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Standard XII
Physics
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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  1. GM(R+a)2⎛⎜ ⎜ ⎜ ⎜ ⎜⎝1−18(1−R2a)2⎞⎟ ⎟ ⎟ ⎟ ⎟⎠
  2. GMa2⎛⎜ ⎜ ⎜ ⎜ ⎜⎝1−18(1−R2a)2⎞⎟ ⎟ ⎟ ⎟ ⎟⎠
  3. GM(R−a)2⎛⎜ ⎜ ⎜ ⎜ ⎜⎝1−18(1−2aR)2⎞⎟ ⎟ ⎟ ⎟ ⎟⎠
  4. GMR2⎛⎜ ⎜⎝1−18(1−2Ra)2⎞⎟ ⎟⎠

A
GM(R+a)2⎛⎜ ⎜ ⎜ ⎜ ⎜⎝1−18(1−R2a)2⎞⎟ ⎟ ⎟ ⎟ ⎟⎠
B
GM(R−a)2⎛⎜ ⎜ ⎜ ⎜ ⎜⎝1−18(1−2aR)2⎞⎟ ⎟ ⎟ ⎟ ⎟⎠
C
GMR2⎛⎜ ⎜⎝1−18(1−2Ra)2⎞⎟ ⎟⎠
D
GMa2⎛⎜ ⎜ ⎜ ⎜ ⎜⎝1−18(1−R2a)2⎞⎟ ⎟ ⎟ ⎟ ⎟⎠
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Solution
Verified by Toppr

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=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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144413.jpg
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