Question

A metallic sphere of radius $r$ and specific heat $S$ is rotated about an axis passing through its centre of mass at an angular speed of $n$ 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. $\frac{2{\pi }^2{n}^2{r}^2}{5S}$
• B. $\frac{{\pi }^2{n}^2}{10r^{2}S}$
• C. $\frac{7}{8}\pi r^2{n}^{2}S$
• D. $\frac{5(\pi rn{)}^2}{14S}$

Answer 1

The correct option is A $\frac{2{\pi }^2{n}^2{r}^2}{5S}$

Moment of Inertia of the solid metallic sphere, $I=\frac{2}{5}M{r}^2$
Given , $\omega =n$ rotations per second or $2\pi n\text{rad/sec}$
Rotational kinetic energy, $\left(K\right)=\frac{1}{2}I{\omega }^2=\frac{1}{2}\left(\frac{2}{5}M{r}^2\right)(2\pi n{)}^2$
or $\left(K\right)=\frac{4}{5}M{\pi }^2{r}^2{n}^2$
Half of this kinetic Energy is converted into heat.
$dQ=\frac{K}{2}=\frac{2}{5}M{\pi }^2{r}^2{n}^2$
But we know that,
$dQ=MSdT$
$\Rightarrow dT=\frac{dQ}{MS}=\frac{\frac{2}{5}M{\pi }^2{r}^2{n}^2}{MS}$
$\Rightarrow dT=\frac{2{\pi }^2{r}^2{n}^2}{5S}$
Thus, option (a) is the correct answer.