A metallic solid sphere is rotating about its diameter as axis of rotation. If the temperature is increased…
A metallic solid sphere is rotating about its diameter as axis of rotation. If the temperature is increased by $200^{\circ} \mathrm{C}$, the percentage increase in its moment of inertia is (Coefficient of linear expansion of the metal $=10^{-5} /{ }^{\circ} \mathrm{C}$ )
$0.1 \%$
$0.2 \%$
$0.3 \%$
$0.4 \%$
Solution
The moment of inertia of a solid sphere about the axis along its diameter is
$
\begin{array}{rlrl}
I & =\frac{2}{5} m R^2 \Rightarrow I \propto R^2 \\
& \therefore \quad \frac{\Delta I}{I} \times 100 =2\left[\frac{\Delta R}{R}\right] 100 \\
\text { But } & \alpha =\frac{\Delta R}{R \times \Delta t} \Rightarrow \frac{\Delta R}{R}=\alpha \Delta t \\
\therefore \quad & \frac{\Delta I}{I} \times 100 =2(\alpha)(\Delta t) 100 \\
& =2\left(10^{-5}\right)(200)(100)=0.4 \% .
\end{array}
$