The terminal velocity (vₜ) of the spherical rain drop depends on the radius (r) of the spherical rain drop as :

Solution

Physics sir · 3 steps

Balance the forces: weight grows as r³, drag only as r, so terminal speed goes as r².

1. Forces balance. At terminal velocity the drop no longer accelerates. So the weight is exactly balanced by the buoyant force plus the viscous drag.
2. How each scales. Weight and buoyancy both depend on volume, 4/3π r³, so they grow as r³. Stokes drag is 6πη r v, which grows only as r for a given speed.

   6πη r vₜ = 4/3π r³(ρ - σ)g ⇒ vₜ = 2r²(ρ - σ)g/9η

3. So. One power of r cancels from both sides, leaving vₜ ∝ r². The answer is (c).

The idea to keep. For terminal velocity, set Stokes drag 6πη r v equal to the net weight 4/3π r³(ρ-σ)g. Volume gives r³, drag gives r, so vₜ ∝ r².

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