A solid sphere of mass M and radius R is divided into two unequal parts. The smaller part having mass M/8 is converted into a sphere of radius r and the larger part is converted into a circular disc of thickness t and radius 2R . If I₁ is moment of inertia of a sphere having radius r about an axis through its centre and I₂ is the moment of inertia of a disc about its diameter, the ratio of their moment of inertia I₂/

Step-by-step solution

Let the mass of the original solid sphere be M and its radius be R . Mass of the smaller sphere, m₁ = M 8 . Since the material remains the same, the density is constant. The volume is directly proportional to the mass. 4 3 r^3 = 1 8 ( 4 3 R^3 ) r^3 = R^3 8 r = R 2 Moment of inertia of the smaller sphere about an axis through its centre is: I₁ = 2 5 m₁ r^2 = 2 5 ( M 8 ) ( R 2 )^2 = 2 5 M 8 R^2 4 = M R^2 80 Mass of the larger part (which is converted into a disc) is: m₂ = M - M 8 = 7M 8 The radius of the disc is give