StreakPeaked· Practice

ExamsJEE MainPhysics

A solid sphere falls at terminal velocity 20 m/s through air where g = 9.8 m/s². The same sphere is then taken into a gravity-free chamber containing air at the same pressure and is given a downward push at 20 m/s. (Treat the density of air as negligible.) Which statement is correct?

  1. Its initial acceleration will be 9.8 m/s² downward.
  2. Its initial acceleration will be 9.8 m/s² upward.
  3. The magnitude of acceleration will decrease as time passes.
  4. It will eventually stop.

Correct answer: Its initial acceleration will be 9.8 m/s² upward.

Solution

On Earth, at terminal velocity the upward viscous drag balances gravity: F_drag = m*g. In the gravity-free chamber gravity is absent, so the only force at 20 m/s is the same-magnitude viscous drag, now directed opposite to the velocity (upward, since motion is downward). Thus initial acceleration = F_drag/m = g = 9.8 m/s², directed upward (decelerating the sphere). As speed drops, drag drops, so deceleration decreases, and the sphere asymptotically slows but never abruptly stops; the correct listed statement is that the initial acceleration is 9.8 m/s² upward.

Related JEE Main Physics questions

⚔️ Practice JEE Main Physics free + battle 1v1 →