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(C) Let $r$ be the radius of each small drop and $R$ be the radius of the big drop.Since the total volume remains constant,the volume of $8000$ small

Vedclass · Accepted Answer

$1:20$

Vedclass · Answer

(C) Let $r$ be the radius of each small drop and $R$ be the radius of the big drop.Since the total volume remains constant,the volume of $8000$ small drops equals the volume of the big drop:$8000 	imes (\frac{4}{3} \pi r^3) = \frac{4}{3} \pi R^3$$R^3 = 8000 r^3$$R = 20r$Surface energy $E = T 	imes A$,where $T$ is surface tension and $A$ is surface area.Initial surface energy $E_i = 8000 	imes (4 \pi r^2 T)$Final surface energy $E_f = 4 \pi R^2 T$The ratio of final surface energy to initial surface energy is:$\frac{E_f}{E_i} = \frac{4 \pi R^2 T}{8000 	imes 4 \pi r^2 T} = \frac{R^2}{8000 r^2}$Substituting $R = 20r$:$\frac{E_f}{E_i} = \frac{(20r)^2}{8000 r^2} = \frac{400 r^2}{8000 r^2} = \frac{1}{20}$Thus,the ratio is $1:20$.

$8000$ identical water drops are combined to form a big drop. Then the ratio of the final surface energy to the initial surface energy of all the drops together is

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