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

The moment of inertia of a rod of mass M and length L is given by 

                $I= kM{L}^2$                                  (1)

where $k=\frac{1}{12}$ if the axis of rotation is through the centre of rod ·and $k=\frac{1}{3}$ if the axis of rotation is through the end of the rod. Partially differentiating (1), we have

$$\begin{gathered} ∆I=2kML∆L \\ Now ∆L=L\alpha ∆t. Therefore, \\ ∆I=2kML\times L\alpha ∆t \\ = 2\left(kM{L}^2\right)\alpha ∆t=2I\alpha ∆t \\ or \frac{∆I}{I}=2\alpha ∆t \end{gathered}$$