When the temperature of a rod increases from t to $\mathrm{t}+\Delta \mathrm{t}$, its moment of iner

Question

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

Professor PC Thomas Classes · Answer

Moment of inertia of a rod $$\begin{equation*} \mathrm{I}=\frac{1}{12} \mathrm{ML}^{2} 	ag{i} \end{equation*}$$ Where M is the mass of the rod and L is the length of the rod\ $	herefore \Delta \ell=\frac{1}{12} 2 \mathrm{ML} \Delta \mathrm{L}(\because \mathrm{M}$ isconstant $)$\ Divide (ii) by (i), we get $$\begin{equation*} \frac{\Delta \mathrm{I}}{\mathrm{I}}=2 \frac{\Delta \mathrm{~L}}{\mathrm{~L}} 	ag{iii} \end{equation*}$$ As $\Delta \mathrm{L}=\mathrm{L} \alpha \Delta \mathrm{t}$ or $\frac{\Delta \mathrm{L}}{\mathrm{L}}=\alpha \Delta \mathrm{t}$\ Substituting the value $\frac{\Delta \mathrm{L}}{\mathrm{L}}$ in (iii), we get\ $\frac{\Delta \mathrm{I}}{\mathrm{I}}=2 \alpha \Delta \mathrm{t}$\