the liquid is approximately (a) 1.8 (c) 1.3

(b) 1.5 (d) 1.6

47. The wavelength of visible light in air can be (a) 5 nm (c) 50 µm

(b) 5 mm (d) 500 nm

48. A double slit, S1 – S2 is illuminated by a light source S emitting light of wavelength . The slits are separated by a distance d. A plane mirror is placed at a distance D in front of the slits and a screen is placed at a distance 2D behind the slits. The screen receives light reflected only by the plane mirror. The fringe-width of the interference pattern on the screen is

MARK YOUR RESPONSE

(a)

D d

(b)

2D d

(c)

3D d

(d)

4D d

49. A possible means for making an airplane invisible to radar is to coat the plane with an antireflective polymer. Radar waves have a wavelength of 3.00 cm and the index of refraction of the polymer is n = 1.50. How thick would you make the coating? (a) 1.50 cm (b) 3.00 cm (c) 0.50 cm (d) None of these 50. In a two-slit experiment, with monochromatic light, fringes are obtained on a screen placed at some distance from the slits. If the screen is moved by 5 × l 0 –2 m towards the slits, the change in fringe width is 10–3 m. Then the wavelength of light used is (given that distance between the slits is 0.03 mm) (a) 4000 Å (b) 4500 Å (c) 5000 Å (d) 6000 Å 51. Specific rotation of sugar solution is 0.5 deg m2/kg. 200 kgm–3 of impure sugar solution is taken in sample polarimeter tube of length 20 cm and optical rotation is found to be 19º. The percentage of purity of sugar is (a) 20% (b) 80% (c) 95% (d) 89% 52. If white light is used in the Newton’s rings experiment, the colour observed in the reflected light is complementary to that observed in the transmitted light through the same point. This is due to (a) 90º change of phase in one of the reflected waves (b) 180º change of phase in one of the reflected waves (c) 145º change of phase in one of the reflected waves (d) 45° change of phase in one of the reflected waves

44.

45.

46.

47.

49.

50.

51.

52.

48.

IIT-JEE PHYSICS Challenger

660

A vessel ABCD of 10 cm width has two small slits S1 and S2 sealed with identical glass plates of equal thickness. The distance between the slits is 0.8 mm. POQ is the line perpendicular to the plane AB and passing through O, the middle point of S1 and S2. A monochromatic light source is kept at S, 40 cm below P and 2 m from the vessel, to illuminate the slits as shown in the figure below. Calculate the position of the central bright fringe (in cm) on the other wall CD with respect to the line OQ.

S1 P

O S2

40 cm S

10 cm B

In YDSE a light containing two wavelengths 500 nm and 700 nm are used. Find the minimum distance (in mm) where maxima of two wavelengths coincide. Given D/d = 103, where D is the distance between the slits and the screen and d is the distance between the slits.

Screen

S h A B Calculate the ratio of the minimum to the maximum intensities in the interference fringes formed near the point P (shown in the figure). The Youngâ&#x20AC;&#x2122;s double slit experiment is done in a medium of refractive index 4/3. A light of 600 nm wavelength is falling on the slits having 0.45 mm separation. The lower slit S2 is covered by a thin glass sheet of thickness 10.4 m and refractive index 1.5. The interference pattern is observed on a screen placed 1.5 m from the slits as shown in Figure.

A point source S emitting light of wavelength 600 nm is placed at a very small height h above a flat reflecting surface AB (see figure). The intensity of the reflected light is 36% of the incident intensity. Interference fringes are observed on a screen placed parallel to the reflecting surface at a very large distance D from it.

O S2

Find the location (inÂľm) of the central maximum (bright fringe

1. MARK YOUR RESPONSE

661

WAVE OPTICS

with zero path difference) on the y-axis. [All wavelengths in this problem are for the given medium of refractive index 4/3. Ignore dispersion] A coherent parallel beam of microwaves of wavelength = 0.5 mm falls on a Young’s double slit apparatus. The separation between the slits is 1.0 mm. The intensity of microwaves is measured on a screen placed parallel to the plane of the slits at a distance of 1.0 m from it as shown in Fig.

thickness (in µm) of the glass plate. (Absorption of light by glass plate may be neglected.) In a modified Young’s double slit experiment, a monochromatic uniform and parallel beam of light of wavelength 6000 Å and intensity (10/ ) W m–2 is incident normally on two circular apertures A and B of radii 0.001 m and 0.002 m respectively. A perfectly transparent film of thickness 2000 Å and refractive index 1.5 for the wavelength of 6000 Å is placed in front of aperture A, see fig.

y A F

d = 1.0 mm

x B

D = 1.0m Screen

If the incident beam falls normally on the double slit apparatus, find the sum of the magnitudes (in m) of the y-coordinates of all the interference minima on the screen. In Young’s experiment, the source is red light of wavelength 7 × 10–7 m. When a thin glass plate of refractive index 1.5 at this wavelength is put in the path of one of the interfering beams, the central bright fringe shifts by 10–3 m to the position previously occupied by the 5th bright fringe. Find the thickness (in µm) of the plate. In Young’s experiment, the upper slit is covered by a thin glass plate of refractive index 1.4 while the lower slit is covered by another glass plate, having the same thickness as the first one but having refractive index 1.7. Interference pattern is observed using light of wavelength 5400 Å. It is found that the point P on the screen where the central maximum (n = 0) fell before the glass plates were inserted now has 3/4 the original intensity. It is further observed that what used to be the fifth maximum earlier, lies below the point P while the sixth minimum lies above P. Calculate the

Calculate the power (in micro-watt) received at the focal spot F of the lens. The lens is symmetrically placed with respect to the apertures. Assume that 10% of the power received by each aperture goes in the original direction and is brought to the focal spot. Screen S is illuminated by two point sources A and B. Another source C sends a parallel beam of light towards point P on the screen (see figure). Line AP is normal to the screen and the lines AP, BP and CP are in one plane. The distance AP, BP and CP are in one plane. The distance AP, BP and CP are 3 m, 1.5 m and 1.5 m respectively. The radiant powers of sources A and B are 90 watts and 180 watts respectively. The beam from C is of intensity 20 watt/m2. Calculate the intensity (in W/m2)at P on the screen. C

60° × A

5. MARK YOUR RESPONSE

60°

IIT-JEE PHYSICS Challenger

662

1. 2.

1 2 3 4 5 6 7 8 9 10

11 12 13 14 15 16 17 18 19 20

(d) (c) (c) (b) (d) (c) (d) (a) (b) (a)

21 22 23 24 25 26 27 28 29 30

(a) (a) (b) (b) (b) (c) (b) (d) (d) (a)

31 32 33 34 35 36 37 38 39 40

(d) (d) (d) (a) (c) (a) (a) (c) (c) (b)

41 42 43 44 45 46 47 48 49 50

(a) (a) (a) (b) (a) (a) (d) (d) (c) (d)

1 2 3 4

(a) (b) (c) (a)

5 6 7 8

(b) (c) (a) (c)

9 10 11 12

(a) (c) (d) (b)

13 14 15 16

(d)

1 2

(d) (c)

3 4

(b) (a)

5 6

(a) (c )

1 2

(a, b, c, d) (b, d)

3 4

(b) (a, c)

A-q, r, s; B-p; C-s; D-r A-s; B-r; C-q; D-p

1 5 9

2 2.78 13.9

5 6

(b) (a, b)

3. 4.

2 6

3.5 7

7 8

51 52 53 54 55 56 57

(a) (a, b)

A-p, q; B-r, s; C-s, t; D-p A-q; B-s; C-r; D-p

3 7

0.06 9.3

4 8

4330 7

663

WAVE OPTICS

(c) Referring to the figure, the path difference between the two waves starting from S1 and S2 turns out to be (2 cos ) = n where n is taken as 1 to get the point of maximum intensity which is the same as a point O. Therefore, the above relation gives cos = 1/2 so that = 60°

(a)

path difference

O screen

(d) The fringe width in Young's double slit experiment is

D d where is the wavelength of light used. D is the distance between slit and screen. d is the distance between the slit. =

3 5 , path differences. 2 2 2 total number of minima = 2nmax = 6. (c) In absence of film or for m = 0 intensity is maximum at screen. As the value of m is increased, intensity shall decrease and then increase alternately. Hence the correct variation is.

(a)

(

5 =

1)t

5 5890 10 0.45

= 2.4 × 10 – 3 m = 2.4 mm

(0.45)t 5890 10

= 6.544 × 10–4 cm

6 +1 8 6 1 8

I max 49 = I min 1

I( )

I 0 cos 2

where

d tan

I ( ) = I 0 cos 2

dy 2 d tan = D

2 ÷ = I 0 cos

I 0 cos 2 5.

tan ÷

I max =

(

I min

(

I1 + I 2

)2 = (

) (

I + 4I I

)2 = (3 I )2 = 9I )

d/2 2

(a) a1 = 6 units, a2 = 8 units

I max = I min a1 1 a2 10.

2D 2 2 600 10 = a 10 3

a1 +1 a2

........(2)

2 51 10

(2 D ) D '= =4 =4 d /2 d (c) There can be three minima from central point to corresponding to

........(1)

= 5.1 × 10–4 m 2 10 3 (d) The distance between the first dark fringe on either side of the central maximum = width of central maximum

D 2.

2 10 –2 m 10 D = 2m, d = ? From eqs. (1) and (2)

According to question, = 5100 × 10–10 m

3 , giving PO = D 3 .

and tan = PO/D =

d/2

S1 D S2

150 tan 3 108 /106

IIT-JEE PHYSICS Challenger

664

= 30°; I( ) = Io cos2

For

2 3

(a + b) 1–

For = 90° ; I( ) = Io cos2 ( ) For = 0° I( ) = I0 I( ) is not constant. 11.

(d) Shift of fringe pattern = (

30 D (4800 10 d 30 4800 10 t=

12.

30 4800 10 0.6

14.

)

(a + b) =

tD d

= (0.6)t

D d

= 24 10

5.0 10 –5 5

2.5 10

20.

= 10 (given) d No. of fringes within 0.12° will be

[2.09] 360 10 3 The number of bright spots will be three 15. (d) For diffraction pattern to be observed, the dimension of slit should be comparable to the wave length of light. 16. (c) Given d = 1mm, = 6.5 × 10–7m, D = 1m. For n th dark fringe yn=

y3 =

2 3 1 1 6.5 10 ÷ 2 1 10 3

(d) We have,

= 1.625 mm.

d n = d (a b) cos

The angular separation for IIIrd order is given by

d =

3.d (a b) 1 – sin 2

We also have for IIIrd order (a + b) sin = 3

sin =

3 ( a + b)

1 2 0.034 10 3 1 = + = 100.5 2 2 680 10 9 mmax = 100 (a) Optical path difference between the waves = (n3 – n2) t ( n3

n2 ) t

( n3

(Vacuum )

21.

n2 ) t n1

(a) When slits are of equal width. Imax µ (a + a)2 ( = 4a2) Imin µ (a – a)2 ( = 0) When one slit's width is twice that of other

I1 W1 a 2 = = I 2 W2 b 2 Imax µ (a + 22.

1 10 –10

Phase difference = 2

2n 1 D ÷ 2 d

For third fringe,

1–

6°

9 4 4 10 –14

2t (m 1/ 2) . The minimum value for m = is m = 1, the maximum value is the integer portion of

0.12 2

–5

(a) Locus of equal path difference are lines running parallel to axis of the cylinder. Hence straight fringes will be observed. (b) Constructive interference happens when

17.

19.

= 2.

(b) Angular width =

3 200 10 –9 1 10

18.

1.44 10 0.6

(a + b)2

1 10 –2 = 1 10 –5 m / line 1000

So, d

= 0.6t =

Here, = 400 × 10–9, d = 200 × 10–9.

13.

3.d

So, d

W a2 = 2 2W b

b = 2a

2a )2 = (5.8 a2)

Imin µ ( 2a – a)2 = (= 0.17 a2) (a) Suppose the mth bright fringe of 6500 Å coincides with the nth bright fringe of 5200Aº.

Xn =

m 1D n 2 D = d d

m 6500 D n 5200 D = d d m 5200 4 = = n 6500 5 m 1D y = 0.156 cm. d 23. (b) When a mica sheet of thickness t and refractive index is introduced in the path of one of the interfering beams in a biprism arrangement, the shift in interference pattern is given by distance y is y =

The ratio of frequencies of the shortest wave lengths of Balmer and Lyman series of hydrogen atom is (a) 4 : 1 (b) 1 : 4 (c) 27 : 5 (d) 5 : 27 The first member of Balmer series of hydrogen has a wavelength of 6563 Å the wavelength of its second member will be (a) 4861 Å (b) 6563 Å (c) 3561 Å (d) 1215 Å An atom emits a photon of wavelength = 600 nm by transition from an excited state of life time 8 × 10–9 s. If v represents the minimum uncertainty in the frequency of the v photon, the fractional width of the spectral line is of v the order of (a) 10–4 (b) 10–6 –8 (c) 10 (d) 10–10 The half life of radioactive Radon is 3.8 days. The time at the 1 th of the radon sample will remain end of which 20

is (R = 1.1 × 10–7 m–1, b = 1 and 5 / 33 = 0.39 ) (a) Co, Z = 27 (b) Iron, Z = 26 (c) Mn, Z = 25 (d) Ni, Z = 28

MARK YOUR RESPONSE

25 25 (b) 1019 J, 6.2 amp 1019 J, 0.4amp 4 3 25 (c) (d) none of these 1019 J, 0.8amp 2 Electrons are bombarded to excite hydrogen atoms and six spectral lines are observed. If Eg is the ground state energy of hydrogen, the minimum energy the bombarding electrons should posses is

(a)

(c)

undecayed is (given log10 e = 0.4343 ) (a) 3.8 days (b) 16.5 days (c) 33 days (d) 76 days. The wavelength of Ka X-rays produced by an X-ray tube is 0.76 Å. Find the atomic number of the anode material of the tube ? (a) 40 (b) 30 (c) 20 (d) 10 The element which has a K x-rays line of wavelength 1.8 Å

Find the number of photon emitted per second by a 25 watt source of monochromatic light of wavelength 6600 Å. What is the photoelectric current assuming 3% efficiency for photoelectric effect ?

8Eg

(b)

9 35Eg

(d)

15 Eg 16 48Eg

36 49 In Bohr theory of hydrogen atom, let r, v and E be the radius of orbit, speed of electron and the total energy of the electron respectively. Which of the following quantities is proportional to the quantum number n? (a) vr (b) r E (c) r/E (d) r/v

10. If the nuclear radius of 27Al is 3.6 Fermi, the approximate nuclear radius of 64Cu in Fermi is (a) 4.8 (b) 3.6 (c) 2.4 (d) 1.2 11. If the wavelength of the first line of the Balmer series in the hydrogen spectrum is , then the wavelength of the first line of the Lyman series is (a) (27/5) (b) (5/27) (c) (32/27) (d) (27/32)

10.

11.

IIT-JEE PHYSICS Challenger

676

12. The difference between the longest wavelength line of the Balmer series and shortest wavelength line of the Lyman series for a hydrogenic atom (atomic number Z) equal to . The value of the Rydberg constant for the given atom is : (a)

5 31

(c)

31 5

(b)

.Z 2

(c)

5 Z2 36

(a)

4 2+ 2 He

+ 2e – + 26 MeV represents

-decay

(b)

19. Calculate binding energy of 92U238.

13. A radioactive material decays by simultaneous emission of two particles with respective half-lives 1620 and 810 years. The time, in years, after which one-fourth of the material remains is) (a) 1080 (b) 2430 (c) 3240 (d) 4860 14. The energy difference between the first two levels of hydrogen atom is 10.2 eV for another element of atomic number 10 and mass number 20, this will be (a) 2040 eV (b) 0.201 eV (c) 510 eV (d) 1020 eV 15. Light of wavelength 200 Å fall on aluminium surface. Work function of aluminium is 4.2 eV. What is the kinetic energy of the fastest emitted photoelectrons? (a) 2 eV (b) 1 eV (c) 4 eV (d) 0.2 eV 16. In an a-decay the kinetic energy of a-particle is 48 MeV and Q-value of the reaction is 50 MeV. The mass number of the mother nucleus is (Assume that daughter nucleus is in ground state) (a) 96 (b) 100 (c) 104 (d) 110 17. The equation 4 11 H +

(d)

(d) none of these

.Z 2

-decay

Given M (U238) = 238.050783 amu, mn = 1.008665 amu and mP = 1.007825 amu (a) 801.7 MeV

(b) 18.7 MeV

(d) 1801.7 MeV

20. At radioactive equilibrium, the ratio between the atoms of two radioactive elements (X) and (Y) was found to be 3.2 × 109 : 1 respectively. If half-life of the element (X) is 1.6 × 1010 years, then half-life of the element (Y) would be (a) 3.2 × 109 years (c)

(b) 5 × 109 years

1.6 × 1010 years

(d) 5 years

Pb200

21. A stationary nucleus emits an alpha -particle with kinetic energy T . The fraction of recoil energy of the daughter nucleus to the total energy liberated is : (a) 1/196

(b) 4/196

(d) 1/50

22. Half life of a radioactive substance is 20 minute. Difference between points of time when it is 33% disintegrated and 67% disintegrated is approximately (a) 40 minute

(b) 10 minute

(d) 20 minute

23. In the photoelectric experiment, if we use a monochromatic light, the I-V curve is as shown. If work function of the metal is 2eV, estimate the power of light used. (Assume efficiency of photo emission = 10–3%, i.e., number of photoelectrons emitted are 10–3% of number of photons incident on metal)

(c) fusion (d) fission 18. A radioactive nucleus X decays to a stable nucleus ‘Y’. Then the graph of rate of formation of ‘Y’ against time ‘t’ will be

I 10µA

–5 volt

(a)

(b)

MARK YOUR RESPONSE

(a) 2W

(b) 5W

(d) 10W

12.

13.

14.

15.

16.

17.

18.

19.

20.

21.

22.

23.

677

MODERN PHYSICS 24.

25.

26.

27.

The ionization energy of a hydrogen-like Bohr atom is 4 Rydbergs. Find the wavelength of radiation emitted when the electron jumps from the first excited state to the ground state: [1 Rydberg = 2.2 × 10–18 joule (h = 6.6 × 10–34 Js, c = 3 × 108 m/s.) Bohr radius of hydrogen atom = 5 × 10–11m] (a) 400 Å (b) 300 Å (c) 500 Å (d) 600 Å Suppose potential energy between electron and proton at separation r is given by U = K ln (r), where K is a constant. For such a hypothetical hydrogen atom, the ratio of energy difference between energy levels (n = 1 and n = 2) and (n = 2 and n = 4) is (a) 1 (b) 2 (c) 3 (d) 4 Consider particles, particles and - rays, each having an energy of 0.5 MeV. In increasing order of penetrating powers, the radiations are: (a)

, ,

(b)

, ,

(c)

, ,

(d)

, ,

A hydrogen atom is in an excited state of principal quantum number (n), it emits a photon of wavelength ( ), when it returns to the ground state. The value of n is (a) (c)

28.

29.

30.

(b)

R 1

(d)

( R 1)

( R 1) R

processes 1 H 2 +1 H 2 1H

+1 H3

4 2 He + n . 10 16 W, the

H3 + p and If the average power radiated by

the star is deuteron supply of the star is exhausted in a time of the order of (The masses of nuclei are:

m(H 2 ) = 2.014amu, m(p) = 1.007amu , m(n) = 1.0084amu, m(He 4 ) = 4.001amu ) (a) 106 s (b) 108 s 12 (c) 10 s (d) 1016 s 32. A radioactive source in the form of metal sphere of diameter 10–3 m emits beta particle at a constant rate of 6.25 × 1010 particles per second. If the source is electrically insulated, how long will it take for its potential to rise by 1.0 volt, assuming that 80% of the emitted beta particles escape from the source? (a) 6.95 µ sec (b) 0.95 µ sec (c) 1.95 µ sec (d) 2.15 µ sec 33. There are two radioactive substances A and B. Decay constant of B is two times that of A. Initially, both have equal number of nuclei. After n half lives of A, rate of disintegration of both are equal. The value of n is (a) 4 (b) 2 (c) 1 (d) 5 2

34. The binding energy of deuteron ( 1 H ) is 1.15 MeV per 4

nucleon and an alpha particle ( 2He ) has a binding energy

R 1 An electron in a hydrogen atom makes a transition from n = n1 to n = n2. The time period of electron in the initial state is eight times that in the final state. Then which of the following statement is true ? (a) n1 = 3n2 (b) n1 = 4n2 (c) n1 = 2n2 (d) n1 = 5n2 A neutron travelling with a velocity v and kinetic energy E has a perfectly elastic head-on collision with a nucleus of an atom of mass number A at rest. The fraction of total energy retained by the neutron is approximately (a) [(A – 1)/(A + 1)]2 (b) [(A + 1)/(A – 1)]2 (c) [(A – 1)/A]2 (d) [(A + 1)/A]2 Determine the power output of a 92U235 reactor if it takes 30 days to use 2kg of fuel. Energy released per fission is 200 MeV and N = 6.023 × 1026 per kilomole. (a) 63.28 MW (b) 3.28 MW (c) 0.6 MW (d) 50.12 MW

MARK YOUR RESPONSE

31.

A star initially has 1040 deutrons. It produces energy via the

of 7.1 MeV per nucleon. Then in the reaction 2 1H

2 H 1

2 2 He + Q

the energy released Q is : (a) 5.95 MeV (b) 26.1 MeV (c) 23.8 MeV (d) 289.4 MeV 35. An energy of 24.6 eV is required to remove one of the electrons from a neutral helium atom. The energy (in eV) required to remove both the electrons from a neutral helium atom is (a) 38.2 (b) 49.2 (c) 51.8 (d) 79.0 36. A horizontal beam of thermal neutrons with v = 2.2 × 10 3 meters/sec is directed to hit a target 1.1 meter away. If gravity is the only force, the beam would miss the target approximately by : (a) 1.01 × 10– 6 meter (b) 1.25 × 10– 6 meter – 6 (c) 1.1 × 10 meter (d) 4.4 × 10– 6 meter

24.

25.

26.

27.

28.

29.

30.

31.

32.

33.

34.

35.

36.

IIT-JEE PHYSICS Challenger

678

37. X-rays are produced in an X-ray tube operating at a given accelerating voltage. The wavelength of the continuous X-rays has values from (a) 0 to (b) (c) (d)

min

0 to

to max

min to

where where max

min max

>0 <

where 0 <

min

max

43. An particle passes rapidly through the exact centre of a hydrogen molecule, moving on a line perpendicular to the internuclear axis. The distance between the nuclei is b. Where on its path does the particle experience the greatest force? (Assume that the nuclei do not move much during the passage of the particle. Also neglect the electric field of the electrons in the molecule.)

38. The electric field of a light wave at a point is E = (100 N/c) sin [(3 × 1015 s–1)t] sin [(6 × 1015s–1)t] where t is time in seconds. This light falls on a metal surface having work function of 2 eV, then maximum possible kinetic energy of photoelectrons is about (a) 16 eV (b) 7 eV (c) 6 eV (d) 4 eV 39. In the hydrogen atom spectrum 3–1 and 2–1 represent wavelengths emitted due to transition from second and first excited states to the ground state respectively. The value of 3–1

(a) 27/32 (b) 32/27 (c) 4/9 (d) 9/4 40. A light of wavelength and intensity I is incident normally on the surface. If reflection coefficient of surface is r, the pressure exerted by light on the surface is equal to

2I c

(b)

I r c

I I (d) (1 + r ) (1 – r ) c c 41. A regular hexagonal lamina of side a made up of perfectly absorbing material is kept in a region where a parallel beam of light with intensity I having a large aperture falls on it. If the area normal of the hexagon makes an angle of 30° with the beam then the force experienced by the hexagon will be (c)

(a)

5a 2 I 4c

(b)

a2 I c

(d)

MARK YOUR RESPONSE

H2 molecule (a)

b 2

(b)

b 2 2

(d) none 2 44. At t = 0, light of intensity 1012 photons/s–m2 of energy 6eV per photon start falling on a plate with work function 2.5 eV. If area of the plate is 2 ×10–4 m2 and for every 105 photons one photoelectron is emitted, charge on the plate at t = 25s is (a) 8 × 10 –15 C (b) 4 × 10 –14 C (c) 12 × 10 –14 C (d) 16 × 10 –14 C 45. When an electron in a hydrogen atom makes a transition from 2nd excited stated to ground state it emits a photon of frequency f. The frequency of photon emitted when an electron of Li++ makes a transition from Ist excited state to ground state is

9a 2 I 4c

6a 2 I c 42. The transition from the state n = 4 to n = 3 in a hydrogen-like atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition (a) 2 1 (b) 3 2 (c) 4 2 (d) 5 4 (c)

b +e

(c)

2–1

(a)

+2e -particle

(a)

243 f 32

(b)

141 f 32

(b)

81 f 32

(d)

63 f 32

46. The activity of a radioactive sample is A1 at time t1 and A2 at time t2. If is average life of sample then the number of nuclei decayed in time (t2 – t1) is

( A1

A2 )

(a) A1t1 – A2t2

(b)

(d) (A1 – A2) .

37.

38.

39.

40.

41.

42.

43.

44.

45.

46.

695

MODERN PHYSICS

(a) The shortest wavelength of Balmer series is for n2 =

1.097 10

=–

1 2

t~h

4 1.097 10 10 7 = B = 1.097 4 For Lyman series, shortest wavelength is for n2 = 1

7 = 1.097 10

1.097 107

10 7 1.097

~ 10 –6 4.

The ratio is

4 /1.097 10 1/1.097 10

(b) t1/2 = 3.8 day

4 1

If the initial number of atom is a = A0 then after time t the number of atoms is a/20 = A. We have to find t.

(a) Wave length of spectral line of Balmer series is

n22

t =

1 4

1 5 = R 9 36

1 1 4 16

3 16

5R 36

16 3R

2 = 1 × 0.746 = 6563 × 0.746

(b)

2 = 4861 Å E. t ~ h

= RZ2

6. . t~h

Now, c = v + =0 D

Z2 = =

1 n12

= RZ 2

A0 2.303 a = log A 0.182 a / 20

1 n22 ÷ , Where R is the Rydbergg

constant. The Ka line characteristic of an element is produced due to transition from the L-shell (n2 = 2) to the K-shell (n1 = 1). Thus

1 1

log

2.303 log 20 = 16.46 days 0.182 (a) The wavelength of X-ray lines is given by

For second line n2 = 4 1

2.303

For 1st line, n2 = 3

0.693 0.693 = = 0.182 t1/ 2 3.8

600 10 –9 h . ~ = t hc tc 8 10 –9 3 108

(a)

FG 1 H1

1 2

IJ = 3 RZ K 4

4 3R 4 3 x (1097 . x 10 7 ) x (0.76 x 10

Z2 = 1600

)

= 1599.25

Z = 40.

3R ( Z 1) 2 4 ( Z 1) =

4 3R

4 3 1.1 10

200 5 78 = = 26 3 33 3 Z = 26 + 1 = 27 =

1.8 10

IIT-JEE PHYSICS Challenger

696

= 6600 Å = 6600 × 10–10 m

(a) Pin = 25W, nhv = P Number of photons emitted/sec, n=

P hc

25 6600 10 34

6.64 10

when No is initial number of atoms Here 1 =

N 1 = No 4

3 108

25 1019 3 3% of emitted photons are producing current = 8.28 × 1019 =

3 ne 100

3 25 × 1019 × 1.6 × 10–19 = 0.4 A 100 3 (b) To obtain 6 spectral line, as electron must be excited to =

fourth orbit with energy

15 Eg 9. 10.

(a) Nuclear radius, r µ A1/ 3 where A is mass number 1/ 3

64 ÷ 27

1/ 3

12.

(c)

= – 3.4 × 100 = –340 eV 22 E2 – E1 = – 340 – (– 1360) = 1020 eV 15. (a) By Einstein’s equation of photo-electric effect, the maximum kinetic energy of emiited photo-electrons is given by

(a)

(2) 2

(3)2

log e

R [1

Z2 31 5

16.

0] Z 2

N No

A 4 .Q A

17.

(c)

18.

(c)

A 4 .50 4

A = 100

4 2+ 4 11H + + 2e – + 26 MeV 2 He represent a fusion reaction.

N0 e

N 0 (1 e

Rate of formation of Y =

)

dN = + N0e dt

At t = 0, R 2

eV – 4.2eV

my + m

. Z2

–(

(b) We have K =

X 1N

–W

2000 10 –10 1.6 10 –19 So, Ek = 2eV

of the Lyman

1 1

–dN dt

6.6 10 –34 3 108

Ek =

On solving , R = 13.

13.6 102

Where, h = Planck’s constant v = frequency of incident light W = work function of metal = wavelength of incident light

3 27 = 4 5

= – 1360 eV

E2 =

5 27

1–

810 + 1620 t 1620 810

Z 2E

Ek = h – W or Ek =

36 5 Wavelength of the first line , series is given by

0.693 ; 810

1 0.693 0.693 =– + ÷t 4 1620 810

13.6 10 2

E1 =

1 = R 1 – ÷ = 36 4 5

n For Z = 10,

4 ÷ = 4.8 Fermi 3 (b) For first line of Balmer series 1 1 1 =R – ÷ 4 9

13.6

rCu = 3.6

11.

2 1620 810 = t = 1080 years 2430 14. (d) For H atom E2 – E1 = 10.2 eV

, so that the difference is

16 (a) Note that v is inversely proportional to n whereas r is directly to n2.

log e

2.303 [ – 2 × (.3010)] = –0.693

rCu A = Cu ÷ rAl AAl

0.693 and 1620

t , R=0 Ny = N0 (1 – e– t)

t=0

N00

t=t