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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 :
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energy = 4VT 
1 - 1 
is released r R -
energy = 3VT 1 + 1 is absorbed r R -
energy = 3VT 1 - 1 is released r R - energy is neither released nor absorbed
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Correct Option: C
As surface area decreases so energy is released. Energy released = 4πR2T[n1 / 3 – 1] where R = n1 / 3r
| = 4πR3T |  | - |  | = 3VT | | - | | ||||
| r | R | r | R |
