Physics

Physics

Problem

A drop of liquid of radius R=10−2 mR = 10^{-2} \text{ m} having surface tension S=0.14π Nm−1S = \frac{0.1}{4\pi} \text{ Nm}^{-1} divides itself into KK identical drops. In this process the total change in the surface energy ΔU=10−3 J\Delta U = 10^{-3} \text{ J} . If K=10αK = 10^\alpha then the value of α\alpha is

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