A metal sphere of radius $r$ and specific heat $S$ is rotated about an axis passing through its centre at a speed of $n$ rotations per second. It is suddenly stopped arid $50\%$ of its energy is used in increasing its temperature. Then the rise in temperature of the sphere is:

Moment of inertia pf the sphere $I=\frac{2}{5}M{r}^2$. Given $\omega$= n rotations per second= 2$\mathrm{\pi } \mathrm{n} \mathrm{rad} {\mathrm{s}}^{-1}$• The kinetic energy is

$$\begin{gathered} KE=\frac{1}{2}I{\omega }^2=\frac{1}{2}\times \frac{2}{5}M{r}^2\times {\left(2\mathrm{\pi n}\right)}^2 \\ =\frac{4}{5}{\mathrm{M\pi }}^2{\mathrm{r}}^2{\mathrm{n}}^2 \end{gathered}$$

Since half of KE is converted into heat energy, we have 

$Q=\frac{1}{2}\times KE=\frac{2}{5}M{\mathrm{\pi }}^2{\mathrm{r}}^2{\mathrm{n}}^2$

Now $Q=MS∆T$ which gives

$∆T=\frac{Q}{MS}=\frac{{\displaystyle \frac{2}{5}}M{\mathrm{\pi }}^2{r}^2{n}^2}{MS}=\frac{2{\mathrm{\pi }}^2{\mathrm{r}}^2{\mathrm{n}}^2}{5S}$