Question

A metal sphere of radius $r$ and specific heat $S$ is rotated about and 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. $\frac{7}{8}{\pi }^2{n}^2{r}^{2}S$
• D. $\frac{5{\left(\pi rn\right)}^2}{14S}$

Answer 1

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

Given :A metal sphere of radius rr and specific heat SS is rotated about and axis passing through its centre at a speed of nn, 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

Solution :

$\omega =2\pi n$ I = moment of inertia

Total energy =$E=\frac{1}{2}I{\omega }^2$ $\Rightarrow$ $E=\frac{1}{2}\times \frac{2}{5}m{r}^2\times {\omega }^2$

$E=\frac{1}{2}\times \frac{2}{5}m{r}^2\times (2\pi n{)}^2$

And $\frac{E}{2}=mS\Delta T$

$\frac{1}{2}\times \frac{1}{2}\times \frac{2}{5}m{r}^2\times (2\pi n{)}^2$=$mS\Delta T$

On solving we get $\Delta T=\frac{2{\pi }^2{n}^2{r}^2}{5S}$

The Correct Opt=A