The temperature of uniform rod of length L having a coefficient of linear expansion `alpha_(L)` is changed by `Delta T` . Calculate the new moment of inertia of the uniform rod about axis passing through its center and perpendicular to an axis of the rod.

Explanation

M. I of uniform rod of mass M. Length / about its center & perpendicular is
` I = (Ml^(2))/(12)`
When rod is heated by temperature `Delta T `, there is increase in length of rod `Delta l`
`Delta l = l alpha_(L) Delta T`
`( alpha _(L) -` coefficient of linear expression )
New M. I ` I ' = (M)/(12) (1 + Delta l)^(2)`
`rArr (M)/(12) (1+ 1 alpha _(L) Delta T)^(2)`
We get substituting `Delta l` from (1)
`I ' = (Ml^(2))/(12) (1+ alpha_(L) Delta T)^(2)`
`I' = (1+ alpha _(L) Delta T)^(2)`