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 (about central axis) of the ring formed by bending the rod, then

Options

A. `I_(1):I_(2)=1:1`
B. `I_(1):I_(2)=pi^(2):3`
C. `I_(1):I_(2)=pi:4`
D. `I_(1):I_(2)=3:5`

Answer

B

Explanation

Here `I_(1)=(ml^(2))/(12)" "...(i)`
and `I_(2)=m[(l)/(2pi)]^(2)" "(because 2pir=l)`
`=(ml^(2))/(4pi^(2))" "...(ii) [r=l//2pi]`
`therefore (I_(1))/(I_(2))=(ml^(2))/(12)xx(4pi^(2))/(ml^(2))=(pi^(2))/(3)`