WBBSE Solutions For Class 8 School Science Chapter 1 Physical Environment Force Active Without Contact Solved Numerical Problems

Chapter 1 Physical Environment Solved Numerical Problems

Problem 1. Let the force of attraction between two bodies of masses M, and M, separated by a distance r, be F1. What will happen to the force of attraction if

1. only the distance between the two bodies is halved;
2. mass of the body M, is doubled?

Solution:

Given:

The force of attraction between two bodies of masses M, and M, separated by a distance r, be F1.

According to the given problem,

WBBSE Solutions for class 8 Physical Environment Section-2 Forces active without contact 4

1. In this case let the force be F2,

WBBSE Solutions for class 8 Physical Environment Section-2 Forces active without contact-3

Hence, the force becomes 4 times the initial force, F1

2. In this case let the force be F3.

= 4F.

So, F3 =

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So the force is twice the initial value.

This is also important to understand that the gravitational forces between two particles are an action-reaction pair.

The first particle exerts a force on the second particle and the second particle also exerts a force on the first particle.

These forces are equal in magnitude but oppositely directed. At this point, some of us may raise a question.

If the gravitational force is attractive in nature and if it is applicable to any object in this universe,

then why don’t we see the objects to be approaching each other on their own? and colliding between themselves? Before, answering it, let us do some mathematics.

 

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Problem 2. Let us calculate the magnitude of the gravitational force between two particles of mass 1 kg each which are 1 metre apart.
Solution:

Given:

The magnitude of the gravitational force between two particles of mass 1 kg each which are 1 metre apart.

Here, m¹ = m, = 1 kg, r = 1 metre,

G=6.67/10¹¹ N.m²/kg² Hence Force of gravitation

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Note: This value is very small. In fact, the force of gravitation (not gravity) is probably the weakest known force in nature.

This means that the law of gravitation is applicable to all the objects on the earth’s surface, but the force of attraction due to gravitation is so small that no two objects collide with each other on the earth’s surface.

Problem 3. Let us now calculate the gravitational force or gravity with which the earth pulls an object of mass 1 kg. [Given: mass of the earth is 5.96 x 1024 kg]
Solution:

Given:

Gravity with which the earth pulls an object of mass 1 kg.

Here, the mass of the object, m¹ = 1 kg

Mass of the earth, m² = 5.96 x 1024 kg

Distance of the object from the earth’s centre,

r = average radius of the earth

= 6370 kilometre

= 6370×103 metre

So, gravitational force (gravity),

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From the above two results, we find that F, (i.e. force of attraction between the earth and an object of mass 1 kg) is much greater than F, (i.e. force of attraction between the two objects of mass 1 kg each).

This means that the gravitational force exerted by the earth on an object is very much greater than the force of attraction between two objects of mass 1 kg each.

Problem 4. Calculate the magnitude of the force of gravitation between two objects of mass 15 kg and 40 kg separated by a distance of 20 cm. Given that the value of G is 6.67/1011 N.m² /kg².
Solution:

Given:

Two objects of mass 15 kg and 40 kg separated by a distance of 20 cm.

From the law of universal gravitation, we know that, IGE

Hence, G= 6.67/1011 N.m² /kg²

m¹=15kg;m² =40kg;r=20cm=0.2m hence, a force of gravitation,

Problem 5. Two bodies of masses 10,000 kg and 100 kg are separated by a distance of 1m. At which point on the line joining them does a third body, when placed, experience no net gravitational force?
Solution:

Given:

Two bodies of masses 10,000 kg and 100 kg are separated by a distance of 1m. At which point on the line joining them does a third body.

The situation is

WBBSE Solutions for class 8 Physical Environment Section-2 Forces active without contact two bodies

Let the third body of mass m be placed x m away from the body of mass 100 kg.

Distance between mass m and that of 10,000 kg mass = (1-x) m.

Let F1 and F2 be the gravitational forces of attraction of 10,000 kg and 100 kg mass on m respectively. In order that m should experience no net gravitational force, F1 and F2 should be equal in magnitude.

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Since the Resultant gravitational force will be zero at a distance of 1/11 m from the mass of 100kg (or 1 m from the mass of 10,000 kg).

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