In Millikan's oil drop experiment, what is the terminal speed of a speed of a drop of radius `2.0 xx 10^(5)m` and density `1.2xx10^(3)m^(-3)`? Take the viscosity of air at the temperature of the experimental to be `1.8xx10^(-5)Nsm^(2)`. How much is the viscous force on the drop at that speed? Neglect buoyancy of the drop due to air.

Explanation

To solve the problem, we need to calculate the terminal speed of the oil drop and the viscous force acting on it. Let's break this down step by step:

### Step 1: Understand the Formula for Terminal Velocity
The terminal velocity \( V_T \) of a sphere falling through a viscous medium can be calculated using the formula:
\[ V_T = \frac{2R^2 (ρ - σ) g}{9η} \]
Where:
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