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Question

A metal sphere has radius R and mass M. A spherical hollow of diameter R is made in this sphere such that its surface passes through the centre of the metal sphere and touches the outside surface of the metal sphere. A unit mass is placed at a distance from the centre of metal sphere. The gravitational field at that point is
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A
GMR2⎜ ⎜118(12Ra)2⎟ ⎟
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B
GMa2⎜ ⎜ ⎜ ⎜ ⎜118(1R2a)2⎟ ⎟ ⎟ ⎟ ⎟
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C
GM(R+a)2⎜ ⎜ ⎜ ⎜ ⎜118(1R2a)2⎟ ⎟ ⎟ ⎟ ⎟
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D
GM(Ra)2⎜ ⎜ ⎜ ⎜ ⎜118(12aR)2⎟ ⎟ ⎟ ⎟ ⎟
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Solution

The correct option is B GMa2⎜ ⎜ ⎜ ⎜ ⎜118(1R2a)2⎟ ⎟ ⎟ ⎟ ⎟
Let ρ be the mass density.
mass of metal sphere M=ρ.43πR3
mass of hollow sphere , m=ρ.43π(R/2)3=M/8
The field at that point = field due to metal sphere field due to hollow sphere.
f=GMa2Gm(aR2)2=GMa2GM8(aR2)2=GMa2GM8a2(1R2a)2=GMa2118(aR2a)2

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