A uniform solid cylinder of radius R and length L has a moment of inertia I₀ about its central axis. A cylindrical hole of radius r is drilled along its entire central axis. The moment of inertia of the resulting hollow cylinder about the same axis is found to be 80 81 I₀ . The value of the ratio R r is __________.

Step-by-step solution

The moment of inertia of a uniform solid cylinder of radius R , length L , and mass M about its central axis is I = 1 2 MR^2 . The mass of the cylinder is M = R^2 L . Substituting this into the moment of inertia formula gives: I = 1 2 ( R^2 L)R^2 = 1 2 L R^4 Thus, for a cylinder of fixed length and constant density, the moment of inertia is proportional to the fourth power of its radius ( I R^4 ). Let the moment of inertia of the original solid cylinder be I₀ = kR^4 , where k = 1 2 L . The moment of inertia of the