A small spherical ball of radius r, falling through a viscous medium of negligible density has terminal velocity 'v'. Another ball of the same mass but of radius 2 r, falling through the same viscous medium will have terminal velocity:

Solution

Physics sir · 3 steps

Same mass means the same weight, so only the drag side changes when the radius doubles.

1. Forces at terminal velocity. The medium has negligible density, so there is no buoyant force. At terminal velocity the weight mg is exactly balanced by the Stokes drag 6πη r v.

   mg = 6πη r v ⇒ v = mg/(6πη r)

2. Fix the mass. Both balls have the same mass m and fall through the same medium, so m, g and η are all the same. That leaves v ∝ 1/r.
3. So. Doubling the radius to 2r halves the terminal velocity: v' = v/2, option (d).

The idea to keep. Do not reach for v ∝ r² by habit. That result assumes the same density. Always balance weight against 6πη r v and see what is actually held fixed.

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