science 10

Revised and Enlarged

Approved by Curriculum Development
Centre, Government of Nepal as an
additional material for schools.

Times’ Crucial

SCIENCE
10Grade
Authors
Kamal Prasad Sapkota
Rajan Kumar Shrestha

Banasthali, Kathmandu, Nepal
Tel.: 01-4385227, 4355052
E-mail: [email protected]
Website: timespublication.com.np

Times’ Crucial

SCIENCE

Published by:
Times International Publication Pvt. Ltd.

Authors:
Kamal Prasad Sapkota
Rajan Kumar Shrestha

Edition:
First: 2070 B.S.
Revised Edition: 2074 B.S.
Revised Edition: 2075 B.S.
Revised Edition: 2077 B.S.

Layout and Design:
Times Desktop

ISBN 993758046-3

© Copyright: Publisher

Printed in Nepal

Preface

Human life is always in progress. Newer technologies and equipmen are developed or
discovered in every second. Lengthy and time consuming work of past are now done
within a few moments. People can travel far off distances within short time. Several
fatal diseases have been eradicated. The achievements, that we are enjoying today
are the results of advancement in science and technology. Hence, science has become
an integral part of our education system.

Times' Crucial Science is a series of text books for the school level students of grades
LKG to class Ten. The series has been prepared for the young learners emphasizing
on student-centred teaching techniques, Learning Circle Based Activities, practicable
activities, scientific approaches and innovative learning techniques. The text of each
lesson is preceded by a warm up activity that encourages students to take part actively
in the learning process. The series includes the teaching techniques and methods for
the teachers under the title Note to the Teacher. It is based on the latest syllabus
prescribed by CDC Government of Nepal. Hence, this series acts as the foundation
of science for the curious and inquisitive young minds.

Each book of this series covers the syllabus of Science . The first part contains chapters
of Science and the second part, chapters of Environment Science. All the chapters
of Science and Environment are provided with a wide variety of exercises which
encourage the students in learning and sharpening their mind. A Project Work is
asked at the end of each lesson so that student can apply their knowledge to solve the
problems of their day-to-day life. In the beginning of each lesson, thought provoking
questions are given under the heading Mind Openers. This activity will encourage
students to use their brain.

We feel delighted to extend our sincere gratitude to all Principals of the schools,
teachers and students for their contribution in revising the book and making it
simpler and more crucial. We are equally thankful to the publisher Mr. Kamal Prasad
Pokharel without whom this series would not have been possible in the present form.
We are also thankful to Times Desktop for providing this smart form to the book.

Constructive suggestions and recommendations from teachers, students and well
wishers for the further improvement of the books are the most welcome.

- Authors

Contents Page No.

Chapter 1-134
Physics
1
1 Force 30
58
2 Pressure 74
3 Energy 88
4 Heat 109
5 Light 135-262
6 Current Electricity and Magnetism 135
156
Chemistry 171
190
7 Classification of Elements 206
8 Chemical Reaction 223
9 Acid, Base and Salt 238
10 Some Gases 263-359
11 Metals 263
12 Hydrocarbon and Its Compounds 273
13 Materials Used in Daily Life 287
302
Biology 311
336
14 Invertebrates 351
15 Human Nervous and Glandular System 360-402
16 Blood Circulatory System in Human Beings 360
17 Chromosomes and Sex Determination 371
18 Asexual and Sexual Reproduction 388
19 Heredity 403-420
20 Environmental Pollution and Its Management

Geology and Astronomy
21 History of the Earth
22 Climate Change and Atmosphere
23 The Universe

Specification Grid / SEE Question Paper

Chapter

1 Force

Sir Isaac Newton

He discovered Newtonian mechanics, Universal
gravitation,Calculus, Newton's laws of motion,
Optics, Binomial series, Newton's method, etc.

Estimated Periods: 10 (8T+2P)

Objectives

At the end of the lesson, students will be able to:
• explain Newton’s Universal law of Gravitation;
• differentiate between gravity and gravitation;
• differentiate between mass and weight with their units;
• measure the mass of different objects;
• explain freefall and weightlessness.

Force

In our daily life, we apply force for doing various activities. We can

apply force either by pulling or pushing. When we apply force to an

object, its state of rest or motion gets changed. For example, when

we kick a ball lying on the ground, it comes to motion. Here, the ball

changes its state of rest into motion. Similarly, we catch a rolling ball

by applying force. However, if we apply force to a big tree or to a wall,

the force cannot move them. Here the force is not sufficient to move

them.

Thus, force can be defined as an external agency which changes or

tends to change the state of rest or uniform motion of a body along a

straight line.

Force is measured by spring balance. Its SI unit is Newton and CGS

unit is dyne. There are different types of forces such as muscular

force, magnetic force, frictional force, electrical force, gravitational

force, nuclear force, etc. In this chapter, we discuss the gravitational

force and gravity. Moon

Gravitation

In this universe, a body attracts another body
with a certain force. The force of attraction
between any two bodies of the universe is
due to their masses. Such force is called
gravitation. The existence of gravitational Earth
force was discovered by sir Isaac Newton in 1687.

1 Times' Crucial Science Book - 10

According to Newton, the bodies in this universe attract each other
by a certain force. Such force is called gravitational force. The earth
attracts the moon and the moon attracts the earth too. In the same
way the sun attracts the earth and the earth attracts the sun with
a certain force. Such force is called gravitational force. Similarly, a
stone attracts the earth and the earth attracts the stone. The effect of
attraction of a big body on a small body can be observed but the effect
of attraction of a small body on a big body cannot be observed easily. It
is because the gravitational force increases with the increase in mass
of bodies. The effect of gravitation can be observed more on liquid
than on solid. Thus, we can see tides in the ocean due to gravitation
of the sun and the moon.

Thus, gravitation is defined as the force of attraction between any two
objects of the universe due to their masses.
Newton's universal law of Gravitation

The great physicist, Sir Issac Newton, propounded the law of
gravitation in 1687 AD which is also known as Newton's universal
law of gravitation. Newton's universal law of gravitation states that
the force of gravitation between any two bodies is directly proportional
to the product of their masses and inversely proportional to the square
of distance between their centres.

Let two bodies of masses m1 and m2 be separated by a distance d

m m1 2

F

d

According to Newton's Universal Law of Gravitation, the force of
gravitation between these bodies F is

i. directly proportional to the product of their masses, i.e.

F α m1 . m2 …… (i)

ii. inversely proportional to the square of distance between their
centres, i.e.

F α 1 ………. (ii)
d2
Combining equations (i) and (ii)

F α m1.m2 × 1
d2

Times' Crucial Science Book - 10 2

Or, F α m1m2
d2

Or, F=G m1m2
d2
Where G is the universal gravitational constant whose value

is 6.67×10-11Nm2/kg2. Its value remains constant throughout the

universe. Therefore, it is called universal gravitational constant.

Newton's law of gravitation is called universal law because it is true

throughout the universe.

The value of G was measured for the first time by Henry Cavendish
(1731 – 1810) using sensitive balance.

Universal gravitational constant (G)

According to Newton's Universal Law of Gravitation,
m1m2
F= G d2

If m1 = 1kg, m2 = 1kg and d = 1m, then F = G×1×1 F=G
1

m1 = 1kg F=G m2 = 1 kg
d = 1m

Thus, universal gravitational constant (G) can be defined as the force
of gravitation between any two bodies each of unit mass separated by
unit distance from their centres. It is a scalar quantity.

SI unit of G

According to Newton's Universal Law of Gravitation,

SI F =ofGfomrcd1em2, d2 ist∴anceGa=nFdm×1mmda22ss are Newton (N), metre (m) and
units

kilogram (kg) respectively. Therefore, the SI unit of

G= N × m2 = Nm2/kg2.
kg.kg

Therefore, SI unit of universal gravitational constant (G) is Nm2/kg2
or Nm2kg–2.

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Explanation of Newton's universal law of gravitation

1. When distance remains same but mass changes

Let two bodies each of mass 'm' be separated by a distance 'd'. Then,

according to Newton's universal law of gravitation, the force of

gravitation between these bodies 'F' is given by formula

F= Gm1m2 m1 m2
d2 F1

Gm1m2 d
or, F1 = d2 …...…. (i)

When mass of one body is doubled keeping the distance between
them constant as shown in the figure, then the new force of

gravitation (F2) is calculated as, m1 2m2

F2 = G m1×2m2 2F1
d2

Gm1m2 d
d2
F2 = 2 …………. (ii)

Substituting the value of Gm1m2 from equation (i) in equation
(ii), we have d2

F2 = 2F1
Thus, the force of gravitation increases two times when the mass
of a body is doubled keeping the distance between them constant.

2. When mass remains same but the distance changes.

Let two bodies of masses m1 and m2 be separated by a distance 'd'.
Then, according to Newton's universal law of gravitation, force of

gravitation between these bodies 'F' is given by m2

m1

F1 = Gm1m2 …….. (i) F1
d2 d

When distance between them is doubled keeping their masses

constant as shown in the figure, then new force of gravitation (F2)
is m1 m2

m1.m2 1 F1
F2 = G (2d)2 4

2d

Times' Crucial Science Book - 10 4

m1.m2
Or, F2 = G 4d2

Or, F2 = 1 Gm1.m2 ……..... (ii)
4 d2

Substituting the value of Gm1.m2 from equation (i) in equation
d2

(ii), we have
1

F2 = 4 F1

Thus, the force of gravitation between the two bodies decreases by
one-fourth when the distance between them is doubled keeping
their masses constant.

Consequences of gravitational force

a. The existence of solar system and constellations is due to the
gravitation between the sun and other heavenly bodies.

b. Atmosphere on a planet is due to gravitational force between
the planet and the atmosphere.

c. The revolution of the earth and other planets around the sun in fixed
orbits is due to gravitational force between the sun and the planets.

d. The tides in the sea or ocean are due to gravitational force
between the water and the sun or the moon.

Solved Numerical Problem 1.1

What is the force of gravitation between two bodies having masses 10kg and 30kg and
separated by 12 meters?

Solution:
Given,Mass of one body (m1) = 10kg

Mass another body (m2) = 30kg

Distance between them (d) = 12m

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

Force of gravitation (F) =?

We have, 6.67 × 10–11 × 10×30
122
F = G m1m2 Or, = = 13.89 × 10-11N
d2
∴ Hence, the force of gravitation between these bodies is 13.89×10–11N

5 Times' Crucial Science Book - 10

Solved Numerical Problem 1.2

Masses of two heavenly bodies are 2×1030kg and 3×1020kg respectively.
The force of gravitation between them is 1025N. Find the distance between
them.

Solution:

Given, Mass of one body (m1) = 2×1030kg
Mass of another body (m2) = 3×1020kg
Force of gravitation (F) = 1025N

Distance between them (d) = ?

We have,

F= Gm1 m2 Or, d2 = Gm1m2
d2 F

Or, d2 = 6.67 × 10–11 × 2 × 1030 × 3 × 1020
1025 = 40 × 1014

∴ d = 6.32 ×107m

∴ Hence, the distance between them is 6.32×107m. 

Gravity

When you throw a ball in upward direction, it
returns downward after sometime. Similarly, fruits

fall downward from a tree, water flows downward in Nepal
rivers and so on. Why is it so? These all are due to
pulling force of the earth. It is called gravity of the Earth

earth. When a person is in Nepal and standing on the 

ground, his head faces towards the sky. Similarly, a America
person standing on the ground of America which is just

opposite to Nepal also faces his head towards the sky.
The reason behind it is that both persons are pulled by the gravity of the earth
towards its centre.

Gravity is the force with which the earth or any heavenly body attracts other
objects towards its centre. Its SI unit is Newton and CGS unit is dyne. It is a
vector quantity.

All planets and heavenly bodies have their own gravity. The body having more
mass has more gravity. Thus, the earth has more gravity than the moon. The
effect of gravity of a heavenly body is more on a larger body than on a smaller

Times' Crucial Science Book - 10 6

body. Therefore, it is difficult to lift a larger stone than a smaller stone from

the ground. m

Calculation of value of gravity

Let a body of mass m lie on the surface of the earth whose R M
mass is M and radius is R. According to Newton’s universal Earth
law of gravitation, the force of gravitation (F) between the

earth and the body is calculated as
Mm
F=G R2 ............(i)

Since the force applied by the body m to the earth is negligible with

comparison to the force applied by the earth to the body, the force of

gravitation F calculated in equation (i) is almost equal to the gravity
of the earth to that body. Therefore, gravity acting on a body of mass

m by a heavenly body of mass M and radius R is calculated by using

formula.

F= GMm
R2

The gravity applied by a heavenly body to an object is also called its

weight on the heavenly body. Therefore, weight

F= GMm
R2
Since value of G is always constant and mass of a particular body

remains constant everywhere, the gravity of a heavenly body depends

upon:

(i) mass and

(ii) radius of the heavenly body.

The gravity of the moon is one -sixth the value of the earth.The value
of acceleration due to gravity on the surface of the moon is only 1.67
m/s2. Therefore, a person can jump 6 times higher on the moon than
on the earth. Similarly, if a person can lift 100kg on the earth, he can
lift 600 kg on the moon.

Mass of the Jupiter is 319 times more than the mass of the earth. The
gravity of the Jupiter does not increase highly even its mass is 319
times. Its gravity is reduced by the radius. Thus, the gravity of the
Jupiter is only 2.5 times more than that of the earth.

7 Times' Crucial Science Book - 10

Solved Numerical Problem 1.3

The mass of the earth is 6×1024kg and its radius is 6380km. What is the force of
gravity of the earth on 1kg object ?

Solution:

Given, Mass of the earth (M) = 6×1024kg
Mass of the object (m) = 1kg

Radius of the earth (R) = 6380km = 6380×1000m = 6380000m
Force of gravity (F) = ?

GMm
F = R2

Or, F = 6.67 × 10–11 × 6 × 1024 × 1

(6380000)2

6.67 × 6 × 1013
Or, F = 4.07 × 1013

= 9.8N
∴ The force of gravity of the earth on 1kg object is 9.8N.

Solved Numerical Problem 1.4

If the mass of the earth is 6×1024kg and its radius is 6380km, what
will be the weight of an object having mass 150kg on the surface of
the earth?

Solution:

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

Radius of the earth (R) = 6380km = 6380×1000m = 6380000m

Mass of the object (m) = 150kg

Weight of the object (F) = ?

We have, F = GMm
Or, R2
6.67 × 10-11× 6×1024×150

F = (6380000)2

Or, F = 6.67 × 6 × 150 × 1013

4.07 × 1013

Or, F = 1475N
∴ The weight of the body on the earth surface is 1475N..

Times' Crucial Science Book - 10 8

Effects of gravity

We can stand, run and walk on the ground due to gravity. Fruits fall from the
trees due to gravity and so on. Some of the effects of gravity are as follows:

a. An object thrown toward the sky returns to the ground.
b. Water flows in the rivers.
c. Atmosphere surrounds the earth and other planets
d. It causes acceleration to a falling object
e. Buildings, bridges, towers, etc remain fixed and stable due to gravity.

Acceleration due to gravity

You have studied that acceleration can be produced on a moving
object when force is applied to an object continuously in the direction
of its motion. In the same way, when a body falls toward the ground,
acceleration is produced on the falling object.

The acceleration developed in an object falling freely towards the
earth or any heavenly body due to the influence of the gravity is called
acceleration due to gravity.

It is denoted by g. Its SI unit is m/s2.

When a body is dropped from a certain height on the 0 m/s
earth surface, its velocity increases. In the beginning, 9.8 m/s
the velocity of the body when it starts falling is 0m/s. 1s
After 1s, it gains a velocity of 9.8m/s. After 2s, it gains
a velocity of 19.6m/s. After 3s, it gains a velocity of
29.4 m/s and so on. Therefore, change in velocity of 1s
the falling body (acceleration) is 9.8 m/s2.

Ancient people used to think that heavier bodies fall 19.6 m/s

faster than the lighter bodies when they are dropped

together. But a famous Italian scientist Galileo

proved in 1690 AD that such ancient thought was 1s
wrong. He conducted an experiment in which he

dropped two objects one larger having more mass

and another smaller having less mass from the top

of Leaning Tower of Pisa simultaneously. Both the 29.4 m/s

objects reached the earth surface at the same time.

This proved that the acceleration due to gravity of a falling object does

not depend upon the mass of the falling object. The reason for falling

heavier object faster and feathery object slower is due to resistance

given by the air.

9 Times' Crucial Science Book - 10

Feather and Guinea experiment

Robert Boyle performed an Feather
experiment to verify the idea of
Galileo. In this experiment, he Air Vacuum
took a long glass tube fitted with
a vacuum pump. He kept a feather Guinea
and a guinea (a coin of England)
inside the glass tube then closed its Feather and guinea experiment
both ends. The tubes were inverted
quickly. He observed that the coin
reached the bottom faster than
the feather. Then he removed all
air from the tube by using vacuum
pump. He repeated the same
experiment. In this case, he found
that both the guinea and feather
reached the bottom simultaneously.

Conclusion: In the first case, the air resistance slowed the falling
of the feather. So, the feather fell slowly. But when air was removed,
the feather and the guinea fell with same acceleration. Thus, this
experiment concludes that the acceleration of freely falling bodies
remains same for all bodies irrespective of their masses.

Activity 1 .1 To observe the effect of air resistance on the bodies of different

surface area

Materials required:
Two pieces of papers of almost equal size

Procedure:
a. Take two pieces of paper of almost equal size.
b. Make a ball from one of the papers.
c. Drop both papers from the top of a building. What do you

observe?

Observation:
The ball made of paper falls faster than the unfolded paper.

Conclusion:
The air offers more resistance to the falling object having more
surface area.

Times' Crucial Science Book - 10 10

Activity 1 .2 To show that acceleration due to gravity does not depend on

mass

Materials required:
Two pieces of stones of unequal sizes.

Procedure:
a. Take two stones of unequal sizes, i.e. one larger and another
smaller.
b. Drop the stones from the first storey of your school building.
Tell one of your friends to stay on the ground and observe the
falling of stones.
c. Go to the top of your school building and drop the same stones.
What will be the observation?

Observations:
Both the stones reach the ground together.

Conclusion:
Acceleration due to gravity developed on a falling object is
independent to the mass of the falling object.

Calculation of value of acceleration due to gravity (g)

Let an object having mass ‘m’ is placed on the surface m

of the earth having mass 'M' and radius ‘R’. R
According to the Newton’s universal law of Earth

gravitation, the force of gravitation between the M

earth and the object is calculated by

GMm
F = R2 ............(i)

When the object is allowed to fall freely towards the earth, it falls
with acceleration equal to ‘g’. Then, according of Newton’s second law
of motion, the force applied by the earth ‘F’ is calculated by using
formula

F = mg …………(ii)

Force of equation (i) and equation (ii) are almost equal. From
equations (i) and (ii), we have:

11 Times' Crucial Science Book - 10

Or, mg = GMm
R2

∴ g= GM
R2

This equation clarifies that the acceleration of a falling object does not

depend upon the mass of the falling object but depends upon the mass

and radius of the pulling object.

Acceleration due to gravity on the surface of the earth

Wdueectaongruasveittyhoefeaqlluhaetaiovnengly=bodGRiMe2s.Wfoerctahneucaselctuhleaftoiromn uoflathge=aGcRcM2elerfoarttiohne
calculation of acceleration due to gravity of the earth, where

G = Gravitational constant
M = Mass of the earth.
R = Radius of the earth.

Value of ‘G’ and ‘M’ are constant but the value of ‘R’ differs from place

to place on the earth surface. Therefore, value of ‘g’ of the earth is
1
inversely proportional to the square of its radius, i.e. g α R2 .

The earth is not perfectly round. It is flattened at the pole and bulged

at the equator. Thus, the radius of the earth at the pole is less whereas

it is more at the equator. Accordingly, value of ‘g’ of the earth is more

at the pole and less at the equator. pole

Value of ‘g’ at the pole gpole = 9.83m/s2. Rp Equator
Value of ‘g’ at the equator gequator = 9.78m/s2. Re
Average value of ‘g’ of the earth (g) = 9.8m/s2.

Value of ‘g’ at the top of the hill m
h
Let an object having mass ‘m’ is lying at the top of a
hill of height ‘h’ from the sea level of the earth. Mass R M
and radius of the earth are ‘M’ and ‘R’ respectively. Earth

According to Newton’s universal law of gravitation,
the force of gravitation between the earth and the
object can be calculated by using formula,

Times' Crucial Science Book - 10 12

F= GMm ........(i)
(R + h)2

If the object is allowed to fall freely to the surface of the earth, it
falls with an acceleration g1. Then, according to Newton’s second law
of motion, the force of gravity applied by the earth to that object is
calculated by using formula

F = mg1 ………..(ii)
From equations (i) and (ii)

GMm
mg1 = (R + h)2

GM
∴g1 = (R + h)2 ...............(iii)

Since, the distance of the top of the hill from the centre of the earth
is more than that at the bottom, the value of g1 is smaller than that
of g. Therefore, value of g decreases when altitude from the surface of
the earth increases.

We know that acceleration due to gravity on the earth surface 'g' is
calculated by using formula

g= GM ................ (iv)
R2

Dividing equation (iii) by equation (iv)

GM R2
+ h)2
g1 = (R + h)2 Or, g1 = (R R2 Or, g1= (R × g
g GM g + h)2

R2

Value of ‘g’ at a depth

Value of ‘g’ at a certain depth is calculated by using formula

gd = [1 – d ] g.
R

Where, d = depth from the surface of the earth

R = Radius of the earth, and

g = Acceleration due to gravity of the earth on the surface.

13 Times' Crucial Science Book - 10

The above formula shows that when the depth increases, the value of
‘g’ decreases. For the centre, depth (d) = Radius (R). In such situation

gd = 0
Hence, the value of ‘g’ is zero at the the centre of the earth.

Acceleration due to gravity of the moon

Moon also has gravity. Therefore, when a body falls towards the
surface of the moon, it falls with acceleration called acceleration due
to gravity of the moon.

Calculation of acceleration due to gravity of the moon

Formula gmoon = GM can be used to calculate the acceleration due to
gravity where, R2

gmoon = Acceleration due to gravity of the moon
G = Gravitational constant

M = Mass of the moon

R = Radius of the moon.

Solved Numerical Problem 1.5

Mass of the moon is 7.2×1022kg and its radius is 1.7×106m. Find
acceleration due to gravity of the moon.

Solution:

Given,G = 6.67×10–11 Nm2/kg2

M = 7.2×1022kg

R = 1.7×106m

gmoon = ?

We have,

GM = 6.67 × 10–11 × 7.2 × 1022
gmoon = R2 (1.7 × 106)2

gmoon = 1.67 m/s2

∴ The acceleration due to gravity of the moon is 1.67 m/s2.

Times' Crucial Science Book - 10 14

Acceleration due to gravity of a planet and weight of a body

Every planet or heavenly body has its own acceleration due to
gravity. The acceleration due to gravity of a heavenly body can be
calculated by the same formula,

g = GM
R2

Where

G = Gravitational constant,

M = Mass of the planet,

R = Radius of the planet

Thus, acceleration due to gravity of a planet is inversely
proportional to the square of its radius and directly proportional to
its mass.

Weight of a body on any planet is the amount of gravity applied by
the planet to that body. According to Newton’s second law of motion,
the amount of gravity applied can be calculated by the formula

F = mg.

Thus, the weight of a body on a planet is the product of mass of the
body and acceleration due to gravity of the planet.

Jupiter is the largest planet. Although it is 319 times bigger than
the earth and its radius is 11 times more than that of the earth, its
force of gravity is only 2.5 times bigger than that of the earth.

Solved Numerical Problem 1.6

Mass of the Jupiter is 1.9×1027 kg and its radius is 71×106 m. What
will be its acceleration due to gravity? What will be the weight of a
man of mass 70 kg on the Jupiter?

Solution:

Given, Mass of the Jupiter (M) = 1.9×1027kg

Radius of the Jupiter (R) = 71×106m

Acceleration due to gravity (g) = ?

Weight on the Jupiter (F) = ?

15 Times' Crucial Science Book - 10

We have,

g= GM = 6.67 × 10–11 × 1.9 × 1027
R2 (7.1 × 107)2

= 25m/s2

Again, mass of the man (m) = 70 kg.
According to formula,

Weight (F) = mg

= 70×25

= 1750N.
Thus, acceleration due to gravity of the Jupiter is 25m/s2 and the
weight of the man on the Jupiter is 1750N.

Differences between gravity and gravitation

Gravity Gravitation

1. Gravity is the force 1. G r a v i t a t i o n i s t h e
with which the earth or any f o r c e o f gravitation between
heavenly body attracts other any two bodies of the universe
objects towards its centre. due to their masses.

2. Gravity of a planet 2. Gravitational force between
depends upon its mass and any two objects depends upon
its radius. the product of their masses
and the distance between
them.

Differences between gravity and acceleration due to gravity

Gravity Acceleration due to gravity

1. Gravity is the force 1. The acceleration produced
with which the earth or on a freely falling body due
any heavenly body attracts to the influence of gravity
other bodies towards its is called acceleration due to
centre. gravity.

2. Its SI unit is Newton. 2. Its SI unit is m/s2.

3. It is the cause for the 3. It is the effect of gravity.
acceleration due to gravity

Times' Crucial Science Book - 10 16

Differences between g and G

Acceleration due to gravity (g) Gravitational constant (G)

1. It is the acceleration 1. It is the force of
produced on a freely falling gravitation between any two
object towards the earth or objects each of unit mass
planet. separated by unit distance.

2. Its SI unit is m/s2. 2. Its SI unit is Nm2/kg2.

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

4. Its value differs from place 4. Its value remains

to place. constant in all places.

5. Its value on the surface of 5. Its value is
6.67×10–11Nm2/kg2.
the earth is 9.8m/s2.

Gravitational field

The space around a planet or any heavenly body upto where its force
of gravity can be experienced is called gravitational field. The planet
having more mass has large gravitational field.

Gravitational field intensity of a planet at a point is defined as the
force of gravity experienced by a unit mass placed at that point.

Mathematically,

Gravitational field intensity, (I) = Force (F)
mass (m)
F
Or, I = m .......(i)

Solved Numerical Problem 1.7

The mass of the earth is 6×1024kg and the radius of the moon is
1.7×106m. Calculate the acceleration due to gravity of the new earth
which is formed by the compression of the earth to the size of the moon.

Solution:

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

although it is compressed)

Radius of the new earth (R) = 1.7×106m

Acceleration due to gravity (g) = ?

We have,

g= GM = 6.67×10–11×6×1024 = 6.67×6×1013
R2 (1.7×106)2 2.89×1012
= 138.5m/s2

∴ The acceleration due to gravity of the new earth is 138.5 m/s2.

17 Times' Crucial Science Book - 10

Solved Numerical Problem 1.8

What mass can a weight lifter lift on the moon if he can lift 100kg on
the earth? [Given: acceleration due to gravity of the earth (ge) = 9.8m/
s2 and acceleration due to gravity of the moon (gm) = 1.67m/s2]

Solution:

Given, Acceleration due to gravity of the earth (ge) = 9.8m/s2.
Acceleration due to gravity of the moon (gm) = 1.67m/s2.
Mass that can be lifted on the earth (me) = 100kg
Mass that can be lifted on the moon (mm)=?

We have,

Weight that can be lifted on the earth = weight that can be
lifted on the moon.

Or, ge×me = gm×mm Or, 9.8×100 = 1.67×mm

Or, mm = 980 =600kg.
1.67

∴ The weight lifter can lift 600kg load on the moon.

Solved Numerical Problem 1.9

Calculate the height above the earth’s surface at which the value of
acceleration due to gravity is 4.9 m/s2. The mass and radius of the
earth are 6×1024 kg and 6400 km respectively.

Solution:

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

Radius of the earth (R) = 6400km = 64,00,000m

Acceleration due to gravity at a height (gh) = 4.9m/s2

Height of the place (h) = ?

We have,

Acceleration due to gravity at a height

gh = GM Or, 4.9 = 6.67×10-11×6×1024
(R+h)2 (6400000+h)2

Or, (6400000+ h)2= 6.67×6×1013
4.9

(6400000 + h)2 = 2(6.4×106)2

Times' Crucial Science Book - 10 18

Taking square root on both sides

6400000 + h = √2 (6.4×106) Or, h = 2650km

∴ The height above the earth’s surface is 2650km.

Mass and weight

Total quantity of matter contained in an object is its mass. Mass
and weight of an object are not the same thing. Sometimes, the term
weight is used for the mass. Objects having equal masses have equal
weights in a particular place. Therefore, the use of the term weight
for mass does not make very much difference. Physical balance is
used for measuring the mass whereas a spring balance is used for
measuring the weight.

Mass

Its SI unit is kilogram. It is a scalar quantity. It is measured by a
beam balance or physical balance. Mass of a body remains constant
throughout the universe whether it is measured on the earth or in
the moon or in the space.

Mass of a body depends upon the (i) size of atoms or molecules and (ii)
number of atoms or molecules contained in the body.

Activity 1 .3 To observe the relation between size of particles and total

mass.

Materials required:
Two equal sized plastic bags, 25 small
marbles, 25 big marbles, beam balance,
etc.

Procedure:
a. Take two plastic bags of almost equal
size.
b. Put 25 small marbles in one of the plastic bags.

c. Put this bag in one pan of the beam balance.
d. Put another plastic bag in another pan and put large marbles in
this bag until both sides of the pans get balanced.
e. Count the number of large marbles in the plastic bag.

Observation:
The number of small sized marbles is more than that of large sized
marbles.

19 Times' Crucial Science Book - 10

Conclusion:
The mass of both bags is equal but the number of marbles differ.
Here, the bag is matter and marbles are particles. The greater the
size of the particles,the more the mass is.

Weight

Weight of an object is the amount of gravity by which the object is pulled
by the earth or any planet towards its centre. Its SI unit is Newton.
It is measured by a spring balance. The spring balance measures the
direct pulling of an object by the earth or a planet. If an object is
pulled by the earth with more gravity, it shows more reading.

Weight of an object is calculated by using formula,

weight = mass × acceleration due to gravity

F=m×g

Here, mass of the object is constant but the value of ‘g’ differs from
place to place on the surface of the earth. The value of ‘g’ at the pole
of the earth is more but it is less at the equator. As a result, a body
weighs more at the pole than at the equator. Value of ‘g’ also differs
from planet to planet. Due to this, weight of an object differs from
planet to planet.

Differences between mass and weight

Mass Weight

1. It is the total quantity of matter 1. It is the amount of gravity

contained in an object. by which an object is pulled

by the earth or any planet

towards its centre.

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

3. Its SI unit is kilogram. 3. Its SI unit is Newton.

4. It is a scalar quantity. 4. It is a vector quantity.

5. Its value is constant throughout 5. Its value differs from place to

the universe. place.

Free fall and falling of parachute

When a body falls towards the centre of the earth or a planet, it falls
with a certain acceleration. If there is no any air to resist the falling

Times' Crucial Science Book - 10 20

of a body, it falls with acceleration equal to the acceleration due to
gravity of the earth or a planet. It is called free fall. Falling of a body in
vacuum is an example of free fall. Similarly, falling of a body towards
the surface of the moon is an example of free fall. It is because there
is no air or atmosphere on the moon to disturb the fall of an object.

Thus, falling of a body under the effect of gravity without any external
resistance is called free fall.

Complete free fall is not possible on the earth. It is because the
atmosphere present on the surface of the earth disturbs the falling of
objects. The resistance given by the air depends upon the shape and
size of the falling objects. When you drop a paper and a stone together
from a certain height, the stone reaches the ground faster than the
paper. It is because the paper is resisted more by the air than the
stone. The fall of stone is almost free fall. But the fall of the paper is
not free fall.

When an object falls towards the surface of the earth, its velocity
increases with a certain rate. When the velocity increases, the
resistance given by the air to the body also increases thereby decreasing
the acceleration. The magnitude of the resistance due to air depends
upon the shape and size of the falling objects. When a parachutist
jumps out from a flying plane with a parachute, the resistance given
by the air to the parachute will be considerably high due to its large
size. During its fall, a condition comes at which the downward force
of the parachute and the upthrust given by the air will be equal. In
such situation, the acceleration of the parachute remains zero, i.e. it
falls with a constant velocity. As a result the velocity of the falling
parachute will be low and balanced. Thus, parachutist lands safely.

Activity 1 .4 To observe the free fall of a parachute.

Materials required: 2200ccmm Plastic Plastic
A large sheet of plastic, Thread
hread, a piece of small Hole
Thread
stone

Procedure: Stone

a. Take a large sheet of plastic and cut it in the form of a circle of

radius 20cm.

b. At the edge of the sheet, make 8 small holes at equal distance

as shown in figure.

c. Tie a thread of 15 cm at each hole.

21 Times' Crucial Science Book - 10

d. Collect all the free ends of the thread and tie a small stone on
it as shown in the figure. Now parachute is ready.

e. Drop the parachute from the top of your school building. What
do you observe?

Observation:

The parachute falls with a constant and balanced velocity.

Conclusion:
Due to special shape of the parachute, the air gives high resistance
to the falling parachute. Thus, the parachute falls slowly.

Weightlessness

The weight of a body

is measured by a

spring balance. When

a stone is suspended

to the hook of a spring

balance, it shows

certain reading.

When the spring

balance is released

along with the stone, the spring balance shows zero reading. It means

the stone has zero weight during free fall. In this condition, the stone

seems weightless. Similarly, suppose a girl is standing inside an

elevator (lift). She is throwing a book upward inside the elevator. If

the elevator can be made to fall under the effect of gravity only, the

acceleration of the elevator, the girl and the book will be equal to the

acceleration due to gravity of the earth, i.e. 9.8m/s2. In this situation,

the book seems to be suspended in the air, or the effect of the gravity

of the earth seems to be zero. Even the person standing inside the

elevator seems weightless because the acceleration of the elevator

and person is equal to the acceleration due to gravity. All the three

things, the elevator, the book and the person are in free fall in this

situation and they seem weightless.

When a spaceship is orbiting around the earth in a circular path,
astronauts inside it feel weightless. They hang in the room of spaceship.
When the spaceship is orbiting around the earth, both the spaceship
and astronaut are in continuous state of free fall towards the earth
surface with acceleration equal to acceleration due to gravity. In such
situation, no reaction force is exerted by the spaceship to the astronaut
and s/he feels weightless.

Times' Crucial Science Book - 10 22

Weightlessness of a body is a state in which the body experiences
having no weight.

Conditions for weightlessness.

a. A body feels weightlessness during free fall due to lack of
reaction force.

b. A body experiences weightlessness when it is at null point in
the space.

c. When a body is inside spaceship which is orbiting around a
heavenly body, it feels weightlessness.

Motion under the influence of gravity

When a body falls towards the surface of the earth, the acceleration of
the falling body will be equal to the acceleration due to gravity if the
air resistance is neglected. In such situation, acceleration (a) can be
replaced by acceleration due to gravity (g) and distance travelled (s)

by height (h), i.e. a = g and s=h. Then, the equations of motions can

be modified as

v = u+at ⇒ V = u + gt

s= ut + 1 at2 ⇒ h = ut + 1 gt2
2 2
v2 = u2+2as ⇒ v2 = u2 + 2gh.

When a body falls freely from a height its initial velocity will be 0m/s.

i.e. u = 0m/s.

When a body is thrown vertically upwards, its final velocity will
be

0m/s, i.e. v = 0m/s.

Learn and Write

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

Newton’s law of gravitation is true throughout the universe. It is
true not only on the earth, but on the moon, Jupiter or in space
too. Therefore, Newton’s law of gravitation is called universal
law of gravitation.

2. A body weighs more at the pole than at the equator. Why?

Weight of a body is calculated by using a formula, weight (F)
mass (m) acceleration due to gravity (g). Mass of a body remains

23 Times' Crucial Science Book - 10

constant but value of ‘g’ differs from place to place. Value of ‘g’ is
inversely proportional to the square of radius of the earth. Since
the polar radius of the earth is less than that of equator, the
value of ‘g’ is more at the pole than at the equator. Therefore, a
body weighs more at the pole than at the equator.

3. The moon does not have atmosphere. Why?

The moon has no atmosphere because the moon has less gravity.
The moon is not massive enough to have a gravitational field
sufficient to keep an atmosphere over the long term. As the gas
molecules are always in random motion of high speed, they can
easily escape out of the gravitational field of the moon. That
is why all the molecules of gases have escaped and there is no
atmosphere on the moon.

4. Parachutists are not hurt when they jump out from an aeroplane. Why?

When parachutists jump out from an aeroplane, the resistance
given by the air to the parachute will be considerably high due
to large size of parachute. During its falling, a condition comes
at which downward force of parachute and upthrust of air will
be equal. In such situation the acceleration remains zero i.e. the
parachute falls with constant velocity. As a result, the parachute
falls with low and balanced velocity. Thus, the parachutists are
not hurt.

5. A man can jump higher on the moon than on the earth. Why?

Gravity of the moon is six times less than that of the earth.
Therefore, a man is attracted by the moon with six time lesser
gravity than by the earth. Thus, a man can jump six times higher
on the moon than on the earth.

6. Tides occur in an ocean but not in a lake. The gravitational
force between the moon and water body on the earth is directly
proportional to the product of their masses. Because the oceans
have large amount of water (huge mass), they create pressure to
overpower the pull of earth’s gravity. So, the gravitational pull
of the moon is able to produce tidal bulges in large water bodies.
However, the smaller water bodies such as lakes and pools do not
produces noticeable tidal bulges. It is due to the lack of enough
liquid (large mass) to create pressure so that it can visibly overcome
the pull of earth’s gravity and get attracted toward the moon.

7. Astronauts experience weightlessness inside an artificial
satellite but not in a natural satellite such as moon. The mass
of the artificial satellite is negligibly small it cannot have detectable
gravity. So, it cannot pull astronomers towards its center. The

Times' Crucial Science Book - 10 24

artificial satellite that is revolving round the earth is in the state of
continuous free fall due to centripetal and centrifugal forces. Such
a free fall pulls the satellite down and enables it to continue motion
on the elliptical orbit. Therefore, the astronauts inside an artificial
satellite feel weightlessness due to free fall. On the other hand,
the moon has mass large enough to have its own gravity. As the
astronomers are pulled by the gravity of the moon, they experience

their weight.

Main points to remember

1. Force is an external agency which changes or tends to change

the state of rest or uniform motion of a body along a straight

line.

2. Gravitation is the force of attraction between any two objects of

the universe due to their masses.

3. Newton’s universal law of gravitation states that the force of

gravitation between any two bodies is directly proportional to

the product of their masses and inversely proportional to the

square of the distance between their centres.

4. Universal gravitational constant can be defined as the force of

gravitation between two bodies each of unit mass separated by

unit distance.

5. Gravity is the force with which the earth or any heavenly body

attracts other objects towards its centre.

6. The acceleration developed on an object falling freely towards

the earth or any heavenly body due to the influence of gravity is

called acceleration due to gravity.

7. The acceleration due to gravity of the earth or any planet is

directly proportional to its mass and inversely proportional to

the square of its radius.

8. The space around a planet or any heavenly body up to which its

gravity can be felt is called gravitational field.

9. Gravitational field intensity of a planet at a point is defined as

the force of gravity experienced by a unit mass placed at that

point.

10. Total quantity of matter contained in an object is called its

mass.

11. Weight of an object is the amount of gravity by which the object

is pulled by the earth or any planet towards its centre.

12. Falling of a body under the effect of gravity without any

external resistance is called free fall.

13. Weightlessness of a body is a state in which a body experiences

having no weight. Times' Crucial Science Book - 10

25

Exercise

A. Choose the best alternative.

1. What is the CGS unit of force?

a. Newton b. Erg c. m/s d. Dyne

2. What is the value of g at the equator of the earth?

a. 9.8 m/s2 b. 9.83 m/s2 c. 9.78 m/s2 d. 9.75 m/s2

3. The value of acceleration due to gravity (g) on the earth is zero at

a. Equator b. Centre c. Pole d. None of these

4. What is the unit of universal gravitational constant?

a. Nm2/kg2 b. kgm/s2 c. Nm/kg2 d. m/s2

5. Who experience weightlessness?

a. Astronauts in spaceship orbiting round the earth

b A stone released along with spring balance from a height

c. A ball thrown upward inside an elevator which is moving
down with the acceleration equal to g

d. All of the above

B. Answer these questions in brief.
1. Define the terms gravity and gravitation.
2. State Newton's universal law of gravitation.
3. Write down the relation of gravitational force with masses of two
bodies and their distance.
4. What is gravitational constant? Write down its value.
5. What do you mean by acceleration due to gravity?
6. The acceleration due to gravity on the moon is 1.67m/s2. What does it mean?
7. Name the two factors that affect gravitational force.
8. What is the value of acceleration due to gravity at the equatorial region
of the earth?
9. How does 'g' vary from place to place?
10. What are the factors that affect gravity?
11. Is there any place in the earth where the value of 'g' is zero?
12. What will be the change in gravitational force if the distance between
the bodies is halved?
13. What will happen to the gravitational force if the distance between two
bodies is halved and the mass of each body is doubled?

Times' Crucial Science Book - 10 26

14. Under what condition, a coin and a feather fall together? What is the
acceleration of the feather and the coin at that instant?

15. Define gravitational field and gravitational field intensity.

16. What is meant by weightlessness?

17. What is free fall? What will be the weight of a body at this condition?

18. What is the difference between the fall of parachute on the earth and
that on the moon? What is the effect of gravity on the falling of objects?

19. In which conditions does an object become weightless?

20. What do you understand by 'weightlessness due to free fall'?

C. Differentiate between:

1. Gravity and gravitation 2. 'g' and 'G'.

3. Gravity and acceleration due to gravity 4. I and g

5. Mass and weight 6. Weightlessness and free fall

D. Prove the following relations.

m1m2 2. g = GM 3. I = GM
1. F = G d2 R2 R2
E. Give reasons.

1. Newton's law of gravitation is known as universal law.

2. A body falls faster at polar region than at equatorial region of the earth.

3. A stone is dropped from the same height at Terai and at the top of the
mountain. Where does it fall faster? Why?

4. The force of gravitation between any two ordinary masses is not
remarkable. Justify with an example.

5. The value of 'g' varies from place to place on the surface of the earth.

6. Weight of an object is greater at the polar region than the equatorial
region of the earth.

7. The weight of an object at the top of a mountain is less than its weight
at the bottom of the mountain.

8. The probability of getting hurt is more when a person jumps from a
large height.

9. The moon does not have atmosphere.

10. It is difficult to lift a large stone but easy to lift a smaller one on the
surface of the earth.

11. It is difficult to drink water inside an artificial satellite in the space.

12. A person can jump higher distance on the moon's surface than on the earth.

13. If a coin and a feather are dropped from the same height at the same

27 Times' Crucial Science Book - 10

time on the moon's surface, they fall together.
14. The mass of the Jupiter is 319 times more than that of the earth but its

gravity is only 2.5 times more than that of the earth.
15. The parachutists are not hurt when they jump out of an aeroplane.
16. When a person jumps from aeroplane with a parachute, he can land

safely on the earth surface but it is not possible on the moon.
C. Numerical Problems

1. Calculate the gravitational force between two bodies of masses
10kg and 55kg, if they are placed at a distance of 2m apart.

2. The mass of the Jupiter is 19×1026kg and that of the earth is
6×1024kg. If the distance between them is 6.29×108km, what
will be the force of gravitation between them?

3. The masses of the earth and the moon are 6×1024kg and
7.1×1022kg respectively and the gravitational force between them
is 1.682×1020N. Calculate the distance between them.

4. Mass of the earth is 6×1024kg and its radius is 6400km. Find
the acceleration due to gravity of the earth and weight of an
object of mass 50kg kept on the surface of the earth.

5. The mass of the earth is 6×1024 kg and its radius is 6400
km. What is the mass of a man weighing 777N in a spring
balance? (Universal gravitational constant 6.67 × 10–11 Nm2/
kg2. Use all data.

6. What is the weight of a body having mass 50 kg at a height of
3200km from the surface of the earth, if the mass of the earth
is 6×1024 kg and its radius is 6400 km?

7. The mass of the earth is 6×1024 kg and radius of the moon is
1.7×106 m. Calculate the acceleration due to gravity of the
new earth which is formed by the compression of the earth
equal to the size of the moon.

8. What mass of a body can a weight lifter lift on the Jupiter if
he can lift 100 kg mass on the earth?

9. Mount Everest is 8848 m above the sea level. Find the
acceleration due to gravity at this height. The value of
acceleration due to gravity at the earth’s surface is 9.8 m/s2
and the radius of the earth is 6400km.

10.What is the change in the force of gravitation between any
two bodies when mass of each body is doubled keeping the
distance between them?

Times' Crucial Science Book - 10 28

11.What is the change in the force of gravitation between any

two bodies when mass of each body is increased two times

and the distance between them is increased three times?.

Answers C.1. 9.17×10–9N 2. 1.92×1018N 3. 4.11×108m

4. 9.8m/s2, 490N 5. 79.5kg 6. 217.1N

7. 138.47 m/s2 8. 40 kg 9. 9.77 m/s2

10. 4 times 11. 4/9 times

Project Work

How to find the centre of gravity?

Try to find the centre of gravity of an object by performing the following
experiment Materials required:

A stick of length 1m.

Procedure:
Take a stick and hold between the thumb and
index finger as shown in the figure. Hold both
ends of the stick in position and move both hands
closer to each other. The point where both the
hands meet, is the centre of gravity of the stick. The centre of gravity is
the point where a small support balances a body without making it lean
or fall down.

Glossary : effect
: not changing, same throughout
• Influence : non uniform
• Uniform : a coin of England
• Random : a person who jumps with a parachute from a
• Guinea
• Parachutist height
: an imaginary circular line that divides the
• Equator
earth into two equal hemispheres
• Intensity : strength, concentration, power

29 Times' Crucial Science Book - 10

Chapter

2 Pressure
Blaise Pascal

Blaise Pascal was a French mathematician,
physicist, inventor and writer.The SI unit of
pressure is Pascal in his honour.

Estimated Periods: 10 (8T+2P)

Objectives

At the end of the lesson, students will be able to:
• define pressure and demonstrate liquid pressure;
• state and prove Pascal’s law;
• state Archimedes’ principle and describe its application with examples;
• demonstrate the law of floatation;
• Solve simple numerical problems related to Archimedes’ principle, Pascal’s law
and liquid pressure.

Introduction

There are many questions in our mind related to the pressure. Why
are tyres of tractors made wider? Why is a nail made pointed? Why
do foods cook faster in a pressure cooker than in an open pot? Why do
buckets get filled faster at the ground floor than at upper floors even
though taps are opened equally?

The answers to these questions can be obtained after getting concept
of pressure.

Pressure can be defined as the force acting perpendicularly on unit
area.

Pressure = Force
Area

The SI unit of pressure is N/m2 or Pascal. It is a scalar quantity
because pressure is always normal to the surface area, i.e. it does not
have direction.

One Pascal pressure

We know that

P = F and A = 1 m2 then,
If F =A1N

Times' Crucial Science Book - 10 30

P = 1 = 1 Pascal
1
Thus, 1 Pascal is defined as the pressure created when 1 Newton

force acts on the area of 1m2.

Atmospheric pressure

Our earth is surrounded by layers of the air called atmosphere. It is
extended to 9600km above the earth surface. The atmosphere has
weight due to which it exerts pressure. The pressure exerted by the
atmosphere is called atmospheric pressure. Atmospheric pressure
decreases when altitude from the sea level increases. The atmospheric
pressure at the sea level is 105 Pascal. It is also called standard or
normal atmospheric pressure.

1 atmosphere = 760mm of Hg = 105 Pascal.

Differences between force and pressure

Force Pressure

1. It is an external agency which 1. Force acting perpendicularly
changes or tends to change on unit area is called pressure.
the state of rest or motion of
a body in a straight line.

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

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

4. It is the cause of pressure. 4. It is the effect of force.

Thrust

The total perpendicular force exerted by a body on the surface is
called thrust. It is measured in Newton. It is a vector quantity.

Solved Numerical Problem 2.1

Ram has mass of 50kg. When he stands on his sandals, the surface
area occupied is 200cm2. Find the pressure exerted on the ground.

Solution:

Given, Mass (m) = 50kg

Force (F) = mg = 50×9.8 = 490N

Area (A) = 200cm2

31 Times' Crucial Science Book - 10

= 200 ×(1010)2 m2 = 0.02m2
Pressure (P) = ?

According to formula,

P= F Or, = 490 Or, = 49000 = 24,500 Pascal
A 0.02 2

∴ The pressure exerted is 24,500 Pascal.

Solved Numerical Problem 2.2

Calculate the pressure exerted by the brick in the following case.

6 cm

Solution 20N 4 cm
Given, Force (F) = 20N 12 cm
Contact surface area of the brick (A)

= 12cm×4cm = 0.12m×0.04 = 0.0048m2

Pressure (P) = ?

We have,

P = F Or, = 20 Or, = 416.67 Pascal
A 0.0048

∴ Pressure exerted on the ground is 4166.67 Pascal.
Uses of pressure in practical life

a. A knife contains sharp edge.

When edge of a knife is made sharp, the force applied by the knife
acts on a small area thereby creating more pressure. Thus, the
knife cuts easily when the edge is sharp.

b. The foundation of a building is made wider than the walls.

The foundation has to withstand the whole load exerted by the
walls and roofs of the building. When the foundation is wider, the
whole load of the building exerts on large area thereby creating
less pressure. Thus, the building is saved from collapsing.

c. Why are tyres of tractors made wider?

The tractors have to carry heavy loads or they have to move in
muddy fields. When they have wide tyres, the huge force exerted

Times' Crucial Science Book - 10 32

by the heavy load acts on large area of the ground through wide
tyres. This reduces the pressure on the ground and prevents the
tyres dipping in the road.

d. Nail is made pointed at end. Why ?

Pointed end of the nail has less area. When force is applied at the
flat end, the more pressure can be created by the pointed end on
the wood. So, the nail can easily penetrate inside the wood.

Liquid pressure

Liquids do not have fixed shape but have fixed weight. Due to weight,
liquid exerts pressure. When a liquid is kept in any container, it
exerts pressure on the base as well as on the walls of the container.

The pressure exerted by a liquid is called liquid pressure. Its SI unit is
Pascal. It can be measured in the unit mm of Hg also.

Properties of liquid pressure

a. The liquid pressure is directly proportional to the depth from the
free surface of the liquid.

b. The liquid pressure is independent of the shape of the container
in which it is kept.

c. A liquid determines its own level.
d. Pressure applied to an enclosed fluid transmits equally in all

directions.
e. Liquid pressure increases with density.
f. The liquid pressure at the same depth is same in all directions.

Pascal’s law

French scientist Blaise Pascal (1623 AD) propounded a law related to
liquid pressure in 1647 AD which is popularly known as Pascal’s law.

Pascal’s law states that “the pressure applied at a place on a liquid
contained in a closed container transmits equally perpendicularly in
all directions”.

Pascal's law is also called the principle of transmission of liquid
pressure. A number of devices such as hydraulic press, hydraulic
garage lift, etc work on the principle of Pascal's law.

Pascal's law is also applicable for the gases.

33 Times' Crucial Science Book - 10

Activity 2 .1 Experimental verification of Pascal’s law .

Materials required:
A balloon or plastic bag, needle, water.

Procedure:
a. Take a balloon or a plastic bag.
b. Make about 20 small holes on the balloon

or plastic bag with a needle.
c. Fill the balloon with water and close its

mouth.
d. Squeeze the balloon slowly from upper end. Observe the outflow

of water from the balloon.

Observation:
The water comes out from all holes with equal force.

Conclusion:
When the balloon is squeezed, pressure is applied to the water at a
point. The applied pressure is transmitted equally to all sides.

Therefore, the water comes out from all holes with equal force.

Activity 2 .2 Experimental verification of Pascal’s law.

Materials required: A
A spherical container fitted with four pistons
D
of equal cross sections and water.

Procedure: B

a. Take a spherical container fitted with C

four pistons A, B, C and D of equal cross

sections.

b. Fill the container with water

c. Push the piston A inward with a force. What happens to the

other pistons? Observe.

Observation:
All the other pistons are pushed outwards when the piston ‘A’ is
pushed inward. The outward distance moved by all the pistons will
be equal to each other as well as to the inward distance moved by

the piston A.

Conclusion:
Pressure given to the water of the container through piston ‘A’ is
transmitted to all directions. So all the other pistons moved
outwards with equal distance. It proves that the liquid pressure
transmits equally in all directions when pressure is applied to a
liquid kept in a closed container.

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Application of Pascal’s law F1 F2
A1 A2
Pascal’s law can be used in
hydraulic lift, hydraulic brake,
hydraulic press, etc.

Hydraulic press

Hydraulic machine is a device Hydraulic lift
which works on the principle of

Pascal’s law and magnifies the applied force to a greater extent.

Construction:

It consists of two cylindrical tubes of different diameters connected
with horizontal tube ‘H’ as shown in the diagram. The tubes are filled
with water. Small cylinder is fitted with an airtight piston having
cross sectional area A1 and large cylinder is fitted with an airtight
piston having cross sectional area A2.

Working mechanism

When a small force F1 is applied to the smaller piston, pressure P1
is exerted to the liquid. This pressure is transmitted to the larger
piston. Suppose, the pressure created on the larger piston is P2.

According to Pascal’s law,

Pressure exerted by small Piston (P1) = Pressure created on large
piston (P2)

Force exerted on the small piston (F1) = Load kept on the large piston (F2)

Cross sectional area of the small piston (A1) Cross sectional area of the large piston (A2)

Or, F1 = F2
A1 A2

Or, A2 = F2
A1 F1

Since, A2 > A1 then F2 >F1. ThOur,sF, 2a=smF1aA×l1lAf2orce can be multiplied by a
hydraulic machine.

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Solved Numerical Problem 2. 3

A hydraulic press has a small cylinder fitted with a piston having
cross section area of 4cm2 and a large cylinder fitted with a piston
having cross sectional area 20cm2. If the force applied on the small
piston is 20N, calculate the load that can be lifted on large piston.

Solution: 20N

Given, A1 (4cm2) F2
Area of small piston (A1) A2 (20cm2)

= 4cm2

= 4 × 1 = 0.0004m2
(100)2

Force on small piston (F1) =20N
Area of large piston (A2) = 20cm2

= 20 × 1 m2 = 0.002m2
(100)2

Load on large piston (F2) = ?

We have,

F2 = F1
A2 A1

Or, F2 = F1 × A2 = 20 × 0.002
A1 0.0004

= 100N.

∴ 100N load can be lifted on the large piston

Hydraulic garage lift

Hydraulic garage lift is a device which is used to lift light vehicles in
the garage during their maintenance or service.

Construction:

It consists of three metallic cylinders of different diameters. They are
connected together by a horizontal connecting tube and whole system
is filled with water. The piston of small cylinder is fitted with a lever
whereas the piston of large cylinder is fitted with a platform to rest
vehicles for maintenance. The reservoir and the large cylinder are
also connected by another tube which is provided with stopper ‘s’
as shown in figure. The horizontal connecting tube is provided with
valves X and Y.

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Handle

Reservoir Effort Small Platform
X Piston Larger Piston

Y Oil or water

stopper
Hydraulic garage lift

Working mechanism

When effort is applied to the small piston with the help of lever, the
valve X gets closed and valve Y opens. The pressure exerted to the
liquid by the small piston is transmitted to the large piston. The effort
is magnified by the large piston due to its large area. Thus, vehicle
can be lifted. When stopper is opened, the water flows from large
cylinder to the reservoir thereby bringing the vehicle to the previous
position.

Hydraulic brakes

Hydraulic brakes are the brakes which work on the principle of
Pascal’s law. They are mostly used in heavy vehicles.

Master cylinder

Lever system Brake pads

Water or oil
Brake pedal

Hydraulic brake Wheel

Construction:

It contains a tube containing brake oil. Its one end is connected to
a master cylinder and another end with slave cylinder. A piston
connected to pedal is fitted to the master cylinder. The slave cylinder
is fitted with two pistons each connected to the brake shoes. The
pistons of slave cylinder have larger diameter than that of the master
cylinder.

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Working mechanism:

When the brake pedal is pressed, piston of the master cylinder exerts
pressure to the brake oil. This pressure is transmitted to the pistons
of slave cylinder. The force will be multiplied by many times due to
large area of these pistons. The large force hence produced presses the
brake shoes which in turn press the rim of wheel. Thus, the vehicle
stops when brake pedal is pressed.

Upthrust

When an object is placed in a liquid, the liquid exerts force to the
object in upward direction. For example, when you dip a cork inside
the water and release, it rises up to the surface immediately as if the
water is pushing it upwards forcefully. Similarly, when you pull a
bucket of water from a well, it is easier when it is inside the water.
The bucket becomes heavier as soon as it comes out of water. Boats,
ships, etc float on water. These all are due to upward force exerted by
the water surface called upthrust.

The resultant force with which a liquid pushes an object in upward
direction when it is kept in the liquid is upthrust. It is also called
buoyant force. Its SI unit is Newton.

Activity 2 .3 Measurement of upthrust.

Materials required: Spring Spring
A piece of stone, a spring balance, a Balance Balance

thread,beaker, water, etc. Stone Beaker
W1 Water
Stone (W2)
Procedure:
a. Take a beaker and fill it with water.

b. Tie a stone with a thread and suspend it to the hook of a spring

balance.

c. Measure its weight in air. Suppose, its weight is w1=15N
d. Immerse the stone in the water of beaker and measure its

weight.

Suppose its weight in water is w2 = 10N

Observation:
The weight of stone in the water is less than that in air by 5N.

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Conclusion:
When the stone is dipped in water, the water pushes the stone in
upward direction. Due to which, its weight is decreased. The
decrease in weight is equal to the resultant upward force exerted

by the liquid (uptrust).

∴Upthrust = Decrease in weight of object
= Weight of the object in air–weight of the object in liquid

∴ Upthrust = w1 – w2 = 15 – 10
∴Upthrust = 5N

Density and relative density

There are various types of substances around us. Some are heavier
and some are lighter. Two substances having equal volume may not
be equally heavier. If you take a piece of iron and a piece of wood of
equal size, the iron piece will be heavier than the wood. The lightness
or heaviness can be explained easily by the term density.

Density is defined as the mass present in a unit volume of a substance.

Mathematically, Density (D) = Mass (m) Or, D = M
Volume (v) V
The SI unit of density is kilogram per cubic meter (kg/m3). It is

measured in g/cc (CGS unit) also.

1 g/cc = 1000 kg/m3

A substance having higher density is heavier than the substance
having lesser density. The following table shows the density of some
substances:

Substance Density Relative
density
Gold Kg/m3 g/cc
Mercury 19.3
Lead 19300 19.3 13.6
Iron
Water 13600 13.6 11
Aluminium 8
Wood 11000 11 1
Kerosene 2.7
8000 8 0.6-0.8
0.8
1000 1

2700 2.7
600-800 0.6-0.8

800 0.8

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Relative density

When density of a substance is compared with density of water at 4°C,
it is called relative density of the substance.

Relative density of a substance is defined as the ratio of the mass of

a certain volume of the substance to the mass of the equal volume of

water at 4°C.

Relative density = Mass of a certain volume of a substance
Mass of the same volume of water at 4°C.

In other words,

It can be defined as the ratio of density of a substance to the density
of the water at 4°C, i.e.

Relative density of a substance = Density of the substance
Density of water at 4°C

Relative density is a simple ratio of two quantities. Therefore, it has
no unit.

Activity 2 .4 To show the relation of upthrust with density .

Materials required:

Two beakers, an egg, salt and water. Fresh Egg

Procedure: Water Salt
Solution
a. Take two beakers and fill two- Egg

thirds of both of them with water. ‘A’ ‘B’

b. Dissolve enough salt in water of

one of the beakers to make saturated salt solution.

c. Put an egg in fresh water then the same egg in the saturated

salt solution. Observe what happens?

Observation:
The egg sinks on pure water but floats on saturated salt solution.

Explanation:
The salt solution has more density than the pure water. Therefore,
the upthrust provided to the egg by the salt solution is more than
that by the pure water. Thus, the egg floats on salt solution due to
sufficient upthrust but sinks in pure water due to lack of sufficient

upthrust.

Conclusion:
Upthrust increases when the density of the liquid increases.

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Density of liquid and upthrust

The upthrust exerted by a liquid to an object depends upon the density
of the liquid. The higher the density of the liquid, the greater will be
the upthrust on the object. And the lesser the density, the lesser will
be upthrust on the same object.

Archimedes’ principle

When an object is kept in a liquid its weight decreases. Greek scientist
Archimedes carried out an experiment to find out the magnitude by
which weight is decreased. The result obtained from the experiment
was incorporated in a principle named Archimedes’ principle. It
states that “when an object is wholly or partially immersed in a
liquid, the upthrust experienced by it is equal to the weight of the
liquid displaced”. This principle is applied for gas too.

Theoretical verification of Archimedes’ principle

Let a cylindrical object of height ‘h’ and h2 h1 A B
cross sectional area ‘A’ be immersed in a D h
liquid of density ‘d’ as shown in the figure.
The horizontal thrusts given by the liquid to C
the cylinder from various sides balance each
other because they are equal in magnitude
due to the same depth. But, they are opposite
in direction.

Let the depth of the upper and lower surface of the cylinder are h1
and h2 respectively.
The liquid pressure acting on the upper face of the body,

P1= d×g×h1 ............................(i)
The liquid pressure acting on the lower face of the body,

P2= d×g×h2.........................(ii)
Similarly,

The force acting on the upper face is given by

F1= P1×A...........(iii)
The force acting on the lower face,

F2= P2×A....................(iv)

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Since h2> h1, the force on the lower face (F2) is greater than the force
on the upper fall (F1). So,
Net resultant thrust (upthrust) on the cylinder,

F = F2 – F1=P2A–P1A
Or, F = dgh2A – dgh1A

= dg A (h2– h1)...........(v)
Let h2– h1 = h is the thickness of the rectangular body.
From equation (v), we have

F = d×g×A×h

∴ F = d×g×v ...........(vi)

(∴v= A× h = Volume of body = Volume of liquid displaced)

Equation (vi) is the mathematical form of Archimedes principle.

Here, the quantity v× d×g represents the weight of the liquid displaced.

∴ v×d×g = m×g = weight

∴ Upthrust = Weight of the liquid displaced

Thus, Archimedes’ Principle is verified. The Archimedes' principle
is used to operate hot air balloon submarine, hydrometer, sailing of
ship, etc.

Activity 2 .5 Experimental verification of Archimedes’ principle.

Materials required:

A eureka can, spring Spring
balance, small stone, a Balance

piece of thread, pan

balance, water, beaker, etc. Thread

Procedure Stone Water
a. Take a eureka can W2(Weight) Small beaker (W4)

and pour water into Top Pan balance

it upto spout level.

b. Place a beaker over Wooden block
pan balance just

below the spout of the can as shown in the figure.

c. Note the weight of the empty beaker as w1.

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d. Take a piece of stone and measure its weight with spring
balance. Note its weight in air as w2.

e. Now, immerse the stone in the water of the Eureka can and
measure its weight in water as w3.

f. The water displaced by the stone gets collected in the beaker
kept over the pan balance. Note the weight of the displaced
water and beaker as w4.

Observation and calculation
Loss in weight of stone in water

= Weight of the stone in air – Weight of the stone in water
= w2 – w3.
Weight of displaced water
= Weight of the beaker + displaced water – Weight of the empty
beaker
= w4 – w1

Conclusion:
The decrease in weight of the stone (upthrust) is equal to the weight
of the displaced water.
Thus, Archimedes’ principle is verified.

Law of floatation
When a body is kept in a liquid, two types of forces exist here.

i. the weight of the body which acts vertically downward.
ii. the upthrust on the body acting vertically upwards.

Depending upon the magnitude of these forces, following three
conditions may exist:

QR

P
ab c

1. Condition I: When the weight of a body is greater than the
upthrust, the resultant force is vertically downward. Then the body
sinks as shown in the figure (a). This condition exists when density of
a body is greater than the density of the liquid.

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2. Condition II: When the weight of a body is equal to the upthrust
acting on it, the resultant force acting on the body is zero. Then, the
body floats just in the liquid as shown in figure (b).

This condition exists when density of a body is equal to the density of
liquid.
3. Condition III: When the weight of the body is less than the upthrust
acting on it, the resultant force acting on the body is directed upward.
As a result the body floats with its some part in the air and little part
in the water as shown in figure (c). In this case, the weight of the body
is equal to the weight of the liquid displaced by the immersed part of
the body. This condition exists when density of a body is less than the
density of the liquid.

If you keep a rubber cork and a wooden cork in water, both of them
float. But more part of the rubber cork remains under the water and
less part remains in air whereas more part of the wooden cork remains
in the air and less part remains in the water. Here, both the corks
displace the water equal to their weight. But, the wooden cork (less
density) is light. When it displaces less water, it becomes equal to its
weight. On the other hand, the rubber cork is heavy (more density). It
has to displace more water by submerging its more part to make the
weight of the displaced water equal to its weight. Thus, weight of the
displaced liquid is the main factor for the floatation of an object. It is
explained by law of flotation.

The law of floatation states that weight of a floating object displaces
liquid equal to its weight.

Activity 2 .6 Experimental verification of law of floatation

Materials required: 10N
A eureka can, a beaker, a top pan
balance, a dry wooden block, a
piece of thread, and a spring

balance.

Procedure: 2N 12N
a. Take a euraka can and fill it

with water up to its spout level.

b. Place an empty beaker over top pan balance below the spout of

the euraka can. Note the weight of the empty beaker as w1 = 2N

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c. Take a dry wooden block and measure its weight with the help
of a spring balance. Note the weight of the wooden block as w2 = 10N

d. Place the wooden block in the water of eureka can slowly, Now,
the water is displaced by the wooden block and is collected in
the beaker kept under the spout.

e. Measure the weight of the beaker with displaced water. Note it

as w3 = 12N

Observation and calculation
Weight of the empty beaker (w1) = 2N

∴ Weight of the displaced water = w3 – w1 = 12 – 2 = 10N
∴ Weight of the dry wooden block (w2) = Weight of the displaced water.

Conclusion:
The weight of a floating object is equal to the weight of the displaced
liquid.

Difference between Archimedes principle and the law of floatation

S.N. Archimedes' Principle S.N. Law of flotation

1. When an object is immersed 1. An object that floats

fully or partially in a liquid, it on liquid displaces

experiences an upthrust which the liquid equal to its

is equal to weight of liquid own weight.

displaced by it.

2. It is equally applicable for 2. It is applicable for the

floating as well as sinking floating objects only.

objects.

3. It can be applied to the working 3. It cannot be applied
of submarine, ship, hydrometer to the working of
etc. submarine because it
floats as well as sinks
Atmospheric pressure in water.

The envelope of air that surrounds the earth is called atmosphere.
It is denser near the earth and is rarer at high altitude. The air is
a mixture of different gases. The air is found near the earth due to
its force of gravity. The earth pulls air with a force of gravity so the
atmosphere cannot escape from the earth's surface. Due to the force
of gravity, air experiences weight, which acts towards the earth's
surface. The atmospheric pressure is due to the weight of air column
acting on a specified area at any place.

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The pressure of air that acts upon the unit area on the surface of the
earth is called atmospheric pressure.

The atmospheric pressure at the sea level is 105 N/m2 or 760 mm of
Hg or 1 atmosphere. This pressure is known as standard atmospheric
pressure. There are different other units for the measurement of
pressure.

1 atm = 760 mm mercury

1 atm = 760 torr

1 atm = 101325 Pascal

1 atm = 1.01 x 105 N/m2

1 atm = 1.01325 bar

The pressure is 1,000 times greater at the depth of 10,000 m from the
surface.

The atmospheric pressure decreases at high altitude because the air
column high above the earth surface is rare. We cannot live in the place
with very high or very low atmospheric pressure. The aeroplanes fly
at the high altitude but we do not feel difficulty in respiration and feel
normal. It is because the air pressure inside the aeroplane is controlled
so that the comfortable air pressure for our body is maintained.

The atmospheric pressure affects each and every thing on the surface
of the earth. But we do not feel the atmospheric pressure because our
blood pressure is slightly greater than the atmospheric pressure. Our
body also exerts equal pressure against the atmospheric pressure.
Hence, we do not feel atmospheric pressure. We can perform the
following activity to experience the atmospheric pressure.

Activity 2 .7 To demonstrate that air exerts pressure.

Materials required:
A cylindrical tin can,
water, burner or stove,
water tap, etc.

Procedure:
1. Take a cylindrical tin
can and pour some
water into it.

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