A large number of droplets,each of radius r,c | vedclass

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(B) When small droplets coalesce to form a bigger drop,the change in surface area is $\Delta A = n(4\pi r^2) - 4\pi R^2$. Since the volume is conserve

Vedclass · Accepted Answer

${\left[ {\frac{6T}{\rho }\left( {\frac{1}{r} - \frac{1}{R}} \right)} \right]^{1/2}}$

Vedclass · Answer

(B) When small droplets coalesce to form a bigger drop,the change in surface area is $\Delta A = n(4\pi r^2) - 4\pi R^2$. Since the volume is conserved,$n(\frac{4}{3}\pi r^3) = \frac{4}{3}\pi R^3$,so $n = \frac{R^3}{r^3}$.The energy released is $\Delta E = T 	imes \Delta A = T(n 4\pi r^2 - 4\pi R^2) = 4\pi T (\frac{R^3}{r} - R^2) = 4\pi R^3 T (\frac{1}{r} - \frac{1}{R})$.According to the problem,this energy is converted into the kinetic energy of the big drop: $\frac{1}{2} M v^2 = \Delta E$.Here,$M = ho 	imes 	ext{Volume} = ho (\frac{4}{3}\pi R^3)$.Substituting the values: $\frac{1}{2} (\frac{4}{3}\pi R^3 ho) v^2 = 4\pi R^3 T (\frac{1}{r} - \frac{1}{R})$.Simplifying: $\frac{2}{3} \pi R^3 ho v^2 = 4\pi R^3 T (\frac{1}{r} - \frac{1}{R})$.$v^2 = \frac{4 	imes 3}{2} \frac{T}{ho} (\frac{1}{r} - \frac{1}{R}) = \frac{6T}{ho} (\frac{1}{r} - \frac{1}{R})$.Thus,$v = {\left[ {\frac{{6T}}{ho }\left( {\frac{1}{r} - \frac{1}{R}} ight)} ight]^{\frac{1}{2}}}$.

$A$ large number of droplets,each of radius $r$,coalesce to form a bigger drop of radius $R$. An engineer designs a machine so that the energy released in this process is converted into the kinetic energy of the drop. The velocity of the drop is ($T=$ surface tension,$\rho =$ density)

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