Physics for You - April 2016

Volume 24 Managing Editor Mahabir Singh

CONTENTS

Editor Anil Ahlawat (BE, MBA)

No. 4

April 2016

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Physics Musing Problem Set 33

8

AIPMT Practice Paper

12

Core Concept

21

AIIMS Special : Assertion & Reason

28

JEE Advanced Practice Paper

31

Thought Provoking Problems

40

Olympiad Problems

44

Brain Map

46

Exam Prep 2016

48

AIPMT Model Test Paper 2016

54

CBSE Board Solved Paper 2016

63

BITSAT Practice Paper

72

Physics Musing Solution Set 32

79

Live Physics

83

You Ask We Answer

81

Crossword

85

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PHYSICS FOR YOU | APRIL ‘16

7

P

hysics Musing was started in August 2013 issue of Physics For You with the suggestion of Shri Mahabir Singh. The aim of Physics Musing is to augment the chances of bright students preparing for JEE (Main and Advanced) / AIIMS / Other PMTs with additional study material. In every issue of Physics For You, 10 challenging problems are proposed in various topics of JEE (Main and Advanced) / various PMTs. The detailed solutions of these problems will be published in next issue of Physics For You. The readers who have solved five or more problems may send their 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.

SINGLE OPTION CORRECT TYPE

1. A spherical insulator of radius R is charged uniformly with a charge Q throughout its volume Q located at its centre. and contains a point charge 16 Which of the following graphs best represents qualitatively, the variation of electric field intensity E with distance r from the centre?

(b)

(a)

(c)

(d)

(c) σ(r2 – r1) B

(a) zero

(b)

(d) σ r22 − r12 B

3. Two long straight cylindrical conductors with resistivities ρ1 and ρ2 respectively are joined together as shown in figure. The radius of each of the conductor is a. If a uniform total current I flows through the conductors, the magnitude of the total free charge at the interface of the two conductors is

(ρ1 − ρ2 ) I ε0

2 (d) ε0I(ρ1 + ρ2 )

(c) ε0I(ρ1 – ρ2)

4. In which material do the conduction electrons have the largest mean time between collisions? (a) Copper (b) Aluminium (c) Nichrome (d) Tungsten 5. A capacitor of capacitance 5 μF is connected to a source of constant emf of 200 V for a long time, then the switch was shifted to contact 1 from contact 2. The amount of heat generated in the 500 Ω resistance is H. Find 3200 H (in joule).

2. A particle of specific charge σ (q / m) moving with a certain velocity v enters a uniform magnetic field of strength B directed along the negative Z-axis extending from x = r1 to x = r2. The minimum value of v required in order that the particle can just enter the region x > r2 is (a) σr2B (b) σr1B

(a) 800 J (c) 200 J

(b) 400 J (d) 100 J

ONE OR MORE OPTION CORRECT TYPE 6. Figure shows three spherical shells in separate situations, with each shell having the same uniform positive net charge. Points 1, 4 and 7 are at the same radial distances from the centre of their respective shells; so are points 2, 5 and 8 ; and so are points 3, 6 and 9. With the electric potential taken equals to zero at an infinite distance, choose correct statement(s).

By Akhil Tewari, Author Foundation of Physics for JEE Main & Advanced, Senior Professor Physics, RAO IIT ACADEMY, Mumbai.

8


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

Point 3 has highest potential. Point 1, 4 and 7 are at same potential. Point 9 has lowest potential. Point 5 and 8 are at same potential.

7. A simple harmonic oscillator consists of a mass sliding on a frictionless surface, attached to an ideal spring. Choose the correct statement(s). (a) Quadrupling the mass will double the period. (b) Doubling the amplitude will change the frequency. (c) Doubling the amplitude will double the total energy of the system. (d) Doubling the amplitude will quadruple the total energy of the system.

COMPREHENSION TYPE For questions 8, 9 and 10 A small caterpillar crawls in the direction of electron drift along bare copper wire that carries a current of 2.56 A. It travels with the drift speed of the electron in the wire of uniform cross section area 1mm2. Number of free electrons for copper = 8 × 1022 cc–1 and resistivity of copper = 1.6 × 10–8 Ω m. 8. How much time would the caterpillar take to crawl 1.0 cm if it crawls at the drift speed of the electrons in the wire? (a) 50 s (b) 5 s (c) 5000 s (d) None of these 9. What is the order of the average time of collision for free electrons of copper? (a) 10–14 s (b) 10–16 s –11 (c) 10 s (d) 10–8 s 10. If the caterpillar starts from the point of zero potential at t = 0, it reaches a point of _____ potential after 10 s. (a) 80 μV (b) –80 μV (c) 160 μV (d) –160 μV 

NASA Astronaut Scott Kelly Returns Safely to Earth after One-Year Mission

N

ASA astronaut and Expedition 46 Commander Scott Kelly and his Russian counterpart Mikhail Kornienko returned to Earth Tuesday after a historic 340-day mission aboard the International Space Station. They landed in Kazakhstan at 11:26 p.m. EST (10:26 a.m. March 2 Kazakhstan time). “Scott Kelly’s one-year mission aboard the International Space Station has helped to advance deep space exploration and America’s Journey to Mars,” said NASA Administrator Charles Bolden. “Scott has become the first American astronaut to spend a year in space, and in so doing, helped us take one giant leap toward putting boots on Mars.” During the recordsetting One-Year mission, the station crew conducted almost 400 investigations to advance NASA’s mission and benefit all of humanity. Kelly and Kornienko specifically participated in a number of studies to inform NASA’s Journey to Mars, including research into how the human body adjusts to weightlessness, isolation, radiation and the stress of long-duration spaceflight. Kelly’s identical twin brother, former NASA astronaut Mark Kelly, participated in parallel twin studies on Earth to help scientists compare the effects of space on the body and mind down to the cellular level. One particular research project examined fluid shifts that occur when bodily fluids move into the upper body during weightlessness. These shifts may be associated with visual changes and a possible increase in intracranial pressure, which are significant challenges that must be understood before humans expand exploration beyond Earth’s orbit. The study uses the Russian Chibis device to draw fluids back into the legs while the subject’s eyes are measured to track any changes. NASA and Roscosmos already are looking at continuing

10


the Fluid Shifts investigation with future space station crews. The crew took advantage of the unique vantage point of the space station, with an orbital path that covers more than 90 percent of Earth’s population, to monitor and capture images of our planet. Kelly and Kornienko saw the arrival of six resupply spacecraft during their mission. Kelly ventured outside the confines of the space station for three spacewalks during his mission. The International Space Station is a convergence of science, technology and human innovation that enables us to demonstrate new technologies and make research breakthroughs not possible on Earth. It has been continuously occupied since November 2000 and, since then, has been visited by more than 200 people and a variety of international and commercial spacecraft. For more information about the one-year mission, visit: http://www.nasa.gov/oneyear 

*K P Singh

1. An electron enters the space between the plates of a charged parallel plate capacitor as shown in the figure. The charge density on the plate is σ. Electric intensity in the space between the plates is E. A uniform magnetic field B also exists in the space perpendicular to the direction of E. The electron moves perpendicular to both E and B without any change in direction. The time taken by the electron to travel a distance L in the space is (a)

σL ε0 B

(b)

σB ε0 L

(c)

ε0 LB ε L (d) 0 σ σB

2. Two capacitors of capacitance 2 μF and 4 μF respectively are connected in series. The combination is connected across a potential difference of 10 V. The ratio of energies stored by capacitors will be (a) 1 : 2

(b) 2 : 1

(c) 1 : 4

(d) 4 : 1

3. An unknown resistance R1 is connected in series with a resistance of 10 Ω. This combination is connected to one gap of meter bridge while, a resistance R2 is connected in the other gap. The balance point is at 50 cm. Now, when the 10 Ω resistance is removed the balance point shifts to 40 cm. The value of R1 (in ohm) is (a) 20

(b) 10

(c) 60

(d) 40

4. A magnet of length 14 cm and magnetic moment M is broken into two parts of lengths 6 cm and 8 cm. They are put at right angle to each other with opposite poles together. The magnetic moment of the combination is M M (a) (b) M (c) (d) 2.8 M 10 1. 4

5. A circuit area 0.01 m2 is kept inside a magnetic field which is normal to its plane. The magnetic field changes from 2 T to 1 T in 1 ms. If the resistance of the circuit is 2 Ω, the amount of heat evolved is (a) 0.05 J

(b) 50 J


(d) 500 J

6. An LC circuit contains a 20 mH inductor and a 50 μF capacitor with an initial charge of 10 mC. The resistance of the circuit is negligible. Let the instant the circuit is closed be t = 0. At what time is the energy stored completely magnetic? (a) t = 0 (b) t = 1.57 ms (c) t = 3.14 ms (d) t = 6.28 ms 7. An alternating voltage e = 200 sin100t V is applied to a series combination R = 30 Ω and an inductor of 400 mH. The power factor of the circuit is (a) 0.01

(b) 0.2

(c) 0.05

(d) 0.6

8. A particle of mass 1 × 10–26 kg and charge 1.6 × 10–19 C travelling with a velocity 1.28 × 106 m s–1 along the positive X-axis enters a region in which a uniform electric field E and a uniform magnetic field of induction B are present. If ^ ^ E = –102.4 × 103 k N C–1 and B = 8 × 10–2 j Wb m–2, the direction of motion of the particle is (a) along the positive X-axis (b) along the negative X-axis (c) at 45° to the positive X-axis (d) at 135° to the positive X-axis 9. An object is placed at a distance of 40 cm in front of a concave mirror of focal length 20 cm. The nature of image is (a) real, inverted and of same size (b) virtual, erect and of same size (c) real, erect and of same size (d) virtual, inverted and of same size

*A renowned physics expert, KP Institute of Physics, Chandigarh, 09872662552

12

(c) 0.50 J

10. A ray of light falls on a transparent glass slab of refractive index 1.62. If the reflected ray and the refracted ray are mutually perpendicular, the angle of incidence is ⎛ 1 ⎞ (a) tan–1(1.62) (b) tan −1 ⎜ ⎝ 1.62 ⎟⎠ (c) tan–1(1.33)

⎛ 1 ⎞ (d) tan −1 ⎜ ⎝ 1.33 ⎟⎠

11. A ray PQ incident on the refracting face BA is refracted in the prism BAC as shown in the figure and emerges from the other refracting face AC as RS, such that AQ = AR. If the angle of prism A = 60° and the refractive index of the material of prism is 3 , then the angle of deviation of the ray is (a) 60° (b) 45° (c) 30° (d) None of these 12. The head lights of a car are 1.2 m apart. If the pupil of the eye of an observer has a diameter of 2 mm and light of wavelength 5896 Å is used, what should be the maximum distance of the car from the observer if the two head lights are just separated? (a) 33.9 km (b) 33.9 m (c) 3.34 km (d) 3.39 m 13. In a Young’s double slit experiment, the two slits act as coherent sources of waves of equal amplitude A and wavelength λ. In another experiment with the same arrangement the two slits are made to act as incoherent sources of waves of same amplitude and wavelength. If the intensity at the middle point of the screen in the first case is I1 and in the second I case I2, then the ratio 1 is I2 (a) 4 (b) 2 (c) 1 (d) 0.5 I1 16 14. In a Young’s double slit experiment, = . Ratio I2 9 of maximum to minimum intensity is (a) 1 : 49 (b) 9 : 16 (c) 16 : 9 (d) 49 : 1 15. Two polaroids are placed in the path of unpolarized beam of intensity I0 such that no light is emitted from the second polaroid. If a third polaroid whose polarization axis makes an angle θ with the polarization axis of first polaroid, is placed between these polaroids, then the intensity of light emerging from the last polaroid will be

⎛I ⎞ (b) ⎜ 0 ⎟ sin2 θ ⎝4⎠

I (a) ⎛⎜ 0 ⎞⎟ sin2 2 θ ⎝8⎠ ⎛I ⎞ (c) ⎜ 0 ⎟ cos 4 θ ⎝2⎠

(d) I0cos4θ

16. The energy that should be added to an electron to reduce its de-Broglie wavelength from 1 nm to 0.5 nm is (a) four times the initial energy (b) equal to the initial energy (c) twice the initial energy (d) thrice the initial energy 17. An electron and a neutron have same momentum. Which of the following statements is correct? (a) Both neutron and electron have same kinetic energy (b) Both neutron and electron have same de-Brolie wavelength. (c) Both neutron and electron have same speed. (d) Both neutron and electron have different de-Broglie wavelength. 18. The energy of a photon is equal to the kinetic energy of a proton. The energy of the photon is E. Let λ1 be the de-Broglie wavelength of the proton and λ2 be the wavelength of the photon. The ratio (λ1/λ2) is proportional to (a) E0

(b)

E

(c) E–1

(d) E–2

19. The product of linear momentum and angular momentum of an electron of the hydrogen atom is proportional to nx, where x is (a) 0

(b) 1

(c) –2

(d) 2

20. A nucleus disintegrates into two nuclear parts which have their velocities in the ratio 2 : 1. The ratio of their nuclear sizes will be (a) 21/3 : 1 (b) 1 : 31/2 (c) 31/2 : 1 (d) 1 : 21/3 21. A radioactive material decays by simultaneous emission of two particles with half-lives 1620 yr and 810 yr respectively. The time in year after which one-fourth of the material remains, is (a) 4860 (b) 3240 (c) 2340 (d) 1080 22. A radioactive sample S1 having an activity of 5 μCi has twice the number of nuclei as another sample S2 which has an activity of 10 μCi. The half lives of S1 and S2 can be PHYSICS FOR YOU | APRIL ‘16

13

(a) 20 yr and 5 yr, respectively (b) 20 yr and 10 yr, respectively (c) 10 yr each (d) 5 yr each 23. A common emitter amplifier gives an output of 3 V for an input of 0.01 V. If β of the transistor is 100 and the input resistance is 1 kΩ, then the collector resistance is (a) 1 kΩ (b) 3 kΩ (c) 10 kΩ (d) 30 kΩ 24. The output of given logic circuit is

(a) A + B + C (c) A⋅(B ⋅ C)

(b) (A + B)·(A + C) (d) A⋅(B + C)

25. A small spherical ball falling through a viscous medium of negligible density has terminal velocity v. Another ball of the same mass but of radius twice that of the earlier falling through the same viscous medium will have terminal velocity v v (c) (d) 2v (a) v (b) 4 2 26. The excess pressure inside one soap bubble is three times that inside a second soap bubble, then the ratio of their surface areas is (a) 1 : 9 (b) 1 : 3 (c) 3 : 1 (d) 1 : 27 27. Two rods of different materials having coefficients of thermal expansions α1 and α2 and Young’s moduli Y1 and Y2 respectively are fixed between two rigid walls. The rods are heated, such that they undergo the same increase in temperature. There is no α 2 bending of rods. If 1 = and stresses developed α2 3 Y in the two rods are equal, then 1 is Y2 1 3 2 (a) (b) 1 (c) (d) 2 2 3 28. 1 g of steam at 100 °C and equal mass of ice at 0 °C are mixed. The temperature of the mixture in steady state will be (latent heat of steam = 540 cal g–1, latent heat of ice = 80 cal g–1) (a) 50°C (b) 100°C (c) 67°C (d) 33°C 14


29. A black body emits radiations of maximum intensity for the wavelength of 5000 Å when the temperature of the body is 1227 °C. If the temperature of the body is increased by 1000 °C, the maximum intensity would be observed at (a) 1000 Å (b) 2000 Å (c) 5000 Å (d) 3000 Å 30. Two solid spheres A and B made of the same material have radii rA and rB respectively. Both the spheres are cooled from the same temperature under the conditions valid for Newton’s law of cooling. The ratio of the rate of cooling of A and B is r2 (d) B rA2 rB2 31. A gas is suddenly expanded such that its final volume becomes 3 times its initial volume. If the specific heat at constant volume of the gas is 2R, then the ratio of initial to final pressure is nearly equal to (a) 5 (b) 6.5 (c) 7 (d) 3.5 (a)

rA rB

r (b) B rA

(c)

rA2

32. An ideal refrigerator has a freezer at a temperature of –13 °C. The coefficient of performance of the engine is 5. The temperature of the air (to which heat is rejected) will be (a) 325°C (b) 325 K (c) 39°C (d) 320°C 33. In a Carnot engine, when T2 = 0 °C and T1 = 200 °C, its efficiency is η1 and whenT1 = 0°C and T2 = –200 °C, η its efficiency is η2, then what is 1 ? η2 (a) 0.577

(b) 0.733

(c) 0.638

(d) 0.95

34. A container with insulating walls is divided into two equal parts by a partition fitted with a valve. One part is filled with an ideal gas at a pressure p and temperature T, whereas the other part is completely evacuated. If the valve is suddenly opened, the pressure and temperature of the gas will be T p T p (a) , T (b) , (c) p, T (d) p, 2 2 2 2 35. If universal gas constant is R, the essential heat to increase the temperature of 4 mol monoatomic ideal gas from 273 K to 473 K at constant volume is (a) 200R (b) 400R (c) 800R (d) 1200R 36. A particle at the end of a spring executes SHM with a period t1 while the corresponding period for another spring is t2. If the period of oscillation with the two springs in series is T, then (a) T = t1 + t2 (b) T 2 = t12 + t22 (c) T–1 = t1–1 + t2–1 (d) T–2 = t1–2 + t2–2

37. A hollow pipe of length 0.8 m is closed at one end. At its open end a 0.5 m long uniform string is vibrating in its second harmonic and it resonates with the fundamental frequency of the pipe. If the tension in the wire is 50 N and the speed of sound is 320 m s–1, the mass of the string is (a) 5 g (b) 10 g (c) 20 g (d) 40 g 38. Ultraviolet light of wavelength 300 nm and intensity 1.0 W m–2 falls on the surface of photoelectric metal. If one percent of incident photons produce photoelectrons, then the number of photoelectrons emitted from an area of 1.0 cm2 of the surface is nearly (a) 2.13 × 1011 s–1 (b) 1.5 × 1012 s–1 (c) 3.02 × 1012 s–1 (d) none of these 39. According to Bohr’s theory of hydrogen atom, for the electron in the nth allowed orbit, the (i) linear momentum is proportional to 1/n (ii) radius is proportional to n 1 (iii) kinetic energy is proportional to n2 (iv) angular momentum is proportional to n Choose the correct option from the codes given below. (a) (i), (iii), (iv) are correct (b) (i) is correct (c) (i), (ii) are correct (d) (iii) is correct 40. Consider the nuclear reaction X200 → A120 + B80. If the binding energy per nucleon for X, A and B are 7.4 MeV, 8.2 MeV and 8.3 MeV respectively, then the energy released in the reaction is (a) 168 MeV (b) 200 MeV (c) 190 MeV (d) 188 MeV 41. An atomic power nuclear reactor can deliver 300 MW. The energy released due to fission of each nucleus of uranium atom U238 is 170 MeV. The number of uranium atoms fissioned per hour will be (a) 30 × 1025 (b) 4 × 1022 20 (c) 10 × 10 (d) 5 × 1015 42. The equation of a wave on a string of linear mass density 0.04 kg m–1 is given by ⎡ ⎛ t ⎞⎤ x y = 0.02 (m)sin ⎢2 π ⎜ − ⎥. ⎟ ⎢⎣ ⎝ 0.04 (s) 0.50 (m) ⎠ ⎥⎦ The tension in the string is (a) 1.25 N (b) 0.5 N (c) 6.25 N (d) 4.0 N

43. Two strings A and B are slightly out of tune and produce beats of frequency 5 Hz. Increasing the tension in B reduces the beat frequency to 3 Hz. If the frequency of string A is 450 Hz, calculate the frequency of string B. (a) 460 Hz (b) 455 Hz (c) 445 Hz (d) 440 Hz 44. If a source emitting waves of frequency υ moves towards an observer with a velocity v/4 and the observer moves away from the source with a velocity v/6, the apparent frequency as heard by the observer will be (v = velocity of sound) 14 10 14 2 υ (b) υ (c) υ (d) υ 9 9 15 3 45. Two bodies are in equilibrium when suspended in water from the arms of a balance. The mass of one body is 36 g and its density is 9 g cm–3. If the mass of the other is 48 g, its density in g cm–3 is (a)

(a)

4 3

(b)

3 2

(c) 3

(d) 5

SOLUTIONS

1. (c) : For no change in the velocity of electron, magnetic force = electrostatic force qvB = qE E σ v= = B ε0 B The time taken by electron to travel a distance L in that space with uniform motion ε LB L L t= = = 0 v σ / ε0 B σ 2. (b) : U =

q2 2C

For series combination of the capacitors, q = constant 1 ⇒ U∝ C U1 C2 4 = = =2 U 2 C1 2 3. (a) : The balance condition of a meter bridge experiment R l = X 100 − l R + 10 50 Case (i) : 1 ... (i) = R2 50 PHYSICS FOR YOU | APRIL ‘16

15

R 40 Case (ii) : 1 = R2 60

... (ii)

^

Using R2 from eqn. (ii) in (i), we get R1 + 10 3 = 1 ⇒ R1 + 10 = R1 ⇒ R1 = 20 Ω 60 2 R 40 1 M 4. (c) : Pole strength of original magnet, m = 14 M M ∴ M1 = .6 and M2 = .8 14 14 Magnetic moment of the combination, M 2 2 10M M 6 +8 = M = M12 + M22 = = 14 14 1. 4 5. (a) : Induced emf in coil dB 1 ∴ |e | = A = 0.01 × = 10 V dt 1 × 10−3 Current produced in coil, | e | 10 i= = = 5A R 2 Heat evolved = i2Rt = (5)2 × (2) × 1 × 10–3 = 0.05 J 6. (b) : For LC circuit, the time period is T = 2 π LC T At time t = , energy stored is completely magnetic. 4 So the time, t = or t =

2 π LC 4

2 π 20 × 10−3 × 50 × 10−6 = 1.57 ms 4

7. (d) : Power factor, cos φ = =

R

= (1.6 × 10−19 )[(−102.4 × 103 k ) ^

^

=0 F =0 m Hence, the particle will move along positive x-axis.

Acceleration of the particle, a = 9. (a) : From mirror formula 1 1 1 1 = − =− v −20 (−40) 40

v = –40 cm The image is on the same side of the object. v (−40) = −1 Now, magnification m = − = − u (−40) i.e., the image is real, inverted and of same size. 10. (a) : Brewster’s law, μ = tan θp θp = θi = tan–1(1.62) 11. (a) : Ray QR travels parallel to base BC, this is the case of minimum deviation thus ⎛ A + δmin ⎞ ⎛ 60° + δmin ⎞ sin ⎜ sin ⎜ ⎟ ⎟⎠ ⎠ ⎝ ⎝ 2 2 3= μ= ⇒ ⎛ A⎞ ⎛ 60° ⎞ sin ⎜ sin ⎜ ⎟ ⎝2⎠ ⎝ 2 ⎟⎠ 3 ⎛ 60° + δmin ⎞ = sin ⎜ ⎟⎠ ∴ δmin = 60° ⎝ 2 2

⇒

12. (c) : x = distance of car from eye D = diameter of eye lens, d = separation between sources.

v = 1.28 × 106 i m s −1 E = −102.4 × 103 k N C −1 B = 8 × 10−2 j Wb m −2 Force on a charged particle in a uniform electric and magnetic field is PHYSICS FOR YOU | APRIL ‘16

2

8. (a) : Here, m = 1 × 10–26 kg q = 1.6 × 10–19 C

^

= (1.6 × 10−19 )[(−102.4 × 103 k + 102.4 × 103 k )]

30 2

^

+ (1.28 × 106 i × 8 × 10−2 j )]

R2 + ω2 L2

(30) + (100) × (400 × 10−3)2 30 30 = = = 0. 6 900 + 1600 50

16

F = qE + q (v × B) = q (E + v × B)

d 1.22λ = x D D×d 2 × 10−3 × 1.2 = 3337 m ⇒ x= = 1.22 λ 1.22 × 5896 × 10−10

dθ =

x = 3.34 km 13. (b) : I = Ia + Ib + 2 Ia Ib cos φ For incoherent sources, (cosφ)av = 0

Linear momentum × angular momentum ∝ nx mcZ nh ∴ × ∝ nx 137 n 2 π n0 ∝ nx ⇒ x = 0

⇒ Iics = Ia + Ib = I2 I is maximum for coherent sources I cs = Ia + Ib + 2 Ia Ib = I1 For Ia = Ib = I0 I1 = 4I0 and I2 = 2I0 I So, 1 = 2 I2 ⎛ A1 ⎞ ⎜⎝ A + 1⎟⎠ 2

2

20. (d) : Using law of conservation of momentum m1v1 = m2v2 m1 v2 ⇒ = m2 v1

⎛4 ⎞ ⎜⎝ + 1⎟⎠ 3

m ∝ r3 for a spherical nucleus of uniform density 1/3 m1 r13 v2 r ⎛1⎞ ∴ = 3= ⇒ 1 =⎜ ⎟ m2 r v1 r2 ⎝ 2 ⎠

2

I 49 14. (d) : max = = = 2 2 I min ⎛ A 1 ⎛4 ⎞ 1 − 1⎞ − 1⎟ ⎜ ⎜⎝ A ⎟⎠ ⎝3 ⎠ 2 15. (a) : For P1, I = (I0)(cos2θ)av =

21. (d) : Effective half-life 1 1 1 1 1 = + = + T T1 T2 1620 810

I0 2

⇒ T = 540 yr

3

1 0

2

n

(90° – )

2

⎛I ⎞ For P3, I = ⎜ 0 ⎟ cos2 θ ⎝2⎠ I For P2, I = ⎛⎜ 0 cos2 θ ⎞⎟ cos2(90° – θ) ⎠ ⎝2 I I 2 I 2 = 0 (cos θ sin θ) = 0 (2 cos θ sin θ) = 0 sin2 2 θ 8 2 8 h 16. (d) : de-Broglie wavelength, λ = 2 mE ∴

E2 E2 λ1 1 × 10−9 = = ⇒ 9 − E1 E1 λ2 0.5 × 10

E2 E ⇒ 2 = 4 ∴ E2 = 4E1 E1 E1 ∴ Energy to be added = E2 – E1 = 4E1 – E1 = 3E1 ⇒ 2=

17. (b) : de-Broglie wavelength, λ =

h h h = = p 2 mK mv

λ1 h / 2 mE λ or 1 ∝ E1/2 = λ2 λ2 hc / E mcZ 19. (a) : Linear momentum, mv = 137 n nh Angular momentum = 2π

18. (b) : Required ratio,

t ⎛1⎞ Fraction left after n half lives is ⎜ ⎟ , n = 540 ⎝2⎠ According to question, 1 ⎛1⎞ =⎜ ⎟ 4 ⎝2⎠

n

⇒ n=2 ∴ 2=

t 540

⇒ t = 1080 yr

1 (Activity of S2) 2 λ1 N 2 1 = or λ1 N1 = (λ2 N 2 ) or λ2 2 N1 2 T 2 N1 0.693 ⎤ ⎡ or 1 = As T = ⎢ T2 N 2 λ ⎥⎦ ⎣ Given N1 = 2N2 ∴ T1 = 4 T2 23. (b) : Voltage gain = current gain × resistance gain V0 R R or AV = β × 0 or =β 0 Ri Vi Ri R0 3 30 = 100 × or or R0 = = 3 kΩ 3 0.01 0 .01 1 × 10

22. (a) : Activity of S1 =

24. (b) : Here A + B = G1 (OR) A + C = G2 (OR) and G1·G2 = Y (AND) 25. (c) : Terminal velocity of the ball falling through a viscous medium of negligible density (σ ≈ 0) is 2 2 v= r ρg 9η ⎞ ⎛ 2 2⎜ m ⎟ v= r g 9 η ⎜ 4 π r3 ⎟ ⎟⎠ ⎜⎝ 3 PHYSICS FOR YOU | APRIL ‘16

17

For constant m, η and g 1 v∝ r Because radius of second ball first ball v ∴ v2 = 1 2 4T 4T 26. (a) : Given, = 3× ⇒ r1 r2 Ratio of surface areas will be A1 = A2

4π r12 4π r22

=

is twice that of the

r1 1 = r2 3

1 9

27. (a) : Thermal stress = YαΔT where Y is Young’s modulus, α the coefficient of linear expansion and ΔT the change in temperature. For no bending, thermal stress in each rod should be equal so as to cancel other. Since, ΔT1 = ΔT2 Y1 α2 3 = = Y2 α1 2 28. (b) : Heat taken by ice to raise its temperature to 100°C Q1 = 1 × 80 + 1 × 1 × 100 = 180 cal Heat given by steam when condensed Q2 = m2L2 = 1 × 540 = 540 cal As Q2 > Q1, hence, temperature of mixture will remain 100°C. 29. (d) : According to Wien’s displacement law, 1 (λm)1 T2 λm ∝ ⇒ = T (λm)2 T1 5000 2227 + 273 ∴ = ⇒ (λm)2 = 3000 Å (λm)2 1227 + 273

32. (c) : Given that, the temperature of freezer, T2 = –13 °C T2 = –13 + 273 = 260 K Coefficient of performance, β = 5 T2 260 β= or 5 = T1 − T2 T1 − 260 ∴ T1 − 260 =

or T1 – 260 = 52 or T1 = (52 + 260) K = 312 K or T1 = (312 – 273)°C = 39°C 33. (a) : Take temperature in Kelvin 273 200 η1 = 1 − = = 0.423 473 473

area dT (T − TS ) ∝ = mass dt mc For given surrounding and object temperature −

dT R2 1 ∝ = dt R3 R H r Ratio of rates of cooling, A = B H B rA 31. (a) : Suddenly expanded ⇒ adiabatic process, i.e., pVγ = constant γ p1V1 = p2(3V)γ Cp 3R p1 = 3γ = 31.5 ≈ 5 γ= = = 1. 5 ⇒ p CV 2 R 2 −

18


... (i)

T 73 200 η2 = 1 − 2 = 1 − = = 0.732 T1 273 273

... (ii)

Dividing eqn. (i) by (ii), η1 0.423 = = 0.577 η2 0.732 34. (a) : Internal energy of the gas remains constant, hence T2 = T Using p1V1 = p2V2 V p p. = p2 V ⇒ p2 = 2 2 35. (d) : Specific heat for a monoatomic gas fR 3 R CV = = 2 2 Required heat is ΔH = nCV ΔT 3 = 4 × R × 200 = 1200R 2 36. (b) : For series springs, equivalent spring constant

30. (b) : Rate of cooling, 4 e A σ TS3

260 5

2

is given by, 2

1⎛T ⎞ 1 1 1 1 = + . Also ⎜ ⎟ = m k π 2 ks k1 k2 ⎝ ⎠ 2

1⎛T ⎞ 1⎛t ⎞ 1⎛t ⎞ = ⎜ 1 ⎟ + ⎜ 2 ⎟ ⎟ ⎜ m ⎝2π⎠ m ⎝2π⎠ m ⎝2π⎠

2

∴ T2 = t12 + t22 37. (b) : According to question, 2 × fundamental frequency of string = fundamental frequency of pipe ⎛ v ⎞ v T / μ 320 2⎜ 1 ⎟ = 2 ⇒ = 4 L2 L1 ⎝ 2 L1 ⎠ 4 L2 (μ = mass per unit length of wire)

50 / μ 320 or, μ = 0.02 kg m–1 = 0. 5 4 × 0. 8 ∵ length of string, l = 0.5 m ∴ Mass of string = μ × l = 0.02 × 0.5 = 10 × 10–3 kg = 10 g hc 38. (b) : Energy of each photon, E = λ −34 8 6.6 × 10 × 3 × 10 = 6.6 × 10−19 J = −9 300 × 10 Power of source is, P = intensity × areav = 1.0 × 1.0 × 10–4 = 10–4 W Number of photons per second (N) fall on the surface, P 10−4 = = E 6.6 × 10−19 Now number of electrons emitted = 1 % of N 1 10−4 = × = 1.5 × 1012 per second − 19 100 6.6 × 10 h 39. (a) : Angular momentum, L = n 2π or

Radius of the orbit, r = 0.52

n2 Z

Z2 Kinetic energy = –E = +13.6 2 eV n 40. (a) : For X, binding energy = 200 × 7.4 = 1480 MeV For A, binding energy = 120 × 8.2 = 984 MeV For B, binding energy = 80 × 8.3 = 664 MeV Therefore, energy released = (984 + 664) – 1480 = 168 MeV energy 41. (b) : Power = = 300 × 106 W = 3 × 108 J s–1 time 170 MeV = 170 × 1.6 × 10–13 J = 27.2 × 10–12 J Number of atoms fissioned per second (N) 3 × 108 = 27.2 × 10−12 Number of atoms fissioned per hour 3 × 108 × 3600 = 4 × 1022 = N × 3600 = −12 27.2 × 10 42. (c) : Compare given equation with y = A sin(ωt – kx) 2π 2π ⇒ ω= and k = 0.04 0.50 ω 0. 5 ∴ v= = = 12.5 m s −1 k 0.04 20


T ⇒ T = v2 μ μ ∴ T = (12.5)2 × 0.04 = 6.25 N 43. (c) : beat frequency υ(A) υ(B) (i) 450 5 υ (ii) 450 3 υ′ (> υ) But v =

(i) ⇒ υ = 455 Hz or 445 Hz (ii) υ′ – 450 = ±3 (iii) Also υ′ > υ′ (slightly) Only 445 Hz satisfies condition (ii) and (iii) 44. (c) : When source and observer both are moving in the same direction and observer is ahead of source, then apparent frequency is given by v v− ⎛ v − vo ⎞ 6 × υ = 10 υ υ′ = ⎜ υ= v ⎝ v − v s ⎟⎠ 9 v− 4 45. (c) : For equilibrium Fnet (Apparent weight) on each pan should be same. Fnet = W – U = mg – σVg σm or m − = constant ρ 36 48 ⇒ 36 − 1 × = 48 − 1 × 9 ρ 2 1 ⇒ = 1− ⇒ ρ=3 3 ρ 

EXAM DATES 2016 JEE Main

:

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: : : : : : : : : : : : : : : :

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A current which periodically changes direction while its magnitude may or may not change is alternating current (ac). Until and unless specified otherwise, an ac is a sinusoidal function of time, of the form i(t) = i0 sin(ωt + φ)

= root of mean (average) of square of current. x2

∫

Note : t2

∫i

∴

A current which does not change its direction, while its magnitude may or may not change is dc (direct current). A dc in which magnitude changes is a pulsating dc and until and unless we specify otherwise, a dc is assumed to be constant dc.

RMS Value of a Variable Current It represents that value of a constant dc current which when allowed to pass through the same resistor for same time interval as the variable current then it produces same amount of heat energy.

f (x)dx

x1

x2 − x1

= average (mean) value of f(x) in the interval x2 – x1.

2

dt

t1

= average value of square of current in the interval t2 – t1 Let us try and find rms value of sinusoidal current. For this we use, sin2θ + cos2θ = 1 ∴ If < f (x) > indicates average of f (x), then < sin2θ > + < cos2θ > = < 1 > = 1 but, sinθ and cosθ are identical functions, just with a phase difference of π/2 between them. 1 ∴ < sin2 θ > = < cos2 θ > = 2 ∴ For sinusoidal current, i = i0sin(ωt) whose time 2π period is T = ω π T but, i2 = i02sin2(ωt) has a period of = ω 2 t2 − t1

t2

2 AC : H ac = ∫ i Rdt t1

2 2 DC : H dc = idc RΔt = irms RΔt ... Hdc = Hac [according to definition] t2 t2

∫i

t1

2

dt

2 ⇒ irms RΔt = ∫ i 2Rdt ⇒ irms = (t2 − t1) t1

∴ A=

π/ω

∫

(i0 sin ωt )2 dt

0

Contributed By: Bishwajit Barnwal, Aakash Institute, Kolkata


21

∴ irms = < square of current > 1 = < i02 sin2 ωt > = i02 × 2 i peak value irms = 0 = 2 2 This result is applicable for entire cycle or only within the crest part or trough part of the cycle. Using this result, now we can easily obtain the area A marked in the graph. T /2

∫

irms =

i02 sin2 ωt dt

0

T /2

i = 0 2

i02

Ti 2 A = ⇒ A= 0 T /2 2 4 We should remember this result since using this we can easily calculate rms current in many cases without integrating. For example, say for i = i0cos(ωt), we are to find rms value in the interval π π (i) t = 0 to t = (ii) t = 0 to t = 2ω ω ⇒

∴ (i) irms =

∴ irms = < (a + b sin ωt )2 > = < a2 > + < (b sin ωt )2 > + 2ab < sin ωt >

22

∴ ε0 = (220) 2 V and ω = 2πf = 2π(50) = 100π rad s–1 Now, let us see the effect of imposing an alternating emf on various circuit components. In general, we would have a combination of R, L and C connected to an ac source and we would be finding current in the circuit. For this we draw phasor diagram where we represent emf and current as projections of rotating vectors (phasors).

Ti02 A A + 2 2 = A = 4 = i0 π T T 2 ω 2 2

A i (ii) irms = 2 = 0 T 2 4 Let us try finding the rms value of current in which dc is superimposed over sinusoidal ac such as i = a + bsin(ωt)

= a2 +

Note that, in ac circuits we cannot employ the voltmeter and ammeter used in dc circuits since they are based on the torque experienced by the coil of the galvanometer and in ac circuits the magnitude as well as direction of torque would keep on oscillating. Therefore we use hot wire ammeters and voltmeters in ac circuits which are based on heating effect of current. Hence they measure rms values. So whenever the reading of a voltmeter or ammeter in ac is given, they are rms values. Remember this. For alternating emf, two forms of representation are used : (i) ε = ε0sin(ωt) 2π Here ε0 = peak value of emf and T = . ω (ii) 220 V, 50 Hz If emf is given as separate value with frequency then it is rms value

b2 b2 + 0 = a2 + 2 2


Both i0 and φ are dependent upon the combination of R, L and C. But in such situations in general we say that the current leads (for '+') or lags (for '–') by φ radians to the applied emf and in phasor diagram, it is represented as :

Here (ωt) is the common phase angle and their relative phase does not change with respect to time hence for convenience we can rotate both the current and voltage phasors by ωt in clockwise direction, as below.

Now, we will apply alternating emf to R, L and C and see how the current changes. 1. Resistor Applying KVL, εs – iR = 0 ε ε ∴ i = s = 0 sin(ωt ) R R

ε0 ∴ i(t) = i0sin(ωt), where i0 = R ∴ The current passing through a resistor is in phase with the applied emf.

2. Capacitor Applying KVL, q εs − = 0 C ⇒ q = Cε0 sin(ωt ) dq = Cε0ω cos(ωt ) dt ε0 π⎞ sin ⎜⎛ ωt + ⎟ ⇒ i(t ) = ⎝ (1 / ωC) 2⎠ 1 = XC = capacitive reactance where ωC It has the same role of play in capacitive circuits which is played by resistance. ε π⎞ ∴ i(t ) = 0 sin ⎜⎛ ωt + ⎟ ⎝ XC 2⎠ π⎞ = i0 sin ⎛⎜ ωt + ⎟ ⎝ 2⎠ ⇒ i(t ) =

The equation clearly shows that the current leads the potential difference across the terminals of capacitor π by rad. 2 3. Inductor Applying KVL, di εs − L = 0 dt ⇒ Ldi = ε0sinωt dt ⇒ L ∫ di = ε0 ∫ sin ωtdt ε ⇒ i = 0 (− cos ωt ) ωL ε π⎞ ∴ i(t ) = 0 sin ⎜⎛ ωt − ⎟ ⎝ XL 2⎠ where XL = ωL = inductive reactance π ∴ i = i0 sin ⎜⎛ ωt − ⎟⎞ ⎝ 2⎠ ε0 where i0 = = peak value of current XL Clearly, the current through the inductor lags in phase by π/2 radians with respect to the potential difference across its terminals. 1 Note : XC = shows that for high frequency, XC ωC is low which means it offers negligible resistance for the passage of such a quickly changing current. XL = ωL indicates for high frequency, XL is high which means it offers high resistance for the passage of such quickly changing current.

With these concepts learnt, let us try finding current in some combinations of R, L and C. Series RC circuit

where i0 = peak value of current ε peak value of potential difference across it = 0 = XC XC PHYSICS FOR YOU | APRIL ‘16

23

i0 = 0 for ω = 0 ε i0 = 0 for ω = ∞ R

Applying KVL, q ε s − iR − = 0 C dq q ⇒ ε0 sin(ωt ) − R − = 0 dt C Now, with our limited knowledge of differential equations, we clearly can see that since we cannot separate the variables q and t so wont be able to find by this method. So now, we use alternate methodphase diagram method. Through both R and C, a common current passes, so we keep current along reference line and hence draw the voltage phasors with respect to it. Here VR and VC indicate the peak value of potential difference across resistor and capacitor. The vector sum of VR and VC will give source emf. ∴ ε20 = VC2 + VR2 2 2 2 ⎛ ε ⎞ ⎛V ⎞ ⎛V ⎞ ⇒ ⎜ 0 ⎟ =⎜ C ⎟ +⎜ R ⎟ ⎝ 2⎠ ⎝ 2⎠ ⎝ 2⎠ ⇒ ε2s = VC2 + VR2 rms rms rms 2

2

∴ In general, ε2s = VC + VR can be used for both peak as well as rms values but not for instantaneous values. From phasor diagram, V i X X 1 tan φ = C = 0 C = C = VR i0R R ωCR

Series RL circuit

Again keeping current phasor along the reference line, we plot the phasors VR and VL. ∴ ε20 = VR2 + VL2 ∴ ε20 = (i0R)2 + (i0 X L )2 ε0 ε ⇒ i0 = = 0 2 2 R + XL Z where Z = R2 + X L2 = impedance of series LR circuit ⎧ ε0 for ω = 0 ε0 ⎪ ∴ i0 = =⎨ R 2 2 ⎪⎩ 0 for ω = ∞ R + (ωL)

ε20 = VC2 + VR2 = (i0 XC )2 + (i0R)2 ε0 ε ⇒ i0 = = 0 R2 + X 2 Z C

2

where Z = R + XC2 = impedance of RC circuit ε0 ⎛ ⎛ X ⎞⎞ ∴ i(t ) = sin ⎜ ωt + tan −1 ⎜ C ⎟ ⎟ ⎝ R ⎠⎠ ⎝ 2 2 R + XC ⎛ ⎛ X ⎞⎞ ∴ VR (t ) = (i0R)sin ⎜ ωt + tan −1 ⎜ C ⎟ ⎟ ⎝ R ⎠⎠ ⎝ ⎛ ⎛ X ⎞ π⎞ VC (t ) = i0 XC sin ⎜ ωt + tan −1 ⎜ C ⎟ − ⎟ ⎝ R ⎠ 2⎠ ⎝ ε0 i0 = clearly shows that 2 1 ⎞ ⎛ R2 + ⎜ ⎝ ωC ⎟⎠ 24


V X From phasor diagram, tan φ = L = L VR R ε0

∴ i(t ) = R

2

+ X L2

⎛ ⎛ X ⎞⎞ sin ⎜ ωt + tan −1 ⎜ L ⎟ ⎟ ⎝ R ⎠⎠ ⎝

Series RLC Circuit (Acceptor Circuit)

Now, either amongst VC = i0XC and VL = i0XL will be larger and will therefore decide whether current will lead the applied emf or it will lag the applied emf. Let VC > VL 2 2 ∴ ε20 = VR2 + (VC − VL )2 = (i0R) + (i0 XC − i0 X L ) ε ∴ i0 = 0 , where Z

Z = R2 + (XC − X L )2 ⎞ ⎛ 1 = R2 + ⎜ − ωL ⎟ ⎠ ⎝ ωC

If the circuit is operated at resonant frequency, Z = R hence it readily allows the current to pass through as if L and C were not present, hence is known as acceptor circuit. Finally, ε0 ⎛ ⎛ X − XL ⎞ ⎞ i(t ) = sin ⎜ ωt + tan −1 ⎜ C ⎟⎠ ⎟⎠ ⎝ ⎝ 2 2 R R + (X − X ) C

L

Parallel RLC Circuit

2

⎧∞ for ω = 0 ⎪⎪ 1 when XL = XC = ⎨R for ω = ω0 = LC ⎪ ⎪⎩∞ for ω = ∞ V − VL XC − X L tan φ = C = VR R if XL = XC ⇒ φ = 0 This condition when the current through the source is found to be in phase with the source emf, is said to be condition of resonance and the frequency at which this 1 ⎞. ⎛ happens is resonant frequency ⎜ ω0 = ⎟ ⎝ LC ⎠ This condition for a series LCR circuit also means that the impedance becomes minimum and hence current maximum but this is not the general definition, since it might happen that when φ = 0, current will be minimum, as will be shown in this article further.

In parallel circuit, the potential dif ference is common for all. So we keep the voltage phasor along the reference line for relative phase angle measurements of current. ε ε Now, iC > iL if 0 > 0 , i.e., XC < X L XC X L and iL > iC if XL < XC iL = iC if XL = XC Let iC > iL then the vector sum of the 3 current phasors gives the resultant current phasor. ∴ i02 = iR2 + (iC − iL )2 2 2 ⎛ ε0 ⎞ ⎛ ε0 ⎞ ⎛ ε0 − ε0 ⎞ = ⎜⎝ ⎟⎠ ⎜⎝ ⎟⎠ + ⎜ ⎝ XC X L ⎟⎠ R Z

1 ⎞ 1 1 ⎛ 1 − ⇒ = +⎜ ⎟ 2 X X Z R ⎝ C L⎠

2

2

1 is admittance. Z ε0 ε0 − iC − iL XC X L R R Clearly, tan φ = = = − ε iR XC X L 0 R Again, current through the source is in phase with source emf if XL = XC 1 i.e. at ω = ω0 = (resonant frequency) LC Inverse of impedance, i.e.,

For XC > XL nature of circuit is overall capacitive. XC = XL nature of circuit is overall resistive. XL > XC nature of circuit is overall inductive.


25

But at this frequency

For a series LCR circuit, 2

1 1 ⎛ 1 ⎞ 1 = + ⎜ ωC − = = minimum ⎟ 2 ⎝ ⎠ Z R ωL R Hence Z = maximum ε ∴ i0 = 0 = minimum R

Power Dissipation in AC circuits In any circuit if we have source emf εs = ε0 sin (ωt) then current is of the form i(t) = i0sin(ωt + φ) where φ may have positive or negative value depending upon the circuit and operational frequency ω. Instantaneous power supplied by source, Pinst = ε0sin(ωt) × i0 sin(ωt + φ) ε i = 0 0 [2 sin(ωt )sin(ωt + φ)] 2 ε0i0 = [cos(φ) − cos(2ωt + φ)] 2 ε i ε i = 0 0 cos φ − 0 0 cos(2ωt + φ) 2 2 Here φ is time independent parameter and cos(2ωt + φ) is a periodic function which is changing very quickly, generally, φ = 50 Hz, therefore their instantaneous values wont be noticeable. What we get to feel is their average value. ∴ Average power supplied by source ε i cos φ ε i P = Pinst = 0 0 − 0 0 cos(2ωt + φ) 2 2 (... = 0) ε i ε i ∴ P = 0 0 cos φ = 0 0 cos φ 2 2 2 P = εrms irms cosφ 26


X − XL tan φ = C R R ∴ cos φ = = power factor Z ⎞ R ⎛ε ∴ P = εrmsirms = ⎜ rms ⎟ irms R Z ⎝ Z ⎠ ε 2 P = irms R where irms = rms Z This result gives a very important conclusion, that the average power supplied by source is equal to the power consumed by resistor alone inspite of the fact that there could have been capacitors and inductors in the circuit, which means in a complete cycle there is no net power consumption by either capacitor or inductor. This is due to the fact that in one half of the cycle, work is done on them while in other half work is done by them. Therefore inductors and capacitors are said to be lossless components of circuit. Often virtual current or virtual emf terms are also used to indicate rms values. ∴ εrms irms = Virtual power Actual power ∴ Power factor= Virtual power Half Power Frequencies (ω1 and ω2) These are those frequencies at which if the circuit is operated, the power consumption becomes half of the maximum possible power consumption. (i )2 R P 2 ∴ P = max ⇒ irms R = rms max 2 2 (i ) (i ) ⇒ irms = rms max ⇒ i0 = 0 max 2 2 Now, for a series LCR circuit, ε0 ε0 / R = Z 2 ⇒ Z = 2R ⇒ R2 + ( XC − X L )2 = 2R2 ⇒ (XC – XL)2 = R2 ⇒ XC – XL = ±R 1 ∴ XC > XL for ω < ω0 ⇒ − ω1L = R ω1C XL > XC for ω > ω0 ⇒

1 − ω2 L = −R ω 2C

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

Adding (i) and (ii), ⎛ ω1 + ω2 ⎞ 1 ⎜ ⎟ − (ω1 + ω2 )L = 0 ⎝ ω1ω2 ⎠ C

Q = 2π ×

1 = ω20 LC Subtracting (ii) from (i), ⎛ ω2 − ω1 ⎞ 1 ⎜⎝ ω ω ⎟⎠ C − (ω1 − ω2 )L = 2R ⇒ ω1ω2 =

1 2

1 ⎤ ⎡ 1 ⇒ (ω2 − ω1 ) ⎢ × + L ⎥ = 2R ⇒ (ω2 – ω1)2L = 2R 1 C ⎥ ⎢ ⎦ ⎣ LC R ⇒ ω2 − ω1 = = Δω = bandwidth of operation L

If ω1 < ω < ω2, current is greater than

Energy stored in oscillators

Energy lost in dissipation Higher Q indicates a lower rate of energy loss relative to the stored energy in the resonator. In electrical systems, the stored energy is the sum of the energy stored in lossless inductors and capacitors. In each cycle, the energy is alternately stored in capacitor and inductor. The maximum energy stored in either of them is equal to their sum of energy at any instant. The lost energy is the sum of the energies dissipated in the resistor per cycle. 1 2 1 2 Li0 Li U osc 2π 2 0 2 = π × = × 2 ∴ Q = 2π 2 U lost irms RT0 T0 i02 R ω L ω 2 Q= 0 = 0 Δω R Hence it can also be defined as the ratio of resonant frequency and the bandwidth of operation.

ε0

, hence power 2R consumption will be greater than half the maximum possible power consumption. Quality factor (Q – factor) Q–factor is 2π times the ratio of the energy stored in the oscillator to the loss in energy per cycle in the circuit at resonant frequency.

∴ The graph with sharp peak has more selectivity. 


27

PRACTICE QUESTIONS ON

Assertion & Reason

Directions : In the following questions, a statement of Assertion is followed by a statement of Reason. Mark the correct choice as : (a) If both assertion and reason are true and reason is the correct explanation of assertion. (b) If both assertion and reason are true but reason is not the correct explanation of assertion. (c) If assertion is true but reason is false. (d) If both assertion and reason are false. 1. Assertion : A gas can be liquified at any temperature by increase of pressure alone.

Reason : On increasing pressure, the temperature of gas decreases. 2. Assertion : In the case of a stationary wave, a person hears a loud sound at the nodes as compared to the antinodes.

Reason : In a stationary wave, all the particles of the medium vibrate in phase. 3. Assertion : When we change the unit of measurement of a quantity, its numerical value changes.

Reason : Smaller the unit of measurement smaller is its numerical value. 4. Assertion : Identical springs of steel and copper are equally stretched. More work will be done on the steel spring.

Reason : Steel is more elastic than copper. 5. Assertion : If in a projectile motion, we take air friction into consideration, then tascent < tdescent. 28


Reason : During ascent magnitude of retardation is greater than magnitude of acceleration during descent. 6. Assertion : The energy stored in the inductor of 2 H, when current of 10 A flows through it is 100 J.

Reason : Energy stored in an inductor is directly proportional to its inductance. 7. Assertion : Thin films such as soap bubble or thin layer of oil on water show beautiful colours when illuminated by white light.

Reason : It happens due to the interference of light reflected from the upper surface of the thin film. 8. Assertion : A charged particle moves perpendicular to a magnetic field. Its kinetic energy remains constant, but momentum changes.

Reason : Force acts on the moving charged particles in the magnetic field. 9. Assertion : If the length of the conductor is doubled, the drift velocity will become half of the original value (keeping potential difference unchanged).

Reason : At constant potential difference, drift velocity is inversely proportional to the length of the conductor. 10. Assertion : A rocket works on the principle of conservation of linear momentum.

Reason : Higher the velocity, smaller is the pressure and vice-versa. 11. Assertion : No work is done when an electron completes a circular orbit around the nucleus of an atom.

Reason : Work done by a centripetal force is always zero. 12. Assertion : A person standing on a rotating platform suddenly stretched his arms. The platform slows down.

Reason : This happens as angular momentum is conserved. 13. Assertion : A planet moves faster, when it is closer to the sun in its orbit and vice-versa.

Reason : Orbital velocity in an orbit of planet is constant. 14. Assertion : A needle placed carefully on the surface of water may float, whereas a ball of the same material will always sink.

Reason : The buoyancy of an object depends both on the material and shape of the object. 15. Assertion : Coefficient of thermal conductivity of a metal rod is a function of length of the rod.

Reason : Longer the rod, larger is the amount of heat conducted. 16. Assertion : Two electric field lines never intersect one another at any point in space.

Reason : Electric field lines always start from a positive charge and end on a negative charge. 17. Assertion : The focal length of the objective of the telescope is larger than that of eyepiece.

Reason : The resolving power of telescope increases when the aperture of objective is small. 18. Assertion : When base region has larger width, the collector current decreases.

Reason : Electron hole combination in base increases base current. 19. Assertion : The relative velocity of two photons travelling in opposite direction is the velocity of light.

Reason : The rest mass of photon is zero. 20. Assertion : Television signals are received through sky-wave propagation.

Reason : The ionosphere reflects electromagnetic waves of frequencies greater than a certain critical frequency.

SOLUTIONS 1. (d) : Gas and vapour are two distinct state of matter. Critical temperature is the distinguishing feature between the two. A vapour above the critical temperature is a gas and a gas below the critical temperature for the substance is a vapour. A gas cannot be liquified by the application of pressure alone, howsoever large the pressure may be while vapour can be liquified under pressure alone. To liquify a gas it must be cooled upto or below its critical temperature. 2. (c) : The person will hear the loud sound at nodes than at antinodes. We know that at antinodes the displacement is maximum and strain is minimum while at nodes the displacement is zero and strain is maximum. The sound is heard due to variation of pressure. Further, P = – E(dy/dt), where E is elasticity and dy/dt is strain. As strain is maximum at nodes, hence there is maximum variation of pressure and loud sound is heard. 3. (c) : If u1, u2 be the units to measure a quantity Q and n1, n2 be the numerical values respectively then we know that Q = n1u1 = n2u2. Since the quantity Q does not change irrespective of the units used to measure 1 it i.e., Q = constant. So nu = constant ⇒ n ∝ i.e., u smaller the unit of measurement, greater is the corresponding numerical value. 4. (a) : As, work done 1 1 = × stress × strain = × Y × (strain)2 2 2 Since elasticity of steel is more than that of copper hence more work has to be done in order to stretch the steel. 5. (a) 6. (b) : The energy stored in the inductor is given by 1 1 U = × LI 02 = × 2 × (10)2 = 100 J 2 2

It is clear that energy stored is directly proportional to its inductance. 7. (c) : The beautiful colours are seen on account of interference of light reflected from the upper and the lower surface of the thin films. Since, condition for constructive and destructive interference depends upon the wavelength of light therefore, coloured interference fringes are observed.


29

8. (b) : The charged particle would move along a circle. Magnitude of velocity remains constant, but its direction goes on changing. That is why ⎛ 1 ⎞ kinetic energy ⎜ = mv 2 ⎟ remains constant but 2 ⎝ ⎠ momentum (= mv) goes on changing. 9. (a) : Drift velocity of free electron is given by eE vd = τ m V Here, E = l 1 ⎛ eV eV ⎞ τ or v d ∝ τ is constant ⎟ ∴ vd = ⎜ ml l ⎝m ⎠ 10. (b) : Assertion and reason, both are correct. But the reason refers to Bernoulli’s theorem, which has nothing to do with the assertion. 11. (a) : W = Fs cos θ = Fs cos 90° = 0 Centripetal force is along the radius, which is at 90° to the direction of motion (along the tangent). 12. (a) : On stretching arms, his moment of inertia (I) increases. As no torque has been applied, therefore Iω = constant. As I increases, ω decreases. 13. (c) : A planet revolves around the sun in such a way that its areal velocity is constant. That is why, it moves faster, when it is closer to the sun and vice-versa. 14. (c) : A needle placed carefully on the surface of water may float due to surface tension, as upward forces due to surface tension balances the weight of the needle. But these upward forces due to surface tension are very small as compared to weight of ball, also the weight of liquid displaced by the ball immersed in liquid is less than the weight of the ball, hence ball sinks into the liquid. 15. (d) : Coefficient of thermal conductivity depends only on nature of material of the rod and the length of rod is inversely proportional to the amount of 1 heat conducted, ΔQ ∝ . Δx 16. (b) : Two electric field lines do not intersect one another because if they do so then at the point of intersection there will be two possible directions of electric field which is impossible. Electric field lines always start from a positive charge and end on a negative charge. 30


f 17. (c) : The magnifying power of telescope, m = 0 . fe So, for high magnification, the focal length of objective should be larger than that of eyepiece. D Resolving power of a telescope = 1.22λ For high resolving power, diameter (D) of objective should be higher. 18. (a) 19. (b) : Velocity of first photon = u = c Velocity of second photon = v = – c Now, relative velocity of first photon with respect to second photon u−v c − (−c) = = uv (c)(−c) 1− 2 1− 2 c c

=

2c 2

c 1+ 2 c

=

2c 2c = =c 1+1 2

Also the rest mass of photon is zero. 20. (d) : In sky wave propagation the radio waves which have frequency between 2 MHz to 30 MHz, are reflected back to the ground by the ionosphere. But radio waves having frequency greater than 30 MHz cannot be reflected by the ionosphere because at this frequency they penetrate the ionosphere. It makes the sky wave propagation less reliable for propagation of TV signal having frequency greater than 30 MHz. Critical frequency is defined as the higher frequency that is returned to the earth by the ionosphere. Thus, above this frequency a wave whether it is electromagnetic will penetrate the ionosphere and is not reflected by it.

Ev Everything is determined, the beginning as well as the end, by forces over which we have no control. It is determined for insects as well as for the stars. Human beings, vegetables or cosmic H dust, we all dance to a mysterious tune, intoned in the distance by an invisible piper.

Exam on 22nd May 2016

PAPER-2

SECTION 1 (MAXIMUM MARKS : 32) • •

This section contains EIGHT questions The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 to 9, both inclusive

1. Four point charges + 8 μC, –1 μC, –1 μC, and + 8 μC 27 3 3 m, − m, + m, 2 2 2 27 and + m respectively on the y-axis. A particle 2 of mass 6 × 10–4 kg and charge + 0.1 μC moves along the x-direction. Its speed at x = + ∞ is v0. Find the least value of v0 (in m s–1) for which the particle will cross the origin. Assume that space is gravity free. 1 = 9 × 109 N m2 C–2. Given 4 πε0 2. A narrow monochromatic beam of light of intensity I is incident on a glass plate as shown in figure. Another identical glass plate is kept close to the first one and parallel to it. Each glass plate reflects 25 percent of the light incident on it and transmits the remaining. Find the ratio of the maximum and the minimum amplitudes in the interference pattern formed by the two beams obtained after one reflection at each plate. are fixed at the points −

according to the empirical relation s = A + BT, where A = 100 cal kg–1 K–1 and B = 2 × 10–2 cal kg–1 K–2. The final temperature of the container is 27°C, the mass of the container is x × 10–1 kg. What is the value of x? (Latent heat of fusion of water = 8 × 104 cal kg–1, Specific heat of water = 103 cal kg–1 K–1). 4. A current I = 10 A flows in a ring of radius r0 = 15 cm made of a very thin wire. The tensile strength of the wire is equal to T = 1.5 N. The ring is placed in a magnetic field, which is perpendicular to the plane of the ring so that the forces tend to break the ring. Find B(in T) at which the ring is broken. 5. There is a stream of neutrons with a kinetic energy of 0.0327 eV. The half life of neutrons is 700 s. Fraction of neutrons that decay before they travel distance of 10 m is 3.9 × 10–p, find p. 6. Figure shows a part of a bigger circuit. The capacity of the capacitor is 6 mF and is decreasing at the constant rate 0.5 mF s–1. The potential difference across the capacitor at the shown moment is changing as follows: dV d 2V 1 = 2 V s −1 , 2 = V s −2 2 dt dt – + 6mF

2H 4

3. An ice cube of mass 0.1 kg at 0°C is placed in an isolated container which is at 227°C. The specific heat s of the container varies with temperature T

3A

The current in the 4 Ω resistor is decreasing at the rate of 1 mA s–1. What is the potential difference (in mV) across the inductor at this moment? PHYSICS FOR YOU | APRIL ‘16

31

7. Consider the situation shown in the figure. Mass of block A is 6 kg and that of block B is 12 kg. The force constant of –1 spring is 50 N m . Friction is absent everywhere. System is released from rest with the spring unstretched. Find the speed of block A (in m s–1) when the extension in x the spring is x = m , xm = maximum extension of 2 spring. 8. A solid ball of density half that of water falls freely under gravity from a height of 19.6 m and then enters water. How much time (in s) will it take to come again to the water surface? Neglect air resistance and viscosity effects in water. (Take g = 9.8 m s–2) SECTION 2 (Maximum Marks : 32) • •

This section contains EIGHT questions Each question has FOUR options (a), (b), (c) and (d). ONE OR MORE THAN ONE of these four option(s) is(are) correct

9. A straight copper wire of length 1000 m and crosssectional area 1.0 mm2 carries a current 4.5 A. Assuming that one free electron corresponds to each copper atom, (Density of copper = 8.96 × 103 kg m–3, Atomic mass of copper = 63.5 g, Resistivity of copper wire = 1.69 × 10–8 Ω m) (a) the time taken by an electron to displace from one end of the wire to the other is 4 × 106 s. (b) the sum of electric force acting on all free electrons in the given wire is 1 × 106 N. (c) the time taken by an electron to displace from one end of the wire to the other is 3 × 106 s. (d) the sum of electric force acting on all free electrons in the given wire is 2 × 106 N. 10. Determine the period of small oscillations of a mathematical pendulum, that is a ball suspended by a thread l = 20 cm in length, if it is located in a liquid whose density is three times less than that of the ball. The resistance of the liquid is to be neglected. (a) 2.2 s (b) 1.1 s (c) 0.5 s (d) 3.1 s 11. The minimum value of d so that there is a dark fringe at O is dmin, the distance at which the next bright fringe is formed is x. Then 32


(a) dmin =

λD

(b) dmin =

λD 2

dmin (d) x = dmin 2 12. If dimensions of length are expressed as Gxcyhz, where G, c and h are the universal gravitational constant, speed of light and Planck's constant respectively, then 1 1 1 1 (a) x = , y = (b) x = , z = 2 2 2 2 3 3 1 1 (c) y = – , z = (d) y = , z = 2 2 2 2 (c) x =

13. A tuning fork vibrating with a frequency of 512 Hz is kept close to the open end of a tube filled with water, as shown in the figure.

The water level in the tube is gradually lowered. When the water level is 17 cm below the open end, maximum intensity of sound is heard. The room temperature is 20°C. Then (a) Speed of sound in air at room temperature is 360 m s–1. (b) Speed of sound in air at 0°C is 336 m s–1. (c) If the water in the tube is replaced with mercury, then the intensity of reflected sound increases. (d) If the water in the tube is replaced with mercury, then the intensity of reflected sound decreases.

14. A steel rod of length 2l, cross-sectional area A and mass M is set rotating (with angular speed ω) in a horizontal plane about an axis passing through the centre. If Y is the Young's modulus for steel, then the extension in the length of the rod will be (Assume the rod is uniform) Mω2l 2 Mω2l 2 (b) (a) 2YA 6YA 2 2 M ω2l 2 Mω l (c) (d) 3YA 4YA 15. The gap between the plates of a parallel-plate capacitor is filled with isotropic dielectric whose permittivity ε varies linearly from ε1 to ε2(ε2 > ε1) in the direction perpendicular to the plates. The area of each plate equals A, the separation between the plates is equal to d. Then the capacitance of the capacitor will be given by (ε2 − ε1 )ε0 A d ln(ε2 / ε1 ) ε2 ε0 A (c) d (a)

(ε2 − ε1 )ε0 A d (ε1 − ε2 )ε0 A (d) d ln(ε1 / ε2 )

(b)

16. A particle moves along a closed trajectory in a centripetal field of force where the particle's potential energy U = kr2 (k is a positive constant, r is the distance of the particle from the centre O of the field). The mass of the particle if its minimum distance from the point O equals r1 and its velocity at the point farthest from O equals v2 will be 2

⎛r ⎞ (a) 2k ⎜ 1 ⎟ ⎝ v2 ⎠

⎛r ⎞ (b) 2k ⎜ 1 ⎟ ⎝ v2 ⎠

⎛ r2 ⎞ (c) 2k ⎜ 1 ⎟ ⎝v ⎠

⎛r ⎞ (d) 4k ⎜ 1 ⎟ ⎝ v2 ⎠

2

2

17. Total magnetic flux in this loop is μ Ia 2μ0 Ia (b) (a) 0 ln 2 ln 2 π π μ Ia 4μ0 Ia (c) (d) 0 ln 2 ln 2 2π π 18. The instantaneous current in the circuit will be 2μ0 I 0 ωa ln 2 sin(ωt − θ) (a) π R2 + ω 2 L2 (b) (c) (d)

2μ0 I 0 ωa ln 2 π R2 + ω2 L2 2μ0 I 0 ωa ln 2 π R2 + ω 2 L2

sin(ωt + θ) sin ωt

ωL ⎞ ⎛ sin(ωt − θ) ⎜ where tan θ = R ⎟⎠ ⎝ π R +ω L μ0 I 0 ωa ln 2 2

2 2

PARAGRAPH 2 A parallel beam of light falls successively on a thin convex lens of focal length 40 cm and then on a thin convex lens of focal length 10 cm as shown in the figure (a).

SECTION 3 (Maximum Marks : 16) • • •

This section contains TWO paragraphs Based on each paragraph, there will be TWO questions Each question has FOUR options (a), (b), (c) and (d). ONE OR MORE THAN ONE of these four options(s) is(are) correct

PARAGRAPH 1 In the shown figure, a square loop consisting of an inductor of inductance L and resistor of resistance R is placed between two long parallel wires. The two long straight wires have time-varying current of magnitude I = I0 cos ωt but the directions of current in them are opposite.


33

In the figure (b), the second lens is an equiconcave lens of focal length 10 cm and made of a material of refractive index 1.5. In both the cases, the second lens has an aperture equal to 1 cm. 19. The ratio of the areas illuminated by the beam of light on the screen, which passes through the second lens in the two cases, i.e., (A2/A1) will be (a) 72/5 (b) 81/1 (c) 56/3 (d) 29/2 20. Now, a liquid of refractive index μ is filled to the right of the second lens in figure (b) such that the area illuminated in both the cases is the same. Determine the refractive index of the liquid. (a) 1 (b) 2.5 (c) 3 (d) 1.5

27 15 or 4 ⎜⎛ x 2 + 3 ⎟⎞ = x 2 + or 3x 2 = ⎝ 2 2 2⎠ 5 m 2 The least value of kinetic energy of the particle at infinity should be enough so that particle can reach or x = ±

at x = + it is zero.

5 5 m because at x = + m net force on 2 2

5 m, net force on the particle is repulsive 2 (towards positive x-axis). For x >

5 m, net force on the particle is attractive 2 (towards negative x-axis). For x <

SOLUTIONS

1. (3) : Let the particle be, at some instant, at a point P distant x from the origin. At this point electric field will be zero. Net force towards origin = 2 F ' cos β Net force away from origin = 2 F cos α For net force at point P to become zero,

Let electric potential at x = ∴

V=

5 m be V. 2

2 × (9 × 109 ) × (8 × 10−6 ) 5 27 + 2 2 −

2 × (9 × 109 ) × (1 × 10−6 )

+8 μC, –1 μC,

27 2 3 2

–1 μC, –

3 2

+8 μC, –

27 2

+0.1μC

2 F cosα = 2 F ' cos β 2k × (8 × 10−6 ) × (0.1 × 10−6 )cos α ⎛ 2 27 ⎞ ⎜⎝ x + ⎟⎠ 2 −6 −6 1 = 2k × (1 × 10 ) × (0.1 × 10 )cosβ where k = 4 πε0 ⎛ 2 3⎞ ⎜⎝ x + ⎟⎠ 2 x x 8 or × = / 1 2 1/2 ⎛ 2 27 ⎞ ⎛ 2 27 ⎞ ⎛ 2 3⎞⎛ 2 3⎞ ⎜⎝ x + ⎟⎠ ⎜ x + ⎟ x + x + ⎟⎜ ⎟ 2 ⎝ ⎝⎜ 2⎠ 2⎠⎝ 2⎠ ∴

3⎞ ⎛ 8 ⎜ x2 + ⎟ ⎝ 2⎠ 34

3/2

27 ⎞ ⎛ = ⎜ x2 + ⎟ ⎝ 2⎠

3/2


⎡8 1⎤ or V = 2 × 9 × 109 × 10–6 ⎢ − ⎥ ⎣4 2⎦ 4 or V = 2.7 × 10 V

5 3 + 2 2

5 Using energy conservation at x = m and x = ∞. 2 1 q × V + 0 = 0 + mv02 2 1 –6 or (0.1 × 10 ) × (2.7 × 104) = (6 × 10–4)v02 2 or v02 = 9 ... v0 = 3 m s–1 2. (7) : Two plates are parallel to each other. Each plate reflects 25% of the incident light and transmits the remaining 75%. Incident beam at A has an intensity I. This beam gets refracted and reflected multiple number of times between the two parallel plates. The intensities of various reflected and transmitted beams are shown in the figure. Beam 1 and 2, from A and B, undergo interference.


35

Intensity of beam 1 at A, I1 = I/4 Intensity of beam 2 at B, I2 = 9I/64 ∴

I min ⎛ I1 − I2 ⎞ =⎜ ⎟ I max ⎝ I1 + I2 ⎠

⎛ dθ ⎞ or, 2T sin ⎜ ⎟ = BIr0dθ ⎝ 2 ⎠ T 1. 5 = B= =1T Ir0 (10)(0.15)

2

2

or

⎛1 3⎞ 2 2 I min ⎜ 2 − 8 ⎟ 1 ⎛ 1/ 8 ⎞ ⎛1⎞ = = =⎜ ⎜⎝ ⎟⎠ = ⎜⎝ ⎟⎠ 1 3⎟ 49 I max 7 8 / 7 ⎜ + ⎟ ⎝2 8⎠

I A 7 ⇒ max = max = = 7 Amin I min 1 3. (5) : Heat lost by container = msdT or dQ = m(A + BT)dT or

Q 300 ∫0 dQ = m∫500 ( A + BT )dT

or

⎡ BT 2 ⎤ Q = m ⎢ AT + ⎥ ⎣ 2 ⎦ 500

300

300

−2 ⎡ ⎤ or Q = m ⎢100T + 2 × 10 T 2 ⎥ ⎣ ⎦ 500 2 ⎡ (300)2 − (500)2 ⎤ or Q = m ⎢100(300 − 500) + ⎥ 100 ⎣ ⎦ or Q = m[–20000 – 1600]cal or Q = –21600 m calorie …(i) Heat gained by ice to become water at 0°C = mL ∴ Q1 = 0.1 × 80000 = 8000 cal Heat gained by water to raise its temperature from 0°C to 27°C = msΔT ∴ Q2 = 0.1 × 1000 × 27 = 2700 cal ∴ Total heat gained = 8000 + 2700 Q1 + Q2 = 10700 cal According to principle of calorimetry, Heat lost = Heat gained or 21600 m = 10700 10700 or m = = 0.495 kg 21600 ∴ Mass of container = 0.495 kg = 4.95 × 10–1 ≈ 5 × 10–1 kg

4. (1) : For the equilibrium of a small part of semicircular arc subtending an angle of dθ at the centre,

36


(dl = r0 dθ)

5. (6) : Kinetic energy of neutrons = 0.0327 eV or K = 0.0327 × 1.6 × 10 –19 J or 1 mv2 = 0.0327 × 1.6 × 10–19 2 2 × 0.0327 × 1.6 × 10−19 or v2 = 1.675 × 10−27 or v2 = 0.0625 × 108 or v = 0.25 × 104 m s–1 distance 10 = ∴ time taken = velocity 0.25 × 104 –3 or t = 4 × 10 s N = (1 – e–λt) ∴ Fraction that decays = N0 ⎧ ⎛ 0.693 −3 ⎞ ⎫ ⎪ − ⎜⎝ 700 × 4 ×10 ⎟⎠ ⎪ –6 –p = 1 – ⎨e ⎬ = 3.96 × 10 =3.96 × 10 ⎪⎩ ⎪⎭ ∴ p = 6. 6. (4) : q = CV, I =

dC dV dq =C +V , IL = I0 – I dt dt dt – +

= 2H

0 = 3 A

dI L dI 0 ⎡ d 2V dV dC dV dC d 2C ⎤ = – ⎢C 2 + + +V 2 ⎥ dt dt ⎣ dt dt dt dt dt dt ⎦ –3 = –1 × 10 – ⎡ ⎤ −3 1 −3 −3 ⎢⎣6 × 10 × 2 + 2(−0.5 × 10 ) + 2(−0.5 × 10 ) + 0⎥⎦ = –2 × 10–3 A s–1 dI L = 2 × 2 × 10–3 = 4 × 10–3 V = 4 mV dt 7. (4) : At maximum extension in the spring vA = vB = 0 (momentarily) Therefore, applying conservation of mechanical energy : Decrease in gravitational potential energy of block B = increase in elastic potential energy of spring. VL = L

1 2 kx 2 m 1 or 2mgxm = kxm2 2 4mg ∴ xm = k x 2mg At x = m = 2 k Let vA = vB = v(say) Then, decrease in gravitational potential energy of block B = increase in elastic potential energy of spring + increase in kinetic energy of both the blocks. 1 1 ∴ mBgx = kx2 + (mA + mB)v2 2 2 2 ⎛ 2mg ⎞ 1 ⎛ 2mg ⎞ 1 or (2m)(g) ⎜ k = + (m + 2m)v2 ⎝ k ⎟⎠ 2 ⎜⎝ k ⎟⎠ 2 or

∴

mBgxm =

v = 2g

6 m = 4 m s–1 = 2 × 10 3 × 50 3k

8. (4) : v = 2 gh = 2 × 9.8 × 19.6 = 19.6 m s–1 Let ρ be the density of ball and 2ρ the density of water. Net retardation inside the water, upthrust − weight a= mass V (2ρ) g − V (ρ) g = (V = volume of ball) V (ρ) –2

= g = 9.8 m s

...

n= =

N Ad ; M

d = density of copper

6.02 × 1023 × 8.96 × 103 63.5 × 10−3

= 8.49 × 1028 electron per m3 l We know, I = ne A vd = neA t neAl 8.49 × 1028 × 1.6 × 10−19 × 10−6 × 103 t= = I 4. 5 = 3 × 106 s Sum of electric forces F = (nV )eE = nAle ρJ I = nAleρ = nleρI A = 8.49 × 1028 × 103 × 1.6 ×10–19 × 1.69 × 10–8 × 4.5 = 1.03 × 106 N 10. (b) : Let us depict the forces acting on the oscillating ball at an arbitrary angular position θ, relative to equilibrium position where FB is the force of buoyancy.

The equation of motion for ball –mgl sinθ + FBl sin θ = ml2 θ 4 4 Using m = πr3σ, FB = πr3ρg and sin θ 3 3 small θ, in eqn (i), we get

...(i) θ for

g ⎛ ρ⎞ ⎜1 − ⎟ θ l ⎝ σ⎠ Thus the time period of the ball θ=−

Hence, the ball will go upto the same depth 19.6 m below the water surface. Further, time taken by the ball to come back to water surface is, ⎛v ⎞ ⎛ 19.6 ⎞ t = 2⎜ ⎟ = 2⎜ =4s ⎝a⎠ ⎝ 9.8 ⎟⎠ 9. (b,c): Every copper atom has one free electron. So, number of electrons per unit volume = number of atoms per unit volume.

T = 2π

= 6.28

1 g ⎛ ρ⎞ ⎜1 − ⎟ l ⎝ σ⎠

= 2π

l/g 1 1− 3

0 .2 / 9 .8 = 1 .1 s 2/3

11. (b, d) : There is a dark fringe at O if the path λ difference Δx = ABO – AO'O = 2 PHYSICS FOR YOU | APRIL ‘16

37

1/2

⎛ d ⎞ Δx = 2D ⎜1 + ⎟ ⎝ D2 ⎠ 2

14. (d) : Let us consider an element of length dx on half portion OA of the rod at a distance x from the axis of rotation,

− 2D

⎛ 1 d2 ⎞ d2 2D ⎜1 + 2 D − = ⎟ ⎝ 2 D2 ⎠ D ∴

2 λ dmin = D 2

λD 2 The bright fringe is formed at P if the path difference = AO'P – ABP = 0 or, dmin =

⇒ D+ ⇒

D 2 + x 2 − D 2 + d 2 − D 2 + (x − d )2 = 0

x 2 d 2 (x 2 + d 2 − 2 xd ) =0 − − 2D 2D 2D

Given d = dmin On solving, x = dmin =

λD 2

12. (b, c) : [L] = [Gxcyhz] whence, x = 1/2, y = –3/2, z = 1/2. 13. (b, c) : The tube filled with water behaves as a closed organ pipe of length, L = 17 cm = 0.17 m v As υ = 20 4L v20 = 4υL = 4 × 512 × 0.17 = 348.16 m s–1 As

v20 T 273 + 20 = = v0 T0 273 v0 = 348.16 ×

273 = 336 m s–1 293

The resonance will still be observed for 17cm length of air column above mercury. However, due to more complete reflection of sound waves at mercury surface, the intensity of reflected sound increases. 38


Since the rod is uniform, mass of the element = (mass/unit length) dx M = ⎜⎛ ⎞⎟ dx ⎝ 2l ⎠ Centripetal force acting on this element, ⎛M ⎞ ⎛M⎞ dF = ⎜ dx ⎟ xω2 = ⎜ ⎟ ω2xdx ⎝ 2l ⎠ 2 l ⎝ ⎠ Here, dF is provided by the tension in the rod due to elasticity. The tension (F) in half of the portion OA of the rod at a distance x from its axis of rotation is due to the centripetal force acting on all the elements from x to l, i.e., M ω2 2 2 M ω2 = xdx (l – x ) 2l ∫x 4l If dδ is the extension in the element of length dx at position x, then stress Fdx dδ = [as Y = ] longitudinal strain YA Extension in half the portion OA (of length l) of the rod, i.e., l

F=

Fdx M ω 2 2 2 ∫0 YA = 4YAl ∫0 (l – x )dx l

δ= =

l

M ω 2 ⎡ 3 l 3 ⎤ ⎛ M ω 2 ⎞ ⎛ 2l 3 ⎞ ⎛ M ω 2 l 2 ⎞ = ⎢l − ⎥ = ⎜ 4YAl ⎣ 3 ⎦ ⎝ 4YAl ⎟⎠ ⎜⎝ 3 ⎟⎠ ⎜⎝ 6YA ⎟⎠

Extension in the entire rod of length 2l, i.e., M ω 2l 2 2δ = 3YA

15. (a, d) : We point the x-axis towards right and place the origin on the left hand side plate. The left plate is assumed to be positively charged. Since ε varies linearly, we can write, ε = a + bx where a and b can be determined from the boundary conditions. We have, ε = ε1 at x = 0 and ε = ε2 at x = d, ⎛ ε −ε ⎞ Thus, ε = ε1 + ⎜ 2 1 ⎟ x ⎝ d ⎠ Now potential difference between the plates d d σ V+ – V– = ∫ E ⋅ dr = ∫ dx ε 0 0 0ε ε σ σd ln 2 dx = (ε2 − ε1 )ε0 ε1 ⎛ ⎛ε −ε ⎞ ⎞ 0 ε0 ⎜ ε1 + ⎜ 2 1 ⎟ x ⎟ ⎝ d ⎠ ⎠ ⎝

d

=

∫

Hence, the required capacitance, (ε − ε )ε A σA C= = 2 1 0 V+ − V− ⎛ ε2 ⎞ ⎜⎝ ln ε ⎟⎠ d 1 16. (b) : For the centripetal force, r × F = 0 so angular momentum is conserved. Also at maximum and minimum separations r is perpendicular to v . Hence from the conservation of angular momentum about the centre v v …(i) mv1r1 = mv2r2 or, 1 = 2 r2 r1 As potential energy, U = kr2 dU = −2kr So, Fr = − dr From Newton's law in projection form, towards centre, at the minimum and maximum separations: 2k r1 = mv21/r1 …(ii) …(iii) and 2k r2 = mv22/r2 From equations (ii) and (iii) v21/r12 = v22 /r22 …(iv) Hence from equations (i) and (iv), r2 = r1 2k r12 Putting this value of r2 in equation (iii), m = v22 17. (a) : dφ = BdA μ0 I ⎡μ I ⎤ dφ = ⎢ 0 + adx ⎥ ⎣ 2πx 2π(3a − x ) ⎦

φ=

μ0 Ia μ0 I ⎡2a dx 2a dx ⎤ ln 2 +∫ ⎢∫ ⎥ a; φ = π 2π ⎣ a x a (3a − x ) ⎦

18. (d) : Magnitude of emf in this circuit ε=

dφ μ0 a(ln 2) dI = dt π dt

μ0 a ln 2 I0ω sin ωt π μ I ωa ln 2 sin(ωt – θ) ac current, I = 0 02 π R + ω 2 L2 =

19. (b) : In case (a), the incident parallel beam emerges as a parallel beam. So area illuminated, 2

π ⎛1⎞ A1 = π ⎜ ⎟ = cm2 ⎝2⎠ 4 In case (b), let x be the diameter of the area illuminated. Then, x 1 = ⇒ x = 9 cm 45 5

2

⎛ 9 ⎞ 81 A2 = π ⎜ ⎟ = π cm2 4 ⎝2⎠ A2 81 = A1 1

20. (c) : When liquid of refractive index μ is filled to the right of this (concave) lens, the first surface of the lens (radius of curvature = 10 cm) forms the image at the object only. Considering the refraction at the second surface. μ 1 . 5 μ − 1. 5 − = 10 ∞ −10 (∵ same area ⇒ v → ∞) ⇒ μ=3 PHYSICS FOR YOU | APRIL ‘16

39

By : Prof. Rajinder Singh Randhawa*

1.

A block-spring pendulum is shown in figure. The system is hanging in equilibrium. A bullet of mass m moving at a speed of u hits the block from 2 downward direction and gets embedded in it. Find the amplitude of oscillation of the block now.

2.

A homogeneous cylinder of mass m and radius r is connected to the springs of spring constant k as shown in figure. Find the time period of oscillation of the cylinder if it rolls without slipping.

3.

One end of wire is fixed to a wall and a heavy block is suspended from its other end as shown in figure.

The pulley is free to rotate about a fixed horizontal axis. Transverse oscillations in a particular tone are excited in the wire. Another particular tone is excited in a closed pipe made of the same material. At temperature T = 300 K, both are found in unison with a source of frequency υ0 = 100 kHz. If coefficient of linear thermal expansion of the material of wire and pipe is α = 10–5 °C–1, calculate the difference in frequencies produced by wire and organ pipe when temperature increases by ΔT = 0.6 K if the same particular tones are excited in wire and pipe. 4.

A train A crosses a station with a speed of 40 m s–1 and whistles a short pulse of natural frequency υ0 = 596 Hz. Another train B is approaching towards the same station with the same speed along a parallel track. Two tracks are separated by a distance d = 99 m. When train A whistles, train B is 152 m away from the station as shown in figure. If velocity of sound in air is v = 330 m s–1, calculate the frequency of the pulse heard by driver of train B.

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

40


5.

A uniform rope of mass 0.1 kg and length 2.45 m hangs from a ceiling. (Take g = 9.8 m s–2). (a) Find the speed of transverse wave in the rope at a point 0.5 m distant from lower end. (b) Calculate the time taken by a transverse wave to travel the full length of the rope.

6.

A diver is at a depth h from the ship. If a siren of the ship emits a sound, find the time after which the diver hears the sound. Assume ρm = density of mercury.

7.

2.

x . r The point C gets displaced to N on the cylinder connected to the springs by x ...(i) y = (r + h)θ = (r + h) r 1 Potential energy stored in springs, U = 2 × ky 2 2 U = ky2 The angle by which the cylinder rotates is θ =

The tension on the wire is 3000 N. The wire is 3 m long and the fundamental frequency is 300 Hz. What is the mass of the wire?

U =k

Extension in spring in equilibrium position, x0 = mg/k. After the bullet hits, extension of spring in m⎞ ⎛ ⎜⎝ m + 2 ⎟⎠ g 3mg equilibrium position, x ′ = = 2k k The distance of new equilibrium position from previous equilibrium position is 3mg mg mg ...(i) x = x ′ − x0 = − = k 2k 2k According to conservation of linear momentum; m 3m u= v 2 2 u ...(ii) ⇒ v= 3 where v is combined velocity. Also, angular velocity is ω=

k 2k = 3m ⎛3 ⎞ ⎜⎝ 2 m ⎟⎠

...(iii)

Since v = ω A2 − x 2 ⇒ A =

v2

+ x2

...(iv) ω Putting equations (i), (ii) and (iii) in equation (iv), we get A =

mu2 ⎛ mg ⎞ +⎜ ⎝ 2k ⎟⎠ 6k

2

2

(r + h)2 x 2

r2 Kinetic energy of cylinder,

SOLUTIONS 1.

Let the centre of cylinder move by a small distance OM = x towards right as shown in figure.

(using (i))

2 1 1 ⎛ mr 2 ⎞ ⎛ v ⎞ 3 = mv 2 K = mv 2 + ⎜ ⎟ 2 2 ⎝ 2 ⎠ ⎜⎝ r ⎟⎠ 4 Now, total energy, E = K + U = constant

k(r + h)2

3 mv 2 = constant 4 r Differentiating with resepect to time, we get ⇒

2

x2 +

2k(r + h)2

3 xv + mva = 0 2 r 4k(r + h)2 4k(r + h)2 ⇒ a=− x ⇒ ω2 = 3mr 2 3mr 2 2

∴ T = 2π 3.

3mr 2 4k(r + h)2

Let n1 and n2 be the harmonics excited in the wire and organ pipe respectively. Also let l1 and l2 be the respective lengths of wire and organ pipe. T n2 v = μ 4l2 It is given that n1 and n2 will not change. Also, after the increase in temperature, T and l1 will remain the same. However, μ, l2 and v will change. As mass of wire is μl = constant Δμ Δl Δμ Δl ⇒ + =0 ⇒ = − = − αΔT μ l μ l New value of mass per unit length is μ′ = μ(1 – αΔT). n Now, 105 = 1 2l1


41

New frequency produced by wire, n υ1′ = 1 2l1

n T = 1 μ(1 − αΔT ) 2l1 ⎛ α ⎞ ⇒ υ1′ = 105 ⎜1 + ΔT ⎟ ⎝ ⎠ 2 Also, v =

−1 ∴ v = 9.8 × 0.5 = 2.21 m s

T (1 − αΔT )−1/2 μ

(b) Let t is the time taken to reach the top of the rope of length l. From (i)

...(i)

dx = gx ⇒ dt

γRT Δv 1 ΔT 1 0.6 ⇒ = = × = 10−3 M v 2 T 2 300

0

0

l

0

∴ t =2

n2v(1 + 10−3 )

4l2 (1 + αΔT ) nv υ2′ = 2 (1 + 10−3 )(1 − αΔT ) 4l2

6.

⇒ υ′2 105(1 + 10–3 – αΔT) ...(ii) From equations (i) and (ii), we get 3 ⎛ ⎞ υ2′ − υ1′ = 105 ⎜10−3 − × 10−5 × 0.6 ⎟ = 100 − 0.9 ⎠ ⎝ 2 = 99.1 Hz 4.

t

−1/2 ∫ x dx = g ∫ dt ,

l 1 ⎡ x1/2 ⎤ ⇒ t= ⎥ =2 ⎢ g g ⎢⎣ 1 / 2 ⎥⎦

New velocity, v′ = v (1 + 10–3) New length of pipe, l′2 = l2 (1 + αΔT). New frequency produced by organ pipe, υ2′ =

l

Let t be the time after which the pulse is heard by the driver of train B. In time t, train B and the sound of the whistle from train A reaches point C.

2.45 =1s 9. 8

B , ρ where B (Bulk modulus) = P (Pressure at any point of water) P + ρgy P then v = 0 = 0 + gy ρ ρ Speed of sound in water is v =

⇒

ρ gh dy = m m + gy ρ dt

∴

⎛ρ h ⎞ dy = ⎜ m m + y⎟ g dt ⎝ ρ ⎠ h

⇒

∫ 0

t

dy (a + y ) g

= ∫ dt 0

ρm hm ⎞ ⎛ ⎜ where, a = ⎟ ρ ⎠ ⎝

h

⎡ (a + y ) ⎤ ⇒ t = ⎢2 ⎥ g ⎥⎦ ⎢⎣ 0 Now, (330t)2 = (99)2 + (152 – 40t)2 On solving, we get t = 0.5 s 152 − 40t 152 − (40 × 0.5) ∴ cos θ = = = 0. 8 330t 330 × 0.5 v + v0 cos θ 330 + 40 × 0.8 = 596 × Therefore, υ ′ = υ0 v − v s cos θ 330 − 40 × 0.8 υ′ = 724 Hz 5.

42

T = gx μ


...(i)

(a + h) g

−2

⎞ 1 ⎛ ρm a =2 hm + h ⎟ ⎜ ⎠ g g⎝ ρ −2

ρm hm ρg

1 ⎡ (ρmhm − hρ) − ρmhm ⎤⎦ ρg ⎣ Since the fundamental frequency is ∴ t =2

7.

(a) Tension at a point at a distance x from the free end is T = μxg where, μ is the mass per unit length of the rope. Speed of transverse wave at this point, v=

⇒ t =2

υ=

1 T 1 TL = 2L μ 2L m

∴ m=

T 2

4υ L

=

3000 4 × (300)2 × 3

kg ≈ 2.80 gm

OLYMPIAD PROBLEMS 1. Consider a system of two thin lenses as shown in figure. An object of height 1 cm is placed at 40 cm from convex lens. Mark the correct option related to final image formed by the two lens system. (a) Final image is formed at 32 cm on right of concave lens and is 0.45 cm in size. (b) Final image is formed at 32 cm on left side of convex lens and is 1 cm in size. (c) Final image is formed at 14.5 cm on the left side of concave lens and is 0.45 cm in size. (d) None of these.

2. A thermostated chamber at a height h above earth’s surface maintained at 30°C has a clock fitted with uncompensated pendulum. The maker of the clock for chamber mistakenly designed it to maintain correct time at 20°C. It is found that when the chamber is brought to earth’s surface the clock in it clicked correct time. Re is the radius of earth. The linear coefficient of the material of pendulum is Re 5 Re h h (a) (b) (c) (d) 5 Re h Re h 3. A light ray enters a glass slab at an angle of incidence θ1. What can be the minimum value for the index of refraction of the glass if the ray does not emerge at face BC, irrespective of angle θ1? (a) 1 (b) 2 3 (c) (d) 2 4. A pendulum bob is suspended on a flat car that moves with velocity v0. The flat car is stopped by a bumper. If the swing angle of pendulum is 60° and 44


l = 10 m, what was the initial speed of the flat car? (a) 10 m s–1 (b) 15 m s–1 (c) 10 3 m s–1 (d) 20 2 m s–1

5. The figure shows a uniform rod lying along the X-axis. The locus of all the points lying on the XYplane, about which the moment of inertia of the rod is same as that about O is (a) an ellipse (b) a circle (c) a parabola (d) a straight line 6. A uniformly charged ring of radius R is rotated about its axis with constant linear speed v of each of its particles. The ratio of electric field to magnetic field at a point P on the axis of the ring at distance x = R from centre of ring is (c is speed of light) 2 2 (a) c (b) v v c (c) c (d) v v c SOLUTIONS

1. (c) : Let image formed by 1st lens be at distance v1 from convex lens. Now, 1 − 1 = 1 v1 − 40 20 ⇒ v1 = 40 cm h v 40 Size of intermediate image is, I = 1 = hO u − 40 ⇒ hI = –1 cm Negative sign tells that image formed is inverted. The image formed by 1st lens would be treated as a virtual object for 2nd lens, let final image forms at a distance v from concave lens.

1 1 1 − = ⇒ v = −14.5 cm v 32 − 10 Negative sign tells that image is formed on left side of concave lens. The magnification produced by 2nd lens is, v 14.5 m2 = = − u 32 So, size of final image is h = 14.5 × 1 = 0.45 cm 32 So, final image is virtual, erect and diminished. 2. (b) : Variation of acceleration due to gravity with ⎛ 2h ⎞ altitude gh = g ⎜1 − ⎟ ⎝ Re ⎠ 2hg Δg = Re Variation of l with temperature, Δl = αlΔθ Coefficient of linear expansion, α = Δl l(Δθ) l l ⎛ Δg ⎞ Th = 2 π = 2π 1− g − Δg g ⎜⎝ g ⎟⎠

−1/2

⎛ Δg ⎞ ≈ T ⎜1 + ⎝ 2 g ⎟⎠

1/2

l + Δl l ⎛ Δl ⎞ ⎛ Δl ⎞ = 2π ⎜1 + ⎟⎠ ≈ T ⎜⎝1 + ⎟⎠ 2l g g ⎝ l The clock shows correct time if Th = Tθ Δl Δg = 2l 2 g Δl Δg 2h h = = = ∴ α= 10l 10 g 10 Re 5 Re Tθ = 2 π

3. (b) : The condition for total internal reflection at the second surface is nsinθ3 ≥ 1. As θ2 + θ3 = 90°, So nsin(90° – θ2) ≥ 1 or n cosθ2 ≥ 1 or n2(1 – sin2θ2) ≥ 1 ...(i) sin θ1 From Snell’s law, sin θ2 = ...(ii) n From eqns (i) and (ii), ⎛ sin2 θ1 ⎞ n2 ⎜1 − ⎟ ≥1 ⎝ n2 ⎠ n2 – sin2θ1 ≥ 1 or n2 ≥ 1 + sin2θ1 The value of n corresponds to θ1 = 90°. Hence n ≥ 2 . 4. (a) : When the flat car collides with the bumper, due to inertia of motion the bob swings forward. No work is done by tension of string on the bob, therefore energy is conserved.

KEA + PEA =KEB + PEB 1 2 mv + 0 = 0 + mg (l − l cos θ) 2 0

θ or v02 = 2 gl(1 − cos θ) = 4glsin2 2 θ or v0 = 2 gl sin 2 On substituting θ = 60°, l = 10 m, g = 10 m s–2 we obtain, v0 = 2 (10) × (10) sin 30° = 10 m s −1 5. (b) : Consider a point P(x, y) in x-y plane where we assume the = moment of inertia of the rod about P is same as that about at point O IP = ICM + Mr2 =

2 – + 2 2

,0 2

2 ⎡⎛ ⎤ ML2 L⎞ + M ⎢⎜ x − ⎟ + y2 ⎥ ; 12 2⎠ ⎣⎝ ⎦

Now, I P = IO ⇒ ⇒

IO =

ML2 3

2 ⎡⎛ ⎤ ML2 ML2 L⎞ + M ⎢⎜ x − ⎟ + y2 ⎥ = 12 2⎠ 3 ⎣⎝ ⎦

2

2 L⎞ ⎛ 2 L =⎛L⎞ ⎜⎝ x − ⎟⎠ + y = ⎜ ⎟ 2 4 ⎝2⎠

2

This is equation of a circle with radius L and centre 2 at ⎛⎜ L , 0 ⎞⎟ . ⎝2 ⎠ 6. (a) : Let Q be the charge on the ring.

P

The electric field at point P, QR 1 Qx 1 = E= 4 πε0 (2R2 )3/2 4 πε0 (x 2 + R2 )3/2 The charged ring is equivalent to a ring in which current I flows, such that I = Qv 2πR The magnetic field at point P, μ μ QvR 2 πIR2 B= 0 = 0 / 2 2 3 2 4 π (x + R ) 4 π (2R2 )3/2 1 E c2 ∴ = = B μ 0 ε0 v v PHYSICS FOR YOU | APRIL ‘16

45

CHAPTERWISE MCQs FOR PRACTICE Useful for All National and State Level Medical/Engg. Entrance Exams MAGNETISM AND MATTER

1. The intensity of magnetic field at a point X on the axis of a small magnet is equal to the field intensity at another point Y on its equatorial axis. The ratio of distance of X and Y from the centre of the magnet will be (a) (2)–3 (b) (2)–1/3 (c) 23 (d) 21/3 2. The length of a magnet is large compared to its width and breadth. The time period of its oscillation in a vibration magnetometer is 2 s. The magnet is cut along its length into three equal parts and these parts are then placed on each other with their like poles together. The time period of this combination will be 2 2 (a) s (b) s (c) 2 3 s (d) 2 s 3 3 3. An iron rod of length 20 cm and diameter 1 cm is placed inside a solenoid on which the number of turns is 600. The relative permeability of the rod is 1000. If a current of 0.5 A is passed in the solenoid, then the magnetisation of the rod will be (a) 2.997 × 102A m–1 (b) 2.997 × 103A m–1 (c) 2.997 × 104A m–1 (d) 2.997 × 105A m–1 4. A long magnet is placed vertically with its S-pole resting on the table. A neutral point is obtained 10 cm from the pole due geographic north of it. If H = 3.2 × 10–5 T, then the pole strength of magnet is (a) 8 ab–A cm (b) 16 ab–A cm (c) 32 ab–A cm (d) 64 ab–A cm 5. A coil of 50 turns and area 1.25 × 10–3m2 is pivoted about a vertical diameter in a uniform horizontal magnetic field and carries a current of 2 A. When the coil is held with its plane in N-S direction, it 48


experiences a couple of 0.04 N m, and when its plane is east-west, the corresponding couple is 0.03 N m. The magnetic induction is (a) 0.2 T (b) 0.3 T (c) 0.5 T (d) 0.4 T 6. Two magnets are suspended by a given wire one by one. In order to deflect the first magnet through 45°, the wire has to be twisted through 540° whereas with the second magnet, the wire requires a twist of 360° for the same deflection. Then the ratio of magnetic moments of the two magnets is (a) 11/7 (b) 3/2 (c) 4/3 (d) 7/6 7. At a certain place, a magnet makes 30 oscillations per minute. At another place where the magnetic field is double, its time period will be (a) 4 s (b) 2 s (c) 1/2 s (d) 2 s 8. A copper rod is suspended in a non-homogeneous magnetic field region. The rod when in equilibrium will align itself (a) in the region where magnetic field is strongest (b) in the region where magnetic field is weakest and parallel to direction of magnetic field there (c) in the direction in which it was originally suspended (d) in the region where magnetic field is weakest and perpendicular to the direction of magnetic field there. 9. Two bar magnets of the same mass, same length and breadth but having magnetic moments M and 2M are joined together by pole and suspended by a string. The time period of assembly in a magnetic field of strength H is 3 s. If now the polarity of one of the magnets is reversed and the combination is again made to oscillate in the same field, the time period of oscillation is

(a) 3 s

(b) 3 3 s (c) 3 / 3 s (d) 6 s

10. A magnet performs 10 oscillations per minute in a horizontal plane at a place where the angle of dip is 45° and the total intensity is 0.707 CGS units. The number of oscillations per minute at a place where dip angle is 60° and total intensity is 0.5 CGS units will be (a) 5 (b) 7 (c) 9 (d) 11 11. A magnetising field of 1600 A m–1 produces a magnetic flux of 2.4 × 10–5 Wb in an iron bar of cross-sectional area 0.2 cm2. The susceptibility of the iron bar is (a) 298 (b) 596 (c) 1192 (d) 1788 12. A bar magnet has a magnetic moment 2.5 J T –1and is placed in a magnetic field of 0.2 T. The work done in turning the magnet from parallel to antiparallel position relative to field direction is (a) 1 J (b) 2 J (c) 2 J (d) 4 J 13. A solenoid has core of a material with relative permeability 500 and its windings carry a current of 1 A. The number of turns of the solenoid is 500 per metre. The magnetization of the material is nearly (a) 2.5 × 103A m–1 (b) 2.5 × 105 A m–1 3 –1 (c) 2.0 × 10 A m (d) 2.0 × 105A m–1 14. Choose the correct statement. (a) A paramagnetic material tends to move from a strong magnetic field to weak magnetic field. (b) A magnetic material is in the paramagnetic phase below its Curie temperature. (c) The resultant magnetic moment in an atom of a diamagnetic substance is zero. (d) Typical domain size of a ferromagnetic material is 1 nm. 15. Materials suitable for permanent magnets, must have which of the following properties? (a) High retentivity, low coercivity and high permeability. (b) Low retentivity, low coercivity and low permeability. (c) Low retentivity, high coercivity and low permeability. (d) High retentivity, high coercivity and high permeability.

ELECTROMAGNETIC INDUCTION

16. The magnetic field in a certain region is given by B = (40 i − 18 k) G. How much flux passes through a 5.0 cm2 area loop in this region, if the loop lies flat on the xy plane? (a) –900 nWb (b) 900 nWb (c) 0 (d) 9 Wb. 17. A fully charged capacitor C with initial charge q0 is connected to a coil of self inductance L at t = 0. The time at which the energy is stored equally between the electric and the magnetic field is π (a) (b) 2π LC LC 4 (c) LC (d) π LC 18. The radius of the circular conducting loop shown in figure is R. Magnetic field is decreasing at a constant rate α. Resistance per unit length of the loop is ρ. Then current in wire AB is (AB is one of the diameters) Rα (a) from B to A 2ρ

Rα (b) from A to B 2ρ 2Rα from A to B (c) ρ

(d) zero

19. A coil having n turns and resistance R Ω is connected with a galvanometer of resistance 4R Ω. This combination is moved in time t seconds from a magnetic field W1 weber to W2 weber. The induced current in the circuit is (W − W1 ) (W − W1 ) (a) − 2 (b) − 2 Rnt 5Rnt n (W2 − W1 ) n (W2 − W1 ) (c) − (d) − Rt 5Rt 20. A metallic ring is dropped down, keeping its plane perpendicular to a constant and horizontal magnetic field at t = 0. The ring enters in the region of magnetic field at t = 0 and completely emerges out at t = T s. The current in the ring varies with time as

(a)

(b)

(c)

(d)


49

21. A vertical ring of radius r and resistance R falls vertically. It is in contact with two vertical rails which are joined at the top, as shown in the figure. The rails are without friction and resistance. There is a horizontal uniform magnetic field of magnitude B perpendicular to the plane of the ring and the rails. When the speed of the ring is v, the current in the section PQ is 2Rrv (a) zero (b) R 8Brv 4Rrv (c) (d) R R 22. The magnet shown in figure rotates on a pivot through its center. At the instant shown, what are the directions of the induced currents?

(a) A to B and C to D (b) B to A and C to D (c) A to B and D to C (d) B to A and D to C 23. Two coils A and B have coefficient of mutual inductance M = 2 H. The magnetic flux passing through coil A changes by 4 Wb in 10 s due to change in current in B. Then (a) the change in current in B in this time interval is 0.5 A (b) the change in current in B in this time interval is 8 A (c) the change in current in B in this time interval is 2 A (d) the change in current of 1 A in coil A will produce a change in flux passing through B by 4 Wb. 24. Two concentric and coplanar circular coils have radii a and b as shown in figure. Resistance of the inner coil is R. Current in the other coil is increased from 0 to i, then the total charge circulating the inner coil is μ iab μ iaπ (a) 0 (b) 0 2R 2ab μ ia2 (c) 0 Rb 50

μ ia2 (d) 0 2Rb


25. The charge which will flow through a 200 Ω galvanometer connected to a 400 Ω circular coil of 1000 turns wound on a wooden stick 20 mm in diameter, if a magnetic field B = 0.012 T parallel to the axis of the stick decreased suddenly to zero is (a) 6.3 μC (b) 63 μC (c) 0.63 μC (d) 630 μC 26. A circular ring of diameter 20 cm has a resistance of 0.01 Ω. The charge that will flow through the ring if it is turned from a position perpendicular to a uniform magnetic field of 2.0 T to a position parallel to the field is about (a) 63 C (b) 0.63 C (c) 6.3 C (d) 0.063 C 27. The equivalent inductance of two inductors is 2.4 H when connected in parallel and 10 H when connected in series. What is the value of inductances of the individual inductors? (a) 8 H, 2 H (b) 6 H, 4 H (c) 5 H, 5 H (d) 7 H, 3 H 28. A small square loop of wire of side l is placed inside a large square loop of wire of side L(>>l). The loops are coplanar and their centres coincide. What is the mutual inductance of the system? μ l2 μ l2 (a) 2 2 0 (b) 8 2 0 π L π L 2 μ l μ L2 (c) 2 2 0 (d) 2 2 0 πl 2π L 29. A circular coil with a cross-sectional area of 4 cm2 has 10 turns. It is placed at the centre of a long solenoid that has 15 turns cm–1 and a cross-sectional area of 10 cm2, as shown in the figure. The axis of the coil coincides with the axis of the solenoid. What is their mutual inductance?

(a) 7.54 μΗ (c) 9.54 μΗ

(b) 8.54 μΗ (d) 10.54 μΗ

30. A conducting ring of radius 1 m is placed in a uniform magnetic field B of 0.01 T oscillating with frequency 100 Hz with its plane at right angle to B. What will be induced electric field? (a) π V m–1 (b) 0.5 V m–1 –1 (c) 10 V m (d) 62 V m–1

SOLUTIONS 1. (d) : If d1is distance of point X on axial line and d2 is distance of point Y on equatorial line, μ 2M μ M ,B = 0 then B1 = 0 4 π d13 2 4 π d23 As B1= B2 ∴ d13 = 2d23

μ0 2 M μ0 M = 4 π d13 4 π d23

d1 = 21/3 d2 I MB When the magnet is cut along its length into three equal parts, moment of inertia of each part becomes I/3 and magnetic moment of each part also becomes M/3. As the three parts are placed on each other with their like poles together, therefore for the combination,

2. (d) : T = 2 s = 2 π

I /3+ I /3+ I /3 I = 2π =2s (M / 3 + M / 3 + M / 3)B MB 1 3. (d) : Here, l = 20 cm, r = cm, n = 600, 2 μr = 1000, i = 0.5 A, I = ?, As 1 + χm = μr χm = μr – 1 = 1000 – 1 = 999 H = ni = 600 × 0.5 = 300 I = χmH = 999 × 300 = 2.997 × 105 A m–1 T = 2π

4. (c) : As the magnet is long, we assume that the upper north pole produces no effect on the table. At the neutral point, magnetic field B due to south pole of the magnet is equal and opposite to horizontal component of earth’s magnetic field. m i.e., B = ⎜⎛ μ0 ⎟⎞ = H ⎝ 4π ⎠ r2 μ In CGS system, 0 = 1 4π m ∴ 1× = 3.2 × 10−5 × 104 gauss 102 m = 32 ab–A cm 5. (d) : As, M = niA = 50 × 2 × 1.25 × 10–3 = 0.125 A m2 If normal to the face of the coil makes an angle θ with the magnetic induction B, then in first case, torque = MB cos θ = 0.04 ...(i)

and in second case, when plane of coil is turned through 90°; torque = MB sin θ = 0.03 ...(ii) Square (i) and (ii) and add MB = (0.04 )2 + (0.03)2 = 0.05 0.05 0.05 ∴ B= = = 0. 4 Τ M 0.125 6. (a) : If C is torque per unit angular twist of the wire, then for a twist φ, τ = C φ = MB sin θ. In the 1st case, φ1 = 540° – 45° = 495°, θ1 = 45° In the 2nd case, φ2 = 360° – 45°= 315°, θ2 = 45° ∴ C (495°) = M1 B sin 45° ...(i) and C (315°) = M2 B sin 45° ...(ii) Dividing (i) by (ii), we get M1 495 11 = = M2 315 7 7. (d) : As, T = 2π ∴

T1 =

1 2

I MB

T

Initial time period, T = 2

= 2s 2 8. (d) : Copper is a diamagnetic material, therefore its rod align itself where magnetic field is weaker and perpendicular to the direction of magnetic field in that region. ∴

T1 =

60 = 2s 30

9. (b) : M1 = 3 M and M2 = M T2 = ? In pole combination, T2 M1 3M = = = 3 or T2 = 3T1 = 3 3 s T1 M2 M 10. (b) : Here, n1 = 10 oscillations per minute δ1 = 45o, R1 = 0.707 CGS units n2 = ?, δ2 = 60o, R2 = 0.5 CGS units n2 H2 R2 cos δ2 ∴ = = n1 H1 R1 cos δ1 n 0.5 cos 60° 0. 5 × 1 / 2 1 ⇒ 2= = = 10 0.707 cos 45° 0. 5 × 2 × 1 / 2 2 10 = 7.07 or n2 = 2 PHYSICS FOR YOU | APRIL ‘16

51

11. (b) : Here, Magnetising field, H = 1600 A m–1 Magnetic flux, φ = 2.4 × 10–5 Wb Area, A = 0.2 cm2 = 0.2 × 10–4 m2 φ 2.4 × 10−5 Wb = = 1.2 Wb m–2 A 0.2 × 10−4 m2 B Magnetic permeability, μ = H 1.2Wb m −2 = = 7.5 × 10−4 Wb A −1m −1 −1 1600 A m μ −1 Magnetic susceptibility, χ = μ0 ∴

=

B=

(7.5 × 10−4 Wb A −1m −1 ) (4 π × 10−7 Wb A −1m −1 )

− 1 596

12. (a) : Work done in changing the orientation of a dipole of moment M in magnetic field B from position θ1 to θ2 is given by W = MB (cos θ1 – cos θ2) Here, θ1= 0° and θ2 = 180° So, W = 2MB = 2 × 2.5 × 0.2 = 1 J 13. (b) : Here, n = 500 turns m–1 I = 1 A, μr = 500 Magnetic intensity, H = nI = 500 m–1 × 1 A = 500 A m–1 Magnetisation, M = χH where χ is the magnetic susceptibility of the material = (μr−1)H = (500 –1) × 500 A m–1 (∵ χ = (μr – 1)) = 499 × 500 A m–1 = 2.495 × 105 A m–1 ≈ 2.5 × 105 A m–1

17. (a) : As ω2 =

1 LC

or ω =

1 LC

2

1 q0 2 C Let at an instant t, the energy be stored equally between electric and magnetic field. Then energy stored in electric field at instant t is 2 1 Q 2 1 ⎡ 1 q02 ⎤ 2 q0 = ⎢ = or Q ⎥ 2 C 2 ⎣2 C ⎦ 2 Maximum energy stored in capacitor =

or

q q Q = 0 , q0 sin ωt = 0 2 2

or

t=

or ωt =

π 4

π π π LC = = 4ω 4 × (1 / LC ) 4

18. (d) : As per Lenz’s law, e.m.fs of the same magnitude are induced in the two loops in the clockwise direction. Thus current is induced in the clockwise direction in the outer boundary, but obviously, there is zero current in wire AB. ε −ndφ / dt 19. (c) : The induced current, I = = R R′ ′ −n(W2 − W1 ) / t I= R + 4R −n(W2 − W1 ) I= 5Rt Note that W1, W2 are in weber. Therefore, they represent magnetic flux and not magnetic field.

15. (d) : Materials suitable for permanent magnets should have high retentivity, high coercivity and high permeability.

20. (c) : At t = 0, the ring enters the region of magnetic field. Magnetic flux linked with the ring varies till the entire ring is in the field. For this time only, e.m.f. is induced in the ring, in one direction. When the entire ring is moving in the magnetic field, the magnetic flux linked with the ring is constant. Therefore, ε = 0. When the ring starts moving out of the magnetic field, magnetic flux linked with the ring is decreasing. Therefore, e.m.f. is induced in the ring in opposite direction, only till the entire ring is out of the magnetic field.

16. (a) : As loop is in xy plane, only z component of magnetic field is effective B = –18 G = –18 × 10–4 T A = 5 × 10–4 m2 φ = BA cos 0o = –18 × 10–4 × 5 × 10–4 = – 90 × 10–8 Wb = – 900 × 10–9 Wb = – 900 nWb

21. (d) : When a ring moves in magnetic field in a direction perpendicular to its plane, we replace the ring by a diameter (2r) perpendicular to the direction of motion. The e.m.f., is induced across this diameter. Current flow in the ring will be through the two semicircular portions in parallel. Induced e.m.f. = B(2r)v.

14. (c) : Diamagnetic substances are those substances in which resultant magnetic moment in an atom is zero. A paramagnetic material tends to move from a weak magnetic field to strong magnetic field. A magnetic material is in the paramagnetic phase above its Curie temperature. Typical domain size of a ferromagnetic material is 1 mm.

52


Resistance of each half of ring = R/2 As the two halves are in parallel, therefore, equivalent resistance = R/4 B(2r )v 8Brv ∴ Current in the section = = R/4 R 22. (a) : As per Lenz’s law, N pole should develop at the end corresponding to C. Induced current flows from C to D. Again S pole should develop at the end corresponding to B. Therefore, induced current in the coil flows from A to B. 23. (c) : Here, M = 2 H, dφ = 4 Wb, dt = 10 s Αs φ = Mi dφ = M di dφ 4 or di = = =2A M 2 Also, dφ = M(di) = 2(1) = 2 Wb

Δφ NA(B2 − B1 ) N πr 2 (B2 − B1 ) = = R R R

1000 × π × 10−4 × (0.012 − 0) (200 + 400)

= 6.3 × 10−6 C = 6.3 μC −dφ BA(cos 0° − cos 90°) 26. (c) : As q = = R R =

= (4π × 10–7 T m A–1) (1500 m–1)(10) (4 × 10–4m2) = 7.54 × 10–6 H = 7.54 μH

Bπr 2 (1 − 0) Bπr 2 = R R

30. (b) : Here, r = 1 m, B = 0.01 T, 1 1 t= = s υ 100 dφ Induced e.m.f, ε = dt

2 × 3.14 × (10−1)

2

=

= 6.28 C

0.01 6.3 C

27. (b) : L1 + L2 = 10 H L1L2 = 2. 4 H and (L1 + L2 )

28. (a) : Let the current I be flowing in the larger loop. The larger loop is made up of four wires each of length L, the field at the centre i.e., at a distance from each wire, will be μ0 I (sin 45° + sin 45°) B=4× 4 π(L / 2) μ 2I 2 μ I =4× 0 =2 2 0 π L 4π L 2 Flux linked with smaller loop

μ l2 φ Hence, M = 2 ⇒ M = 2 2 0 I π L 29. (a) : Let us refer to the coil as circuit 1 and the solenoid as circuit 2. The field in the central region of the solenoid is uniform, so the flux through the coil is φ12 = B2A1 = (μ0n2I2)A1 where n2 = N2/l = 1500 turns m–1, The mutual inductance is Nφ M = 1 12 = μ0n2 N1 A1 I2

2 μ ia2 ⎛ μ i ⎞ πa = 0 =⎜ 0 ⎟ ⎝ 2 πb ⎠ R 2Rb

=

...(iii)

μ I φ2 = BA2 = 2 2 0 × l 2 π L

μ i 24. (d) : Change in flux, dφ = ⎜⎛ 0 ⎟⎞ πa2 ⎝ 2 πb ⎠ dφ Αs dq = R ∴ Τotal charge circulating the inner coil is

25. (a) : q =

(L1 – L2)2= (L1 + L2)2 _ 4L1L2 L1 – L2 = [102– 4 × 24]1/2 = 2 H Solving (i) and (iii), we get L1 = 6 H, L2 = 4 H

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

Substituting the value of (L1 + L2) from (i) into (ii), we get L1 L2 = (2.4)(L1 + L2) = 2.4 × 10 = 24

=

BA Bπr 2 0.01 × π(1)2 =πV = = t t ⎛ 1 ⎞ ⎜⎝ ⎟⎠ 100

Induced electric field, E =

π ε ε = = l 2 πr 2 × π × 1

= 0.5 V m–1 PHYSICS FOR YOU | APRIL ‘16

53

Exam on st 1 May

MODEL TEST PAPER 2016 1. A body of mass 1 kg is projected with velocity 50 m s–1 at an angle of 30° with the horizontal. At the highest point of its path a force of 10 N acts on body for 5 s vertically upward besides gravitational force. What is the horizontal range of the body? (g = 10 m s–2) (a) 125 3 m (c) 500 m 2. The kinetic energy along a circle of the distance s as constant. The force s2 R

(a) 2a (c) 2as

(b) 200 3 m (d) 250 3 m K of a particle moving radius R depends upon K = as2, where a is a acting on the particle is ⎡ s2 ⎤ (b) 2as ⎢1 + 2 ⎥ ⎣ R ⎦ (d) 2a

1/2

3. A hemispherical bowl of radius R is kept on a horizontal table. A small sphere of radius r(r <
4 gR / 3

(d)

10 gR / 7

4. Three plates A, B, C each of area 50 cm2 have separation 3 mm between A and B and 3 mm between B and C. The energy stored when the plates are fully charged is (a) 1.6 × 10–9 J (b) 2.1 × 10–9 J –9 (c) 5 × 10 J (d) 7 × 10–9 J 5. In the circuit shown in the figure, ammeter and voltmeter are ideal. If ε = 4 V, R = 9 Ω and r = 1 Ω , then readings of ammeter and voltmeter are 54

PHYSICS FOR YOU

| APRIL ‘16

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

1 A, 3 V 2 A, 3 V 1 A, 4 V 4 A, 4 V

E

6. Some magnetic flux is changed through a coil of resistance 10 Ω. As a result an induced current is developed in it, which varies with time as shown in the figure. The magnitude of change in the flux through the coil in Wb is (a) 2 (b) 4 (c) 6 (d) none of these 7. An AC source is connected with a resistance (R) and an uncharged capacitance (C), in series. The potential difference across the resistor is in phase with the initial potential difference across the capacitor for the first time at the instant (assume that at t = 0, emf is zero). 3π 2π π π (a) (b) (c) (d) 2ω ω 2ω 4ω 8. The bob of a simple pendulum of length 1.2 m has a velocity of 7 m s–1 when it is at the lowest point. The bob would leave the circular path above the centre at a height (a) 1.0 m (b) 0.867 m (c) 0.652 m (d) 0.512 m 9. What is the height at which the weight of body will be the same as at the same depth from the surface of the earth? Radius of earth is R. (a) R (b) 5 R − R 2 3R−R 5R−R (c) (d) 2 2

10. A steel wire of length 20 cm and uniform cross sectional area 1 mm2 is tied rigidly at both the ends. The temperature of the wire is altered from 20°C to 40°C. Coefficient of linear expansion of steel is α = 1.1 × 10–5 per °C and Y for steel is 2.0 × 1011 N m–2; the tension in wire is (a) 2.2 × 106 N (b) 16 N (c) 8 N (d) 44 N 11. The dimensional formula of a physical quantity is [M–1 L–3 T2]. The respective errors in measuring M, L, T are 2%, 3%, 4%. The maximum percentage error in measuring the quantity is (a) 2% (b) 9% (c) 8% (d) 19%

16. Two identical magnetic dipoles of magnetic moment 2 A m2 are placed at a separation of 2 m with their axes perpendicular to each other in air. The resultant magnetic field at a midpoint between the dipoles is (a) 4 5 × 10 −5 T (b) 2 5 × 10 −5 T −7 (c) 4 5 × 10 T (d) 2 5 × 10 −7 T 17. An object is kept at a distance of 16 cm from a thin lens and the image formed is real. If the object is kept at a distance of 6 cm from the same lens, the image formed is virtual. If the sizes of the image formed are equal then focal length of the lens will be (a) 21 cm (b) 11 cm (c) 15 cm (d) 17 cm

(c) 10 i + 10 2 j + 10 k (d) 5 i + 5 2 j + 10 k 13. The given graph (m s–1) shows the variation of velocity of a 1000 rocket with time. Find the time of 110 120 0 10 (s) burning of fuel from the graph. (a) 10 s (b) 110 s (c) 120 s (d) cannot be estimated from the graph

18. A charged oil drop falls with terminal velocity v0 in the absence of electric field. An electric field E keeps it stationary. The drop acquires charge 3q, it starts moving upwards with velocity v0. The initial charge on the drop is q 3q (a) (b) q (c) (d) 2q 2 2 19. Which one of the following statements regarding photo-emission of electrons is correct? (a) Kinetic energy of electrons increases with the intensity of incident light. (b) Electrons are emitted when the wavelength of the incident light is above a certain threshold wavelength. (c) Photoelectric emission is instantaneous with the incidence of light. (d) Photo electrons are emitted whenever a gas is irradiated with ultraviolet light.

14. In Young’s double slit experiment, the y-coordinates of central maxima and 10th maxima are 2 cm and 5 cm respectively. When the YDSE apparatus is immersed in a liquid of refractive index 1.5, the corresponding y-coordinates will be (a) 2 cm, 7.5 cm (b) 3 cm, 6 cm (c) 2 cm, 4 cm (d) 4/3 cm, 10/3 cm

20. A ball of mass m is thrown vertically upwards. Another ball of mass 2m is thrown at angle θ with the vertical. Both of them stay in air for the same period of time. The heights attained by the two balls are in the ratio of (a) 2 : 1 (b) cos θ : 1 (c) 1 : cos θ (d) 1 : 1

15. Light of wavelength 330 nm falling on a piece of metal ejects electrons with sufficient energy which requires voltage V0 to prevent them from reaching a collector. In the same setup, light of wavelength 220 nm, ejects electrons which require twice the voltage V0 to stop them from reaching a collector. The numerical value of voltage V0 is 8 15 15 16 V (a) V (d) V (c) V (b) 15 8 16 15

21. A block of mass 0.5 kg is moving with a speed of 2.0 m s–1 on a smooth surface. It strikes another mass of 1.0 kg at rest and then they move together as a single body. The energy loss during the collision is (a) 0.16 J (b) 1.00 J (c) 0.67 J (d) 0.34 J

12. A bird moves with velocity 20 m s–1 in a direction making an angle of 600 with the eastern line and 600 with vertical upward line. The velocity vector of bird in cartesian coordinates is (a) 10 i + 10 j + 10 k

(b) 10 i + 5 2 j + 10 k

22. The temperature of equal masses of three different liquids A, B and C are 12 °C, 19 °C and 28 °C respectively. The final temperature when A and B are mixed is 16 °C and when B and C are mixed is 23 °C. PHYSICS FOR YOU

| APRIL ‘16

55

The final temperature when A and C are mixed is (a) 18.2 °C (b) 22.2 °C (c) 20.2 °C (d) 24.2 °C 23. The specific heat of the mixture of two gases at constant volume is 13 R/6. The ratio of the number of moles of the first gas to the second is 1 : 2. The respective gases may be (a) He, Ne (b) He, N2 (c) N2, O2 (d) N2, He 24. A spring has force constant K and a mass m is suspended from it. The spring is cut in two parts in the ratio 1 : 3, and the same mass is suspended from the larger part. If the frequency of oscillation in the first case is υ, then the frequency in the second case will be 3 2υ υ (a) 2 υ (b) 3υ (c) (d) 2 3 25. A string is under tension so that its length is 1 increased by times its original length. The ratio of n fundamental frequency of longitudinal vibrations and transverse vibrations will be (a) 1 : n (b) n2 : 1 (c) n :1 (d) n : 1 26. The driver of a car travelling with speed 30 m s–1 towards a hill sounds a horn of frequency 600 Hz. If the velocity of sound in air is 330 m s–1, the frequency of reflected sound as heard by driver is (a) 500 Hz (b) 550 Hz (c) 555.5 Hz (d) 720 Hz 27. When forward bias is applied to a p-n junction, then what happens to the potential barrier VB' and the width of charge depleted region x ? (a) VB increases, x decreases (b) VB decreases, x increases (c) VB increases, x increases (d) VB decreases, x decreases 28. The half-life of radium is 1620 yr and its atomic weight is 226 g mol–1 The number of atoms that will decay from its 1 g sample per second will be (Avogadro's number N = 6.023 × 1023 atom per mole) (a) 3.61 × 1010 Bq (b) 3.61 × 1012 Bq (c) 3.11 × 1015 Bq (d) 31.1 × 1015 Bq 29. Which of the following statements is incorrect regarding the given LCR circuit ? (a) Voltage will lead the current. 56

PHYSICS FOR YOU

| APRIL ‘16

(b) rms value of current is 20 A. (c) Power factor of the circuit is

1

. 2 (d) Voltage drop across resistance is 200 V.

30. The potential difference across 100 Ω resistance in the circuit is measured by a voltmeter of 900 Ω resistance. The percentage error in reading the potential difference is 10 (a) 9 (b) 0.1 (c) 1.0 (d) 10.0 31. A charge q is moving with a velocity v1 = 1i m s–1 at a point in a magnetic field and experiences a force F = q[− j + k] N . If the charge is moving with a velocity v2 = 1 j m s–1 at the same point, it experiences a force F2 = q(i − k ) N. The magnetic induction at that point is −2 −2 (a) (i + j + k) Wb m (b) (i − j + k) Wb m −2 −2 (c) (−i + j − k ) Wb m (d) (i + j − k) Wb m

32. A light string passing over a smooth light pulley connects two blocks of masses m1 and m2 (vertically). If the acceleration of the masses is (g/8), then the ratio of masses is (a) 8 : 1 (b) 9 : 7 (c) 4 : 3 (d) 5 : 3 33. The velocity of a body moving in a vertical circle of radius r is 7gr at the lowest point of the circle. What is the ratio of maximum and minimum tension ? (a) 4 : 1 (b) 7 : 1 (c) 3 : 1 (d) 2 : 1 34. A coin of mass m and radius r having moment of inertia I about the axis passes through its centre and perpendicular to its plane. It is beaten uniformly to form a disc of radius 2r. What will be the moment of inertia about the same axis? (a) I (b) 2I (c) 4I (d) 16I 35. A spherical soap bubble of radius 1 cm is formed inside another bubble of radius 3 cm. The radius of a single soap bubble which maintains the same pressure difference as inside the smaller and outside the larger soap bubble is (a) 0.75 cm (b) 0.75 m (c) 7.5 cm (d) 7.5 m

36. A heat engine has an efficiency η. Temperatures of source and sink are each decreased by 100 K. Then, the efficiency of the engine (a) increases (b) decreases (c) remains constant (d) becomes 1 37. Two masses m1 and m2 are suspended together by a massless spring of force constant k. When the masses are in equilibrium, m1 is removed without disturbing the system. The amplitude of oscillation is mg m g (b) 2 (a) 1 k k (m1 − m2 ) g (m1 + m2 ) g (c) (d) k k 38. A beam of light of wavelength 600 nm from a distant source falls on a single slit 1.00 mm wide and the resulting diffraction pattern is observed on a screen 2 m away. The distance between the first dark fringes on either side of the central bright fringe is (a) 1.2 cm (b) 1.2 mm (c) 2.4 cm (d) 2.4 mm 39. The instantaneous magnitudes of the electric field (E) and the magnetic field (B) vectors in an electromagnetic wave propagating in vacuum are related as B (a) E = (b) E = cB c B (c) E = 2 (d) E = c2B c 40. The threshold frequency for a photosensitive metal is 3.3 × 1014 Hz. If light of frequency 8.2 × 1014 Hz is incident on this metal, the cut-off voltage for the photoelectron emission is nearly (a) 1 V (b) 2 V (c) 3 V (d) 5 V 41. Half life of a radioactive substance A is two times the half life of another radioactive substance B. Initially, the number of nuclei of A and B are NA and NB respectively. After three half lives of A, number N of nuclei of both are equal. Then the ratio A is NB 1 1 1 1 (b) (c) (d) (a) 3 6 4 8 42. The transition in He+ ion that will give rise to a spectral line having the same wavelength as that of some spectral line in hydrogen atom is (a) n = 3 to n = 1 (b) n = 3 to n = 2 (c) n = 4 to n = 2 (d) n = 4 to n = 3 43. One way in which the operation of an n-p-n transistor differs from that of a p-n-p transistor is that

(a) the emitter junction is reverse biased in the n-p-n and forward biased in the p-n-p (b) the emitter injects minority carriers into the base region of the p-n-p and majority carriers in the base region of the n-p-n (c) the emitter injects holes into the base region of the p-n-p and electrons into the base region of the n-p-n (d) the emitter injects electrons into the base region of the p-n-p and holes into the base region of the n-p-n. 44. An AC source of variable frequency υ is connected to an L-C-R series circuit. Which one of the graphs in figure represents the variation of current I in the circuit with frequency υ ?

(a)

(b) U

U

(d)

(c) U

U

45. The insulation property of air breaks down at E = 3 × 106 V m–1. The maximum charge that can be given to a sphere of diameter 5 m is approximately (in coulombs) (a) 2 × 10–2 (b) 2 × 10–3 –4 (c) 2 × 10 (d) 2 × 10–5 SOLUTIONS 1. (d) : For 5 s, weight of the body is balanced by the given force. Hence, it will move in a straight line as shown in the figure.

R= =

u 2 sin 2 θ g

+ (u cos θ)(5)

(50)2 ⋅ sin 60° + (50 × cos 30°)(5) = 250 3 m 10 PHYSICS FOR YOU

| APRIL ‘16

57

1 2 2 2. (b) : Here,K = mv = as ∴ mv 2 = 2as 2 2 Differentiating with respect to time t dv ds dv 2mv = 4as = 4asv ⇒ m = 2as dt dt dt ∴ Tangential force, Ft = 2as mv 2 2as 2 Centripetalforce, Fc = = R R ∴ Net force acting on the particle 2

F=

Ft2

+ Fc2

⎛ 2as 2 ⎞ = 2as 1 + s 2 / R2 = (2as) + ⎜ ⎝ R ⎟⎠ 2

3. (d) : As, it is clear from figure, on reaching the bottom of the bowl, loss in potential energy = mgR, and gain in kinetic energy 1 1 = mv 2 + Iω 2 2 2 1 2 1 ⎛ 2 2 ⎞ v2 = mv + × ⎜ mr ⎟ 2 ⎠r 2 2 ⎝5 1 2 1 2 7 = mv + mv = mv 2 2 5 10 As, gain in KE = loss in PE 7 ∴ mv 2 = mgR 10 10 gR ⇒ v= 7 4. (b) : Here circuit is equivalent to two capacitors in parallel. ε A ε A 2ε A ∴ Ceq = C1 + C2 = 0 + 0 = 0 d d d 1 1 ⎛ 2ε A ⎞ ∴ Energystored = Ceq V 2 = ⎜ 0 ⎟ V 2 2 2⎝ d ⎠ =

8.86 × 10 −12 × 50 × 10 −4 × 12 × 12

= 2.1 × 10–9 J 3 × 10 −3 5. (a) : The given three resistors are in parallel

⎛ 1⎞ i V = R = ⎜⎝ ⎟⎠ (9) = 3 V 3 3 6. (a) : Charge induced in coil dφ = i dt = Area under i-t graph dq = R ∴ dφ = (Area under i-t graph) R 1 = × 4 × 0.1 × 10 = 2 Wb 2 (as V = 0 at t = 0) 7. (d) : Let V = V0 sin ωt Then VR = V0 sin ωt VC = V0 sin(ωt − π / 2) and V and VR are in same phase. While VC lags V (or VR) by 900. Now VR is in same phase with initial potential difference across the capacitor for the first time when, π 3π ωt = − + 2 π = 2 2 3π ∴ t= 2ω 8. (b) : Here, r = l = 1.2 m, u = 7 m s–1 2 gr = 2 × 9.8 × 1.2 = 4.85 m s −1

Now, and

5 gr = 5 × 9.8 × 1.2 = 7.67 m s −1

The condition 2 gr < u < 5 gr is satisfied. Therefore, the bob will leave the vertical circle at P, where component of weight along PO is just equal to centripetal force (mv2/r). mv 2 i.e., mg cos θ = r

∴ i=

58

ε 4 = = 1A r + R / 3 1+ 3

PHYSICS FOR YOU

| APRIL ‘16

or v2 = rg cos θ = gh ...(i) Height of P above the bottom = r + h At P, loss in KE = gain in PE 1 1 mu2 − mv 2 = mg (r + h) 2 2 or u2 – v2 = 2g (r + h) using (i), u2 = gh + 2gh + 2gr = g (3h + 2r) ⎤ ⎤ 1 ⎡ 72 1 ⎡ u2 − 2 × 1.2⎥ = 0.867 m h = ⎢ − 2r ⎥ = ⎢ 3⎣ g ⎦ ⎦ 3 ⎣ 9. 8 9. (c) : Weight of a body at height h = weight of the body at depth h g R2

E

Q

⎛ h⎞ = g ⎜1 − ⎟ ⎝ R⎠ ( R + h) ⎛ h ⎞ ⎛ h2 2h ⎞ or ⎜ 1 − ⎟ ⎜ 1 + 2 + ⎟ = 1 ⎝ R⎠ ⎝ R R⎠ 2

h3

h ⎛ h2 h ⎞ + − 1⎟ = 0 ⎜ R ⎝ R2 R ⎠

h2

or

h + 2 − = 0 or 3 R R R

or

5 −1 h −1 ± 1 + 4 = = or 2 2 R

5R − R h= 2

10. (d) : Increase in length due to rise in temperature, ΔL = αL Δ T FL As, Y = ; AΔL so F =

YA ΔL L

=

YA × αL ΔT L

eV0 = e (2V0 ) =

hc 330 × 10 −9 hc

−W

...(i)

−W ...(ii) 220 × 10 −9 Solving these two equations, we get 109 × h × c 109 × 6.6 × 10 −34 × 3 × 108 15 V0 = = volt = 110 × e × 6 8 110 × 1.6 × 10 −19 × 6 16. (d) :

= YA α ΔT

∴ F = 2 × 1011 × 10 −6 × 1.1 × 10 −5 × 20 = 44 N 11. (d) : Here, x = [M L T ] Δx ΔM ⎛ ΔT ⎞ ⎛ ΔL ⎞ + 2⎜ = + 3⎜ ⎝ T ⎟⎠ ⎝ L ⎟⎠ x M ∴ % error in the quantity x = 2% + 3 (3%) + 2 (4%) = 19%

Let point P be the midpoint between the dipoles. The point P will be in end-on position with respect to one dipole and in broad-side on position with respect to the other. μ 2 M 10 −7 × 2 × 2 ∴ B1 = 0 3 1 = = 4 × 10 −7 T 4 π r1 (1)3

12. (c) : Let eastern line be x-axis, northern line as → y-axis and vertical upward line as z-axis. Let v make angles α, β and γ with x, y and z-axis respectively. Then α = 60°, γ = 60°

μ 0 M2 10 −7 × 2 = = 2 × 10 −7 T 3 3 4 π r2 (1) As B1 and B2 are perpendicular to each other, therefore the resultant magnetic field at point P is

–1

–3

2

As cos2 α + cos2 β + cos2 γ = 1

B2 =

B = B12 + B22 = (4 × 10 −7 )2 + (2 × 10 −7 )2

∴ cos2 60° + cos2 β + cos2 γ = 1 or (1 / 2)2 + cos2 β + (1 / 2)2 = 1 or cos β =

and

−7

1 2

→

Now v = (v cos α) i + (v cos β) j + (v cos γ )k = (20 cos 60°) i + (20 × 1 / 2 ) j + (20 cos 60°)k = 10i + 10 2 j + 10k 13. (a) : When the fuel is burning, velocity of the rocket is increasing. After the fuel is exhausted, velocity starts decreasing. From the graph, time of burning of fuel = 10 s. Dλ , i.e., β ∝ λ d So, wavelength λ and hence fringe width β decreases 1.5 times when immersed in liquid. The distance between central maxima and 10th maxima is 3 cm in vacuum. When immersed in liquid it will reduce to 2 cm. Position of central maxima will not change while 10th maxima will be obtained at y = 4 cm.

14. (c) : Fringe width, β =

15. (c) : Let W be the work function of metal. Then

= 10 −7 16 + 4 = 10 −7 20 = 2 5 × 10 T 17. (b) : As a convex lens alone can form a real image as well as a virtual image, therefore, the lens in the present question is a convex lens. Let, f be the focal length of the lens and m be the magnification produced. In the first case, when image is real, u = –16 cm, v = (m × 16) cm 1 1 1 − = As v u f 1 1 1 1 16 ...(i) ∴ + = or 1 + = 16m 16 f m f In the second case, when image is virtual. u = – 6 cm, v = (–6m) cm 1 1 1 − = From v u f 1 1 1 1 6 + = or 1 − = ...(ii) −6m 6 f m f Adding eqn (i) and eqn (ii), we have 22 22 = 11 cm 2 = or f = f 2 PHYSICS FOR YOU

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59

18. (c) : When drop is stationary, then q1E = Weight –Upthrust q1E = 6πηrv0 or q1 = 6πηrv0 / E When drop moves upwards, then 6πηr (v0 + v0 ) ⎛ 6πηrv0 ⎞ 3q = =2×⎜ ⎟ = 2q1 ⎝ E E ⎠ 3 ∴ q1 = q 2 19. (c) 20. (d) : As per question, time of flight for both the balls is equal. 2u1 2u2 cos θ = or u1 = u2 cos θ ...(i) g g 1 For first ball ; mu12 = mgh1 2 1 For second ball ; 2m (u2 cos θ)2 = 2mgh2 2 1 2 or mu1 = mgh2 [From (i)] 2 ∴ h1 = h2

...(ii) ...(iii)

21. (c) : Here, m1 = 0.5 kg, u1=2.0 m s–1 m2 = 1.0 kg, u2 = 0 If v is velocity of the system after collision, then according to the principle of conservation of linear momentum, (m1 + m2) v = m1 u1 + m2 u2 = m1 u1 m1 u1 0. 5 × 2. 0 = v= m1 + m2 0.5 + 1.0 1. 0 2 = m s −1 1. 5 3 Energy loss = initial energy – final energy 1 1 = m1 u12 − (m1 + m2 )v 2 2 2 2 1 1 ⎛ 2⎞ 2 = × 0.5 × (2.0) − (0.5 + 1.0) ⎜ ⎟ ⎝ 3⎠ 2 2 1. 5 4 = 1. 0 − × = 0.67 J 2 9 v=

22. (c) : Let m1 = m2 = m3 = m Let s1, s2, s3 be the respective specific heats of the three liquids. (i) When A and B are mixed, temperature of mixture = 160 C As heat gained by A = heat lost by B ∴ ms1 (16 –12) = ms2 (19 – 16) 60

PHYSICS FOR YOU

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4s1 = 3s2 ...(i) (ii) When B and C are mixed, temperature of mixture = 230C As heat gained by B = heat lost by C, ms2 (23 – 19) = ms3 (28 – 23) ∴ 4s2 = 5s3 ...(ii) 3 15 From eqns (i)and (ii), s1 = s2 = s3 4 16 When A and C are mixed, suppose temperature of mixture = t heat gained A = heat lost by C ms1 (t – 12) = ms3 (28 – t) 15 s3 (t − 12) = s3 (28 − t ) 16 15t – 180 = 448 – 16t 31t = 448 + 180 = 628 ∴ t = 20.2°C 23. (b) : As CV = or

n1 CV 1 + n2 CV 2 n1 + n2

1C + 2 CV 2 13 13 R = V1 or CV 1 + 2 CV 2 = R 6 1+ 2 2

3 If first gas is monoatomic, then CV1 = R and 2 5 second gas is diatomic, then CV 2 = R . 2 5 13 3 ∴ CV 1 + 2CV 2 = R + 2 × R = R 2 2 2 It means the option (b) is true. 24. (c) : Let k be the force constant of smaller piece of spring. Then the longer piece is a combination of three smaller pieces in series. Their effective force constant k1 = k/3. In a bigger spring, the smaller pieces are connected in series, so 1 1 3 4 = + = or k = 4 K K k k k 4K ∴ k1 = 3 1 k1 1 4K / 3 2υ 1 K and υ′ = = = υ= m 2π m 2π 2π m 3 25. (c) : Velocity of longitudinal waves, v1 = velocity of transverse waves, v2 =

T m

If a is area of cross-section of string mass mass = × area = ρ a then m = length volume

Y ρ

∴ v2 = As Y = ∴

v1 = v2

T v1 Ya Y ρa , = ⋅ = ρ a v2 T ρ T F T = aΔ l / l a(Δ l / l ) T a ⎛ Δl⎞ =⎜ ⎟ ⎛ Δl⎞ T ⎝ l ⎠ a⎜ ⎟ ⎝ l ⎠

we are given, ∴

v1 ⎛ 1 ⎞ =⎜ ⎟ v2 ⎝ n ⎠

Δl l

=

−1/2

1 n

= n

26. (d) : As the source is moving towards the hill (the listener), therefore, apparent frequency of horn striking the hill is v 330 × 600 = 660Hz υ′ = ×υ= v − vs 330 − 30 For the reflected sound, driver acts as listener moving towards source (the hill) (v + v L )υ (330 + 30) 660 ∴ υ′′ = = = 720H z v 330 27. (d) : In forward biasing, both potential barrier VB and the width of charge depleted region x decreases. 28. (a) : Rate of change of radioactive nuclei, ⎛ dN ⎞ ⎟ = λN ⎜⎝ − dt ⎠ 0.693 0.693 ∵ λ= = 1620 × 365 × 24 × 60 × 60 T 12/ 6.023 × 1023 and N = 226 0.693 × 6.023 × 1023 ⎛ dN ⎞ = ∴⎜− ⎝ dt ⎟⎠ 1620 × 365 × 24 × 60 × 60 × 226 = 3.61 × 1010 Bq 29. (a) : XC > XL. Hence, current will lead the voltage. Z = R2 + ( XC − X L ) = 10 2 Ω 2

I rms =

10 ⎛ 100 ⎞ 30. (c) : V V = V1 = V actual = ⎜ ⎝ 110 ⎟⎠ 11 Equivalent resistance of 100 Ω and 900 Ω is 90 Ω 9 ⎛ 90 ⎞ Vmeasured = ⎜ V = V = V2 ⎟ ⎝ 90 + 10 ⎠ 10 V − V2 % error = 1 × 100 V1

−1/2

If υ1 and υ2 are the fundamental frequencies of longitudinal and transverse waves then υ v1 = υ1 λ and v2 = υ2 λ. ∴ 1 = n υ2

∴

1 R = Z 2 VR = Irms R = (20)(10) = 200 V cos φ =

Vrms Z

=

400 / 2 10 2

= 20 A

=

⎛ 10 ⎞ ⎛ 9 ⎞ ⎜⎝ V ⎟⎠ − ⎜⎝ ⎟⎠ V 11 10 ⎛ 10 ⎞ ⎜⎝ ⎟⎠ V 11

× 100 = 1%

31. (a) : Let the magnetic field is B = B1 i + B2 j + B3 k Applying F m = q(v × B) we have, q[− j + k] = q[(i) × (B1 i + B2 j + B3 k)] = q[B2 k − B3 j] Comparing two sides, we get, B2 = 1 and B3 = 1 Further, q[i − k] = q[( j) × (B1 i + B2 j + B3 k)] = q[− B1 k + B3 i] Again comparing we get, B1 = 1 and B3 = 1 ∴ B = (i + j + k) Wb m −2 32. (b) : In the given system, a =

(m1 − m2 ) g g = 8 m1 + m2

m1 − m2 1 = m1 + m2 8 8m1 – 8m2 = m1 + m2 m1 9 = 7m1 = 9m2 or m2 7 33. (a) : Tension is maximum at the lowest point and minimum at the highest point. ∴

Tension at the lowest point,

TL = mg +

mv L2 7mgr = mg + = 8mg r r

(∵ v

L

= 7gr

Tension at the highest point,

mv H2 m(v L2 − 4 gr ) − mg = − mg r r m(7gr − 4 gr ) = − mg ∵ v L2 − v H2 = 4 gr r = 3mg – mg = 2mg T TL 4 4 ∴ = or max = TH 1 Tmin 1 TH =

(

PHYSICS FOR YOU

| APRIL ‘16

)

61

)

34. (c) : Moment of inertia of coin of mass m and radius r about the axis passing through the centre of mass and perpendicular to its plane is mr 2 I= ...(i) 2 Moment of inertia of disc of mass m and radius 2r about the axis passing through the centre of mass and perpendicular to its plane is m (2r )2 (Using (i)) I′ = = 4I 2 35. (a) : Pressure outside the bigger drop = P1 Pressure inside the bigger drop = P2 Radius of bigger drop, r1 = 3 cm 4S 4S Excess pressure = P2 − P1 = = r1 3 Pressure inside small drop = P3 4S 4S Excess pressure = P3 – P2 = = 1 r2 Pressure difference between inner side of small drop and outer side of bigger drop 4S 4S 16S = P3 − P1 = + = 3 1 3 This pressure difference should exist in a single drop of radius r. 4S 16S 3 or r = cm = 0.75 cm ∴ = r 3 4 36. (a) : Efficiency of the heat engine, T T −T η =1− 2 = 1 2 T1 T1 where T1 and T2 are the temperatures of source and sink respectively. When T1 and T2 both are decreased by 100 K each, (T1 – T2) stays constant. T1 decreases. ∴ η increases. (m + m2 ) g 37. (a) : For equilibrium of (m1 + m2 ), x1 = 1 k m2 g and for equilibrium of m2 , x2 = k ∴ Amplitude of oscillation will be (m + m2 ) g m2 g m1 g A = x1 − x2 = 1 − = k k k 38. (d) : In case of diffraction at a single slit, the position of minima is given by dsinθ = nλ y If θ is small, sinθ = θ = D So, the position of first minimum relative to centre will be given by d(y/D) = λ, i.e., y = (D/d)λ 62

PHYSICS FOR YOU

| APRIL ‘16

Here, D = 2 m; d = 1 × 10–3 m and λ = 6 × 10–7 m 2 × 6 × 10 −7 So, y = = 1.2 mm 1 × 10 −3 ∴ Distance between first minima on either side of central maxima, Δy = 2y = 2.4 mm. 39. (b) : At every instant, the ratio of the magnitude of the electric field to that of the magnetic field in an electromagnetic wave equals the speed of light. 40. (b) : According to Einstein’s photoelectric equation eVs = hυ – h υ0 where, υ = Incident frequency υ0 = Threshold frequency Vs = Cut-off or stopping potential h or Vs = (υ − υ0 ) e Substituting the given values, we get Vs =

6.63 × 10 −34 (8.2 × 1014 − 3.3 × 1014 ) 1.6 × 10 −19

≈2V

41. (c) : Three half lives of A are equivalent to six half lives of B. As number of nuclei left are equal in the two cases. 3 6 ⎛ 1⎞ ⎛ 1⎞ Therefore, NA ⎜⎝ 2 ⎟⎠ = NB ⎜⎝ 2 ⎟⎠ 3

NA (1 / 2)6 ⎛ 1 ⎞ 1 = =⎜ ⎟ = 3 8 NB (1 / 2) ⎝ 2 ⎠

42. (c) : To have the same wavelength of some spectral lines from different hydrogen like atoms, one must have 1⎤ 1⎤ ⎡1 2 ⎡ 1 ⎢ 2 − 2⎥=Z ⎢ 2 − 2⎥ ⎢⎣ n f ni ⎥⎦ ⎢⎣ p f pi ⎥⎦ where nf and ni are principal quantum numbers of final and initial orbits for hydrogen atom and pf and pi are those for He+ ion. Also, Z = 2. This gives pf = 2nf and pi = 2ni. 43. (c) 44. (c) : At resonance frequency, current is maximum. 45. (b) : Electric field on the surface of a conducting sphere q 1 is E = ⋅ 2 4 πε0 r ∴ q = Er 2 ⋅ 4πε0 3 × 106 × (2.5)2 = = 2.08 × 10 −3 C 9 × 109

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 aĴempt only one of the choices in such questions.

SECTION-A 1. Write the underlying principle of a moving coil galvanometer. 2. Why are microwaves considered suitable for radar systems used in aircraft navigation? 3. Define 'quality factor' of resonance in series LCR circuit. What is its SI unit? 4. A point charge +Q is placed at point O as shown in the figure. Is the potential difference VA – VB positive, negative or zero ? 5. How does the electric flux due to a point charge enclosed by a spherical Gaussian surface get affected when its radius is increased ? SECTION-B 6. A nucleus with mass number A = 240 and BE/A = 7.6 MeV breaks into two fragments each of A = 120 with BE/A = 8.5 MeV. Calculate the released energy. OR Calculate the energy in fusion reaction : , where BE of = 2.23 MeV

and of

= 7.73 MeV.

7. Two cells of emfs 1.5 V and 2.0 V having internal resistances 0.2 Ω and 0.3 Ω respectively are

connected in parallel. Calculate the emf and internal resistance of the equivalent cell. 8. State Brewster's law. The value of Brewster angle for a transparent medium is different for light of different colours. Give reason. 9. Explain the terms (i) Attenuation and (ii) Demodulation used in Communication System. 10. Plot a graph showing variation of de-Broglie 1 wavelength λ versus , where V is accelerating V potential for two particles A and B carrying same charge but of masses m1, m2 (m1 > m2). Which one of the two represents a particle of smaller mass and why? SECTION-C 11. (i) Define mutual inductance. (ii) A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the other coil ? 12. Two parallel plate capacitors X and Y have the same area of plates and same separation between them. X has air between the plates while Y contains a dielectric of εr = 4. PHYSICS FOR YOU | APRIL ‘16

63

(i) Calculate capacitance of each capacitor if equivalent capacitance of the combination is 4 μF. (ii) Calculate the potential difference between the plates of X and Y. (iii) Estimate the ratio of electrostatic energy stored in X and Y. 13. Two long straight parallel conductors carry steady current I1 and I2 separated by a distance d. If the currents are flowing in the same direction, show how the magnetic field set up in one produces an attractive force on the other. Obtain the expression for this force. Hence define one ampere. 14. A charge is distributed uniformly over a ring of radius 'a'. Obtain an expression for the electric intensity E at a point on the axis of the ring. Hence show that for points at large distances from the ring, it behaves like a point charge. 15. Write three characteristic features in photoelectric effect which cannot be explained on the basis of wave theory of light, but can be explained only using Einstein's equation. 16. (a) Write the expression for the magnetic force acting on a charged particle moving with velocity v in the presence of magnetic field B. (b) A neutron, an electron and an alpha particle moving with equal velocities, enter a uniform magnetic field going into the plane of the paper as shown. Trace their paths in the field and justify your answer.

17. (a) Calculate the distance of an object of height h from a concave mirror of radius of curvature 20 cm, so as to obtain a real image of magnification 2. Find the location of image also. (b) Using mirror formula, explain why does a convex mirror always produce a virtual image. 64


18. (i) State Bohr's quantization condition for defining stationary orbits. How does de Broglie hypothesis explain the stationary orbits ? (ii) Find the relation between the three wavelengths λ1, λ2 and λ3 from the energy level diagram shown below.

19. Draw a schematic ray diagram of reflecting telescope showing how rays coming from a distant object are received at the eye-piece. Write its two important advantages over a refracting telescope. 20. How are em waves produced by oscillating charges ? Draw a sketch of linearly polarized em waves propagating in the Z-direction. Indicate the directions of the oscillating electric and magnetic fields. OR Write Maxwell's generalization of Ampere's Circuital Law. Show that in the process of charging a capacitor, the current produced within the plates of the capacitor is dΦ i = ε0 E dt where ΦE is the electric flux produced during charging of the capacitor plates. 21. (a) Explain any two factors which justify the need of modulating a low frequency signal. (b) Write two advantages of frequency modulation over amplitude modulation. 22. (i) Write the functions of three segments of a transistor. (ii) Draw the circuit diagram for studying the input and output characteristics of n-p-n transistor in common emitter configuration. Using the circuit, explain how input, output characteristics are obtained. SECTION-D 23. Meeta's father was driving her to the school. At the traffic signal she noticed that each traffic light was made of many tiny lights instead of a single bulb. When Meeta asked this question to her father, he explained the reason for this.

Answer the following questions based on above information : (i) What were the values displayed by Meeta and her father ? (ii) What answer did Meeta's father give ? (iii) What are the tiny lights in traffic signals called and how do these operate ? SECTION-E 24. (i) Define the term drift velocity. (ii) On the basis of electron drift, derive an expression for resistivity of a conductor in terms of number density of free electrons and relaxation time. On what factors does resistivity of a conductor depend ? (iii) Why alloys like constantan and manganin are used for making standard resistors ? OR

(a) number of turns in secondary (b) current in primary (c) voltage across secondary (d) current in secondary (e) power in secondary 26. (i) In Young's double slit experiment, deduce the condition for (a) constructive, and (b) destructive interference at a point on the screen. Draw a graph showing variation of intensity in the interference pattern against position 'x' on the screen. (ii) Compare the interference pattern observed in Young's double slit experiment with single slit diffraction pattern, pointing out three distinguishing features. OR

(i) Plot a graph to show variation of the angle of deviation as a function of angle of incidence for light passing through a prism. Derive an expression for refractive index of the prism in terms of angle of minimum deviation and angle of prism. (ii) What is dispersion of light ? What is its cause ? (iii) A ray of light incident normally on one face of a right isosceles prism is totally reflected as shown in figure. What must be the minimum value of refractive index of glass ? Give relevant calculations.

(i) State the principle of working of a potentiometer. (ii) In the following potentiometer circuit AB is a uniform wire of length 1 m and resistance 10 Ω. Calculate the potential gradient along the wire and balance length AO (= l ).

25. (i) An a.c. source of voltage V = V0 sinωt is connected to a series combination of L, C and R. Use the phasor diagram to obtain expressions for impedance of the circuit and phase angle between voltage and current. Find the condition when current will be in phase with the voltage. What is the circuit in this condition called ? (ii) In series LR circuit XL = R and power factor of the circuit is P1. When capacitor with capacitance C such that XL = XC is put in series, the power factor becomes P2. Calculate P1/P2. OR (i) Write the function of a transformer. State its principle of working with the help of a diagram. Mention various energy losses in this device. (ii) The primary coil of an ideal step up transformer has 100 turns and transformation ratio is also 100. The input voltage and power are respectively 220 V and 1100 W. Calculate

SOLUTIONS 1.

When a current carrying coil is suspended in a uniform magnetic field, a torque acts on it, magnitude of which depends on the strength of current. This torque tends to rotate the coil about the axis of suspension, so that the magnetic flux passing through the coil is maximum.

2.

Microwaves have short wavelengths so they are suitable for radar systems used in aircraft navigation. They can penetrate through clouds also.

3.

The quality factor (Q) of resonance in series LCR circuit is defined as the ratio of voltage drop across inductor (or capacitor) to the applied voltage, I X ω L V 1 i.e., Q = L = 0 L = 0 = ω0CR VR I0 R R It is an indicator of sharpness of the resonance. Quality factor has no unit. PHYSICS FOR YOU | APRIL ‘16

65

4.

εeq = ∴ Potential difference due to a point charge Q at a distance r is given by 1 Q V= 4 πε0 r ∴ From the given figure 1 Q 1 Q , VB = VA = 4 πε0 rA 4 πε0 rB 1 Q 1 Q ∴ VA − VB = − 4 πε0 rA 4 πε0 rB =

5.

According to question when radius of spherical Gaussian surface is increased, its surface area will be increased but point charge enclosed in the sphere remains same. Hence there will be no change in the electric flux. 6.

For a big nucleus, A = 240, BE/A = 7.6 MeV Initial binding energy = 240 × 7.6 = 1824 MeV For two small nuclei, A = 120, BE/A = 8.5 MeV Final binding energy = 2 × 120 × 8.5 = 2040 MeV Energy released during fission = (final BE) – (Initial BE) = 2040 – 1824 = 216 MeV OR Fusion reaction, Energy released = Final BE – Initial BE = 7.73 – (2.23 + 2.23) = 3.27 MeV

7.

66

Brewster's law : The tangent of the polarizing angle of incidence of a transparent medium is equal to its refractive index, i.e., μ = tan(ip) Brewster angle, ip = tan–1(μ) Refractive index of a transparent medium depends on the wavelength of light which falls on the medium. So a transparent medium has different values of refractive index for light of different colours. Hence the value of Brewster angle for a transparent medium is different for light of different colours.

9.

(i) Attenuation : The loss of strength of signal during its propagation through the transmission medium is called attenuation. Repeater is used to compensate the attenuation. (ii) Demodulation : The process of recovering the original information signal from the modulated wave at the receiver end is called demodulation. It is the reverse process of modulation.

rB > rA ⇒

According to Gauss's law, the electric flux passing through a closed surface is given by qenclosed ∫ E . ds = ε0


1.5 × 0.3 + 2 × 0. 2 0. 3 + 0. 2 0.45 + 0.4 0.85 = = = 1. 7 V 0. 5 0. 5 0.2 × 0.3 0.06 req = = = 0.12 Ω 0 . 2 + 0 . 3 0. 5 εeq =

8.

Q ⎡1 1⎤ ⎢ − ⎥ 4 πε0 ⎣ rA rB ⎦

⎛1 1⎞ 1 1 < ⇒⎜ − ⎟>0 rB rA ⎝ rA rB ⎠ Hence, (VA – VB) > 0 i.e., potential difference (VA – VB) is positive. ∵

ε1r2 + ε2r1 rr , req = 1 2 r1 + r2 r1 + r2

10. When a charge q at rest is accelerated through a

potential difference V then its kinetic energy is increased by qV. i.e., K = qV Momentum, p = 2m K = 2mqV de-Broglie wavelength, λ= λ=

h h = p 2mqV h 2mq

×

1 V

Since m1 > m2 and q1 = q2 = q ∴ (slope)(q, m ) < (slope)(q, m ) 1 2 Line 2 represents particle of smaller mass because its slope is more than that of line 1. 11. (i) Mutual inductance : When an emf is produced in a coil because of change in current in a coupled

coil, the effect is called mutual inductance. The most common application of mutual inductance is the transformer. (ii) Here, M = 1.5 H, ΔI1 = 20 A, Δt = 0.5 s, Δφ = ? We know, emf induced in the second coil, M ΔI1 (Δφ)2 ε=− =− Δt Δt ∴ (Δφ)2 = MΔI1 = 1.5 × 20 = 30 Wb ε A 12. Here, C x = 0

d ε0 εr A = εr Cx = 4 Cx Cy = d (i) Cx and Cy are in series, so equivalent capacitance is given by Cx × C y C= Cx + C y ⇒ 4=

Cx × 4 Cx

(∵ C = 4 μF) Cx + 4 Cx 4 Cx ∴ Cx = 5 μF ⇒ 4= 5 and Cy = 4 Cx = 20 μF (ii) Charge on each capacitor, Q = CV Q = 4 × 10–6 × 15 = 60 × 10–6 C Potential difference between the plates of X, Q 60 × 10−6 Vx = = = 12 V Cx 5 × 10−6 Potential difference between the plates of Y, Vy = V – Vx = 15 – 12 = 3 V. (iii) Ratio of electrostatic energy stored, Q2 U x 2Cx C y 4Cx = 2 = = =4 Uy Cx Cx Q 2Cy 13. When two parallel infinite straight wires carrying currents I1 and I2 are placed at distance d from each other, then current I1 produces magnetic field, which at any point on the second current carrying wire is μ0 I1 directed 2 πd inwards perpendicular to plane of wires.

B1

So, this current (I2) carrying wire then experiences a force due to this magnetic field which on its length l is given by F21 = I2 (l × B1 ) μ I F21 = F12 = I2lB1 sin 90° = I2l × 0 1 2πd μ 0 I1I 2 l or F21 = F12 = 2πd The vector product (l × B1 ) has a direction towards the wire carrying current I1. Hence, both the wires attract each other. So, force per unit length that each wire exerts on the other is μ II f = 012 2πd If I1 = I2 = 1 A and d = 1 m and l = 1 m μ then f = 0 = 2 × 10 −7 N m −1 2π Thus, electric current through each of two parallel long wires placed at distance of 1m from each other is said to be 1 ampere, if they exert a force of 2 × 10–7 N m–1 on each other. 14. Suppose total charge on ring of radius a is q. Charge

q is uniformly distributed. We want to find electric field at point P on the axis of the charged ring. Consider a small element of the ring carrying charge dq. Electric field due to this small element is dE .

dE can be resolved into two components as (i) dE cosθ along PX and (ii) dE sinθ along PY. Due to symmetry of ring all components of electric fields of small elements along y-axis cancel out. Resultant electric field at point P, E = ∫ dE cos θ dq 1 dq 1 = 4 πε0 r 2 4 πε0 (x 2 + a2 ) x x cos θ = = 2 r x + a2 dq x 1 × 2 2 × E=∫ 4 πε0 (x + a ) ( x 2 + a2 )

Here, dE =

=

∴


67

x 1 dq 2 4 πε0 (x + a2 )3/2 ∫ qx 1 E= 2 4 πε0 (x + a2 )3/2 =

For large x as x >> a, so a2 can be neglected, 1 qx 1 q = ∴ E= 3 4 πε0 x 4 πε0 x 2 which is the electric intensity due to a point charge at a distance x. Hence charged ring behaves as a point charge for points at large distances from it. 15. According to Einstein's photoelectric equation,

kinetic energy of photoelectrons is given by Kmax = hυ – hυ0 Three characteristic features explained by this equation : i. Photoelectrons are not ejected unless the frequency of incident light is above a certain threshold frequency. The threshold frequency depends on the work function of material. ii. If the frequency of incident radiation is greater than threshold frequency, even a light of very weak intensity will cause photoelectrons to be emitted. If the frequency of incident radiation is less than the threshold frequency, even the most intense light will not cause photoelectrons to be emitted. iii. The kinetic energy of the ejected electrons was proportional to the frequency of the illuminating light. This showed that whatever was knocking the electrons out had an energy proportional to light frequency. The remarkable fact that the ejection energy was independent of the total energy of illumination showed that the interaction must be like that of a particle which gave all of its energy to the electron . iv. The emission of photoelectrons is almost instantaneous. i.e. there is no time lag between the emission of electrons and switching on of the light source. 16. (a) Magnetic force acting on a charged particle q

moving with a velocity v in a uniform magnetic field B is given by F = q (v × B) (b) Magnetic force on α-particle Fα = qv × B = 2 e v B upward So, curve will bend upwards as force is perpendicular to the velocity. 68


Magnetic force on neutron, F = 0 (as q = 0) So, neutron will move along straight line. Magnetic force on electron Fe = q v × B = | −e v B | downwards So, curve will bend downwards as force is perpendicular to the velocity, For a charged particle moving in a uniform magnetic field B perpendicular to velocity, mv 2 mv qvB = ⇒r = r qB r is the radius of curved path. Here vα = vn = ve = v Radius of path traced by 4mev 2mev α-particle, rα = = 2e B eB Radius of path traced by mv electron, re = e eB

17. (a) Here, R = – 20 cm, f = R/2 = –10 cm

m = –2 (image is real) u → object distance v → image distance v m = − ⇒ v = 2u u 1 1 1 Using mirror formula, + = v u f 1 1 1 3 1 + = ⇒ = 2u u −10 2u −10 ∴ u = – 15 cm Hence, v = 2u = –30 cm. (b) For convex mirror : f > 0, u < 0 Using mirror formula, 1 + 1 = 1 v u f 1 1 1 1 1 1 1 1 f ×u = − = − ⇒ = + ⇒v = f +u v f u f (−u) v f u ∴ v>0 This implies that image of object placed in front of a convex mirror is always formed behind the mirror which is virtual in nature. 18. (i) Bohr's quantization condition :

The electron revolves around the nucleus only in those orbits for which the angular momentum is some integral multiple of h/2π. h ; n = 1, 2, 3, .... i.e., L = mvr = n 2π de Broglie hypothesis may be used to derive Bohr's formula by considering the electron to be a wave spread over the entire orbit, rather than as a particle which at any instant is located

at a point in its orbit. The stable orbits in an atom are those which are standing waves. Formation of standing waves require that the circumference of the orbit is equal in length to an integral multiple of the wavelength. Thus, if r is the radius of the orbit nh h⎞ ⎛ 2π r = nλ = ⎜∵λ = p ⎟ p ⎝ ⎠ which gives the angular momentum quantization h L = pr = n 2π

(ii) The use of parabolic mirror reduces spherical aberration. 20. Oscillating charge produces an oscillating electric

field in space which produces an oscillating magnetic field, which in turn, is a source of oscillating electric field and so on. The oscillating electric and magnetic fields thus generate each other as the wave propagates through the space. Hence em waves are produced by oscillating charges. A plane electromagnetic wave is said to be linearly polarized. The transverse electric field wave accompanied by a magnetic field wave is illustrated.

(ii)

Clearly, from energy level diagram, EC – EA = (EC – EB) + (EB – EA) (On the basis of energy of emitted photon). hc hc hc = + λ 3 λ1 λ 2 ⇒

1 1 1 λλ = + ⇒ λ3 = 1 2 λ 3 λ1 λ 2 λ1 + λ 2

which is the required relation between the three given wavelengths. 19. Reflecting telescope : Schematic ray diagram

OR Maxwell's generalization of Ampere's circuital law, dφ ⎞ ⎛ ∫ B . dl = μ0 (i + id ) = μ0 ⎜⎝ i + ε0 dtE ⎟⎠

In the process of charging the capacitor there is change in electric flux between the capacitor plates. dφ E d = (EA) dt dt E → Electric field between the plates = A → Area of the plate dφ E d ⎛ q ⎞ 1 dq id = ⎜ × A⎟ = = So, dt dt ⎝ Aε0 ⎠ ε0 dt ε0 ∴

id = i = ε0

q Aε0

dφ E dt

21. (a) Need of modulating a low frequency signal:

Two important advantages of reflecting telescope over a refracting telescope : (i) A concave mirror of large aperture has high gathering power and absorbs very less amount of light than the lenses of large apertures. The final image formed in reflecting telescope is very bright.

(i) Low energy : The audio/video signals when converted into em waves do not have sufficient high energy to travel upto long distances, because of their lower frequency. Hence these signals are modulated with high frequency carrier waves, before being sent and are demodulated or separated from the carrier waves at the receiving end. PHYSICS FOR YOU | APRIL ‘16

69

(ii) Size of antenna : For the effective transmission by an antenna, the size of the antenna should be at least of the size λ/4, where λ is wavelength of signal to be sent. For an em wave of the frequency of the order of audio signal i.e., 20 kHz, we need an antenna of size λ/4 i.e., 3.75 km high, which is practically impossible. Hence these low frequency signals are superimposed with high frequencies or radio frequencies before transmission. (b) Advantages of frequency modulation over amplitude modulation : (i) Frequency of a wave does not change while travelling through different media, while amplitude of a wave changes while travelling through different media. An amplitude modulated wave carries information in terms of variation of amplitude, which can get disturbed. This is why FM signal is less susceptible to noise than AM signal. (ii) In FM transmission, all the transmitted power is useful, whereas in AM transmission most of the power wastes in transmitting carrier wave, with no useful information.

varying VBE. For silicon diode we have knee voltage around 0.7 V . After overcoming the knee voltage, current will rise sharply. The input characteristic will be different if we go on increasing the VCE. It will be shifting right, means for the same VBE we will be getting lower input current IB.

Similarly, for different values of IB, the IC versus VCE graph (output characteristic) is shown below.

22. (i) Functions of three segments of a transistor :

Emitter : It supplies a large number of majority charge carriers for the flow of current through the transistor. Base : It controls the flow of majority charge carriers from emitter to collector. Collector : It collects a major portion of the majority carriers supplied by emitter for the circuit operation. (ii)

Let us first consider the input characteristic. Input characteristic means we have to plot the graphical representation between IB and VBE. VBE is the emitter to base voltage or the forward bias voltage and IB is the base current. In this forward biasing, E is at lower potential than B. We will be plotting IB versus VBE because base is at higher potential than emitter, so that will be reflected here. Now go on 70


23. (i) Values displayed by Meeta, are curiosity to learn

and good observation. Values displayed by her father are patience and knowledgeable. (ii) Meeta's father most probably explained her the benefits of using tiny bulbs (LEDs) over a single bulb. (a) Tiny lights are semiconductor devices which consume very less power than a single bulb. (b) Tiny lights are very cheap. (c) If some of these tiny lights are not working, then traffic system will not be affected. But if a single bulb is fused, traffic system will be disturbed. (iii) Tiny lights in traffic signals are called LEDs. LEDs are operated in forward biased and emits spontaneous radiation. 24. (i) Refer point 2.1(3) page no. 92 (MTG Excel in Physics)

(ii) Refer points 2.1(4) and 2.2(4, 9(a)) page no. 92-94 (MTG Excel in Physics) (iii) Refer point 2.2(9(b)), page no. 94 (MTG Excel in Physics) OR (i) Refer point 2.5(7) page no. 101 (MTG Excel in Physics)

(ii) Here AB = 1 m, RAB = 10 Ω, Potential gradient, k = ?, AO = l = ? Current passing through AB, 2 I= 15 + R AB 2 2 = = A 15 + 10 25 2 4 × 10 = V VAB = I × RAB = 25 5 VAB 4 − 1 ∴ k= = Vm AB 5 Current in the external circuit, 1.5 1. 5 I′ = = =1A 1 . 2 + 0 . 3 1. 5 For no deflection in galvanometer, Potential difference across AO = 1.5 – 1.2 I' ⇒ k (l) = 1.5 – 1.2 × I′ 4 0 .3 × 5 ⇒ l = 0.3 or, l = = 0.375 m 5 4 ∴ l = 37.5 cm

PO 1100 1 = = A εS 22000 20 (e) PS = PO = PI = 1100 W. (d) IS =

(∵ PO = PI)

26. (i) Refer point 6.13(6) page no. 447

(MTG Excel in Physics) (ii) Refer point 6.14(7) page no. 452 (MTG Excel in Physics) OR (i) Refer point 6.7(3, 5) page no. 378 (MTG Excel in Physics) (ii) Refer point 6.7(6, 7) page no. 379 (MTG Excel in Physics) (iii) At point B, for total internal reflection, μ sin i ≥ 1 1 μ≥ sini 1 μ≥ = 2 sin 45° (∵ i = 45°) ∴ μ≥ 2 μ min = 2 .



25. (i) Refer point 4.6(6, 7) page no. 269, 270

(MTG Excel in Physics) (ii) For LR circuit, XL = R Power factor, R 1 R = = P1 = cos φ = 2 2 2 R +R R 2 + X L2

SOLUTION OF MARCH 2016 CROSSWORD

For LCR circuit, as C is put in series with LR circuit Also, XL = XC R Power factor, P2 = cos φ' = R 2 + (X L − XC )2 =

Required ratio =

R R + (X L − X L ) 2

2

=

R =1 R

P1 1 = P2 2

OR (i) Refer point 4.8(1) page no. 274 (MTG Excel in Physics) N (ii) Here NP = 100, S = 100 NP εi = εP = 220 V, PI = 1100 W (a) NP = 100 NS = 10000 P 1100 =5A (b) IP = I = ε P 220 N (c) εS = S × εP = 100 × 220 = 22000 V NP

     

WINNERS (March 2016) Manjit Bakshi (Punjab) Rohini Rani (Bihar) Anirban Das (WB) Solution Senders (February 2016) Sreehari K (Kerala) Vishal Saxena (Rajasthan) Satwik Jain (Rajasthan) PHYSICS FOR YOU | APRIL ‘16

71

Exam from 14th to 28th May 2016

1. In a race for 100 m dash, the first and the second runners have a gap of one metre at the mid way stage. Assuming the first runner goes steady, by what percentage should the second runner increase his speed just to win the race? (a) 2% (b) 4% (c) more than 4% (d) less than 4% 2. A sand bag of mass m is suspended from a m moving with a long string. A bullet of mass 20 horizontal velocity v strikes it and gets embedded into it. Calculate the velocity gained by the bag in this process and fraction of energy lost in the process. v 20 v 18 ; ; (b) (a) 21 21 24 20 v 20 v 20 ; ; (c) (d) 21 22 19 21 3. The bodies situated on the surface of earth at its equator, become weightless, when the kinetic energy of rotation of earth about its axis is (a) MgR (b) 2MgR/5 (c) MgR/5 (d) 5MgR/2 4. Two linear SHMs of equal amplitude A and angular frequencies ω and 2ω are impressed on a particle along the axes x and y respectively. If the initial phase difference between them is π/2, the resultant path followed by the particle is (a) y2 = x2(1 – x2/A2) (b) y2 = 2x2(1 – x2/A2) (c) y2 = 4x2(1 – x2/A2) (d) y2 = 8x2(1 – x2/A2) 5. A calorie is a unit of heat and equals 4.2 J. Suppose we employ a system of units in which the unit of mass is α kg, the unit of length is β m and the unit of time is γ s. In this new system, 1 calorie will be (a) α–1β–2γ2 (b) 4.2αβ2γ2 2 2 (c) αβ γ (d) 4.2α–1β–2γ2 6. Which of the following graph represents the variation of magnetic flux density B with distance r for a straight long wire carrying an electric current? 72


(a)

(b)

(c)

(d)

7. In an L – R circuit, the value of L is (0.4/π) H and the value of R is 30 Ω. If in the circuit, an alternating emf of 200 V at 50 cycle per second is connected, the impedance of the circuit and current will be (a) 11.4 Ω, 17.5 A (b) 30.7 Ω, 6.5 A (d) 50 Ω, 4 A (c) 40.4 Ω, 5 A 8. A microscope has an objective of focal length 1.5 cm and eyepiece of focal length 2.5 cm. If the distance between objective and eyepiece is 25 cm, what is the approximate value of magnification produced for relaxed eye? (a) 75 (b) 110 (c) 140 (d) 25 9. Two small conducting spheres of equal radius have charges +10 μC and –20 μC and placed at a distance R from each other experience force F1. If they are brought in contact and separated to the same distance, they experience force F2. The ratio of F1 to F2 is (a) 1 : 2 (b) –8 : 1 (c) 1 : 8 (d) –2 : 1 10. Two masses 40 kg and 30 kg are connected by a weightless string passing over a frictionless pulley as shown in the figure. The tension in the string will be (a) 188 N (b) 368 N (c) 288 N (d) 168 N 11. At constant temperature, the volume of a gas is to be decreased by 4%. The pressure must be increased by (a) 5.34% (b) 4.16% (c) 2.96% (d) 3.86%

12. In a Young's double slit experiment, one of the slits is covered with a transparent sheet of thickness 3.6 × 10–5 m due to which position of central bright fringe shifts to a position originally occupied by 30th fringe. The refractive index of the sheet, if λ = 6000 Å, is (a) 1.5 (b) 1.9 (c) 1.3 (d) 1.7 13. At time t = 0, activity of a radioactive substance is 1600 Bq, at t = 8 s activity becomes 100 Bq. Find the activity at t = 2 s. (a) 200 Bq (b) 400 Bq (c) 600 Bq (d) 800 Bq 14. Find the ratio of minimum to maximum energy of radiation emitted by electron in ground state of Bohr's hydrogen atom. 4 2 3 1 (a) (b) (c) (d) 3 3 4 2 15. In an experiment with sonometer, a tuning fork of frequency 256 Hz resonates with a length of 25 cm and another tuning fork resonates with a length of 16 cm. Tension in the string remains constant, the frequency of the second tuning fork is (a) 204 Hz (b) 160 Hz (c) 400 Hz (d) 320 Hz 16. Refer to the arrangement of logic gates. For A = 0, B = 0 and A = 1, B = 0, the values of output Y are, respectively

(a) 0 and 1 (c) 1 and 1

(b) 1 and 0 (d) 0 and 0

17. If the length of stretched string is shortened by 40% and the tension is increased by 44 %, then the ratio of the final and initial fundamental frequencies is (a) 2 : 1 (b) 3 : 2 (c) 3 : 4 (d) 1 : 3 18. A 220 V input is supplied to a transformer. The output circuit draws a current of 2.0 A at 440 V. If the efficiency of the transformer is 80%, the current drawn by the primary windings of the transformer is (a) 3.6 A (b) 2.8 A (c) 2.5 A (d) 5.0 A 19. When a metal surface is illuminated with light of wavelength λ, the stopping potential is V0. When the same surface is illuminated with light of V0 . If the wavelength 2λ, the stopping potential is 4 velocity of light in air is c, the threshold frequency of photoelectric emission is c c 2c 4c (a) (b) (c) (d) 6λ 3λ 3λ 3λ

20. A launching vehicle carrying an artificial satellite of mass m is set for launch on the surface of the earth of mass M and radius R. If the satellite is intended to move in a circular orbit of radius 7R, the minimum energy required to be spent by the launching vehicle on the satellite is (Gravitational constant = G) 13 GMm GMm (b) (a) 14 R R (c)

GMm

(d)

GMm

7R 14 R 21. An inductance coil is connected to an ac source through a 60 Ω resistance in series. The source voltage, voltage across the coil and voltage across the resistance are found to be 33 V, 27 V and 12 V respectively. Therefore, the resistance of the coil is (a) 30 Ω (b) 45 Ω (c) 105 Ω (d) 75 Ω 22. The cylindrical tube of spray pump has a crosssection of 8 cm2, one end of which has 40 fine holes each of area 10–8 m2. If the liquid flows inside the tube with a speed of 0.15 m min–1, the speed with which the liquid is ejected through the holes is (a) 50 m s–1 (b) 5 m s–1 –1 (c) 0.05 m s (d) 0.5 m s–1 23. A body is fired vertically upwards. At half the maximum height, the velocity of the body is 10 m s–1. The maximum height raised by the body is (Take g = 10 m s–2) (a) 5 m (b) 10 m (c) 15 m (d) 20 m 24. If the masses of deuterium and helium are 2.0140 amu and 4.0026 amu, respectively and 22.4 MeV energy is liberated in the reaction 6 2 4 4 6 3Li + 1H → 2He + 2He, the mass of 3Li is (a) 6.015 amu (c) 5.980 amu

(b) 6.068 amu (d) 6.00 amu

25. The maximum range of projectile fired with some initial velocity is found to be 1000 m, in the absence of wind and air resistance. The maximum height reached by the projectile is (a) 250 m (b) 500 m (c) 1000 m (d) 2000 m 26. In the adjoining figure, the equivalent resistance between A and B is ⎛ 17 ⎞ (a) ⎜ ⎟ Ω ⎝ 24 ⎠ ⎛ 24 ⎞ (c) ⎜ ⎟ Ω ⎝ 17 ⎠

⎛4⎞ (b) ⎜ ⎟ Ω ⎝3⎠ ⎛3⎞ (d) ⎜ ⎟ Ω ⎝4⎠ PHYSICS FOR YOU | APRIL ‘16

73

27. A satellite in a circular orbit of radius R has a period of 4 hours. Another satellite with orbital radius 3R around the same planet will have a period (in hours) (a) 16 (b) 4 (c) 4 27 (d) 4 8 28. A body of mass M suspended from two springs separately executes simple harmonic motion. During oscillation the maximum velocity is equal A1 in both cases. The ratio of amplitude is A2 k12 k1 k2 k2 (b) (c) (d) 2 (a) k2 k1 k1 k2 29. A uniform thin bar of mass 6m and length 12L is bent to make a regular hexagon. Its moment of inertia about an axis passing through the centre of mass and perpendicular to the plane of hexagon is (a) 20mL2 (b) 6mL2 12 2 mL (d) 30mL2 (c) 5 30. An object is displaced from position vector ^

^

^ ^ r1 = (2 i + 3 j ) m to r2 = (4 i + 6 j ) m under a force ^

^

F = (3x 2 i + 2 y j ) N. The work done by this force is (a) 63 J (b) 73 J (c) 83 J (d) 93 J 31. A charge Q is enclosed by a gaussian spherical surface of radius R. If the radius is doubled, then the outward electric flux will (a) increase four times (b) be reduced to half (c) remain the same (d) be doubled 32. A small bulb emits 100 W of electromagnetic radiation uniformly in all directions. What is the maximum energy density in the electric field at a point 1.0 m from the source? (a) 10.5 × 10–8 J m–3 (b) 5.30 × 10–8 J m–3 (c) 2.65 × 10–8 J m–3 (d) 20.1 × 10–8 J m–3 33. In an interference pattern produced by two identical slits, the intensity at the site of the central maximum is I. The intensity at the same spot when either of the two slits is closed is I0. Therefore, (a) I = I0 (b) I = 2I0 (c) I = 4I0 (d) I and I0 are not related to each other. 34. In an L-R ciruit shown in figure, switch S is closed at time t = 0. If ε denotes the induced emf across inductor and I, the current in the 74


circuit at any time t, then which of the following graphs, shows the variation of ε with I ?

(a)

(b)

(c)

(d)

35. The time of vibration of a dip needle vibrating in the vertical plane in the magnetic meridian is 3 s. When the same magnetic needle is made to vibrate in the horizontal plane, the time of vibration is 3 2 s. Then the angle of dip is (a) 30° (b) 45° (c) 60° (d) 90° 36. A current of 2 A flows in the system of conductors as shown in the figure. The potential difference VP – VR will be nearly (a) – 2 V (b) – 1 V (c) + 1 V (d) + 2 V

37. When the reverse potential in the semiconductor diode are 10 V and 15 V, the corresponding reverse currents are 10 μA and 35 μA respectively. The reverse resistance of junction diode will be (a) 20 kΩ (b) 200 kΩ (c) 2000 kΩ (d) none of these 38. A metal rod of Young’s modulus Y and coefficient of thermal expansion α is held at its two ends such that its length remains invariant. If its temperature is raised by t °C, the linear stress developed in it is 1 αt Y (c) Yαt (d) (a) (b) (Y αt ) Y αt 39. The deflection in a moving coil galvanometer falls from 50 divisions to 10 divisions when a shunt of 12 ohm is applied. What is the resistance of the galvanometer? (a) 12 Ω (b) 24 Ω (c) 36 Ω (d) 48 Ω 40. The angular momentum of an electron in the 3h . Here, h is Planck’s constant. hydrogen atom is 2π The kinetic energy of this electron is (a) 4.35 eV (b) 1.51 eV (c) 3.4 eV (d) 6.8 eV

SOLUTIONS 1. (c) : Let v1, v2 be the initial speeds of first and second runners. Let t be the time taken by them when the first runner has completed 50 m. During this time, the second runner has covered a distance = 50 – 1 = 49 m. 50 49 So, t = = ...(i) v1 v2 Suppose the second runner increases his speed to v3 so that he covers the remaining distance (51 m) in time t. So 51 49 51 2 ⎞ ⎛ t= = or v3 = × v2 = ⎜1 + ⎟ v2 ⎝ 49 ⎠ v3 v2 49 v3 − v2 2 v3 2 = −1 = or or v2 49 v2 49 2 or % increase = × 100% = 4.1% 49 i.e. more than 4% 2. (a) : Applying principle of conservation of linear momentum, 21m m m⎞ ⎛ × v = ⎜m + ⎟ v ′ = v′ ⎝ ⎠ 20 20 20 v 20 v v′ = × = 20 21 21 This is the velocity gained by the bag. 1⎛m⎞ 2 Initial kinetic energy of bullet, E1 = ⎜ ⎟ v 2 ⎝ 20 ⎠ Final kinetic energy of bag and bullet, 2

1⎛ 1 ⎛ 21m ⎞ ⎛ v ⎞ m⎞ E2 = ⎜ m + ⎟ v ′2 = ⎜ ⎟⎜ ⎟ 2⎝ 2 ⎝ 20 ⎠ ⎝ 21 ⎠ 20 ⎠ Loss of energy 20 1⎛m⎞ ⎡ 1 ⎤ 1⎛m⎞ = E1 − E2 = ⎜ ⎟ v 2 ⎢1 − ⎥ = ⎜ ⎟ v 2 × 21 2 ⎝ 20 ⎠ ⎣ 21 ⎦ 2 ⎝ 20 ⎠ E1 − E2 20 = Fraction of energy lost = E1 21 3. (c) : When there is a weightlessness in the body at the equator, then g′ = g – Rω2 = 0 or ω = g /R and linear velocity ωR = ( g /R )R = gR 1 2 ∴ Kinetic energy of rotation of earth = Iω 2 1 2 1 1 = × MR2 × ω2 = M (ωR)2 = MgR 2 5 5 5 4. (c) : x = A sin(ωt + π/2) = A cos ωt ∴

y2 = 4A2 sin2 ωt cos2 ωt ⎛ x2 ⎞ x 2 ⎛ A2 − x 2 ⎞ 2 = x 4 1− = 4 A2 × ×⎜ ⎜ ⎟ ⎟ ⎝ A2 ⎠ A2 ⎝ A2 ⎠ 5. (d) : 1 calorie = 4.2 J ∴ [calorie] = [ML2T–2]. Comparing with general dimensional formula [MaLbTc], we get a = 1, b = 2, c = –2 or

a

∵

b

⎡M ⎤ ⎡L ⎤ ⎡T ⎤ n2 = n1 ⎢ 1 ⎥ ⎢ 1 ⎥ ⎢ 1 ⎥ ⎣ M2 ⎦ ⎣ L2 ⎦ ⎣ T2 ⎦

c

−2 ⎡ 1 kg ⎤ ⎡ 1 m ⎤ ⎡ 1 s ⎤ –1 –2 2 ⇒ n2 = 4.2 ⎢ ⎢ ⎥ ⎢ ⎥ = 4.2 α β γ ⎥ ⎣ α kg ⎦ ⎣ β m ⎦ ⎣ γ s ⎦ 6. (c) : Magnetic field induction at a point due to a long current carrying wire is related with distance r by relation B ∝ 1/r. Therefore graph (c) is correct. 7. (d) : Here, XL = ωL = 2πυL 0. 4 = 2 π × 50 × = 40 Ω π R = 30 Ω 2

1

Z = R2 + X L2 = 302 + 402 = 50 Ω V 200 irms = rms = =4A Z 50 8. (c) : Length of the tube is L = v0 + fe v0 = L – fe = 25 – 2.5 = 22.5 cm 1 1 1 Now applying − = , we have v0 u0 f0 1 1 1 − = 22.5 u0 1.5 ∴ |u0| ≈ 1.6 cm v0 D ⎛ 22.5 ⎞ ⎛ 25 ⎞ × =⎜ ∴ M = ⎟ ⎜ ⎟ = 140 u0 fe ⎝ 1.6 ⎠ ⎝ 2.5 ⎠ ∴

9. (b) : Here, F1 =

k(+10)(−20) 2

=

−k × 200

R R2 As spheres are of equal radius, their capacities are same. On touching, the net charge = + 10 – 20 = –10 μC is shared equally between them i.e, each sphere carries –5 μC charge. k(−5)(−5) k × 25 F −8 F2 = ∴ 1= = 2 2 F2 1 R R

10. (d) :

cos ωt = x/A and sin ωt = 1 − (x 2 /A2 ) y = A sin 2ωt = 2A sin ωt cos ωt PHYSICS FOR YOU | APRIL ‘16

75

m1gsin30° – T = m1a T – m2gsin30° = m2a Adding (i) and (ii), =

1

sin 30° −

sin 30°

2

1+

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

2

From eqn. (ii), we get T = m2g sin30° + m2a = =

2

2

sin 30° + 1 2

2

1+

sin 30°

1+

2

=

(

1

sin 30° −

2

sin 30°)

2

2 × 40 × 30 × 9.8 × (1 / 2) 40 + 30

1200 = × 9.8 = 168 N 70

11. (b) : At constant temperature, pV = constant p1 V2 = p1V1 = p2V2 or p2 V1 Fractional change in volume V1 − V2 4 1 = = V1 100 25 V2 24 V2 1 1− = or, = V1 25 V1 25 p1 V2 24 p2 25 ∴ or = = = p2 V1 25 p1 24 p2 − p1 25 1 = −1 = p1 24 24 100 % increase in pressure = = 4.16 % 24 30 λD 12. (a) : The position of 30th bright fringe y30 = d 30 λD New position of central fringe is y0 = d But we know , y0 = shift due to transparent sheet D = (μ – 1)t d 30 λD D = (μ − 1)t So, d d 30 λ 30 × (6000 × 10−10 ) (μ − 1) = = = 0 .5 t (3.6 × 10−5 ) ∴ μ = 1.5 n

⎛1⎞ 13. (d) : Activity, R = R0 ⎜ ⎟ ⎝2⎠ where n is the number of half-lives. At

⎛1⎞ t = 8 s, 100 = 1600 ⎜ ⎟ ⎝2⎠ 1 ⎛1⎞ =⎜ ⎟ 16 ⎝ 2 ⎠

76

n

n

or

n=4


four half-lives are equivalent to 8 s. Hence, 2 s is equal to one half-life. So, in one half-life activity will fall half of 1600 Bq i.e., 800 Bq. 14. (c) : Energy of radiation corresponding to transition between two energy levels n1 and n2 is given by 1 ⎞ ⎛ 1 E = 13.6 − eV . ⎜ n2 n2 ⎟ ⎝ 1 ⎠ 2 E is minimum when n1 = 1 and n2 = 2. 3 ⎛1 1 ⎞ Hence, Emin = 13.6 ⎜ − ⎟ eV =13.6 × eV ⎝1 4 ⎠ 4 E is maximum when n1 = 1 and n2 = ∞ (when the atom is ionised). Hence E 3 ⎛ 1⎞ Emax = 13.6 ⎜1 − ⎟ = 13.6 eV ∴ min = ⎝ ∞⎠ Emax 4 15. (c) : For sonometer, frequency of vibration of string, provided its tension and mass per unit length 1 remain same, is given by υ ∝ l υ2 l1 υ2 25 = ∴ = or υ1 l2 256 16 256 × 25 υ2 = = 400 Hz 16 16. (b) :

17. (a) : Initial fundamental frequency of a stretched string is 1 T …(i) υ= 2L μ where the symbols have their usual meanings. When the length of a stretched string is shortened by 40% and the tension is increased by 44%, then its length and tension become 40 3 44 36 L′ = L − L = L , T′ = T + T= T 100 5 100 25 Then, final fundamental frequency is 1 36 T 2 T 1 T′ = = υ′ = …(ii) ⎛ 3 ⎞ 25 μ 2L μ 2L ′ μ 2 ⎜ L⎟ ⎝5 ⎠ υ′ 2 = Dividing (ii) by (i), we get υ 1

18. (d) : Here, Vp = 220 V, Is = 2 A, Vs = 440 V η = 80%, Ip = ? VI VI η = s s or I p = s s Vp I p ηV p

19. 20.

21. 22.

23.

24.

Substituting the given values, we get 440 × 2 Ip = =5 A 80 × 220 100 (b) (b) : The energy of the satellite on the surface of the earth is GMm ⎛ GMm ⎞ Es = KE + PE = 0 + ⎜ − ⎟⎠ = − ⎝ R R The energy of the satellite in an orbit of radius r is 1 ⎛ GMm ⎞ Eo = mvo2 + ⎜ − ⎟ ⎝ 2 r ⎠ ⎡ GM ⎤ 1 ⎛ GM ⎞ GMm = m⎜ As vo = − ⎢ ⎥ ⎟ r ⎦ 2 ⎝ r ⎠ r ⎣ GMm =− 2r The minimum energy required to be spent by the vehicle is ΔE = Eo − Es GMm ⎡ GMm ⎤ =− − − (∵r = 7R) 2(7 R) ⎢⎣ R ⎥⎦ GMm GMm 13 GMm =− + = 14 R R 14 R (b) (b) : According to equation of continuity, a1v1 = a2v2 ⎛ 0.15 ⎞ ∴ (40 × 10−8 ) × v1 = 8 × 10−4 × ⎜ ⎝ 60 ⎟⎠ −4 8 × 10 × 0.15 or v1 = = 5 m s −1 −8 40 × 10 × 60 (b) : Let h be maximum height reached by the body. Taking motion of the body from half the maximum height upto the highest point, we have h u =10 m s–1, a = – g = – 10 m s–2, v = 0, S = 2 As v2 = u2 + 2aS h ∴ 0 = 102 + 2 (– 10) × or h = 10 m 2 (a) : In the given reaction, B.E. = [(m(63Li) + m(21H) – 2m(42He)] × 931 = 22.4 22.4 – 2.0140 + 2 × 4.0026 ∴ m(63Li) = 931 = 6.015 amu

u2 sin 2θ g For maximum range, angle of projection θ is 45°. u2 sin (2 × 45°) u2 = ∴ Rmax = = 1000 m …(i) g g

25. (a) : Range, R =

∴

u2 sin2 45° Maximum height, H = 2g 2

1 ⎛ 1 ⎞ × 1000 × ⎜ = 250 m (Using (i)) ⎝ 2 ⎟⎠ 2 26. (b) : The equivalent circuit is as shown in figure. H=

Resistance of arm AGB 8×2 4×6 = + = 1.6 + 2.4 = 4 Ω 8+2 4+6 For equivalent resistance between A and B, the resistance of arms CD, AGB and FE are in parallel. Thus 1 1 1 1 4+3+2 9 = + + = = RAB 3 4 6 12 12 12 4 = Ω 9 3 27. (c) : According to Kepler’s third law T2 ∝ R3 or

RAB =

3/2

3/2

⎛R ⎞ ⎛ 3R ⎞ T2 = T1 ⎜ 2 ⎟ = 4 ⎜ ⎟ = 4 27 hours ⎝R⎠ ⎝ R1 ⎠ 28. (c) : For SHM, Maximum velocity, vmax = Aω k For 1st case, vmax1 = A1ω1 = A1 1 M k2 For IInd case, vmax 2 = A2 ω2 = A2 M According to question, vmax1 = vmax2 ∴ A1

k1 k = A2 2 M M

or

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

A1 k = 2 A2 k1

29. (a) : Length of each side of hexagon = 2L Mass of each side = m Let O be centre of mass of hexagon. Therefore, perpendicular distance of O from each side PHYSICS FOR YOU | APRIL ‘16

77

r = Ltan60° = L 3 The desired moment of inertia of hexagon about O is I = 6 [Ione side]

I . …(i) MR where R is resultant intensity of earth’s field I t2 = 3 2 = 2π …(ii) MH Dividing eqn.(i) by eqn.(ii)

35. (c) : t1 = 3 = 2 π

⎤ ⎡ m (2 L) =6⎢ + mr 2 ⎥ ⎥⎦ ⎢⎣ 12 2

⎤ ⎡ mL2 + m(L 3 )2 ⎥ = 6 =6⎢ ⎥⎦ ⎢⎣ 3 30. (c)

⎤ ⎡ m L2 ⎢ + 3 mL2 ⎥ = 20 mL2 ⎥⎦ ⎢⎣ 3

31. (c) : According to Gauss’s law, the total outward electric flux linked with gaussian surface 1 φE = × charge enclosed by surface. ε0 If the radius of the gaussian surface is doubled the total outward electric flux will remain the same as charge enclosed by the guassian surface is unchanged. 32. (c) : Here, P = 100 W, r = 1.0 m P Intensity of radiation at distance r, I = 4 πr 2 Let maximum energy density in the electric field = uEmax P Then I = uE c = max 4 πr 2 P or uE = max 4 πr 2c 100 = = 2.65 × 10–8 J m–3 2 8 4 × 3.14 × (1) × (3 × 10 ) 33. (c) : When one slit is closed, amplitude becomes half and intensity becomes 1/4th 1 i.e., I0 = I or I = 4I0 4 34. (c) : In L–R circuit, current at any time t is given by I=

−R ⎡1 − e L t ⎤ = V

V R ⎢⎣

dI V = dt R

−R t eL

⎥⎦

R

−

⎛R⎞ V ⎜⎝ ⎟⎠ = L L

V R

−R t eL

...(i)

...(ii)

From (i), IR = V Using (ii), IR = V – ε or ε = V – IR Therefore, graph between ε and I is a straight line with negative slope and positive intercept. PHYSICS FOR YOU | APRIL ‘16

50

5

5

From figure, (I – IG)S = IGG ( −

1 1 ) × 12 = 5 5

or, G = 4 × 12 = 48 Ω

−R

−R t − Ve L

=

5 × 10 10 = Ω 36. (b) : Resistance between Q and S, R ′ = 5 + 10 3 Potential difference across Q and S, 2 × 10 20 VQ − VS = = V 3 3 20 4 Current through arm QPS, I1 = = A 3×5 3 4 8 VQ − VP = × 2 = V 3 3 20/3 2 = A Current through arm QRS, I2 = 10 3 2 VQ − VR = × 3 = 2 V 3 VP – VR = (VQ – VR) – (VQ – VP) 8 −2 =2− = ≈ −1 V. 3 3 ΔV 15 − 10 = 37. (b) : Reverse resistance = ΔI (35 − 10) × 10−6 3 = 200 × 10 Ω = 200 kΩ 38. (c) : Due to change in temperature t °C, increase in length, Δl Δl = l α t or = αt l Δl stress stress ∴ Stress = Y × = Yαt Y = = l strain Δl / l 39. (d) : In case of a galvanometer, ∝ θ 10 1 1 So, = = = . .,

. .,

−R t eL

t dI Induced emf ε = L = Ve L dt

78

H R cos δ = = cos δ R R 2 1 cos δ = , δ = 60° 2 1

3h ⎛ h ⎞ = n⎜ ⎟ ∴ n = 3 ⎝ 2π ⎠ 2π The kinetic energy of the electron in nth orbit is 13.6 K n = 2 eV n 13.6 13.6 ∴ K 3 = 2 eV = eV = 1.51 eV 9 3 

40. (b) :

SOLUTION SET-32

1. (b): Since the wires are infinite, so the system of these two wires can be considered as a closed rectangle of infinite length and breadth equal to d. Flux through the strip of area l dr, due to current flowing in one wire is given by d −a μ0 I μ Il d −a ⎞ φ=∫ ( ldr ) = 0 ln ⎛⎜ ⎟ a 2πr 2π ⎝ a ⎠

The other wire produces the same result, so the total flux through the dotted rectangle is μ Il ⎛ d − a ⎞ φtotal = 0 ln ⎜ ⎟ π ⎝ a ⎠ The total inductance of length l, φ μ l ⎛ d −a ⎞ L = total = 0 ln ⎜ ⎟ I π ⎝ a ⎠ L μ ⎛ d −a ⎞ Inductance per unit length = = 0 ln ⎜ ⎟ l π ⎝ a ⎠ 2. (d): Current density, J = ⇒

J=

4I

I πR2 − π ( R / 2 )2

3πR2 Current in smaller cylinder,

μ=

In ΔOPQ, ∠OQP = 90° + θ − α Applying sine rule, r R = 0 sin ( 90° + θ − α ) sin α R R sin 30° sin 30° = r0 = sin ( 60° + 45° ) sin ( 90° + 45° − 30° ) 1⎞ ⎛ R⎜ ⎟ 2R ⎝2⎠ = = ⎡ 1 ⎤ ( 3 1 1 3 + 1) + × ⎥ ⎢ × 2 2⎦ ⎣ 2 2 4. (c): Since they move under mutual attraction and no external force acts on them, their momentum and energy are conserved. GM1M2 1 1 ...(i) 0 = M1v12 + M2v22 − s 2 2 It is zero because in the beginning, both kinetic energy and potential energy are zero. 0 = M1v1 + M2v2 ...(ii) Solving the two equations (i) and (ii),

2

I ⎛R⎞ I1 = Jπ ⎜ ⎟ = ⎝2⎠ 3 For A, BA = Bwhole-cylinder – Bsmall-cylinder μ ( I / 3 ) −μ0 I BA = 0 − 0 = 2π ( R / 2 ) 3πR For B, BB = Bwhole-cylinder – Bsmall-cylinder ⇒

μ ( I + I / 3 )( R / 2 ) μ I = 0 −0 = 0 2 3πR 2πR 3. (c): Let θ be angle of incidence and be the angle of refraction for the extreme rays. 1 ⎛ R ⎞1 ⇒ θ = 45° sin θ = ⎜ ⎟R = 2 ⎝ 2⎠

sin θ 1 1 = = 2 ⇒ sin α = ∴ α = 30° sin α 2 2 sin α

v12 =

2GM22 s ( M1 + M2 )

2 and v2 =

2GM12 s ( M1 + M2 )

Solution Senders of Physics Musing SET-32 1.

Spandan Senapati (Odisha)

2.

Mikhail Joseph (WB)

3.

Subrata Dutta (WB)

4.

Parvathi Nair (Tamil Nadu) SET-31

1.

Harsh Mehta (New Delhi)

2.

Sri Krishna Sahoo (Rajasthan) PHYSICS FOR YOU | APRIL ‘16

79

Velocity of approach = v1 –(–v2) = v1 + v2 =

2G ( M1 + M2 ) s

5. (d): Rate of loss of heat ∝ difference in temperature with the surroundings. dQ At 50°C, = k ( 50 − 20 ) = 10 , where k = constant dt 1 ∴ k= 3 At an average temperature of 35°C, dQ 1 = ( 35 − 20 ) = 5 J s −1 dt 3 Heat lost in 1 minute, dQ = × 60 = 5 × 60 J = 300 J = −Q dt Fall in temperature = 0.2°C = Δθ As Q = CΔθ Heat capacity, C = Q = 300 = 1500 J °C −1 Δθ 0.2 6. (a): According to question, angular acceleration ∝ θ dω ω = −c θ dθ

α = −c θ ⇒ or

ωf

∫ω

i

θ

ω dω = − ∫ c θ dθ 0

1 1 1 or ωi2 − ω2f = c θ2 2 2 2 1 2 1 2 1 or I ωi − I ω f = ΔE = Ic θ2 ∴ ΔE ∝ θ2 2 2 2 7. (b): By definition, r

r

⎛ −2α β ⎞ + ⎟ dr U r − U ∞ = − ∫ F ⋅ dr = − ∫ ⎜ 3 ⎝ r r2 ⎠ ∞ ∞

r

r

r

r ⎡ 2α ⎤ ⎡1 ⎤ =∫ dr − ∫ dr = ⎢ β + ⎥ 2 ⎢⎣ r ⎥⎦ 3 2 ∞ ⎣ ( −2 ) r ⎦ ∞ ∞r ∞r α β ⎞ ⎛1 ⎞ ⎛ 1 = −α ⎜ − 0 ⎟ + β ⎜ − 0 ⎟ ⇒ U r = − 2 + r ⎠ r ⎠ ⎝r ⎝ r2

2α

β

8. (c): At equilibrium, F = 0 2α β 2α ⇒ − + = 0 ⇒ r0 = 3 2 β r0 r0 Ionization energy, E0 = –ΔU = (U –Ur0) ⎡ ⎛ α β ⎞⎤ = ⎢0 − ⎜⎜ − 2 + ⎟⎟ ⎥ ⎢⎣ ⎝ r0 r0 ⎠ ⎥⎦ E0 =

α

( 2α / β )

⇒ E0 =

2

−

β ( 2α / β )

β2 β2 β2 β2 − =− ⇒ E0 = 4 α 2α 4α 4α

9. (b): Maximum possible charge on capacitor of capacitance C = CV0 and that on capacitor of capacitance 2C = 2CV0. So possible change through each capacitor is CV0 as they are in series. Since same charge will be on both capacitors, so 3V CV0 CV0 potential difference = = 0 = 2 Ceq C × 2C C + 2C 10. (c): Both the capacitors are in parallel means voltage across each is equal. Breakdown potential of both is also equal. Therefore, they will undergo breakdown at the same moment. 

Dr. A.P.J. Abdul Kalam Technical University Uttar Pradesh, Lucknow (Formerly U.P. Technical University, Lucknow) U.P. State Entrance Examination (UPSEE- 2016) Dr. A.P.J. Abdul Kalam Technical University Uttar Pradesh, Lucknow would conduct State Entrance Examination known as UPSEE- 2016. A. 1st Year of B.Tech/ B. Tech. (Biotech)/B. Tech. (Ag)/ B. Arch./ B. Pharm./ B. HMCT/ B.FAD/ B.FA/ MBA/ MCA/ MAM (5 years Dual Degree) B. Lateral Entry: Direct Admission to 2nd year of B.Tech/ B. Pharm/ MCA SCHEME OF ENTRANCE EXAMINATION Courses with Paper Code Mode of Examination B.Tech (Paper- 1), B. Tech (Biotech) (Paper 1 or Paper 2), B.Tech. (Ag) (Paper 1/Paper 3), OMR based Test B.Arch. (Paper 4), B. Pharm. (Paper 1 or Paper 2) April 23,2016 (Saturday) B.HMCT/ B.FAD/ B.FA (Paper 5), Lateral Entry: Direct Admission to 2nd year of B.Tech (Paper6/ Paper 8), Computer Based Test B.Pharm (Paper 7) April 24, 2016 (Sunday) MBA (Paper 9), MCA (Paper 10), MAM (5 years Dual Degree) (Paper 11), Lateral Entry: Direct Admission Computer Based Test to 2nd year MCA (Paper 12) Date of Exam April 17,2016 (Sunday)

The application form can be filled ONLINE only through the website http://www.upsee.nic.in. The link for filling up the ONLINE Application Form will be opened from February 24, 2016 at 11:00 AM. The last date and time for filling up the Application Form is March 27,2016, 05:00 PM.

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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. What happened to the neutrino and positron produced in nuclear fusion in the Sun? –Pinaki Chattopadhyay

Ans. Nuclear fusion in the Sun provides huge energy. It takes place dominantly by proton-proton cycle as follows:

Overall view, 4 41H + 2e– He + 2ν + 6γ + 26.7 MeV In the first reaction, released positron (e+) very quickly encounters a free electron (e–) in the Sun and both particles annihilate. And their rest energies appear as two gamma ray photons (γ). Internal energy inside the Sun is not all the energy produced in nuclear fusion. About 0.5 MeV is associated with two neutrinos that are produced in each cycle. Neutrinos escape from the sun carrying this energy with them, because they are so penetrating. Some are intercepted by the Earth, bringing us our only direct information about the Sun’s interior.

Q2. If a helium atom loses its two electrons, it becomes an alpha particle which means it contains two protons and two neutrons in its nucleus. Is it stable? –John Gokul (Tamil Nadu)

Ans. Alpha particle emission is modeled as a barrier penetration process. The alpha particle is the nucleus of the helium atom and is the nucleus of highest stability. The nuclear binding energy of the alpha particle is extremely high 28.3 MeV. It is an exceptionally stable collection of nucleons, and those heavier nuclei which can be viewed as collections of alpha particles (carbon-12, oxygen16, etc.) are also exceptionally stable. This contrasts with a binding energy of only 8 MeV for helium-3, which forms an intermediate step in the protonproton fusion cycle. Q3. What is the exact definition of system and surroundings in thermodynamics? –Santanu Chatterjee, Kolkata (West Bengal)

Ans. System: We are free to define our system in any convenient way, as long as we are consistent and can account for all energy transfer to or from the system.

For example, consider a calorimeter consisting water and ice. Then we may assume two systems depending on the situation (i) ice + water (calorimeter does not take any heat.) (ii) ice + water + calorimeter (calorimeter is not ideal.) Surrounding: Region outside the system is known as surrounding. We define boundary between system and surrounding. In the shown figure, environment is surrounding. Q4. Mechanical transverse waves cannot travel in gas. Then how does light travel in air? –Sivanujhaa, Coimbatore (Tamil Nadu)

Ans. Mechanical transverse waves cannot be produced in the gas because of very low elasticity. Hence they cannot travel in the gas. Light is an electromagnetic wave which is produced by accelerating charge. It requires no medium to travel. Also electromagnetic waves are transverse in nature.  PHYSICS FOR YOU | APRIL ‘16

81

Nasa tests inflatable heat shield for its spacecraft to Mars

N

asa has su successfully tested its donut-shaped inflatable heat shield technology that works like a parachu parachute and will enable a spacecraft to land safely when it descends through the high temp peratu atmosphere of a planet such as Mars. Before Nasa uses its new inflatable technology temperature fo or slow for slowing spacecraft that are entering the atmosphere of other planets, it will first need to be packed into the tight confines of a rocket. Engineers at Nasa’s Langley Research Cen Centre in US, put the technology to the test by packing a donut-shaped test article with d a diameter of 9 feet and also known as a torus, to simulate what would happen before a space mission, called the Hypersonic Inflatable Aerodynamic Decelerator, or HIAD, i works like a parachute, using the drag of a planet’s atmosphere to slow the space it vehicle as it descends towards the surface, researchers said. Slowing the spacecraft protects it from the intense heat of atmospheric entry, and allows it to land more softly. “During testing, we used a vacuum pump to compress the test article into a s small space,” said Keith Johnson, a lead engineer for the project. “We packed and u unpacked it and did thorough inspections to check for leaks and damage to the Zy and Teflon materials. We repeated this three times,” said Johnson. Zylon The technology will enable the delivery of heavy cargo, science instruments and people oth worlds. According to test engineer Sean Hancock, HIAD was packed the same to other waay each eac time to see how the material would handle folding, packing, and compressing. way h Doing so helps engineers understand how it would perform after exposure to handling, dep storage and deployment during a space flight mission, researchers said. “The test included all the components for the latest inflatable torus design, so it was a good final check to prove that the materials can tolerate packing,” Johnson said. After successful testing, Nasa engineers can move forward in the development of creating a larger HIAD that can withstand the high temperatures experiences when it descends through the atmosphere of a planet such as Mars.

Oz scientists create world’s thinnest lens

No more a dwarf? Pluto may get back its `planet’ status

A

P

ustralian scientists have claimed that they have developed a lens which is 2,000 times thinner than human hair, a breakthrough set to revolutionise nanotechnology. The finished lens is 6.3 nanometre nanometres in size as compared to the previous ssmallest lens which was 50 nanom nanometres thick. According to ABC News, the lens, has been created by a team of rese researchers led by Yuerui Lu fro from Australian National U University (ANU). Scientists ssaid the lens could have revolutionary applications in medicine, science and technology and it could be used to create bendable ccomputer screens. While th the new lens has already bee been experimented by techn technology companies with prototy prototype TV and computer screens th that can be rolled up or folded, the ma mass production at cheaper i iis yet to b i price be d devised. “This type of material is the perfect candidate for future flexible displays,” Lu said. Another application of the Australian-made lens could be arrays of micro-lenses that mimic the compound eyes of insects. Lu said a crystal called molybdenum disulphide was the special ingredient.

luto could be about to become a planet again, after scientists spotted what appears to be clouds on its surface. As part of the New Horizons mission which has already found far more complexity on Pluto than had otted cloud like parts been expected -scientists have spotted cloud-like of the images released from the mission. That ven richer could mean that the planet has an even ought, atmosphere than had been thought, according to the New Scientist. he Scientists aren’t yet sure that the images show clouds. But emailss seen by the magazine seem to indicate that the now dwarf planet might have clouds made up of the same things in its general atmosphere, which is mostly made up of nitrogen. The discovery could lead to further calls for Pluto to be reinstated as a ady planet -a question that has already been asked more and more since New Horizons sent back detailed imagess of the dwarf planet’s rich surface. Pluto still won’t cal Union’s rules satisfy the International Astronomical for what is and isn’t a planet. It was excluded on the basis of a criterion that it still doesn’t pass, whether or not the pictures show clouds: that a planet should have a clear neighbourhood of orbiting bodies. But that IAU definition has proven controversial with scientists arguing that the definition is complicated. Courtesy : The Times of India

PHYSICS FOR YOU |

APRIL ‘16

83

10 TIPS to DRIVE OUT EXAM PHOBIA If you are one of the many people who gets stressed out when it comes to taking exams then we have a few tips for you that will help you to overcome this and concentrating on achieving good grades.

1

LOCATION, LOCATION, LOCATION...

If you’re serious about getting work done, find a place that is relatively free of distractions. Establish guidelines with roommates for quiet times or use the libraries, study rooms, or empty classrooms.

2

MAKE IT A HABIT: WORK EVERY DAY

Avoid all-night cram sessions in which you (unsuccessfully) try to understand and retain large amounts of information. Spend time on your studies each day, and you can stay on top of your courses and still have time for fun. Use small blocks of time - you’ll be amazed what you can get done between study sessions.

3

HELP EXISTS! SEEK IT OUT AND IMPROVE YOUR GRADES

Whether you’re an ‘A’ student or a ‘D’ student, you can strengthen your learning skills. Whenever a doubt crops up get it cleared. Never hesitate to ask for help from your teachers/parents.

4

WRITE IT DOWN

Remember important dates. It’s up to you to remember due dates for sending applications and test dates. A wall calendar of important dates is also a good idea.

5

GET ENERGIZED - EAT, EXERCISE, SLEEP

Not understanding? Trouble remembering? Comprehension and memory are affected by stress and fatigue. When you’re hungry, tense, or tired your brain can’t function at its full potential. It’s especially crucial to eat well, exercise, and get adequate sleep during exam periods.

6

BE A KEENER : GO TO CLASS PREPARED AND TAKE GOOD NOTES

Don’t fall into the habit of missing class. Someone else’s notes aren’t going to be as good as having gone to the lecture yourself.

In class, listen for emphasis and examples. Take a thorough set of notes; you’ll be thankful at test time.

7

LECTURES AND TEXTBOOKS: WHAT'S THE BIG PICTURE?

Many unsuccessful students see a course as “a lot of stuff to memorize.” School/College learning requires understanding how pieces of information fit together to form a “BIG picture.”Use course outlines, tables of content, headings and subheadings to organize the information in each of your courses. Routinely ask yourself, “What’s the purpose of this detail?”and Does it make sense?

8

DO SOMETHING (ANYTHING!) TO REMEMBER KEY INFORMATION

Capture your understanding of course material in an active way. Generate examples, create mnemonics, make summary notes, identify key words, highlight textbooks or add margin notes. Be creative and interested and you’ll certainly do good at test time. Also no matter how well you understand something, without practice some forgetting will occur. Before a test, make sure that you can recall important information from memory. Self-test by recalling information without looking at notes or textbooks and by doing practice exams if available.

9

BE TEST SMART

Don’t lose marks because of test-writing errors such as misreading a question or running out of time. Also, carefully read instructions, budget time to marks, and do less difficult questions first to build confidence.

10

GET A MENTOR

And finally have a mentor– someone you know who has succeeded in Exam. It can even be your mother, father, brother or teacher. Idolize your mentor and keep asking him for advices. Run to him when you have a problem. He can act as a single source of motivation and problem solver for you.

At the end, once exams are over, don't forget to reward yourself. The reward gives you the break you deserve after all of your studying. Treating yourself also helps you stop thinking about the test and analyzing every little mistake you may have made. 84


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 1. Lens designed so as to minimize both spherical and coma aberration. (9,4) 4. An imaginary line around the earth parallel to the equator. (8) 10. A massive astrophysical compact halo object, a kind of astronomical body that might explain the apparent presence of dark matter in galaxy halos. (5) 14. A period of exponential

expansion thought to have occurred around 10–36 s after the universe began. (9) 15. A CGS unit of pressure equals to 1 dyne per square centimetre. (5) 17. Energy equivalent to the mass of a particle at rest. (4, 6) 22. Unit of mass in FPS system. (5) 23. The progressive decrease of a property due to repeated stress. (7) 25. A unit prefix in the metric system denoting a factor of 10–24. (5) 26. Product of force and the lever arm. (6) 27. The quasiparticle associated with spin waves in a crystal lattice. (6) 28. A device which controls the flow of electricity. (6) DOWN 1. A non dimensional, unitless quantity that indicates how well a surface reflects solar energy. (6) 2. Unit of magnetic flux in CGS system. (7) 3. The upper layer of the ionosphere at about 200 km altitude by day and at 300 km by night. (8) 5. A prefix used in metric system to denote multiple of 10–1. (4) 6. The speed less than that of the speed of sound in medium. (8) 7. A type of galvanometer designed to measure brief flow of charges through it. (9) 8. A temporary connection between electrical or electronic circuits or a temporary communications channel. (4, 2) 9. The production of shadow photographs of the internal structure of bodies, opaque to visible light. (11)

11. A line that just touches a curve at one point, without cutting across it. (7) 12. An organization founded by inventor Dean Kamen in 1989 in order to develop ways to inspire students in engineering and technology fields. (5) 13. A means of making measurements, in which the measured quantity is distant from the recording apparatus and the data is sent over a particular telecommunication system from the measuring position to the recording position. (9) 16. Timbre or tone quality that is not characterized by frequency or amplitude. (4, 5) 18. A superconducting quantum interference device. (5) 19. A stable, isolated wave that travels at a constant speed. (7) 20. The curve or surface formed by the reflection of parallel rays of light in a large aperture concave mirror. (7) 21. An alloy of iron and other elements which is magnetic. (5) 23. An elementary particle with half integer spin. (7) 24. Particles in the standard model that mediate strong interactions. (6)  PHYSICS FOR YOU | APRIL ‘16

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Physics for You - April 2016

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