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}.

#### Solution

Given:

Initial radius of mercury drop R = 2 mm = 2 × 10^{−3} m

Surface tension of mercury T = 0.465 J/m^{2}

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 × 4πr^{2} − 4πR^{2}) 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.