Shown in the figure is a hollow icecream cone (it is open at the top). If its mass is M, radius of its top, R and height, H, then its moment of inertia about its axis is :

Solution

Physics sir · 3 steps

Squash the cone flat along its axis: every ring keeps its radius, and you get a uniform disc of radius R.

1. Slice into rings. Cut the cone's surface into thin rings, each centred on the axis. A ring of radius r and mass dm adds r² dm to I, however high up the cone it sits.
2. Mass of a ring. Let L be the slant length. A ring at slant distance l from the tip has radius r = R/Ll and area 2π r dl. With σ = M/(π R L), its mass is dm = σ 2π r dl, so dm ∝ r dr. That is exactly how mass is spread over a uniform flat disc.

   I = ∫₀^L (R/Ll)² σ 2π R/Ll dl = (2πσ R³)/L³·L⁴/4 = M/(π R L)·(π R³ L)/2 = MR²/2

3. So. The height H cancels out completely, and the hollow cone has the same moment of inertia as a uniform disc: I = MR²/2, option (b).

The idea to keep. Moment of inertia about an axis depends only on distances from that axis. Slide mass parallel to the axis and I does not change, so a cone surface flattens into a disc.

Answer from the NTA final answer key. Solution by PrepWiser.