Question

From 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 mr². Then the value of k is

Answer 1

Answer:

Explanation:

Mass of remaining portion is M/4

Mass of ring is M

Inertia of ring is MR^2

Inertia of remaining portion is MR^2/4