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A drop of water of radius 0.0015 mm is f...

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^(-5)kg//ms`, what will be the terminal velocity of the drop? Density of water `= 1.0 xx 10^(3) kg//m^(3)` and `g = 9.8 N//kg`. Density of air can be neglected.

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To find the terminal velocity of a water drop falling in air, we can use Stokes' law, which states that the terminal velocity \( V_t \) of a sphere falling through a viscous fluid is given by the formula: \[ V_t = \frac{2}{9} \cdot \frac{R^2 \cdot (\rho - \sigma) \cdot g}{\mu} \] Where: - \( R \) = radius of the drop ...
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