Question

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 $)$

• A. $4\pi r\Delta rS$
• B. $8\pi r\Delta rS$
• C. $2\pi r\Delta rS$
• D. $10\pi r\Delta rS$

Answer 1

Answer: (B)

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 $=(SA{)}_f-(SA{)}_i$
where, $S=$ surface tension and $A$ is the surface area.
$=\left|4\pi (r+\Delta r{)}^2-4\pi r^2\right|S$
$=\left(4\pi r^2+4\pi r^2+8\pi r\Delta r-4\pi r^2\right)S=8\pi r\Delta rS$
(neglecting $\Delta r^2$ as it is very small)