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Question

A solid sphere of uniform density and radius R applies a gravitational force of attraction equal to F1 on a particle placed at a distance 2R from the centre of the sphere. A spherical cavity of radius R2 is now made in the sphere as shown in the figure. The sphere with the cavity now applies a gravitational force F2 on the same particle. The ratio F1F2 is


A
12
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B
34
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C
78
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D
97
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Solution

The correct option is D 97
Let the density of the sphere be ρ and the mass of sphere be M and let the mass of the particle at point A be m.

Mass of complete sphere =M

Force on the mass m due the solid sphere is F1=GMm(2R)2F1=GMm4R2

Mass of cavity MC=ρ×Volume of cavity

MC=M43πR3×43π(R2)3

MC=M8

By principle of super position, force acting the particle of mass m by the sphere with cavity is

F2=FcompleteFcavity

Where,
Fcomplete=F1= Force due to whole solid sphere.
Fcavity=Force due to the mass of spherical cavity.

Substituting the values,

F2=GMm(2R)2GMm8(3R2)2

F2=GMm4R2GMm18R2

F2=736GMmR2

F2=736GMmR2



Ratio of the forces will be,

F1F2=14×367

F1F2=97
Why this question: To make the student familier with use of the principle of superposition when the mass of body is changed. This can be applied for any kind of body shape.

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