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(C) The volume of the liquid remains constant when two drops coalesce. Let $r$ be the radius of each smaller drop and $R$ be the radius of the bigger

Vedclass · Accepted Answer

$2^{2/3} : 1$

Vedclass · Answer

(C) The volume of the liquid remains constant when two drops coalesce. Let $r$ be the radius of each smaller drop and $R$ be the radius of the bigger drop.Volume of two smaller drops = Volume of the bigger drop$2 	imes (\frac{4}{3} \pi r^3) = \frac{4}{3} \pi R^3$$R^3 = 2r^3 \implies R = 2^{1/3} r$Surface energy $W$ is given by $W = T 	imes A$,where $T$ is surface tension and $A$ is the surface area.Surface energy of the bigger drop,$W_1 = T 	imes (4 \pi R^2) = 4 \pi T (2^{1/3} r)^2 = 2^{2/3} (4 \pi r^2 T)$.Surface energy of one smaller drop,$W_2 = T 	imes (4 \pi r^2) = 4 \pi r^2 T$.The ratio of the surface energy of the bigger drop to the smaller drop is:$\frac{W_1}{W_2} = \frac{2^{2/3} (4 \pi r^2 T)}{4 \pi r^2 T} = 2^{2/3} : 1$.

Two drops of equal radius coalesce to form a bigger drop. What is the ratio of the surface energy of the bigger drop to that of a smaller one?

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