Physics for You March 2015

edit

rial

Training for research starts in schools Vol. XXIII

No. 3

March 2015

Corporate Office: Plot 99, Sector 44 Institutional area, Gurgaon -122 003 (hr), Tel : 0124-4951200

Regd. Office 406, Taj apartment, Near Safdarjung hospital, ring road, New Delhi - 110029. e-mail : [email protected] website : www.mtg.in

Managing Editor Editor

: :

mahabir Singh anil ahlawat (BE, mBa)

contents Physics Musing (Problem Set-20)

8

AIPMT Special Practice Paper 2015

12

Thought Provoking Problems

23

Core Concept

26

You Asked We Answered

30

JEE Foundation Series

31

Brain Map

50

CBSE Board Practice Paper 2015

58

JEE Main Practice Paper 2015

67

JEE Advanced Practice Paper 2015

75

Physics Musing (Solution Set-19)

91

25 Must Know Facts

92

Crossword

93

Owned, Printed and Published by Mahabir Singh from 406, Taj Apartment, New Delhi - 29 and printed by Personal Graphics and Advertisers (P) Ltd., Okhla Industrial Area, Phase-II, New Delhi. Readers are adviced to make appropriate thorough enquiries before acting upon any advertisements published in this magazine. Focus/ Infocus features are marketing incentives MTG does not vouch or subscribe to the claims and representations made by advertisers. All disputes are subject to Delhi jurisdiction only. Editor : Anil Ahlawat Copyright© MTG Learning Media (P) Ltd. All rights reserved. Reproduction in any form is prohibited.

W

hen should students start thinking, is a difficult question to answer. It is like asking when should children start learning music. children start learning music when they are in primary school stage, at home. By the time they are grown up children, it is amazing to see the advance they have made, when they have not yet reached 14 or 15 years of age. research is only an attitude of mind which drives a person to think deeper and deeper. But to avoid the mistakes of repeating what others have done, a lot reading is also advised. Before starting something new, one should know what others have done earlier and what the great scientists are thinking about the problem. In modern books, written for graduate levels, one finds first a short history of the work and the thinking of the great scientists in about half a page. We are happy that in the NcErT books, particularly for high schools, the system of historic introduction and the thinking of the great scientists are also given. One may not gain extra marks for learning the history of science, but this gives extra inputs for the development of mind. reading the biography of scientists is as interesting as reading a novel. To keep the attention of the students, after every heavy derivation, one should give a short digression on the scientists. as the editorial has a wider readership, every teacher would also like to know about the methods of research and also teach their students, how to succeed in research. For success in one’s career, one must learn simultaneously how to concentrate on microproblems as well as the art of increasing a wide vision. Anil Ahlawat Editor subscribe online at www.mtg.in

individual subscription rates combined subscription rates 1 yr. 2 yrs. 3 yrs. 1 yr. 2 yrs. 3 yrs. Mathematics Today 300 500 675 PCM 800 1200 1700 300 500 675 PCB 800 1200 1700 Chemistry Today Physics For You 300 500 675 PCMB 900 1500 2100 Biology Today 300 500 675 Send D.D/M.O in favour of MTG Learning Media (P) Ltd. Payments should be made directly to : MTG Learning Media (P) Ltd, Plot No. 99, Sector 44 Institutional Area, Gurgaon - 122003 (Haryana) We have not appointed any subscription agent.

physics for you | march ‘15

7

MUSING

PHYSICS

P

hysics Musing was started in August 2013 issue of Physics For You with the suggestion of Shri Mahabir Singh. The aim of Physics Musing is to augment the chances of bright students preparing for JEE (Main and Advanced) / AIIMS / Other PMTs with additional study material. In every issue of Physics For You, 10 challenging problems are proposed in various topics of JEE (Main and Advanced) / various PMTs. The detailed solutions of these problems will be published in next issue of Physics For You. The readers who have solved five or more problems may send their solutions. The names of those who send atleast five correct solutions will be published in the next issue. We hope that our readers will enrich their problem solving skills through "Physics Musing" and stand in better stead while facing the competitive exams. By : Akhil Tewari

single option correct type

1. A point object O is placed at a distance of 20 cm in front of a equiconvex lens (amg = 1.5) of focal length 10 cm. The lens is placed on a liquid of refractive index 2 as shown in the figure. Image will be formed at a distance h from lens. The value of h is (b) 10 cm (a) 5 cm (c) 20 cm (d) 40 cm

O

(a)

^

3^ i + j 2 2

(c) −

3^ ^ i+ j 2

(b)

^

3^ i − j 2 2

(d) −

3 ^ 1^ i− j 2 2 ^

^

3. A ray of light moving along the vector ( i − 2 j ) undergoes refraction at an interface of two media, which is x-z plane. The refractive index 8

physics for you | march ‘15

^

^

−3 i − 5 j 34

^

(c)

2. For a certain reflecting surface, the unit vector along the incident ray is ^i and that along the  ^  3 ^ i j . outward normal of the surface is  − −  2 2  The unit vector along the reflected ray will be (a)

5 for y > 0 is 2 while for y < 0 it is . The unit 2 vector along the refracted ray is

^

−3 i − 4 j 5

^

(b) (d)

^

−(4 i − 5 j) 5 ^

^

4i −3 j 5

4. A soap bubble of radius r is blown up to form a bubble of radius 2r under isothermal conditions. If s is the surface tension of soap solution, the energy spent in doing so is (a) 3psr2 (b) 6psr2 2 (c) 12psr (d) 24psr2 5. Find the minimum vertical force required to pull a thin wire ring up as shown in figure, if it is initially resting on a horizontal water surface. The circumference of the ring is 20 cm and its weight is 0.1 N. The surface tension of water is 75 dyne cm–1. (a) 0.125 N (b) 0.225 N (c) 0.115 N (d) 0.130 N 6. A point mass m is welded to a ring of mass m and radius R as shown in the figure. Assuming that the ring does not slip and initially the

system is released from rest. What would be the speed of the point mass as seen from the ground after the ring has turned through an angle of 90° ? m

m

R

(a)

gR

(b)

gR 2

(c)

2gR

(d)

gR 3

7. Two infinitely long conducting parallel rails are connected through a capacitor of capacitance C as shown in the figure. A perfect conductor of length l is moved with constant speed v0. Which of the following graph truly depicts the variation of current through the conductor with time ?

9. Pulley and strings as shown in figure are massless. The force acting on the block of mass m is (a) 2F (b) F F (c) 2 (d) 4F 10. A particle of mass m moves along a circle of radius R. The modulus of average value of force acting on particle over the distance equal to a quarter of circle, if the particle moves uniformly with velocity v is 2mv 2 2 2mv 2 (a) (b) pr pr 2 (c)

2 2mv 2 pr

(d)

mv 2 pr nn

solution of february 2015 crossword

B v0

l

(a)

(b)

(c)

(d)

8. An organ pipe of cross-sectional area 100 cm2 resonates with a tuning fork of frequency 1000 Hz in fundamental tone. The minimum volume of water to be drained so the pipe again resonates with the same tuning fork is (Take velocity of wave = 320 m s–1) (a) 800 cm3 (b) 1200 cm3 3 (c) 1600 cm (d) 2000 cm3 10 physics for you |

march ‘15

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Winner (February 2015) riya Kataria Solution Senders (January 2015)



Divya Acharya

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Shubhneet Bhatia



Akash Kashyap

PRACTICE PAPER 2015

1. A 15 g ball is shot from a spring gun whose spring has a force constant 600 N m–1. The spring is compressed by 5 cm. The greatest possible horizontal range of the ball for this compression (Take g = 10 m s –2) (a) 6 m (b) 8 m (c) 10 m (d) 12 m 2. Two weights w1 and w2 are suspended from the ends of a light string over a smooth fixed pulley. If the pulley is pulled up with acceleration g, the tension in the string will be 4w1w2 2w1w2 (a) (b) w1 + w2 w1 + w2 w − w2 w1w2 (c) 1 (d) w1 + w2 2(w1 + w2 ) 3. A gas bubble from an explosion under water oscillates with a time period T, depends upon static pressure p, density of water r and the total energy of explosion E. Find the expression for the time period T. k is a dimensionless constant. (a) T = kp–5/6r1/2E1/3 (b) T = kp–4/7r1/2E1/3 (c) T = kp–5/6r1/2E1/2 (d) T = kp–4/7r1/3E1/2 4. A charged particle of mass m and charge q is released from rest in an electric field of constant magnitude E. The kinetic energy of the particle after time t is 2 E 2t 2 E 2 q 2t 2 Eqm Eq2m (b) (c) (d) qm 2m 2t 2t 2 5. The intensity of magnetic field at a point X on the axis of a small magnet is equal to the field intensity at another point Y on its equatorial (a)

12 physics for you |

march ‘15

Exam on 3rd May

axis. The ratio of distances of X and Y from the centre of the magnet will be (a) (2)–3 (b) (2)–1/3 (c) 23 (d) 21/3 6. 5 mole of an ideal gas with g = 7/5 initially at STP are compressed adiabatically so that its temperature becomes 400°C. The increase in the internal energy of gas in kJ is (a) 21.55 (b) 41.55 (c) 65.55 (d) 50.55 7. A circular platform is mounted on a frictionless vertical axle. Its radius R = 2 m and its moment of inertia about the axle is 200 kg m2. It is initially at rest. A 50 kg man stands on the edge of the platform and begins to walk along the edge at the speed of 1 m s–1 relative to the ground. Time taken by the man to complete one revolution with respect to disc is 3p p s (c) 2p s (d) s (a) p s (b) 2 2 8. A vessel contains oil (density = 0.8 g cm–3 over mercury (density = 13.6 g cm–3). A uniform sphere floats with half its volume immersed in mercury and the other half in oil. The density of the material of sphere in g cm–3 is (a) 3.3 (b) 6.4 (c) 7.2 (d) 12.8 9. Maxwell’s velocity distribution curve is given for the same quantity two different temperatures. For the given curves (a) T1 > T2 T2 N T (b) T1 < T2 1 (c) T1 ≤ T2 (d) T1 = T2 v 10. Two capacitors of 25 mF and 100 mF are connected in series to a source of 120 V. Keeping

physics for you

| march ‘15 13

their charges unchanged, they are separated and connected in parallel to each other. Find out energy loss in the process. (a) 5.2 J (b) 52 J (c) 50.2 J (d) 0.052 J 11. The steady state current in a 2 W resistor when the internal resistance of the battery is negligible and the capacitance of the condenser is 0.1 mF is A

2

B

3

0.1 F

4

6V

2.8 

(a) 0.6 A (b) 0.9 A (c) 1.5 A (d) 0.3 A 12. In an experiment, a magnet with its magnetic moment along the axis of a circular coil and directed towards the coil, is withdrawn away from the coil and parallel to itself. The current in the coil, as seen by the withdrawing magnet, is (a) zero (b) clockwise (c) anticlockwise (d) first (a) then (b) 13. A luminous object is placed at a distance of 30 cm from the convex lens of focal length 20 cm. On the other side of the lens, at what distance from the lens a convex mirror of radius of curvature 10 cm be placed in order to have an inverted image of the object coincident with it? (a) 12 cm (b) 30 cm (c) 50 cm (d) 60 cm 14. Two slits separated by a distance of 1 mm are illuminated with red light of wavelength 6.5 × 10–7 metre. The interference fringes are observed on a screen placed one metre from the slits. The distance between the third dark fringe and fifth bright fringe on the same side of centre is equal to (a) 0.65 mm (b) 1.63 mm (c) 3.25 mm (d) 4.8 mm 15. An electric bulb is marked 100 W, 230 V. If the supply voltage drops to 115 V, what is the heat and light energy produced by the bulb in 20 min? (a) 10 kJ (b) 15 kJ (c) 20 kJ (d) 30 kJ 14 physics for you |

march ‘15

16. A body hanging from a spring stretches it by 1 cm at the earth’s surface. How much will the same body stretch the spring at a place 16400 km above the earth’s surface? (Radius of the earth = 6400 km) (a) 1.28 cm (b) 0.64 cm (c) 3.6 cm (d) 0.12 cm 17. A resistor R and 2 mF capacitor in series are connected through a 200 V direct supply. Across the capacitor is a neon bulb that lights up at 120 V. Find the value of R to make the bulb light up 5 s after the switch has been closed. (Take log10 2.5 = 0.4) (a) 1.7 × 105 W (b) 2.7 × 106 W 7 (c) 3.3 × 10 W (d) 1.3 × 104 W 18. A coil of resistance 400 W is placed in a magnetic field. If the magnetic flux f (Wb) linked with the coil varies with time t (s) as f = 50t2 + 4. The current in the coil at t = 2 s is (a) 0.5 A (b) 0.1 A (c) 2 A (d) 1 A 19. An electromagnetic wave of frequency 3 MHz passes from vacuum into a dielectric medium with permittivity er = 4, then (a) the wavelength and frequency both remain unchanged. (b) the wavelength is doubled and the frequency remains unchanged. (c) the wavelength is doubled and the frequency becomes half (d) the wavelength is halved and the frequency remains unchanged. 20. A polyster fibre rope of diameter 3 cm has a breaking strength of 150 kN. If it is required to have 600 kN breaking strength. What should be the diameter of similar rope? (a) 12 cm (b) 6 cm (c) 3 cm (d) 1.5 cm 21. A thin uniform rod of mass m moves translationally with acceleration a due to two antiparallel forces of lever arm l. One force is of magnitude F and acts at one extreme end. The length of the rod is mal 2(F + ma)l (a) (b) ma + F ma F   (c) l  l +   ma 

(d)

(F + ma)l 2ma

22. The amount of work done in stretching a spring from a stretched length of 10 cm to a stretched length of 20 cm is (a) equal to the work done in stretching it from 20 cm to 30 cm (b) less than the work done in stretching it from 20 cm to 30 cm (c) more than the work done in stretching it from 20 cm to 30 cm (d) equal to the work done in stretching it from 0 to 10 cm.

XC = 40  220 V 20 Hz

R1 = 40 

XL = 100  R2 = 40 

(a) 0.2

(b) 0.4

(c) 0.8

(d) 0.6

25. A compound microscope has an eye piece of focal length 10 cm and an objective of focal length 4 cm. Calculate the magnification, if an object is kept at a distance of 5 cm from the objective, so that the final image is formed at the least distance of distinct vision 20 cm. (a) 12 (b) 11 (c) 10 (d) 13

28. At ordinary temperature, the molecules of an ideal gas have only translational and rotational kinetic energies. At high temperatures they may also have vibrational energy. As a result of this at higher temperatures (CV = molar heat capacity at constant volume) 3 (a) CV = R for a monoatomic gas 2 3 (b) CV > R for a monoatomic gas 2 5 (c) CV > R for a diatomic gas 2 5 (d) CV = R for a diatomic gas 2 29. A body is projected vertically upwards with a velocity of 10 m s–1. It reaches the maximum height h in time t. In time t/2, the height covered is h 2 3 5 (a) (b) h (c) h (d) h 2 5 4 8 30. A wheel is subjected to uniform angular acceleration about its axis. Initially, its angular velocity is zero. In the first 2 s, it rotates through an angle q1, in the next 2 s, it rotates through an q angle q2. The ratio of 2 is q1 (a) 1 (b) 2 (c) 3 (d) 5

26. In a galvanometer 5% of the total current in the circuit passes through it. If the resistance of the galvanometer is G, the shunt resistance S connected to the galvanometer is G G (a) 19G (b) (c) 20G (d) 19 20 27. The power factor of the circuit as shown in figure is

31. A uniform chain of mass m and length l is l lying on a table with of its length hanging 4 freely from the edge of the table. The amount of work done in dragging the chain on the table completely is mgl mgl mgl mgl (a) (b) (c) (d) 4 8 32 16

23. The rms value of the electric field of the light coming from the sun is 720 N C–1. The average total energy density of the electromagnetic wave is (a) 3.3 × 10–3 J m–3 (b) 4.58 × 10–6 J m–3 (c) 6.37 × 10–9 J m–3 (d) 81.35 × 10–12 J m–3 24. In Young’s double slit experiment, one of the slits is wider than the other, so that the amplitude of the light from one slit is double that from the other slit. If Im be the maximum intensity, the resultant intensity when they interfere at phase difference f is given by (a)

Im  I  f f 1 + 2 cos2  (b) m  1 + 4 cos2   3  2 5  2

(c)

Im  I  f f 1 + 8 cos2  (d) m  8 + cos2   9  2 9  2

physics for you

| march ‘15 15

32. The diode used in the circuit shown in the figure has a constant voltage drop at 0.5 V at all currents and a maximum power rating of 100 milliwatts. What should be the value of the resistor R, connected in series with diode, for obtaining maximum current? R

0.5 V

1.5 V

(a) 6.76 W (b) 20 W (c) 5 W

(d) 5.6 W

33. The half-life of a radioactive isotope X is 50 years. It decays to another element Y which is stable. The two elements X and Y were found to be in the ratio of 1 : 15 in a sample of a given rock. The age of the rock was estimated to be (a) 100 years (b) 150 years (c) 200 years (d) 250 years

34. On shining light of wavelength 6.2 × 10–6 m on a metal surface photo-electrons are emitted. The work function of the metal is 0.1 eV. Find the kinetic energy of a photo-electron (in eV) (a) 0.1 (b) 0.2 (c) 0.3 (d) 0.4 35. A mass of 0.2 kg is attached to the lower end of a massless spring of force constant 200 N m–1, the upper end of which is fixed to a rigid support. Which of the following statement is not true? (a) The frequency of oscillation will be nearly 5 Hz. (b) In equilibrium, the spring will be stretched by 2 cm. (c) If the mass is raised till the spring is unstretched state and then released, it will go down by 2 cm before moving upward (d) If the system is taken to a planet, the frequency of oscillation will be the same as on the earth. 36. The equation of a wave is represented by x  Y = 10−5 sin 100t −  m, then the velocity of 10   wave will be (a) 100 m s–1 (b) 4 m s–1 (c) 1000 m s–1 (d) zero 16 physics for you |

march ‘15

37. Force on a 1 kg mass on earth of radius R is 10 N. Then the force on a satellite revolving around the earth in the mean orbital radius 3R/2 will be (mass of satellite is 100 kg) (a) 4.44 × 102 N (b) 3.33 × 102 N (c) 500 N (d) 6.66 × 102 N 38. The far point of a near sighted person is 6.0 m from her eyes, and she wears contacts that enable her to see distant objects clearly. A tree is 18.0 m away and 2.0 m high. How high is the image formed by the contacts? (a) 1.0 m (b) 1.5 m (c) 0.75 m (d) 0.50 m 39. You drive a car at a speed of 70 km h–1 in a straight road for 8.4 km and then the car runs out of petrol. You walk for 30 min to reach a petrol pump at a distance of 2 km. The average velocity from the beginning of your drive till you reach the petrol pump is (a) 16.8 km h–1 (b) 35 km h–1 –1 (c) 64 km h (d) 18.6 km h–1

40. A fork A has frequency 2% more than the standard fork and B has a frequency 3% less than the frequency of same standard fork. The forks A and B when sounded together produced 6 beats s–1. The frequency of fork A is (a) 116.4 Hz (b) 120 Hz (c) 122.4 Hz (d) 238.8 Hz 41. When a wire of length 10 m is subjected to a force of 100 N along its length, the lateral strain produced is 0.01 × 10–3. The Poisson’s ratio was found to be 0.4. If the area of cross-section of wire is 0.025 m2, its Young’s modulus is (a) 1.6 × 108 N m–2 (b) 2.5 × 1010 N m–2 (c) 1.25 × 1011 N m–2 (d) 16 × 109 N m–2 42. In an experiment on photoelectric emission from a metallic surface, wavelength of incident light is 2 × 10–7 m and stopping potential is 2.5 V. The threshold frequency of the metal is approximately (Charge of electron e = 1.6 × 10–19 C, Planck’s constant h = 6.6 × 10–34 J s) (a) 12 × 1015 Hz (b) 9 × 1015 Hz (c) 9 × 1014 Hz (d) 12 × 1013 Hz

43. Two cells of emf e1 and e2 (e1 > e2) are connected as shown in figure. When a potentiometer is connected between A and B, the balancing length of the potentiometer wire is 300 cm. On connecting the same potentiometer between A and C, the balancing length is 100 cm. The ratio is e1 : e2 is 1

A

(a) 3 : 1

(b) 1 : 3

2

B

C

(c) 2 : 3

(d) 3 : 2

44. A body is thrown horizontally from the top of a tower of 5 m height. It touches the ground at a distance of 10 m from the foot of the tower. The initial velocity of the body is (Take g = 10 m s–2) (a) 2.5 m s–1 (b) 5 m s–1 –1 (c) 10 m s (d) 20 m s–1 45. Two bodies of 6 kg and 4 kg masses have their velocity 5i − 2 j + 10k and 10i − 2 j + 5k respectively. Then the velocity of their centre of mass is (a) 5i + 2 j − 8k (b) 7i + 2 j − 8k (c) 7i − 2 j + 8k (d) 5i − 2 j + 8k solutions

1. (c) : Here, Rmax =

u2 1 2 2 = mu × g 2 mg

1 1 But mu2 = kx 2 2 2 kx 2 1 2 2 \ Rmax = kx × = mg mg 2 2 600 × (0.05) = = 10 m 0.015 × 10

m2

m1

w2 m2a0 m1a0 w1

( a0 = g) ... (ii)

4. (a) : Here, u = 0, a = qE t m

F qE = m m

mq2 E 2t 2 E 2q2t 2 1 KE = mv 2 = = 2 2m 2m2

T

a

( a0 = g) ... (i)

3. (a) : Time period, T ∝ parbEc or T = kparbEc k is a dimensionless constant. According to homogeneity of dimensions, LHS = RHS \ [T] = [ML–1T–2]a[ML–3]b[ML2T–2]c [T] = [Ma+b+c][L–a–3b+2c][T–2a–2c] Comparing the powers, we obtain a+b+c=0 –a – 3b + 2c = 0 –2a – 2c = 1 On solving, we get 5 1 1 a=− ,b= , c= 6 2 3

v = u + at = 0 +

2. (a) : For solving the problem, we assume that observer is situated in the frame of pulley (noninertial reference frame). m1g = w1 m2g = w2 T

From force diagram, T – m2a0 – w2 = m2a or T – m2g – w2 = m2a or T – 2w2 = m2a From force diagram, m1a0 + w1 – T = m1a or m1g + w1 – T = m1a or 2w1 – T = m1a From eqs. (i) and (ii), we get 4w1w2 T= w1 + w2

a

5. (d) : If d1 is the distance of point X on axial line and d2 is distance of point Y on equatorial line, then m 2M m M , B2 = 0 B1 = 0 4 p d13 4 p d23 As B1 = B2 m0 2 M m0 M \ = 4 p d13 4 p d23

d d13 = 2d23 ; 1 = 21/3 d2 physics for you

| march ‘15 17

6. (b) : Here, n = 5, g = 7/5, T1 = 0°C, T2 = 400°C nR dT dU = g −1 5 × 8.31 × (400 − 0) = = 41550 J = 41.55 kJ (7 / 5) − 1 7. (c) : Using angular momentum conservation, Li = 0, Lf = mvR – Iw, so, mvR = I w mvR 50 × 1 × 2 1 w= = = I 200 2 For one complete revolution, (v + wR)t = 2pR  1   1 + 2 × 2  t = 2p × 2 ⇒ t = 2p s.   8. (c) : Weight = Buoyant force V V V rm g = rHg g + roil g 2 2 rHg + roil rm = 2 13.6 + 0.8 = 2 14.4 = = 7. 2 2

13. (c) : For the lens,

Oil Mercury

9. (b) : Higher is the temperature, greater is the most probable velocity. 1 1 1 1 1 5 1 = + = + = = Cs C1 C2 25 100 100 20 Cs = 20 mF = 20 × 10–6 F 1 1 U1 = CsV 2 = (20 × 10−6 )(120)2 = 144 × 10–3 J 2 2 Charge on each capacitor,

10. (d) :

q1 = q2 = Cs ⋅ V = 20 × 120 =2400 mC In parallel, Cp = C1 + C2

= 25 + 100 = 125 mF = 125 × 10–6 F \ U2 =

Q 2 [(2400 + 2400) × 10−6 ]2 = 2C p 2 × 125 × 10−6

= 92.16 × 10–3 J Loss of energy = U1 – U2 = (144 – 92.16) × 10–3 J = 51.84 × 10–3 J = 0.052 J 11. (b) : Capacitor will work as open key. Therefore no current flows through resistance 18 physics for you |

march ‘15

4 W. The total resistance of circuit 2×3 = 2. 8 + = 2. 8 + 1. 2 = 4 W 2+3 6 3 \ Main current, I = = A 4 2 Potential difference across A and B 3 = × 1.2 = 1.8 V 2 1.8 \ Current through 2 W = = 0.9 A 2 12. (b) : As magnet is withdrawn from the coil, field into the coil decreases. To increase this field, current induced in the coil must be clockwise as seen by the withdrawing magnet. 1 1 1 1 1 1 = + = − = v f u 20 30 60

v = 60 cm. Therefore, to have an inverted image of the object, coincident with it, image should tend to form at centre of curvature of convex mirror. Therefore, distance of convex mirror from the lens = 60 – 10 = 50 cm. 14. (b) : For dark fringes, lD ⇒ Y3 = 5 lD Yn = (2n − 1) 2d 2 d For bright fringes, 5lD nlD ⇒ Y5 = Yn = d d Y = Y5 – Y3 Y=

5l D 5 × 6.5 × 10−7 × 1 = 2 d 2 × 10−3

= 1.625 × 10–3 m = 1.63 mm 15. (d) : Here, power of the bulb, P = 100 W Supply voltage, e = 230 V Let R be the resistance of the bulb. e2 (230)2 e2 ⇒ R= = = 529 W P 100 R Changed supply voltage, e′ = 115 V Heat and light energy produced by the bulb in 20 min. As P =

=

2 e′2t 115 × 20 × 60 = = 30,000 J = 30 kJ R 529

16. (b) : In equilibrium, weight of the suspended body = stretching force. \ At the earth’s surface, mg = k × x At a height h, mg′ = k × x′

21. (b) : Let L be the length of the rod of mass m, with centre of mass at C. Suppose F1 is the magnitude of other force. Let F1 > F. \ F1 – F = ma or F1 = F + ma

Re2 g ′ x′ (6400)2 = = = g x (Re + h)2 (6400 + 1600)2

F1 L

2

 6400  16 = =  8000  25 16 16 x ′ = × x = × 1 cm = 0.64 cm 25 25 17. (b) : As VC = e(1 – e–t/RC) 120 3 V or 1 – e–t/RC = C = = e 200 5

or et/RC = 2.5 or log et/RC = loge 2.5 t or = 2.3026 log10 2.5 = 0.92 RC 5 t or R = = = 2.7 × 106 W 0.92C 0.92 × 2 × 10−6

df d 18. (a) : e = − = − (50t 2 + 4) = −100t dt dt When t = 2 s, |e| = 200 V Induced current at t = 2 s, | e | 200 I= = = 0.5 A R 400 19. (d) : Frequency remains unchanged with change of medium. c 1 / e0 m 0 = er m r m (refractive index) = = v 1 / em Since, mr is very close to 1, m = er = 4 = 2 l l Thus, lmedium = = m 2 F L × A ∆L The breaking strength F ∝ A.

20. (b) : Y =

\

F2 A2 p(D22 / 4) D22 = = = F1 A1 p(D12 / 4) D12 1/2

F  or D2 = D1  2   F1 

1/2

 600  = 3   150 

= 6 cm

F

l C

As the rod moves translationally and there is no rotation, therefore, net torque about C must be zero. L L  L  \ F   = F1  − l  = (F + ma)  − l  2 2  2  L L L F   = F   − Fl + ma − mal 2 2 2 2(F + ma)l L ma = l(F + ma) \ L = ma 2

(

)

1 22. (b) : W = K x22 − x12 2 1 = K(202 − 102 ) = 150 kJ 2 which is less than work done in stretching it from 20 cm to 30 cm. 1   2 2  viz. 2 K(30 − 20 ) = 250 kJ    23. (b) : Total average energy density of electromagnetic wave is 1 1 2 2 < u > = e0 Erms + Brms 2 2m0 =

2 1 1  Erms 2  e0 Erms + 2 2m0  c 2

  E    Brms = rms    c  

1 1 2 2 e0 Erms + Erms e0m0 2 2m0 1 1 2 2 2 = e0 Erms + e0 Erms = e0 Erms 2 2 = (8.85 × 10–12) × (720)2 = 4.58 × 10–6 J m–3 24. (c) : Here, A2 = 2A1  Intensity ∝ (Amplitude)2 =

2

\

2

I2  A2   2 A1  =  =  =4 I1  A1   A1  physics for you

| march ‘15 19

I2 = 4I1 Maximum intensity, Im =

(

I1 + I2

Z = R2 + ( X L − XC )2 = (80)2 + (100 − 40)2

)

2

I = I1 + 4 I1 = 3 I1 = 9 I1 or I1 = m ...(i) 9 Resultant intensity, I = I1 + I2 + 2 I1I2 cos f

(

) (

)

2

2

= I1 + 4 I1 + 2 I1 (4 I1 ) cos f = 5I1 + 4I1cosf = I1 + 4I1 + 4I1cosf = I1 + 4I1(1 + cosf)  2 f f = I1 + 8 I1 cos2 1 + cos f = 2 cos 2    2 f  = I1  1 + 8 cos2  2  Putting the value of I1 from eq. (i), we get I  f I = m  1 + 8 cos2  9  2 25. (a) : Here, uo = –5 cm, fo = 4 cm fe = 10 cm, D = 20 cm According to lens formula, 1 1 1 1 1 1 = + = − = vo 4 −5 4 5 20 vo = 20 cm v  D Magnification, M = o  1 +  uo  fe  20  20  =  1 +  = 12 5  10 

I

IG

27. (c) : Resistance of the circuit, R = R1 + R2 = 40 W + 40 W = 80 W Impedance of the circuit, 20 physics for you |

march ‘15

Power factor, cos f =

R 80 = = 0.8 Z 100

3 28. (c) : Monoatomic gas CV = R 2 This value is same for high temperature also. In case of diatomic gas 5 CV = R (at low temperature) 2 5 Also, CV > R (at high temperature due to 2 vibrational kinetic energy) 29. (c) : As v2 – v02 = 2gh, 0 – (10)2 = 2(–10) h or h = 5 m Also, v = v0 + at, 0 = 10 + (–10) t or t = 1 s Height covered in time t/2, i.e., (1/2 s), 2

1 1 1 1 h′ = v0t + (− g )t 2 = 10 × − × 10 ×   2 2 2 2 = 3.75 m = (3/4) h

26. (b) : As shunt is a small resistance S in parallel with a galvanometer (of resistance G) as shown in figure. (I – IG)S = IGG S I G (I – IG) S= G ( I − IG ) G 5 I Here, IG = 100 5 IG G \ S = 100 = 5 19 I− I 100

= (80)2 + (60)2 = 100 W

I

1 1 30. (c) : As q1 = w0t + at 2 = 0 + a(2)2 = 2a 2 2 1 1 (q1 + q2 ) = w0t + at 2 = 0 + a(4)2 = 8a 2 2 q Thus, q2 = 6a or 2 = 3 q1 l m 31. (c) : Mass of   length of the chain = 4 4 The weight of this part of the chain acts as its l CG which is at a distance   from the edge 8 of the table.  m   l  mgl Work done =   g   =  4   8  32

2 V 2 (0.5 V) = 2.5 W 32. (c) : As RD = D = PD 0.1 W

V I D = D = 0. 2 A RD Total resistance required in the circuit,

V 1.5 = = 7.5 W I D 0.2 Resistance of the series resistor, R = Req – RD = 7.5 – 2.5 = 5 W NX 1 = , N + NY = 16N X 33. (c) : As NY 15 X Req =

NX 1 = N X + NY 16 1 1 or N X = (N X + NY ) = (N X + NY ) 16 24 Age of the rock = number of half-lives of isotope X passed = 4 = 4 × 50 years = 200 years Thus,

34. (a) : Here, l = 6.2 × 10–6 m, f0 = 0.1 eV hc Energy of the incident photon, E = hu = l (6.6 × 10−34 )(3 × 108 ) or E = J 6.2 × 10−6 =

6.6 × 3 × 10−26

(6.2 × 10−6 )(1.6 × 10−19 ) As E = K + f0 , K = E − f0

eV = 0.2 eV

= 0.2 eV – 0.1 eV = 0.1 eV 1 k 1 200 = = 5 Hz. 2 p m 2 p 0. 2 In equilibrium, kx = mg mg 0.2 × 10 or x = = = 0.01 m k 200 When mass is raised till the spring is unstretched, 1 the work = kx 2 = mgx 2 When the mass is released from the unstretched position of spring, then total work done 1 mgx ′ = (mgx ) + kx 2 = 2mgx 2

35. (b) : u =

or x′ = 2x = 2 × 0.1 = 0.02 m As u of spring is independent of g so that the frequency of oscillation will be the same as that on the earth. x  36. (c) : Here, Y = 10−5 sin 100t −  10   Comparing it with, standard equation of wave motion

 2p 2p  Y = r sin  t − x  l  T 2p 2p p 2p 1 = , l = 20p = 100, T = = s ; 100 50 l 10 T l 20p Velocity, v = = = 1000 m s −1 T p / 50 37. (a) : On the surface of earth, the force on a mass of 1 kg is GMm GM × 1 ... (i) F= = = 10 R2 R2 When the radius of the satellite, r = 3R/2, the force on the satellite is GMm′ GM × 100 F= = r2 (3 / 2)2 R2 10 × 4 × 100 = = 4.44 × 102 N (Using (i)) 9 38. (d) : The far point of 6.0 m tell us that the focal length of the lens is f = – 6.0 m, u = – 18 m and h=2m 1 1 1 = − f v u 1 1 1 1 1 ⇒ = + = − v f u −6.0 18.0

Using,

⇒ v = – 4.5 m \ The image size, v   −4.5  h′ = h  = 2 ×  = 0.50 m u  −18.0  39. (a) : Here, displacement = 8.4 + 2 = 10.4 km 1 8. 4 Total time taken = + = 0.62 h 2 70 Displacement Average velocity = Total time taken 10.4 km = = 16.8 km h −1 0.62 h 40. (c) : Let u be frequency of standard fork. The 2 frequency of A, uA = u + u 100 3 u and the frequency of B, uB = u − 100 According to question, uA – u B = 6 physics for you

| march ‘15 21

2   3   \ u+ u −  u − u =6 100   100  

u0 =

5 600 u = 6 or u = = 120 Hz 100 5 The frequency of A 2  2  uA =  u + u  = 120 + × 120 = 122.4 Hz 100  100  or

Lateral strain

41. (a) : Poisson’s ratio = Longitudinal strain = = Young’s modulus, Y = Y=

Longitudinal strain Lateral strain Poisson’s ratio −3

0.01 × 10 ... (i) 0.4 Normal stress

42. (c) : Energy of incident photon, E = E=

6.6 × 10−34 × 3 × 108

hc l

2 × 10−7

= 9.9 × 10–19 J =

e e 100 1 or 1 − 2 = or 2 = 1 − e1 300 e1 3 or

9.9 × 10

−19

where the symbols have their usual meanings. hc or f0 = − K max = 6.2 eV – 2.5 eV = 3.7 eV l f0 Threshold frequency, u0 = h

march ‘15

e2 2 e 3 = or 1 = e1 3 e2 2

44. (c) :

u 5m Tower Ground

10 m

Let t be time taken by the body to reach the ground. \

1 2H 2×5 = =1 s H = gt 2 or t = 2 g 10

R = ut or 10 = u × 1 or u = 10 m s–1 45. (c)

nn Form IV

eV = 6.2 eV 1.6 × 10−19 Kmax = eVs = e × 2.5 V = 2.5 eV According to Einstein’s photoelectric equation hc K max = − f0 l

22 physics for you |

43. (d) : When potentiometer is connected between A and B, then it measures only e1 and when connected between A and C, then it measures e1 – e2. e1 l e −e l \ = 1, 1 2 = 2 e1 − e2 l2 e1 l1

(Using (i))

 0.01 × 10−3  A×    0.4   100 × 0.4 = N m −2 = 1.6 × 108 N m–2 −3 0.025 × 0.01 × 10

6.6 × 10−34

= 0.9 × 1015 Hz = 9 × 1014 Hz

Longitudinal strain

F

3.7 × 1.6 × 10−19

1. Place of Publication 2. Periodicity of its publication 3. Printer’s and Publisher’s Name Nationality Address

: : : : :

4. Editor’s Name Nationality Address

: : :

New Delhi Monthly Mahabir Singh Indian Physics for You, 406, Taj Apartment, New Delhi - 110029. Anil Ahlawat Indian Physics for You, 19, National Media Centre, Gurgaon Haryana - 122002 Mahabir Singh 406, Taj Apartment New Delhi

5. Name and address of : individuals who own the newspapers and partners or shareholders holding more than one percent of the total capital I, Mahabir Singh, hereby declare that particulars given above are true to the best of my knowledge and belief. Mahabir Singh Publisher

1. A sample of hydrogen gas in its ground state is irradiated with photons of 10.2 eV energies. The radiation from the above sample is used to irradiate two other samples of excited ionised He+ and excited ionised Li2+ respectively. Both the ionised samples absorb the incident radiation. (i) How many lines are obtained in the He+ and Li2+ emission spectra? (ii) What are the smallest and biggest wavelengths in their spectra? 2. The half-value thickness of an absorber is defined as the thickness that will reduce the intensity of a beam of particles by a factor of 2. Calculate the half-value thickness for lead, assuming an X-ray beam of wavelength 20 pm. Total linear attenuation coefficient, m = 55 cm–1 for X-rays in lead at wavelength l = 20 pm. 3. When an electron beam interacts with atoms on the surface of a solid, by studying the angular distribution of the diffracted electrons, one can indirectly measure the geometrical arrangement of atoms. Assume that the electrons strike perpendicular to the surface of a solid as shown in figure, and that their energy is low, K = 100 eV, so that they interact only with the surface layer of atoms. If the smallest angle at which a diffraction maximum occurs is at 24°, what is the separation d between the atoms on the surface?

By : Prof. Rajinder Singh Randhawa*

4. Initial activity of a b– emitter isotope 90Sr is 10 mCi. How many decays per second will be taking place after 84 years. The half life of 90Sr is 28 years. 5. Nuclei of a radioactive element A are being produced at a constant rate a. The element has a decay constant l. At time t = 0, there are N0 nuclei of the element. (a) Calculate the number N of nuclei of A at time t. (b) If a = 2lN0, calculate the number of nuclei of A after one half life of A, and also the limiting value of N as t → ∞. 6. Find the decay constant and mean life time of 55Co radio nuclide if its activity is known to decrease by 4% per hour. The decay product is non-radioactive. SOLUTIONS

1. After absorbing photons of energy 10.2 eV, hydrogen atom would reach the first excited state of –3.4 eV, since energy difference

*Randhawa Institute of Physics, S.C.O. 208, First Fl., Sector-36D & S.C.O. 38, Second Fl., Sector-20C, Chandigarh, Ph. 09814527699

PHYSICS FOR YOU | March ‘15

23

corresponding to n = 1 and n = 2 is 10.2 eV. When this excited hydrogen atom deexcites, it would release 10.2 eV, which is absorbed by He+ and Li2+. Energy of nth state of a hydrogen like atom with atomic number Z is given by, En = −

13.6Z 2 n2

eV

1242 = 10.4 nm n = 6 → n = 1 13.6 9 − 9   36  l min

=

2. The intensity varies with distance travelled in the medium, according to relation I I ( x ) = I 0 e − mx \ 0 = I 0 e − mx 2 − mx 1 mx e = or 2 = e 2 Taking log both sides, mx = ln 2 x=

ln 2 m

or x =

ln 2 55

=

0.693 55

= 1.26 × 10 −2 cm = 0.126 mm Hence, we conclude that lead is a very good absorber for X-rays. 3. The path difference is dsinq. For constructive interference, dsinq = nl. For the smallest value of q, dsinq = 1 × l The kinetic energy is After absorbing 10.2 eV, He+ electron moves from n = 2 to n = 4 and Li2+ electron moves from n = 3 to n = 6.

K= \

l= =

p2 h2 = 2me 2me l 2

h   p =  l

h 2me K

6.63 × 10 −34

2 × (9.1 × 10 −31 ) × 100 × 1.6 × 10 −19

= 0.123 nm In the spectrum of He+ there would be 4C = 6 lines. 2 hc 1242 l max = = = 470 nm E ∆ 4 4  n=4→n=3 13.6  −   9 16  hc 1242 = = 24.4 nm n = 4 → n = 1 ∆E 13.6  4 − 4   16  Similarly, in spectrum of Li2+ there will be 6C = 15 lines. 2 l min

=

1242 = 830.2 nm n = 6 → n = 5 13.6  9 − 9   25 36  l max

=

24 PHYSICS FOR YOU |

March ‘15

On the surface, interatomic spacing of atoms is l 0.123 d= = = 0.30 nm sin q sin 24° 4. As T1/2 = 28 year, there will be three half lives in 84 years. 1  1 1 1 1 So the activity will be only  as × × =  . 8  2 2 2 8 So the activity of source will be, \

A = 10 ×

1 = 1.25 mCi = 1.25 × 10 −3 Ci 8

(Q 1 Ci = 3.70 × 1010 decays/sec) \

A = (1.25 × 10 −3 ) × (3.70 × 1010 ) = 4.63 × 107 decays per second

5. (a) The rate of production of A is a while its rate of decay is (–lN). Thus the net rate of change of the nuclei of A is given by dN dN = (a −lN ) or = dt dt a − lN

∫

N0

t

−1 dN N = ∫ dt or log e (a − lN ) =t N0 (a − lN ) l 0

or (a − lN ) = (a − lN 0 )e − lt which gives the number of nuclei at time t, 1 …(i) N = [a − (a − N 0 l)e − lt ] l (b) For a = 2lN0, eqn. (i) becomes N = 2N 0 − N 0e − lt

…(ii)

To obtain the number of nuclei after one half life of A, put t = T1/2 = N = 2N 0 − N 0 e

 0.693  −l  l 

N = 2N 0 −

{

N0 2

 e −0.693 =

3 = N0 2 The limiting value of N as t → ∞ is

1 2

}

N = lim 2N 0 − N 0e − lt  = 2N 0 − N 0e −∞ = 2N 0 t →∞

Integrating, we get N

\

0.693 in eqn. (ii) l

= 2N 0 − N 0e −0.693

6. Initial activity, A0 = lN0 Activity of nuclei at time t, A = lN = lN0e–lt = A0e–lt …(i) Since activity decreases at h = 4% per hour, so activity of 55Co radio nuclide at t = 1 h, A = A0 – hA0 = A0 (1 – h) …(ii) Taking natural log of equation (i), we get A ln = − lt A0 A (1 − h) ln 0 =−l A0 or l = –ln(1 – h) ≈ –(–h) \ l=h Put the value, we get l ≈ 1.1 × 10–5 s–1 Mean life time, t =

ATTENTION COACHING INSTITUTES:

1 = 9 × 104 s l

(using (ii))

nn

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PHYSICS FOR YOU | March ‘15

25

The apparent changes in frequency detected due to relative motion between source and observer is termed as Doppler’s effect. Now, the question is how does one layman understand whether there has been any change in frequency or not? The answer is very easy. If the sound appears to be more or less shrill, its frequency has either increased or decreased. Shrillness of sound is directly related to frequency. This is a common phenomenon experienced in day to day life. Supposedly, a bike is stationary and it blows its horn. The sound of the horn will be more shrill if the bike starts approaching us. Now, let us explore the details of the “how and why” of Doppler’s effect.

1 f0 Let, a wave be emitted at t = 0, hence, the wave would travel a distance l0 = vwT in a time t = T. But the source itself has moved towards observer by a distance vST as shown in figure.

Case-I : Source(S) moving, observer(O) stationary

Clearly, the waves appear to have been compressed, hence wavelength decreases for the observer. 1 \ lapp = vwT − vST = (vw − vS ) f0

T=

⇒ vw = wave velocity f0 = frequency of source towards observer vS = velocity of source towards observer In this case, the waves, once emitted, the propagation speed is only medium dependent, since the observer is stationary too. But what about wavelength? Wavelength is defined as the shortest distance between two points oscillating in same phase. Had the source been stationary, this distance surely would not have changed. Let me explain. If f0 is the frequency of the wave emitted, its time period,

vw 1 = (vw − vS ) fapp f0

 vw  ⇒ fapp =  f0  vw − vS  where, fapp = apparent frequency detected by observer. Clearly, fapp > f0, where source approaches observer. Similarly, had the source been moving away (receding) from the observer, the waves would have expanded, i.e., wavelength increased, hence frequency decreased. In general, due to motion of source towards/away from observer, the apparent frequency detected would be

Contributed By: Bishwajit Barnwal, Aakash Institute, Kolkata

26 physics for you |

march‘15

physics for you

| march ‘15 27

 v  fapp =  w  f0  vw  v s  Adjust the – or + sign using the simple logic, that when separation between S and O decreases, fapp should be more than f0, hence obviously – sign has to be used here. Case-II : Source(S) stationary, observer(O) moving

vO = velocity of observer away from source. In this case, as the source is stationary, there is no scope of wavelength appearing to have changed. But here, as the observer is moving away, the wave appears to be coming at a slower rate, hence the wave velocity changes in this case. \ vapp = vw – vO = apparent wave velocity for observer v \ vw − vO = fapp l0 = fapp w f0 v −v  \ fapp =  w O  f0  vw 

Clearly, the apparent frequency is lesser than the frequency emitted by source. A more generalised result when the observer moves towards/away from stationary source, would be v ±v  fapp =  w O  f0  v  w

Again, the same rule for using + or – sign, i.e., separation decreases then frequency should increase and vice versa. Case-III : Source(S) as well as observer(O) moving

This case taken is just case-I and II combined, i.e. due to the motion of source, wavelength changes whereas due to motion of observer, wave velocity changes, and due to cumulative effect of both, frequency of wave appears to have changed. A more 28 physics for you |

march‘15

generalised expression for finding out the apparent frequency would be v ±v  fapp =  w O  f0  v v  w

S

Let us learn, how to apply this formula for the above case shown where the source move towards observer and observer moves away from source. Due to motion of source, the separation decreases, hence frequency should increase (therefore, negative sign in denominator) whereas due to the motion of observer, the separation increases, hence frequency should decrease (hence negative sign in numerator too). v −v  \ fapp =  w O  f0  v −v  w

S

This is the frequency detected by observer for the case shown. Typical Examples

(1) Source(S) and observer(O), both moving with same velocity in same direction v

v

S

O

In this case, since there is no relative motion between source and observer, the observer would not detect any change in frequency but this does not mean that wavelength or wave velocity does not appear to have changed for the observer. Infact, the wavelength has decreased due to motion of source whereas the wave speed decreases due to the motion of observer and their cumulative effect is that there is no change in frequency. v −v  fapp =  w f0 = f0  v − v  w

(2) If the direction of motion of source/observer does not match with the line joining them. In this case, break the components of the velocity of the source and observer along the line joining them and use the same generalised formula.

 v + v cos θO  \ fapp =  w O f0  vw − vS cos θS  (3) One among the source or observer is stationary while the other moves perpendicular to the line joining them

Now, clearly there is no component of velocity along OS, there would not be any change in frequency. \ fapp = f0 (4) Frequency detected after reflection from a rigid boundary (wall/building/cliff)

We have two observers here, A and B. A will get to hear two frequencies, one of the wave which has been emitted from the source directly and the other after reflection from the wall. Hence he can also hear beats if the difference of frequencies is less than 10 Hz (due to limitation of resolution). The frequency received by the wall,  v  frec =  w  f0 ,  vw − v  since the wall becomes a stationary observer and the source is approaching the observer. The received frequency will be equal to the reflected frequency (fref).  v  \ fref = frec =  w  f0 v −v  w

Now the wall behaves as a stationary source of sound of frequency fref whereas A is the observer (moving with speed v towards wall). v +v  \ fapp =  w  fref  v w

 v + v   vw  f0 = w  vw   vw − v 

v +v  fapp =  w f0  vw − v  where fapp is the apparent frequency detected by the observer A after reflection from wall. \ Beat frequency detected by him (A) is fb = fref – f0 v +v  = w f0 − f0  vw − v   2v  = f0  vw − v  Now, what about B? Since B is stationary, he will receive the same frequency as received by wall, and the reflected frequency being equal to the received frequency, both the frequencies received by B, directly from the source as well as after reflection are identical and hence he would not hear any beats. A shortcut can also be used to find the reflected frequency as detected by A, where we create a virtual source S′ of S by taking reflection on the wall as shown in figure.

v +v  \ fapp =  w f0  vw − v  nn IMPORTANT EXAMINATION DATES 2015 4 april

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physics for you

| march ‘15 29

Y U ASKED

WE ANSWERED Do you have a question that you just can’t get answered? Use the vast expertise of our mtg team to get to the bottom of the question. From the serious to the silly, the controversial to the trivial, the team will tackle the questions, easy and tough. The best questions and their solutions will be printed in this column each month.

Q1. Why does a mobile phone blast while attending call during charging ? How can it be prevented ? –Sachin Vats (New Delhi)

Ans. It is a fact that mobile phones when answered while charging can sometimes lead to electrocution of the person and other fire and explosion hazards, but it is important to note that such accidents happen rarely and that to because of faulty mobile phone manufacturing, battery problem and low quality chargers. As a matter of safety, it is important to follow some precautions. Make sure the battery and charger are of the same brand as mobile phone. Avoid using phone on charging. Don’t tamper with the battery or bring it to contact with other metal objects outside the phone. Plug off as soon as mobile is fully charged. Avoid over heating of battery. Most importantly, follow the instructions of manufacturer for battery usage, storage, and recharging. Q2. What are auroras and how are they form?

-Aditya Prabhakar Warke

Ans. The bright dancing lights of aurora are caused by collisions between fast-moving particles (electrons) from space and the oxygen and nitrogen gas in our atmosphere. These electrons originate in the magnetosphere, the region of space controlled by Earth’s magnetic field. As they enter into the atmosphere, the electrons impact energy to oxygen and nitrogen molecules, making them excited. When the molecules return to their normal state, they release photon, small bursts of energy in the form of light. When billions of these collisions occur and enough photons are released, the oxygen and nitrogen in the atmosphere emit enough light for the eye 30 physics for you |

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to detect them. This ghostly glow can light up the night sky in a dance of colours. But since the aurora is much dimmer than sunlight, it cannot be seen from the ground in the daytime. The colour of the aurora depends on which gas is being excited by the electrons and on how much energy is being exchanged. Oxygen emits either a greenish-yellow light or a red light, nitrogen generally gives off a blue light. Auroras usually occur in ring-shaped areas centered around the magnetic poles of Earth. The brighter the colour, the more intense the aurora. The crescent of colour on the left is from sunlight scattered over the upper atmosphere. Q3. Why magnetic field intensity at the end of long solenoid is half than at the centre of the solenoid? – Suraj Gohel (Rajkot) Ans. A solenoid is made out of a current carrying wire which is coiled into a series of turns. In a solenoid, a large field is produced parallel to the axis of the solenoid. Components of the magnetic field in other directions are cancelled by opposing fields from neighbouring coils. Outside the solenoid the field is also very weak due to this cancellation effect and for a solenoid which is long in comparison to its diameter, the field is very close to zero. Inside the solenoid the fields from individual coils add together to form a very strong field along the center of the solenoid. The magnetic field at any point in space can be computed by summing over the magnetic fields produced by each turn of wire in the solenoid. It turns out that for an infinitely long solenoid, with the same number of turns per unit length of the solenoid, the magnetic field is constant in strength everywhere inside. If solenoid has ends, then you can think of it as an infinitely long solenoid minus the end parts that stretch off to infinity. The magnetic field strength on the axis going right through the solenoid, in the place on the end of the solenoid is then the field of an infinitely long solenoid minus half of it because half is missing, and so the field strength is half as big on the ends (but right in the middle). The field strength in the middle of a long solenoid is almost exactly that of an infinitely long solenoid, or twice that on the ends. nn

8

Optics and Modern Physics

reflection of light

When a light ray strikes the surface between two media, a part of it get return back in the initial medium. It is known as reflection. laws of reflection

• The incident ray, the Normal reflected ray and the normal i r to the surface, all lie in the same plane. • The angle of incidence is equal to the angle of reflection, i = r reflection of light at plane surface

In case of a reflection from plane surface such as plane mirror, the image is always erect, virtual and of same size as the object. It is also at the same distance behind the mirror as the object in front of it. When two plane mirrors are inclined at an angle q and an object is placed between them, multiple images of the object are formed as a result of multiple successive reflections.  360°  If  is an even integer, then the number of q   360°  images (n) is given by n =  −1 q   360°  If  is an odd integer, then the number of q  images (n) is decided according to the following two situations :

• If

the object lies symmetrically, then  360°  n= − 1. q  • If the object lies unsymmetrically, then  360°  n= . q  360° If is a fraction, the number of images formed q will be equal to its integral part. KEY POINT

• When two plane mirror are placed parallel to each other and an object is placed between them, the number of image formed will be infinite. reflection of light at spherical surface

A spherical mirror is a part of a spherical reflecting surfaces. They are of two types : • Concave mirror : If the reflection occurs from the inner surface of the spherical mirror, the mirror is called a concave mirror. • Convex mirror : If the reflection occurs from the outer surface of the spherical mirror, the mirror is called a convex mirror. C

F

P f

Concave Mirror

P

F

C

f Convex Mirror physics for you

| march ‘15 31

Here, P = pole of mirror, F = principal focus f = focal length, C = centre of curvature New cartesian sign conventions : All distances have to be measured from the pole of the mirror. Distances measured in the direction of incident light are taken as positive, while those measured in opposite direction are taken as negative. Heights measured upwards and normal to the principal axis of the mirror are taken as positive, while those measured downwards are taken as negative. the Mirror equation

1 1 1 where u is the distance of object from + = u v f the pole of the mirror and v is the distance of image from the pole of the mirror. f = R/2 where R is the radius of curvature of mirror. KEY POINT

• f or R is negative for concave mirror and positive for convex mirror Linear Magnification m=

size of image(I ) f −v v f =− = = size of object(O) u f −u f

m is positive for erect image and m is negative for inverted image. Axial Magnification 2

v  max = −   u Areal Magnification area of image mar = area of object Newton’s formula is f 2 = xy, where x is distance of object from the focus and y is distance of image from the focus of the mirror. refraction of light

Refraction of light is the change in the path of light due to change in velocity, when it goes from one medium to another. laws of refraction

• The incident ray, the refracted ray and the 32 physics for you |

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normal to the interface at the point of incidence, all lie in the same plane. • The ratio of sine of angle of incidence to i Medium 1 the sine of angle of Medium 2 refraction for any two r media is constant. sin i 1 i.e. = m2 sin r where 1m2 is the refractive index, of the medium 2 with respect to medium 1. This is also known as Snell’s law. • If 1m2 > 1, r < i the refracted ray bends towards the normal. In such a case medium 2 is said to be optically denser in comparison to medium 1. If 1m2 < 1, r > i the refracted ray bends away from the normal. In such a case medium 2 is said to be optically rarer in comparison to medium 1. Absolute refractive index : Refractive index of a medium with respect to vacuum (or in practice air) is known as absolute refractive index of the medium m=

c speed of light in vacuum = v speed of light in medium

General expression for Snell’s law  c   v  v sin i m 1 m2 = 2 = 2 = 1 = m1  c  v 2 sin r  v  1 where c is the speed of light in air, v1 and v2 be the speeds of light in medium 1 and medium 2 respectively. Principle of Reversibility : 1

lateral shift

1 m2 = 2 m1

When the medium is same on both sides of a glass slab, then the deviation of the emergent ray is zero. That is the emergent ray is parallel to the incident ray but it does suffer lateral shift with respect to the incident ray and is given by

Lateral shift, d = t

sin (i − r) cos r

where t is the thickness of the slab.

Air (2) Water (1) Incident ray

i2 i 1 i1

Refracted ray Reflected ray

When the angle of incidence is equal to the critical angle iC, the angle of refraction is 90°. i2 = 90°

2

real Depth and apparent Depth

When one looks into a pool of water, it does not appear to be as deep as it really is. Also when one looks into a slab of glass, the material does not appear to be as thick as it really is. This all happens due to refraction of light. If a beaker is filled with water and a point lying at its bottom is observed by someone located in air, then the bottom point appears raised. The apparent depth is less than the real depth. It can be shown that apparent depth =

real depth refractive index (m)

total internal reflection

It is the total reflection of light from a boundary between two mediums. It occurs when the angle of incident is greater than the critical angle for the two surfaces involved. Critical angle is the angle of incidence for which the angle of refraction is 90°. It exists only when light passes from a denser to a rarer medium. 1 Critical angle, sin iC = R mD If the rarer medium is air or vacuum, then sin iC =

1 m

When light travels from a higher refractive medium (water) into a lower refractive index medium (air), the refracted ray is bent away from the normal.

1

i C iC

If i1 is greater than iC, there is no refracted ray, and total internal reflection occurs. 2 1

iC

Total internal reflection

KEY POINT

• Critical angle depends on nature of media in contact and on the wavelength of light. application of total internal reflection

• The brilliance of diamond : The critical angle for diamond air interface is 24.4°. The diamond is cut suitably so that light entering the diamond from any face falls at an angle greater than 24.4°, and suffers multiple total internal reflections which results in sparkling of diamond. • Mirage : It is an optical illusion which occurs in hot, sunny days. The object such as a tree appears to be inverted, as if the tree is on the bank of a pond of water. • Optical fiber : It is a thin tube of transparent material that allows light to pass through, without being refracted into the air. Light undergoes successive total internal reflections as it moves through an optical fibre. physics for you

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refraction through a prism

Prism is a homogeneous, transparent medium enclosed by two plane surfaces inclined at an angle. These surfaces are called the refracting surfaces and angle between them is known as the refracting angle or the angle of prism. The angle between the incident ray and the emergent ray is known as the angle of deviation.

colours turn through different angles on passing through the prism. This is the cause of dispersion.

Red Yellow Violet

angular Dispersion

The difference in deviation between any two colours is known as angular dispersion. Angular dispersion dV – dR = (mV – mR)A where mV and mR are the refractive index for violet and red rays.

For refraction through a prism it is found that d = i + e – A where A = r1 + r2

d + dR . Mean deviation d = V 2 Dispersive power, w=

When A and i are small \ d = (m – 1) A

In a position of minimum deviation d = dm, i = e, and r1 = r2 = r  A + dm  A  and r = \ i =  2  2 The refractive index of the material of the prism is  (A + d m )  sin    2 m= A sin 2 This is known as prism formula, where A is the angle of prism and dm is the angle of minimum deviation.

()

Dispersion of light

It is the phenomenon of splitting of white light into its constituent colours on passing through a prism. This is because different colours have different wavelengths (lR > lV). According to Cauchy’s formula B

C

m=A+ 2 + 4 l l where A, B, C are arbitrary constants. Therefore, m of material of prism for different colours is different (mV > mR). As d = (m – 1) A, therefore different 34 physics for you |

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angular dispersion (dV − d R ) mean deviation (d)

m − mR , w= V (m − 1) m + mR where m = V = mean refractive index 2 Dispersion without Deviation Suppose we combine two prisms of refracting angles A and A′, and dispersive powers w and w′ respectively in such a way that their refracting angles are reversed with respect to each other. For no deviation, the condition is

d + d′ = 0

(m − 1) A (m – 1) A + (m′ – 1) A′ = 0 or A′ = − (m′ − 1) Under this condition, net angular dispersion produced by the combination = (dV − d R ) + (dV ′ − d′R )

= (mV − m R ) A + (mV ′ − m′R ) A′ Deviation without Dispersion

The condition for no dispersion is

(mV − m R ) A + (mV′ − m′R ) A′ = 0

or A′ = −

(mV − m R ) A (mV′ − m′R )

Under this condition, net deviation produced by the combination is

R1 R2

= d + d′ = (m – 1) A + (m′ – 1) A′ Rainbow : The rainbow is an example of the dispersion of sunlight by the water drops in the atmosphere. It is due to combined effect of dispersion, refraction and reflection of sunlight by spherical water droplets of rain. scattering of light

As sunlight travels through the earth’s atmosphere, it gets scattered (changes its direction) by the atmospheric particles. Light of shorter wavelengths is scattered much more than light of longer wavelengths. The amount of scattering is inversely proportional to the fourth power of the wavelength. This is known as Rayleigh scattering. refraction froM a spherical surface

A spherical refracting surface is a portion of a refracting medium whose curved surface is a part of a sphere. Spherical refracting surface are of two types : • Convex refracting surface • Concave refracting surface For refraction from rarer to denser medium m − m1 m m − 1+ 2= 2 u v R where u, v and R be the object distance, image distance and radius of curvature from the spherical surface respectively. When refraction occurs from denser to rarer medium we interchange m1 and m2 m m m − m2 − 2+ 1= 1 u v R refraction by a lens A lens is a portion of a transparent refracting medium bound by two spherical surfaces, or one spherical surface and a plane surface. Lenses are of two types: • Convex lens is a lens that is thicker in the middle than at the edges as shown in figure (a). • Concave lens is a lens that is thinner in the middle than at the edges as shown in figure (b).

Biconvex R1

Plano convex (a)

Concavo convex

R2

Biconcave

R2

R1

Plano concave Convexo concave (b)

KEY POINT

• The sign convention for thin lenses are same as those of spherical mirrors except that instead of pole of the mirror, we take use of optical centre of the lens. lens Maker’s formula

1 1 1 = (m − 1) − f R1 R2

when R1 and R2 are radii of curvature of the two surfaces of the lens and m is the refractive index of material of lens with respect to surrounding medium. When the refractive index of the material of the lens is greater than that of the surroundings, then biconvex lens acts as a converging lens and a biconcave lens acts as a diverging lens as shown in the figure.

When the refractive index of the material of the lens is smaller than that of the surrounding medium, then biconvex lens acts as a diverging lens and a biconcave lens acts as a converging lens as shown in the figure.

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thin lens formula

1 1 1 − = v u f

SELF CHECK

For a convex lens, P is positive, and for a concave lens, P is negative.

1. A thin convex lens made from crown glass 3   m = 2  has focal length f. When it is measured in two different liquids having refractive 4 5 indices and , it has the focal lengths f1 and 3 3 f2 respectively. The correct relation between the focal lengths is (a) f1 and f2 both become negative (b) f1 = f2 < f (c) f1 > f and f2 becomes negative (d) f2 > f and f1 becomes negative (JEE Main 2014) 2. An object 2.4 m in front of a lens forms a sharp image on a film 12 cm behind the lens. A glass plate 1 cm thick, of refractive index 1.50 is interposed between lens and film with its plane faces parallel to film. At what distance (from lens) should object be shifted to be in sharp focus on film? (a) 2.4 m (b) 3.2 m (c) 5.6 m (d) 7.2 m (AIEEE 2012)

combination of thin lenses in contact

optical instruMents

where u is the distance of the object from the optical centre of the lens, v is the distance of the image from the optical centre of the lens, f is the focal length of lens. KEY POINT

• f is positive for converging or convex lens and f is negative for diverging or concave lens. linear Magnification

m=

size of image (I ) v = . size of object (O) u

m is positive for erect image and m is negative for inverted image. power of a lens

P=

1 . focal length in metres

The SI unit of power of lens is dioptre (D). 1 D = 1 m–1.

When a number of thin lenses of focal length f1, f2, ...etc. are placed in contact coaxially, the equivalent focal length F of the combination is given by 1 1 1 1 = + + + .... F f1 f 2 f 3 The total power of the combination is given by P = P1 + P2 + P3 + ... The total magnification of the combination is given by m = m1 × m2 × m3 .... When two thin lenses of focal lengths f1 and f2 are placed coaxially and separated by a distance d, the focal length of a combination is given by d 1 1 1 = + − . F f1 f 2 f1 f 2 In terms of power P = P1 + P2 – dP1P2. 36 physics for you |

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simple Microscope

It is used for observing magnified images of tiny objects. It consist of a converging lens of small focal length. Magnifying power of a simple microscope when the image is formed at infinity, M=

D f

where D is the least distance of distinct vision and f is the focal length of convex lens. When the image is formed at near point (at D), D M = 1+ . f compound Microscope

A compound microscope consist of two convex lenses coaxially separated by some distance. The

lens nearer to the object is called the objective and the lens through which the final image is viewed is called the eyepiece. Magnifying power of a compound microscope

When the final image is formed at infinity (normal adjustment), v D M=− o   uo  f e 

In normal adjustment, f Magnifying power, M = o fe Length of the tube, L = fo + 4f + fe

Length of tube, L = vo + fe When the final image is formed at least distance of distinct vision, v  D M = − o 1 +  uo  fe  where uo and vo represent the distance of object and intermediate image from the objective lens, fe is the focal length of an eye lens.  f D  Length of the tube, L = v o +  e  f e + D  astronomical refracting telescope

It is used for observing astronomical bodies. Refracting telescope use lenses as their main components. The one facing the object is called objective or field lens and has large focal length, while the other facing the eye is called eye-piece or ocular has small focal length. Magnifying power of a astronomical telescope

When the final image is formed at infinity (normal adjustment), f M=− o fe Length of tube, L = fo + fe When the final image is formed at least distance of distinct vision, f  f  M = − o 1 + e  fe D Length of tube, L = f o +

final image must be erect with respect to the object. To achieve it, an inverting convex lens (of focal length f) is used in between the objective and eye piece of astronomical telescope. This lens is known as erecting lens.

fe D fe + D

terrestrial telescope

It is used for observing far off objects on the ground. The essential requirement of such a telescope is that

reflecting type telescope

Reflecting type telescope was designed by Newton in order to overcome the drawbacks of refracting type telescope. In a reflecting type telescope, a concave mirror of large aperture is used as objective in place of a convex lens. It possesses a large light gathering power and a high resolving power. Due to this, it enables us to see even faint stars and observe their minute details. In normal adjustment

()

R fo Magnifying power, M = = 2 fe fe

where R is the radius of curvature of concave mirror. Reflecting type telescope is free from chromatic aberration because light does not undergo refraction. wave optics

Wave optics deals with the theories of the nature of light and provides an explanation for different phenomena like reflection, refraction, interference, diffraction and polarisation. wavefront

Particles of light wave which are equidistant from the light source and vibrate in the same phase constitute a wavefront. Depending on the type of light source, wavefronts are of three types: • Spherical wavefront : It is formed by a point source of light. • Cylindrical wavefront : It is formed by a linear source of light. physics for you

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• Plane wavefront : It is obtained from a point source of light when the observation point is far away from the light source.

I = I1 + I 2 + 2 I1I 2 cos f If I1 = I2 = I0, then I = I0 + I0 + 2I0 cos f = 4I0 cos2 I = I1 +I 2 + 2 I1I 2 cos f

huygens principle

According to this principle, each and every point on the given wavefront called primary wavefront, acts as a source of new disturbances, called secondary wavelets, that travel in all directions with the velocity of light in the medium. A surface touching these secondary wavelets tangentially in the forward direction at any instant gives a new wavefront at that instant, which is known as secondary wavefront. coherent sources

Sources of light which emit continuous light waves having the same wavelength, same frequency and in same phase or having a constant phase difference are known as coherent sources of light. interference of light

It is the phenomenon of redistribution of energy on account of superposition of light waves from two coherent sources. Interference pattern produce points of maximum and minimum intensity. Points where resultant intensity is maximum, interference is said to be constructive and at the points of destructive interference, resultant intensity is minimum. conditions for sustained interference of light

The two sources should continously emit waves of the same wavelength or frequency. The amplitudes of waves from two sources should preferably be equal. The waves emitted by the two sources should either be in phase or should have a constant phase difference. The two sources must lie very close to each other. The two sources should be very narrow. intensity Distribution

If a, b are the amplitudes of interfering waves due to two coherent sources and f is constant phase difference between the two waves at any point P, then the resultant amplitude at P will be R = a 2 + b2 + 2ab cos f If a2 = I1, b2 = I2, then resultant intensity, I = R2 = a2 + b2 + 2 ab cos f 38 physics for you |

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When cos f = 1, I max = I1 + I 2 + 2 I1I 2 =

When cos f = − 1, I min = or

I max = I min

( (

(

I1 − I 2

)2

(

f 2

I1 + I 2

)2

)2 = (a + b)2 2 (a − b)2 I2 )

I1 + I 2 I1 −

If I1 = I2 = I0, then Imax = 4I0, Imin = 0. f

Resultant intensity, I = 4I0 cos2 2 If the sources are incoherent, I = I1 + I2 young’s double slit experiment

Young’s double slit experiment was the first to demonstrate the phenomenon of interference of light. Using two slits illuminated by monochromatic light source, he obtained bright and dark bands of equal width placed alternately. These were called interference fringes. • For constructive interference (formation of bright fringes) For nth bright fringe, d Path difference = xn = nl D where n = 0 for central bright fringe n = 1 for first bright fringe, n = 2 for second bright fringe and so on d = distance between the two slits D = distance of slits from the screen xn = distance of nth bright fringe from the centre. D \ xn = nl d • For destructive interference (formation of dark fringes). For nth dark fringe, d l = (2n − 1) D 2 where, n = 1 for first dark fringe, n = 2 for 2nd dark fringe and so on. path difference = xn

xn = distance of nth dark fringe from the centre lD \ xn = (2n −1) 2 d Fringe width : The distance between any two consecutive bright or dark fringes is known as fringe width. lD Fringe width, β = d β l Angular fringe width, q = = D d If W1, W2 are widths of two slits, I1, I2 are intensities of light coming from two slits; a, b are the amplitudes of light from these slits, then W1 I1 a 2 = = W2 I 2 b2

2 I max (a + b) = I min (a − b)2

I −I Fringe visibility, V = max min I max + I min When entire apparatus of Young’s double slit experiment is immersed in a medium of refractive index m, then fringe width becomes l′D lD β β′ = = = d md m When a thin transparent plate of thickness t and refractive index m is placed in the path of one of the interfering waves, fringe width remains unaffected but the entire pattern shifts by D β = (m − 1) t d l This shifting is towards the side in which transparent plate is introduced. Dx = (m −1) t

KEY POINT

• For constructive interference at a point, the phase difference between the two waves reaching that point should be zero or an even integral multiple of p. • For destructive interference at a point the phase difference between the two waves reaching that particular point should be an odd integral multiple of p.

Diffraction

It is the phenomenon of bending of light around the corners of an obstacle placed in its path, on account of which it penetrates into the region of geometrical shadow of the obstacle. Diffraction of light at a single slit

In this case, the diffraction pattern obtained on the screen consists of a central bright band, having alternate dark and weak bright bands of decreasing intensity on both sides. Condition for nth secondary maximum in terms of l path difference = a sin qn = (2n − 1) 2 where n = 1, 2, 3,....... Condition for nth secondary minimum in terms of path difference = asinqn = nl where n = 1, 2, 3,....... Width of secondary maxima or minima lD lf = a a where, a is the width of slit, D is the distance of screen from the slit, f is the focal length of lens for diffracted light. 2lD 2 f l = . Width of central maximum = a a β=

Angular fringe width of central maximum =

2l . a

Angular fringe width of secondary maxima or l minima = . a resolving power

It is the ability of an optical instrument to produce distinctly separate images of two close objects i.e. it is the ability of the instrument to resolve or to see as separate, the images of two close objects. limit of resolution

The minimum distance between two objects which can just be seen as separate by the optical instrument is known as the limit of resolution of the instrument. Smaller the limit of resolution of the optical instrument, greater is its resolving power and vice-versa. physics for you

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resolving power of a Microscope

It is defined as the reciprocal of the minimum distance d between two point objects, which can just be seen through the microscope as separate. 1 2m sin q Resolving power = = d l where m is refractive index of the medium between object and objective lens, q is half the angle of cone of light from the point object, d represents limit of resolution of microscope and msinq is known as the numerical aperture. resolving power of a telescope

It is defined as reciprocal of the smallest angular separation (dq) between two distant objects, whose images are just seen in the telescope as separate. Resolving power =

1 D = dq 1.22 l

where D is diameter or aperture of the objective lens of the telescope, dq represents limit of resolution of telescope. polarisation

The phenomenon of restricting vibrations of light to a single plane is known as polarisation of light. Angle of polarisation is the angle of incidence for which an ordinary light is completely polarised in the plane of incidence when it gets reflected from a transparent medium. Plane of vibration

Unpolarised light

Plane polarised light

Plane of vibration

The plane in which the vibrations of polarised light are confined is known as plane of vibration and plane perpendicular to the plane of vibration is known as plane of polarization. Brewster’s law

According to this law, when unpolarised light is incident at polarising angle on the interface separating air from a medium of refractive index m, the reflected light is fully polarised, provided 40 physics for you |

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the refractive index of the medium is equal to the tangent of the polarising angle. m = tan ip Malus’ law

According to this law, when a beam of plane polarised light is incident on the analyzer, the intensity of light transmitted from the analyser is directly proportional to the square of the cosine of the angle between the planes of transmission of the polariser and analyser. I a cos2q If the intensity of plane polarised light incident on analyser is I0, then intensity of light emerging from analyser is I = I0 cos2q polaroids

A polaroid is a type of plastic sheet which polarises light. It can be used to control the intensity of light in sunglasses, windowpanes, photographic cameras and 3D movie cameras. KEY POINT

• When light is incident at polarising angle, the reflected and refracted rays are perpendicular to each other.

SELF CHECK 3. Two beams, A and B, of plane polarized light with mutually perpendicular planes of polarisation are seen through a polaroid. From the position when the beam A has maximum intensity (and beam B has zero intensity), a rotation of polaroid through 30° makes the two beams appear equally bright. If the initial intensities of the two beams are IA and IB I respectively, then A equals IB 1 3 (a) (b) 3 (c) (d) 1 3 2 (JEE Main 2014) 4. A beam of unpolarised light of intensity I0 is passed through a polaroid A and then through another polaroid B which is oriented so that its

(b) I0

(c) I0/2

(d) I0/4

(JEE Main 2013) Dual nature of Matter anD raDiation

Phenomena like interference, diffraction and polarisation can be explained only on the basis of wave nature of radiation whereas phenomena like black body radiation, photoelectric effect, compton effect can be explained only on the basis of particle (quantum) nature of radiation. Thus, radiation has dual nature i.e. particle and wave. photoelectric effect

It is the phenomenon of emission of electrons from the surface of metals, when light radiations of suitable frequency fall on them. The emitted electrons are known as photoelectrons and the current so produced is known as photoelectric current. work function

The minimum energy needed by an electron to come out from a metal surface is known as work function of the metal. It is denoted by f0 or W0 and measured in electron volt (eV). The work function depends on the properties of the metal and the nature of its surface. laws of photoelectric emission

The laws of photoelectric effect are as follows : For a given metal and frequency of incident radiation, the number of photoelectrons ejected per second is directly proportional to the intensity of the incident light. For a given metal, there exists a certain minimum frequency of the incident radiation below which no emission of photoelectrons takes place. This frequency is known as threshold frequency. Above the threshold frequency, the maximum kinetic energy of the emitted photoelectron is independent of the intensity of incident light but depends only upon the frequency (or wavelength) of the incident light. The photoelectric emission is an instantaneous process.

Hertz found that high voltage sparks across a detector loop were enhanced when an emitter plate was illuminated by ultraviolet light from an arc lamp. Hallwach’s and Lenard found that when ultraviolet radiation was allowed to fall on the emitter plate of an evacuated glass tube enclosing two metal plates, current flowed in the circuit. After the discovery of electrons, it became evident that the incident light causes electrons to be emitted from the emitter plate. It was also observed that no electrons were emitted at all when the frequency of the incident light was smaller than a certain minimum value. experimental features and observations of photoelectric effect

Experimental features and observations of photoelectric effect are as follows : • For a given photosensitive material and frequency of incident radiation (above the threshold frequency) the photoelectric current is directly proportional to the Intensity of light intensity of incident light. • For a given photosensitive material and frequency of incident radiation, saturation current (the maximum value of photoelectric current) is found to be proportional to the intensity of incident radiation whereas the stopping potential is independent of its intensity. Photoelectric current

(a) I0/8

hertz’s and lenard’s observation

Photoelectric current

principal plane makes an angle of 45° relative to that of A. The intensity of the emergent light is

I3 > I2 > I1 I3 I2 I1

Stopping potential –V0 Retarding potential

O Collector plate potential

• For a given photosensitive material, there exists a certain minimum cut-off frequency physics for you

| march ‘15 41

of the incident radiation, called the threshold frequency, below which no emission of photoelectrons takes place, no matter how intense the incident light is. Above the threshold frequency, the stopping potential or equivalently the maximum kinetic energy of the emitted photoelectrons increases linearly with the frequency of incident radiation, but is independent of its intensity. Stopping potential (V0)

O

Metal A  > 0

Metal B  > 0

0 0 ( ) Frequency of incident radiation

The photoelectric emission is an instantaneous process without any apparent time lag (~ 10–9 s or less), even when the incident radiation is made exceedingly dim. particle nature of light : the photon

Einstein proposed that electromagnetic radiation is quantized and exists in elementary amounts we now call photons. The photon picture of electromagnetic radiations and the characteristic properties of photons are as follows : • In the interaction of radiation with matter, radiation behaves as if it is made of particles like photons. • Each photon has energy E (= hu = hc/l) and  hu h  momentum p  = = , where h is Planck’s c l constant, u and l are the frequency and wavelength of radiation and c is the velocity of light. • Irrespective of the intensity of radiation, all the photons of a particular frequency have the same energy and same momentum. • The photon energy is independent of the intensity of radiations. • All the photons emitted from a source of radiations travel through space with the same speed c. 42 physics for you |

march ‘15

• The frequency of photon gives the radiation, a definite energy (or colour) which does not change when photon travels through different media. • The velocity of photon in different media is different which is due to change in its wavelength. • The rest mass of a photon is zero. According to theory of relativity, the mass m of a particle moving with velocity v, comparable with the velocity of light c is given by m0 or m0 = m 1 − v 2 /c 2 ...(i) m= 2 2 1 − v /c where m0 is the rest mass of particle. As a photon moves with the speed of light, v = c, so from (i), m0 = 0. • Photons are not deflected by electric and magnetic fields. This shows that photons are electrically neutral. • In a photon-particle collision (such as photoelectron collision), the energy and momentum are conserved. However the number of photons may not be conserved in a collision. Matter waves

The waves associated with the moving material particles are called matter waves or de Broglie waves. de Broglie wavelength

The wavelength associated with moving particle is called de Broglie wavelength and it is given by l=

h h = p mv

where m is the mass, p is the momentum and v is the velocity of particle and h is the Planck’s constant. de Broglie wavelength is independent of the charge and nature of the material particle. In terms of kinetic energy K, de Broglie wavelength is given by l =

h 2mK

If a particle of charge q is accelerated through a potential difference V, its de Broglie wavelength is given by

h 2mqV 1/2

 150  l= ¯ For an electron, V  For a gas molecule of mass m at temperature T kelvin, its de Broglie wavelength is given by h l= , where k is the Boltzmann constant. 3mkT Davisson anD gerMer eXperiMent

The wave nature of electrons was first experimentally verified by C.J. Davisson and L.H. Germer in 1927 and independently by G.P. Thomson, in 1928, who observed diffraction effects with beams of electrons scattered by crystals.

SELF CHECK 5. The surface of a metal is illuminated with the light of 400 nm. The kinetic energy of the ejected photoelectrons was found to be 1.68 eV. The work function of the metal is (hc = 1240 eV nm) (a) 3.09 eV (b) 1.41 eV (c) 1.51 eV (d) 1.68 eV (AIEEE 2009) atoMs

The first model of atom was proposed by J J Thomson in 1898. According to this model, the positive charge of the atom is uniformly distributed through out the volume of the atom and the negatively charged electrons are embedded in it like seeds in a watermelon. This model was called as plum pudding model of atom. alpha particle scattering eXperiMent anD rutherforD’s MoDel of atoM

Rutherford’s alpha scattering experiment led to the discovery of the nucleus. Rutherford and his co-workers directed a beam of alpha particles at a thin metal foil made of gold. If the plum pudding model were correct, then the alpha particles would be expected to pass nearly straight through the foil. But in actual, alpha particles were scattered in different directions.

A graph was plotted, between the scattering angle q and the number of alpha particles scattered at ∠q for a very large number of alpha particles, as shown in figure. The following observations can be made from the graph. • Most of the alpha particles passed straight through the gold foil undeflected. • Some of the alpha particles were deflected at small angles. • Very few (one in thousands) alpha particles retraced their paths after passing through the gold foil. Number of alpha particles scattered

l=

107 106 105 104 103 102 10

0

20 40 60 80 100 120 140 160 180 Scattering angle (°)

alpha particle trajectory

The perpendicular distance between the initial velocity vector of an alpha particle from a central line passing through the centre of nucleus, when the alpha particle is far away from the nucleus, is known as its impact parameter. For larger impact parameters, the force F experienced by the alpha particle is weak because 1 . F∝ (distance)2 The scattering angle q of the alpha particle and impact parameter b are related as b=

Ze 2 cot(q / 2) 4 pε0 K

where K is the kinetic energy of alpha particle and Z is the atomic number of the nucleus. Smaller the impact parameter, larger the angle of scattering q. Distance of closest approach

At the distance of closest approach, whole of the kinetic energy of the alpha particles is converted into potential energy. physics for you

| march ‘15 43

Distance of closest approach, r0 =

Bohr developed a theory of hydrogen and hydrogen-like atoms which have only one orbital electron. His postulates are as follows : • An electron can revolve around the nucleus only in certain allowed circular orbits of definite energy and in these orbits it does not radiate. These orbits are called stationary orbits. • Angular momentum of the electron in a stationary orbit is an integral multiple of h/2p. i.e., L = nh or, mvr = nh 2p 2p where m is the mass of the electron, v is the velocity of the electron, r is the radius of the orbit and n is a positive integer called principal quantum number. This is called as Bohr’s quantisation condition. This postulate is equivalent to saying that in a stationary state, the circumference of a circular orbit contains integral numbers of de Broglie wavelength. nh nh 2pr = nl = i.e. L = mvr = . mv 2p • The emission of radiation takes place when an electron makes a transition from a higher to a lower orbit. The frequency of the radiation is given by E − E1 u= 2 h where E2 and E1 are the energies of the electron in the higher and lower orbits respectively. Since the centripetal force for circular orbit is provided by the Coulomb’s force, we have 1 Ze 2 mv 2 = r 4pε0 r 2 where Z is the atomic number of the element and e is the electronic charge. Bohr’s formulae

• Radius of rn =

orbit

4pε0n2h 2

ε0n2h 2 = 4p2mZe 2 pmZe 2

44 physics for you |

ε n2h 2 rn = 0 2 = a0n2 pme

2 Ze 4 pε0 K

Bohr’s MoDel of atoM

nth

For hydrogen atom, Z = 1

2

march ‘15

h 2ε 0

= 0.53 × 10 −10 m pme 2 is called Bohr’s radius • Velocity of electron in nth orbit where a0 =

1 2pZe 2 acZ Ze 2 c Z = = = 4pε0 nh 2ε0nh 137 n n

vn =

1 e2 = 2ε0hc 137 a is called fine structure constant and is a pure number. • Frequency of electron in nth orbit where a =

un =

2

vn  1  4p2Z 2e 4m me 4 Z 2 = = 2 3 3 2prn  4pε0  n3h 3 4ε0n h

• Time period of revolution of electron in nth orbit Tn =

2prn n3h 3(4pε0 )2 4ε20n3h 3 = = vn me 4 Z 2 4p2Z 2e 4m

• Kinetic energy of electron in nth orbit Kn = =

2

1 Ze 2  1  2p2me 4 Z 2 =  4pε0 2rn  4pε0  n2h 2

13.6Z 2

eV. n2 • Potential energy of electron in nth orbit Un = − =

2

1 Ze 2  1  4p2me 4 Z 2 = − 4pε0 rn  4pε0  n2h 2

−27.2Z 2

eV n2 • Total energy of electron in nth orbit 2

 1  2p2me 4 Z 2 En = U n + K n = −   4pε0  n2h 2 =−

13.6Z 2

eV. n2 • When an electron makes a transition from initial state ni to final state nf (ni > nf), then the frequency of emitted radiation is given by

1   1 −  n2 n2   f i  • Wavelength of emitted radiation is given by 1 1   1 = RZ 2 2 − 2 n  l  f ni  where R is called Rydberg’s constant. u = RcZ 2

2

 1  2p2me 4 = 1.097 × 107 m −1. R= 3  4pε0  ch 1 is called wave number and is denoted by u. l This relation holds for radiation by hydrogen like atoms i.e. H (Z = 1), He+ (Z = 2), Li++ (Z = 3) and Be+++ (Z = 4). ionization energy and ionization potential

• Ionization energy =

13.6Z 2

• Ionization potential =

n2

eV.

13.6Z 2 n2

V.

SELF CHECK 6. In a hydrogen like atom electron makes transition from an energy level with quantum number n to another with quantum number (n – 1). If n > > 1, the frequency of radiation emitted is proportional to 1 1 (a) 3 (b) n n (c)

1

n2

(d)

1

n

3/2

(JEE Main 2013)

spectral series of hyDrogen atoM

When the electron in a H atom jumps from higher energy level to lower energy level, the difference of energies of the two energy levels is emitted as radiation of particular wavelength, known as spectral line. lyman series

Emission spectral lines corresponding to the transition of electron from higher energy levels (n2 = 2, 3, ...,∞) to first energy level (n1 = 1) constitute Lyman series.

1 1 1 = R  2 − 2  where n2 = 2, 3, 4, ......,∞ l 1 n2  Balmer series Emission spectral lines corresponding to the transition of electron from higher energy levels (n2 = 3, 4, ....∞) to second energy level (n1 = 2) constitute Balmer series. 1 1 1 = R  2 − 2  where n2 = 3, 4, 5...........,∞ l n2  2 paschen series

Emission spectral lines corresponding to the transition of electron from higher energy levels (n2 = 4, 5, .....,∞) to third energy level (n1 = 3) constitute Paschen series. 1 1 1 = R  2 − 2  where n2 = 4, 5, 6.........,∞ l n2  3 Brackett series Emission spectral lines corresponding to the transition of electron from higher energy levels (n2 = 5, 6, 7,.....,∞) to fourth energy level (n1 = 4) constitute Brackett series. 1 1 1 = R  2 − 2  where n2 = 5, 6, 7..........,∞ l n2  4 pfund series

Emission spectral lines corresponding to the transition of electron from higher energy levels (n2 = 6, 7, 8,.......,∞) to fifth energy level (n1 = 5) constitute Pfund series. 1 1 1 = R  2 − 2  where n2 = 6, 7,...........,∞ l n2  5 Number of spectral lines due to transition of electron from nth orbit to lower orbit is N=

n(n − 1) . 2

energy quantisation

In quantum mechanics, the energies of a system are discrete or quantised. The energy of a particle of mass m confined to a box of length L can have only discrete values of energy given by the relation En =

n2 h 2

8mL2

where n = 1, 2 , 3 , ......... physics for you

| march ‘15 45

The radius of the nucleus is given by

SELF CHECK (1H1),

H2),

7. Hydrogen Deuterium (1 singly ionised Helium (2He4)+ and doubly ionised lithium (3Li6)++ all have one electron around the nucleus. Consider an electron transition from n = 2 to n = 1. If the wavelengths of emitted radiation are l1, l2, l3 and l4 respectively then approximately which one of the following is correct? (a) l1 = 2l2 = 3l3 = 4l4 (b) 4l1 = 2l2 = 2l3 = l4 (c) l1 = 2l2 = 2l3 = l4 (d) l1 = l2 = 4l3 = 9l4 (JEE Main 2014) 8. Hydrogen atom is excited from ground state to another state with principal quantum number equal to 4. Then the number of spectral lines in the emission spectra will be (a) 3 (b) 5 (c) 6 (d) 2 (AIEEE 2012)

R = R0 A1/3 where R0 is a constant. nuclear Density

Nuclear density, ρ =

Mass of the nucleus Volume of the nucleus

Nuclear density is independent of A and is in the order of 1017 kg m–3. KEY POINT

• The sizes of nuclei are different, the heavier nuclei being bigger than the lighter ones. But density of nuclear matter is same for all nuclei. isotopes, isoBars anD isotones

Isotopes : The atoms of same element which have same atomic number but different mass numbers are called isotopes.

nuclei

e.g. (1H1, 1H2 and 1H3) and (8O16, 8O17 and 8O18) are isotopes.

A tiny central core in which the entire positive charge and almost entire mass of the atom are concentrated is called as nucleus.

Isobars : The atoms of different elements which have same mass number but different atomic numbers are called isobars.

atoMic Masses anD coMposition of nucleus

e.g. (8O17 and 9F17) and (11Na24 and isobars.

A nucleus is made up of protons and neutrons. The number of protons in a nucleus (called the atomic number or proton number of the nucleus) is represented by the symbol Z, the number of neutrons (the neutron number) is represented by the symbol N. A neutron is a nucleon with no charge. The total number of neutrons and protons in a nucleus is called its mass number A. Thus A=Z+N Neutrons and protons, when considered collectively, are called nucleons. The atomic mass unit is one-twelfth of the mass of a 12 atom of carbon. It is represented by the symbol 6C u and it is the average mass of a nucleon.

24 12Mg )

are

Isotones : The atoms of different elements of which the nuclei have the same number of neutrons but different number of protons are called isotones. e.g. (11Na23 and 12Mg24) and (19K39 and 20Ca40) are isotones. Mass Defect

The mass of a nucleus is always less than the total mass of its constituents. The difference in mass of a nucleus and its constituents is called mass defect and is given by DM = [Zmp + (A – Z)mn – M]

properties of nucleus

where mp is the mass of the proton and mn is the mass of the neutron and M is the mass of the nucleus.

nuclear size

BinDing energy

The size of the nucleus is of the order of fermi. 1 fm = 10–15 m 46 physics for you |

march ‘15

The energy equivalent of mass defect of a nucleus is called the binding energy of the nucleus.

The binding energy of nucleus is given by Eb =

DMc2

= [Zmp + (A – Z)mn –

M]c2

Mean life or average life of a radioactive substance is given by

= [Zmp + (A – Z)mn – M] × 931.49 MeV/u.

τ=

The binding energy per nucleon of a nucleus = Eb/A The greater the binding energy per nucleon, the more stable is the nucleus. Binding Energy curve

It is curve drawn between binding energy per nucleon and mass number as shown in the figure.

1 T1/2 = = 1.44T1/2 l 0.693

activity

The number of disintegrations occurring in a radioactive substance per second is called activity and it is given by R = –dN/dt. The SI unit of activity is becquerel. 1 becquerel = 1 Bq = 1 decay/sec.

Binding energy per nucleon (MeV)

The traditional unit of activity is the curie. 1 curie = 1 Ci = 3.70 × 1010 decays/s = 37 GBq. activity law

R(t) = R0e–lt, where R0 = lN0 is the decay rate at t = 0 and R = Nl. Fraction of nuclei left undecayed after n half lives is n

The main features of the curve are as follows : • The binding energy per nucleon is practically constant, i.e. independent of the atomic number for nuclei of middle mass number (30 < A < 170). The curve has a maximum of about 8.75 MeV for A = 56 and has a value of 7.6 MeV for A = 238. • The binding energy per nucleon is lower for both light nuclei (A < 30) and heavy nuclei (A > 170). radioactivity

A nuclear phenomenon in which an unstable nucleus undergoes decay with the emission of alpha particles, beta particles, or gamma rays. Law of radioactive decay dN = −lN (t ) or N (t ) = N 0e − lt dt where l is the decay constant or disintegration constant, N is the number of nuclei left undecayed at the time t, N0 is the number of radioactive nuclei at t = 0. Half-life of a radioactive substance is given by T1/2 =

ln 2 0.693 = l l

N 1 1 = =  N0  2  2

t /T1/2

or t = nT1/2

SELF CHECK 9. The half life of a radioactive substance is 20 minutes. The approximate time interval 2 (t2 – t1) between the time t2 when of it has 3 1 decayed and time t1 when of it had decayed 3 is (a) 7 min (b) 14 min (c) 20 min (d) 28 min (AIEEE 2011) 10. A radioactive nucleus (initial mass number A and atomic number Z) emits 3a-particles and 2 positrons. The ratio of number of neutrons to that of protons in the final nucleus will be (a)

A−Z −4 Z −2

(b)

A−Z −8 Z−4

(c)

A−Z −4 Z −8

(d)

A − Z − 12 Z−4 (AIEEE 2010)

physics for you

| march ‘15 47

nuclear reaction

A nuclear reaction represents the transformation of one stable nucleus into another nucleus by bombarding the former with suitable high energy particles. It is represented by A + a → B + b + Q where A is the target nucleus, a is the impinging particle, B and b the products, Q is the energy released in the process. Q value of nuclear reaction, Q = (mA + ma – mB – mb)c2 If Q is positive, the reaction is termed as exothermic and if Q is negative the reaction is termed as endothermic. Every nuclear reaction obey following laws : • Conservation of charge • Conservation of mass • Conservation of linear momentum • Conservation of energy nuclear fission

It is the phenomenon of splitting a heavy nucleus into two or more smaller nuclei. The

nuclear fission of 235 92 U is represented as 235 1 141 92 1 92 U + 0 n → 56 Ba + 36 Kr + 3 0 n + Q

The value of Q is 200 MeV per fission reaction. nuclear chain reaction

Under suitable conditions, the three secondary neutrons may cause further fission of U235 nuclei

and start what is known as nuclear chain reaction. The nuclear chain reaction is controlled by Neutron reproduction factor (K) rate of production of neutrons = rate of loss of neutrons Uncontrolled nuclear chain reaction is the basis of an atom bomb. Controlled nuclear chain reaction is the basis of a nuclear reactor. nuclear reactor

Nuclear reactor uses nuclear energy for peaceful purposes. It is based on the phenomenon of controlled nuclear chain reaction. Moderators like heavy water, graphite, paraffin and deuterium slow down neutrons. Rods of cadmium or boron serve as control rods. Ordinary water and heavy water serve as coolants. nuclear fusion

It is the phenomenon of fusing two or more lighter nuclei to form a single heavy nucleus. The nuclear fusion reaction of two deutrons is represented as 2 1H

+ 21H → 42He + 24 MeV Temperature ≈ 107 K are required for fusion to take place. Nuclear fusion is a basis of hydrogen bomb. nn

answer keys (self check) 1. (c)

2. (c)

3. (a)

4. (d)

5. (b)

6. (a)

7. (d)

8. (c)

9. (c)

10. (c)

IMAX 3-D MOVIES An exciting application of crossed polarisers that is, two polarisers which are perpendicular to each other is in viewing IMAX 3-D movies. These movies are recorded on two separate rolls of film, using a camera that provides images from two different perspectives that correspond to what is observed by human eyes and allow us to see in three dimensions. The camera has two apertures or openings located at roughly the spacing between our eyes. The 3-D films are projected using a projector with two lenses. Each lens has its own polariser, and the two polarisers are crossed. Viewers watch the action on-screen using glasses with corresponding polarisers for the left and right eyes, as the drawing shows. Because of the crossed polarisers the left eye sees only the image from the left lens of the projector, and the right eye sees only the image from the right lens. Since the two images have the approximate perspectives that the left and right eyes would see in reality, the brain combines the images to produce a realistic 3-D effect.

48 physics for you |

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1. Consider an equiconvex lens of radius of curvature R and focal length f. If f > R, the refractive index of the material of the lens is (a) greater than zero but less than 15 (b) greater than 1.5 but less than 2.0 (c) greater than one but less than 1.5 (d) none of these 2. A parallel beam of light of intensity I0 is incident on a glass plate, 25% of light is reflected by upper surface and 50% of light is reflected from lower surface. The ratio of maximum to minimum intensity in interference region of reflected rays is 5+2 6 5+ 6 (a) (b) 5− 6 5−2 6 5 8 (c) (d) 8 5 3. An electron of mass m and charge e initially at rest gets accelerated by a constant electric field E. The rate of change of de-Broglie wavelength of this electron at time t (ignoring relativistic effects) is −h −mh −h −eEt (a) (b) (c) (d) 2 2 eEt h eEt eEt

4. Two radioactive sources A and B of half lives 1 h and 2 h respectively initially contain the same number of radioactive atoms. At the end of two hours, their rates of disintegration are in the ratio of (a) 1 : 4 (b) 1 : 3 (c) 1 : 2 (d) 1 : 1 5. A ray of light undergoes a deviation of 30° when incident on an equilateral prism of refractive index 2. What is the angle subtended by the ray inside the prism with the base of the prism? (a) 0° (b) 45° (c) 60° (d) 90° 6. Young’s double slit experiment is performed with two wavelengths simultaneously,

l1 = 480 nm and l2 = 600 nm. The distance

between the slits is 5.0 mm and the slits are 1.0 m from the screen. What is the separation of the screen between third order (n = 3) bright fringes of the two interference patterns? (a) 7.2 × 10–9 m (b) 7.2 × 10–5 m –9 (c) 5.2 × 10 m (d) 5.2 × 10–5 m 7. A convex lens of focal length 20 cm and another plano-convex lens of focal length 40 cm are placed co-axially. The plano-convex lens is silvered on plane surface. What should be the distance d (in cm) so that final image of the object O is formed on O itself ? d O 10 cm f = + 20 cm

(a) 10

(b) 15

(c) 20

(d) 25

8. Photons of energy 5 eV are incident on cathode. Electrons reaching the anode have kinetic energies varying from 6 eV to 8 eV. Then which of the following is true? h

5V

(a) Work function of the metal is 2 eV . (b) Work function of the metal is 3 eV. (c) Current in the circuit is equal to saturation value. (d) Both (a) and (c). 9. Ultraviolet light of wavelengths 800 Å and 700 Å when allowed to fall on the hydrogen atoms in their ground state is found to liberate electrons with kinetic energy 1.8 eV and physics for you

| march ‘15 49

4.0 eV respectively. The value of the Planck’s constant is (a) 1.66 × 10–19 J s (b) 5.67 × 10–19 J s –34 (c) 7.57 × 10 J s (d) 6.57 × 10–34 J s 10. Hydrogen atom in its ground state is excited by means of monochromatic radiation of wavelength 970.6 Å. How many different wavelengths are possible in the resulting emission spectrum? (a) 4 (b) 6 (c) 8 (d) 10 11. A sample of isotope 131I which has a half-life of 8.04 days has an activity of 5 mCi at the time of shipment. Upon receipt in a medical laboratory, the activity is 4.2 mCi. How much time has elapsed between the two measurements? (a) 2 days (b) 3 days (c) 4 days (d) 5 days 12. Assuming that in a star, three alpha particles join in a single reaction to form 126 C nucleus. Find the energy released in this reaction. Given mass of 24 He = 4.002604 u and that of 12 = 12.000000 u. 6C (a) 12.2 MeV (b) 10.2 MeV (c) 7.2 MeV (d) 3.2 MeV 13. For the given incident ray as shown in figure, the condition of total internal reflection of ray will be satisfied if the refractive index of block will be 45°

(a)

3 +1 (b) 2

Incident ray

2 +1 (c) 2

3 2

(d)

7 6

14. The plane face of a planoconvex lens is silvered. If m be the refractive index and R, the radius of curvature of curved surface, then the system will behave like a concave mirror of radius of curvature R (a) mR (b) (m −1) R  m +1 (c) (d)  R m  m − 1  52 physics for you |

march‘15

15. In Young’s double slit experiment, the distance

between the two slits is 0.1 mm and wavelength of light used is 4 × 10–7 m. If the width of fringe on screen is 4 mm, the distance between screen and slits is (a) 0.1 mm (b) 1 cm (c) 0.1 cm (d) 1 m

16. The size of the image of an object, which is at infinity, as formed by a convex lens of focal length 30 cm is 2 cm. If a concave lens of focal length 20 cm is placed between the convex lens and the image at a distance of 26 cm from the convex lens, the new size of the image is (a) 1.25 cm (b) 2.5 cm (c) 1.05 cm (d) 2 cm 17. Interference fringes are produced in Young’s double slit experiment using light of wavelength 5000 Å. When a film of material 2.5 × 10–6 m thick was placed over one of the slits, the fringe pattern shifted by a distance equal to 2 fringe widths. The refractive index of the material of the film is (a) 1.25 (b) 1.33 (c) 1.4 (d) 1.5 18. Two radioactive materials X1 and X2 contain same number of nuclei. If 6l s–1 and 4l s–1 are the decay constants of X1 and X2 respectively, then the ratio of number of nuclei, undecayed 1 of X1 to that of X2 will be   after a time  e 1 1 1 1 s (b) (a) s (c) s (d) s 2l 10l 5l l 19. The ratio of maximum and minimum intensities in the interference pattern of two sources is 4 : 1. The ratio of their amplitudes is (a) 1 : 2 (b) 3 : 1 (c) 1 : 9 (d) 1 : 16 20. Li nucleus has three protons and four neutrons. Mass of Li nucleus is 7.016005 amu, mass of proton is 1.007277 amu and mass of neutron is 1.008665 amu. Mass defect of lithium nucleus in amu is (a) 0.040486 amu (b) 0.040500 amu (c) 0.040524 amu (d) 0.040555 amu 21. A man is trying to start a fire by focusing sunlight on a piece of paper using an equiconvex lens of focal length 10 cm. The diameter of the sun is

1.39 × 109 m and its mean distance from the earth is 1.5 × 1011 m, the diameter of the sun’s image on the paper is (a) 3.1 × 10–4 m (b) 6.5 × 10–5 m –4 (c) 6.5 × 10 m (d) 9.3 × 10–4 m

27. The de Broglie wavelength and kinetic energy of a particle is 2000 Å and 1 eV respectively. If its kinetic energy becomes 1 MeV, then its de Broglie wavelength becomes (a) 1 Å (b) 5 Å (c) 2 Å (d) 10 Å

22. The power of a lens having refractive index 1.25 is +3 D. When placed in a liquid its power is –2 D. The refractive index of the liquid is (a) 1.2 (b) 1.4 (c) 1.5 (d) 1.6

28. The work function of a certain metal is 3.31 × 10–19 J. Then, the maximum kinetic energy of photoelectrons emitted by incident radiation of wavelength 5000 Å is (a) 2.48 eV (b) 0.42 eV (c) 2.07 eV (d) 0.82 eV

23. The focal length of the lens of refractive index 1.5 in air is 10 cm. If air is replaced by water of 4 m = , its focal length is 3 (a) 20 cm (b) 30 cm (c) 40 cm (d) 25 cm 24 Half-lives of two radioactive substances A and B are respectively 20 minutes and 40 minutes. Initially, the samples of A and B have equal number of nuclei. After 80 minutes, the ratio of the remaining numbers of A and B nuclei is (a) 1 : 16 (b) 4 : 1 (c) 1 : 4 (d) 1 : 1 25. An equiconvex lens of focal length f is cut into two halves along (i) XOX′ and (ii) YOY′ as shown in the figure. Let f ′, f ′′ be the focal lengths of complete lens, of each half in case (i), and of each half in case (ii), respectively.

29. A ray is incident at an angle of incidence i on one surface of a prism of small angle A and emerges normally from opposite surface. If the refractive index of the material of prism is m, the angle of incidence i is nearly equal to mA (a) A (b) A (c) mA (d) 2 2m m 30. If the wavelength of the first line of the Balmer series of hydrogen is 6561 Å, the wavelength of the second line of the series should be (a) 13122 Å (b) 3280 Å (c) 4860 Å (d) 2187 Å solutions

1.

Y

X

O

Choose the correct statement from the following. (a) f ′ = 2f and f ′′ = f (b) f ′ = f and f ′′ = f (c) f ′ = 2f and f ′′ = 2f (d) f ′ = f and f ′′ = 2f 26. A radioactive substance decays at the rate of 5000 disintegration per minute. After 5 minutes it disintegrates at 1250 disintegration per minutes. The decay constant is (a) 0.2ln2 min–1 (b) 0.4ln2 min–1 –1 (c) 0.6ln2 min (d) 0.8ln2 min–1

1 R 2 = (m − 1)   or f = R f 2(m − 1)

Now, f > R R 1 \ > R or > 1 or 2(m − 1) < 1 2(m − 1) 2(m − 1) 1  1 or (m − 1) < or m < 1 +  or m < 1.5.  2 2

X

Y

(c) : Here,

2.

(a) : The intensity of light reflected from upper surface is I1 = 25% of I0 25 I0 = I0 × = 100 4 The intensity of transmitted light from upper surface is 3I I I = I0 − 0 = 0 4 4 The intensity of reflected light from lower surface is physics for you

| march ‘15 53

I2 =

3I0 50 3I0 × = 4 100 8

As m = 2, the ray suffers minimum deviation through the prism. Thus A r1 = r2 = r = = 30° 2 Inside the prism, the ray makes an angle 60° with the face AB, so it is parallel to base.

2 I max ( I1 + I2 ) = I min ( I1 − I2 )2

\

 I0 3I 0   4 + 8   

2

6.

2

 2+ 3 5+2 6 = = = 2  2 − 3  5−2 6  I0 3I 0  −  4 8   3.

(a) : Here, u = 0, a =

eE m

eE t m de-Broglie wavelength, \

v = u + at = 0 +

l=

h h h = = mv m(eEt / m) eEt

Rate of change of de-Broglie wavelength d l h  1  −h = = − dt eE  t 2  eEt 2 4.

(c) : Rate of disintegration ∝ number of atoms left

8.

2

In case of source A,

N 1 1 =  =   N0 2 4

In case of source B,

N 1 1 = = N 0  2  2

\ 5.

7.

1

RA N 0 / 4 1 = = RB N 0 / 2 2

(a) : The given parameters are d = 30° and A = 60°. Let us test whether the prism is in the position of minimum deviation.  30° + 60°  sin   sin 45°  2 m= = sin 30°  60°  sin   2  =

1 2

×2= 2

54 physics for you |

march‘15

A 60° 30° B

60°

C

(b) : Condition for maxima is dsinq = nl nl or dq  nl ⇒ q = d The angular separation of two maxima associated with different wavelength and same order is n(l2 − l1 ) ∆q = d The separation on a screen at a distance D away is nD(l2 − l1 ) ∆y = D tan(∆q)  D∆q = d (as tanq ≈ q for small angles)  3 × 1. 0  –9 –5 =  (600 – 480) × 10 = 7.2 × 10 m  5.0 × 10−3  (c) (a) : Maximum kinetic energy, Kmax = (5 – f) eV When these electrons are accelerated through 5 V, they will reach the anode with maximum energy = (5 – f + 5) eV \ 10 – f = 8 or f = 2 eV Current is less than saturation current because even if slowest electron reaches the plate it will have 5 eV energy at the anode, but there it is given that the minimum energy is 6 eV.

(d) : Here, l1 = 800 Å = 8 × 10–8 m l2 = 700 Å = 7 × 10–8 m Kinetic energy of electron,  hc  = Incident energy   – Binding energy (13.6 eV) l hc In the first case, 1.8 eV= − 13.6 eV ... (i) l1 9.

In the second case, 4 eV =

hc − 13.6 eV ... (ii) l2

From eqns. (i) and (ii), we get  1  l − l2  1 2.2 eV = hc  −  = hc  1   l2 l1   l1l2  h=

or =

(2.2 eV)l1l2 c(l1 − l2 )

2.2(1.6 × 10−19 J)(8 × 10−8 m)(7 × 10−8 m) 8

−1

(3 × 10 m s )(8 × 10

−8

−8

− 7 × 10 ) m

= 6.57 × 10–34 J s 10. (b) : Energy of the monochromatic radiation, E=

−15 eV s)(3 × 108 m s −1 ) hc (4.14 × 10 = l 970.6 × 10−10 m

= 12.79 eV Energy of the hydrogen atom is given by 13.6 En = − eV n2 In the ground state, n = 1, E1 = –13.6 eV Energy of the hydrogen atom after excitation of the state En = E1 + E = –13.6 eV + 12.79 eV = –0.81 eV 13.6 As En = − eV n2 13.6 − 0.81 eV = − eV n2 13.6 or n2 = = 16.8 0.81 \n  4 Thus, when the electron is excited to state n = 4, six different wavelengths in the emission spectrum correspond to n = 4 to n = 3, 2, 1 (3 lines) n = 3 to n = 2, 1 (2 lines) and n = 2 to n = 1 (1 line) transitions. 11. (a) : As R = lN and R0 = lN0. 5 mCi N 0 R0 = = = 1.190 N R 4.2 mCi Also, T1/2 = 8.04 days N0 = 2t /T1/2 ⇒ 1.190 = 2t /T1/2 As N t t /T log 2 or log 1.190 = log 2 1/2 = T1/2

 log 1.190  0.0755 or t =  × 8.04  2 days  T1/2 = 0.3010  log 2  12. (c) : According to the assumption, as three alpha particles form a 12 nucleus, 6C

324 He →126 C

Mass of three alpha particles = 3(4.002604 u) = 12.007812 u Mass of carbon nucleus = 12 u Mass defect = 12.007812 u – 12 u = 0.007812 u Energy released in the reaction = 0.007812 × 931.5 MeV = 7.277 MeV 13. (c) : C

iC

B r

45° A

As r + iC = 90°, r = (90° – iC) sin i sin 45° 1 m= = = sin r sin(90° − iC ) 2 cos iC 1 , sin iC = 2 cos iC As m = sin iC 2 or tan iC = 2 , ⇒ sin iC = 3 1 3 = siniC 2 14. (b) : For planoconvex lens (without its plane surface silvered) 1  1 1  m −1 R = (m − 1)  −  = or f L = R ∞ fL R (m − 1) or m =

When an object is placed in front of the planoconvex lens with its plane face silvered, light rays are : (i) refracted at the convex surface (ii) reflected at the silvered surface and (iii) refracted again at convex surface. If F is the effective focal length of the combination, then 1 1 1 1 2 = + + = (as f M = ∞) F fL fM fL fL or

f R F= L = 2 2(m − 1) physics for you

| march ‘15 55

Radius of curvature of the concave mirror R = 2F = (m − 1) lD bd or D = 15. (d) : As, b = d l Here, d = 0.1 mm = 0.1 × 10–3 m, l = 4 × 10–7 m, b = 4 mm = 4 × 10– 3m 4 × 10 −3 × 0.1 × 10 −3 \ D= =1m 4 × 10 −7 16. (b) : In figure, I1 is image formed by convex lens. This acts as a virtual object for concave lens. As for concave lens

1 1 1 1 1 1 − = \ − =− v u f v 4 20 1 1 1 1 \ = − = or v = 5 cm v 4 20 5 Magnification produced by concave lens v 5 m = = = 1.25 u 4 As size of image I1 is 2 cm, therefore, size of image I2 = 2 × 1.25 = 2.5 cm 17. (c) : Here, l = 5000 Å = 5 × 10–7 m, t = 2.5 × 10–6 m, x = 2b D As x = (m − 1)t d lD D b But b = \ = d d l \ x = (m − 1)t

b l

(m − 1)(2.5 × 10 −6 )b 5 × 10 −7 m – 1 = 0.4 or m = 1.4 2b =

18. (a) : After a time t, N X 1 = N 0e −6lt , N X2 = N 0e −4lt \

N X1 e −6lt = = e −2lt N X2 e −4lt

56 physics for you |

march‘15

or

1 1 1 = or 2lt = 1 or t = s e e 2lt 2l

2 19. (b) : I max = (a + b) = 4 I min (a − b)2 1 a+b 2 ⇒ = a −b 1

⇒ 2a – 2b = a + b or a = 3b \

a 3 = b 1

20. (a) : Here mp = 1.007277 amu mn = 1.008665 amu, mLi = 7.016005 amu, Sum of the masses of three protons and four neutrons = 3mp + 4mn = 3 × 1.007277 + 4 × 1.008665 = 3.021831 + 4.03466 = 7.056491 amu \ Mass defect = 7.056491 – 7.016005 = 0.040486 amu −f f −u Here, f = 10 cm = 10 × 10–2 m, u = 1.5 × 1011 m −10 × 10−2 \ m= = 6.67 × 10−13 10 × 10−2 − 1.5 × 1011 \ Diameter of the image d = m × 1.39 × 109 = 6.67 × 10–13 × 1.39 × 109 = 9.3 × 10–4 m

21. (d) : Magnification, m =

  22. (c) : 1 = (m − 1) 1 − 1 = 3  fa  R1 R2  \

1 1  3 = (1.25 − 1)  −   R1 R2 

and

 1.25   1 1 1  = −2 =  − 1  −   m   R1 R2  fl 3 2

\ − =

0.25m ⇒ −0.5m = 3.75 − 3m 1.25 − m

or m = 3.75/2.5 = 1.5 23. (c) : According to lens maker’s formula The focal length of the lens in air is 1  ml   1 1 − = −1 f air  m a   R1 R2 

1  3/2   1 = − 1  −  1 R R  1 2

\

1 1 1 1 = − f air 2  R1 R2 

2mK Since mass of the particle remains constant \ l∝ 1

1  3/2   1 = −1 − 4 / 3   R1 R2  1 1 1 1 = − f water 8  R1 R2 

K

...(ii)

Divide (i) by (ii), we get f water =4 f air fwater = 4 × fair = 4 × 10 cm = 40 cm n

 2 where n is the number of half-lives. nA

4 N 1 = N0   = 0  2 16

t 80   nA = T = 20 = 4  A   1 For B, N B = N 0    2

nB

2 N 1 = N0   = 0  2 4

t 80   nB = T = 40 = 2 B  

\ 25. (d)

h

27. (c) : As l =

1 1  1  ml − = −1 f water  m w   R1 R2 

1 For A, NA = N 0    2

2 = ln 2 = 0.4 ln 2 min −1 5

...(i)

The focal length of the lens in water is

24. (c) : N = N 0  1 

1  5000  1 = ln 4 l = ln  5  1250  5

NA 1 = or N A : N B = 1 : 4 NB 4

26. (b) : The rate of disintegration R is given by R = R0e–lt where R0 is the initial rate at t = 0. R0 \ = e lt R Taking the natural logarithm on both sides, we get R  1 R  ln  0  = lt ; l = ln  0  R t R Here, R0 = 5000 dpm, R = 1250 dpm, t = 5 min

1 eV l′ K 1 = = = 6 l K′ 1 × 10 eV 103 or l ′ =

l 3

10

=

2000 103

Å=2Å

28. (b) : Here, work function, f0 = 3.31 × 10–19 J Wavelength, l = 5000 Å = 5000 × 10–10 m Energy of the incident photon,

hc 6.63 × 10−34 × 3 × 108 = = 3.98 × 10–19 J −10 l 5000 × 10 According to Einstein’s photoelectric equation Kmax = hu – f0 = 3.98 × 10–19 J – 3.31 × 10–19 J = 0.67 × 10–19 J E=

=

0.67 × 10−19 1.6 × 10−19

eV = 0.42 eV

29. (c) : As refracted ray emerges normally from opposite surface, r2 = 0 As A = r1 + r2 \ r1 = A sin i1 i1 i ≈ = or i = mA sin r1 r1 A 30. (c) : For Balmer series, n1 = 2, n2 = 3 for 1st line and n2 = 4 for second line. Now, m =

1  1 − 2  2 l1  2 4  = 3 / 16 = 3 × 36 = 27 = l2  1 1  5 / 36 16 5 20 −  22 32  l2 =

20 20 ° l = × 6561 = 4860 A 27 1 27 nn physics for you

| march ‘15 57

Time Allowed : 3 hrs

Maximum Marks : 70 GENERAL INSTRUCTIONS

(i) All questions are compulsory. There are 26 questions in all. (ii) This question paper has five sections: Section A, Section B, Section C, Section D and Section E. (iii) Section A contains five questions of one mark each, Section B contains five questions of two marks each, Section C contains twelve questions of three marks each, Section D contains one value based question of four marks and Section E contains three questions of five marks each. (iv) There is no overall choice. However, an internal choice has been provided in one question of two marks, one question of three marks and all the three questions of five marks weightage. You have to attempt only one of the choices in such questions. (v) You may use the following values of physical constants wherever necessary.

c = 3 × 108 m/s, h = 6.63 × 10–34 J s, e = 1.6 × 10–19 C, μo = 4p × 10–7 T m A–1, e0 = 8.854 × 10–12 C2 N–1 m–2, 9 2 –2 1/4pe0 = 9 × 10 N m C , m = 9.1 × 10–31 kg, mass of neutron = 1.675 × 10–27 kg, mass of proton = 1.673 × 10–27 kg e



Avogadro’s number = 6.023 × 10

23

per gram mole, Boltzmann constant = 1.38 × 10

section-A 1. Identify the parts X and Y in the block diagram

of a generalised communication system. X

Transmitter

Y

2. Twelve wires of equal length

C

D

B G F

3. An unsymmetrical double convex thin lens

forms the image of a point object on its axis. Will the position of the image change if the lens is reversed?

4. A TV tower has a height of 100 m. How much

population is covered by TV transmission if the

58 physics for you |

March ‘15

JK

–1

average population density around the tower is 1000 km–2 ? (Take Radius of the Earth = 6.37 × 103 km). 5. There are materials which absorb photons of

Receiver

are connected to form a A skeleton cube as shown in the figure, which moves with a v H  velocity v perpendicular to  E the magnetic field B. What B will be the induced emf in each arm of the cube?

–23

shorter wavelength and emit photons of longer wavelength. Can there be stable substances which absorb photons of larger wavelength and emit light of shorter wavelength. section-B

6. Power P is to be delivered to a device via

transmission cables having resistance RC. If V is the voltage across the device and I the current through it, find the power wasted. How can it be reduced?

7. A convex lens of focal length 20 cm has a point

object placed on its principal axis at a distance of 40 cm from it. A plane mirror is placed 30 cm behind the convex lens. Locate the position of the image formed by this combination.

8. Deduce the expression for the magnetic field at

the centre of circular electron orbit of radius r where angular speed of orbiting electron is w. 9. Voltages across L and C in the series are 180° out of phase while for LC in parallel, currents in L and C are 180° out of phase. Explain. 10. Two charges q and –3q are placed fixed on X-axis separated by distance d. Where should a third charge 2q be placed such that it will not experience any force? OR Find the work done to dissociate the system of three charges each of value 1.6 × 10–9C placed on the vertices of a triangle as shown in the figure. q

10

cm

cm

10

–4q

10 cm

+2q

section-c 11. At what angle should a ray of light be incident

on the face of a prism of refracting angle 60° so that it just suffers total internal reflection at the other face? The refractive index of the material of the prism is 1.524. 12. A student performs an experiment on photoelectric effect, using two materials A and B. Plot of Vstop versus u is given in the figure. (a) Which material A or B has a higher work function? (b) Given the electric charge on an electron = 1.6 × 10–19 C, find the value of h obtained from the experiment for both A and B. Comment on whether it is consistent with Einstein’s theory. Vstop(V)

A

3.0 2.5 2.0 1.5 1.0

B

13. (a) Two circular metal plates, each of radius

10 cm, are kept parallel to each other at a distance of 1 mm. What kind of the capacitor do they make? Mention one application of this capacitor. (b) If the radius of each of the plates is increased by a factor of 2 and their distance of separation is reduced to half of its initial value. Find the ratio of the capacitances in the two cases. (c) Suggest any one possible method by which the capacitance in the second case be increased by n times. 14. A telephone cable at a place has four long straight horizontal wires carrying a current of 1.0 A in the same direction east to west. The earth’s magnetic field at the place is 0.39 G, and the angle of dip is 35°. The magnetic declination is nearly zero. What is the resultant magnetic field at points 4.0 cm below the cable? 15. Define the two current gains of a transistor and deduce a relation between them. OR

The current in the forward bias is known to be more (~mA) than the current in the reverse bias (~μA). What is the reason then, to operate the photodiodes in reverse bias? 16. Explain what is meant by radioactive decay.

A radioactive nucleus is represented by the symbol abV. How is the new nucleus represented after the emission of (i) an alpha particle (ii) a beta particle (iii) a gamma ray? The activity of a source undergoing a single type of decay is R0 at time t = 0. Obtain an expression in terms of the half-life T1/2 for the activity R at any subsequent time t.

17. A beam of protons enters a uniform magnetic

5

10

15

 (1014 Hz)

field of 0.3 T with a velocity of 4 × 105 m s–1 at an angle of 60° to the field. Find the radius of the helical path taken by the beam. Also find the pitch of the helix. physics for you

| March ‘15 59

18. An amplitude modulated wave is as shown in

V

the figure. Find: (a) the percentage modulation (b) peak carrier voltage (c) peak value of information voltage.

23. Teena went out for shopping with her mother.

100 V 20V

t

19. Five charges, q each, are placed at the corners

of a regular pentagon of side a as shown in the figure. Aq E q

B q

r q

a

q

C

(i) What will be the electric field at O, the centre of the pentagon? (ii) What will be the electric field at O if the charge from one of the corners (say A) is removed? (iii) What will be the electric field at O if the charge q at A is replaced by –q? 20. Using Bohr’s postulates, derive an expression for the radii of the permitted orbits in the hydrogen atom. 21. In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0 × 1010 Hz and amplitude 48 V m–1. (a) What is the wavelength of the wave? (b) What is the amplitude of the oscillating magnetic field? (c) Show that the average energy density of the E field equals the average energy density of the B field. 22. Can reflection result in plane polarised light if the light is incident on the interface from the side with higher refractive index? 60 physics for you |

March ‘15

During purchase of vegetables, she noticed that the vendor used a digital weighing machine. On another shop, she noticed that the vendor was using an ordinary weighing machine. She remembered having studied about logic gates where, digital codes are used. (i) What do you mean by logic gate? Mention the basic universal gates. (ii) Draw symbols for OR, AND and NOT gates. (iii) What is the value, in your opinion, that Teena created by the above incident? section-e

O

D

section-D

24. (a) State

the working principle of a potentiometer. With the help of the circuit diagram, explain how a potentiometer is used to compare the emf ’s of two primary cells. Obtain the required expression used for comparing the emfs. (b) Write two possible causes for one sided deflection in a potentiometer experiment. OR State Kirchhoff ’s rules for an electrical network. Explain their use by drawing a simple circuit diagram. Find the expression of net emf if two cell of emf ’s e1 and e2 and internal resistance r1 and r2 are combined in parallel. 25. Explain with the help of a labelled diagram, the principle, construction and working of an a.c. generator. OR Explain with the help of a labelled diagram, the principle, construction and working of a transformer. Why is the core of transformer laminated? Give few energy losses in transformer. 26. With the help of a ray diagram, explain the formation of image in an astronomical telescope for a distant object. Define the term magnifying power of a telescope. Derive an expression for its magnifying power when the final image is formed at the least distance of distinct vision.

OR

(a) In Young’s double slit experiment, derive the condition for (i) constructive interference and (ii) destructive interference at a point on the screen. (b) In a Young’s double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. Determine the wavelength of light used in the experiment. solutions

If P is the power to be delivered to the device, then P P = VI or I = ...(ii) V where V is the voltage across the device and I is the current through it. From eqns. (i) and (ii), 2

P 2 RC P PC =   RC = V  V2 In order to reduce PC , power should be transmitted at high voltage. 7. For the lens : u = – 40 cm, f = 20 cm L

1. X is the information source,

10 cm

Y is the communication channel.

2. Induced emf is set up in a conductor moving

  with a velocity v in a magnetic field B only if   it is perpendicular to both v and B . Since, only the arms AD, BC, EH and FG of the cube are   perpendicular to both v and B , induced emf (e = Blv) is set up in these arms (l being the length of each arm). 3. According to the principle of reversibility of light, if a reflected or refracted ray is reversed in direction, it will retrace its original path. Thus, the position of the image will not change if the lens is reversed. 4. Given hT = 100 m = 10–1 km, population density = 1000 km–2 Radio horizon of the transmitting tower dT = 2RhT Area covered, A = pdT2 = 2pRhT Population covered = A × population density = 2 × 3.14 × 6.37 × 103 × 10–1 × 1000 = 40 × 105 = 40 lakhs 5. If a material absorbs a photon of larger wavelength (less energy) and emits a photon of shorter wavelength (more energy), energy has to be supplied by the material. This is not possible in case of a stable material. 6. Power wasted in the cables carrying current

I and having resistance RC, i.e., PC = I2RC

M

...(i)

O

40 cm

I 30 cm

I 10 cm

40 cm

As

1 1 1 1 1 1 − = , − = v u f v (−40) 20

or v = +40 cm In the absence of mirror M, the image of the object O would have been formed at I′ which lies at a distance of 10 cm (40 cm – 30 cm) behind M. For the mirror : As I′ acts at a virtual object for M, its real image is formed at I at a distance of 10 cm in front of the mirror. 8. The circular electron orbit can be considered to be a circular coil carrying current I, where e e ew I= = = T 2p / w 2p Magnetic field at the centre of circular electron orbit of radius r, i.e., μ I μ e w μ 0e w = B= 0 = 0 × 2r 2r 2 p 4 pr 9. When L and C are in series, the current everywhere in the circuit has the same phase. But the voltage in L leads current by p/2 and in C it lags the current by p/2. Therefore, the same phase difference between the voltages in L and C is p. When L and C are in parallel, the voltage across both L and C has the same phase. But physics for you

| March ‘15 61

current in L lags the voltage by p/2 and in C it leads the voltage by p/2 and as such the phase difference between the current in L and C is p. 10.

FCA

FCB

C

A +q

+2q x

B –3q

X

or (d + x)2 = 3x2 or 2x2 – 2dx – d2 = 0 d d 3 ± 2 2 Neglecting –ve sign because it means that the point C is between charges + q and –3q, which is not possible as per the given condition, x=

d d 3 d + = (1 + 3 ) 2 2 2 OR Here, q1 = q, q2 = –4q, q3 = +2q and x1 = x2 = x3 = r = 10 cm = 10–1 m Initial potential energy q q q q  q q Ui = k  1 2 + 2 3 + 3 1  x2 x3   x1 x=

 q(−4q) (−4q)(2q) (2q)q  = k + + r r   r = −k

r

=−

9 × 109 × 10 (1.6 × 10−9 )2 10−1

= –2.3 × 10–6 J Final potential energy, Uf = 0 Thus, work to be done to dissociate the system of three charges, W = Uf – Ui = –Ui = 2.3 × 10–6 J 11. The beam should be incident at critical angle

or more than critical angle, for total internal reflection at second surface of the prism.

62 physics for you |

March ‘15

i

60° 90 °– 90°–r ic r ic

d

It is obvious that charge +2q must be placed at C (farther from charge –3q at B than from charge +q at A) so that FCA (force on charge +2q due to charge +q) = FCB (force on charge + 2q due to charge – 3q), 2q (q) 2q (3q) i.e., k =k 2 x (d + x )2

10 q2

A

Let us first find critical angle for air glass interface. 1 We know, sin ic = a μg  1   1  ic = sin −1   = sin −1  a  1.524   μ g  Critical angle ic = 41° In the smaller triangle, 60° + (90° – r) + (90° – ic) = 180° or r = 19° Using Snell’s law, required angle of incidence i at first surface can be calculated. sin i sin i a μg = , 1.524 = sin r sin 19° sin i = 1.524 (sin 19°) i = sin–1 (0.4962) = 29.75° 12. (a) Threshold frequency (when Vstop = 0)

for material A, i.e., uA = 5 × 1014 Hz Threshold frequency (when Vstop = 0)

for material B, i.e., uB = 10 × 1014 Hz Work function for material A, i.e., fA = h(5 × 1014 Hz) Work function for material B, i.e., fB = h(10 × 1014 Hz) \ fB > fA h (b) (i) As = slope of A e 2−0 2 = = 14 5 × 1014 (10 − 5) × 10

  1.6 × 10−19 × 2 2 h = e = 14  5 × 1014  5 × 10  = 6.4 × 10–34 J s

(ii) As

in direction east to west. We want resultant magnetic field 4.0 cm below.

h = slope of B e 2.5 − 0 2.5 = = , 14 (15 − 10) × 10 5 × 1014

4 cm

 2.5  1.6 × 10−19 × 2.5 = h = e  5 × 1014  5 × 1014 –34

= 8.0 × 10 J s Since the values of h (which is a constant) obtained for material A and B are different, the experiment is not consistent with the Einstein’s theory. 13. (a) The two circular metal plates form a

parallel-plate capacitance. A capacitor is used in an LC circuit (oscillation circuit) along with an inductance (L). LC circuit is an important part of radio circuits.

∈ A ∈ ( pr 2 ) (b) Original capacitance, C = 0 = 0 d d ∈ [ p (r 2 )2 ] Changed capacitance, C ′ = 0 d/2 =

4 ∈0 pr 2 d

= 4C

C 1 = =1: 4 C′ 4 (c) By inserting a dielectric of dielectric constant n between the plates of the capacitor. Thus,

14. Let us first decide the directions which can best

represent the situation. East ×

S

Up

West

N BV Down

I=4A

×

BH

Here, BH = B cos d = 0.39 × cos 35° G BH = 0.32 G and BV = B sin d = 0.39 × sin 35° G BV = 0.22 G Telephone cable carry a total current of 4.0 A

Bwire

BH

BV

μ 2I 2×4 Now, Bwire = 0 = 10−7 × 4p r 4 × 10−2 –5 = 2 × 10 T = 0.2 G Net magnetic field Bnet =

( BH − Bwire )2 + BV2

Bnet =

(0.12 )2 + (0.22 )2

= 0.25 G

15. Usually two types of current gains are defined

for a transistor. (a) Common base current amplification factor or a.c. current gain (a): It is defined as the ratio of the small change in the collector current to the small change in the emitter current when the collector-base voltage is kept constant.  DI  Thus, a =  C   DI E  V = constant CB

(b) Common emitter current amplification factor or d.c. current gain (b) : It is defined as the ratio of the small change in the collector current to the small change in the base current when the collector-emitter voltage is kept constant.  DI  Thus, b =  C   DI B  VCE = constant Relation between a and b: For both n-p-n and p-n-p transistors, we have IE = IB + IC For small changes, we can write DIE = DIB + DIC Dividing both sides by DIC DI E DI B = +1 DI C DI C physics for you

| March ‘15 63

or

1 1 b a = + 1, a = and b = a b 1+b 1− a OR

Consider the case of an n-type semiconductor. Obviously, the majority carrier density (n) is considerably larger than the minority hole density p (i.e., n >>p). On illumination, let the excess electrons and holes generated be Dn and Dp, respectively: n′ = n + Dn p′ = p + Dp. Here n′ and p′ are the electron and hole concentrations at any particular illumination and n and p are carrier concentrations when there is no illumination. Remember Dn = Dp and n >>p. Hence, the fractional change in the majority carriers (i.e., Dn/n) would be much less than that in the minority carriers (i.e., Dp/p). In general, we can state that the fractional change due to the photo-effects on the minority carrier dominated reverse bias current is more easily measurable than the fractional change in the forward bias current. Hence, photodiodes are preferably used in the reverse bias condition for measuring light intensity. 16. Radioactive decay is the spontaneous disintegration of the nucleus of an atom with the emission of one or more penetrating radiations like a, b and g-rays. (i) a-decay :

a bV

→

a−4 b − 2X

+

4 2 He

+Q

(ii) b-decay : ba V → b + a1 X + −01e + u + Q (iii) g-decay :

a bV Excited state

→ ba V

+g

Ground state

dN dt According to radioactivity decay law, Activity, R = − dN − = lN dt

As N = N0 e–lt R = R0 e–lt 64 physics for you |

\ R = lN \ R = lN0 e–lt

March ‘15

where l N0 = R0 = activity of the sample at t= 0. 0.693 Also l = T1/ 2 \

R = R0 e

−

0.693 t T1/2

17. The components of the proton’s velocity parallel

and perpendicular to the magnetic field are 1 v|| = v cos 60° = 4 × 105 × = 2 × 105 m s−1 2 v⊥ = v sin 60° = 4 × 105 ×

3 2

= 3.464 × 105 m s–1 The component v|| makes the electron move along the field B while v⊥ makes the proton move along a circular path. Hence the path of the proton is a helix. The radius r of the helix is given by qv⊥ B =

2 mv⊥ r

or r =

× 3.464 × 10 mv⊥ 1.67 × 10 = − qB 1.6 × 10 19 × 0.3

−27

5

\ r = 12 × 10–3 m Period of revolution of the electron is −3 2 pr 2 × 3.14 × 12 × 10 T= = v⊥ 3.464 × 105

= 21.75 × 10–8 s Pitch of the helix is p = v|| × T = 2 × 105 × 21.75 × 10–8 = 43.5 × 10–3 m = 4.35 cm 18. Vmax =

100 V

= 50 V, Vmin =

2 (a) Modulation index,

20 V = 10 V 2

V − Vmin   50 − 10  2 = μ =  max =  Vmax + Vmin   50 + 10  3 Percentage modulation = μ × 100 = (2/3)100 =66.67% (b) Peak carrier voltage, V + Vmin 50 V + 10 V VC = max = = 30 V 2 2

(c) Peak information voltage, 2 Vm = μVC =   (30 V) = 20 V 3      19. (i) Let EAO , EBO , ECO , EDO and EEO represent the electric fields at O due to the five charges (q each) placed at A, B, C, D and E respectively. Since we can represent these fields by the sides of another regular pentagon taken in the same order,       EAO + EBO + ECO + EDO + EEO = 0 ...(i) (ii) When charge on one of the corners (say A)   is removed, EAO = 0 Thus, resultant electric field at O, i.e.,       ER = EBO + ECO + EDO + EEO = − EAO   or ER = EOA

[From eqn. (i)]

  Thus, the resultant electric field ER (= EOA )  acts along OA  1 q Also, | EOA | = 4 pe0 r 2 (iii) If the charge q at A is replaced by –q, then resultant electric field at O, i.e.,       ER′ = − EAO + ( EBO + ECO + EDO + EEO )  (negative sign with EAO is due to charge q being replaced by –q)    or E R′ = − EAO − EAO [from eqn. (i)]   i.e., E ′R = − 2 EAO Thus, the resultant electric    ER′ (= 2 EOA ) acts along OA .   1 2q Also | E ′R | = | 2 EOA | = 4 p ∈0 r 2

field

20. According to Bohr’s theory, a hydrogen atom

consists of a nucleus with a positive charge Ze, and a single electron of charge –e, which revolves around it in a circular orbit of radius r. Here Z is the atomic number and for hydrogen Z =1. The electrostatic force of attraction between the nucleus and the electron is

F=

k Ze . e r2

=

k Ze 2 r2 +Ze r Nucleus

e–, m

To keep the electron in its orbit, the centripetal force on the electron must be equal to the electrostatic attraction. Therefore, 2 mv 2 k Ze = r r2

or mv 2 = or r =

k Ze 2

k Ze 2

...(i)

r

...(ii) mv 2 where m is the mass of the electron, and v, its speed in an orbit of radius r. Bohr’s quantisation condition for angular momentum is nh nh L = mvr = or r = ...(iii) 2p 2p mv From equation (ii) and (iii), we get k Ze 2 mv

2

=

or v =

nh 2 p mv 2 p k Ze 2

...(iv) nh Substituting this value of v in equation (iii), we get r=

nh nh n 2 h2 . \ r = 2 pm 2 p k Ze 2 4p2mk Ze 2

21. (a) Wavelength of electromagnetic wave can

be calculated as c = ul 3 × 108= 2 × 1010 × l or l = 1.5 × 10–2 m (b) Amplitude of magnetic field E E0 = c ; B0 = 0 c B0

physics for you

| March ‘15 65

48

B0 =

3 × 108

There are two basic universal gates NAND and NOR. (ii) OR Gate,

= 16 × 10−8 T

(c) Energy density of electric field 1 U E = e0 E 2 2 1 or U E = e0c 2 B2 2 As speed of em wave, 1 c= μ 0e0 1 e0 2 So, U E = B 2 μ 0e0 UE =

A

E    B = c 

22. Rarer medium (a)

(i)

90 ic

Rarer medium (a) Denser medium (b)

(ii)

Polarisation by reflection occurs when tan ip = bμa where ip = Brewster’s angle (or polarising angle) From figure (ii) For light travelling from medium (b) to medium (a) sin ic = bμa We know that for large angles, |tan ip| > |sin ic| (e.g., tan45° = 1 and sin 45° = 0.707) This implies that ip < ic, i.e., polarisation by reflection always occurs when light is incident on the interface from the side with higher refractive index (i.e., denser medium). 23. (i) A gate is digital circuit that is designed for

performing a particular logical operation. As it works according to some logical operation between input and output voltages, so it is generally known as a logic gate.

66 physics for you |

March ‘15

A

Y = A·B

B

Y=A

A

B2 = UB 2μ 0

Denser medium (b)

AND Gate

NOT Gate

= Energy density of magnetic field

ip

Y=A+B

B

(iii) Concentration and observation in the class room, retaining capacity, co-relating of what was taught with the real life incident. 24. (a) Refer point 2.5 (7), MTG Excel in Physics (b) (i) The emf of the cell connected in main circuit may not be more than the emf of the primary cells whose emfs are to be compared. (ii) The positive ends of all cells are not connected to the same end of the wire. OR Refer points 2.5 (1, 2, 3) and 2.3 (13), MTG Excel in Physics 25. Refer point 4.8 (2), MTG Excel in Physics OR Refer point 4.8 (1), MTG Excel in Physics 26. Refer point 6.9 (2), MTG Excel in Physics. OR (a) Refer point 6.13 (6), MTG Excel in Physics (b) Here d = 0.28 mm, D = 1.4 m Distance of fourth bright fringe from center = 1.2 cm Linear position of nth bright fringe nD l yn = d Linear position of 4th bright fringe 4 Dl y4 = d 1.2 × 10−2 =

4 (1.4) l 0.28 × 10−3

⇒ l = 6000 Å nn

1. A boat which has a speed of 5 km h–1 in still water crosses a river of width 1 km along the shortest possible path in 15 minutes. The velocity of the river water in km h–1 is (a) 1 (b) 3 (c) 4 (d) 41 2. A planet is at an average distance d from the sun, and its average surface temperature is T. Assume that the planet receives energy only from the sun, and loses energy only through radiation from its surface. Neglect atmospheric effects. If T ∝ d–n, the value of n is 1 1 (a) 2 (b) 1 (c) (d) 2 4 3. A glass prism of refractive index 1.5 is immersed in water (refractive index 4/3). A light beam incident normally on the face AB is totally reflected to reach on the face BC if 8 (a) sin q ≥ B A 9 θ 2 8 (b) < sin q < 3 9 C 1 8 (c) < sin q < 2 9 (d) none of these 4. A rod of length L lies along the axis of a concave mirror of focal length f. The near end of the rod is at a distance L > f from the mirror. Its image will have a length (a) (c)

Lf 2 (L − f )(2L − f )

2Lf 2 (L − f )(2L − f )

(b)

L2 f (L − f )(2L − f )

(d)

2L2 f (L − f )(2L − f )

5. Between two infinitely long λ –λ wires having linear charge A B densities l and –l there are two points A and B as shown a a a in the figure. The amount of work done by the electric field in moving a point charge q0 from A to B is equal to 2 lq0 lq0 ln 2 (a) (b) − ln 2 pe0 2 pe0

lq0 2 lq0 (d) ln 2 ln 2 pe0 pe0 6. A uniform chain of length L and mass M is lying on a smooth table and one third of its length is hanging vertically down over the edge of the table. If g is acceleration due to gravity, the work required to pull the hanging part on the table is MgL MgL MgL (a) MgL (b) (c) (d) 18 9 3 7. A flexible wire loop in the shape of a circle has a radius that grows linearly with time. There is a magnetic field perpendicular to the plane of the loop that has a magnitude inversely proportional to the distance from the centre of 1 the loop, B(r ) ∝ . How does the e.m.f. e vary r with time? (a) e ∝ t2 (b) e ∝ t (c) e ∝ t (d) e is constant (c)

8. A person of mass 60 kg wants to lose 5 kg by going up and down a 10 m high stairs. Assume he burns twice as much fat while going up than coming down. If 1 kg of fat is burnt on expending 7000 kilo calories, how many times Physics for you | March ‘15

67

must he go up and down to reduce his weight by 5 kg? (a) 25300 (b) 9989 (c) 8756 (d) 16334 9. Consider the spectral line resulting from the transition n = 2 → n = 1 in the atoms and ions given below. The shortest wavelength is produced by (a) hydrogen atom (b) deuterium atom (c) singly ionised helium (d) doubly ionised lithium 10. A hypothetical experiment conducted to determine Young’s modulus, gave the formula, cos qT x t Y= . If T = time period, t = torque l3 and l = length, then the value of x is (a) zero (b) 1 (c) 2 (d) 3 11. Voltmeter reads the potential difference between the terminals of an old battery as 1.4 V while a potentiometer reads its voltage to be 1.55 V. The voltmeter resistance is 280 W. Then, (a) the battery of the cell is of 1.4 V (b) the battery of the cell is of 1.5 V (c) the internal resistance of the battery is 30 W. (d) the internal resistance of the battery is 5 W. 12. Two bodies M and N of equal masses are suspended from 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 k k2 (a) 2 (b) k1 k1 (c)

k1k2

(d)

k1 k2

13. A light emitting diode (LED) has a voltage drop of 2 V across it and passes a current of 10 mA when it operates with a 6 V battery through a limiting resistor R. The value of R is (a) 40 kW (b) 4 kW (c) 200 W (d) 400 W 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 68 Physics for you |

March ‘15

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. A pure inductor L, a capacitor C and a resistance R are connected across a battery of e.m.f. e and internal resistance r as shown in the figure. The switch Sw is closed at t = 0, select the correct alternative. L R C ε

r Sw

(a) Current through resistance R is zero all the time. (b) Current through resistance R is zero at t = 0 and t → ∞. (c) Maximum charge stored in the capacitor is Ce. (d) Maximum energy stored in the inductor is equal to the maximum energy stored in the capacitor. 16. A block P of mass m is placed on a horizontal frictionless plane. A second block Q 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 maximum value of frictional force between the two blocks? k

Q

µs

P

kA (b) kA (c) msmg (d) zero 2 17. Two radioactive materials X1 and X2 have decay constants 10l and l respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of X1 to that of X2 will be 1/e after a time 1 1 11 1 (a) (b) (c) (d) 10 l 11l 10 l 9l (a)

18. Imagine a light planet revolving around a very massive star in a circular orbit of radius R with a period of revolution T. If the gravitational force of attraction between the planet and the star is proportional to R–5/2, then T2 is proportional to (a) R3 (b) R7/2 (c) R3/2 (d) R7/3 19. A wall of dimensions 2.00 m by 3.50 m has a single-pane window of dimensions 0.75 m by 1.20 m. If the inside temperature is 20°C and the outside temperature is –10ºC, effective thermal resistance of the opaque wall and window are 2.10 m2 K W–1 and 0.21 m2 K W–1 respectively. Find the heat flow through the entire wall. (a) 215 W (b) 205 W (c) 175 W (d) 110 W 20. In the given circuit, the potential difference between A and B is

(a) 0

(b) 5 V

(c) 10 V (d) 15 V

C1 21. Two capacitors C1 = 2 mF and C2 = l mF are charged + – to same potential V = 100 V, S1 S2 C2 but with opposite polarity – + as shown in the figure. The switches S1 and S2 are closed. The ratio of final energy to the initial energy of the system is 1 1 1 (a) 1 (b) (c) (d) 2 9 4 22. 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 while 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.

23. A proton has kinetic energy E = 100 keV which is equal to that of a photon. The wavelength of photon is l2 and that of proton is l1. The ratio l2/l1 is proportional to (a) E2 (b) E1/2 (c) E–1 (d) E–1/2 24. Find the pressure at which temperature attains its maximum value if the relation between pressure and volume for an ideal gas is P = P0 + (1 – a )V2 . 4 P0 P 2P0 (a) (b) 0 (c) P0 (d) 3 3 3 25. A small metallic ball is charged positively and negatively in a sinusoidal manner at a frequency of 106 cps. The maximum charge on the ball is 10–6 C. What is the displacement current due to this alternating current? (a) 6.28 A (b) 3.8 A (c) 3.75 × 10–4 A (d) 122.56 A 26. A body of mass 1 kg, initially at rest, explodes and breaks into three fragments of masses in the ratio 1 : 1 : 3. The two pieces of equal mass fly off perpendicular to each other with a speed of 30 m s–1 each. What is the velocity (in m s–1) of the heavier fragment? (a) 6 2 (b) 10 2 (c) 8 (d) 12 27. A particle of charge q and mass m starts moving from the origin under the action of an electric   field E = E0 i and a magnetic field B = B0 i with  a velocity v = v0 j. The speed of the particle will become 2v0 after a time 2B q 2mv0 (a) t = (b) t = 0 qE0 mv0 (c) t =

3B0q mv0

(d) t =

3mv0 qE0

28. A rectangular block of mass m and area of cross-section A floats in a liquid of density r. If it is given a small vertical displacement from equilibrium, it undergoes oscillation with a time period T. Which one is not possible? (a) T ∝ m (b) T ∝ r 1 1 (c) T ∝ (d) T ∝ g A Physics for you | March ‘15

69

29. In Young’s double slit experiment of equal width slits, if intensity at the centre of screen is I0, then intensity at a distance of b/4 from the central maxima is (b is the fringe width) I0 I0 I (a) I0 (b) (c) (d) 0 2 4 3 30. Sound of wavelength l passes through a Quincke’s tube, which is adjusted to give a maximum intensity I0. Through what distance should the sliding tube be moved to give an I intensity 0 ? 2 l l l l (a) (b) (c) (d) 2 3 4 8 solutions 1. (b) : v →

vB →

R

v

→

1 km vR →

Resultant velocity of boat, 1 km 1 km v= = = 4 km h −1 15 min (1 / 4) h Velocity of river, v R = v 2B − v 2 = 52 − 42 = 3 km h −1 2. (c) : Let P = power radiated by the sun, R = radius of planet. P Energy received by planet = × pR2 . 2 4 pd Energy radiated by planet = (4pR2)sT 4. For thermal equilibrium, P × pR2 = 4 pR2sT 4 4 pd 2 1 1 or T ∝ or T ∝ d −1/2 or T 4 ∝ 1 /2 2 d d 3. (a) : Total internal B A reflection occurs when θ the ray of light travels θ from denser medium T.I.R. P to rarer medium and the angle of incidence is greater than the critical C angle. 70 Physics for you |

March ‘15

At P, glass is denser and water is rarer. g

mw =

a

a

mw mg

=

4/3 8 = 3/2 9

8 where iC = critical angle 9 \ q = iC or sin q = 8/9 for critical angle. 8 For total internal reflection, sin q ≥ 9 \

sin iC =

4. (a) : For the near end of the rod, uf Lf 1 1 1 = + ⇒ v= = f u v u− f L− f For far end of the rod, 2L − f 1 1 1 1 1 1 = − = = + ⇒ f u + L v1 v1 f 2L ( f )(2L) ( f )(2L) 2L − f Lf (2L)( f ) Length of image = v − v1 = − L − f (2L − f ) \ v1 =

= =

Lf (2L − f ) − 2Lf (L − f ) (L − f )(2L − f )

2L2 f − Lf 2 − 2L2 f + 2Lf 2 Lf 2 = (L − f )(2L − f ) (L − f )(2L − f )

5. (d) : Electric field at P l 1 1  E=  +  2 pe0 x 3a − x  W=

2a

∫ q0 E dx a

2a 2a lq0  dx dx  lq0 = ∫ +∫ ln 2 2 pe0  x 3a − x  pe0  a a 6. (d) : Weight of length L of the chain = Mg L 1 Weight of length of the chain = Mg 3 3 As the centre of gravity of the hanging part lies L at its midpoint i.e. at a distance equal to 6 below the edge of the table, so the work required to pull the hanging part on the table is 1 L MgL W = Force × distance = Mg × = 3 6 18

=

Physics for you | March ‘15

71

1 c 7. (d) : Here, B ∝ = r r A = pr2 c φ = BA = ⋅ pr 2 = pcr r dφ dr \ e= = pc = pck dt dt So, e is constant.

[... r = r0 + kt]

8. (d) : Work done to burn 5 kg of fat = 5 × 7000 × 103 = 35 × 106 cal = 147 × 106 J Work done towards burning of fat in one trip (up and down the stairs) = mgh × mgh/2 = 60 × 10 × 10 + (1/2) × 60 × 10 × 10 = 9 × 103 J (as only half the work done while coming down is useful in burning fat) 147 × 106 Number of trips required = = 16334 9 × 103 1 1   1 − 9. (d) : = RZ 2  2 2 l  n2 n1  1 \ ∝ Z2 l l is shortest if Z is largest. Z is largest for doubly ionised lithium atom (Z = 3) among the given elements. Hence wavelength for doubly ionised lithium will be the least. 10. (a) : As T x = or

x = zero.

Yl 3 [ML−1T−2 ][L3 ] = =1 cos qt [ML2 T−2 ]

11. (c) : Potentiometer reads the voltage of battery and voltmeter reads the potential across the terminals of battery \ battery of cell is of 1.55 V VAB = 1.4 V 1.55 1.55 V r B A I= 280 + r I VAB = IR 1.55 280 Ω 1.4 = 280 × 280 + r r × 1.4 = 280 (1.55 – 1.4) r × 1.4 = 280 × 0.15 2. 8 r= × 15 = 30 W 1. 4 72 Physics for you |

March ‘15

12. (b) : In case of SHM, maximum velocity = aw a × 2p \ v M = aM w M = M TM 2p vN = aN w N = aN × TN But vM = vN (given) a M × 2 p aN × 2 p a T or M = M ...(i) \ = TM TN aN TN As

TM m 1 k2 = 2p × TN k1 2 p m

or

TM k = 2 TN k1

...(ii)

a k From (i) and (ii), M = 2 aN k1 13. (d) : As the LED is connected in series with the limiting resistor R, the potential difference across R = Battery voltage – voltage drop across LED =6–2=4V 4V 4V V \ R= = = = 400 W I 10 mA 10 × 10−3 A 14. (a) : For open pipe, N in fundamental mode, A A l1 l= l 2 v v v \ Frequency = = \ u1 = ...(i) 2l l1 2l For closed pipe, vibrating in third harmonic, 3l l= 2 N 4 N A A \ Frequency l v 3v = = l 2 4l 3v or u2 = ... (ii) 4l u2 – u1 = 100 (given) 3v v v \ − = 100 or = 100 or v = 400l 4l 2l 4l v 400l \ u1 = = = 200 Hz 2l 2l \ Fundamental frequency of open pipe = 200 Hz. 15. (b) : At t = 0, charge on C is zero, so potential difference across C is zero, so also across R the

potential difference is zero. Hence there is no current in R. At t = ∞, current through L is maximum and constant, so potential difference across L is zero, therefore potential difference across R is zero. Hence no current in R. 16. (a) : The block Q oscillates but does not slip on P. This indicates acceleration is same for both Q and P. A force of friction acts between the two blocks but the horizontal plane is frictionless. The system P-Q oscillates with angular frequency, k k w= = m+m 2m Maximum acceleration of the system will be kA amax = w2 A = 2m This acceleration is provided to the lower block by the force of friction. kA kA \ f max = mamax = m = . 2m 2 N 1 17. (d) : Given : 1 =  N = N 0e − lt N2 e \

N 0e −10 lt − lt

=

1 1 1 or = or 9 lt = 1 9 l t e e e

N 0e 1 or t = 9l 18. (b) : For motion of planet in a circular orbit, Centripetal force = Gravitational force GmM or mRw2 = or R7/2 w2 = GM R5/2 GM GM T 2 × GM or R7/2 = = = w2 (2 p / T )2 4 p2 or T2 ∝ R7/2 19. (a) : The wall and the window are in parallel arrangement; so net heat flow is the sum of the heat flow through the wall and the window. The temperature difference, TH – TC = 30 K Area of window, A1= (0.75)(1.20) = 0.90 m2 Heat flow through window pane, 30  dQ  TH − TC = (0.90) = 128.6 W   = 0.21 dt 1 R1 The area of the wall, A2= (2.00)(3.50) – (0.75)(1.20) = 6.10 m2

Heat flow through the wall,  dQ  TH − TC (6.10)(30) = = 87 W   = dt  2 R2 (2.10) dQ = 128.6 + 87 = 215.6 W dt 20. (c) : The forward biased p-n junction does not offer any resistance. 10 × 10 \ RAB = = 5 kW 10 + 10 Total resistance, R = 10 + 5 = 15 kW Current in the circuit, V 30 I= = = 2 × 10−3 A 3 R 15 × 10 I Current through each arm = = 10−3 A 2 \ VAB = 10 × 103 × 10–3 = 10 V. 21. (c) : Q = total charge shared by the capacitors = C1V – C2V = (C1 – C2)V V1 = Voltage across each capacitor Q  C − C2  = = 1 V C1 + C2  C1 + C2  100  2 −1 = × 100 = V  2 + 1  3 Ei = Initial energy of the system 1 1 1 = C1V 2 + C2V 2 = (C1 + C2 )V 2 2 2 2 Ef = Final energy of the system 1 = (C1 + C2 )V12 2 E f V12  100  2 1 1 = = =  Ei V 2  3  (100)2 9 Net heat flow,

22. (b) : Frictional force acts at the point of contact of the two surfaces. The frictional force acts in a direction opposite to the direction of net acceleration of point of contact.

C denotes centre of mass and A is point of contact. The acceleration of centre of mass is gsinq down Physics for you | March ‘15

73

the plane whether the cylinder rolls up or rolls down. Accordingly, the point of contact A has an acceleration down the plane, similar to that of C. Point of contact A moves down in both the cases The frictional force, acting on the cylinder, is always up the incline, while ascending as well as descending. hc hc or l2 = l2 E 1 2 For proton, kinetic energy K = m p v p 2 or 2m p K = m2p v 2p or 2mpK = p2

23. (d) : For photon, E =

... (i)

2

h  or 2m p K =   (by de Broglie equation)  l1  h h = or l1 = ... (ii) 2m p K 2m p E From (i) and (ii), 2m p E l2 hc = × l1 E h

l2 c × 2m p = = c 2m p × E −1/2 l1 E l2 or ∝ E −1/2 l1 or

24. (a) : Ideal gas equation PV = nRT (P0 +(1 – a)V 2)V = nRT

(P + (1 − a)V 2 )V T= 0 nR dT P0 3V 2 (1 − a) = + =0 dV nR nR [For maximum temperature] P0 = 3(a – 1)V2 P0 or V 2 = 3(a − 1) P0 \ P = P0 + (1 − a) 3(a − 1) P0 2P0 = P0 − = 3 3 25. (a) : Charge oscillating sinusoidally is given by q = q0sinwt Displacement current, dq Id = = q0 w cos wt dt 74 Physics for you |

March ‘15

(Id )max = q0 w = q0 × 2 pu = 10–6 × 2 × 3.14 × 106 = 6.28 A 26. (b)    27. (d) : E is parallel to B and v is perpendicular to both. Therefore, path of the particle is a helix with increasing pitch. Speed of particle at any time t is v = v 2x + v 2y + v z2

... (i)

qE Here, v 2y + v z2 = v02 , v x = 0 t m and v = 2v0 Substituting the values in eqn. (i), we get 3 mv0 t= qE0 28. (b) : In equilibrium, let l = length of the block immersed. \ mg = Alrg If the block is given a further downward displacement x and its downward acceleration becomes a then mg – A(l + x)rg = ma  Arg  or a = −  x  m  Arg Put w2 = . m \ a = – w2x. This is SHM with time period T = 2p

m . Arg

29. (b) : Let the intensity of individual waves be I, then I I0 = 4 I ; I = 0 4 dy Also, ∆x = d sin q = D d b d lD l ∆x = × = × = D 4 D 4d 4 2p l p ∆φ = × = l 4 2 So required intensity I p I ′ = I + I + 2 I 2 cos = 2 I = 0 2 2 30. (d) nn

paper-1 Section-1 One or More Than One Options Correct Type This section contains 10 multiple choice questions. Each question has four choices (a), (b), (c) and (d) out of which ONE or MORE THAN ONE are correct.

1. A planet moves round the Sun in an elliptical orbit such that its kinetic energy is k1 and k2 when it is nearest to the Sun and farthest from the Sun respectively. The smallest distance and the largest distance between planet and the Sun are r1 and r2 respectively. (a) If total energy of the planet is U, then r2 U − k1 = . r1 U − k2 (b) If total energy of the planet is U, then r2 U − k2 = . r1 U − k1 (c) If r2 = 2r1, the total energy of planet in terms of k1 and k2 is 2k1 – k2. (d) If r2 = 2r1, the total energy of planet in terms of k1 and k2 is 2k2 –k1. 2. One mole of monoatomic gas is taken through cyclic process shown below. TA = 300 K. Process AB is defined as PT = constant. Select the correct statements. P 3P0 P0

C

B A

T

(a) Work done in process AB is –400 R. (b) Change in internal energy in process CA is 900 R. (c) Heat transferred in the process BC is 2000 R. (d) Change in internal energy in process CA is –900 R. 3. Three concentric conducting spherical shells have radii r, 2r and 3r and charges q1, q2 and q3 respectively. Innermost and outermost shells are earthed as shown in figure. Select the correct alternatives. q3 q2 (a) q1 + q3 = –q2 q1 − q2 (b) q1 = 4 q3 =3 (c) q1 q 1 (d) 3 = − q2 3 4. A source is moving across a circle given by the equation x2 + y2 = R2, with constant speed 330 p m s −1 , in anti-clockwise direction. A 6 3 detector is at rest at point (2R, 0) with respect to the center of the circle. If the frequency emitted by the source is u and the speed of sound is 330 m s–1, then (a) the position of the source when the detector records the maximum frequency R 3  is  , − R . 2  2 PHYSICS FOR YOU

| March ‘15 75

(b) the co-ordinate of the source when the detector records minimum frequency is (0, R). (c) the minimum frequency recorded by the detector is

6 3 p+6 3

A

u.

(d) the maximum frequency recorded by the 6 3

u. 6 3−p 5. Seven identical rods of material of thermal conductivity K are connected as shown in figure. All the rods are of identical length L and cross sectional area A. If one end A is kept at 100°C and the other end B is kept at 0°C, what would be the temperature of the junctions C, D and E (TC, TD and TE) in the steady state? detector is

K

C K A

K K

D K

K E

K

B

(a) TC > TE > TD (b) TC = TD = 37.5°C, TE = 50°C (c) TC = 62.5°C, TD = 37.5°C, TE = 50°C (d) TC = 60°C, TD = 40°C, TE = 50°C 6. It is observed that only 0.39% of the original radioactive sample remains undecayed after eight hours. Select the correct options. (a) The half-life of that substance is 1 hour. 1 (b) The mean life of the substance is hour. ln 2 (c) Decay constant of the substance is ln2 per hour. (d) If the number of radioactive nuclei of this substance at a given instant is 108 then the number left after 30 min would be

2 × 107. 7. Block B in figure weighs 700 N. The coefficient of static friction between block and table is 0.25, assume that the cord between B and the knot is horizontal. Choose the correct options. 76 PHYSICS FOR YOU |

March ‘15

30°

B

(a) If the weight of A is 100 N, block B will start sliding. (b) In the static equilibrium condition, the tension in the inclined string is more than tension of other two strings. (c) If the weight of A is 50 N, friction acting on block B is 175 N. (d) If the system is in equilibrium, weight of A cannot be equal to the force of friction acting on B. 8. In displacement method, the distance between object and screen is 96 cm. The ratio of length of two images formed by a convex lens placed between them is 4.84. (a) Ratio of the length of object to the length of 11 shorter image is . 5 (b) Distance between the two positions of the lens is 36 cm. (c) Focal length of the lens is 22.5 cm. (d) Distance of the lens from the shorter image is 30 cm. 9. Which of the following statements is/are correct? (a) Average speed of a particle in a given time period is never less than the magnitude of average velocity. (b) It is possible to have situations in which   dv d|v | ≠ 0, but = 0. dt dt (c) It is possible to have situations in which   d |v | dv ≠ 0, but = 0. dt dt (d) If the average velocity of a particle is zero in a time interval, then it is possible that the instantaneous velocity is never zero in that interval.

PHYSICS FOR YOU

| March ‘15 77

10. The potential energy of a particle in a certain a b field has the form U = − , where a and b r2 r are positive constants, r is the distance from the center of the field. Then 2a (a) At r = , particle is in steady equilibrium. b 2a (b) At r = , particle is in unsteady equilibrium. b (c) Maximum magnitude of force of attraction is

b3

.

27a2 (d) Maximum magnitude of force of attraction is

27 b3 a2

. Section-2

One Integer Value Correct Type This section contains 10 questions. Each question, when worked out will result in one integer from 0 to 9 (both inclusive). q3

5

3m q1

m

11. Three charges q1 = 3 mC, q2 = –3 mC and q3 are kept at the vertices of a triangle as shown in the figure.



4m



 F

q2

 If the net force acting on q1 is F, the charge q3 is 2  1 then given as 1 +  mC. Find the value of n.  n

12. One end of a spring of natural length l0 = 0.1 m and spring constant k = 80 N m–1 is fixed to the ground and the other end is fitted with a smooth ring of mass m = 2 g, which is allowed to slide on a horizontal rod fixed at a height h = 0.1 m. 0.1 m

{

37°

Initially the spring makes an angle of 37° with the vertical when the system is released from rest. When the spring becomes vertical, if the speed 4 of ring is v m s–1, find v. (Given cos 37° = ) 5 78 PHYSICS FOR YOU |

March ‘15

13. A steady current I goes through a wire loop PQR having shape of a right angle triangle with PQ = 3x, PR = 4x and QR = 5x. If the magnitude of the magnetic field at P due to this loop is  m I  k  0  , find the value of k.  48 px  14. In brass, the velocity of longitudinal wave is 100 times the velocity of the transverse wave. If Y = 1 × 1011 N m–2, then stress in the wire is x × 107 N m–2. What is the value of x ? 15. A glass vial containing 16 g sample of an enzyme is cooled in an ice bath. The bath contains water and 0.120 kg of ice. The sample has specific heat capacity 2250 J kg–1 K–1; the glass vial has mass 6 g and specific heat capacity 2800 J kg–1 K–1. How much ice (in g) melts in cooling the enzyme sample from room temperature (19.5°C) to the temperature of ice bath? Lf (ice) = 80 cal g–1. 16. A silver sphere of radius 1 cm and work function 4.7 eV is suspended from an insulating thread in free space. It is under continuous illumination of 200 nm wavelength light. As photoelectrons are emitted, the sphere gets charged and acquires a potential. The maximum number of photoelectrons emitted from the sphere is A × 10z (where 1 < A < 10), the value of z is 17. A 2.5 kg wooden block is initially at rest on a fix horizontal rough table of height 1 m. The block is initially 2 m away from edge. It is pushed with a constant force of 50 N for a distance of 1 m and then let go. The block falls off the 2m 1m edge and lands 2 m from the bottom of the table. 2m The coefficient of kinetic friction between the n block and the table is . Find n. 10 In physics, you don’t have to go around making trouble for yourselfnature does it for you. – Frank Wilczek

18. A series RC combination is connected to an ac voltage of angular frequency w = 500 rad s–1. If the impedance of the RC circuit is R 1.25 , the time constant (in milli second) of the circuit is

source is 3 m away from A, a person standing at a point O on a road perpendicular to the track hears a sound of frequency u′. The distance of O from A at that time is 4 m. If the original frequency is 640 Hz, then the value of u′ is 340 × x Hz. What is the value of x? (Given : Velocity of sound = 340 m s–1)

19. An annular ring with inner and outer radii R1 = 1.2 cm and R2 = 4.8 cm is rolling without slipping with a uniform angular speed. What is the ratio of the forces experienced by the two particles of same mass situated on the outer and inner parts of the annular ring?

A

3m

100 m s–1 3 S 

4m

100 m s−1 3 along a road, towards a point A. When the

20. A source of sound is travelling at

O

paper-2 Section-1 Only One Option Correct Type This section contains 10 multiple choice questions. Each question has four choices (a), (b), (c) and (d) out of which ONLY ONE is correct.

1. A ball is dropped from a certain height on a horizontal floor. The coefficient of restitution between the ball and the floor is 1/2. The displacement time graph of the ball will be s

(a)

(b) t s

s

(c)

(d) t

t

2. In the circuit shown in figure, I1 and I2 are the steady state currents in the coils after the switch S is closed. Then

(a) I1 =

I1

L1

I2

L2

E

R

EL2 R(L1 + L2 )

S

(b) I1 =

EL1 R(L1 + L2 )

(c) I 2 =

EL2

(d) I 2 =

E L1L2 RL2

R L1L2 3. A point object is placed at a distance of 12 cm on the axis of a convex lens of focal length 10 cm. On the other side of the lens, a convex mirror is placed at a distance of 10 cm from the lens such that the image formed by the combination coincides with the object itself. What is the focal length of convex mirror? (a) 25 cm (b) 50 cm (c) 10 cm (d) 20 cm 4. One end of a string of length l is tied to the ceiling of a lift accelerating upwards with an acceleration g/2. The linear mass density of the string is m(x) = m0x1/2 where, x is measured from the bottom. The time taken by a pulse to reach from bottom to top is (a)

3l g

(b) 2

l g

(c)

l g

(d)

l 3g

5. Magnetic field at the centre of a Bohr’s hypothetical hydrogen atom in the nth orbit of the electron is (a) directly proportional to charge of electron e (b) directly proportional to e2 (c) inversely proportional to n5 (d) directly proportional to n5 PHYSICS FOR YOU

| March ‘15 79

6. Two point charges +q and –q are held fixed at (–d, 0) and (d, 0) respectively of a x-y coordinate system. Then (a) the electric field E at all points on x-axis has the same direction. (b) electric field at all points on y-axis is parallel to x-axis. (c) work has to be done in bringing a test charge from ∞ to the origin. (d) the dipole moment is 2qd along the x-axis. 7. A hollow sphere of outer radius R is rolling down an inclined plane without slipping and attains a speed v0 at the bottom. Now the inclined plane is made smooth and the sphere is allowed to slide without rolling. Now it attains a speed 5v0 . What is the radius of gyration of sphere? 4 3 4 2 2 (a) R (b) R (c) R (d) R 4 5 5 3 8. Two point monochromatic and coherent sources of light of wavelength l are each placed as shown in the figure below. The initial phase difference between the sources is zero. Select the incorrect statement. S1

S2

D

d (D > > d)

O Screen

7l (a) If d = , O will be minima. 2 (b) If d = l, only one maxima can be observed on screen. (c) If d = 4.8 l, then total 10 minimas would be there on screen. 5l (d) If d = , then intensity at O would be 2 maximum. 9. A particle with charge Q, moving with a momentum p, enters a uniform magnetic field normally. The magnetic field has magnitude B and is confined to a region of width d, where p d< . The particle is deflected by an angle BQ q in crossing the field. Then 80 PHYSICS FOR YOU |

March ‘15

× × × p Q

× × × B × × × × × × d

p BQd (b) sin q = BQd p Bp pd (c) sin q = (d) sin q = Qd BQ 10. A thin uniform rod of length l is pivoted at its upper end. It is free to swing in a vertical plane. Its time period for oscillations of small amplitude is (a) sin q =

(a) 2p

l g

(b) 2 p

2l 3g

(c) 2 p

3l 2g

(d) 2 p

l 2g

Section-2 Comprehension Type (Only One Option Correct) This section contains 3 paragraphs, each describing theory, experiments, data etc. Six questions relate to the three paragraphs with two questions on each paragraph. Each question has only one correct answer among the four given options (a), (b), (c) and (d).

Paragraph for questions 11 and 12 When a charged capacitor is connected to an inductor, both the current and the charge on the capacitor oscillate. The angular frequency of the oscillations depends solely on the inductance (L) and the capacitance (C) of the circuit and is given 1 . The variation of charge (Q) on by w = LC the capacitor with time (t) is given by Q = Qmax cos wt where Qmax is the maximum charge on the capacitor. V = 12 V

C = 9 pF L = 2.81 mH

S2 S1

(c) (–6.79 × 10–4 A) sin wt

14. Find the minimum height h1 (in situation-1), for which the block just starts to move up. 2h 5h (a) (b) 3 4 5h 5h (c) (d) 3 2 Paragraph for questions 15 and 16 Figure shows a rod of length l = 1 m and mass m = 8 kg. It is hinged at the end P (on wall). At the end Q, a block of mass M = 50 kg is suspended. At the point R, which is at a distance equal to 25 cm from end Q, a string is connected whose other end is connected to the wall.

(d) (12.6 × 10–6 A) sin wt Paragraph for questions 13 and 14

4 3   Given : sin 37° = 5 and cos 37° = 5 

In figure, a capacitor is initially charged when S1 is open and S2 is closed. Then S1 is thrown closed at the same instant that S2 is opened so that the capacitor is shorted across the inductor. 11. The maximum value of current (in A) in the circuit is (a) 6.79 × 10–3

(b) 6.79 × 10–4

(c) 6.79 × 10–6

(d) 12.6 × 10–6

12. The equation of current as a function of time is (a) (– 6.79 × 10–3 A) sin wt (b) (– 6.79 × 104 A) sin wt

(Take g = 10 m s–2)



h1 h2

4r /3 Hole

h

Q

2r

A cylinder tank has a hole of diameter 2r in the bottom. The hole is covered by wooden cylindrical block of diameter 4r, height h and density r/3. Situation 1 : Initially, the tank is filled with water of density r to a height such that the height of water above the top of block is h1 (measured from the top of the block). Situation 2 : The water is removed from the tank to a height h2 (measured from the bottom of the block) as shown in figure. The height h2 is smaller than h (height of the block) and thus the block is exposed to the atmosphere. 13. In situation-2, if h2 is further decreased, then (a) block will not move up and remains at its original position. h (b) for h2 = , block starts moving up. 3 h (c) for h2 = , block starts moving up. 4 h (d) for h2 = , block starts moving up. 5

ing Str 37° R

Wall m

P

l

M

15. The tension in the string is (a) 600 N (b) 80 N (c) 1200 N (d) 800 N 16. The horizontal component Nx and vertical component Ny of the reaction at end P are given by (a) Nx = 1300 N, Ny = 960 N (b) Nx = 960 N, Ny = 140 N (c) Nx = 960 N, Ny = 960 N (d) Nx = 1300 N, Ny = 1300 N Section-3 Matching List Type (Only One Option Correct) This section contains four questions, each having two matching lists. Choices for the correct combination of elements from List-I and List-II are given as options (a), (b), (c) and (d), out of which one is correct.

17. List-II is depicting a standard situation in which a modification is introduced. Match the effect on variable x with appropriate entry in List-I. PHYSICS FOR YOU

| March ‘15 81

List I Must increase

R. If a transparent paper 3. Bright fringe (m = 1.45) of thickness of order 62 t = 0.02 mm is pasted on S1 i.e. one of the slits, the nature and order of interference at P S. After inserting the 4. Bright fringe transparent paper of order 280 (m = 1.45) of thickness t = 0.02 mm in front of slit S1, the nature and order of interference at O Code : (a) P - 1, Q - 4, R - 2, S - 3 (b) P - 3, Q - 1, R - 2, S - 4 (c) P - 2, Q - 4, R - 1, S - 3 (d) P - 4, Q - 1, R - 2, S - 3 19. The spectral lines of hydrogen-like atom fall with in the wavelength range given in List II. Match the following.

18. In Young’s double slit experiment, the point source S is placed slightly off the central axis as shown in figure. If l = 500 nm, then match the following.

Q. If it is atomic 2. l = 134 Å and it hydrogen atom and corresponds to energy E = –3.4 eV, transition from then 2 to 1

2 mm

P

S1

S

y = 10 mm

List II 1. Two tuning forks of slightly different frequency. Wax is put on tuning fork of higher frequency. Beat frequency = x Q. Must 2. An oscillating simple decrease pendulum, positive charge is given to the bob and an electric field is switched on in vertically downward direction. Time period of oscillation = x R. Must 3. A pulse is sent down a taut remain string. The tension is suddenly same decreased but the string is still taut. The time taken by the pulse to reach the other end = x S. Cannot 4. In a LC oscillation circuit be dielectric slab is introduced predicted when capacitor is discharged. Value of maximum current in inductor = x Code : (a) P - 1, Q - 4, R - 2, S - 3 (b) P - 3, Q - 2, R - 4, S - 1 (c) P - 3, Q - 4, R - 2, S - 1 (d) P - 1, Q - 3, R - 4, S - 2 P.

O

20 mm S2 1m

2m

List I List II P. Nature and order of 1. Bright fringe interference at the point P of order 80 Q. Nature and order of 2. Bright fringe interference at point O of order 262 82 PHYSICS FOR YOU |

March ‘15

List I P. If it is atomic hydrogen atom and energy E = –0.85 eV, then

List II 1. l = 1212 Å and it corresponds to transition from 2 to1

R. If it is doubly 3. l = 303 Å and it ionized lithium corresponds to atom, then transition from 2 to 1 S. If it is singly ionized 4. l = 970 Å and it helium, then corresponds to transition from 4 to 1 Code : (a) P - 1, Q - 4, R - 3, S - 2 (b) P - 4, Q - 1, R - 2, S - 3 (c) P - 3, Q - 2, R - 2, S - 4 (d) P - 2, Q - 1, R - 2, S - 3

20. Match the entries of List-I and List-II.

PA TB 1 T = ⇒ = B PB TA 3 TA

List II 1. Bifocal lens

Now,

Q. Hypermetropia

2. Cylindrical lens

⇒ TB =

R. Presbyopia

3. Concave lens

S.

4. Convex lens

P.

List I Myopia

Astigmatism

\ WAB =

TA/TC = PA/PB ⇒ TA/TC = 1/3 ⇒ TC = 3TA

SolutionS

or TC = 900 K

PaPer-1

1. (a, d) : Apply energy conservation maximum and minimum distances.

k1 +

− GmM r1

From (i),

sun M r1

r2

= k2 +

− GmM

−

− GmM r2

=U

...(i)

...(iii)

⇒ 2U – 2k2 = U –k1 ⇒ U = 2k2 –k1

2. (a, c, d) : Process AB : PT = constant = k dV 2nRT nRT 2 = =k ⇒ dT k V B Bk WAB = ∫ PdV = ∫ dV A AT

⇒

k 2nRT dT k

∫TA T .

DU = nCVDT 3 3 = (1) R × (TA − TC ) = R × (300 − 900) 2 2 =

GmM = U − k1 r1

r2 U − k1 = r1 U − k2 2r U − k1 If r2 = 2r1 ; 1 = r1 U − k2

TB

m k1

...(ii)

\

WAB =

for

= U − k2

r2

100

∫300 2nR dT = 2nR (100 − 300)

⇒ WAB = –400nR = – 400 R (Q n = 1 mole) Process CA : P/T = constant \ TA/TC = PA/PC

Code : (a) P - 3, Q - 4, R - 1, S - 2 (b) P - 4, Q - 3, R - 1, S - 2 (c) P - 2, Q - 3, R - 4, S - 2 (d) P - 4, Q - 2, R - 1, S - 4

k2

300 = 100 K 3

3 R × (− 600) = − 900R 2

Process BC : Isobaric Q = nCPDT 5 \ Q = (1) R × (TC − TB ) 2 Q=

5 5 R × (900 − 100) = R × 800 2 2

⇒ Q = 2000 R 3. (a, b, c) :

q3 q2 q1

A

r 2r

B

C

3r

As shell A is earthed kq kq kq \ VA = 1 + 2 + 3 = 0 2r 3r r q2 q3 ⇒ q1 + + =0 2 3 kq kq kq Also VC = 1 + 2 + 3 = 0 3r 3r 3r PHYSICS FOR YOU

...(i)

| March ‘15 83

⇒ q1 + q2 + q3 = 0 ⇒ q1 + q3 = –q2

H7 =

...(ii)

Solving (i) and (ii), we get q3 − 3 q3 q −1 = , = 3 and 1 = q2 q2 4 q1 4

Q

\

4. (a, c, d) : Sound emitted by source at A will result in maximum frequency while sound emitted by source at B will result in minimum frequency.

H2 100°C A

TD D

H4 H1

H6 E TE



H7 0°C B

H5

Let H1, H2, H3, H4, H5, H6 and H7 be the rate of heat flow through AE, AC, CD, CE, EB, ED and DB respectively. Then KA(100 − TE ) KA(100 − TC ) H1 = , H2 = L L KA(TC − TD ) KA(TC − TE ) H3 = , H4 = L L H5 =

KA(TE − 0) L

84 PHYSICS FOR YOU |

, H6 =

March ‘15

KA(TE − TD ) L

L

(a, b, c) : Here, t = 8 hours

5. (a, c) : This problem can be solved similar to electric circuit problem. H3

KA(TE − 0)

6.

        330 330 umax =  u  u, umin =   330 + 330 p   330 − 330 p     6 3 6 3 

TC C

L

=



A

Time taken by sound to reach from A to 2R sin 60° detector = 330 Angle travelled by source 330 p 2R sin 60° p = × = 330 6 6 3R

H1 = H5 KA(100 − TE )

or TC + TD = 100 ...(ii) Q H2 = H3 + H4 = H7 KA(100 − TC ) KA(TC − TD ) \ = L L KA(TC − 50) KA(TD ) (Using (i)) + = L L or 2TC – 2TD = 50 ...(iii) Solving (ii) and (iii), we get TC = 62.5°C, TD = 37.5°C ⇒ TC > TE > TD

Detector

60°

L

or TE = 50°C ...(i) Q H4 = H6 KA(TC − 50) KA(50 − TD ) \ = (Using (i)) L L

B

2R

KA(TD − 0)

As N = N0e–lt N = e −lt N0

0.0039 = e–l8 1 = 256 or e l8 = 28 el8 = 0.0039 Taking natural logarithm on both sides, we get 8l = 8ln2 l = ln2 per hour Option (c) is correct. ln 2 T1/ 2 = = 1 hour l Option (a) is correct. 1 1 Mean time, t = = hour l ln 2 Option (b) is correct. 1   8  1  2 

N = (10)   2

=

1

2 or N = 5 2 × 107 Option (d) is incorrect.

× 108

7. (b, d) : T cos30° = T2 = fs T sin30° = T1 = mA g From eqn (i) and (ii), T=

f s2

T1

2

+ (mA g )

fs

m g tan 30° = A fs \ mA g = WA = =

...(i) ...(ii)

fs 3

0.25 × 700 N 3

=

B

T2

m mB g

A mA g

3

In uniform circular motion,   dv d|v | = 0, ≠0 dt dt In circular motion, A from point A to point A again, average velocity = 0 Instantaneous velocity ≠ 0 (at any time)

= 101 N

I 8. (a, b, d) : Given, D = 96 cm, 2 = 4.84 I1 Let I1 = a and I2 = 4.84 a \ O = I1I 2 = 2.2 a 2.2a 11 (a) Required ratio = = 5 a v1 11 = (b) Also, ...(i) u1 5 v1 + u1 = 96 ...(ii) 5v1 \ v1 + = 96 11 16v1 ⇒ = 96 or v1 = 66 cm 11 \ u1 = 30 cm Distance between two positions = v1 – v2 = 66 – 30 = 36 cm (Q u1 = v2) 1 1 1 30 + 66 (c) = + = f 66 30 30 × 66 30 × 66 330 ⇒ f = = = 20.625 cm 96 16 (d) u1 = 30 cm 9. (a, b, d) : Average speed =

t otal distance

total time displacement Average velocity = time

Q distance ≥ displacement \ Average speed ≥ average velocity  d|v | = tangential acceleration dt  dv = net acceleration dt

dU  −2a b  = +  dr  r 3 r2 

10. (a, c) : For

dU b 2a 2a = 0, = ⇒ r= 2 3 dr b r r

d 2U dr

2

=

At r = d 2U dr 2

=

+6a r

4

−

2b r

3

=

2  3a  − b 3  r  r

2a , b  2 b b 2  3a × b − b = = >0   r3 2 r3 r 3  2a

i.e., U is minimum. So, it is a position of stable (steady) equilibrium. dU 2a b F=− = − dr r3 r2 2 dF − d U For maximum force, = =0 dr dr 2 2  3a 3a  ⇒ − b = 0 ⇒ r = 3  r  b r F=

2a

−

b

=

2ab3

−

b3

=−

b3

3 2 27a3 9a2 27a2  3a   3a   b   b    11. (8) : Let F1 be the force on q1 due to q2 and F2 be that due to q3. As q1 > 0, q2 < 0, the charge q3 should be positive  to get the net force F .

PHYSICS FOR YOU

| March ‘15 85

kq q F kq q tan q = 2 , F1 = 1 2 , F2 = 1 3 16 9 F1 1 3 q3 16 and k = \ = × 4 pε0 4 9 q2 2 27 81  9  \ q3 = × q2 = =   mC 64 64  8 

14. (1) : The velocity of longitudinal wave is Y vL = r The velocity of transverse wave is vT =

2

 1 = 1 +  mC  8 Hence, n = 8.

\

12. (5) : Loss in elastic potential energy = Gain in kinetic energy 1 1 2 i.e., k(l − l0 ) = mv 2 2 2 k  k  l v = (l − l0 ) =  0 − l0   m  cos 37° m  1  k = l0  −1  cos 37°  m 80 5  = (0.1)  − 1  4  2 × 10−3 1 = 0.1 × × 2 × 102 = 5 m s−1 4 13. (7) : Using the concept of area of triangle 1 1 × PD × 5x = × 3x ×Q4 x D 2 2 5x 1 1 3x 1 × PD × 5x = × 3x × 4 x 2 2 2 R P 4x 12 x \ PD = 5 144 x 2 9 x 2 2 = 9x 2 − = QD = (PQ ) − (PD ) 25 5 9 x 16 x and DR = 5x − = 5 5 Magnetic field at P due to current elements PQ and PR is zero as the point P is on the conductor. Therefore, magnetic field at P due to current element QR is m0 I B= (sin φ1 + sin φ2 ) 4 pPD m I × 5  (9 x / 5) (16 x / 5)  + B= 0 4 p × 12 x  3x 4 x  =

m 0 I 5  3 4  7m 0 I + = 48 px  5 5  48 px

86 PHYSICS FOR YOU |

March ‘15

\ k=7

T = m

T

pr 2 r

vL Y pr 2 r Y = × = = vT T r T / pr 2

\ Stress =

Y

(v L / vT )2

=

Y Stress

1 × 1011 (100)2

= 1 × 107 N m–2 Hence x = 1 15. (3) : The heat lost by the sample (and vial) melts a mass m of the ice. Q \ m= = Lf (16 × 10−3 )(2250)(19.5) + (6 × 10−3 )(2800)(19.5) 80 × 4 . 2 × 103

or m =

702 + 327.6 3

336 × 10

= 3.06 × 10−3 kg ≈ 3 g

hc − φ, l0 1242 eVnm − 4.7 eV = 6.2 eV − 4.7 eV eV0 = 200 nm

16. (7) : As eV0 =

or eV0 = 1.5 eV or V0 = 1.5 V

−2 V r 1.5 × 10 q As V0 = k , q = 0 = C k r 9 × 109 Since q = ne,

1.5 × 10−2 C q = = 1.04 × 107 e (9 × 109 ) (1.6 × 10−19 C) \ z=7 n=

17. (5) : Since, y = x tan 0° − −1 =

1 gx 2 2 v 2 cos2 0°

−1 10 × 4 2 × , v = 20 m2 s–2 , 2 2 v

WF + Wfr = DK

I L or L1I1 = L2I2 , 1 = 2 I 2 L1 E Steady current I passing through R is, I = R Also, I = I1 + I2

1 × 2.5 v 2 2 ⇒ 50 – 50m = 25 ⇒ 50 – 25 = 50m 5 or m = \ n=5 10 50 × 1 − m × 2.5 × 10 × 2 =

2

 1  18. (4) : As Z = R 2 +  = R 1.25 ,  wC   1  R 2 (1.25) = R 2 +   wC 

2

2

1  1  = 0.25R 2 or = 0.5 R or    wC  wC or CR =

1 = 0.004 s = 4 ms 0.5 × 500

F2

R2

F1 R1 19. (4) : As the annular ring is rolling without slipping, therefore, its angular velocity w is constant, so is its linear velocity v. Therefore, no net force or net torque is acting on the ring. The force experienced by the two particles, one on outer part and other on inner part is only centripetal force directed towards the centre of ring as shown in figure. 2 R F2 m R2 w 4.8 = = 2 = =4 2 R1 1.2 F1 m R1 w

20. (2) : Effective value of velocity of source, 100 vs = cos q 3m S 3 100 3 = × = 20 m s−1 4m 3v 5 5m u′ = u v − vs 340 × 640 Hz = 340 × 2 Hz u′ = 340 − 20 \ x=2 PaPer-2

1. (c) : The ball will stop after a long time. The final displacement of the ball will be equal to the height. The motion is first accelerated, then retarded, then accelerated and so on. 2. (a) : L1

dI dI1 = L2 2 or L1dI1 = L2dI2 dt dt

 L2  EL2 I= \ I1 =  R(L1 + L2 )  L1 + L2   L1  EL1 and I 2 =  I=  L + L R(L1 + L2 )  1 2 3. (a) : For convex lens, 1 1 1 − = or, v = 60 cm v −12 10 In the absence of convex mirror, convex lens will form the image I1 at a distance of 60 cm behind the lens. Since, the mirror is at a distance of 10 cm from the lens, I1 will be at a distance of 60 – 10 = 50 cm from the mirror, i.e., MI1 = 50 cm. I2 L

O

I1

M

10 cm

50 cm 60 cm

12 cm

Now as the final image I2 is formed at the object O itself, the rays after reflection from the mirror retraces its path, i.e., rays on the mirror are incident normally, i.e., I1 is the centre of the mirror, so that R = MI1 = 50 cm R 50 Hence f = = = 25 cm 2 2 Solution Senders of Physics Musing SET-19 1. Anjali Sharma (New Delhi) 2. Sattwik Sadhu (W.B.) 3. Koushik Das (W.B.) SET-18 1. 2. 3. 4. 5.

Soumik Chandra (W.B.) Ronal Gaonkhowa (Assam) Satyajeet Kumar (Kerala) Shubhneet Bhatia (Ambala Cantt) Rushikesh Joshi PHYSICS FOR YOU

| March ‘15 87

4. (b) : Consider a small element dx at a distance x from the bottom of string. Weight of this element g = [m( x )dx ]( g + ) 2 Tension at this element, x 3 Tx = ∫ m( x ) gdx 2 0

dx

3

m en  me 2  m pm 2 e 7 = 0 = 0   4 pm  4 pε0 2n2  8ε30h5n5 h   Q  = 2p 

g 2

x

x

3 g m x1/ 2dx 2 ∫ 0 0 3 g m 0 x 3/ 2 = = g m 0 x 3/ 2 2 (3 / 2)

Tx =

m e  v  m e (n / mrn ) B= 0  n = 0 2rn  2 prn  4 p rn2

6. (b) : (a) E will not have same direction along entire x-axis. Option (a) is incorrect. –q

+q (–d, 0)

(d, 0)

(0,0) E(+q)

y P

E E(–q)

Velocity of transverse wave at this element v= dt =

T , v = dx = m dt dx

g m 0 x 3/ 2 m 0 x1/ 2

=

gx

gx

Integrate both sides, we get t

l

dx

1

l

l x −1/ 2dx ⇒ t = 2 ∫ gx g g 0 0 0 5. (c) : An electron revolving in a circular orbit in a hydrogen atom is equivalent to a current of eu. Magnetic field at the centre of Bohr’s hypothetical orbit m I m eu B= 0 = 0 ...(i) 2rn 2rn where rn is the radius of nth orbit. The electric force between the nucleus and electron in the nth orbit provides the centripetal force for circular motion.

∫ dt =

∫

t=

mvn2 e2 or rn = ...(ii) rn 4 pε0rn2 4 pε0mvn2 According to Bohr’s quantisation condition n ...(iii) mvnrn = n or vn = mrn From equations (ii) and (iii), we get \

e2

rn =

=

2 2

4 pε0  n

me 2 Then equation (i) becomes 88 PHYSICS FOR YOU |

March ‘15

–q

+q (–d, 0)

(0, 0)

(d, 0)

x

For – d ≤ x ≤ d, the field is along + x-axis For all other points, E is along negative x-axis (b) Electric field at P, a point on y-axis is parallel to x-axis. (c) Electric potential at origin = zero. No work has to be done in bringing a test charge from infinity to the origin. Option (c) is incorrect. (d) The dipole moment is directed from – q charge to + q charge, along negative direction of x-axis. \ Dipole moment = – 2 qd along x-axis. The option is incorrect. Hence option (b) is correct answer. 7. (b) : Let the sliding acceleration be a. So rolling a , acceleration = 1 + K 2 / R2 where K is the radius of gyration Using v2 – u2 = 2as, 2

5   v0  = 2aS 4

and v02 =

...(i) 2aS 2

... (ii)

2

1+ K /R From (i) and (ii), we get

K2 2

=

25 or 16

K2

25 −1 16 R R 9R 2 or K = 3R K2 = 4 16

1+

2

=

PHYSICS FOR YOU

| March ‘15 89

8. (d) : Path difference, Dx at O = d. 7l So, if d = , O will be a minima. 2 If d = l, O will be maxima. 5l , O will be minima and hence intensity If d = 2 is minimum. If d = 4.8 l, then total 10 minimas can be observed on screen, 5 above and 5 below O, which correspond to l 3l 5l 7l 9l Dx = ± , ± ,± ,± ,± 2 2 2 2 2 9. (a) : A to D is part of circle with centre C and radius CD = r.

C

mv = p = BQr p or r = BQ



×

×

×

ED d BQd sin q = = = CD r p

×

E A ×

d

D



12. (c) : I = –Imax sin wt = –(6.79 × 10–4 A) sin wt 13. (a) : When h2 is decreased, the upward force (buoyancy) on the block decreases while the downward force (weight) remains the same. Thus, the block does not move up and remains at its original position. 14. (c) :

P0 (acting through hole) [P0 + (h1+ h)g]

For the equilibrium of the block, (P0 + h1rg) [p(2r)2] + p(2r)2 h(r/3)g =[P0 + (h1 + h) rg] [p(2r)2 – pr2] +P0(pr2) 5h On solving, we get h1 = 3 15. (c) : Tsin37°

×

mg

3g 2l This represents an angular SHM with the time period 2 Comparing with a = –w2q, w =

2l 3g

11. (b) : The initial charge on the capacitor equals the maximum charge, i.e., Qmax = CV

= (9 × 10–12 F) (12 V) = 1.08 × 10–10 C dQ As Q = Qmax cos wt, I = = −wQmax sin wt , dt w=

1 LC

=

1 2.81 × 10

−3

× 9 × 10

−12

= 6.28 × 106 s −1

Imax = wQmax = 6.28 × 106 × 1.08 × 10–10 = 6.79 × 10–4 A

90 PHYSICS FOR YOU |

March ‘15

T

37° Tcos37°

Q

 3g  or a = −   q  2l 

2p = 2p w

Block

W

l  1 10. (b) : T = mg  sin q   mg lq = − I a 2  2 1 1 or mg lq = − ml 2a  2 3

T=

(P0 + h1g)

P

R

500 N

80 N

Nx

Ny

25 cm 50 cm

For equilibrium, about P : StP = 0 \ [(500)(1) + 80(0.5) – (Tsin37°)(0.75)] = 0 Torque produced by Tcos37°, Nx and Ny about the point P is equal to zero. 3 500 + 40 − T   (0.75) = 0 5 540(5) T= = 1200 N 3(0.75) 16. (b) : For equilibrium, SFx = 0 \ T cos37° – Nx = 0 4 4 or N x = T = (1200) = 960 N 5 5 For equilibrium, SFy = 0 \ Ny – Tsin37° + 500 + 80 = 0 3 or N y = (1200)   − 500 − 80 = 140 N 5 17. (b)

18. (d)

19. (b)

20. (b) nn

2 2 As, Q = V t1 = V t 2 ⇒ R1 = t1 = 15 = 1

R1

solution set-19

1.

  F F0 (b) : As, a = = [cos(t ) ^i + sin(t ) ^j ] m m t F0   ^ v = ∫ a dt = [sin(t ) i^+ (1 − cos t ) j] m 0

⇒ Q=

2.

F02 (1 − cos t ) m

⇒

where n is number of half lives elapsed for species A. N′ 5  1 1  + = =  n + 1 n N 32 +1  2  22  1  r1

−

1 r2 

2(0.07)  1 1  2T  1 1  = − h= −   rg  r1 r2  103 × 9.8  1 × 10−3 1.5 × 10−3 

5 2 When it dissociates,

R2 t = 10 minutes R1 + R2 1

1 2

3

2

3 2 mr , I 2 = I 3 = mr , 2 2 7 \ I = I1 + I 2 + I 3 = mr 2 2 I1 =

6

1 Now, DW = I w 2 2 1 ml 2 2 1 2 2 = × w = ml w 2 2 4 DU = –qlE \

4. (c) : Initial CV = R

+q B A� q

qE ml 2 2 w = qlE ⇒ w = 2 4 ml

9. (b) : By conservation of linear momentum mv1 + m(v1 + wR) = mv0  ⇒ 2v1 + wR = v0 …(i) By applying conservation of angular momentum

n f +n f 1 . 2 × 3 + 0. 4 × 5 7 f av = 1 1 2 2 = = n1 + n2 1. 2 + 0. 4 2 f R 7 initial CV 10 \ Final CV = av = R \ = 2 4 final CV 7

C = 2CV = 2 ×

⇒ t 3 = 3t1 = 45 minutes

2

= 4.76 × 10–3 m = 4.76 mm

\

2

V 2t 4 (R1 + R2 ) V 2t1 = R1R2 R1

8. (b) : As DW = – DU

T = 0.07 N m–1, r1 = 1mm, r2 = 1.5 mm,

5. (c) : From first law of thermodynamics DQ = DU + DW Here, DQ = Q, DW = Q/2 So, DU = Q –Q/2 = Q/2 = DW DU = DW = nCVDT From eqn (i) DQ = 2nCVDT nCDT = 2nCVDT

30

\ 3mk 2 = 7 mr 2 ⇒ k = 7 r

Here, r = 103 kg m–3, g = 9.8 m s–2, \

t2

Also, moment of inertia, I = 3mk2 , where k is radius of gyration.

On solving, we get, n = 4 3. (b) : As, rgh = 2T 

t4 =

R2

7. (a): Moment of inertia of the system about the given axis, I = I1 + I2 + I3

N  N (a): As, N ′ =  2 + 2   n  2 2n/2  Now

V 2 t 3 V 2 t1 = Req R1

In parallel, Q =

\ Kinetic energy at time t,

 1   1  F2 = m (v ⋅ v ) = m  0 (2 − 2 cos t ) = 2 2  m2  

R2

or 2R1 = R2 In series, Req = R1 + R2.

v1

O

… (i)

f R = 3R 2

(Q f = 3 for monoatomic gas)

6. (b) : Let R1 and R2 are the resistances of the first and the second coils and V be mains voltage.

about O, Li = 0 (where O is a point of ground frame near the point of collision) Lf = – mv1R + mR2w [only due to ring applying

   rcm × p + I cm w considering outward normal as positive] Li = Lf ⇒ v1 = Rw …(ii) v0 v Solving (i) and (ii), we get w = = 3R 3R 10. (a): As, DV = –E Dr \ VA –VB = –E(a – b) Here, E =

σ σ ⇒ VA − VB = (b − a) ε0 ε0

nn

physics for you | march ‘15

91

System of Particles and Rotational Motion ª Every particle of a rigid body rotating about a fixed axis moves in a circle, which lies in a plane perpendicular to the axis and has its centre on the axis. ª A rigid body fixed at one point or along a line can have only rotational motion. ª When a rigid body rotates about a fixed axis, every part of the body has the same angular velocity and angular acceleration. ª If all the external forces acting on a body or system of bodies are applied on the center of mass of the body, the state of rest or motion of the body or system of bodies remains unaffected. ª The centre of mass of a system of two particles lies on the straight line joining the two particles. ª The centre of mass of all homogenous bodies of regular shape coincides with the geometrical centre of the bodies. ª To determine the motion of the center of mass of a system, the knowledge of external forces acting on the body is required. ª The position of centre of mass of a system is independent of the choice of coordinate system. ª The system as a whole can change its shape or orientation due to internal forces, but it will have no effect on the trajectory of center of mass of isolated system. ª The centre of gravity of an external body is that point where the total gravitational torque on the body is zero. ª The centre of gravity of a body coincides with its centre of mass only when the gravitational field does not vary from one part of the body to the other. ª The total torque on a system is independent of the origin if the total external force is zero. 92 Physics for you |

March ‘15

ª The concept of torque accounts for the motion of our arms and legs. ª The vector sum of all the torque acting on a particle is equal to the time rate of change of the angular momentum of that particle. ª In case of pure rotation, a vector defines an axis of rotation and not the direction in which an object moves. ª The linear momentum of the body must be zero for static equilibrium. ª Angular acceleration of a body in rotational equilibrium will be zero. ª If algebraic sum of moments of all forces acting on the body about a fixed point is zero, the body will be in rotational equilibrium. ª Moment of inertia plays the same role in rotational motion as mass plays in translational motion. ª The moment of inertia of a body about a given axis is equal to twice the kinetic energy of rotation of the body rotating with unit angular velocity about the given axis. ª According to sign convention, anticlockwise moments are taken as positive and clockwise moments are taken as negative. ª The radius of gyration of a body about an axis is equal to the root mean square distance of the various particles constituting the body from the axis of rotation. ª Moment of inertia of circular ring about a given axis is greater than the moment of inertia of circular disc of same mass and same size about the same axis. ª Rolling motion of a body is a combination of translatory as well as rotatory motion. ª In rolling motion, there is no loss of mechanical energy because the contact point is at rest relative to the surface at any instant. nn

Readers can send their answer with complete address by 15 th of every month to win exciting prizes. Winners' name with their valuable feedback will be published in next issue.

down 1. 2. 3. 6. 7. 8. 11. 13. 18. 21. 22. 23. 25. 27. 28. 29. 31. 34.

ACRoSS 1. 4. 5.

9. 10. 12. 14. 15.

–15

A unit equal to 10 metre. (5) Biggest moon of saturn. (5) Force of attraction between like molecules. (8) Alloy involving the metal mercury. (7) A fold in rocks with younger layers closer to the centre of the structure. (8) Gradual loss in intensity of any flux through a medium. (11) An electron in combination with a hole in a crystalline solid. (7) Wave that obeys superposition principle. (6, 4) An angle whose value is 0° at the equator and 90° at the poles. (8) Instrument used for navigation and orientation. (7) The condition of a body in free fall. (14) A body that absorbs all the radiation incident upon it. (5, 4) Branch of physics dealing with the flow of air and gases. (12) Displacement of a vibrating body at zero time. (5) Tendency of a material to undergo permanent deformation. (10) Instrument for determining the angle of dip. (3, 6) Technique which enables 3D images to be made. (10) An isolating circuit used to minimize the reaction between a driving and driven circuit. (6) Reciprocal of dynamic viscosity. (8) Shortest path between points in the space. (8) Fuel material that comes from living or recently living organism. (7) Solidification curve. (3, 4) Sound with frequency higher than the human audible range. (10) Production of transverse electric field in a conductor carrying current when it is placed in a magnetic field perpendicular to the current. (4, 6) Repetitive layering in metamorphic rocks. (9) An area of closed circular fluid motion rotating in the same direction as the earth. (7)

16. Zero rank tensor. (6) 17. Temperature scale designed in 1717. (10) 19. A theory that is not proven but leads to further study. (10) 20. A material used to reduce the frictional force between two surfaces in contact. (9) 24. Ratio of angular momentum to angular velocity. (6, 2, 7) 26. Residual periodic variation of dc output which has been derived from an ac source. (6) 30. It happens when wavelength of light from an object gets increased. (8) 32. Rupture in the earth’s crust where molten lava and gases from below the earth’s crust escape into the air. (7) 33. Invented by scientist William Sturgeon in 1824. (13) 34. A small area on the retina that is insensitive to light. (5, 4) 35. Emission of EM radiation from a body as a result of its temperature. (13) 36. Inverse of collision frequency. (4, 4, 4) 37. A type of collision where kinetic energy is conserved.(7) 38. Speed of fluid defined on the basis of Torricelli’s theorem. (6, 5) g g

physics for you

| march ‘15 93

94 physics for you |

march ‘15