Question

If the temperature of a uniform rod is slightly increased by $\Delta t$, its moment of inertia I about a perpendicular bisector increases by -

• A. zero
• B. $\alpha I\Delta t$
• C. $2\alpha I\Delta t$
• D. $3\alpha I\Delta t$

Answer 1

The correct option is C $2\alpha I\Delta t$

We know that moment of inertia of a road along its perpendicular bisector is -

$I$ = $\frac{m{l}^2}{12}$.

As $\Delta l$ is small, $\Delta I$ will be a much smaller quantity,

$\therefore$ $\Delta I$ $\sim$ $dI$.

And,

$dI$ = $d{l}^2\left(\frac{m}{12}\right)$

$\Rightarrow \Delta I$ $\approx$ $dI$ = $2\times l\times \Delta l\times \frac{m}{12}$. ..........(1)

Now,

$\frac{\Delta l}{l}$ = $\alpha \Delta t$

$\Rightarrow \Delta l$ = $\alpha l\Delta t$. ................(2)

From (1) & (2),

$\Delta I$ = $2\times \left(\frac{m{l}^2}{12}\right)\times \alpha \Delta t$

$\Rightarrow$ $\Delta I$ = $2\alpha I\Delta t$