tạp chí physics for you số tháng 11 năm 2015

If a very small, positive point charge Q is placed at any point in an electric field and it experiences a force F, then the electric field at that point is defined as The magnitude of

Physics Musing Problem Set 28 8

JEE Advanced Practice Paper 2016 58

Ace Your Way CBSE XII 74

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Volume 23 No 11 November 2015

single oPtion correct tyPe

1 A metal ring of initial radius r and cross-sectional

area A is fitted onto a wooden disc of radius R > r If

Young’s modulus of the metal is Y, then the tension

in the ring is

(a) AYR

(c) AY R r(r − ) (d) Y R r(Ar− )

2 A piece of pure gold (r = 19.3 g cm–3) is suspected

to be hollow from inside It weighs 38.250 g in air

and 33.865 g in water The volume of the hollow

portion in gold is

3 A thermally insulated vessel contains an ideal

gas of molecular mass M and ratio of specific

heats g It is moving with speed v and is suddenly

brought to rest Assuming no heat is lost to the

surroundings, its temperature increases by

4 In the figure shown, there are

10 cells each of emf e and

internal resistance r The

current through resistance

2 T perpendicular to the plane of loop Resistance

3 Shiekh Md Shakeel Hassan (Assam)

4 Swati Shah (Rajasthan)

set-26

1 Md Samim Jahin (Assam)

2 Deep Anand Basumatary (Assam)

3 Harsimran Singh (Punjab)

4 Sayantan Bhanja (WB)

7 The figure shows several

points A and B, choose the

best possible statement

(a) The electric field has a greater magnitude at

point A and is directed to left.

(b) The electric field has a greater magnitude at

point A and is directed to right.

(c) The electric field has a greater magnitude at

point B and is directed to left

(d) The electric field has a greater magnitude at

point B and is directed to right.

8 A light wire AB of length

10 cm can slide on a

vertical frame as shown

in figure There is a film

of soap solution trapped

between the frame and the wire

Find the mass of the load W that should be

suspended from the wire to keep it in equilibrium

Neglect friction Surface tension of soap solution

Consider the situation shown in

figure in which a block A of mass

2 kg is placed over a block B of

mass 4 kg

The combination of the blocks are placed on an inclined plane of inclination 37° with horizontal The system is released from rest

(Take g = 10 m s–2 and sin 37° = 0.6)

9 The coefficient of friction between block B and

inclined plane is 0.4 and in between the two blocks

is 0.5 Then

(a) Both blocks will move but block A will slide over the blocks B.

(b) Both blocks will move together

(c) None of them will move

(d) Only block A will move.

10 The frictional force acting between the blocks will

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1 Assume ABCDEF to be a regular hexagon Choose

the correct statements

(a) ED DB BE   +   +  = 0 

(b) FE BC  = 

C F

(c) AD  = 2 FE 

(d) DC  = −AF 

2 A spring mass system is hanging

from the ceiling of an elevator in

equilibrium as shown in figure The

elevator suddenly starts accelerating

upwards with acceleration a,

consider all the statements in the

reference frame of elevator and

choose the correct one(s)

(a) The frequency of oscillation is 1

with a piston which is free to

move

Mass of the frictionless piston is

9 kg Initial volume of the gas is

0.0027 m3 and cross-section area of the piston is

0.09 m2 The initial temperature of the gas is 300 K

m k

Atmospheric pressure P0 = 1.05 × 105 N m–2 An amount of 2.5 × 104 J of heat energy is supplied to the gas, then

(a) Initial pressure of the gas is 1.06 × 105 N m–2.(b) Final temperature of the gas is 1000 K.(c) Final pressure of the gas is 1.06 × 105 N m–2.(d) Work done by gas is 9.94 × 103 J

4 A man has fallen into a ditch of width d and two of

his friends are slowly pulling him out using a light rope and two fixed pulleys as shown in figure

Assume both the friends apply forces of equal magnitude Choose the correct statements

(a) The force exerted by both the friends decreases

as the man moves up

(b) The force applied by each friend is

mg

(c) The force exerted by both the friends increases

as the man moves up

(d) The force applied by each friend is mg h d2+h2

when the man is at depth h.

5 Two balls are thrown from an inclined plane at angle

of projection a with the plane, one up the incline and other down the incline as shown in figure

(Here, T stands for total time of flight).

d

Which of the following are correct?

(c) R2 – R1 = g(sinq) T12

(d) v t1 =v t2

6 Two identical buggies move one after other due to

inertia (without friction) with the same velocity v0

A man of mass m rides the rear buggy At a certain

moment, the man jumps into the front buggy with

a velocity u relative to his buggy If mass of each

buggy is equal to M and velocity of buggies after

jumping of man are vrear and vfront Then

7 A spherical body of radius R rolls on a horizontal

surface with linear velocity v Let L1 and L2 be the

magnitudes of angular momenta of the body about

centre of mass and point of

contact P respectively

Then (here K is the radius

of gyration about its

geometrical axis)

(a) L2 = 2L1 if radius of gyration K = R

(b) L2 = 2L1 for all cases

(c) L2 > 2L1 if radius of gyration K < R

(d) L2 > 2L1 if radius of gyration K > R

8 Two solid spheres A and B of equal volumes but of

different densities d A and d B are connected by a

string They are fully immersed in a

fluid of density d F They get arranged

into an equilibrium state as shown in

the figure with a tension in the string

The arrangement is possible only if

(a) d A < d F (b) d B > d F

(c) d A > d F (d) d A + d B = 2d F

9 A body of mass m is attached to a spring of spring

constant k which hangs from the ceiling of an

elevator at rest in equilibrium Now the elevator

starts accelerating upwards with its acceleration

varying with time as a = pt + q, where p and q are

A B

positive constants In the frame of elevator,

(a) The block will perform SHM for all value of p and q.

(b) The block will not perform SHM in general

for all value of p and q except p = 0.

(c) The block will perform SHM provided for all

value of p and q except p = 0.

(d) The velocity of the block will vary simple

harmonically for all value of p and q.

10 A string of mass m is fixed at both its ends The

fundamental mode of string is excited and it has an angular frequency w and the maximum

displacement amplitude A Then

(a) The maximum kinetic energy of the string is

R 2 and is filled with

honey of density 2r The upper zone of the bottle is filled with the water of density r and has a cross-

sectional radius R The

height of the lower zone is

H while that of the upper

zone is 2H If now the

honey and the water parts are mixed together to form a homogeneous solution, then

(Assume that total volume does not change)(a) The pressure inside the bottle at the base will remain unaltered

(b) The normal reaction on the bottle from the ground will remain unaltered

(c) The pressure inside the bottle at the base will increase by an amount 12rgH

(d) The pressure inside the bottle at the base will

4



 rgH

12 A particle moving with kinetic energy 3 J makes an

elastic collision (head-on) with a stationary particle

which has twice its mass During impact

(a) The minimum kinetic energy of system is 1 J

(b) The maximum elastic potential energy of the

system is 2 J

(c) Momentum and total energy are conserved at

every instant

(d) The ratio of kinetic energy to potential energy of

the system first decreases and then increases

by a massless spring of natural length L and spring

constant k The blocks are initially resting on a

smooth horizontal floor with the spring at its

natural length as shown in figure A third identical

block C, moving on the floor with a speed v along

the line joining A and B, collides with A Then

(a) The maximum compression of the spring is

v m k/

(b) The maximum compression of the spring is

v m k/2

(c) The kinetic energy of A-B system at maximum

compression of the spring is zero

(d) The kinetic energy of A-B system at maximum

compression of the spring is mv2/4

14 Three planets of same density and with radii R1, R2

and R3 such that R1 = 2R2 = 3R3, have gravitational

fields on the surfaces E1, E2, E3 and escape velocities

2 = 2 (d) v v1

3

13

=

15 Water is flowing smoothly through a closed pipe

system At one point A, the speed of the water is

3.0 m s–1 while at another point B, 1.0 m higher, the

speed is 4.0 m s–1 The pressure at A is 20 kPa when

the water is flowing and 18 kPa when the water flow

k m

From the reference frame of elevator, A ma= k

m

ma k

32

,

∆

∆

W Q

nC

P V p

5

25

4 (b, c) : From figure, sinq =

mg d/2 h

where a = angle of projection from inclined plane

q = angle of inclination of surface

v t1 and v t2 are the velocities of the particles at their

maximum height Let the particles reach their

maximum heights at time t1 and t2 respectively

Hence, 0 = (v0 sin a) – (g cos q) t1

8 (a, b, d) :Let V be the volume of

each sphere and T is the tension in

d V F g

(b) is correctFor an equilibrium

k m

dx dt

mp k

10 (a, d) : Let the displacement of the string be given by

same instant of time, where y = 0 for all x Since

dm = mdx where m =m

L is the mass per unit

length of the uniform string

The maximum kinetic energy,

The integral sin2

0

πx

L dx

L

the average value of L2

14

2 2w m 2 2w

\ (a) is correct

The mean kinetic energy of the string averaged

over one periodic time is obtained by integrating

the time dependent factor sin2 (wt + d) over one

The mean kinetic energy of the string averaged over

one periodic time is

18

92

12 (a, b, c, d) : In a head on elastic collision between

two particles, the kinetic energy becomes minimum

and potential energy becomes maximum at the

instant when they move with a common velocity

The momentum and energy are conserved at every

instant

Let m and u be the mass and initial velocity of the

first particle, 2m be the mass of second particle and

v be the common velocity

Maximum potential energy of system = 2 J

13 (b, d) : After collision of C with A, let velocity

acquired by A and B be v′ and spring gets compressed

by length x Using law of conservation of linear

12

12

12

At maximum compression of the spring, the kinetic

energy of A-B system will be

R R

R R

1 2

1 2

2 2

= = = \ (a) is not correct.

E E

R R

R R

1 3

1 3

3 3

v

R R

1 2

1

2 2

v v

R R

1 3

P1 gh1 1 v12 P2 gh2 v22

2

12

Pressure inside a liquid

At a depth h below the free surface, pressure = P.

h

P0

P

To find P, we choose a liquid

column of height h and

The additional pressure with respect to atmospheric

pressure is known as Gauge pressure

Hence we conclude that pressure changes by an amount

rgh on moving through a distance h vertically.

Note that this result has been derived from equilibrium

of the liquid column Hence if the container or liquid

was vertically accelerated, it would not be applicable

In such cases if the container is vertically accelerated,

say upward with a, then

(Fnet)upward direction = ma

So, it is interesting to see that we can also have a

situation that all points inside a liquid irrespective of

their location, will have same pressure as atmospheric if

geff = 0, as in case of free fall

Measurement of atmospheric pressure (P0 )

We take a tub filled partially with mercury and a tube completely filled with mercury We seal the mouth of the tube and invert it upside down with the mouth inside the mercury in the tub

The liquid in the tube drops down a little, creating almost vacuum in the upper closed end of the tube as shown

At equilibrium,

P A + rgh = P B = P C = P0

Hence, measuring the length of the liquid column in

the tube, P0 can easily be calculated

Supposedly, we keep this set-up in an upward

accelerating frame, then how will h change?

Clearly, in such case also, atmospheric pressure does

not change, we need to change g with geff

\ h′ < h ⇒ height of liquid column decreases.

Contributed By: Bishwajit Barnwal, Aakash Institute, Kolkata

Pressure difference in a horizontally accelerated

container

In such case, clearly, the pressure at same horizontal

level would not be same To know the exact relation we

consider a thin horizontal liquid column of length l as

drawn Hence from free body diagram,

Vertically pressure increases in the direction of gravity,

horizontally acceleration increases opposite to the

This could also have been found out using the fact that

liquids cannot tolerate tangential force on its surface

Hence the free surface should be perpendicular to g eff

g eff = + −g ( )a

[We revert the acceleration of container and add it

vectorially to g ]

Let us apply this to a more

complicated situation Assume

a container falling down on a

smooth inclined plane

Archimedes’ principle

If an object is submerged inside a liquid, partially or completely, it experiences an upward force by the liquid due to pressure difference along the vertical column

of the liquid, which is equal to the weight of liquid displaced by the object To prove this, let us imagine an

object of cross-sectional area A and height H partially submerged till height h as shown here.

H h

Note : This result is applicable only if the container is

vertically unaccelerated, else we need to replace g with

g eff in the result

Now, let us see what would the condition of floatation

for the object be

\ If rs ≤ rl, the object floats, else it sinks Hence it

does not matter how heavy an object is for floating,

what matters is how dense the object is!

Note here, that the fraction of submerged portion, h

H is

independent of g, hence even in accelerated containers,

this same fraction will be submerged

Now, let us seen an application

Suppose a helium filled balloon

He air

is floating in air (their densities

given as rHe and rair) with a

string tied to a box as shown

here

Now, if the box is accelerated towards right with

acceleration a, we have to find the direction and angle

with the vertical in which the string gets deflected at

equilibrium

One would be tempted to say that the string will deflect

towards left due to the pseudo force

But there is a basic point, one is missing here As we

saw force of upthrust being generated due to pressure

difference along vertical column of liquid, similar to it

a side thrust can also be generated if pressure difference

is created along horizontal column and on similar

approach it can be proved

Fside thrust = rl V sub a

Hence, since, rair > rHe,



so side thrust will be greater

than the pseudo force, so the

balloon deflects towards right

a g

Force exerted by a liquid on a vertical surface

Let us assume, that the wall of a dam of width w stops a liquid of height h from flowing It obviously is

experiencing force due to the liquid’s pressure

To find this, let us consider a horizontal strip of height

dy at a depth y below the free surface.

The force experienced by this surface due to pressure

= pressure at centroid of submerged portion ×

area of submerged portion

Force exerted by liquid on a horizontal surface

Let us consider a horizontal face of area A at a depth h

below the free surface

We have to find the force exerted by the liquid only

Net vertical force = PA

= (rgh)A = r(Ah)g

= weight of liquid column above its surface

nn

Solution Set-27

1 (d) :Here, d = 0.5 mm = 0.5 × 10–3 m

D = 0.5 m

l = 500 nm = 500 × 10–9 m

The distance of third maxima from the second

minima on the other side is

= 2.25 × 10–3 m = 2.25 mm

2 (c) : During motion of the particle, total mechanical

energy remains constant

At the surface of earth, total mechanical energy is

0

or −2gR v+ 02= −gR v ∴+ 2 v= v02−gR

3 (a) : If we consider the cylindrical surface to be a

ring of radius R, there will be an induced emf due

As the field is increasing being directed inside the

paper, hence there will be anticlockwise induced

current (in order to oppose the cause) in the ring

(assumed) Hence there will be a force towards left

on the electron

4 (c) : Total time of flight is T = 4 s and if u is its

initial speed and q is the angle of projection Then

g g

sincos

When lift is accelerated upwards with acceleration

a, let time period becomes T2 Then

a g

Accelerations of the block down the two planes are

a1 = g sinq1 and a2 = g sinq2

As l1 1a t1 12and l2 a t2 22

2

12

a l

a l

g g

1 2

1 2

sin

sinsin

qq

qq

t21 12

sinsin

qq

7 (a) : Given : f = at2 + bt

The magnitude of induced emf is

at b dt

a b

0 0

0

2 0

25

where a is the angle made by the vector (A B + with ) A.

where b is the angle made by the vector(A B − )with A.

Note that the angle between Aand ( )−B is (180° – q)

Adding (i) and (ii), we get

cos

sincos

q

qq

aa

=

−

221

2

2

sincos

sincossin( cos )

qq

qqqq1

2212

2 2

⋅+

−+

On solving, cosq = − 1

2

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solution of october 2015 crossword

1

S K E W R A Y

K A O N I

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A T

D F R M S

Z E N I T H

U C L E A

C A N L R A Y S

Y C T O

C T

F L I P F L O P

D R Y C E L L P

F L A V O R

R I P P L E

I T S U N

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T E N S I M E T E R

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W E I G H T

O E A N E C H O I C

G A T E

8

9 10

Mechanical ProPerties of fluids

1 A hemispherical bowl just floats without sinking

in a liquid of density 1.2 × 103 kg m–3 If outer

diameter and the density of the bowl are 1 m and

2 × 104 kg m–3 respectively, then the inner diameter

of the bowl will be

2 A body of density r is dropped from rest at a height

h into a lake of density s, where s > r Neglecting

all dissipative forces, calculate the maximum depth

to which the body sinks before returning to float on

the surface

3 Water in a vessel of uniform cross-section escapes

through a narrow tube at the base of the vessel

Which of the following graphs represents the

variation of the height h of the liquid with time t?

4 Water rises to a height of 10 cm in a capillary tube

and mercury falls to a depth of 3.42 cm in the

same capillary tube If the density of mercury is

13.6 g cm–3 and the angle of contact of mercury and water are 135° and 0° respectively, the ratio of surface tension of water and mercury is

5 A metallic sphere of mass M falls through glycerine with a terminal velocity v If we drop a ball of mass 8M of same metal into a column of glycerine, the

terminal velocity of the ball will be

6 A cylindrical drum, open at the top, contains 15 L

of water It drains out through a small opening at

the bottom 5 L of water comes out in time t1, the

next 5 L in further time t2 and the last 5 L in further

time t3 Then

(a) t1 < t2 < t3 (b) t1 > t2 > t3(c) t1 = t2 = t3 (d) t2 > t1 = t3

7 A sealed tank containing a liquid of density r moves

with a horizontal acceleration a, as shown in figure The difference in pressure between the points A and

chapterwise McQs for practice

Useful for All National and State Level Medical/Engg Entrance Exams

atmospheric pressure is 10 m of water, the depth of

the water in the tank is

What is the density of liquid?

10 A spherical ball is dropped in a long column of

viscous liquid Which of the following graphs

represent the variation of

(i) gravitational force with time

(ii) viscous force with time

(iii) net force acting on the ball with time?

F

P Q

R

t

11 Water is flowing through a horizontal pipe If at one

point pressure is 2 cm of Hg and velocity of flow of

the liquid is 32 cm s–1 and at another point, velocity

of flow is 40 cm s–1, the pressure at this point is

12 The rate of flow of glycerine of density

1.25 × 103 kg m–3 through the conical section

of a pipe, if the radii of its ends are 0.1 m and

0.04 m and the pressure drop across its length is

10 N m–2, is

(a) 5.28 × 10–4 m3 s–1 (b) 6.28 × 10–4 m3 s–1

(c) 7.28 × 10–4 m3 s–1 (d) 8.28 × 10–4 m3 s–1

13 Water rises in a capillary tube to a height h Choose

the false statement regarding a capillary rise from

the following

(a) On the surface of Jupiter, height will be less

than h.

(b) In a lift, moving up with constant acceleration,

height is less than h.

(c) On the surface of the moon, the height is more

than h.

(d) In a lift moving down with constant acceleration,

height is less than h.

14 A candle of diameter d is floating on a liquid

in a cylindrical container of diameter D(D>>d)

as shown in figure If it is burning at the rate of

2 cm h–1, then the top of the candle will

L L d D

(a) remain at the same height(b) fall at the rate of 1 cm h–1 (c) fall at the rate of 2 cm h–1(d) go up at the rate of 1 cm h–1

15 A frame made of metallic wire enclosing a surface

area A is covered with a soap film If the area of the

frame of metallic wire is reduced by 50%, the energy

of the soap film will be changed by(a) 100% (b) 75% (c) 50% (d) 25%

therMal ProPerties of Matter

16 The plots of intensity versus wavelength for

three black bodies at temperatures T1, T2 and T3

respectively are as shown Their temperatures are such that

T1

T3

T2I



(a) T1 > T2 > T3 (b) T1 > T3 > T2(c) T2 > T3 > T1 (d) T3 > T2 > T1

17 A solid copper sphere (density r and specific heat

capacity c) of radius r at an initial temperature

200 K is suspended inside a chamber whose walls are at almost 0 K The time required (in ms) for the temperature of the sphere to reach 100 K is

(a) 807 r crs (b) 807 r csr(c) 277 r crs (d) 277 r crs

18 A clock with an iron pendulum keeps correct time

at 15°C What will be the error in time per day, if

the room temperature is 20°C?

(The coefficient of linear expansion of iron is

0.000012°C–1.)

10 min If room temperature is 25°C,

temperature of body at the end of next

10 min will be

20 Two rods of same length and material transfer a

given amount of heat in 12 s, when they are joined

end to end (i.e., in series) But when they are joined

in parallel, they will transfer same heat under same

conditions in

21 A spherical black body with a radius of 12 cm

radiates 450 W power at 500 K If the radius were

halved and the temperature doubled, the power

radiated in watt should be

22 A body in laboratory takes 4 min to cool from 61°C

to 59°C If the laboratory temperature is 30°C, then

the time taken by it to cool from 51°C to 49°C is

23 The wavelength of maximum intensity of radiation

emitted by a star is 289.8 nm The radiation intensity

for the star is (Take s = 5.67 × 10–8 W m–2 K–4,

Wien’s constant, b = 2898 mm K)

(a) 5.67 × 108 W m–2 (b) 5.67 × 1012 W m–2

(c) 5.67 × 107 W m–2 (d) 5.67 × 1014 W m–2

with that of Fahrenheit thermometer in a liquid

The temperature of the liquid is

25 For a black body at temperature 727 °C, its radiating

power is 60 W and temperature of surrounding

is 227°C If the temperature of the black body is changed to 1227°C, then its radiating power will be

27 A 2 kg copper block is heated to 500°C and then it

is placed on a large block of ice at 0°C If the specific heat capacity of copper is 400 J kg–1°C–1 and latent heat of fusion of water is 3.5 × 105 J kg–1, the amount

of ice that can melt is

28 The maximum wavelength of radiation emitted

at 2000 K is 4 mm What will be the maximum wavelength emitted at 2400 K?

(d) nature of its surface

30 We plot a graph, having temperature in °C on x-axis and in °F on y-axis If the graph is straight line,

then it(a) passes through origin

(b) intercepts the positive x-axis (c) intercepts the positive y-axis (d) intercepts the negative axis of both x-and

y-axis

solutions

1 (c) : Let D1 be the inner diameter of the hemispherical bowl As bowl is just floating so4

3

1

43

2 (c) : The speed of the body just before entering

the liquid is u = 2 The buoyant force F gh B of

the lake, i.e., upward thrust of liquid on the body

is greater than the weight of the body W, since

s > r If V is the volume of the body and a is the

acceleration of the body inside the liquid, then

3 (a) : Let dV be the decrease in volume of water in

vessel in time dt Therefore rate of decrease of water in

vessel = rate of water flowing out of narrow tube

dt

P P r l

ph

Integrating it within the limits, as time changes

from 0 to t, volume changes from V0 to V.

Thus, the variation of h and t will be represented by

exponential curve as given by (a)

1 2

coscosθθ rr

= −( )103 42 ×coscos1350° ×° 13 61.

= 1 ×3 42.0 0 70713 6. . =6 51.

5 (b) : As, M=43p rr3 and 8M=43p rR3 ,

So, R3 = 8r3 ⇒ R = 2r Now v ∝ r2 so, v

v

R r

r r

=   =  =

or v1 = 4 v

6 (a) : If h is the initial height of liquid in drum

above the small opening, then velocity of efflux,

v = 2 As the water drains out, h decreases, gh

hence v decreases This reduces the rate of drainage

of water Due to which, as the draining continues,

a longer time is required to drain out the same

volume of water So, clearly, t1 < t2 < t3

7 (a) : Since points A and C are in the same horizontal line but separated by distance l and liquid tank is moving horizontally with acceleration a, hence

Points B and C are vertically separated by h

1

pp

Using Boyle’s law, we have

P1V1 = P2V2

V

h kA kA

10 (c) : Gravitational force remains constant on the

falling spherical ball It is represented by straight

line P The viscous force (F = 6 phrv) increases

as the velocity increases with time Hence, it is

represented by curve Q Net force = gravitational

force – viscous force As viscous force increases,

net force decreases and finally becomes zero Then

the body falls with a constant terminal velocity It

is thus represented by curve R.

11 (b) : As per Bernoulli’s theorem,

P1 1 v12 P2 v22

2

12

0 1

0 04

254

pp

1

Since, candle is burning at the rate of 2 cm h–1, then

after an hour, candle length is 2L – 2

L x L

2 3

21 (d) : For a spherical black body of radius r at T K,

Power radiated = energy radiated per second

If T is the temperature of star, then

according to Wien’s law, lm T = b

32180

If the temperature is T at which the readings of two

scales coincide, then from

2 1

\ Thermal conductivity of material

27 (c) : Let x kg of ice melts.

Using law of calorimetry,

heat lost by copper = heat gained by ice

max max

2 1

1 2

Thus the graph between °C and °F is a straight line

with positive intercept on y-axis as shown in the

figure above



There are plenty of technologies a

vailable in the market to treat water But most

of them are capable of treating only tap water a

nd are either membrane-based or use chemicals

In remote villages

or areas hit by natural disasters, po

nds and wells still remain the primary sou

rce of water, but this is not fit for drinking.

ust be disposed of

in a safe manner At the Flexible Ele

ctronics Lab at the IISc, we have been working on the id

ea of developing a more efficient water tre

atment technology since 2010;

a small, portable, maybe even a ha

nd-held device that could convert very dirty w

ater into drinkable water. It was only a year later that we seriously sta

rted working on the

project We used electric fields to achieve the objective.

When an electric field is established

in a container with water, the impurities and small pa

rticles get polarised and are then attracted to each oth

er Smaller particles coalesce into larger granu

les while remaining suspe

nded

in water These granules can the

n be filtered easily, using a normal tea filter or

a clean cloth.

For our tests, we used water from

Mavalipura, near Bangalore, where the w

ater quality is really bad, and the results we achieved were

satisfactory.

There was no new phys

ics to the experiment that we performed, as the problem

was essentially an engineering

issue The key was to u

nderstand how electric

fields permeate ion rich fluids and how

they interact with neutral impurities This understa

nding helped us identify the field strengths neede

d to maximise this interaction and then use this conc

ept to improve the

filtration process.

We have managed to develop a t

echnology that can convert very dirty water — from pon

ds, lakes, wells and

even waste water — into drinkable quality Unlike in the traditional filtration technologies, there is zero wastage

of water In fact, even the waste w

ater can be reused

The entire instrument can be encased

in a one-litre bottle and needs very little po

wer to run, either a battery or hand-held dynamo will serve the

purpose.

The instrument can filter one liter of w

ater in a maximum time of about 10 minu

tes, though there have been occasions where one litre of water

has been treated

within three minutes.

So far, we have tested the newly

developed water purifying system in labo

ratories and now we are trying

to take the technology to the field.

We estimate that a bottle filter could cost a

bout Rs 1,000, but the prices will come down substantially as we in

crease the capacity

of the instrument The new filter and

our ideas on how

to introduce the technology to so

ciety was recently awarded the first place at a compe

tition organised by Google in Zurich, Switzerland. Courtesy : Indian

to filter very dirty water

and make it fit for drinking

Electrostatics is the branch of science that deals with the

study of electric charges at rest Here we study the forces,

fields and potentials associated with static charges

Electric charges

Charges are of two types, positive charge and negative

charge The charge developed on a glass rod when rubbed

with silk is positive charge The charge developed on a

plastic rod when rubbed with wool is negative charge

Basic properties of charge

Charge is a scalar quantity

•

Charge is transferable :

in contact with an uncharged body, the uncharged

body becomes charged due to transfer of electrons

from one body to the other

Charge is always associated with mass,

can not exist without mass though mass can exist

without charge So, the presence of charge itself is a

convincing proof of existence of mass

Quantization of charge :

always an integral multiple of a basic unit of charge

denoted by e and is given by q = ne where n is any

integer, positive or negative and e = 1.6 × 10–19 C

The basic unit of charge is the charge that an electron

•

or proton carries By convention the charge on

electron is –e (–1.6 × 10–19 C) and charge on proton is

+e (1.6 × 10–19 C)

Additivity of charge :

the algebraic sum (i.e sum is taking into account

with proper signs) of all individual charges in the system

Conservation of charge :

isolated system remains unchanged with time In other words, charge can neither be created nor be destroyed Conservation of charge is found to hold good in all types of reactions either chemical or nuclear

Methods of charging : A body can be charged by

bodies without any loss of charge If q be the source

of charge, then charge induced on a body of dielectric

constant K is given by ′ = − − q q1 1K

For metals, K = ∞ \ q′ = –q

i.e., charges induced are equal and opposite only in case

of conductors In general, magnitude of induced charge

is less than that of inducing charge

coulomb’s law

It states that, the electrostatic force between two

stationary charges is proportional to the product of

magnitude of charges and inversely proportional to the

square of the distance between them

e0 = 8.854 × 10–12 C2 N–1 m–2 is permittivity of free space

Vectorially Coulomb’s law can be written as

(The force on charge q2 due to charge q1)

where the position vectors of charges q  1 and q2 are

medium=41 1 2=4 1

2

0

1 2 2

affects and is affected by a medium

comparison between coulomb force and

gravitational force are as follows :

Coulomb force and gravitational force follow the

•

same inverse square law

Coulomb force can be attractive or repulsive while

•

gravitational force is always attractive

Coulomb force between the two charges depends

•

on the medium between two charges while

gravitational force is independent of the medium

between the two bodies

The ratio of coulomb force to the gravitational

Superposition theorem : The interaction between any

two charges is independent of the presence of all other charges

Electrical force is a vector quantity therefore, the net force on any one charge is the vector sum of all the forces exerted on it due to each

of the other charges interacting

with it independently i.e, Total force on charge q,    

F F F F= + + +1 2 3

continuous charge Distribution

Linear charge density : Charge per unit length is known

as linear charge density It is denoted by symbol l

l = ChargeLengthIts SI unit is C m–1

Surface charge density : Charge per unit area is known

as surface charge density It is denoted by symbol s

s = ChargeAreaIts SI unit is C m–2

Volume charge density : Charge per unit volume

is known as volume charge density It is denoted by symbol r

r = ChargeVolumeIts SI unit is C m–3

SELFCHECK

1 Two charges, each equal to q, are kept at x = – a and

x = a on the x-axis A particle of mass m and charge

q0= is placed at the origin If charge q q2 0 is given a

small displacement (y < < a) along the y-axis, the

net force acting on the particle is proportional to

(JEE Main 2013)

–Q inside the cavity of a spherical shell The charges

are kept near the surface of the cavity on opposite sides of the centre of the shell If s1 is the surface

charge on the inner surface and Q1 net charge on it and s2 the surface charge on the outer surface and

Q2 net charge on it then

An electric field is a region where an electric charge

experiences a force If a very small, positive point

charge Q is placed at any point in an electric field and it

experiences a force F, then the electric field at that point

is defined as

The magnitude of E is the force per unit charge and its

direction is that of F(i.e., of the force which acts on a

positive charge) Thus, electric field is a vector quantity

If F is in newton (N) and Q is in coulomb (C) then the

unit of E is newton per coulomb (N C–1)

Electric field due to a Point charge

The magnitude of E due to an isolated positive point

charge +q0 at a point P distance r away, in a medium of

permittivity e, can be calculated by imagining a very small

E is directed away from +q0 as shown If a point charge

–q0 is replaced by +q0, E would be directed towards

–q0 since unlike charges attract

Electric field lines

An electric field can be represented and so visualized

by electric field lines These are drawn so that, the field lines at a point, (or the tangent to it if it is curved) gives the direction of E at that point, i.e., the direction in

which positive charge would move and the number of

lines per unit cross-section area is proportional to E

The field lines are imaginary but the field it represents

is real

The friction caused by rapidly rising and falling currents of moving air creates electrical charges within

a cloud Water droplets and ice pellets fall, carrying charged electrons to the lower portion of the cloud, where a negative charge builds A positive charge builds up near the top of a cloud most of the electrical energy in a thunderstorm is dissipated within the clouds, as lightning hops between the positively and negatively charged areas Lightning becomes dangerous, though, when it reaches earth When the negative charge in the cloud becomes great enough, it seeks an easy path to the positively charged ground below The current looks for a good conductor of electricity, or a tall structure anchored to the ground.

As negative charges collect at the bottom of a storm cloud, a change happens on the ground below electrons on the ground feel the power of the cloud’s negative charges The electrons are pushed away from the area underneath the cloud The ground and the objects on it are left with a positive charge.

If you were standing on the ground below a storm cloud, you wouldn’t be able

to see electrons move but you might feel your skin tingle or your hair stand

on end.

As the ground becomes positively charged, the attraction between the cloud and the ground grows stronger Suddenly, electrons shoot down from the cloud They move in a path that reaches out in different directions—like the branches

of a tree each branch, or step, is about 45 m long This branching path is called

a stepped leader After the first electrons have blasted their way through the air, other electrons from the cloud follow and make new branches A stepped leader cuts through the air very quickly Its average speed is about 1.2 × 105 m s–1 As the stepped leader nears the ground, a positive streamer reaches up for it only then, once this channel is made, does the visible lightning happen A return stroke runs from the ground to the clouds in a spectacular flash Though the bolt appears continuous, it is actually a series of short bursts most lightning strikes occur in less than a half second and the bolt is usually less than 5 cm in diameter.

The electric field due to a positive point charge is

represented by straight lines originating from the charge

as shown in figure (i)

The electric field due to a negative point charge is

represented by straight lines terminating at the charge

as shown in figure (ii)

The lines of force for a charge distribution containing

more than one charge, is such that from each charge we

can draw the lines isotropically The lines may not be

straight as one moves away from a charge

The shape of lines for some charge distribution is shown

below

The lines of force are purely a geometrical construction

which help us to visualise the nature of electric field in

the region They have no physical existence

The number of lines originating or terminating on

a charge is proportional to the magnitude of charge

In rationalised MKS system (1/e0) electric lines are

associated with unit charge So if a body encloses a

charge q, total lines of force associated with it (called

flux) will be q/e0

Lines of force per unit area normal to the area of a

point represents magnitude of intensity, crowded lines

represent strong field while distant lines represent weak

field

Electric flux

If the lines of force pass through a surface then the

surface is said to have flux linked with it It is given by

dφ = ⋅E dS 

where dS is the area vector of the small area element

The area vector of a closed surface is always in the

direction of outward drawn normal

The total flux linked with whole of the body

φ = ∫E dS ⋅where ∫ represents closed integral done for a closed surface

The SI unit of electric flux is N m2 C–1 and its dimensional formula is [ML3T–3A–1]

ElEctric DiPolE

It is a pair of two equal and opposite charges separated

by a small distance

Electric Dipole Moment

It is a vector quantity whose magnitude is equal to product of the magnitude of either charge and distance

between the charges i.e | | p q a= 2

By convention the direction of dipole moment is from negative charge to positive charge

The SI unit of electric dipole moment is C m and its dimensional formula is [M0LAT] The practical unit of electrical dipole moment is debye

Electric field intensity on axial line (End on Position)

of the Electric Dipole

At the point at a distance r from the centre of the electric

moment (i.e from –q to q).

Electric field intensity on Equatorial line (broad on Position) of Electric Dipole

At the point at a distance r from the centre of electric

Electric field intensity at any Point due to an Electric

Electric field intensity due to a charged ring

At a point on its axis at distance r from its centre,

At the centre of the ring, i.e r = 0, E = 0.

torque on an Electric Dipole Placed in a uniform Electric field

When an electric dipole of dipole moment p is placed

in a uniform electric field E, it will experience a torque and is given by

List of formula for electric field intensity due to various types of charge distribution :

q

r

charge to the test point

outwards due to +ve charges and

from line charge

r

^

the charge to test point

E r

3 A long cylindrical shell carries positive surface

charge s in the upper half and negative surface

charge –s in the lower half The electric field lines

around the cylinder will look like figure given in

(Figures are schematic and not drawn to scale)

(d)

(JEE Main 2015)

4 A thin disc of radius b = 2a has

a concentric hole of radius

a in it (see figure) It carries uniform surface charge s on

it If the electric field on its axis

at height h (h < < a) from its centre is given as Ch then value

of C is

(a) se

(JEE Main 2015)

5 A wire, of length L(= 20 cm), is bent into a

semi-circular arc If the two equal halves of the arc, were

each to be uniformly charged with charges ±Q,

[|Q| = 103 e0 Coulomb where e0 is the permittivity

(in SI units) of free space] the net electric field at the

centre O of the semi-circular arc would be

where q1 is the initial angle between p and and qE 2 is

the final angle between pand E

Gauss’s law

It states that the total electric flux through a closed

surface S is 1/e0 times the total charge enclosed by S.

 

 ∫ E dS q⋅ =e0

where q is charge enclosed by the closed surface S.

The total electric flux through a closed surface is zero if

no charge is enclosed by the surface

In the situation when the surface is so chosen that there are some charges inside and some outside, the electric field (whose flux is calculated) is due to all the charges, both inside and outside the surface However, the term

(q) represents only the total charge inside the closed

surface

KEYPOINT

Gauss’s law is true for any closed surface, regardless

•

of its shape or size

The surface that we choose for the application of

of potential is volt and its dimensional formula is [ML2T–3A–1]

Electric potential due a point charge q at a distance r

from the charge is

V= 4peq0rElectric potential due to various charge distributions are given in table below :

Point charge

V kq= r ff q r is the distance of the point is source charge.

from the point charge

of sphere to the point

Q

Uniformly charged

solid non-conducting

sphere

For r R V kQ≥ , = R

conducting thin

sheet

defined Potential difference between

potential V0 (measured with respect to ∞) on its

surface For this sphere the equipotential surfaces

R1, R2, R3 and R4 respectively Then

An equipotential surface is a surface with a constant

value of potential at all points on the surface

Properties of an equipotential surface

Electric field lines are always perpendicular to an

equipotential surface Work done in moving an electric

charge from one point to another on an equipotential surface is zero Two equipotential surfaces can never intersect one another

Relationship between EandV

Electric Potential Energy

Electric potential energy of a system of charges is the total amount of work done in bringing the various charges to their respective positions from infinitely large mutual separations The SI unit of electrical potential energy is joule

Electric potential energy of a system of two charges is

U= 14 q q r

0

1 2 12

pe

where r12 is the distance between q1 and q2.

Electric potential energy of a system of n point charges

jk

=4pe10 ∑

all pairs

Note in this summation, we should include only one

term for each pair of charges

conductors and insulators

On the basis of conductivity, all bodies can be divided

in two classes, conductors and insulators

In conductors, electric charges are free to move

throughout the volume Insulators do not have free

charges to move

Basic electrostatics properties of a conductor

Inside a conductor, electric field is zero

•

At the surface of a charged conductor, electric field

•

must be normal to the surface at every point

The interior of a conductor can have no excess

•

charge in the static situation

Electric potential is constant throughout the volume

where s is the surface charge density and n^ is a unit

vector normal to the surface in the outward direction

Electrostatic shielding : It is the phenomenon of

protecting a certain region of space from external

electric field

Polar and non-Polar Molecules

Dielectrics : Dielectrics are non conducting substances

In contrast to conductors, they have no (or negligible

number of) charge carriers

Polar molecule : A polar molecule is one in which the

centres of positive and negative charges are separated

(even when there is no external field) A polar molecule

has a permanent dipole moment e.g., water (H2O) and

HCl

Non-polar molecule : A non-polar molecule is one

in which the centres of positive and negative charges

coincide A non-polar molecule has no permanent

dipole moment e.g., oxygen (O2) and hydrogen (H2)

Polarisation : The dipole moment per unit volume

is called polarisation and is denoted by P For linear 

isotropic dielectrics P= c ,e E

where ce is a constant characteristic of the dielectric

and is called the electric susceptibility of the dielectric

medium

capacitor

A capacitor is a device that stores electrical energy It consists of conductors of any shape and size carrying charges of equal magnitudes and opposite signs and separated by an insulating medium

Capacity of capacitor (Capacitance)

Capacitance (C) of a capacitor is the ratio of charge(Q) given and the potential (V) to which it is raised.

C Q=VThe SI unit of capacitance is farad (F)

Capacitance of spherical conductor

Capacitance of a spherical conductor of radius R is

C = 4pe0R

Taking earth to be a conducting sphere of radius

6400 km, its capacitance will be

Capacity of capacitor depends upon

Total outer surface area

Parallel plate capacitor : C= ed0A

(when air is between the plates)

C=e0K A d (when dielectric is between the plates)

Here, A is area of each plate and d is separation between

the two plates

Spherical capacitor :C=4pe0b a ab−

Here, a and b are the radii of inner and outer coatings

(spherical shells) of the spherical capacitor

Cylindrical capacitor :

a

l b a e

Here, a and b are the radii of the inner and outer

coatings (cylindrical shells) and l is the length of the

either curved surface of the cylindrical capacitor

Same charge flows through each capacitor

Different potential difference exists across each

capacitor joined in series The ultimate effect is to

reduce the capacitance in a circuit

Capacitors in parallel

Across each capacitor, potential difference is same

Different charges flow through each capacitor

q = q1 + q2 + q3

C = C1 + C2 + C3

The resultant capacity C is greater than the greatest

capacitance joined in parallel Thus the ultimate effect

is to enhance the capacitance in a circuit

Energy stored in a capacitor

Work done in charging a capacitor gets stored in the

capacitor in the form of its electrostatic potential

energy

U=12CV2=12QV=12Q C2

Electric Energy Density (u E )

The energy stored per unit volume in the electric field

between the plates is called energy density

called common potential (V), where

V=Total capacityTotal charge =Q Q C C1++ 2 =C V C V C C++

When a dielectric slab of dielectric constant K is

introduced between the plates of a charged parallel plate capacitor and the charging battery remains connected, then, potential difference between the plates remains

constant i.e., V = V0

Capacitance C increases i.e., C = KC0

Charge on a capacitor increases i.e., Q = KQ0

Electric field between the plates remains unchanged i.e.,

E = E0

Energy stored in a capacitor increases

i.e., U = KU0

When a dielectric slab of dielectric constant K is

introduced between the plates of a charged parallel plate capacitor and the charging battery is disconnected,

then, charge remains unchanged i.e., Q = Q0

Capacitance increases i.e., C = KC0

Potential difference between the plates decreases i.e.,

Van de Graaff generator

A Van de Graaff generator consists of a large spherical conducting shell (a few metres in diameter) By

means of a moving belt and suitable brushes, charge

is continuously transferred to the shell, and potential

difference of the order of several million volts is built up,

which can be used for accelerating charged particles

SELFCHECK

7 In the given circuit, charge Q2 on the 2 mF capacitor

changes as C is varied from 1 mF to 3 mF Q2 as a

function of ‘C’ is given properly by (figures are

drawn schematically and are not to scale)

8 In figure is shown a system of four capacitors

connected across a 10 V battery Charge that will

flow from switch S when it is closed is

(JEE Main 2015)

currEnt ElEctricity

Electric current

It is the amount of charge flowing across any section of

wire per unit time If the moving charges are positive,

the current is in the direction of motion of charges If

they are negative the current is opposite to the direction

of motion of charges

Instantaneous current is the current at any point of time

= dq dt and average current = q t

The unit of current is ampere (A)

1A=1CsThe dimentional formula of current is [M°L°T°A]

Drift Velocity

It is defined as the average velocity with which free electrons get drifted towards the positive end of the conductor under the influence of an external electric field Drift velocity of electrons is given by

v d= −eE m t

where –e is the charge and m is the mass of an electron, 

E is the electric field applied and t is known as relaxation

time The value of drift velocity of an electron is about

10–4 m s–1 and value of relaxation time is about

10–14 second

The direction of drift velocity of electrons in a metal conductor is opposite to that of electric field applied E.

Drift velocity depends on electric field as v d ∝ E So

greater the electric field, larger will be the drift velocity.The thermal speed or rms speed of electrons at room temperature is about 105 m s–1, which is very large as compared to the drift velocity of electrons

Relaxation time =rms speed of electronsmean free path

relationship between current and drift velocity

I = nAev d

where n is the number density of electrons or number

of electrons per unit volume of the conductor and A is

the area of cross-section of the conductor

Mobility

It is defined as the magnitude of drift velocity per unit electric field It is denoted by symbol m

m=v E d =qE mtE/ =q mt

where q, t and m are charge, relaxation time and mass

of a charge carrier respectively

Mobility is positive for both electrons and holes although their drift velocities are opposite to each other

The SI unit of mobility is m2 V–1 s–1 and its dimensional formula is [M–1L0T2A]

ohm’s law

It was discovered by German physicist Georg Simon

Ohm in 1828 It states that the current (I) flowing

through a conductor is directly proportional to the

potential difference (V) across the ends of the conductor,

provided physical conditions of the conductor such as

temperature, mechanical strain etc are kept constant

V ∝ I or V = RI

where the constant of proportionality R is called

resistance of the conductor.

The graph between potential



V

I O

slope = tan

difference (V) and current (I)

through a metallic conductor is

a straight line passing through

the origin as shown in figure

The slope of V-I graph gives resistance

R VI= = tan (q slope of - graphV I )

Electrical resistance

The resistance of a conductor is the obstruction posed

by the conductor to the flow of current through it

Resistance of a conductor (R) is defined as the ratio

of potential difference (V) applied across the ends of

conductor to the current (I) flowing through it

R VI=

Resistance is a property of the conductor and is not related

to the circuit in which the conductor is connected The

equation R VI= can be used to calculate the current in

a conductor if we know the resistance and the potential

difference across the conductor The SI unit of resistance

is ohm (W) and its dimensional formula is [ML2T–3A–2]

The resistance of a conductor not only depends on the

material of the conductor but also on the dimensions of

the conductor

The resistance of a conductor is proportional to its

length and inversely proportional to its area of

cross-section

R∝A l or R=rA l

where constant of proportionality r is called resistivity

The SI unit of resistivity is W m and its dimensional

where m is the mass and e is the magnitude of charge of

an electron, n is the number density of electrons, t is the

relaxation time

The resistivity of a conductor is independent of its

dimensions but depends on the material of the conductor

A perfect conductor would have zero resistivity and a

perfect insulator would have infinite resistivity

If the conductor is in the form of wire of length l and radius r, then its resistance is R l

If the area of cross-section of the given metallic wire

of resistance R becomes n times, then its resistance becomes (1/n2)R.

A cylindrical tube of length l has inner and outer radii r1 and r2 respectively The resistance between its end faces

current density, conductance and conductivity

Current density at a point inside the conductor is defined as the amount of current flowing per unit area around that point of the conductor, provided the area is held in a direction normal to the current It is denoted

by symbol J.

J IA=

If area A is not normal to the current but makes an angle

q with the direction of current, then

J=AcosI q or I JA= cosq= ⋅J A 

The SI unit of current density is A m–2 and its dimensional formula is [M0L–2T0A]

Current density is a vector quantity Its direction is that

of the flow of positive charge at the given point inside the conductor

Current density is a characteristic property of a

•

particular point inside the conductor and not of

the whole conductor

Current is the flux of current density

•

relationship between current density and drift

velocity

J = nqv d where n is the number density of charge carriers

each of charge q, and v d is the drift velocity of the charge

carriers For electrons q = –e

Conductance : The reciprocal of resistance is called

conductance It is denoted by symbol G.

G R= 1

The SI unit of conductance is ohm–1 which is called mho

and is represented by the symbol ( ) The unit mho is

also called siemen (S) and its dimensional formula is

[M–1L–2T3A2]

Conductivity : The reciprocal of resistivity is called

conductivity or specific conductance It is denoted by

symbol s

s r= =1 ne m2t=nem As m=v E d =e mt

The SI unit of conductivity is W–1 m–1 or S m–1

[M–1L–3T3A2]

relationship between J, s and E

J = sE

It is a microscopic form of Ohm’s law

ohmic and non-ohmic conductors

Ohmic conductors : Those conductors which obey

Ohm’s law are called ohmic conductors, e.g metals For

Ohmic conductors, V-I graph is a straight line passing

through the origin

Non-ohmic conductors : Those conductors which do

not obey Ohm’s law are called non-ohmic conductors

e.g diode valve, junction diode For non-ohmic

conductors V-I graph is non-linear.

Effect of temperature on resistance

The resistance of a metallic conductor increases with

increase in temperature

The resistance of a conductor at temperature T°C is

given by

R T = R0 (1 + aT + bT2)

where R T is the resistance at T°C, R0 is the resistance at

0°C and a and b are the characteristics constants of the

material of the conductor If the temperature T°C is not

sufficiently large, b is negligible, the above relation can

be written as

R T = R0 (1 + aT)

where a is the temperature coefficient of resistance Its unit is K–1 or °C–1 For metals, a is positive, therefore resistance of a metal increases with rise in temperature.For insulators and semiconductors, a is negative therefore their resistance decreases with rise in temperature For alloys like manganin and constantan, the value of a is very small as compared to that for metals Due to high resistivity and low temperature coefficient of resistance, these alloys are used in making standard resistance coils

thermistors

A thermistor is a heat sensitive device whose resistivity changes very rapidly with change of temperature

colour code for resistors

A colour code is used to indicate the resistance value and its percentage accuracy Every resistor has a set of coloured rings on it The first two coloured rings from the left end indicate the first two significant figures of the resistance in ohms The third colour ring indicates the decimal multiplier and the last colour ring stands for the tolerance in percent The colour code of a resistor is

as shown in the table

Colour Number Multiplier Tolerance

Suppose a resistor has green, red, orange and gold rings

as shown in the figure below The resistance of the resistor is (52 × 103 W) ± 5%

combination of resistors

resistors in series

The various resistors are said to be connected in series if

they are connected as shown in the figure

The current through each resistor is the same

The equivalent resistance of the combination of resistors

Series combinations of resistors are used in resistance

box and decorative bulbs

resistors in parallel

The various resistors are said to be connected in parallel

if they are connected as shown in figure below

The potential difference is same across each resistor

• n wires each of resistance R are connected in

series, the equivalent resistance is nR.

If

• n wires each of resistance R are connected in

parallel, the equivalent resistance is R/n.

(JEE Main 2015)

10 Suppose the drift velocity v d in a material varied

with the applied electric field E as v d∝ E Then

V-I graph for a wire made of such a material is best

given by (a)

I

V

(b)

I V

(JEE Main 2015)

Electric cell

An electric cell is a device which maintains a continuous flow of charge in a circuit by a chemical reaction

Electromotive force (emf) of a cell

It is defined as the potential difference between the

two terminals of a cell in an open circuit i.e., when no

current flows through the cell It is denoted by symbol e.The SI unit of emf is J C–1 or volt and its dimensional formula is [ML2T–3A–1]

The emf of a cell depends upon the nature of electrodes, nature and the concentration of electrolyte used in the cell and its temperature

terminal potential difference of a cell

It is defined as the potential difference between two

terminals of a cell in a closed circuit i.e when current

is flowing through the cell It is generally denoted by

symbol V and is measured in volt.

internal resistance of a cell

It is defined as the resistance offered by the electrolyte and electrodes of a cell when the current flows through it Internal resistance of a cell depends upon the following factors:

Distance between the electrodes

relationship between e, V and r

When a cell of emf e and internal resistance r is

connected to an external resistance R as shown in the

figure



I I

R r

Current in the circuit, I R r= +e

Terminal potential difference, V IR= =(R re+R )

Internal resistance of a cell,

r=eV−V =R Ve −1R

During discharging of a cell, terminal potential

difference = emf of a cell – voltage drop across the

internal resistance of a cell i.e terminal potential

difference across it is lesser than emf of the cell The

direction of current inside the cell is from negative

terminal to positive terminal

During charging of a cell, terminal potential difference

= emf of a cell + voltage drop across internal resistance of

a cell i.e terminal potential difference becomes greater

than the emf of the cell The direction of current inside

the cell is from positive terminal to negative terminal

When the cell is short circuited, i.e., R = 0, the

maximum current is drawn from a cell whose value is

given byImax= er

Grouping of cells

Cells can be grouped in the following three ways:

Series grouping

If n identical cells each of emf e and internal resistance

r are connected to the external resistor of resistance R

as shown in the figure, they are said to be connected in

series grouping

I I

n identical cells, each of emf e, it will reduce the

total emf by 2e i.e effective emf = ne – 2e.

The total internal resistance of cells =

no effect on the total internal resistance of the cells

In series grouping of cells their emf’s are additive

R

m

emf e and internal

resistance r are connected

to the external resistor of

resistance R as shown in

figure, they are said to

be connected in parallel grouping

Equivalent emf of the cells is eeq = e Equivalent internal resistance of the cells is r eq= m r

Current in the circuit, I

R m r

=+  

n cells in series in one row and m such rows of cells in

parallel Suppose all the cells are identical Let each cell

be of emf e and internal resistance r.