Find the increase in the moment of inertia $I$ of a uniform rod (coefficient of linear expansion $\alpha$) about its perpendicular bisector when its temperature is slightly increased by $\Delta T$.

(N/A) Let the mass of the rod be $M$ and its length be $l$. The moment of inertia of the rod about its perpendicular bisector is given by:
$I = \frac{Ml^2}{12}$
When the temperature increases by $\Delta T$,the change in length $\Delta l$ is given by:
$\Delta l = l \alpha \Delta T$
The new length of the rod is $l' = l + \Delta l$. The new moment of inertia $I'$ is:
$I' = \frac{M(l + \Delta l)^2}{12}$
Expanding the term:
$I' = \frac{M}{12}(l^2 + 2l\Delta l + (\Delta l)^2)$
Since $\Delta l$ is very small,we can neglect the $(\Delta l)^2$ term:
$I' \approx \frac{M}{12}(l^2 + 2l\Delta l) = \frac{Ml^2}{12} + \frac{2Ml\Delta l}{12} = I + \frac{Ml\Delta l}{6}$
The increase in moment of inertia $\Delta I = I' - I$ is:
$\Delta I = \frac{Ml\Delta l}{6}$
Substituting $\Delta l = l \alpha \Delta T$:
$\Delta I = \frac{Ml(l \alpha \Delta T)}{6} = \frac{Ml^2 \alpha \Delta T}{6} = 2I \alpha \Delta T$