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:

$\left(G = universal g r a v i t a t i o n a l \, c o n s t a n t\right)$

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

Answer 1

Answer: (C)

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