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Physics

Complete Course at a Glance for Class X

[Prepared strictly according to the latest syllabus issued by the Central Board of

Secondary Education, New Delhi for 2020 Examination of Class X]

ISBN 978-81-943778-2-5

By

Marks Adda Team of Experts

Ultra Creations Publications

e-Crash Course

https://www.marksadda.com/


Preface to the First Edition

This book is an effort to provide basic concepts and explanations in a concise way.

The topics have been explained through the examples.

Some main features of this book are:

1. The language used in this particular book is simple, lucid and easily

understandable by the students.

2. Essential graphs and diagrams have been correctly drawn and labeled well.

3. The concepts and explanations are provided in a concise manner.

Improvement is a consistent process to make the things better. Ultra Creations will

add more values to this particular work regarding betterment and upgradation of

content and quality in further editions. Suggestions and feedbacks are always

welcomed from teachers and students on [email protected]



Copyright © 2019 Marks Adda

All rights reserved. No part of this publication may be reproduced, distributed or transmitted in

any for or by any means, including photocopying, recording, uploading, scanning, sharing on

social media or other electronic or mechanical methods, without the written permission of the

publisher. For permission request write to [email protected]

Disclaimer:

The information provided in this book is to help the students preparing for the examination. All

efforts have been done to terminate errors in the content provided in this book. Neither the

publisher nor the author shall be responsible for any errors, omissions or damages arising out of

this information. Neither the publisher nor the author do not take any guarantee for any success

in the examination.

Jurisdiction:

The jurisdiction area for any legal dispute will be Nainital, Uttarakhand, India only.


mailto:[email protected]


CONTENTS

CHAPTER 1 – LIGHT-REFLECTION AND REFRACTION …………………5

CHAPTER 2 – HUMAN EYE AND THE COLOURFUL WORLD…….…48

CHAPTER 3 – ELECTRICITY………………………………………………….…….61

CHAPTER 4 – MAGNETIC EFFECT OF ELECTRIC CURRENTS………..89

CHAPTER 5 – SOURCE OF ENERGY …………………………..…………….108



ELECTRICI

Electronic Current & Electric Circuit

Electric Current

An electric current is define as the amount of charge flowing through any cross

section of a conductor. In unit time.

qI

t

Where Q is coulomb and ‘t’ is second. If t = 1 sec. I = Q

“Flow of charge in a conductor is known as electric current and the electrons which are

responsible for the conductivity is known as conduction electron.

Unit of Electric Current

SI unit of electric current is Ampere.

1 Ampere

QI

t

1coulombIAmp

1sec

&

1 Amp = 1C/sec = C sec1.

1 ampere is define as if 1 coulomb charge flows through any cross section of the

conductor in 1 sec.

1 mA = 103 AMP

3 ELECTRICITY



1 A = 106 AMP

Currents in terms of number of electrons.

We know that Q

It

19

Q ne

neI

t

i.e.,n no.of electrons.

e 1.6 10 C

[Wheree ismagnitude]

Electric current is a scalar quantity

Ammeter

Ammeter is a device which is used to measure the electric current. Ammeter should be

connected in series i.e., positive terminal is connected with positive & negative terminal

is connected with negative terminal of the body.

Electric Circuit

An electric circuit is a closed conducting patch containing a source of energy i.e., cell or a

battery and a device or a element or load which utilising the electric energy.

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These are of two types:

OPEN CIRCUIT

CLOSED CIRCUIT

Open Circuit

An electric circuit through which no electric current flows is known as open electric

circuit.

Closed Circuit

An electric circuit through which electric current flows continuously is knows as closed

electric circuit.

Numerical

Calculate the number of electrons constituting 0.1 coulomb of charge.

Given.

19

19

19

17

Q 0 1C

N ?

e 1.6 10 C

Q ne

0.1 n 1.6 10

0.1 Cn

1.6 10 C

n 6.25 10

6.25 ×1018 electrons flow from one end to another end of the conductor in 5sec. Find the

current flowing through the conductor?

n = 6.25 ×1018

t = 5 sec

I = ?, e = 1.6 × 1019 C






18 19

1

Q neI

t t

6.25 10 1.6 10I

5

2.000 10

I 0.2Amp

In a particular television tube a beam of electrons is emitted. If the beam of current is 80

A. How many electrons strike the screen of television every second? Also find the total

charge striking the screen in two minutes?

6

2

6 19

I 80 A 80 10 Amp

n ?, t 1sec

Weknow that

Q API

t t

80 10 n 1.6 10 C

1sec

6

19

6

13

6

6

80 10n

1.6 10

50 10 19

n 50 10

Case II Whent 2minutes

2 60

120sec.

FindQ ?

QI

t

80 10 Q/120

9600 10 Q

Q 9600 C

Calculate the amount of charge that flows in 2 hours through an element of an electric bulb

drawing a current of 0.25 amp?



Here,

I = 0.25 A

T = 2 hours × 60 × 60

Q = ?

QI

t

So,

Q 0.25 2 60 60

25 2 3600Q

100

Q 1800C.

A current of 0.5 amp passes through a conductor in 2 sec. How many electrons flow

through this conductor from one end to the another end during this internal of time?

Charge on each electron is 1.6 × 1019C.

19

18

19

18

I 0.5 Amp

t 25

E 1.6 10 C

neI ,Weget

t

It 0.5 2n 6.25 10

e 1.6 10

6.25 10 electron

Electric Potential & Potential Difference

Electric potential is define as the amount of work done in caring a unit positive charge

from infinite to a point against any electric field i.e.,



WV

q

If q IC

V W

Where v is electric potential, W is work done & q is charge.

[SI unit of potential is Valt]

One Volt

1. We know that

WorkdoneWV

Charg eq

1Jolue1Volt

1columb

1 volt = 1J/C or JC1

Thus, electric potential is said to be one volt if one Joule work is done in moving one

coulomb charge .

[Electric potential is a scalar quantity]

Electric Potential Difference

If VA&VB are the two electric potential of the conductor or wire then, The difference

between two electric potential of the conductor is known as potential difference i.e.,

WV

q

WVA VB

q

[SI unit of potential difference is volt V]

It is a scalar quantity

Voltmeter

A B VA VB






It is a device which is used to measure electric potential difference & a volt meter is

always connected in parallel across a conductor.

Resistance of an ideal volt meter is infinite [high resistance device]

Ohm’s Law

This law states that “The electric current flowing in a conductor is directly proportional

to the potential difference across the end of the conductor at constant temperature &

other physical conditions.

So, We know that

I V i.e., V I

So,

V = RI

i.e.,

V = IR

Resistance

Resistance of a conductor is the property of a conductor to oppose the flow of charge

through it.

Since,

V = IR



Therefore, VR

I

SI unit of resistance is ohm

1 resistance

Therefore,

V = I R

& VR

I

Now, 1 = 1Volt

1Amp

So, v = 1 volt/ Amp = Vot Amp1

Resistance of a conductor is said to be 1 if a potential difference of 1. Volt across the

end of a conductor makes a current of 1 Amp to flow through it”.

Resister

Acomponent in an electric circuit which offers resistance (opposition) to a flow of

electrons constituting electric current is known as resistor. For example: Metal wire &

conductor.

Variable Resistance

Since, sometime current has to be increased or decreased. A component used in an

electric circuit to change the current without changing the potential difference across

the circuit is called variable resistance.

Rheostat

It is a device used in an electric circuit to change the resistance hence current in the

circuit. It act as a variable resistance.

Good Conductor

A conductor is that a which offers low resistance to the flow of electrons in an electric

circuit is known as good conductor example : Silver.

Poor / Bad Conductor






A Conductor or a material which offer high resistance to the flow of electrons in an

electric circuit is known as poor or bad conductor. Example : Iron as compared to silver.

Insulator

A conductor or a material with very high resistance to flow of electrons in an electric

circuit is known as Insulator. Example : Rubber, Plastic etc.

“Electric current does not flow through a conductor.”

Ohmic & Non Ohmic Material

Those material which obeys ohm’s law are called Ohmic material & those material which

doesn’t obeys ohm’s law are called non Ohmic material.

Cause of Resistance

Since, every metal (conductor) contains large numbers of electrons, when these

electrons during in a movement collide with atoms then these electrons during the

collision loose energy they are the cause of resistance.

[More the collocation more will be the resistance]

Factors effecting on Resistance or resistivity or specific resistance

Resistance of a conductor is directly proportional to the length

R l _____________(1)

Resistance of a conductor is inversely proportional to the area.

1R _____(2)

A






From 1 & 2 relations.

lR

A

lR

A

Where is proportionalityconstant known as resistivity or specific resistance of a

conductor.

Resistivity [Specific resistance]

l

lR

A

RA

If A = 1 unit area

L = 1 unit length

So, it conclude that

= R

Resistance of a conductor is said to be resistance if length & the cross sectional area of

the conductor must be 1 unit.

[SP unit of resistivity is m]

In C.G.S system ( Cm)

2

R.A

l

m

m

m



Question

We have a copper wire of resistance R. This wire is pulled so that its length is doubled.

Find the new resistance of the wire in terms of its original resistance.

Let be the resistance of original wire

l

l

l l

l l

l

l

l ll

ll

l

1

1

1

1

1

2 1

2 1

2 2 1 1

2 1

212 1

2 1

2 1 1 2 2

2 1 21

1

2 1

2 1

Wire R

&Length

let Area A

RA1 __________(1)

After pulledso that its length isdoubled

2

R A R A

AR R

A

A A2

R R A

A

R 4R

If the resistance of a aluminium wire is 4 . This wire is pulled so that its radius is half

with respect to original wire. Calculate new resistance of wire.

Resistance of original wire R1 = 4



l

l l

l

l

l

l

2

2 1

1

1 2

2 21 1

1 2

21 12

2

211

2 1

length1r r

Area A2

From

R AR A

Now,

R AR

1 A

AR

A



l l

l

l

1 1 2 2

1 11 1 1

2 2

2 2

2

11

2

22

12 1 2

2

22

2

1 12 2

22

4

1

2

4

1

1

4

2

A AA A

R AA A

A

AR

A

rR R

r

r rR 4 4

rr

r4r

r41r

2

4[2] 4 16

R 64

A cylinder of a material is 10cm long and has a cross section of 2cm2. If of 2cm2. If its

resistance along the length be 20 ohms, what will be its resistivity?

L = 10cm, A = 2cm2

R = 20 Ohms

l

RAResistivity

Now,

20 2

10

4ohms

= 4 ohms cm



0.04 ohms m

Resistors Connected In series

Two or more conductors [resistors] are said to be connected in series if they are

connected one after the other such that same current flows through each resistor when

some polential difference is applied across the combination.






Equivalent Resistance of the Series Combinations

In above circuit, R1, R2& R3. Are three resistors which is connected in series. If current I

is same flows throw each resistor then potential difference of each resistors.

V = V1 + V2 + V3______________(1)

By using ohms law.

V = IR

From eq. I

IRs = I1 R1 + I2R2 + I3 R3

IRs = I [R1 + R2 + R3] 1 2 3

ISome

I I I

Rs = R1 + R2 + R3

If n resistors are connected in series having equal resistance

Rs=R1 + R2 + R3 .......

=nR

To achieve the maximum resistance in a circuit. Resistors are connected in series.

Or

Resistance of a series combination is very high (maximum)

Parallel Combinations

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Equivalent Resistance in Parallel.

V (potential difference) across each resistors are same but I (current) is distributed

across each R.

1 2 3

2 31

P 1 2 3

1 2 3

P 1 2 3

P 1 2 3

I I I I ____________(1)

By using ohms law

VV IR I

R

V VVV

R R R R

V 1 1 1V V V V V

R R R R

1 1 1 1

R R R R

For parallel combination

P 1 2

1 1 1..........n

R R R

p

p

1 n,

R R

RR

n

Equivalent Resistance of a parallel combination is less than that individual resistance.

Advantages of Parallel Combination

If anyone of the electric appliances or devices connected in parallel doesn’t work [fuses],

then the working of other devices will not be affected.

When different electric devices are connected in parallel then potential difference across

each conductor are same and hence they work properly.



Disadvantages of Series Combinations

If anyone resistor is fused which is connected series then all the appliances doesn’t work

properly.

Questions:

Three conductor of resistance 1, 2& 3 are connected in

i series, ii in parallel

Calculate effective resistances.

For Series.

Rs = R1 + R2 + R3

Rs = 1, 2 + 3 = 6

For Parallel.

p 1 2 3

p

p

s

1 1 1 1

R R R R

1 1 1 6 3 2 11

1 2 3 6 6

1 11

R 6

6R

11

&

R 6

If a 12 volt battery is connected to the arrangement of resistance shown in fig. Calculate

i total effective resistance

ii total current flowing in the circuit.



R1= 10 , R2 = 20

R3 = 5, R4 = 25

R1& R2 are connected in series.

Rs = R1 + R2 = 10 + 20

Rs = 30

Similarly R3& R4 are converted in series.

R’s = R3 + R4 = 5 + 25

'

's

'

s 's

1

P s s

P

R 30

R & R areconnected inParallel

1 1 1

R R R

1 1

30 30

30R 15 R

2

(ii) Now

V 12

R 15

I ?

By can

V IR

12V I 15

12I 0.8amp15

I 0.8amp

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A Wire of resistance 20 is bent in the foam of a closed circle. What is the effective

resistance between two points at the ends of any diameter of the circle.

20 is divided into two equal parts for diameter.

R1 = 10

R2 = 10

Now,

P 1 2

P

1 1 1

R R R

1 1 1

R 10 10



P

P

2

1 2

R 10

1 10

R 2

Rp 5

A copper wire having resistance R is cut into four equal parts.

Find the resistance of each part? And calculate effective resistance of the combination if

these four parts are connected in parallel.

Original resistance = R

R is cut into four equal parts.

New resistance of each part =R

4

R

4four parts are connected in parallel

1 1 1 1 1

Rp (R /4) (R /4) (R /4) (R /4)

4 4 4 4

R R R R

1 16

Rp R

RRp

16

Show diagrammatically, how you would connect three resistors each of resistance 6

so that the combination has a resistance of 4 & 9 ?

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R1& R2 are connected in II.

P 1 2

P

P

1 1 1

R R R

1 1

6 6

1 2

R 6

6R 3

2

Then 6Ω & 3Ω would be in series

Reff = 3 + 6 Ω

= 9

(II) R1& R2 are in series

Rs = 6 + 6 = 12

Then 12 & 6 are in 11.



p

p

p

1 1 1

R 12 16

1 2

12

1 3

R 12

12R 4

3

How can three resistors of resistance is 2, 3& 6 be connected to get a resistance

of:-

i 4

ii 1

If we connect R1 in series with respect to R2& R3 in parallel






p 1 2

I.

1 1

R R R

1 1

3 6

2 1 3

6 6

6R 2

3

Now, 2Ω & 2Ω would be in series

Rs = 2 + 2 = 4

II. Now, If we connect R1, R2& R3 are connected in parallel

Then

p 1 2 3

1 1 1

R R R R

1 1 1

2 3 6

3 2 1 61

6 6

1






How many 440 resistors [in parallel] are required to carry 5 Amp. In a 220 volt line

[important]

p 1 2

p

p

R 440

1 1 1..........n

R R R

RR

n

440R _________(1)

n

V 220,I 5A,Rp ?

By ohm's law

V IRp

VRp

I

From(1)

440 220

n 5

440 5

220

n 10

Heating Effect of Electric Current.

Since a metallic conductor contain large amountof free electron these electrons move

randomly in a conductor. Due to randomly movement of electrons they collide with

atoms ions & with each other. As a result the Kinetic energy of free electrons is transfer

to the atoms or ions. Then these atoms or ions vibrate with large amplitude about the

mean positions. Hence kinetic energy is converted into heat energy & it increases in

increasing temperature.



Amount of Heat Producing in conductor : through which electricity flows.

wv

q

w vq

qI q It

t

W VIt

Now we know that V = IR

W = IR × IT

2

2

W I Rt

H I Rt

Joule’s Law of Hating

Joule’s law can be stated as the amount of heat produced in a conductor is depends on

electric current, Resistance and time according to this law heat energy. i.e.,

2

2

2

2

H i

H R

H t

H i Rt

H Ki Rt

if K 1

Then

H i Rt

Different forms of Joule’s law.






2

2

2

2

2

H i Rt

By ohm's law

V IR

VIR

vThen,H i t

i

H ivt _______(i)

Again,

VV IR I

R

VH Rt

R

Then,

v tH _________(ii)

R

Also,

q It

H i it R

Then,

H qiR ________(iii)

Now,

VR

I

vH qi

i

Then,

H qV ________(iv)

Practical Applications of Heating of current.

Electric Heater, Water Heater, etc., Work on the heating effect of current.

An electric appliance is connected to the electricity (main supply) then appliances

become red & hot but connecting wires remains cold (reason)

Because the element of electric heater is made of michrome. Michrome as a high

resistance & heat is directly. Proportional to resistance (by jouel’s law) so filament of

electric heater becomes red hot.

Reason of connecting wire remains cold.



Since, Metal wire is made up of copper or aluminum it has very law resistance. So, a very

small heat is produced in a connecting wire which is made up of copper or aluminum.

Electric Bulb Glows when electric current flows through the filament.

Since, Filament [tungsten] has very high melting points [about 3380°C] & very high

resistance so, when electric current flows through filament then large amount of heat

energy is produced & filament becomes white hot. [Due to presence of inactive gases]

Like nitrogen & carbon in a glass envelope. Hence the filament of the emits light & heat.

Electric fuse in the electric circuit melts when large current flows in the circuit.

Electric fuse is a circuit device connected in series with the electric circuit. Electric fuse

is a wire made fo a material whose melting point is very low like copper. Tin, lead, alloy.

When large current flows through the circuit & hence through a fuse wire then large.

Amount of heat is produced due to low melting point fuse wire melts, breaks the circuit

by which current stops flowing in the circuit. This saves the electric circuit from burning.

Electric Energy

The work done by a source of electricity to maintain a current in an electric circuit, is

known as electric energy.

WV

q

W Vq

I q /tE Vq

q it

Therefore,

W= E = VIT

Electric Power



The rate which electric energy is consumed or “The amount of energy is consume in a

circuit per unit time”

So

WP

t

Where, W is a electric energy

VItW

t

P VI

Thus, electric power is defined as the product of potential difference applied across the

circuit & current flowing through it.

Other forms of Electric Power

We know that

P = VI

2

2

2 2

2

i. By 's law V IR

P IR I I R

Vii.Now,I V IR

R

V VP .R

RR

q Vq qiii.P V. I

t t t

Units of Power

SI unit of power is Watt [W]

1 WATT

Since, W

Pt

Therefore,

1 Watt = 1Joule

1Sec

Or



1 Watt = 1 J/S

and,

P = VI

1 W = 1 volt × 1 AMP

1 W = 1 volt AmP

Kilo Watt Hour [commercial unit of electrical energy]

It is represented by KWh the department ofelectricitysales the electric energy to the

consumers in units called Kilo watt hour.

Example.

1 unit = 1KWh

A KWh is the amount of electric energy used by 1000 watt. Electric appliance when it

operator for one hour.

Kwh is also known as boarding of trade unit hot.

Relation Between KWh & Joule.

1 KWh = 1000 Wh

= 1000 × 60 × 60 sec [ 1 h = 60 × 60 sec.

= 36 × 105 J

1 KWh = 3.6 × 106 J

Practical Unit of Power

1 h.p = 746 Watt.

h.p = horse power

How to Calculate Electricity Bill

Number of units of electricity consumed in household = No of KWh = Watt hrs

1000

Total units of electricity consumed in a month = Number of units × Number of days in

a month.

Total cost of electricity = Total units × Cost per unit of electricity






Example:- Suppose electric appliances consumed 100 KWh of electric energy in a month &

the cost of 1 unit is 50 paisa. Then Calculate the electricity bill?

Total bill = 100 × 50 p

= 500 p

In rupees = 500

25100

If a consumer consumed 2000 W In 24 hrs. If this energy consumes continuously

for 25 day then calculate the electricity bill of these days. If cost of per unit Rs .2

Water hourNoof units

1000

2000 24

1000

48

TotalNumberof unit in25days

48 25

1200unit

Bill 1200 2

Rs.2400



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