Two small drops of mercury,each of radius R,c | vedclass

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(B) Let $r$ be the radius of the large drop formed after coalescence.Since the volume remains constant,the volume of the large drop equals the sum of

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$2^{1/3} : 1$

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(B) Let $r$ be the radius of the large drop formed after coalescence.Since the volume remains constant,the volume of the large drop equals the sum of the volumes of the two small drops:$\frac{4}{3} \pi r^{3} = 2 	imes \frac{4}{3} \pi R^{3}$$r^{3} = 2R^{3} \implies r = 2^{1/3} R$The surface energy of a drop is given by $E = T 	imes A$,where $T$ is the surface tension and $A$ is the surface area $(4 \pi R^{2})$.Surface energy before coalescence $(E_{i})$ = $2 	imes (4 \pi R^{2} T) = 8 \pi R^{2} T$.Surface energy after coalescence $(E_{f})$ = $4 \pi r^{2} T = 4 \pi (2^{1/3} R)^{2} T = 4 \pi 2^{2/3} R^{2} T$.The ratio of surface energies before and after is:$\frac{E_{i}}{E_{f}} = \frac{8 \pi R^{2} T}{4 \pi 2^{2/3} R^{2} T} = \frac{2}{2^{2/3}} = 2^{1 - 2/3} = 2^{1/3}$.Thus,the ratio is $2^{1/3} : 1$.

Two small drops of mercury,each of radius $R$,coalesce to form a single large drop. The ratio of the total surface energies before and after the change is

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