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_(2)=(I_(1))/(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)`