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A drop of mercury of radius 2 mm is split into 8 identical droplets. Find the increase in surface energy. Surface tension of mercury = 0.465 J m-3.

A drop of mercury of radius 2 mm is split into 8 identical droplets. Find the increase in surface energy. Surface tension of mercury = 0.465 J m-3.

Grade:11

1 Answers

Kevin Nash
askIITians Faculty 332 Points
7 years ago
Sol. 4/3 πR3 = 4/3 πr2 ⇒ r = R/2 = 2 Increase in surface energy = TA’ – TA

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