Moment of inertia of a thin uniform rod rotating about the perpendicular axis passing through its center is l. If the same rod is bent into a ring and its moment of inertia about its diameter is I. then the ratio `(I)/(I)` is

Options

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

Answer

D

Explanation

`I= (Ml^(2))/(12) rArr Ml^(2)= 12I`
`I. = (MR^(2))/(2), ["As " l= 2pi R rArr R= (l)/(2pi)]`
`= (M)/(2) ((l)/(2pi ))^(2)= (Ml^(2))/(8pi^(2))`
`therefore I. = (12I)/(8pi^(2)) rArr (I)/(I.)= (8pi^(2))/(12)= (2pi^(2))/(3)`