The moment of inertia of ring about an axis passing through its diameter is `I`. Then moment of inertia of that ring about an axis passing through its centre and perpendicular to its plane is

Options

A. `(I)/(2)`
B. `2I`
C. `(I)/(4)`
D. `4I`

Answer

B

Explanation

For a ring, `I_(z) =MR^(2)`
From perpendicular axis theorem,
`I_(x)+I_(y)=I_(z)`
Given, `I_(x)=I_(y)=I`
From Eq. (i), we get
`2I-I_(z)` .