# TEXTBOOK ANSWERS AND SOLUTIONS OF CBSE CLASS IX SCIENCE Chapter 10 Gravitation

**1) How does the force of gravitation between two objects change when the distance between them is reduced to half?**

ANSWER:-According to Universal
Law of gravitation, the gravitational
force of attraction between any two objects of mass M and m is proportional to the
product of their masses and
inversely proportional to the square of distance between them

So, force is given by

**2) Gravitational force acts on all objects in proportion to their masses. Why then, a heavy object does not fall faster than a light object?**

ANSWER:-All objects fall on
ground with constant acceleration, called acceleration due to gravity (in the
absence of air resistances). It is constant and does not depend upon the mass of an object. Hence, heavy objects
do not fall faster than light objects.

**3) What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface? (Mass of the earth is 6 × 1024 kg and radius of the earth is 6.4 × 10 6 m).**

ANSWER:-Mass of the body, m= 1 kg

Mass of the Earth, M= 6 x1024 kg

Radius of the earth, R = 6.4 x106
m

Now magnitude of the
gravitational force (F) between the
earth and the body can be given
as,

**4) 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?**

ANSWER:-According to universal
law of gravitation tow objects attract each other with equal force but in
opposite directions. The Earth and the moon both attract each other with equal
force. Which means that the magnitude of force with which earth attracts moon is equal to
the magnitude of force with which moon attracts earth but their directions are
opposite to each other.

**5) If the moon attracts the earth, why does the earth not move towards the moon?**

ANSWER:-The Earth and the moon
experience equal gravitational forces from each other. However, the mass of the
Earth is much larger than the
mass of the moon. Hence, it accelerates
at a rate lesser than the acceleration rate of the moon towards the Earth.

**6) 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?**

ANSWER : i)-

ii)The force F is inversely
proportional to the distance
between the objects. So if the
distance between two objects is
doubled then the gravitational force of attraction between them is reduced to
one fourth of its original value. Similarly f the distance between two objects
is tripled, then the gravitational force of attraction becomes one ninth of its
original value.

(iii) Again from Universal law of
attraction from equation 1 force
F is directly proportional to the product of both the masses. So if both the
masses are doubled then the gravitational force of attraction becomes four
times the original value.

**7)What is the importance of universal law of gravitation?**

ANSWER:-Universal law of Gravitation
is important because it tells
us about the force that is responsible for binding us to Earth. The motion of
moon around the earth, the motion of planets around the sun the tides formed by rising and
falling of water level in the ocean are due to the gravitational force exerted
by both sun and moon on the earth.

**8) What is the acceleration of free fall?**

ANSWER:-Acceleration of free fall
is the acceleration produced when a body falls under the influence of the force
of gravitation of the earth alone. It is denoted by g and its value on the
surface of the earth is 9.8 ms-2.

**9) What do we call the gravitational force between the Earth and an object?**

ANSWER:-Gravitational force
between the earth and an object is known as the weight of the object.

**10) Amit buys 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? [Hint: The value of gis greater at the poles than at the equator].**

ANSWER:-W= mg

Where,

m= Mass of the body

g= Acceleration due to gravity

The value of gis greater at poles
than at the equator. Therefore, gold at the equator weighs less than at the
poles. Hence, Amit's friend will not agree with the weight of the gold bought.

**11) Why will a sheet of paper fall slower than one that is crumpled into a ball?**

ANSWER:-When a sheet of paper is
crumbled into a ball, then its density
increases. Hence, resistance to its motion through the air decreases and it
falls faster than the sheet of paper.

**12) Gravitational force on the surface of the moon is only 1/6 as strong as gravitational force on the Earth. What is the weight in Newton’s of a 10 kg object on the moon and on the Earth?**

ANSWER:-Weight of an object on
the moon = 1/6 x Weight of an object on the Earth

Also,

Weight = Mass x Acceleration

Acceleration due to gravity, g
= 9.8 m/s2

Therefore, weight of a 10 kg
object on the Earth = 10 x 9.8 =
98 N

And, weight of the same object
on the moon= 1.6 x 9.8 = 16.3 N.

**13) A ball is thrown vertically upwards with a velocity of 49m/s. Calculate**

**(i) The maximum height to which it rises.**

**(ii) The total time it takes to return to the surface of the earth.**

ANSWER:-equation of motion under
gravity:

v 2 - u2= 2 gs

Where,

u= Initial velocity of the ball

v = Final velocity of the ball

s = Height achieved by the ball

g= Acceleration due to gravity

At maximum height, final velocity of the ball is zero,
i.e.,

v = 0

u= 49 m/s

During upward motion, g= - 9.8

m s -2

Let h be the maximum height
attained by the ball.

Hence,

Let t be the time taken by the ball to reach the
height 122.5 m, then according to the equation of motion:

v = u + gt

We get,

0= 49 + t x (- 9.8)

9.8t= 49

t= 49 / 9.8= 5s

But,

Time of ascent = Time of
descent

Therefore, total time
taken by the ball to return = 5 +
5 = 10s

**14) A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity just before touching the ground.**

ANSWER:-According to the equation of motion under
gravity:

v 2 − u2 = 2 gs

Where,

u = Initial velocity of the stone= 0

v = Final velocity of the stone

s = Height of the stone = 19.6m

g = Acceleration due to gravity

= 9.8 m s −2

∴ v2 − 0 2 = 2 × 9.8 × 19.6

v 2 = 2 ×
9.8 × 19.6 = (19.6) 2

v = 19.6 m s − 1

**15) A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g = 10 m/s 2, find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?**

ANSWER:-According to the equation of motion under
gravity:

v 2 − u2 = 2 gs

Where,

u = Initial velocity of the stone= 40m/s

v = Final velocity of the stone= 0

s = Height of the stone

g = Acceleration due to gravity= −10 m s −2

Let h be the maximum height attained by the stone.

Therefore,

0 - (40)2 = 2 x h x (-10)

h= 40 x 40 / 20 = 80 m

Therefore, total distance covered by the stone during
its upward and downward journey= 80 + 80 = 160 m

Net displacement of the stone during its upward and downward journey = 80 + (−80) = 0

**16) Calculate the force of gravitation between the earth and the Sun, given that the mass of the earth = 6 × 1024 kg and of the Sun = 2 × 1030 kg. The average distance between the two is 1.5 × 1011 m.**

ANSWER:-MSun = Mass of the Sun = 2 × 1030 kg

MEarth = Mass of the Earth = 6 ×1024 kg

R = Average distance between the Earth and the Sun =
1.5 ×1011 m

From Universal law of
gravitation,

**17) 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.**

ANSWER:-Let t be the point at which two stones meet
and let h be their height
from the ground

**18) A ball thrown up vertically returns to the thrower after 6s. Find**

**(a) The velocity with which it was thrown up,**

**(b) The maximum height it reaches, and**

**(c) Its position after 4 s.**

ANSWER:-Acceleration due to gravity, g = −9.8 m s −2

Equation of motion, v = u + gt will give,

0 = u + (−9.8
× 3)

u = 9.8 × 3
= 29.4 ms− 1

Hence, the ball was thrown upwards with a velocity of
29.4 m s −1.

(b) Let the maximum height attained by the ball be h. Initial
velocity during the upward journey, u = 29.4 m s−1

Final velocity, v = 0

Acceleration due to gravity, g = −9.8 m s −2

From the equation of motion, s= ut + 1/2 at 2

h= 29.4 x 3 + 1/2 x -9.8 x (3) 2 = 44.1 m

(c) Ball attains the maximum height after 3 s. After
attaining this height, it will start falling downwards.

In this case,

Initial velocity, u = 0

Position of the ball after 4 s of the throw is given by the distance travelled by it during its downward journey in 4 s − 3

s = 1 s.

Equation of motion, s= ut + ½ gt 2 will give,

s= 0 x t + 1/2 x 9.8 x 12 = 4.9m

Total height = 44.1 m

This means that the ball is 39.2 m (44.1 m −
4.9 m) above the ground after 4
seconds.

**19) In what direction does the buoyant force on an object immersed in a liquid act?**

ANSWER:-An object immersed in a liquid experiences
buoyant force in the upward direction.

**20) Why does a block of plastic released under water come up to the surface of water?**

ANSWER: Because of gravitational force and buoyant.

**21) The volume of 50 g of a substance is 20 cm 3. If the density of water is 1 g cm −3, will the substance float or sink?**

ANSWER:-density of the substance

= Mass of the substance /Volume of the substance

= 50 / 20

= 2.5 g cm -3

The density of the substance is more than the density of water (1 g cm −3).
Hence, the substance will sink in
water.

**22) 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 −3? What will be the mass of the water displaced by this packet?**

ANSWER:-Density of the 500 g sealed packet= Mass of the Packet /Volume of the Packet

= 500 / 350

= 1.428 g cm −3

The density of the substance is more than the density
of water (1 g cm−3) Hence, it will sink in water.

The mass of water displaced by the packet is equal to
the volume of the packet, i.e., 350g.