When the temperature of a uniform rod is slightly raised by Δ t ,...

The moment of inertia of a body about any axis is given by $I = \int r^2 dm$, where $r$ is the perpendicular distance of the mass element $dm$ from the axis of rotation.

When the temperature is raised by $\Delta t$, the linear dimensions of the rod increase due to thermal expansion. The new perpendicular distance of each mass element from the axis becomes:
$r' = r(1 + \alpha \Delta t)$
where $\alpha$ is the coefficient of linear expansion.

The new moment of inertia is:
$I' = \int (r')^2 dm = \int r^2 (1 + \alpha \Delta t)^2 dm$

Since $\alpha$ is very small, we can neglect the higher-order term $\alpha^2 (\Delta t)^2$ and use the binomial approximation $(1 + \alpha \Delta t)^2 \approx 1 + 2\alpha \Delta t$.

Thus, the new moment of inertia becomes:
$I' = \int r^2 (1 + 2\alpha \Delta t) dm$
$I' = (1 + 2\alpha \Delta t) \int r^2 dm$
$I' = I(1 + 2\alpha \Delta t)$

The increase in the moment of inertia is:
$\Delta I = I' - I = 2\alpha I \Delta t$