n number of liquid drops each of radius r coa | vedclass

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Question

(D) Conservation of volume: $\frac{4}{3} \pi R^3 = n 	imes \frac{4}{3} \pi r^3 \implies R^3 = n r^3$.Energy released due to decrease in surface area:

Vedclass · Accepted Answer

$\sqrt{\frac{6T}{\rho}\left[\frac{1}{r}-\frac{1}{R}\right]}$

Vedclass · Answer

(D) Conservation of volume: $\frac{4}{3} \pi R^3 = n 	imes \frac{4}{3} \pi r^3 \implies R^3 = n r^3$.Energy released due to decrease in surface area: $\Delta U = T 	imes (n 	imes 4 \pi r^2 - 4 \pi R^2)$.Since $n = \frac{R^3}{r^3}$, we have $\Delta U = 4 \pi T \left( \frac{R^3}{r} - R^2 ight) = 4 \pi T R^3 \left( \frac{1}{r} - \frac{1}{R} ight)$.This energy is converted into kinetic energy $(K.E.)$ of the big drop: $K.E. = \frac{1}{2} M v^2$, where $M = ho 	imes \frac{4}{3} \pi R^3$.Equating energy: $\frac{1}{2} (ho 	imes \frac{4}{3} \pi R^3) v^2 = 4 \pi T R^3 \left( \frac{1}{r} - \frac{1}{R} ight)$.Simplifying: $\frac{2}{3} ho \pi R^3 v^2 = 4 \pi T R^3 \left( \frac{1}{r} - \frac{1}{R} ight)$.$v^2 = \frac{4 \pi T R^3 	imes 3}{2 \pi ho R^3} \left( \frac{1}{r} - \frac{1}{R} ight) = \frac{6 T}{ho} \left( \frac{1}{r} - \frac{1}{R} ight)$.Therefore, $v = \sqrt{\frac{6 T}{ho} \left( \frac{1}{r} - \frac{1}{R} ight)}$.

$n$ number of liquid drops each of radius $r$ coalesce to form a single drop of radius $R$. The energy released in the process is converted into the kinetic energy of the big drop so formed. The speed of the big drop is [$T = \text{surface tension of liquid}, \rho = \text{density of liquid}$.]

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