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A certain number of spherical drops of a liquid of radius 'r' coalesce to form a single drop of radius 'R' and volume 'V'. If 'T' is the surface tension of the liquid, then:

1. energy = 4VT (1/r - 1/R) is released
2. energy = 3VT (1/r + 1/R) is absorbed
3. energy = 3VT (1/r - 1/R) is released
4. energy is neither released nor absorbed

Correct answer: energy = 3VT (1/r - 1/R) is released

Solution

When smaller drops coalesce into a larger drop, the total surface area decreases, leading to a reduction in surface energy. The energy released is proportional to the change in surface area, which is given by 3VT(1/r - 1/R).

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