Two mercury drops, each of radius r, coalesce to form a single larger drop. If the surface tension is 1/pi (in SI units), the surface energy of the bigger drop is:

2^(5/3) r²
4r²
2r²
2^(8/3) r²

Correct answer: 2^(8/3) r²

Solution

Volume conservation: R³ = 2 r³, so R = 2^(1/3) r. Surface energy E = T * 4*pi*R² = (1/pi)*4*pi*R² = 4*R² = 4*(2^(1/3) r)² = 4*2^(2/3) r² = 2² * 2^(2/3) r² = 2^(8/3) r².

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