Eight mercury drops, each of radius r , coalesce to form a bigger drop. The surface energy released in this process is _______. ( S is the surface tension of mercury).

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Correct answer: B. 16π r^2 S

Let the radius of the bigger drop be R . Since the total volume remains conserved during the process: 8 × (4)/(3) π r^3 = (4)/(3) π R^3 R^3 = 8r^3 R = 2r Initial surface energy of the 8 drops is: U_i = 8 × (4 π r^2 S) = 32 π r^2 S Final surface energy of the bigger drop is: U_f = 4 π R^2 S = 4 π (2r)^2 S = 16 π r^2 S The surface energy released in the process is: Δ U = U_i - U_f = 32 π r^2 S - 16 π r^2 S = 16 π r^2 S Answer: 16π r^2 S

Eight mercury drops, each of radius r , coalesce to form a bigger...

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