Question

A small mass $m$ is transferred from the centre of a hollow sphere of mass $M$ and radius $R$ to infinity. During this process $W_1$ is the amount of work done. If instead of a hollow sphere, a solid sphere of mass $\frac{M}{2}$ and radius $\frac{R}{3}$ were there, then $W_2$ is the amount of work done. The ratio of $\frac{W_1}{W_2}$ will be

• A. $\frac{1}{3}$
• B. $\frac{4}{3}$
• C. $\frac{4}{9}$
• D. $-\frac{3}{2}$

Answer 1

The correct option is C $\frac{4}{9}$

The gravitational potential is taken as zero at infinity.

Thus, if $V_i$ be the gravitational potential at the centre of hollow sphere, then

Work done by external force = Change in potential energy.

For hollow sphere,

Gravitational potential at the centre of the hollow sphere, $V_i=-\frac{GM}{R}$

Gravitational potential at infinity, $V_f=0$

Workdone by external force, $W_1=m(V_f-V_i)$
$\Rightarrow W_1=m(0-(-\frac{GM}{R}\left)\right)$
$\Rightarrow W_1=\frac{GMm}{R}$

For solid sphere,

Gravitational potential at the centre of the solid sphere, $V_i=-\frac{3G\left(M/2\right)}{2\left(R/3\right)}$

Gravitational potential at infinity, $V_f=0$

Workdone by external force, $W_2=m(V_f-V_i)$
$\Rightarrow W_2=m(0-(-\frac{9GM}{4R}))$
$\Rightarrow W_2=\frac{9GMm}{4R}$

According to question ,
$\frac{W_1}{W_2}=\frac{\frac{GMm}{R}}{\frac{9GMm}{4R}}$
$\Rightarrow \frac{W_1}{W_2}=\frac{4}{9}$

Hence, option (c) is correct.

Why this question? We can see that the work required to transfer the mass from the centre of the solid sphere is higher than hollow spherical case.