If `I_(1)` is the moment of inertia of a thin rod about an axis perpendicular to its length and passing through its centre of mass and `I_(2)` is the moment of inertia of the ring about an axis perpendicular to plane of ring and passing through its centre formed by bending the rod, then

Options

A. `(I_(1))/(I_(2))=(3)/(pi^(2))`
B. `(I_(1))/(I_(2))=(2)/(pi^(2))`
C. `(I_(1))/(I_(2))=(pi^(2))/(2)`
D. `(I_(1))/(I_(2))=(pi^(2))/(3)`

Answer

D

Explanation

According to the question, `l = 2piR`
`R=(l)/(2pi)rArrI_(1)=(Ml^(2))/(12)`
`I_(2)=MR^(2)=(Ml^(2))/(4pi^(2))rArr(I_(1))/(I_(2))=(pi^(2))/(3)`.