The moment of inertia of a uniform rod about a perpendicular axis passing through one end is `I_(1)`. The same rod is bent into a ring and its moment of inertia about a diameter is `I_(2)`. Then `I_(1)//I_(2)` is

Options

A. `(pi^(2))/(3)`
B. `(2pi^(2))/(3)`
C. `(4pi^(2))/(3)`
D. `(8pi^(2))/(3)`

Answer

D

Explanation

`I_(1)= (ML^(2))/(3)` ....(i)
`2pi R = L rArr R= (L)/(2pi)`
`I_(2)= (MR^(2))/(2)= (M)/(2) (L^(2))/(4pi^(2))= (ML^(2))/(8pi^(2))` ...(ii)
From (i) & (ii)
`(I_(1))/(I_(2))= (1)/(3) xx 8pi^(2)= (8pi^(2))/(3)`