Question

A drop of water of radius 0.01 m is falling through a medium whose density is 1.21 kg/m\(^3\) and co-efficient of viscosity \( \eta = 1.8 \times 10^{-5} \, \text{Ns/m}^2 \). Then the terminal velocity of the drop is:

• A. \( 120.0 \times 10^{-2} \, \text{m/s} \)
• B. \( 0.012 \, \text{m/s} \)
• C. \( 1.2 \, \text{m/s} \)
• D. \( 1.2 \times 10^{-2} \, \text{m/s} \)

Hint:

Stokes' law is used to calculate the terminal velocity of small spheres falling in a fluid. The velocity depends on the radius, the densities, and the viscosity of the fluid.

Answer 1

The Correct Option is B

The terminal velocity \( v_t \) of a falling sphere is given by Stokes' law: \[ v_t = \frac{2r^2 (\rho_{\text{liquid}} - \rho_{\text{air}}) g}{9 \eta} \] where:
- \( r = 0.01 \, \text{m} \),
- \( \rho_{\text{liquid}} = 1000 \, \text{kg/m}^3 \) (density of water),
- \( \rho_{\text{air}} = 1.21 \, \text{kg/m}^3 \),
- \( \eta = 1.8 \times 10^{-5} \, \text{Ns/m}^2 \),
- \( g = 9.81 \, \text{m/s}^2 \). Substituting the given values: \[ v_t = \frac{2(0.01)^2(1000 - 1.21) \times 9.81}{9 \times 1.8 \times 10^{-5}} \] Solving for \( v_t \): \[ v_t \approx 0.012 \, \text{m/s} \] Thus, the terminal velocity is \( 0.012 \, \text{m/s} \).