A uniform solid sphere of mass M and radius R has a spherical cavity of radius R 2 created in it. The cavity touches the center of the sphere and the outer surface, as shown in the figure. What is the moment of inertia of the remaining portion of the sphere about an axis PQ which is tangent to the sphere at a point diametrically opposite to the cavity?

Step-by-step solution

Let the mass of the original solid sphere be M and its radius be R . The moment of inertia of the original sphere about the tangent axis PQ is given by the parallel axis theorem: I_ original = I_ CM + M R^2 = 2 5 M R^2 + M R^2 = 7 5 M R^2 = 224 160 M R^2 The cavity is a sphere of radius r = R 2 . Since the original sphere is uniform, the mass of the removed cavity is proportional to its volume: m_ cavity = M ( r R )^3 = M ( 1 2 )^3 = M 8 The cavity touches the center of the sphere and the outer surface, so its cent