Moment of inertia of a ring of mass M and radius R about a tangent to the circle of the ring is

Options

A. `(5)/(2) MR^(2)`
B. `(3)/(2) MR^(2)`
C. `(1)/(2) MR^(2)`
D. `MR^(2)`

Answer

B

Explanation

Moment of inertia of a ring of mass M and radius R about the diameter AB is
`I_(AB) = (1)/(2) MR^(2)`
XY is the tangent to the circular ring in the plane of the ring .
According to theorem of parallel axes ,
`I_(XY) = I_(AB) + MR^(2) = (1)/(2) MR^(2) + MR^(2) = (3)/(2) MR^(2)`