Question

Inside a fixed sphere of radius $R$ and uniform desnity $\rho$, there is a sphericfal cavity of radius R/2 such that the surface of cavity passes through the center of the sphere as shown in the figure. A particle of mass $m_0$ is released from rest from centre B of cavity which strikes the center A of sphere with a velocity which is equal to $\frac{\sqrt{2\pi G\rho R^2}}{\sqrt{\beta }}$. Then the value of $\beta$ will be.

Answer 1

$V_A$ $=$ [Potential due to the complete sphere at A]$-$[Potential due to the spherical cavity at A]
$\Rightarrow V_A=$$\frac{-3GM}{2R}-\frac{-G{M}^{{}^{\prime }}}{r}$
$M=\frac{4}{3}\pi R^3\rho$
$M^{{}^{\prime }}=\frac{4}{3}\pi (\frac{R}{2}{)}^3\rho$
$=$$\frac{1}{8}\left(\frac{4}{3}\pi R^3\rho \right)=\frac{M}{8}$
Therefore
$V_A=\frac{G}{R}[\frac{\pi \rho R^3}{3}-2\pi \rho R^3]=-\frac{5}{3}\pi G\rho R^2$
Now
$V_B$$=$[Potential due to the complete sphere at B]$-$[Potential due to the spherical cavity at B]
$=\frac{-GM}{2R^3}(3R^2-r^2)-[-\frac{3G{M}^{{}^{\prime }}}{2r}]$
or
$V_B=-\frac{11GM}{8R}+\frac{3G{M}^{{}^{\prime }}}{R}=-\frac{GM}{R}$
Now,
$\frac{1}{2}{m}_0{v}^2=U_B-U_A=m_0(V_B-V_A)$
$v=\sqrt{2(V_B-V_A)}$
$v=\sqrt{\frac{2\pi G\rho R^2}{3}}$
$\beta =3$