M.I. about its axis along perpendicular bisector
= \(\frac{1}{12}\) ml2
When temperature increased by ∆T, length of rod increases.
∆l = lα∆T
∴ New M.I., I1 = \(\frac{M}{12}\) (l + ∆l)2 = \(\frac{M}{12}\) (l2 + ∆l2 + 2l∆l)
Neglecting (∆l)2 (very very small quantity) –
I1 = \(\frac{M}{12}\) (l2 + 2l∆l)
= \(\frac{Ml^2}{12}+\frac{Ml∆l}{6}\) = I + \(\frac{Ml∆l}{6}\) .
Therefore, moment of inertia, increase.
∆I = I1 − I = \(\frac{M_l∆l}{6}\)
= 2( \(\frac{Ml^2}{12}\) )\(\frac{∆l}{l}\)
∆l = 2l ∝ ∆T.