Question 150 of 223

From a solid sphere of mass $M$ and radius $R\text{,}$ a spherical portion of radius $\left(\dfrac{R}{2}\right)$ 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$\dfrac{-2GM}{R}$B$\dfrac{-GM}{2R}$C$\dfrac{-GM}{R}$D$\dfrac{-2GM}{3R}$

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Answer: (C) $\dfrac{-GM}{R}$

Central idea is, consider the cavity as negative a mass and apply the superposition of gravitational potential.
Consider the cavity formed in a solid sphere as shown in the figure.
The gravitational potential at infinite distance,
$V=\dfrac{-GM}{r}$
$\Rightarrow V\left(\infty \right)=0$

According to the question, we can write potential at an internal point $P$ due to complete solid sphere,
${V}_{s}=-\dfrac{GM}{2{R}^{3}}\left[3{R}^{2}-{\left(\dfrac{R}{2}\right)}^{2}\right]$
$=\dfrac{-GM}{2{R}^{3}}\left[3{R}^{2}-\dfrac{{R}^{2}}{4}\right]$
$=\dfrac{-GM}{2{R}^{3}}\left[\dfrac{11{R}^{2}}{4}\right]=\dfrac{-11GM}{8R}$
Mass of removed part
$=\dfrac{M}{\dfrac{4}{3}\times \pi {R}^{3}}\times \dfrac{4}{3}\pi {\left(\dfrac{R}{2}\right)}^{3}=\dfrac{M}{8}$
The potential at a point $P$ due to removed part
${V}_{C}=\dfrac{-3}{2}\times \dfrac{\left(\dfrac{GM}{8}\right)}{R}=\dfrac{-3GM}{8R}$
Thus, potential due to the remaining part at the point $P\text{,}$
${V}_{P}={V}_{s}-{V}_{c}=\dfrac{-11GM}{8R}-\left(-\dfrac{3GM}{8R}\right)$
$\dfrac{\left(-11+3\right)GM}{8R}=\dfrac{-GM}{R}$