Question

What is the terminal velocity of a raindrop of radius \(0.01\,\text{mm}\), where the coefficient of viscosity is \(1.8\times10^{-5}\,\text{N-s/m}^2\) and its density is \(1.2\,\text{kg/m}^3\), density of water \(=1000\,\text{kg/m}^3\)? (Take \(g=10\,\text{m/s}^2\))

• A. \(1.2\,\text{cm/s}\)
• B. \(2.4\,\text{cm/s}\)
• C. \(2.1\,\text{m/s}\)
• D. \(2.1\,\text{cm/s}\)

Hint:

Use Stokes' law \(v_t=\dfrac{2r^2(\rho-\sigma)g}{9\eta}\); the drop is water, the medium is air.

Answer 1

The Correct Option is A

Terminal velocity comes from balancing the net downward weight (weight minus buoyancy) against the viscous drag on the drop.

Weight minus buoyancy for a sphere: $\tfrac{4}{3}\pi r^3(\rho-\sigma)g$. Viscous drag (Stokes): $6\pi\eta r v_t$. At terminal speed these are equal: \[\tfrac{4}{3}\pi r^3(\rho-\sigma)g=6\pi\eta r v_t\] Solving for $v_t$ cancels one $\pi r$ and gives \[v_t=\frac{2r^2(\rho-\sigma)g}{9\eta}\] With $r=10^{-5}$ m, $\rho-\sigma=998.8\approx10^3$ kg/m$^3$, $g=10$, $\eta=1.8\times10^{-5}$: \[v_t=\frac{2\times10^{-10}\times10^{3}\times10}{9\times1.8\times10^{-5}}=\frac{2\times10^{-6}}{1.62\times10^{-4}}\approx0.0123\ \text{m/s}\] Converting, $v_t\approx1.2$ cm/s. The air density $1.2$ kg/m$^3$ is tiny next to water, so it barely changes the answer. \[\boxed{v_t\approx1.2\,\text{cm/s}}\]