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. `2l`
B. `(l)/(2)`
C. `(3)/(2)l`
D. `l`

Answer

A

Explanation

By the theorem of perpendicular axes, the momet of inertia about the central axis `l_(C)`, will be equal to the sum of its moments of inertia about two mutually perpendicular diameters lying in its plane.
`"Thus, "l=(1)/(2)MR^(2)`
`therefore" "l_(C)=l+l=(1)/(2)MR^(2)+(1)/(2)MR^(2)=l+l=2l`