JEE Main 2026 · 23 January · morning · Single correct · +4 / −1

Two small balls with masses m and 2 m are attached to both ends of a rigid rod of length d and negligible mass. If angular momentum of this system is L about an axis (A) passing through its centre of mass and perpendicular to the rod then angular velocity of the system about A is :

  1. (a) 4/3L/md²
  2. (b) 3/2L/md²
  3. (c) 2L/5md²
  4. (d) 2L/md²
Answer and solution

Answer (b) 3/2L/md²

Solution

Physics sir · 3 steps

Find where the centre of mass sits first, then the moment of inertia, and divide L by it.

  1. Locate the centre. The centre of mass is closer to the heavier ball. Measured from m, it is at (2m × d)/3m = 2d/3. So m is 2d/3 from the axis and 2m is d/3 from it.

    I = m(2d/3)² + 2m(d/3)² = 4md²/9 + 2md²/9 = 2md²/3

  2. Use L = Iω. ω = L/I = L/(2md²/3) = 3L/2md².
  3. So. The angular velocity about A is 3/2L/md², which is option (b).

The idea to keep. For two point masses on a light rod, the moment of inertia about their centre of mass is μ d², with reduced mass μ = (m₁ m₂)/(m₁ + m₂).

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

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Asked about 180 times in JEE Main since 2020. This question is from 2026.