P-9-T1 - 04 Gravitation PDF

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GRAVITATION & FLUIDS IX -Physics
INTRODUCTION
When a body is held in hand and then released, the body falls vertically downwards. Let us think as to why
the body falls towards the earth? Why is it not going up? It was Sir Issac Newton who posed this question
and he himself answered it.
It is said that when Newton was sitting under a tree, an apple fell on him. The fall of the apple made Newton
think. He thought that if the earth can attract an apple can it not attract the moon? Is the force the same in
both cases? He conjectured that the same type of force is
responsible in both the cases.
Newton was sitting under an apple tree, when an apple fell on him.
He thought that the apple fell due to downward pull of the Earth
on the apple.
When a body is thrown up, it reaches a certain height and then
falls down. The downward pull of the Earth, on the body,
decreases its velocity in the upward direction to zero at some
height. The body cannot rise further. This is, therefore, the
maximum height attained by that body. The same downward pull
of the Earth on the body makes it fall downwards from the
maximum height.
Consider a piece of stone tied to one end of a string. Hold the other end of the string in hand and whirl it
around. The stone moves in a circular path with a certain speed, but its direction of motion changes
continuously. The change in direction of motion involves change in velocity or acceleration of the stone.
The external force F that causes this acceleration and keeps the stone moving uniformly along the circular
path is acting towards the centre of the circular path. This is called centripetal force, i.e., centre-seeking
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force. At any instant, if we release the string, the stone flies along the tangent to the circular path at that
instant. This is because, the moment the string is released, centripetal force is no longer provided. The stone
is free to fly off along the tangent. Based on this activity, we can perceive that the motion of the Moon
around the Earth is due to centripetal force, provided by force of attraction of the Earth on the Moon.
Gravitation is the phenomenon of attraction between any two objects in the universe. The objects may be
terrestrial (which are on the Earth) or celestial (which are in outer space such as stars, planets, satellite etc.).
Further, the objects may be of any size, shape or mass and they may be any distance apart (small or large),
with any medium between them. The force gravitation is always the force of attraction and it is never
repulsive.
NEWTON’S UNIVERSAL LAW OF GRAVITATION

Newton gave the universal law that gave the relationship between the force of attraction between any two
bodies lying at certain distance.
According to Newton’s universal law of gravitation:
“Every object in the universe attracts every other object with a force. This force of attraction between any
two objects is directly proportional to the product of their masses and inversely proportional to the square of
the distance between them”.
The direction of force is along the line joining the centers of two objects.
A
B
Let two objects A and B of masses M and m lie at a distance d from each other (distance between their
centers) as shown in figure. Let the force of attraction between two objects be F. According to the universal
law of gravitation, the force between two objects is directly proportional to the product of their masses. That
is,
F M  m …(i)
And the force between two objects is inversely proportional to the square of the distance between them, that
is,
1
F …(ii)
d2
Combining equation (i) and (ii)

M m
F 
d2
M m
or F=G
d2
Where G is the constant of proportionality and is called as universal gravitation constant.
M m
F=G
d2
F  d 2 = GM  m
F d2
G= …(iii)
M m
The SI unit of G can be obtained by substituting the unit of force, distance and mass in equation (iii).
Nm 2
G=
kg  kg
G = Nm 2 / kg 2 or G = Nm 2 kg −2
The value of ‘G’ was found out by Henry Cavendish (1731 – 1810) by using a sensitive balance.
The accepted value of ‘G’ is 6.673 x 10-11 Nm2/kg2.

Note : As the value of G is extremely small, therefore, gravitational force between any two ordinary objects
will be really weak. However, the gravitational force becomes appreciable when even one of the objects has
a large mass.
Importance of the universal law of gravitation

The universal law of gravitation successfully explained several phenomena which were believed to be
unconnected.
1. It is the gravitational force between the sun and the earth which keeps the earth in uniform
motion around the sun.
2. The gravitational force between the earth and the moon makes the moon revolve at uniform
speed around the earth. Thus the gravitational force is responsible for the existence of our
solar system.
3. The tides in the sea formed by the rising and falling of water level in the sea, are due to the
force of attraction which the sun and the moon exert on the water surface in the sea.
4. The gravitational force of the earth is responsible for holding the atmosphere above the earth.
5. It is also responsible for rain falling on the earth and for the flow of rivers.
6. It is also the gravitational force of the earth which keeps us firmly on the ground.
Gravitational Force between Light Objects and Heavy Object

The formula applied for calculating gravitational force between light objects and heavy objects is the same,
Gm1m 2
i.e., F = . Let us take three cases:
r2
(a) When two bodies of mass 1 kg each are 1 metre apart.
i.e., m1 = m 2 = 1 kg, r = 1 m
Taking G = 6.67  10−11 Nm2 / kg2, we obtain gravitational force of attraction,
Gm1m 2 6.67  10 −11  1 1

F = 2
= 2
= 6.67  10 −11 N
r (1)
This is extremely small. Hence, we conclude that though every pair of two objects exertsgravitational
pull on each other, yet they cannot move towards each other because this gravitational pull is too
weak.
(b) When a body of mass 1 kg is held on the surface of Earth.

Here, m 1 = 1 kg
m 2 = mass of Earth = 6  1024 kg
r = distance of body from centre of Earth
= radius of Earth = 6400 km = 6.4  103 km = 6.4  106 m

Gravitational force of attraction between the body and Earth,
Gm1m 2 6.67  10 −11  1 6  10 24

F = = = 9 .8 N
r2 (6.4  10 6 )2
It means that the Earth exerts a gravitational force of 9.8 N on a body of mass 1 kg. This force is
much larger as compared to the force when both the bodies are lighter. That is why when a body is
dropped from a height, it falls to the Earth.
(c) When both the bodies are heavy.

Let us calculate gravitational force of attraction between Earth and the Moon.
Mass of Earth, m1 = 6  1024 kg
Mass of Moon, m 2 = 7.4  1022 kg
Distance between Earth and Moon, r = 3.84  105 km = 3.84  108 m
Gravitational constant, G = 6.67  10−11 Nm2 / kg2
The gravitational force between Earth and Moon,
Gm1m 2 6.67  10 −11  (6  10 24 )  (7.4  10 22 )
F = = = 2.01  10 20 N
r 2
(3.84  10 )8 2
This is really large. It is this large gravitational force exerted by Earth on Moon, which makes
the Moon revolve around the Earth.
Gravitation and Newton's Third Law of Motion

As we have read Newton's third law of motion states that, to every action, there is always an equal and
opposite reaction. It means if an object A exerts some force on another object B, then the object B exerts an
equal and opposite force on the object A at the same instant. This law is applied to the force of gravitation
also. We say that an apple falls towards the Earth due to gravitational pull of the Earth on it. As a reaction,
the apple also exerts an equal force of attraction on Earth in the opposite direction (i.e., towards the apple).
But then why do we not see the Earth rising towards the apple? The answer to this question is provided by
Newton's second law of motion, according to which
force = mass × acceleration

force
or acceleration =
mass
i.e., for a given force, acceleration produced varies inversely as the mass. Now, force on Earth (exerted by
apple) = force on apple (exerted by Earth)
But mass of Earth is too large compared to the mass of apple. Therefore, acceleration produced in
Earth is too small to be seen. That is why the Earth is not seen rising towards the apple.
We know that acceleration produced in a body due to gravitational pull of Earth on it is 9.8 m/s 2. As this
acceleration is very large, we can see the body falling towards Earth. We shall show that when gravitational
pull of same magnitude acts on Earth (whose mass is 6  1024 kg ), the acceleration produced in Earth is
1.63  10−24 m/s 2 . As this value of acceleration is too small, we cannot see the Earth moving towards the
falling body.
Free Fall
When we drop a body say a stone, we observe that its speed increases as it falls
towards the earth. We have already seen that the body falls towards earth due
to gravitational force between body and the earth. Whenever objects fall
towards the earth under this force alone, we say that the objects are under free
fall. While falling, there is no change in the direction of motion of the objects.
But due to the earth’s attraction, there will be a change in the magnitude of the
velocity. Any change in velocity involves acceleration.
The question is now whether we can measure its acceleration? Will this
acceleration be more for heavier bodies?
If we drop stone and paper from a top of a house, we find that stone reaches the
earth earlier than the paper. It was therefore thought that the acceleration for
heavier bodies is more than lighter bodies.
Galileo decided to test this common belief. He dropped two stones of different masses from the top of
leaning Tower of Pisa. It was found that both reached the earth in almost same time. He concluded that all
the bodies fall towards the earth with equal acceleration. The acceleration with which the bodies fall towards
the earth is independent of the masses of the bodies. Reason for slowing down of lighter bodies was
attributed to the resistance or friction offered by the air.
Robert Boyle took a vacuum pump to remove air from a tube containing a heavy coin and a piece of paper.
When the tube was inverted, both the coin and the paper hit the bottle at the same time. Thus, Galileo’s
prediction that all bodies fall with the same acceleration towards earth stands confirmed.
Gravity
As we have read that gravitation is the phenomenon of attraction between any two bodies of the universe.
Gravity is one particular case of gravitation when one of the two attracting bodies is Earth.
Hence, gravity is the phenomenon of attraction between Earth and any other body. The force of
attraction exerted by Earth on the body is called force of gravity.
The force of gravity is calculated using Universal Law of Gravitation given by Newton as,
GmM
F=
r2
Where m = mass of the body
M = mass of earth
r = distance of the body from the centre of Earth
F = force of gravity exerted by Earth on the body
ACTIVITY
Fill in the blanks

If the separation distance between the object and the Earth is ...
increased by a factor of 2, then the Fgrav is ___________ by a factor of _____.
If the mass of the object is ...
increased by a factor of 2, then the Fgrav is increased by a factor of _____.
decreased by a factor of 4, then the Fgrav is ___________ by a factor of _____.
If the mass of the Earth is ...

increased by a factor of 2, then the Fgrav is increased by a factor of _____.
decreased by a factor of 4, then the Fgrav is ___________ by a factor of _____.
Acceleration Due To Gravity

When a body is dropped from a certain height, it falls with a constant acceleration.
This uniform acceleration produced in a freely falling body due to the gravitational pull of the earth is
known as acceleration due to gravity and it is denoted by the letter ‘g’.
Although g varies very slightly from place to place but its average value is taken to be 9.8 m/s2. This means
that the velocity of a body increases by 9.8 m/s every second. Say, if the body is dropped with zero velocity,
its velocity becomes 9.8 m/s after 1s; 19.6 m/s after 2s; 27.4 m/s after 3s and so on. Similarly, if a body is
projected upwards, its velocity decreases by 9.8 m/s after every second.
Relation between ‘g’ and ‘G’

Case I: If we drop a body (say, a stone) of mass ‘m’ from a distance ‘d’ from the center of the earth
of mass
M, then the force exerted by the earth on the stone is given by Newton’s law of gravitation
as:
M m
F =G … (i)
d2
We also know from the second law of motion that force is the product of mass and
acceleration. We already know that there is acceleration involved in falling objects due to the
gravitational force and is denoted by g.
Therefore, the magnitude of the gravitational force F will be equal to the product of mass and
acceleration due to the gravitational force, that is,
F = mg …(ii)
From equation (i) and (ii)
M m
mg =G
d2
M
g =G 2
d
Case II : Let an object be on or near the surface of the earth. The distance d will be equal to R , the
radius of the earth. Thus, for objects on or near the surface of the earth,
M m
mg = G
R2
M
g =G
R2
The earth is not a perfect sphere. As the radius of the earth increases from the poles to the
equator, the value of ‘g’ becomes greater at the poles than at the equator.
Value of g on earth
To calculate the value of g, we should put the values of G = 6.7  10 −11 Nm 2 /kg 2 ; mass of the earth
M = 6  10 24 kg and radius of the earth (R) = 6.4  10 6 m , in the formula
M
g =G 2
R
6.7  10 −11Nm 2 /kg 2  6  10 24 kg
=
( 6.4  10 6 m) 2
= 9.81m/s 2
Thus, value of acceleration due to gravity of the earth, g = 9.81m/s2 .
Value of g on moon
Mass of moon = 7.4  10 22 kg
Radius of moon = 1,740 km
= 1,740,000 m = 1.74  10 6 m
M 6.67  10 −11  7.4  10 22
g =G = = 1.63 m/s 2
R2 (1.74  10 6 ) 2
g m 1.63 1
Therefore, = =
ge 9.8 6
Acceleration due to gravity (g) does not depend upon mass of the body
GM
The relation between g and G is g =
R2
Where, G = gravitational constant,
M = mass of Earth, and
R = radius of Earth.
We observe from this relation that value of g depends on mass of Earth and radius of Earth. Mass of the
body (m) is nowhere involved in this relation. It means that the value of acceleration due to gravity does
not depend upon mass of the body. Hence, all bodies when drop will falls towards the centre of earth with
the same acceleration – whatever be their masses. It means that a big stone and a small stone when dropped
from a particular height will move with the same acceleration due to gravity. Both will strike the ground at
the same time whenever they fall under gravitational pull alone (vacuum).
Variation in Acceleration due to Gravity (G)

GM
The relation between g and G is g = . It shows that value of g depends on gravitational constant G,
R2
mass of Earth M and radius of the Earth R. Now, G and M both are constants. But radius of earth R is not
constant as Earth is not a perfect sphere. Therefore, value of g changes from place to place on the surface of
earth. These changes are briefly discussed here
(a) Effect of shape of Earth on 'g'
As is known, Earth is flattened at the poles and bulged out at the equator. Therefore, polar radius of
Earth= R p is minimum, and equatorial radius of Earth = R p is maximum.
Acceleration due to gravity at poles, g p = GM = maximum, as R p is minimum.
2
Rp
Acceleration due to gravity at equator, g e = GM2 = minimum, as R e is maximum.

Re
(b) Effect of height above the surface of Earth on 'g'
As we move above the surface of the Earth, the distance (r) from the centre of earth increases. As
GM
g= , therefore, value of acceleration due to gravity decreases with height above the surface
r2
of Earth.
Gravity at height h is given by

g
gh = 2
 h
 1 + 
 R
for small heights,
 h 
g h  g 1 −  ( when h  R )
 R
(c) Effect of depth below the surface of Earth on 'g'

Most of you might be thinking that as we go down the surface of Earth,
the value of radius of Earth R decreases. Therefore, g must be
increasing. But it is not true. You will learn in higher classes that the
value of acceleration due to gravity decreases as we go down inside
the Earth.
So much so that at the centre of Earth, g = 0.
 d
g d = g 1 − 
 R
Hence, we conclude that value of acceleration due to gravity is

maximum at the surface of Earth. It decreases as we move above the
surface of Earth or go inside the surface of Earth.
Equations of Motion of Freely Falling Bodies

When the bodies are falling under influence of gravity, they experience acceleration g i.e. 9.8 m/s2 .
However, when these are going up against gravity they move with retardation of 9.8 m/s2 .
For vertically upward motion, ‘a’ is replaced by g. Here ‘s’ is also replaced by h, the height.
v = u + at changes to v = u + gt
1 2 1 2
s = ut + at changes to h = ut + gt
2 2
v 2 = u 2 + 2as changes to v 2 = u 2 + 2gh
Mass and Weight

The similarities and differences between mass and weight are discussed as follows
MASS
• The mass of a body is the quantity of matter contained in it.
• Mass is a scalar quantity. The unit of mass is kilogram (kg).
• A body contains the same quantity of matter whether it be on the earth, moon or even in
outer space. Thus, the mass of a body is constant and does not change from place to place.
• Mass of a body is usually denoted by the small letter ‘m’.
• Mass of a body is a measure of inertia of the body and hence it is also known as inertial
mass.
The mass of a body cannot be zero.
WEIGHT
We know that the earth attracts every object with a certain force and this force depends on the mass (m) of
the object and the acceleration due to gravity (g). The weight of an object is the force with which it is
attracted towards the earth.
We know that, F = m  a
That is, F = mg
• The force of attraction of the earth on an object is known as the weight of the object. It is denoted by
W.
So we have, W = m  g
• As the weight of an object is the force with which it is attracted towards the earth, the S.I. unit of
weight is the same as that of force i.e. Newton (N).
• The weight is a force acting vertically downwards; it has both magnitude and direction, so it is a
vector quantity.
The value of g is constant at a given place. Therefore at a given place, the weight of an object is directly
proportional to the mass, say m, of the object, that is, W  m . It is due to this reason that at a given place,
we can use the weight of an object as a measure of its mass.
The mass of an object remains the same everywhere, that is, on the earth or on any planet whereas its weight
depends on its location.
Weight of a freely falling body

Let us suppose that a body is placed on a lift, the weighing machine will show the weight of the body on its
scale. Now, the lift is made to fall freely due to gravity, both the weighing machine as well as the body will
fall with same acceleration i.e., with g in the downward direction. The body will, therefore, not press the
weighing machine with any force and hence show zero weight. Thus a body is weightless during free fall.
Weight of an object on the moon

We have learnt that the weight of an object on the earth is the force with which the earth attracts the object.
In the same way, the weight of an object on the moon is the force with which the moon attracts that object.
The mass of the moon is less than that of the earth. Due to this, the moon exerts lesser force of attraction on
objects. Let the mass of an object be m. Let its weight on the moon be Wm. Let the mass of the moon be Mm
and its radius be Rm.
By applying the universal law of gravitation, the weight of the object on the moon will be
Mm  m
Wm = G
R m2
If the weight of the same object on the earth be We . The mass of the earth is M and its radius is R.
M m
Now, We = G
R2
Mass of earth = 5.98  10 24 kg
Radius of earth = 6.37  10 6 m
Mass of moon = 7.36  10 22 kg
Radius of moon = 1.74  10 6 m
Substituting the values:
7.36  10 22 kg  m
Wm = G
(1.74  10 6 m ) 2
Wm = 2.431  10 10 G  m …(i)
And We = 1.474  10 G  m
11
…(ii)
Dividing equation (i) by (ii), we get
Wm 2.431 1010
=
We 1.474  1011
Wm 1
or = 0.165 
We 6
Weight of the object on the moon 1
=
Weight of the object on the earth 6
 Weight of the object on the moon = (1/6) ×
weight of the object on the earth.
Kepler’s Laws of Planetary Motion

Johannes Kepler derived three laws, which govern the motion of planets. These are called Kepler’s laws.
These are:
1. LAW OF ORBITS: The orbit of a planet is an ellipse with the sun at one of the foci, as shown in
the figure given below. In this figure O is the position of the sun.
2. LAW OF AREAS: The line joining the planet and the sun sweeps equal areas in equal intervals of
time. Thus, if the time of travel from A to B is the same as that from C to D, then the areas OAB and
OCD are equal.
3. LAW OF PERIODS: The cube of the mean distance of a planet from the Sun is proportional to the
square of its orbital period T. or, r3/T2 = constant.
major axis
b
F F'
a
minor axis
F, F are foci
FLUIDS, THRUST AND PRESSURE

The force acting on an object perpendicular to the surface is called thrust. Let us understand the
meaning of thrust and pressure practically.
Situation-1:
To fix a poster on a notice board and while doing so you need to press drawing pins with your thumb. So
pressing drawing pins means applying force on the surface area of the head of the pin. This force is directed
perpendicular to the surface area of the board.
Why is the tip of a needle sharp?
Situation-2:
When you stand on loose sand your feet go deep into the sand. But when you lie down on the sand, you will
find that your body will not go deep on the sand.
This is because when you stand on loose sand, the force i.e., the weight of your body is acting on an area
equal to area of your feet. When you lie down, the same force acts on an area equal to the contact area of
your whole body. Which is larger than the area of your feet.
Thus, the effect of thrust on sand is larger while standing than while lying.
The thrust on unit area is called pressure. Thus,

thrust
Pressure = .
area
S.I. unit of pressure is N/m 2 or Nm −2 .
In honour of scientist Blaise Pascal, the S.I. unit of pressure is called
Pascal, denoted as Pa .
Pressure depends on two factors :
(i) Force applied (ii) Area over which force acts.
PRESSURE IN FLUIDS
All liquids and gases are called as fluids. A solid exerts pressure on a surface due to its weight.
Similarly, fluids have weight, and they also exert pressure on the base and walls of the container in which
they are enclosed. A fluid exerts pressure in all directions (even upwards) always perpendicular to the
surface of the container.
Density
We describe the lightness or heaviness of different substances by using the word density.
The density of a substance is defined as mass of the substance per unit volume. That is,
mass of the substance
Density =
volume of the substance
SI unit is kg/m 3 or kg m −3
The density of a given substance, under specified conditions remains the same. Therefore, the density of a
substance is one of the characteristic properties of a substance.
For example, density of gold is 19300kg/m3 , while that of water is 10 00 kg/m 3 .
The density of a given sample of a substance can help us to determine its purity.
Relative Density
The relative density of a substance is the ratio of its density to that of water.
Density of the substance

Relative density of a substance =
Density of water
Since, the relative density is a ratio of similar quantities, it has no unit.
The relative density of a substance expresses the heaviness of the substance in comparison to water. For
example, the relative density of iron is 7.8. This means that iron is 7.8 times as heavy as an equal volume of
water.
The relative density of water is 1.
Now, if the relative density of a substance is more than 1, then it will be heavier than water and hence it will
sink in water.
On the other hand, if the relative density of a substance is less than 1, it will be lighter than water and hence
will float in water.
Buoyancy
To understand the term buoyancy, let us perform an activity
Take an empty plastic bottle. Close the mouth of the bottle with an air tight stopper. Put it in a bucket filled
with water.
(a) We will see that the bottle floats.
(b) Now push the bottle into the water. We will feel an upward push. If we push it further
down, we will find it difficult to push deeper and deeper.
(c) This indicates that water exerts a force on the bottle in the upward direction.
(d) This upward force exerted by the water goes on increasing as the bottle is pushed deeper till it
is completely immersed.
(e) If we release the bottle, it bounces back to the surface.
Explanation
The force due to the gravitational attraction of the earth acts on the bottle in the downward direction. So the
bottle is pulled downwards. But the water exerts an upward force on the bottle. Thus, the bottle is pushed
upwards.
The weight of an object is the force due to gravitational attraction of the earth. When the bottle is immersed,
the upward force exerted by the water on the bottle is greater than its weight therefore it rises up, when
released.
To keep the bottle completely immersed the upward force on the bottle due to water must be balanced. This
can be achieved by an externally applied force acting downwards. This force must at least be equal to the
difference of upward force and the weight of the bottle.
The net upward force exerted by the liquid acting on an object immersed in a liquid is called buoyant
force or upthrust and this phenomenon of exerting buoyant force is called buoyancy.
Factors affecting buoyant force

The magnitude of buoyant force acting on an object immersed in a liquid depends on two factors:
(a) Volume of the object immersed in the liquid.
The buoyant force exerted by a liquid depends on the volume of the solid object immersed in the
liquid. As the volume of solid object immersed inside the liquid increases, the upward buoyant force
also increases. And when the object is completely immersed in the liquid, the buoyant force becomes
the maximum and remains constant. Also, the magnitude of buoyant force acting on a solid object
does not depend on the nature of the solid object.
It depends only on its volume. For e.g. If two balls made of different metals having different
weights but equal volumes are fully immersed is a liquid, they will experience an equal upward
‘buoyant force’.
(b) Density of the liquid.

The buoyant force exerted by a liquid depends on the density of the liquid in which the object is
immersed. As the density of liquid increases, the buoyant force exerted by it also increases, for
example sea water has higher density then fresh water therefore sea water will exert more buoyant
force on an object immersed in it than the fresh water. Therefore it is easier to swim in sea water
because sea water exerts a greater buoyant force on the swimmer due to its higher density.
Note : Even a very heavy material like an iron block floats in mercury because mercury exerts a
very high buoyant force on iron block due to its very high density.
Archimedes’ Principle
Archimedes was a Greek scientist. He discovered the principle, subsequently named after him, after noticing
that the water in a bath tub overflowed when he stepped into it. He ran through the streets shouting
“Eureka”! which means “I have got it”. This knowledge helped him to determine the purity of the gold in the
crown made for the king. This work in the field of geometry and mechanics made him famous. His
understanding of levers, pulleys, wheels–axle helped the Greek army in its war with roman army.
“Archimedes’ principle states that when a body is partially or wholly immersed in a liquid, it
experiences a net upward force that is equal to the weight of the fluid displaced by it.”
Activity to understand Archimedes’ principle

(a) Take a piece of stone and tie it to one end of a rubber string or a spring balance.
(b) Suspend the stone by holding the balance or the string as shown in figure below.
(c) Note the elongation of the string or the reading on the spring balance due to the weight of
the stone.
(d) Now, slowly dip the stone in the water in a container. You will find that the elongation of
the string or the reading of the balance decreases as the stone is gradually lowered in the
water.
(e) However no further change is observed once the stone gets fully immersed in the water.
Explanation:
The elongation is produced in the string or the spring balance due to the weight of the stone. Since the
extension decreases once the stone is lowered in water, it means that some force acts on the stone in upward
direction. As a result, the net force on the string decreases and hence the elongation also decreases. This
upward force exerted by water is known as the force of buoyancy.
Applications of Archimedes’ principle

(a) Archimedes’ principle is used in determining the relative density of a substance.
(b) The hydrometers used for determining the density of liquids are based on Archimedes’
principle.
(c) The lactometers used for determining the purity of milk are based on Archimedes’ principle.
(d) Archimedes’ principle is used in designing ships and submarines.
Why objects float or sink in a liquid?

When an object is put in a liquid, then two forces act on it:
• Weight of object acting downwards due to the gravitational pull of
the earth on the object.
• Buoyant force acting upwards which tends to push the object up.
(a) Sinking of an object in water

If we place an iron nail on the surface of water in a beaker then
the nail sinks. The force due to the gravitational attraction of the earth on the iron nail pulls it
downwards. There is an upthrust of water on the nail, which pushes it upwards. But the downward
force acting on the nail is greater than the upthrust of water on the nail, so it sinks.
(b) Floating of an object in water

If we place a piece of cork on the surface of water in a beaker, then the cork floats. This happens
because of the difference in their densities. The density of a cork is less than the density of water.
This means that the upthrust of water on the cork is greater than the weight of the cork. So, it
floats.
Therefore, objects of density less than that of a liquid float on the liquid. The objects of density
greater than that of a liquid sink in the liquid.
SOLVED EXAMPLE
1. Why is Newton’s law of gravitation called as universal law of gravitation?

Sol: The law of gravitation is universal in the sense that it is applicable to all the bodies, whether the
bodies are big or small, whether they are celestial or terrestrial.
2. If the distance between two objects is increased by a factor of 6, how is the force altered?
Sol: Saying that F is inversely proportional to the square of d means that if d gets bigger by a factor of 6,
1
F becomes times (smaller).
36
3. The mass of the earth is 6  10 24 kg and that of the moon is 7.4  10 22 kg . If the distance
between the earth and the moon is 3.84  10 5 km , calculate the force exerted by the earth on the
moon. [ G = 6.7  10 −11 Nm 2 kg −2 ]
Sol: The mass of the Earth, M = 6  10 24 F
The mass of the Moon, m = 7.4  10 22 kg
The distance between the earth and the moon,
d = 3.84  10 5 km
d = 3.84  10 5  1000 m
d = 3.84  10 8 m
G = 6.7  10 −11 Nm 2 kg −2
The force exerted by the earth on the moon is given as :
M m
F =G
d2
6.7  10 −11 Nm 2 /kg 2  6  10 24 kg 7.4  10 22 kg
= = 2.01  10 20 N
(3.84  10 m )
8 2
Thus, the force exerted by the earth on the moon is2.01 x 1020 N.
4. Calculate the force of gravitation due to a child of mass 25 kg on his mother of mass 75 kg
standing at a distance of 1 m from each other.
20
(Given: G =  10 −11 Nm 2 kg )
3
20
Sol: Here, m1 = 25 kg; m2 = 75 kg; d = 1m; G =  10 −11Nm 2 /kg 2
3
20  10 −11  75  25
F =
3  (1) 2
F = 12,500  10 −11
F = 1.25  10 −7 N
Since, this force is too small i.e., 0.000,000,125 N, hence the child or the mother is attracted
by a negligible force due to gravitation and is not noticeable.
5. A mass of 50 kg attracted by a mass of 20 kg lying at a distance of 2m with a force of

1.67  10 −8 N . Find the value of G.
Sol: F = 1.67  10 −8 N
d = 2m
m1 = 50 kg
m2 = 20 kg
G=?
G(50 ) ( 20 ) G(1000 )
F = =
22 4
G (1000 )
1.67  10 −8 =
4
−8
4  1.67  10
G=
1000
G = 6.68  10 −11 Nm 2 / kg 2
6. Calculate the force of gravitation due to earth on a child weighing 10 kg standing on the
ground. (Mass of earth = 6  10 kg ; Radius of earth
24
= 6.4  10 km &
3
G = 6.7  10 −11 Nm 2 /kg 2 )

Sol: The force of gravitation is calculated by using the formula
m1  m2
F =G
d2
G = 6.7  10 −11Nm 2 / kg 2
Mass of earth, m1(M ) = 6  10 24 kg
Mass of child, m2 (m ) = 10 kg
Distance between center, R (d) = Radius of earth,
= 6.4  10 3 km
= 6.4  10 3  1000 m = 6.4  10 6 m
Now, putting these values in the above formula, we get
6.7  10 −11  6  10 24  10
F =
(6.4  10 6 ) 2
F = 98.1 N .
7. A planet whose mass and radius both are half as that of the earth, the acceleration
due to gravity at the planet's surface would be
(a) 19.6 m/s2 (b) 9.8 m / s 2 (c) 4.9 m / s 2 (4) 2.45 m / s 2
M
Sol: We know g e = G for earth.
R2
In case of Planet M is replaced by M/2 and R by R/2.
M
G
gp = 2 = 2 GM ;
2
R R2
 
2
g p = 2 ge
g e = 9.8 m/s 2
So, g p = 2  9.8 = 19 .6 m/s 2
8. If a planet existed whose mass was twice as that of the earth and whose radius 3times greater, a
10 kg mass on its surface will weigh:
(a) 21.7 N (b) 4.4 N (c) 6.7 N (d) 13.3 N
Choose the correct answer.
M
Sol: For earth g e = G …(i)
R2
For planet M is replaced by 2 M and R by 3R
G (2M ) 2GM
gp = 2
= …(ii)
(3R ) 9R 2
From equation (i) and (ii)
gp 2
=
ge 9
2
gp =  ge
9
2
g p =  9.8 = 2.17 m/s2
9
Weight = mgp = 10 x 2.17 = 21.7 N
9. Mass of an object is 10 kg. What is its weight on the earth?

Sol: Mass, m = 10 kg
Acceleration due to gravity, g = 9.8 m/s 2
W = mg
W = 10  9.8 = 98N
Thus, the weight of the object is 98 N.
10. What is the mass of object whose weight is 49 N?

Sol: W = mg
49 490
 49 = m  9.8  m= = = 5 kg
9 .8 98
11. A force of 100 N is applied to an object of area 2 m2 . Calculate the pressure.

Force 100 N
Sol: Pressure = = = 50 N/m 2
Area 2 m2
= 50 Pa
12. A man weighs 600N on the earth, what is its mass? If it was taken to the moon, his weight
would be 100 N. What is his mass on moon? What is his accelerations due to gravity on the
moon.
Sol: (i) Let m be the mass of body on earth. We know,
W = mg
600 = m  10
m = 60 kg
(ii) Mass will be same on Moon i.e., 60 kg
[ Mass of the body does not change from place to place]
(iii) Weight on moon = mass × acceleration due to gravity on moon.
100 = 60  gm
100
= gm
60
g m = 1.67 m/s 2
13. A car falls off a ledge and drops to the ground in 0.5s. Let g = 10 m/s 2 .
(i) What is its speed on striking the ground?
(ii) How high is the ledge from the ground?
(iii) What is its average speed during the 0.5 s?
1
Sol: Time, t = second
2
Initial velocity, u = 0 m/s
Acceleration due to gravity, g = 10 m/s 2
Acceleration of the car, a = +10 m/s 2 (downward)
(i) Speed v = u + gt
v = 0 + (10  0.5)
v = 5 m/s
1
(ii) Distance traveled (S ) = ut + at 2
2
1
= 0  0 .5 +  10  (0.5) 2
2
= 0 + (5  0.25 ) = 1.25 m
(iii) Average speed = total distance / total time
= 1.25 / 0.5
= 2.5 m/s
14. An object is thrown vertically upwards and rises to a height of 10m. Calculate
(i) the velocity with which the object was thrown upwards and
(ii) the time taken by the object to reach the highest point.
Sol: Distance traveled (h) = 10 m
Final velocity (v ) = 0 m/s
Acceleration due to gravity, g = 9.8 m/s 2
Acceleration of the object, a = −9.8 m/s 2
(i) v 2 = u 2 + 2gh
0 = u 2 + 2( −9.8)  10
0 = u 2 − 196
u 2 = 196
u = 196 = 14 m/s
(ii) v = u + gt 0 = 14 − 9.8  t => t = 1.43 s
15. To estimate the height of a bridge over a river, a stone is dropped freely in the river from the
bridge. The stone takes 2 seconds to touch the water surface in the river.
Calculate the height of the bridge from the water level (g = 9.8 m/s 2 ) .
Sol: Now, initial velocity of stone, u = 0
Time taken, t = 2s
Acceleration due to gravity,
And, height of the bridge, h = ? (To be calculated)
We know that for a freely falling body
1 2
Height, h = ut + gt
2
Putting the above values in this formula, we get
16. A block of wood is kept on a table top. The mass of wooden block is 5 kg and its dimensions are
40 cm × 20 cm × 10 cm. Find the pressure exerted by the wooden block on the table top if it is
made to lie on the table top with its sides of dimensions
(a) 20 cm × 10 cm (b) 40 cm × 20 cm
Sol: The mass of the wooden block = 5 kg
The dimensions = 40 cm × 20 cm × 10 cm
Here, the weight of the wooden block applies a thrust on the table top.
That is,
Thrust, F = mg
= 5 kg  9.8 m/s 2
= 49 N
Area of a side = l  b = 20 cm 10 cm
= 200 c m 2 = 0.02 m 2
F 49
Pressure = = = 2450 N/m 2 or 2450 Pa
A 0.02
When the block lies on its side of dimensions 40 cm × 20 cm, it exerts the same thrust.
Area = l  b = 40  20
F 49
Pressure = = = 612 .5 N/m 2 or 612.5 Pa
A 0.08
17. When a ball is thrown vertically upwards, it goes through a distance of 19.6 m. Find the initial
velocity of the ball and the time taken by it to rise to the highest point. (Acceleration due to
gravity, g = 9.8 m/s 2 )
Sol: Here, Initial velocity of ball, u = ? (To be calculated)
Final velocity of ball, v =0
Acceleration due to gravity, g = −9.8 m/s 2
And height, h = 19.6 m
Now, putting all these values in the formula:
v 2 = u 2 + 2gh
We get (0) 2 = u 2 + 2  ( −9.8)  19 .6
0 = u 2 − 19 .6  19 .6
u 2 = (19 .6) 2
So, u = 19 .6 m/s
Here, Final velocity, v = 0 (The ball stops)
Initial velocity, u = 19 .6 m/s (Calculated above)
And, Time t = ? (To be calculated)
So, putting these values in the equation, v = u + gt
0 = 19.6 + ( −9.8)  t
0 = 19 .6 − 9.8t
19 .6
9.8t = 19.6 t = t = 2s
9 .8
Thus, the ball takes 2 seconds to reach the highest point of its upward journey. The ball will take an
equal time that is 2 seconds to fall back to the ground. So the ball will take a total of 2 + 2 = 4
seconds to reach back to the thrower.
18. A ball is thrown up with a speed of 15 m/s. How high will it go before it begins to
fall? (g = 9.8 m/s 2 ) .
Sol: Here, initial speed of ball, u = 15 m/s
Final speed of ball, v = 0 (The ball stops)
And, Height, h = ? (To be calculated)
Now, putting all these values in the formula,
v 2 = u 2 + 2gh
We get, (0)2 = (15 )2 + 2  ( −9.8)  h
0 = 225 − 19 .6 h
225
19.6h = 225 or h = = 11.48 m
19 .6
Thus, the ball will go to a maximum height of 11.48 metres before it begins to fall.
19. Give reason for the following:

(a) Why school bags have wide straps?
(b) Why a sharp knife cuts better than a blunt knife?
(c) Why a nail has a pointed tip?
(d) Why buildings have wide foundations?
Sol: (a) A school bag has wide straps made of thick cloth so that the weight of bag may fall over a
large area of the shoulder of the child producing less pressure on the shoulder. Due to less
pressure, it would be more comfortable to carry the heavy school bag. On the other hand, if
the school bag has a straps made of thin strings, then the weight of school bag will fall over a
small area of the shoulder. This will produce a large pressure on the shoulder of the child and
it will become very painful to carry the heavy school bag.
(b) A sharp knife has a very thin edge of its blade. A sharp knife cuts objects better because due
to its very thin edge, the force of our hand falls over a very small area of the object producing
a large pressure. And this large pressure cuts the object easily. On the other hand, a blunt
knife has a thicker edge. A blunt knife does not cut an object easily because due to its thicker
edge, the force of our hand falls over a large area of the object and produces lesser pressure.
This lesser pressure cuts the object with difficulty.
(c) A nail has a pointed tip, so that when it is hammered the force of the hammer falls on a very
small area of the wood or the wall creating a large pressure which pushes the nail into the
wood or the wall.
(d) The foundation of buildings and dams are laid on a larger area of ground so that the weight of
the building or dam produces less pressure on ground and they may not sink into the ground.
20. The mass of 2 m3 of steel is 15600 kg. Calculate the density of steel in S.I. units.
mass
Sol: We know that density =
volume
15600 kg
= = 7800 kg/m 3
2 m3
21. When an aluminium object is immersed in water it displaces 5 kg of water. How much is the
buoyant force acting on the aluminium object in Newton? (g = 10 m/s 2 )
Sol: According to Archimedes’ principle, the buoyant force acting on the aluminium object will be equal
to the weight of water displaced by this aluminium object.
Now, W = m  g
m = 5 kg , g = 10 m/s 2
W = 5  10 = 50 N
The weight of water displaced by the aluminium object is 50 Newton, therefore, the buoyant force
acting on the aluminium object will also be 50 N.
22. An object of mass 50 g has a volume of 20 cm3. Calculate the density of the object. If the
density of water be 1g/cm 3 . State whether the object will float or sink in water.
mass
Sol: (a) Density =
volume
Mass of the object = 50 g
Volume of the object = 20 cm3
50 g
Density = = 2.5 g/cm 3
20 cm3
density of the object
(b) Relative density =
density of water
2.5 g/cm 3
= 3
= 2.5 g/cm3
1g/cm
Since, the relative density of the object is more than 1, then it will be heavier than water and hence
it will sink in water.
23. The relative density of silver is 10.8. If the density of water be 1 10 3 kg/m 3 . Calculate the
density of silver is S.I. units.
Density of the substance
Sol: Relative Density =
Density of water
Relative density of silver = 10.8
Density of silver = ?
Density or water = 1 10 3 kg/ m 3

Density of silver
10 .8 =
1 10 3
Density of silver = 10 .8  10 3 kg/m 3
24. The volume of a solid mass 500g is 350 cm 3 .

(a) What will be the density of this solid?
(b) What will be the mass of water displaced by this solid?
(c) What will be the relative density of the solid?
(d) Will it float or sink in water?
Mass 500
Sol: (a) Density = = = 1.42 g/cm 3
Volume 350
(b) The solid will displace water equal to its volume. Since the volume of solid is
350 cm3 so it will displace 350 cm3 of water. Now volume of water displaced is
350 cm3 and the density of water is 1g/cm3 .
mass of water
Density of water =
volume of water
mass of water
1g/cm 3 =
350 cm3
Mass of water = 1g/cm3  350 cm3 = 350 g
Density of solid 1.42 g/cm 3

(c) Relative density = = = 1.42
Density of water 1 g/cm 3
(d) Since the relative density of this solid is greater than relative density of water i.e.,
1.42 > 1, therefore this solid is heavier than water and hence it will sink in water.
CROSSWORD PUZZLE
1 2
3 4
6 7
8 9
10
11
12 13
14
Across: Down:
1 All objects with ______________ are 2 The gravitational force is ALWAYS

affected by gravity. _____________.
6 Newton's Law of Universal __________ 3 The force due to gravity between

states that all objects with mass are the earth and an object is referred
attracted to all other objects with mass. to as the object's _________when
the object is on the earth.
7 If you double the distance between two
objects without changing the mass of 4 If we double the masses of two
either object, the force due to gravity stationary objects, the
between the objects decreases by a gravitational force between the two
factor of _________. objects will ___________.
8 Newton did not discover gravity; he 5 If we double the mass of one

discovered that gravity is object without changing the mass
__________________. of another and without changing
the distance between the two
10 The gravitational constant G is equal to objects, the gravitational force
6.67 x 10^-11 N*m^2/kg^2 and so is a(n) between the two objects will
_______________ number. _______________.
11 As the distance between two objects 9 As the masses of two stationary

increases, the gravitational force objects increase, the gravitational
between them _________________, if force between them
the masses of the objects remain __________________.
constant.
13 According to Newton's
12 If two objects have constant masses and _______________ Law of Motion,
remain at a constant distance apart, the the gravitational forces exerted by
gravitational force between them two objects on each other must be
remains _________________. equal in strength and opposite in
direction.
14 The force due to gravity follows the
__________________ Square Law
because as the distance between two
objects increases, the gravitational
force between them decreases as the
square of the distance between their
centers.
EXERCISE
OBJECTIVE QUESTIONS:
LEVEL – I
1. The weight of an object on the moon will be about ___________ of what it is on the earth.
(a) 1/7 (b) 1/10 (c) 1/6 (d) 1/16
2. The S.I. unit of mass is

(a) kg (b) Newton (c) m/s (d) m/s 2
3. Thrust per unit area is called

(a) Force (b) Pressure (c) Man (d) Weight
4. A fluid exerts pressure

(a) In all directions (b) In upward direction only
(c) In downward direction only (d) None of these
5. The maximum loss in weight of an object takes place when

(a) It is ¼th immersed in water (b) It is partially immersed in water
(c) It is ⅓rd immersed in water (d) it is completely immersed in water
6. The magnitude of buoyant force acting on an object immersed is a liquid depends on

(a) Volume of object and density of liquid (b) Volume of liquid and density of object
(c) Only volume of object (d) Only density of liquid
7. If the buoyant force exerted by the liquid is less than the weight of the object, then the object will
(a) Sink in the liquid (b) Float in the liquid
(c) Rise in the liquid and then float (d) None of the above
8. A ship made of iron and steel floats in water because

(a) Average density of ship is more than that of water
(b) Average density of ship is less than that of water
(c) Average density of ship is equal to that of water
(d) None of the above
9. Density of water is
(a) 10 kg/ m 3 (b) 100 kg/ m 3 (c) 1000 kg / m 3 (d) 10000 kg/ m 3
10. The S.I. unit of relative density is

(a) g/cm3 (b) kg/m3 (c) Newton (d) None
11. A stone dropped from a building top takes 4 s to reach the ground. The height of the building is
(a) 78.4 m (b) 19.6 m (c) 156.8 m (d) 39.2 m
12. A stone is dropped from a cliff. Its speed after it has fallen 400 m is
(a) 9.8 m/s (b) 88.4 m/s (c) 19.6 m/s (d) 98 m/s
13. When an object is thrown up, the force of gravity

(a) is opposite to direction of motion (b) is in the same direction of motion
(c) becomes zero at the highest point (d) increases as it rises up.
14. A ball is thrown up and attains a maximum height of 100 m. Its initial speed was
(a) 9.8 m/s (b) 44.2 m/s (c) 19.8 m/s (d) 24.6 m/s
15. The weight of an object

(a) is the gravity of the matter it contains
(b) refers to its inertia
(c) is the same as its mass but expressed in different units
(d) is the force with which it is attracted to the earth
16. The SI unit of gravitational constant is

(a) N (b) ms/2 (c) J (d) Nm2 kg-2
17. The mass of a body is measured to be 12 kg on Earth. If it is taken to Moon, the mass will be
(a) 2 kg (b) 72 kg (c) 2 N (d) 12 kg
18. Pascal is a unit of

(a) pressure (b) force (c) linear momentum (d) energy
19. Which of the following solids is the densest of all?

(a) Lead (b) Gold (c) Silver (d) Tungsten
20. Which of the following gas is the densest of all?

(a) Air (b) Chlorine (c) Ozone (d) Argon
LEVEL – II
1. The escape velocity of a body projected vertically upwards from the surface of earth is ‘v’ if the
body is projected at an angle of 30 o with the horizontal, the escape velocity would be
3
(a) v / 2 (b) v (c) 2v (d) v
2
2. If both the mass and radius of the earth decrease by 1%, the acceleration due to gravity on the surface
of the earth will
(a) decrease by 1% (b) increase by 1% (c) increase by2% (d) remain unchanged
3. A satellite revolves around a planet with a speed ‘v’ in circular orbit of radius ‘r’. if ‘R’ is the radius
of the planet, the acceleration due to gravity on its surface is
(a) g = v 2 r / R 2 (b) g = v 2 R / r (c) g = v2 / r (d) g = v 2 / R
4. The time of revolution of satellite is ‘T’. Its kinetic energy is proportional to

(a) 1/T (b) 1 2 (c) 1 3 (d) T −2/3
T T
5. A satellite is launched into a circular orbit of radius ‘R’ around the earth. A second satellite is
launched into an orbit of radius 1.01 R. The period of the second satellite is longer than that of first
by approx.
(a) 0.5% (b) 1% (c) 1.5% (d) 3%
6. a comet is moving in a highly elliptical orbit round the sun. when it is closest to the sun, its distance
from the sun is ‘r’ and its speed is ‘v’ when it is farthest from the sun, its distance from the sun is ‘R’
and its speed will be
(a) v ( r / R ) (b) v ( r / R ) (c) v ( r / R )
1 3
(d) v ( r / R )
2 2 2
7. A satellite is revolving around the earth in a circular orbit of radius ‘r’ then which is wrong statement
(a) kinetic energy varies as 1/r (b) angular momentum varies as 1/ r
(c) linear momentum varies as 1 (d) frequency of revolution varies as 1 3
r r 2
8. A body weights 63 N on the surface of the earth. How much will it weight at a height equal to half
the radius of the earth?
(a) 63 N (b) 32.5 N (c) 28 N (d) None
9. A block floats in a liquid contained in a beaker. The beaker is placed on the floor of an elevator. If
the elevator descends with acceleration a (<g) the upthrust on the block due to the liquid
(a) is equal to the weight of the liquid displaced

(b) is greater than weight of the liquid displaced
(c) is less than the weight of the liquid displaced
(d) becomes equal to the zero
10. The stone of relative density ‘k’ is released from rest on the surface of lake. If viscous effects are
ignored the stone sinks in water with an acceleration of
(a) g (1 − k ) (b) g (1 + k ) (c) g (1 − 1/ k ) (d) g 1 + 1
k ( )
SUBJECTIVE QUESITONS - I
1. What is the value of acceleration due to gravity in S.I. units?
2. What is the value of ‘G’?
3. What is the S.I. unit of weight?
4. What is the relation between mass and weight?
5. What is the weight of body of man with mass 1kg?
6. Is weight a vector or a scalar quantity?
7. What is the relation between g and G?
8. Where does a body weigh more, at poles or at equator? Why?
9. The earth’s gravitational force causes an acceleration of 5 m/s 2 in a 1 kg man somewhere in

space. How much would be the acceleration of a body of man 3 kg at a place on earth.
10. Define gravitation and gravity.
11. Distinguish between mass and weight of a body.
12. What is weightlessness in space? Discuss the weightlessness in spaceship.
13. State Newton’s law of gravitation. Give its mathematical expression.
14. Explain why buildings and dams have wide foundation.
15. What happens to the buoyant force as more and more volume of a solid object is immersed in a
liquid? When does the buoyant force become maximum?
16. Why does a ship made of iron and steel floats in water whereas a small piece of iron sinks is it?
17. Explain why, a piece of glass sinks in water but it floats in mercury.
18. Does the acceleration produced in a freely falling body depend on the mass of the body? Explain.
19. Why objects float or sink when placed on the surface of water?
20. List some phenomenon explained on the basis of universal law of gravitation.
SUBJECTIVE QUESITONS - II
1. State the universal law of gravitation.
2. Write the formula to find the magnitude of the gravitational force between the earth and an
object on the surface of the earth.
3. How does the force of gravitation between two objects change when the distance between them is
reduced to half?
4. What is the magnitude of the gravitational force between the earth and a 1kg object on its
surface? (Mass of earth is 6  10 24 kg and radius of the earth is 6.4  10 6 m )
5. What do you mean by free fall? Explain
6 Define thrust, what is its unit?
7. Explain why a wide steel belt is provided over the wheels of an army tank.
8. Explain why it is easier to walk in soft sand if you have flat shoes rather than shoes with sharp
heels.
9. Why the tractors have broad tyres?
10. An object weighs 10N when measured on the surface of the earth. What would be its weight
when measured on the surface of the moon?
11. How much would a 70 kg man weight on the moon? What would be his mass on the earth?
12. What is the difference between the mass of an object and its weight?
13. Why is the weight of an object on the moon 1/6th its weight on the earth?
14. A stone is dropped from a building and it reaches the ground 2.5s later. How high is the building?
(g = 9.8 m/s 2 )
15. When a stone is thrown vertically upwards, it reaches a maximum height of 5m.
(a) What was the initial speed of the stone?
(b) How much time is taken by the stone to reach the highest point? (g = 10 m/s 2 )
16. A rectangular wooden block has mass of 4 kg. The length, breadth and height of this wooden
block are 50 cm, 25 cm and 10 cm respectively. Find the pressure on the table.
(a) When the wooden block is kept with its surface measuring 50 cm × 25 cm on the table.
(b) When the wooden block is kept with its surface measuring 25 cm × 10 cm on the table.
(g = 10 m/s 2 )
17. A cricket ball is dropped from a height of 20 metres.

(a) Calculate the speed of the ball when it hits the ground.
(b) Calculate the time it takes to fall through this height. (g = 10 m/s 2 )
18. Define density. What is its S. I. unit?
19. Define relative density. What is its S.I. unit?
20. What is the value of relative density of water?
21. Define buoyant force? Name two factors an which buoyant force depends.
22. State Archimedes’ Principle.
23. The density of turpentine is 840 kg/m 3 . What will be its relative density? (Density of water
= 1000 kg/m 3 )
24. When a golf ball is lowered into a measuring cylinder containing water, the water level rises by
30 cm3 , when the ball is completely submerged. If the mass of ball in air is 33 g, find its density.
25. When a boat is partially immersed in water it displaces 600 kg of water. How much is the buoyant
force acting on the boat in Newton’s. (g = 10 m/s 2 )
26. A mug full of water appears light when immersed is a bucket full of water than when it is outside
water. Why?
CONCEPT MAP
The force with which earth pulls The falling of a body from a
the objects towards it is called height towards the earth under
the gravitational force of earth the gravitational force of earth
or gravity. is called free fall.
Newton’s law of gravitation

states that every object in the
universe attracts every other
object with a certain force The uniform acceleration
which is directly proportional to produced in a freely falling
body due to the gravitational
the product of their masses and
GRAVITATION
force of the earth is known as
inversely proportional to the acceleration due to gravity.
square of the distance between
g = 9.8 m/s2
them
m m
F =G 1 2 2
R
Equation of motion for freely
falling bodies
‘G’ is the universal gravitational
v = u + gt
constant whose value is equal
to 6.67  10 −11Nm 2 / kg 2
h = ut + ½gt 2
v 2 = u 2 + 2gh
Archimedes’ principle states
that when a body is partially or
wholly immersed in a liquid, it
experiences an upthrust which
Force
is equal to weight of liquid Pressure =
displaced by the object. Area
Density of a substance is mass The weight of a body is the

per unit volume. Its unit is force with which it is attracted
(g/cm 3 ) or (kg/m3) towards the earth.
Mass of a body is the quantity Relative density of a substance

of matter contained in it. is the ratio of density of
substance to that of water.

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112

GRAVITATION & FLUIDS IX -Physics

NEWTON’S UNIVERSAL LAW OF GRAVITATION

Combining equation (i) and (ii)

The accepted value of ‘G’ is 6.673 x 10-11 Nm2/kg2.

Importance of the universal law of gravitation

Gravitational Force between Light Objects and Heavy Object

Taking G = 6.67  10−11 Nm2 / kg2, we obtain gravitational force of attraction,

Gm1m 2 6.67  10 −11  1 1

(b) When a body of mass 1 kg is held on the surface of Earth.

m 2 = mass of Earth = 6  1024 kg

r = distance of body from centre of Earth

= radius of Earth = 6400 km = 6.4  103 km = 6.4  106 m

Gm1m 2 6.67  10 −11  1 6  10 24

(c) When both the bodies are heavy.

Gravitation and Newton's Third Law of Motion

force = mass × acceleration

Fill in the blanks

increased by a factor of 2, then the Fgrav is ___________ by a factor of _____.

increased by a factor of 3, then the Fgrav is ___________ by a factor of _____.

increased by a factor of 4, then the Fgrav is ___________ by a factor of _____.

If the mass of the object is ...

increased by a factor of 2, then the Fgrav is increased by a factor of _____.

increased by a factor of 3, then the Fgrav is ___________ by a factor of _____.

decreased by a factor of 4, then the Fgrav is ___________ by a factor of _____.

If the mass of the Earth is ...

increased by a factor of 3, then the Fgrav is ___________ by a factor of _____.

decreased by a factor of 4, then the Fgrav is ___________ by a factor of _____.

Acceleration Due To Gravity

Relation between ‘g’ and ‘G’

Variation in Acceleration due to Gravity (G)

(a) Effect of shape of Earth on 'g'

Acceleration due to gravity at equator, g e = GM2 = minimum, as R e is maximum.

(b) Effect of height above the surface of Earth on 'g'

Gravity at height h is given by

(c) Effect of depth below the surface of Earth on 'g'

Hence, we conclude that value of acceleration due to gravity is

Equations of Motion of Freely Falling Bodies

Mass and Weight

Weight of a freely falling body

Weight of an object on the moon

Kepler’s Laws of Planetary Motion

FLUIDS, THRUST AND PRESSURE

Why is the tip of a needle sharp?

The thrust on unit area is called pressure. Thus,

Density of the substance

Factors affecting buoyant force

(b) Density of the liquid.

Activity to understand Archimedes’ principle

Applications of Archimedes’ principle

Why objects float or sink in a liquid?

(a) Sinking of an object in water

(b) Floating of an object in water

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

5. A mass of 50 kg attracted by a mass of 20 kg lying at a distance of 2m with a force of

G = 6.7  10 −11 Nm 2 /kg 2 )

9. Mass of an object is 10 kg. What is its weight on the earth?

10. What is the mass of object whose weight is 49 N?

increased by a factor of 2, then the Fgrav is _______ by a factor of _.

increased by a factor of 3, then the Fgrav is _______ by a factor of _.

increased by a factor of 4, then the Fgrav is _______ by a factor of _.

increased by a factor of 3, then the Fgrav is _______ by a factor of _.

decreased by a factor of 4, then the Fgrav is _______ by a factor of _.

increased by a factor of 3, then the Fgrav is _______ by a factor of _.

decreased by a factor of 4, then the Fgrav is _______ by a factor of _.