Inside a fixed sphere of radius $R$ and uniform density $\rho$, there is spherical cavity of radius $\frac{R}{2}$ such that surface of the cavity passes through the center of the sphere as shown in figure. A particle of mass $m_0$ is released from rest at center $B$ of the cavity. Calculate velocity with which particle strikes the center $A$ of the sphere. Neglect earth's gravity. Initially sphere and particle are at rest.

Applying conservation of mechanical energy, Increase in kinetic energy = decrease in gravitational potential energy

or  $\frac{1}{2}{m}_0{v}^2=U_3-U_A=m_0\left(Y_3-V_A\right)$ 

$\therefore v=\sqrt{2\left(V_z-V_A\right)}$

Potential at $A$

$V_A=$ potential due to complete sphere - potential due to cavity

     $=-\frac{1.5GM}{R}-\left[-\frac{Gm}{R/2}\right]=\frac{2Gm}{R}-\frac{1.5GM}{R}$

Here, $m=\frac{4}{3}\pi {\left(\frac{R}{2}\right)}^3\rho =\frac{\pi \rho R^3}{6}$ and $M=\frac{4}{3}\pi R^3\rho$

Substituting the values, we get

$V_A=\frac{G}{R}\left[\frac{\pi \rho R^3}{3}-2\pi \rho R^3\right]=-\frac{5}{3}\pi G\rho R^2$

Potential at $B$

$V_B=-\frac{GM}{R^3}\left[1.5R^2-0.5{\left(\frac{R}{2}\right)}^2\right]+\frac{1.5Gm}{R/2}$

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

   $=\frac{G}{R}\left[\frac{\pi \rho R^3}{2}-\frac{11}{6}·\pi \rho R^3\right]=-\frac{4}{3}\pi G\rho R^2$

$\therefore V_s-V_A=\frac{1}{3}\pi G\rho R^2$

So, from Eq. (i) $v=\sqrt{\frac{2}{3}}\pi G\rho R^2$ Ans.