From a circular ring of mass M and radius R, an arc corresponding to a `90^(@)` sector is removed. The moment of inertia of the remaining part of the ring about an axis passing through the centre of the ring and perpendicular to the plane of the ring is k times `MR^2`. Then the value of k is

Options

A. `3//4`
B. `7//8`
C. `1//4`
D. `1//8`

Answer

A

Explanation

Moment of inertia of a ring about an axis passing through the centre and perpendicular to the ring is I= `MR^(2)`
Mass of the remaining portion of the ring as shown in figure (ii) is `M - (M)/(4)= (3M)/(4)`
Moment of inertia of the remaining portion of the ring about the given axis is
`I. = (3)/(4) MR^(2)` But `I. = k MR^(2)` (given)
`therefore k = (3)/(4)`