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Question

A drop of mercury of radius 2 mm is split into 8 identical drops. Find the increase in surface energy. Surface tension of mercury =0.465 Jm2

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Solution

R=2mm 2×103m

The volume of drop is given as:
43πR3=8×43×r3
R3=8r3

R38=r3 (R2)3=r3
r=1mm=103m

A1=4πR2
4π(2×102)2=4π×4×106=16π×106

A2=8×4πr2=32π(102)3=32π106

ΔA=A1A2
(32π16π)106=16π×106

the increase in surface energy will be:
Es=TΔA=23.376

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