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Simran Bhatia Grade: 11
         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.
3 years ago

Answers : (1)

Kevin Nash
askIITians Faculty
332 Points
										Sol. 4/3  πR3 = 4/3 πr2

⇒ r = R/2 = 2
Increase in surface energy = TA’ – TA
3 years ago
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