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

Answer 1

Answer: (A)

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