Approved by the Government of Nepal, Ministry of Education, Science and Technology,
Curriculum Development Centre (CDC), Sanothimi, Bhaktapur, as an additional material for schools
MODERN GRADED
SCIENCE
Class 10
Jai Prakash Srivastav By
Bijay Shankar Mishra
Khaga Raj Ghimire
Mahendra Bahadur Thapa
VIDYARTHI PUSTAK BHANDAR
Publisher and Distributor
Kamalpokhari/Bhotahity, Kathmandu
MODERN GRADED SCIENCE CLASS 10
Publisher : VIDYARTHI PUSTAK BHANDAR
Kamalpokhari/Bhotahity, Kathmandu
Phone : 01-4423333, 4245834, 4227246
E-mail : [email protected]
Web : www.vpb.com.np
Copyright : Authors
Edition : First Edition, 2000
Seventh Revised Edition, 2020
ISBN : 978-99937-699-08-2
Layout : Vidyarthi Desktop
Printed at :
PREFACE
Advancement of science and technology has reached its apex in the 21st century.
Therefore, understanding of facts, concepts and principles of science has become
a dynamic force in our life. Development of science has brought many changes in
the present civilization. So, science has become one of the major disciplines in our
education system.
Modern Graded Science 10 is a part of recently revised and updated edition
of Modern Graded Science series, brought out for the students of Grade 10 who
are going to appear Secondary Education Examination in order to enhance their
knowledge, understanding, application and ability as per the requirements of the
current curriculum of science. The immense popularity of the previous series among
our students and teachers has provided us additional energy and encouragement to
make this edition more practical, more user-friendly and more comprehensive. Based
on the latest approaches to teaching science, the series envisages learner-centred
teaching method and activity-based learning techniques.
Heavy emphasis on developing concepts rather than providing mere information,
special efforts to stimulate learners’ creative thinking encouraging them to get
involved in learning activities, presentation of scientific knowledge in a very logical
and sequential manner in simple and lucid language to keep up with the ability of the
young learners are the main features of this series.
Each chapter of the textbook includes Competencies of the chapter to guide
students and teachers towards the right directions, Some Reasonable Facts to enhance
the conceptual knowledge, Things To Know as the summary of the entire lesson and
Things to Do to develop the skills of project work. Similarly, Test Yourself to test the
learning achievement and understanding, Glossary as a list of technical terminologies
and their meanings and Do You Know to impart the knowledge of new facts are
given in the same order. Plenty of exercise questions of different levels of behaviours
including MCQS, are also introduced in the book. Moreover, the book also incorporates
SEE Specification Gird and two sets of Model Questions.
The authors of this series would like to acknowledge Asso. Prof. Yadab Prasad
Adhikari and Vidyarthi Desktop for their contribution in editing and layout designing
respectively.
Further constructive suggestions and recommendations from subject experts,
teachers, students and well-wishers will be highly appreciated.
Authors
Contents
Chapter-1 FORCE 1
Chapter-2 PRESSURE 31
Chapter-3 ENERGY 55
Chapter-4 HEAT 69
Chapter-5 LIGHT 83
Chapter-6 ELECTRICITY AND MAGNETISM 102
Chapter-7 CLASSIFICATION OF ELEMENTS 124
Chapter-8 CHEMICAL REACTION 146
Chapter-9 ACID, BASE AND SALT 157
Chapter-10 SOME GASES 171
Chapter-11 METALS 184
Chapter-12 CARBON AND ITS COMPOUNDS 195
Chapter-13 MATERIALS USED IN DAILY LIFE 214
Chapter-14 INVERTEBRATE 235
Chapter-15 HUMAN NERVOUS AND GLANDULAR SYSTEMS 246
Chapter-16 BLOOD CIRCULATION IN HUMAN BODY 260
Chapter-17 CHROMOSOME AND SEX DETERMINATION 276
Chapter-18 REPRODUCTION 290
Chapter-19 HEREDITY 305
Chapter-20 ENVIRONMENTAL POLLUTION AND ITS MANAGEMENT 317
Chapter-21 HISTORY OF THE EARTH 327
Chapter-22 ATMOSPHERE AND CLIMATE CHANGE 342
Chapter-23 THE UNIVERSE 357
Specification Grid (Science) SEE 371
Model Question Sets 374
Chapter FORCE
1 Total estimated periods: 10 (T 8 + P 2)
describe Newton's law of gravitation and its application.
differentiate between gravity and gravitation.
differentiate between mass and weight.
express the units of mass and weight.
measure the mass of different substances.
describe free fall and weightlessness.
Force
We frequently observe many surprising and interesting events in our daily life. For
instance, when a body is released from a certain height, it always falls towards the surface
of the earth. Fruits from a tree fall towards the earth after they are ripened. Why do they
fall towards the earth? Why do they not move upwards? Likewise, the planets revolve
around the sun in their own oval orbits. The satellites of the planets also revolve around
them in their own oval orbits like the planets round the sun. Why do they not go away?
We usually take the events mentioned above as simple ones and most of them go
unnoticed. But there are some scientific reasons to cause these events. We are going to
discuss gravitation, Newton's universal law of gravitation, gravity, acceleration due to
gravity, etc. to get the answer of the above questions.
Gravitation
Before the sixteen century, the earth was considered to be the centre of the solar system.
Other heavenly bodies like the sun, the planets and their satellites were also regarded to
revolve around the earth with a uniform circular motion. This earth-centred model of the
solar system is called geocentric or Ptolemaic theory. In the 16th century, Nicholas Copernicus,
an astronomer put forward the heliocentric theory about the solar system. In this heliocentric
model, the sun is assumed to be the centre of the solar system and all the planets are believed
to revolve around the sun in their own fixed orbits. In the seventeenth century, Galileo,
observing through his telescope, found that there were four satellites. Also he observed that
these satellites were revolving around the planet, the Jupiter. Thereafter, later Newton (1642-
1727) tried to think and find out the reasons why planets revolve around the sun and began
to investigate. In 1687 AD, Newton propounded that the earth attracts every object towards
FORCE CLASS - 10 MODERN GRADED SCIENCE 1
its centre. So, when a body is released from a
height, it falls towards the earth because of the sun
force of attraction of the earth. He also found out
that not only the earth but also every object in moon
this universe attracts other bodies towards them. gravitation
He called this force of attraction between any two gravitation
bodies in the universe the force of gravitation
or simply gravitation. In this way, Newton
propounded the law of gravitation. On this basis, earth’s orbit
the force of gravitation could be measured.
moon’s orbit
According to Newton, the mutual force of Fig: 1.1 celestial bodies
attraction between any two bodies in the universe is called gravitation. It is also called
gravitational force. For example, the earth attracts the moon and the moon also attracts the
earth. Similarly, the earth attracts the Jupiter and the Jupiter attracts the earth. The effect
of force of attraction of a body with big mass on a body of small mass can be observed. But
the effect of a body with small mass on a body with big one cannot be observed. The sun
is so large in size that even the sum of the masses of all the planets and satellites is only
about 0.0015th part of the sun.
The earth's revolution around the sun is possible due to the force of gravitation
between the sun and the earth. Similarly, the force of gravitation between planets and
their satellites causes the satellites to revolve around their respective planets in their fixed
orbits. Thus, the force of gravitation is responsible for the existence of the solar system.
The effect of the gravitational force can be observed more on liquid than on solid. Thus,
the tides in seas and oceans are due to the gravitation of the sun and the moon on the
earth when both of them come on the same straight line.
Newton’s universal law of gravitation
The universal law of gravitation between two masses was formulated by Newton
in 1687 AD. Now, it is known as Newton's universal law of gravitation. Newton's law of
gravitation is called universal law because this law holds truth for all the bodies, whether
microscopic or massive placed everywhere in the universe.
Newton's universal law of gravitation states that "every object attracts every other
object in the universe 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."
The direction of force is along the line joining the centre of two masses.
Verification of Newton’s law of gravitatiomn1 m2
B
Suppose two bodies A and B having their masses
m1 and m2 respectively are separated by a distance d A F
from their centres. If the force of attraction between Fig:d1.2
them is F then according to Newton's universal law
of gravitation, we have:
F ∝ m1 × m2 ....... (i) [keeping the distance constant]
and F ∝ 1 ..................... (ii) [keeping the masses m1 and m2 constant]
d2
2 MODERN GRADED SCIENCE CLASS - 10 FORCE
Combining equations (i) and (ii), we have,
F ∝ m1.m2 where G is a constant of proportionality which is
d2 known as universal gravitational constant and its
value is 6.67 × 10–11 Nm2kg–2.
or, F=G m1.m2 ............. (iii)
d2
Hence, the equation (iii) gives the measure of the gravitational force between two bodies.
Universal gravitational constant ‘G’
It is found that the value of G for any pair of masses in the universe is the same. On
the other hand, the value of G does not depend on the nature of the mutually attracting
bodies (terrestrial or celestial) and the properties of the medium (rarer or denser) in which
the bodies are placed. Its value is 6.67 × 10–11 Nm2kg–2. It was first calculated by Henry
Cavendish. The value of G is constant everywhere in the universe. That is why, it is called
universal gravitational constant. It is expressed as,
F × d2 m1 = 1kg m2 = 1kg
G = m1 × m2
F=G
Definition of G d = 1m
We have,
F=G m1.m2 Fig: 1.3
d2
If m1 = m2 = 1kg and d = 1m. Put the value of m1, m2 and d in the given relation,
F=G 1×1
(1)2
∴ F = G
Thus, the universal gravitational constant 'G' can be defined as the force of gravitation
produced between two bodies of unit mass each, separated by unit distance between their
centres.
Unit of G,
F × d2 A scientist named Henry Cavendish (1731-
We have, G = m1.m2 1810) determined the value of G experimentally
by using an extra sensitive apparatus called
In the SI system, the unit of force (F) is torsion balance. The value of G was found to be
newton (N), distance (d) is metre (m) and extremely small. The approximate value of G is
mass (m) in kilogram (kg). Then, the unit of 6.67×10–11Nm2kg–2.
G is:
G = newton × (metre)2 = Nm2
kg×kg kg2
Therefore, the SI unit of G is Nm2 kg–2.
The value of G is very small. Thus, it is known as the weakest force in the nature. Due
to the small value of G, the force of gravitation between two small masses is extremely
small and it becomes unnoticeable. But it becomes noticeable when at least one of the two
bodies is huge like the heavenly bodies.
FORCE CLASS - 10 MODERN GRADED SCIENCE 3
Relation of gravitation with its factors
The force of gravitation depends upon the product of the masses of two bodies
and the square of the distance between their centres. The relation between the force of
gravitation, masses of two bodies and distance between them in various conditions can
be illustrated as follows.
1. When the masses of bodies change keeping the distance between them same or constant.
a) If the mass of one of the bodies is doubled by keeping the distance constant.
A B
Let us consider, two bodies, A and B have the
masses 'm1' and 'm2'. The distance between O1 F O2
them from their centres is 'd'. Then the force
of gravitation 'F' between them is given by,
Gm1.m2 m1 m2
d2 d
F= ……………… eqn (1)
Fig: 1.4
According to the statement, the mass of one of the bodies is doubled i. e. m2 is
doubled (2m2) keeping the distance constant, then,
A B
F1 = Gm1.2m2 O1 F1 O2
d2
or, F1 = 2 Gm1.m2 ……………… eqn (2)
d2
m1 m2
From the equation (1) and (2) d
∴ F1 = 2F Fig: 1.5
Hence, the force of gravitation between two bodies is doubled if the mass of one of
the bodies is doubled.
b) If the mass of both the bodies is increased by two times each by keeping the
distance between their centres constant:
The forces of gravitation between two bodies is given by, B
A B
Gm1×m2
Here, F = d2 ……………… eqn (1)
According the statement, the mass of the F
body A is 2m1 and the mass of the body B d
is 2m2, then, Fig: 1.6
F1 = G2m1×2m2 A
d2
F1 = 4 Gm1×m2 ……………… eqn F1
(2) d2
From the equation (1) and (2) 2m1 2m2
∴ F1 = 4F d
Fig: 1.7
Hence, the force of gravitation between the bodies becomes four times of the initial
force if the mass of both the bodies is increased by two times each.
4 MODERN GRADED SCIENCE CLASS - 10 FORCE
From these examples, we can say that the force of gravitation between two bodies
changes according to the product of their masses keeping the distance constant. Thus,
the gravitational force between two bodies is directly proportional to the product of their
masses. That is,
F ∝ m1 × m2
2. When the mass of the bodies is constant but the distance between their centres
changes:
a) If the distance between two bodies, A and B from their centres is halved,
keeping their masses constant,
Suppose, two bodies, A and B have the masses A B
m1 and m2 lying apart from each other from F
their centres by a distance 'd'. The force of
attraction between them is F.
The force of gravitation (F1) between them is m1 m2
given by d
Gm1.m2 Fig: 1.8
d2
F= ……………… eqn (1)
According to the statement, the distance between the two bodies from their
d
centres is made half, i. e. is 2 , keeping the masses constant
G.m1.m2
Now, F1 = d2
2 AB
or, F1 = 4 [G.m1.m2] ………… eqn (2) F1
d2
From equation (1) and (2)
or, F1 = 4 Gm1×m2 m1 1 d m2
d2 2
∴ F1 = 4F Fig: 1.9
Thus, while decreasing the distance between two bodies by two times, the
gravitation becomes four times of the initial force.
b) If the distance between two bodies from their centre is increased by 2 times
(2d) by keeping the masses of the bodies constant:
Suppose the masses of two bodies A and B be m1 and m2 respectively and the
distance between their centres be d. Then the force of the gravitation is given by
F = G.m1.m2 ……………… eqn (1) A B
d2 F
According to the statement, distance
between the centre of the bodies is made
double i. e. 2d by keeping the masses
constant. Now, 2d
F1 = G.(m2d1.)m2 2 Fig: 1.10
FORCE CLASS - 10 MODERN GRADED SCIENCE 5
or, F1 = G4.dm2 1 A F2 = f41 B
m1 2d m2
1 G.m1.m2 ………….(ii) Fig: 1.11
or, F1 = 4 d2
From equation (1) and (2)
1
∴ F1 = 4 F
Thus, if the distance between any two bodies is increased double keeping their masses
constant, the force of attraction between them becomes one fourth of the initial force.
From these illustrations, we can say that the force of gravitation between two
bodies changes as the distance between them from their centres varies. Thus the force
of gravitation between two bodies is inversely proportional to the square of the distance
between their centres keeping the masses constant. That is,
1
F ∝ d2
Example 1: Two bodies with the mass of 200 kg each are kept 2 m apart from their centre
on the surface of the earth. Calculate the force of attraction produced between them.
Solution:
Here, Masses of m1 = m2 = 200kg
Distance (d) = 2m
Gravitational constant (G) = 6.67 × 10–11 Nm2/kg2
Force of attraction (F) = ?
We know that
F = G.m1.m2
d2
= 6.67 × 10–11 × 200×200
22
= 6.67 × 10–11 × 4×104
4
∴ F = 6.67 × 10–7 N
Thus, the force of attraction between these two bodies is 6.67 × 10–7 N.
Example 2: What will be the force of attraction if the distance between these bodies is
halved in the example 1 above?
Solution:
Here, Mass of the bodies m1 = m2 = 200 kg
Distance (d) = 2m = 1m ( distance is halved)
2
Gravitational constant (G) = 6.67 × 10–11 Nm2/kg2
6 MODERN GRADED SCIENCE CLASS - 10 FORCE
Force of attraction (F) = ?
We know that,
F = G.m1.m2
d2
= 6.67 × 10–11 200×200
1
6.67 × 10–11 × 4×104
= 1 = 26.68 × 10–7
∴ F = 2.67 × 10–6 N
Thus, the force of attraction between two bodies is 2.67×10–6 N
Example 3: What will be the force if the mass of one of the bodies is tripled keeping the
distance between them unaltered in the first example?
Solution:
Here, mass of a body (m1) = 200kg
Mass of another body (m2) = 200kg × 3 = 600 kg [ mass of one of the bodies is tripled]
Distance (d) = 2m
Gravitational constant (G) = 6.67 × 10–11 Nm2/kg2
Force of attraction (F) = ?
We know that,
F = G m1.m2
d2
= 6.67 × 10–11 × 200×600
22
= 6.67 × 10–11 × 1.2×105
4
∴ F = 2 × 10–6 N
Thus, the force of attraction between two bodies is 2 × 10–6 N
Example 4: The mass of Earth is 6 × 1024 kg. The mass of the Jupiter is 319 times more than
that of Earth and the distance between Earth and Jupiter from their centres is 9.68 × 1011 m.
Calculate the force of attraction between Earth and Jupiter.
Solution:
Mass of the earth (Me) = 6 × 1024 kg
Mass of the Jupiter (Mj) = 319 Me
= 319 × 6 × 1024 = 1.91 × 1027 kg
Distance between Earth and Jupiter (d) = 9.68 × 1011 m
Gravitational constant (G) = 6.67 × 10–11 Nm2/kg2
Force of attraction (F) = ?
According to the formula,
FORCE CLASS - 10 MODERN GRADED SCIENCE 7
F = G.Me.Mj
d2
= 6.67 × 10–11 × 6 × 10–11 × 1.91 × 1027
(9.68 × 1011)2
= 6.67 × 10–11 × 11.46×1051
93.70 × 1022
= 76.44 × 1040
93.70 × 1023
∴ F = 8.16 × 1017 N
Hence, the force of gravitation between Earth and Jupiter is 8.16 × 1017 N
Example 5: An apple with the mass of 200 g falls from a tree. What is the acceleration
of the apple towards the earth? What is the acceleration of the earth towards the apple?
(Given: mass of the earth = 6 × 1024 kg, radius of the earth = 6.4 × 106 m, G = 6.67 × 10–11 Nm2
kg–2 and neglecting height of the tree).
Solution:
Here, mass of an apple (m) = 200g = 200 kg = 0.2 kg
1000
Mass of the earth (M) = 6 × 1024 kg
Radius of the earth (R) = 6.4 × 106m
Universal gravitational constant (G) = 6.67 × 10–11 Nm2 kg–2
The force of attraction between the earth and the apple is given by
F = G.M.m
R2
6×1024×0.2
= 6.67 × 10–11 (6.4×106)2 = 1.954N.
If a1 is the acceleration of the apple towards the earth, then
F = m . a1 [ m = mass of the apple]
or, a1 = 1.954 = 9.77 m/s2
0.2
If a2 is the acceleration of the earth towards the apple then
F = M . a2 [ M = mass of the earth]
or, a2 = 1.954
6×1024
= 3.25 × 10–25 m/s2
Thus, the apple accelerates towards the earth by 9.77 m/s2 while the earth accelerates
towards the apple by 3.25 × 10–25 m/s2. Such small acceleration of the earth towards the
apple is not detectable. That is why, we do not see the earth rising towards the apple
although the apple exerts an upward force on the earth.
8 MODERN GRADED SCIENCE CLASS - 10 FORCE
Special features of gravitational force
1. The gravitational force between light bodies is extremely small and hence such
smaller force is not felt in practice. However, it becomes noticeable in case of massive
bodies. This is clear from the following example:
Suppose, you and your friend have the same mass of 40 kg each and both of you are
standing at a distance of 1 m from each other. The mutual force of attraction between
you is given by,
F = G m1.m2
d2
= 6.67 × 10–11 40×40 = 1.067 × 10–7N
(1)2
This is the force of negligible magnitude. Thus, the gravitational force between you
and your friend does not stick you and your friend standing nearby.
The distance between the sun and the earth is about 1.5 × 107 km. The gravitational
force between them is 1027N, which cannot be ignored. Thus, the earth revolves around
the sun due to the gravitational force of such a large magnitude.
Gravity
We know that the earth is spherical like an orange. It may be a great surprise to think how
people can stay on the spherical earth. Nepal and America are located just nearly opposite on
the surface of the earth. A person living in Nepal will have their head directed towards the sky
while standing on their feet on the surface. Likewise, a person living in America, just opposite
to Nepal, will have their head faced towards the sky too while standing on their feet on the
surface. The reason behind this effect is that the earth pulls both of them towards the centre of
the earth with a force. This is called the force of the gravity of the earth.
If, out of the two involved bodies, in Newton's law centre
of gravitation, one body is the earth (or planets/satellite)
then the gravitational force is called force of gravity or
simply gravity. Thus, gravity is merely a special case of
gravitation in which there is involvement of a heavenly
body. For example, the gravitational force between the
earth and a body near or on its surface is called the earth's
gravity. Similarly, the attraction of Jupiter and a body
near or on its surface is called Jupiter's gravity.
Thus, the force of attraction with which the earth earth
pulls a body towards its centre is called gravity of the Fig: 1.12
earth or simply gravity. The force of gravity on a body
is its weight. The factors that affect gravity are mass and
radius of the planets and satellites.
On the basis of Newton's universal law of gravitation, the gravity of a planet or a
satellite can be calculated.
Let a body of mass 'm' be placed on the surface of the earth of mass M as shown in the
figure. The distance between them can be taken to be equal to the radius 'R' of the earth. The
weight 'W' of the body on the surface of the earth is given by,
FORCE CLASS - 10 MODERN GRADED SCIENCE 9
Mm m
W = G R2 ............. (i) ( F = W)
R=d
If m is kept constant, M
M O
W ∝ R2 ............. (ii)
1. Equation (ii) shows that the weight (W) of a body on the surface of a Fig: 1.13
heavenly body is directly proportional to the mass (M) of the heavenly
body. Greater the mass of the heavenly body, larger is its gravity i. e. greater will be the
weight on the heavenly body.
2. The weight (W) of a body on the surface of a heavenly body is inversely proportional
to the square of the radius (R) of that body. The weight of the body will be less if the
radius of heavenly body is longer.
It is found that the gravity of the moon is 1/6 of the gravity of the earth. That's why, a
person, who can jump 0.5 m high on the earth's surface, can jump 3m high on the surface
of the moon. Similarly, a person, who can lift a mass of 10 N on the surface of the earth can
lift a mass of 60 N on the surface of the moon.
The Jupiter is about 319 times greater than the earth. Due to its larger radius (about
11 times) than the earth, the gravity of the Jupiter is only about 2.5 times more than the
1
gravity of the earth as F ∝ R2 . It indicates the gravity of Jupiter does not increase highly.
Also, the weight of the body weighs 2.5 times more on the Jupiter than on the earth. This
means, if a person can lift a body of mass 50 kg on the earth's surface, then he/she would
lift 20 kg only on the surface of the Jupiter.
Example 1: The earth has the mass of 6 × 1024 kg and its radius 6380 km. What will be the
gravity of the earth to a body of mass 1 kg on the surface of the earth?
Solution: [ 1 km = 1000m]
Here,
Mass of the earth (M) = 6 × 1024kg
Radius of the earth (R) = 6380 km = 6.38 × 106 m
Mass of the body (m) = 1 kg
Gravitational constant (G) = 6.67 × 10–11Nm2/kg2
Gravity (F) = ?
We have,
G.M.m
F = R2
= 6.67 × 10–11 × 6 × 1024 × 1
(6.38 × 106)2
40.02 × 1013
= 40.7 × 1012 = 9.8 N
Therefore, the gravity of the earth on the body is 9.8 N
10 MODERN GRADED SCIENCE CLASS - 10 FORCE
Example 2: The mass of the moon is 7.2 × 1022 kg and its radius 1.7 × 106 m. What will be
the force of gravity of the moon to a body of mass 1 kg on the surface of the moon?
Solution:
Here,
Mass of the moon (M) = 7.2 × 1022 kg
Radius of the moon (R) = 1.7 × 106 m
Mass of the body (m) = 1 kg
Gravitational constant (G) = 6.67 × 10–11 Nm2/kg2
Gravity of the moon (F) = ?
We have,
G.M.m
F = R2
= 6.67 × 10–11 × 7.2 × 1022×1
(1.7 × 106)2
48.02 × 10–11+22
= 2.89 × 1012 = 9.8 N
= 16.61 × 10–11+12
= 1.66 N
Therefore, the gravity of the moon on the body is 1.66 N.
Example 3: The mass of a heavenly body is one third of the mass of the earth and its radius
is just half of the radius of the earth. If a stone weighs 200 N on the earth's surface, find its
weight on the heavenly body.
Solution:
Consider the mass of the earth is 'M' and its radius is 'R'.
Now,
Mass of the a heavenly body = 1/3 M
Radius of the heavenly body = 1/2 R
Weight of the body on the earth's surface (W1) = 200 N
Weight of the body on the heavenly body (W2) = ?
We have, for the earth,
G.M.m
W1 = R2 …………..(i) [ W = F]
For the heavenly body,
W2 = G. 1 M.m …………..(ii)
3
1 2
2
R
By dividing eq. (i) by eq. (ii)
WW12 GM.m 1 R2
R2 4
= × 1
3
G Mm
FORCE CLASS - 10 MODERN GRADED SCIENCE 11
11
Or, 4 W2 = 3 W1
4
Or, W2 = 3 W1
4 800
∴ W2 = 3 × 200 = 3 = 266.67 N
Therefore, weight of the body on that heavenly body is 266.67 N.
Effects of gravity
Gravity plays a vital role in our everyday life as well. Some of the effects of gravity
are as follows:
1. We are able to stand up, run and perform other activities freely due to the gravity of
the earth.
2. Construction of buildings and bridges is possible due to the gravity of the earth.
3. The existence of atmosphere around the earth is due to the force of gravity of the
earth. The atmosphere makes life possible on the earth. As the gravity of the moon is
very small, it does not have its atmosphere.
4. If a body is thrown upwards, its motion is opposed by the gravity of the earth. So, it comes
back to the earth.
5. Tides occur due to the gravity of the moon on ocean water. As the mass of ocean
water is greater, there is more force of gravity between the water and the moon. It
results in the pull of the water due to which the ocean level increases and tides occur.
6. The flowing property of liquid i. e. flowing of water in rivers and streams is also one
of the effects of gravity.
Acceleration due to gravity
When a body is released from a height, it freely falls towards the surface of the earth
due to the gravity of the earth. While falling, its velocity goes on increasing so that it
possesses uniform acceleration. Thus, when a body falls freely, it has acceleration equal to
the acceleration due to gravity of the earth. Hence, acceleration due to gravity is defined as
the acceleration produced in a freely falling body due to the force of the gravity of the
earth. It is denoted by g and its SI unit is ms–2. Its average value is 9.8ms–2.
The acceleration due to gravity near the surface of the earth velocity = 0
is about 9.8 ms–2. It means that the velocity of a body rises by 9.8
ms–1 each second if the air resistance is neglected. In the figure, velocity = 9.8 ms–1
a small ball is kept at a height, so it has zero initial velocity. If
it is allowed to fall freely, after one second its velocity becomes velocity = 19.6 m–1
9.8 ms–1; after further one second, its velocity is 19.6 ms–1 and velocity = 29.4 m–1
so on. In this way, the velocity is increased by 9.8 ms–1 each
second. Thus, acceleration of the freely falling stone is 9.8 ms–2. Ground
Similarly, if a body is thrown upward, its velocity decreases by Fig: 1.14
9.8 ms–2 every second.
12 MODERN GRADED SCIENCE CLASS - 10 FORCE
Falling bodies and acceleration due to gravity
According to the legend, Galileo Galilee
simultaneously released several iron balls of different
masses from the top of the Leaning Tower of Pisa. All
these balls reached the ground simultaneously which
proved that all the bodies of different masses fall with
the same acceleration.
Galileo did this experiment to demonstrate it in Fig: 1.15 leaning tower of Pisa
1590 AD. Before Galileo, it was thought that the speed
with which a body falls to the ground depends on its
mass. Higher the mass, more speedily it falls to the
ground. This assertion was Aristotle's thought which
proved to be wrong.
If a stone and a feather are allowed to fall from a
height, the former falls to the ground much more quickly than the later. This may give a
wrong conclusion supporting Aristotle's idea. The actual fact is that the air resistance on the
large surface area of the light feather is very large compared with the force of gravity on it.
Air resistance as well as gravity determines the rate at which a body falls towards the earth.
In other words, the acceleration of freely falling bodies does not depend on mass; it is found
to be equal for all bodies.
It means two bodies, one heavier and another lighter, fall freely at the same rate if
there is no air resistance. That is, they have the same acceleration due to gravity equal
to 'g' at every point during the motion, irrespective of their masses. If air resistance is
excluded, the feather falls with the same acceleration as the stone.
Activity 1.1
Take two paper sheets of equal size. Squeeze one of the papers into a small paper ball
and another remains as unfolded. Drop both of them from the same height and at the
same time. Which one, paper ball or unfolded paper sheet will reach the ground faster?
Why? What conclusion can be drawn from this activity? The paper ball reaches the
ground faster than the unfolded sheet of paper. This is because the air resistance on
the small paper ball is less than that on the unfolded sheet of paper. Due to greater air
resistance on the unfolded sheet of paper, it falls more slowly than the small paper ball
even though both of them have the same mass. This activity confirms that the speed of
a falling body does not depend on its mass.
Activity 1.2
Take two pieces of stones of 0.2 kg and 3 kg masses, both having about equal density.
Keep a plain sheet of tin at the base of the height where the stones strike. Drop both the
stones from a height simultaneously. Listen to the sound produced by both the stones
while they strike on the tin. Will both the stones having different masses hit the tin sheet
at the same time? Why? The stones with both small and large masses hit the ground at
the same time. It is because the magnitude of acceleration due to gravity is equal to all
the freely falling bodies.
FORCE CLASS - 10 MODERN GRADED SCIENCE 13
Feather and coin experiment
It is the experimental fact that the fall of an to vacuum pump
object does not depend on its mass. feather
Sir Isaac Newton put a coin and a feather coin
into a closed long and wide glass tube. The coin
and the feather, initially, were at the bottom of air Fig: 1.16 vacuum
the tube. When the tube was inverted, it was a b
observed that the coin fell much more quickly
than the feather, as in the figure (a). The tube was again inverted so that the coin and the
feather were at the bottom of the tube. The tube was evacuated using a vacuum pump as
far as possible. On inverting the evacuated tube, this time, it was observed that both the
coin and feather fell together and reached the bottom of the tube simultaneously, as shown
in the figure (b). This showed that they fall at the same rate in the absence of air resistance.
This experiment confirms that acceleration due to
gravity of freely falling bodies, whatever their masses, Parachute has no use on the moon.
is the same in the absence of air resistance. Larger the This is because the parachute falls
surface area, more will be air resistance. This explains freely with acceleration due to
why a small paper ball falls more quickly than a sheet gravity on the moon due to the
of paper. absence of atmosphere.
The astronauts travelling on Apollo-12 performed an experiment on the moon. They
took a mallet and a feather and dropped them from the same height at the same time.
They found that both the mallet and the feather fall at the same rate and reached the
surface of the moon at the same time. What is the reason behind it? There is absence of
air on the moon and hence no air resistance to slow down the speed of the feather while
falling. Thus, a feather accelerates at the same rate as the mallet on the surface of the
moon.
Activity 1.3
Take a measuring tape, a piece of stone and a sheet of tin. Place the sheet of tin on the ground
below from the height of the school's building which will produce sound while the falling
stone hits it. Measure the height of the building from its rooftop to the ground using a
measuring tape. Take a stone in one hand and a stopwatch in another hand. Drop the stone
and start the stopwatch to record the falling time of the stone at the same time. Record
the time period until you listen to the sound produced while the stone hits the tin. Repeat
this experiment 4 times and find out the average acceleration due to gravity using the
observation table. Compare the average value of acceleration due to gravity with 9.8 m/s2.
14 MODERN GRADED SCIENCE CLASS - 10 FORCE
Observation table Time (t) Acceleration due to
S. Height of the school's building 2h
No. from the ground (h) gravity (g = t2 )
1
2
3
4
Therefore, the average acceleration due to gravity = …………….m/s2
The acceleration due to gravity in all the falling bodies will be the same at a given
place if there is no air resistance.
Relation of acceleration due to gravity with the earth's radius
Let 'M' be the mass of the earth and 'R' its radius; then the force of attraction on a body
of mass 'm' on its surface is as shown in the figure.
According to Newton's law of gravitation, the force of attraction between them is
given by,
G.M.m
F = R2 ........ (i)
The force with which the body is attracted by the earth towards its centre is given by,
F = mg ........... (ii) m
From equation (i) and (ii) R
G.M.m MO
mg = R2
G.M Fig: 1.17
or, g = R2 ........... (iii)
This formula does not contain the mass of the body (m). Hence the acceleration due
to gravity (g) is independent of the mass of the freely falling body (m).
Since the values of G and M are constant but R is variable on the earth's surface due
to its orange shape then,
1
g ∝ R2 ……..(iv)
This equation (iv) is also the expression for acceleration due to gravity on a heavenly
body. The value of acceleration due to gravity (g) on the surface of the planet or satellite
depends on its radius.
The acceleration due to gravity of the earth is also inversely proportional to the square
of its radius.
FORCE CLASS - 10 MODERN GRADED SCIENCE 15
Variations in the value of g
(1) Variation of g due to the shape of the earth
The value of acceleration due to gravity on the surface Rp polar radius
of the earth is given by: Re equatorial
radius
G.M
g = R2
Since G and M are constants, we have, Fig: 1.18
1
g ∝ R2 ......... (i)
The shape of the earth is not completely spherical. Its polar regions are flattened
little and it bulges out at the equator. The radius of the earth at the poles is smaller in
comparison to the radius of the earth at the equator (Rp < Re). Therefore, the acceleration
due to gravity is more at the poles than at the equator of the earth. The equation (i) shows
that the value of 'g' at a place on the surface of the earth is inversely proportional to
the square of the radius of the earth at that place. From this fact, we can say that the
acceleration due to gravity produced in any objects differs from place to place.
At equatorial region of the earth acceleration due to gravity (ge) = 9.78 m/s2
At polar region of the earth acceleration due to gravity (gp) = 9.83 m/s2
On the average of the earth acceleration due to gravity (g) = 9.8 m/s2
Value of acceleration due to gravity (g) of some heavenly bodies
Acceleration due to gravity (g) Earth Moon Jupiter
Poles Equator Average 1.67m/s2 25 m/s2
9.83m/s2 9.78m/s2 9.8m/s2
(2) Variation of g with height from the surface of the earth h
R
Let M and R be the mass and radius of the earth respectively. The
value of acceleration due to gravity at the surface of the earth is given by: Fig: 1.19
GM
g = R2 ............ (i)
Where, G is the universal gravitational constant. If g1 is the value of
acceleration due to gravity at a height h from the surface of the earth, we
have
GM
g1 = (R+h)2 ............ (ii)
Dividing equation (ii) by equation (i)
g1 GM × R2 R2
g = (R+h)2 GM = (R+h)2
R2
∴ g1 = (R+h)2 × g ............ (iii)
16 MODERN GRADED SCIENCE CLASS - 10 FORCE
R2 2
Since R+h < 1, hence g1 < g
Therefore, the value of acceleration due to gravity decreases with increase in height
from the surface of the earth.
(3) Variation of g with depth from the earth's surface x
Suppose that the mass of the earth is M and its radius is Rm
'R'. A body is at depth x from the earth's surface. Now, forth R–x=d
acceleration due to gravity on the earth's surface (g).
C
We have,
GM Gρ 4 π(R)3 4 M
g = R2 = 3 = Gρ 3 π R ………………..(i)
(R)3
Now, for the acceleration due to gravity at depth x Fig: 1.20
g1 = GM = Gρ 4 π(R–x)3 4
(R–x)2 3 = Gρ 3 π (R – x) ………………….(ii)
(R–x)2
44
Here, ρ (Rho) is density and 3 π R3 is volume of the earth and 3 π (R – x)3 is volume
of the earth's part below the depth, as we know that:
m4
d = v and V = 3 πr3
Now from eq (i) and (ii)
g Gρ 4 πR R
g1 = 3 R–x
4 =
Gρ 3 π(R–x)
R–x
∴ g1 = R × g
R–x
We know that R is always lesser than 1, therefore g1 is lesser than g.
It proves that if depth increases, acceleration due to gravity decreases.
Again, at the earth's centre x = R, thus R – R = 0 i. e. at the earth's centre g is zero.
Acceleration due to gravity on the surface of the moon
The moon is smaller than the earth. Its acceleration due to gravity equals to 1/6th times
of the acceleration due to gravity of the earth. The value of acceleration due to gravity on
the moon's surface is 1.67 m/s2.
Acceleration due to gravity and weight of a body on planets/satellites
Generally, the weight of a body is defined as the force with which the earth pulls a
body towards its centre. According to Newton's second law of motion,
F = m.a
or, F = m.g ( a = g)
FORCE CLASS - 10 MODERN GRADED SCIENCE 17
To solve a numerical related to weight, the acceleration due to gravity of the planets/
satellites is used. Thus, we can use the following formula for the calculation of weight–
W = m.g
The weight of a given mass of a body on a heavenly body is as;
W = m.g ......................... eq(i)
The value of g for a heavenly body is given as;
GM
g = R2 ......................... eq(ii)
GMm
From equation (i) and (ii), W = R2 ......................... eq(iii)
Here, M and R are the mass and radius of the planet respectively. Since the values of
M and R are different for different heavenly bodies, the value of 'g' is variable on them.
It makes the weight 'W' of a body different on different heavenly bodies even though the
body has the same mass. It is found out that the acceleration due to gravity 'gm' on the
surface of the moon is 1/6th the value of 'g' on the earth.
If the earth is compressed to a volume equal to that of the moon, its acceleration due
to gravity will increase by 14 times more than that of its present value. This is because the
radius of the earth is decreased but the mass of the earth remains the same and we have
1
known that g ∝ M and g ∝ R2 .
Therefore, the weight of an object on the moon is one sixth (1/6th) of its weight on the
earth. Similarly, on the surface of the Jupiter, the weight of a body is 2.5 times more than
that on the earth's surface. We know that value of 'g' on the Jupiter is about 25 ms–2.
Example 1: How much would a 70 kg man weigh on the moon? What would be his mass
on the earth and on the moon?
Solution:
Here,
Weight on the moon (Wm) = 1/6 times the weight on the earth
Weight on the earth (We) = mg
= 70 × 9.8 = 686 N
Therefore,
(Wm) = 1/6th of the weight on the earth
1 FORCE
or, Wm = 6 × We
1
= 6 × 686 = 114.33 N
Thus, the weight of the person on the moon will be 114.33N
Mass of an object remains the same on any planet,
Then, mass on the earth (Me) = 70 kg and Mass on the moon (Mm) = 70 kg
18 MODERN GRADED SCIENCE CLASS - 10
Example 2: The mass of Jupiter is 1.92 × 1027 kg and it radios is 7.1 × 107 m. What is the
acceleration due to gravity of Jupiter. Also, calculate the weight of a stone of mass 15 kg
on the surface of Jupiter.
Solution:
Here,
Mass of Jupiter (M) = 1.92 × 1027 kg
Radius of Jupiter (M) = 7.1 × 107 m
Gravitational constant (G) = 6.67 × 10–11 Nm2/kg2
Mass of a stone (m) = 15kg
Acceleration due to gravity (g) = ?
Weight of the stone (W) = ?
According to formula,
G.M
g = R2
6.67 × 10–11 × 1.92 × 1027
= (7.1 × 107)2
128.06 × 1015
= 50.41 × 1014
∴ g = 25.4 m/s2
Again,
We have,
W = mg
= 15 × 25.4 = 381 N
∴ W = 381 N
Therefore, acceleration due to gravity on Jupiter is 25.4 m/s2 and the weight of a stone
on Jupiter is 381 N.
Example 3: What is the acceleration due to gravity on the surface of the earth when its
mass is 6×1024 kg and radius is 6400 km. Given, G = 6.67×10–11 Nm2/kg2
Solution:
Here,
Mass of the earth (M) = 6 ×1024 kg
Radius of the earth (R) = 6400 km = (6400×1000)m = 6400000 = 6.4 × 106 m
Gravitational constant (G) = 6.67 × 10–11 Nm2/kg2
Acceleration due to gravity of the earth (g) = ?
We have,
GM
g = R2
FORCE CLASS - 10 MODERN GRADED SCIENCE 19
or, g = 6.67×10–11×6×1024
(6.4×106)2
40.02×10–11+24
or, g = 40.96×1012
∴ g = 0.977×10 = 9.77 m/s2
Therefore, the mean value of acceleration due to gravity is 9.77 m/s2.
Example 4: A person can lift a mass of 100 kg on the surface of the earth. What mass can
he/she lift on the surface of the moon?
Solution:
Here, on the earth,
Mass of the body on the earth (me) = 100 kg
Acceleration due to gravity of the earth (g) = 9.8 m/s2
Weight (W) = ?
We have,
W = me × ge The radius of the Jupiter is much
= 100 × 9.8 = 980 N greater (about 11 times) than the
The person can lift 980 N on the earth. radius of the earth. Thus, the gravity
On the moon, we have of the Jupiter is not so large; even the
Again, mass of the Jupiter is 319 times more
than that of the earth.
Weight (W) = 980 N
Acceleration due to gravity of the moon (gm) = 1.67 m/s2
Mass can be lifted on the moon (mm) = ?
We know that,
W = mm × gm
or, mm = W
gm
∴ mm = 980×1.67 = 586.8 kg
Thus, the person can lift 586.8 kg mass on the moon.
Difference between acceleration due to gravity (g) and universal gravitational constant (G)
Acceleration due to gravity (g) Universal gravitational constant (G)
1. It is the acceleration produced on a 1. It is the gravitational force between
body due to gravity. two unit masses separated by unit
distance between their centres.
2. Its unit is m/s2. 2. Its unit is Nm2kg–2.
3. It is a vector. 3. It is a scalar.
4. It is not universal. 4. It is universal.
5. Its value changes from place to place. 5. Its value is 6.67 × 10–11 Nm2kg–2.
However, on the earth's surface, its
average value is 9.8 m/s2.
20 MODERN GRADED SCIENCE CLASS - 10 FORCE
Mass
The amount of matter contained
in a body is called mass of that body. The space around a heavenly body over which it can
It remains the same everywhere in exert gravitational force on other masses is called the
the universe. The mass of a body is gravitational field of that heavenly body. Theoretically,
measured by using a physical balance the gravitational field extends up to infinity. In
or beam balance. When a physical practice, the gravitational field may become too weak
balance is in the state of equilibrium to be measured beyond a particular distance.
with its one pan with 2 kg standard
mass and other pan with 2 kg sugar The gravitational field intensity (I) at a point in
as shown in the figure, the force of the gravitational field is defined as the force (F)
gravity on both of the pans is the experienced by unit mass (m) placed at that point.
same.
F
Hence, the mass of the body is That is, I = M
determined by comparing the force
of gravity on the standard mass and the body to be weighed
using a physical balance. Its unit is kg in the SI system. The
mass of a body is dependent upon the number of atoms and
average mass of atoms present in the body.
Weight
The weight of a body is defined as the force with which Fig: 1.21 physical balance
it is attracted towards the centre of a planet. A spring balance
measures it. Its unit is Newton (N) in the SI system and dyne in the spring
CGS system. Its value changes from place to place as it depends balance
upon the value of acceleration due to gravity of planets (g) and
mass of the body. It is a vector quantity. string
stone
The weight of a given mass 'm' on a planet is written as, Fig: 1.22 spring balance
W = m.g (where, g is acceleration due to gravity)
As g depends on radius, depth and height from the earth's
surface, weight is also affected by the same factors. Where there is
more value of g, there is more weight and where there is less value
of g, there is less weight. For example, the weight of an object in
the Terai is more than that at the top of Mount Everest. You have
already studied about the variation of 'g' on the earth's surface.
Difference between mass and weight
Mass Weight
1. It is the amount of matter contained in 1. It is the force with which a body
a body. is attracted towards the centre of a
planet/satellite.
2. Its value is the same everywhere in the 2. Its value differs from place to place.
universe.
3. It is a scalar quantity. 3. It is a vector quantity.
FORCE CLASS - 10 MODERN GRADED SCIENCE 21
4. Its SI unit is kilogram (kg). 4. Its SI unit is newton (N).
5. It is measured by using a physical 5. It is measured by using a spring
balance. balance.
Free fall
When a body is falling freely towards the centre of the earth under the influence of
the gravitational pull of the earth, the motion of the body is called free fall. In other words,
if a body is falling with acceleration equal to the acceleration due to gravity (i. e. a = g),
thus the body is said to have a free fall. In fact, a body falling in vacuum or in space is free
fall.
One of the examples of free fall could be a body falling on the surface of the moon.
This is because the moon has no atmosphere around it. The acceleration of the body in the
free fall is equal to the acceleration due to gravity 'g'. On the earth, there is no prefect free
fall of any falling body due to air resistance. Free fall is a type of motion on a body which
is moving under the influence of gravity. In this condition, the body is in weightlessness.
This is because the body feels zero reaction force.
Weightlessness
Normally, when we stand or walk, we feel reaction of the floor and that makes us
conscious of our weight. Thus, it causes the sensation of weight. If this reaction becomes zero
due to some reason, one will feel weightless. This will be clear from the following activity.
Activity 1.3
Take a stone and tie it with a string. Suspend it in the hook of the 200 g 0 reading
spring balance as shown in the figure (a). Now, note the weight
of the stone in the spring balance. Then, release the spring
balance with the stone from your hand, and allow them to fall
freely. In this situation, the spring balance shows zero reading as
in the figure (b). This state of the stone is called weightlessness.
Fig: 1.23
Therefore, weightlessness is the condition of a body when its weight seems to be zero.
In other words, weightlessness is the state of the body in which it feels that the body is not
influenced by any force. A body with a fixed mass will be weightless under the following
conditions:
1. When a body falls freely under the influence of the force of gravity or when the
acceleration of a falling body is equal to the acceleration due to gravity.
2. When a body is at the earth's centre or in the space at null point (g = 0 → w = mg = 0)
3. The body in the artificial satellite (rocket) becomes weightless when the artificial
satellite is orbiting around a heavenly body.
Example 1
Suppose, a person standing in a lift moving down along with it near the surface of the
earth. The person feels 'weightless', which is explained as follows. If the lift is falling
freely with acceleration 'g' as seen in the figure, the floor of the lift cannot exert any
22 MODERN GRADED SCIENCE CLASS - 10 FORCE
upward force to prevent it from falling. Newton's third law of motion states that the
person is not pressing any force on the floor and hence no reaction force takes place.
In this situation– g
The apparent weight will be given by–
R = m (g - a)
Where, R = Apparent weight
m = mass of the falling body
g = Acceleration due to gravity
a = Acceleration of the falling body
But, for a freely falling object, a = g
Thus, R = m (g - g) Fig: 1.24 lift in free fall condition
or, R = m × o = o
Example 2
An artificial satellite revolves around the centripetal B
earth in the space in an elliptical path. For force centrifugal
this, two forces are in action on the spacecraft. C
The forces are centripetal and centrifugal. The force
force of gravity of the earth on the spaceship is
centripetal force which keeps the spaceship in earth A
the elliptical orbit as shown in the diagram. The artificial
other force is centrifugal which is against the satellite
gravity. Study the diagram given:
Fig: 1.25 an artificial satellite revolving around
In the given diagram, a satellite is shown at position A. It should go to B due to its
centrifugal force due to inertia. But the force of gravity of the earth brings the satellite at
place 'C' i. e. due to free fall caused by centripetal force or gravity. It causes weightlessness
of the objects inside the satellite.
Falling of a parachute
A parachute is used for jumping out from a flying aeroplane. It
opens and expands when a paratrooper jumps out from the aeroplane.
Due to large surface area of the parachute, the air resistance will be
more. As a result, the acceleration of the parachute decreases. The
decreased acceleration due to gravity causes the paratrooper to fall
down slowly. As a result of slow motion, the paratrooper can balance
their body and land safely without any hurt.
The use of a parachute is not safe on the moon. Can you say why?
It is because of the absence of the atmosphere on the moon, hence no air
pressure is felt. Thus, the parachute gets attracted by the gravitational Fig: 1.26 falling parachute
force of the moon and the parachute falls forcibly onto the surface of the moon.
At what condition is the acceleration of a falling parachute approximately zero while
getting down from a parachute? When air resistance is equal to the force of gravity of
the earth on the falling parachute, the acceleration of the falling parachute is nearly zero
while getting down from the parachute.
FORCE CLASS - 10 MODERN GRADED SCIENCE 23
S me Reasonable Facts
1. If the earth is compressed so as to make its volume equal to that of the moon, the
gravity of the earth increases by about 14 times more than that of its present value.
The gravity at the surface of the earth is given by:
GM
g = R2 .......... (i)
Where, M and R are the mass and the radius of the earth respectively.
The radius of the earth (6400 km) is about 3.74 times that of the moon (1711.23 km). If
the earth is compressed so as to make its volume equal to that of the moon, the radius
of the earth would be reduced to 1/3.74 of the present radius (R) of the earth. In such
a case, its acceleration due to gravity would be,
g1 = GM
R2
3.74
or, g1 = 13.98GM .......... (ii)
R2
Dividing (i) by (ii), we get,
g GM R2
g1 = R2 × 13.98 GM
∴ g1 = 14g (Approx)
Thus, the earth's gravity would be increased by about 14 times more than that of the
present value at its surface.
2. The weightlessness in free fall does not mean the acceleration due to gravity is zero. This
is because whether a body is in the state of free fall or not, it is constantly attracted by
the earth towards its centre which is called weight of a body and the weight is directly
proportional to the acceleration due to gravity. For example, if a body is suspending on
the spring balance during free fall, its indicator reads zero but it is falling freely towards
the centre of the earth because of acceleration due to gravity (g).
3. An astronaut in the spaceship is orbiting the earth. The astronaut and the spaceship are
in the continuous state of free fall towards the earth and both of them are in the condition
of weightless. This is because there will be acceleration equal to g (acceleration due to
gravity) of the spaceship towards the earth. Since the astronaut and the spaceship both
fall with the same acceleration (a), the astronaut feels weightless as there is no action and
reaction of the forces between the astronaut and the spaceship.
4. A person in an artificial satellite of the earth feels weightless. But the same person on
the moon has weight though the moon is also a satellite of the earth. When a spaceship
(or the moon) orbits around the earth, necessary centripetal force is provided by the
gravitational force between the earth and the spaceship (or the moon). The only force
that the astronaut experiences while orbiting the spaceship is due to the gravitational
force exerted by the spaceship. This force is negligibly small force and the astronaut
does not experience any gravity. However, when he is on the moon, he will experience
the force of gravity of the moon. As the mass of the moon is much more relative to the
spaceship, the person has weight there.
24 MODERN GRADED SCIENCE CLASS - 10 FORCE
5. Tide occurs in oceans but not in Lake Fewa. Tide is an effect of gravity on ocean water
that causes the sea level to rise. As gravity is directly proportional to the product of
two masses, it shows its effect on the ocean. Due to less mass of the water of Lake
Fewa, the gravity of the moon is not much active there.
6. The weight of a body reduces when height or depth increases from sea level (earth's
surface). It is because there is maximum value of 'g' at sea level. As the depth or
height increases from sea level, the 'g' decreases, due to which the weight of a body
also decreases.
7. It is not safe to use parachute on the moon. The moon has no atmosphere thus, there is no
air resistance while a parachute falls and the velocity of the parachute is not controlled.
Due to this reason, the parachute falls forcibly onto the surface of the moon.
Things To Know
1. The mutual force of attraction between two bodies is called gravitation.
2. The force with which the earth attracts an object towards its centre is called gravity of
the earth.
3. Gravitation and gravity both are denoted by F and their SI unit is N.
4. The gravitational force produced between two bodies of unit mass each, when they
are kept at a unit distance from their centres, is called gravitational constant. Its value
is 6.67 × 10–11 Nm2/kg2.
5. Newton's universal law of gravitation states that "every body in this universe attracts
another 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 centres."
6. When a body is thrown up or dropped down, its velocity changes. The rate of change
of velocity due to gravity is called acceleration due to gravity (g).
7. Value of g at the poles is 9.83m/s2, at the equator 9.78 m/s2 but average value of g is
9.8m/s2.
8. The mass and the radius of the planets and satellites are the factors upon which the
acceleration due to gravity depends.
9. If a body falls freely under the influence of the gravity excluding air resistance, the
motion is called free fall. During free fall, acceleration of the falling body is equal to
acceleration due to gravity. But the acceleration due to gravity is independent of the
mass of the falling body.
10. The quantity of matter contained in a body is called mass. Its SI unit is kilogram (kg).
11. The force with which a planet attracts any body on its surface towards its centre is
called weight. Its SI unit is Newton (N).
12. Weightlessness of a body is the state in which the body experiences that the body is
not being attracted by any force.
13. Weightlessness occurs during free fall and at the null point.
14. The space between any two heavenly bodies at which the resultant gravity is zero, is
known as the null point.
15. In a falling parachute, when a weight of the parachute becomes equal to the air
resistance, its acceleration is zero.
FORCE CLASS - 10 MODERN GRADED SCIENCE 25
Things To Do
Take a thick plastic sheet and cut it
in the form of a circle with its radius plastic plastic
18 cm. Make small holes at as equal
distance towards the edge of the hole thread
sheet. Tie small threads at each hole
and collect all the other ends of the thread
threads and bind a small stone as
shown in the figure. Now, a parachute stone
is made. Take this parachute to a
height and drop it from there. Why does the parachute fall down gradually? Is it free fall? At
what condition will acceleration of the parachute be zero? The fall of the parachute towards
the earth is not a free fall. This is because air resistance forcibly influences the parachute.
Test Yourself
1. Multiple choice questions (MCQs).
a. What is the acceleration due to gravity of the moon?
A. 1.67 m/s2 B. 9.83 m/s2
C. 9.87 m/s2 D. 2.57 m/s2
b. What is the value of gravitational constant?
A. 6.67×10–11 Nm2/kg B. 6.67×10–10 Nm2/kg2
C. 6.67×10–11 Nm/kg2 D. 6.67×10-–11 Nm2/kg2
c. If the distance between two bodies is doubled, the force of attraction F between
them will be:
1 1 C. 2 F D. 4F
A. 2 F B. 4 F
d. What is the value of gravitational constant on the surface of the moon?
A. 9.78 m/s2 B. 9.8 m/s2
C. 6.67×10–11 Nm2/kg2 D. 1.67 Nm2/kg2
e. The acceleration due to gravity is zero at:
A. the centre of the earth B. poles
C. sea level D. the equator
f. A feather and a coin released at the same time from the same height do not reach
the ground at the same time because of the:
A. force of gravity B. resistance of the air
C. force of gravitation D. difference in mass
g. How much would a person whose mass is 60kg weigh on the moon?
A. 9.8 N B. 500 N
C. 50 N D. 98 N
26 MODERN GRADED SCIENCE CLASS - 10 FORCE
h. What is the relation between the weight of a body and acceleration due to gravity?
A. W = g 1
B. W ∝ g
C. W ∝ g D. W = 2g
2. Answer the following questions.
a. What is gravitation? What are the factors that affect gravitation? Write the SI unit
of gravitation.
b. State Newton's universal law of gravitation.
c. What will happen to the gravitational force between two bodies if the distance
between them is halved keeping their masses constant?
d. What will be the gravitational force between two heavenly bodies if the masses
of both are tripled keeping the distance between them constant?
e. What will be the gravitational force between two bodies if the mass of each is
doubled and the distance between them is halved?
f. Define G. Also write its value and SI unit.
GM 1
g. In this formula, g = R2 , what do G, M, and R stand for? Prove that: g ∝ R2 .
h. What will be the effect on the acceleration due to gravity of the earth if it is
compressed to a the size the moon?
i. The acceleration due to gravity of Jupiter is 25 m/s2. What does it mean?
j. A coin and a feather are dropped in vacuum. Which one will reach the ground
first? Why?
k. Where does a body have more weight: at the pole or at the equator of the earth?
Explain with reason.
l. When an apple falls towards the earth, the earth moves up to meet the apple. Is
this true? If yes, why is the earth's motion not noticeable?
m. Show with the help of an experiment that the acceleration of a freely falling body
does not depend on the mass of that body.
n. An astronaut is said to be weightless when he/she travels in a satellite. Does it
mean that the earth does not attract him/her?
o. What is the importance of gravitational force? Write any two points.
p. What is free fall? What happens to the weight of a body when it is falling freely
under the action of gravity?
q. What is the effect of gravity on a falling object? Write the conclusion obtained
from the coin and feather experiment.
r. Write the difference between the fall of a parachute on the earth and its fall on the
moon? On what condition does a body have free fall?
s. Write any two examples of weightlessness.
t. Define:
i. acceleration due to gravity ii. weightlessness
iii. free fall iv. null point
FORCE CLASS - 10 MODERN GRADED SCIENCE 27
3. Distinguish between:
a. Gravity and gravitation
b. Gravity and acceleration due to gravity
c. Mass and weight
d. g and G
e. Free fall and weightlessness
f. Gravitational force and acceleration due to gravity
g. Weightlessness in space and weightlessness on the earth
4. Give reasons.
a. Newton's law of gravitation is called a universal law.
b. The weight of a body decreases with increase in distance from the earth's surface.
c. The weight of a stone is less at the top of a mountain than that at its bottom.
d. The weight of a body is less at the moon than its weight on the earth.
e. The moon has no atmosphere.
f. The value of g is greater at the poles than at the equator.
g. There is more role of the moon than the sun to cause tides in oceans.
h. Tides occur in oceans but not in lakes.
i. It is difficult to lift a large stone on the surface of the earth but it is easy to lift a
small one.
j. Mass of the Jupiter is 319 times more than that of the earth but its acceleration
due to gravity is only 2.5 times more than the earth.
k. The weight of a body will be zero during free fall.
l. A paratrooper does not get injured while jumping on the earth from a flying
areoplane.
m. When a coin and a feather are dropped from the roof of a building, the coin
reaches the ground earlier.
5. Diagrammatic questions
1. The earth revolves around the sun in its sun
orbit as shown in the figure. Is there any A B
difference in the amount of gravitational
force of the sun towards the earth when the earth 1 earth 2
earth moves from a point A to point B on its
orbit? Explain.
2. Study the diagram of an experiment to prove a fact and to vacuum pump
answer the following questions. feather
a. Name the experiment given in diagram. coin
b. What conclusion can be drawn from the experiment?
c. What effect is seen if the tube is filled with air? vacuum
28 MODERN GRADED SCIENCE CLASS - 10 FORCE
3. Study the given diagram and find the gravitational force 6m 4m
between two spherical objects. (Ans: 3.335×10–7N)
400kg
6. Numerical Problems. 800kg
(Take g = 10m/s2 and G = 6.67 × 10–11 Nm2 kg–2 if required)
a. A sphere of mass 40 kg is attracted by another sphere of mass 15 kg with a force
of 9.8 × 10–7 N. Find the value of the universal gravitational constant if the centres
of spheres are 20 cm apart.
b. A heavenly body has a mass equal to half of the mass of the earth and its radius
half as that of the earth. If a stone weighs 100 N on the surface of the earth, find
the weight of the stone on the heavenly body.
c. Calculate the force of attraction between two bodies with their mass 100 kg each and
they are 1m apart on the surface of the earth. Will the force of attraction be different
if the same bodies are taken on the moon, their separation remaining constant?
d. The acceleration due to gravity at the surface of the moon is 1.6 m/s2. If the radius
of the moon is 1.7 × 106 m, calculate the mass of the moon.
e. The acceleration due to gravity at a place on the surface of the earth is 9.8 m/s2.
Find the mass of the earth. (R = 6400 km)
f. The masses of the sun and the earth are 2 × 1030 kg and 6×1024 kg respectively.
If the distance between their centres is 1.5 × 1011 m, find the gravitational force
between them.
g. A body weighs 63 N on the surface of the earth. What is the gravitational force on
it due to the earth at a height of 3200 km from the earth's surface? (Radius of the
earth is 6400 km)
h. A man can lift a mass of 200 kg on the surface of the earth. What is the amount of
mass he can lift on the surface of the moon?
i. The mass of the Jupiter is 2 × 1027 kg and its radius 6.5×107 m. What is the
acceleration due to gravity on the Jupiter? Also, calculate the weight of a person
having the mass of 70 kg on the Jupiter.
j. The mass of the earth is 6×1024 kg and radius of the moon is 1.7×106 m. Calculate the
acceleration due to gravity of the new earth which is formed by the compression
of the earth equal to the size of the moon.
k. The mass of the Jupiter is 1.9×1027 kg and that of the sun is 2×1030 kg. If the
magnitude of the gravitational force between these two masses is 4.166×1023N,
find the distance between them.
l. A man who weighs 75 kg is on the surface of the earth whose mass is 6×1024 kg. If
the radius of the earth is 6380 km, calculate the force of attraction between them.
(Given, G = 6.67×10–11 Nm2/kg2)
FORCE CLASS - 10 MODERN GRADED SCIENCE 29
a. 6.37 × 10–11 Nm2 kg–2 b. 200N c. 6.67 × 10–7 N
d. 6.93 × 1022 kg e. 6 × 1024 kg f. 3.56 × 1022 N
g. 28 N h. 1173.6 kg i. 31.57 m/s2, 2209.9 N
j. 138.48m/s2 k. 7.8 × 1011 m l. 737.38N
Microscopic : too small to be seen without a microscope
Microscopy : use of microscopes
Massive : large, heavy
Satellite : natural body in space orbiting round a planet
Equator : the imaginary line around the earth at an equal distance from the
Astronaut north and south poles
Mallet : a person who travels in a spacecraft
: a tool resembling a hammer but with a large head
30 MODERN GRADED SCIENCE CLASS - 10 FORCE
Chapter PRESSURE
2 Total estimated periods: 10 (T 8 + P 2)
demonstrate liquid pressure with an experiment (hydrostatic pressure).
verify Pascal's law.
interpret the application of Archimedes' principle in our daily life with examples.
demonstrate the law of floatation.
describe atmospheric pressure and its use (barometer, syringe, water pump and
air pump).
Liquid pressure [Hydrostatic Pressure]
Liquid is a state of matter that does not have a definite shape but it takes the shape
of the container in which it is kept. When any liquid is placed in a container, it exerts
pressure on the base as well as on the walls of the container.
Suppose that a liquid with density 'd' is kept in a beaker of cross-sectional area 'A'.
Let 'h' be the depth of the liquid from its free surface as shown in the figure.
According to the definition of pressure:
F
P = A
m×g h
or, P = A [ F = mg]
A
d×v×g
or, P = A [ d = mv or m = d×v] Fig: 2.1
d×A×v×g
or, P = A [ V = A×h]
∴ P = dhg
Where P = liquid pressure
d = density of liquid and
h = depth of liquid and
g = acceleration due to gravity
PRESSURE CLASS - 10 MODERN GRADED SCIENCE 31
Thus, the factors influencing liquid pressure are density of the liquid (d), depth of the
liquid (h) and acceleration due to gravity (g).
For a given liquid at a given place, the density of the liquid and acceleration due to
gravity are constant. Hence, the pressure exerted by a liquid is directly proportional to the
depth of the liquid from the free surface of the liquid.
i. e. P ∝ h.
Though pressure is measured in Nm–2 in the SI system, the liquid pressure is also
measured in mm of Hg. The pressure exerted by a mercury column of height 760 mm is
called 760 mm of Hg or 1 atmospheric pressure.
Pascal's law
This law is formulated by the French scientist Blaise Pascal in 1647 AD. This is one
of the basic principles of hydrostatics. Hydrostatics is that branch of physics which deals
with the properties of fluids (liquids and gas) at rest.
Pascal's law states that "when pressure is applied on a liquid enclosed in a vessel, it is
transmitted equally in all the directions." The pressure acts at right angles to the surface
of the vessel exposed to the liquid.
Experimental verification of Pascal's law S P force = F
water
Suppose, a vessel is filled with water and fitted cylinder
with pistons P, Q, R, and S in different positions as
shown in the figure. These four pistons P, Q, R and S A
have the same cross-sectional area 'A'. If the piston 'P' Q
is pushed inward with a force F, the pressure exerted
by the piston P is F/A. spherical
vessel
This pressure is transmitted throughout the
liquid and then it acts at right angles to the surfaces piston
of the pistons. It causes pistons Q, R and S to push R
outwards equally i. e. the outward distance moved Fig: 2.2 verification of Pascal's law
by these pistons is the same.
This experiment proves that liquid pressure is
transmitted equally in all the directions by the liquid
enclosed in a vessel. Pascal's law is also known as the
principle of transmission of fluid pressure.
Applications of Pascal's law
Pascal's law tells that if pressure at any point in a liquid is changed, there will be
equal change in the pressure at other points in the liquid. This fact is used in hydraulic
machines such as hydraulic press (used to make bales of cotton), hydraulic brakes (used in
four wheelers), hydraulic garage lift (used at service stations to lift car, van, etc.) hydraulic
cranes (used to lift a very heavy load), etc.
Principle of hydraulic machine
Hydraulic machine is a machine which works under the principle of Pascal’s law and
converts a small force into a larger force.
32 MODERN GRADED SCIENCE CLASS - 10 PRESSURE
Construction: Suppose, two cylindrical tubes of a cross-sectional area A1 and A2
connected by a horizontal tube T as shown in the diagram. The apparatus is filled with a
liquid (say, water). There are two watertight pistons P1 and P2.
Working: If a force ‘F1’ is applied at piston ‘P1’, the pressure exerted at any point on
given by: .........................(i)
it is F1 load F2 P2
P1 = F1/A1 P1
This pressure is transmitted unchanged to the piston
‘P2’. Therefore, the upward force ‘F2’
exerted on piston ‘P2’ is given by A1 T A2
P2 = F2/A2 .........................(ii)
From eq (i) and (ii) Fig: 2.3 a hydraulic machine
F1 = F2 or, [ P1= P2 due to Pascal’s law)]
A1 A2
∴ A2 = F2
A1 F1
If A2 > A1, then F2 > F1. Thus, a small force can be used to exert a much larger force.
Thus, a hydraulic machine is a force multiplier.
The principle of hydraulic machine can be summed up as given below:
1. Any liquid cannot be compressed.
2. Pressure applied on an enclosed liquid is transmitted equally to its each part (Pascal’s
law).
3. The ratio of the cross-sectional area of a big cylinder (A2) to that of a small cylinder
(Al) is equal to the ratio of the load overcome (F2) to the effort applied (F1).
i. e. AA21 = F2
F1
Example 1: Find out x in the given diagram. x 5000 25 cm2
Solution: 4 cm2
33
In the diagram, a hydraulic machine
Cross sectional area of the big cylinder (A) = 25 cm2
Cross sectional area of the small cylinder (a) = 4 cm2
Load overcome by the big piston (F) = 5000 N
Effort applied in the small piston (f) = ?
We have,
AF
a = f
25 5000
or, 4 = f
PRESSURE CLASS - 10 MODERN GRADED SCIENCE
or, 25f = 20000
20000
∴ f = 25 = 800 N
Thus, 800 N effort will be required on the small piston.
Example 2: Find the values of X, Y, Z in the given diagram.
Solution: P 20N
10 cm2
For the value of X, we have,
A
AA = 10 cm2 4 cm2 X
AB = 80 cm2 Z B
FA = 20 N
FB = ? D
According to formula, C 80 cm2
AAAB = FA
FB
10 20 Y 30N
or, 80 = FB
or, 10 FB = 80×20
∴ FB = 80×2 = 160 N
Again for Y, we have,
FA = 20 N
FC = 30 N
AA = 10 cm2
AC = ?
According to formula,
AAAc = FA
Fc
10 20
or, Ac = 30
or, 20 AC = 30×10
30×10 30 PRESSURE
∴ AC = 20 = 2 = 15 cm2
Finally for Z, we have,
AA = 10 cm2
AD = 4 cm2
FA = 20 N
FD = ?
34 MODERN GRADED SCIENCE CLASS - 10
According to formula,
AADA = FA
FD
10 20
or, 4 = FD
or, 10 FD = 80
∴ FD = 8 N
Thus value of X is 160 N, Y is 15 cm2 and of Z is 8 N.
Example 3: In a hydraulic machine the ratio of the cross sectional area of a big cylinder to
small cylinder is 30:1. Calculate the load overcome by the big piston by applying the effort
of 600 N at the small piston.
Solution:
Consider the value of the cross sectional area of the small cylinder is x.
Now,
Cross sectional area of the big cylinder = 30 x
Load overcome by the small piston (f) = 600 N
Load overcome by the big piston (F) = ?
We have,
AF
a = f
60 x F
or, x = 600
or, 36000 = F
Thus, a load of 36000 N will be lifted by the effort.
This principle is applicable in different hydraulic machines. Some of such hydraulic
machines are explained below.
Hydraulic garage lift
Hydraulic garage lift is a machine used to lift light four wheelers like cars, jeeps, vans
etc. in service stations during their servicing.
Construction: A hydraulic garage lift consists of three metallic cylinders of different
diameters. They are connected together by connecting tubes and whole the system is
filled with water. The connecting tubes have valve ‘A’ and valve ‘B’ as shown in the figure.
The piston of the small cylinder is connected to a lever to apply force. The piston of the
big cylinder is connected to a platform on which the vehicle, which is to be lifted up, rests.
Reservoirs and big cylinders are also connected by another tube, which has a stopper ‘T’.
PRESSURE CLASS - 10 MODERN GRADED SCIENCE 35
lever vehicle
plateform
reservoir small
cylinder big piston
big cylinder
small
piston
water
valves
stopper
Fig: 2.4 hydraulic garage left
Working: To lift a vehicle by using it, the effort is applied at the small piston with the help
of the lever. The applied force exerts a pressure that closes valve ‘A’ and opens valve ‘B’.
The pressure exerted by the small piston is transmitted to the big piston. Because of more
area of the big piston, the force magnifies and it can lift the vehicle by using less effort.
To bring the vehicle at the initial position, the stopper ‘T’ is opened. Since liquid
maintains its level itself, the platform comes back to its original place due to the flow of
water in the reservoir.
Hydraulic brake
A hydraulic brake is a braking system is usually used in four wheelers and other
heavy automobiles.
Construction: It consists of tube ‘T’ containing oil called brake oil. One end of this tube
is connected to the master cylinder 'C' fitted with a piston that is connected to the brake
pedal. The other end of the tube is connected to a slave cylinder 'C1' having two pistons
connected to the brake shoes. The cylinder 'C1' has a larger diameter than that of 'C'. The
cylinders also contain brake oil.
pedal
tube (T) brake oil
lever
master cylinder (C) master PRESSURE
piston
inner rim of wheel
brake piston
brake shoe
slave cylinder (C1)
Fig: 2.5 hydraulic brake
36 MODERN GRADED SCIENCE CLASS - 10
Working: When the brake pedal is pressed, the piston in the master cylinder ‘C’ exerts
pressure in the brake oil of the tube ‘T’. This pressure is transmitted through the oil to
the pistons in the slave cylinder ‘C’1. Being the diameter of ‘C’1 larger than that of ‘C’, the
force applied at the brake pedal is magnified. Such a large force in the pistons of C1 pushes
the brake shoes. When the brake shoes are pressed against the rim of a wheel, the brake is
jammed on retarding the motion of the automobile.
Upthrust
When a bucketful of water is lifted from a well, we feel it lighter until it is under water.
But it is felt heavier if it comes out of the water surface. When a solid object is immersed in
water, it experiences pressure due to the water around it. These examples show that when
a body is wholly or partially immersed in water, the water pushes the body up.
Hence, when a body is partially or wholly immersed in a fluid, the resultant force
acted on it by the fluid is called upthrust or buoyant force.
Measurement of upthrust
Activity 2.1
Materials required: spring balance, overflow
can, beaker, stone, string, water
Method: W1 W2
1. Take a stone tied in a string and suspend it string
from a spring balance as shown in the figure
(a). stone in air overflow can
displaced
2. Record the weight (W1) of the stone in air. water
Let the weight of the stone in air (W1) = 50 N.
(a) water (b) stone in
3. Now, immerse the stone completely in a water
beaker containing water as shown in the
figure (b). Let its weight in water be W2 = Fig: 2.6 measuring upthrust
30N.
Now, the loss of the weight by the stone is W1 - W2 (= 50 - 30 = 20 N). This 20 N force
acts upwards on the stone in water, which causes the weight of the stone in water
to decrease. The upthrust on the stone immersed in water is 20 N. Thus, we feel the
stone lighter in water than in air.
Therefore, when a body is immersed in a liquid, it experiences two forces: its weight (W1)
acting downward and upthrust (U) due to the liquid acting upward. The effective weight
or apparent weight of the body in the liquid (W2) is the difference between the weight of
the body in air (W1) (real weight) and upthrust (U).
∴ Apparent weight = real weight - upthrust
(weight in air)
Density and relative density
There are different substances around us. The equal volume of two substances varies
in mass. For example, if we take the equal volume of stone and cotton, we feel the stone
PRESSURE CLASS - 10 MODERN GRADED SCIENCE 37
to be heavier than the cotton. The lightness or the heaviness of different substances is
because of their different densities.
The density of a substance is defined as the mass per unit volume of that substance.
Mathematically,
Mass(m)
Density (d) = Volume(v)
In SI system, density is measured in kg/m3 and in CGS system, it is measured in g/
cm3. 1000kg/m3 = 1g/cm3.
The following table presents the density of different substances:
Substance Density
Gold SI (kg/m3) CGS (g/cm3)
Mercury
Lead 19300 19.3
Iron 13,600 13.6
Common salt 11,000 11.0
Aluminium 8,000 8.0
Sugar 2170 2.17
Pure milk 2700 2.7
Water 1590 1.59
Ice 1030 1.03
Wood 1000 1
Kerosene 920 0.92
600-800 0.6-0.8
800 0.8
The substances having less density are said to be lighter ones while the substances
having greater density are said to be heavier ones. If a substance has more density than
that of the fluid, it sinks in the fluid and if a substance has less density than that of the
fluid, it floats on the fluid.
When the density of any substance is compared with the density of pure water at
4°C, it is called relative density of that substance. Therefore, when the mass of a particular
volume of substance is divided with the mass of the same volume of pure water at 4°C, it
is called relative density of that substance.
Density of the substance
Relative density of a substance = Density of pure water at 4°C
Thus, the ratio of the density of a substance to the density of pure water at 4°C is
called relative density of that substance.
Relative density is a simple ratio of two substances. That’s why, it does not have any
unit. Density of a substance in CGS system, without any unit, is numerically equal to the
relative density of that substance. It is also synonymized by specific gravity. For example,
relative density/ specific gravity of aluminium is 2.7.
38 MODERN GRADED SCIENCE CLASS - 10 PRESSURE
Interrelation between density and upthrust of liquid
The density of a liquid directly influences the upthrust. As the density of a liquid
increases, the pressure given by the liquid also increases. Upthrust is also increased in this
condition. The following activity will help to understand the relation between density of
liquid and upthrust.
Activity 2.2
To study the relation of density and upthrust of liquid with upthrust
Materials required:
beakers, egg, water, salt, spoon. floating egg
beaker
Method:
water
1. Take a beaker with water in it. saturated salt
2. Keep an egg in the water and solution
observe. sinking egg
Fig: 2.7 measuring upthrust
3. Now make a saturated solution
by dissolving enough salt in
the water.
4. Repeat the activity 2.
You can do the activity by taking potato, tomato etc. instead of an egg. It is found that the
egg sinks in water but it floats in a saturated salt solution. How is it possible? It is due to
more density of the salt solution, which exerts more upthrust on the egg. Due to the same
reason, a cargo-loaded ship floats more in oceans than in rivers. You may get surprised
that in the dead seas, which contains about 270 gram of salt in one litre of water, a person
does not sink due to its greater upthrust.
Archimedes’ principle
Greek scientist Archimedes first carried out experiments to measure the upthrust
of a liquid. It states that “when a body is partially or wholly immersed in a liquid, it
experiences an upthrust which is equal to the weight of the liquid displaced by it.” This
principle is also applicable in terms of gases,
or upthrust (u) = weight of displaced fluid (w2)
Theoretical proof of Archimedes’ principle free surface
d
Let’s consider a cylinder of height ‘h’ and its cross-sectional h1
area ‘A’ be completely immersed in a liquid of density ‘d’ as AM h2
shown in the figure. The horizontal thrusts on the cylinder
(shown by the arrows pointing towards the sides) balance each hh
other because they are equal in magnitude due to the same N beaker
depth but opposite in direction.
Let the top face M of the cylinder be at a depth h1, and the Fig: 2.8 a cylinder inside liquid
bottom face N at the depth h2 below the free surface of the liquid.
The thrust acted on the upper face of the cylinder due to liquid = Pressure × Area
i. e. F1 = P1 × A
PRESSURE CLASS - 10 MODERN GRADED SCIENCE 39
or, F1 = dgh1 × A ........... eqn (1) [ P = dgh]
The thrust acted on the lower face of the cylinder due to liquid = Pressure × Area
i. e. F2 = P2 × A
or, F2 = dgh2 × A .................. (2)
Therefore, the net upward thrust on the cylinder
F2 - F1 = dgh2 A - dgh1 A
or U = dg A [h2 - h1] [F2 - F1= upthrust]
∴ U = dgV [ V = A × h]
Thus, the upthrust acting on a body immersed in a liquid depends on the following
factors:
(a) the volume of the body ‘v’ (U ∝ V)
(b) the density of the liquid ‘d’ (U ∝ d)
(c) the acceleration due to gravity ‘g’(U ∝ g).
This is the mathematical form of Archimedes’ principle and the quantity dgV
represents the weight of the liquid displaced.
i. e. Upthrust (U) = weight of liquid displaced (W)
∴ Upthrust = mg ( w = m.g)
Experimental verification of Archimedes’ principle
Activity 2.3
To verify Archimedes’ principle
Materials required: spring balance, solid piece of denser material, eureka can, beaker,
tripod stand, string, water, top pan balance
Method:
spring balance W2
W1
eureka can
solid
beaker
displaced
water
top pan
balance
Fig: 2.9 verification of Archimedes' principle
40 MODERN GRADED SCIENCE CLASS - 10 PRESSURE
1. Weigh out a stone in air using a spring balance as shown in the figure (a).
2. Take an overflow can (eureka can) and pour water into it up to the spout.
3. Keep a beaker over the top pan-balance, just below the spout of the can as shown in
the figure (b).
4. Note the reading shown by the top pan-balance which gives the weight of the
beaker.
5. Immerse the solid in the eureka can.
The water displaced by the stone is collected in the beaker over the top pan balance.
Record the weight shown by the spring balance. It is the weight of the beaker and
displaced water.
The following observations are taken:
Weight of the stone in air = W1 gram weight
Weight of the stone in water = W2 gram weight
∴ Loss in weight of the stone in water = (W1 - W2) gram weight
Weight of the empty beaker = W3 gram weight
Weight of the beaker + displaced water = W4 gram weight
∴ Weight of the displaced water = W4 - W3 gram weight
After calculation, it is observed that W1 - W2 = W4 - W3. That is, loss in weight of
a body (i. e. upthrust) in water is equal to the weight of the displaced water. Thus,
Archimedes’ principle is verified.
Principle/Law of flotation
When a body is allowed to immerse completely in a liquid, the following two forces
act on it:
(1) the weight of the body acting vertically downward and
(2) the upthrust on the body acting vertically upward.
Depending on the magnitude of these forces, the following three cases may arise:
Case I: When the weight of the body is greater than the upthrust acting on it, the resultant force
will be vertically downwards. As a result, the body gets sunk in the liquid as shown in the
figure. In this case,
Weight of the body (w1) > weight of the liquid displaced (w2). Fig: 2.10 solid
or, m1g > m2g [ w = m × g] immersed in water
or, m1 > m2
m
or, d1 v1 > d2 v2 [ d = v or m = d × v)
d1 > d2 [ v1 = v2]
Thus, a body sinks in a liquid only if the density of the body (d1) is greater than the
density of the liquid (d2) in which the body is kept. Due to it, a piece of iron, being denser
than water, sinks completely in water.
PRESSURE CLASS - 10 MODERN GRADED SCIENCE 41
Case II: When the weight of a body is just equal to the upthrust acting on it, the resultant force
acting on it is zero. In this case, the body floats in the liquid as shown in the figure.
or, w1 = w2
∴ or, m1g = m2g [ w = m × g]
m
or, m1 = m2 [ d = v or m = d × v]
or, d1v1 = d2v2 Fig: 2.11 solid floats
Since, the volume of the body (v1) is equal to the volume of maintaining similar upper
displaced liquid (v2) i. e., v1 = v2, d1 is also equal to d2. surface with water level
∴ d1 = d2
Thus, a body just floats on a liquid when the density of the body (d1) is equal to the
density of the liquid (d2) in which the body is kept.
Case III: When the weight of the body is less than the upthrust acting on it, the resultant force
will be directed vertically upwards. As a result, the body comes at rest with its some part
in air and some part in water as shown in the figure. In this case, the weight of the body
will be equal to the weight of the liquid displaced by the immersed portion of the body.
We have for floating bodies:
∴ Weight of the body (W1) = weight of the liquid displaced (W2) Fig: 2.12 solid
or, m1g = m2g [ w = m.g] floating on liquid
or, m1 = m2
m
or, d1v1 < d2v2 [ d = v or m = d × v)
As the body is immersed partially, the volume of the body > the volume of the liquid
displaced i. e. v1 > v2 and hence we have d1 < d2
Thus, a body floats keeping some parts of it out when its density of the body (d1) is
lesser than that of the liquid (d2).
From the above discussion, we conclude that a body sinks in a liquid if the density of
the body is greater than that of the liquid. In this case, the weight of the liquid displaced
is lesser than the weight of the body. If, anyhow, the body is given a proper shape so that
the weight of the liquid displaced by the body is equal to the weight of the body, the body
can be made to float on the liquid. A floating boat ship is its best example.
The principle of floatation is, in fact, a special case of the Archimedes’ principle.
The law of floatation states that “weight of a floating body is equal to the weight of
the fluid displaced by the body”.
or, wt. of floating body (w1) = wt. of the displaced fluid (w2).
Experimental verification of law of floatation
Activity 2.4
To verify the law of floatation
Materials required: eureka can, beaker, top pan balance, dry wooden block, water,
tripod stand.
42 MODERN GRADED SCIENCE CLASS - 10 PRESSURE
Method: eureka can floating
spout body
1. An overflow can is
kept over a top pan water empty beaker
balance. top pan beaker displaced
balance water
2. The can is filled with tripod
water up to its spout stand
and an empty beaker
is kept below the spout Fig: 2.13 verification of law of floatation
of the can.
3. The weight of water with the overflow can [Eureka can] is noted.
4. Now, a dry wooden block is kept in the water of the can.
It displaces water into the beaker but the top pan balance does not show any change.
It proves that the weight of the displaced water is equal to the weight of the block. It
verifies the law of floatation.
Example 1: A solid object weighs 24 N in air and 10 N in water. What is the upthrust and
weight of the liquid displaced?
Solution:
Here,
Weight of the object in air (W1) = 24 N
Weight of the object in water (W2) = 10 N
Upthrust (U) = ?
Weight of the water displaced (W2) = ?
We know that,
Upthrust = W1 - W2
= 24 - 10 = 14 N
Again,
We know,
Upthrust (U) = Weight of the liquid displaced (W2)
∴ The weight of the liquid displaced = 14 N
Thus, upthrust on the body is 14 N and the weight of displaced fluid is 14 N.
Example 2: The mass of a brick is 2 kg. Find the mass of water displaced by it when it is
completely immersed in water. (Density of the brick is 2.5 g/cm3).
Solution:
Here, mass of a brick (m1) = 2 kg
2.5 × 10–3 kg
Density of the brick (d1) = 2.5 g/cm3 = 10–6 m3 = 2500 kg/m3 = 2500kg/m3
Density of water (d2) = 1000 kg/m3
Mass of water displaced = (m2) = ?
PRESSURE CLASS - 10 MODERN GRADED SCIENCE 43
(For sinking bodies)
V1 = V2
or, dm11 = m2 [ d = mv ]
d2
dd21 = m1
m2
2500 2
1000 = m2
1000×2 20 4
∴ m2 = 2500 = 25 = 5 = 0.8 kg
∴ Mass of the displaced water is 0.8 kg.
Example 3: An ice cube of volume 0.03 m3 floats on water. The densities of ice and water
are 900 kg/m3 and 1000 kg/m3 respectively. What fraction of the volume of the ice would
be in air? What would be the height of the ice cube in air if its total height is 0.2 m?
Solution:
Here,
Volume of ice (v1) = 0.03 m3
Density of ice (d1) = 900 kg/m3
Density of water (d2) = 1000 kg/m3
Height of the ice cube (h) = 0.2 m
Fraction of ice in air (v2/v1) = ?
Height of ice-cube above water = ?
We have, for floating bodies-
w1 = w2 [According to law of flotation]
or, m1g = m2g [ w = mg]
m
or, d1v1 = d2v2 [ d = v or m = dv]
d1 v2
∴ d2 = v1
900 v2 [ v2 is the fraction inside water]
1000 = v1 v1
∴ v2 9
v1 = 10
Again, for the fraction of ice in air-
9 10–9 1
= 1 – 10 = 10 = 10
For the height of ice in air-
= Fraction of ice × total height
1
= 10 × 0.2
44 MODERN GRADED SCIENCE CLASS - 10 PRESSURE
21
= 100 = 50 m = 0.02 m
1
∴ The fraction of ice cube in air is 10 and 0.02 m of the ice in air.
Example 4: A block of wood floats on a liquid with four-fifths of its volume submerged. If
the density of the wood is 800 kg/m3, find the density of the liquid.
Solution:
Here,
Fraction of block inside liquid v2 4
v1 =5
Density of wood (d1) = 800 kg/m3
Density of liquid (d2) = ?
We have, for floating bodies Air pressure is maintained artificially for the
passengers' convenience inside aeroplanes.
W1 = W2 [law of floatation]
or, m1g = m2g [ w = m × g] At very high altitudes, nose, ears and eyes may
or, d1v1 = d2v2 [d = mv or m = d × v] bleed due to less atmospheric pressure.
∴ d1 = v2
d2 v1
800 4
d2 = 5
800×5
∴ d2 = 4 = 1000 kg/ m3
Thus, density of the liquid is 1000 kg/m3
Example 5: Different weights of a piece of stone on weighing in three different media- air,
water and salt solution are shown in the table given.
(i) Which one is water and which one is the salt solution out of Medium Weight
the three media A, B and C? Write with reasons. A 15 N
B 18 N
(ii) If the weight of 1kg of mass in air is 10 N, find out the mass C 16 N
of the piece of stone.
(iii) Find out the mass of water displaced by the piece of stone.
Solution:
(i) Since the density of water is higher than the density of air but less than the
density of a salt solution, water exerts more upthrust on the body than air but it
exerts less upthrust than the salt solution. So, the apparent weight of the body is
highest in air and least in the salt solution. From the given table, we conclude that
the medium C is water and the medium A is salt solution.
(ii) The weight of the stone in air is 18 N.
W 18
Hence, the mass of the stone (m) = g = 10 = 1.8 kg. [ w = m.g]
(iii) We know that‚ upthrust due a liquid = Wt. of the liquid displaced
PRESSURE CLASS - 10 MODERN GRADED SCIENCE 45
or, loss in wt. of the stone in the liquid = wt. of the liquid displaced
∴ wt. of the stone in air - wt. of the stone in water = wt. of the liquid displaced (w2)
or, 18-16 = m2 × g
2
or, m2 = 10 = 0.2
Therefore, the mass of water displaced = 0.2 kg
Atmospheric pressure
The earth is surrounded by air. The air around the earth is called atmosphere. It is
estimated that thickness of the atmosphere is 9600 km i. e. about 10,000 km. We know
that air is a matter and it has weight. The weight of air exerts pressure on the bodies on
the earth's surface that is called atmospheric pressure. The value of atmospheric pressure
on sea level is one atmosphere. It is also termed as standard atmospheric pressure or
normal pressure. Thus, the normal pressure is the pressure exerted by 760 mm long
column of mercury at the temperature of 0°C at sea level. It is equivalent to 1.01×105 Nm–2.
Atmosphere, torr and bar are some other units used to express atmospheric pressure.
Their interrelation is shown below:
1 atmosphere = 760 mm of Hg = 1.01×105 Pa
1 torr = 1 mm of Hg = 132.322 Pa
1 bar = 106 dynes/cm2 = 105 Pa (Approx 750 mm of Hg) = 0.987 atm
1 millibar = 100 Pa
The value of atmospheric pressure decreases as the height increases from sea level. We
cannot survive at very high and very low atmospheric pressure. We do not feel the normal
atmospheric pressure on our body as the pressure of air and the blood pressure inside our
body are almost equal.
We can demonstrate atmospheric pressure with the help of the following activity.
Activity 2.5
To prove that atmospheric pressure occurs steam
Materials required: a tin can with lid, water, spirit cold water
lamp, tripod stand, wire gauze
partially
Method: vacuum
atmospheric
1. Take a tin can with its lid and fill less than one Pressure
third part of the can with water.
2. Keep the can on a tripod stand with its lid opened.
3. Now heat the water till the water begins to boil. (a) (b)
4. Now put off the fire and tighten the lid in the can. Fig: 2.14
5. Keep the can in air or water to cool.
Observation:
When the water boils, the water vapour expels all the air of the can out. But when the can
is cooled, the vapour turns in to water. Due to it, partial vacuum is formed inside the can.
In this situation, the atmospheric pressure crushes the tin can. It shows that every object
on the earth's surface receives atmospheric pressure.
46 MODERN GRADED SCIENCE CLASS - 10 PRESSURE
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