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.

Answer 1

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

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