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JEE Advance 2017
Paper-1
Question
A drop of liquid of radius $R=10^{-2} m$ having surface tension $S=\frac{0.1}{4 \pi} \mathrm{Nm}^{-1}$ divides itself into $K$ identical drops. In this process the total change in the surface energy $\Delta \mathrm{U}=10^{-3} \mathrm{~J}$. If $\mathrm{K}=10^\alpha$ then the value of $\alpha$ is
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JEE Advance
Physics
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