The moment of Inertia  $I$ of uniform rod about a perpendicular bisector increases to  $I+\Delta I,$  If the temperature is increased slightly by  $\Delta T.$  If the coefficient of linear expansion is '$\alpha$' then $\frac{\Delta I}{I}$ is (Assume  $\frac{\Delta T}{T}<<1$ )

Moment of inertia of rod about perpendicular bisector before heating 

$I=\frac{1}{12}M{L}^2$ , now moment of inertia after heating  $I+\Delta I=\frac{1}{12}M{(L+\Delta L)}^2$

$I(1+\frac{\Delta I}{I})=\frac{1}{12}M{L}^2{(1+\frac{\Delta L}{L})}^2$  $1+\frac{\Delta I}{I}={(1+\frac{\Delta L}{L})}^2$

(Apply binomial expansion ${(1+x)}^n\approx 1+nxifx<<1$  )

$1+\frac{\Delta I}{I}=1+2\frac{\Delta L}{L}$   $\frac{\Delta I}{I}=2\alpha \Delta T(\Delta L=L\alpha \Delta T)$