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A mental 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 raise in temperature of the sphere is
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Verified Answer
The correct answer is:
$\frac{2 \pi^2 n^2 r^2}{5 S}$
Rotational KE of sphere $=\frac{1}{2} I \omega^2$
For sphere, $\quad I=\frac{2}{5} m r^2$
$$
\begin{aligned}
& \text { KE } \\
& =\frac{1}{2} \times \frac{2}{5} m r^2(2 \pi n)^2=\frac{1}{5} m r^2 \times 4 \pi^2 n^2 \\
& =\frac{4}{5} m r^2 \pi^2 n^2 \mathrm{~J} \\
& \therefore \text { Heat produced } \Delta Q=\frac{1}{2} \mathrm{KE} \\
& =\frac{1}{2}\left(\frac{4}{5} m r^2 \pi^2 n^2\right) \\
& =\frac{2}{5} m r^2 \pi^2 n^2 \\
& \therefore \quad \Delta Q=m s \Delta t \\
& \therefore \text { In temperature } \\
&
\end{aligned}
$$
$$
\begin{aligned}
\Delta t & =\frac{\Delta Q}{m s}=\frac{2}{5} \frac{m r^2 \pi^2 n^2}{m \cdot s} \\
& =\frac{2}{5} \frac{r^2 \pi^2 n^2}{s}
\end{aligned}
$$
For sphere, $\quad I=\frac{2}{5} m r^2$
$$
\begin{aligned}
& \text { KE } \\
& =\frac{1}{2} \times \frac{2}{5} m r^2(2 \pi n)^2=\frac{1}{5} m r^2 \times 4 \pi^2 n^2 \\
& =\frac{4}{5} m r^2 \pi^2 n^2 \mathrm{~J} \\
& \therefore \text { Heat produced } \Delta Q=\frac{1}{2} \mathrm{KE} \\
& =\frac{1}{2}\left(\frac{4}{5} m r^2 \pi^2 n^2\right) \\
& =\frac{2}{5} m r^2 \pi^2 n^2 \\
& \therefore \quad \Delta Q=m s \Delta t \\
& \therefore \text { In temperature } \\
&
\end{aligned}
$$
$$
\begin{aligned}
\Delta t & =\frac{\Delta Q}{m s}=\frac{2}{5} \frac{m r^2 \pi^2 n^2}{m \cdot s} \\
& =\frac{2}{5} \frac{r^2 \pi^2 n^2}{s}
\end{aligned}
$$
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