Question

A cavity of radius $R/2$ is made inside a solid sphere of radius $R$. The centre of the cavity is located at a distance $R/2$ from the centre of the sphere. The gravitational force on a particle of mass $m$ at a distance $R/2$ from the centre of the sphere on the line joining both the centres of the sphere and the cavity is (opposite to the centre of the cavity)
[Here $g=\left(GM\right)/R^2$, where $M$ is the mass of the sphere]

• A. $\frac{mg}{2}$
• B. $\frac{3mg}{8}$
• C. $\frac{mg}{16}$
• D. $noneofthese$

Answer 1

The correct option is B $\frac{3mg}{8}$

The case can be considered as a sphere of density $\rho$ and another sphere of density $-\rho$ of the exact dimensions and position as the cavity.

Attractive force due to the sphere of density $\rho =\frac{G\left(\rho \frac{4}{3}\pi \right(R/2{)}^3)m}{(R/2{)}^2}=\frac{mg}{2}$(Since $M=\rho \frac{4}{3}\pi R^3$)

Force due to sphere of density $-\rho =-\frac{G\left(\rho \frac{4}{3}\pi \right(R/2{)}^3)}{R^2}=\frac{mg}{8}$

Hence net force $=\frac{mg}{2}-\frac{mg}{8}=\frac{3mg}{8}$