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# 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 isGM(R+a)2⎛⎜ ⎜ ⎜ ⎜ ⎜⎝1−18(1−R2a)2⎞⎟ ⎟ ⎟ ⎟ ⎟⎠GMa2⎛⎜ ⎜ ⎜ ⎜ ⎜⎝1−18(1−R2a)2⎞⎟ ⎟ ⎟ ⎟ ⎟⎠GM(R−a)2⎛⎜ ⎜ ⎜ ⎜ ⎜⎝1−18(1−2aR)2⎞⎟ ⎟ ⎟ ⎟ ⎟⎠GMR2⎛⎜ ⎜⎝1−18(1−2Ra)2⎞⎟ ⎟⎠

A
GM(R+a)2⎜ ⎜ ⎜ ⎜ ⎜118(1R2a)2⎟ ⎟ ⎟ ⎟ ⎟
B
GM(Ra)2⎜ ⎜ ⎜ ⎜ ⎜118(12aR)2⎟ ⎟ ⎟ ⎟ ⎟
C
GMR2⎜ ⎜118(12Ra)2⎟ ⎟
D
GMa2⎜ ⎜ ⎜ ⎜ ⎜118(1R2a)2⎟ ⎟ ⎟ ⎟ ⎟
Solution
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#### Let ρ be the mass density. mass of metal sphere M=ρ.43πR3mass of hollow sphere , m=ρ.43π(R/2)3=M/8The field at that point = field due to metal sphere − field due to hollow sphere.f=GMa2−Gm(a−R2)2=GMa2−GM8(a−R2)2=GMa2−GM8a2(1−R2a)2=GMa2⎛⎝1−18(a−R2a)2⎞⎠

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