A large number of liquid drops each of radius | vedclass

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(D) When $n$ drops of radius $r$ coalesce to form a single drop of radius $R$,the volume is conserved: $n(\frac{4}{3}\pi r^3) = \frac{4}{3}\pi R^3$,so

Vedclass · Accepted Answer

$\sqrt{\frac{6T}{\rho}\left(\frac{1}{r} - \frac{1}{R}\right)}$

Vedclass · Answer

(D) When $n$ drops of radius $r$ coalesce to form a single drop of radius $R$,the volume is conserved: $n(\frac{4}{3}\pi r^3) = \frac{4}{3}\pi R^3$,so $n = (R/r)^3$.The energy released during the process is equal to the decrease in surface area multiplied by the surface tension $T$:$E = T 	imes (A_{initial} - A_{final}) = T 	imes (n \cdot 4\pi r^2 - 4\pi R^2)$.Substituting $n = R^3/r^3$:$E = 4\pi T (\frac{R^3}{r^3} \cdot r^2 - R^2) = 4\pi T R^3 (\frac{1}{r} - \frac{1}{R})$.This energy is converted into the kinetic energy of the big drop:$K.E. = \frac{1}{2} M v^2 = E$,where $M$ is the mass of the big drop,$M = ho \cdot V = ho \cdot \frac{4}{3}\pi R^3$.Equating the two:$\frac{1}{2} (\frac{4}{3}\pi R^3 ho) v^2 = 4\pi T R^3 (\frac{1}{r} - \frac{1}{R})$.$\frac{2}{3} \pi R^3 ho v^2 = 4\pi T R^3 (\frac{1}{r} - \frac{1}{R})$.$v^2 = \frac{4 	imes 3}{2} \frac{T}{ho} (\frac{1}{r} - \frac{1}{R}) = \frac{6T}{ho} (\frac{1}{r} - \frac{1}{R})$.$v = \sqrt{\frac{6T}{ho} (\frac{1}{r} - \frac{1}{R})}$.

$A$ large 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 kinetic energy of the big drop so formed. The speed of the big drop is (given,surface tension of liquid $T$,density $\rho$):

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