Moment of inertia of a thin uniform rod rotating about the perpendicular axis passing through its centre is 'I'. If the same rod is bent in the form of ring, its moment of inertia about the diameter is ' I ₁ ^ . If I₁=x I , then the value of ' x ' is

Step-by-step solution

The moment of inertia of a thin uniform rod about its center is I = 1 12 ML^2 . When bent into a ring of radius R , the circumference equals the rod length: L = 2 R , so R = L 2 . The moment of inertia about a diameter becomes I₁ = 1 2 MR^2 = 1 2 M ( L 2 )^2 = ML^2 8 ^2 . With I₁ = xI , we substitute: ML^2 8 ^2 = x ( 1 12 ML^2 ) . Canceling ML^2 yields 1 8 ^2 = x 12 , so x = 12 8 ^2 = 3 2 ^2 . Final answer: B