Explanation

(a) Moment of inertia of a rod,
$I = \frac{1}{12}ML^{2}$ $\ldots(i)$
Where $M$ is the mass of the rod and $L$ is the length of the rod
$\therefore\mathrm{\Delta}I = \frac{1}{12}2ML\mathrm{\Delta}L\ (\because M\ is\ a\ constant)$ $\ldots(ii)$
Divide (ii) by (i), we get
$\frac{\mathrm{\Delta}I}{I} = 2\frac{\mathrm{\Delta}L}{L}$ $\ldots(iii)$
As $\mathrm{\Delta}L = L\ \alpha\mathrm{\Delta}t$
Or $\frac{\mathrm{\Delta}L}{L} = \alpha\mathrm{\Delta}t$
Substituting the value of $\frac{\mathrm{\Delta}L}{L}$ in (iii), we get
$\frac{\mathrm{\Delta}I}{I} = 2\alpha\mathrm{\Delta}t$