Question

A solid sphere of mass M and radius R is divided into two unequal parts. The first part has a mass of $\frac{7M}{8}$ and is converted into a uniform disc of radius $2\text{R}$. The second part is converted into a uniform solid sphere. Let $I_1$ be the moment of inertia of the disc about its axis and $I_2$ be the moment of inertia of the new sphere about its axis. The ratio $\frac{I_1}{I_2}$ is given by

Answer 1

The given situation is shown in the figure given below

Density of given sphere of radius R is
$\rho =\frac{M}{V}=\frac{M}{\frac{4}{3}\pi R^3}$
Let radius of sphere formed from second part is r, then
mass of second part = $V\times \rho$
$\frac{1}{8}M=\frac{4}{3}\pi r^3\times \frac{M}{\frac{4}{3}\pi R^3}$
$\therefore r^3=\frac{R^3}{8}\Rightarrow r=\frac{R}{2}$
Now, $I_1$= moment of inertia of disc (radius is 2R)
$=\frac{\text{Mass}\times R^2}{2}=\frac{\frac{7}{8}M\times (2R{)}^2}{2}=\frac{7}{4}M{R}^2$
and $I_2$ = moment of inertia of sphere
$\left(\text{radius}\frac{R}{2}\text{and mass}\frac{1}{8}M\right)\text{about its axis}=\frac{2}{5}\times \text{Mass}\times (\text{Radius}{)}^2$
$=\frac{2}{5}\times \frac{1}{8}M\times {\left(\frac{R}{2}\right)}^2=\frac{M{R}^2}{80}$
$\therefore \frac{I_1}{I_2}=\frac{\frac{7}{4}M{R}^2}{\frac{1}{80}M{R}^2}=140$