SMART EDUCATIONS
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GRAVITATION
Q. What is gravitation?
Ans. Every object in the universe attracts each other with a force and the force is called
gravitation. When one of the objects is the earth then the gravitation is called just
gravity.
(Gravitational force can only be seen in the case of an object having huge mass
and in the other cases there is gravitation but canโ€™t be seen or felt)
Q. What is gravity?
Ans. Gravity is an attraction force between an object and the earth.
Q. What is the reason behind the revolving of all the planets around the sun?
Ans. It is the gravitational force that makes all the planets revolve around the sun. In
this case it is also known as centripetal force.
Q. What is centripetal force?
Ans. Centripetal force is a force that makes an object revolve around in a circular path
and the force is applied at the center of the circular path.
Q. States the universal law of gravitation.
Ans. Universal law of gravitation states that
โ€œ Every object in the universe attracts every other object with a force which is
proportional to the product of their masses and inversely proportional to the square of
the distance between them. The force is along the line joining the centers of two
objects. โ€
๐น =
๐บ๐‘€๐‘š
๐‘‘
2
Where F is the gravitational force, G is the universal gravitational constant, M is the
mass of the 1
st
object, m is the mass of the 2
nd
object and d is the distance between the
two objects.
Q. Derive the expression of the universal law of gravitation.
Ans. According to the universal law of gravitation
1. Every object in the universe attracts every other object with a force which is
proportional to the product of their masses
๐น โˆ ๐‘€๐‘š ( ๐‘– )
2. and inversely proportional to the square of the distance between them. The force
is along the line joining the centers of two objects.
๐น โˆ
1
๐‘‘
2
( ๐‘–๐‘– )
๐‘†๐‘œ , ๐‘“๐‘Ÿ๐‘œ๐‘š ( ๐‘– ) ๐‘Ž๐‘›๐‘‘ ( ๐‘–๐‘– )
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We get:-
๐น โˆ
๐‘€๐‘š
๐‘‘
2
Or,
๐น =
๐บ๐‘€๐‘š
๐‘‘
2
( ๐‘คโ„Ž๐‘’๐‘Ÿ๐‘’ ๐บ ๐‘–๐‘  ๐‘๐‘œ๐‘›๐‘ ๐‘ก๐‘Ž๐‘›๐‘ก ๐‘œ๐‘“ ๐‘๐‘Ÿ๐‘œ๐‘๐‘œ๐‘Ÿ๐‘ก๐‘–๐‘œ๐‘›๐‘Ž๐‘™๐‘–๐‘ก๐‘ฆ )
Derived
๐น =
๐บ๐‘€๐‘š
๐‘‘
2
๐‘–๐‘  ๐‘กโ„Ž๐‘’ ๐‘’๐‘ฅ๐‘๐‘Ÿ๐‘’๐‘ ๐‘ ๐‘–๐‘œ๐‘› ๐‘œ๐‘“ ๐‘กโ„Ž๐‘’ ๐‘ข๐‘›๐‘–๐‘ฃ๐‘’๐‘Ÿ๐‘ ๐‘Ž๐‘™ ๐‘™๐‘Ž๐‘ค ๐‘œ๐‘“ ๐‘”๐‘Ÿ๐‘Ž๐‘ฃ๐‘–๐‘ก๐‘Ž๐‘ก๐‘–๐‘œ๐‘› .
Q. What is G?
Ans. G is the universal gravitation constant. SI unit of G is . The value of G is
๐‘ ๐‘š
2
๐‘˜๐‘”
โˆ’ 2
measured by Henry Cavendish and the value of G is . The value
6 . 673 ร— 10
โˆ’ 11
๐‘ ๐‘š
2
๐‘˜๐‘”
โˆ’ 2
of G is constant i.e does not change.
Q. The mass of the earth is and that of the moon is . If the
6 ร— 10
24
๐‘˜๐‘” 7 . 4 ร— 10
22
๐‘˜๐‘”
distance between the earth and the moon is , calculate the force exerted
3 . 84 ร— 10
5
๐พ๐‘š
by the earth on the moon.
๐บ = 6 . 67 ร— 10
โˆ’ 11
๐‘ ๐‘š
2
๐‘˜๐‘”
โˆ’ 2
Ans.
๐‘€ = 6 ร— 10
24
๐‘˜๐‘”
๐‘š = 7 . 4 ร— 10
22
๐‘˜๐‘”
๐‘‘ = 3 . 84 ร— 10
5
๐พ๐‘š
= 3 . 84 ร— 10
5
ร— 10
3
๐‘š
= 3 . 84 ร— 10
8
๐‘š
๐บ = 6 . 67 ร— 10
โˆ’ 11
๐‘ ๐‘š
2
๐‘˜๐‘”
โˆ’ 2
๐น =
๐บ๐‘€๐‘š
๐‘‘
2
๐น =
6 . 67 ร— 10
โˆ’ 11
ร— 6 ร— 10
24
ร— 7 . 4 ร— 10
22
3 . 84 ร— 10
8
( )
2
๐น =
6 . 67 ร— 6 ร— 7 . 4 ร— 10
22
ร— 10
โˆ’ 11
ร— 10
24
3 . 84 ร— 10
8
ร— 3 . 84 ร— 10
8
๐น =
6 . 67 ร— 6 ร— 7 . 4 ร— 10
22 +โˆ’ 11 + 24 โˆ’ 8 โˆ’ 8
3 . 84 ร— 3 . 84
๐น =
6 . 67 ร— 3 ร— 3 . 7 ร— 10
19
1 . 92 ร— 1 . 92
๐น =
74 . 037 ร— 10
19
3 . 6864
๐น = 20 . 0838 ร— 10
19
๐น = 2 . 0083 ร— 10
20
๐น = 2 . 01 ร— 10
20
๐‘
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Q. Write the importance of the universal law of gravitation?
Ans. The importance of the universal law of gravitation are as follows
1. It helped us to know why we are bound to the earth.
2. It helped us to know the reason behind the motion of the celestial bodies.
3. It helped us to know the reason behind the production of the tides.
Q. Write the formula to find the magnitude of the gravitational force between the earth
and an object on the surface of the earth.
Ans.
๐น =
๐บ๐‘€๐‘š
๐‘Ÿ
2
( ๐‘Ÿ = ๐‘Ÿ๐‘Ž๐‘‘๐‘–๐‘ข๐‘  ๐‘œ๐‘“ ๐‘กโ„Ž๐‘’ ๐‘’๐‘Ž๐‘Ÿ๐‘กโ„Ž )
Q. What is Free Fall?
Ans. When an object falls due to only gravity then the fall is called free fall.
Q. What is gravitational acceleration?
OR
What do you mean by acceleration due to gravity?
Ans. During the free fall there is a constant acceleration on the object due to the gravity
and the acceleration is called the gravitational acceleration and is represented as โ€œgโ€.
Q. What is โ€œgโ€?
Ans. โ€œgโ€ is the gravitational acceleration. The SI unit of the g is . The value of g is
๐‘š / ๐‘ 
2
different for different places. For the earth it is . Even on the earth the value of
9 . 8 ๐‘š / ๐‘ 
2
g is not constant. On the pole end the value of g increases and vice-versa, the higher
the height of an object from the earth the lower is the value of g and vice-versa and this
is because the value of g is inversely proportional to the distance between the objects.
๐‘” =
๐บ๐‘€
๐‘Ÿ
2
M is the mass of the object of which g is to be calculated and r is the radius of the
object.
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Q. What are the differences between โ€œGโ€ and โ€œgโ€?
Ans.
G
g
It is the universal gravitation constant.
It is the gravitational acceleration.
๐บ =
๐น ๐‘‘
2
๐‘€๐‘š
๐‘” =
๐บ๐‘€
๐‘Ÿ
2
SI unit is
๐‘ ๐‘š
2
๐‘˜๐‘”
โˆ’ 2
SI unit is
๐‘š ๐‘ 
โˆ’ 2
Value =
6 . 673 ร— 10
โˆ’ 11
๐‘ ๐‘š
2
๐‘˜๐‘”
โˆ’ 2
.
Value = on the earth.
9 . 8 ๐‘š / ๐‘ 
2
Its value is constant.
Its value is variable.
Q. Find the value of g on the earth.
Ans.
๐‘€ = 6 ร— 10
24
๐‘˜๐‘” ( ๐‘š๐‘Ž๐‘ ๐‘  ๐‘œ๐‘“ ๐‘กโ„Ž๐‘’ ๐‘’๐‘Ž๐‘Ÿ๐‘กโ„Ž )
๐‘Ÿ = 6 . 4 ร— 10
6
๐‘š ( ๐‘Ÿ๐‘Ž๐‘‘๐‘–๐‘ข๐‘  ๐‘œ๐‘“ ๐‘กโ„Ž๐‘’ ๐‘’๐‘Ž๐‘Ÿ๐‘กโ„Ž )
๐บ = 6 . 67 ร— 10
โˆ’ 11
๐‘ ๐‘š
2
๐‘˜๐‘”
โˆ’ 2
๐‘” =
๐บ๐‘€
๐‘Ÿ
2
๐‘” =
6 . 67 ร— 10
โˆ’ 11
ร— 6 ร— 10
24
6 . 4 ร— 10
6
( )
2
๐‘” =
6 . 67 ร— 6 ร— 10
โˆ’ 11 + 24
6 . 4 ร— 10
6
ร— 6 . 4 ร— 10
6
๐‘” =
6 . 67 ร— 6 ร— 10
โˆ’ 11 + 24 โˆ’ 6 โˆ’ 6
6 . 4 ร— 6 . 4
๐‘” =
6 . 67 ร— 6 ร— 10
6 . 4 ร— 6 . 4
๐‘” =
6 . 67 ร— 3 ร— 10
6 . 4 ร— 3 . 2
๐‘” = 9 . 77
๐‘” = 9 . 8 ๐‘š ๐‘ 
โˆ’ 2
Q. A car falls off a ledge and drops to the ground in 0.5 s. Let (for simplifying the
๐‘” = 10 ๐‘š ๐‘ 
โˆ’ 2
calculations). (i) What is its speed on striking the ground? (ii) What is its average speed during
the 0.5 s? (iii) How high is the ledge from the ground?
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Ans.
๐‘ข = 0 ๐‘š / ๐‘ 
๐‘ก = 0 . 5 ๐‘ 
๐‘” = 10 ๐‘š ๐‘ 
โˆ’ 2
๐‘ฃ = ?
๐ด๐‘ฃ๐‘’๐‘Ÿ๐‘Ž๐‘”๐‘’ ๐‘ ๐‘๐‘’๐‘’๐‘‘ = ?
๐‘  = ?
๐‘ฃ = ๐‘ข + ๐‘”๐‘ก
๐‘ฃ = 0 + 10 ร— 0 . 5
Ans(i)
๐‘ฃ = 5 ๐‘š / ๐‘ 
๐ด๐‘ฃ๐‘’๐‘Ÿ๐‘Ž๐‘”๐‘’ ๐‘ ๐‘๐‘’๐‘’๐‘‘ =
๐‘ข + ๐‘ฃ
2
๐ด๐‘ฃ๐‘’๐‘Ÿ๐‘Ž๐‘”๐‘’ ๐‘ ๐‘๐‘’๐‘’๐‘‘ =
0 + 5
2
Ans(ii)
๐ด๐‘ฃ๐‘’๐‘Ÿ๐‘Ž๐‘”๐‘’ ๐‘ ๐‘๐‘’๐‘’๐‘‘ = 2 . 5 ๐‘š / ๐‘ 
๐‘  = ๐‘ข๐‘ก +
1
2
๐‘” ๐‘ก
2
๐‘  = 0 ร— 0 . 5 +
1
2
ร— 10 ร— 0 . 5 ร— 0 . 5
๐‘  = 0 + 5 ร— 0 . 5 ร— 0 . 5
Ans(iii)
๐‘  = 1 . 25 ๐‘š
Q. An object is thrown vertically upwards and rises to a height of 10 m. Calculate (i) the
velocity with which the object was thrown upwards and (ii) the time taken by the object
to reach the highest point.
Ans.
๐‘ฃ = 0 ๐‘š / ๐‘ 
๐‘  = 10 ๐‘š
๐‘ข = ?
๐‘ก = ?
๐‘ฃ
2
โˆ’ ๐‘ข
2
= 2 ๐‘”๐‘ 
0
2
โˆ’ ๐‘ข
2
= 2 ร— โˆ’ 9 . 8 ( ) ร— 10
โˆ’ ๐‘ข
2
=โˆ’ 196
๐‘ข = 196
Ans(i)
๐‘ข = 14 ๐‘š / ๐‘ 
๐‘ฃ = ๐‘ข + ๐‘”๐‘ก
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0 = 14 + โˆ’ 9 . 8 ( ) ร— ๐‘ก
0 = 14 โˆ’ 9 . 8 ๐‘ก
14 โˆ’ 9 . 8 ๐‘ก = 0
โˆ’ 9 . 8 ๐‘ก = 0 โˆ’ 14
๐‘ก =
โˆ’ 14
โˆ’ 9 . 8
๐‘ก = 1 . 428
Ans(ii)
๐‘ก = 1 . 43 ๐‘ 
Q. Define the mass of an object?
Ans. Inertia of an object is the mass of the object. Mass of an object is represented by
โ€œmโ€. Its SI unit is kg.
Q. Define the weight of an object?
Ans. The gravitational force between the place (on which weight is to be calculated)
and an object (of which the weight is to be calculated) is called its weight. Weight is
represented as โ€˜Wโ€™. SI unit of weight is N.
๐‘Š = ๐‘š๐‘”
For the earth:- โ€œ Weight of an object is the force by which it is attracted towards
the earth.โ€
Q. What are the differences between mass and weight of an object?
Ans.
Mass
Weight
It is the inertia.
It is the force.
Its SI unit is kg.
Its SI unit is N.
It is constant.
It is variable.
It is represented as โ€˜mโ€™.
It is represented as โ€˜Wโ€™.
Q. How to calculate the weight of an object on the moon?
Ans . To calculate the weight of an object in any place we must know the value of โ€˜gโ€™ in
that place and to find g we must know the mass of the body(moon) and its radius.
In case of moon:-
๐‘š = 7 . 36 ร— 10
22
๐‘˜๐‘”
๐‘Ÿ = 1 . 74 ร— 10
6
๐‘š
๐‘” =
๐บ๐‘€
๐‘Ÿ
2
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๐‘” =
6 . 67 ร— 10
โˆ’ 11
ร— 7 . 36 ร— 10
22
1 . 74 ร— 10
6
( )
2
๐‘” =
6 . 67 ร— 7 . 36 ร— 10
โˆ’ 11 + 22
1 . 74 ร— 10
6
ร— 1 . 74 ร— 10
6
๐‘” =
6 . 67 ร— 7 . 36 ร— 10
โˆ’ 11 + 22 โˆ’ 6 โˆ’ 6
1 . 74 ร— 1 . 74
๐‘” =
6 . 67 ร— 7 . 36 ร— 10
โˆ’ 1
1 . 74 ร— 1 . 74
๐‘” =
6 . 67 ร— 1 . 84
0 . 87 ร— 0 . 87 ร— 10
๐‘” =
6 . 67 ร— 1 . 84
0 . 87 ร— 0 . 87 ร— 10
๐‘” = 1 . 621 ๐‘š ๐‘ 
โˆ’ 2
The value of g on the moon is approximately of the value of g on the earth and hence
1
6
weight of an object on the moon is approximately of the weight of that object on the
1
6
earth.
๐‘Š
๐‘š
=
๐‘Š
๐‘’
6
Q. Mass of an object is 10 kg. What is its weight on the earth?
Ans.
๐‘š = 10 ๐‘˜๐‘”
๐‘Š = ?
๐‘Š = ๐‘š๐‘”
๐‘Š = 10 ร— 9 . 8
Ans
๐‘Š = 98 ๐‘
Q. An object weighs 10 N when measured on the surface of the earth. What would be
its weight when measured on the surface of the moon?
Ans.
๐‘Š
๐‘’
= 10 ๐‘
๐‘Š
๐‘š
= ?
๐‘Š
๐‘š
=
๐‘Š
๐‘’
6
๐‘Š
๐‘š
=
10
6
Ans.
๐‘Š
๐‘š
= 1 . 67 ๐‘
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Q. Why is the weight of an object on the moon its weight on the earth?
1
6
( )
๐‘กโ„Ž
Ans. Weight of any object is the product of its mass and gravitational acceleration.
Because mass is constant so Weight of any object is directly proportional to the
gravitational acceleration. Gravitational acceleration on the moon is of the
1
6
( )
๐‘กโ„Ž
earthโ€™s gravitational acceleration, so the weight of an object on the moon its
1
6
( )
๐‘กโ„Ž
weight on the earth.
Q. What is Thrust?
Ans. Thrust is a perpendicular force on any surface. Its SI unit is N.
Q. What is Pressure?
Ans. Pressure is the force acting on a per unit area of the surface.
๐‘ƒ๐‘Ÿ๐‘’๐‘ ๐‘ ๐‘ข๐‘Ÿ๐‘’ =
๐‘‡โ„Ž๐‘Ÿ๐‘ข๐‘ ๐‘ก
๐ด๐‘Ÿ๐‘’๐‘Ž
It SI unit is or just Pascal(Pa) . Pressure is inversely proportional to the surface
๐‘ / ๐‘š
2
area.
Q. A block of wood is kept on a tabletop. The mass of the wooden block is 5 kg and its
dimensions are 40 cm ร— 20 cm ร— 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
ร— 10 cm and (b) 40 cm ร— 20 cm.
Ans.
(a)
๐‘š = 5 ๐‘˜๐‘”
๐ด๐‘Ÿ๐‘’๐‘Ž = 20 ร— 10 ๐‘๐‘š
2
= 200 ๐‘๐‘š
2
=
200
100 ร— 100
๐‘š
2
= 0 . 02 ๐‘š
2
๐‘ƒ๐‘Ÿ๐‘’๐‘ ๐‘ ๐‘ข๐‘Ÿ๐‘’ =
๐‘‡โ„Ž๐‘Ÿ๐‘ข๐‘ ๐‘ก
๐ด๐‘Ÿ๐‘’๐‘Ž
๐‘ƒ๐‘Ÿ๐‘’๐‘ ๐‘ ๐‘ข๐‘Ÿ๐‘’ =
๐‘š๐‘”
0 . 02
๐‘ƒ๐‘Ÿ๐‘’๐‘ ๐‘ ๐‘ข๐‘Ÿ๐‘’ =
5 ร— 9 . 8
0 . 02
๐‘ƒ๐‘Ÿ๐‘’๐‘ ๐‘ ๐‘ข๐‘Ÿ๐‘’ =
49
0 . 02
๐‘ƒ๐‘Ÿ๐‘’๐‘ ๐‘ ๐‘ข๐‘Ÿ๐‘’ = 2450 ๐‘ƒ๐‘Ž
(b)
๐‘š = 5 ๐‘˜๐‘”
๐ด๐‘Ÿ๐‘’๐‘Ž = 40 ร— 20 ๐‘๐‘š
2
= 800 ๐‘๐‘š
2
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=
800
100 ร— 100
๐‘š
2
= 0 . 08 ๐‘š
2
๐‘ƒ๐‘Ÿ๐‘’๐‘ ๐‘ ๐‘ข๐‘Ÿ๐‘’ =
๐‘‡โ„Ž๐‘Ÿ๐‘ข๐‘ ๐‘ก
๐ด๐‘Ÿ๐‘’๐‘Ž
๐‘ƒ๐‘Ÿ๐‘’๐‘ ๐‘ ๐‘ข๐‘Ÿ๐‘’ =
๐‘š๐‘”
0 . 08
๐‘ƒ๐‘Ÿ๐‘’๐‘ ๐‘ ๐‘ข๐‘Ÿ๐‘’ =
5 ร— 9 . 8
0 . 08
๐‘ƒ๐‘Ÿ๐‘’๐‘ ๐‘ ๐‘ข๐‘Ÿ๐‘’ =
49
0 . 08
๐‘ƒ๐‘Ÿ๐‘’๐‘ ๐‘ ๐‘ข๐‘Ÿ๐‘’ = 612 . 5 ๐‘ƒ๐‘Ž
Q. What is Buoyancy ?
Ans. Buoyancy is a thrust which is exerted by a fluid to an object which is on the
surface of the fluid or immersed in the fluid. It always acts in the upward direction so it is
also known as upthrust.
Q. Define density?
Ans. Density is the mass per unit volume. Its SI unit is
๐‘˜๐‘” / ๐‘š
3
๐ท๐‘’๐‘›๐‘ ๐‘–๐‘ก๐‘ฆ =
๐‘€๐‘Ž๐‘ ๐‘ 
๐‘‰๐‘œ๐‘™๐‘ข๐‘š๐‘’
Q. Why do objects float or sink when placed on the surface of water?
Ans. When the density of an object is more than that of the water then the object will
sink but if the density of an object is less than that of the water then the object will float.
OR
When the gravitational force exerted by the earth in the object is more than that of the
buoyancy force exerted by the water in the object then the object will sink and
vice-versa.
Q. Why is it difficult to hold a school bag with a strap made of a thin and strong string?
Ans. Strap of the bag is in contact with the holderโ€™s shoulder so it applies pressure on
the holderโ€™s shoulder. A thin strap has less surface area and as we know that the less
the surface area is more the pressure so a school bag with a strap made of a thin and
strong string applies more pressure in the holderโ€™s shoulder and it is difficult for the
holder to hold it.
Q. States the Archimedesโ€™ principle.
Ans. Archimedes' principle states that โ€œWhen a body is immersed fully or partially in a
fluid, it experiences an upward force that is equal to the weight of the fluid displaced by
it.โ€
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Q. Write the applications of the Archimedesโ€™ principle.
Ans. Following are the applications of the Archimedesโ€™ principle.:-
1. It is used in designing ships and submarines.
2. It is used in designing lactometers.
3. It is used in designing hydrometers.
Q. You find your mass to be 42 kg on a weighing machine. Is your mass more or less
than 42 kg?
Ans. All objects on the earth are immersed in the atmospheric air so all the body feel
the buoyancy force exerted by the air so when the mass of an object is measured by a
weighing machine it indicates the less mass than actual mass so if the weighing
machine is indicating 42 kg then my mass is actually more than 42 kg.
Q. What is relative density?
Ans. Relative density of is the ratio of the density of an object to that of the water.
๐‘…๐‘’๐‘™๐‘Ž๐‘ก๐‘–๐‘ฃ๐‘’ ๐‘‘๐‘’๐‘›๐‘ ๐‘–๐‘ก๐‘ฆ =
๐ท๐‘’๐‘›๐‘ ๐‘–๐‘ก๐‘ฆ ๐‘œ๐‘“ ๐‘Ž๐‘› ๐‘œ๐‘๐‘—๐‘’๐‘๐‘ก
๐ท๐‘’๐‘›๐‘ ๐‘–๐‘ก๐‘ฆ ๐‘œ๐‘“ ๐‘ค๐‘Ž๐‘ก๐‘’๐‘Ÿ
It has no SI unit.
Q. Relative density of silver is 10.8. The density of water is . What is the
10
3
๐‘˜๐‘” ๐‘š
โˆ’ 3
density of silver in SI unit?
Ans.
๐‘…๐‘’๐‘™๐‘Ž๐‘ก๐‘–๐‘ฃ๐‘’ ๐‘‘๐‘’๐‘›๐‘ ๐‘–๐‘ก๐‘ฆ ๐‘œ๐‘“ ๐‘ ๐‘–๐‘™๐‘ฃ๐‘’๐‘Ÿ = 10 . 8
๐ท๐‘’๐‘›๐‘ ๐‘–๐‘ก๐‘ฆ ๐‘œ๐‘“ ๐‘ค๐‘Ž๐‘ก๐‘’๐‘Ÿ = 10
3
๐‘˜๐‘” ๐‘š
โˆ’ 3
๐ท๐‘’๐‘›๐‘ ๐‘–๐‘ก๐‘ฆ ๐‘œ๐‘“ ๐‘ ๐‘–๐‘™๐‘ฃ๐‘’๐‘Ÿ = ?
๐‘…๐‘’๐‘™๐‘Ž๐‘ก๐‘–๐‘ฃ๐‘’ ๐‘‘๐‘’๐‘›๐‘ ๐‘–๐‘ก๐‘ฆ =
๐ท๐‘’๐‘›๐‘ ๐‘–๐‘ก๐‘ฆ ๐‘œ๐‘“ ๐‘Ž๐‘› ๐‘œ๐‘๐‘—๐‘’๐‘๐‘ก
๐ท๐‘’๐‘›๐‘ ๐‘–๐‘ก๐‘ฆ ๐‘œ๐‘“ ๐‘ค๐‘Ž๐‘ก๐‘’๐‘Ÿ
10 . 8 =
๐ท๐‘’๐‘›๐‘ ๐‘–๐‘ก๐‘ฆ ๐‘œ๐‘“ ๐‘ ๐‘–๐‘™๐‘ฃ๐‘’๐‘Ÿ
10
3
Ans
๐ท๐‘’๐‘›๐‘ ๐‘–๐‘ก๐‘ฆ ๐‘œ๐‘“ ๐‘ ๐‘–๐‘™๐‘ฃ๐‘’๐‘Ÿ = 10 ร— 10
3
๐‘˜๐‘” ๐‘š
โˆ’ 3
Q. How does the force of gravitation between two objects change when the distance
between them is reduced to half ?
Ans.
For previous gravitational force For new gravitational force
๐‘€ = ๐‘€ ๐‘€ ' = ๐‘€
๐‘š = ๐‘š ๐‘š ' = ๐‘š
๐‘‘ = ๐‘‘ ๐‘‘ ' =
๐‘‘
2
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11 SMART EDUCATIONS
๐ถโ„Ž๐‘Ž๐‘›๐‘”๐‘’ ๐‘–๐‘› ๐‘กโ„Ž๐‘’ ๐‘”๐‘Ÿ๐‘Ž๐‘ฃ๐‘–๐‘ก๐‘Ž๐‘ก๐‘–๐‘œ๐‘›๐‘Ž๐‘™ ๐‘“๐‘œ๐‘Ÿ๐‘๐‘’ =
๐‘๐‘’๐‘ค ๐‘”๐‘Ÿ๐‘Ž๐‘ฃ๐‘–๐‘ก๐‘Ž๐‘ก๐‘–๐‘œ๐‘›๐‘Ž๐‘™ ๐‘“๐‘œ๐‘Ÿ๐‘๐‘’
๐‘ƒ๐‘Ÿ๐‘’๐‘ฃ๐‘–๐‘œ๐‘ข๐‘  ๐‘”๐‘Ÿ๐‘Ž๐‘ฃ๐‘–๐‘ก๐‘Ž๐‘ก๐‘–๐‘œ๐‘›๐‘Ž๐‘™ ๐‘“๐‘œ๐‘Ÿ๐‘๐‘’
=
๐บ๐‘€ ' ๐‘š '
๐‘‘ '
2
๐บ๐‘€๐‘š
๐‘‘
2
=
๐บ๐‘€๐‘š
๐‘‘
2
( )
2
๐บ๐‘€๐‘š
๐‘‘
2
=
๐บ๐‘€๐‘š
๐‘‘
2
4
ร—
๐‘‘
2
๐บ๐‘€๐‘š
= 4
New gravitational force will be 4 times that of previous gravitational force. Ans
Q. Gravitational force acts on all objects in proportion to their masses. Why then, a
heavy object does not fall faster than a light object?
Ans. All objects fall with a constant acceleration, called gravitational acceleration
whether it is heavy or light because the gravitational acceleration is directly proportional
to the mass of the body on which it is falling and not to the mass which is falling.
Q. What is the magnitude of the gravitational force between the earth and a 1 kg object
on its surface? (Mass of the earth is kg and radius of the earth is 6 ร— 10
24
6 . 4 ร— 10
6
m.)
Ans.
๐‘€ = 6 ร— 10
24
๐‘˜๐‘” ( ๐‘š๐‘Ž๐‘ ๐‘  ๐‘œ๐‘“ ๐‘กโ„Ž๐‘’ ๐‘’๐‘Ž๐‘Ÿ๐‘กโ„Ž )
๐‘š = 1 ๐‘˜๐‘”
๐‘Ÿ = 6 . 4 ร— 10
6
๐‘š ( ๐‘Ÿ๐‘Ž๐‘‘๐‘–๐‘ข๐‘  ๐‘œ๐‘“ ๐‘กโ„Ž๐‘’ ๐‘’๐‘Ž๐‘Ÿ๐‘กโ„Ž )
๐บ = 6 . 67 ร— 10
โˆ’ 11
๐‘ ๐‘š
2
๐‘˜๐‘”
โˆ’ 2
๐น =
๐บ๐‘€๐‘š
๐‘‘
2
๐น =
6 . 67 ร— 10
โˆ’ 11
ร— 6 ร— 10
24
ร— 1
6 . 4 ร— 10
6
( )
2
๐น =
6 . 67 ร— 6 ร— 10
โˆ’ 11 + 24
6 . 4 ร— 10
6
ร— 6 . 4 ร— 10
6
๐น =
6 . 67 ร— 6 ร— 10
โˆ’ 11 + 24 โˆ’ 6 โˆ’ 6
6 . 4 ร— 6 . 4
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12 SMART EDUCATIONS
๐น =
6 . 67 ร— 6 ร— 10
6 . 4 ร— 6 . 4
๐น =
6 . 67 ร— 3 ร— 10
6 . 4 ร— 3 . 2
๐น = 9 . 77
๐น = 9 . 8 ๐‘
Q. The earth and the moon are attracted to each other by gravitational force. Does the
earth attract the moon with a force that is greater or smaller or the same as the force
with which the moon attracts the earth? Why?
Ans. They both attract each other with the same force. Because according to the third
law of motion:- โ€œAction is always equal to the reaction but in the opposite direction.โ€
Q. If the moon attracts the earth, why does the earth not move towards the moon?
Ans. They both attract each other with the same force but the force produces greater
acceleration in an object which is lower in mass. The mass of the earth is greater than
that of the mass of the moon so it feels less acceleration which is not sufficient to make
the earth move toward the moon.
Q. What happens to the force between two objects, if
(i) the mass of one object is doubled?
(ii) the distance between the objects is doubled and tripled?
(iii) the masses of both objects are doubled?
Ans.
(i) For previous gravitational force For new gravitational force
๐‘€ = ๐‘€ ๐‘€ ' = 2 ๐‘€
๐‘š = ๐‘š ๐‘š ' = ๐‘š
๐‘‘ = ๐‘‘ ๐‘‘ ' = ๐‘‘
๐ถโ„Ž๐‘Ž๐‘›๐‘”๐‘’ ๐‘–๐‘› ๐‘กโ„Ž๐‘’ ๐‘”๐‘Ÿ๐‘Ž๐‘ฃ๐‘–๐‘ก๐‘Ž๐‘ก๐‘–๐‘œ๐‘›๐‘Ž๐‘™ ๐‘“๐‘œ๐‘Ÿ๐‘๐‘’ =
๐‘๐‘’๐‘ค ๐‘”๐‘Ÿ๐‘Ž๐‘ฃ๐‘–๐‘ก๐‘Ž๐‘ก๐‘–๐‘œ๐‘›๐‘Ž๐‘™ ๐‘“๐‘œ๐‘Ÿ๐‘๐‘’
๐‘ƒ๐‘Ÿ๐‘’๐‘ฃ๐‘–๐‘œ๐‘ข๐‘  ๐‘”๐‘Ÿ๐‘Ž๐‘ฃ๐‘–๐‘ก๐‘Ž๐‘ก๐‘–๐‘œ๐‘›๐‘Ž๐‘™ ๐‘“๐‘œ๐‘Ÿ๐‘๐‘’
=
๐บ๐‘€ ' ๐‘š '
๐‘‘ '
2
๐บ๐‘€๐‘š
๐‘‘
2
=
๐บ ร— 2 ๐‘€ ร— ๐‘š
๐‘‘
2
๐บ๐‘€๐‘š
๐‘‘
2
=
2 ๐บ๐‘€๐‘š
๐‘‘
2
ร—
๐‘‘
2
๐บ๐‘€๐‘š
= 2
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13 SMART EDUCATIONS
Hence the force will also be double Ans
(ii)(a) For previous gravitational force For new gravitational force
๐‘€ = ๐‘€ ๐‘€ ' = ๐‘€
๐‘š = ๐‘š ๐‘š ' = ๐‘š
๐‘‘ = ๐‘‘ ๐‘‘ ' = 2 ๐‘‘
๐ถโ„Ž๐‘Ž๐‘›๐‘”๐‘’ ๐‘–๐‘› ๐‘กโ„Ž๐‘’ ๐‘”๐‘Ÿ๐‘Ž๐‘ฃ๐‘–๐‘ก๐‘Ž๐‘ก๐‘–๐‘œ๐‘›๐‘Ž๐‘™ ๐‘“๐‘œ๐‘Ÿ๐‘๐‘’ =
๐‘๐‘’๐‘ค ๐‘”๐‘Ÿ๐‘Ž๐‘ฃ๐‘–๐‘ก๐‘Ž๐‘ก๐‘–๐‘œ๐‘›๐‘Ž๐‘™ ๐‘“๐‘œ๐‘Ÿ๐‘๐‘’
๐‘ƒ๐‘Ÿ๐‘’๐‘ฃ๐‘–๐‘œ๐‘ข๐‘  ๐‘”๐‘Ÿ๐‘Ž๐‘ฃ๐‘–๐‘ก๐‘Ž๐‘ก๐‘–๐‘œ๐‘›๐‘Ž๐‘™ ๐‘“๐‘œ๐‘Ÿ๐‘๐‘’
=
๐บ๐‘€ ' ๐‘š '
๐‘‘ '
2
๐บ๐‘€๐‘š
๐‘‘
2
=
๐บ๐‘€๐‘š
2 ๐‘‘ ( )
2
๐บ๐‘€๐‘š
๐‘‘
2
=
๐บ๐‘€๐‘š
4 ๐‘‘
2
ร—
๐‘‘
2
๐บ๐‘€๐‘š
=
1
4
Hence the force will be Ans.
1
4
( )
๐‘กโ„Ž
(ii) (b) For previous gravitational force For new gravitational force
๐‘€ = ๐‘€ ๐‘€ ' = ๐‘€
๐‘š = ๐‘š ๐‘š ' = ๐‘š
๐‘‘ = ๐‘‘ ๐‘‘ ' = 3 ๐‘‘
๐ถโ„Ž๐‘Ž๐‘›๐‘”๐‘’ ๐‘–๐‘› ๐‘กโ„Ž๐‘’ ๐‘”๐‘Ÿ๐‘Ž๐‘ฃ๐‘–๐‘ก๐‘Ž๐‘ก๐‘–๐‘œ๐‘›๐‘Ž๐‘™ ๐‘“๐‘œ๐‘Ÿ๐‘๐‘’ =
๐‘๐‘’๐‘ค ๐‘”๐‘Ÿ๐‘Ž๐‘ฃ๐‘–๐‘ก๐‘Ž๐‘ก๐‘–๐‘œ๐‘›๐‘Ž๐‘™ ๐‘“๐‘œ๐‘Ÿ๐‘๐‘’
๐‘ƒ๐‘Ÿ๐‘’๐‘ฃ๐‘–๐‘œ๐‘ข๐‘  ๐‘”๐‘Ÿ๐‘Ž๐‘ฃ๐‘–๐‘ก๐‘Ž๐‘ก๐‘–๐‘œ๐‘›๐‘Ž๐‘™ ๐‘“๐‘œ๐‘Ÿ๐‘๐‘’
=
๐บ๐‘€ ' ๐‘š '
๐‘‘ '
2
๐บ๐‘€๐‘š
๐‘‘
2
=
๐บ๐‘€๐‘š
3 ๐‘‘ ( )
2
๐บ๐‘€๐‘š
๐‘‘
2
=
๐บ๐‘€๐‘š
9 ๐‘‘
2
ร—
๐‘‘
2
๐บ๐‘€๐‘š
=
1
9
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14 SMART EDUCATIONS
Hence the force will be Ans.
1
9
( )
๐‘กโ„Ž
(iii) For previous gravitational force For new gravitational force
๐‘€ = ๐‘€ ๐‘€ ' = 2 ๐‘€
๐‘š = ๐‘š ๐‘š ' = 2 ๐‘š
๐‘‘ = ๐‘‘ ๐‘‘ ' = ๐‘‘
๐ถโ„Ž๐‘Ž๐‘›๐‘”๐‘’ ๐‘–๐‘› ๐‘กโ„Ž๐‘’ ๐‘”๐‘Ÿ๐‘Ž๐‘ฃ๐‘–๐‘ก๐‘Ž๐‘ก๐‘–๐‘œ๐‘›๐‘Ž๐‘™ ๐‘“๐‘œ๐‘Ÿ๐‘๐‘’ =
๐‘๐‘’๐‘ค ๐‘”๐‘Ÿ๐‘Ž๐‘ฃ๐‘–๐‘ก๐‘Ž๐‘ก๐‘–๐‘œ๐‘›๐‘Ž๐‘™ ๐‘“๐‘œ๐‘Ÿ๐‘๐‘’
๐‘ƒ๐‘Ÿ๐‘’๐‘ฃ๐‘–๐‘œ๐‘ข๐‘  ๐‘”๐‘Ÿ๐‘Ž๐‘ฃ๐‘–๐‘ก๐‘Ž๐‘ก๐‘–๐‘œ๐‘›๐‘Ž๐‘™ ๐‘“๐‘œ๐‘Ÿ๐‘๐‘’
=
๐บ๐‘€ ' ๐‘š '
๐‘‘ '
2
๐บ๐‘€๐‘š
๐‘‘
2
=
๐บ ร— 2 ๐‘€ ร— 2 ๐‘š
๐‘‘
2
๐บ๐‘€๐‘š
๐‘‘
2
=
4 ๐บ๐‘€๐‘š
๐‘‘
2
ร—
๐‘‘
2
๐บ๐‘€๐‘š
= 4
Hence the force will be 4 times.
Q. What do we call the gravitational force between the earth and an object?
Ans. Weight of that object.
Q. Amit buys a few grams of gold at the poles as per the instruction of one of his
friends. He hands over the same when he meets him at the equator. Will the friend
agree with the weight of gold bought? If not, why?
Ans. No, because the gold was purchased at the poles where the gravitational
acceleration is higher as compared to the equator so the weight of the gold will also be
higher at the poles and lower at the equator. So the friend will not agree with the weight
of gold bought but will agree with the mass of gold.
Q. Why will a sheet of paper fall slower than one that is crumpled into a ball?
Ans. Mass of both the papers is the same but the difference between the two papers is
their surface area. The crumpled paper has less surface area so it feels less friction
force which is exerted by the air as compared to the sheet of paper. So the crumpled
paper will fall faster and the sheet of paper slower.
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15 SMART EDUCATIONS
Q. Gravitational force on the surface of the moon is only as strong as gravitational
1
6
force on the earth. What is the weight in newtons of a 10 kg object on the moon and on
the earth?
Ans.
๐‘š = 10 ๐‘˜๐‘”
๐‘Š
๐‘š
= ?
๐‘Š
๐‘’
= ?
๐‘Š
๐‘š
=
๐‘š ๐‘”
๐‘’
6
๐‘Š
๐‘š
=
10 ร— 9 . 8
6
Ans
๐‘Š
๐‘š
= 16 . 33 ๐‘
๐‘Š
๐‘’
= ๐‘š๐‘”
๐‘Š
๐‘’
= 10 ร— 9 . 8
Ans
๐‘Š
๐‘’
= 98 ๐‘
Q. A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate (i) the
maximum height to which it rises, (ii) the total time it takes to return to the surface of the
earth.
Ans.
๐‘ข = 49 ๐‘š / ๐‘ 
๐‘ฃ = 0 ๐‘š / ๐‘ 
๐‘” = โˆ’ 9 . 8 ๐‘š ๐‘ 
โˆ’ 2
๐‘  = ?
๐‘ก = ?
2 ๐‘”๐‘  = ๐‘ฃ
2
โˆ’ ๐‘ข
2
2 ร— โˆ’ 9 . 8 ( ) ร— ๐‘  = 0
2
โˆ’ 49
2
๐‘  =
โˆ’ 49 ร— 49
2 ร—โˆ’ 9 . 8 ( )
Ans(i)
๐‘  = 122 . 5 ๐‘š
๐‘ฃ = ๐‘ข + ๐‘”๐‘ก
0 = 49 +โˆ’ 9 . 8 ๐‘ก
0 + 9 . 8 ๐‘ก = 49
๐‘ก =
49
9 . 8
Ans(ii)
๐‘ก = 5 ๐‘ 
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16 SMART EDUCATIONS
Q. A stone is released from the top of a tower of height 19.6 m. Calculate its final
velocity just before touching the ground.
Ans. ๐‘ข = 0 ๐‘š / ๐‘ 
๐‘  = 19 . 6 ๐‘š
๐‘” = 9 . 8 ๐‘š ๐‘ 
โˆ’ 2
๐‘ฃ = ?
2 ๐‘”๐‘  = ๐‘ฃ
2
โˆ’ ๐‘ข
2
2 ร— 9 . 8 ร— 19 . 6 = ๐‘ฃ
2
โˆ’ 0
2
19 . 6 .ร— 19 . 6 = ๐‘ฃ
2
๐‘ฃ = 19 . 6 .ร— 19 . 6
Ans
๐‘ฃ = 19 . 6 ๐‘š / ๐‘ 
Q. A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g =
, find the maximum height reached by the stone. What is the net displacement
10 ๐‘š ๐‘ 
โˆ’ 2
and the total distance covered by the stone?
Ans.
๐‘ฃ = 0 ๐‘š / ๐‘ 
๐‘ข = 40 ๐‘š / ๐‘ 
๐‘” = โˆ’ 10 ๐‘š ๐‘ 
โˆ’ 2
๐‘  = ?
๐‘๐‘’๐‘ก ๐‘‘๐‘–๐‘ ๐‘๐‘™๐‘Ž๐‘๐‘’๐‘š๐‘’๐‘›๐‘ก = ?
๐‘‡๐‘œ๐‘ก๐‘Ž๐‘™ ๐‘‘๐‘–๐‘ ๐‘ก๐‘Ž๐‘›๐‘๐‘’ = ?
2 ร— โˆ’ 10 ( ) ร— ๐‘  = 0
2
โˆ’ 40
2
โˆ’ 20 ๐‘  = 0 โˆ’ 1600
๐‘  =
โˆ’ 1600
โˆ’ 20
Ans
๐‘  = 80 ๐‘š
Ans
๐‘๐‘’๐‘ก ๐‘‘๐‘–๐‘ ๐‘๐‘™๐‘Ž๐‘๐‘’๐‘š๐‘’๐‘›๐‘ก = 0 ( ๐ต๐‘’๐‘๐‘Ž๐‘ข๐‘ ๐‘’ ๐‘–๐‘ก ๐‘Ÿ๐‘’๐‘ก๐‘ข๐‘Ÿ๐‘›๐‘’๐‘‘ ๐‘ก๐‘œ ๐‘กโ„Ž๐‘’ ๐‘ ๐‘Ž๐‘š๐‘’ ๐‘๐‘™๐‘Ž๐‘๐‘’ )
๐‘‡๐‘œ๐‘ก๐‘Ž๐‘™ ๐‘‘๐‘–๐‘ ๐‘ก๐‘Ž๐‘›๐‘๐‘’ = 180 + 180
Ans
๐‘‡๐‘œ๐‘ก๐‘Ž๐‘™ ๐‘‘๐‘–๐‘ ๐‘ก๐‘Ž๐‘›๐‘๐‘’ = 160 ๐‘š
Q. Calculate the force of gravitation between the earth and the Sun, given that the mass
of the earth kg and of the Sun kg. The average distance
= 6 ร— 10
24
= 2 ร— 10
30
between the two is m.
1 . 5 ร— 10
11
Ans.
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17 SMART EDUCATIONS
๐‘€ = 6 ร— 10
24
๐‘˜๐‘”
๐‘š = 2 ร— 10
30
๐‘˜๐‘”
๐‘‘ = 1 . 5 ร— 10
11
๐‘š
๐บ = 6 . 67 ร— 10
โˆ’ 11
๐‘ ๐‘š
2
๐‘˜๐‘”
โˆ’ 2
๐น =
๐บ๐‘€๐‘š
๐‘‘
2
๐น =
6 . 67 ร— 10
โˆ’ 11
ร— 6 ร— 10
24
ร— 2 ร— 10
30
1 . 5 ร— 10
11
( )
2
๐น =
6 . 67 ร— 6 ร— 2 ร— 10
โˆ’ 11 + 24 + 30
1 . 5 ร— 10
11
ร— 1 . 5 ร— 10
11
๐น =
6 . 67 ร— 6 ร— 2 ร— 10
43 โˆ’ 11 โˆ’ 11
1 . 5 ร— 1 . 5
๐น =
667 ร— 6 ร— 2 ร— 10 ร— 10 ร— 10
21
15 ร— 15 ร— 100
๐น =
667 ร— 2 ร— 2 ร— 10
21
5 ร— 15
๐น =
2668 ร— 10
21
75
๐น = 35 . 5733 ร— 10
21
Ans
๐น = 3 . 56 ร— 10
22
๐‘
Q. A stone is allowed to fall from the top of a tower 100 m high and at the same time
another stone is projected vertically upwards from the ground with a velocity of 25 m/s.
Calculate when and where the two stones will meet.
Ans.
Let the 1
st
stone was released from
A and the 2
nd
stone was projected
from D and let after they meet at
๐‘ฅ ๐‘ 
C.
1
st
stone 2
nd
stone
๐‘ข = 0 ๐‘š / ๐‘  ๐‘ข = 25 ๐‘š / ๐‘ 
๐‘” = 9 . 8 ๐‘š ๐‘ 
โˆ’ 2
๐‘” =โˆ’ 9 . 8 ๐‘š ๐‘ 
โˆ’ 2
๐‘  = ๐ด๐ถ ๐‘  = ๐ท๐ถ
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18 SMART EDUCATIONS
For the 1
st
stone For the 1
st
stone
๐‘  = ๐‘ข๐‘ก +
1
2
๐‘” ๐‘ก
2
๐‘  = ๐‘ข๐‘ก +
1
2
๐‘” ๐‘ก
2
๐ด๐ถ = 0 ร— ๐‘ฅ +
1
2
ร— 9 . 8 ร— ๐‘ฅ
2
๐ท๐ถ = 25 ร— ๐‘ฅ +
1
2
ร— โˆ’ 9 . 8 ( ) ร— ๐‘ฅ
2
๐ด๐ถ = 0 + 4 . 9 ๐‘ฅ
2
๐ท๐ถ = 25 ๐‘ฅ โˆ’ 4 . 9 ๐‘ฅ
2
๐ด๐ถ = 4 . 9 ๐‘ฅ
2
( ๐‘– ) ๐ท๐ถ = 25 ๐‘ฅ โˆ’ 4 . 9 ๐‘ฅ
2
( ๐‘–๐‘– )
According to the figure:-
๐ด๐ถ + ๐ท๐ถ = ๐ด๐ท
4 . 9 ๐‘ฅ
2
+ 25 ๐‘ฅ โˆ’ 4 . 9 ๐‘ฅ
2
= 100 [ ๐‘“๐‘Ÿ๐‘œ๐‘š ( ๐‘– ) ๐‘Ž๐‘›๐‘‘ ( ๐‘–๐‘– )]
25 ๐‘ฅ = 100
๐‘ฅ =
100
25
๐‘ฅ = 4 ๐‘ 
Now
๐ท๐ถ = 25 ๐‘ฅ โˆ’ 4 . 9 ๐‘ฅ
2
๐ท๐ถ = 25 ร— 4 โˆ’ 4 . 9 ร— 4
2
๐ท๐ถ = 100 โˆ’ 4 . 9 ร— 16
๐ท๐ถ = 100 โˆ’ 78 . 40
๐ท๐ถ = 21 . 60 ๐‘š
So the both stones will meet after 4s at the height of 21.60m from the ground. Ans.
Q. A ball thrown up vertically returns to the thrower after 6 s. Find (a) the velocity with
which it was thrown up, (b) the maximum height it reaches, and (c) its position after 4 s.
Ans.
๐ฟ๐‘’๐‘ก ๐‘กโ„Ž๐‘’ ๐‘๐‘Ž๐‘™๐‘™ ๐‘ค๐‘Ž๐‘  ๐‘กโ„Ž๐‘Ÿ๐‘œ๐‘ค๐‘› ๐‘“๐‘Ÿ๐‘œ๐‘š ๐ท
๐‘‡๐‘œ๐‘ก๐‘Ž๐‘™ ๐‘ก๐‘–๐‘š๐‘’ = 6 ๐‘ 
โ„Ž๐‘’๐‘›๐‘๐‘’ ๐‘ก๐‘–๐‘š๐‘’ ๐‘ก๐‘Ž๐‘˜๐‘’๐‘› ๐‘–๐‘› ๐‘œ๐‘›๐‘’ ๐‘ค๐‘Ž๐‘ฆ ( ๐ด๐ท ) =
6
2
= 3 ๐‘ 
๐‘คโ„Ž๐‘–๐‘™๐‘’ ๐‘”๐‘œ๐‘–๐‘›๐‘” , ' ๐‘” ' = โˆ’ 9 . 8 ๐‘š / ๐‘ 
2
๐‘Ž๐‘›๐‘‘ ๐‘ฃ = 0 ๐‘š / ๐‘ 
๐‘†๐‘œ , ๐‘คโ„Ž๐‘–๐‘™๐‘’ ๐‘Ÿ๐‘’๐‘ก๐‘ข๐‘Ÿ๐‘›๐‘–๐‘›๐‘” ' ๐‘” ' = 9 . 8 ๐‘š / ๐‘ 
2
๐‘€๐‘Ž๐‘ฅ๐‘–๐‘š๐‘ข๐‘š โ„Ž๐‘’๐‘–๐‘”โ„Ž๐‘ก ' ๐‘  ' = ? ( ๐ด๐ท )
๐‘ƒ๐‘œ๐‘ ๐‘–๐‘ก๐‘–๐‘œ๐‘› ๐‘Ž๐‘“๐‘ก๐‘’๐‘Ÿ 4 ๐‘  = ? ( ๐ถ๐ท )
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19 SMART EDUCATIONS
๐‘Šโ„Ž๐‘–๐‘™๐‘’ ๐‘”๐‘œ๐‘–๐‘›๐‘” ( ๐ด๐ท ):โˆ’
๐‘ฃ = ๐‘ข + ๐‘”๐‘ก
0 = ๐‘ข + โˆ’ 9 . 8 ( ) ร— 3
0 = ๐‘ข โˆ’ 29 . 4
๐‘ข = 29 . 4 ๐‘š / ๐‘ 
๐‘  = ๐‘ข๐‘ก +
1
2
๐‘” ๐‘ก
2
๐ด๐ท = 29 . 4 ร— 3 +
1
2
ร— โˆ’ 9 . 8 ( ) ร— 3
2
๐ด๐ท = 88 . 2 โˆ’ 4 . 9 ร— 9
๐ด๐ท = 88 . 2 โˆ’ 44 . 1
๐ด๐ท = 44 . 1 ๐‘š
After 4s the stone was at C because in 3 s the stone will reach at its maximum height A
and after 3s it starts falling and in 1s it reaches at C. So after 4s the position of the
stone is DC from the ground.
๐‘†๐‘œ , ๐‘คโ„Ž๐‘–๐‘™๐‘’ ๐‘Ÿ๐‘’๐‘ก๐‘ข๐‘Ÿ๐‘›๐‘–๐‘›๐‘” ' ๐‘” ' = 9 . 8 ๐‘š / ๐‘ 
2
๐‘ข = 0 ๐‘š / ๐‘ 
๐‘ก = 1 ๐‘ 
๐‘  = ๐‘ข๐‘ก +
1
2
๐‘” ๐‘ก
2
๐ด๐ถ = 0 ร— 1 +
1
2
ร— 9 . 8 ร— 1
2
๐ด๐ถ = 0 + 4 . 9
๐ด๐ถ = 4 . 9 ๐‘š
๐ท๐ถ = ๐ด๐ท โˆ’ ๐ด๐ถ
๐ท๐ถ = 44 . 1 โˆ’ 4 . 9
๐ท๐ถ = 39 . 2 ๐‘š
So after 4s the position of the stone is 39.2 m from the ground. Ans.
Q. In what direction does the buoyant force on an object immersed in a liquid act?
Ans. The buoyant force always acts in the upward direction, so it is also called as
upthrust.
Q. Why does a block of plastic released under water come up to the surface of water?
Ans. A block of plastic released under water comes up to the surface of water because
of the buoyancy force exerted by the water.
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20 SMART EDUCATIONS
Q. The volume of 50 g of a substance is 20 cm
3
. If the density of water is 1 g cm
โ€“3
, will
the substance float or sink?
Ans.
๐ท๐‘’๐‘›๐‘ ๐‘–๐‘ก๐‘ฆ ๐‘œ๐‘“ ๐‘กโ„Ž๐‘’ ๐‘ ๐‘ข๐‘๐‘ ๐‘ก๐‘Ž๐‘›๐‘๐‘’ =
๐‘€๐‘Ž๐‘ ๐‘ 
๐‘‰๐‘œ๐‘™๐‘ข๐‘š๐‘’
๐ท๐‘’๐‘›๐‘ ๐‘–๐‘ก๐‘ฆ ๐‘œ๐‘“ ๐‘กโ„Ž๐‘’ ๐‘ ๐‘ข๐‘๐‘ ๐‘ก๐‘Ž๐‘›๐‘๐‘’ =
50
20
๐ท๐‘’๐‘›๐‘ ๐‘–๐‘ก๐‘ฆ ๐‘œ๐‘“ ๐‘กโ„Ž๐‘’ ๐‘ ๐‘ข๐‘๐‘ ๐‘ก๐‘Ž๐‘›๐‘๐‘’ = 2 . 5 ๐‘” ๐‘๐‘š
โˆ’ 3
๐ต๐‘’๐‘๐‘Ž๐‘ข๐‘ ๐‘’ ๐ท๐‘’๐‘›๐‘ ๐‘–๐‘ก๐‘ฆ ๐‘œ๐‘“ ๐‘กโ„Ž๐‘’ ๐‘ ๐‘ข๐‘๐‘ ๐‘ก๐‘Ž๐‘›๐‘๐‘’ > ๐ท๐‘’๐‘›๐‘ ๐‘–๐‘ก๐‘ฆ ๐‘œ๐‘“ ๐‘กโ„Ž๐‘’ ๐‘ค๐‘Ž๐‘ก๐‘’๐‘Ÿ
Hence the substance will sink. Ans.
Q. The volume of a 500 g sealed packet is 350 cm
3
. Will the packet float or sink in
water if the density of water is 1 g cm
โ€“3
? What will be the mass of the water displaced
by this packet?
Ans.
๐ท๐‘’๐‘›๐‘ ๐‘–๐‘ก๐‘ฆ ๐‘œ๐‘“ ๐‘กโ„Ž๐‘’ ๐‘ ๐‘ข๐‘๐‘ ๐‘ก๐‘Ž๐‘›๐‘๐‘’ =
๐‘€๐‘Ž๐‘ ๐‘ 
๐‘‰๐‘œ๐‘™๐‘ข๐‘š๐‘’
๐ท๐‘’๐‘›๐‘ ๐‘–๐‘ก๐‘ฆ ๐‘œ๐‘“ ๐‘กโ„Ž๐‘’ ๐‘ ๐‘ข๐‘๐‘ ๐‘ก๐‘Ž๐‘›๐‘๐‘’ =
500
350
๐ท๐‘’๐‘›๐‘ ๐‘–๐‘ก๐‘ฆ ๐‘œ๐‘“ ๐‘กโ„Ž๐‘’ ๐‘ ๐‘ข๐‘๐‘ ๐‘ก๐‘Ž๐‘›๐‘๐‘’ = 1 . 43 ๐‘” ๐‘๐‘š
โˆ’ 3
๐ต๐‘’๐‘๐‘Ž๐‘ข๐‘ ๐‘’ ๐ท๐‘’๐‘›๐‘ ๐‘–๐‘ก๐‘ฆ ๐‘œ๐‘“ ๐‘กโ„Ž๐‘’ ๐‘ ๐‘ข๐‘๐‘ ๐‘ก๐‘Ž๐‘›๐‘๐‘’ > ๐ท๐‘’๐‘›๐‘ ๐‘–๐‘ก๐‘ฆ ๐‘œ๐‘“ ๐‘กโ„Ž๐‘’ ๐‘ค๐‘Ž๐‘ก๐‘’๐‘Ÿ
Hence the substance will sink. Ans.
Because the volume of the substance is 350 cm
3
hence according to the Archimedes'
Principle the volume of the water displaced by it is also 350 cm
3
.
Hence:-
๐‘€๐‘Ž๐‘ ๐‘  ๐‘œ๐‘“ ๐‘กโ„Ž๐‘’ ๐‘‘๐‘–๐‘ ๐‘๐‘™๐‘Ž๐‘๐‘’๐‘‘ ๐‘ค๐‘Ž๐‘ก๐‘’๐‘Ÿ = ๐‘‰๐‘œ๐‘™๐‘ข๐‘š๐‘’ ๐‘œ๐‘“ ๐‘กโ„Ž๐‘’ ๐‘‘๐‘–๐‘ ๐‘๐‘™๐‘Ž๐‘๐‘’๐‘‘ ๐‘ค๐‘Ž๐‘ก๐‘’๐‘Ÿ ร— ๐ท๐‘’๐‘›๐‘ ๐‘–๐‘ก๐‘ฆ ๐‘œ๐‘“ ๐‘กโ„Ž๐‘’ ๐‘ค๐‘Ž๐‘ก๐‘’๐‘Ÿ
๐‘€๐‘Ž๐‘ ๐‘  ๐‘œ๐‘“ ๐‘กโ„Ž๐‘’ ๐‘‘๐‘–๐‘ ๐‘๐‘™๐‘Ž๐‘๐‘’๐‘‘ ๐‘ค๐‘Ž๐‘ก๐‘’๐‘Ÿ = 350 ร— 1
Ans. ๐‘€๐‘Ž๐‘ ๐‘  ๐‘œ๐‘“ ๐‘กโ„Ž๐‘’ ๐‘‘๐‘–๐‘ ๐‘๐‘™๐‘Ž๐‘๐‘’๐‘‘ ๐‘ค๐‘Ž๐‘ก๐‘’๐‘Ÿ = 350 ๐‘”