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Two small drops of mercury each of radius R coalesce to form a single large drop. The ratio of total surface energy before and after the change is:

1. 2^(1/3): 1
2. 1: 2^(1/3)
3. 2: 1
4. 1: 2

Correct answer: 2^(1/3): 1

Solution

Volume conservation gives R' = 2^(1/3) R. Surface energy is proportional to area (r^2). Before/after = (2*4piR^2)/(4pi*2^(2/3)R^2) = 2/2^(2/3) = 2^(1/3). So the ratio is 2^(1/3):1.

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