First slide

Surface tension and surface energy

Question

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 $0 .465 J / m^{2}$ )

Moderate

A

$23.4 μ J$
B

$18.5 μ J$
C

$26.8 μ J$
D

$16.8 μ J$

Solution

Increase in surface energy or work done in splitting a big drop $= 4 π R^{2} T (n^{1 / 3} - 1)$
$\Rightarrow W = 4 π \times {(2 \times 10^{- 3})}^{2} \times 0.465 (8^{1 / 3} - 1) = 23.4 μ J$

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