A small steel ball is dropped into a long cylinder containing glycerine. Which one of the following is the correct representation of the velocity time graph for the transit of the ball?

Solution

Physics sir · 4 steps

Viscous drag grows with speed until it balances the weight, so the ball levels off at terminal velocity.

1. Forces on the ball. Weight mg acts down. Buoyancy B and the Stokes drag F = 6πη r v act up. The drag is zero at the start and grows as v grows.

   mdv/dt = (mg - B) - 6πη r v

2. Shape of the curve. At t = 0 the drag is zero, so the acceleration is largest and the graph starts steep. As v rises the drag grows, the acceleration shrinks, and the slope keeps decreasing.
3. Terminal velocity. When 6πη r v equals mg - B, the net force is zero. The velocity stops changing and stays at the terminal value, so the curve flattens out instead of falling or rising forever.
4. So. The graph rises from zero with a decreasing slope and levels off at the terminal velocity. That is option (b).

The idea to keep. Whenever a resisting force grows with speed, the velocity rises steeply at first, then levels off at the speed where the forces balance. It never drops back.

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