Question

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

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

Answer 1

Answer: (D)

Potential at point O (Centre of cavity) before removing the spherical portion,

$V_{1}=\frac{- G M}{2 R^{3}}\left(3 R^{2} - \left(\frac{R}{2}\right)^{2}\right)=\frac{- G M}{2 R^{3}}\left(3 R^{2} - \left(\frac{R}{4}\right)^{2}\right)=\frac{- 11 G M}{8 R}$
Mass of spherical portion to be removed,
$M'=\frac{M V '}{V}=\frac{m \frac{4 \pi }{3} \left(\frac{R}{2}\right)^{3}}{\frac{4 \pi }{3} R^{3}}=\frac{M}{8}$
Potential at point P due to spherical portion to be removed
$V_{2}=\frac{- 3 G M '}{2 R '}=\frac{- 3 G \left(\right. M / 8 \left.\right)}{2 \left(\right. R / 2 \left.\right)}=\frac{- 3 G M}{8 R}$
Potential at the centre of cavity formed
$V_{P}=V_{1}-V_{2}=\frac{- 11 G M}{8 R}-\left(\frac{- 3 G M}{8 R}\right)=\frac{- G M}{R}$