Thank you for reporting, we will resolve it shortly
Q.
A spherical drop of radius $r$ is in equilibrium. The extra surface energy, if radius of bubble is increased by $\Delta r$, is ( $S=$ surface tension $)$
Mechanical Properties of Fluids
Solution:
Suppose a spherical drop of radius $r$ is in equilibrium. If its radius increases by $\Delta r$, then the extra surface energy is
$\Delta E_{S}=$ final surface energy - initial surface energy $=(S A)_{f}-(S A)_{i}$
where, $S=$ surface tension and $A$ is the surface area.
$=\left|4 \pi(r+\Delta r)^{2}-4 \pi r^{2}\right| S $
$=\left(4 \pi r^{2}+4 \pi r^{2}+8 \pi r \Delta r-4 \pi r^{2}\right) S=8 \pi r \Delta r S$
(neglecting $ \Delta r^{2}$ as it is very small)