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 $r=\in fty,$ the potential at the centre of the cavity thus formed is:

$\left(G=universalgravitationalconstant\right)$

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

Answer 1

Answer: (C)

Potential due to whole sphere if cavity is not there at distance $=\frac{-GM}{R^3}{\left(\frac{3}{2}{R}^2-0.5r^2\right)}_{r=\left(\frac{R}{2}\right)}$
$=\frac{-GM}{R^3}\left(\frac{3}{2}{R}^2-\frac{R^2}{8}\right)$
$=\frac{-GM}{R^3}\left(\frac{12R^2-R^2}{8}\right)$
$=\frac{-11GM}{8R}$
Let mass of Sphere Of radius $\frac{R}{2}$ is M'
$\therefore \therefore$ M 4 3 π R 3 = M 1 4 3 π R 2 3
$\Rightarrow {\text{M}}^{{}^{\prime }}=\frac{\text{M}}{8}$
Potential due to the sphere of $\frac{\text{R}}{2}$ radius at its centre is
$=\frac{-3}{2}\frac{{\text{GM}}^{{}^{\prime }}}{\frac{\text{R}}{2}}$ $=\frac{-3}{2}\frac{\text{GM}\times 2}{8R}$ $=\frac{-3}{8}\frac{\text{GM}}{R}$ ____(2)
∴ Net Potential at $\text{r}=\frac{\text{R}}{2}$ is = (1) - (2)
$=\frac{-11}{8}\frac{\text{GM}}{\text{R}}+\frac{3}{8}\frac{\text{GM}}{\text{R}}=\frac{-\text{GM}}{\text{R}}$