Question

A rain drop of radius 0.3 mm has a terminal velocity in air is 1 m/s. The viscosity of air is \( 8 \times 10^{-5} \) poise. The viscous force on it is:

• A. \( 45.2 \times 10^{-4} \) dyne
• B. \( 101.73 \times 10^{-5} \) dyne
• C. \( 16.95 \times 10^{-5} \) dyne
• D. \( 16.95 \times 10^{-6} \) dyne

Hint:

Use Stokes’ law to calculate the viscous force on small spherical objects moving through a fluid.

Answer 1

The Correct Option is C

Step 1: Formula for viscous force.
The viscous force \( F \) is given by Stokes' law: \[ F = 6 \pi \eta r v \] where \( \eta \) is the viscosity, \( r \) is the radius of the drop, and \( v \) is the velocity. Step 2: Substituting the values.
Given: \( \eta = 8 \times 10^{-5} \, \text{poise} \), \( r = 0.3 \, \text{mm} = 0.3 \times 10^{-3} \, \text{m} \), and \( v = 1 \, \text{m/s} \), we calculate the force: \[ F = 6 \pi (8 \times 10^{-5}) (0.3 \times 10^{-3}) (1) = 16.95 \times 10^{-5} \, \text{dyne} \]
Final Answer: \[ \boxed{16.95 \times 10^{-5} \, \text{dyne}} \]