Coalescence of falling drops – Rankers Physics
Topic: Solid and Fluids
Subtopic: Surface Tension and Viscosity

Coalescence of falling drops

Eight drops of equal radii are falling through air with a steady velocity of \(3\text{ cm/s}\). If the eight drops combine to form a single drop, then its steady velocity will be
\(3\text{ cm/s}\)
\(12\text{ cm/s}\)
\(6\text{ cm/s}\)
\(24\text{ cm/s}\)

Solution:

Terminal velocity \(v \propto r^2\). Since volume remains constant, \(\frac{4}{3}\pi R^3 = 8 \times \frac{4}{3}\pi r^3 \implies R = 2r\). Thus, the new terminal velocity is \(v' = \left(\frac{R}{r}\right)^2 v = 2^2 \times 3 = 12\text{ cm/s}\).

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