Question

A drop of mercury of radius 2 mm is split into 8 identical droplets. Find the increase in surface enery. Surface tension of mercury = $0.465J{m}^{-2}$.

Answer 1

r = 2 mm = $2\times {10}^{-3}m$,

T = 0.465 $J{m}^2$

Vol. of initial drop = 8 $\times$ (vol of small drops)

$\left(\frac{4}{3}\right)\pi R^3=\left(\frac{4}{3}\right)\pi r^3\times 8$

$r=\frac{R}{2}=2$

Increase in surface energy

= $T{A}^{\prime }-TA=8\times 4\pi r^{2}T-4\pi R^{2}T$

$=4\pi T[8\times \left(\frac{R^2}{4}\right)-R^2]$

= $4\pi T{R}^2$

= $4\times \left(3.14\right)\times \left(0.465\right)\times (4\times {10}^{-6})$

= $23.36\times {10}^{-6}$

= 23.4$\mu J$