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)`
`2piR=LimpliesR=(L)/(2pi)`
`I_(2)=(MR^(2))/(2)=(M)/(2)(L^(2))/(4pi^(2))=(ML^(2))/(8pi^(2))`
From (i) & (ii)
`(I_(1))/(I_(2))=(1)/(3)xx8pi^(2)=(8pi^(2))/(3)`