Gravitation
Class 9th Physics Part I Kerala Board Solution
- If the distance between two bodies that attract each other is trebled, how many times will…
- A body, the mass and the weight of which were already determined as the Equator, is now…
- What is meant by the terms mass and weight?
- Are they vector or scalar quantities? Why?
- The mass of a body is 30 kg. What is its weight on earth? (g = 9.8 ms-2)…
- What is its weight on the moon? (g = 1.62 ms-2)
- If a body of mass 40 kg is kept at a distance of 0.5 m from a body of mass 60 kg, what is…
- Observe the figure and complete the table.
Let Us Assess
Question 1.If the distance between two bodies that attract each other is trebled, how many times will their mutual force of attraction be?
(9 times, 3 times, 1/3, 1/9).
Answer:
The correct option is 
.
Explanation:
The mutual force of attraction between two bodies is inversely
proportional to the square of the distance between them.
where r = distance between the two bodies.
Thus, if the distance is tripled (3×), the mutual force of attraction will
become 1/9th of its original value.
Using the formula:
Now, r becomes 3r after tripling and F becomes F’ ,
as 
Question 2.
A body, the mass and the weight of which were already determined as the Equator, is now placed at the Pole. In this context, choose the correct statement from the following:
a. Mass does not change, weight is maximum
b. Mass does not change, weight is minimum
c. Both mass and weight are maximum
d. Both mass and weight are minimum.
Answer:
(a. Mass does not change, weight is maximum)
Due to the shape of the earth, the radius at the poles is less than the radius of earth at equator.
The value of acceleration due to gravity is calculated as
where
g=acceleration due to gravity
M= Mass of earth
R=radius of earth
G=Universal gravitational constant
More the radius of the earth, lesser is the acceleration due to gravity. Thus, 
is minimum at equator
and maximum at the poles.
The weight(W) of a body is given by
W=mass(m)×acceleration due to gravity(g)
⇒W=mg
At the equator, due to minimum value of acceleration due to gravity, the weight is minimum.
At the poles due to maximum value of acceleration due to gravity, the weight is maximum.
Tip: The mass of a body does not depend on acceleration due to gravity.
Thus, the mass remains constant at both the equator and the poles.
Question 3.
What is meant by the terms mass and weight?
Answer:
The mass is the fundamental property of a body that is a measure of its resistance to change in the state of motion. It also describes the strength of its mutual gravitational force of attraction to other bodies. It does not depend on the value of g. It is measured by a beam balance.
The weight of a body is the force with which earth pulls the body towards its centre.
Weight(W)=mass(m)×Acceleration due to gravity(g)
W=mg
It changes with the change in the value of g. It is measured by a spring balance.
Question 4.
Are they vector or scalar quantities? Why?
Answer:
Weight is a force. It has both magnitude and a direction. The direction of weight is towards the centre of the earth. Thus, weight is vector quantity.
Mass has no direction but only magnitude. Thus, mass is scalar quantity.
Question 5.
The mass of a body is 30 kg. What is its weight on earth?
(g = 9.8 ms-2)
Answer:
Mass of body = 30 kg
Acceleration due to gravity = 9.8 ms-2
Formula used:
Weight(W) = mass(m)×Acceleration due to gravity(g)
⇒ W = 30 kg × 9.8 ms-2
=294 kgms-2
Question 6.
What is its weight on the moon?
(g = 1.62 ms-2)
Answer:
Mass of the body=30 kg
Acceleration due to gravity on moon=1.62 ms-2
Formula used:
Weight(W) = mass(m)×Acceleration due to gravity(g)
W1=30×1.62=48.6 kgms-2
Question 7.
If a body of mass 40 kg is kept at a distance of 0.5 m from a body of mass 60 kg, what is the mutual force of attraction between them?
Answer:
Given:
Mass of one body(M)=60 kg
Mass of other body(m)=40 kg
Distance between the bodies(R)=0.5 m
Formula Used:
where
F=Mutual force of attraction between two bodies
G=Universal Gravitational Constant=6.67×10-11 Nm2kg-2
M=Mass of one body
m=Mass of other body
R=Distance between the bodies
Putting these values,
Question 8.
Observe the figure and complete the table.
Answer:
Given:
Mass of ball A=20 kg
Mass of ball B=30 kg
Mass of ball C=60 kg
Formula Used:
where
F=Mutual force of attraction between two bodies
G=Universal Gravitational Constant=6.67×10-11 Nm2/kg2
M=Mass of one body
m=Mass of other body
r=Distance between the bodies
Tip: We have to consider two attracting bodies at a time to calculate the mutual force between them.
For A and B,
For B and C,
For C and A,
Extended Activities
Question 1.Collect different types of lamina and determine their centres of gravity.
Answer:
Question 2.
Prepare a table of things that can keep their stability by overcoming great disturbances.
Answer:
For a body to remain stable during great disturbances, the vertical line through its centre of gravity must pass through its base. This condition is known as stable equilibrium.
Question 3.
The values of g on different planets are given. Determine the weight of a body of mass 100 kg on these planets.
Answer:
Given:
Mass of body(m)=100 kg
Formula Used:
Weight(W)=Mass(m)×Acceleration due to gravity(g)
W=mg
