Question

When the temperature of a rod increases from $t$ to $t+\Delta t$, the moment of inertia of the rod increases from $I$ to $I+\Delta I$. If the coefficient of linear expansion of the rod is $\alpha$, the ratio $\frac{\Delta I}{I}$ is

• A. $\alpha \Delta t$
• B. $2\alpha \Delta t$
• C. $\frac{\alpha \Delta t}{t}$
• D. $\frac{2\alpha \Delta t}{t}$

Answer 1

The correct option is B

$2\alpha \Delta t$

The moment of inertia of a rod of mass $m$ and length $l$ is given by $I=km{l}^2$ where $k$ is a constant.
Partially differentiating, $\Delta I=2kml\Delta l$ ..... (1)
Now, $\Delta l=\alpha l\Delta t$
Substituting in equation (1), $\Delta I=2kml\alpha l\Delta t=2km{l}^2\alpha \Delta t=2I\alpha \Delta t\Rightarrow \frac{\Delta I}{I}=2\alpha \Delta t$
Hence, the correct choice is (b).