Question

# A uniform ring of mass m and radius a is placed directly above a uniform sphere of mass (M) and of equal radius. The centre of the ring is at a distance √3a from the centre of the sphere. The gravitational force (F) exerted by the sphere on the ring is

A
3GMm8a2
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B
2GMm3a2
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C
7GMm2a2
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D
3GMma2
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Solution

## The correct option is D √3GMm8a2Given data :Mass of the ring - mRadius of the ring - aMass of the sphere - MRadius of sphere - aDistance between center of ring and center of sphere - r=√3aThe gravitational field, F′ of the sphere is given by:F′=GMr(r2+a2)32 ........(1)where, G is gravitational constant.Using the value of r in equation (1), we get:F′=GM(√3a)((√3a)2+a2)32F′=√3GMa(3a2+a2)32F′=√3GMa(4a2)32F′=√3GMa(23)(a3)F′=√3GM8(a2)Now, the force, F exerted by the sphere on the ring is,F=F′×mF=(√3GM8(a2))mF=√3GMm8(a2)

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