Moment of Inertia (M.I.) of four bodies having same mass 'M' and radius '2R' are as follows:
I₁ = M.I. of solid sphere about its diameter
I₂ = M.I. of solid cylinder about its axis
I₃ = M.I. of solid circular disc about its diameter
I₄ = M.I. of thin circular ring about its diameter
If 2(I₂ + I₃) + I₄ = x . I₁, then the value of x will be __________.

Solution

Physics sir · 3 steps

All four share the same Mr², so write each as a fraction of Mr² and compare the fractions.

1. Common factor. Every body has mass M and radius r = 2R, so each moment of inertia is a number times Mr². The Mr² cancels in the end. That is why the value of 2R does not matter.

   Body	Axis	Moment of inertia
   Solid sphere, I₁	diameter	2/5Mr²
   Solid cylinder, I₂	its own axis	1/2Mr²
   Solid disc, I₃	diameter	1/4Mr²
   Thin ring, I₄	diameter	1/2Mr²

2. Diameter axes. For the disc and the ring, use the perpendicular axis theorem: the moment about a diameter is half the moment about the central axis. So the disc gives 1/2 × 1/2 = 1/4 and the ring gives 1/2 × 1 = 1/2.

   2(1/2+1/4)Mr² + 1/2Mr² = 2Mr² = x · 2/5Mr²

3. So. x = 2/(2/5) = 5, so the answer is 5.

The common slip: 6.25. Using 1/2Mr² for the disc about its diameter, which is the formula for its central axis, gives 6.25 instead of 5.

The idea to keep. Always check the axis before writing a formula. For a flat disc or ring, the diameter moment is half the central-axis moment, by the perpendicular axis theorem.

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