If `I_(1)` is moment of inertia of a thin rod about an axis perpendicular to its length and passing through its centre and `I_(2)` is its moment of inertia when it is bent into a shape of a ring then (Axis passing through its centre and perpendicular to its plane)

Options

A. `I_(1) = (I_(2))/(4pi^(2)`
B. `I_(2)=I_(1)/(pi^(2))`
C. `(I_(2))/(I_(1)) = (pi^(2))/3`
D. `(I_(2))/(I_(1)) = 3/(pi^(2))`

Answer

D

Explanation

` I_(1) = (ML^(2))/(12) , I_(2) = MR^(2) , R = L/(2pi)`