Question

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 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 a sphere,
$I=\frac{2}{5}m{r}^2$
Given : $\omega =2\pi nrad/sec$
$\therefore KE=\frac{1}{2}I{\omega }^2=\frac{1}{2}.\frac{2}{5}m{r}^2\times (2\pi n{)}^2$
$=\frac{4}{5}m{\pi }^2{r}^2{n}^2$
Half of this energy is converted into heat.
$i.e.,dQ=\frac{4}{5}m{\pi }^2{r}^2{n}^2\times \frac{1}{2}$
$\therefore$ Specific heat, $S=\frac{1}{m}\frac{dQ}{dT}$
or $dT=\frac{dQ}{mS}$ or $dT=\frac{\frac{2}{5}m{\pi }^2{r}^2{n}^2}{mS}$
or $dT=\frac{2{\pi }^2{r}^2{n}^2}{5S}$
Hence, the correct answer is option (a).