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edit Vol. XXIII

No. 8

rial

solar Energy and

August 2015

its Wonderful Applications

Corporate Office : Plot 99, sector 44 Institutional area, gurgaon -122 003 (HR). tel : 0124-4951200 e-mail : [email protected] website : www.mtg.in

t

he use of solar energy to produce electricity is not new to India. In Jammu, in the rest-house complex of Mata Vaishno Devi temple,

Regd. Office 406, taj Apartment, Near safdarjung Hospital, Ring Road, New Delhi - 110029.

Managing Editor Editor

: :

the need for electricity is completely met by solar energy. A large open terrace is completely filled with solar panels and are used for charging

Mahabir singh Anil Ahlawat (BE, MBA)

batteries which in turn is serving all the needs of the complex. In the city of Coimbatore, many housing complexes and individual

contents Physics Musing (Problem Set-25)

houses are tapping solar energy for the purpose of producing electricity.

8

Thought Provoking Problems

10

J & K CET

16

Solved Paper 2015 Ace Your Way CBSE XII

Companies distributing hydroelectricity are also supplementing by a large use of wind-energy to convert it to electrical energy for turning transformers. Andre Borschberg, in his solar Impulse 2, an airplane, powered by the sun, has flown for three consecutive days from Japan to Hawaii. the actual

23

Series 3

time of flight was more than 100 hours. Borschberg’s collaborator monitoring from the ground was happy that this

JEE Accelerated Learning Series

31

was done without any other fuel and was able to fly on its own a longer

Brain Map

46

distance than any jet plane, without refuelling. the wings of the carbon

AMU Engg.

57

Solved Paper 2015 Exam Prep

68

Physics Musing (Solution Set-24)

75

Core Concept

77

You Ask We Answer

84

Crossword

85

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.

fibre aircraft have move than 17,000 solar cells. the plane flies up to 28, 000 feet during the day to recharge the solar cells and descends to under 10,000 feet at night to minimise the power consumption. Anil Ahlawat Editor

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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, Gurgaon - 122003 (Haryana) We have not appointed any subscription agent.

Physics for you | August ‘15

7

P

PHYSICS

MUSING

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 detailed solutions with their names and complete address. 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.

Set 25 subjective type

1. A plane mirror of circular shape with radius r = 20 cm is fixed to the ceiling. A bulb is to be placed on the axis of the mirror. A circular area of radius R = 1 m on the floor is to be illuminated after reflection of light from the mirror. The height of the room is 3 m. What is the maximum distance between the centre of the mirror and the bulb so that the required area is illuminated ?  C 2. Some gas  P = 1.25  follows V B C C   V

A the cycle ABCDA shown in the D figure. Determine the ratio of T the energy given out by the gas to its surroundings during the isochore section of the cycle to the expansion work done during the isobaric section of the cycle. 3. An equilateral prism provides the least deflection angle 45º in air. Find the refractive index of an unknown liquid in which same prism gives least deflection angle of 30°. 4. The final image I of the object O shown in the figure is formed at a point 20 cm below a thin biconcave lens, which is at a depth of 65 cm from principal axis. From the given geometry, calculate the radius of curvature in cm of lens kept at A.

glasses have refractive index 1.5. Now they jump into a swimming pool and look at each other. B appears to be present at distance 2 m (from A) to A. A appears to be present at distance 1 m (from B) to B. If the refractive index of water in the swimming X pool is then find the value of X. 10 6. A planet revolves about the sun in elliptical orbit of semi-major axis 2 × 1012 m. The areal velocity of the planet when it is nearest to the sun is 4.4 × 1016 m s–1. The least distance between planet and the sun is 1.8 × 1012 m. Find the minimum speed of the planet in km s–1. 7. We would like to increase the length of a 15 cm long copper rod of cross-section 4 mm2 by 1 mm. The energy absorbed by the rod if it is heated is E1. The energy absorbed by the rod if it is stretched slowly E is E2. Then find 1 . E2 [Various parameters of Copper are : Density = 9 × 103 kg m–3; Thermal co-efficient of linear expansion = 16 × 10–6 K–1, Young’s modulus = 135 × 109 Pa, Specific heat = 400 J kg–1 K–1] Contd. on page 30

Solution Senders of Physics Musing

f = 30 cm

set-24 1. Sameer Joshi (Jharkhand)

45°

O 36 cm

2. Anjali Saxena (New Delhi) 65 cm

1m A 20 cm

 = 3/2 I

5. Two persons A and B wear glasses of optical powers (in air) P1 = + 2 D and P2 = + 1 D respectively. The

3. Shikha Vats (MP) 4. Rihan Saifi (UP) set-23 1. Akansha Arora (Punjab) 2. Varsha Jain (Haryana)

By Akhil Tewari, author Foundation of Physics for JEE Main & advanced, senior Professor Physics, RaO IIt aCaDEMY, Mumbai.

8

physics for you | august ‘15

By : Prof. Rajinder Singh Randhawa*

1. An electron accelerated through a potential difference of 2.5 kV, moves horizontally into a region of space in which there is a downward electric field of magnitude 10 kV m–1. (a) In what direction must a magnetic field be applied so that the electron moves undeflected? What is the magnitude of the smallest magnetic field? (b) What happens if the charge is a proton that passes through the same combination of fields? 2. A particle of mass m and charge +q enters a region of magnetic field with a uniform velocity v as shown in figure. (a) Find the angle subtended by the circular arc described by it in the magnetic field. (b) How long does the particle stay inside the magnetic field? (c) If the particle enters at A, what is the intercept AB? 3. A charged particle of mass m and charge q is projected on a rough horizontal x-y plane surface with z-axis in the vertically upward direction. Both electric and magnetic fields are acting in the region   ^ ^ and given by E = − E0 k and B = − B0 k respectively. z The particle enters B0 into the field at E0 y (a0, 0, 0) with  ^ velocity v = v0 j . ^ v0 j The particle starts (a0, 0, 0) x moving into a circular path on the plane. If the coefficient of friction between particle

and the plane is m. Then calculate the (a) time when the particle will come to rest. (b) time when the particle will hit the centre. (c) distance travelled by the particle when it comes to rest. 4. Charge Q is uniformly distributed over the same surface of a right circular cone of semi-vertical angle q and height h. The cone is uniformly rotated about its axis at angular velocity w. Calculate the associated magnetic dipole moment. 5. A long horizontal wire AB which is free to move in a vertical plane and carries a steady current of 20 A, is in equilibrium at a height of 0.01 m over another parallel long wire CD, which is fixed in horizontal plane and carries a steady current of 30 A as shown in figure. B

A

0.01 m D

y x

C

Show that when AB is slightly depressed and released, it executes S.H.M. Find the time period of oscillations. 6. A wire loop carrying a current is placed in the x-y plane as shown in figure. If a particle with charge q and mass m is placed at the centre P and given a velocity v along NP. Find its instantaneous acceleration. If an external uniform magnetic ^ induction B = B i is applied, find the force and torque acting on the loop.

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

10

physics for you | AUGUST ‘15

y M 

v

I

120°

P

60°

x

N

soLuTioNs

1. (a) experienced by the electron is  The Lorentz   force  F = − e[E + (v × B)] Since the net force is zero as the electron moves undeflected, E ...(i) E = vB sin q or B = v sin q The magnetic field is to be smallest E \ q = 90°, then B = v

(b) The length of arc traced by the particle, l = R(180° – 2q) = R(p – 2q) Time spent in the field, l R(p − 2q) t= = v v mv (p − 2q) m(p − 2q)  mv  = = Using R =  Bq v Bq Bq   2pm T As time period T = ,t= (p − 2q) Bq 2p (c) Intercept AB = 2Rcosq (from figure) 3. (a) Normal by the surface on the particle, N = mg + qE0 Centrifugal force on the particle, qvB0 =

mv 2 R

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

dv = mN dt From equation (ii), mv R= B0q

...(iii)

−m

...(iv)

From equations (i) and (iii), we get dv −m = m(mg + qE0 ) dt K.E. of electron = eV = 2.5 KeV, 1 or mv 2 = eV 2 2 × 2.5 × 103 × 1.6 × 10−19 2eV \ v= = m 9.1 × 10−31 = 2.96 × 107 m s −1 3 E 10 × 10 \ B= = = 3.37 × 10−4 T v 2.96 × 107 (b) Proton also moves undeflected through the given E and B as eE = evB. 2. (a) The particle circulates under the influence of magnetic field. As the field is uniform the charge comes out symmetrically. The angle subtended at the centre is (180° – 2q).

⇒

0

dt

v0

0

−m ∫ dv = m(mg + qE0 ) ∫ dt

mv0 ⇒ t = m(mg + qE0 ) (b) From equation (iv), m(mg + qE0 )dt m dv dR = =− B0q qB0 0 −m(mg + qE 0 ) t dR = ∫ ∫ dt qB0 Ri

⇒

Ri =

qB0 mv0 \ t= m(mg + qE0 )

(c) −m  

B Path of particle

O (180° – 2) FB

  v

12

physics for you | AUGUST ‘15

0

×t  mv0  Using Ri =  B0q  

dv dv = −mv = m(mg + qE0 ) dt dl

0

l

v0

0

or − m ∫ v dv = m(mg + qE0 )∫ dl \

A

m(mg + qE 0 )

mv02 l= 2m0 (mg + qE0 )

4. Charge q rotating along a ring of radius r with angular velocity w is equivalent to a dipole of magnetic moment

Compare (iii) with equation of S.H.M. a = – w2y, we get d g 2p \ T = 2p w= = g d T

w A 2p Dipole moment of differential ring, Q dx w dm = 2pr ⋅ ⋅ pr 2 cos q 2p pRl m = IA = qυA = q

r

x

R



Here, d = 0.01 m, g = 10 m s–2 0.01 = 2 p 10−3 = 0.199 s ≈ 0.2 s \ T = 2p 10

l

6. The magnetic field at P due to arc  m I 2p / 3 ^ B1 = 0 × (k) 2a 2p The magnetic field at P due to straight segment MN  m I I ^ B2 = 0 × (sin 60° + sin 60°)(− k) 4p a cos 60° The resultant magnetic field at P due to the given loop,    m I  p ^ BR = B1 + B2 = 0  3 −  (− k)  2pa 3

h O

From similar triangles, R h we have, = r h−x 3 h − x  Q cos q dx wp  R \ dm =   h pRh cos q Integrating both sides, we get 4 0

y

QwR h Qw 2 QwR  (h − x) h tan2 q × =   = 4  4  4 4 4 h h h dF m0 I1I2 = 5. Since 2pd dl m=

2

2

4

M 

v

I

120°

FB

d

mg

D

I2

I1

A

x y

(d – y) C

For antiparallel currents, force is repulsive. In equilibrium position, mg ^ m0 I1I2 ^ (− j) + ( j) = 0 2pd L m0 I1I2 mg = or ...(i) 2pd L If upper wire is displaced slightly from its equilibrium position, the resultant force per unit length on it  m0 I1I2 ^ mg ^ ...(ii) j− j FR = L 2p (d − y)  m II  1 y 1 ^ m I I ^ j FR = 0 1 2  − j= 0 1 2 2p  d − y d  2p (d − y)d [Using eqn (i)] For small displacement y, y < < d,  m II ^ FR = 0 1 22 y j 2p d m II L From (i), m = 0 1 2 2p dg F g Acceleration of wire, a = R = − y ...(iii) d (m/L) ( y and a are in opposite directions) 14

physics for you | AUGUST ‘15

60°

x

acos60°

y

B

P

N

^

^

Velocity of charge at P, v = v cos 60° i + v sin 60° j Force experienced by a moving charge in a magnetic field is given by   F = qv × B  v ^ 3 ^ m0 I  p ^ F =q i + v j ×  3 −  (− k)  2 2 2pa 3  m0qvI  p ^ ^ ^ F=  3 −  (i + 3 j)(− k) 4pa  3 p   m0qvI  3 − 3  ^ ( j − 3 ^i ) F= 4pa So, acceleration of the charge particle  m qvI  3 − p  0  F 3  ( ^j − 3 ^i ) = 4pam m Net magnetic force on a current carrying coil of any shape in an external magnetic field is zero. Torque on the loop     τ = m × B = I ( A × B) 1 Here, A = (area of circle) − area of ∆MNP 3 1 1 a = pa2 − × × 3a = 0.614a2 3 2 2  B = Bi  \ τ = I (0.614a2 k × Bi) = 0.614a2 IB j nn

J & K CET

Jammu & Kashmir Common Entrance Test

1.

Which of the following is not an electromagnetic wave? (a) Sound wave (b) Thermal radiation (c) Microwave (d) Gamma ray

2.

The 220 V a.c. line voltage that we receive in our homes is (a) rms value (b) peak value (c) average value (d) none of the above

3.

A solid sphere is rolling down an inclined plane. Then the ratio of its translational kinetic energy to its rotational kinetic energy is (a) 2.5 (b) 1.5 (c) 1 (d) 0.4

4.

The dimension of magnetic flux is (a) MLT–1A–1 (b) ML–1TA–2 –2 2 –2 (c) ML T A (d) ML2T–2A–1

5.

A bullet fired from a rifle loses 20% of its speed while passing through a wooden plank. Then minimum number of wooden planks required to completely stop the bullet is (a) 3 (b) 5 (c) 15 (d) 25

6.

The energy per mole per degree of freedom of an ideal gas is 3 1 3 1 (a) kBT (b) kBT (c) RT (d) RT 2 2 2 2 Two copper spheres having same radii, one solid and other hollow, are charged to the same potential. Which of the following statements is correct? (a) Hollow sphere will hold more charge. (b) Solid sphere will hold more charge. (c) Solid sphere will have uniform volume charge density. (d) Both spheres will hold same charge.

7.

8.

16

The path of a charge particle after it enters a region of a uniform electrostatic field with velocity perpendicular to the field will be (a) straight line (b) circular (c) helical (d) parabolic physics for you | august ‘15

9.

In a cyclic process, the change in the internal energy of a system over one complete cycle (a) depends on the path (b) is always negative (c) is always zero (d) is always positive

10. Dimensions of Planck's constant are (a) ML2T–1 (b) ML2T–3 –1 (c) MLT (d) ML3T–3 11. Which of the following is not an example of primary cell? (a) Voltaic cell (b) Lead-acid cell (c) Daniel cell (d) Leclanche cell 12. In a transformer the number of primary turns is four times that of the secondary turns. Its primary is connected to an a.c. source of voltage V. Then (a) current through its secondar y is about four times that of the current through its primary. (b) voltage across its secondary is about four times that of the voltage across its primary. (c) voltage across its secondary is about two times that of the voltage across its primary. 1 (d) voltage across its secondary is about 2 2 times that of the voltage across its primary. 13. A magnet makes a single pass through a coil. Then across the ends of the coil it produces (a) d.c. voltage (b) sinusoidal voltage (c) single voltage pulse (d) two voltage pulses 14. An observer standing near the sea-coast counts 48 waves per min. If the wavelength of the wave is 10 m, the velocity of the waves will be (a) 8 m/s (b) 12 m/s (c) 16 m/s (d) 20 m/s

15. In an n-p-n transistor, p is (a) intrinsic semiconductor (b) emitter (c) collector (d) base 16. The carbon resistor has the colour band sequence of green, orange, blue and silver. The value of resistance will be (a) 64 × 107 ± 20% W (b) 53 × 106 ± 20% W (c) 64 × 107 ± 10% W (d) 53 × 106 ± 10% W 17. A series LCR circuit is connected to an a.c. source and is showing resonance. Then (a) VR = 0 (b) VL = VR (c) VC = VR (d) VL = VC 18. The wave nature of electrons is demonstrated by the (a) photoelectric effect (b) Rutherford's experiment (c) Doppler's effect (d) Davisson and Germer experiment 19. A ball is projected up at an angle q with horizontal from the top of a tower with speed v. It hits ground at point A after time tA with speed vA. Now this ball is projected at same angle and speed from the base of the tower (located at point P) and it hits ground at point B after time tB with speed vB. Then (a) PA = PB (b) tA < tB (c) vA > vB (d) ball A hits the ground at an angle (–q) with horizontal 20. Consider the two cells having emf E1 and E2 (E1 > E2) connected as shown in the figure. A potentiometer is used to measure potential difference between P and Q, and the balancing length of the potentiometer wire is 0.8 m. Same potentiometer is then used to measure potential difference between P and R, and the balancing length is 0.2 m. Then the ratio E1/E2 is Q P

(a)

4 3

E1

(b)

5 4

E2

(c)

5 3

R

(d)

4 1

21. A particle is undergoing uniform circular motion with angular momentum L. While moving on the same path if its kinetic energy becomes four times,

then its angular momentum will be L L (a) (b) (c) L (d) 2L 2 4 22. Radius of Earth is 6400 km and that of Mars is 3200 km. Mass of Mars is 0.1 that of Earth's mass. Then the acceleration due to gravity on Mars is nearly (a) 1 m/s2 (b) 2.5 m/s2 2 (c) 4 m/s (d) 5 m/s2 23. A ball is dropped from the top of 80 m high tower. If after 2 sec of fall the gravity (g = 10 m/s2) disappears, then time taken to reach ground since the gravity disappeared is (a) 2 sec (b) 3 sec (c) 4 sec (d) 5 sec 24. Assuming density d of a planet to be uniform, we can say that the time period of its artificial satellite is proportional to 1 1 (a) d (b) d (c) (d) d d 25. A charge particle having charge 1 × 10–19 C revolves in an orbit of radius 1 Å such that the frequency of revolution is 1016 Hz. The resulting magnetic moment in SI units will be (a) 1.57 × 10–21 (b) 3.14 × 10–21 –23 (c) 1.57 × 10 (d) 3.14 × 10–23 26. Consider a bi-convex lens and a plano-convex lens with radii of curvature of all the curved surfaces being same. If f is focal length of bi-convex lens then the focal length of the plano-convex lens is (a) 4f (b) 2f (c) f (d) 0.5f 27. Consider a ray of light travelling from a denser to a rarer medium. If it is incident at the critical angle then (a) it will emerge out into the rarer medium (b) it will undergo total internal reflection (c) it will travel along the interface separating the two media (d) it will retrace its path 28. A concave mirror has focal length f. A convergent beam of light is made incident on it. Then the image distance v is (a) zero (b) less than f (c) equal to f (d) more than f 29. A 1 m long solenoid containing 1000 turns produces a flux density of 3.14 × 10–3 T. The current in the solenoid will be (a) 2.0 A (b) 2.5 A (c) 3.0 A (d) 3.5 A physics for you | august ‘15

17

30. Values for Brewster's angle can be (a) only less than 45° (b) only greater than 45° (c) any value in the range 0° to 90° except 45° (d) any value in the range 0° to 90° including 45°

38. A person carrying a whistle emitting continuously a note of 272 Hz is running towards a reflecting surface with a speed of 18 km/h. If the speed of sound is 345 m/s, the number of beats heard by him are (a) 4 (b) 6 (c) 8 (d) 10

31. The dielectric constant of a perfect conductor is (a) +1 (b) 0 (c) infinite (d) –1

39. Bulk modulus is defined by (a) increase in length per unit length per unit applied stress (b) increase in volume per unit volume per unit applied stress (c) lateral displacement per unit length per unit applied stress (d) change in cross-sectional area per unit area per unit applied stress

32. Newton's law of cooling applies when a body is losing heat to its surroundings by (a) conduction (b) convection (c) radiation (d) conduction as well as radiation 33. A block of mass m is placed on an inclined plane having coefficient of friction m. The plane is making an angle q with horizontal. The minimum value of upward force acting along the incline that can just move the block up is (a) mgcosq (b) mmgcosq (c) mgsinq (d) mmgsinq 34. Which of the following is incorrect about sky waves? (a) Sky waves are not used in long distance communication. (b) Their propagation takes place by total internal reflection. (c) Sky waves support the so-called AM band. (d) The frequency of sky waves ranges typically from 3 MHz to 30 MHz. 35. Consider boiling water converting into steam. Under this condition, the specific heat of water is (a) less than zero (b) zero (c) slightly greater than zero (d) infinite 36. Consider an electric dipole placed in a region of non-uniform electric field. Choose the correct statement out of the following options. (a) The dipole will experience only a force. (b) The dipole will experience only a torque. (c) The dipole will experience both force and the torque. (d) The dipole will neither experience a force nor a torque. 37. Conductivity of semiconductors (a) is maximum at 0 K (b) decreases with increase in temperature (c) increases with increase in temperature (d) is maximum at 300 K 18

physics for you | august ‘15

40. Unpolarized light is travelling from a medium of refractive index 2 to a medium of index 3. The angle of incidence is 60°. Then (a) reflected light will be partially polarized. (b) reflected light will be plane polarized in a plane perpendicular to plane of incidence. (c) refracted light will be plane polarized in a plane perpendicular to plane of incidence. (d) refracted light will be plane polarized in a plane parallel to plane of incidence. 41. Which of the following is correct statement about the magnitude of the acceleration a of the particle executing simple harmonic motion? (a) a will be maximum at the equilibrium position. (b) a will be maximum at the extreme position. (c) a will be always constant. (d) a will always be zero. 42. The ratio of mass defect of the nucleus to its mass number is maximum for (a) U238 (b) N14 (c) Si28 (d) Fe56 43. Consider a region of uniform magnetic field directed along positive x-axis. Now a positive rest charge Q, located at origin O (0, 0) inside the field is released from rest position. The particle will (a) remain stationary at origin O (b) move along positive x-axis (c) move along negative x-axis (d) undergo a circular motion in the x-y plane. 44. If R is Rydberg's constant, the series limit of the wavelength of Balmer series for hydrogen atom is given by 1 4 16 9 (a) (b) (c) (d) R R R R

45. A body is travelling east with a speed of 9 m/s and with an acceleration of 2 m/s2 acting west on it. The displacement of the body during the 5th second of its motion is (a) 0.25 m (b) 0.5 m (c) 0.75 m (d) zero 46. In the fringe pattern of a Young's double slit experiment the ratio of intensities of maxima and minima is 25 : 9. Then the ratio of the amplitudes of interfering waves is (a) 4 : 1 (b) 5 : 3 (c) 4 : 3 (d) 25 : 9 47. The combination of gates as shown in the figure forms the (a) AND gate (b) OR gate (c) NOR gate (d) NOT gate 48. Metal alloys are used for making standard resistance coils because (a) they have high thermal conductivity. (b) their resistance depend weakly on temperature. (c) they have low thermal conductivity. (d) their resistance depend strongly on temperature. 49. If the forward bias voltage in a p-n diode is decreased, the length of depletion region will (a) increase (b) decrease (c) not change (d) initially increase and then decrease 50. A block of mass 3 kg starts from rest and slides down a curved path in the shape of a quarter-circle of radius 2 m and reaches the bottom of path with a 4 m/s speed . If g is 10 m/s2, the amount of work done against friction is (a) 60 J (b) 36 J (c) 24 J (d) 12 J 51. The length of antenna to transmit waves of 1 MHz will be (a) 3 m (b) 15 m (c) 30 m (d) 300 m 52. In which of the following both transverse and longitudinal waves propagate? (a) Heat transfer (b) Elastic wave motion in a solid (c) Microwave communication (d) X-ray motion 53. A person is standing on a weighing-scale and observes that the reading is 60 kg. He then suddenly jumps up and observes that reading goes to 70 kg. Then his maximum upward acceleration is (a) zero (b) 1.4 m/s2 2 (c) 1.63 m/s (d) 9.8 m/s2

54. What amount of original radioactive material is left after 3 half-lives? (a) 6.5% (b) 12.5% (c) 25.5% (d) 33.3% 55. An ideal gas is heated at constant volume until its pressure doubles. Which one of the following statements is correct? (a) The mean speed of the molecules doubles. (b) Root mean square speed of the molecules doubles. (c) Mean square speed of the molecules doubles. (d) Mean square speed of the molecules remains unchanged. 56. A parallel narrow-beam of light is falling normally on a glass sphere. It will come to a focus (a) inside the sphere (except at its center) (b) on the surface of the sphere (c) outside the sphere (d) exactly at the center of the sphere 57. Smallest division on the main scale of given vernier callipers is 0.5 mm. Vernier scale has 25 divisions and these coincide with 24 main scale divisions. The least count of vernier callipers is (a) 0.001 cm (b) 0.002 cm (c) 0.01 cm (d) 0.02 cm 58. The electric field of an electric dipole at a point on its axis at a distance d from the centre of the dipole varies as 1 1 1 1 (a) (b) (c) (d) 3/2 d d d2 d3 59. Red, blue, green and violet colour lights are one by one made incident on a photocathode. It is observed that only one colour light produces photoelectrons. That light is (a) red (b) blue (c) green (d) violet 60. Two pendula oscillate with a constant phase difference of 45° and same amplitude. If the maximum velocity of one of them is v and that of other is v + x, then the value of x will be v v (a) 0 (b) (c) (d) ( 2 )v 2 2 solutions 1. (a) : Among the given waves, sound wave is not an electromagnetic wave whereas all others are electromagnetic waves. 2. (a) : The 220 V a.c. line voltage that we receive in our homes is an rms value. 3. (a) physics for you | august ‘15

19

4. (d) : As magnetic flux = magnetic field × area \ [magnetic flux] = [MT–2A–1][L2] = [ML2T–2A–1] 5. (a) 6. (d) : According to the law of equipartition of energy, the energy per mole per degree of freedom of an ideal gas  1 1 R  = kB N AT = RT as kB =  2 2 N  

8  \ v =  s −1  (10 m) = 8 m/s  10  15. (d) : In an n-p-n transistor, p is base. 16. (d) : The number for green is 5, for orange is 3 and that for blue is 6. For silver, tolerance is 10%.

A

7. (d) : The capacitance of a spherical conductor is C = 4pe0R, where R is its radius. As both spheres have same radii, so their capacitances will be same. Further, when they are charged to the same potential, both spheres will hold same charge (as Q = CV). 8. (d) : After entering the region of a uniform electrostatic field with velocity perpendicular to the field, the path of the charged particle will be parabolic. 9. (c) : The change in the internal energy of a system is path independent and depends only on the initial and final states of the system. i.e. DU = Uf – Ui In a cyclic process, the system returns to its initial state, so Uf = Ui \ DU = Uf – Ui = 0 10. (a) : As Planck's constant = \ [h] =

[ML2 T−2 ] −1

[T ]

energy frequency

= [ML2 T−1 ]

11. (b) : Lead-acid cell is an example of secondary cell whereas all other given cells are the examples of primary cell. 12. (a) 13. (d) 14. (a) : Standing near the sea-coast the observer counts 48 waves per min, so the frequency of the wave is 48 48 8 u= = = s −1 1 min 60 s 10 The velocity v, wavelength l and frequency u of the wave are related as v = ul 8 −1 Here, u = s and l = 10 m 10 20

physics for you | august ‘15

Green Orange

Blue

Silver

\ The resistance of the resistor shown in the figure is R = 53 × 106 ± 10% W 17. (d) : At resonance, voltage across inductor = voltage across capacitor i.e. VL = VC 18. (d) : The wave nature of electrons is demonstrated by the Davisson and Germer experiment and was first experimentally verified by C.J. Davisson and L.H. Germer in 1927. 19. (c) 20. (a) :

Q P

E1

E2

R

As emf ∝ balancing length of the potentiometer wire \ When the potentiometer is connected between P and Q, E1 ∝ 0.8 ...(i) and when it is connected between P and R, E1 – E2 ∝ 0.2 ...(ii) (As E1 and E2 are in opposition and E1 > E2) Dividing eqn. (ii) by eqn. (i), we get E1 − E2 0.2 1 E or 1 − 2 = 1 = = E1 E1 4 0.8 4 E2 1 3 = 1− = E1 4 4

or

E1 4 = E2 3

21. (d) : The kinetic energy K and angular momentum L of a particle undergoing uniform circular motion are related as K=

L2

or L = r 2mK ... (i) 2mr 2 where m is its mass and r is the radius of circle. While moving on the same path, (i.e. r remains the same) if its kinetic energy becomes four times, then its new angular momentum will be

L ′ = r 2m(4 K ) = 2(r 2mK ) = 2L

(using (i))

22. (c) 23. (b) 24. (c) : The time period of an artificial satellite revolving very close to a planet's surface is R3 ... (i) GM where M is the mass of the planet and R its radius. Assuming the planet to be of uniform density d, so its mass is 4 mass   M = pR3d as density =   3 volume  T = 2p

\ T = 2p

1 3p R3 or T ∝ = d Gd 4  G  pR3d  3 

25. (d) : The current due to orbital motion of the charged particle having charge q is I = qu where u is the frequency of revolution. If r is the radius of the orbit, then the resulting magnetic moment is M = IA = qu(pr2) = pqur2 Here, q = 1 × 10–19 C, u = 1016 Hz, r = 1 Å = 10–10 m \ M = (3.14)(1 × 10–19 C)(1016 Hz)(10–10 m)2 = 3.14 × 10–23 Am2 26. (b) 27. (c) : When a ray of light Rarer travelling from a denser medium medium to a rarer medium is incident at ic Denser the critical angle ic, then medium it will travel along the interface separating the two media as shown in the figure. 28. (b) 29. (b) : The flux density produced by a long solenoid m NI Bl or I = is B = 0 l m0 N where N is the number of turns in the solenoid and l its length and I is the current in the solenoid. Here, B = 3.14 × 10–3 T, l = 1 m m0 = 4p × 10–7 Tm/A, N = 1000

\ I=

(3.14 × 10−3 T)(1 m) (4 p × 10

−7

10 A = 2.5 A Tm/A)(1000) 4 =

30. (b) : According to Brewster's law, the Brewster's angle iB and refractive index m of a material are related as taniB = m or iB = tan–1(m) As m > 1 \ iB > 45° Thus the values for Brewster's angle iB can be only greater than 45°. 31. (c) : The dielectric constant of a perfect conductor is infinite. 32. (b, c) : Newton's law of cooling applies when a body is losing heat to its surroundings by convection and radiation. 33. (*) : The various forces acting on the block are shown in the figure. N

n

si mg

f

F

 mg

s

co mg



When the block moves up the inclined plane, the force of friction f acts down the plane. So the minimum force required to just move the block up the inclined plane is F = mgsinq + f = mgsinq + mN = mgsinq + mmgcosq (Q N = mgcosq) = mg(sinq + mcosq) * None of the given options is correct. 34. (a) : S k y w av e s a re u s e d i n l on g d i s t an c e communication. All other statements are correct about sky waves. Q 35. (d) : As specific heat, s = mDT Since boiling of water is an isothermal process, so DT = 0 Q \ s= = ∞ m×0 36. (c) : When an electric dipole is placed in a region of non-uniform electric field, it will experience both force and the torque. 37. (c) : Conductivity of semiconductors increases with increase in temperature and vice versa. Semiconductors behave as insulators at 0 K. physics for you | august ‘15

21

38. (c) 39. (*) : Bulk modulus B is defined as the ratio of normal stress to the volumetric strain and is given by DP B=− DV / V –ve sign shows that with increase in pressure the volume decreases. * None of the given options is correct. 40. (a) : As unpolarized light is travelling from the medium 1 of refractive index m1 (= 2) to the medium 2 of refractive index m2 (= 3), so the refractive index of the medium 2 with respect to the medium 1 is m 3 1 m2 = 2 = m1 2 According to Brewster's law, the polarizing angle ip for given two media is 3 tani p = 1m2 = 2 3 i p = tan −1   = 56.3° 2 But angle of incidence i = 60° (given) i.e. i > ip For i > ip, both reflected and refracted light will be partially polarized. 41. (b) : The acceleration of the particle executing simple harmonic motion at the displacement x from the equilibrium position is a = –w2 x where w is the angular frequency. Its magnitude will be minimum at the equilibrium position (x = 0) and maximum at the extreme position (x = A(amplitude)). 42. (d) : The ratio of mass defect of the nucleus to its mass number is maximum for Fe56. 43. (a) 44. (b) : According to Rydberg formula, the wavelength l of a spectral line in Balmer series is 1 1 1  = R  −  ; n = 3, 4, 5,... 2  2 n2  l For series limit, n = ∞ 1 1 1  R \ = R − =  22 ∞2  4 l

or l =

4 R

45. (d) : Taking west to east as positive, then u = 9 m/s and a = –2 m/s2 As S

n

22

th

a = u + (2n − 1) 2

physics for you | august ‘15

W

+ve

E

\ S

5th

46. (a) 47. (a) :

2 = 9 − (2 × 5 − 1) = 9 − 9 = 0 2

A

B

A

B

Y

The Boolean expression of the given combination of gates is Y = A + B = A ⋅ B (using de Morgan's theorem) = A⋅B which is Boolean expression for AND gate. Thus the given combination of gates forms AND gate. 48. (b) : Metals are used for standard resistance coils because their resistance depend weakly on temperature. 49. (a) : In a p-n diode when the forward bias voltage is decreased the length of depletion region will increase and vice versa. m

50. (b) : Here, mass of the block, m = 3 kg Initial speed of the block, h u = 0 (as it starts from rest) Final speed of the block, v v = 4 m/s Height, h (in this case the radius of quarter circle) =2m The change in kinetic energy of the block is 1 1 1 DK = mv 2 − mu2 = mv 2 − 0 2 2 2 1 2 = (3 kg)(4 m/s) = 24 J 2 The work done by the gravitational force is Wg = mgh = (3 kg)(10 m/s2)(2 m) = 60 J If W f is the work done by the friction, then according to work energy theorem, Wg + Wf = DK or Wf = DK – Wg = 24 J – 60 J = –36 J As work done against friction is equal and opposite to work done by the friction, \ The amount of work done against friction is 36 J. ...contd. on page 83

Series 3 CHAPTERWISE PRACTICE PAPER : Electromagnetic Induction and Alternating Currents | Electromagnetic Waves Time Allowed : 3 hours

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.

section-A 1. For circuits used for transporting electric power,

2.

3.

4. 5.

a low power factor implies large power loss in transmission. Explain. On the basis of electromagnetic theory, show how does the refractive index of a material medium depend on its relative permeability and dielectric constant. Consider a metallic pipe with an inner radius of 1 cm. If a cylindrical bar magnet of radius 0.8 cm is dropped through the pipe, it takes more time to come down than it takes for a similar unmagnetised cylindrical iron bar dropped through the metallic pipe. Explain. Mention the two characteristic properties of the material suitable for making core of a transformer. Can the voltage drop across the inductor or the capacitor in a series LCR circuit be greater than the applied voltage of the AC source? Justify your answer. section-b

6. Two identical loops, one of copper and the other

of aluminium, are rotated with the same angular speed in the same magnetic field. Compare:

7.

8.

9.

10.

(a) the induced emf, and (b) the current produced in the two coils. Justify your answer. Electric field at a given point in space is oscillating with an angular frequency of 9p × 1012 rad s–1 and an amplitude of 7.5 V m–1 along Y-axis. Write the equation of electric field and magnetic field. Given that the electromagnetic wave is propagating along x-direction. OR How would you establish an instantaneous displacement current of 1.0 A in the space between the plates of 1 µF capacitor? Although there is no direct electric connection between the two coils of a transformer, yet energy is being transferred from primary coil to secondary coil. How? An 80 V, 800 W heater is to be operated on a 100 V, 50 Hz supply. Calculate the inductance of the choke required. Ordinary moving coil galvanometer used for direct current cannot be used to measure an alternating current even if its frequency is low. Explain, why? physics for you | august ‘15

23

section-c 11. (a) Obtain the expression for the magnetic energy

12.

13. 14.

15.

16.

17.

18.

24

stored in a solenoid in terms of magnetic field B, area A and length l of the solenoid. (b) How does this magnetic energy compare with the electrostatic energy in a capacitor? Name the constituent radiation of electromagnetic spectrum which: (a) is used in satellite communication, (b) is used for studying crystal structure, (c) is similar to the radiations emitted during decay of radioactive nuclei, (d) has its wavelength range between 390 nm and 700 nm, (e) is absorbed from sunlight by ozone layer, and (f) produces intense heating effect. Describe the use of a series resonant circuit in the tuning of a radio receiver. A circuit contains two inductors in series with selfinductances L1 and L2 and mutual inductance M. Obtain a formula for the equivalent inductance in the circuit. Show that in the free oscillations of an LC circuit, the sum of energies stored in capacitor and the inductor is constant in time. Name the SI unit of magnetic flux and show that it equals volt-second. Give three possible ways of producing an induced emf in a coil giving an example in each case. OR How does the mutual inductance of a pair of coils change when: (i) the distance between the coils is increased? (ii) the number of turns in each coil is decreased? (iii) a thin iron sheet is placed between the two coils, other factors remaining the same. Justify your answer in each case. (i) The primary of a transformer has 400 turns while the secondary has 2000 turns . If the power output from the secondary at 1100 V is 12.1 kW, calculate the primary voltage. (ii) If the resistance of the primary is 0.2 W and that of the secondary is 2.0 W and the efficiency of the transformer is 90%, calculate the heat losses in the primary and the secondary coils. Light with an energy flux of 18 W cm–2 falls on a non-reflecting surface at normal incidence. If the surface has an area of 20 cm2, find the average force exerted on the surface during a 30 minute time span. physics for you | august ‘15

19. Three students X, Y and Z performed an experiment

for studying the variation of alternating currents with angular frequency in a series LCR circuit and obtained the graphs shown in the figure. They all used AC sources of the same rms value and inductances of the same value. I

X Y

Z 



What can we (qualitatively) conclude about the: (a) capacitance values, and (b) resistance values used by them? (c) In which case, will the quality factor be maximum? (d) What can be concluded about the nature of the impedance of the set-up at frequency w0? 20. Derive an expression for the mutual inductance of two long coaxial solenoids. 21. A coil of inductance 0.50 H and resistance 100 W is connected to a 240 V, 50 Hz AC supply. (a) What is the maximum current in the coil? (b) What is the time lag between the voltage maximum and the current maximum? 22. Prove that an ideal inductor connected to an AC source does not dissipate any power. section-D 23. One day Kapil requested his grandfather to show

him the working of his big size radio set. He was amazed to observe that radio set was catching the radio stations on tuning, by rotating the knob. His grandfather, who was a retired scientist, explained the working of tuning circuits and also explained how the similar circuits work in tuning the TV for different channels and FM radio installed in the car. Kapil was very happy to learn that all these electronic equipments were working on the fundamental concepts which he learned in his physics class recently. (a) What are the major components of tuning circuit in a radio? (b) Name the phenomena involved in tuning. (c) What values can be associated with this?

section-e 24. (a) What are eddy currents? Write their two

applications. (b) Figure shows a rectangular conducting loop PQSR in which arm RS of length ‘l’ is movable. The loop is kept in a uniform magnetic field ‘B’ directed downward perpendicular to the plane of the loop. The arm RS is moved with a uniform speed ‘v’.

l

×P × × × ×Q

× × × × ×

×R × × × ×S

× × × × ×

× × ×v × ×

Deduce an expression for (i) the emf induced across the arm ‘RS’, (ii) the external force required to move the arm, and (iii) the power dissipated as heat. OR State Faraday’s law of electromagnetic induction and explain three methods for producing induced EMF. 25. Explain with the help of a labelled diagram, the principle, construction and working of an AC generator. OR Draw a schematic diagram of a step-up transformer. Explain its working principle. Deduce the expression for the secondary to primary voltage in terms of the number of turns in the two coils. In an ideal transformer, how is this ratio related to the currents in the two coils? How is the transformer used in large scale transmission and distribution of electrical energy over long distances? 26. Derive an expression for the impedance of a series LCR circuit connected to an AC supply of variable frequency. Plot a graph showing variation of current with the frequency of the applied voltage. OR Derive an expression of power in an LCR AC circuit. What is wattless current?

solutions 1. We know that power in an AC circuit is given by the

relation, P = VrmsIrms cos f, where cos f is the power factor. To supply a given power to an AC circuit at a given voltage if power factor cos f is small, then we have to increase the value of current accordingly. However, increase in current means increased power loss (power dissipation due to Joule’s heating) being given by I2R. Thus, transmission line loss is more. 2. As per electromagnetic theory, speed of light in a given material medium, ...(i) c = v mr ⋅ K and we know that refractive index n of a medium is given by c ...(ii) n= v On putting the value of c in equation (ii), we get v mr ⋅ K = mr ⋅ K v Thus, refractive index of a medium is proportional to square root of its relative permeability and dielectric constant. 3. When the bar magnet is dropped into the metallic pipe, eddy currents are produced in the pipe. As these currents oppose the motion of the magnet, its acceleration is less than that due to gravity (g). Since no eddy currents are produced in the metallic pipe when the unmagnetised bar is dropped into the pipe, its acceleration is g. Obviously, the bar magnet takes more time to come down through the pipe than the time taken by the unmagnetised bar. n=

4. The material used for making the core of a

transformer should have high value of magnetic permeability but a low value of magnetic hysteresis so that magnetic flux linked with each turn of primary or secondary coil is large and energy loss due to hysteresis phenomenon is the least possible. 5. Yes, the voltage drop across L or C in a series LCR circuit can be greater than the applied voltage of the AC source. This is due to the fact that these voltages are not in phase and cannot be arithmetically added up. In fact, VL and VC are in mutually opposite phase and try to balance each other. 6. (a) When two identical loops, one of copper and the other of aluminium, are rotated with same angular speed w in the same magnetic field B, the induced emf in both coils will be same physics for you | august ‘15

25

because rate of change of magnetic flux of both loops is exactly the same. e (b) Induced current, I = R where e is the induced emf and R is the resistance offered by the loop. As copper is a better conductor than aluminium, resistance offered by copper loop is less and consequently, more induced current is produced in the copper loop. 7. As per question, w = 9p × 1012 rad s–1 and E0 = 7.5 V m–1 12 w 9p × 10 \ k= = = 3p × 10 4 m −1 c 3 × 108 E 7. 5 and B0 = 0 = = 2.5 × 10 −8 T c 3 × 108 Hence, equation of electric field, Ey = E0 sin (wt – kx) = 7.5 sin (9p × 1012t – 3p × 104 x) V m–1 Similarly, Bz = B0 sin (wt – kx) = 2.5 × 10–8 sin (9p × 1012 t – 3p × 104 x) T OR As per question, ID = 1.0 A and C = 1 µF = 10–6 F We know that, df d I D = e0 ⋅ = e 0 (EA) dt dt = e0

d  V  e 0 A dV dV ⋅ =C⋅ A = dt  d  d dt dt

where A is the area of capacitor plate and d is the distance between the plates. dV I D 1. 0 = = = 106 V s −1 ⇒ dt C 1 × 10 −6 Thus, to establish a displacement current of 1.0 A, the potential difference between the plates of capacitor must change at a rate of 106 V s–1. 8. Although there is no electrical connection between the primary and secondary coils of a transformer, still energy is transferred from the primary circuit to secondary circuit. Owing to an AC voltage applied across the primary coil an alternating current flows in primary coil and an alternating magnetic flux per unit turn is created. Through the iron core, this alternating magnetic flux is also linked with each turn of secondary coil, due to which, an induced emf is set up in the secondary coil. In this manner, electrical energy is continuously being transferred from primary to secondary coil circuit. 26

physics for you | august ‘15

9. As P = VI

P 800 V 80 = = 10 A and R = = = 8 W V 80 I 10 As the choke is connected in series with the heater, the current should remain same for the impedance adjusted. Veff Veff \ I eff = = 2 2 2 2 R +w L R + 4p 2u2L2 \ I=

100

or

10 =

or

64 + 10000 p2L2 = 100

or

L2 =

2

8 + 4p 2 × 502 × L2 36 6 or L = = 0.019 H 100 p 10000 p 2

10. Ordinary moving coil galvanometer cannot be used

to measure AC. Ordinary moving coil galvanometer is based on magnetic effect of current which, in turn, depends on direction of current. So it cannot be used to measure AC. During one half cycle of AC, its pointer moves in one direction and during next half cycle, it will move in the opposite direction. Now the average value of AC over a complete cycle is zero. Even if we measure an alternating current of low frequency, the pointer, will appear to be stationary at the zero position due to persistence of vision. We can measure AC by using a hot-wire ammeter which is based on heating effect of current and this effect is independent of the direction of current. To measure AC, we define the mean value of AC over half a cycle or its root mean square value. 11. (a) l

I

A

I

In a current carrying solenoid the energy is stored in the form of magnetic field. If current I is flowing in the solenoid, the magnetic field energy associated is 1 UB = LI2 2 2 1  B  or UB = L  (Q B = m0nI) 2  m 0n  As coefficient of self induction L = m0 n2 Al  B2  1 1 2 \ UB = m0 n2 Al  2 2  = B Al 2  m 0n  2 m 0

Let Leq be the equivalent inductance of the two self-inductances L1 and L2 connected in series. For the series combination, the emfs induced in the two coils get added up. Thus eeq = e1 + e2 ...(i) If the rate of change of current in the series dI circuit is , then dt dI dI e1 = − L1 − M , ...(ii) dt dt

(b) Magnetic field energy density in solenoid U B 2 Al B2 uB = B = = 2m 0 Al 2m 0 V Electric field energy density in capacitor 1 uE = e0E2 2 So both energy density are proportional to square of corresponding fields. 12. (a) Microwaves are used in satellite communication. (b) X-rays are used for studying crystal structure. (c) Gamma rays are similar to the radiations emitted during decay of radioactive nuclei. (d) Visible light radiation has its wavelength range between 390 nm and 700 nm. (e) Ultraviolet radiations are mainly absorbed from sunlight by ozone layer. (f) Infrared rays produce intense heating effect when fall on a material. 13. The tuning circuit of a radio or TV is an example of LCR resonant circuit. Signals are transmitted by different stations at different frequencies. These frequencies are picked up by the antenna and corresponding to these frequencies, a number of voltages appear across the series LCR circuit. But maximum current flows through the circuit for that AC voltage which has frequency equal to 1 fr = . 2p LC

e 2 = − L2

dI dI −M dt dt

...(iii) dI and e eq = − Leq ...(iv) dt The negative sign throughout indicates that both self and mutual induced emfs are opposing the applied emf. Using eqns. (ii), (iii) and (iv) in eqn. (i), we get dI dI − Leq = − (L1 + M + L2 + M ) dt dt Leq = L1 + L2 + 2M

or

(ii) Let the series combination be such that the current flows in opposite senses in the two coils, as shown in figure.

I

L1

I

I

I

L2

I

Antenna L

C

The emfs induced in the two coils will be dI dI dI dI e1 = − L1 + M , e 2 = − L2 + M dt dt dt dt

To receiver

Series resonant circuit

If Q-value of the circuit is large, the signals of the other stations will be very weak. By changing the value of the adjustable capacitor C, the signal from the desired station can be tuned in. 14. (i) Let the series connection be such that the current flows in the same sense in the two coils as shown in figure.





Here the mutual emfs act in the direction of applied emf and hence positive. For this series combination also, the emfs induced in the two coils get added up. Here, e eq = e1 + e 2 = − [L1 − M + L2 − M] dI dt dI But e eq = − Leq dt dI dI \ −Leq = − [L1+ L2 − 2M] dt dt or Leq = L1 + L2 – 2M.

15. Just as the sum of potential and kinetic energies

remains constant in SHM, the sum of energies stored in the capacitor and inductor remains constant during LC oscillations. Suppose at time physics for you | august ‘15

27

t = 0, capacitor is fully charged with a battery of voltage V0. 1 So, U E = CV02 2 It is connected with an inductor. So initial energy in the inductor in zero. 1 Total initial energy of the LC circuit, U v = CV02 2 At any instant, the electrostatic energy stored in the capacitor is 1 q2 UE = 2C The magnetic energy stored in the inductor at any instant is 1 U B = LI 2 2 If there is no (resistive) loss of energy, then the total energy of the LC circuit at any instant will be 1 q2 1 2 + LI 2C 2 dq But q = q0 cos w 0t and I = dt = − q0w 0 sin w 0t 1 2 1 \ U= ⋅ q cos 2 w 0t + Lw 20 q02 sin 2 w 0t 2C 0 2 U =UE +UB =

=

1 2 1 1 2 2 ⋅ q cos 2 w 0t + L q sin w 0t 2C 0 2 LC 0

=

1 1 q2 ⋅ q02 [cos 2 w 0t + sin 2 w 0t] = ⋅ 0 2C 2 C

1 = CV02 = Initial energy 2 16. SI unit of magnetic flux is weber (Wb).

By Faraday’s law df |e| = dt 1 weber \ 1 volt = or1 weber = 1 volt second 1 second Magnetic flux linked with a closed coil is given by f = BA cos q. Whenever f changes, emf is induced. Hence emf may be produced by three methods: (i) By changing the magnetic field B, e.g., by changing the relative separation between a closed coil and magnet. (ii) By changing the area A of the closed coil, e.g., by moving a closed loop into and out of a region of magnetic field. (iii) By changing the relative orientation q between a closed coil and a magnetic field, e.g., by rotating

28

physics for you | august ‘15

a closed coil about an axis perpendicular to the magnetic field. OR (i) The mutual inductance of two coils decreases when the distance between them is increased. This is because the flux passing from one coil to another decreases. mNN A (ii) Mutual inductance, M = 0 1 2 l i.e., M ∝ N1N2 Clearly, when the number of turns N1 and N2 in the two coils is decreased, the mutual inductance decreases. (iii) When an iron sheet is placed between the two coils the mutual inductance increases, because M ∝ permeability (µ). 17. (i) Here, N1 = 400, N2 = 2000, e2 = 1100 V N 400 e1 = e 2 ⋅ 1 = 1100 × = 220 V . N2 2000



(ii) Resistance of primary, R1 = 0.2 W Resistance of secondary, R2 = 2.0 W Output power = e2I2 = 12.1 kW = 12100 W \ Current in the secondary, e I 12100 I2 = 2 2 = = 11 A e2 1100 Output power As Efficiency = Input power 12100 W 90 = 100 Input power or Input power 12100 × 100 = 13.44 × 103 W e1I1 = 90 Current in the primary, 3

e1I1 13.44 × 10 = = 61.1 A e1 220 Power loss in the primary = I12R1 = (61.1)2 × 0.2 = 746.64 W Power loss in the secondary = I22R2 = (11)2 × 2.0 = 242 W 18. Here, Energy flux = 18 W cm–2 = 18 J s–1 cm–2 Area = 20 cm2, Time = 30 min = 1800 s Total energy falling on the surface U = Energy flux × time × area = (18 J s–1 cm–2) × 1800 s × 20 cm2 = 6.48 × 105 J The total momentum delivered to the surface, I1 =

S1 but zero in the annular region between the two solenoids. Hence B1 = µ0n1I1 where n1 = N1/l = the number of turns per unit length of S1. Total flux linked with the outer solenoid S2 is f 2 = B1AN 2 = m 0n1I1 × AN 2

6.48 × 105 J U = = 2.16 × 10–3 kg m s–1 c 3 × 108 m s −1 The average force exerted on the surface, −3 p 2.16 × 10 F= = = 1.2 × 10 −6 N 1800 t p=

19. (a) As per question, resonant frequency w0 is same

for all three students X, Y and Z. 1 Since, w 0 = LC and LX = LY = LZ Hence, CX = CY = CZ (b) At resonance condition, V V I= = Z R As per graph IX > IY > IZ, hence we conclude that RX < RY < RZ. (c) Q-factor = wL/R, hence Q-factor is maximum for X and minimum for Z. It means that QX > QY > QZ. (d) At resonant frequency w0, impedance is purely resistive in nature, i.e., Z = R. 20. As shown in figure, consider two long co-axial solenoids S1 and S2, with S2 wound over S1. I2

S2

r1

I1 S1

r2

I1

\ Mutual inductance of coil 2 with respect to coil 1 is f mNN A M 21 = 2 = 0 1 2 I1 l Clearly M12 = M21 = M(say) mNN A \ M = 0 1 2 = m 0n1n2 Al = m 0n1n2pr12l l 21. For an LR circuit, if V = V0 cos wt, then

I=

V0

2

R + w 2L2

where tan f =

cos (wt − f),

wL . R

(a) Maximum current in the coil is V0 V0 = I0 = 2 2 2 2 R +w L R + 4p 2u2L2 Given L = 0.50 H, R = 100 W, Veff = 240 V and u = 50 Hz.



\ I0 =

2 × 240 (100)2 + 4p 2 × (50)2 × (0.50)2

V0    Veff = 2   

l

Let l = length of each solenoid r1, r2 = radii of the two solenoids A = pr21 = area of cross-section of inner solenoid S1 N1, N2 = number of turns in the two solenoids. Let a time varying current I2 pass through S2. The magnetic field set up inside S2 due to I2 is B2 = µ0n2I2 where n2 = N2/l = the number of turns per unit length of S2. Total magnetic flux linked with the inner solenoid S1 is f1 = B2AN1 = µ0n2I2×AN1 \ Mutual inductance of coil 1 with respect to coil 2 is mNN A f M12 = 1 = m 0n2 AN 1 = 0 1 2 I2 l We now consider the flux linked with the outer solenoid S2 due to the current I1 in the inner solenoid S1. The field B1 due to I1 is constant inside

A



1.414 × 240

1.414 × 240 = 1.82 A 186.2 10000 + 24674 f (b) V is maximum at t = 0, I is maximum at t = w (i.e., when wt – f = 0). If f is positive, this means current maximum lags behind voltage maximum by time lag, f ∆t = w 2puL 2p × 50 × 0.5 Now, tan f = = = 1.571 R 100 57.5 p \ f = tan −1 (1.571)  57.5° = rad 180 57.5 p f Time lag, ∆t = = s w 180 × 2p × 50 =

=

= 3.19 × 10–3 s  3.2 ms physics for you | august ‘15

29

22. When AC is applied to an ideal inductor, current

lags behind the voltage in phase by p/2 radian. So we can write the instantaneous values of voltage and current as follows: V = V0 sin wt p and I = I 0 sin  wt −  2  p  = − I 0 sin  − wt  = − I 0 cos wt 2  Work done in small time dt is dW = P dt = –V0I0 sin wt cos wt dt VI = − 0 0 sin 2 wt dt 2

Excel in Physics).

T

W 1 = dW T T∫ 0

Physics).

0

V I  cos 2wt  VI = 0 0 =+ 0 0  2T  2w  4T w 0

T

4p   cos T t   0

V0I 0 VI [cos 4p − cos 0] = 0 0 [1 − 1]= 0 4T w 4T w Thus the average power dissipated per cycle in an inductor is zero. =

PHYSICS

MUSING

Contd. from page 8

8. A direct-vision prism is made out of three prisms,

each with a refracting angle of 60°, attached to each other as shown in the figure. Light of a certain wavelength is incident on the first prism. The angle of incidence is 30° and the ray leaves the third prism parallel to the direction of   incidence. The refractive  index of the glass of the first and third prisms is 1.5. Find the refractive index of the material of the middle prism. (Take 6 = 2.45) 9. A light ray parallel to the principal axis is incident (as shown in the figure) on a thin planoconvex lens with radius of curvature of its curved part equal to 10 cm. Assuming that the refractive index of the material of the lens is 4/3 and medium on both sides of the lens is air, the distance of the point from the lens where this ray meets the principal axis is 30

physics for you | august ‘15

(b) Refer point 4.1(7(e)) page no. 248 (MTG Excel in Physics). OR Refer point 4.1(5, 7(a, b, c)) page no. 247, 248 (MTG Excel in Physics). 25. Refer point 4.8(2) page no. 275 (MTG Excel in

T

VI = − 0 0 ∫ sin 2 wt dt 2T T

converts electromagnetic waves into electrical signals and an LCR series resonant circuit with variable capacitor whose value varies by rotating the knob. (b) Phenomena involved in tuning is resonance. (c) Awareness, scientific thinking, good communication between old generation and new generation is important to take advantage of the hard earned learning by elders.

24. (a) Refer point 4.3(1, 4) page no. 252, 253 (MTG

The average power dissipated per cycle in the inductor is Pav =

23. (a) A tuning receiver circuit has antenna which

OR Refer point 4.8(1) page no. 274 (MTG Excel in Physics). 26. Refer point 4.6(6, 8) page no. 269, 270 (MTG Excel

in Physics).

OR Refer point 4.6(9) page no. 271 (MTG Excel in Physics). nn

X cm. Find the value of X. 7 6 cm

10. One mole of an ideal monoatomic gas undergoes

the cyclic process as shown in the figure . Find out efficiency (in percent) of the cycle. [Take ln 2 = 0.7]

nn

kInematICS

Kinematics is the branch of mechanics which describes the motion of points, bodies and system of bodies without consideration of the cause of motion. Rest and motion

Rest or motion is relative term. An object is said to be at rest if its position does not change with time with respect to a reference point. An object is said to be in motion if its position changes with respect to its immediate surroundings. For example : A car moving on a highway is in motion with respect to surrounding (tree) but the same car is at rest with respect to its driver. Point object

If an object is small enough so that we can treat it mentally similar to a point then we call such an object a point object or point particle. Position, path length and displacement

• • •

Position : Position of a particle is its location with respect to reference point (origin of mutually perpendicular axes). Path length : Path length (distance) is a measure of the length of the path taken during the change in position of an object. Displacement : The measure of change in position of an object with time is known as displacement. It is a vector quantity having both magnitude as well as direction. Displacement along x-axis, Dx = x – x0

KEY POINT • The magnitude of displacement may or may not be equal to the path length traversed by an object. Uniform and non-Uniform motion

If an object moving along the straight line covers equal distance in equal intervals of time, it is said to be a uniform motion along a straight line. If an object moving along the straight line covers equal distances in unequal intervals of time or unequal distance in equal intervals of time, it is said to be in non-uniform motion along a straight line. average Velocity and average Speed

The average velocity is defined as the ratio of the change in position or displacement (Dx) of the object to the time interval (Dt). The SI unit of the average velocity is m s–1 and it is a vector quantity. Mathematically, the average velocity is expressed as Displacement Dx x 2 − x1 v avg = . = = Time interval Dt t 2 − t1 Average speed is defined as the ratio of the distance traveled by a particle to the time taken. The average speed involved the total distance covered (for example, the number of meters moved). It is independent of direction and is given by Total distance savg = . Total time taken Instantaneous Velocity and Speed The velocity of an object at any given instant of time is PHYSICS FOR YOU | august ‘15

31

known as instantaneous velocity (or simply velocity) v. The SI unit of instantaneous velocity is m s–1 and it is a vector quantity. Mathematically, the instantaneous velocity is represented as Dx dx v = lim = . dt x →∞ Dt Instantaneous speed or simply speed is the magnitude of velocity. KEY POINT • Average velocity of a body can be positive or negative but average speed is always positive.

• If a moving object reverses its direction then in slow motion it would seem to stop and then begin moving in the opposite direction. There is an instant of reversal. If the velocity is positive before this instant then after it, the velocity is negative. In changing sign, it must pass through zero. At this instant, the instantaneous velocity is zero. Displacement-time graph

•

•

•

If the graph is a straightline parallel to time-axis, shown by line AB, it means that the body is at rest i.e. velocity = zero.

A

Displacement

•

O

The acceleration of an object is defined as the ratio of change of velocity of the object and time taken Change in velocity

Time taken Acceleration is a vector quantity. Acceleration is positive, if the velocity is increasing and is negative if velocity is decreasing. The negative acceleration is called retardation or deceleration. Uniform acceleration and variable acceleration : If the velocity of an object changes by equal amounts in equal intervals of time, it is said to be moving with a PHYSICS FOR YOU | august ‘15

(i) v = u + at (iii) v 2 = u2 + 2as

1 2 (ii) s = ut + at 2 a (iv) sn = u + (2n − 1) 2

KEY POINT • When a body is accelerated with accelerations   a1 in time t1 and a2 in time t2, then the average acceleration is given as   a t +a t  aavg = 1 1 2 2 t1 + t 2

• The free-fall acceleration near the Earth’s surface is a = – g = –9.8 m s–2, and the magnitude of the acceleration is g = 9.8 m s–2. Do not substitute – 9.8 m s–2 for g.

Time

acceleration

32

Formulae for uniformly accelerated motion along a straight line

D C B E

If the graph is a straight line inclined to time-axis (such as OC) shows that body is moving with a constant velocity. If the graph obtained is a curve like OD whose slope decreases with time, the velocity goes on decreasing, i.e., motion is retarded. If the graph obtained is a curve like OE whose slope increases with time, the velocity goes on increasing, i.e. motion is accelerated.

i.e., Acceleration =

uniform acceleration. If the velocity of an object changes by unequal amounts in equal intervals of time, it is said to be moving with a variable acceleration. Instantaneous acceleration : The acceleration of an object at a given instant or at a given point of motion is called its instantaneous acceleration. It is defined as the first time derivative of velocity at a given instant or it is also equal to second time derivative of the position of the object at a given instant i.e., Instantaneous acceleration, Dv dv d 2 x a = Lt = = dt dt 2 Dt →0 Dt

SELF CHECK

1. From a tower of height H, a particle is thrown vertically upwards with a speed u. The time taken by the particle, to hit the ground, is n times that taken by it to reach the highest point of its path. The relation between H, u and n is (a) gH = (n – 2)u2 (b) 2 gH = n2u2 2 2 (c) gH = (n – 2) u (d) 2 gH = nu2(n – 2) (JEE Main 2014) 2. An object moving with a speed of 6.25 m s–1, is dv decelerated at a rate given by = −2.5 v , where dt v is the instantaneous speed. The time taken by the object, to come to rest, would be (a) 1 s (b) 2 s (c) 4 s (d) 8 s (AIEEE 2011) Velocity-time graph

•

If the graph is a straight line parallel to time axis shown by line AB, it means that the body is moving with a constant velocity or zero acceleration.

(x1 – x2)

A

Velocity

D C E

O

• • • •

If the graph is a straight line and inclined to the time-axis with +ve slope (line OC) it means that the body is moving with constant acceleration. If the graph obtained is a curve like OD whose slope decreases with time, the acceleration goes on decreasing. If the graph obtained is a curve like OE whose slope increases with time, the acceleration goes on increasing. The area of velocity-time graph with time axis represents the displacement of that body.

The velocity with which an object moves with respect to another object is known as relative velocity of that object. The relative velocity of an object A with respect to object B when both are in motion, is the time rate of change of position of object A with respect to that of B.    v AB = v A − v B When vA and vB are inclined to an angle q, v AB = v 2A + v 2B + 2v A v B cos q

•

If two objects are moving in same direction along straight line, the magnitude of relative velocity of one object with respect to the other object will be equal to difference in magnitude of the velocities of the two objects. If two objects are moving in opposite direction along straight line, the magnitude of relative velocity of one object with respect to the other object will be equal to the sum of the magnitude of the velocities of the two objects.

SELF CHECK

3. A body is at rest at x = 0. At t = 0, it starts moving in the positive x-direction with a constant acceleration. At the same instant another body passes through x = 0 moving in the positive x-direction with a constant speed. The position of the first body is given by x1(t) after time t and that of the second body by x2(t) after the same time interval. Which of the following graphs correctly describes (x1 – x2) as a function of time t?

(b) O

Time

Relative Velocity

•

(a)

B

(c)

t

(x1 – x2)

O

(x1 – x2)

O (x1 – x2)

t

(d) t

O

t

(AIEEE 2008) SCalaRS anD VeCtORS

Scalars : The physical quantities which have only magnitude and no direction are called scalars e.g., mass, length, time, speed, work, power, etc. Vectors : The physical quantities which have both magnitude and direction are called vectors e.g., displacement, velocity, acceleration, force momentum, etc. • Position vector : A vector which gives position of an object with reference to the origin of a coordinate system is called position vector. • Displacement vector : It is that vector which tells how much and in which direction an object has changed its position in a given time interval. • Polar vectors : These are the vectors which have a starting point or a point of application e.g., displacement, force, velocity, etc. • Axial vectors : The vectors which represent rotational effect and act along the axis of rotation in accordance with right hand screw rule are called axial vectors e.g., torque, angular momentum, etc. • Equal vectors : Two vectors are said to be equal if they have the same magnitude and direction. • Negative vector : The negative of a vector is defined as another vector having the same magnitude but having an opposite direction. • Zero vector : A vector having zero magnitude and an arbitrary direction is called a zero or null vector. • Collinear vectors : The vectors which either act along the same line or along parallel lines are called collinear vectors. • Coplanar vectors : The vectors which act in the same plane are called coplanar vectors. • Modulus of a vector : The magnitude or length of a vector is called its modulus.   Modulus of vector A = A = A • •

Fixed vector : The vector whose initial vector is fixed is called a fixed vector or localised vector. Unit vector : A unit vector is a vector of unit PHYSICS FOR YOU | august ‘15

33

magnitude drawn in the direction of a given vector.  A unit vector in the direction of A is given by  A  A=  A • • •

•

Co-initial vectors : The vectors which have the same initial point are called co-initial vectors. Co-terminus vectors : The vectors which have the common terminal point are called co-terminus vectors. Properties of zero vector : A zero vector has the following properties :        A + O = A; l O = O; 0 A = O Multiplication of a vector by a real number :  When a vector A is multiplied by a real number  n, it becomes another vector nA . Its magnitude  becomes n times the magnitude of A.

addition of Vector

Addition of two collinear vectors   Suppose A and B are two collinear vectors.  A   O P A B Q  O  P R B O Q    Now the resultant vector R = A + B Suppose a body is displaced through 4 m due west and then is further displaced through 6 m due west. Then the resultant displacement of the body = (4 m + 6 m) = 10 m due west. Triangle law of vector addition : If two vectors are represented both in magnitude and direction by the two sides of a triangle taken in the same order, then the resultant of these vectors is Q represented both in magnitude and direction by the third side of   R the triangle taken in reverse order B as shown.        or O R = A+B = B+A P A Parallelogram law of vector   addition : If the non-zero vectors A and B are represented by the two adjacent sides of a parallelogram then the resultant is given by the diagonal of the parallelogram passing through the point of intersection of the two vectors.  The magnitude of R is 34

PHYSICS FOR YOU | august ‘15

 R = R = A2 + B 2 + 2 AB cos q   where q is the angle between A and B . Here, B sin q tan a = A + B cos q A sin q and tan b = B + A cos q Special cases If q = 0°, Rmax = A + B q = 180°, Rmax = A – B

...(i)

...(ii)

and q = 90°, R = A2 + B 2 In all other cases equation (i) and (ii) can be used to  calculate magnitude and direction of R.  Polygon law of vector addition : E U  If a number of vectors are D represented both in magnitude and T direction by the sides of a polygon   taken in the same order, then the R C resultant vector is represented both S in magnitude and direction by the  B closing side of polygon taken in the  P Q opposite order. A       R=A+B+C+D+E Properties of vector addition (i) Vector addition is commutative. i.e.,     A+B = B+A (ii)

Vector addition is associative i.e.,       ( A + B) + C = A + ( B + C )

(iii) Vector addition is distributive. i.e.,     l( A + B) = lA + lB Subtraction of Vectors

 Negative of a vector (− A) is a vector of the same  magnitude as vector A but pointing in a direction  opposite to that of A.     Therefore, A − B = A + (− B)   Let the angle between vectors A and B be q, then the   angle between A and − B will be 180° – q.   A − B = A2 + B 2 − 2 AB cos q  B



 B

180° – 

 A

Rectangular Components of a Vector in a Plane

When a vector is splitted into two component vectors at right angles to each other, the component vectors are called rectangular components of a vector. If it makes an   angle q with x-axis and Ax and A y be the rectangular  components of A along x-axis and y-axis respectively, then    A = A + A = A i + A j x

y

x

y

Here, Ax = A cos q and Ay = A sin q. \ A2(cos2q + sin2q) = Ax2 + Ay2 or

(

A = Ax2 + A2y

)

1/2

and tan q =

Ay Ax

Addition of vectors in rectangular coordinates   If A = Ax i + A y j + Az k and B = Bx i + B y j + Bz k   Then, A + B = ( Ax i + A y j + Az k ) + (Bx i + B y j + Bz k ) = ( A + B )i + ( A + B )j + ( A + B )k x

y

y

z

z

Dot Product or Scalar Product

The dot product or scalar  product of two vectors A and    by A . B (read B represented  A dot B ) is defined as the product of the magnitudes of   A and B and the cosine of the angle between them.   A . B = AB cos q Cross Product or Vector Product

 B 

PROjeCtIle mOtIOn

Any body projected into space, such that it moves under the effect of gravity alone is called a projectile. The path followed by a projectile is called its trajectory which is always a parabola. A projectile executes two independent motions simultaneously: (i) uniform horizontal motion and (ii) uniform accelerated downward motion. Projectile Fired Horizontally

KEY POINT • Two vectors can be added if both of them are of same nature, for example, a displacement vector cannot be added to a force vector but can be added to displacement vector only. • Vector subtraction is neither commutative nor associative by nature. • Resolution of a vector into two component vectors along the directions of two given vectors is unique.

x

where n is a unit vector indicating the direction of   A × B.

 A

  The cross product of two vectors A and B is a vector        C = A × B (read as A cross B ). The magnitude of A × B   is defined as the product of the magnitudes of A and B and the sine of the angle between them. The direction of     the vector C = A × B is perpendicular to the plane A     and B and such that A, B and C form a right handed system.   A × B = AB sin q n .

Suppose a body is projected horizontally with velocity u from a height h above the ground. Let it reach the point (x, y) after time t. Then position of the projectile after time t, 1 x = ut, y = gt 2 2 g 2 Equation of trajectory, y = 2 x 2 u Velocity after time t,  gt  v = u 2 + g 2 t 2 ; b = tan −1   u 2h Time of flight, T = g 2h Horizontal range, R = u × T = u . g Projectile Fired at an angle with the Horizontal

Suppose a projectile is fired with velocity u at an angle q with the horizontal. Let it reach the point (x, y) after time t. Then Components of initial velocity, ux = u cos q, uy = u sin q Components of acceleration at any instant, ax = 0, ay = – g Position after time t, 1 x = (u cos q) t, y = (u sin q) t − gt 2 2 Equation of trajectory, g y = x tan q − x2 2 2 2u cos q Maximum height, H =

u2y

u 2 sin 2 q

= 2g 2g 2u y 2u sin q = Time of flight, T = g g Horizontal range, R =

2u x u y g

=

u 2 sin 2q g

PHYSICS FOR YOU | august ‘15

35

Maximum horizontal range is attained at q = 45° and its value is u2 Rmax = g Velocity after time t, vx = u cos q, vy = u sin q – gt KEY POINT • When the motion is two-dimensional, the time variable t has the same value for both the x-and y-directions.

• In projectile motion, the horizontal motion and the vertical motion are independent of each other; that is, neither motion affects the other. • The horizontal range is the same if an object is projected at angles q and 90° – q. • A projectile moves under the combined effect of two velocities (i) uniform horizontal velocity which would not change if there is no air resistance (ii) uniform changing vertical velocity due to gravity.

SELF CHECK

4. A projectile is given an initial velocity of  (i + 2 j) m s −1 , where i is along the ground and j s–2,

is along the vertical. If g = 10 m the equation of its trajectory is (a) 4y = 2x – 25x2 (b) y = x – 5x2 (c) y = 2x – 5x2 (d) 4y = 2x – 5x2

(JEE Main 2013) 5. A boy can throw a stone up to a maximum height of 10 m. The maximum horizontal distance that the boy can throw the same stone up to will be (a) 10 m

(b) 10 2 m

(c) 20 m

(d) 20 2 m (AIEEE 2012)

Uniform Circular motion

When a body moves along a circular path with uniform speed, its motion is said to uniform circular motion. Angular displacement : It is the angle swept out by a radius vector in a given time interval. Arc s q (rad) = = Radius r Angular velocity : The angle swept out by the radius vector per second is called angular velocity. q −q q w= or w = 2 1 t t 2 − t1 36

PHYSICS FOR YOU | august ‘15

Relationship between v and w : It is given by v = rw i.e. Linear velocity = Radius × angular velocity. Angular acceleration and its relation with linear acceleration : The rate of change of angular velocity is called angular acceleration. It is given by w − w1 a= 2 t 2 − t1 Also, a = r a i.e. Linear acceleration = Radius × angular acceleration Centripetal acceleration : A body moving along a circular path is acted upon by an acceleration directed towards the centre along the radius. This acceleration is called centripetal acceleration. It is given by a=

v2 = rw 2 . r

KEY POINT • A centripetal force accelerates a body by changing the direction of the velocity of body without changing the speed of body.

SELF CHECK

6. For a particle in uniform circular motion, the  acceleration a at a point P(R, q) on the circle of radius R is (Here q is measured from the x-axis) (a)

v2 ^ v2 ^ i+ j R R

(b) −

^ v2 ^ v2 cos q i + sin q j R R

(c) −

^ v2 ^ v2 sin q i + cos q j R R

(d) −

^ v2 ^ v2 cos q i − sin q j R R

(AIEEE 2010)

FORCe anD IneRtIa

Force is defined as an agency (a push or a pull) which changes or tends to change the state of rest or of uniform motion or the direction of motion of a body. Inertia is a property of the body due to which body opposes the change of its state itself. Inertia of a body is measured by mass of the body i.e., Inertia ∝ Mass Heavier the body greater is the force required to change its state and hence greater is inertia. The reverse is also true. i.e., lighter body has less inertia.

types of Inertia

• • •

Inertia of rest : It is the inability of a body to change its state of rest by itself. Inertia of motion : It is inability of a body to change its state of uniform motion by itself. Inertia of direction : It is the inability of a body to change its direction of motion by itself.

KEY POINT • Force on a body can be exerted from a distance by the external agencies of force. Examples are gravitational and magnetic forces. newtOn’S lawS OF mOtIOn

Sir Isaac Newton (1642-1727) made a systematic study of motion and extended the ideas of Galileo. He arrived at three laws of motion which are called Newton’s laws of motion.

duration the total effect of force is called impulse. This type of force is sometimes called impulsive force. Impulse, I = F × Dt     I = mDv = m(v − u)     = mv − mu = p f − pi    \ I = D p = p f − pi

SELF CHECK

7. A particle of mass m is at rest at the origin at time t = 0. It is subjected to a force F(t) = Foe–bt in the x direction. Its speed v(t) is depicted by which of the following curves? (a)

Impulse

In many cases, the forces involved in collisions and other interactions that act a very short time are known explicitly. When a large force act for an extremely short

(b)

v(t) t

First law

A body continues in its state of rest or uniform motion in a straight line unless compelled by an external force to change that state. The first law defines inertia hence it is also called law of inertia. momentum It is the product of the mass and velocity of a body, i.e. p = mv The concept of momentum was introduced by Newton to measure the quantitative effect of force. Second law The rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction in which the force acts.   dp Mathematically, F = dt It can be shown experimentally that the acceleration is  (i) proportional to the net force F net on the object and (ii) inversely proportional to the mass m of the object. 1 a ∝ Fnet and a ∝ . m Fnet kF Thus, a ∝ ⇒ a = net , m m The value of constant k is 1. F \ a = net , Fnet = ma m

F0b m

(c)

F0 mb

(d)

v(t)

t

F0 mb

v(t)

t

F0 mb

v(t)

t

(AIEEE 2012)

8. The figure shows the position - time (x-t) graph of one-dimensional motion of a body of mass 0.4 kg. The magnitude of each impulse is 2 x(m) 0

(a) 0.2 N s (c) 0.8 N s

2

4

6 8 t(s)

10

12

14

(b) 0.4 N s (d) 1.6 N s

16

(AIEEE 2010)

third law

For every action there is an equal and opposite reaction. Meaning of action and reaction : Suppose that a body A  experiences a force FAB due to other body B. Also body  B will experience a force FBA due to A. According to Newton’s third law if two forces are equal in magnitude and opposite mathematically we write it as  in direction,  FAB = − FBA   Here we can take either FAB or FBA as action force and the other will be the reaction force. Another important thing is these two forces always act on different bodies, and so, they never cancel each other. apparent weight of Body in a lift (i) When a lift moves upwards with uniform acceleration, apparent weight of a body in the lift increases. PHYSICS FOR YOU | august ‘15

37

From Fig. (a), R – mg = ma or R = m (g + a) R m mg (a)

a

m

R

mg (b)

a

R m

a=0

mg ( c)

(ii) When a lift moves downwards with acceleration, a the the apparent weight of a body in the lift decreases. From Fig. (b) mg – R = ma or R = m (g – a) (iii) When a lift is at rest or moves with uniform velocity, a = 0, the apparent weight of the body is equal to its true weight. From Fig. (c), R = mg (iv) When a lift fall freely, (a = g) the apparent weight of a body in the lift becomes zero. R = m(g – g) = 0 KEY POINT • If the elevator is at rest or is moving uniformly in the upward/downward direction, the apparent weight is equal to actual weight of the person. COnSeRVatIOn OF lIneaR mOmentUm

When no external force acts on a system of several interacting particles, the total linear momentum of the system is conserved. The total linear momentum is the vector sum of the linear momenta of all the particles of the system. Derivation of the law of conservation of momentum from Newton’s second law of motion Suppose the n particles have masses m1, m2, m3 ... mn and    are moving with velocities, v1 , v2 ..... vn respectively. Then total linear momentum of the system is     p = m1v1 + m2v2 ..... mnvn     p = p1 + p2 + ....... pn  If F is the external force acting on the system, then according to Newton’s second law  dp F= dt For an isolated system    dp F = 0 or = 0 or p = constant dt    or p1 + p2 + ....... + pn = constant Thus in the absence of any external force, the total linear momentum of the system is constant. This is the law of conservation of linear momentum. application of the law of Conservation of momentum

(i) Recoil of a gun. Let M be the mass of the gun and 38

PHYSICS FOR YOU | august ‘15

m be the mass of the bullet. Before firing, both the gun and the bullet are at rest. After firing, the bullet  moves with velocity v and the gun moves with  velocity V . As no external force acts on the system, so according to the principle of conservation of momentum, Total momentum before firing = Total momentum after firing   or 0 = mv + MV   V M m v or MV = − mv  m or V = − v M   The negative sign shows that V and v are in opposite directions i.e., the gun gives a kick in the backward direction or the gun recoils with  velocity V . Further, as M > > m, so V < < v i.e., the recoil velocity of the gun is much smaller than the forward velocity of the bullet. (ii) When a man jumps out of a boat to the shore, the boat slightly moves away from the shore. Initially, the total momentum of the boat and the man is zero. As the man jumps from the boat to the shore, he gains a momentum in the forward direction. To conserve momentum, the boat also gains an equal momentum in the opposite direction. So the boat slightly moves backwards. (iii) An astronaut in open space, who wants to return to the spaceship, throws some object in a direction opposite to the direction of motion of the spaceship. By doing so, he gains a momentum equal and opposite to that of the thrown object and so he moves towards the spaceship. Rocket propulsion. It is an example of momentum conservation in which the large backward momentum of the ejected gases imparts an equal forward momentum to the rocket. Due to the decrease in mass of the rocketfuel system, the acceleration of the rocket keeps on increasing. Let u = velocity of exhaust gases v0, v = initial velocity and velocity of the rocket at any instant t m0, m, me = initial mass, mass of the rocket at any instant t and mass of empty rocket respectively. dm Thrust on rocket, F = −u dt    dm  u Acceleration of rocket, a =  dm  dt   mo − t  dt 

Velocity of rocket, v = v0 + u log e

m0 m

Burnt-out speed of rocket, vb = v0 + u log e

SELF CHECK

m0 me

9. A machine gun fires a bullet of mass 40 g with a velocity 1200 m s–1. The man holding it can exert a maximum force of 144 N on the gun. How many bullets can he fire per second at the most? (a) one (b) four (c) two (d) three. (AIEEE 2004) FRee BODY DIagRam

A single body or a subsystem of bodies isolated from its surroundings showing all the forces acting on it is its free body diagram. Steps for Free Body Diagram • Step 1: Identify the object or system and isolate it from other objects clearly, specify its boundary. • Step 2 : First draw non-contact external force in the diagram. Generally it is weight. • Step 3 : Draw contact forces which acts at the boundary of the object or system. Contact forces are normal, friction, tension and applied force. In free body diagram, internal forces are not drawn, only external are drawn. Free body diagram of a block of mass m is kept on the ground as shown in figure. tenSIOn

When we pull the two ends of a string in opposite directions then a force is developed in the body of the string which is called tension.

tension in a String

Features of an ideal string • Ideal string is massless. • Ideal string is perfectly smooth. • Ideal string is inextensible. Length of the strings remains constant. • Ideal string is perfectly flexible. Rules of marking tension force on a body by a string (i) The tension force should be along the length of the string. (ii) The tension force should be away from the body. Tensions are of two different types : • Extensional

• Compressional In a string or in a chain tension is only extensional. In a rod tension can be extensional or compressional or both. Tension in a string is the force by which one part of the string pulls the other part. An ideal string has zero mass then the net force acting on that string will always be equal to zero. Fnet = 0 × a \ Fnet = 0 In an ideal string the tension at all the points of the string will be same.

We had to find out the tension in the string between the blocks by drawing the free body diagram of both the blocks and the string. m1F T = m1a ⇒ T = m1 + m2

Motion of connected bodies. Suppose two bodies of masses M and m(M > m) are tied at the ends of an inextensible string passing over a frictionless pulley. Then M −m g Acceleration of the masses, a = M +m 2 Mm g Tension of the string, T = M +m Clearly, a < g. Equilibrium of concurrent forces: A number of concurrent forces acting on a body are said to be in equilibrium if their vector sum is zero or if these forces can be completely represented by the sides of a closed polygon taken in the same order.     F1 + F 2 + F 3 + ......F n = 0 Lami’s theorem : It states that if three forces acting on a particle keep it in equilibrium, then each force is proportional to the sine of the angle between other two    forces. If a, b, g be the angles between F 2 and F 3 ; F 3    and F1 ; F1 and F 2 respectively, then according to Lami’s theorem : F F1 F = 2 = 3 . sin a sin b sin g PHYSICS FOR YOU | august ‘15

39

SELF CHECK

10. A rocket with a lift-off mass 3.5 × 104 kg is blasted upwards with an initial acceleration of 10 m s–2. Then the initial thrust of the blast is (a) 3.5 × 105 N (b) 7.0 × 105 N 5 (c) 14.0 × 10 N (d) 1.75 × 105 N. (AIEEE 2003) FRICtIOn

Friction is a contact force that opposes the relative motion or tendency of relative motion of two bodies. Consider a block on a horizontal table as shown in the figure.

flim ∝ N ⇒ flim = msN Here ms is a constant, the value of which depends on nature of surfaces in contact and is called as coefficient of static friction. (ii) The magnitude of limiting frictional force is independent of area of contact between the surfaces. kinetic Frictional Force

Once relative motion starts between the surfaces in contact, the frictional force is called as kinetic frictional force. The magnitude of kinetic frictional force is also proportional to normal force. f k = m kN From the previous observation we can say that mk < ms. angle of Friction

If we apply a force, acting to the right, the block remains stationary if F is not too large. The force that counteracts F and keeps the block at rest is called frictional force. If we keep on increasing the force, the block will remain at rest and for a particular value of applied force, the body comes to state of ‘about to move’. Now if we slightly increase the force from this value, block starts its motion with a jerk and we observe that to keep the block moving we need less effort than to start its motion. If we draw the graph between applied force and frictional force for this observation its nature is as shown in figure. Static Frictional Force

When there is no relative motion between the contact surfaces, frictional force is called static frictional force. It is a self-adjusting force, it adjusts its value according to requirement (of no relative motion). In the taken example static frictional force is equal to applied force. Hence one can say that the ab portion of graph will have a slope of 45°. (fs ≤ msN) limiting Frictional Force

This frictional force acts when body is about to move. This is the maximum frictional force that can exist at the contact surface. We calculate its value using laws of friction. laws of Friction

(i) The magnitude of limiting frictional force is proportional to the normal force at the contact surface. 40

PHYSICS FOR YOU | august ‘15

The resultant of normal reaction    R and the frictional force f is S which makes an angle l with R. Now, the angle l is called the angle of friction. f mR tan l = = =m R R angle of repose

The minimum angle of inclination of plane with the horizontal, at which the body placed on the plane just begins to slide down the incline, is known as angle of repose. a = tan–1m Rolling Friction

The opposing force that comes into existence when one object rolls over the surface of another object is known as rolling friction.

SELF CHECK 11. A block of mass m is placed on a surface with a x3 vertical cross section given by y = . If the 6 coefficient of friction is 0.5, the maximum height above the ground at which the block can be placed without slipping is 1 2 1 1 (a) m (b) m (c) m (d) m 6 3 3 2 (JEE Main 2014) 12. Two fixed frictionless inclined planes making an angle 30° and 60° with the vertical are shown in the figure. Two blocks A and B are placed on the two

planes. What is the relative vertical acceleration of A with respect to B?

end of the string is attached as shown in figure, then vH

H

A O

B

60°

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

P  vL L mgsin mgcos mg

30°

4.9 m s–2 in vertical direction 4.9 m s–2 in horizontal direction 9.8 m s–2 in vertical direction zero (AIEEE 2010)

•

CIRCUlaR mOtIOn Centripetal Force

Centripetal force is the force required to move a body uniformly in a circle. Magnitude of this force mv 2 is F = = mrw2 where v is the linear velocity, w r is angular velocity of the body and r is radius of the circular path. a Vehicle taking a Circular turn on a level Road

If the coefficient of friction between the tyres and the road is ms, the maximum velocity with which a vehicle can take a circular turn of radius r without slipping is given by v = m s rg A cyclist move along a circular level track of radius r with speed v. The angle through which the cyclist bends himself is given by tanθ = v2/rg Banking of Roads

Vehicles move on a curved track of radius r with maximum speed v, the track is banked through an angle q is given by tanθ = v2/rg (frictionless road) A car is moving on a circular track of radius R and angle q. ms is the coefficient of friction between the wheels of the car and the road. The optimum speed to avoid skidding is given by 1/2

vmax

 Rg (m s + tan q)  =   1 − m s tan q 

motion in a Vertical Circle

• • • •

Bending of a Cyclist

When a small body of mass m is attached to an inextensible light string of length r and whirling in a vertical circle about a fixed point O to which the other

T

Tension at any position of angular displacement, (q) along a vertical circle is given by mv 2 T= + mg cos q r Thus, tension at the lowest point (q = 0) is given by mv L2 TL = + mg r and tension at the highest point (q = 180°) is given by 2 mv H TH = − mg r Minimum speed at the highest point, v H = gr Minimum speed at the lowest point for looping the loop, v L = 5 gr . When the string is horizontal, q = 90°, minimum velocity, v = 3 gr . Height through which a body should fall for looping the vertical loop or radius r is, h = 5r/2. wORk, eneRgY anD POweR

work

Work is said to be done whenever a force acts on a body and the body moves through some distance in the direction of the force. This work is done on a body only if the following conditions are satisfied. (i) A force acts on the body (ii) The point of application of the force moves in the direction of the force. work Done by a Constant Force (i) Measurement of work done when the force acts along the direction of motion. Work done (W) = Force (F) × distance moved in the direction of force (s) (ii) Measurement of work done when force and displacement are inclined to each other. Work done (W) = component of force in the direction of displacement Fcosq) × magnitude of displacement (s)   = Fs cos q = F ⋅ s PHYSICS FOR YOU | august ‘15

41

Positive and negative work

Work done is a scalar quantity. It can be positive or negative. F  F  F

q < 90°, W is +ve

s

s

q = 90°, W is zero



s q > 90°, W is –ve. If the force applied to a body does a positive work, the velocity of the body increases. In this case the force, and hence the acceleration are directed along velocity.

If however, the force does a negative work, the acceleration is directed against the velocity and the velocity of the body decreases.

gravitational force, electrostatic force, elastic force in a spring etc. Forces acting along the line joining the centres of two bodies are called central force. Gravitational and electrostatic forces are two important examples. Central forces are conservative forces. A force is said to be non-conservative if work done by or against the force in moving a body depends upon the path. e.g., frictional forces, air resistance, viscous force, induced electric force etc.

SELF CHECK 13. When a rubber-band is stretched by a distance x, it exerts a restoring force of magnitude F = ax + bx2 where a and b are constants. The work done in stretching the unstretched rubber-band by L is 1  aL2 bL3  (a)  (b) aL2 + bL3 + 2 2 3  aL2 bL3 1 + (c) (aL2 + bL3 ) (d) 2 3 2 (JEE Main 2014)

work Done by Variable Force

 A variable force F, dependent on the position of the particle, is acting on a particle. If the particle moves  through an incremental displacement ds .  ^ ^ ^ F = Fx i + Fy j + Fz k  ^ ^ ^ ds = dx i + dy j + dz k   increment of work done on the dW = F ⋅ ds is the   particle by force F during the displacement ds .   ^ ^ ^ ^ ^ ^ dW = F ⋅ ds = (Fx i + Fy j + Fz k) ⋅ (dx i + dy j + dz k) dW = Fxdx + Fydy + Fzdz xf

yf

zf

xi

yi

zi

W = ∫ dW = ∫ Fx dx + ∫ Fy dy + ∫ Fz dz (xi, yi, zi) is initial position and (xf, yf, zf) is final position. KEY POINT • Work is energy transferred to or from a system by means of an external force acting on that system.

• Work done by a variable force is numerically equal to the area under the force curve and the displacement axis. Conservative Force and non-Conservative Force

A force is said to be conservative if work done by or against the force in moving a body depends only on the initial and final positions. It is independent of the path followed. Work done in a round trip is always zero. e.g., 42

PHYSICS FOR YOU | august ‘15

eneRgY

Energy is the capacity of a body to do work. It is measured in joule, like work. mechanical energy

There are two types of mechanical energy • Kinetic energy • Potential energy kinetic energy

Kinetic energy of body of mass m moving with velocity v is given by 1 K = mv 2 2 Two points regarding kinetic energy are as follows : • Since both m and v2 are always positive, K is always positive and does not depend on the direction of motion of the body. • Kinetic energy depends on the frame of reference. e.g., the KE of a person of mass m sitting in a train moving with speed v is zero in the frame of train 1 2 but mv in the frame of earth. 2

SELF CHECK

14. An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range (a) 2,000 J - 5,000 J (b) 200 J - 500 J (c) 2 × 105 J - 3 × 105 J (d) 20,000 J - 50,000 J. (AIEEE 2008)

Relation between kinetic energy (K) and linear momentum (p) p2 or p = 2mK K= 2m K

The graph between K and p is a parabola as shown in the figure. The graph between K and p is a straight line as shown in figure. The graph between K and 1/p is a rectangular hyperbola as shown in figure. work energy theorem

p K

where R is the radius of earth. If h > m1 then in equations (iii) and (iv), m1 can be neglected as compared to m2. m2u1 = −u1 , v2 = 0. m2 Therefore, when a light body A collides with a heavy body B at rest, the body A should start moving with same velocity just in opposite direction while the body B should practically remain at rest.

v1 = −

According to conservation of linear momentum, m1u1 + m2u2 = m1v + m2v or

v=

m1u1 + m2u2 (m1 + m2 )

...(i)

Kinetic energy of the system before collision is 1 1 K I = m1u12 + m2u22 2 2 Kinetic energy of the system after collision is 1 K F = (m1 + m2 )v 2 2 Loss in kinetic energy during collision, DK = KI – KF 1 1  1 =  m1u12 + m2 u22  − (m1 + m2 )v 2 2 2  2

...(ii)

Substituting the value of v from eqn. (i), we get DK =

1 m1m2 (u − u )2 2 (m1 + m2 ) 1 2

SELF CHECK

17. A block of mass 0.50 kg is moving with a speed of 2.00 ms–1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is (a) 0.34 J (b) 0.16 J (c) 1.00 J (d) 0.67 J. (AIEEE 2008) elastic Collision in two Dimensions or Oblique Collision

Let us consider two bodies A and B of masses m1 and m2 moving along X-axis with velocities u1 and u2 respectively. When u1 > u2, the two bodies collide. After collision, body A moves with velocity v1 at an angle q1 with X-axis and body B move with a velocity v2 at an angle q2 with X-axis as shown in the figure.

Perfectly Inelastic Collision in One Dimension Consider two bodies A and B of masses m1 and m2 moving with velocities u1 and u2 (u2 < u1) respectively along the same line collide head on and after collision they have same common velocity v. PHYSICS FOR YOU | august ‘15

45

BRAIN

ELECTROMAGNETIC INDUCTION Michael Faraday

Energy Consideration in Motional emf

B

l × B 2l 2v F R R  Power required to move the wire B 2l 2v 2 P R It is dissipated as Joule heat.

v

F

mo

f em al n tio

nge in flux produ Cha ces

Eddy Current  The currents induced in surface of bulk pieces of conductors when the magnetic flux linked with the conductor changes are known as Eddy currents. Slotting the conductor reduces the eddy current.  Applications : Electromagnetic damping, Induction furnace, Electric power meter, Magnetic braking in trains.

 It is produced by change in magnetic field in a region. This is nonconservative in nature.

is g

 The direction of the induced current is such that it opposes the change that has induced it.  If a current is induced by an increasing(decreasing) flux, it will weaken (strengthen) the original flux.  It is a consequence of the law of conservation of energy.

se

mf

nt

f

by ved hie c ea





dI Emf induced in the coil/conductor,   L dt  Coefficient of self induction N  L  B  I dI / dt 1 2  Self inductance of a coil L  0 N R 2  0 r N 2 A 2  Self inductance of a long solenoid L  0r n Al  l N 2 2 2 1  Mutual inductance, M    I1 (dI1/dt ) (dI 2 /dt )  Mutual inductance of two closely wound circular coils, M  N1N 2

Mutual inductance of two long coaxial solenoids  NN A M   0 r r12n1n2l  0 r 1 2 1 l M  Coefficient of coupling, k  L1L2



Combination of Inductors





For perfect coupling, k = 1 so,

em

Electric Generator Mechanical energy is converted into electrical energy by virtue of electromagnetic induction.  Induced emf,  = NAB sint  0sint

uc e

cu rre of ge

co il i nd

the

ch an

in

of Ra te

Inductance

This is also known as integral form of Faraday’s law.



b

d B dB  A 0 dt dt

Lenz's Law

al

n ca



by Lenz's l aw iven

   B  B  A  BA cos   Faraday’s Laws : It states that whenever magnetic flux linked with a coil changes, an emf is induced in the coil. d Induced emf,    N B dt dB / dt   Induced current, I   N R R  Induced charge flow , Q  I t   N B R Ro  The polarity of induced emf is such that it tat ion B tends to produce a current which opposes pr of od co uc il i the change in magnetic flux that es n mo has produced it. tio n

f ie ld )

net ic

(mag



 Magnetic flux

UB B2  V 2 0

M  L1L2

Induced Electric Field

 E  dl  

MAGNETIC FLUX AND FARADAY’S LAW

Magnetic Energy  Energy stored in an inductor 1 U B  LI 2 2  Energy stored in the solenoid, 1 2 UB  B Al 20  Magnetic energy density,



field



d behin

uB 

induced electric

produces emf





A wire moving in B

Emf in the wire = Bvl  Bvl Induced current, I   R R I Force exerted on the wire,

Motional emf On a straight conducting wire,  = Bvl On a rotating conducting wire about one end Bl 2  2    Here, B, v( rvˆ)and l are perpendicular to each other. 





A magnetic field can produce an electric field that can drive a current. This link between a magnetic field and the electric field is now known as Faraday's law of induction. The observations by Michael Faraday and other scientists which led to this law were at first just basic science. Today, however, applications of this basic science is everywhere.

Current, NBA I sin t  I 0 sin t R

Inductors in series, LS  L1  L2  2 M

Inductors in parallel, LP 

L1L2  M 2 L1  L2  2 M

If coils are far away, then M = 0 LL So, LS  L1  L2 and LP  1 2 L1  L2

L – R Circuit Current growth in L–R circuit I  I 0 (1  e t / L ) 



Growth /decay of

Current decay in L–R circuit I  I 0 (e t / L ) L Here,  L  Time constant  R  I0  R

Since the collision is elastic, kinetic energy is conserved. 1 1 1 1 ...(i) m1u12  m2u22  m1v12  m2v22 2 2 2 2 Also momentum along X-axis before collision = momentum after collision along X-axis  m1u1  m2u2  m1v1 cos 1  m2v2 cos 2 ...(ii)

Similarly along Y-axis ...(iii) 0  m1v1 sin 1  m2v2 sin 2 Thus from these three equations (i), (ii) and (iii) we can find the required quantities. Coeffi cient of Restitution It is defined as the ratio of relative velocity of separation after collision to the relative velocity of approach before collision. It is represented by e. relative velocity of separation (after collision) e relativee velocity of approach (before collision) v2  v1 e u1  u2 where u1, u2 are velocities of two bodies before collision, and v1, v2 are their respective velocities after collision.  For perfectly elastic collision, e = 1.  For perfectly inelastic collision, e = 0.  A ball falls from a height h, it strikes the ground with a velocity u  2 gh . Let it rebound with a

velocity v and rise to a height h1. 2 gh1 h v e   1 or h  e h 1 u h 2 gh  





or h1  e 2h.

A ball dropped from a height h and attaining height hn after n rebounds. Then hn = e2nh. A ball dropped from height h and travelling a total distance S before coming to rest. 1  e2  Then S  h  2 1  e  A ball dropped from a height h and rebounding. The time taken by the ball in rising to height h1 and 2h1 2h coming back is 2  2e . g g Note that h1 = e2h. Total time taken by the ball in coming to rest is t

1  e 2h . 1 e g ANSWER KEYS (SELF CHECK)

1. 6. 11. 16.

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

2. 7. 12. 17.

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

3. (c) 8. (c) 13. (d)

4. (c) 9. (d) 14. (a)

5. (c) 10. (a) 15. (b) 

APPLICATION OF NEWTON’S LAWS– THE ROLLER COASTER

S

ir Isaac Newton in 1687, published the Principia. This book contained his three laws of motion which changed mechanics forever. These three laws became the foundation of classical mechanics. Newton’s three laws of motion have been proven over and over again in the last three centuries. These are the basis of many of our modern inventions. One among these is a roller coaster. It is a modern invention that uses the laws of motion to thrilling ends. Roller coasters, with their twists, turns, and loops seem to defy everything we know about how people and objects move. Roller coaster designers, however, use Newton’s laws to push people past their usual limits. Newton’s first law is the law of inertia. Most roller coasters run by the Law of Inertia. Since an object at rest stays at rest, all roller coasters have to be pushed or pulled to get started. Most are pulled up a large hill called as lift hill. The first hill is usually the biggest hill. As the cars go uphill, it store potential energy. At the top, the cars have the largest amount of gravitational potential energy. These are about to be put into motion. Roller coaster cars will gain enough energy from the lift hill to be powered through the rest of the ride. Once put into motion, they will not stop until the brakes are applied at the end of the ride. Newton’s second law is the law of force and acceleration. You feel this second law when you start going down the hills. The coaster cars and your body have mass. The gravity provides acceleration.

48

PHYSICS FOR YOU | AUGUST ‘15

That causes force. The rider feels the force as it moves the cars along the coaster track. The track directs the force and the cars. The mass of your body and the coaster cars is constant. The amount of force changes due to changes in acceleration. The changes in acceleration are mostly due to friction. The thrill of acceleration on a roller coaster comes from Newton’s second law. Newton’s third law is about action and reaction. As you push down on the seat, the seat pushes back at you. This law really comes into play with newer roller coasters that expose riders to high gravitational force. Gravitational force relates an acceleration on a body to the acceleration due to gravity. Your body will experience an acceleration twice as strong as the pull of gravity. Older roller coasters did not expose riders to gravitational force. Newer roller coasters, however, can expose riders to very high gravitational forces. Gravitational force like all others forces, is not just acting on the things we can see. It is also acting on our blood and our blood vessels. Just as they push your body into the seat, it pushes your blood back away from your brain and toward your feet. As we use the laws of physics to create more exciting roller coasters, it becomes more important to keep in mind the limits of our human bodies. 

1. An aeroplane requires for take off a speed of 80 km h–1, the run on the ground being 100 m. The mass of the aeroplane is 104 kg and the coefficient of friction between the plane and the ground is 0.2. Assume that the plane accelerates uniformly during the take off. What is the maximum force required by the engine of the plane for take off? (a) 2.51 × 104 N (b) 4.43 × 104 N 2 (c) 4.42 × 10 N (d) 8.22 × 102 N 2. A particle of mass m is moving in a horizontal circle of radius r, under a centripetal force equal k  to −   , where k is constant. What is the total  r2  energy of the particle? k k (a) − (b) 3r 2r k k (c) − (d) 2r r 3. A particle is moving with uniform acceleration along a straight line AB. Its speed at A and B are 2 m s–1 and 14 m s–1 respectively. Then which one is incorrect? (a) Speed of particle at the mid-point of AB is 10 m s–1. (b) Speed of particle at a point P such that AP : PB = 1:5 is 5 m s–1. (c) The time taken by particle to go from A to the mid-point of AB is double to go from mid-point to B. (d) None of these 4. What is the velocity of ring as shown in the figure when spring becomes horizontal? m 3m 37° m = 10 kg k = 400 N m–1 l0 = 4 m

(a) 10 m s–1 (c) 30 m s–1

(b) 20 m s–1 (d) 5 m s–1

5. A ball is projected with velocity v0 and at an angle of projection a. After what time is the ball moving at right angles to the initial direction? v v (a) 0 sin a (b) 0 cosec a g 2g v v (c) 0 cos a (d) 0 cot a g 2g 6. Two bars of masses m1 and m2 connected by a nondeformed light spring rest on a horizontal plane. The coefficient of friction between the bars and the surface is equal to m. If F is the minimum constant force that has to be applied in the horizontal direction to the bar of mass m1 in order to shift the other bar. Then, the value of F is m  m    (a) mg  m1 + 2  (b) mg  m2 + 1    2  2  mg (c) mg (m1 + m2) (d) (m1 + m2 ) 2 7. The motion of a particle of mass m is given by x = 0 for t < 0 s, x(t) = A sin 4pt for 0 < t < (1/4)s (A > 0), and x = 0 for t > (1/4) s. Which of the following statements is incorrect? (a) The force at t = (1/8) s on the particle is – 16p2Am. (b) The particle is acted upon by an impulse of magnitude (4p2 Am) at t = 0 s and t = (1/4) s. (c) The particle is not acted upon by any force. (d) The particle is not acted upon by a constant force. 8. Two particles A and B, of masses m and 2m, are moving along the X-axis and Y-axis respectively with the same speed v. They collide, at the origin and coalesce into one body after the collision. What is the loss of energy during this collision? 1 4 (a) mv 2 (b) mv 2 3 3 3 2 2 2 (c) mv (d) mv 2 3 9. In the arrangement shown in figure, the ends P and Q of an inextensible string move downwards with uniform speed u. Pulleys A and B are fixed. The mass M moves upwards with a speed Physics for you | august ‘15

49

A

wedge. What magnitude F must have if the block is to remain at constant height above the table top?

B

 M

P

(a) 2u cosq

Q

(b)

M

u cos q

2u (d) u cosq cos q 10. What is the maximum compression in the spring, if the lower block is shifted to rightwards with acceleration a as shown in the given figure? All the surfaces are smooth. (c)

k

ma (a) 2k

2ma (b) k

m

ma (c) k

a

4ma (d) k

11. A chain is held on a frictionless table with 1/nth of its length hanging over the edge. If the chain has a length l and a mass m, how much work is required to pull the hanging part back on the table? mgl mgl mgl mgl (a) (b) (c) (d) 2 2 3 3n 2n2 4n 2n 12. A car accelerates from rest at a constant rate a for some time after which it decelerates at a constant rate b and comes to rest. If the total time elapsed is t, the maximum velocity acquired by the car will be a +b (a)  t  ab 

 ab  (b)  t  a + b 

 a 2 − b2  (c)  t  ab 

2  2 (d)  a + b  t  ab  13. A heavy particle hanging from a string of length l is projected horizontally with speed gl. Find the speed of the particle at the point where the tension in the string equals weight of the particle.

gl gl (a) (b) (c) 2gl (d) 3gl 2 3 14. A wedge with mass M rests on a frictionless horizontal surface. A block with mass m is placed on the wedge. There is no friction between the block and the wedge. A horizontal force F is applied to the 50

m

F

Physics for you | august ‘15

(a) Mg tanq (c) (M + m)g cosq



(b) (M – m)g tanq (d) (M + m)g tanq

15. In a car race, car A takes time t less than car B and passes the finishing point with a velocity v more than the velocity with which car B passes the point. Assuming that the cars start from rest and travel with constant accelerations a1 and a2. Which of the equations is correct? (a) v = a2t (b) v = a1t (c) v = t a1a2

(d) v = 2t a1a2

16. A 0.5 kg particle moves along the x-axis from x = 5 m to x = 17.2 m under the influence of a force 200 F (x ) = N. The total work done by this force 2x + x 3 during this displacement will be (a) 6.72 J (b) 2.54 J (c) 3.51 J (d) 3.00 J 17. A particle is moving in a circle of radius R in such a way that at any instant the normal and tangential components of the acceleration are equal. If its speed at t = 0 is u0, the time taken to complete the first revolution is u R (a) (b) 0 u0 R R R −2 p (1 − e −2 p ) e (c) (d) u0 u0 18. Suppose the average mass of raindrops is 3.0 × 10–5 kg and their average terminal velocity is 9 m s–1. The energy transferred by rain to each square metre of the surface at a place which receives 100 cm of rain in a year is (a) 2.05 × 102 J (b) 4.05 × 102 J 4 (c) 9.10 × 10 J (d) 4.05 × 104 J 19. A particle is observed from two frames S1 and S2. The frame S2 moves with respect to S1 with an acceleration a. Let F1 and F2 be the pseudo forces on the particle when seen from S1 and S2 respectively. Which of the following are not possible? (a) F1 = 0, F2 ≠ 0 (b) F1 ≠ 0, F2 = 0 (c) F1 ≠ 0, F2 ≠ 0 (d) F1 = 0, F2 = 0

20. The speed of a motor launch with respect to still water is 7 m s–1 and the speed of stream is 3 m s–1. When the launch began travelling upstream, a float was dropped from it. The launch travelled 4.2 km upstream, turned about and caught up with the float. How long is it before the launch reaches the float? (a) 35 min(b) 36 min (c) 34 min (d) 33 min 21. A particle is tied to an ideal string and whirled in a vertical circle of radius L, where L is the length of the string. If the ratio of the maximum to minimum tension in the string throughout the motion is 2 : 1, then the maximum possible speed of the particle will be (a) 11 gL (b) 5 gL (c) 10 gL (d) 3 gL 22. In the arrangement, shown in figure, pulleys are massless and frictionless and threads are in-extensible, block of mass m1 will remain at rest if 4 1 1 (a) = + m1 m2 m3 (b) m1 = m2 = m3 (c)

1 1 1 = + m1 m2 m3

(d)

A B m1

m2 m3

1 2 3 = + m3 m2 m1

23. In shown figure, l the trolley starts m accelerating with acceleration a. The maximum angle deflected by thread from vertical will be a  2a  (a) tan −1   (b) tan −1   g  g 

U(x)

(a)

U(x) x

(b) U(x)

U(x)

(c)

x

x

(d)

x

   26. A force F = v × A is exerted on a particle in addition  to the force of gravity, where v is the velocity of the  particle and A is a constant vector in the horizontal direction. The minimum speed of projection for a particle of mass m so that it continues to move with a constant velocity is given by mg mg mg (a) (b) (c) (d) mg A 2A 3A 27. If the potential energy of two molecules is given A B by U = − , then at equilibrium position, its r 6 r 12 potential energy is equal to B2 A2 B2 2B (b) (c) (d) − 4A 4B 4A A 28. A particle P is sliding down a frictionless hemispherical bowl. It passes the point A at t = 0. At this instant of time, the horizontal component of its velocity is v. A bead Q of the same mass as P is ejected from A at t = 0 along the horizontal string AB with speed v. Friction Q B between the bead and A the string may be P neglected. Let tP and tQ be the respective times C taken by P and Q to reach the point B. Then (a) tP < tQ (b) tP = tQ (c) tP > tQ Length of arc ACB t (d) P = t Q Length of chord AB (a)

a

a  a  (c) 2 tan −1   (d) tan −1   g    2g  24. A cyclist is riding with a speed of 27 km h–1. As he approaches a circular turn on the road of radius 80 m, he applies brakes and reduces his speed at the constant rate 0.5 m s–2. What is the magnitude of the net acceleration of the cyclist on the circular turn? (a) 0.74 m s–2 (b) 0.86 m s–2 –2 (c) 0.50 m s (d) 0.25 m s–2 25. A particle, which is constrained to move along the x-axis, is subjected to a force in the same direction which varies with the distance x of the particle from the origin as F(x) = –kx + ax3. Here k and a are positive constants. For x ≥ 0, the functional form of the potential energy U(x) of the particle is

29. A string of negligible mass going over a clamped pulley of mass m supports a block of mass M as shown in the figure. What is the force exerted on the pulley by the clamp? (a) 0 mg (b) 2 Mg

m M

2 2 2 2 (c) g M + (m + M ) (d) g m + (M + m) Physics for you | august ‘15

51

30. A block is projected with an initial velocity vBlock on a long conveyor belt moving with velocity vBelt (at that instant) having constant acceleration aBelt. Mark the correct option regarding friction after long time (friction coefficient between block and belt = m). If P. Q. R. S. (a) (b) (c) (d)

Column-I v Block = 2v Belt and aBelt = 0 v Block = 2v Belt and aBelt > mg v Block = 2v Belt and abelt = mg v Block = 2v Belt and aBelt < mg

(i)

Column-II zero

f s static friction (0 < fs < f l) (iii) f l limiting friction (ii)

(iv) f k kinetic friction

P-(i), Q-(ii), R-(iii), S-(iv) P-(i), Q-(iv), R-(iii), S-(ii) P-(iv), Q-(i), R-(iii), S-(ii) P-(ii), Q-(iv), R-(iii), S-(i)

1. (b) : Here u = 0, s = 100 m, 5 200 m s −1 v = 80 km h–1 = 80 × = 18 9 2 2 As v – u = 2as 2

 200    − 0 = 2a × 100 9  40000 200 = m s −2 or a = 81 × 200 81 Force required to produce acceleration a, 200 F1 = ma = 10 4 × = 2.47 × 10 4 N 81 Force required to overcome friction, F2 = mR = mmg = 0.2 × 104 × 9.8 = 1.96 × 104 N Maximum force required by the engine for take off, F = F1 + F2 = 2.47 × 104 + 1.96 × 104 = 4.43 × 104 N. 2. (c) : As the particle is moving in horizontal circle, so Centripetal force, F =

mv 2 k = r r2

r

r

r

∞

∞

∞

 k  k U = − ∫ Fdr = − ∫  −  dr = k ∫ r −2 dr = −  r2  r 52

Physics for you | august ‘15

k k k − =− . 2r r 2r d

A 2 m s–1

B 14 m s–1

P

142 = 22 + 2 × a × d At mid-point, let velocity is v d 2

⇒ v2 = 22 + 2 × a ×

...(i) 14 2 − 22 2 [Using eqn. (i)]

⇒ v2 = 4 +

192 = 100 ; v = 10 m s–1 2 AP 1 d If = then AP = PB 5 6 Let velocity at P is v1 v2 = 4 +

d 14 2 − 22 =4+ = 36 6 6

⇒ v1 = 6 m s–1 Let time taken to reach mid-point from A is t1, and t2 be time taken to reach B from mid-point. 6 = 2 + at1 ; t1 = 4/a 14 = 6 + at2 ; t2 = 8/a t1 4 1 = = ⇒ t 2 = 2t1 t2 8 2

4. (a) : m = 10 kg, k = 400 N m–1 Natural length of spring =4m 5m Decrease in PE = Increase in KE 37° C

A 3m=h

4m

B

1 1 k × (Dl )2 + mgh = mv 2 2 2 1 1 400 × 12 + 10 × 10 × 3 = × 10 × v 2 2 2 2 ⇒ 200 + 300 = 5v ⇒ 5v2 = 500 \

k r 1 k \ K.E. of the particle, K = mv 2 = 2 2r dU As F = − dr \ Potential energy, This gives mv 2 =

3. (b) :

v12 = 22 + 2 × a ×

SolutionS

\

Total energy = K + U =

v = 100 = 10 m s −1

5. (b) : If initial velocity v0 and velocity at time t are perpendicular, then the final velocity will be at angle a with the vertical as shown in the figure.

v0 

 90° 

v

Horizontal component of velocity is unchanged throughout the motion. Therefore v0 cos a = v sin a or v = v0 cot a Vertical component of velocity after time t = –v cos a From the equation vy = v0 sin a – gt –v cos a = v0 sin a – gt v 0 sin a + v cos a or t = g v sin a + v 0 cot a cos a = 0 g 2 v  sin a + cos 2 a  v 0 = 0   = cosec a g  sin a g  6. (a) : m m 2

1

F

For mass m2 to move, kx = mm2g ...(i) Using work energy theorem on m1, 1 Fx – mm1gx – kx2 = 0 2 1 or, F = mm1 g + kx 2 mm2 g F = mm1 g + 2 m   = mg  m1 + 2  [using eqn. (i)]  2  7. (c) : At t = (1/8) s, as x = A sin 4pt, dx v= = A (cos 4 pt ) 4 p dt dv = 4 pA (− sin 4 pt )(4 p) = −16 p2 A and a = dt Thus, F = ma = –(16p2Am) 1  Impulse, I = FDt = (–16p2Am) ×  s − 0 s  = – 4p2Am   4 | I | = 4p2Am For t > (1/4) s, x = 0, v = 0, a = 0, F = 0. Hence option (c) is incorrect. 8. (c) :

Y

A v m B 2m

O

 5 5v 2 , hence V =   v 9  3  Total kinetic energy before collision, i.e. V2 =

or

1 1 3mv 2 K i = mv 2 + (2m)v 2 = 2 2 2 Total kinetic energy after collision, i.e. 1 5 35 K f = (3m)V 2 =     mv 2 = mv 2 29 2 6 Loss of kinetic energy during the collision 2 3 5 = K i − K f =  −  mv 2 = mv 2 2 6 3 2 2 2 9. (b) : l = x + b v dl dx db ⇒ 2l = 2 x + 2b dt dt dt  dx  ⇒ 2l (u) = 2 x   + 0  dt  u  dx  l ⇒   = u= u  dt  x cos q 10. (b) : Pseudo force on mass m = ma Spring force developed = kx where x = compression in spring Using Newton’s second law of motion, mv

dv = (ma − kx ) dx

0

V

x′

∫0 mv dv = ∫0 ⇒ 0 = max ′ −

b l

v m

ma

x



FBD of mass m

M

kx

(ma − kx ) dx (Q x′ = maximum compression)

kx ′ 2

2

⇒

x′ =

2ma k

m l Let length of the hanging part of the chain = y m Mass of the hanging part of the chain = . y l Force required to be applied = Weight of the hanging part mg m  or F =  y  g = .y l l 

11. (d) : Mass per unit length of the chain =

 3m

v

Using law of conservation of linear momentum, For the X-components : mv = 3mV cosa ...(i) For the Y-components : 2mv = 3mV sina ...(ii) From eqns. (i) and (ii), 2 2  v   2v  V 2 cos 2 a + V 2 sin 2 a =   +   3  3 

X

 Let V be the velocity of the combined mass and let its direction make an angle a with the positive X-axis after the collision.

Physics for you | august ‘15

53

The work done in pulling the chain through a small distance dy is mg dW = − . y dy l Here negative sign indicates that the weight and displacement are oppositely directed. Total work done in pulling the 1/nth length of the chain is W = ∫ dW = − mg =− l

mg l 0

y =0

∫

ydy

l y= n

 y2  mg  l 2  mgl   =− 0 − 2  = 2 . 2l  n  2n  2  l /n

2 \ mv = mg − mg cos q

l

1 2 mgl or, 2 mv = 2 (1 − cos q) 1 2 mgl mv = (1 − cos q) ...(iii) 2 2

l cos 

12. (b) : For accelerated motion for time t 1 , from v = v0 + at1, we get v = 0 + at1 (as v0 = 0) v or, t1 = a For the retarded motion for time t2, from v = v0 + at2, we get v 0 = v – bt2 or t 2 = (as v0 = v, v = 0) b a +b v v Total time, t = t1 + t2 = + = v   ab  a b  ab  t or v =   a + b  13. (a) : Speed at bottom = gl < 2 gl Using energy conservation principle, 1 1 mgl (1 − cos q) = mgl − mv 2 ...(i) 2 2 mv 2 Also T − mg cos q = ...(ii) l According to question, h = l(1 – cos) T = mg  T h  v 0  ( gl )

From eqn. (i),

v  mg

1 1 mgl(1 − cos q) = mgl − mgl(1 − cos q) 2 2 1 2 1 − cos q = ⇒ cos q = 3 3 From eqn. (iii), gl v= 3

54

Physics for you | august ‘15

mg cos

14. (d) : This problem can be solved very easily if we analyse the block in the reference frame of wedge instead of analysing it in ground reference frame. Reference frame of wedge is non-inertial, therefore we must apply a pseudo force on block m. System

M

F

N Y

mA cos  Pseudo force  mA  is n  A A  m os c g m mg

m 

X

m gs

in



According to condition of problem the block m remains at constant height h, i.e., it does not slip downwards along the incline. For an observer on the wedge the block will be stationary. SFx = mg sin q – mA cos q = 0 ...(i) SFy = N – mA sin q – mg cos q = 0 ...(ii) From eqn. (i), A = g tan q From eqn. (ii), we get N = mg/cos q If the block is kept on a scale its reading will be N = mg/cos q. We have to consider block and wedge as a single body (because block does not slip). Therefore F = (M + m)A = (M + m) g tan q 15. (c) : Let s be the distance covered by each car. Let the times taken by the two cars to complete the journey be t1 and t2, and their velocities at the finishing point be v1 and v2 respectively. According to the problem, v1 – v2 = v and t2 – t1 = t 0+v v When u = 0, s = × t = .t 2 2 v t v t \ s= 1 1 = 2 2 2 2 1 2 1 Also s = a1t1 = a2 t 22 2 2 Hence 2s 2s − 2s(t 2 − t1 ) v v1 − v 2 t1 t 2 2s = = = = t t 2 − t1 t 2 − t1 t1 t 2 (t 2 − t1 ) t 1 t 2 =

4s 2 t12 t 22

=

v = t a1 a2 .

2s 2s . = a1 a2 t12 t 22

16. (c) : W=

xf

17.2

xi

5

∫ F dx = ∫

200 2x + x 3

dx =

17.2

∫ 5

19. (d)

200 x (x 2 + 2)

dx

Put t = x2 + 2, dt = 2 xdx. Then dt 200 W= ∫ 2 (t − 2) t 1  1 = 100 ∫  −  dt  2(t − 2) 2t  t − 2  = 50  log e (t − 2) − log e t  = 50  log e t   17.2

 x2  = 50 × 2.303  log  x 2 + 2  5  = 50 × 2.303 log

295.84 27 × 297.84 25

= 50 × 2.303 log 1.07275 = 50 × 2.303 × 0.0305 = 3.51 J. 17. (c) : Given, tangential acceleration = centripetal acceleration dv v 2 = dt R v

⇒

∫ v2

u0

⇒

dv

...(i) t

=∫ 0

 1 1 R −  =t  u0 v 

...(ii)

⇒

∫

u0

dv = v

2 pR

∫ 0

ds ⇒ v = u0e 2 p R

R From (ii) and (iii), t = (1 − e −2 p ) u0

⇒

T = mg +

mv 2 L

12mg = mg +

mv 2 L

v = 11 gL

22. (a) : FBD of masses m1, m2 and m3 T

dv ds v 2 dv v 2 . = ⇒ v = ds R ds dt R

v

21. (a) : We know Tmax – Tmin = 6mg and Tmax = 2Tmin(given) \ Tmin = 6mg, Tmax = 12 mg Since tension is maximum at lowest points

1 dt R

Again from eqn. (i) Q

20. (a) : For upstream motion of launch : Relative velocity = 7 – 3 = 4 m s–1 Distance moved = 4.2 km = 4200 m 4200 Time taken, t1 = = 1050 s 4 For downstream motion of launch: Distance moved downstream by float in 1050 s = 3 × 1050 = 3150 m Distance between float and launch turned about = 4200 + 3150 = 7350 m This distance is to be covered by launch with its own velocity (7 m s–1) because stream velocity is being shared by both. 7350 \ Time taken, t 2 = = 1050 s 7 Total time taken, t = t1 + t2 = 1050 + 1050 = 2100 s = 35 min

...(iii)

18. (d) : Here; m = 3 × 10–5 kg, v = 9 m s–1, Rain received in a year, h = 100 cm = 1 m, A = 1 m2 Volume of the rain falling, V = Ah = (1 m2) (1m) = 1m3 Mass of rain, M = Vr = (1 m3) (103 kg m–3) = 103 kg Energy transferred by the rain, 1 1 E = Mv 2 = (103 kg )(9 m s −1 )2 2 2 = 40.5 × 103 J = 4.05 × 104 J

T

m1

m2

m1g

a

m2g

T m3

a

m3g

T′ = m1g m2g − T ′ = m2a 2 T′ − m3 g = m3a 2 From eqns. (i) and (ii) (m − m3 ) g a= 2 m2 + m3 Putting value of a in eqn. (i),  m − m2  T ′ = 2m2 1 + 3 g  m2 + m3  Physics for you | august ‘15

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

55

2m2 × 2m3 g m2 + m3 4 1 1 = + m1 m2 m3

23. (c) : By work energy theorem from trolley, maL sin q = mgL(1 – cosq) or, a sin q = 2gsin2q/2 q a ⇒ tan =  2 g ma a mg ⇒ q = 2 tan −1   g 24. (b) : Here r = 80 m 27 × 5 v = 27 km h −1 = m s −1 = 7.5 m s −1 , 18 Centripetal acceleration, ac =

v 2 (7.5)2 = = 0.7 m s −2 r 80

Suppose the cyclist v applies brakes at the point A of the ac A O circular turn, then, a at  tangential acceleration aT (negative) will act opposite to velocity. Given at = 0.5 m s–2 As the accelerations ac and at are perpendicular to each other, so the net acceleration of the cyclist is a=

ac2

+ at2

2

= (0.7) + (0.5)

x

U (x ) = − ∫ (−kx + ax 3 ) dx 0

or

U (x ) =

kx 2 ax 4 x 2  ax 2  − = k −   2 4 2  2 

2k a 2k For x > , U(x) will be negative. a Clearly, U(x) = 0 at x = 0 and x =

56

28. (a) : As the particle P moves from A to C, its horizontal velocity increases from v. Again the horizontal velocity decreases to v as P moves from C to B. But the horizontal velocity of P remains greater than or equal to v. But the horizontal velocity of bead Q remains constant equal to v. For same horizontal displacement AB, P takes smaller time than Q i.e., tP < tQ. 29. (c) : The free body diagram for the pulley is shown in figure. Fy

2

= 0.49 + 0.25 = 0.74 = 0.86 m s −2 If q is the angle between the total acceleration and the velocity of the cyclist, then, a 0. 7 tan q = c = = 1.4 or q = 54°28 ′ . at 0.5 dU 25. (d) : F = − dx or dU = – F dx \

dU =0 dx i.e., slope of U-x graph is zero at x = 0. Hence, the most appropriate option is (d).  26. (b) : For moving with constant velocity Fnet = 0   ⇒ F + mg = 0  A is in x-direction  \ For net force to be zero, F should be in +ve y-direction.   Now | v × A | = mg mg vA sinq = mg or v = A sin q For minimum v, sin q should be maximum mg \ v min = A 27. (a) At x = 0, F = −

m1 g =

Physics for you | august ‘15

Mg

Fx mg

Mg

For the equilibrium of the pulley, Fx = Horizontal component of the force by the clamp on the pulley = Mg Fy = Vertical component of the force by the clamp on the pulley = (M + m) g The net force exerted on the pulley by the clamp, F = Fx2 + Fy2 = g (m + M )2 + M 2 . 30. (b) : Maximum acceleration due to friction is mg. So a > mg → f k a < mg → fs a = mg → f l a=0→f=0 nn

1.

AMU

An oil drop of n excess electrons is held stationary under a constant electric field E in Millikan's oil drop experiment. The density of oil is r. The radius of the drop is 1/2

1/2

 3nrg  (b)    2 peE 

1/3

3nrg  (d)    4 peE 

  (a)  3neE   2 prg 

2.

The potential at a point x (measured in mm) due to some charges situated on the x-axis is given by V(x) = 20/(x2 – 4) volt The electric field E at x = 4 mm is given by (a) (10/9) volt/mm and in the +ve x direction (b) (5/3) volt/mm and in the –ve x direction (c) (5/3) volt/mm and in the +ve x direction (d) (20/9) volt/mm and in the –ve x direction

3.

In the following circuit, if potential difference between A and B is VAB = 4 V, then the value of X will be 10 

6.

B 2V

7.

8.

5V

A

(a) 5 W

total resistance is P. If S = nP, then the minimum possible value of n is (a) 4 (b) 3 (c) 2 (d) 1

1/3

  (c)  3nEe   4 prg 

X

(b) 10 W (c) 15 W

(d) 20 W

4.

A wire when connected to 220 V mains supply has power dissipation P1. Now the wire is cut into two equal pieces which are connected in parallel to the same supply. Power dissipation in this case is P2. Then P2 : P1 is (a) 1 (b) 4 (c) 2 (d) 3

5.

The resistance of the series combination of two resistors is S. When they are joined in parallel the

ENGG.

9.

A triangular loop of side l carries a current i. It is placed in a magnetic field B such that the plane of the loop is in the direction of B. The torque on the loop is (a) iBl (b) i2Bl 3 2 (c) (d) infinity Bil 4 A long wire carries a steady current. It is bent into a circle of one turn and the magnetic field at the centre of the coil is B. It is then bent into a circular loop of n turns. The magnetic field at the centre of the coil will be (a) nB (b) n2B (c) 2nB (d) 2n2B A magnetic needle lying parallel to a magnetic field requires W units of work to turn it through 60°. The torque needed to maintain the needle in this position will be (a) 3W (b) W  3 (c)  (d) 2W W  2  The self inductance of the motor of an electric fan is 10 H. In order to impart maximum power at 50 Hz, it should be connected to a capacitance of (a) 1 mF (b) 2 mF (c) 4 mF (d) 8 mF

10. An alternating current is given by i = i1 coswt + i2 sinwt. The rms current is given by i +i |i + i | (a) 1 2 (b) 1 2 2 2 (c)

i12 + i22 2

(d)

i12 + i22 2

physics for you | AUGUST ‘15

57

11. If the wavelengths of light used in an optical instrument are l1 = 4000 Å and l2 = 5000 Å, then ratio of their respective resolving powers (corresponding to l1 and l2) is (a) 16 : 25 (b) 9 : 1 (c) 4 : 5 (d) 5 : 4 12. Refractive index of glass is 1.520 for red light and 1.525 for blue light. Let D1 and D2 be angles of minimum deviation for red and blue light respectively in a prism of this glass. Then (a) D1 > D2 (b) D1 < D2 (c) D1 = D2 (d) D1 can be less than or greater than depending upon angle of prism 13. The maximum number of possible interference maxima for slit separation equal to twice the wavelength in Young's double slit experiment is (a) infinite (b) five (c) three (d) zero 14. The work function of a substance is 4.0 eV. The longest wavelength of light that can cause photoelectron emission from this substance is approximately (a) 540 nm (b) 400 nm (c) 310 nm (d) 220 nm 15. A proton when accelerated through a potential difference of V volts has a wavelength l associated with it. An a particle in order to have the same wavelength l, must be accelerated through a potential difference (in volts) V V (d) 8 4 16. Frequency of the series limit of Balmer series of hydrogen atom in terms of Rydberg constant R and speed of light c is Rc 4 (a) Rc (b) 4Rc (c) (d) 4 Rc (a) 2V

(b) V

(c)

17. In the following nuclear reaction, how many a and b particles are emitted? → 82 Pb206 (a) 8a, 6b (c) 8a, 8b 92 U

238

(b) 6a, 10b (d) 12a, 6b

18. The circuit shown in figure contains two diodes, each with a forward resistance of 50 W and with infinite reverse resistance. If the battery voltage is 6 V, the current through 100 W resistance is 58

physics for you | AUGUST ‘15

150  50 

6V

(a) zero

100 

(b) 0.02 A (c) 0.03 A (d) 0.036 A

19. Which of the following statements is not true for zener diode? (a) It is used as a voltage regulator. (b) It is fabricated by heavily doping both p and n sides of the junctions. (c) It depletion region is very thin. (d) The electric field of the junction is very low. 20. For an amplitude modulated wave the maximum amplitude is found to be 10 V while the minimum amplitude is found to be 2 V, the modulation index is 3 2 (a) 5 (b) 0.2 (c) (d) 2 3 21. The dimension of magnetic field in M, L, T and C (coulomb) is given as (a) MLT–1C–1 (b) MT–2C–1 2 –2 (c) MT C (d) MT–1C–1 22. A particle located at x = 0 at time t = 0, starts moving along the positive x-direction with a velocity v that varies as v = a x where a is dimensionless constant. The displacement of the particle varies with time as (a) t3 (b) t2 (c) t (d) t1/2 23. From a building two balls A and B are thrown such that A is thrown upwards and B downwards with the same speed (both vertically). If vA and vB are their respective velocities on reaching the ground then, (a) vB > vA (b) vA = vB (c) vA > vB (d) their velocities depend on their masses 24. A shell fired from a gun at sea level rises to a maximum height of 5 km when fired at a ship 20 km away. The muzzle velocity should be (a) 7 m/s (b) 14 m/s (c) 28 m/s (d) 56 m/s 25. Two racing cars of masses m1 and m2 are moving in circles of radii r1 and r2 respectively. Their speeds are such that each makes a complete circle in the same time t. The ratio of the angular speeds of the first to the second car is (a) r1 : r2 (b) m1 : m2 (c) 1 : 1 (d) m1m2 : r1r2

26. A block of mass M is pulled along a horizontal frictionless surface by a rope of mass m. If a force P is applied at the free end of the rope, the force exerted by the rope on the block is Pm Pm (a) (b) (M − m) (M + m) PM (M + m) 27. A man fires a bullet of mass 200 g at a speed of 5 m/s. The gun is of one kg mass. By what velocity the gun rebounds backward? (a) 1 m/s (b) 0.01 m/s (c) 0.1 m/s (d) 10 m/s (c) P

(d)

28. A man weighs 80 kg. He stands on a weighing scale in a lift which is moving upwards with a uniform acceleration of 5 m/s2. What would be the reading on the scale? (g = 10 m/s2) (a) zero (b) 400 N (c) 800 N (d) 1200 N 29. A wire of length 100 cm is connected to a cell of emf 2 V and negligible internal resistance. The resistance of the wire is 3 W, the additional resistance required to produce a potential difference of 1 mV/cm on the wire is (a) 297 W (b) 60 W (c) 57 W (d) 35 W 30. The only force acting on a 2.0 kg body as it moves along a positive x-axis has an x-component Fx = –6x with x in meters. The velocity at x = 3.0 m is 8.0 m/s. The velocity of the body at x = 4.0 m is (a) 6.6 m/s (b) 46.6 m/s (c) 60 m/s (d) 96.6 m/s 31. A quarter horse power motor runs at a speed of 600 rpm. Assuming 40% efficiency, the work done by the motor in one rotation will be (a) 7.46 J (b) 74.6 J (c) 7400 J (d) 7.46 erg 32. A rocket is launched vertically upward from the surface of the earth with an initial velocity of 10 km/s. If the radius of the earth is 6400 km and atmospheric resistance is negligible, find the distance above the surface of the earth that the rocket will go. (a) 2.5 × 104 km (b) 3.0 × 104 km 3 (c) 4.0 × 10 km (d) 3.0 × 103 km 33. Consider a two particle system with particles having masses m1 and m2. If the first particle is pushed towards the centre of mass through a distance d, by what distance should the second particle be moved, so as to keep the centre of mass at the same position.

m2 d m1 m1 d (d) m2

(a) d (c)

(b)

m1 d m1 + m2

34. A body of mass M while falling vertically downwards under gravity breaks into two parts ; a 1 2 body B of mass M and body C of mass M. The 3 3 centre of mass of bodies B and C taken together shifts compared to that of body A towards (a) body C (b) body B (c) does not shift (d) depends on height of breaking 35. A metre stick is balanced on a knife edge at its centre. When two coins, each of mass 5 g are put one on top of the other at the 12.0 cm mark, the stick is found to be balanced at 45.0 cm. The mass of the metre stick is (a) 13 g (b) 33 g (c) 66 g (d) 77 g 36. A particle performing uniform circular motion has angular momentum L. If its angular frequency is doubled and its kinetic energy halved, then new angular momentum is L L (b) 2L (c) 4L (d) 4 2 37. Suppose the gravitational force varies inversely as the nth power of distance. Then the time period of a planet in circular orbit of radius R around the sun will be proportional to (a)

(a) R (c)

 n + 1  2 

(b) R

Rn

(d) R

 n − 1  2 

n − 2  2 

38. A geo-stationary satellite is in an orbit of radius 36000 km. Approximately what would be the time period of a spy satellite orbiting a few hundred kilometers above the surface of the earth? (Earth radius = 6400 km) (a) One hour (b) Two hour (c) Four hour (d) Eight hour 39. If S is the stress and Y is the Young's modulus of material of a wire, the energy stored in the wire per unit volume is (a)

2Y S

(b)

S 2Y

(c) 2S2Y

(d)

physics for you | AUGUST ‘15

S2 2Y

59

40. Two identical cylindrical vessels with their bases at the same level, each contains a liquid of density 1.3 × 103 kg/m3. The area of each base is 4.00 cm2, but in one vessel, the liquid height is 0.854 m and in the other it is 1.560 m. Find the work done by the gravitational force in equalizing the levels when the two vessels are connected. (a) 0.0635 J (b) 0.635 J (c) 6.35 J (d) 63.5 J 41. A wire extends by 1 mm when a force is applied. Double the force is applied to another wire of the same material and length but half the radius of cross-section. The elongation of the wire in mm will be (a) 8 (b) 4 (c) 2 (d) 1 u 42. The energy density of an ideal gas is related to V its pressure P as u u 3 = 3P = P (a) (b) V V 2 u 2 u 1 = P = P (c) (d) V 3 V 3 43. One kg of a diatomic gas is at a pressure of 8 × 104 N/m2. The density of the gas is 4 kg/m3. What is the energy of the gas due to its thermal motion? (a) 3 × 104 J (b) 5 × 104 J 4 (c) 6 × 10 J (d) 7 × 104 J 44. A refrigerator is to maintain eatables at 9°C. If room temperature is 36°C, then the coefficient of performance is (a) 8.6 (b) 10.4 (c) 11.2 (d) 12.5 45. According to Newton's law of cooling, the rate of cooling of a body is proportional to (Dq)n, where Dq is the difference of the temperature of the body and surroundings, and n is equal to (a) four (b) three (c) two (d) one 46. Two spheres of the same material have radii 1 m and 4 m and temperatures 4000 K and 2000 K respectively. The ratio of energy radiated per second by the first sphere to the second is (a) 1 : 1 (b) 16 : 1 (c) 4 : 1 (d) 1 : 9 47. A whistle producing sound waves of frequency 9500 Hz and above is approaching a stationary person with speed v m/s. The velocity of sound in air is 300 m/s. If the person can hear frequencies upto a maximum of 10000 Hz, the maximum value of v upto which he can hear the whistle is 60

physics for you | AUGUST ‘15

(a) 30 m/s (b) 15 2 m/s 15 m/s (c) (d) 15 m/s 2 48. The displacement of an object attached to a spring and executing simple harmonic motion is given by x = 2 × 10–2 cos pt metre. The time at which the maximum speed first occurs is (a) 0.25 s (b) 0.50 s (c) 0.75 s (d) 0.125 s 49. Two concentric spherical shells of radii r1 and r2 have similar charges and equal surface charge densities (s). What is the potential at the common centre? s s (r − r ) (r1 + r2 ) (a) (b) ∈0 1 2 ∈0 (c)

s r12 ∈0 r2

(d)

s ∈0

r22 r1

  A  50. The electric field in a region is given by E =   i .  x3  An expression for the potential in the region, assuming the potential at infinity to be zero, is A 2x 2 (a) (b) 2x 3 A (c)

A2 2x

(d)

A 2x 2

solutions 1. (c) : Let r be radius of the drop. Then the mass of the drop is 4 m = volume × density = pr 3r 3 Since the drop has n excess electrons, therefore it is negatively charged and the charge on the drop is q = ne As the drop is held stationary under the constant electric field E, \ weight of the drop = force on the drop due to electric field mg = qE 4 3 + + + + + + pr rg = neE qE 3 3 3neE – r = E 4prg mg 1/3

 3neE  r=   4prg 

2. (a) : The electric field E and potential V at a point on the x-axis are related as dV E=− dx 20 Here, V (x) = 2 volt = 20(x 2 − 4)−1 volt (x − 4) d 2 d −1 20 (x 2 − 4)−1 = − 20 (x − 4) dx dx 40x = – 20(–1)(x2 – 4)–2 (2x) = 2 (x − 4)2 At x = 4 mm 40(4) 160 10 E= = = volt/mm 2 2 144 9 ( 4 − 4) \ E=−

Thus the electric field E at x = 4 mm is (10/9) volt/mm and in the +ve x direction. 3. (d) 4. (b) : Let R be resistance of the wire. When it is connected to 220 V mains supply, then power dissipation is  (220 V)2 V2  ...(i) P1 =  P =  R R   When the wire is cut into two equal pieces, the resistance of each piece becomes R/2. Now they are connected in parallel, their equivalent resistance is RR     R 2 2 = Req = R R 4 + 2 2 As this parallel combination is connected to the same supply, so power dissipation is P2 =

(220 V)2 Req

=

(220 V)2 (R/4)

...(ii)

Dividing eqn. (ii) by eqn. (i), we get (220 V)2 P2 (R/4) = =4 P1 (220 V)2 R

5. (a) : Let R1 and R2 be the two given resistances. Then RR S = R1 + R2 and P = 1 2 R1 + R2 But S = nP (given) RR \ R1 + R2 = n 1 2 R1 + R2

(R1 + R2)2 = nR1R2 R12 + R22 + 2R1R2 = nR1R2 (R1 – R2)2 + 4R1R2 = nR1R2 (R1 – R2)2 = R1R2(n – 4) For n to be minimum, R1 = R2, \ n – 4 = 0 or n = 4 6. (c) : Area of the loop is 1 A = × base × height 2 1 = × l × l sin 60° 2

l

i 60°

i i

l

l

1 3 3 2 ...(i) = ×l×l = l 2 2 4 As the loop carries current i and placed in a magnetic field B such that plane of the loop is in the direction of B i.e. q = 90°, so the torque on the loop is t = iABsin90° = iAB = i =

3 2 l B 4

(using (i))

3 2 Bil 4

7. (b) : Let l be the length of the wire carrying steady current I. When it is bent into the circle of one turn of radius r, then l ...(i) 2pr = l or r = 2p \ The magnetic field at the centre of the coil is m I m0 I m Ip (using (i)) ...(ii) B= 0 = = 0  l  l 2r 2  2p  When the same wire is bent into the circular loop of n turns of radius r′, then l ...(iii) n(2pr ′) = l or r ′ = 2pn Now the magnetic field at the centre of the coil is m nI m0nI n2m0 I p B′ = 0 = = (using (iii))  l  l 2r ′ 2   2pn  = n2 B (using (ii)) 8. (a) : The work done to turn the magnetic needle of magnetic moment M lying parallel to a magnetic field B through an angle q is W = MB(1 – cosq) physics for you | AUGUST ‘15

61

Here, q = 60°

 1  MB \ W = MB (1 − cos 60°) = MB 1 −  =  2 2 or MB = 2W ...(i) The torque needed to maintain the needle in this position is t = MB sin q = MB sin 60° =

3 3 MB = (2W ) = 3W 2 2

(using (i))

9. (a) : Maximum power is imparted at resonance frequency. 1 As resonance frequency ur = 2p LC 1 \ C= 2 2 4p ur L Here, ur = 50 Hz, L = 10 H 1 \ C= 2 4 p (50 Hz)2 (10 H) 1 = 5 2 F = 10−6 F = 1 mF 10 p 10. (d) : The alternating current is i = i1coswt + i2sinwt The rms current is T 2

∫0 i dt T ∫0 dt

irms =

=

1 T (i cos wt + i2 sin wt )2 dt T ∫0 1

1 T 2 (i cos2 wt + i22 sin2 wt = T ∫0 1 + 2i1i2 cos wt sin wt )dt i12 T i22 T 2 2 + cos w sin wt dt t dt ∫ T ∫0 = T 0 2i i T + 1 2 ∫ cos wt sin wt dt T 0 T

T wt dt = , 2

∫0 cos

and

∫0 cos wt sin wt = 0

T

\ irms = = 62

2

But

T

∫0 sin

2

wt dt =

T 2

i12  T  i22  T  2i1i2 (0)  +  + T 2 T 2 T i12 + i22 i2 + i2 i12 i22 + = 1 2 = 2 2 2 2

physics for you | AUGUST ‘15

1 wavelength (l) 5000 Å 5 = = 4000 Å 4

11. (d) : As resolving power (RP) ∝ \

RP for l1 l2 = RP for l2 l1

12. (b) : The angle of minimum deviation is D = (m – 1)A where m is the refractive index of the material of prism and A is the angle of the prism. If mr and mb be refractive indices of glass for red and blue light respectively, then D1 = (mr – 1)A and D2 = (mb – 1)A D1 (mr − 1) \ = D2 (mb − 1) As mr = 1.520 and mb = 1.525 (given) D1 (1.520 − 1) 0.520 \ = = 0 [Q 0° < qA < 90°] At B, slope mB = tanqB < 0 [Q 90° < qB < 180°] Also remember that tanq is a non decreasing function within its continuous limits, which means if q increases, tanq will also increase. Following are examples of increasing and decreasing slopes. Increasing slope y

y

1

2 1

2

2

1 (A)

2

1

x

(B)

x

Clearly, if q2 > q1 ⇒ tanq2 > tanq1 in both cases. But note that in case B, tanq < 0, hence |tanq1| > |tanq2|. Decreasing slope y

y

2

1

1

2

1

2

x

2

1

x

Here q2 < q1 ⇒ tanq2 < tanq1 Therefore, if acceleration is negative, it means velocity decreses, not necessarily speed. Imagine velocity to be changing from –5 m s–1 to –6 m s–1, speaking mathematically velocity has decreased, since Dv = –6 – (–5) = –1 m s–1, hence acceleration is negative but speed increased. Keep in mind that if velocity and acceleration have same sign (in same direction) then speed will increase. Synopsis of learning

€ Slope of position-time graph at any point gives us instantaneous velocity. 80

physics for you | august ‘15

€ Slope of velocity-time graph gives instantaneous acceleration. € Area of velocity-time graph gives – distance travelled if area above and below time axis are added. – displacement if area below time axis is subtracted from the area above it. € Area under acceleration time graph gives us the change in velocity. Suppose the acceleration is known to be constant, say ‘a’ m s–2. Then we can use standard formulae to find various quantities related to motion as derived below: Assumptions: Initial position at t = 0, xi = x0 Initial velocity at t = 0, vi = u Final position t sec later, xf = x Final velocity t sec later, vf = v Displacement, s = xf – xi = x – x0 Results: v t dv a= ⇒ ∫ dv = ∫ adt dt u 0 but since acceleration a is a constant, we can take it out of integration. \ v – u = at ⇒ v = u + at ... (i) x t dx Again, v = ⇒ ∫ dx = ∫ (u + at )dt dt x0 0 1 ⇒ x − x0 = ut + at 2 2 1 2 ... (ii) ⇒ s = ut + at 2 Note here that s indicates displacement, which may or may not be equal to distance travelled. v x dv Again, a = v ⇒ ∫ vdv = ∫ adx dx u x 0

v 2 u2 ⇒ − = a(x − x0 ) = as 2 2 ⇒ v2 = u2 + 2as ...(iii) Each of the equation derived above (i), (ii) and (iii) are applicable only for constant acceleration. If the acceleration is not constant, and we are given a relation between v, x, t and a to describe the motion and one or the other is to be found out then we use either of the three relations dv dx vdv (i) a = (ii) v = (iii) a = dt dt dx will help us, as used, while deriving equations (i), (ii) and (iii).

Now, let us begin with the applications of whatever we learnt. Graphical Application

1. The position of a particle moving along x-axis varies as below. x (m) 15 10 5

10

t (s)

Find (i) distance travelled and displacement in the interval 0 to10 second (ii) average acceleration in 0 to10 second interval Soln.: (i) As long as the position increases, particle is travelling in +ve x-direction but as soon as the position starts decreasing (after 5 s), the direction of motion has reversed, and continues to move in –x direction. Hence distance ≠ displacement. \ Distance travelled = |x(5) – x(0)| + |x(10) – x(5)| = 5 + 15 = 20 m Displacement = x(10) – x(0) = 0 – 10 = –10 m v(10) − v(0) (ii) aavg = 10 − 0  dx   dx  −15 5  dt  −  dt  − 5 10 0 = = 5 10 10 = – 0.4 m s–2 2. The position of a particle moving along x-axis is given by x(t) = 3t2 – 12t + 6, where t in seconds gives x in metres. Find the distance travelled by particle in the duration (0 - 6) s. Soln.: One incorrect method to do this would be that distance = x(6) – x(0) Since we don’t know whether the particle changed its direction of motion or not. To determine this always remember that at the instant when it will change the direction of motion, its instantaneous velocity will definitely be zero. Here let us find velocity first. dx v= = 6t − 12 dt \ If v = 0 ⇒ t = 2 s, hence at t = 2 s it has changed its direction. \ Distance travelled = |x(2) – x(0)| + |x(6) – x(2)|

= |(3(2)2 – 12(2) + 6 – (6)| + |(3(6)2 – 12(6) + 6) – (3(2)2 – 12(2) + 6)| = |–6 – 6| + |42 – (–6)| = 12 + 48 = 60 m Alternatively, this could have been done quickly v(m s–1) from v-t graph. v = 6t – 12 (linear) 24 v6 tan q = =6 v6 6−2 A2 2 ⇒ v6 = 24 m s −1 6 A1 t(s) \ Distance travelled –12 = A1 + A2 1 1 = (2 × 12) + (4 × 24) = 60 m 2 2 3. The velocity-time graph of a particle moving along x-axis is shown in the figure. v (m s–1) 10 15

t(s)

10

Find the instant of time when the particle reaches the initial position which was at t = 0. Soln.: Since xf = xi ⇒ Dx = 0 ⇒ area under v-t graph = 0 v (m s–1) tc

10 A1 10

45°

15 5 A2

t(s)

A1 = A2 1 1 ⇒ × 10 × 10 = (Dtc + Dtc + 5)(5) ⇒ Dtc = 7.5 s 2 2

\ Required time = 15 + 7.5 = 22.5 s

4. The acceleration-time graph of a particle moving along x-axis is shown. If the particle starts moving with a speed of 6 m s–1 along negative x-direction then find its velocity at t = 8 s. a (m s–2)

4

t(s)

physics for you | august ‘15

81

Soln.: Th e graph clearly shows that

tan 37 

 (m s–2)

 y 2 1 4

 

7. If the v-x graph of a particle moving along x-axis varies as shown. Predict the nature of acceleration.

A1 = A2  v = 0  v2 – v1 = 0  v2 = v1 = –6 m s–1



5. The acceleration-time graph of a particle is shown which starts from origin from rest. Find the distance travelled by it till the journey of 15 s.  (m s–2)

5 10

 

1 1 15 170 85  t  10    10    s 2 2 2 4 2



8

15 2

3 y  4 10

15

5

(s)

–5



dv = v × slope of v-x graph dx Since v increases but slope decreases, therefore, we cannot predict with certainty about the nature. It might be constant, increasing or decreasing as well.

Soln.: We have a  v

8. If the position (x) varies with respect to time (t) as shown in the figure. Justify whether force acts in the direction of motion or not? 

Soln.: The corresponding v-t graph is as follows  (m s–1)  0 =  = 25

 5

10

(s)

15

 Distance travelled = A 1  (15  5)(25)  250 m 2 6. (1/v) is plotted against x for a particle moving along x-axis. Find the time taken to move from x = 0 to x = 10 m? (v is in m s–1 .)

Soln.: If force acts in the direction of motion speed will increase, hence acceleration and velocity, both should either be +ve or both –ve. Slope < 0  v < 0 and v2 < v1  a < 0  Both are negative. Hence force acts in the direction of motion.    



   

Soln.: The area under the graph 1 1   dx    dx   dt  t v (dx / dt ) 82

PHYSICS FOR YOU | AUGUST ‘15

The greatest enemy of knowledge is not ignorance, it is illusion of knowledge. -Stephen Hawking

J & K CET

...contd. from page 22

51. (d) : For transmission frequency, u = 1 MHz = 106 Hz

c 3 × 108 m/s = = 300 m u 106 Hz To transmit a signal the length of an antenna is comparable to the wavelength of the signal (atleast l/4). Thus the length of the antenna is 300 m. 52. (b) : Both transverse and longitudinal waves propagate in a solid. 53. (c) : Before jumping the reading of the weighing scale on which the person is standing is 60 kg and after jumping on it, the reading goes to 70 kg. If a be his maximum upward acceleration, then (70 kg)g = (60 kg)g + (60 kg)a (70 kg)g − (60 kg)g \ a= 60 kg The wavelength is l =

=

g 9.8 m/s2 = = 1.63 m/s2 6 6

54. (b) : If N0 is the original amount of a radioactive material, then the amount of substance left after n half-lives is n n N 1 1 N = N 0   or =  2 N0  2  Here, n = 3 \

3

N 1 1 1 = = = × 100 = 12.5% N 0  2  8 8

55. (c) 56. (c) : When a parallel narrow beam of light is falling normally on a glass sphere, it will come to focus outside the sphere as shown in the figure. 57. (b) : 25 VSD = 24 MSD 24 ...(i) 1 VSD = MSD 25 The least count of vernier callipers is LC = 1 MSD – 1 VSD 24 1 = 1 MSD − MSD = MSD (using (i)) 25 25 As 1 MSD = 0.5 mm (given) 1 \ LC = (0.5 mm) = 0.02 mm = 0.002 cm 25

58. (c) 59. (d) : The energy of incident light is E = hu where h is the Planck's constant and u is the frequency of incident light. As uviolet > ublue > ugreen > ured \ Eviolet > Eblue > Egreen > Ered Since the incident energy is maximum for violet colour, therefore violet light produces photoelectrons. 60. (a) : Since the phase difference is constant therefore angular velocity is same for both pendula. Let it be w. If A be amplitude of each pendulum, then the maximum velocity of one of them is v = Aw ... (i) and that of other is v + x = Aw ... (ii) Substituting the value of v from eqn. (i) in eqn. (ii), we get Aw + x = Aw or x = 0 nn Refer to MTG J&K CET Explorer for complete solutions solution of July 2015 crossword C O L L I D 2E 3R A 4D I I Y 5 N P L U T O N 7 S C R E E 10 S E X T 11A N T T 12 E T D I G L E I 15 GA L V A N I S M 17 O S F 19 R E A 20 21 22 R P G I L B E R T A 24 R M O T D C 25 G T R O C H I L I L L O S R A 28 N CO R E M E E A W B E T I T 29 30 HO L E FO C I M E N R 1

U M 6

N

8

T E I T R O A 16 P R E 23 C E L I C S I O N T R Y

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Winners (July 2015) Pratyusha Das (Odisha)

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ravinder singh (Punjab)



Arnab Banerjee (West Bengal)



Vivek Giri (Maharashtra)

C H A R G E R N 13 T R 14A C M 18 A P L Y E R I T Y O H 26 E L V E 27S O A P A G R Y K

E B O R E

9

solution senders (June 2015) 

Parth Kapoor (Delhi)



saundarya P (Kerala) physics for you | august ‘15

83

Y U ASK

WE ANSWER 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 is the sky dark at night? Why is there no light due to diffraction from the Earth’s surface? – Pooja Yadav (UP)

Ans. Light from the illuminated side of the Earth is not scattered by the atmosphere to a significant extent, so as to illuminate the side darkened by night. This does happen at dawn and dusk, the land is bright even before sunrise and after sunset. But as the Earth turns and the Sun dips further away, this light disappears. Our atmosphere does not extends very far (less than a hundred kilometres). So there is nothing above us that could receive the scattered light from the sunlit side. The situation is somewhat different near the polar regions. During the summer months, the Sun does not dip much below the horizon and the night stays illuminated by either direct or scattered light. Therefore, mid-summer nights are never complete dark at the poles, sunlight or a scattered twilight keeps the land or the sea bright. Q2. Many roadway construction sites have flashing yellow lights to warn motorists of possible dangers. What causes the lightbulbs to flash? – Divya Nair (Kerala)

Ans. A typical circuit for such R a flasher is shown in the  L C figure. The lamp L is a gasfilled lamp that acts as an k open circuit until a large potential difference causes an electric discharge in the gas, which gives off a bright light. During this discharge, charge flows through the gas between the electrodes of the lamp. After switch k is closed, the battery charges up the capacitor of capacitance C. At the beginning, the current is high and the 84

physics for you | August ‘15

charge on the capacitor is low, so that most of the potential difference appears across the resistance R. As the capacitor charges, more potential difference appears across it, reflecting the lower current and, thus, lower potential difference across the resistor. Eventually, the potential difference across the capacitor reaches a value at which the lamp will conduct, causing a flash. This discharges the capacitor through the lamp and the process of charging begins again. The period between flashes can be adjusted by changing the time constant of the RC circuit. Q3. If you sit in front of a fire with your eyes closed, you can feel significant warmth in your eyelids. If you now put on a pair of eyeglasses and repeat this activity, your eyelids will not feel nearly so warm. Why?

– Subash Basu (WB)

Ans. Much of the warmth you feel is due to electromagnetic radiation from the fire. A large fraction of this radiation is in the infrared part of the electromagnetic spectrum. Your eyelids are particularly sensitive to infrared radiation. On the other hand, glass is very opaque to infrared radiation. Thus, when you put on the glasses, you block much of the radiation from reaching your eyelids, and they feel cooler. Q4. If an orchestra does not warm up before a performance, the strings go flat and the wind instruments go sharp during the performance. Why ? – Shruti Roy (New Delhi)

Ans. Without warming up, all the instruments will be at room temperature at the beginning of these concert. As the wind instruments are played, these are filled with warm air from the player’s exhalation. The increase in temperature of the air in the instrument causes an increase in the speed of sound, which raises the resonance frequencies of the air columns. As a result, the wind instruments go sharp. The temperature of the strings on the stringed instruments also increase due to the friction of it rubbing with the bow. This results in thermal expansion, which causes a decrease in the tension in the strings. With a decrease in tension, the wave speed on the strings drops, and the fundamental frequencies decrease. Thus, the stringed instruments go flat. nn

Readers can send their responses at [email protected] or post us with complete address by 25th of every month to win exciting prizes. Winners’ name with their valuable feedback will be published in next issue.

across

6.

12. 14. 16. 17. 19. 24. 25. 26. 27. 28. 29. 30.

2. 3. 4. 5. 7. 8. 9.

2

A piece of solid matter which retains a permanent electron polarization like a permanent magnet. (8) The measurement of force used in doing work. (11) An instrument for measuring the bulk flow of plasma in space. (5,5) Any material that stops ionizing radiation. 17 (8) The science, technology and application of 22 nuclear energy. (10) A phenomenon in which wave velocity changes with its wavelength. (10) The transformation of one element into another by a nuclear reaction. (13) A particle postulated to move with velocity 28 greater than that of electromagnetic radiation. (7) The movement of one atomic plane over another in a crystal.(5) A telescope designed to observe the outer portions of the solar atmosphere. (11) The abbreviated term for an occurrence of atmospheric radio noise. (6) A unit of nuclear cross section. (4) The ratio of charge stored per increase in potential difference. (11)

Down

1.

Cut Here 1

Particle that carries or transmits the weak interaction of force. (4,5) A combination of two or more tones that sound unpleasant together. (10) A reference equipotential surface around a planet where the gravitational potential energy is defined to be zero. (5) A hypothetical entity postulated as the building blocks of leptons and quarks. (5) Acronym for microwave amplification by stimulated emission of radiation. (5) Dates on which the day and night are of equal length. (7) A term used in the study of water waves on a free surface, such as waves on the surface of the ocean. (8) A thin bar of metal or cane clamped at one end and set into transverse vibration generally by a flow of air. (4)

3

4

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7

8

5

9

10

12 13 14

15 16

18

19 20 21 23

24 25 26

27 29 30

10. The first legal unit of voltage for the United States was based on this cell. (5,4) 11. The determination of the absolute values of arbitrary indications of an instrument. (11) 13. A meteoroid that explodes or breaks up during its passage through the atomsphere as a meteor. (6) 15. An electronic instrument that indicates, on a meter, the number of radiation induced pulses per minutes from radiation detectors. (4,5) 18. The study of technology of optical instruments and apparatus. (12) 20. The fraction of the incident power at a prescribed wavelength that is scattered within a volume. (11) 21. A device to accelerate particles in which particles move in circular path. (11) 22. The area of study that focuses on precise measurements of the positions and movements of stars and other celestial bodies. (10) 23. The production of photographs of the internal structure of bodies, opaque to visible light by the radiation from X-rays or gamma rays from radioactive substances. (11)  Physics for you | august ‘15

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Physics for you | august ‘15