Question

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

• A. $\frac{2{\pi }^2{f}^2{r}^2}{5S}$
• B. $\frac{{\pi }^2{f}^2}{10r^{2}S}$
• C. $=\frac{7}{8}\pi r^2{f}^{2}S$
• D. $\frac{5(\pi rf{)}^2}{14S}$

Answer 1

Answer: (A)

Rotational kinetic energy of the sphere
$KE=\frac{1}{2}I{\omega }^2$ where $I=\frac{2}{5}M{r}^2$
$=\frac{1}{2}\times \frac{2}{5}M{r}^2{\omega }^2$
$=\frac{1}{5}M{r}^2(2\pi f{)}^2$
$=\frac{4}{5}{\pi }^2{r}^{2}M{f}^2$
Since half of the KE converts into heat,
$\therefore A=\frac{1}{2}\left[\frac{4}{5}{\pi }^2{r}^{2}M{f}^2\right]=M.S.\Delta t$
$MS.\Delta t=\frac{2}{5}{\pi }^2{r}^{2}M{f}^2$
$\therefore \Delta t=\frac{2}{5S}{\pi }^2{r}^2{f}^2$