Approved by Government of Nepal, Ministry of Education, Curriculum
Development Centre, Sanothimi, Bhaktapur as an additional material
Green
10
Editor
Dr. Deepak Chand
M.Sc. (TU, Kirtipur, Kathmandu)
Ph.D. (University of Idaho, USA)
Author
Bishnu Prasad Bhatt
M.Sc. (TU, Kirtipur, Kathmandu)
Lalitpur, Nepal, Tel: 977-1-5529899
e-mail: [email protected]
www.greenbooks.com.np
Publisher: Green Books 10
Copyright: Author (2074 BS)
All rights reserved. No part of this book may
be reproduced, stored in a retrieval system,
or transmitted in any form or by any means
without prior permission in writing from the
author and editor.
Edition
First : B.S. 2074 (2017 AD)
Reprint : B.S. 2075 (2018 AD)
Revised : B.S. 2076 (2019 AD)
Revised : B.S. 2077 (2020 AD)
Illustrator
Prakash Samir
Layout
The Focus Computer
[email protected]
Printed in Nepal
Preface
It gives me an immense pleasure in presenting this book
Green Science for Class 10. This book is written specially to
meet the requirements of the new syllabus introduced by the
Government of Nepal, Ministry of Education, Curriculum
Development Centre, Sano Thimi, Bhaktapur, Nepal.
My aim and effort while writing this book has been to help
students understand, enjoy and appreciate the fascinating
subject of Science and Environment by making the process of
learning enjoyable and stimulating. I have attempted to present
the subject matter covering the entire prescribed syllabus in a
simple language and interesting style with a large number of
illustrative examples for easy understanding and application
of the fundamental principles of science. Each unit of the book
has been carefully planned to make it student-friendly and
present the subject matter in an interesting, understandable and
enjoyable manner. A Structural Programme Learning Approach
(SPLA) has been followed and exhaustive exercises are given
at the end of each unit to test knowledge, understanding and
applications of concepts taught/learnt.
The text is supplemented with weighting distribution, learning
objectives, word power, teaching instructions, sample test
papers and a large number of well-labelled accurate pictures. I
sincerely hope that this book will serve its intended purpose and
be received enthusiastically by both the students and teachers
concerned.
I wish to express my sincere gratitude to Green Books Team for
publishing this book. My hearty thank goes to Focus Computer
for excellent type setting and layout.
I also wish to acknowledge my great indebtedness to many
teachers for their valuable suggestions and advice concerning
the textbook. I am confident that as result of their suggestions
this book will be more useful than any other textbooks. However,
sympathetic criticisms and constructive suggestions for further
improvement of the book, if any, will be welcomed and with
warm regards incorporated in the subsequent editions.
Author
Kathmandu, Nepal
Contents
Physics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1. Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
2. Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
3. Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53
4. Heat . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67
5. Light . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83
6. Current Electricity and Magnetism . . . . . . . . . . . . . . 107
Chemistry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127
7. Classification of Elements . . . . . . . . . . . . . . . . . . . . . . 127
8. Chemical Reaction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149
9. Acid, Base and Salt . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168
10. Some Gases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 188
11. Metals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204
12. Hydrocarbon and Its Compounds . . . . . . . . . . . . . . 219
13. Materials Used in Daily Life . . . . . . . . . . . . . . . . . . . . 242
Biology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263
14. Invertebrates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263
15. Human Nervous and Glandular System . . . . . . . . . 280
16. Blood Circulation in Human body . . . . . . . . . . . . . . 299
17. Chromosomes and Sex Determination . . . . . . . . . . . 316
18. Asexual and Sexual Reproduction . . . . . . . . . . . . . . 328
19. Heredity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 346
20. Environmental Pollution and Management . . . . . . 361
Geology and Astronomy . . . . . . . . . . . . . . . . . . . . . . . . . 377
21. History of the Earth . . . . . . . . . . . . . . . . . . . . . . . . . . . 377
22. Climate Change and Atmosphere . . . . . . . . . . . . . . . 391
23. The Earth in the Universe . . . . . . . . . . . . . . . . . . . . . . 407
Specification Grid . . . . . . . . . . . . . . . . . . . . . . . . . . . . 424
Models Questions (Issued by CDC) . . . . . . . . . . . . 425
Model Question 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 427
Model Question 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 430
Physics
1UNIT Force
Weighting Distribution Theory : 8 Practical: 2
Before You Begin
Force is a very essential physical quantity to perform different types of
works. We are all familiar with the concept of force from our common
experiences. For example, while playing football, opening a door, pushing
a table, pulling a cart, beating a drum, ringing a bell, lifting a bucket of
water, etc. We have to act in different ways. A book kept on a table does
not move on its own. To move it, we must either lift it, push it or pull it.
The effort required to do so it is generally called force. So, force can be
defined as a pull or push which changes or tends to change the position
(rest or motion), the direction or the shape and size of a body. In SI system,
force is measured in newton (N). It is a vector quantity. There are different
types of forces, viz muscular force, frictional force, intermolecular force,
magnetic force, electrostatic force, gravitational force, etc. In this unit, we
will study about gravitational force. The force of attraction between any
two bodies due to their masses is called gravitational force.
Learning Objectives Syllabus
After completing the study of this unit, students will be able to: • Gravitation
• Newton’s law of gravitation
i. introduce force and its types. • Gravity and gravitational field
• Effects of gravity
ii. introduce gravitational force and state and prove • Acceleration due to gravity
Newton’s law of gravitation. • Feather and coin experiment
• Mass and weight
iii. introduce gravity, gravitational field and acceleration • Free fall and weightlessness
due to gravity. • Simple numerical problems
iv. differentiate between gravity and acceleration due to
gravity.
v. differentiate between mass and weight.
vi. introduce free fall and weightlessness and
differentiate between them.
Glossary: A dictionary of scientific/technical terms
force : the pull or push that changes or tends to change the position of a body
gravitation : the force of attraction between any two bodies due to their masses
acceleration : the rate of change in velocity of a moving body
gravity : the force which pulls an object towards the centre of a heavenly body
weightlessness : the condition in which the weight of a body of a certain mass becomes
zero
GREEN Science (Physics) Book-10 5
Concept of Gravitation
When a body is dropped from a certain height, it falls towards
the surface of the earth. Similarly, fruits fall down from trees.
It shows that the earth attracts various objects towards its
centre with a certain force. This force is called gravity.
Sir Isaac Newton came to the conclusion that it is not only the Fig.
earth which attracts the other objects, in fact, every object in Fig.
Fig.
this universe attracts every other object with a certain force. 1.1 Gravitational force
This force is called gravitation or gravitational force. So,
gravitational force can be defined as the force of attraction
between any two bodies due to their masses. Gravitational force is measured in newton (N).
According to Newton, every body in this universe attracts another body. For example, the earth
attracts the sun and the sun attracts earth too. Similarly, the moon attracts the earth and the
earth also attracts the moon. All the planets of the solar system attract each other. The effect of
gravitation between a very big body and a small body can be observed but the effect of a small
body on big body cannot be observed easily. The gravitational force acts even if two bodies
are not connected by any means. For example, the earth and the moon are not connected by
any means but gravitational force exists between these heavenly bodies. Similarly, gravitational
force exists between the earth and the sun. The planets of the solar system revolve around the
sun due to gravitational force. The sun is extremely large so that the sum of masses of all the
planets of the solar system is only about 0.0015th part of the solar mass.
Before 1542AD, it was believed that the earth remains at the centre and planets and the
sun revolve around it. This model is called geocentric model. Later Nicholas Copernicus
propounded a theory stating that all planets revolve around the sun and sun remains at
the centre. This model is called heliocentric model. Galileo watched the space through
telescope and found that there were four satellites revolving around the Jupiter. It helped
to prove the hypothesis of Nicholas Copernicous that all planets revolve around the sun.
Sir Isaac Newton propounded the famous law of gravitation in 1687 AD. The effect of
gravitation can be seen more on liquid than on solid. So, we see tides in the sea and ocean
due to gravitation of the sun and the moon.
Newton’s Universal Law of Gravitation
According to Newton’s law of gravitation, “Every object in the 1.2
universe attracts every other object with a certain force which is
directly proportional to the product of their masses and Newton
inversely proportional to the square of the distance between
their centres.” The direction of the force is along the line joining
the centres of the two bodies.
Derivation of the formula F = Gm1m2 m1 m2 B
d2 AF
Suppose two objects A and B of masses m1 and m2 1.3 d
are lying at a distance 'd' from each other. Let the Gravitational force
force of attraction between these two objects be ‘F’.
Now, according to Newton’s law of gravitation,
6 GREEN Science (Physics) Book-10
i. The force between two objects is directly proportional to the product of their masses.
That is,
F ∝ m1 × m2 ........................ (1)
ii. The force between two objects is inversely proportional to the square of the distance
between them from their centres. That is,
F ∝ 1 ........................ (2)
d2
Combining (1) and (2), we get
F ∝ m1 × m2 ........................ (3)
d2
F = Gm1 × m2 Where, G is a constant called “Newton’s universal gravitational constant”.
d2
\ Gravitational force, F=G m1 × m2 proved.
d2
The value of gravitational constant ‘G’ does not depend on the medium between two
bodies. It also does not depend on the masses of the bodies or the distance between them.
The above formula gives the force of attraction ‘F’ between two bodies of masses m1 and
m2 which are at a distance ‘d’ from one another. This formula is applicable anywhere in
this universe, and it is a mathematical expression of Newton’s law of gravitation. Since
the gravitational force between two bodies is inversely proportional to the square of the
distance between them, therefore, if we double the distance between two bodies, the
gravitational force becomes one-fourth and if we halve the distance between two bodies,
1
then the gravitational force becomes four times, q F ∝ r.
d2
Unit of Gravitational Constant (G)
According to Newton’s law of gravitation, the force ‘F’ between two bodies of masses m1
and m2 placed at a distance ‘R’ apart is given by,
F = G m1 × m2
R2
This can be rearranged to get an expression for the gravitational constant G as follows:
R2
G=F×
m1 × m2
Now, the unit of force F is newton (N), the unit of distance R is metre (m), and the unit of
masses m1 and m2 is kilogram (kg). So, the SI unit of gravitational constant (G) becomes
newton (metre)2 or Nm2 or Nm2/kg2 or Nm2kg–2.
(kilogram)2 kg2
GREEN Science (Physics) Book-10 7
Value of Gravitational Constant (G)
If the masses m1 and m2 of the two bodies are 1kg each and the distance R between them
is 1m, then putting m1 = 1kg, m2 = 1kg and R = 1m in the above formula,
We get, Fig. 1kg
F = Gm1 × m2 1kg
AB
R2
1×1 1m
or, F = G 1.4 F=G
12
or, F = G [when m1 = m2 = 1kg and R = 1m]
Thus, the gravitational constant (G) is numerically equal to the force of gravitation which
exists between two bodies of unit masses kept at a unit distance from each other. The
value of universal gravitational constant ‘G’ has been found to be 6.67 × 10–11 Nm2/kg2.
The value of gravitational constant
(G) has been determined by Do You Know
performing careful experiments
with two gold balls. Two heavy gold Gravitational constant (G) is a scalar quantity. Its
balls were suspended near each value remains the same throughout the universe.
other by strong but delicate threads. Newton’s law of gravitation is applicable for any
The force between the two gold balls two objects present in the universe whether those
were measured by using a sensitive objects are small or big, celestial or terrestrial.
balance called torsion balance. Therefore, Newton’s law of gravitation is called a
Knowing the force, the masses of universal law.
gold balls, and the distance between
them, the value of ‘G’ was calculated by using Newton’s gravitation formula. The
experiment with gold balls gave the value of gravitational constant (G) as 6.67×10–11 Nm2/
kg2.
Solved Numerical: 1
The mass of the earth is 6×1024kg and that of a satellite is 2×103kg. If the distance between
them is 15km, calculate the gravitational force between them. Take G = 6.67×10–11 Nm2/kg2.
Solution:
Given, Mass of the earth (m1) = 6 × 1024 kg
Mass of the satellite (m2) = 2 × 103 kg
Distance (d) = 15 km
= 15 × 1000 m
= 15000m
8 GREEN Science (Physics) Book-10
Gravitational constant (G) = 6.67 × 10–11 Nm2/kg2
Gravitational force (F) =?
We know,
F = G m1 × m2
d2
6.67×10–11×6×1024×2×103
=
(15000)2
6.67×6×2×10–11+24+3
=
2.25×108
80.04×1016
=
2.25×108
= 35.573 × 1016–8
= 35.573 × 108
= 3.557×109N
∴ The gravitational force between them is 3.557×109N.
Gravitational Force between Objects of Small Size and Big Size
Suppose two objects of 1kg each are kept at 1m apart from their centres, then they attract
each other with a force given by.
F = G m1 × m2
R2
Putting G = 6.67×10–11 Nm2/kg2, m1 = 1kg, m2 = 1kg and R = 1m.
We get,
F = 6.67×10–11×1×1 N
12
= 6.67×10–11 N
It is obvious that the gravitational force of attraction between two bodies of 1kg each and
1m apart is 6.67 × 10–11 N and this is a very small force. Though various objects on the earth
attract each other constantly, they do not cause any motion because the force of attraction
between them is very small. However, if at least one of the bodies is large (like the earth
or the sun) the gravitational force becomes very large. For example, the earth has a very
large mass and this force is quite large between the earth and other objects. Therefore,
objects when thrown up, fall back on the earth.
GREEN Science (Physics) Book-10 9
Consequences of Gravitation
Since the masses of the sun and the earth are extremely large, they exert very large force
on one another. It is the gravitational force between the sun and the earth which keeps
the earth in uniform circular motion around the sun. Similarly, the gravitational force
between the earth and the moon makes the moon revolve at uniform speed around the
earth. Thus, the gravitational force is responsible for the existence of the solar system. The
tides in the sea formed by rising and falling of water level in the sea are due to the force of
attraction which the sun and the moon exert on the water surface in the sea.
The gravitational force of the earth is responsible for holding the atmosphere above the
surface of the earth, for the rain falling to the earth and for the flow of rivers. It is also the
gravitational force of the earth which keeps us firmly on the ground.
Variation of Gravitational force with Mass and Distance
1. What change in gravitational force can be observed when the mass of a body is
tripled by keeping the distance between them unchanged?
Solution: We know,
F = Gm1 × m2
d2
According to the question, when the mass of a body is tripled, then
Fnew = G3m1 × m2
d2
d
Gm1 × m2
or Fnew = 3 d2 3m1 m2
or Fnew = 3F
∴ The gravitational force between any two bodies is tripled when the mass of a body is
tripled by keeping the distance between them unchanged.
2. What change in gravitational force can be noticed when the distance between
two bodies is doubled by keeping their masses unchanged?
Solution: We know,
F = Gm1 × m2
d2
According to the question, when the distance between two bodies is doubled, then
Fnew = Gm1 × m2
(2d)2
or Fnew = Gm1 × m2 2d
4d2 m1 m2
10 GREEN Science (Physics) Book-10
or Fnew = 1 Gm1 × m2
4 d2
or Fnew = 1 [F]
4
1
∴ The gravitational force between two bodies becomes 4 th of the original force when
the distance between them is doubled by keeping their masses unchanged.
Gravity
If we raise a body above the ground and Do You Know
then release it, the body falls down towards
the earth. Similarly, fruits fall down from the The area around a planet or a heavenly
trees. It shows that there must be a force body upto where the gravity of the planet
which pulls every object towards the centre or heavenly body has its influence on
of the earth. This force is called gravity. So, a body, is called gravitational field. It
gravity can be defined as the force which depends on the mass and radius of the
pulls an object towards the centre of the planet or heavenly body.
earth or a heavenly body. Gravity is a vector
quantity. In SI system, gravity is measured in newton (N). The direction of the gravity is
towards the centre of the planet or heavenly body.
All planets and satellites have their own gravity. Sky
The gravity of the earth is more on a large body than
a smaller one. Therefore, we can lift a small stone Sky
but it becomes very difficult to lift a large stone on
the surface of the earth. The weight of a body is the Sky
measure of the gravity acting on the body.
Earth
Even while we walk, run and jump, the gravity of the
earth affects us. The gravity of the earth also affects Fig. Sky Sky
the construction of house, building, bridge, etc. The
gravity of the earth also affects the movement of 1.5 Sky
vehicles, ships, aeroplanes, etc. We all know that the
earth is spherical. It may be surprising to think how people can stay on the spherical
earth. A person living in Nepal will have his head faced out to the sky and the same case
will be on the various parts on the earth like America, China, Australia, Canada, etc. The
reason behind this fact is that the gravity of the earth pulls all of them towards the centre
of the earth.
On the basis of Newtons’ law of gravitation, the gravity of a planet or a heavenly body can
be calculated by the given formula.
F = W = GMm
R2
Where,
F = Gravity
GREEN Science (Physics) Book-10 11
W = Weight of a body
G = Gravitational constant
M = Mass of the earth/planet
m = Mass of the body kept on the surface of the planet
R = Radius of the planet/heavenly body
The radius of the planet/earth is taken as the distance between the centre of the planet and
that of the object.
The Jupiter has its own gravity like that of the earth. Similarly, the moon has its own gravity
like that of the earth. According to the scientists, the gravity of the Jupiter is 2.5 times more
than that of the earth. Similarly, the gravity of the moon is six times less than that of the earth.
Solved Numerical: 2
Calculate the force of gravity of the earth that acts on 1kg object. The mass and radius of
the earth are 6×1024kg and 6380km respectively.
Solution:
Given,
Mass of the body (m) = 1 kg
Mass of the earth (M) = 6 × 1024 kg
Radius of the earth (R) = 6380 km = 6380 × 1000 m = 6.38×106m
Gravity (F) = ?
According to the formula, [ G = 6.67×10–11Nm2/kg2]
We know,
GMm
F =
R2
6.67×10–11×6×1024×1
=
(6.38×106)2
6.67×6×10–11+24
=
40.70×1012
= 40.02 = 0.98×101 = 9.8m/s2.
40.70
∴ The gravity of the earth that acts on an object of 1 kg is 9.8N.
12 GREEN Science (Physics) Book-10
Differences between Gravitation and Gravity
Gravitation Gravity
1. Gravitation is the force of attraction 1. Gravity is the force with which a body
between any two bodies due to their is pulled towards the centre of the
masses. earth/ planet.
2. It depends on the mass of two bodies 2. It depends on the mass and radius of
and distance between them from their the earth/ heavenly body.
centres.
Acceleration due to gravity t = 0s
u=0
An object thrown to the sky returns to the surface of
the earth due to the gravity of the earth. The velocity of 4.9 m
the body falling towards the earth’s surface increases
at a constant rate. In other words, when a body is 19.6 m t = 1s,
dropped from a certain height, a uniform acceleration g = 9.8m/s²,
is produced in it by the gravity of the earth which is d = 4.9m
known as acceleration due to gravity. So, acceleration
due to gravity (g) can be defined as the uniform
acceleration produced in a freely falling object due
to the gravity of the earth/ heavenly body. Its SI unit
is m/s² or ms–² (metre per second per second). It is a
vector quantity. -
The acceleration due to gravity of all the falling objects t = 2s,
towards the surface of the earth will be the same in the g = 9.8 m/s²,
absence of external resistance. d = 19.6m
Fig.
Feather and Coin Experiment Fig. 1.6
When an object falls towards the Feather Vacuum
surface of the earth under the effect Air
of gravity, its velocity increases at Coin Feather
the constant rate of 9.8 m/s in every Coin
second of time it is falling. It means
that when a body is dropped freely, 1.7 Feather and coin experiment
it falls with an acceleration of 9.8 m/
s². So, the velocity of a body thrown
vertically upwards will decrease at
the rate of 9.8m/s² until it reaches
zero. The body then falls back to
the earth due to the effect of earth’s
gravity.
GREEN Science (Physics) Book-10 13
Robert Boyle performed and experiment in a vacuum to prove that the acceleration
produced on every freely falling body remains the same and it does not depend on the
mass of the falling body. This famous experiment is commonly known as feather and coin
experiment.
In feather and coin experiment, a feather and coin are placed inside a transparent glass
tube as shown in the given figure. Both ends of the glass tube are closed and the glass tube
is inverted. It is observed that the coin reaches the bottom earlier than the feather when
the tube contains air.
Then the air is taken out of the glass tube with the help of a vacuum pump and the experiment
is repeated. In this case, both the feather and coin reach the ground simultaneously. When
the tube contains air, the feather falls down slowly due to air resistance as the surface area
of the feather is larger than that of the coin.
From the feather and coin experiment, it can be concluded that, “the acceleration produced
on freely falling bodies remains the same for all the bodies and it is independent of the
mass of these bodies.”
Activity 1
Take a whole brick and another small piece of brick.
Select a suitable place (porch or ceiling) to drop the brick and the piece of the brick.
Place a sheet of tin on the ground where the bricks strike.
Drop the brick and the piece of brick at the same time from the porch or ceiling.
Listen carefully to the sound produced by the bricks when they strike the sheet of tin.
Which object will strike the sheet faster?
Now, take a sheet of paper and a ball made of paper. Drop them on the sheet of tin
simultaneously. Which one will reach the ground faster?
What conclusion can be drawn from this activity?
Calculation of value of acceleration due to gravity (g)
If we drop a body of mass 'm' from a distance of 'R' from the centre of the earth of mass 'M',
then the force exerted by the earth on the body is given by Newton's law of gravitation as:
F = GMm ................. (1)
R²
This force exerted by the earth produces acceleration in the body due to which the body
moves downwards. According to Newton's second law of motion,
F = mg ....................... (2)
From equation (1) and (2), we get, Fig. R
mg = GMR²m M
or, g = GRM²mm 1.8 Earth
14 GREEN Science (Physics) Book-10
or, g = GM
R²
or, g ∝ 1 [∵ G and M are constants.]
R2
From the above relation, it becomes clear that the mass of the falling body does not affect
the acceleration due to gravity. The acceleration due to gravity of the falling bodies is
inversely proportional to the square of the radius of the earth or a planet.
Solved Numerical: 3
The mass of the earth is 6 × 1027 g and its radius is 6400km. Calculate the acceleration due
to gravity on the surface of the earth. Take G = 6.67 × 10 – 11 Nm²/kg².
Solution: = 6 × 1027 g = 6 × 1024 kg [∵ 1000g = 1kg]
Given, Mass of the earth (M) = 6400 km
Radius of the earth (R)
= 6400 × 10³ m [∵ 1km = 1000m]
Gravitational constant (G) = 6.67 × 10–11 Nm²/kg²
Acceleration due to gravity (g) = ?
We know,
g = GM
R²
= 6.67 × 10–11 × 6 × 1024
(6400 × 10³)²
= 6.67 × 6 × 10– 11 + 24
4.096 × 1013
= 40.02 × 1013
4.096 × 1013
= 9.77m/s²
\ The acceleration due to gravity on the surface of the earth (g) = 9.77m/s².
Acceleration due to gravity of the earth
The value of acceleration due to
gravity depends on the values of G, Do You Know
M and R. As G and M are always
constant, the value of acceleration The value of acceleration due to gravity (g) is
due to gravity (g) depends on the maximum on the surface of the earth. It decreases
radius of the earth/ planet. The value on going above the surface of the earth or on going
of 'g' remains constant at a certain inside the surface of the earth.
place but its value changes from
place to place on the surface of the earth. This is due to the fact that the earth is not a
perfect sphere. So, the value of its radius (R) is not same at all the places on its surface. The
earth is flat at poles and bulging out at equator, all the places on its surface are not at the
same distance from its centre, so the value of 'g' varies with latitude. Since the radius of
GREEN Science (Physics) Book-10 15
the earth at the poles is minimum, the value of 'g' is maximum (9.83m/s²) at poles. Again,
the radius of the earth is maximum at the equator, the value of 'g' is minimum (9.78m/s²)
at the equator. However, the average value of 'g' on the earth's surface is taken as 9.8m/s².
The value of acceleration due to gravity (h) at a certain height (h) from the earth’s surface
is calculated by the given formula:
g = GM
(R + h)2
Where 'h' is the height from the surface of the earth.
The value of 'g' is inversely proportional to the square of the distance from the centre of
1
the earth o g ∝ R² p. As we go upward from the surface of the earth, the distance from
the centre of the earth increases, and hence the value of 'g' decreases.
The above formula suggests that the value of 'g' should increase on going down inside the
earth because the value of 'R' decreases then we go down from the earth’s surface. This,
however, is not true. Actually, this formula for 'g' is not applicable at any point inside the
surface of the earth. Please note that the value of 'g' also decreases as we go down inside
the earth, and it becomes zero at the centre of the earth.
value of 'g' decreases
while going upwards
h from the surface
g (maximum
at poles)
Do You Know
value of g decreases
R while going The value of acceleration due to gravity
(g) is maximum on the surface of the
downwards from
the surface earth and it decreases on going above
the surface of the earth or on going
Fig. inside the surface of the earth.
Earth The value of ‘g’ is zero at the centre of
1.9 the earth.
Acceleration due to gravity on the jupiter
Jupiter is the largest planet of the solar system. Its radius is 11 times and mass is 319 times
than that of the earth. The value of acceleration due to gravity on the surface of the jupiter
is 25.13 m/s². It shows that the value of acceleration due to gravity on the surface of the
jupiter is about 2.5 times than that on the earth.
The acceleration due to gravity does not increase significantly though its mass is 319 times
that of the earth because the radius of the jupiter is 11 times that of the earth. On the other
hand, if the earth is compressed to the size of the moon, the value of acceleration due to
gravity will increase 14 times. o g ∝ M, g ∝ 1 p.
R²
16 GREEN Science (Physics) Book-10
Differences between Gravity (F) and Acceleration due to gravity (g)
Gravity (F or W) Acceleration due to gravity (g)
1. It is the force with which a body is 1. It is the acceleration produced in a
pulled towards the surface of the earth freely falling body due to the effect of
or a heavenly body. gravity.
2. Its SI unit is newton (N). 2. Its SI unit is m/s² or ms–2.
3. It produces acceleration in the falling 3. It is the effect of gravity.
bodies.
Solved Numerical: 4
The mass of the earth is 6 × 1024 kg and its radius is 6380km. Calculate the acceleration
due to gravity at the top of Mt. Manaslu of height 8163m. What will be the weight of a
mountaineer of mass 75 kg on the top of Mt. Manaslu?
Solution:
Given,
Mass of the earth (M) = 6 × 1024 kg
Radius (R) = 6380 km
= 6380 × 1000 m = 6380 × 103 m
Height of Mt. Manaslu (h) = 8163 m
Acceleration due to gravity (g) = ?
We know,
g = GM
(R + h)²
= 6.67 × 10–11 × 6 × 1024 [∵ G = 6.67 × 10 –11 Nm²/kg]
(6380 × 10³ + 8163)²
= 6.67 × 6 × 10– 11 + 24
4.08 × 1013
= 40.02 × 1013
4.08 × 1013
= 9.8m/s²
Now,
\ Acceleration due to gravity (g) = 9.8m/s²
Mass of a mountaineer (m) = 75kg
Weight of the mountaineer (W) = ?
GREEN Science (Physics) Book-10 17
We know,
W = m × g
= 75 × 9.8 N
= 735 N
\ The weight of the mountaineer at the top of Mt. Manaslu is 735N.
Acceleration due to gravity on the surface of the moon
The mass (7.2 × 10²² kg) and radius (1.7 × 106 m) of the moon are less than that of the earth.
So the acceleration due to gravity (g) of the moon is less than that of the earth. It is about
1 th of the earth, i.e. 1.67m/s². From this fact, it becomes clear that a body weighing 600 N
6
on the earth weighs only 100 N on the moon. Similarly, a person who can jump 1 m high
on the earth’s surface can jump 6 m high on the surface of the moon.
Solved Numerical: 5
A weightlifter can lift 150kg mass on the earth. How much mass can he lift on the surface
of the moon? The value of 'g' on the earth is 9.8m/s² and that on the moon is 1.67m/s²?
Solution:
Given, On the earth,
The weight that can be lifted by a weightlifter (W) = m × g
= 150 × 9.8
= 1470 N
\ The weight that a weightlifter can lift on the moon (W) = 1470 N
On the moon,
Weight that can be lifted on the moon (W) = 1470 N
Acceleration due to gravity on the moon (g) = 1.67m/s²
Mass that can be lifted on the moon(m) = ?
We know,
W = m × g
m = W = 1470 = 880.23kg
g 1.67
\ The mass that a weightlifter can lift on the moon (m) = 880.23 kg.
18 GREEN Science (Physics) Book-10
Mass and Weight
In physics, mass and weight are two different physical quantities. However, mass is also
called weight in practice. The objects with equal masses have equal weight on a certain
place of a certain planet.
Mass
Mass of an object is defined as the total amount
of matter present in a body. Its SI unit is kilogram
(kg). Mass is also measured in gram (g), milligram
(mg), metric ton, etc. It is measured by using a
physical balance or a beam balance. Mass is a
constant quantity, i.e. it does not change from
place to place. It is a scalar quantity.
Factors that affect the mass of a body
i. Size of atoms or molecules of the body.
Fig.1.10
Fig.
Beam balance
ii. Number of atoms or molecules of the body.
Weight
The weight of a body can be defined as the measure of gravity acting on the body. It is the
force with which a body is pulled towards the centre of the earth/planet. The SI unit of
weight is newton(N). It is measured by using a spring balance. Weight is a variable quantity.
The weight of a body differs from place to place on the surface of the earth because of
the difference in the value of acceleration due to gravity from place to place. The value
of acceleration due to gravity (g) is more at polar region than that in the equator. So, the
weight of a body is more at polar region than that in the equator. Similarly, the weight of
a body is more in the Terai than that in the hilly region. The weight of a body decreases
when shifted to a higher altitude from the surface of the earth.
1.11 Do You Know
Spring balance
The value of 'g' of the Jupiter is 2.5 times
more than that of the earth. So the weight
of a body on the surface of Jupiter is 2.5
times more than that on the surface of the
earth.
Factors that affect the weight of a body
i. Mass of the body
ii. Value of acceleration due to gravity (g) where the body is kept
GREEN Science (Physics) Book-10 19
Differences between Mass and Weight
Mass Weight
1. Mass of a body is the total amount of 1. Weight of a body is the force with
matter present in the body. which the body is pulled towards the
centre of the earth/planet.
2. It is measured by using a pan balance 2. It is measured by using a spring
or a beam balance. balance.
3. Its SI unit is kilogram (kg). 3. Its SI unit is newton (N).
4. Its value does not change from place 4. Its value changes from place to place.
to place.
Activity 2
Take a beam balance and a spring balance.
Measure the mass and weight of the following objects using those devices:
i. Science book ii. Pen iii. Duster
iv. Geometry set v. Calculator vi. Science notebook
Free Fall
Take a piece of stone and a sheet of paper. Drop them simultaneously from the same
height. What do you observe? The stone reaches the ground faster than the sheet of the
paper. Due to air resistance, the velocity of the sheet of the paper is reduced but the effect
of air resistance on the stone is negligible. So the piece of stone reaches the ground faster.
The acceleration produced in the piece of stone is equal to the acceleration due to gravity
of that place. So, falling of the stone can be considered as free fall.
However, it is not called a complete free fall due to the presence of air resistance on the
surface of the earth. But falling of the paper is not a free fall because acceleration produced
in the sheet of paper is less than the acceleration due to gravity. When a body falls freely
only under the effect of gravity, the fall of the body is called free fall. When a body falls
downwards in a vacuum, the fall of the body is considered a free fall. Similarly, the fall of
a body towards the surface of the moon is called a free fall as there is no atmosphere on
the surface of the moon.
The spacecrafts revolve around the Do You Know
earth in a circular path. Centripetal
force is essential for an object to There is no perfect free fall on the earth due to
revolve in a circular path. This force air resistance.
should be equal to the gravitational
force between the earth and the An abject appears to be weightless as a result of
free fall.
object. Free fall is also experienced
by the astronauts while revolving around the earth inside the spacecraft.
Equation of motion for freely falling objects
When a body falls freely towards the surface of the earth, a uniform acceleration is
produced due to the effect of earth's gravity. Three equations of motion are modified to
get equation of motion for freely falling bodies as follows:
20 GREEN Science (Physics) Book-10
General equation of motion Modified equation of motion
1. v = u + at 1. v = u + gt
2. s = ut + 1 at² 2. h = ut + 1 gt²
2 2
3. v² = u² + 2as 3. v² = u² + 2gh
For freely falling bodies, the acceleration due to gravity is 'g', so we replace the acceleration
'a' of the equation with 'g' and since the vertical distance of freely falling body is 'h', we
replace the distance 's' in the equation with the height 'h'.
The points that we should remember while solving numerical problems of falling bodies
are as follows.
1. When a body is dropped freely from a certain height, its initial velocity (v) is zero.
2. When a body is thrown vertically upwards, its final velocity (v) is zero.
3. When a body is falling vertically downwards, its final velocity increases. So the
acceleration due to gravity (g) is taken as positive.
4. When a body is thrown vertically upwards, its final velocity decreases. So the
acceleration due to gravity (g) is taken as negative.
Solved Numerical: 6
A ball is thrown vertically downwards from the top of a tower. If the ball takes 3 seconds
to reach the ground, calculate the height of the tower.
Solution:
Given, Initial velocity (u) = 0 [∵ The ball is thrown from rest.]
Time taken (t) = 3s
Acceleration due to gravity (g) = 9.8m/s²
Height of the tower (h) = ?
We know,
h = ut + 1 gt²
2
= 0×3+ 1 × 9.8 × 3²
2
= 12 × 9.8 × 9
= 44.1m
\ The height of the tower (h) = 44.1m
Weightlessness
When a body falls freely, its weight appears to be zero. This condition is called
weightlessness. A body becomes weightless during free fall, at null point and at the centre
of the earth. So, weightlessness can be defined as the condition at which a body of certain
mass becomes weightless.
GREEN Science (Physics) Book-10 21
Conditions for weightlessness
A body becomes weightless under the following conditions:
a. When a body is falling freely.
b. When a body is at the centre of planet/heavenly body.
c. When a body is in space at null point.
The astronauts inside the revolving aircraft also experience weightlessness. They can hang
anywhere in the room of the aircraft. During this condition, the reaction force is zero, so
astronauts feel weightless inside the aircraft. While jumping out of an aeroplane with
parachute, the air resistance will be considerably high. As the velocity increases, there
comes a condition when the downward force of falling object is equal to the upthrust of
the air. In this condition, the acceleration due to gravity is zero. The falling object has no
acceleration and it falls with a uniform velocity. As a result, the velocity of the paratrooper
will be low and s/he can balance the body and land safely.
Differences between Free fall and Weightlessness
Free fall Weightlessness
1. When a body falls only under the 1. Weightlessness is the condition in
effect of gravity, it is called free fall. which a body of a certain mass seems
to be weightless.
2. Free fall is not possible in the absence 2. Weightlessness is possible in the
of gravity. absence of gravity.
Activity 3
Take a few apples in a
polythene bag and measure
their weight by holding a
spring balance in one hand.
Note down the weight of these 0N
apples.
Now, release the spring 0.8 N
balance and note down the
reading of the spring balance.
What is the weight of these
apples?
The pointer of the spring balance Polythene bag
shows zero weight during its fall. Apples
It shows that a body becomes
weightless during free fall. When the spring balance When the spring
Fig. is held by hand, it balance is released, it
shows a certain weight, shows zero weight.
1.12 i.e. 0.8 N
22 GREEN Science (Physics) Book-10
Key Concepts
1. Gravitational force can be defined as the force of attraction between any two bodies
due to their masses. Gravitational force is measured in newton (N).
2. According to Newton’s law of gravitation, “Every object in the universe attracts every
other object with a force which is directly proportional to the product of their masses
and inversely proportional to the square of the distance between their centres.”
3. The gravitational constant (G) is numerically equal to the force of gravitation which
exists between two bodies of unit masses kept at a unit distance from each other. The
value of universal gravitational constant ‘G’ has been found to be 6.67 × 10–11 Nm2/kg2.
4. Newton’s law of gravitation is applicable to any two objects present in the universe
whether those objects are small or big, celestial or terrestrial. Therefore, Newton’s
law of gravitation is called a universal law.
5. Gravity can be defined as the force which pulls an object towards the centre of the
earth or a heavenly body.
6. The acceleration due to gravity of all the falling objects towards the surface of the
earth will be the same in the absence of external resistance.
7. When an object falls towards the surface of the earth under the effect of gravity, its
velocity increases at the constant rate of 9.8 m/s in every second of time it is falling.
8. The value of acceleration due to gravity (g) is maximum on the surface of the earth
and it decreases on going above the surface of the earth or on going inside the
surface of the earth.
9. Mass of an object is defined as the total amount of matter present in a body. Its SI
unit is kilogram (kg).
10. The weight of a body can be defined as the measure of gravity acting on the body. It
is the force with which a body is pulled towards the centre of the earth/planet. The
SI unit of weight is newton(N).
11. When a body falls freely only under the effect of gravity, the fall of the body is
called free fall. When a body falls downwards in a vacuum, the fall of the body is
considered free fall.
12. Weightlessness can be defined as the condition at which a body of certain mass
becomes weightless.
Sequential General Exercise 1
1. Choose the best answer from the given alternatives.
a. The force of attraction between any two bodies due to their masses is called
..............................
gravity gravitation mass weight
b. The SI unit of gravitational constant is .......................
Nm² kg² Nm²/kg² Nm–2 kg–2 N/kg
GREEN Science (Physics) Book-10 23
c. The value of acceleration due to gravity on the surface of the moon is
..............................
9.8m/s² 1.67m/s² 25.13m/s² 6.17m/s²
d. .............................. is measured by using a beam balance.
Weight Force Mass Gravity
2. Answer the following questions.
a. What is force? Write its SI unit.
b. Define gravitational force with its SI unit.
c. Write down the factors on which gravitational force depends.
d. State Newton's law of gravitation and derive the mathematical expression for it.
e. State any two effects of gravitational force.
f. Prove that : F = Gm1 × m2 where the symbols have their usual meanings.
R²
g. What is Newton's gravitational constant ? Write down its value and SI unit.
h. What is gravity? Write its SI unit.
i. State any three effects of gravity.
j. What is acceleration due to gravity? Write its SI unit.
k. Prove that :
i. g = GM ii. g ∝ 1
R² R²
l. What is meant by the statement that the acceleration due to gravity on the surface
of the moon is 1.67m/s² ?
m. Write down the conclusion of feather and coin experiment.
n. What is mass? Write its SI unit.
o. What is weight? Write its SI unit.
p. What is free fall? In which condition does a body fall freely?
q. What is weightlessness? Write down any two conditions for weightlessness.
3. What change in gravitational force between two bodies can be observed when:
i. the mass of a body is doubled by keeping the distance between them constant?
ii. the distance between two bodies is doubled?
iii. the distance between two bodies is halved?
iv. the mass of both bodies is doubled and distance between them is halved?
v. the distance between them made one-third keeping their masses constant?
24 GREEN Science (Physics) Book-10
4. Differentiate between:
a. Gravity and Gravitation
b. Gravitation and Acceleration due to gravity
c. Gravity and Acceleration due to gravity
d. Mass and Weight
e. Free fall and Weightlessness
f. Weightless in space and weightlessness on the earth
5. The earth's orbit is oval in shape. Explain how the magnitude of the gravitational
force between the earth and the sun changes as the earth moves from position P to
Q as shown in the figure.
P
Q
6. The mass of the Jupiter is 319 times than that of the earth, but the acceleration due
to gravity of the jupiter is only 2.5 times than that of the earth. Explain the reason.
7. At what condition do a coin and a feather fall together? What is the acceleration of
the coin and feather at that instant? Justify your answer.
8. If a body is dropped from the same height once in the equator and then in the
polar region, in which place will it fall faster? Why?
9. Write down the conclusion of feather and coin experiment. At what condition the
value of acceleration due to gravity is approximately zero while getting down from
a parachute?
10. Give reason:
a. It is easier to lift a small stone but difficult to lift a heavy one on the surface of the
earth.
b. Planets revolve around the sun.
c. The value of acceleration due to gravity changes from place to place on the
surface of the earth.
d. The weight of a body changes from place to place on the earth.
GREEN Science (Physics) Book-10 25
e. The weight of a body is more at polar region than that in the equator.
f. Complete free fall cannot be observed on the surface of the earth.
g. Parachutists do not get hurt while jumping out of an aeroplane towards the
surface of the earth.
h. The probability of getting hurt is more while jumping from a significant height.
i. A body falls faster at polar region than that in the equator.
j. The apparent weight of a freely falling body is zero.
k. A feather and a coin of equal mass fall at different speed in air.
11. Numerical Problems:
a. A man of 100 kg mass is standing on the surface of the earth of mass 6 × 1024 kg and
radius 6400km. Calculate the gravitational force between them. [Ans: 977.05 N]
b. The mass of the Mars is 6 × 1023 kg and that of the Earth is 6 × 1024 kg. If the
gravitational force between them is 6.67 × 1016 N, calculate the distance between
them. [Ans: 6 × 106m]
c. The distance between any two bodies is 10m and the gravitational force between
them is 3.2 × 10–9 m. If the mass of one object is 40kg, calculate the mass of another
body. [Ans: 119.9kg]
d. The mass of the Jupiter is 1.9 × 1030 g and its radius is 7.1 × 104km. Calculate the
acceleration due to gravity on the surface of the Jupiter. What will be the weight
of the person of mass 85 kg on the Jupiter? [Ans: 25.13 m/s2, 2136 N]
e. The radius of the earth is 6380km and the value of acceleration due to gravity on
the earth's surface is 9.8m/s². Calculate the acceleration due to gravity at the top
of Mt. Everest. The height of Mt. Everest is 8848m from the earth's surface.
[Ans: 9.7 m/s²]
f. The weight of a body kept at a distance of 6.4 × 106m from the centre of the earth
is 40 N. Calculate its weight when it is shifted to a distance of 1.28 × 107 m from
the centre of the earth. [Ans: 10N]
g. A person can jump 12m high on the surface of the moon. Calculate the height
that he can jump on the earth's surface. [Ans: 2m]
h. A weightlifter can lift 586.8 kg mass on the moon. Calculate the mass that he can
lift on the earth. [Ans: 100 kg]
i. A ball is dropped freely from a tower of 78.4m height. The ball reaches the ground
in 4 seconds. Calculate the acceleration due to gravity on the ball. [Ans: 9.8m/s²]
j. An iron ball is dropped freely from a tower and it strikes the ground after 5
seconds. Calculate the height of the tower from the ground. (take g = 9.8m/s²)
[Ans: 122.5m]
26 GREEN Science (Physics) Book-10
Grid-based Exercise 2
Group ‘A’ (Knowledge Type Questions) (1 Mark Each)
1. Define gravitational force. Write down its SI unit.
2. What is acceleration due to gravity ?
3. State Newton’s universal law of gravitation.
4. The bodies in this universe attract one another. Name the scientist who propounded
this statement.
5. What is gravity ? In which unit is it measured ?
6. What is the weight of an object ?
7. On which factors does the mass of an object depend?
8. What is the average value of acceleration due to gravity on the polar region of the
earth?
9. What is a free fall ?
10. What is weightlessness ?
11. Write any two conditions at which a body becomes weightless.
12. What are the factors affecting gravity ?
13. Write down the value of gravitational constant.
14. Write any two effects of earth’s gravity.
15. What is the value of acceleration due to gravity at the equatorial region of the earth ?
Group ’B’ (Understanding Type Questions) (2 Marks Each)
16. Write any two differences between gravitation and gravity.
17. Why is Newton’s law of gravitation called universal law ?
18. The value of universal gravitational constant is 6.67 x 10–11 Nm2/kg2. What does it
mean?
19. The weight of a body is more at the polar region of the earth than that at the equatorial
region. Why ?
20. The fall of a parachute towards the earth’s surface is not a free fall. Why ?
21. The weight of a body of certain mass becomes zero in space. Why ? Write with reason.
22. Astronauts feel them weightless inside spacecrafts. Why ?
23. The fall of a body on the earth’s surface cannot be a complete free fall. Why ?
24. A satellite does not need any energy to revolve around the earth, why ?
25. Write any two differences between acceleration due to gravity and gravity.
GREEN Science (Physics) Book-10 27
26. What happens to the weight of an object when it is brought from the moon to the
earth? Why?
27. Why does a person feel weightlessness during free fall?
Group ‘C’ (Application Type Questions) (3 Marks Each)
28. Prove that: F = Gm1m2
d2
29. Radius of the earth is 6.4 × 103 km and value of acceleration due to gravity on its
surface is 9.8 m/s2. Find the value of acceleration due to gravity produced on a meteor
at a distance of 9850 m from the earth surface. (Ans: 9.76 m/s2)
30. Mention any three consequences of gravitational force.
31. The mass of the jupiter is 1.9 × 1027 kg and its radius is 7.1 × 104 km. Calculate the
acceleration due to gravity on the surface of the jupiter. What will be the weight of the
person of mass 65 kg on that planet ? (Ans: 25.13m/s2)
32. Write any two applications of gravity. A stone is dropped freely from 45 m height of
the tower, it reaches the ground in 3 seconds. Calculate the acceleration due to gravity
of that stone. (Ans: 10 m/s2)
Group ‘D’ (Higher Abilities Type Questions) (4 Marks Each)
33. How would a parachute fall on the surface of the moon? Describe. The mass of the
Jupiter is 1.9 × 1027 kg and that of the sun is 2 x 1033 kg. If the gravitational force acting
between them is 4.166 × 1023 N, calculate the distance between their centres.
(Ans: 7.8 × 1011m) 4
34. In which condition does the value of gravitation force become equal to that of the
gravitational constant ? The radius of the earth is 6380 km and the height of Mt.
Everest is 8848 m. If the value of acceleration due to gravity on the top of Mt. Everest
is 9.77 m/s², calculate the value of acceleration due to gravity on the surface of the
earth. What is the weight of a body of mass 100 kg on the top of Mt. Everest ?
(Ans: 9.8 m/s2, 980.5 N) 1 + 3
35. What do you understand by ‘weightlessness due to free fall’ ? The mass of the earth
is 6 × 1024 kg and its radius is 6400 km, what will be the acceleration due to gravity at
a distance of 3600 km far from the earth’s surface ? (Ans: 4.002 m/s2) 1+3
36. If a spring balance with the iron ball is released, what reading does the spring balance
record and why? The mass of the sun is 2 x 1030 kg, that of the earth is 6 x 1024 kg and
the distance between them is 1.5 x 1011 m. What is the gravitational force produced
between them ? (G = 6.67 × 10–11 Nm2/kg2) (Ans: 3.557 × 1022N) 1+3
37. Atmosphere is present around the surface of the earth but absent around the surface
of the moon. Why ? The mass of the jupiter is 1.9 × 1027 kg and its radius is 7.1 × 104
km. Calculate the acceleration due to gravity on the surface of the jupiter. What will
be the weight of the person of mass 85 kg on that planet ?
(Ans: 25.13 m/s2,2136 N) 1+ 3
28 GREEN Science (Physics) Book-10
UNIT Pressure
2
Weighting Distribution Theory : 8 Practical: 2
Before You Begin
When a body is placed on a surface, the body exerts a force equal to
its weight on the surface. The total perpendicular force acting on the
surface of contact is called thrust. The effect of thrust depends on the
area of the surface over which the thrust is applied. The thrust per
unit area of the surface is called pressure. The SI unit of pressure is
pascal (Pa) or N/m2. Pressure is a scalar quantity. In this unit, we will
study about liquid pressure, Pascal’s law of liquid pressure, hydraulic
press, density of liquid and upthrust, Archimedes’ principle, law of
floatation, atmospheric pressure, mercury barometer, syringe, air
pump and water pump.
Learning Objectives Syllabus
After completing the study of this unit, students will be able to: • Introduction to pressure
• Liquid pressure
i. demonstrate liquid pressure. • Pascal's law of liquid pressure
• Hydraulic press
ii. verify Pascal's law of liquid pressure. • Density of liquid and upthrust
• Archimedes' principle
iii. explain the application of Archimedes' principle in • Law of floatation
our daily life. • Atmospheric pressure
• Mercury barometer, syringe,
iv. demonstrate the law of flotation.
air pump and water pump
v. introduce atmospheric pressure and describe its
application.
vi. introduce mercury barometer, syringe, air pump and
water pump with their working mechanism.
Glossary: A dictionary of scientific/technical terms
pressure : the thrust acting per unit area of a surface
thrust : the total perpendicular force acting on the surface in contact
atmospheric : related to atmosphere
density : the mass per unit volume of a substance
upthrust : upward force experienced by a body immersed in a fluid
barometer : a device which is used to measure atmospheric pressure
GREEN Science (Physics) Book-10 29
Liquid Pressure
Any substance that has weight exerts
pressure. Liquids also have weight and
exert pressure on the walls and bottom of
the vessel in which they are stored. The
thrust exerted by a liquid per unit area of
the surface is called liquid pressure. The SI
unit of liquid pressure is Pa or Nm–2.
Ships, boats, etc. float on water due to
Fig.
Fig.liquid pressure. In service stations, heavy
Fig.vehicles are lifted easily using hydraulic 2.1
machines. It is possible due to pressure
Liquid pressure increases with depth
exerted by liquid. A body immersed in a
liquid is felt lighter due to the upthrust of the fluid. Due to the upthrust of water, ships
and boats float. Like liquids, air also exerts pressure. The force exerted by air on per
unit area of a surface is called atmospheric pressure. Atmospheric pressure is very useful
for human beings. Due to atmospheric pressure, we can fill ink in fountain pen, inject
medicine in the body of the patient, draw underground water using hand pump, inflate
tyres of vehicles and so on.
2.2
Pascal's Law
The French scientist Blaise Pascal Piston A Pressure in piston A Piston B
propounded a law regarding liquid = Pressure in piston B
pressure which is known as Pascal's
law of liquid pressure. Pascal was born Piston head Piston head
in 1623 AD. He was a famous physicist, 1.0 in2 100 in2
mathematician and philosopher of his
time.
Pascal's law of liquid pressure states that, 2.3 Fluid
"The pressure exerted on a confined liquid
is transmitted equally and undiminished Hydraulic press
in all directions throughout the mass of
the liquid."
30 GREEN Science (Physics) Book-10
When pressure is applied to a liquid kept in an enclosed vessel, the pressure is transmitted
equally throughout the enclosed liquid. This fact was first recognized by Blaise Pascal.
In science laboratory, following activities can be performed to verify Pascal's law of liquid
pressure.
Activity 1
Take a large balloon and fill it with water. Fig.2.4
Fig.
Tie the mouth of the balloon with a thread.
Take a needle and make a number of holes in the
balloon.
Squeeze the balloon and observe the pressure of
water in each hole.
When the balloon is squeezed, water comes out with
the same pressure. This activity verifies Pascal's law
of liquid pressure.
Activity 2
In the figure, a glass vessel having four pistons A, B, C and D of the same area are
shown. B
20 Pa
This vessel is filled with water and the pistons
fitted in this vessel are frictionless and water
tight.
When a pressure of 20 Pa is exerted in piston 20 Pa 20 Pa
A, same pressure is transmitted equally in all A
pistons. As a result, pistons B, C and D exert a Liquid (water) C
pressure of 20 Pa in outward direction.
Various equipment are constructed on the 2.5 20 Pa
basis of Pascal's law of liquid pressure. Some D
of them are hydraulic press, hydraulic lift or
jack and hydraulic brakes.
Application of Pascal’s Law
Pascal’s law is applied to construct hydraulic machines like hydraulic brake, hydraulic
press, hydraulic lift, etc.
Hydraulic Press
Hydraulic press is an equipment which is constructed on the basis of Pascal's law of liquid
pressure.
GREEN Science (Physics) Book-10 31
Hydraulic press is used for extracting F1 = 10 N F2 = 40 N
oil from seeds, pressing cotton, pressing
paper, etc.
In a hydraulic press, two pipes, one A1 = 20 cm2 A2 = 80 cm2
having small cross-sectional area (A1) and
another having large cross-sectional area Fig.
(A2) are fitted with a third pipe. Liquid
Suppose, the cross-sectional area of the
small piston A1 is 20 cm2 and that of A2 is 2.6
80 cm2. Both pipes have frictionless and watertight pistons. The hydraulic press is filled
with water or oil and when the small piston is pressed with a certain force F1, the large
piston moves upwards and vice-versa.
When an effort of 10 N is applied in the small piston, the piston exerts pressure in
downward direction. The pressure exerted in the liquid transmits equally in all directions
and the same pressure is exerted on the large piston and the large piston moves upward.
In order to balance the pressure exerted by the small piston, certain weight should be
kept on the large piston to balance both the pistons. According to Pascal's law of liquid
pressure, when both pistons are in balanced condition, equal pressure is exerted on in
both pistons. Therefore, 40 N force on large piston can balance 10 N force on small piston.
Here,
Cross-sectional area of small piston (A1) = 20 cm2
= 20 m2
100 × 100
= 2 × 10–3 m2
Cross–sectional area of large piston (A2) = 80 cm2
= 80 m2
100 × 100
= 8 × 10–3 m2
According to Pascal's law,
Pressure on small piston (P1) = Pressure on large piston (P2)
or, AF11 = AF22 [ P = AF]
or, 2 ×1100–3 = 8 40
× 10–3
or, 5 × 103 N/m2 = 5 × 103 N/m2
32 GREEN Science (Physics) Book-10
From above calculation, it becomes clear that the pressure exerted on both pistons is the
same, i.e. 5 × 103 N/m2 or Pa. It verifies Pascal's law of liquid pressure.
The principle of hydraulic machine states that, when a small force is applied on the small
piston, a large force is exerted on the large piston.
On the basis of this principle, hydraulic press, hydraulic lift or jack and hydraulic brakes
are constructed. Hydraulic machine is commonly known as force multiplier vessel.
Hydraulic Lift
Hydraulic lift is a device constructed on the basis of Pascal’s law which is used for raising
automobiles like car, truck, bus, motorcycle, etc. in service stations. In a hydraulic lift,
when a small force is applied in a small piston, a large force develops on the large piston
and heavy vehicles can be lifted easily.
Let us study the following figure.
Let us consider a Car
hydraulic lift having Effort (F1) Load (F2)
two vessels, viz. A1
having small cross–
sectional area and A2 Small Large
having large cross– piston piston (A2)
sectional area. Those
vessels are fitted with (A1) Water or oil
frictionless and water-
tight pistons. When Fig. V1 V2
an effort F1 is applied Valve 1 Valve 2
on the piston A1, a
pressure P1 is exerted 2.7
on the piston A1.
According to Pascal's law of liquid pressure, the same pressure is transmitted to the larger
piston too, i.e.
Pressure in small piston (P1) = Pressure in large piston (P2)
or, P1 = P2
or, AF11 = AF22 [ P = AF]
or, F1A2 = F2A1
Since A2 > A1 When there is no friction
Thus, F1 < F2 and no leakage of the liquid
Therefore, hydraulic machine is called a force multiplier vessel.
In the above figure, valve (V2) prevents the reverse flow of fluid on the small piston.
Similarly, valve (V1) prevents the reverse flow of fluid in the tank.
GREEN Science (Physics) Book-10 33
Solved Numerical: 1
Study the given figure and answer the following questions.
i. Calculate the pressure exerted on pistons A, C and D.
ii. Calculate the weight that can be balanced on piston A.
iii. Calculate the area of piston C.
Solution: A 20cm2
i. Given,
Area of piston B (A2) = 10 cm2 10cm2
B 125N
= 10
100 × 100
D
= 10–3 m2
Force on piston B (F2) = 125 N Fig.
\ Pressure on piston B (P2) = F2 C 375N
A2
2.8
125
= 10–3
= 125 × 103 Pa
\ Pressure on piston A, C and D = 125 × 103 Pa [ Pascal's law]
ii. Given,
Area of piston A (A1) = 20 cm2 = 20 m2 = 2 × 10–3 m2
100 × 100
Pressure on piston A (P1) = 125 × 103 Pa
Force (weight) on piston A (F1) = ?
We know,
P1 = F1
A1
or, F1 = P1 × A1
= 125 × 103 × 2 × 10–3
= 250 N
\ The weight that can be balanced on piston A = 250 N.
iii. Given,
Force on piston C (F3) = 375 N
Pressure on piston C (P3) = 125 × 103 Pa
Area of piston C (A3) = ?
34 GREEN Science (Physics) Book-10
We know,
P3 = F3
A3
or, A3 = F3
P3
= 375 m2
125 × 103
= 3 × 10–3 m2
\ Area of piston C = 3 × 10–3 m2
= 30 cm2
Density of liquid and Upthrust
Density of a liquid is the ratio of mass per unit volume of the liquid, i.e.
Density (d) = Mass (m)
Volume (V)
\ d = m
V
The SI Unit of mass is kilogram (kg) and that of volume is cubic metre (m3). So the SI unit
of density is kilogram per cubic metre (kg/m3) or kgm–3.
If we try to lift a large block of stone in water, we will notice that it may be easily lifted
but it seems to become much heavier when it comes out of water. Every liquid exerts an
upward force on objects immersed in it. Like liquids, gases also exert an upward force
when an object is immersed in them. For example, a balloon filled with hydrogen gas
experiences an upward force when it is released in the air. This upward force exerted by
liquids or gases is called upthrust. It is also called buoyant force and this phenomenon is
called buoyancy. So, upthrust can be defined as the upward force experienced by a body
when partially or wholly immersed in a fluid.
Since upthrust is an upward force, its SI unit is newton (N).
Factors affecting upthrust
The upthrust acting on a body depends on following factors:
1. The volume of the body submerged in a fluid
Larger the volume of the body submerged in a fluid, greater is the upthrust and vice-
versa. If we push a small piece of cork into the water, our fingers will experience some
upthrust. Now, if we push a large piece of cork into the water, we will experience a
great upthrust in our fingers.
In short, Upthrust (U) ∝ the volume of the body submerged in a liquid (V)
GREEN Science (Physics) Book-10 35
2. The density of the liquid in which the body is submerged
Larger the density of the liquid, greater is the upthrust and vice-versa. This is the
reason that an iron nail floats in mercury, but it sinks in water because the density of
mercury is 13.6 times greater than that of water. Hence, for the same volume, upthrust
provided by mercury is much greater than that provided by water.
In short, Upthrust (U) ∝ the density of the liquid in which a body is immersed (V)
3. Acceleration due to gravity
More the value of acceleration due to gravity (g), more is the upthrust and vice-
versa. Therefore, same liquid exerts more upthrust in the polar region than that in
the equatorial region because the value of acceleration due to gravity is slightly more
in the polar region than that in the equatorial region.
In short, Upthrust (U) ∝ Acceleration due to gravity (G)
Formula to calculate upthrust
U = Vdg
Where, U = upthrust
V = volume of the body submerged in a liquid
d = density of the liquid in which a body is submerged/immersed
g = acceleration due to gravity
Cause of upthrust
When a body (for example, a piece of metal or wood) is completely immersed in a liquid,
the pressure acting on the sides of the body called lateral pressure, mutually cancels each
other. The pressure acting upward on the lower surface of the body is more than the
pressure acting downward on the upper surface of the body due to its greater depth.
Therefore, there is net upward pressure acting on the body. As force = pressure × area of cross-
section, there is a net upward force acting on the body. It is the force which is called upthrust.
Container
metalFig. Lateral pressure
piece Liquid
2.9
36 GREEN Science (Physics) Book-10
Activity 3
Take an empty sealed ‘tin can’ and place it in a bucket about three-fourths filled
with water. The ‘tin can’ floats freely.
Tin can
Water Water
Bucket
‘Tin can’ floats freely
2.10 Water
The ‘tin can’ experiences an upward force
when pushed in water
Fig. Fig.
Fig. Push the ‘can’ into the water. You feel an upward force (push) and you find it difficult
to push it deeper. Obviously, the ‘can’ experiences an upward force (upthrust). The
upward force exerted by water goes on increasing as the ‘can’ is pushed deeper.
Now, if you release the can in this position, you notice that the can bounces back. It
implies that water exerts an upward force (upthrust) on the can which pushes it upwards.
Activity 4
Take a beaker and fill it with fresh water. Now, keep a fresh egg in water. Does the egg
sink in water?
When a fresh egg is kept in water, it tries to go
down due to its weight but the upthrust due to
water pushes the egg upward. Since the weight of Fresh water
the egg is more than the upthrust due to water, the Egg
egg sinks in fresh water. An egg sinks in fresh water
Now, add some salt in the beaker to make 2.11 Egg
concentrated salt solution. When salt is added, Concentrated
salt solution
the density of water increases and upthrust also
increases. As a result, the egg floats in water. In
this condition, the weight of egg is equal to the
upthrust. When some more salt is added, upthrust
increases which pushes some part of egg slightly
above the surface of water. In this condition too,
the weight of egg is equal to the upthrust. 2.12
An egg floats in concentrated
salt solution
The upthrust due to a liquid in a body does not depend on the weight of the body. The
upthrust acting on the body differs according to volume of the immersed part of the
body. The upthrust acting on a body is more when the body is wholly immersed in a
liquid than it is partially immersed. These facts are described by Archimedes’ principle.
GREEN Science (Physics) Book-10 37
Archimedes' Principle
It is a common experience that a body appears to be heavier when it is lifted above the
surface of water. The apparent change of weight may be followed if a large piece of stone
is hung from a spring balance and slowly lowered into the water. As it enters the water,
the reading of the spring balance will decrease and this decrease will continue as more
and more of the stone is submerged. Once the stone is completely covered by the water,
the spring balance reading will remain constant as the stone is lowered more deeply. On
raising the stone upward, reverse changes of the spring balance reading take place until
the stone is completely out of water. When the original reading will be given once more,
the apparent loss in weight is equal to the upthrust on the body. This is the Archimedes'
principle.
Archimedes’ was a mathematician, Do You Know
physicist, engineer, inventor as well
as astronomer. He was born in 287 Archimedes' principle is equally applicable
BC in Syracuse Sicily. He performed to gases as well. So the word 'liquid' may be
many experiments on upthrust and replaced by fluid.
finally propounded a principle which
is commonly known as Archimedes’
principle.
Archimedes' principle states that, "When a body is partially or wholly immersed in a
liquid (fluid), it experiences an upthrust (or apparent loss in weight) which is equal to the
weight of the liquid (fluid) displaced by it."
Experimental Verification of Archimedes' Principle
Take a ‘displacement can’ with a beaker under its spout as shown in the figure. Water is
poured until it runs from the spout. When the water has ceased dripping, the beaker is
removed and replaced by another beaker which has been previously dried and weighed.
Any suitable solid, a piece of stone or metal, is suspended by a thin thread from the
hook of a spring balance and the weight of the body in air is measured. The body still
attached to the balance is then carefully lowered into the displacement can and when it is
completely immersed, its new weight is noted.
Spring Weight in air (W1) 7N
balance 10 N weight in water (W3)
Stone Water
Fig. 3N
Top pan balance
2.13
38 GREEN Science (Physics) Book-10
The displaced water is taken in the weighed beaker. When no more water drips from the
spout, the beaker and water are weighed. Results obtained are as follows:
Weight of the body in air = W1
Weight of the empty beaker = W2
Weight of the body in water = W3
Weight of the beaker with displaced water = W4
Apparent loss in weight of the body or upthrust = (W1 – W3)
Weight of displaced water = W4 – m2
It is found that W1 – W3 = W4 – W2
or, Upthrust = Weight of liquid displaced
This gives the approximate verification of Archimedes' principle.
Solved Numerical: 2
Study the given figure and answer the following questions. 7N
3N
i. Which principle is the experiment based on?
ii. What is the upthrust given by the liquid?
iii. What is the weight of the stone in air?
Ans: Fig.
i. The experiment is based on Archimedes' principle.
ii. The upthrust given by the liquid is 3N. 2.14
iii. The weight of the stone in air = Wt. in water + upthrust
= (7 + 3) N
= 10 N
Applications of Archimedes' Principle
1. Archimedes’ principle is applicable to construct sub-marines, hydrometer, etc.
2. This principle is applicable to fly hot air balloons, to float ships in sea, etc.
Law of Floatation
Law of floatation states that , "When a body floats in a liquid, the weight of the floating
body is equal to the weight of the liquid, displaced by it."
When a body is immersed in a liquid, the following two forces act on it.
i. The weight of the body (W1) acting vertically downward at the centre of gravity of the
body. This force has a tendency to sink the body in a liquid.
GREEN Science (Physics) Book-10 39
ii. The upthrust (W2) due to weight of the liquid displaced, acting vertically upwards at
the centre of buoyancy.
The magnitude of upthrust is always equal to the weight of the liquid displaced by the
immersed part of the body. It has a tendency to push the body out of the given liquid with
respect to above forces.
Conditions for floatation
It is evident from three cases discussed in the law of floatation that for a body to float, the
weight of the liquid displaced by it must be equal to the weight of the body. When this is
the case, the resultant force acting on the body is zero and so there will be no motion of
translation, but it is possible that the two forces may not act along the same line, will form
a couple and produce rotation. Hence, for equilibrium the two forces must also act along
the same vertical line, so as the centre of gravity of the body and the centre of gravity of
the liquid displaced i.e., the centre of buoyancy must lie in the same vertical line.
Thus, the conditions for the equilibrium of a floating body in a liquid are:
i. The weight of the liquid displaced by the portion of the body below the liquid surface
must be exactly equal to the weight of the floating body.
ii. The centre of gravity of the body and centre of buoyancy of the displaced liquid must
lie in the same vertical line.
Experiment to verify law of floatation
i. Take a small piece of a wooden block that floats in water.
ii. Measure the weight of the wooden block in air and note it down. Let it be W1.
iii. Now, take a Eureka (overflow) can and fill it completely with water and wait until it
stops dripping.
iv. Take a top pan balance and keep a clean beaker on the pan of the balance.
v. Now, keep the top pan balance just below the spout of the water.
vi. Immerse the wooden block in the Eureka can. The block displaces some water which
is collected in the beaker kept below the spout of the over flow can.
Fig. Eureka can
W1
Displaced water
2.15 Wooden block Top pan balance
vii. Now, measure the weight of the water displaced by the wooden block. Let it be W2.
Compare the weight of the wooden block and the weight of the liquid displaced. You
will find that,
Weight of the floating body = Weight of the liquid displaced
40 GREEN Science (Physics) Book-10
viii. Now, repeat the above experiment with wooden blocks of different weights and fill
in the given table.
Weight of object (W1)
Weight of displaced water (W2)
In each case, you will find that the weight of the floating body is equal to the weight
of water displaced.
In this way, law of floatation can be verified.
Activity 5
Take a beaker and fill it with water as shown in the figure.
Take a wooden cork and push it in the water Fig. Cork
kept in a beaker and release it. What do you Beaker
experience?
Water
Repeat the same activity with a small rubber
ball filled with air. What do you experience?
What is the difference between floatation of
a cork and a rubber ball.
Now, take a piece of stone and keep it in 2.16
water. The stone will sink.
The density of a cork is more than the density of the ball. So more part of the cork
immerses in water to float so that the weight of the cork is equal to the weight of the
water displaced. Since the ball is lighter, less portion of the ball sinks in water to float.
In this case also the weight of the ball is equal to the weight of the water displaced. But
the stone cannot float in water because the weight of the stone is more than the weight
of the water displaced because the density of stone is more than that of water.
Atmospheric Pressure
The gaseous envelope surrounding the earth is called the atmosphere. It consists of various
gases like nitrogen, oxygen, carbon dioxide, etc. The atmosphere extends upto a height
of 1000 km from the surface of the earth. The mass of atmosphere is about 5.15 × 1018 kg.
Atmosphere has weight. So it exerts pressure. The pressure exerted by atmosphere due to
its weight is called atmospheric pressure. It is also known as barometric pressure.
The atmospheric pressure varies in different parts of the earth. At the sea level, the atmospheric
pressure is about 101300 N/m2 or 760 mmHg. Different altitudes have different atmospheric
pressure. Due to this, air blows from one place to another. The atmospheric pressure decreases
as the altitude increases. Therefore, atmospheric pressure is maximum at the sea level and
minimum at the top of Mt. Everest on the earth’s surface.
GREEN Science (Physics) Book-10 41
Do You Know
Barometer is used to measure the
atmospheric pressure.
Fig. Simple barometer was designed by
Fig. Torricelli in 1643 AD.
2.17
Barometer
One atmospheric pressure is defined as the standard atmospheric pressure at the sea level
which is about 101300 Pa or 760 mmHg.
Experiment: 1
To demonstrate atmospheric pressure
Take a ‘tin can’ and keep Steam
some water into it. Heat the
tin can until steam issues
(forms) freely, driving out
the air from the can.
Remove the flame and cork Cork
up the can quickly. Hold
it under the cold water Partial Air
tap. The steam inside will vacuum pressure
condense and the can will
collapse under the external
pressure of the atmosphere. 2.18 a b
When the water in the can boils, steam is formed in the tin can. The steam pushes most
of the air out of the can. When the corked up can is cooled by pouring cold water,
the steam inside it condenses and creates a partial vacuum in the can. As a result, the
atmospheric pressure crushes the can inwards.
This experiment shows the presence of large atmospheric pressure around us.
Activity 6
Take an empty glass and fill it completely with water in such a way that there is no
space for air.
Take a thick cardboard and cover the glass gently and press the cardboard. Now,
invert the glass pressing the cardboard with your hand.
Now, remove the hand from the cardboard gently and observe it carefully.
42 GREEN Science (Physics) Book-10
Glass
Water Water
Cardboard
Fig.
Fig. 2.19 Pressure
The cardboard does not fall down for a while because the atmospheric pressure holds
the water in the glass by pressing the cardboard upward. This activity shows the
presence of atmospheric pressure.
Importance of Atmospheric Pressure
We can do various works and use various equipment due to the presence of atmospheric
pressure. The importance of atmospheric pressure is given below.
1. Atmospheric pressure helps to fill ink in the ink pen.
2. It helps to fill medicine in the syringe.
3. It helps to fill air in the tube of vehicles.
4. It helps to lift water using a water pump.
5. It helps to drink soft drinks using a straw.
Atmospheric pressure
2.20
Medicine is being filled in a syringe Drawing soft drink through straw
The working of fountain pen, syringe, suction pad, straw, water pump, etc. is based on
atmospheric pressure. These equipment cannot be used in the absence of atmospheric
pressure. Therefore, atmospheric pressure is very important for human beings.
Measurement of atmospheric pressure
We can measure atmospheric pressure. It can be measured with a device called barometer.
It was invented by Evangelist Torricelli in 1643. Barometers are of different types such as
mercury barometer, aneroid barometer, etc. In this unit, we will study mercury barometer
only.
GREEN Science (Physics) Book-10 43
Mercury Barometer Do You Know
The instrument which is used to measure Mercury barometer was designed by
atmospheric pressure is called barometer. Torricelli in 1643 AD.
There are different types of barometer. The
barometer having mercury is called mercury
barometer.
To construct a mercury barometer, a glass Glass Vacuum
tube of about 1m length is taken and it is tube Scale
completely filled with mercury. The open Mercury
end of the tube is pushed with a thumb and
the tube is inverted on a trough filled with
mercury. Then the tube is supported with
a stand.
All air bubbles should be tapped out while Air Air
filling the mercury in the tube before Pressure Pressure
inverting it on a trough. The mercury
falls by a few centimeters creating empty Fig. Reservoir
space above the level of mercury which is Fig.
commonly known as Torricellian vacuum. 2.21
The tube is then graduated to take the
reading of atmospheric pressure.
It is found that the level of mercury in the tube is 76 cm above the level of mercury in the
bowl at sea level which shows that mercury in the tube is supported by the pressure of air
on the surface of mercury in the bowl.
Pressure of atmosphere = Pressure of mercury column = 76 cm
When the atmospheric pressure decreases, the level of mercury also decreases and vice-
versa.
Syringe Piston
Syringe is a medical instrument Storage cylinder
which is used to extract blood (barrel)
from the patient's body and to
inject medicines into the body of
the patient.
A syringe consists of three parts. Needle
They are storage cylinder (barrel), 2.22
piston and needle.
In order to fill medicine into the syringe, the piston is pushed into the barrel in order to
take air out of it. Then, the needle is inserted in the container of medicine. When piston is
pulled out, it creates a partial vacuum inside the barrel. As a result, medicine moves into
the storage cylinder (barrel) since liquids always flow from high pressure to low pressure.
In the same way, blood is also drawn into the barrel of the syringe from the patient's
44 GREEN Science (Physics) Book-10
body. When the piston of the syringe Atmospheric 1 Piston
is pushed in, medicine can be injected pressure 2
into the body of the patient through 3 Low pressure
blood vessel. 4 Barrel
5
When the piston is pulled up, the Needle
atmospheric pressure inside the barrel Fig. Liquid (medicine)
will decrease and the atmospheric Fig.
pressure outside pushes the liquid
into the barrel of syringe.
2.23
Air Pump
Air pump is an instrument which is used to inflate
tyres of bicycle, motor cycle, car, etc. We should
maintain proper air pressure inside the tyres of
vehicles. Otherwise, they cannot move properly.
We use pressure gauze to measure the air pressure
inside the tyres of vehicles.
In the given figure, air pump which is used to
inflate the tyres of bicycle is shown. It is a simple 2.24
pump which is used manually to inflate the tyres. It works on the principle of atmospheric
pressure. A manual air pump consists of three parts. They are cylinder, piston and nozzle.
In order to inflate the tyre of bicycle, we should move the piston up and down regularly.
When the piston of the air pump is moved up, it creates a partial vacuum in the cylinder
and air is filled in the cylinder. When the piston is forced down, air enters inside the tube
of bicycle tyre through the nozzle by opening the valve of the tube. When the piston is
moved up, the valve of the tube closes itself and the air inside tube does not come out.
When up and down movement of the piston is continued several times, the tube of the
bicycle is inflated.
Piston going up
Air in Piston going down
Inlet valve openFig. Inlet valve shut
Outlet valve open
Outlet valve shut
Air out
When the piston is pulled
When the piston is pulled down, air is
2.25 up, air enters the pump pushed out of the pump
GREEN Science (Physics) Book-10 45
Water Pump
Water pump is an instrument which is used to lift underground water. It is based on
the principle of atmospheric pressure. A water pump consists of two main parts. They
are barrel and piston. Both barrel and piston are fitted with one valve each. The piston
is provided with piston valve (V1). The piston valve opens when piston moves down
and closes when piston moves up. The valve also moves up and down along with the
piston. A valve called foot valve (V2) is fitted at the base of the barrel. The foot valve (V2)
opens when the piston moves up and it closes when the piston moves down. The barrel is
connected to a pipe deep in the ground down to the level of water.
Water outlet Force rod
Piston rod
Water suction line
Water Cylinder
Piston
PSeisatloinngvOal-vrein(gV1)
Foot valve (V2)
Fig.
2.26
Upstroke
When the handle of the water pump is pushed down, the piston moves upward. It creates
a partial vacuum between the valves V1 and V2. In this condition, air pressure decreases in
the barrel, the valve V1 closes and water rises up in the barrel through the valve V2. Here,
water rises up due to atmospheric pressure. This process is called upstroke.
Downstroke
When the handle of the water pump is pulled upward, the piston moves downwards. It
decreases the distance between valves V1 and V2. It increases pressure in the foot valve
(V2) in the barrel and the foot valve (V2) is closed. The piston valve opens due to water
pressure. As a result, water rises upward and water comes out of the barrel. This process
is called down stroke.
46 GREEN Science (Physics) Book-10
Key Concepts
1. The thrust exerted by a liquid on per unit area of the surface is called liquid pressure.
2. The force exerted by air on per unit area of a surface is called atmospheric pressure.
3. Pascal’s law of liquid pressure states that, “The pressure exerted on a confined
liquid is transmitted equally and undiminished in all directions throughout the
mass of the liquid.”
4. Hydraulic press is an equipment which is constructed on the basis of Pascal’s law
of liquid pressure.
5. The principle of hydraulic machine states that, “When a small force is applied on
the small piston, a large force is exerted on the large piston.”
6. Density of a liquid is the ratio of mass per unit volume of the liquid.
7. The SI unit of mass is kilogram (kg) and that of volume is cubic metre (m3). So the
SI unit of density is kilogram per cubic metre (kg/m3) or kgm–3.
8. Upthrust can be defined as the upward force experienced by a body when partially
or wholly immersed in a fluid.
9. Archimedes’ principle states that, “When a body is partially or wholly immersed in
a liquid, it experiences an upthrust (or apparent loss in weight) which is equal to the
weight of the liquid displaced by it.” Archimedes’ principle is equally applicable to
gases as well.
10. The gaseous envelope surrounding the earth surface is called the atmosphere.
11. The pressure exerted by atmosphere due to its weight is called atmospheric
pressure.
12. The atmospheric pressure varies in different parts of the earth. At the sea level, the
atmospheric pressure is about 101300 N/m2 or 760 mmHg.
13. The instrument which is used to measure atmospheric pressure is called barometer.
There are different types of barometers. The barometer having mercury is called
mercury barometer.
14. Syringe is a medical instrument which is used to extract blood from the patient’s
body and to inject medicines into the body of the patient.
15. Air pump is an instrument which is used to inflate tyres of bicycle, motor cycle, car,
etc.
16. We use pressure gauze to measure the air pressure inside the tyres of vehicles.
17. Water pump is an instrument which is used to lift underground water. It is based
on the principle of atmospheric pressure.
GREEN Science (Physics) Book-10 47
Sequential General Exercise 1
1. Choose the best answer from the given alternatives.
a. In SI system, pressure is measured in ..............................
N Pa Nm2 m3
b. Which of the given equipment is based on Pascal's law?
Hydraulic press Barometer
Water pump Syringe
c. Which of the given equipment is based on Archimedes' principle?
Submarine Hydraulic brake
Water pump Barometer
d. Which of the given equipment is used to measure atmospheric pressure?
Hydrometer Thermometer
Barometer Syringe
2. Answer the following questions.
a. What is pressure? Write its SI unit.
b. State Pascal's law of liquid pressure.
c. What is hydraulic press? Write down its principle.
d. Name any three equipment based on Pascal's law of liquid pressure.
e. What is density? Write its formula and SI unit.
f. What is upthrust? Write down its SI unit.
g. State Archimedes’ principle.
h. Name any two equipment constructed on the basis of Archimedes' principle.
i. Write any two applications of Archimedes' principle.
j. What is atmospheric pressure? What is the atmospheric pressure at the sea level?
k. Write any three applications of atmospheric pressure.
l. What is barometer? Draw a neat figure showing mercury barometer.
m. What is syringe? Name its main parts.
n. Describe the working mechanism of a syringe in brief.
o. What is air pump? Draw a neat figure showing the structure of air pump.
48 GREEN Science (Physics) Book-10
p. What is water pump? Draw a neat and labelled figure showing the structure of
a water pump.
3. Explain Pascal’s law with a neat figure.
4. Explain Archimedes’ Principle with a neat and labelled figure.
5. Explain the law of floatation with a neat figure.
6. Describe the working mechanism of a water pump in brief.
7. How does an air pump work? Describe in brief.
8. Differentiate between:
a. Force and Pressure
b. Pressure and upthrust
c. Upthrust and Density
d. Archimedes' principle and Law of floatation
e. Air pump and Water pump
f. Upstroke and Downstroke
9. Give reason:
a. An egg sinks in fresh water but floats in concentrated salt solution.
b. An iron nail sinks in water but floats on mercury.
c. The piece of ice floats on water, why?
c. When a ball is pushed into water, it moves upward.
d. An iron nail sinks in water but the ship made of iron floats.
e. A balloon filled with air bursts when it rises up in the sky.
10. Numerical problems: A 20cm2
10cm2
a. Study the given figure and answer the following
questions. B 125N
C 375N
i. Calculate the pressure exerted on pistons A, C
7N
and D. [Ans: A = C = D = 125 × 103 Pa] D
3N
ii. Calculate the weight that can be balanced on
piston A. [Ans: 250 N]
iii. Calculate the area of piston C. [Ans: 30 cm2]
b. Study the given figure and answer the following
questions.
i. Which principle is the experiment based on?
ii. What is the upthrust given by the liquid?
[Ans: 3 N]
iii. What is the weight of the stone in air?
[Ans: 10 N]
GREEN Science (Physics) Book-10 49
c. In a hydraulic machine, the area of small piston is 50 cm2 and that of the large
piston is 5 m2. If an effort of 20 N is applied on the small piston, how much load
can be lifted on the large piston? Calculate.
[Ans: 2 × 104 N]
d. Study the given figure and calculate the cross-sectional area of the large piston.
300 N [Ans: 150 cm2]
30 N
Small piston Large piston
15 cm2
e. The weight of an object is 25 N in air. If the weight of the body is seen 22 N in the
water:
i. Calculate the upthrust on the object by water. [Ans: 3 N]
ii. What is the weight of the liquid displaced? [Ans: 3 N]
Grid-based Exercise 2
Group ‘A’ (Knowledge Type Questions) (1 Mark Each)
1. What is thrust? Write down its SI unit.
2. What is pressure ?
3. Name the two factors on which the pressure exerted by a body depends.
4. What is hydraulic pressure ?
5. What is hydraulic press?
6. On which factors does hydraulic pressure depend? Write.
7. State Pascal’s law.
8. State Archimedes’ principle.
9. Name any two instruments based on Archimedes’ principle.
10. What is barometer?
11. What is one pascal pressure?
12. What is hydraulic brake ?
13. What is upthrust due to liquid ?
14. Weight of an object is decreased when immersed in water, which law does this
statement represent?
15. What is relative density ?
50 GREEN Science (Physics) Book-10
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