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

Step 1: A small sphere falling through a viscous medium reaches a steady terminal velocity given by Stokes' law: \[v_t=\frac{2r^2(\rho-\sigma)g}{9\eta}\] where \(r\) is the radius of the drop, \(\rho\) is the density of the drop (water), \(\sigma\) is the density of the medium (air), \(\eta\) is the coefficient of viscosity, and \(g\) is the acceleration due to gravity.

Step 2: Convert the radius to SI units: \[r=0.01\,\text{mm}=1\times10^{-5}\,\text{m}\]
Step 3: Put in the values \(\rho=1000\), \(\sigma=1.2\), \(g=10\), \(\eta=1.8\times10^{-5}\): \[v_t=\frac{2(1\times10^{-5})^2(1000-1.2)(10)}{9(1.8\times10^{-5})}\]
Step 4: Numerator \(=2\times10^{-10}\times998.8\times10=1.9976\times10^{-6}\). Denominator \(=1.62\times10^{-4}\). \[v_t=\frac{1.9976\times10^{-6}}{1.62\times10^{-4}}\approx1.23\times10^{-2}\,\text{m/s}\]
Step 5: That is \(1.23\,\text{cm/s}\), which rounds to \(1.2\,\text{cm/s}\). \[\boxed{v_t\approx1.2\,\text{cm/s}}\]