JEE Main 2025 · 8 April · evening · Single correct · +4 / −1
A rod of linear mass density 'λ' and length 'L' is bent to form a ring of radius 'R'. Moment of inertia of ring about any of its diameter is.
- (a) (λ L³)/(8 π²)
- (b) (λ L³)/(16 π²)
- (c) (λ L³)/(4 π²)
- (d) (λ L³)/12
Answer and solution
Answer (a) (λ L³)/(8 π²)
Solution
Physics sir · 3 steps
Bending does not change the mass, so write M and R in terms of L, then use the diameter formula.
- Mass and radius. The whole rod becomes the ring, so the mass stays M = λ L. The rod's length becomes the circumference, so 2π R = L, which gives R = L/2π.
- Axis along a diameter. About the axis through the centre and perpendicular to the ring, I_z = MR². By the perpendicular axis theorem, I_z equals the sum of the moments about two perpendicular diameters, and these two are equal, so each one is 1/2MR².
I_(dia) = 1/2MR² = 1/2(λ L)(L/2π)² = (λ L³)/8π²
- So. The moment of inertia about any diameter is (λ L³)/8π², which is option (a).
The idea to keep. When a rod is bent into a shape, the mass and the length carry over. Rewrite M and R in terms of λ and L first, then use the standard formula for the new shape.
Answer from the NTA final answer key. Solution by PrepWiser.