Explanation

(d) We know that surface tension is related to work as
$W = T \times \mathrm{\Delta}A$
Since, surface area of sphere is
$4\pi R^{2}$ and there are two free surfaces, we have
$W = T \times 8R^{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots\ldots(i)$
$and\ volume\ of\ sphere = \frac{4}{3}\pi R^{3}$
$ie,\ V = \frac{4}{3}\pi R^{3}$
$\Longrightarrow \ R = \left( \frac{3V}{4\pi} \right)^{1/3}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots..(ii)$
From Eqs. (i) and (ii), we get
$W = T \times 8\pi \times \left( \frac{3V}{4\pi} \right)^{2/3}$
$\Longrightarrow \ W \propto V^{2/3}$
$\therefore\ \ W_{1} \propto V_{2}^{2/3}$
$and\ \ W_{2} \propto V_{2}^{2/3}\ $
$\ \frac{W_{2}}{W_{1}} = \left( \frac{{2V}_{1}}{V_{1}} \right)^{2/3}$
$\Longrightarrow \ W_{2} = 2^{2/3}W_{1} = 4^{1/3}W$