Find out the increases 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 `DeltaT`.

Explanation

Suppose mass of rod is M and moment of inertia is I.
Moment of inertia of rod about perpendicular bisector,
`I=(Ml^(2))/(l^(2))`

Increase in length due to increase in temperature `DeltaT`,
`Deltal=l alphaDeltaT` . . . (1)
`:.` Now, new moment of inertia of rod,
`I.=(M)/(12)(l+Deltal)^(2)`
`=(M)/(12)(l^(2)+2l+Deltal^(2))`
But, `Deltal` is very small, `Deltal^(2)` is neglected.
`I.=(M)/(12)(l^(2)+2lDeltal)`
`=(Ml^(2))/(12)+(MlDeltal)/(6)`
`=I+(MlDeltal)/(6)`
Increase in moment of inertia,
`I.-I=(MlDeltal)/(6)`
`:.DeltaI=2xx(Ml^(2))/(12)*(Deltal)/(l)" "[because" Multiplying and dividing by 2"]`
`:.DeltaI=2Ialpha DeltaT" "[because(Deltal)/(l)=alphaDeltaT]`