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

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(C) Surface energy $E$ is proportional to the surface area $A$,where $E = T 	imes A$ ($T$ is surface tension).Initial surface area of two drops: $A_{

Vedclass · Accepted Answer

$2^{1/3} : 1$

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(C) Surface energy $E$ is proportional to the surface area $A$,where $E = T 	imes A$ ($T$ is surface tension).Initial surface area of two drops: $A_{	ext{before}} = 2 	imes (4 \pi R^2) = 8 \pi R^2$.When they coalesce into a single large drop of radius $R_0$,the volume remains conserved:$2 	imes (\frac{4}{3} \pi R^3) = \frac{4}{3} \pi R_0^3 \Rightarrow R_0^3 = 2R^3 \Rightarrow R_0 = R(2)^{1/3}$.Final surface area: $A_{	ext{after}} = 4 \pi R_0^2 = 4 \pi (R \cdot 2^{1/3})^2 = 4 \pi R^2 \cdot 2^{2/3}$.The ratio of surface energies is the ratio of surface areas:$\frac{E_{	ext{before}}}{E_{	ext{after}}} = \frac{8 \pi R^2}{4 \pi R^2 \cdot 2^{2/3}} = \frac{2}{2^{2/3}} = 2^{1 - 2/3} = 2^{1/3} : 1$.

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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