`I` is moment of inertia of a thin circular ring about an axis perpendicular to the plane of ring and passing through its centre. The same ring is folded into `2` turns coil. The moment of inertia of circular coil about an axis perpendicular to the plane of coil and passing through its centre is

Options

A. 2I
B. 4I
C. `I/2`
D. `I/4`

Answer

D

Explanation

Mass is same. `l = 2pirn rArr ralpha 1/n ` where n is number of turns `I prop r^(2)` and ` (I_(1))/(I_(2)) = ((r_(1))/(r_(2)))^(2) = ((n_(2))/(n_(1)))^(2)`