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Q.

A drop of mercury of radius 2 mm is split into 8 identical droplets. Find the increase in surface energy. (Surface tension of mercury is 0.465 J/m2 )

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a

23.4μJ

b

18.5μJ

c

26.8μJ

d

16.8μJ

answer is A.

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Detailed Solution

Increase in surface energy or work done in splitting a big drop =4πR2T(n1/3−1)⇒W=4π×(2×10−3)2×0.465(81/3−1)=23.4 μJ
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