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

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$2^{-1/3} : 1$

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(C) Let the radius of each small drop be $r$ and the radius of the big drop be $R$.Since the volume remains conserved, the volume of the big drop equals the sum of the volumes of the two small drops:$2 	imes (4/3 \pi r^3) = 4/3 \pi R^3$$2r^3 = R^3 \implies R = 2^{1/3}r$.The surface energy $E$ is given by $E = T 	imes A$, where $T$ is surface tension and $A$ is the surface area.$E_{before} = 2 	imes (T 	imes 4\pi r^2) = 8\pi r^2 T$.$E_{after} = T 	imes 4\pi R^2 = 4\pi T (2^{1/3}r)^2 = 4\pi T r^2 2^{2/3}$.The ratio of the total surface energies after and before the change is:$E_{after} / E_{before} = (4\pi T r^2 2^{2/3}) / (8\pi r^2 T) = 2^{2/3} / 2 = 2^{2/3 - 1} = 2^{-1/3}$.Thus, the ratio is $2^{-1/3} : 1$.

Two small drops of liquid of same radius coalesce to form a big drop. The ratio of the total surface energies after and before the change is

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