Moment of interia of a thin uniform rod rotating about the perpendicular axis passing through its center is I. If the same rod is bent into a ring and its moment of inertia about its diameter is

Options

A. `8pi^(2)//3`
B. `5pi^(2)//3`
C. `3pi^(2)//2`
D. `2pi^(2)//3`

Answer

D

Explanation

`I=(1)/(12)ml^(2)`
when rod is bent into a ring then radius of ring is r
`therefore 2pir=limpliesr=(l)/(2pi)`
Moment of inertia of ring about its diameter is
`I.=(1)/(2)mr^(2)=(1)/(2)m(l^(2))/(4pi^(2))`
`therefore (I)/(I.)=(1)/(12)ml^(2)//(ml^(2))/(8pi^(2))=(2pi^(2))/(3)`