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Page | 1 CBSE Physics Class 10th https://www.marksadda.com/
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
Page | 2 CBSE Physics Class 10th 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]
Page | 3 CBSE Physics Class 10th https://www.marksadda.com/
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.
Page | 4 CBSE Physics Class 10th https://www.marksadda.com/
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
Page | 5 CBSE Physics Class 10th https://www.marksadda.com/
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
Page | 6 CBSE Physics Class 10th https://www.marksadda.com/
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
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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?
Page | 9 CBSE Physics Class 10th https://www.marksadda.com/
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.,
Page | 10 CBSE Physics Class 10th https://www.marksadda.com/
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
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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
Page | 12 CBSE Physics Class 10th https://www.marksadda.com/
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
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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
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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
Page | 15 CBSE Physics Class 10th https://www.marksadda.com/
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
Page | 16 CBSE Physics Class 10th https://www.marksadda.com/
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
Page | 17 CBSE Physics Class 10th https://www.marksadda.com/
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
Page | 18 CBSE Physics Class 10th https://www.marksadda.com/
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.
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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.
Page | 21 CBSE Physics Class 10th https://www.marksadda.com/
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.
Page | 22 CBSE Physics Class 10th https://www.marksadda.com/
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
Page | 24 CBSE Physics Class 10th https://www.marksadda.com/
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.
Page | 26 CBSE Physics Class 10th https://www.marksadda.com/
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
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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
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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.
Page | 29 CBSE Physics Class 10th https://www.marksadda.com/
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.
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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.
Page | 31 CBSE Physics Class 10th https://www.marksadda.com/
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
Page | 32 CBSE Physics Class 10th https://www.marksadda.com/
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
Page | 33 CBSE Physics Class 10th https://www.marksadda.com/
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
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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
Page | 35 CBSE Physics Class 10th https://www.marksadda.com/
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