A drop of water of radius 0.001 mm is falling in air. If the coeffcient of viscosity of air is `1.8 xcx 10^(-5) kg m^(-1) s^(-1)`, what will be the terminal velocity of the drop? Neglect the density of air.

Explanation

To find the terminal velocity of a falling drop of water in air, we can use the formula for terminal velocity derived from Stokes' law:

\[ V_t = \frac{2r^2(\rho_d - \rho_a)g}{9\eta} \]

Where:
- \( V_t \) = terminal velocity
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