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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 = 0.465 J m−2.
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Solution
Given:
Initial radius of mercury drop R = 2 mm = 2 × 10−3 m
Surface tension of mercury T = 0.465 J/m2
Let the radius of a small drop of mercury be r.
As one big drop is split into 8 identical droplets:
volume of initial drop = 8 × (volume of a small drop)
\[ \left( \frac{4}{3} \right)\pi R^3 = \left( \frac{4}{3} \right)\pi r^3 \times 8\]
Taking cube root on both sides of the above equation: \[r\]\[=\]\[\frac{R}{2}\]\[=\]\[10\]\[\]-3 m
Surface energy = T × surface area
∴ Increase in surface energy = TA' − TA
= (8 × 4πr2 − 4πR2) T
\[= 4\pi T\left[ 8 \times \left( \frac{R^2}{4} \right) - R^2 \right]\]
\[ = 4\pi T R^2\]
= 4 × (3.14) × (0.465) × (4 × 10−6)
= 23.36 × 10−6
= 23.4 μJ
Hence, the required increase in the surface energy of the mercury droplets is 23.4 μJ.
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