find out the increase in moment of inertia I of a uniform rod (coefficient of linear expansion `alpha`) about its perpendiuclar bisector when itsw temperature is slightly increased by `Delta T.`

Explanation

Moment of inertial of a uniform rod of mass M length l about its perpendiuclar bisector is `I = M l^2 //12.` When rod is heated by temperature `Delta t` (quite small value) there is increase in length of rod `Delta l` given by `Delta l = l alpha Delta T `..... (i)
New Moment of inertia, `I' (M)/(12) [I + Delta l]^2 = (M)/(12) [l^2 + Delta l^2 + 2l Deltal]`
Since `Delta l` is a small quantity, the term with `Delta l^2` being very very small can be neglected, `:. I' = (M)/(12) [ I^2 + 2 l Delta l] =(Ml^2)/(12) + (M)/(6) I DeltaI = I +(M)/(6)l xx (l alpha Delta T)` [from (i)]
` = I +(Ml^2)/(6)alpha Delta T`
Increase in moment of inertia ` = l' - l = (Ml^2)/(6) alpha Delta T = 2((Ml^2)/(12)) alpha Delta T = 2 I alpha Delta T`