Moment of inertia (M. I.) of four bodies, having same mass and radius, are reported as;
I₁ = M.I. of thin circular ring about its diameter,
I₂ = M.I. of circular disc about an axis perpendicular to disc and going through the centre,
I₃ = M.I. of solid cylinder about its axis and
I₄ = M.I. of solid sphere about its diameter.
Then :

Solution

Physics sir · 3 steps

Write all four as a number times MR², and the comparison does itself.

1. Standard results. Each body has the same M and R, so every moment of inertia is some fraction times MR². Recall each fraction and compare only the fractions.

   Body and axis	Moment of inertia	As a decimal of MR²
   I₁: ring about a diameter	1/2MR²	0.5
   I₂: disc about the central perpendicular axis	1/2MR²	0.5
   I₃: solid cylinder about its axis	1/2MR²	0.5
   I₄: solid sphere about a diameter	2/5MR²	0.4

2. Ring about diameter. About its own axis a ring has MR². By the perpendicular axis theorem, its two equal diameters share that value, so each diameter gets 1/2MR².
3. So. I₁ = I₂ = I₃ = 0.5 MR² and I₄ = 0.4 MR², so I₁ = I₂ = I₃ > I₄. The answer is (a).

The idea to keep. When the bodies share the same M and R, compare only the fractions in front of MR². Remember that the ring about a diameter is 1/2MR², not MR².

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