Blooming Science-10-2077- final press

Approved by Government of Nepal, Ministry of Education, Curriculum Development Center (CDC).

Blooming

SCIENCE

Book

10

Authors
Raj Kumar Dhakal
Purushottam Devkota

Shubharambha Publication Pvt. Ltd.

Published by:

Shubharambha Publication Pvt. Ltd.

Kathmandu, Nepal
URL: www.shubharambhapublication.com
E-mail: [email protected]

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Blooming Science Book-10

Authors : Raj Kumar Dhakal

Purushottam Devkota

Video Content : Laxmi Nand Dhakal

Layout Design : Ram Malakar

Language Editor : Krishna Prasad Regmi

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Edition : 2077

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Preface

The series ‘Blooming Science’ has been brought out as an indispensable
resource for school level students and has intended to provide concise and
comprehensible explanation of key concepts, facts and principles across
science disciplines. Organized around the National Science curriculum
prescribed by Curriculum Development Centre, Sanothimi, Bhaktapur, the
series presents solid overviews of the most commonly encountered school
science topics with sound academic and fun activities.

The clear and accessible definitions, concise language, helpful diagrams and
illustrations and other science activities offered in this series will nonetheless
help teachers understand science concepts to the degree to which they can
develop rich and exciting inquiry approaches to exploring these concepts with
students in the classroom. As the series has been brought out considering
the age and other psychological factors of children, the learning materials
in this series appeal to the sense of the children and they are related to the
world of young learners. Activities with varieties of questions in this series are
meant to assess and evaluate the level of students’ inquisitiveness.

As each unit begins with its objectives and estimated teaching periods which
help teachers to complete the course in time. Moreover, each lesson in the
series ends with Let's Learn, Points to Remember and Relevant Exercises;
these sections are meant to provide good review to students and enhance
their ability to solve the exercise questions. Each lesson has Multiple Choices
Questions and Project Work that are meant to arouse more creativity and
interest in the students for better understanding and adjustment with their
scientific world.

We are grateful to teachers, students and principals who shared their valuable
suggestions in materializing this series. Any constructive suggestions and
recommendations for the betterment of this series will be highly acknowledged.

AUTHORS

Contents

PHYSICS

1. Force.......................................................................................................................... 7
2. Pressure................................................................................................................... 28
3. Energy...................................................................................................................... 48
4. Heat......................................................................................................................... 61
5. Light......................................................................................................................... 76
6. Electricity and Magnetism........................................................................................ 93

CHEMISTRY

7. Classification of Elements...................................................................................... 114
8. Chemical Reaction................................................................................................. 129
9. Acid, Base and Salt................................................................................................ 141
10. Some Gases.......................................................................................................... 152

Ammonia ............................................................................................................... 168
11. Metals.................................................................................................................... 177
12. Hydrocarbon and its compounds ...........................................................................193
13. Materials Used in Daily Life................................................................................... 213

BIOLOGY

14. Invertebrates.......................................................................................................... 223
15. Human Nervous System and Glandular System................................................... 237
16. Circulatory System in Human................................................................................ 255
17. Chromosome and Sex Determination.................................................................... 268
18. Reproduction.......................................................................................................... 283
19. Heredity.................................................................................................................. 295
20. Environmental Pollution and Management............................................................ 302

ASTRONOMY & GEOLOGY

21. History of the Earth................................................................................................ 314
22. Atmosphere and Climate Change.......................................................................... 330
23. The Universe.......................................................................................................... 328
Class X Examination Specification Grid - 2074..................................................... 347
Model Question ..................................................................................................... 348
Practical work sheet...............................................................................................353
List of Video Experiments ......................................................................................357

Chapter

1 Force Sir, Issac Newton
Jan 4, 1643-Mar 31, 1727

Learning Outcomes Estimated Periods: 8+2

On the completion of this unit, the students will be able to:
 describe and use universal law of gravitation.
 differentiate between gravity and gravitation.
 identify mass and weight with their units.
 define free fall and weightlessness.

Introduction

Force is used in most of the activities in our daily life. We use force to walk on the ground, to
lift the objects, to throw cricket ball or to move a given body. Newton’s first law of motion
gives the definition of force. The second law of motion gives the measurement of force and the
third law of motion clears that forces exist in pair as action and reaction. There are different
types of forces, but the word force generally denotes push or a pull. For example, a duster
on a table is at rest. If we apply a push or a pull, the duster moves on the table. Similarly, a stone
is on the ground. If we want to raise the stone, we apply a lift. During the pushing or pulling
or lifting, we have to spend something that we have in us as an expense. The thing that we spend
as expense is called energy. The push or the pull or the lift is called force.

Duster
Duster

Table Table Stone
Ground

Similarly, suppose a body is moving and we want to stop it. For that we have to apply a push on
the body in a direction opposite to its motion. The opposition which stops the motion of the body
is called opposing ( retarding) force.

However, sometimes, it happens that we apply the force on a body, but the
body remains as it were. For example, a huge rock is resting on the ground.
If the push or pull applied on the rock by a man is very small as compared
to the mass of the rock, the rock does not come into motion. Similarly, a
moving body may not stop if the opposing force applied on it is very weak.

Thus, the force can be defined as: “Force is an external agent which
changes or tends to change the position of a body”. Simply the pulling
or pushing effort to an object is called force.

Blooming Science Book 10 7

Gravitation

Sir Isaac Newton, in 1687 AD, discovered that there exists a force between two masses. For
example, two people sitting in a room, a table and a rock lying in a room, the sun and the earth,
the earth and the moon and so on exert force on each other. The force between two masses is
called gravitational force. The gravitational force is always attractive. That is each body (or
mass) attracts evey other body (or the other mass) towards it. This force exists universally among
all bodies in the universe. This force of attraction between any two bodies in the universe is
called gravitational force or simply gravitation.

Newton’s Universal Law of Gravitation

Newton was able to formulate the universal law of gravitation A m1 m2 B
when he observed the motion of the planets around the sun.
O F O’
Newton’s universal law of gravitation states that, ''everybody
in this universe attracts every other body with a force which d
is

(a) directly proportional to the product of their masses and

(b) inversely proportional to the square of the distance between their centres."

Let m1 and m2 are masses of two bodies A and B are separated by a distance. If the force of attraction
between them is 'F', then according to Newton’s law of gravitation, the attractive force (F) between
the two bodies is:

F ∝ m1m2 ............(i) and Scan for practical experiment

F∝ 1 ............ .......(ii) visit: csp.codes/c10e01

d2

Combining (i) and (ii), we have

F∝ m1m2

d2

Removing the proportionality sign, we get as:

F = Gm1m2 ............ (iii)
d2


Where G is the proportionally constant called universal gravitational constant and its
value is 6.67× 10-11 Nm2/kg2.

The Universal Gravitational Constant (G)

The constant 'G' appearing in the equation (iii) given d = 1m

above is called universal gravitational constant. The

value of G for any pair of masses in the universe is the

same. On the other hand the value of 'G' is independent m1 = 1 kg m2 = 1 kg
of the medium in which the masses are placed. This is

why 'G' is called universal gravitational constant.

8 Blooming Science Book 10

Definition of 'G'

We have,

F= Gm1m2
d2

Consider two unit masses separated by a unit distance. That is let m1 = 1 kg m2 = 1 kg and d = 1m.

G×1×1
Then, F = (1)2 = G

∴ F = G

Hence, the Universal gravitational constant (G) can be defined as the gravitational force between
two unit masses which is separated by a unit distance.

Unit

We have F= Gm1m2
d2
F × d2
G = m1m2

In SI system, m1 and m2 are expressed in kg, d is expressed in metre and F is expressed in Newton
(N).

Hence,

N × m2
G = (kg)2

= Nm2 = Nm2kg-2
kg2

Hence, the SI unit of 'G' is Nm2kg-2.

Value of G: The value of 'G' was first determined experimentally by Henry Cavendish. The value
of 'G' measured experimentally by him was G = 6.6734 × 10-11 Nm2/kg2

Thus, approximately, we accept the value of 'G' as

G = 6.67 × 10-11Nm2/kg2.

Newton'’s Gravitational Law is Universal Law

Newton’s law of gravitation is valid everywhere in the universe. It is valid for everybody in the
universe. In other words, it holds true for both heavenly bodies and terrestrial bodies. Also, it
holds true for both micro bodies like electrons, protons in an atom and macro bodies like stars,
galaxies, etc.

Evidences of Newton’s Law of Gravitation/ Consequences of Gravitation force

Following are the evidences in support of the gravitational law:

1. All the planets revolve round the sun and all the satellites revolve round the planets. This
is possible because of the force of attraction existing between them.

Blooming Science Book 10 9

2. Due to the gravitational attraction of the Sun and the Moon on earth, sea tides occur and are
maximum on full moon day.

3. Human has been able to go into space only with the help of the principle of gravitation.

4. The presence of atmosphere on the earth is due to the gravitational force.

5. Artificial satellites are orbiting round the earth because of the gravitational force between
them so they don’t need extra energy to revolve around the planets.

Relation between the Force of Gravitation with Mass and Distance

1. Force of gravitation between two bodies when A F1 B
their masses change and distance remains the m1 m2
same.

a. If the mass of one of the bodies is doubled, d
keeping their distance same.

Consider the masses of two bodies ‘A’ and ‘B’ be m1 and m2. They are placed at the
distance 'd' from their centres and gravitational force between them be 'F1'.

The gravitational force (F1) between them will be

F1 = Gm1m2 ............. (i)
d2

If the mass of one body is doubled and the distance between them is kept constant.

F2 = Gm12m2
d2

F2 = 2 Gm1m2 A F2 B
d2 m1 2m2

F2 = 2F1 [from (i)] d

Then the new gravitational force (F2) between them also becomes double.

b. If the masses of both bodies are made double without changing the distance.

F2 = G2m12m2
d2



F2 = 4 Gm1m2
d2

F2 = 4F1

Then, the new gravitation force (F2) between them will increase by 4 times. B
Thus, the gravitation force between two bodies A F1 m2

changes according to the product of their m1

masses. The gravitational force between two bodies is

directly proportional to the product of their masses. d
F ∝ m1m2 (keeping the distance constant)

10 Blooming Science Book 10

2. Force of gravitation between two bodies of same masses when kept at different distance.

a. If the distance between their centres is made half, keeping their masses same.

Consider the masses of two bodies ‘A’ and ‘B’ be ‘m1’ and ‘m2’ respectively. They are
placed at a distance 'd’ from their centres. Then the force of gravitation between them

will be F1.

F1 = Gm1m2 A F2 B
d2 m1 m2

If the distance between them is made half without

changing their masses, then the new gravitational force F2 d
will be increased by 4 times. 2



F2 = Gm1m2 = Gm1m2 =4 Gm1m2 = 4F1
d2
1 d 2 1 d 2
2 4

Thus, decreasing the distance between the two bodies by one half increase the force by
four times. Similarly decreasing the distance between them by one third, the force increases nine
times.

Relation between distance and gravitational force

Distance made Gravitational force increased by
1 4 times
9 times
2 16 times
25 times
1

3

1

4

1

5

b. If the distance between their centres is made double, keeping their masses same.

If the distance between two bodies is made double without changing their masses,
then the new gravitational force F2 will decrease by four times.

F2 = Gm1m2 = 1 Gm1m2 = 1 F1 F2 = 1 × F1

(2d)2 4 d2 4 4

A F2 B
m1 m2

2d

Blooming Science Book 10 11

Thus, increasing the distance between the two bodies by double decreases the force
by one fourth. The force of gravitation between two bodies decreases with the increase in the
distance between them.

Relation between distance and gravitational force

Distance increased by Gravitational force decreased by
2 4 times
3 9 times
4 16 times
5 25 times

Thus, the gravitational force between two bodies is inversely proportional to the square of the
distance between their centres.

F ∝ 1 (keeping the masses constant)

d2

Solved Numerical problems

Example 1: Calculate the force of gravitation between two bodies having masses 20 kg and 30
kg which are kept 20 m apart.

Solution

Given,

Mass of first body (m1) = 20 Kg
Mass of second body (m2) = 30 Kg
Distance between them (d) = 20 m

Gravitational constant (G) = 6.67 × 10-11 Nm2/Kg2

Force of gravitation (F) = ?

We know that,

Gm1m2
F= d2

= 6.67 × 10-11 ×20 ×30

(20)2
6.67 × 10-11 ×20 ×30
=
400

= 10.005 ×10-11 N

Therefore, the force of gravitation between them is 10.005×10-11N.

Example 2:

The mass of the earth is 6 ×1024Kg and its radius is 6300K. An object of mass 300Kg is kept on

its surface. Calculate the gravitational force between the earth and the object.

12 Blooming Science Book 10

Solution,

Given,

Mass of the earth (m) = 6×1024 Kg

Mass of object (m)= 300 Kg

Radius of the earth (R)= 6380 Km = 6380 ×103m

= 6.38 × 106 m

Gravitation force (F) = ?

We know that,

G Mm

F = R2 ( d = R, as radius of object neglisible compared to earth)

= 6.67×10-11×6×1024×300

(6.38×106)2



6.67×6×3×10-11+24+2
=

6.38×6.38×106+6
= 2949.5N

∴ Gravitational force between the earth and given object is 2949.5 N.

Gravity

In Newton’s law of gravitation, we talked about the force of attraction between two bodies. These
bodies could be any. Now, out of these bodies, if one of the body is the earth (or a planet), the
gravitational force is called force of gravity. Thus, the force of gravity is the special case of the
force of gravitation.

When a stone is released from the top of a tower, it falls towards the ground. Similarly, if a
body is thrown upward, its speed slows down. The speed becomes zero when the body reaches
a certain height. After the maximum height, the body begins falling down. Finally, the body
reaches (returns) to the ground. If a stone is released at the top of a deep well, it goes into the
earth. This implies that the earth attracts every body lying near its surface and on its surface
towards its center. The force of attraction of the earth on the bodies which lie on its surface (or
near its surface) is called gravity or force of gravity.

Not only the earth but also all the planets or other heavenly bodies have gravity. The gravity of a
heavenly body depends upon its mass and radius of the planets. The gravity of the moon is about
1/6 time the gravity of the earth.
Effects of Gravity
1. We can walk on earth due to gravity
2. Atmosphere present around the earth is due to gravity.
3. Construction of infrastructure like bridges is possible due to gravity.
4. Tides in seas and oceans are caused due to gravity of earth.

Blooming Science Book 10 13

Acceleration due to Gravity

When a body is released from a height, it falls towards the earth. When m
the body falls down, its velocity goes on increasing. This means that small body
there is a change in velocity every time. The change in velocity per falling freely
second is the acceleration. Thus, when a body falls freely, it has an
acceleration. The acceleration produced in a freely falling body due to r
gravity of earth (heavenly body) is called acceleration due to gravity.
The acceleration due to gravity is uniform. It acts in the direction of line M
joining the body to the center of the earth. It is denoted by 'g'.
R

Expression for 'g': Consider a small body of mass 'm' falling freely from Earth
a small height. Let a body is at a distance 'r' from the center of the earth as
shown in figure. If M is the mass of the earth and G is the universal gravitational constant, the force
of gravity on the body due to the earth is given by

GMm
F = r2 ........................................ (1)

G Mm
mg = r2 ( F = mg)

GM
\ g = r2 .............................................. (2)
This is acceleration due to gravity at a point near earth’s surface which is at a distance r from the
center of the earth.

On the surface of the earth r = R

GM
\g = R2

1
\g ∝ R2 , (where G and M are constant.)
Value of g near Earth surface: Near the earth surface, the distance (r) of the body from the
centre of the earth can be taken as the radius (R) of the earth, i.e. r = R.

We have,

(G) = 6.67 × 10-11 Nm2kg-2

Mass of the earth (M) = 6.0 × 1024 kg.

Radius of the earth (R) = 6.4 × 106m.

∴ The value of g on the earth surface is

GM 6.67 × 10-11 × 6.0 × 1024
g = R2 = (6.4 × 106)2
= 9.8 m/s2

It means that velocity of any falling body on the earth increases by 9.8m/s in every second.

The SI unit of acceleration due to gravity is meter per second square (m/sec2).

Also, F = mg

g = F
m

14 Blooming Science Book 10

The force is measured in Newton (N) and mass is measured in kg. Hence, an alternative unit of 'g'
is N/kg. Acceleration due to gravity of any body depends on (i) mass of heavenly body (ii) radius of
heavenly body (iii) height above the heavenly body.

Activity -1

To determine acceleration due to gravity of the earth and show that it is independent of the mass

of the failing body.

Materials needed:

Ball bearing or stone of various size (10g-100gm), metal sheet, digital or spring

balance, stopwatch, measuring tape, etc.

Method

1. Choose at least two storey building for the experiment.

2. Keep the metal square sheet on the floor.

3. Measure the height of the building and mass of each body.

4. Let the mass to fall one at a time and record the time interval between it starts to fall

and hits the metal sheet.

5. Repeat this process for remaining masses.

Calculation of ‘g’

When a body falls from certain height,

h = ut + 1 gt2 h= 1 gt2
2 2
1 2h
h = O×t + 2 gt2 ∴ g = t2

Observation:

S.N. Height of building (m) Mass of falling body Time interval ( 2h Mean 'g'
(g) g = t2
Sec.)

Result:
The average acceleration due to gravity is ………………..

Conclusion:

The value of ‘g’ for different masses is same or close to each other.
The mass of Jupiter is 319 times of earth but its acceleration due to gravity is only 2.5 times of
earth because Jupiter has radius 11 times of the earth.

Blooming Science Book 10 15

Solved Numerical Problems

Example 3: The mass of the Jupiter is 1.9 × 1027 Kg and its radius is 71×106m. Calculate the value

of acceleration due to gravity on its surface.

Solution

Given,

Mass of Jupiter (m) = 1.9 × 1027 Kg.

Radius of Jupiter (R) = 71 × 106 m

Gravitation constant (G) = 6.67 × 10-11 Nm2/Kg2

Acceleration due to gravity (g) = ?

∵ We know that,

GM
g = R 2

6.67 ×10-11 ×1.9 ×1027
=

(71 ×106)2

= 25.13 m / s2

∴ Acceleration due to gravity on the surface of Jupiter is 25.13 m/s2.

Example 4: Calculate the acceleration due to gravity on the surface of the moon. (mass of the
moon = 7.35×1022Kg, radius of the moon = 1737 Km)

Solution

Given,

Mass of the moon (M) = 7.35 ×1022 Kg

Radius of the moon (R) = 1737 Km

= 1737 ×103m = 1.737 × 106m

Acceleration due to gravity (g) =?
∵ we know that,

GM
g = R 2

6.67 ×10-11 ×7.35 ×1022 49.02 ×1011
= =

(1.737 ×106)2 3.017 ×1022

= 16.247×10-1

∴ g = 1.62 m/s2

The acceleration due to gravity on the surface of the moon is 1.62 m/s2. It is about 1 part of the
acceleration due to gravity produced by the earth.
2

16 Blooming Science Book 10

Variation of Acceleration Due to Gravity (g): Pole of
the earth
(a) Variation of g due to radius of the earth: We know Equator of
that the earth is not a perfect sphere. Its shape is like the earth Rp
an orange as shown in figure. It is therefore, the radius
of the earth varies on its surface. The value of g on the O Re
surface is given by

GM 1
g= R2 That is g ∝ R 2

This means the value of g at a place on the surface of the earth is inversely proportional to
the square of the distance of the place from the center of the earth.

Now, the radius of the earth at its equator (Re) is greater than the radius at the pole (Rp). That
is Re > Rp. Therefore, the value of g at the pole is greater than the value of g at the equator.
That is gp > ge. The g is maximum at the pole and minimum at the equator.

i.e. gp = 9.83 m/s2 and ge = 9.78 m/s2.

(b) Variation of g due to height (altitude): Let a mass m is placed at a point P which is at a

height h from the earth’s surface as shown in figure. The acceleration due to gravity (g’) at

the point is given by the relation. Pm
R2 h

g’ = g oR+hp ................................ (1)

GM
Proof: We know, on the surface of the earth, g = R 2 .......(i)
GM M R
and at height (h) from the surface of the earth, g’ = (R + h) 2 ....(ii) O

R2
Dividing equation (ii) and (i) we get, g’ = g oR+hp
Earth

The value of g decreases gradually when we approach above the earth from its surfaces.
This also explains why the value of g at the top of Mount Everest is less than the value of

g at sea level (or at a place in Terai).

(c) Variation with depth: As we go deep inside the earth, the value of acceleration due to
gravity decreases. It is because, the mass of the earth goes on decreasing as we go inside
R–d
the earth. At certain depth ‘d’, acceleration due to gravity is given by: g’ = R ×g

The value of ‘g’ at the center of earth is zero so the weight of the body is zero.

Galileo’s Idea of Falling Bodies

Aristotle’s idea of falling bodies continued until sixteenth century. In 1590 AD, Galileo, a
great scientist of that time, challenged the idea of Aristotle. Galileo argued that all bodies when
dropped from the same height should reach the ground together. To prove his idea, he performed
experiments at the Leaning Tower of Pisa. A large crowd of people was present at the Leaning
Tower to see Galileo’s experiment. In front of the gathering, Galileo took bodies of different

Blooming Science Book 10 17

masses and dropped them from the top of the tower. All the bodies irrespective of their masses
reached the ground at the same time. However, when a very heavy body (a stone) and a very
light body (a paper) were dropped simultaneously from the tower, certainly the stone reached
the ground earlier than the paper. Galileo explained that the difference in time is due to the
air resistance. The air resistance on the paper was greater than the air resistance on the stone.
Therefore, the paper fell slower than the stone. Galileo further predicted that if the stone and the
paper are dropped from the same height in vacuum, they reach the ground together.

Guinea (coin) and Feather Experiment (Robert Boyle Experiment)

Robert Boyle performed an

experiment to verify Galileo’s

idea that a light body and a

heavy body fall together in Feather Air
vacuum. He took a guinea (a Vacuum

coin of England) and a feather.

He put them in a long glass tube Feather Coin Coin
fitted with a vacuum pump as

shown in figure. Initially, the

coin and the feather were at Feather Coin
the bottom of the tube. The air

inside the tube was removed

with the help of the vacuum

pump. A vacuum was created

in the tube. The tube was (i) (ii) (iii)
inverted quickly. In doing so,

the coin and the feather were

found to be falling together as in figure no (ii). Finally, they reached the other end of the tube

simultaneously.

The tube was reinverted so that the coin and the feather were on the bottom. Now, air was allowed
to enter into the tube. With the air inside, the tube was again inverted quickly. In doing so, it was
observed that the coin fell much faster than the feather as in figure no (iii).

From these experiments, it is concluded that if there is no air resistance, all bodies, irrespective of
their masses, fall with the same acceleration. The acceleration of the body at that instant is equal
to acceleration due to gravity of that place.

Weight

The force of attraction of the earth on a body is called force of gravity. It is also
called weight of a body. So, weight of a body can be defined as the force with
which it is attracted towards the centre of earth. Weight is measured by spring
balance and its unit is Newton in SI system. It is a vector quantity. It varies from
place to place. It is calculated by using formula,

W = mg ;where ‘g’ is acceleration due to gravity.

18 Blooming Science Book 10

Solved Numericals

Example 5: Calculate the acceleration due to gravity on the surface of the earth. And what is
the weight of an object having mass 40 Kg on its surface. (mass of the earth is 6 × 1024 Kg and
redius of the earth is 6380 Km.)

Solution

Given,

Mass of the earth = 6 × 1024Kg

Radius of the earth (R) = 6380 Km
= 6380×103m
Mass of object (m) = 40Kg = 6.38 ×106m

Acceleration due to gravity (g) = ?

Weight of object (w) = ?

We know that,

GM
g = R 2

6.67 ×10-11 ×6.35 ×1024
=

(6.38 ×106)2

40.02
= 40.7 × 101

= 0.98×101

= 9.8 m / s2

∴g = 9.0 m / s2

Again, W = mg

= 40 ×9.8 = 392 N

Here, acceleration due to gravity on the surface of the earth is 9.8m/s2 and weight of the object
is 392 N.

Example 6 : If a man can lift 80kg on earth, how much could he lift on the moon surface?

(gmoon = 1.67m/s2).

On the earth, mass (m) = 80kg

The acceleration due to gravity (g) = 9.8 m/s2

Weight (w) = m×g

= 80 × 9.8

= 784N

On moon surface mass (m) =?

The acceleration due to gravity (g) = 1.6 m/s2

Blooming Science Book 10 19

Weight (w) = m×g
or, 784 N
= m × 1.67 [It is because, the weight he can lift on
earth, can be lifted on moon]

Therefore, mass on moon (m) = 469.4 kg

Free Fall

If a body moves under the influence of the gravity alone without air resistance, the motion is
called free fall. In fact, a body falling in vacuum is the free fall. The best example of free fall
could be a body falling onto the surface of the moon. It is, because, the moon has no atmosphere
around it. The weight of body during free fall is zero or the body becomes weightless during free
fall. The falling of parachute on the moon is a free fall but if a person jump with parachute on
earth he can land safely and it is not the free fall. In the earth the velocity of parachute is affected
air resistance so parachute falls slowly at uniform velocity. When weight of parachute is equal to
upthrust given by air at that condition parachute has zero acceleration.

If a body is falling freely under the action of gravity alone neglecting air resistance, the motion
is called free fall. The acceleration of a body in free fall is equal to acceleration due to gravity
‘g’ of the place.

Apparent Weight and Weightlessness

Apparent Weight

The weight of a body is the force with which the earth attracts the body. Hence, the actual weight
of a man is the force exerted by the earth on him.

We measure our weight by standing on a weighing machine. When we stand on the weighing
machine, we exert a force on it. The force we exert on the machine is equal to the force exerted
by the earth on us. Hence, if we stand on a weighing machine which is resting on the ground, the
weighing machine shows reading which is equal to our weight.

When a man is standing on a weighing machine, he exerts a force equal to his weight (mg) on
the weighing machine. Now, according to Newton’s third law, the weighing machine also exerts
an equal and opposite force called reaction on the man. Thus, in fact, the weight of the man read
by the weighing machine is the reaction of the machine on the man. The reaction of the weighing
machine on the man who is standing on it is equal to the actual weight (the force of earth’s
attraction on the man) of the man if the weighing machine is not moving up or down.

However, if the weight of machine on which the man is standing is moving (up or down), the
reaction of the machine on the man may not be equal to the actual weight (mg) of the man. In
such a situation, the weighing machine gives reading which may not be equal to the mg.

Apparent Weight: The weight of the man measured by a moving weighing machine is called the
apparent weight of the man. In other words, the apparent weight of a body is the reaction of a
moving weighing machine on the body.

Apparent Weight in a Lift: Let a man is standing on a weighing machine which is in a moving
lift as shown in figure. In the figure, W = mg is the actual weight of the man and 'R' is the

reaction of the weighing machine (or the lift) on the man. Then

20 Blooming Science Book 10

R R R

v = constant a
v = 0 and
a=0

a

mg mg mg

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

i. his actual weight mg acting downward.
ii. the reaction (R) on him acting upward.
If 'a' is the net acceleration of the man, then we have, m x a = Net force
Case I: The lift Moving with Uniform Speed
If the lift is moving uniformly with a speed either up or down, the acceleration a = 0. Hence,
R = mg = m × 0 = 0
∴ R = mg ; This is also true in the case when the lift is at rest.
In this case, the apparent weight (R) = actual weight as in figure no (b).
Case II: The Lift Moving Upward with Acceleration (a)
If the lift is moving upward with a uniform acceleration (a), then we have
R - mg = ma
∴ R = mg + ma = m (g + a)
Hence, in this case, the apparent weight (R) of the man is greater than its actual weight as
illustrated in figure no. (c).
Case III: The Lift Moving Downward with Acceleration (a)
If the lift is moving downward with acceleration a, then we have, mg - R = ma
∴ R = m (g - a)
Hence, in this case, the apparent weight of the man is less than the actual weight as in figure no. (d).

Weightlessness (i) (ii)

When a stone tightened with a string is hanged by the hook of a
spring balance which is hanged by a hand as in figure no (i). In this
case, the spring shows some reading which gives the actual weight
of the stone. If the spring balance is released from hand, the spring
balance as well as the stone falls freely. The spring balance shows
zero readings in figure no. (ii). In other words, the apparent weight
on the stone is Zero. This state of the stone is called weightlessness.

Thus, the weightlessness of a body is the state in which the body
feels that it is not being attracted by any force.

Blooming Science Book 10 21

Activity -2 400g
Aim: To observe weightlessness in a freely falling object. 0

Material needed: (i) (ii)

Spring balance, a piece of stone, thread, etc.

Method:
1. Take a spring balance and tie a stone with a piece of
thread to measure its weight.
2. Release the spring balance with stone from certain height
and observe the reading shown by spring balance.
Observation:
When the spring balance is falling freely along with
stone, it shows a ‘zero’ reading.
Conclusion:
When a body falls freely from certain height it falls
freely and shows the state of weightlessness. (There
should be no or negligible air resistance to the falling
body.)

Weightlessness in the lift: In case of the man moving in a lift, consider that the rope of the lift is
broken. Then, the lift also moves down with downward acceleration a = g. Hence, the apparent
weight of the man

R = m(g - a) = m (g - g) = 0.

Hence, a person in a freely falling lift feels weightlessness. It occurs when acceleration of a
falling body is equal to acceleration due to gravity.

Weightlessness in space: When the astronaut in the space ship is orbiting the earth, the
astronaut and space ship are in the continuous state of free fall towards the earth. That is they
fall downward with the same acceleration (g). Since the astronaut and the space ship both
fall with the same acceleration, the astronaut exerts no force on the space ship. Since the
astronaut exerts no force on the ship, there is no reaction on the astronaut. Hence, the astronaut
feels as if he/she does not have weight. Any weighting machine used to measure the weight of
the astronaut in one space ship shows zero reading. It occurs when acceleration due to gravity
is zero.

True Weightlessness/Condition for Weightlessness

We know that the weight of a body is given by mg. The mass of the body cannot be zero. Thus, a
body will be truly weightless only if g = 0. In the outer space, the value of g = 0. Hence, a body
feels truly weightless if it is in the outer space. This is called true weightlessness. During free fall
and at the centre of earth also the body will be weightless.

22 Blooming Science Book 10

Points to Remember

1. The mutual force of attraction between any two bodies is called force of gravitation or
simply gravitation.

2. The mass of the sun is bigger than all the remaining masses in the solar system so all
planets and moons revolve round the sun.

3. The gravitational force between two masses depends on the product of their masses and
the square of the distance between them.

4. The gravitational force exerted between two bodies is directly proportional to the product
of their masses, and inversely proportional to the square of their central distance.

5. Newton’s universal law of gravitation states that “the gravitational force produced
between two bodies is directly proportional to the product of their masses and inversely
proportional to the square of the distance between their centres”.

6. Gravitational constant is the force of gravitation between two bodies each with a mass of
1 kg placed at a distance of 1 meter apart.

7. The gravitational force between a planet or satellite and a body on or near its surface is
called the gravity of the planet or satellite.

8. Gravity of a planet is directly proportional to the mass (M) of the planet and when the
mass of the planet is constant and its radius is large, the gravity will be low and if the
radius is short, the gravity will inverse.

9. Gravitational field is defined as an area around a planet up to where the gravity of the
planet has its influence on an object.

10. Gravitational field intensity of a planet or a satellite is defined as the gravity that applies to
an object of unit mass present on the surface of the planet or satellite. Its SI unit is N/kg.

11. As the value of R is greater at equator, the value of g is the least at equator.

12. As the value of R is less at pole, the value of g is the greatest at the pole.

13. Due to gravity we are staying on this spherical earth atmosphere exists, river flows, wind
blows and there is change in the weather.

14. The acceleration produced on a body due to the force of gravity is called acceleration due
to gravity. Its unit is meter per second per second (m/s2).

15. Acceleration due to gravity is inversely proportional to the square of the radius of earth.
So, g is greater at the poles than at equator.

16. Value of g and E are numerically equal at the same place.

17. In every second, the velocity increases by 9.8m/s i.e. the body is accelerating down
towards the centre of the earth at the rate of 9.8m/s2.

Blooming Science Book 10 23

18. The falling of an object without any external air resistance is called free fall. An object
becomes weightless during free fall.

19. A state of apparent loss of weight is called weightlessness.

Let's Learn

1. It is easy to lift small stone than the big stone on the earth.

It is because mass of a small stone is less so the weigh of the small stone is less but a big
stone has more mass so it has more weight as w∝m.

2. The gravitational force between ordinary sized bodies is not noticeable.

The gravitational force between ordinary sized bodies is very small. For example, the
force of attraction between two 1 kg, masses placed at 1m apart is 6.67 × 10-11N. Such
a small force cannot be measured ordinarily. Therefore, the gravitational force between
ordinary sized bodies is not noticeable.

3. The value of g varies from place to place on the surface of the earth.
On the surface of the earth, g ∝ 1 where R is the radius of earth. The earth is not a

R2
perfect sphere. That is, different points on the earth’s surface are at different distance from
the center of the earth. That is value of R vary from place to place on the earth surface
Therefore, the value of g vary from place to place on the earth surface.

4. The value of g is greater at the poles then at the equator.

The value of g at a place on the earth is inversely proportional to the square of the radius
of the earth at the place. The radius of the earth at the poles is less than that at the equator.
Therefore, the value of g at the poles is greater than the value of g at the equator.

5. An iron ball and a feather are dropped simultaneously from the same height in a
vacuum both will strike the ground together.

The acceleration produced by the earth’s attraction on all bodies is the same irrespective of
their masses if there is no air resistance. Therefore, if an iron ball and a feather are dropped
simultaneously in vacuum, their speeds increase with the same rate. As a result, they reach
the ground together.

6. The probability of getting hurt increases when a person jump from great height.

It is because the velocity of falling body increases while falling to the earth from the
greater height. Since f∝v, the force of impact on falling body increases during collision on
the ground and gets more hurt.

24 Blooming Science Book 10

Project Work
 Make a model of parachute by using a plastic sheet or wind proof cloth, thread and
small plastic basket or can.
Method
Take a plastic sheet or wind proof cloth and cut it into a circular shape with radius
20cm. Make few small holes near the edge of the circle at equal distance. Tie a
thread at each hole in such a way that the length of each thread is equal. Collect all
the free ends of the thread and tie a small basket with load or a can with load. Now,
parachute is ready. Now, drop it from a certain height and observe its falling.

Plastic sheet

Plastic basket/can with

Exercise
1. Give short answer to the following questions.

a) What is force of gravitation? What are the two main factors that affect the force of
gravitation?

b) Why is acceleration due to gravity low in the space?
Gm1 m2

c) State Newton’s law of gravitation and show F = d2
d) In what condition is there weightlessness?
e) What is free fall?
f) At what condition the acceleration of falling parachute zero?
g) What is acceleration due to gravity? Write the factors which affect it.
h) Prove g ∝ 1 on the surface of the earth.

R2
2. What is the relationship between the weight of an object and the radius of the earth?
3. What mass of a body can a weight lifter lift on the Jupiter if he can lift 100 kg mass on the earth?

Blooming Science Book 10 25

4. What change in gravitational force is seen in the following conditions? Show with
calculation.
a) Masses are doubled and distance is constant
b) Masses are half and distance is constant
c) Masses are constant and distance is half
d) Masses are half and distance is double

5. Differentiate the following.
a) Gravity and gravitational force
b) Mass and weight
c) Acceleration due to gravity (g) and Gravitational constant (G)
d) Weightlessness in free fall and weightlessness in the space
d) Falling a parachute on earth and on the moon

6. Give reason:
a) The weight of an object is more at poles than at equator.
b) The parachutists are not hurt when they jump out of an aeroplane.
c) Miners feel less weight inside mine.
d) The effect of gravitation force is more in liquid than in solid.

e) A coin falls faster than feather on earth.

7. Numerical Problems

a. Calculate the gravitational force between two masses each of 30 kg separated by 2 meter
distance. Given G = 6.67 × 10-11 Nm2/kg2.

Also calculate the acceleration produced by each on the other. [1.5 × 10-8N, 5 × 10-10 ms-2]

b. The mass of the earth is 6 × 1024 kg. The mass of the moon is 7.4 × 1022 kg and the distance
between their centers is 3.84 × 105 km. Calculate the force exerted between them [2.008
× 1021N]

c. At what condition is the force of gravitation equal to 6.67 × 10–11N? Calculate.

d. Calculate the value of g on the surface of the earth. Given that mass of the earth = 6 ×
1024kg, radius of the earth = 6.4 × 106 meter and G = 6.67 × 10-11 Nm2/kg2. [9.8 ms-2]

e. The height of Mt. Everest is 8848m from sea level. Find acceleration due to gravity at its
height, if acceleration due to gravity at earth’s surface is 9.8m/s2 and radius of earth is 6.4
× 106m. Also find weight of person whose mass is 80kg at that height. [9.77m/s2, 781.6N]

f. A heavenly body has mass equal to half of the mass of the earth and its radius is half as

that of earth. If a stone weigh 100N on the surface of earth, find the weight of stone on the

heavenly body. [200N)

26 Blooming Science Book 10

g. A man of mass 75 kg is standing on the ground. What is the force of gravity on him? What

is his weight ? ( Given R= 6400km, M = 6×1024kg) [735 N, 735 N]

h. Two spheres of equal masses are separated by a distance of 1 km. The gravitational force
between them is 100 N. What is their mass?

(G = 6.67 × 10-11 Nm2kg-2). [1.22 × 109 kg]

i. What will be the acceleration due to gravity on a heavenly body if its mass and radius

are 4 times of the earth. [2.45 m/s2]

j. If earth is squeezed to the size of moon, calculate the weight of 100kg mass on the
squeezed earth. [Rmoon = 1.75×105m] [13840N]

8. Choose the correct alternative from the following options.

i. The SI unit of ‘G’ is?

a. Nm/Kg b. Nm2/Kg2 c. Nm-2 Kg2 d. NKg2/m2

ii. The value of ‘g’ at centre of earth is:

a. 9.8m/s2 b. 0m/s2 c. 10m/s2 d. 1.67m/s2

iii. How is gravitational force related with product of masses?

a. remains same b. directly proportional

c. Inversely proportional d. None of the above

iv. If a person can jump 1 m on earth, he jumps …….. on moon.

a. 2m b. 1/6 m c. 6 m d. 1 m

v. What is the acceleration due to gravity on the moon?

a. 10m/s2 b. 1.67m/s2 c. 0m/s2 d. All of them

Blooming Science Book 10 27

Chapter Pressure

2 Blaise Pascal
1623 AD-1662 AD

Learning Outcomes Estimated Periods: 8+2

On the completion of this unit, the students will be able to:
 identify liquid pressure.
 prove the law of liquid pressure (Pascal’s law).
 describe Archimedes Principle and its application in daily life.
 demonstrate law of floatation.
 introduce atmospheric pressure and describe its uses.
 solve simple numerical related to pressure.

Introduction

Pressure is a derived quantity which plays wide role in the study of hydrostatics. Pressure is a
quantity which tells how effective a force is. It is defined as the force acting normally in per unit
area.

It is calculated by using the formula; Scan for practical experiment

Pressure = Force
Area

P = F

A

The SI unit of pressure is N/m2 which is also called as Pascal (Pa). visit: csp.codes/c10e02

Liquid Pressure

Liquids do not have fixed shape but have fixed weight. The liquids exert pressure due to their
weight. When a liquid is kept in a container, it experts pressure in all directions.

The pressure exerted by a liquid is called liquid pressure. It is calculated as P = dgh; where ‘d’ is
density of liquid, ‘g’ is acceleration due to gravity and ‘h’ is height of liquid column. The SI unit of
liquid pressure is Pascal (pa)

Properties of liquid pressure
1. The pressure in liquid is directly proportional to the depth from the free surface of liquid

or the height of liquid column.
2. The pressure at the same depth, in a liquid is the same in all directions.
3. The pressure of liquid is independent of the shape of container in which it is kept.
4. The pressure at any point in a liquid depends on its density.

5. A liquid finds its own level.

28 Blooming Science Book 10

Pascal’s law

Blaise Pascal was born in 1623 AD in France. He was a French scientist, famous mathematician
and also a philosopher. He predicted a law on how pressure applied to an enclosed fluid transmits
in 1650AD. This law later became famous from his name as Pascal’s law. Liquid cannot be
compressed, so pressure is equally exerted on all sides.

Pascal’s law states that ‘‘the pressure exerted at any point in an enclosed liquid is transmitted
equally in all directions throughout the liquid ’’.

Verification of Pascal's Law

Activity -1
To demonstrate that liquid transmits pressure equally in all directions
Material required: a plastic bag, needle, water
Method:
a. Take a plastic bag filled with water.
b. Make several pores covering all area in it.
c. Squeeze the bag gently and observe the water coming out
from these pores.
Observation: Water comes out with equal force from all pores.
Conclusion: Liquid exerts pressure equally in all directions.

Activity -2

To demonstrate that liquid transmits pressure equally in all directions

Materials required: a spherical container fitted with four pistons of equal cross sectional

area and water.

Method: F2
B
a. Take a spherical container fitted with four piston
A, B, C and D of equal cross sectional area.

b. Fill the container with water.

c. Push the piston A inward with a force and observe F1 A C
other pistons. F3

Observation:

All other pistons are pushed outwards when the piston D
'A' is pushed inward. The outward distance moved by F4
all the pistons will be equal to each other as well as to
the inward distance moved by piston 'A'.

Conclusion:

Pressure given to the water of the container through piston 'A' is transmitted equally in all
pistons. It means liquid transmits pressure equally in all directions.

Blooming Science Book 10 29

Applications of Pascal’s Law

Pascal’s law is used to produce a large force by using a small force. Hydraulic machines like
hydraulic press, hydraulic brake, hydraulic jack, etc work in this fact.

Hydraulic Press F1 F2

Hydraulic press is a device based on A B water
Pascal’s law and it is used for pressing or
compressing purpose.

Principle

a. Liquids are incompressible.

b. Liquid transmits pressure equally in Fig: Hydraulic press
all directions.

Construction and working

It consists of two cylinders of different areas of

cross section fitted with pistons and connected by F1 F2
horizontal pipe as shown in figure. The space under B

the pistons and the pipe is filled with liquid, the A A2
column of the liquid remains same if no force acts A1 PB

on the pistons. PA

Consider, cross-sectional area of piston A is A1 and water
that of B is A2. Here, A1 < A2. A weight F1 is placed
on the piston A. It produces a pressure PA which is Fig: Hydraulic lift

given by, PA= F1 .
A1

This pressure PA is transmitted in all directions by the water. It acts on the piston B from
downward. Let a load F2 be kept on the piston B which makes the pistons in equilibrium, and the
upward pressure on the piston B F2
is PB. Now, PB = A2 .

According to Pascal’s law, PA= PB

or F1 = F2
A1 A2

∴ F2 = A2 × F1
A1

Since A2 > A1, F2 > F1 and hence the applied force is multiplied in the system. F2 is as many times
larger than F1 as A2 is larger than A1. Therefore, hydraulic machine is a force multiplier.

30 Blooming Science Book 10

Uses of Hydraulic Press
Hydraulic press is used for the following purposes.
1. to lift heavy loads.
2. to extract oil from cotton seeds by crushing.

3. to press the bales of cotton, book, etc.

Solved Problems

Example 1 : In a hydraulic press, the area of small piston is 0.001m2 and that of large piston is
0.5m2 calculate the weight needed on the larger piston to balance the load of 15N on the small
piston.

Here, Force generated on smaller piston (F1) = 15N

Area of small piston (A1) = 0.001m2

Area of large piston (A2) = 0.5m2 F1 F2

Force generated on lager piston (F2) = ? A B water

We have,

F1 = F2
A1 A2

F1 × A2 Fig: Hydraulic press
or, F2 = A1

= 15×0.5 = 7500 N
0.001

The weight lifted on the larger piston is 7500 N.

Hydraulic Car Brake Cylinder Pivot

Hydraulic car brakes are the brakes which work on the Master Brake pedal
principle of Pascal’s law and are used in car, truck,
plane, etc. Its advantage is that the equal pressure is Disc brake
applied to each wheel. So, all the wheels stop at a time
and a large force can be generated with a small effort. Fig: Car brake
A hydraulic brake is a breaking system that is used in
heavy automobiles.

Blooming Science Book 10 31

Hydraulic Garage Lift

Hydraulic Garage lift is a device used to work on the basis of Pascal’s law. It is used to lift
light vehicles in service stations during their services.

Lever Vehicle
Platform
Reservoir Smaller
X cylinder Big piston
2
Small
piston Big cylinder

valve valve
A B

Stopper
1

Fig: Hydraulic garage lift

Upthrust

While lifting a bucketful of water from a well, we feel it lighter if it is under water but it is felt
heavy if it comes out of the water surface. While swimming, we feel as if water is pushing us
up. It is because water pushes the things up with certain force. This shows that there must be a
force acting on a body immersed in water in the upward direction. Since this force acts against
the gravity, it reduces the weight of an object immersed in water. This force is called upthrust.

The upward force exerted by a liquid on an object immersed in liquid is called the upthrust or
buoyant force. It is the vector quantity and is measured in newton in SI system. Upthrust acting
on a body immersed on a fluid depends on:

• Volume of immersed part ‘V’

• density ‘d’ of liquid and

• value of acceleration due to gravity ‘g’

∴ Upthrust =Vdg

At a particular place, value of g is constant and volume of given liquid can be known so the
upthrust is directly proportional to the density. Upthrust of liquid does not depend upon the shape
of vessel. i.e. upthrust ∝ d

The upthrust given to a body is directly proportional to the density of liquid in which the body
is immersed.

Density of Liquid and Upthrust

The density of liquid directly influences upthrust. As density of liquid increases the pressure
given by the liquid, the upthrust is also affected by the density of the liquid. Its verification is
done by the Archimedes’ principle, as upthrust has very close relationship with the principle. The
following activity will help to understand the relation between density of liquid and upthrust.

32 Blooming Science Book 10

Measurement of Upthrust

Activity -3
To measure upthrust of water

Materials required: Spring balance
a beaker, spring balance, water, thread, a piece of stone etc.

Method:

a. Take a beaker and fill it with water. Water
Stone
b. Tie a stone on a string and suspend it at the spring
balance. Beaker

c. Measure the weight of stone in air.

d. Now, immerse stone gradually inside water contained in Scan for practical experiment
beaker and observe the
weight of stone in spring balance.

Observation:

Let,

Weight of stone in air = 3N

Weight of stone in liquid = 2N visit: csp.codes/c10e03

So, upthrust = weight of stone in air (W1)- weight of stone in liquid (W2)
= (3 - 2) N

=1 N

Conclusion:

The force with which liquid pushes up an object partially or wholly immersed in liquid is
called its upthrust. All liquids exert upthrust.

Activity -4
To show the relation of upthrust and density
Materials required
two beakers, an egg, salt and water
Method
a. Take two beakers and fill about half of both of them with water.
b. Dissolve enough salt in water of beaker 'B' to make saturated solution of salt in
water.
c. Put an agg in fresh water then the same egg in salt solution. observation
what happens?

Blooming Science Book 10 33

Floating egg
Beaker

Water
Salt Solution

Sinking egg

Beaker 'A' Beaker 'B'

Observation

The egg sinks in pure water ( Beaker A) and floats on saturated solution
(Beaker B).

Conclusion:

The density of salty water is more than the density of pure water. Salty water with more
density exerts more upthrust and the egg floats on it.

It is found that the egg sinks in water but floats in saturated salt solution. How is it possible? It
is due to the more density of salt solution, which applies more upthrust on the egg. Due to the
same reason a cargo loaded ship floats more in seas than in rivers. You will surprise that in dead
seas, which contains about 270 gram of salt in one litre of water, a person does not sink, due to
the great upthrust.

Archimedes’ Principle
Archimedes was a Greek scientist. He was born in 287 BC in Sicily. He discovered that an object
apparently loses some of its weight when immersed on water. He developed a principle that
describes the loss of weight in terms of upthrust.

Statement

Archimedes’ principle states that ‘‘when a body, wholly or partially, is immersed in a liquid(fluid),
it experiences an upthrust which is equal to the weight of the liquid (fluid) displaced”.

i.e. Upthrust = weight of displaced liquid

This principle is also applied for gases. A balloon filled with hydrogen gas is raised high in the atmosphere.

Verification of Archimedes’ Principle

Tie a stone to the hook of a spring balance. Take the weight (W1) of the stone in air.
Take an eureka can or overflow can and fill it with water up to the spout. Take a beaker and place
it on the pan balance just below the spout of the eureka can. Take its weight (W2).

34 Blooming Science Book 10

W1 (10N) W3(6N)
Stone

Eureka/ W2 Beaker (W4)
Overflow can

Wooden block Pan balance
Fig: Experimental verification of Archimeds' Principle


Now, slowly lower the stone into the water of the eureka can till it is completely immersed in
it. Take the new weight (W3) of the stone immersed in water. The displaced water is overflowed
through the spout of the eureka can and gets collected in the weighted beaker. Take the weight of
the displaced water and beaker (W4)

Calculation,

Loss in weight of a stone in water = upthrust

= W1 - W3
Weight of water displaced by a stone = Weight of beaker and water - weight of a beaker

= W4 - W2
∴ The loss in weight of a stone (W1 - W3) = Weight of water displaced by a stone (W4 - W2)

Upthrust = Weight of liquid displaced

Thus, it is found that the upthrust is equal to the weight of water displaced by it. This verifies the
Archimedes’ principle.

In the above figure, suppose the stone weighs 10 N in air and 6 N when completely immersed in
water. The weight of the water collected in a beaker is also 4 N.

Weight of stone in air (W1) = 10 N

Weight of stone in water (W3) =6N

Weight of beaker (W2) =1N

Weight of beaker and displaced water (W4) =5 N

The apparent loss in weight of a stone upthrust =W1 - W3 =10N - 6N = 4N

Weight of water displaced = W4 - W2 = 5N - 1N = 4N

Hence, upthrust =Weight of displaced water

Blooming Science Book 10 35

Thus, Archimedes principle is verified. Due to the upthrust of liquid, any object immersed in it
loses some of its weight.

Uses of Archimedes Principle

1. The volume and density of a solid can be determined.

2. Volume of cavity in body can be determined.

3. Impurity in a given material can be determined.

Law of Floatation

A body immersed in a liquid experiences two types of forces:

a. The force of gravity (i.e. weight) directed vertically downward.

b. The upthrust (i.e. buoyant force) directed vertically upward.



[W = weight, U = upthrust] freely floats W<U
(c)

Just floats (b)
W=U

Sinking W>U
(a)



Under the action of these two forces, a body will move in the direction of greater force. Three
cases are possible if an object is immersed in liquid.

i. If weight of an object is greater than upthrust, the body sinks to the bottom as shown in
above figure (a).

ii. If weight of an object is equal to the upthrust, the body can be in equilibrium condition at
any place in liquid as shown in figure (b).

iii. If weight of an object is less than upthrust body rises to the surface and floats on the
surface of the liquid partly submerged as in figure (c)

Law of floatation states that a body floats in a liquid if it displaces the liquid of its own weight.

Weight of floating object = weight of liquid displaced

The bodies having relatively more weight require more upthrust to float in a liquid. Since the
upthrust is directly proportional to the volume of the immersed part of the body, the heavier body
submerges more by displacing more liquid.

The density of iceberg is less than the density of water. The density of ice is 0.917 g/cm3 and that
of water is 1 g/cm3. Therefore huge mass of ice known as iceberg floats in water.

36 Blooming Science Book 10

Activity -5
Verification of Principle of Floatation

1. Take an eureka can and fill it with water up to its spout and place it on the top pan
balance. The pointer in the top of pan balance indicates the weight of the eureka can and
water in it.

2. Place an empty beaker below the spout of the eureka can and take its weight.

3. Take an empty test tube and put lead shots in it and take its weight. (For this test-tube
with rim should be taken. Tie it with string to hang on the hook of the spring balance).

4. Immerse test tube with lead shots Test tube Beaker
into the eureka can containing Lead shots Water
water. Measure the weight of beaker
containing displaced water.

5. Add more lead shots in the tube and
note its weight.

6. Again fill the eureka can with water up to its spout and immerse the test tube with
more lead shots in it. Note the weight of displaced water.

7. Repeat this activity by adding more lead shots and fill the recorded weight of
displaced water in the table given below.

Observation

Weight of test tube and lead Weight of displaced water (= weight of
shots in gm beaker and displaced water – weight of

1. beaker) gm
2.
3.
4.

Result and Conclusion

It is found that test tube with lead shots when float, the weight of the water displaced by
it always equals to its weight. With the increase of weight of the test tube on adding
more lead shots, the weight of displaced liquid also increases. The test tube with lead shots
sinks only when its weight exceeds the weight of the displaced water. In other words when
the weight of displaced water (up-thrust) is less than the test tube with lead shots, it sinks.

Thus, weight of liquid displaced = weight of floating body. Hence, the law of floatation is
verified

Blooming Science Book 10 37

Application of Principle of Floatation

1) Floatation of Iron Ship

A nail of iron dropped in a trough of water sinks. But an empty bowl made of steel floats in the
trough of water. The nail is compact and not hollow. Therefore, it displaces less weight of water
than its own weight and hence sinks in water. However, bowl is hallow within. Thus it has fairly
large volume. This reduces its apparent density to such an extent that it becomes less than the
density of water. Thus, it is able to displace more weight of water than its own weight and hence
floats on the surface of water.

Ships are streamlined and are hollow inside. This reduces its apparent density to such an extent
that it becomes less than that of water. Thus it is able to displace more weight of water than its
own weight and hence floats on the surface of water.

2) Floatation of Balloons

Children commonly use hydrogen filled balloons for playing. These ballons float upward in air.
Because the weight of balloon and fabric is less than the weight of air displaced by it. Thus a net
upthrust act on the balloon and hence it rises upward.

Similarly, hot air balloons are used for sport. The hot air is produced by burning kerosene oil in a
high pressure stove and is filled in giant balloons. The density of hot air is 0.40 kg/m3. Whereas
the density of cold air is 1.30 kg/m3. The upthrust acting on the hot air balloons is far more
than its own weight. Thus, the balloon shoots up in air if released. Conversely the cold air filled
balloons does not rise up. Because the weight of air displaced by it is less than its own weight.

3) Floatation of man in sea water and freshwater

It is easier to swim in sea water than in fresh water because density of sea water is more (around
1.04 gm/cm3) and provides more upthrust as compared to fresh water.

Similarly, ships entering river mouth from sea sink more because of the same reason.

4) Floatation of icebergs

The density of sea water is 1.04 gm/cm3 and that of ice is 0.917 gm/cm3. Thus, when glaciers
push their way into sea, they break and float on the surface of seawater, as they displace more
weight of seawater than their own weight. It is found that iceberg floats in seawater with 11 parts
out of 12 parts of its volume below water and 1 part out of 12 parts of its volume above sea water.

Differences between Force (thrust) and Pressure

Force (thrust) Pressure

1. It is an external agency that changes or 1. Normal force per unit area acting on a
tends to change the state of motion or state surface is called pressure.
of rest of a body in a straight line.

2. Its SI unit is Newton (N). 2. Its SI unit is N/m2 or Pascal (Pa).

3. It is a vector quantity. 3. It is a scalar quantity.

4. It is measured by spring balance. 4. It is measured by manometer, barometer
and pressure gauge, etc.

38 Blooming Science Book 10

Solved Problems

Example 2 : What portion of the ice will remain above water when an ice block of dimensions
20 cm×30 cm×50 cm is put into water ?The density of ice is 0.9g/cm3.

Solution:

The volume of ice block (V) = (20×30×50) cm3

= 30,000 cm3

Density of ice =0.9 gm/cm3

According to the formula,

Mass = Density × volume = 0.9 × 30,000 = 27,000 gm

Weight of the water displaced by the immersed part of iceberg = total weight of the iceberg

= 27,000gm

Mass
∴ Total volume of water displaced = density

27000 gm
= 1gm/cm3

= 27,000 cm3

\ The volume of iceberg submerged = volume of displaced water.

The portion of the iceberg submerged in water = Volume of water displaced

Volume of iceberg

27000cm3 9
10
= 30000cm3 =

91
Thus, the portion of iceberg above the surface of water =1 - 10 = 10

In other words, if 20 cm is the height of an iceberg, the portion of the iceberg above the

1
surface of water = 10 × 20 cm = 2 cm
Example 3 : Density of a brick is 2.5g/cm3 and its mass is 1 kg. How much water does it displace
when it is immersed in water?

Solution:

Alternative method

Density of brick (d) =2.5 gm/cm3 Or d∝m in case of sinking or inside

Mass of brick (m) =1kg =1000g so, db = mb

Density of water =1gm/cm3 dw mwater

We know that, 2.5 = 1000
mass 1000
1 mw
Volume of brick = density = 2.5 = 400cm3
or, mw = 1000 400gm

2.5

Blooming Science Book 10 39

According to Archimedes’s principle,

Volume of displaced water = Volume of brick = 4000cm3

∴ A brick displaces the water of weight 4000 gm i.e. 4000 cm3 because the weight of
1cm3 water is 1gm

Atmospheric Pressure

The air that surrounds the earth is called atmosphere. It is extended about 9,600 km above the
earth surface. The air has weight. Thus, it exerts pressure on the surface of the earth. The pressure
exerted by the atmospheric air is called atmospheric pressure. At sea level, the air pressure is said
to be one atmosphere. This is also termed as standard atmospheric pressure or normal pressure.

The normal pressure is the pressure exerted by 760 mm long column of mercury at 0°C at sea
level. It is equal to 1.01 × 105N/m2.

In short, we can write as,

l atmosphere = 760mm of Hg = 1.01 × 105N/m2 or Pascal.

Barometer

The scientific instrument or device which is used to measure the atmospheric pressure is called
Barometer. Evangelista Torricelli was the first Italian scientist who invented Barometer in
1643 AD. There are different types of barometer such as mercury barometer, aneroid barometer,
etc.

Mercury Barometer

The barometer which contains mercury to measure atmospheric pressure is called mercury
Barometer. It consists of a one end closed glass tube of length 100 cm. with an open mercury
filled reservoir at the base.

Construction Torricellian
vacuum
The glass tube is fully filled with pure and dry mercury and inverted
over the bowl of mercury by putting the thumb on its open mouth so
that no air bubbles left inside the tube.

The weight of mercury inside the tube creates vacuum which is mercury
known as Toricellian vacuum.

The scale of the barometer is so adjusted that the pressure exerted by air is equivalent to the
pressure extorted by 76 cm of vertical height of mercury column.

Working

The rising and falling of mercury inside the glass tube indicates high and low atmospheric
pressure. The place having high atmospheric pressure exerts high force on the mercury of the
bowl which results to raise the mercury at the greater height inside the tube and can be termed
as high atmospheric pressure. Similarly, low atmospheric pressure of air exerts less force on the
bowl and allows the mercury to drop at lower level in the column.

40 Blooming Science Book 10

Factors affecting barometric height

1) Height from sea level
As the height from the sea level increases the force given by air column on the mercury

reservoir or bowl decreases so that there is fall on mercury column indicating lower
atmospheric pressure. Hence barometric height will be less.

2) Temperature of the atmosphere
The temperature and the density of the air are inversely varied with each other i.e. greater

the temperature lower will be the density of air. As the pressure of air directly depends
upon the density of air i.e. pressure decreases with decrease in density so atmospheric
pressure decreases hence the barometric height also decreases.

3) Moisture
If the moisture of water is left on the Toricellian vacuum, it will create pressure on the

mercury column and hence height of mercury column decreases.

Advantages of using mercury as a barometric liquid

1) Mercury doesn’t evaporate under normal temperature so that the reading given by
barometer is correct.

2) Since the density of mercury is higher than other liquids at the normal temperature it helps
to balance the atmospheric pressure.

3) The shining nature of mercury helps to locate the reading in barometer easily.

4) Unlike other liquids, it does not wet the walls of the tube.

But if we use water as a barometric liquid in the barometer there will be following disadvantages:

1) It evaporates at normal temperature due to which the vapor of water deposited inside
the Toricellian vacuum gives pressure on the water filled inside the tube and ultimately
barometer gives wrong measurement.

2) It wets the wall of the glass tube.
3) It is colourless liquid so for locating the reading, it will be quite difficult.
4) If water will be used as a barometric liquid, we need very long glass tube to measure

pressure which is not fruitful to make the barometer.

Some common devices that use air pressure

Syringe

The word syringe is derived from the Greek word ‘syrinx’ which means tube.

The scientific instrument which is used to inject or withdraw the fluid from any object of body is
called injection or syringe.

Blooming Science Book 10 41

Construction

It consists of nozzle of certain length provided with cylindrical transparent barrel made up of
plastic or glass. The nozzle contains a needle while barrel has a rubber piston which can be
moved to and fro inside it.

Working

When piston is pulled out, the volume of air inside the barrel Barrel

increases causing the low pressure inside it. As the needle Piston
Nozzle
of syringe is exposed to the fluid at higher pressure than

the barrel, hence fluid is forced in and reverse phenomenon

takes place as the piston is pressed in. Neddle

Water Pump

The pump is simply a device which is used to move the air or fluid from one point to another
point. In this topic we will discuss about working of water hand pump on the basis of laws of
mechanics, as there are so many other types of water pump which can be run by electricity.

The water hand pump is a device which is used for getting the underground water. It can be used
for irrigation, household purpose etc.

Working of water lifting pump

The water lifting pump is a Water outlet Force rod
type of positive displacement Piston rod
pump i.e. it collects the certain
amount of fluid by reducing Cylinder
pressure inside it and forcing Piston
the fluid to the discharge pipe Check valve
(Outlet).It consists of a piston Sealing O-valve
tightly fitted with a piston valve Check valve
into the cylindrical chamber.
Below the piston there is
another valve which is called
lower valve or foot valve. It
works with downstroke and
upstroke of the handle.

Downstroke

When pump rod is pulled up (handle is pressed down) the piston valve closes the piston and
pressure is decreased or vacuum is created below the piston. Due to the lower pressure or vacuum
the foot valve opens by the force of fluid (water) hence water is collected into the cylindrical
chamber.

Upstroke

Similarly, when pump rod is pressed down (handle is lifted up), due to the incompressible nature
of the water, upper valve or piston valve opens while foot valve closes the mouth of rising water.
As a result, water flows out through the outlet.

42 Blooming Science Book 10

Air Pump

It is also called positive displacement pump because it traps the fixed amount of gas and there
forcing that trapped volume into the discharge tube.

It is a device which is used to pump the air into tyre handle
ball etc. There are different types of air pumps but in
this topic we well only discuss a simple type. body
piston
It consists of cylindrical body with handle connected piston valve
with a piston. The piston contains piston valve or inlet (inlet valve)
valve (washer) below which there is compressing
chamber. The washer must be lubricated because it compressor
works as piston as well as valve. In flexible condition outlet valve
only, it can expand and contract to work properly.
The lubrication also reduces friction. paddle

The compressing chamber has an outlet pipe having outlet valve and bears two or may be one
paddle.

When handle is lifted up, air enters through the external environment into chamber due to which
pressure is acted on the piston valve and opens.

At the same time, due to less pressure inside the compressor, the air through the outlet pipe
presses the outlet valve and remains closed. In the same way when handle is pressed down, the
volume of air tends to decrease as a result pressure increases and piston valve remains closed
while outlet valve opens due to greater pressure inside the chamber than the environment, in this
way air pump can be used to inflate the tyres or balls.

Let's Learn

1. It is easier to move our limbs in water than in air. This is because water exerts upthrust on
the limbs and thus reduces apparently the weight of the limbs. Thus, force required by a
person to move his limbs when immersed in water is smaller than the force for the same
movement in air.

2. The blood pressure in human body is greater at the feet than at the brain. This is because the
depth of the feet from the heart is more than the depth of the brain. Since P = dgh (where d
= density, g = acceleration due to gravity and h = distance), it implies that pressure is more
for the feet where h is more and it is less for the brain where h is less.

3. An empty ship sinks less then the loaded ship on sea-water. Weight of sea-water displaced
by the immersed portion of each ship is equal to the total weight of each ship. The empty
ship is lighter than the loaded Ship. Due to it the loaded ship has to displace more weight
of water to be floated, it sinks more. Thus the loaded ship sinks more than the empty ship.

4. An egg sinks in pure water but floats on a saturated salt solution. Density of an egg is
higher than the density of pure water but is less than saturated salt solution. The salt
solution exerts more upthrust on the egg than the pure water due to its more density.

5. The density of ice is less than that of water. Thus, the upthrust of water on ice is greater
than its weight acting downward. Hence, ice floats on water.

Blooming Science Book 10 43

6. Water in ground floor tap comes with greater force than in top floor. This is because the
height of water to the ground floor tap is more than the height of water in top floor tap. As
the liquid pressure increases with height of the liquid, the water comes forcefully in the
ground floor tap.

7. Water can’t be used in mercury barometer. It is because the density of water is very less
than that of mercury if it is used in the barometer, the height of barometer will be nearly
11 times than the height of mercury barometer, which is impractical.

8. Only water is used in hydraulic machine. It is because water can’t be compressed and it
transmitted pressure equally in all direction.

Points to Remember

1. The force applied perpendicularly on a unit area is called pressure. The SI unit of pressure
is Newton per square meter or Pascal.

2. Pressure of liquid depends upon height (h) of the liquid column, density (d) of the liquid
and acceleration due to gravity (g) i.e. P =hdg.

3. Liquids are incompressible.
4. Pascal’s law states that the pressure is equally exerted perpendicularly on all sides as

pressure is applied at a place on a liquid contained in a closed container.
5. Pascal’s law is used in Hydraulic press, Hydraulic brake, Hydraulic jack, etc.
6. Liquid pushes the things up with a certain force. It is called upthrust.
7. Upthrust changes as per the change of weight of displaced liquid.
8. The upthrust given to an object increases with density of liquid.
9. Upthrust acting on a body immersed in a fluid depends upon volume of body, density of

liquid and acceleration due to gravity of that place.
10. Archimedes’ principle states that when a body wholly or partially immersed in a liquid, it

loses some of its weight. This apparent loss in weight of body is equal to the weight of the
liquid displaced.
11. A floating object displaces the liquid equal to its own weight. This is the law of floatation.
12. Generally substances whose density is more than liquid sink into liquid and less than
liquid float on it.
13. Internal body pressure of human beings is nearly 1,00,000 Pascal.
14. The scientific instrument or device which is used to measure the atmospheric pressure is
called Barometer.
15. The scientific instrument which is used to inject or withdraw the fluid from any object of
body is called injection or syringe.

44 Blooming Science Book 10

Project Work
1. Get a tin can or a tin box with screw cap. Clean the can and take about 2/3rd of water it.
Heat the water for sometime and let it to boil for few minutes. Stop heating and cover
the can tightly with its cap. observe what happens after some time? Discuss, why the
tin collapses on closing the can? Draw the conclusion and submit to your teacher.

2. How is it possible to drink cold drinks or juices from the can or bottle with the help of
straw pipe? Discuss in your classroom and draw the conclusion.

Exercise

1. What is pressure? What factors affect the pressure of an object?

2. What are the factors which affect pressure due to liquid contained in a vessel?

3. What is buoyancy? Draw a Hydraulic brake and label it.

4. As in the given figure, the tank with capacity 1000 liters has greater

cross-sectional area than that with capacity 500 liters. Answer the

following questions with reason. 1000 liters 500 liters

i. Which one exerts more thrust at the bottom?

ii. If both contain equal amount of water, which one exerts more pressure at the bottom?

iii. What happens on the pressure if salt is added?

5. The figure shows a small iron ball sunk in a beaker of fresh water. Fresh water
Now if some table salt is added to the water in the beaker, explain Iron ball
how the upthrust experienced by the ball will be affected?

6. In what condition does a body heavier than water float in water?

7. State Pascal’s law .

8. State Archimedes’ principle and describe an experiment to verify it.

9. Explain the effect of density of liquid on the floatation of an object.

10. Write differences between force and upthrust.

11. Prove that hydraulic machine is a force multiplier.

12. In the given figure, there are three holes A, B and C of equal size in
a water tank. In which hole water pressure is high and from which
hole more water comes out? Give reason.

13. Mention the relationship between the density of liquid and

upthrust? In the given diagram an object ABCD is shown h1 Water
dipped in liquid, on which surface, the pressure is greater in AB Object
AB
face or CD surface? Why? h2 DC

Blooming Science Book 10 45

14 What are the advantages of using mercury as a barometric liquid?

15. Study the diagram and answer the questions. B
i. Which of them are floating and why? A

ii. Which law is used to explain B? State it. C
iii. State the law used to explain C.

16. Write short notes on working principle of the following:
i. Syringe
ii. Air pump
iii. Water pump
17. Explain with reason.

i. An iron nail sinks in water while a ship made up of iron floats in water.
ii. A man exerts more pressure while standing on one foot than by two feet.
iii. Iron nail sinks in water but floats in mercury.
iv. Deep sea divers wear diving suit.
v. A piece of wood floats on water.
vii. Balloon filled with hydrogen rises up in the air.
viii. Swimming is easier in sea water than in river water.
ix. It is easier to lift a heavy stone under water.
x. Pistons are not made up of same size in hydraulic machine.
xi. The bottom surface of dam is made thicker as compared to the upper surface.

18. Solve the following Numerical problems.

a. The water tank of dimension 2m×1m×1m is half filled with water. Calculate pressure

exerted by water at bottom of tank. [4900Pa]

b. A stone is weighed in air, water and salt solution. Answer the questions on the basis of

given table.

i. Name ‘A’, ‘B’ and ‘C’ Media Weight
A 20N

ii. Calculate upthrust given by salt solution. [Ans 4N] B 18N

iii. If 1kg is equal to 10N. Calculate the mass of water displaced. C 16N

B

A

c. In the diagram if area of piston A is 0.05m2 and that of B is 0.15m2. What force

is in B when 20N force is applied in piston A? [60N]

d. In a Hydraulic press, if the cross sectional area of narrow cylinder is 4cm2 and that of wide

cylinder is 0.4m2. What load is necessary on the wide piston to balance 600N load kept on

small piston? [6000N]

46 Blooming Science Book 10

e. If an iron box having a length 3m, breadth 1m and height 2m has mass of 3000kg. Does it

float on water? Show calculations. [D = 500kg/m3]

f. Calculate the pressure exerted by water at the bottom of pond whose depth is 6m. (Take

density of water =1000kg/m3, g =10m/s2) [6 × 104 Pa]

g. An iceberg of volume 100cm×60cm×40cm and density 0.9g/cm3 floats in water. What
1
portion of the iceberg will be above the water surface? [ 10 part]

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

Given that the density of mercury is 13600kg/m3 and g = 9.8m/s2. [101292.8 Pa]

i. Study the diagram and answer the questions.

i. Name the experiment verified by this 20N
experiment.
4N
ii. Calculate the apparent weight of stone.

iii. If some salt is added what happens to the
weight of stone in liquid? Why?

j. The weight of body in air is 100N. How much will it weigh in water if it displaces 400

cm3 of water? [Ans. 96N]

k. The volume of a ballon is 1000m3. It is filled with helium of density 0.18kg/m3. How
much load can it lift if the density of air is 1.29 kg/m3

19. Choose the correct alternatives from the following options.

i. Liquid pressure is directly proportional to:

a. density b. area c. volume d. None of the above

ii. SI unit of pressure is:

a. Pascal b. Newton c. Dyne d. Joule

iii. Upthrust is equal to the weight of liquid displaced. Which law is it?

a. Pascal’s b. Archimedes’ c. Floatation d. Newton’s

iv. Atmospheric pressure is measured by:

a. Thermometer b. Barometer c. Calorimeter d. All of the above

v. Washer is used as piston in:

a. Air pump b. Syringe c. Water Pump d. None of the above

vi. If weight of a body in air is 40 N and that in water is 30N, what is the upthrust
activity on it?

a. 10 N b. 30N c. 40 N d. 70N

Blooming Science Book 10 47

Chapter Energy Albert Einstein
1879 AD-1955 AD
3

Leaning Outcomes Estimated Periods : 5

On the completion of this unit, the students will be able to:
 define and describe energy with examples.
 describe the sources and uses of energy.
 identify the problems of energy crisis and measures to solve energy crisis.
 explain alternative sources of energy.
 find out the major factors of energy crisis.
 list the measures of energy conservation with examples.

Objects are pushed, pulled or lifted while doing different works. Energy is required to do different
types of works. Energy produces force and performs the work. Different types of machines are
used to do the work. Energy is necessary for the operation of all the machines. Some works need
more amount of energy while others need less.

Energy is present in running water, wind, coal, petrol, kerosene, the sun etc. Similarly human
beings get their required energy from food.

According to the principle of physics, energy can neither be created nor be destroyed rather it can
be transformed from one form to another. The capacity of doing work is called energy.

Sources of Energy

The sources of energy which are used in the form of natural availability are called primary
sources of energy. They are of two types:

a. Renewable Sources of Energy

The energy sources which can never be used up fully or if they are once used up they can
be replenished or renewed by natural cycles are called renewable sources of energy. The solar
energy, wind energy, hydropower, biomass energy, tidal energy etc. are the examples of renewable
sources of energy. The energy of the sun can never be used up fully, whereas the wind & water
are being rapidly renewed by natural cycles so they are inexhaustible sources of energy. Biomass
can be reproduced to replace the old ones hence it is renewable but it can be used up and so it is
exhaustible source of energy. The tidal energy is renewed by nature.

b. Non- renewable Sources of Energy

The energy sources which can be used up or exhausted by regular use and cannot be replenished
are called non-renewable energy sources. These energy sources are formed in nature taking very
long duration may be millions of years. Coal, natural gas, petroleum, firewood, coke etc. are the
examples of non-renewable sources of energy.

48 Blooming Science Book 10

1. Major sources of energy

a. The Sun

The major source of energy, the sun is an extremely hot mass of gases. It is estimated that the
temperature of outer surface of the sun is about 5500oC. The diameter of the sun is about 1.39
million km. This diameter is about 109 times than that of the earth. By mass also, the sun is very
large. Even when the masses of all the planets and satellites of the solar system are combined, it
forms only about 0.0015th part of the sun. It has been estimated that the temperature at the center
of the sun is about 15 million degree celsius. Since even iron is vaporized at about 2700oC,
everything is vaporized near the sun at that temperature.

The sun radiates about 27×1023KW of energy per second. The sun is extremely far (15×107km)
from the earth. Even though very small fraction of the energy that the sun radiates (about one
part of 2.2 trillion) reaches to the earth, the earth receives 1.4 KW of energy per square meter in
average. Guess how hot it will be when the sun is there in the daytime. It is difficult for us to stay
for even a short time under the sun. The solar energy is also not equally received in all parts of
the earth. For example, more energy per square meter is received in the Terai belt of Nepal while
less solar energy pre square meter is received in Kathmandu valley. Similarly, more heat is felt in
the Terai belt than in Kathmandu valley, and another thing is that the solar energy is spread over
large area. It is difficult to harness it.

When the temperature reaches 5 millions degree Celsius, an atom ionizes completely into the
nucleus and electrons. Hydrogen gas is found in the sun in abundance, the Hydrogen gas is
completely ionized to Proton (Hydrogen ion) and electron as a result of this high temperature but
there is a powerful repulsion between protons, the free protons fuse together due to extremely
high pressure and temperature in the sun forming a helium atom. These reactions take place in
different steps. A vast quantity of energy is released during each step of the reaction. A fraction
of mass is also lost in the reaction. The mass thus lost is converted into tremendous quantity of
energy. This was formulated by Einstein as E= mc2.

Sun as the Ultimate Source of Energy

Directly or indirectly, the sun supplies most of our energy that we use today. Wind energy,
hydro-energy, fossil fuel energy, tidal energy, geothermal energy are all derived from the sun.

The heat energy of the sun causes the air near the surface of the earth to be heated. Being lighter,
it rises up and at the same time, cooler air from neighboring region flow into take the vacuated
region in the form of wind. Thus, wind energy has been derived from the sun’s energy.

During photosynthesis process, the sun’s energy is stored in plants, in the form of compounds.
Animals eat plants and thus Sun’s energy is indirectly stored in animals. The plants and animals
which get buried under the earth millions of years ago, had sun’s energy stored in them. They were
converted into fossil fuels. Hence, the energy of fossils fuels is also derived from the sun’s energy.

The solar energy evaporates water from oceans and surface of the earth. The water vapour thus
formed rises up in the atmosphere to form cloud. When it gets cooled, it falls back to the earth in
the form of rain and snow. The water is used in the conduction of generators for hydroelectricity.
Thus, real source of hydroelectricity is the solar energy.

Blooming Science Book 10 49

b. Fossil Fuels

Fossils are remains of prehistoric animals or plants buried under the earth millions of years ago.
Fossil fuels are formed from living matter by the decomposition in the earth’s crust, buried long
ago. In the earth’s crust the living matter decomposes into coal, petroleum and natural gas due to
chemical effects condensed by pressure, heat, bacteria and in the absence of oxygen. These fuels
are non-renewable as when once used, they cannot be replenished. Here, we discuss on each
fossil fuels separately.

i. Coal

Coal is a mixture of carbon compounds of free carbon, hydrogen, oxygen and small amount of
nitrogen and sulphur compounds. The slow chemical process of the conversion of wood forms
the coal under the earth’s crust. This process is called carbonization. This carbonization is a very
slow process and may take thousands of years. Depending upon the time taken different types of
coal are formed:

i. Peat:- It contains 60% of carbon:
ii. Lignite: It contains 70% of carbon.
iii. Bituminous: It contains 80% carbon.

iv. Anthracite: It contains 90% carbon.

Anthracite is high quality coal. In Nepal anthracite is not found but some low quality coal mines
are found in Dang district.

ii. Mineral Oil (Petroleum )

Petroleum is non-renewable fossil fuel The word petroleum is derived from Latin words ‘petra’ and
‘oleum’ meaning rock and oil respectively. The crude oil petroleum is a complex mixture of many
solid, liquid and gaseous hydrocarbons containing soil particles, water and salts. Petroleum is formed
by decomposition of the remains of microscopic plants and animals microorganism buried under the
sea millions of years ago. The petroleum thus formed gets trapped between two layers of non-porous
rocks. The natural gas is available above the petroleum. The petroleum is drawn out by drilling holes.
First natural gas comes out, which is also very efficient fuel.

Petroleum is not a pure substance of fuel therefore it has to be processed by the process called
fractional distillation. By processing other useful fuels are obtained such as petroleum gases
LPG (ethane, butane and propane), liquid fuels ( Kerosene, petrol and diesel), paraffin, wax,
asphalt and lubricating oil.

The petroleum fuels are used in automobiles ships and aeroplanes. Petroleum fuels fulfil about
45% of world’s energy requirement.

iii. Natural Gas

Natural gas is non-renewable source of energy. It fulfills about 20% of the world’s energy
requirement. Natural gas is formed under the earth by the decomposition of vegetable matter
lying under water. The decomposition takes place by anaerobic bacteria in the absence of oxygen.
Natural gas consists of mainly methane (CH4) about 95% with small quantities of ethane (C2H6)
and propane (C3H8). Natural gas occurs deep under the earth’s crust either with mineral oil or
separately. Many oil wells produce natural gas as by-product. Some wells produce natural gas
only.

50 Blooming Science Book 10