A uniform solid cone of base radius R and height H has a moment of inertia I₁ about its central axis of symmetry. A smaller solid cone of base radius R / 3 and height H / 3 is carved out from the top of this cone. The moment of inertia of this smaller cone about the same central axis is I₂ . The ratio I₁ I₂ is ________.

Step-by-step solution

The moment of inertia of a uniform solid cone of mass M and base radius R about its central axis is I = 3 10 MR^2 . Since the mass of the cone is M = ( 1 3 R^2 H ) , the moment of inertia can be written as: I = 3 10 ( 1 3 R^2 H ) R^2 = 1 10 R^4 H For the original large cone: I₁ = 1 10 R^4 H For the smaller carved-out cone, the base radius is R/3 and the height is H/3 . Its moment of inertia about the same central axis is: I₂ = 1 10 ( R 3 )^4 ( H 3 ) = 1 10 R^4 81 H 3 = 1 243 ( 1 10 R^4 H ) Thus, I₂ = I₁ 243 , which