The moment of inertia of a circular ring about an axis passing through its diameter is I . This ring is cut then unfolded into a uniform straight rod. The moment of inertia of the rod about an axis perpendicular to its length passing through one of its ends is

Options

A. `4pi^(2)I/3`
B. `8pi^(2)I/3`
C. `16pi^(2)I/3`
D. `2pi^(2)I/3`

Answer

B

Explanation

For the ring `I = (mr^(2))/(2)`
for the rod, `l = 2pi r` and
`I'=(ml^(2))/(3)=(m(4pi^(2)r^(2)))/(3)`
`=(8pi^(2)I)/(3)`