Discussion
The terminal velocity of a steel ball of radius r and density ρ falling through an oil column of density σ and viscosity η is
Answer: D
The terminal velocity is directly proportional to \(r^{2}\).
At terminal velocity the weight minus buoyancy equals Stokes' drag \(6\pi\eta rv\):
\[ \begin{aligned} \tfrac{4}{3}\pi r^{3}(\rho - \sigma)g &= 6\pi\eta r v \\ v &= \frac{2r^{2}(\rho - \sigma)g}{9\eta} \end{aligned} \]
So v grows as \(r^{2}\): a ball twice as large falls four times as fast. It is inversely, not directly, proportional to \(\eta\), and the options in r alone or \(r^{-2}\) give the wrong power.
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