A drop of liquid of radius R=10–2m having surface tension S=0.14πNm−1 divides itself into K identical drops. In this process the total change in the surface energyΔU=10−3J . If K=10α then the value of α is

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answer is 6.

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Detailed Solution

Let radius of small drops =r . 43πR3=K43πr3 R=K13r … (i)S(K4πr2−4πR2)=10−3 0.14π4π(kR2K13−R2)=10−3 R2(K13−1)=10−2 K13−1=100 K13=101 10α3=101 α≈6

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