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Two mercury drops (each of radius r) merge to form bigger drop. The surface energy of the bigger drop, if T is the surface tension, is:
- 4πr²T
- 2πr²T
- 2^(8/3)πr²T
- 2^(5/3)πr²T
Correct answer: 2^(8/3)πr²T
Solution
When two mercury drops merge, their combined volume increases, leading to a larger drop with a new radius. The surface area of the larger drop is greater than the sum of the surface areas of the two smaller drops, resulting in a change in surface energy that can be calculated using the formula for surface energy, which incorporates the new radius and surface tension.
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