One thousand small water drops of equal radii | vedclass

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(D) Let $n = 1000$ be the number of small drops,each of radius $r$. Let $R$ be the radius of the big drop.Since the volume remains constant,the volume

Vedclass · Accepted Answer

$1:10$

Vedclass · Answer

(D) Let $n = 1000$ be the number of small drops,each of radius $r$. Let $R$ be the radius of the big drop.Since the volume remains constant,the volume of the big drop equals the sum of the volumes of $n$ small drops:$\frac{4}{3}\pi R^3 = n 	imes \frac{4}{3}\pi r^3$$R^3 = n r^3 \implies R = n^{1/3} r$For $n = 1000$,$R = (1000)^{1/3} r = 10r$.The initial surface energy $E_i = n 	imes (4\pi r^2 T)$,where $T$ is the surface tension.The final surface energy $E_f = 4\pi R^2 T$.The ratio of final surface energy to initial surface energy is:$\frac{E_f}{E_i} = \frac{4\pi R^2 T}{n 	imes 4\pi r^2 T} = \frac{R^2}{n r^2}$Substituting $R = n^{1/3} r$:$\frac{E_f}{E_i} = \frac{(n^{1/3} r)^2}{n r^2} = \frac{n^{2/3}}{n} = \frac{1}{n^{1/3}}$For $n = 1000$:$\frac{E_f}{E_i} = \frac{1}{(1000)^{1/3}} = \frac{1}{10} = 1:10$.

One thousand small water drops of equal radii combine to form a big drop. The ratio of final surface energy to the total initial surface energy is

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