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

Let the mass and length of a uniform rod be M and l respectively.
Moment of intertia of the rod abour its perpendicular bisector `(I) = (Ml^(2))/(12)` ...(i)

Length in length of the rod when temperature is increased by `Delta T`, is given by
`Delta l = l. alpha Delta T`
`:.` New moment of intertia of the `(I) = (M)/(12) (l+Delta l)^(2) =(M)/(12) (l^(2)+Delta l^(2)+2I Delta l)`
As change in length `Delta l` is very small, therefore, neglecting `(Delta l)^(2)`, we get
`I' = (M)/(12)(l^(2)+2lDelta l)`
`= (Ml^(2))/(12)+(MI Delta l)/(6) = l + (MI Delta l)/(6)`
`:.` Increase in moment of inertia
`Delta I = l'-I = (Ml Delta l)/(6) = 2 xx ((Ml^(2))/(12))(Delta l)/(l)`
`Delta I = 2. I alpha Delta T` [ Using Eq. (i)]