A total of n identical liquid drops, each of radius r, coalesce into one large spherical drop. The liquid surface tension is S. Which expression gives the surface energy released during coalescence?
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Question

A total of n identical liquid drops, each of radius r, coalesce into one large spherical drop. The liquid surface tension is S. Which expression gives the surface energy released during coalescence?

G1 Physics · Accepted Answer

By volume conservation, R³ = nr³. Therefore, R = r∛n. Initial total surface energy: U(initial) = n(4πr²S) = 4πSr²n. Final surface energy: U(final) = 4πSR² = 4πS(r∛n)² = 4πSr²(∛n)². Therefore the released energy is E(released) = U(initial)−U(final) = 4πSr²[n−(∛n)²]. Correct option: A.