Eight identical spherical mercury drops, each of radius r, coalesce to form a single larger spherical drop. If S is the surface tension of mercury, how much surface energy is released during the coalescence?
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Question

Eight identical spherical mercury drops, each of radius r, coalesce to form a single larger spherical drop. If S is the surface tension of mercury, how much surface energy is released during the coalescence?

Options

G1 Physics · Accepted Answer

Conservation of volume gives 8(4πr³ ÷ 3) = 4πR³ ÷ 3. Therefore, R³ = 8r³, so R = 2r. Initial total surface area: A(initial) = 8×4πr² = 32πr². Final surface area: A(final) = 4π(2r)² = 16πr². Surface energy equals surface tension multiplied by area. Hence the energy released is ΔU = S[A(initial)-A(final)] = S(32πr²-16πr²) = 16πr²S. Correct option: B.