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

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answer is 6.

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Detailed Solution

Surface energy U1=4πR2S(for big drop)U2=K4πr2S ( for small drop) R3=Kr3⇒r=RK1/3ΔU=K4πr2S−4πR2S=K4πR2SK2/3−4πR2S=4πR2SK1/3−1≈4πR2SK1/3=4π10−220.14πK1/3=10−3⇒10−5K1/3=10−3K1/3=10210α/3=102α=6
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A drop of liquid of radius R=10−2 m having surface tension S=0.14πNm−1 divides itself into K identical drops. In this process the total change in the surface energy is ΔU=10−3J . If K=10α then the value of α is