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(B) Let the radius of each small droplet be $r$ and the radius of the large drop be $R$.Since the volume remains constant,the volume of the large drop

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$2^{1/3}:1$

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(B) Let the radius of each small droplet be $r$ and the radius of the large drop be $R$.Since the volume remains constant,the volume of the large drop equals the sum of the volumes of the two small droplets: $\frac{4}{3}\pi R^3 = 2 	imes \frac{4}{3}\pi r^3$.This simplifies to $R^3 = 2r^3$,or $R = 2^{1/3}r$.Surface energy $E$ is given by $E = T 	imes A$,where $T$ is surface tension and $A$ is surface area.Initial surface energy of two small droplets: $E_i = 2 	imes (4\pi r^2 T) = 8\pi r^2 T$.Final surface energy of the large drop: $E_f = 4\pi R^2 T = 4\pi (2^{1/3}r)^2 T = 4\pi (2^{2/3}r^2) T$.The ratio of initial to final surface energy 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 droplets combine to form a single large drop. What is the ratio of the surface energy of the small droplets to the surface energy of the large drop?

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