A drop of liquid of radius R=10^ -2 m having surface tension S= 0.1 4 Nm^ -1 divides itself into K identical drops. In this process the total change in the surface energy U=10^ -3 J. If K=10^ then the value of is

JEE Advanced 2017 Physics question

A drop of liquid of radius m having surface tension Nm divides itself into identical drops. In this process the total change in the surface energy J. If then the value of is

Answer: 6

Solution

Volume conservation gives the radius of each small drop . Change in surface energy: (Kr^2R^2)$=4π

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Question ID VK-026599