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

Answer

D

Explanation

Let M be the mass of the rod and L be its length . Then
`I = 1/12 ML^(2)`
When the same rod is bent into the ring of radius R , then
`2piR = L or R = L/(2pi) " " ….(i)` `I' = 1/2 MR^(2) = 1/2 M (L/(2pi))^(2) ` (using (i))
` = 1/(8pi^(2)) ML^(2)`
` :. " " I/(I') = (1/12ML^(2))/(1/(8pi^(2)) ML^(2)) = (8pi^(2))/12 = 2/3 pi^(2)`