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ExamsJEE MainPhysics

Two identically sized balls of different masses m1 and m2 (with m2 > m1) are tied to the two ends of a light thread and dropped together from rest. Air exerts a viscous drag (depending on size and speed) and a buoyant force (equal to the weight of air displaced) on each ball. After a long time both reach a common terminal velocity. Find the tension in the connecting thread at this stage.

  1. 0.5(m2 - m1)g
  2. (m2 - m1)g
  3. 0.5(m2 + m1)g
  4. zero

Correct answer: 0.5(m2 - m1)g

Solution

Both balls have the same size and the same terminal speed, so the combined upward (drag + buoyancy) force F is identical for each. For the heavier ball the thread pulls up (tension T), for the lighter ball the thread pulls down. Writing equilibrium for each and solving gives T = 0.5(m2 - m1)g.

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