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 $W_1=100\text{kJ}$ amount work is done. If instead of a hollow sphere, a solid sphere of mass $\frac{M}{2}$ and radius $\frac{R}{3}$ were there, then $W_2$ amount of work is done. The $W_2$ will be

• A. $200\text{kJ}$
• B. $150\text{kJ}$
• C. $66.67\text{kJ}$
• D. $225\text{kJ}$

Answer 1

The correct option is D $225\text{kJ}$

Given,

Work done for hollow spehere, $W_1=100\text{kJ}$,

The gravitational potential is taken as zero at infinity.

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

$\text{Work done by external force = Change in potential energy}$

$\text{For hollow sphere:}$

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,

$W_1=m(V_f-V_i)$

$\Rightarrow W_1=m(0-(-\frac{GM}{R}\left)\right)$

$\Rightarrow W_1=\frac{GMm}{R}$

According to the problem,

$W_1=\frac{GMm}{R}=100\text{kJ}......\left(1\right)$

$\text{For solid sphere:}$

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,

$W_2=m(V_f-V_i)$

$\Rightarrow W_2=m(0-(-\frac{9GM}{4R}))$

$\Rightarrow W_2=\frac{9GMm}{4R}$

From equation $\left(1\right)$,

$\Rightarrow W_2=\frac{9}{4}\times 100$

$\therefore W_2=225\text{kJ}$

Hence, option (d) 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 sphere of same mass and radius.