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

• A. $\frac{GM}{R^2}(1-\frac{1}{8{(1-\frac{2R}{a})}^2})$
• B. $\frac{GM}{a^2}(1-\frac{1}{8{(1-\frac{R}{2a})}^2})$
• C. $\frac{GM}{(R+a{)}^2}(1-\frac{1}{8{(1-\frac{R}{2a})}^2})$
• D. $\frac{GM}{(R-a{)}^2}(1-\frac{1}{8{(1-\frac{2a}{R})}^2})$

Answer 1

The correct option is B $\frac{GM}{a^2}(1-\frac{1}{8{(1-\frac{R}{2a})}^2})$

Let $\rho$ be the mass density.
mass of metal sphere $M=\rho .\frac{4}{3}\pi R^3$
mass of hollow sphere , $m=\rho .\frac{4}{3}\pi (R/2{)}^3=M/8$
The field at that point $=$ field due to metal sphere $-$ field due to hollow sphere.
$f=\frac{GM}{a^2}-\frac{Gm}{(a-\frac{R}{2}{)}^2}=\frac{GM}{a^2}-\frac{GM}{8(a-\frac{R}{2}{)}^2}=\frac{GM}{a^2}-\frac{GM}{8a^2(1-\frac{R}{2a}{)}^2}=\frac{GM}{a^2}(1-\frac{1}{8(a-\frac{R}{2a}{)}^2})$