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

Answer 1

Answer: (D)

By superposition principle, $v_1=\frac{-GM}{2R^3}[3R^2-(\frac{R}{2}{)}^2]$

$=-\frac{11GM}{8R^3}$

Also, $v_2=-\frac{3}{2}\frac{G\left(M/8\right)}{\left(R/2\right)}=\frac{-3GM}{8R}$

The required potential is, $v=v_1-v_2$

$=-\frac{11GM}{8R}-(-\frac{3GM}{8R})$

$V=-\frac{GM}{R}$