Question

A drop of water of radius 0.0015 mm is falling in air. The coefficient of viscosity of air is \( 1.8 \times 10^5 \, \text{kgm}^{-1} \, \text{s}^{-1} \). If the density of the air is neglected, then what will be the terminal velocity of the drop?

• A. \( 2.72 \times 10^{-5} \, \text{m/s} \)
• B. \( 1.35 \times 10^{-5} \, \text{m/s} \)
• C. \( 2.72 \times 10^{-4} \, \text{m/s} \)
• D. \( 3.45 \times 10^{-4} \, \text{m/s} \)

Hint:

To calculate terminal velocity, remember to apply Stokes' law, which assumes the object is spherical, small, and moving slowly through a viscous medium.

Answer 1

The Correct Option is A

To calculate the terminal velocity of the falling drop, we can use Stokes' Law, which is applicable for small spherical objects moving through a viscous medium. The formula for the terminal velocity \(v_t\) of a spherical drop is given by: \[ v_t = \frac{2 r^2 (\rho_{\text{drop}} - \rho_{\text{air}}) g}{9 \eta} \] Where:
- \(r\) = radius of the drop
- \(\rho_{\text{drop}}\) = density of the drop (water in this case)
- \(\rho_{\text{air}}\) = density of air (which is neglected in this case)
- \(g\) = acceleration due to gravity (\(9.8 \, \text{m/s}^2\))
- \(\eta\) = coefficient of viscosity of air
For the given problem: - Radius of the drop, \(r = 0.0015 \, \text{mm} = 1.5 \times 10^{-6} \, \text{m}\)
- Coefficient of viscosity of air, \(\eta = 1.8 \times 10^{-5} \, \text{kg/m} \cdot \text{s}\)
- Density of the drop (water), \(\rho_{\text{drop}} = 1000 \, \text{kg/m}^3\)
- Density of air is neglected.
The simplified formula becomes:
\[ v_t = \frac{2 r^2 \rho_{\text{drop}} g}{9 \eta} \] Substituting the values:
\[ v_t = \frac{2 \times (1.5 \times 10^{-6})^2 \times 1000 \times 9.8}{9 \times 1.8 \times 10^{-5}} \] \[ v_t = \frac{2 \times 2.25 \times 10^{-12} \times 1000 \times 9.8}{1.62 \times 10^{-4}} \] \[ v_t = \frac{4.41 \times 10^{-8}}{1.62 \times 10^{-4}} \] \[ v_t = 2.72 \times 10^{-5} \, \text{m/s} \] Thus, the terminal velocity of the drop is \( 2.72 \times 10^{-5} \, \text{m/s} \).