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(D) Initial surface area $A_i = 4 \pi r^2 + 4 \pi (2r)^2 = 4 \pi r^2 + 16 \pi r^2 = 20 \pi r^2$.Initial surface energy $E_i = A_i S = 20 \pi r^2 S$.Vo

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$2.7$

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(D) Initial surface area $A_i = 4 \pi r^2 + 4 \pi (2r)^2 = 4 \pi r^2 + 16 \pi r^2 = 20 \pi r^2$.Initial surface energy $E_i = A_i S = 20 \pi r^2 S$.Volume conservation: $\frac{4}{3} \pi r^3 + \frac{4}{3} \pi (2r)^3 = \frac{4}{3} \pi R^3$.$r^3 + 8r^3 = R^3 \implies R^3 = 9r^3 \implies R = 9^{1/3} r$.Final surface area $A_f = 4 \pi R^2 = 4 \pi (9^{1/3} r)^2 = 4 \pi (9^{2/3}) r^2$.Given $9^{2/3} = 4.326$,so $A_f = 4 \pi (4.326) r^2 = 17.304 \pi r^2$.Final surface energy $E_f = 17.304 \pi r^2 S$.Energy released $\Delta E = E_i - E_f = 20 \pi r^2 S - 17.304 \pi r^2 S = 2.696 \pi r^2 S \approx 2.7 \pi r^2 S$.

Two mercury drops of radii $r$ and $2r$ merge to form a bigger drop. The surface energy released in the process is nearly (Surface tension of mercury is $S$ and take $9^{2/3} = 4.326$). (in $\pi r^2 S$)

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