If the temperature of a uniform rod is slightly elevated by Δ t ,...

The moment of inertia of a uniform rod of mass $M$ and length $L$ about its perpendicular bisector is given by

$I = \dfrac{ML^2}{12}$

When the temperature is increased by $\Delta t$, the length of the rod increases due to thermal expansion. The new length is

$L' = L(1 + \alpha \Delta t)$

The new moment of inertia is

$I' = \dfrac{M(L')^2}{12} = \dfrac{M[L(1 + \alpha \Delta t)]^2}{12}$

$I' = \dfrac{ML^2}{12}(1 + \alpha \Delta t)^2$

Since $\alpha \Delta t$ is very small, we can use the binomial approximation $(1 + x)^n \approx 1 + nx$

$I' \approx I(1 + 2\alpha \Delta t)$

$I' = I + 2\alpha I \Delta t$

The increase in the moment of inertia is

$\Delta I = I' - I = 2\alpha I \Delta t$