A ball twice as big, falling through oil.

A small steel ball falls through oil at a terminal speed v. A ball of the same steel with twice the radius has a terminal speed of 4v. At terminal speed, the viscous force on the bigger ball is how many times that on the smaller?

Answer: 8.

At terminal speed the viscous force balances the weight minus the buoyancy, and both of those grow with the volume, which goes as the cube of the radius.

F = 6 π η r v = (ρ - σ) \frac{4}{3} π r^{3} g

\Rightarrow \frac{F'}{F} = \frac{2 r \cdot 4 v}{r \cdot v} = 8 = 2^{3}

So the drag on the bigger ball is 8 times larger. Stokes' law gives the same answer as the volume does: twice the radius times four times the speed. Taking the drag to follow the speed alone, which gives 4, is the trap.

Original practice question · JEE Physics Decoded

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