A uniform solid cylinder with radius R and length L has moment of inertia I₁ about the axis of the cylinder. A concentric solid cylinder of radius R/2 and length L/2 is carved out of the original cylinder. If I₂ is the moment of inertia of the carved out portion of the cylinder then I₁/I₂ is (Both I₁ and I₂ are about the axis of the cylinder)

Step-by-step solution

Let the density of the material of the cylinder be . The mass of the original solid cylinder is M₁ = R^2 L . The moment of inertia of the original solid cylinder about its axis is: I₁ = 1 2 M₁ R^2 = 1 2 ( R^2 L) R^2 = 1 2 R^4 L The carved out portion is a solid cylinder of radius R/2 and length L/2 . Its mass is: M₂ = ( R 2 )^2 ( L 2 ) = 1 8 R^2 L The moment of inertia of this carved out portion about the same axis is: I₂ = 1 2 M₂ ( R 2 )^2 = 1 2 ( 1 8 R^2 L ) ( R^2 4 ) = 1 64 R^4 L Taking the ratio of I₁ to I₂ : I