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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, distant 2R from the center of the sphere. A spherical cavity of radius R/2 is now made on the sphere with cavity now applies a gravitational forceF2. Then F2F1 will be

A
1/2
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B
3/4
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C
7/8
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D
14/9
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Solution

## The correct option is C 7/8Let, M is the mass of the sphere of uniform density ρ and m is the mass of the particle.The gravitational force of attraction F1 is given by,F1=GMm(2R)2 ........................(1)The mass of sphere of volume V, is given byM=ρVM=ρ43πR3 ......................(2)From equations (1) and (2), we getF1=G[43πR3]m(2R)2 ...................(3)When the spherical cavity is made in sphere, the gravitational force F2 will beF2=G[M−ΔM]m(2R)2F2=G[43πR3−43π(R2)3]m(2R)2F2=G[43πR3(1−(12)3)]m(2R)2 .......................(4)Now,F2F1=G[43πR3(1−(12)3)]m(2R)2G[43πR3]m(2R)2 ...............from equations (3) and (4)F2F1=1−(12)3F2F1=1−(18)F2F1=78

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