Volume 24
Managing Editor
Mahabir Singh

CONTENTS

Editor
Anil Ahlawat
(BE, MBA)

No. 4

April 2016

Corporate Office:
Plot 99, Sector 44 Institutional area, Gurgaon -122 003 (HR).
Tel : 0124-4951200 e-mail : website : www.mtg.in
Regd. Office:
406, Taj Apartment, Near Safdarjung Hospital, New Delhi - 110029.

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

Subscribe online at www.mtg.in
Individual Subscription Rates

Combined Subscription Rates

1 yr.


2 yrs.

3 yrs.

1 yr.

2 yrs.

3 yrs.

Mathematics Today

330

600

775

PCM

900

1500

1900

Chemistry Today

330


600

775

PCB

900

1500

1900

Physics For You

330

600

775

PCMB

1000

1800

2300

Biology Today


330

600

775

Send D.D/M.O in favour of MTG Learning Media (P) Ltd.
Payments should be made directly to : MTG Learning Media (P) Ltd,
Plot No. 99, Sector 44, Gurgaon - 122003 (Haryana)
We have not appointed any subscription agent.
Owned, Printed and Published by Mahabir Singh from 406, Taj Apartment, New Delhi - 29 and printed
by Personal Graphics and Advertisers (P) Ltd., Okhla Industrial Area, Phase-II, New Delhi. Readers are
adviced to make appropriate thorough enquiries before acting upon any advertisements published in this
magazine. Focus/Infocus features are marketing incentives. MTG does not vouch or subscribe to the claims
and representations made by advertisers. All disputes are subject to Delhi jurisdiction only.
Editor : Anil Ahlawat
Copyright© MTG Learning Media (P) Ltd.
All rights reserved. Reproduction in any form is prohibited.

PHYSICS FOR YOU | APRIL ‘16

www.pdfgrip.com

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)












(d)

(c)




(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

PHYSICS FOR YOU | APRIL ‘16

www.pdfgrip.com


www.pdfgrip.com






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






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.






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.

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: />

PHYSICS FOR YOU | APRIL ‘16

www.pdfgrip.com


www.pdfgrip.com


*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

PHYSICS FOR YOU | APRIL ‘16

www.pdfgrip.com


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

www.pdfgrip.com

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





(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

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)



(a) A + B + C
(c) A⋅(B ⋅ C)

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

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

PHYSICS FOR YOU | APRIL ‘16

www.pdfgrip.com


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

www.pdfgrip.com

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


=

^

R

^

^

= (1.6 × 10−19 )[(−102.4 × 103 k + 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
⎛ 60° + δmin ⎞
⎛ A + δmin ⎞
sin ⎜
sin ⎜
⎟⎠
⎟
⎠
⎝
⎝
2
2
3=
⇒
μ=
⎛ 60° ⎞
⎛ A⎞
sin ⎜
sin ⎜ ⎟

⎝ 2 ⎟⎠
⎝2⎠
3
⎛ 60° + δmin ⎞
= sin ⎜
⎟⎠ ∴ δmin = 60°
⎝
2
2



R2 + ω2 L2

30
2





2







d 1.22λ

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

dθ =

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

^

+ (1.28 × 106 i × 8 × 10−2 j )]

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

(30) + (100) × (400 × 10−3)2
30

30
=
= = 0. 6
900 + 1600 50

16

= (1.6 × 10−19 )[(−102.4 × 103 k )

⇒

2 π 20 × 10−3 × 50 × 10−6
= 1.57 ms
4

7. (d) : Power factor, cos φ =

F = qE + q (v × B) = q (E + v × B)

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

PHYSICS FOR YOU | APRIL ‘16

www.pdfgrip.com


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
R
V0
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

www.pdfgrip.com

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

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 =

r1 1
=
r2 3

or T1 – 260 = 52 or T1 = (52 + 260) K = 312 K
or T1 = (312 – 273)°C = 39°C

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


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

... (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
2

4 e A σ TS3

−

... (i)

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

30. (b) : Rate of cooling,
−

260
5

is given by,
2


1⎛T ⎞
1
1 1 1
= + . Also ⎜ ⎟ =
m
π
k
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)

PHYSICS FOR YOU | APRIL ‘16

www.pdfgrip.com


www.pdfgrip.com


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

:

VITEEE
MGIMS
AMU (Engg.)
Kerala PET
Kerala PMT
APEAMCET (Engg. & Med.)
AIPMT
Karnataka CET
MHT CET
COMED K
BITSAT
WB JEE
JEE Advanced
AIIMS
AMU (Med.)
JIPMER

:
:
:
:
:
:

:
:
:
:
:
:
:
:
:
:

PHYSICS FOR YOU | APRIL ‘16

www.pdfgrip.com

3rd April (offline),
9th & 10th April (online)
6th to 17th April
17th April
24th April
25th & 26th April
27th & 28th April
29th April
1st May
4th & 5th May
5th May
8th May
14th to 28th May
17th May
22nd May

29th May
1st June
5th June


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

PHYSICS FOR YOU | APRIL ‘16

www.pdfgrip.com

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 =






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
∴ 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)

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 :

∴ irms = < (a + b sin ωt )2 >
= < a2 > + < (b sin ωt )2 > + 2ab < sin ωt >
= a2 +
22

b2
b2
+ 0 = a2 +

2
2

PHYSICS FOR YOU | APRIL ‘16

www.pdfgrip.com


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

www.pdfgrip.com

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 ⎠⎠
⎝

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

2

+ X L2

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

Series RLC Circuit (Acceptor Circuit)

⎛
⎛ X ⎞ π⎞
VC (t ) = i0 XC sin ⎜ ωt + tan −1 ⎜ C ⎟ − ⎟
⎝ R ⎠ 2⎠
⎝

ε0
i0 =
clearly shows that
2
1
⎞
⎛
R2 + ⎜
⎝ ωC ⎟⎠
24

ε0

∴ i(t ) =

PHYSICS FOR YOU | APRIL ‘16

www.pdfgrip.com


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


C

L

Parallel RLC Circuit

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 )

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.

PHYSICS FOR YOU | APRIL ‘16

www.pdfgrip.com

25


But at this frequency

For a series LCR circuit,
2

1

1 ⎛
1 ⎞
1
=
+ ⎜ ωC −
= = minimum
⎟
2
⎝
⎠
Z
ωL
R
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
(... <cos(2ωt + φ)> = 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 ⇒

PHYSICS FOR YOU | APRIL ‘16


www.pdfgrip.com

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
=
2
π
×
=
×
∴ 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.


PHYSICS FOR YOU | APRIL ‘16

www.pdfgrip.com

27