A drop of water of radius `0.0015 mm` is falling in air. If the coefficient of viscosity of air is `1.8 xx 10^(-3)kg//m^(3)`, what will be the terminal velocity of the drop? Density of water `= 1.0 xx 10^(3) kg//m^(2)` and `g = 9.8 N//kg`. Density of air can be neglected.

Explanation

To find the terminal velocity of a water drop falling in air, we can use Stokes' law, which states:

\[ V_t = \frac{2}{9} \cdot \frac{r^2 \cdot (\rho - \sigma)}{\eta} \cdot g \]

Where:
- \( V_t \) = terminal velocity
- \( r \) = radius of the drop
- \( \rho \) = density of the liquid (water)
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