When the temperature of a rod increases from t to `r+Delta t`, its moment of inertia increases from I to `I+Delta I`. If `alpha` is the value of `Delta I//I` is

Options

A. `(DeltaT)/(alpha T) `
B. `(2 alpha Delta T)/(T)`
C. `(alpha Delta T`
D. `2 alpha Delta T`

Answer

D

Explanation

Moment of inertia of a rod of mass M and length L is given by
`I = KML^2` ....(i)
When `K = 1/12` if the axis of rotation is through the centre and `K= 1/3` if the axis of rotation is through one of its ends.
Differentiating (i) partially, we get
`Delta l = 2KML Delta L`
Also , `Delta L = L alpha Delta T`
` therefore Delta I = 2KML xx L alpha Delta t`
` = 2(KML^2) xx alpha Delta T`
` therefore (Delta I)/(I) = 2 alpha Delta T`