Moment of inertia of a ring of mass M and radius R about an axis passing through the centre and perpendicular to the plane is I. What is the moment of inertia about its diameter?

Options

A. I
B. `(I)/(2)`
C. `(I)/(sqrt2)`
D. `I + MR^(2)`

Answer

B

Explanation

Moment of inertia of the ring of mass M and radius R about an axis passing through the centre and perpendicular to the plane ,
`I = MR^(2) " " … (i)`
Moment of inertia of the ring about its diameter
`I_("diameter") = (MR^(2))/(2) = (1)/(2) (MR^(2)) = (I)/(2) " "` (Using (i))