The moment of inertia of a ring about its geometrical axis is I, then its moment of inertia about its diameter will be

Explanation

A thin uniform-density disk of radius R is confined to the xy plane, as shown in Fig.

Calculations: According to perpendicular-axis theorem:
`I_(xO)+I_(yO)=I_(zO)`.
However, by symmetry, we have
`I_(xO)+I_(yO)=I_(d)`.
where `I_(d)` denotes the moment of inertia about any diameter. Therefore, we have
`2I_(d)=I_(zO)`.
The moment of inertia of a solid cylinder or disk of radius R about the symmetry axis is `1//2MR^(2)`. Thus,
`I_(zO)=1/2MR^(2)`
and `I_(d)=1/4MR^(2)`.