Question

From a solid sphere of mass M and radius R, a spherical portion of radius $\frac{R}{2}$ is removed, as shown in the figure. Taking gravitational potential $V=0$ at $r=\infty$, the potential at the centre of the cavity thus formed is (G = gravitational constant)

• A. $-\frac{GM}{R}$
• B. $-\frac{2GM}{3R}$
• C. $-\frac{2GM}{R}$
• D. $\frac{-GM}{2R}$

Answer 1

The correct option is A $-\frac{GM}{R}$

As $M\alpha R^3$ mass of the cavity is M/8.
$V_{required}=V_M-V_{M/8}$
$=-\frac{GM}{W{R}^3}[3R^2-\frac{R^2}{4}]+\frac{\frac{GM}{8}}{2{\left(\frac{R}{2}\right)}^3}\left[3{\frac{R}{2}}^2\right]$
$=-\frac{11GM}{8R}+\frac{3GM}{8R}=-\frac{GM}{R}$