Science and Environment 10

10th Edition

Approved by Government of Nepal, Ministry of
Education, Curriculum Development Centre,

Sanothimi, Bhaktapur as an additional material

Oasis

10Grade

Editor Author
Dr. Rameshwar Adhikari Jayananda Kapadi

Reader M.Sc. (Zoology), B.Ed. (Health Education)
Tribhuvan University, Kirtipur, Kathmandu
Central Department of Chemistry

Tribhuvan University, Kirtipur, Kathmandu

Oasis

10Grade

Publisher and Distributor:

Oasis Publication Pvt. Ltd.

Tel: 014224004, Fax: 4227718

Author:
Jayananda Kapadi

Language Editors:
Ramesh Lamsal
Bhan Dev Kapadi

Edition:
First 2066
Second 2067
Third 2068
Fourth 2069
Fifth 2070
Sixth 2071
Seventh 2072 (Completely revised)
Eighth 2073 (Revised)
Ninth 2074 (Completely revised and updated)
Tenth 2075 (Revised and updated)

Copyright @ All rights reserved. No part of this book may be reproduced, stored
Author in a retrieval system or transmitted, in any form or by any means
without prior permission of the author.

Computer layout:
Oasis Desktop group

Printed in Nepal

Preface

Oasis School Science for Grade 10 is an attempt to make the learning process a joyful
experience. This textbook has been written in strict conformity with the latest syllabus
prescribed by the Curriculum Development Centre, Sano Thimi, Bhaktapur, Nepal. This
book has been designed to help students develop their conceptual thinking and scientific
skills. I think this book is an excellent introduction to experimentation and practical
application of science. I hope it will greatly facilitate the teaching learning process in an
easy and enjoyable manner.

The beauty of this textbook lies in having high resolution pictures an attractive layout,
and clear illustrations with a lucid language. It emphasizes concept building rather than
merely focusing on providing and collecting information without critical thinking. I
expect this book to assist students in making them eager and quizzical learners, which
will reinforce their conceptual learning in the classroom. Besides the learning process of
the students, this book will help in the teaching process of the teachers. Each unit of this
book presents subject matter in an interesting, understandable and enjoyable manner. The
exercise at the end of each unit includes a variety of questions to facilitate the integration
of various concepts taught. Above all, I sincerely believe that this book will be helpful in
the overall understanding of science in an interesting manner.

It is not a hidden fact that the modern era is an era of science and technology. Science is
a part of the world we live in and an avenue to technology. A good textbook in science
should assist the learners to realize different activities and events around us to encourage
them towards further discovery and innovation. I strongly believe that students should
enjoy science and this book will be a means of enjoying and learning science in the modern
era of science and technology.

I wish to express my sincere gratitude to Mr. Harish Chandra Bista, Managing
Director of Oasis Publication Pvt. Ltd., for publishing this book. Similarly, my
hearty thanks go to Oasis Desktop Group for the layout. Thanks are due to Mr.
Deepak Bhatta, Mr. Naresh Budal, Mr. Tek Bahadur Shahi, Mr. Deepak Chand,
Mr. Arjun Shrestha, Mr. Surendra Mishra, Mr. Navneesh Prasad Yadav, Mr.
Ram Maharjan, Mrs. Bimala Shah, Mrs. Jamuna Maharjan, Mr. R.C. Neupane,
Mr. Ujjwol Bhomi, Mr. Shivendra Karki, Mr. Binod Kumar Yadav, Mr. Prakash Bhatta, Mr.
Sudip Bajagain and Mr. Rabindra Agrawal for their valuable help during the preparation
of the book. Likewise, thanks are due to Mr. Ramesh Lamsal and Mr. Bhan Dev Kapadi
for their praiseworthy language editing. I gratefully acknowledge the help of the teachers
across the country as well as my well-wishers for their inspiration and support during
the preparation and publication of the book.

In my opinion, the real judges of a book are the teachers concerned and the students for
whom it is meant. Despite all my efforts, there might be textual as well as technical errors.
Therefore, constructive suggestions for rectification and improvement of the book would
be gratefully acknowledged and incorporated in further editions.

January 2018 Author
Kathmandu, Nepal

Contents

Physics
Unit 1 Force................................................................................. 1
Unit 2 Pressure............................................................................ 25
Unit 3 Energy............................................................................... 52
Unit 4 Heat.................................................................................. 68
Unit 5 Light.................................................................................. 84
Unit 6 Current Electricity and Magnetism.............................. 104

Chemistry

Unit 7 Classification of Elements.............................................. 128
Unit 8 Chemical Reaction.......................................................... 151
Unit 9 Acid, Base and Salt......................................................... 165
Unit 10 Some Gases...................................................................... 179
Unit 11 Metals............................................................................... 194
Unit 12 Hydrocarbon and its Compounds............................... 206
Unit 13 Materials Used in Daily Life......................................... 223

Biology

Unit 14 Invertebrates.................................................................... 249
Unit 15 Human Nervous and Glandular System..................... 262
Unit 16 Blood Circulatory System in Human Body................. 283
Unit 17 Chromosome and Sex Determination......................... 304
Unit 18 Asexual and Sexual Reproduction............................... 313
Unit 19 Heredity............................................................................ 338
Unit 20 Environment Pollution and Management.................. 352

Geology and Astronomy

Unit 21 History of the Earth........................................................ 365
Unit 22 Atmosphere and Climate Change................................ 376
Unit 23 The Universe.................................................................... 392

Teaching Strategies

Science deals with the systematic knowledge of different activities and events that occur in our surroundings.
Therefore, various teaching learning activities can be adopted in science according to the nature of the subject
matter.

Teachers are expected to adopt various teaching methods, like observation, experiment, demonstration,
discovery, invention, discussion, question-answer, field visit, etc. in the teaching learning process in science.
Besides these methods, teachers can adopt the explanation or lecture method in the course of introducing any
event, subject matter or result of something. ‘Student Centered Method’ is supposed to be the most appropriate
in teaching science. In this method, every student gets chance to think critically in solving his/her problems.

In teaching learning activities, teachers are expected to involve of every student in ‘process skills’ like
classification, comparison, putting query, reasoning, keeping record, assessment, etc. Teaching science not only
aims at accumulating knowledge but also at discovering knowledge. Therefore, the teaching learning process
is expected to be centered on / oriented towards discovery and invention.

Students should be encouraged to learn things on their own by discovering, inventing, experimenting or
solving problems. For this purpose, teachers are expected to involve all the students more and more in practical
activities along with the theoretical knowledge.

Specially, for a successful teaching learning process, teachers are expected to keep the following points in mind.

i. Asking students about situations or events happening in their surroundings

ii. Encouraging the students to hypothesize in advance about the result or effect of the events or situations

iii. Encouraging the students to test their hypothesis

iv. Providing an opportunity to every student to reach his/her own conclusion and to rethink of the
significance of his/her conclusion

For an effective teaching learning process, teachers are expected to emphasize the use of teaching learning
materials. It is emphasized that the use of teaching learning materials is helpful to make the concept of each
lesson clear for easy understanding of the students. Teachers are expected to make optimum use of local
teaching learning materials as far as possible according to the nature of the subject matter. For successful
teaching learning process in science, teachers are expected to adopt to the following activities.

1. Figure/Picture Observation: Picture(s) or figure(s) related to the subject matter of the lesson play(s) a vital
role in making the concept of the lesson clear. Teachers are expected to show or demonstrate the picture(s) or
figure(s) related to the lesson and to involve the students in observation. This activity helps the students to take
part in discussions on the basis of their observation.

Weighting Distribution

S.N. Area Weighting Percentage Estimated teaching periods

1 Physics 30.58 Theoretical Practical
2 Chemistry 30.58 42 10
3. Biology 30.58 42 10
4. Astronomy and Geology 8.26 42 10
100 12 2
Total 138 32

2. Project Work: Project work is supposed to be an important activity in enhancing the learning capacity of
the students. Teachers are expected to provide project work to the students individually or in group(s) to be
finished within a limited time frame. After finishing the project work(s), teachers are expected to provide an
opportunity to the students to present the process and result of the work in front of the class. This activity helps
the students to improve further.

3. Practice: It is well known that ‘learning without practice is meaningless.’ Therefore, practice is one of the
major components of a successful teaching learning process. In this activity, teachers are expected to focus
on the process of every finding rather than merely focusing on finding answer(s). This activity provides an
opportunity to the students for further improvement through feedback from the teacher(s).

4. Activities: For the positive change in the concept, skill or attitude of the students, performance of every
activity is supposed to be an essential part in every teaching learning process. Teachers are expected to more
and more involve of the students in different activities that help them to experience themselves for further
improvement.

5. Field Visit: The teaching learning process in science equally focuses on field visits according to the nature
of the subject matter in the lesson. On the one hand, students naturally might feel boring and monotony in
classroom teaching only. On the other hand, nature of the subject matter in science demands for field visit to
make the students familiar with the environment and the different activities happening around us. Therefore,
teachers are expected to take the students outside the classroom to break their monotony as well as make them
familiar with the surroundings. Teachers are expected to make frequent visits to the concerned areas with their
active participation. This activity is supposed to be beneficial to the students that provides them the chance of
self observation and evaluation of the matter to enhance their knowledge in the fields.

Evaluation of the Students

The evaluation process in science is taken as an interlinked part of the teaching learning process of this subject.
Teachers are expected to emphasize the continuous evaluation of the students in terms of achieving the
intended goals rather than merely focusing on the formal written test. Observation of the students’ activities is
supposed to be the best method of evaluation in science.

Teachers are expected to involve the students in the continuous teaching learning process to achieve the
intended goals. The evaluation process is expected to be continued along with the teaching process to identify
students’ problems that helps both teachers as well as students for further improvement. Teachers can evaluate
the students by various means, like evaluation of class work, homework, project work as well as the evaluation
of the change in behaviour of the students. Specially, in science, teachers are expected to evaluate the students’
procedural skills and keep a systematic record of the achievements. The marking distribution of “Theoretical
Exam” of 75 marks is as follows:

S. N. Subject Area Marks (%) Weighting (Marks)

1. Physics 30.00 23
2. Chemistry 30.00 23
3. Biology 30.00 22
4. Astronomy and Geology 10.00 7
100 75
Total

Similarly, the following bases should be taken for practical evaluation.
Drawing, labeling, collection of materials, observation, identification and explaining characteristics
Record of practical work
Construction of materials and their uses
Mini project work
Viva voce
The marking distribution of “Practical Exam” of 25 marks is as follows:

S. N. Particulars Weighting (Marks)
1. Drawing/labeling/explaining characteristics 5
2. Record of practical work 5
3. Materials construction and their uses 5
4. Mini project work 6
5. Viva voce 4
25
Total

Compulsory Science

(Part 1 : Physics)
Scope and sequence of the subject matter

Area Unit Syllabus
Motion and Energy 1. Force
• Gravitation
Physical Phenomena 2. Pressure • Newton’s law of gravitation
Around Us • Gravity
3. Energy • Acceleration due to gravity
Current Electricity and • Mass and Weight
Magnetism 4. Heat • Free fall and weightlessness

5. Light • Liquid pressure
6. Current • Pascal’s law
• Archimedes’ principle and its application
Electricity and • Law of floatation
Magnetism • Simple numericals on Archimedes’ principle and

pressure
• Introduction, application and measurement of

atmospheric pressure (Barometer, syringe, water
pump and air pump)

• Concept of energy
• Sources and utility
• Alternative sources of energy
• Energy conservation
• Biofuel

• Heat and Temperature
• Sources of heat
• Specific heat capacity
• Importance of heat
• Measurement of heat (Simple thermometer, clinical

thermometer, digital thermometer)

• General introduction to lens and its types
• Defects of vision
• Optical instrument

• Effects of current electricity
• Electromagnetic instruments and their use
• Dynamo, transformer, AC and DC, structure of solar

battery
• Battery charger, adaptor and inverter
• Evaluation of electric tariff

(SEE Specification Grid-2074)

S.N. Unit (K) (U) (A) (HA) Total Remarks
1×4 23
1. Force 5×1 4×2 2×3
2. Pressure
3. Energy
4. Heat
5. Light
6. Current Electricity & Magnetism



1UNIT Estimated teaching periods

Theory 8

Practical 2

FORCE Sir Isaac Newton

Objectives (1642–1727 AD)

After completing the study of this unit, students will be able to:
• state Newton’s law of gravitation and prove it.
• differentiate between gravity and gravitation.
• differentiate between mass and weight with their units.
• measure the mass of different bodies.
• describe free fall and weightlessness.

1.1 Introduction

In our daily life, we apply a force to perform various activities. Pull, push, squeeze, stretch,
etc. denote a force. The force of our hands can push or pull objects. A force can change the
position and shape of a body. Similarly, a force can change the direction of motion and speed
of a moving body. However, if a force is applied to move a tree or a fixed wall, the effort will
not be effective. Hence, force is a pull or push which changes or tends to change the state of
rest or of uniform motion or the shape and size of a body. Force is called a vector quantity as
it has both magnitude and direction. In the SI system, force is measured in newton (N) and in
the CGS system, it is measured in dyne [1 N= 105 dyne].

There are different types of forces such as frictional force, gravitational force, electrostatic
force, magnetic force, etc. On the basis of physical proximity, forces can be classified into two
types: contact force and non-contact force.

Contact forces are those which act only in physical contact with each other. For example,
frictional force, collision force, pull, push, etc. Similarly, the forces which do not involve
physical contact between the objects but act through the space between them are called
non-contact forces. For example, gravitational force, electrostatic force, magnetic force,
etc. These forces come into action even though the objects are not in physical contact. The
magnitude of non-contact forces depends on their masses and the distance of separation. These
forces increase with the increase in their masses and decrease with the increase in distance.

In this unit, we will learn about one of the important non-contact forces, i.e., gravitational
force, or gravitation.

proximity /prɒkˈsɪməti/ - the state of being near sb/sth in distance or time
gravitation /ˌɡrævɪˈteɪʃən/ - the force of attraction between any two bodies

PHYSICS Oasis School Science - 10 1

1.2 Gravitation

Gravitation is the force of attraction which exists between any two bodies due to their masses.
If a stone is raised above the ground and released, it falls towards the earth. Since the stone
starts moving downwards, a force must be acting on it. The force is attributed to the attraction
between the earth and the stone. It is called the force of gravity. In fact, the earth attracts all
the objects towards its center. It is due to the gravitational force of the earth that all objects
fall towards the earth when released from a certain height. The gravitational force of the
earth pulls every object towards its center. Therefore, we should apply a force to lift a body
from the earth's surface. It was Newton who said that every object in this universe attracts
every other object with a force called the gravitational force. For a small body, the force of
gravitation is small and cannot be detected easily. Newton concluded that it is not only
the earth which attracts the other objects but every object in this universe attracts every
other object. For example, two stones lying on the ground attract each other. The force of
gravitation between them is very small and we do not notice any motion. However, if one of
the bodies has a very large mass (like the earth), the small body lying near it moves towards
the bigger body.

Since the masses of the sun and the earth are very large, they exert a very large force on one
another. The mass of sun is so large that even the sum of the masses of all its planets and
satellites is only about 0.0015th part of the sun. 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 effect of gravitation can be observed more on liquids than on solids. The tides
in the sea are due to the force of attraction, which the sun and the moon exert on the water
surface in the sea. Approximately twice a month at the time of the new moon and full moon,
the tide's range reaches its maximum due to the effect of the combined gravitational force of
the sun and the moon on the earth when they lie on a straight line.

Sun F F

Moon

Earth

Fig. 1.1. Gravitational force

Thus, gravitation is the force of attraction, which exists between any two bodies of the universe
due to their masses. Newton was the first person who gave the popular law in 1687 AD called
Newton’s universal law of gravitation.

1.3 Newton’s Universal Law of Gravitation

The great physicist Sir Isaac Newton propounded the law of gravitation when he saw an apple

2 Oasis School Science - 10 PHYSICS

falling from a tree. Newton's law of gravitation states, “Every body in the universe attracts
every other body with a force which is directly proportional to the product of their masses and
inversely proportional to the square of distance between their centers.”

Let us consider two bodies of masses 'm1' and 'm2' separated by a distance 'd' from their centers.
If the force of attraction between them is F,

m1 m2

F
d

Fig.1.2
According to Newton’s law of gravitation,

(i) The gravitational force between two bodies is directly proportional to the product
of their masses, i.e.,

F ∝ m1 × m2............................. (i)
(ii) The gravitational force between two bodies is inversely proportional to the square

of the distance between their centers, i.e.,

F ∝ 1 .............................. (ii) Reasonable Fact-1
d2
Why is Newton's law of gravitation called
Combining (i) and (ii), we get the universal law?

F ∝ m1m2 Ans: Newton's law of gravitation holds true
d2 or is applicable to all the objects present in the
universe, whether the objects are terrestrial or
∴ F =G m1m2 celestial. The gravitational force between any
d2 two objects exists everywhere in this universe.
Therefore, Newton's law of gravitation is also
[Where 'G' is a constant called called the universal law of gravitation.
the universal gravitational

constant. Its value is 6.67×10-11

Nm2/kg2.]

1.4 Major Consequences of Gravitational Force

(i) Existence of solar system, galaxy and constellations
(ii) Revolution of planets around the sun
(iii) Revolution of natural satellites around the planets
(iv) Formation of tides in the sea
(v) Rainfall on the earth

terrestrial /təˈrestriəl/ - connected with the planet earth

celestial /səˈlestiəl/ - of the sky or heaven, the sun, moon, stars, etc.

PHYSICS Oasis School Science - 10 3

1.5 Applications of Newton’s Law of Gravitation

(i) This law helps us to determine the mass of the earth and other heavenly bodies.
(ii) This law helps us to calculate the distance between any two heavenly bodies such

as the earth and moon, sun and earth, etc.
(iii) This law helps in discovering new planets, stars and other heavenly bodies.

1.6 Universal Gravitational Constant (G)

Universal Gravitational Constant (G) is the force of attraction which exists between two
bodies of unit masses kept at a unit distance from their centers. 'G' is a scalar quantity. Its
value remains the same throughout the universe and is independent of the nature and size of
the bodies as well as the nature of the medium between them.

According to Newton's law of gravitation, we have

F = G Mm 1 kg 1 kg
d2

If, M = m = 1kg and d = 1m, then 1m

F 1×1 F=G
= G 12

∴ F = G Fig. 1.3

Thus, universal gravitational constant is the force of attraction between two bodies each of 1
kg mass separated at 1 m apart from their centers.

1.7 SI Unit of G

According to Newton's law of gravitation, the gravitational force acting between any two
bodies is given by

F = G Mm
d2

This formula can be rearranged to get the expression for G as follows:

or, =G MF=dm2 Nm2 = Nm2 / kg2
kg.kg

∴ The SI unit of gravitational constant (G) is Nm2/kg2 or Nm2 kg-2.

Fact File -1

The value of gravitational constant (G) is 6.67×10-11 Nm2/kg2. It was calculated by Henry
Cavendish by using a sensitive balance called the torsion balance in 1798 AD.

Fact File -2

If 'G' by some miracle were suddenly multiplied by a factor of 10, we would be crushed to
the floor by the earth's attraction and if 'G' were divided by this factor, the earth's attraction
would be so weak that we would be able to jump over a building easily.

4 Oasis School Science - 10 PHYSICS

1.8 Variation of Gravitation with Mass and Distance

A. When the mass of a body is doubled keeping the distance between two bodies

constant

According to Newton's law of gravitation,

F= GMm
d2

When the mass of a body is doubled, d

M → 2M M Mm m
d2
Then, F = G

Fnew = G(2M)m Fig. 1.4(a)
d2

= 2  GMm  2M d
 d2  Fnew
= G (2M).m m
d2
2F  F GMm 
= = d2  Fig. 1.4(b)



Hence, the gravitation between two masses is doubled when the mass of a body is
doubled keeping the distance between them constant.

B. When the distance between two bodies is halved keeping the masses constant

According to Newton's law of gravitation,

F= GMm
d2

WWhheenn tthheemdaissstaonf acebobdeytwisedeonutbwleod,

bodieMs is→ha2lvMed, d

Then, d G(2M)m M Mm m
d→ =2 d2 d2
F=G
ThenF, new
G GMMdd2mm2
( )Fnew == 2  Fig 1.5(a)


= 2FGMd2Fm2 = GMm  d
= d2 

4 M 2 m
Fnew = G(M2dm)2
= 4  GMm 
 d2 
Fig. 1.5 (b)
 F = GMm 
= 4F d2 



Hence, the gravitation between two masses becomes four times the previous force
when the distance between them is halved keeping their masses constant.

PHYSICS Oasis School Science - 10 5

Worked out Numerical 1

Calculate the gravitational force between two masses of 30 kg and 10 kg when they are kept
500 cm apart.

Solution:

Given,

Mass of one body (m1) = 30 kg
Mass of another body (m2) =
10 kg
Distance (d) =
Gravitational force (F) = 500 cm = 500 =5m [ ∵ 1 m = 100cm]
100

?

We have, m1m2
d2
F = G

= 6.67 × 10−11 × 30 × 10 [ ∵ G = 6.67 × 10-11 Nm2/kg2]
(5)2

= 8 × 10-10 N.

∴ The gravitational force existing between them is 8 × 10-10 N.

Worked out Numerical 2

Calculate the gravitational force acting on a body of mass 50 kg on the surface of the earth.
The mass and radius of the earth are 6×1024 kg and 6400 km respectively.
Solution:

Given,

Mass of the earth (M) = 6 × 10 24 kg

Mass of a body (m) = 50 kg

Radius of the earth (R) = (d)= 6400 km = 6400 × 1000m [ ∵ 1km = 1000 m]

Gravitational force (F) = ?

We have,

F = G Mm
d2

= 6.67 × 10−11 × 6 × 1024 × 50 ( ∵ G = 6.67 × 10-11 Nm2/kg2)
(6400 × 1000)2

= 488.5N

∴ The gravitational force between them is 488.5N.

6 Oasis School Science - 10 PHYSICS

Activ ity 1

• Go to an open ground. Take a ball and throw it upward. What do you observe? It
reaches a certain height and then it starts falling. What is the reason behind it?

1.9 Gravity and Gravitational Field

The earth attracts all the bodies which are near its surface. When a ball is dropped from a roof
of a house, it falls on the surface of the earth. Similarly, fruits fall from a tree. The force that
pulls these objects downwards is the gravity of the earth. Thus, gravity is the force that pulls
a body towards the center of the earth or a planet.

Fig. 1.6

The SI unit of gravity is newton (N). It is a vector quantity. The gravity (F) of the earth or a
planet depends on its mass (M) and radius (R). The gravity of a planet acts towards its center.

All the bodies present inside the gravitational field of the earth or a planet are pulled towards
the center of the earth or the planet. The gravitational field of a planet is the area around the
planet upto where the gravity of the planet has its influence on an object. The gravitational
field of a planet depends on the mass and radius of the planet.

The force of attraction between a planet and a body on the surface or near its surface is called
gravity, or weight of the body on that planet. So gravity of a planet, or a heavenly body, can be
calculated on the basis of Newton's law of gravitation.

Gravity (F) = Weight (W) = GMm [∵ F=W]
R2

Where, F = Gravity, W = Weight of the body, M = Mass of the planet, m = Mass of the body,

R = Radius of the planet and G = Gravitational constant

∴F=W= GMm
R2

PHYSICS Oasis School Science - 10 7

Fact File -3

The moon has its own gravity like that of the earth and other heavenly bodies. According
to scientists, the gravity of the earth is about six times more than that of the moon. Thus, a
body weighing 100 N on the moon weighs 600 N on the earth.

Different heavenly bodies have their own gravity. The effect of the earth's gravity is more on
a body of larger mass than that on a body of a smaller mass. Therefore, it becomes difficult to
lift a large stone than a smaller one. The weight of a body is the measure of the gravity acting
on the body.

Worked out Numerical 3

Calculate the weight of a body of mass 50 kg: (i) on the surface of the earth and (ii) on the
surface of the moon. [Given, mass of the earth = 6×1024 kg, mass of the moon = 7.2 ×1022 kg,
radius of the earth = 6400 km, radius of the moon = 1.7× 10³ km]

Solution:
Given

i. On the surface of the earth,

Mass of the earth (M) = 6 × 1024 kg

Mass of the body (m) = 50 kg

Radius of the earth (R) = 6400 km = 6400 × 1000 m

Weight of the body (F or W) = ?

We have, GMm
F = W = R2

= 6.67 × 10−11 × 6 × 1024 × 50 [ ∵ G = 6.67 × 10-11 Nm2/kg2]
= (6400 × 1000)2

488.5 N

∴ Weight of a body on the earth's surface = 488.5 N.

ii. On the surface of the moon,
Given,
Mass of the moon (M) = 7.2 × 1022 kg

Mass of the body (m) = 50 kg

Radius of the moon (R) = 1.7 ×10³ km

= 1.7 × 10³ ×1000 m

= 1.7 × 106 m

Weight of the body (F or W) = ?

8 Oasis School Science - 10 PHYSICS

We have,

F = W = GMm
R2
6.67 × 10−11 × 7.2 × 1022 × 50
= (1.7 × 106 )2 [ ∵ G = 6.67 × 10-11Nm2/kg2]

= 83.08 N

∴ Weight of the body on the surface of the moon = 83.08 N

Now, compare the weight of a body of mass 50kg on the surface of the earth and that on the
surface of the moon. You notice that the weight of the same body differs on the surface of the
earth and on the moon. It also proves that the weight of a body on the surface of the moon is

1
about th of that on the surface of the earth.

6

1.10 Effects of Gravity

The gravity of the earth affects various activities and objects on the earth. The major effects of
the earth's gravity are given below:

(i) Falling of objects towards the center of the earth
(ii) Stability of large buildings and bridges
(iii) Existence of the atmosphere on the surface of the earth
(iv) Flowing of rivers
(v) Blowing of wind

1.11 Acceleration due to Gravity

When a body is dropped from a certain height, its velocity increases at a constant rate. It
means that a uniform acceleration is produced in the falling body due to the gravitational pull
of the earth. Thus, the acceleration produced in a freely falling body due to the influence of
gravity is called acceleration due to gravity. It is denoted by 'g'. The SI unit of acceleration due
to gravity is meter per second per second (m/s2).

Vacuum

Same acceleration is pro-
duced in all the falling bodies in the
absence of external resistance

Fig. 1.7
acceleration /ək,seləˈreɪʃn/ - rate of change in velocity of a body

PHYSICS Oasis School Science - 10 9

From the above figure, it becomes clear that the acceleration produced in freely falling bodies
will be the same, and it does not depend on the mass of the falling bodies in the absence
of external resistance. Gravity and acceleration due to gravity are two different physical
quantities. The acceleration produced in the bodies falling towards the surface of the earth is
the effect of earth's gravity.

Relation between Acceleration due to gravity and Radius of the earth

Let us consider a body of mass 'm' is falling on the surface of the earth of mass 'M' and radius
'R'. Then the force exerted by the earth on the body is given by Newton’s law of gravitation, i.e.

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

R

M

Earth
Fig. 1.8

The force exerted by the falling body is given Reasonable Fact-2

by F = mg ________ (ii) (Where 'g' is the Why does the value of 'g' vary from place to
acceleration produced in the falling body) place on the surface of the earth? Explain.

From (i) and (ii), we get Ans: The value 'g' is inversely proportional to
the square of the radius of the earth. The earth
mg = GMm is not perfectly spherical. It is flat at the poles
R2 and bulges at the equator. Similarly there are
plains, hills, valleys and mountains of various
or, g = GM heights on the surface of the earth. Due to this,
R2 the distance between the center and various
places of the earth is variable. Therefore, the
or, g∝ 1 value of 'g' varies from place to place on the
R2 surface of the earth.

[Where G and M are constants.]

∴ g∝ 1 is proved.
R2

Thus, acceleration (g) produced on the surface of the earth is inversely proportional to the
square (R) of the radius of the earth. So, when 'R' decreases, 'g' increases and when 'R' increases
'g' decreases in squared value.

GM
From the relation g = R² , it is clear that the value of acceleration due to gravity (g) depends
on the values of M, R and G. Since 'M' and 'G' are always constant, the value of 'g' remains

constant as long as the radius of the earth 'R' remains constant. Therefore, the value of 'g'

remains constant at a given place on the surface of the earth. However, the value of 'g' does not

remain constant at all places on the surface of the earth. The earth is not a perfect sphere. It is

flat at the poles and bulging at the equator. Similarly, there are high mountains, hills, valleys

10 Oasis School Science - 10 PHYSICS

and plains on the surface of the earth. So the value of 'R' differs from place to place on the

surface of the earth. Therefore, the value of 'g' also changes from place to place on the surface

1 Pole (g=9.83 m/s2)
of the earth [ ∵ g ∝ R² ].

R less

R more Equator
(g=9.78 m/s2)
g∝ 1
R²
Earth

Fig. 1.9

Since the radius (R) of the earth at the poles is minimum, the value of acceleration due to
gravity (g) is maximum, i.e. 9.83 m/s2. Similarly, the radius (R) of the earth at the equators is
maximum, so the value of 'g' is minimum, i.e., 9.78 m/s2. However, the average value of 'g' on
the earth's surface is taken as 9.8 m/s2.

m

g = 9.8 m/s²

M

Earth

Fig. 1.10

When a body falls towards the earth's surface only under the action of gravity, its velocity
increases at a constant rate of 9.8 m/s for every second of time it is falling. When a body is
dropped freely, it falls with an acceleration of 9.8 m/s2. When a body is thrown vertically up-
wards, it undergoes a retardation of 9.8 m/s2. So the velocity of a body thrown upwards will
decrease at the rate of 9.8 m/s in every second of time and the velocity decreases until it reach-
es zero. The body falls back to the earth with an acceleration of 9.8 m/s2.

Prove that the acceleration produced on a freely falling body is independent of the
mass of that body.

Let us consider a body of mass 'm' is falling freely towards the m R
surface of the earth. Let the mass and radius of the earth be 'M' and
'R' respectively. M
Earth
According to Newton’s law of gravitation,
F = GMm ............................. (i) Fig. 1.11

R2
From Newton's second law of motion,

F = mg............................. (ii)

From (i) and (ii), we have

PHYSICS Oasis School Science - 10 11

mg = GMm
R2

or, g = GM
R2
Fact File -4
In the above expression, the mass of the falling
All freely falling bodies, small or large, fall
body is cancelled, and there is no mass of the together on the surface of the earth under
the same effect of acceleration due to grav-
falling body. So, the acceleration produced ity (g).

on a freely falling body is independent of the

mass of that body.

Worked out Numerical 4

Calculate the value of acceleration due to gravity on the surface of the earth. The mass and
radius of the earth are 6 × 1024 kg and 6380 km respectively.

Solution:

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

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

Acceleration due to gravity (g) = ?

We have, GM
g = R2
6.67 × 10−11 × 6 × 1024
= [ ∵ G = 6.67 × 10–11 Nm2/kg2]
(6380 × 1000)2

= 9.8 m/s2

Acceleration due to gravity at a certain height (h) from the earth's surface m
Let us consider a body of mass 'm' is placed at a height 'h' above the surface h

of the earth of mass 'M' and radius 'R' as shown in the figure.

According to Newton's law of gravitation, R
Earth
F = GMm Fig. 1.12
(R + h)2

or mg' = GMm [∵ F = mg'
(R + h)2
g' = acceleration due to gravity at the given height]
or g' GM
= (R + h)2



∴ g' = GM
(R + h)2

12 Oasis School Science - 10 PHYSICS

Fact File - 5

Since the distance from the center of the earth to a certain height (R+h) is greater than that
on the surface, the value of g' is less at the given height than that on the surface of the earth.
Similarly, the value of 'g' is less at the top of a hill than at its foot.

Worked out Numerical 5

The mass of the earth is 6×1024 kg and its radius is 6400 km. Calculate the acceleration due
to gravity at the top of Mt. Everest of height 8848 m from the earth's surface. Also, calculate
the weight of a person of mass 60 kg on top of Mt. Everest.

Solution:

Given,

Mass of the earth (M) = 6×1024 kg

Radius of the earth (R) = 6400 km

= 6400×1000 m

= 6.4 ×106 m

Height of Mt. Everest (hF) == (8RG84M+8hmm)2

Acceleration due to groarvitym(gg'') = ?GMm
= (R + h)2

According to the formula, GM
(R + h)2
or g' g ' ==

6.67 × 10–11 × 6 × 1024
= (6.4 × 106 + 8848)²

= 4.002 × 1014
4.1 × 1013

= 9.76 m/s2

∴ The acceleration due to gravity at the top of Mt. Everest is 9.76 m/s2.

Now,

Acceleration due to the gravity (g') = 9.76 m/s²

Mass of the person (m) = 60 kg

Weight of the person (W) = ?

According to the formula,

W = m × g'

= 60 × 9.76

= 585.6N

∴ Weight of the person on the top of Mt. Everest = 585.6 N.

PHYSICS Oasis School Science - 10 13

Differences between Acceleration due to gravity (g) and Gravitational constant (G)

S.N. Acceleration due to gravity (g) S.N. Gravitational constant (G)

1. The acceleration produced on a 1. Gravitational constant is the

freely falling body under the effect force of attraction between two bodies

of gravity is called acceleration due of unit mass each kept at unit distance

to gravity. apart from their centers.

2. It is a variable quantity. Its value 2. It is a constant quantity.
changes from place to place.

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

4. Its SI unit is m/s2. 4. Its SI unit is Nm2/kg2.

Differences between Acceleration due to gravity (g) and Gravity (F)

S.N. Acceleration due to gravity (g) S.N. Gravity (W or F)

1. The acceleration produced on a 1. Gravity is the force by which a body

freely falling body under the effect is attracted towards the center of the

of gravity is called acceleration due earth or any other heavenly body.

to gravity.

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

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

1.12 Coin and Feather Experiment

On the basis of his experiments, Galileo Galilei Feather Air
concluded that the acceleration of an object falling Coin
freely towards the earth does not depend on the Vacuum
mass of the body. To verify this fact, Robert Boyle
performed an experiment in a vacuum. This is a
famous experiment to prove that the acceleration
produced on every freely falling body remains the
same. It does not depend on the mass of the falling
body.

In this experiment, a coin and a feather are kept inside Fig. 1.13 Coin and feather experiment
a tall glass tube and its ends are closed. The tube is
inverted as soon as possible.

The coin reaches the bottom earlier as compared to the feather as the tube contains air. Then
the tube is made free from air by using a vacuum pump. When the tube is quickly inverted,
the coin and the feather reach the bottom simultaneously.

When the tube contains air, the feather falls slower than the coin. Because the resistance offered
by air on the feather is more as the surface area of the feather is larger than that of the coin.

vacuum /ˈvækjuəm/ - a space that is completely empty of all substances, including all air or other gas
feather /ˈfeðər/ - one of the many soft light parts covering a bird's body.

14 Oasis School Science - 10 PHYSICS

Conclusion of feather and coin experiment
The acceleration produced on freely falling bodies remains the same for all the bodies and it
is independent of the mass of those bodies.

Activity 2

• Take a sheet of paper and a stone. Drop them simultaneously from the top of a
building. Observe whether both of them reach the ground simultaneously or not.

When a body falls down, air offers resistance to the falling body due to friction between the
body and air. The resistance offered by air to the paper is more than that to the stone. If this
experiment is conducted in a vacuum, the paper and the stone would fall simultaneously.

Activity 3

• Take two stones of different sizes. Put a sheet of tin on the ground and drop these
stones from the same height simultaneously on the tin. Listen to the sound produced
on their falling. Observe whether they fall simultaneously or not. What is the reason
behind it?

Activity 4
• Take two sheets of paper (one folded and another unfolded) instead of the above
two stones and repeat the activity 3. What do you observe?

When a parachutist jumps from a very high altitude, s/he does
not get hurt. It is because of the resistance offered by the air,
and the acceleration due to gravity is wholly affected by the
upthrust of the air. Since the parachute falls slowly towards
the surface of the earth due to air resistance, the fall of the
parachute is not a free fall.

Activity 5 Fig.1.14 (a) Paragliding

• Take a thick plastic sheet. Cut it in the form of a Plastic
circle of radius 20 cm. Make 8 holes at the edge Thread
of the plastic. Tie a stone with the support of
the thread through these holes. Drop it from the Stone
top of a building and observe the motion of the Fig.1.14 (b) Model of a parachute
model of parachute. Write down the conclusion
of this activity.

PHYSICS Oasis School Science - 10 15

Worked out Numerical 6

A person can lift 50 kg mass on the earth’s surface. Find the mass that he can lift on the
surface of the moon. The value of 'g' on the moon is 1.66m/s².

Solution:
Let 'm' be the mass that the person can lift on the surface of the moon.

Now,

Force applied by the person on the earth’s surface = Force applied by the person on the surface of the moon

or, mearth × gearth = m × gmoon

or, 50 × 9.8 = m × 1.66

or, 50 × 9.8 =m
1.66

m = 295.18 kg

Thus, the person can lift 295.18 kg on the surface of the moon.

1.13 Mass

The total quantity of matter contained in a body is called its mass. It is a constant quantity. The
mass of a body remains the same wherever it be on the earth, moon or even in outer space.
It is measured by a pan balance or a beam balance. It is a scalar quantity. The mass of a body
cannot be zero. It is measured in kilogram (kg), gram (g), milligram (mg), etc.

Mass of a body depends on:
i. size of atoms or molecules and
ii. number of atoms or molecules of the body.

1.14 Weight

Weight of a body is the gravity acting on the body. Thus, the weight of a body is defined as the
force with which it is pulled towards the center of the earth or a planet. It is a variable quantity.
It has direction as well as magnitude. The direction of the weight is towards the center of a
massive body (like the earth). So, weight is called a vector quantity. It is measured by using the
spring balance. The SI unit of weight is newton (N).

The weight of a body at a place depends on :

i. mass of the body, and
ii. value of acceleration due to gravity at that place.

The weight of a body is directly proportional to the mass of that body. The weight of a body is
calculated by W = m×g. The value of g (acceleration due to gravity) differs from place to place.
Therefore, the weight of a body also differs from place to place. The weight of a body becomes
zero where g = 0, such as in interplanetary space. In such places we feel true weightlessness.
We know that acceleration due to gravity of the moon is six times less than that of the earth.
So, the weight of a body on the moon will be about one-sixth of its weight on the earth.

16 Oasis School Science - 10 PHYSICS

Reasonable Fact-3 Reasonable Fact-4

Weight of a body is found less at the top The weight of an object is more at the
of a mountain than at the bottom of it, polar region than that in the equatorial
why? region of the earth. Justify this statement.

Ans: The value of 'g' is more at the bottom Ans: The value of 'g' is more at the poles (i.e.
of a mountain due to less radius and less at
9.83 m/s2) due to less radius and less at the equator

the top of the mountain due to more radius ( )(i.e. 9.78m/s2) due to more radius∵ga1 .
R2
g( )∵a1 . Therefore, the weight of an object Therefore, the weight of an object is more at the
R2
is less at the top of a mountain than at the polar region than in the equatorial region of the

bottom of the mountain. earth.

Differences between Mass and Weight

S.N. Mass S.N. Weight

1. The total quantity of matter present in 1. The weight of a body is the measure of

a body is called its mass. the gravity acting on the body.

2. Its SI unit is kg. 2. Its SI unit is N.

3. It is measured by using a beam 3 It is measured by using a spring

balance. balance.

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

5. It is a constant quantity for a particular 5. It is a variable quantity.
body.

Worked out Numerical 7

The radius of Jupiter is 11 times more than that of the earth and the mass of Jupiter is 319
times more than that of the earth. Now, find out how many times more the gravity of Jupiter
is than that of the earth.
Solution:
Given,

Mass of the earth = me
Mass of Jupiter (mj) = 319 × me
Radius of the earth = re
Radius of Jupiter (rj) = 11 × re
Let the mass of a body be 'm'.

PHYSICS Oasis School Science - 10 17

Gravity acting on a body on the surface of Jupiter (F) = G mjm
rj2

= G 319m em
(11re )2

= G 319m e m
121re ²

= 2.6 G me m
re ²

Hence, the gravity of Jupiter is about 2.6 times more than that of the earth.

1.15 Equations of Motion for Freely Falling Bodies

Freely falling bodies fall with uniform acceleration (g) towards the surface of the earth or a

planet. The three modified equations of motion can be applied to the motion of freely falling

bodies, which are as follows:

i) v = u + gt Where, v = final velocity
u = initial velocity
ii) h = ut + 1 gt2 g = acceleration due to gravity
2 t = time taken
h = height from the earth's surface
iii) v2 = u2 +2gh

These modified equations of motion are used to solve numerical problems related to freely
falling bodies. We should remember the following points for the motion of freely falling bodies:

i) When a body is falling vertically downwards, the acceleration due to gravity (g) is taken
to be positive.

ii) When a body is thrown vertically upwards, the acceleration due to gravity (g) is taken to
be negative.

iii) When a body is dropped freely from a height (h), its initial velocity (u) becomes zero.
iv) When a body is thrown vertically upwards, its final velocity (v) becomes zero.
v) The time taken by a body to rise to the highest point is equal to the time it takes to fall

from the same height.

Worked out Numerical 8

To estimate the height of a bridge over the Mahakali River, a stone is dropped freely in
the river from the bridge. The stone takes 2 seconds to touch the water surface in the river.
Calculate the height of the bridge from the water level. (g = 9.8m/s2)

Solution:

Given, = 0 m/s [∵ The stone is dropped freely.]
Initial velocity (u) = 2s
Time taken (t)

18 Oasis School Science - 10 PHYSICS

Acceleration due to gravity (g) = 9.8 m/s2

Height of the bridge (h) =?

According to the formula,

1
h = ut + 2 gt2

= 0 × 2 + 1 × 9.8 × 22
12
= 2 × 9.8 × 4

= 19.6 m

∴ The height of the bridge above the water level is 19.6 m.

1.16 Free Fall Fig. 1.15 An artificial satellite
revolving around the earth
When a body is falling due to the effect of gravity,
acceleration is produced in it, which is called acceleration
due to gravity. So, if a body is falling only under the effect
of gravity, the fall of the body is called a free fall. Thus,
when a body is falling only under the effect of gravity
without external resistance, the fall of the body is called
a free fall. A body falling in a vacuum is an example of a
free fall.

The value of acceleration due to gravity depends on the
radius of the earth or distance from the center of the earth.
The value of acceleration due to gravity decreases with
the increase in the distance from the center of the earth.

Fact File - 6

A man made satellite can revolve on its own path due to the balance between the force
of gravitation and the outward force due to the velocity of the satellite. Centripetal force
is required to move a body in a circular path. The direction of this force is towards the
center of the circular path.

Reasonable Fact-5

The fall of a parachute towards the earth's surface is not a free fall. Justify this statement.

Ans: The fall of an object towards the earth's surface only under the influence of gravity
without external resistance is called a free fall. But there is the presence of atmosphere around
the surface of the earth, which creates external resistance. Therefore, the fall of a parachute the
towards earth's surface is not a free fall.

satellite /ˈsætəlaɪt/ - a natural/artificial object that revolves around a planet

PHYSICS Oasis School Science - 10 19

Reasonable Fact-6

Parachutists are not hurt when they jump out of a plane from a very high place. Why?

Ans: A parachute opens and expands when a parachutist jumps out of a plane. The larger
the area of the parachute, the more will be the resistance of the air. So, the acceleration of the
parachute decreases. Due to decreased acceleration caused by gravity, a parachutist falls slowly.
As a result, a parachutists can balance his body and is not hurt when he jumps out of a plane.

Activity 6 Spring balance Spring balance
held by hand released
• Take a spring balance, hold it by
one hand and measure the weight Spring balance 5 N Zero
of a stone. Now release the spring
balance along with the stone and Stone weighs Stone weighs 0
note down the reading of the pointer 5N
of the spring balance.
Fig. 1.16 Weight of a body at the time of the free fall
The value of the weight at the time of
the free fall will be zero.

1.17 Weightlessness

The weight of a body on the earth’s surface is the force Fig. 1.17 A man falling
by which the earth attracts it. The weight of a body is freely from a lift
measured by using a weighing machine or a spring
balance. Suppose a weighing machine is placed on the
floor of a lift parked at the top floor of a tall building,
and a man is standing on it. In this condition, the
machine shows a certain weight of the man. If the lift
is allowed to fall freely, both the weighing machine and
the man would fall freely towards the earth with the
same acceleration due to gravity. Under this condition,
it is not possible to get the value of the weight on the
machine. It is to be noted that the weight of the body
shown by the machine is due to its reaction force. When
the reaction force is zero (at the time of the free fall), the
machine shows zero weight. Hence, we can say a man
is weightless in a freely falling lift. Thus, a body is said
to be weightless when it is falling freely only under the
influence of gravity.

When a body is in a state of zero weight, it is called
weightlessness. It can also be defined as the condition at which a body of a certain mass
becomes weightless.

20 Oasis School Science - 10 PHYSICS

Weightlessness in Space

An astronaut floats in space when s/he is in a spaceship orbiting the earth. Consider
an astronaut in a spaceship orbiting the earth about 1000 km above its surface where the
acceleration due to the gravity of the earth is quite strong, i.e., 7.34 m/s². The weight of the
astronaut in the spaceship cannot be zero since the weight = m.g and g cannot be zero. But we
can say that the astronaut is weightless. When the astronaut is in the spaceship orbiting the
earth, both the astronaut and the spaceship are in a continuous state of free fall towards the
earth’s surface with the same acceleration due to gravity. So the astronaut does not exert any
force on the sides of the spaceship and appears to be floating weightlessly. Thus, the astronaut
feels weightlessness due to the zero reaction force. But true weightlessness is experienced in
the region of outer space where the acceleration due to gravity is zero.

Fig. 1.18 An astronaut falling freely

Reasonable Fact-7

A satellite does not need any energy to revolve around the earth, why?
Ans: A satellite does not need any energy to revolve around the earth because of the balanced
centripetal and centrifugal forces provided by the gravitational force.

Conditions for Weightlessness
i. When a body is falling freely (∵ The reaction force is zero during the free fall.)

ii. When a body is in space at null point
(∵ The value of g = 0, weight = m.g. = m.0 = 0.)

iii. When a body is in a rocket which is orbiting around a heavenly body
(∵ The body inside the rocket is in a state of free fall.)

iv. When a body is at the center of a planet (∵ g = 0)

astronaut /ˈæstrənɔːt/ - a person who travels and works in a spacecraft

PHYSICS Oasis School Science - 10 21

SUMMARY

• Force is an external agent, which changes or tries to change the position of
a body.

• Newton’s universal law of gravitation states, “Every body in the universe
attracts every other body with a force which is directly proportional to the
product of their masses and inversely proportional to the square of the
distance between their centers."

• Gravitational constant (G) is defined as the force of attraction between two
bodies of unit mass kept at a unit distance apart from their centers.

• Newton's law of gravitation is called the universal law since it is applicable
to all bodies whether big or small, celestial or terrestrial.

• Gravity is the force by which a body is attracted towards the center of the
earth or any other heavenly body if the body is on or near the surface.

• The gravity acting on a body is called the weight of the body.
• The total quantity of matter contained in a body is called its mass.
• The mass of a body depends on the size and number of the atoms present

in the body.
• The weight of a body depends on the mass of the body and the value of

acceleration due to gravity.
• Acceleration produced on a freely falling body due to the influence of

gravity is called acceleration due to gravity.
• Acceleration produced on the surface of the earth is inversely proportional

to the square of the radius of the earth.
• The acceleration produced on a freely falling body is independent of the

mass of that body.
• The space around a massive body where the gravitational force can be felt

is called the gravitational field.
• When a body is falling freely only under the effect of gravity, it is called a

free fall.
• The state of a body having zero weight is called weightlessness.

22 Oasis School Science - 10 PHYSICS

Exercise

Group-A
1. What is gravitational force? Write down its SI unit.
2. State Newton’s universal law of gravitation.
3. State the factors that affect the gravitational force.
4. What is the relation between the gravitational force acting between any two objects with

their masses and the distance between their centres?
5. What is gravity? Write down its unit in SI system.
6. What are the factors affecting gravity?
7. Write any two effects of gravity.
8. Which force affects the movement of aeroplane and ships?
9. What is acceleration due to gravity? In which unit is it measured?
10. Write down the conclusion of feather and coin experiment.
11. Define 'mass' of an object?
12. On which factors does the mass of an object depend?
13. What is weight? Write down its unit in SI system.
14. What is the average value of acceleration due to gravity on the surface of the earth?
15. What is a free fall?
16. In which condition does an object fall freely?
17. What is meant by weightlessness?
18. In which condition does a body of certain mass become weightless?
19. Write any two conditions at which a body becomes weightless.
20. What is gravitational field? On which factors does gravitational field depend?
21. What is the value of acceleration due to gravity (g) at the centre of the earth?
22. State the two factors on which the mass of an object depends?
23. When a feather and a coin are dropped towards the surface of earth, do they reach the

ground together?
Group-B

1. Why is Newton’s law of gravitation called universal law?
2. What is the relation between acceleration due to gravity and the radius of the earth?
3. Write any two differences between free fall and weightlessness.
4. Write any two differences between acceleration due to gravity and gravity.
5. Write any two differences between mass and weight.

PHYSICS Oasis School Science - 10 23

6. It is easier to lift a small stone but harder to lift a heavy one on the surface of the earth. Why?
7. The weight of a body is slightly more in the Terai and less in the Himalayan region.

Why?
8. The fall of a parachute towards the earth’s surface is not a free fall. Why?
9. The weight of a body of certain mass becomes zero in space. Why? Write with reason.
10. Astronauts feel them weightless inside spacecrafts. Why?

11. The fall of body on the earth’s surface cannot be a complete free fall. Why?

Group-C

1. Mention any three consequences of gravitational force. Sun Earth
2. Prove that: F = Gm1 × m2 A
3. Prove that: g ∝ 1 d2
B
R2
(Ans: 10m/s2)
4. Write any two effects of gravitational force. 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 ‘A’ to ‘B’ as shown in the figure alongside.

5. 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 acceleration due to gravity of that stone.

Group-D

1. If a spring balance with the iron ball is released, what reading does the spring balance

record and why? If mass of the sun is 2×1030 kg, that of the earth is 6 × 1024 kg and the

distance between them is 1.5×1011 m. What is the gravitational force produced between

them? (G = 6.67 × 10-11 Nm2/kg2). (Ans: 3.557×1022N) 1+3

2. What do you understand by ‘weightlessness due to freefall'? The mass of the earth is

6 x 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)

3. In which condition does the value of gravitation 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 50 kg on the top of Mt. Everest? (Ans: 9.8 m/s2, 488.5 N)

4. Atmosphere is present around the surface of the earth but it is not present around

the surface of the moon, Why? The mass of the Jupiter is 1.9 x 1027 kg and its radius is

7.1 x 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)

5. 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 × 1030 kg. If the gravitational force acting between

them is 4.166 x 1023 N, calculate the distance between their centres. (Ans: 7.8 x 1011m)

24 Oasis School Science - 10 PHYSICS

2UNIT Estimated teaching periods

Theory 8

Practical 2

PRESSURE

Objectives Archimedes

After completing the study of this unit, students will be able to: (287–212 BC)

• introduce pressure and demonstrate liquid pressure.

• state and prove Pascal’s law.

• state Archimedes’ principle and describe its application with examples.

• demonstrate the law of flotation.

• solve the problems related to Pascal’s law and Archimedes’ principle.

• introduce atmospheric pressure and explain its applications.

2.1 Introduction

Bags are provided with broad handles so that less pressure is exerted on the hand. Railway
tracks are laid on large sleepers so that the force due to the weight of the train is spread over a
large area. It reduces pressure on the ground. It is painful to walk with bare feet on gravelled
road. It is due to the fact that our body is based on a small area on the ground and there is
large pressure on our feet.

a. b. c.

Fig. 2.1

We find it easy to chop vegetables with a sharp knife as it exerts more pressure on the
vegetables. Food is cooked faster in a pressure cooker than in an open pot due to higher
pressure inside the pressure cooker. Similarly, a helicopter can fly in air because the fans of the
helicopter keep the air pressure below them higher than above during the flight. Water flows
at high speed in the tap downstairs than in upstairs. These examples justify the existence and
utility of pressure in our daily life. The thrust per unit area of a surface is called pressure. The
pressure exerted by air is called atmospheric pressure. It is very useful for us as it helps to
fill ink in pen, to fill medication in a syringe, to lift underground water, to inflate the tyres of
vehicles and so on. In this unit, we will study liquid pressure, various laws related to pressure
(Pascal's law, Archimedes' principle, law of floatation) and atmospheric pressure.

pressure /ˈpreʃər/ - thrust acting per unit area of a surface

PHYSICS Oasis School Science - 10 25

2.2 Thrust Units of thrust

When a body is placed on a surface, it exerts a force on that SI unit newton (N)
surface. The total perpendicular force exerted by a body on
the surface is called thrust. The effect of thrust depends on CGS unit dyne
the area of the surface on which the thrust is applied. Since
thrust is a force (F), it is measured in newton (N), dyne, etc.
It is a vector quantity.

2.3 Pressure

Pressure is defined as the thrust per unit area of a surface. When a thrust (F) is acting on a
surface area (A) in contact, the pressure exerted is given by

Fact File -1

Pr essure (P) = Thrust (F) ∴P= F Pressure is a scalar quantity as it is
Area (A) A taken perpendicular to the surface area.
In case of liquid, the pressure is exerted
equally in all directions, which means
that pressure has no definite direction.

The SI unit of pressure is Nm-2 or pascal (Pa). Any substance having weight can exert pressure.
Like solids, fluids, i.e., liquid and gas also have weight. Fluids exert pressure on the walls and
bottom of the vessel in which they are kept.

Differences between Force and Pressure

S. N. Force S. N. Pressure
1. It is a push or pull acting on a body. 1. It is the thrust acting per unit area.
2. Its SI unit is newton (N). 2. Its SI unit is pascal (Pa).
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.

Activity 1

• Take a brick and foam of thickness about 10cm Brick
to 20cm. Measure the dimension and weight Foam
of the brick. Put the brick on the foam with its
broad surface as shown in fig.2.2(a).

• Observe the depression made by the brick. Put (a) Fig. 2.2 (b)
the same brick on the foam with its narrow
surface as shown in fig. 2.2 (b).

• Observe the depression made by the brick and calculate the pressure in both cases.

Also, write down the conclusion of this activity.

26 Oasis School Science - 10 PHYSICS

Reasonable Fact-1 Reasonable Fact-2

Studs are made on the sole of a football The rear wheels of a tractor are made
player's boot, why? large and flat, why?

Ans: Studs are made on the sole of a football Ans: The rear wheels are made larger and
player's boot to increase the pressure on flat to reduce the pressure. It prevents the
the ground which prevents the player from tyres from descending into the muddy
falling or sliding. land.

2.4 Liquid Pressure

Any substance that has weight does exert pressure. Fluids (gas or liquid) have weight and
exert pressure on the walls and bottoms of the container. If we consider a point within a liquid,
it will be clear that there must be equal upward and downward pressure at that point. The
thrust exerted by a liquid per unit area of the surface is called liquid pressure.

2.5 Properties of Liquid Pressure

1. Liquid pressure increases with depth.
2. Pressure applied on a liquid is transmitted equally in all directions.

3. The pressure at a point in a liquid does not depend on the volume of the liquid in
a vessel.

4. The liquid pressure is independent of the shape of the vessel in which the liquid is kept.

2.6 Pascal’s Law of Liquid Pressure

When a liquid is completely enclosed in a vessel and Fact File - 2
a pressure is applied to it at any part of its surface,
the pressure is transmitted equally throughout the Blaise Pascal (19 June 1623-19 August
liquid. This fact was first propounded by French 1662) was a French mathematician,
scientist Blaise Pascal, which is popularly known physicist, inventor and writer, who
as Pascal's law. propounded the famous law, which is
commonly known as Pascal's law of
According to this law, "Pressure is transmitted liquid pressure.
equally in all directions when pressure is applied
at a place on a liquid kept in a closed container.”

Verification of Pascal's law Polythene
bag
a. Take a plastic bag. Fill the bag with water. Make
a number of holes on the bag with a needle. Now, Water
squeeze the bag as shown in the figure. Does water
come out from all the holes with the same pressure? It 2.3 Plastic bag with water
can be seen that water comes out through all the holes
simultaneously with the same pressure.

PHYSICS Oasis School Science - 10 27

b. Let us consider a spherical glass vessel with four water-tight frictionless pistons P, Q, R
and S having the same cross-sectional area (A). If the piston (P) is pushed by applying
force (F), it exerts a pressure. The pressure exerted on the pistons Q, R and S will be the
same as the pressure exerted on piston P on which force is applied.

These experiments prove Pascal's law of liquid pressure.
The following equipment has been constructed on the basis of Pascal's law of liquid pressure.

1. Hydraulic jack 2. Hydraulic press 3. Hydraulic brakes

Q

P R Fact File - 3

Pascal's law of liquid pressure is
applicable for gases also.

S
2.4 Spherical glass vessel

1. Hydraulic jack: It is used in raising automobiles, like trucks, buses, cars, motorcycles,
etc. for repair works in service stations. The construction of the hydraulic jack is shown
in the figure. This machine works on the basis of Pascal's law. When a small effort is
applied on a small piston, a large force is developed on the large piston. As a result, the
vehicle placed on the large piston is raised upwards. The hydraulic jack is also called a
hydraulic lift.

Applied force

Valve 1 F1 Valve 2 F2
Fig. 2.5 Hydraulic jack Liquid

2. Hydraulic press: Hydraulic press is used for compressing cotton bales, extracting oil
from oil seeds, punching holes in metals, giving specific shapes to metal sheets, etc. It
also works on the basis of Pascal's law.

The construction of the hydraulic press is given in the figure. It has two cylinders X and
Y. The cylinder X is smaller than Y. Those cylinders are fitted with water-tight frictionless
pistons P1 and P2 respectively. The cylinders are filled with a liquid. A rigid ceiling is
placed over the large piston. When a small effort is applied on the small piston, a large
force is developed on the large piston and it is raised upwards. As a result, the substance
placed on the large piston is lifted and gets compressed between the platform of the large
piston and the ceiling.

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Rigid ceiling Pivot
Platform
Material for
pressing

A2 A1 Piston P1

Piston P2

Y F2 F1 X

Liquid

Fig. 2.6 Hydraulic press

3. Hydraulic brakes: Hydraulic brakes are used in heavy automobiles, like trucks, buses,
aeroplanes, cars, motorbikes, etc. to stop them by applying a small effort. The construction
of hydraulic brakes is also based on Pascal's law.

The construction of a hydraulic brake is shown in the following figure. When the
driver presses on the foot pedal, a piston moves inwards, pressing the column of liquid
contained in the master cylinder. This helps to expand the brake holes with the help of
the wheel cylinder. When the pressure on the foot pedal is released, the brake shoe pulls
up due to the force of the spring. The wheel pistons move back into the cylinder and
the liquid returns to the master cylinder. (fig. 2.7). The advantage of this system is that
the pressure set up in the master cylinder is transmitted equally to all wheels so that the
braking effort is equal on all wheels of the vehicle.

Foot pedal

Pipeline Liquid Piston

To other Master cylinder
wheel Wheel cylinder

Brake Brake-shoe
shoe

Hinge Return spring

Fig. 2.7 Hydraulic brake

Hydraulic Machine: A Force Multiplier Vessel
Let us consider a hydraulic machine having two vessels of cross-sectional area 'A1' and 'A2'
respectively. These vessels are fitted with water-tight frictionless pistons. The cross-sectional
area of the small vessel is A1 and that of the large vessel is A2. So A2 > A1.

PHYSICS Oasis School Science - 10 29

F1 F2
A1 A2

Fig. 2.8 Hydraulic machine

When a force F1 is applied on the small piston of area A1­, a pressure P1 is exerted on the small piston.

According to Pascal's law,

Pressure on the small piston = Pressure on the big piston

i.e. P1 = P2
or,
F1 = F2
A1 A2

or, F1×A2 = F2 × A1
Since, A2 > A1

Thus, F1 < F2
Hence, a hydraulic machine is a force multiplier.

Worked out Numerical 4
Study the given figure and answer the following questions:
i) Name the equipment shown in the figure. On the basis of which law does it work?
ii) Calculate the pressure exerted at A.
iii) Calculate the effort (F2) on piston B.

240N F2 = ?

AB

20 cm2 0.4 m2

Ans:
i) The equipment shown in the given figure is a hydraulic machine. It works on the basis

of Pascal's law.

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ii) Solution:

Given,

Force on piston A (F1) = 240 N

Area of piston A (A1) = 20 cm2 = 20 × 10-4 m2 [ ∵ 100 cm = 1m]

We know,

P1 = F1 = 240 = 12 × 104 = 1.2 × 105 Pa
A1 20 × 10−4

∴ Pressure at A ( P1) = 1.2 × 105 Pa

iii) Effort on piston B (F2) = ?

Area of piston B (A2) = 0.4 m²

We know, F1A2 = F2A1

or, F2 = F1A2 = 240 × 0.4 = 4.8 × 104 N
A1 20 × 10−4

∴ Effort on piston B (F2) = 4.8 ×104 N

Application of Pascal's law

Hydraulic machines like the hydraulic jack, hydraulic press and hydraulic brakes are con-
structed on the basis of Pascal's law of liquid pressure.

Principle of a hydraulic machine

The principle of a hydraulic machine states, "A large force is developed on a larger piston
when a small effort is applied on the smaller piston."

2.7 Upthrust

When a body is placed in a liquid, the liquid exerts an upward force on it. For example, when
a piece of cork is held below the surface of water and then released, the cork immediately rises
to the surface. It appears as if some upward force is exerted by water on the cork which pushes
it to the surface. Similarly, if we lift a bucket lying at the bottom of a pond, it appears to be
light as long as it is being lifted inside the water. But as soon as the bucket is lifted out of the
water, the same bucket seems to get heavier. This is because of the upward force of the liquid
on the bucket. In general, whenever an object is partially or wholly immersed in any liquid, it
loses some weight and becomes lighter. When an object is immersed in a fluid (liquid or gas)
the object experiences a resultant force due to the liquid pressure. The resultant upward thrust
exerted by a fluid is called upthrust. Since upthrust is a force acting upward, it is measured
in newton (N).

PHYSICS Oasis School Science - 10 31

Activity 2

• Take an empty plastic bottle. Close the mouth of the bottle with an airtight stopper.
Put it in a bucket filled with water. You see that the bottle floats. Push the bottle into
the water. You feel an upward push. Try to push it further down. You will find that
it is difficult to push it deeper. This indicates that water exerts a force on the bottle
in the upward direction. The force exerted by the water goes on increasing as the
bottle is pushed deeper till it is completely immersed. The upward force exerted by
a liquid is called upthrust.

2.8 Derivation of the Formula of Upthrust

Let us consider a cylinder of height 'h' and uniform cross-sectional area ‘A’ is immersed in a
liquid of density ‘d’. The height of the liquid column above the upper surface of the cylinder
is 'h1' and that of the liquid above the lower surface of the cylinder is 'h2'.
Now,

Force on upper face (F1) = P1 × A = h1dg A [∵ P1 = h1dg]

Force on lower face (F2) = P2 × A = h2dg A [∵ P2 = h2dg] h1
h2 h
[g = acceleration due to gravity]
Fig. 2.9 A cylindrical
Upthrust (U) = F2 – F1 body in a liquid
= h2dgA – h1dgA

∴ U = dgA (h2 – h1)

h2 – h1 = height of the body (h)
U = dgAh

∴ U = V dg [ ∵ V = l × b × h = A × h]

Again, m
U = mg [∵ d = V ]

∴ U = mg where, 'mg' is the weight of the liquid displaced.

Factors affecting upthrust

The upthrust due to a fluid on a body immersed in the fluid depends on the following factors:
1. The volume of the submerged part of a body: Upthrust (U) is directly proportional to the

volume (V) of the submerged part of the body in a fluid, i.e., U ∝ V .
2. Density of the fluid: Upthrust (U) is directly proportional to the density (d) of the fluid

in which the body is submerged, i.e., U ∝ d .

hydraulic /haɪˈdrə:lik/ - operated by liquid moving under pressure

32 Oasis School Science - 10 PHYSICS

3. Acceleration due to gravity: Upthrust (U) is directly proportional to the value of

acceleration due to gravity (g), i.e., U ∝ g .

4. Temperature of the fluid: Upthrust (U) is inversely proportional to the temperature (T)

of the fluid in which the body is immersed, i.e., U∝ 1 .
T

How to calculate upthrust of a liquid

Upthrust of a liquid on an immersed body is measured by finding the difference in the weight
of the body in air and in the liquid.

Upthrust (U) = Weight loss in the liquid

or, U = Weight in air (Wa) – Weight in liquid (Wl)
∴ U = Wa – Wl

Activity 3

• Take a piece of stone and tie it to one end of a rubber Wa Spring Wi
string on a spring balance. Suspend the stone by balance
holding the balance on the string as shown in the
figure.

Note the reading on the spring balance due to the Stone
weight of the stone. Now, slowly dip the stone in Water
a container as shown in the figure. Observe what
happens to the reading on the balance. Fig. 2.10

Reasonable Fact-3 Reasonable Fact-4

An egg sinks in fresh water but floats in A balloon filled with hydrogen gas ris-
a strong solution of salt, why? es up, but a balloon filled with air falls
down, why?
Ans: Fresh water has less density than
that of the egg so the egg cannot displace Ans: The weight of the air displaced by
water equal to its weight due to less the balloon filled with hydrogen gas is
upthrust and it sinks. But salt on mixing more than the weight of the balloon. So the
with water increases the density of water upthrust of the air is also more than the
and hence the upthrust. As a result, the weight of the balloon. As a result, the bal-
egg can displace water equal to its weight loon filled with hydrogen gas rises up. But
and floats. the weight of the balloon filled with air is
more than the upthrust of the air. As a re-
sult, the balloon filled with air falls.

PHYSICS Oasis School Science - 10 33

Activity 4

• Take two beakers, one with salt water Egg
solution and another with pure water.

Put an egg each in both the beakers. Pure Water
In which liquid does the egg float and Salt Water
why?
solution
The density of the salt water solution Egg
is higher than that of the pure water.
Fig. 2.11

So, the upthrust of the salt water solution is more than that of the pure water. Please

note that upthrust is directly proportional to the density of a liquid.

2.9 Archimedes’ Principle

When a solid object is immersed in a fluid, an upward force acts on it. The magnitude of the
force is given by Archimedes' principle. This principle was propounded by Archimedes, a
famous Greek scientist. Archimedes' principle states, "When a body is partially or wholly
immersed in a fluid, it experiences an upthrust which is equal to the weight of the fluid
displaced by it.”

Theoretical Proof of Archimedes’ Principle h1
h2 h
Let us consider a cylinder of height 'h' and uniform cross-sectional area
'A' is immersed in a liquid of density 'd'. If h1 and h2 are the depths of Fig. 2.12
the upper and lower surfaces of the immersed cylinder, then

Force on upper face (F1) = P1 × A = h1dg A

Force on lower face (F2) = P2 × A = h2dg A
[g = acceleration due to gravity]

Now, Upthrust (U) = F2 – F1

= h2dgA – h1dgA

= (h2 – h1) Adg
= hAdg
Let h2 – h1 = h, then,

U

U = Vdg. [∵ V = h × A]

U = mg [∵ V.d = m]

∴ Upthrust (U) = Weight of the liquid displaced

Hence, Archimedes’ principle is proved.

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2.10 Experimental Verification of Archimedes' Principle

a. Place a beaker under the spout of the ureka Spring balance W2
can. Pour water into the can until it starts W1
overflowing through the spout. When the
water stops dripping, keep a beaker under
the spout.

b. Now suspend a stone by the hook of a spring Stone Ureka
balance. Let the reading be W1 on the scale of can
the balance (wt. of the stone in the air).
Water Beaker
c. Gently lower the stone into the water in Top pan
the ureka can. It displaces water which Fig. 2.13 balance
overflows through the spout and collects
in the empty beaker. When the stone is
completely immersed in water and the water
has stopped dripping, let the reading on the
spring balance be W2. You will find that W2 is
less than W1.

d, The water collected in the beaker is the liquid displaced by the stone. Find the
weight of the liquid displaced.

Let

Weight of the stone in air = W1
Weight of the stone in water = W2

∴ Upthrust = W1 – W2 ......................... (i)

Weight of the beaker with displaced water = W3

Weight of the beaker = W4
∴ Weight of the displaced water = W3 – W4 ............. (ii)

The value of (i) and (ii) will be equal. This
verifies Archimedes' principle.

Spring balance

Worked out Numerical 5

Study the given figure and answer the following 20N
questions:

i) What is the upthrust given by the liquid? Stone Ureka
ii) Calculate the weight of the stone in air. Water can
iii) Which principle is this experiment based on?
Solution: Beaker

i) The upthrust given by the liquid (U) = 5 N 5N
[ ∵ Archimedes' principle]

PHYSICS Oasis School Science - 10 35

ii) The weight of stone in air = Weight of stone in water + Weight of water displaced

= (20+5) N

= 25N

iii) This experiment is based on Archimedes' principle.

2.11 Law of Flotation

The law of flotation states, “A body floats on a liquid if it can displace the liquid equal to its
own weight.”
For a floating body,

Wt. of the floating body = Wt. of the liquid displaced

The weight of the solid floating in a liquid is equal to the weight of the liquid displaced by the
immersed part of the solid. In the floating condition, the apparent weight of the body will be
zero, and the body will be weightless.

Verification of the Law of Flotation W1

According to the law of flotation, "The weight of the Wood
liquid displaced is equal to the weight of a floating
body". This can be verified with the help of a simple
experiment.

Pour water into an overflow can to the level of the Wood Wt. of
spout and place a beaker under the spout. Take a solid Water displaced
block (wood) which floats on the liquid and weigh it. water = W2
Let it be W1. Now, immerse the block in the overflow Fig. 2.14
can. It displaces the water which overflows through
the spout and collects in the empty beaker.

Now, weigh the displaced water. Let it be W2. Finally, you will get that
Wt. of the floating body = Wt. of the liquid displaced

∴ Wt. of the floating body = Wt. of the liquid displaced

2. 12 Conditions that Determine the Sinking or Floating of a Body

1. When the weight of a body (solid) is more than the upthrust,
the body sinks in the liquid. It happens so when the density
of the body is greater than that of the liquid (fig. 2.15).

In short,

density of solid > density of liquid

Fig. 2.15

36 Oasis School Science - 10 PHYSICS

2. When the weight of a body (solid) is equal to the upthrust, the body Fig. 2.16
floats below the surface of the fluid. It happens so when the density Fig. 2.17
of the body is equal to the density of the liquid (fig. 2.16).

In short,
density of solid = density of liquid

3. When the weight of a body (solid) is less than the upthrust, the body
floats partially above the surface of the fluid. It happens so when the
density of the body is less than that of the liquid (fig. 2.17).

In short,

density of solid < density of liquid

2.13 Applications of the Law of Flotation

1. Flotation of ships made of iron

An iron nail sinks in water but a ship made of iron floats. The

density of an iron nail is more than that of water. Therefore,

the weight of the iron nail is more than the weight of the

water displaced by it, so it sinks. But a ship made of iron

does not sink because it is made hollow from inside so that

its average density becomes less than that of water. So the Fig. 2.18 A ship made of iron
weight of water displaced becomes equal to the weight of floats on water
the ship, and the ship floats.

2. Flotation of submarines

A submarine displaces water equal to its own weight
when floating on the surface. To make it dive water is
allowed to run into the ballast tanks inside it so that
the weight of the submarine is more than the water
displaced by it.

Fig. 2.19 A submarine dives in water

3. Rising of balloons

When balloons are filled with lighter gases, like hydrogen or
helium, the weight of the air displaced by the balloon becomes
more than the weight of the gas filling the balloon. As a result,
the balloon rises.

As the balloon rises up, the density of air decreases and, hence, Fig. 2.20
the weight of the air displaced also becomes less. Ultimately,
when the weight of the displaced air becomes equal to the
weight of the balloon, it does not rise any further, or it becomes
stationary at a certain height.

PHYSICS Oasis School Science - 10 37

4. Flotation of a man

The density of the human body with empty
lungs is 1.07 g/cm3 while with lungs filled with
air it is 1 g/cm3. A good swimmer with filled
lungs can displace water nearly equal to his
own weight. Because of this s/he can swim
easily in water.

It is easier to swim in sea water than in river Fig. 2.21 A man swimming
water due to higher density of sea water as it
contains dissolved salts, minerals, etc. Greater
upthrust is applied on the swimmer by the sea
water as compared to river water, making it
easier to swim in sea water.

In the Dead Sea, the density of water is 1.16 g/cm3, which offers greater upthrust.
Therefore, chances of drowning in the Dead Sea are negligible.

5. Flotation of icebergs

The density of ice (0.9 g/cm3) is less than that of water (1 g/cm3). So, the weight of water
displaced by ice is more than the weight of the ice. As a result, iceberg floats in water.

Differences between Archimedes’ Principle and Law of flotation

S.N. Archimedes’ Principle S.N. Law of flotation
1. 1. A body floats on a liquid if it can
When a body is partially or wholly displace the liquid equal to its weight.
2. immersed in a liquid, the upthrust 2.
on the body is equal to the weight of It applies only to floating bodies.
the liquid displaced by it.

It applies equally to floating as well
as sinking bodies.

Activity 5 Cork

• Take a beaker and fill it with water. Take a piece of Iron Nail
cork and an iron nail of equal mass. Place them on the Water
surface of the water. Observe what happens.

The cork floats, but the iron nail sinks since the cork
displaces the liquid equal to its weight but the iron
nail cannot displace the liquid equal to its weight.

Fig. 2.22

2.14 Hydrometer

A hydrometer is a device used to measure the density of a liquid or relative density of a liquid.
It works on the basis of the law of flotation.

38 Oasis School Science - 10 PHYSICS

Reasonable Fact-5
An iron nail sinks in water but floats in mercury, why?
Ans: The density of iron is more than that of water. The iron nail cannot displace the water
equal to its weight and it sinks. But the density of mercury is much more than that of the
water. It gives more upthrust, which helps the iron nail to float on mercury.

Reasonable Fact-6

It is easier for a man to swim in sea water than in river water, why?
Ans: The density of sea water (1.026 g/cm3) is more than that of river water (1.00 g/cm3).
Thus, for the same volume, the sea water provides more upthrust as compared to the river
water. Therefore, it is easier for a man to swim in sea water than in river water.

Reasonable Fact-7

A ship can carry more cargo on salty water than on fresh water, why?
Ans: The density of salty water is more than that of fresh water. So salty water provides
more upthrust to the ship than that of fresh water. As a result, a ship can carry more cargo
on salty water than on fresh water.

2.15 Density

Density of a substance is defined as the mass per unit volume of that substance. The SI unit

of density is kg/m³ and its CGS unit is g/cm³. The density of a body is calculated by the given

formula.

Density (d) = Mass (m)
Volume (V)


m

∴ d = V

Worked out Numerical 6
Calculate the density of water if a tank of 3 m3 volume contains 3000 kg mass of water.
Solution:

Given,
Mass (m) = 3000 kg
Volume (V) = 3 m³
Density (d) = ?
We have,

PHYSICS Oasis School Science - 10 39

d =m
V

= 3000
3

= 1000 kg/m3

∴ Density of water is 1000 kg/m3.

2.16 Relative Density

The ratio of density of a substance to the density of water at 40C is called relative density or
specific gravity. It gives an idea about how heavy a body is with respect to water. It can also be
defined as the ratio of the weight of a certain volume of a substance to the weight of the same
volume of water at 40C. Since it is the ratio of two densities, it has no unit. Relative density can
be calculated by the given formulae,

Relative density (R. D.) = Density of substance
Density of water at 40C

Relative density (R. D.) = Weight of certain volume of a substance
Weight of the same volume of water at 40C

Differences between Density and Relative Density

S.N. Density S.N Relative density

1. Mass per unit volume is called 1. The ratio of density of a body to

density. the density of water at 40C is called

relative density.

2. The SI unit of density is kg/m³. 2. Relative density has no unit since it is
the ratio of two densities.

Reasonable Fact-8

An iron ball (nail) sinks in water, but a ship made up of iron floats, why?

Ans: Since the density of iron is greater than that of water, the iron ball (nail) sinks as the
weight of the ball is greater than the weight of the water displaced by it. But a ship made
up of iron is designed in such a way that the relative density of the ship is always less than
that of water. Therefore, the weight of water displaced becomes equal to the total weight
of the ship, and the ship floats.

submarine /ˌsʌbməˈriːn/ - a ship that can travel under water

ballast /ˈbæləst/ - heavy material placed in a ship to make it heavier and keep it steady.

40 Oasis School Science - 10 PHYSICS

Reasonable Fact-9

An iceberg made up of water floats in water, why?

Ans: The density of ice (0.917 g/cm3) is less than that of water (1 g/cm3). Due to this, icebergs
(huge masses of ice) are able to displace the water equal to their weight. Therefore, icebergs
float in water.

Worked out Numerical 7

Calculate the relative density of iron if its density is 7.86 g/cm³. The density of water at 40C
is 1 g/cm³.

Solution: Density of iron = 7.86 g/cm3


Density of water at 40C = 1 g/cm3

We have, density of iron
R.D. = density of water at 40 C

7.86 g / cm3
, = 1g / cm3

= 7.86

∴ Relative density of iron is 7.86

The density and relative density of some substances are tabulated below:

S.N. Substances Density Relative density

1 Wood kg/m³ g/cm³ 0.6-0.8
2 Water 1
3 Milk 600 – 800 0.6 – 0.8
4 Kerosene 1.03
5 Mercury 1000 1 0.8
6 Ice 13.6
7 Lead 1030 1.03 0.92
8 Gold 11
800 0.8 19.3

13600 13.6

920 0.92

11000 11

19300 19.3

2.17 Atmospheric Pressure

The earth is surrounded by a thick layer of air which is called the atmosphere. Air has
weight and it exerts pressure, the pressure exerted by the atmosphere due to its weight is
called atmospheric pressure. Atmospheric pressure at sea level is called standard atmospheric

PHYSICS Oasis School Science - 10 41

pressure, which is about 101300 N/m2 or 760 mmHg. Atmospheric pressure decreases with
increase in altitude. So, atmospheric pressure is less at the top of Mt. Everest than in the Terai.

Atmospheric pressure acts on each and every living organism including different objects on
the earth surface. But we do not feel the atmospheric pressure because it is neutralized by the
blood pressure. Thus, the pressure exerted by the blood in our body is almost equal to the
atmospheric pressure.

Nose bleeding occurs as we go to a higher altitude because the pressure in our body remains
constant but the external atmospheric pressure decreases. Due to this reason, body pressure is
more than that of the atmospheric pressure and hence bleeding of the nose starts.

Different altitudes have different atmospheric pressure. This helps air to blow from one place
to another. The pressure in aeroplanes and jet planes is adjusted so that the passengers feel
comfortable, and they can breathe easily though they fly at high altitude.

Activity 6 Fig. 2.23

• Take a water trough and dip an empty glass as shown in
the figure. Tilt the glass gradually. Observe the activity.

Air bubbles come out of the glass. The glass, which is
dipped into the water trough, contains air though it
seems empty.

Activity 7

• Take a tin can and fill it with water leaving some space. Boil the water for a few

minutes and close the lid. Allow it to cool by pouring cold water over it. What do

you observe? Lid

Water Partial Atmospheric
vapour vaccum pressure

Water

Tin

Fig. 2.24

When the water in the tin can boils, it drives the air out of the can. On cooling the can
by closing the lid, a partial vacuum is formed inside the can. As a result, the tin can
shrinks due to the atmospheric pressure.

42 Oasis School Science - 10 PHYSICS