A certain number of spherical drops of a liqu | vedclass

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Question

(C) Let $n$ droplets each of radius $r$ coalesce to form a big drop of radius $R$.Volume of $n$ droplets = Volume of big drop$n 	imes \frac{4}{3}\pi

Vedclass · Accepted Answer

energy $= 3VT \left( \frac{1}{r} - \frac{1}{R} \right)$ is released.

Vedclass · Answer

(C) Let $n$ droplets each of radius $r$ coalesce to form a big drop of radius $R$.Volume of $n$ droplets = Volume of big drop$n 	imes \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3 \Rightarrow n = \frac{R^3}{r^3}$ $(i)$Volume of big drop,$V = \frac{4}{3}\pi R^3$ $(ii)$Initial surface area of $n$ droplets,$A_i = n 	imes 4\pi r^2 = \frac{R^3}{r^3} 	imes 4\pi r^2 = \frac{4\pi R^3}{r} = \left( \frac{4}{3}\pi R^3 ight) \frac{3}{r} = \frac{3V}{r}$ $(Using (i) 	ext{ and } (ii))$Final surface area of big drop,$A_f = 4\pi R^2 = \left( \frac{4}{3}\pi R^3 ight) \frac{3}{R} = \frac{3V}{R}$ $(Using (ii))$Decrease in surface area,$\Delta A = A_i - A_f = \frac{3V}{r} - \frac{3V}{R} = 3V \left( \frac{1}{r} - \frac{1}{R} ight)$Energy released = Surface tension $	imes$ Decrease in surface areaEnergy released $= T 	imes \Delta A = 3VT \left( \frac{1}{r} - \frac{1}{R} ight)$.

$A$ certain number of spherical drops of a liquid of radius $r$ coalesce to form a single drop of radius $R$ and volume $V.$ If $T$ is the surface tension of the liquid,then

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