Find the radius of gyration of a circular ring of radius r about a line perpendicular to the plane of the ring and passing through one of this particles.

Answer

B

Explanation

Moment of inertia at the centre and perpendicular to the plane of the ring.
So, about la point on the rim of the rind and the axisw perpendicular to the plane of the rign the moment of inertia
`=mR^2+mR^2=2mR^2`
(parallel axis therom)
`rarr mK^2=2mR^2`
`(K=radis of the gyration)`
`rarr K=sqrt(2R^2)=sqrt2R`