Question

When the temperature of a rod increases from $t$ to $t+\Delta t$, its moment of inertia about a perpendicular axis passing through its centre increases from $I$ to $I+\Delta I$. If $\alpha$ be the coefficient of linear expansion of the rod, then the value of $\frac{\Delta I}{I}$ is

• A. $2\alpha \Delta t$
• B. $\alpha \Delta t$
• C. $3\alpha \Delta t$
• D. $4\alpha \Delta t$

Answer 1

The correct option is A $2\alpha \Delta t$

Given:
Temperature change $=\Delta t$
Change in moment of inertia $=\Delta I$
Coefficient of linear expansion of the rod $=\alpha$
To find:
Ratio, $\frac{\Delta I}{I}=?$
We know that, moment of inertia of a rod about a perpendicular axis passing through its centre is given by
$I=\frac{m{L}^2}{12}$ ........(1)
where $m$ is mass and $L$ is the length of the rod.
Differentiating w.r.t. $L$, we get
$\frac{dI}{dL}=\frac{1}{12}\times 2mL$
$\Rightarrow \Delta I=2\times \frac{1}{12}\times mL\Delta L$
$\Rightarrow \Delta I=2\times \frac{1}{12}m\times \frac{L^2}{L}\times \Delta L=\frac{2I\Delta L}{L}$
[ from (1) ]
$\Rightarrow \Delta I=\frac{2I\alpha L\Delta T}{L}$
[ from formula of linear expansion ]
$\Rightarrow \frac{\Delta I}{I}=2\alpha \Delta T$