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

Explanation

The moment of inertia of uniform rod
`I = 1/12 Ml^2`
When heated by temperature `Delta T` of quite small value, the increase in length of the rod is
` Delta l = l alpha Delta T`
the new M.I. of the rod is
` I. = 1/12 M (l + Delta l)^2`
` = 1/12 Ml^2 + 1/12 xx 2M l Delta l + 1/12 M (Delta l)^2`
since `Delta l` is small , the term containing `(Delta l)^2` is neglected
` therefore I. = 1/12 Ml^2 + 1/12 M2l Delta l `
since ` Delta l = l Delta T alpha `
so `I. = (Ml^2)/(12) + M/6 (l Delta l)`
`=I + (M/6) xx l^2 alpha Delta T`
` = I + ( (Ml^2)/(6) ) xx alpha Delta T`
`Delta I = I. - I = ( (Ml^2)/(6) ) xx alpha Delta T`
` = 2 ( (Ml^2)/(12) ) xx alpha Delta T = 2I alpha Delta T`