The moment of inerta of a ring of mass `1kg` about an axis passing through its centre perpendicular to its surface is `4kgm^(2)`. Calculate the radius of the ring.

Options

A. 2 m
B. 4 m
C. 5 m
D. 6 m

Answer

B

Explanation

Moment of inertia of circular ring about an axis passing through its centre of mass and perpendicular of plane,
`l=MR^(2)`
`"Here, "l=4kg-m^(2), m=1kg`
`therefore " "R^(2)=(4)/(1)=4rArr R=2m`
Therefore, diameter of ring = 4m.