Question

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 centre of the sphere as shown in figure. A particle of mass $m$ is released from rest at centre $\text{B}$ of the cavity. Calculate velocity with which particle strikes the centre of the sphere. Neglect earth's gravity. Initially sphere and particle are at rest.

• A. $\sqrt{\frac{2}{3}{\pi }^2{G}^2{\rho }^2{R}^2}$
• B. $\sqrt{\frac{2}{3}\pi G{\rho }^2{R}^2}$
• C. $\sqrt{\frac{2}{3}\pi G\rho R^2}$
• D. $\sqrt{\frac{2}{5}\pi G\rho R^2}$

Answer 1

The correct option is C $\sqrt{\frac{2}{3}\pi G\rho R^2}$

Since internal and non-conservative forces all are absent in this system, So, the mechanical energy of the system (sphere $+m$) will be conserved.


Applying conservation of mechanical energy,

$K.E_B+P.E_B=K.E_A+P.E_A$

Initially the particle is at rest, so $K.E_B=0$

If $v$ is the velocity of mass $m$ at point $\text{A}$, then

$\Rightarrow 0+m{V}_B=\frac{1}{2}m{\nu }^2+m{V}_A$

$\Rightarrow \frac{1}{2}m{v}^2=m(V_B-V_A)...\left(1\right)$

Where,
$V_A$ is the net potential at point $\text{A}$ and $V_B$ is the net potential at point $\text{B}$.

Now,
$V_B=\text{potential due to complete sphere (V)+potential due to cavity}\left(V_C\right)...\left(2\right)$

Potential due to complete solid sphere at distance $r$ from its centre inside is given by, $$V=-\frac{GM}{R^3}\frac{(3R^2-r^2)}{2}$$ For point $\text{B}$, $r=\frac{R}{2}$

$\Rightarrow V=-\frac{GM}{R^3}\frac{(3R^2-(R/2{)}^2)}{2}$

$\Rightarrow V=-\frac{11GM}{8R}$

Now,
Potential at point $\text{B}$ due to cavity of negative mass of $-M_C$

$V_C=-\frac{3G(-M_C)}{2\times \frac{R}{2}}$

$\Rightarrow V_C=\frac{3G{M}_C}{R}$

Here,
$M=\text{Mass of solid sphere}=\frac{4}{3}\pi R^3\rho$
$M_C=\text{Mass of the cavity}=\frac{4}{3}\pi {\left(\frac{R}{2}\right)}^3\rho =\frac{1}{6}\pi R^3\rho$

Substituting the values in equation $\left(2\right)$, we get

$V_B=-\frac{11GM}{8R}+\frac{3G{M}_C}{R}$

$\Rightarrow V_B=-\frac{11G\times \left(\frac{4}{3}\pi R^3\rho \right)}{8R}+\frac{3G\times \left(\frac{1}{6}\pi R^3\rho \right)}{R}$

$\Rightarrow V_B=\pi G{R}^2\rho (-\frac{11}{6}+\frac{1}{2})$

$\Rightarrow V_B=-\frac{4\pi G{R}^2\rho }{3}$

Similarly potential at point $\text{A}$ which is a centre of complete sphere, can be calculated by using the formula,

$V_A=\text{potential due to complete sphere (V)+potential due to cavity}\left(V_C\right)$

$\Rightarrow V_A=-\frac{3GM}{2R}+[-\frac{G(-M_C)}{R/2}]$

Substituting the values of $M$ and $M_C$,

$\Rightarrow V_A=-\frac{3G\times \left(\frac{4}{3}\pi R^3\rho \right)}{2R}+\frac{G\times \left(\frac{1}{6}\pi R^3\rho \right)}{R/2}$

$\Rightarrow V_A=\pi G{R}^2\rho (-2+\frac{1}{3})$

$\Rightarrow V_A=-\frac{5\pi G{R}^2\rho }{3}$

Substituting the values of $V_A$ and $V_B$ in equation $1$, we get

$\frac{1}{2}m{v}^2=m[-\frac{4\pi G{R}^2\rho }{3}-(-\frac{5\pi G{R}^2\rho }{3})]$

$\Rightarrow v^2=\frac{2\pi G{R}^2\rho }{3}$

$\therefore v=\sqrt{\frac{2\pi G{R}^2\rho }{3}}$

Hence, option (a) is the correct answer.

Key Concept: The gravitational potential for solid sphere at an internal point at any distance $r$ from the centre is given by $$V=\frac{GM}{2R^3}(3R^2-r^2)$$ Why this question: To help students apply the principle of superposition for potential of continuous mass distribution.