Class 10 Science Solutions Nepal

Approved by the Government of Nepal, Curriculum Development Centre
(CDC), Sanothimi, Bhaktapur as an additional material for school
NEW
COURSE

NEW CREATIVE

SCIENCE

Rajani Maharjan
Yuwaraj Guragain

Surendra Karki
Sangharsh Raj Khanal

Book NEW CREATIVE

SCIENCE
Class - 10

Publisher JBD Publication Pvt. Ltd.

Bhotahity, Kathmandu, Nepal
Tel: 4252371

Edition 1st, 2065 B.S.
Authors 2nd, 2068 B.S.
3rd Revised, 2073 B.S.
Layout 4th, 2074 B.S. (Completely Revised)
Reprint 2075 B.S.

Rajani Maharjan
Yuwaraj Guragain
Surendra Karki
Sangharsh Raj Khanal

Deltrox IT Solutions Pvt. Ltd.

#9851004706

ISBN 978-9937-544-51-1
Printed at
Mahabir Offset Press, 025-521634

Preface

As this is the age of science and technology, science is a very important subject
to the students of the present days. Scientific inventions and discoveries have
brought great changes in the world. The study of science and technology for
the students of developing countries like Nepal has great importance to make
many achievements. For this, the students should have good knowledge of
science through systematic as well as properly designed textbooks, instructional
materials and good teaching methods. In schools, textbooks are the most
important educational materials. So, the books designed according to the need
and interests of the students are very useful and appropriate.

Keeping these facts in mind this series of textbooks for the students of
classes 1-10 has been prepared with fresh and fascinating approach to the
study of science at school level. During teaching learning process both teachers
and students have to face problems of what is to be taught and learnt, what
should be focused for the examination, how to learn and write to the point, how
to start and how to end.

This series for the first time itself is the best attempt and presentation to
solve these problems. After facing all these problems during teaching, we have
designed and prepared this series as a solution with these salient features.

Each book of this series includes– objectives of the unit in the very beginning,
important questions, clues and memory tips as important things, definitions
in separate colour, answer writing skills as model questions and their answer,
glossary of difficult words, lots of activities as creative activities, summary of the
unit as revision, etc.

Further this series has been designed according to the syllabus prescribed by
CDC of Nepal. Cognitive and practical student-friendly, colourful presentation and
use of lucid and easy language, both knowledge and exam-oriented matters in
sequential order are the main attractions of this series.

We are grateful to Deltrox IT Solutions for computer work and JBD Publication
for publishing this series.

We are highly indebted to those who provide their positive support, suggestions
and comments for the improvement of the series in further editions.

Authors

CONTENTS 1
21
Unit 1. Force 43
Unit 2. Pressure 56
Unit 3. Energy 70
Unit 4. Heat 89
Unit 5. Light 110
Unit 6. Electricity and Magnetism 127
Unit 7. Classification of Elements 140
Unit 8. Chemical Reaction 154
Unit 9. Acid, Base and Salt 164
Unit 10. Some Gases 176
Unit 11. Metals 190
Unit 12. Hydrocarbons and their Compounds 215
Unit 13. Materials Used in Daily Life 224
Unit 14. Invertebrates 244
Unit 15. Human Nervous and Glandular System 262
Unit 16. Blood Circulatory System in Human Body 270
Unit 17. Chromosomes and Sex Determination 290
Unit 18. Reproduction in Animals and Plants 303
Unit 19. Heredity 314
Unit 20. Environmental Pollution and Management 327
Unit 21. History of the Earth 343
Unit 22. Climate Change and Atmosphere 362
Unit 23. The Earth in the Universe 363
Specification Grid
A sample set of model question

UNIT

1 Force

About the Scientist INTRODUCTION

Isaac Newton We are familiar with the fact that a force is needed to change
(1642-1727) shape, size and state of a body and to change speed and
Isaac Newton was born in Wools direction of a moving body. Thus,
Thorpe near Grantham, England.
He is generally regarded as the most Force is an external factor which changes or tends to change the
original and influential theorist state (rest or motion) of the body. More simply, we can say that,
in the history of science. He was pulling or pushing factor is called force.
born in a poor farming family and
was sent to study at Cambridge The force is measured by a spring balance in Newton or
University in 1661. In 1665, a dyne. Newton is the S.I. unit of force, and 1N = 105 dynes
plague broke out in Cambridge and (dyne is the C.G.S. unit of force).
so Newton took a year off. It was
during this year that the incident There are different types of forces such as mechanical,
of the apple falling on him is said muscular, centripetal, frictional, centrifugal, magnetic,
to have occurred. This incident gravitational etc.
prompted Newton to explore the
possibility of connecting gravity Gravitation (Gravitational force)
with the force that kept the moon
in its orbit. This led him to the Before 16th century, astrologists believed that at the center
universal law of gravitation. He of the universe there is the earth and other heavenly bodies
also formulated the well-known like the sun, moon and other planets revolve around the
laws of motion. He worked on earth. It is called geocentric theory. But, this theory had
the theories of light and colour. been modified and heliocentric theory was put forward.
He also designed an astronomical According to heliocentric theory, the sun is at the center of
telescope to carry out astronomical the universe and all other planets and satellites revolve the
observations. He was also a great sun in their own elliptical orbit. Do you know why are the
mathematician and invented a planets and satellites revolving the sun? Newton found its
new branch of mathematics, called answer in 1687. He said that there is a force of attraction
Calculus. Newton transformed the between these heavenly bodies and the sun. So, they are
structure of physical science with revolving around the sun in their own orbit. This force is
his three laws of motion and the called gravitational force.
universal law of gravitation.
Though the gravitational theory We always observe that an object dropped from a height
could not be verified at that time, falls towards the earth. Planets go around the sun, and the
there was hardly any doubt about moon goes around the earth. In all these cases, there must be
its correctness because Newton some force acting on the objects. Isaac Newton could grasp
based his theory on sound scientific that the same force is responsible for all these. This force is
reasoning and backed it with called the gravitational force.
mathematics.

New Creative Science, Class 10 | 1

The motion of the moon around the earth is due to the centripetal force. This centripetal
force is provided by the force of attraction of the earth.

It is seen that a falling apple is attracted towards the earth; the apple also attracts the
earth. As the mass of an apple is negligibly small compared to that of earth, we do not
see the earth moving towards the apple.

From the above facts, Newton concluded that not only does the earth attract an apple
and the moon but also all objects in the universe attract each other with a force called
the gravitational force.

Thus, t he force of attraction between any two objects of the universe due to their masses is
called gravitation.

MEMORY TIPS

● The force of gravitation is always the force of attraction. It is never repulsive.

● The gravitational force between the earth and the apple is very weak due to less mass
of the apple. Due to less mass of the apple it cannot produce noticeable acceleration
in the earth but the earth having more mass produces noticeable acceleration in the
apple. Thus, we see apple falling towards the earth. But we cannot see earth moving
towards the apple.

● The effect of gravitational force is more is liquid than solid. Because the molecules
of the liquid are loosely packed with less intermolecular force of attraction. So, tides
occur in the sea and ocean due to force of attraction of the sun and the earth.

QUESTIONS

# Define gravitational force.
# When the apple falls, does the apple also attract the earth?
# If so, we do not see the earth moving towards an apple, why?

NOTE Teachers are highly requested to create similar types of many questions in each
topic and sub-topic.

NEWTON’S UNIVERSAL LAW OF GRAVITATION

According to Newton's universal law of gravitation every object in the universe attracts
every other object with a force which is called gravitational force. This gravitational force is (i)
directly proportional to the product of their masses and (ii) inversely proportional to the square
of the distance between their centres.

Suppose two objects A and B of masses m1 and m2 respectively are separated by
the distance ‘d’ between their centres. If F is the gravitational force between them then
according to the Newton’s law of gravitation:

2 | Force

F ∝ m1m2 ........................... (1) A B
d
and F ∝ 1 ........................... (2)
d2
Combining eqn (1) and (2), we get,

F ∝ m1m2 ........................... (3) m1 m2
d2 ........................... (4)
m1m2
∴ F = G d2

Where, G is a proportionality constant known as universal gravitational constant.

Eqn (4) is the mathematical expression of Newton’s law of gravitation. This formula
shows that, when m1, m2 are large, F is large and when ‘d’ is large F is less.

Newton’s law of gravitation is called the universal law because it is applicable to all the bodies
either terrestrial or celestial having any shape, size, mass or at any distance apart with any
medium between them, at any time (past, present or future).

Newton’s law can be illustrated by the following examples:

1. When the bodies A and B have the A B

masses m1 and m2 respectively. F
d
Here, the mass of A and B are m1 and m2 and
distance between their centres be r then, from m1 m2
Newton’s law:

F = G m1m2 .......... (A)
d2

2. When the mass of ‘A’ is increased by 2 times keeping mass of ‘B’ and distance

between them constant.

Here, mass of A (m1) = 2m
Mass of B (m2) = m2
Distance between them = d

Then,

F1 = G (2m1).m2 [where, F1 is the new gravitational force]
d2

F1 = 2G m1.m2
d2

∴ F1 = 2F (from equation A)

Hence, the gravitational force increases by two times when the mass of only one body
(A) is doubled.

New Creative Science, Class 10 | 3

3. When the mass of both bodies is doubled keeping distance constant.

Here, mass of A (m1) = 2m1 F 2m2
Mass of B (m2) = 2m2 2m1 d
Distance between them = d

Then,

F1 = G (2m1).(2m2) [where, F1 is the new gravitational force]
d2

F1 = 4G m1.m2
d2

∴ F1 = 4F (from equation A)

Hence, when the mass of both the objects is doubled keeping the distance constant
then the gravitational force will be increased by four times than the initial value.

This proves that the gravitational force is directly proportional to the product of masses when
the distance between them is kept constant.

4. When the distance between their centres is halved keeping the masses of the
objects constant.

Here, mass of A (m1) = m1

Mass of B (m2) = m2

Distance between them = 1 d F

Gm1.m2 2 m1 1 d m2
1d 2 2
( )Then, F1 =

2

F1 = 4G m1.m2
d2

∴ F1 = 4F (from equation A)

Hence, it proves that, the gravitational force is inversely proportional to the distance between

their centres when the masses of both objects are kept constant.

UNIVERSAL GRAVITATIONAL CONSTANT (G)

From Newton’s universal law of gravitation,

We have,

F = G m1.m2 F
d2 1m
1 kg 1 kg

Fr2
∴ G = m1 m2
4 | Force

When, m1 = m2 = 1kg, d = 1 m then, G = F
Thus, universal gravitational constant (G) is equal to the force of gravitation between any two
unit masses (i.e. of 1 kg each) separated by a unit distance (i.e. 1m).

Its value is 6.67 × 10–11 Nm2/kg2.

To find the S.I. unit of ‘G’

We have, from above equation,

Fd2 Nm2 [S.I. unit of F is Newton, d is ‘m’ and that of m1 & m2 is kg]
G = m1 m2 = kg.kg

G = Nm2/kg2

G is called the universal constant because its value does not depend upon the time, masses and
distances of the bodies as well as shape, size and nature of bodies and the medium separating them.

MEMORY TIPS

● Escape velocity is the velocity required for a body to go to the infinite distance away
from the surface of the earth, so that the body does not come back to the earth's
surface. Its value is 11.2 km/s.

● Orbital velocity is the velocity required to put a satellite in its orbit. Its value is 7.2
kms–1.

QUESTIONS

# Derive F = m1.m2 where the symbols have their usual meanings.
d2
# State Newton’s universal law of gravitation. Why is this law called the universal law?

# Define G. Why is it called the universal constant?

Properties of G

a. The value of G is independent to the nature of the masses of the body.
b. It is not affected by the time, temperature and pressure.
c. It is independent to the directions of the gravitational forces between the

masses.
d. The value of G is independent to the chemical composition of the masses

and the medium between them.

Consequences of the gravitational force

Newton has explained successfully several phenomena on the basis of gravitational
forces, e.g.

a. Due to the gravitational force of the earth, all terrestrial objects are bounded
on it and the objects thrown upwards return to the earth’s surface.

New Creative Science, Class 10 | 5

b. The earth holds the atmosphere around it due to its gravitational force.
c. The gravitational force of the earth is responsible for rainfall, snowfall and

flowing of water in rivers.
d. The moon revolves round the earth due to the earth’s gravitational force

and the planets revolve around the sun due to the gravitational pull of the
sun on the planets.
e. The tides formed in the seas and oceans are due to the mutual gravitational
force of attraction between the earth, the sun and the moon.

MEMORY TIPS

● Tide occurs on every new moon day and full moon day when the sun, the moon and
the earth lie on same plane, because of which mutual attraction of the sun and the
moon easily effects the flexible water of seas and oceans.

SOLVEDNumerical

1. Calculate the gravitational force between two bodies of mass 1 kg each
separated by 1m distance apart.

Solution:

Given, mass of one body (m1) = 1 kg
Mass of another body (m2) = 1kg
Distance between their centres (d) = 1m

Gravitational force (F) = ?

Universal gravitational constant (G) = 6.67 × 10–11 Nm2/kg2

We have,

From Newton’s universal law of gravitation,

F = G m1.m2
d2

6.67 × 10–11 × 1 × 1
F = (1)2

∴ F = 6.67 × 10–11 N

Hence, the gravitational force is 6.67 × 10–11N

2. Calculate the gravitational force between the earth and an object of mass 1 kg
held on the surface of the earth. Take G = 6.67 × 10–11 Nm2/kg2, radius of the
earth = 6400 km. (mass of earth = 6 × 1024 kg)

Solution:

Given, Mass of the earth (m1) = 6 × 1024 kg

6 | Force

Mass of the object (m2) = 1 kg
Distance between them (d) = 6400 km = 6.4 × 106m

Gravitational force (F) = ?

We have,

From Newton’s universal law of gravitation,

F = G m1.m2
d2

6.67 × 1024 × 6 × 1024 × 1 = 9.8 N
= (6.4 × 106)2

Therefore, the gravitational force between the earth and the object is 9.8 N.

3. Calculate the gravitational force between the earth and the moon. Given, mass
of the earth = 6 × 1024 kg, mass of the moon = 7.4 × 1022 kg, distance between
the earth and the moon = 3.84 × 105 km.

Solution:

Given, mass of the earth (m1) = 6 × 1024 kg
Mass of the moon (m2) = 7.4 × 1022 kg
Distance between the earth and the moon (d) = 3.84 × 105 km

= 3.84 × 108 m

Force of gravitation (F) = ?

We have,

From Newton’s universal law of gravitation,

F = G m1.m2
d2

6.67 × 1024 × 6 × 1024 × 7.4 × 1022
= (3.84 × 108)2

= 2.01 × 1020 N

Hence, the gravitational force between the earth and the moon is 2.01×1020 N.

MEMORY TIPS

Kepler started the laws of planetary motion which governs the motion of planets around
the sun. But it was Newton who established the real cause of the planetary motion that is
due to gravitational force between planets and the sun.

New Creative Science, Class 10 | 7

QUESTIONS

# What are the consequences of the gravitational force?
# What will happen to the gravitational force between two bodies if their masses are

doubled and distance between them is tripled?

ACTIVITY

Throw a stone to a height and observe.
 The stone falls after reaching a certain height? Why?
 As the stone falls on the earth's surface does the earth also move upwards to meet the

stone. Explain the phenomenon.

GRAVITY OR WEIGHT

As we have discussed, the phenomenon of attraction between any two bodies of the
universe is the gravitation. Gravity is one particular case of gravitation, when one of
the two attracting bodies is the earth or any massive (very large and heavy) object like
stars, planets, satellites, etc.

Thus, the force of attraction exerted by the earth on a body is called force of gravity or weight
of the body.

Each heavenly body has its own force of gravity with the help of which it attracts the
body near on its surface. For example, the gravitational force between the earth and a
body near to its surface is called earth’s gravity and the attracting force between the
moon and a body near to its surface is called the moon’s gravity.

Let, a body of mass ‘m’ is placed on the earth’s surface having the mass (M). The
distance between their centres can be considered to be equal to the radius of the earth.
If ‘F’ be the force of attraction between the object and the earth,

From Newton’s law of gravitation, m
R
F = G m1.m2 O
d2
M
from the given condition,

Mm
or, F = G R2

Mm (where F = W = weight of the body)
or, W = G R2

The S.I. unit of gravity is also Newton (N).

This relation shows that greater the mass of the heavenly body greater will be the force
of gravity given by it and greater will be the weight of the body in that heavenly body.

8 | Force

Similarly, more will be the radius of that heavenly body less will be its force of gravity
and less will be the weight of the object on its surface.

It is found that the moon’s gravity is just 1/6th of the gravity of our earth. Thus, a person
who can lift 10 kg on the earth can lift 60 kg on the moon’s surface and the person who
can jump 1m high on the earth’s surface can jump 6m high on the moon’s surface.

QUESTIONS

# The mass of Jupiter is 319 times greater than that of the earth but its gravity is just 2.5
1
times that of the earth. Why? (Hint: F ∝ m also, F ∝ R2 and R of Jupiter is 11 times that of the earth.)

# Why is it difficult to lift a bigger stone than a smaller one?

MASS

Mass is the total amount of matter contained in a body. It is measured by a beam balance
in kg. It is constant all over the universe. It is a scalar quantity. The amount of mass
never gets changed whenever we go.

WEIGHT

The force of attraction exerted by the earth on a body toward its center is called weight of the
body. It is measured by a spring balance. Its S.I. unit is Newton (N). Its value changes
from place to place according to the acceleration due to gravity. It is a vector quantity.

According to Newton’s second law of motion, weight is the product of mass and
acceleration due to gravity at that place.

Weight = mass × acceleration due to gravity

W=m×g

From this equation it is clear that weight of a body at the polar region of the earth is
more and weight of the same body at the equator is less. This is because acceleration
due to gravity (g) at the polar region of the earth is more and at the equatorial region
is less. Similarly, weight of the object at terai region is more than at hilly region.
Differences between mass and weight.

Mass Weight

1. It is the total amount of matter contained 1. It is the force of gravity acting on a body

in a body. of certain mass.

2. It is measured by a beam balance. 2. It is measured by a spring balance.

3. Its S.I. unit is kilogram (kg). 3. Its S.I. unit is Newton (N).

4. Its value is constant all over the universe. 4. Its value changes from place to place
according to the acceleration due to gravity.

5. It is a scalar quantity. 5. It is a vector quantity.

New Creative Science, Class 10 | 9

SOLVEDNumerical

4. The mass of the moon is 7.2 × 1022 kg and its radius is 1.7 × 106m. Find the
weight of 1kg mass on the moon.

Solution:

Given, Mass of the moon (M) = 7.2 × 1022 kg
Radius of the moon (R) = 1.7 × 106 m
Mass of a body on the surface of the moon (m) = 1 kg
Weight of the body on the moon (W) = ?

We know that,
Mm

W = G R2

6.67 × 10 –11 × 7.2 × 1022 × 1
= (1.7 × 106)2
= 1.67 N

∴ Weight of the body having mass 1 kg on the moon is 1.67 N.

Effects of gravity

a. We can stand, walk, play and perform other activities freely on the earth due to its
gravity.

b. The rivers flow up to down due to the gravity.
c. The earth is covered by the atmosphere.
d. Construction of roads, bridges, building is also possible because of the gravity of

the earth.
e. All types of small and big objects thrown upwards return to the earth’s surface.
f. Acceleration is produced on a body when it falls towards the earth's surface freely

without any resistance.

MEMORY TIPS
The weight of an object on the earth is 6 times more than the weight of the same object
on the moon. However its mass remains the same on the both.

ACCELERATION DUE TO GRAVITY (g)

Everybody on the earth is attracted by the earth towards its centre. So, when the body
is released from a height, it falls towards the earth with an acceleration that is called
the acceleration due to gravity (g).

Thus, acceleration due to gravity is defined as the acceleration produced in a freely falling body
due to the earth’s gravity.

Its S.I. unit is m/s2 and its average value on the earth’s surface is 9.8 m/s2.

10 | Force

Relation between radius and acceleration due to gravity of the earth m
R
Let M be the mass of the earth with its radius R. If a body of mass ‘m’
is on the surface of the earth. Then,

According to the Newton’s law of gravitation, OM

the force of R attraction between them is given by,

Mm .......... (a)
F = G R2

Here, F represents the force by which the body is attracted towards the earth. So, F

represents the weight (W) of the object. Then, eqn (a) becomes

Mm .......... (b)
W = G R2

But we have, from the second law of motion,

W = mg .......... (c)

Hence, from (b) and (c) we get,

Mm
mg = G R2

GM
∴ g = R2
This is the expression for the acceleration due to gravity. This relation shows that, the
value of acceleration due to gravity is,
a. directly proportional to the mass of the earth i.e. g ∝ M

1
b. inversely proportional to the square of radius of the earth i.e. g ∝ R2 and
c. independent to the mass of the object lying on the earth’s surface.

MEMORY TIPS

As an object falls on the ground its velocity goes on increasing per second to produce
more or less uniform acceleration due to gravity.

QUESTIONS
# Show by equation that acceleration due to gravity is independent to the mass of the

object.

ACTIVITY

Take two stones of different masses having almost the same density. Drop these stones from a
height at the same time and observe. Do these stones reach the ground at the same time? Why?
What can you conclude from this experiment?

New Creative Science, Class 10 | 11

Falling bodies and acceleration due to gravity

Before Galileo it was thought that the bodies of higher mass fall on the ground earlier
than the bodies having less mass. Galileo Galilee in 1650 released two iron balls
of different mass at the same time from the top of leaning tower of Pisa. These balls
reached the ground at the same time proving that the acceleration due to gravity is
independent to the masses of the objects.

If a stone and a feather are allowed to fall from the height, the stone falls to the ground
much quicker than the feather. It is due to the large air resistance on the large surface
area of the feather. This leads to the conclusion that the acceleration due to gravity is the
same for all the masses only if the air resistance is neglected.

This was Galileo’s view and it was further proved by Robert Boyle by doing an
experiment on Guinea and feather.

Guinea and feather experiment

In this process a guinea (a coin of England) and a feather

are kept into a long and wide glass tube. Initially both the F eather V acu um
guinea and the feather must be at the bottom of the tube

and the tube should be inverted. It is observed that the

coin falls faster than the feather as shown in the figure Guinea

A. But when the air from the tube is evacuated by using

a vacuum pump and the same process is repeated then

it is observed that both the coin and the feather reach AB
the bottom at the same time. This leads to the conclusion Guinea and feather experiment
that:

The acceleration due to gravity is the same for all kinds of objects. It does not depend upon mass
of the object if air resistance is neglected.

MEMORY TIPS

The value of g does not depend upon the mass of the body. But the gravitational pull of
the earth F = mg, depends on the mass of the body. Heavier the body, greater is the
gravitational pull of the earth on it and vice-versa.

QUESTIONS

# What happens to the acceleration due to gravity when a person jumps with a parachute?
# A parachutist does not get hurt when he jumps from a certain height. Why?

12 | Force

SOLVEDNumerical

5. If the mass of the earth is 6 × 1024 kg and its radius is 6380 km, find the
acceleration due to gravity of the earth.

Solution:

Given, Mass of the earth (M) = 6 × 1024 kg

Radius of the earth (R) = 6380 km = 6.38 × 106 m

Acceleration due to gravity (g) = ?

We have, from the relation = 9.78 m/s2
GM

g = R2
6.67 × 10 –11 × 6 × 1024

= (6.38 × 106)2

Hence, the value of acceleration due to gravity on the earth’s surface is 9.78 m/s2.

Variation in the value of acceleration due to gravity (g)

The relation between g and R is,
GM

g = R2

It shows that the value of g depends on the gravitational constant G, mass of the
earth M and radius of the earth R. It is not constant as earth is not a perfect sphere.
Therefore, the value of g changes from place to place on the surface of the earth. The
variation of ‘g’ according to the various factors is as assumed below.

1. Variation of ‘g’ due to the shape of the earth N pole
Rp
As we know, our earth is not a perfect sphere. It has like the Re > Rp
O Re
shape of an orange. It is flat at poles and bulges out at the equator

equatorial region. Thus, the equatorial radius is longer than S pole

the polar radius and we know that,

GM
g = R2

1 [G and M are constants]
or, g α R2

Thus, the value of g is more at the poles than at the equatorial region.

Since weight = mg, the weight of a person is more at poles than at the equatorial
region.

New Creative Science, Class 10 | 13

MEMORY TIPS

The value of ‘g’ at poles is 9.83 m/s2 and that in the equatorial region is 9.78 m/s2. Hence
the average value of ‘g’ on the earth’s surface is 9.8 m/s2.

QUESTIONS

# Why the weight of a body is more at poles than at the equatorial region?
# Where the value of g is more, at poles or at the equator? Why?

2. Variation of g with the height from the surface of the earth

As we move up from the surface of the earth, the distance between the centre of the
earth to that height (h) becomes (R + h). Hence, from Newton’s universal law,

GMm and F = mg
F = (R+h)2

Where, g’ is the acceleration due to gravity at the height ‘h’ from the earth’s

surface

GMm
∴ mg' = (R+h)2

GM
∴ g' = (R+h)2
Since, with the increase in h, (R + h)2 increases and which in turn decreases the value
of ’g’. So, the value of g at the top of Mt. Everest is less than that in the Terai region.

3. Variation of g with the depth from the earth’s surface

We may have a misconcept that, as the depth increases from the earth’s surface, the

radius of the earth decreases and this must increase the value d
of g. But, it is not true. The value of acceleration due to gravity

(g) at the depth ‘d’ from earth’s surface is given by. m
MR
( )g' = 1 – d .g
R

Where, g = value of acceleration due to gravity at the earth’s Earth
surface.

From this relation, it is clear that as depth (d) increases the value of g' also decreases.

MEMORY TIPS
The value of g at the centre of the earth is zero.

14 | Force

Acceleration due to gravity and weight of a body on planets/satellites

As we have already known that GM
GM

g = R2
So, the value of ‘g’ changes according to the mass and radius of the heavenly bodies.
Different heavenly bodies have different mass and radius. So, the value of ‘g’ in these
bodies changes accordingly.

Similarly, the weight of body = mass of body × g

Mass is a constant quantity but ‘g’ is a variable. The weight of a body is directly
proportional to the acceleration due to gravity of that planet.

SOLVEDNumerical

6. The mass of Jupiter is 1.9 × 1027 kg and its radius is 71 × 106 m. What is the
acceleration due to gravity on the Jupiter? Also calculate the weight of a
person having mass 60kg on the Jupiter.

Solution:

Given, Mass of the Jupiter (M) = 1.9 × 1027 kg
Radius of the Jupiter (R) = 71 × 106 m
Acceleration due to gravity of the Jupiter (g) = ?

We have, from the relation,
GM

g = R2
6.67 × 10 –11 × 1.9 × 1027

= (71 × 106)2

= 25 m/s2
Again, mass of the person (m) = 60 kg

Acceleration due to gravity (g) = 25 m/s2
From the formula,

W = mg = 60 × 25 = 1500 N
Hence, the acceleration due to gravity of the Jupiter is 25 m/s2 and the weight of
a person on it is 1500 N.

New Creative Science, Class 10 | 15

7. If a person can lift 100kg on the surface of the earth, what is the amount of the
mass he can lift on the surface of the moon?

Solution:

Given, weight of an object on the earth = m × g = 100 × 9.8 = 980 N
Now, on the moon,

Mass of the object that the person can lift (M ) = ?
Weight of the object that the person can lift (W) = 980 N
Acceleration due to gravity of the moon (g ) = 1.67 m/s2
From formula,

W = Mm . gm
or,980 = Mm × 1.67

980
Mm = 1.67 = 586.6 kg
Hence, the person can lift 586.6 kg mass on the moon.

GRAVITATIONAL FIELD AND GRAVITATIONAL FIELD INTENSITY

The area around a heavenly body up to which its gravitational force can be felt is called
gravitational field of that body.
Gravitational field intensity at any point is the amount of force experienced by a unit
mass kept at that point in the gravitational field of the heavenly body.

Gravitational field intensity is numerically equal to the acceleration due to gravity.
The SI unit of gravitational field intensity is N/kg.

FREE FALL

When a body falls freely under the effect of gravity only neglecting the air resistance, then the

body is said to be in free fall. 0 m/s
During the time of free fall;
1 sec

Acceleration produced in the falling body = acceleration due to gravity 9.8 m/s
19.6 m/s
a=g 1 sec
1 sec
A body falling in vacuum and falling on the moon are some examples
of free fall. Acceleration produced on the falling body is equal to the
acceleration due to gravity of that place. In the moon there is no air.

So, always there is free fall. Similarly, in the revolving satellite, the 29.4 m/s
astronauts fell free fall.

16 | Force

WEIGHTLESSNESS

Weightless is a condition in which the effective weight of a body becomes zero. It is due to the
zero reaction force of the earth or other heavenly bodies.

For example, when a person falls freely in a lift, he feels weightless.

As the person falls freely, his acceleration (a) will be equal to (g) i.e. a = g and the
reaction force given by the earth to the person also becomes zero. Due to this, the
person feels that he is not being attracted by any force and thus feels zero weight or
weightlessness.

Conditions for weightlessness

a. When a body falls freely.
b. When a body is in the satellite orbiting round the earth or a heavenly body.
c. When a body is at null point outside the gravitational field of a heavenly body.
d. When a body is taken to the centre of the earth.

MEMORY TIPS
Null point is the point in the space where the gravitational forces due to different bodies
cancel out and a body at the null point feels weightlessness.

ACTIVITY

Take the weight of a stone using a spring balance and note its weight. Now release the spring
balance with stone and note the weight of the stone during the fall.
The value of the weight at the time of fall will be zero. Can you explain it?

MODEL QUESTIONS ANSWER
1. Differentiate between g and G.

Gg

1. It is the force of attraction 1. It is the acceleration produced on a

between two bodies of unit masses freely falling body under the effect

kept at unit distance. of gravity.

2. Its value is a universal constant. 2. Its value changes from place to place.

3. Its value is 6.67 × 10 –11 Nm2/kg2 3. Its average value on the earth’s
surface is 9.8 m/s2.

4. Its S.I. unit is Nm2/kg2. 4. Its S.I. unit is m/s2.

5. It is a scalar quantity. 5. It is a vector quantity.

2. The weight of the body is more in the Terai than in the Himalayan region.
Why?

Ü The weight of the body is directly proportional to 'g' as,

Weight = mass × g

New Creative Science, Class 10 | 17

Similarly, 'g' is inversely proportional to the square of distance from the centre
of the earth.

GM
i.e. g = (R+h)2
As the Terai region has less distance from the centre of the earth, there is more
value of 'g' and consequently more weight of the body.
3. The value of ‘g’ at the centre of earth is zero. Why?

Ü The value of 'g' at the centre of the earth is zero because at the centre, depth from
the surface of the earth is equal to the radius of the earth and from the relation,g'

( )d

= 1 – R .g , the ‘g’ becomes zero as, d = R.
4. What is the difference between the falling of parachute on the earth and on

the moon?

Ü When the parachute falls on the earth, it is affected by the resistance of the
air. So, it falls down slowly and constantly. On the other hand, moon has no
atmosphere and no resistance of the air. So, falling of parachute is free fall on the
moon, which accelerates every time until it reaches to the moon's surface.

5. A person falling freely in a lift feels weightlessness. Why?

Ü From the Newton's third law of motion, our weight is balanced by the reactional
force of the earth.

i.e. Weight = reactional force

mg = R

When, the body is falling with an acceleration 'a',

Then, net weight = R

i.e. mg – ma = R

When the person is falling freely, a = g

Thus, R = mg – mg = 0

The person feels weightlessness due to the zero reactional force.
6. Can weightlessness be called as masslessness? Give reason.

Ü No, weightlessness can't be said as masslessness because mass is constant all
over the universe. The body feels weightlessness only due to the zero reactional
force, which may be due to the zero gravity and freefall.

7. Parachutist does not get hurt. Why?

Ü When a person falls with a parachute, it falls with slow and a constant velocity
known as terminal velocity. It is due to the resistance of the air. So, he lands
safely and does not get hurt.

18 | Force

8. The value of ‘g’ is more at poles than in equator. Why?

Ü GM
The value of 'g' is given as, g = (R)2 .

From this relation, it is clear that as the radius increases the value of g decreases.

As we know the earth is not a perfect sphere. It is like an orange. it is flat at poles
and bulges out at the equator. So, polar radius is shorter than equatorial radius.
Hence, the value of g is more at poles than in equator.

SUMMARY

 Newton's universal law of gravitation states that, every object in the universe attracts every
other object with a force which is (i) directly proportional to the product of their masses and
(ii) inversely proportional to the square of the distance between their centres.

 The total amount of matter contained in a body is called its mass.
 The force of gravity acting on a body is called its weight.
 G is called the universal gravitational constant. Its value is 6.67×10 –11 Nm2/kg2 all over the world.
 Acceleration produced on a freely falling body is called acceleration due to gravity.

 The acceleration due to gravity is directly proportional to the mass of the heavenly body and
GM

inversely proportional to the square of the radius.g = (R)2
 The space around a heavenly body up to which its gravitational force can be felt is called

gravitational field.
 The gravitational force experienced by a unit mass kept at a point in the gravitational field of a

heavenly body is called gravitational field intensity.
 When a body falls freely under the effect of gravity only, then the body is said to be in free fall.
 Weightlessness is the condition in which the effective weight of a body becomes zero due to

the zero reactional force of the earth or other heavenly bodies.

EXERCISE

1. State the Newton’s universal law of gravitation. Why is it called the universal law?

2. Prove F = Gm1m2 where the symbols have their usual meaning.
d2

3. Differentiate between g and G.

4. Write down any three effects of gravity.

5. Differentiate between:
(a) Gravity and gravitational force
(b) Mass and weight
(c) Gravitational field and gravitational field intensity
(d) Gravity and acceleration due to gravity
(e) Free fall and weightlessness

New Creative Science, Class 10 | 19

6. Prove, g = GM . The weight of the body is more in the Terai than in the Himalayan
region. WhyR?2

7. The weight of the body is more at poles than in equator. Why?

8. Explain the coin (Guinea) and feather experiment. What conclusion can be drawn
from the experiment?

9. Prove that the acceleration due to gravity is independent to the mass of the object.
10. It is difficult to lift a big stone than a small stone. Why?
11. G is called universal gravitational constant. Why?
12. The value of g varies from place to place. Why?
13. Why does a person feel weightlessness in a spacecraft orbiting around a heavenly body?
14. Why does a person feel weightlessness during free-fall?
15. What happens to the force of gravitation if:

(a) Masses of the bodies are doubled keeping the distance between them constant.
(b) Distances between the bodies are halved keeping masses constant.
(c) Masses are doubled and distance between them is halved?
(d) Moon has no atmosphere. Why?

Numerical problems

16. The mass of Jupiter is 1.9 × 1027 kg and that of the sun is 2 × 1030 kg. If the distance
between them is 78 × 107 km, find the gravitational force between them.(Ans: 4.166 × 1023 N)

17. Calculate the gravitational force between two bodies of masses 750 kg and 1500

kg if they are placed at 500 m distance? (Ans: 3.0 × 10–10 N)

18. A heavenly body has mass one third of the earth and its radius is half as that of

the earth. If a stone weights 200N on the earth’s surface, find its weight on that

heavenly body. (Ans: 266.67 N)

19. The mass of Jupiter is 1.9 × 1027 kg and its radius is 71 × 106 m. Find the gravitational

field intensity at its surface and at a distance of 142 × 106 m. What is the weight of

the body of mass 80 kg at its surface? (Ans: 25.14 N/kg, 2011.2N)

20. How much would a 70 kg man weigh on the moon? What would be his mass on

the earth and on the moon? (Ans: 114.33N, 70 kg)

21. The masses of the sun and the earth are 2 × 1030 kg and 6 × 1024 kg respectively.

Find the gravitational force between them if the distance between their centres is

1.5 × 1011 m. (Ans: 3.56 × 1022N)

A

B GLOSSARY
C

Acceleration : rate of change of velocity
Equator : an imaginary line that divides the earth into two hemispheres
Guinea : a coin of England



20 | Force

UNIT

2 Pressure

About the Scientist INTRODUCTION

Archimedes Have you ever wondered why a camel can run in a desert
Archimedes was a Greek scientist. while a girl with high-heeled sandals feels it difficult? Why
He discovered the principle an army tank weighing more than a thousand tonne rests
subsequently named after him, upon a continuous chain? Why a tractor has much wider
when he noticed that the water tyres? Why cutting tools have sharp edges? Why food cooks
in a bathtub overflowed when he faster in a pressure cooker? In order to find the answers of
steeped into it. He ran through these questions, we should introduce the term pressure.
the streets shouting ‘Eureka’,
which means ‘I have got it’. The relation of force, pressure and area will be clear from
This knowledge helped him to the following activity.
determine the purity of the gold
in the crown made for the king. ACTIVITY

A Take a piece of foam and a brick. Keep the brick as shown in the
B figure (a) with its larger surface area over the foam and observe the
C depression on the foam. Now, keep the brick as shown in the figure
(b) with its smaller surface area over the foam and again observe
the depression. Now, add one similar brick over the foam in the
figure (a) and observe the depression as shown in the figure (c).

Brick

Foam

(a) (b) (c)

QUESTIONS

A # Why the foam gets depressed in the condition (b) than in (a)?

# Why the depression in the foam increases with the increase

D B in the number of bricks?
# What can you conclude from this activity?

From this activity, it is clear that pressure increases if the

C force (weight of the bricks) increases and it decreases if the
surface area increases.

New Creative Science, Class 10 | 21

PRESSURE

The perpendicular force acting on per unit area of the surface is called pressure.

Force (F)
i.e. Pressure (P) = Area (A)
F
∴ P = A

Its S.I. unit is Nm–2 or Pascal (Pa).

From the above relation it is clear that, pressure (P) ∝ F when area is kept constant and
1

(P) ∝ A when force is kept constant.

If 1N force is applied on the surface area of 1m2 then the pressure exerted on that

surface is known as one pascal pressure.

MEMORY TIPS
In honour of scientist Blaise Pascal, the S.I. unit of pressure is called Pascal (Pa).

QUESTIONS
# All the cutting and piercing tools such as scissors, needles, knives, etc. have sharp edges.

Why?
# It is difficult to walk on sand. Why?

SOLVEDNumerical

1. Calculate the pressure exerted by a girl of 30kg if she is standing on only one
leg. (Area of her sole is 100 cm2).

Solution:

150
Given, Area (A) = 150 cm2 = 10000 = 1.5 × 10–2 m2

Mass (m) = 30 kg

Weight/force (F) = m × g = 30 × 9.8 = 294 N

Pressure (P) = ?

We have,

From formula of pressure,
F 294

P = A = 15 × 10–2 = 1.96 × 104 Pa
∴ She exerts the pressure of 1.96 × 104 Pa. on the ground while standing with only
one leg.

22 | Pressure

LIQUID PRESSURE

Liquids do not have fixed shape but they have their weights and can occupy space. So,
the liquids can also exert pressure on the bottom and the walls of the container in which
they are kept.

The force exerted by liquid on per unit area of the wall of the container is called liquid pressure.

Consider a liquid of density (d) is kept in a container having the base area (A) up to
the depth (h) from the free surface of the liquid as shown in the figure.

We know that,

Force (F) mg
Pressure (P) = Area (A) = A

Where, F = mg, m = mass of liquid, g = acceleration due to gravity.

d×v×g [m = d × v, where v is the volume of liquid]
=A [where, v = A × h]

d×A×h×g
=A

∴ Pressure (P) = dgh

This is the expression for the pressure exerted by liquid. From this, it is clear that
liquid pressure depends on three factors:

a. Depth of the liquid from its free surface (h)
b. Density of liquid (d)
c. Acceleration due to gravity (g)

MEMORY TIPS
● For a liquid at a given place, its density and acceleration due to gravity is constant.

So, liquid pressure (P) ∝ depth of the liquid from the free surface (h).

Laws of liquid pressure

a. Liquid pressure is directly proportional to the depth of liquid from its free surface.
b. Liquid finds its own level.
c. Pressure applied on an enclosed liquid transmits equally in all directions.
d. The pressure of liquid doesn’t depend upon the shape and volume of the container

in which it is kept.
e. Liquid pressure depends on its density.

Liquid pressure and its depth

From the relation p = hdg, it is clear that liquid pressure increases with depth.
For its experimental verification, take a can with side holes A, B and C at its different
heights as shown in the figure.

New Creative Science, Class 10 | 23

Fill the can with water by closing all three holes. A Can with holes
When the can is full of water, release the holes Water
simultaneously. Observe the speed of water in all B
three holes. C Stand

Water comes out with more speed from the hole C
than the others because it is the deepest from the free
surface of the water level.

It proves that liquid pressure increases with depth. Due to this reason, dams are
made wider at the base to withstand the greater pressure of water. Similarly, the
blood pressure in human body is greater at the feet than at the brain and to prevent
the enormous pressure, deep sea-divers wear diving suits etc.

QUESTIONS
# Why dams are made wider at the base?
# A bucket can be filled faster at the bottom floor than at the top floor of a building, though

the size of the taps are the same. Why?
# Deep sea divers wear diving suit. Why?

ACTIVITY

Pour some water into a communicating tube in which the vessels of different shapes are
connected with a tube. Observe the water level in each vessel. Is the water level the same? What
can you conclude from this?

PASCAL’S LAW

It is a famous law in hydrostatics that states, “when the pressure is changed in any part of
the enclosed fluid then this change of pressure is transmitted equally and perpendicularly to
all the other parts of the fluid.”

MEMORY TIPS

● Pascal’s law was formulated by a French scientist, Blaise Pascal in 1647 A.D.
● Hydrostatics is a branch of physics that deals with the properties of fluids (liquids and

gases).

Experimental Verification of Pascal’s law D A
B
For this, take a vessel full of water with four openings provided
each with the pistons A, B, C and D as shown in the figure. C

The area of pistons may or may not be equal. If one of the
Pistons say ‘A’ is pushed inward, this pressure is transmitted
equally to all parts of the vessel which causes the piston B, C
and D move outwards equally.

24 | Pressure

The given above water displacement experiment proves that the enclosed liquid transmits
pressure equally to all directions or the Pascal’s law. This law is also known as the principle
of transmission of fluid pressure.

ACTIVITY

Take a balloon and make small holes at its different parts. Fill it with water. Now,
press the balloon by covering its mouth. Observe the speed of water from each
hole. What can you conclude from this experiment?

Application of Pascal’s law

On the basis of the principle of Pascal’s law many hydraulic machines are constructed.

They are: b. hydraulic press
a. hydraulic lift d. hydraulic cranes etc.
c. hydraulic brake

a. Hydraulic lift

It is a machine based on the Pascal’s law. It is used to lift the vehicles in automobile
servicing stations during their service. It is also used to carry people from one floor to
another in tall buildings by multiplying the force.

Valve 1 Valve 2

b. Hydraulic press

It is a simple machine which is used to magnify the applied force. It is also based on
the Pascal’s law which is used to squeeze cotton bundles by magnifying our force.

New Creative Science, Class 10 | 25

c. Hydraulic brakes

By the use of hydraulic brakes, we can easily stop the heavy moving vehicles like
trucks, buses, aeroplanes etc. applying a little force. It is also based on Pascal’s law.

Foot pedal

Pipeline Liquid Piston

To other wheel Master cylinder
Break shoes Wheel cylinder

Break shoes
Return spring

Working Principle of Hydraulic machines

Hydraulic machines are the devices which work on the principle of Pascal’s law
by converting a little force into a larger force, e.g. hydraulic lift, hydraulic brake, etc.

It consists of two cylindrical tubes of different cross sectional areas A1 and A2
connected together with a horizontal tube. The apparatus is filled with water. The tubes

are provided with the water and airtight pistons. The general principle of a hydraulic

machine is given below:

When force ‘F ’ is applied at piston A , the

pressure exerted on the liquid is given by, A B

P1 = F1 ..................... (i)
A1

This pressure is transmitted to another piston

B and pressure exerted on the piston B by the

water is

P2 = F2 ..................... (ii)
A2

From Pascal’s law, we have,

P1 = P2

or, F1 = F2
A1 A2

∴ F2 = F1 × A2 (when the machine is in frictionless condition)
A1

Since A2 > A1 , F2 will be greater than F1 . This proves that the hydraulic machine is a
force multiplier.

26 | Pressure

SOLVEDNumerical

1. Calculate weight that can be lifted in piston A2 when 50 N force is applied on
piston A1.

Solution: = 50 N

Given,

Force in piston A1 (F1) = 50 N 10 cm2 50 cm2
Area in piston A1 = 10 cm2
Force in piston A2 (F2) = ?
Area in piston A2 = 50 cm2

We know that,

F1 = F2
A1 A2

F2 = F1 × A2 = 50 × 50 = 250 N
A1 10

∴ The weight that can be lifted in piston A2 is 250 N.

MEMORY TIPS

Teeth scaling and high-pressure water jet cutting are also the application of Pascal’s law.

On teeth scaling, teeth are hit by fine jet of water at high pressure, and in high pressure
water jet cutting, stones, slates, rubbers, foams, asbestos, etc. are cut by high pressure
of water jet.

QUESTIONS

# State Pascal’s law. Give its two applications.
# How can you show that a hydraulic machine is a force multiplier?

DENSITY AND RELATIVE DENSITY

The density of the substance is defined as the mass of the substance in per unit volume.

i.e. Density (D) = Mass (M)
M
Volume (v)

∴D= v

Its SI unit is kg/m3 and CGS. unit is gm/cm3.

The density of water in SI unit is 1000 kg/m3 and in CGS unit 1 g/cm3

Relative density of a substance is defined as the ratio of density of any substance to the density
of water at 4°C.

New Creative Science, Class 10 | 27

Density of substance
i.e. Relative Density = Density of water at 4ºC
Since, relative density is the ratio of two similar quantities, it has no unit.

Density of substance
Further, relative density = Density of water at 4ºC

mass of substance/volume of substance
= mass of water/volume of water at 4ºC
If the volume of the given substance is equal to the volume of water at 4°C then,

mass of substance
Relative density = mass of equal volume of water at 4ºC

Hence, relative density can also be defined as the ratio of mass of the substance and the mass of
an equal volume of water at 4°C. Relative density is also known as specific gravity.

S.N. Substances Density g/cm3 Relative density

1. Water kg/m3 1.0 1
2. Milk 1000 1.03 1.03
3. Mercury 1030 13.6 13.6
4. Ice 13600 0.92 0.92
5. Gold 920 19.3 19.3
19300
UPTHRUST

The resultant upward force exerted by a fluid on an object which is completely or partially
immersed into the fluid is called the upthrust or buoyancy.

Because of the upthrust, we can easily lift up a heavy object in
water.

Measurement of upthrust

Take the weight of a stone in the air using a spring balance.
Let, its weight be w1 . Now, take its weight by completely
immersing it into water. Let its weight in water be w2 .
The weight of the stone in water (w2) is less than its weight
in the air (w1). It is because of the upthrust given by water.
Thus,

Upthrust given by water = wt. of stone in air – wt. of stone in water.

i.e. upthrust (U) = w1 – w2

28 | Pressure

QUESTIONS
# It is easier to pull a bucket of water from the well until it is inside water. Why?

Density of liquid and upthrust

From the relation of liquid pressure, we have, P = hdg
If depth of liquid column is constant at any place then, p ∝ d.
Similarly, from the relation U = vdg, upthrust is directly proportional to the density
of the liquid. Liquid with more density applies more upthrust. So, egg floats in salt
solution and sinks in pure water.
Hence, pressure increases as the density of the liquid increases. Similarly the upthrust
is also affected by the density of the liquid.

ACTIVITY

1. Take a beaker with fresh tap water.
2. Now put an egg in the above fresh water and observe.
3. The egg sinks in water. Why?
4. Now, take out the egg and make saturated salt solution in above water. Put the egg in it and

observe.
5. Why does the egg float on the solution?

MEMORY TIPS
● In Dead Sea, which contains about 270 gram of salt per litre of water, a person does

not sink due to the greater upthrust.
● Upthrust does not depend upon the weight of the body immersed in a fluid.

QUESTIONS
# Why does an egg float on the salt solution but it sinks in pure water?

ARCHIMEDES’ PRINCIPLE

This principle was formulated by Archimedes. It states, “When a body is wholly or
partially immersed in a fluid, it experiences an upthrust which is equal to the weight of the liquid
displaced.”

New Creative Science, Class 10 | 29

Theoretical proof of Archimedes’s principles

Suppose we have a cylinder of height (h) and cross-sectional area (A) be completely
immersed in liquid of density (d) as shown in the figure. Let
the top face of the cylinder is at depth (h ) and bottom face be
at the depth (h ) below the free surface of liquid. 1

The force acting on the upper face of the cylinder due to

liquid (F1) is,

F1 = Pressure × Area

i.e. F1 = P1 × A

or, F1 = dgh1 × A ..........(i) [p = dgh]

Similarly,

Force acting on lower surface of the cylinder due to liquid (F2) is,
F2 = P2 × A
or, F2 = dgh2 × A ............................ (ii)

Hence, the net upward force on the cylinder (F) is,

F = F2 – F1 (h = h1 – h2)
= dgh2A – dgh1A (v = A – h)
= dgA(h2 – h1)
= dgAh

F = dgv

Since, the net upward force is the upthrust.

Thus,

Upthrust (U) = v × d × g ............................. (iii)

Hence, from this relation it is clear that, upthrust due to the liquid is:
a. directly proportional to the volume of the body immersed or volume of water

displaced
b. directly proportional to the density of liquid
c. directly proportional to the acceleration due to gravity of that place

MEMORY TIPS
When a floating body on water in a beaker falls freely under the gravity with its whole
system then it experience zero upthrust and the body sinks during the free fall.

Experimental verification of Archimedes’s principle

Materials required: Eureka can, top pan balance, stone, thread, spring balance
a. Find out the weight of a stone in the air. Let its weight in the air be w1.

30 | Pressure

b. Fill the Eureka can with water
up to the spout.

c. Keep a beaker over pan balance w3
just below the spout of the can
as shown in the figure.

d. Take the weight of the empty
beaker, let it be w2.

e. Immerse the stone wholly into
the water of Eureka can and
note the weight of the stone
inside the water. Let it be w3.

f. The water overflows when stone
is immersed into it and gets
collected in the beaker. The weight of the beaker with water is taken. Let it be w4.
∴ Loss in weight of the stone = w1 – w3 and
weight of displaced water = w4 – w2
It is observed that w1 – w3 = w4 – w2

Thus, loss in weight of a body (upthrust) in liquid is equal to the weight of liquid displaced.

Hence, Archimedes’s principle is verified experimentally.

QUESTIONS
# An iron nail sinks in water but a ship made of the same material floats on it. Why?

LAW OF FLOATATION

It states that, “the weight of a floating body is equal to the weight of the liquid displaced by it.”

i.e. weight of floating body = weight of displaced liquid

Experimental verification of principle of floatation

Materials required: A block of wood, spring
balance, Eureka can, beaker, top pan balance
a. Take the weight of the block of wood.

Let it be w1 .
b. Fill the Eureka can with water up to

the spout and place a beaker below its
spout. Take the weight of the empty
beaker. Let it be w2.
c. Now put the block of wood into the
Eureka can. It will float on the water
and displace the water.

New Creative Science, Class 10 | 31

d. The water is collected in the empty beaker. Let the weight of beaker with
water be w3.
∴ Weight of displaced water = w3 – w2
It is observed that, w1 = w3 – w2
i.e. Weight of floating body = weight of liquid displaced

Hence, the principle of floatation is verified experimentally.

Conditions for different solids kept in a liquid

Condition-I
When the weight of the body is greater than the upthrust acting in it, then the
resultant force will be vertically downwards and the body sinks in the liquid.
In this case, density of the object > density of the liquid Hence, an iron
piece sinks in water.
Condition-II
When the weight of the body is equal to the upthrust acting on it then the
resultant force acting on the body will be zero and the body floats on the
liquid below its surface at any depth and remains suspending.
In this case, density of the object = density of the liquid.
Condition-III
When the density of the body is less than density of liquid, the body floats on
the surface of that liquid. Similar to the case-II, in this case, upthrust is also
equal to the weight the body.
In this case, density of the body < density of liquid

MEMORY TIPS
Though the density of materials used in ships is greater than the density of water, the
structure of the ship is made such that it displaces more volume of water and therefore
experiences more upthrust which becomes equal to its weight and it floats.

QUESTIONS
# A piece of solid iron sinks, but a ship made of the same material floats on water. Why?
# Loaded ship sinks more than the empty one. Why?

ACTIVITY

Place a cork and an iron nail on the surface of water inside the beaker.
Observe what happens and give reason for it.

32 | Pressure

Applications of principle of floatation

Different types of hydrometers, ships, boats, etc. are made on the principle of floatation.

Hydrometer

It is a device used to measure the density or the relative density of liquids. It is based
on the principle of floatation.

When the hydrometer is placed in a liquid, it sinks until it displaces the
liquid equal to its own weight. The hydrometer sinks less in the liquids having
more density and sinks more in the liquids having less density. So, graduation in
stem is done in such a way that the numbers (showing density) are bigger
at the bottom, decreasing upwards. The stem is graduated in kg/m3. Hence, it
measures density of liquids directly.

A specially designed constant weight hydrometer which is used to measure the purity of milk
is called lactometer.

MEMORY TIPS
The density of pure milk is 1030 kg/m3, density for milk with water is less than that and
for the milk with powder, density is greater than 1030 kg/m3.

ATMOSPHERIC PRESSURE

The earth has its force of attraction called gravity. Due to this force of gravity it binds
the mixture of air around the earth. The mixture of air contains nitrogen, oxygen,
carbon dioxide, water vapour, dust, smoke, etc. Thus, the envelop of air which surrounds
the earth is called atmosphere. The atmosphere extends upto about 9600 km above the
earth’s surface. Air has weight. Due to its weight, it applies pressure on the surface of
the earth called atmospheric pressure.

The pressure exerted by the mixture of different gases on the surface of the earth is called
atmospheric pressure. The atmospheric pressure at the sea level is called normal
atmospheric pressure or standard atmospheric pressure.

Normal atmospheric pressure = 760 mm of Hg or one atmosphere is equal to 105
Pascal. As the amount of air decreases at the heights, the atmospheric pressure also
decreases subsequently. So, the atmospheric pressure decreases with altitude.

The atmosphere also exerts pressure on the human body. The force exerted by the
atmosphere on the human body of area 2m2 is about 2 × 105N as:

F = P × A = 105 × 2 = 2 × 105 N

But, we do not feel it or crushed by it. This is because our internal pressure balances
this pressure. But as we go to the higher altitude, our internal pressure becomes more.
So, bleeding from nose and ear may occur.

New Creative Science, Class 10 | 33

MEMORY TIPS

Jet planes fly at about an altitude of 2000m. At this height, there is less atmospheric
pressure. So, air pressure is controlled inside the planes. This is why, passengers inside
the plane could breathe normally.

Uses of the atmospheric pressure

a. Atmospheric pressure is used to fill ink in the pen.
b. It is used to make syringe.
c. It is used to make air filling pump.
d. It is used to make tubewell.

QUESTIONS

# We feel uneasy at higher altitudes. Why?
# The atmosphere exerts enormous pressure to us but we do not feel it. Why?

ACTIVITY (a) (b)

To show atmospheric pressure
Take a tin can with a little of water in it. Boil it for few minutes
as shown in the figure (a). Close the lid of the can immediately
and cut off the flame. Now, cool the can under a tap as shown
in the figure (b).
At first, the pressure of the air inside the can and outside it was
equal. When it was heated, the air inside the can escaped out
and the space was occupied with the steam. On cooling the
can, the steam condensed creating partial vacuum inside the
can. Hence, the atmospheric pressure becomes more than
the internal pressure and the tin can got crushed inwards.
This proves that atmosphere exerts pressure.

ACTIVITY

To show atmospheric pressure
Take a glass filled with water and cover its mouth with a
cardboard. Invert the glass supporting the cardboard with your
palm. Hold the glass in your hand and remove the other hand
slowly which was supporting the cardboard.
The atmosphere exerts pressure upwards on the cardboard
and the cardboard and the water do not fall. This again proves
that atmosphere exerts pressure from all directions.

34 | Pressure

ACTIVITY

To show air exerts pressure
Take some water in a bottle. Suck it by using a straw pipe. When the
air from the pipe is sucked up, the water enters the mouth through the
straw pipe. It is because when the air from the pipe is sucked out, the
air pressure inside the pipe becomes less than the outside pressure.
Hence, the water enters the straw and comes into the mouth due to the
more atmospheric pressure at the surface of the liquid. This also proves
that atmosphere exerts pressure. We fill ink in the pen on the basis of
the same principle.

BAROMETER

It is an instrument used to measure the atmospheric pressure of the certain place. It is also
used to measure altitudes as the atmospheric pressure changes with altitude. There
are two types of barometers. They are:
a. Mercury barometer
b. Aneroid barometer

a) Mercury barometer

It consists of mercury as the barometric substance. For
its construction, a one meter long calibrated glass tube
is completely filled with mercury and it is inverted
carefully over a disc of mercury by pressing its mouth
with our thumb.

Now, it will be observed that the mercury level inside the tube falls slowly at first
and becomes constant at any point. The reading of this point gives the atmospheric
pressure of that place. The atmospheric pressure at sea level is 760 mm of Hg. As
the altitude increases, the atmospheric pressure decreases. So, the level of mercury in
the barometer also falls. Similarly, as the altitude decreases, the atmospheric pressure
increases. So, the level of mercury in the barometer also rises up.

Advantages of mercury as the barometric substance

i) The density of mercury is very high i.e. 13600 kg/m3 so, only one meter long
glass tube is sufficient.

ii) It doesn’t stick inside the glass tube. So, the reading will be correct.
iii) It is silvery in colour. So, it can be seen easily from outside.
iv) It doesn’t vapourize easily.

But in contrast, water is transparent so it sticks on the wall of the tube, and its
density is also less. So, a tube of about eleven meters is required. Hence, it is not
used as the barometric substance.

New Creative Science, Class 10 | 35

MEMORY TIPS
● Mercury barometer was invented by an Italian scientist E. Torricellie.
● The empty space above the surface of mercury (Hg) in the barometric glass-tube is

called torricellian vacuum.
● Sphygmomanometer is the instrument used to measure the blood pressure of human

body.

QUESTIONS
# The mercury level decreases in barometer as the altitude increases, why?
# We cannot use water as a barometric substance, why?

b) Aneroid barometer

It doesn’t contain any liquid as the barometric substance and is
easy to carry from place to place. It is used as altimeter in aircrafts
and for the mountaineers.

PRESSURE GAUGE

It is an instrument which is used to measure the air pressure inside the
tyres of the vehicles.

More air pressure inside the tyre causes the tyre to burst and less
pressure inside it doesn’t allow the vehicles to run smoothly. So,
the appropriate amount of air should be filled in the tubes of the
tyres. And pressure gauge helps us for this.

SYRINGE

Syringe is a medical device which is used in the hospital and clinic
to inject medicine inside the body and to take out blood from the
body of the patients. Mainly, it has three parts. They are needle,
storage cylinder and piston. During the injection of the liquid
medicine, first of all the needle of the syringe is inserted inside
the medicine bottle. The piston of the syringe is pulled slowly. As we pull the piston,
there occurs vacuum in the storage cylinder. We know that liquid always flows from
the region of the high pressure to the region of the low pressure. So, medicine moves
toward the storage cylinder as there is less pressure. Now, the needle is taken out
from the bottle and inserted in the blood vessel of the patient. As we push the piston,
the medicine moves into the body through blood vessel. Same phenomenon happens
while taking out blood from the body of the patients.

36 | Pressure

AIR FILLING PUMP

To fill air inside the tubes of tyres, tubes of volleyball,
football, etc. we use a pump called air filling pump
or simply pump. It has a cylinder, piston and nozzle.
To fill air inside the tube, the piston must be moved
up and down continuously. While pulling the piston
up, there creates a vacuum inside the cylinder. In
this vacuum air is filled from the atmosphere. As we
push the piston down, air enters into the tube through the nozzle of the cylinder.
Similarly, when we pull the piston up, air from the tube cannot come out as there is a
valve in the tube, which gets closed automatically. This process is continued to fill the
tube completely.

TUBEWELL OR HAND PUMP

Different kinds of tubewells or water pulling pumps
are used to pull the underground water from the
dip well. All these tubewells are constructed on the
principle of atmospheric pressure. It contains two
parts. They are barrel and piston. Both of these parts
are connected with valves. The movable valve (V1)
is connected to the piston. This valve moves up and
down along with the piston. At the bottom of the
barrel there is another valve (V2). Now, the barrel and
the pipe are connected together and send at the dip
well.

UP STROKE

As we push the handle, the piston comes up. At the same time, there occurs vacuum
in between these two valves. Due to low pressure created between these valves, the
valve (V1) gets closed and water rises up in the barrel through valve (V2). Thus, the
pushing down of the handle to rise up the piston is called up stroke.

DOWN STROKE

As we pull up the handle, the piston moves down. As a result the valve (V1) which
is present in the piston and the valve (V2) which is present in the barrel come closer.
Due to high pressure in the valve (V2) it gets closed but the valve (V1) remains open.
Through the valve (V1), water comes up. Thus, the pulling up of the handle to push down
the piston is called down stroke.

New Creative Science, Class 10 | 37

SOLVEDNumerical

1. Calculate the pressure at the bottom of the container if the depth of the water
in the container is 5m.

Solution:

Given, depth of water (h) = 5m
Density of water (d) = 1000 kg/m3
Acceleration due to gravity (g) = 10 m/s2
Liquid pressure (p) = ?

We have, from the relation
P =hdg
= 5 × 1000 × 10
= 5 × 104 Pascal

Hence, the pressure of the water is 5 × 104 Pa.
2. Calculate the density and relative density of the liquid having mass 4000 kg

and volume 4m3.

Solution:

Given, mass of liquid (m) = 4000 kg
Volume of liquid (V) = 4 m3
Density of liquid (D) = ?
Relative density of liquid (R.D.) = ?

We have,from the relation,
M 4000

D = V = 4 = 1000 kg/m3
Again,

Density of liquid 1000
R.D. = Density of water at 4ºC = 1000 = 1
Hence, the density of liquid is 1000 kg/m3 and its relative density is 1.
3. The density of a brick is 2.5 gm/cm3 and its mass is 1 kg. Find the weight of
water displaced when this brick is immersed in water.

Solution:

Given, density of brick (d) = 2.5 gm/cm3
Mass of brick (m) = 1 kg = 1000 gm
Volume of brick (V) = ?
Wt. of water displaced = ?

38 | Pressure

We have,
m 1000

Volume of brick = d = 2.5 = 400 cm3
Volume of brick = volume of water displaced = 400 cm3

Now, mass of water displaced = volume of water displaced × density of water

= 400 × 1 = 400 gm

Hence, the mass of displaced water is 400 gm.

MODEL QUESTIONS ANSWER

1. All cutting instruments have sharp edges. Why?
Ü The sharp edges make the cross-sectional area of cutting instruments small and

as we know,
F

P=A
So, our force will get distributed over small area thereby increasing pressure.
This makes cutting easier. So, cutting instruments are provided with sharp
edges.
2. Deep sea-divers wear special diving suits. Why?
Ü We know that, liquid pressure = h × d × g

Which shows that the liquid pressure increases with depth from its free surface.
So, to withstand the enormous pressure applied by water, the deep sea-divers
wear special diving suits.
3. It is easier to pull a bucket of water from the well until it is inside the water
but difficult when it comes out of water. Why?
Ü The upthrust given by water is more than that given by the air, as water is denser
than the air. Due to this, it is easier to pull a bucket of water until it is inside the
water.

4. An iron nail sinks but a ship made up of iron floats. Why?
Ü A ship is so designed that it displaces water equal to its own weight. Hence it

experiences equal upthrust and floats but the iron nail cannot displace the water
equal to its own weight and thus sinks into the water.

5. A helium filled balloon rises to a certain height and then halts. Why?
Ü Since, the density of helium is lower than that of air. So, due to the upthrust of

air, the balloon rises to a certain height where the density of air is equal to the
density of helium and then it halts there.

6. Why a man will be found lighter when weighted in the air than in vacuum?
Ü Air also gives upthrust due to which the resultant weight of a person in air is

less than in vacuum.

New Creative Science, Class 10 | 39

7. An ice cube floats in water. Will the water level rise if the ice melts completely?
Explain.

Ü No, the water level will not rise if the ice melts completely because the volume of
water obtained from the melted ice is equal to the volume of the ice inside water
level and the water level remains the same.

8. Loaded ship sinks more than the empty one. Why?
Ü The loaded ship has to displace more water (equal to its weight) than the empty

ship to be floated. So, the loaded ship sinks more than the empty one.

9. The weights of a piece of stone on weighing Medium Weight
in three different mediums- air, water and salt 15N
A 18N
solution are shown in the table below. 16N

i. Which one is water and which one is the B

salt solution out of these three mediums? C

Ü C is water and A is the salt solution.

ii. If the weight of 1 kg mass in air is 10N, find out the mass of the piece of
stone.

Ü Given, 10N weight = 1 kg mass
1

∴ 18N weight = 10 × 18 = 1.8 kg
Hence, mass of the stone in air = 1.8 kg

iii. Find out the mass of water displaced by the piece of stone.
Ü Given, weight lost = weight in air – weight in water = 18 – 16 = 2N

∴ Weight lost = weight of water displaced = 2N

∴ Mass of water displaced × g = 2N (since g = 10 m/s2)
2

∴ Mass of water displaced = 10 = 0.2 kg

SUMMARY

 Pressure is defined as the force acting normally on per unit area. It is given by,
Force

pressure = Area .
 The pressure exerted by liquid column is given by hdg. The liquid pressure increases with the

depth of liquid column from its surface level.

 Pascal’s law states that ‘when the pressure is changed in any part of the enclosed liquid, then
this change in pressure is transmitted equally to all the other parts of the fluid.

 Hydraulic brake, hydraulic lift, hydraulic press etc. are constructed on the basis of Pascal’s law.
 The resultant upward force exerted by a fluid on an object which is completely or partially

immersed into the fluid is called upthrust.

40 | Pressure

 Archimedes’ principle states that, ‘when a body is wholly or partially immersed into a
liquid, it experiences an upthrust which is equal to the weight of the displaced liquid.
i.e. upthrust = weight of liquid displaced.

 Law of floatation states that when the weight of the body is equal to the weight of displaced
liquid then the body floats on it. It is the special case of Archimedes’ principle.

 Hydrometer is used to measure the density or relative density of liquid. It is based on principle
of floatation.

 Lactometer is used to measure the purity of milk.

EXERCISE

1. Define pressure. Write its formula and S.I. unit.

2. Prove P = hdg. What are the factors which affect the liquid pressure?

3. State Archimedes’ principle. Verify it experimentally.

4. State Pascal’s law of liquid pressure. Give its two applications.

5. How can hydraulic machines magnify our force?

6. State the law of floatation. Why is it called the special condition of Archimedes’
principle?

7. What do you mean by hydrometer? Write down its uses.

8. What is atmosphere and atmospheric pressure? Write down the value of standard
atmospheric pressure.

9. Describe the structure and working of the syringe with the help of diagram.

10. What is barometer? Describe the structure of the mercury barometer with the help
of diagram.

11. Describe the structure and working of the air filling pump.

12. What is tubewell? Describe its structure and working.

13. Give reasons:

(a) A girl with the pointed heel shoes makes more depression in sand than an
elephant.

(b) Dams are made wider at the bottom.

(c) The bags are provided with broad handles.

(d) Water is filled faster in a bucket in a downstairs tap than in the upstairs tap.

(e) Wooden sleepers are used below the rails.

(f) A steel ball sinks in water but floats on mercury.

(g) Ice floats on water.

(h) Deep sea-divers wear diving suits.

14. Write the difference between:

(a) Force and pressure (b) Density and relative density

(c) Archimedes’s principle and principle of floatation

New Creative Science, Class 10 | 41

15. The weight of a stone in air is 10N and that inside water is 7N. Answer the
following questions.

(a) Find the upthrust due to water and weight of displaced liquid.

(b) Which principle does it verify?

(c) State the principle.

16. Two tanks A and B of different cross-sectional areas
are shown in the figure containing the water upto the
same level.

(a) Which one exerts more pressure at the bottom?

(b) If both contain the equal amount of water, which A B
one exerts more pressure at the bottom?

17. A body ABCD is completely immersed into water as shown in the
figure. Study it and answer the following questions.

(a) What is the resultant pressure exerted on the surfaces BC and AD?

(b) Out of the surfaces AB and CD which one will experience more
pressure? Why?

Numerical problems

18. Find the pressure exerted on the bottom of the pond having depth 1.5m. (Given

g = 9.8 m/s2) [Ans: 14.7 × 103 Pa]

19. Calculate the pressure exerted by a mercury column of 76 cm high at its bottom.

Given that the density of mercury is 13600 kg/m3, g = 9.8 m/s2.[Ans: 1.01 × 105 Pa]

20. If your mass is 45 kg, what weight of water is to be displaced by you to float on

water? Why? [Ans: 450N]

21. The density of ice is 0.92 gm/cm3. A piece of ice has length 50cm, breadth 30cm

and height 20cm. Find out the mass of water displaced by the ice when it is kept

in water. (density of water = 1 gm/cm3) [Ans: 27.6 kg]

22. The weight of a piece of stone when immersed on water is 18N and it displaces

4N of water, what is its weight in air? [Ans: 22N]

23. A hydraulic automobile lift is designed to lift cars with a maximum mass of 3000
kg. The area of cross-section of the piston carrying the load is 425 cm2. What
maximum pressure would the smaller piston have to bear? [Ans: 6.92 × 105 Pa]

A

B GLOSSARY
C

Hydrostatics : a branch of physics which deals with the properties of fluids
Graduated : divided into levels in a scale
Fluid : that can flow freely (liquids and gases)
Thrust : force

Forge : to shape metal by heating and hammering



42 | Pressure

UNIT

3 Energy

About the Scientist INTRODUCTION

Albert Einstein We need energy to do work and make all the movements.
(1879-1955) When we eat, our bodies transform the food into energy,
which we use when we run or walk or do some physical
Albert Einstein is a very famous activities. Vehicles like cars, planes, trains also use diesel
scientist who led to hundreds or kerosene are the source of energy. Similarly, to run
of new discoveries and added industries, agricultural, physical or biological sectors, we
milestones in different branches need various forms of energy. Thus, energy is defined as the
of science. The theory of relativity ability or capacity to do work. Its SI unit is joule (J).
propounded by him (in 1905) is
one of the most famous inventions. SOURCES OF ENERGY
The theory is governed by a simple
equation, A source of energy is that which is capable of providing
enough useful energy at a steady rate over a long period of
E = ∆mc2 time. A good source of energy should be,
∆m is the difference in mass
between the original nucleus and a. safe and convenient to use,
product nuclei in nuclear fission b. easy to transport and
reaction. C is the velocity of light c. easy to store
in vacuum and E is the energy
produced. The sources of energy which can be used in the same form,
In 1774 Lavoises had stated that in which they occur in nature, are called primary energy
matter can neither be created nor sources like coal, wood, crude oil, natural gas, dung-cake,
be destroyed. This law was called solar energy, wind energy, etc.
as the law of conservation of mass.
But according to Einstein, mass Similarly, the sources of energy which are derived from the
can be converted into energy and primary energy sources are called secondary sources like wood-
vice versa. Einstein has written, gas, coke, petrol-diesel, bio-gas, hydroelectricity, wind
“Pre-relativity physics contains energy, nuclear energy, etc.
two conservation of mass and
the law of conservation of CLASSIFICATION OF THE SOURCES OF ENERGY
energy, these two appear there
as completely independent of The energy sources are classified as,
each other. Through relativity a. Non-renewable sources of energy
theory, they melt together into one b. Renewable sources of energy
principle.”
a. Non-renewable sources of energy

The sources of energy which are exhaustible and cannot be replaced
once they have been used are called non-renewable sources of
energy.

New Creative Science, Class 10 | 43

These sources have been accumulated in nature over a very long period of time, e.g.
coal, mineral oil, natural gas and all the other fossil fuels.

Due to the extensive use, these sources are depleting fast and it is difficult to explore
new deposits to these sources.

b. Renewable sources of energy

The sources of energy which are inexhaustible and can be replaced as they have been used are
called renewable sources of energy. They can be used again and again.

These sources are available in an unlimited amount in nature and develop within
a relatively short period of time, e.g. solar energy, wind energy, hydro energy,
geothermal energy, tidal energy, biomass energy, etc.

Renewable sources will last as long as the earth receives light from the sun. They are
freely available in nature.

Differences between renewable and non-renewable sources of energy.

Renewable sources of energy Non-renewable sources of energy

1. These are inexhaustible and can be 1. These are exhaustible and cannot be re-

replaced once they have been used up. placed once they have been used.

2. They can be regained within a short pe- 2. They cannot be regained quickly and take

riod of time. a million of years to form only under very

Example: solar energy, wind energy, suitable condition.

hydro energy, etc. Example: fossil fuels like natural gas,

petroleum, coal, etc

MEMORY TIPS

Wood is a renewable source of energy only if we plant trees in a planned manner to ensure
its continuous supply.

QUESTIONS
# Why is wood called as the renewable but petrol is called the non-renewable sources of

energy?

SOME NON-RENEWABLE SOURCES OF ENERGY

Fossil fuels

Fossil fuels are the remaining parts of prehistoric plants and animals, which got buried
deep inside the earth millions of years ago due to some natural processes and are used
as the sources of energy in the present days.

Coal, petroleum (Mineral oil) and natural gas are the major fossil fuels.

44 | Energy

a. Coal

It is the most abundant fossil fuel on the earth found in deep mines beneath the surface
of the earth. It constitutes mainly carbon and is used as fuel, which provides 27% of
the world’s energy.

When the dead plants and animals were buried inside the earth’s crust, they did not
decompose in a normal way and were protected from oxidizing effect of the air. The
action of certain bacteria released oxygen and hydrogen making the residue richer
and richer in carbon. Pressure and temperature further compressed this residue and it
slowly changed into coal over a period of millions of years. In this way, coal is formed.

According to the amount of carbon content, coal is classified into the following types:
i. Peat: 27% of carbon
ii. Lignite: 28%-30% of carbon
iii. Bituminous: 78%-87% carbon
iv. Anthracite: 94%-98% carbon

MEMORY TIPS
● In Nepal, low quality of coalmine (lignite) is found in Dang district.
● Anthracite is the superior quality of coal.

QUESTIONS
# How is coal formed?

b. Petroleum (Petra-rock oleum-oil)

It is a complex mixture of different types of hydrocarbons. It may also consist of other
elements like oxygen, nitrogen and sulphur in trace amounts.

Petroleum (also called crude oil) is a thick black liquid found trapped beneath
impermeable rocks in the earth’s crust. It is extracted from such reserves through wells
driven in them by cutting through the impervious rocks. The crude oil is then refined
to obtain a number of useful components like petrol (gasoline), diesel, kerosene,
asphalt, lubricating oil, paraffin wax etc.

The refining of crude oil is done in distillation or fractionating companies by the
process of fractional distillation.

i. Petrol is used to run small vehicles and diesel is used to run big and heavy
vehicles.

ii. Kerosene is used as fuel for domestic purposes and pure kerosene is used as
fuel in aeroplanes.

iii. Asphalt is used in road construction.
iv. Paraffin is used for manufacturing candles, books, polish, wax, vaseline, etc.

New Creative Science, Class 10 | 45

v. Lubricating oil is used as a lubricant in moving parts of the machines.
vi. They are also used in fertilizer industries.

MEMORY TIPS
On liquefaction, petroleum gas changes into a liquid, which is called liquefied petroleum gas
(LPG). It is also called cooking gas. The main constituent of LPG is butane.

QUESTIONS
# Why are petroleum products used more than other sources of energies?

c. Natural Gas
It is an important fossil fuel that is found underground near the oil source. It is a
mixture of methane (95%), ethane, propane and butane. It is used as a domestic and
industrial fuel. Compressed Natural Gas (CNG) is in the liquid form, which is used as
a fuel in vehicles. Natural gas is the cleanest burning fossil fuel.

Advantages of fossil fuel

Fossil fuels are used more than other sources of energy due to the following reasons:
a. It is cheaper for short term use and easy to transport.
b. Many machines and engines which are conducted by petroleum are
constructed.
c. It can be used to produce electricity.

SOME RENEWABLE SOURCES OF ENERGY

Hydropower

The kinetic energy of water has been used for a
long time to run watermills. The electricity produced
from the kinetic energy of flowing water is called
hydroelectricity. It is the most versatile source of
energy.

The total hydropower capacity of Nepal is
estimated to be 83,000 mw, out of which 600 mw
of electricity has been produced till today.

Advantages of hydropower

i. It is used to operate different electrical goods like the radio, T.V., computer, heater,
cooker, etc.

ii. It is also used to run vehicles.
iii. Since the initial construction cost is high, it becomes cheaper in the long-term.
iv. It is a pollution-free source of energy.

46 | Energy