Explanation

(a) Surface tension (T )of a liquid is equal to the work (W ) required to increases the surface area $(\mathrm{\Delta}A)$ of the liquid film by unity at constant temperature.
$W = T\mathrm{\Delta}A = surface\ energy$
Also, volume of big drop $= 27 \times volume\ of\ small\ drop$
$ie,\ V^{'} = 27V$
Where $V'$ is volume of big drop of diameter D and V the volume of small drop of diameter $d.$
$\therefore\ \ \frac{4}{2}\pi\left( \frac{D}{2} \right)^{3} = 27 \times \frac{4}{3}\pi\left( \frac{d}{2} \right)^{3}$
$\Longrightarrow \ \ \ \frac{D}{2} = 3 \times \frac{d}{2}$
$\Longrightarrow \ \ d = \frac{D}{3}$
$Radius\ of\ small\ drop,\ r = \frac{d}{2} = \frac{D}{6} \bullet$
$\therefore Change\ in\ surface\ energy = T\left( A_{2} - A_{1} \right)$
$\ \ = T\left\lbrack 27.4\pi r^{2} - 4\pi R^{2} \right\rbrack$
$\ \ = T4\pi\left\lbrack 27\left( \frac{D}{6} \right)^{2} - \left( \frac{D}{6} \right)^{2} \right\rbrack$
$\ \ = 4\pi T\left\lbrack \frac{{3D}^{2}}{4} - \frac{D^{2}}{4} \right\rbrack = 2\pi D^{2}T$