Question

A drop of water and a soap bubble have the same radii. Surface tension of soap solution is half of that of water. Find the ratio of excess pressure inside the drop and bubble.

Answer 1

$r _{1}= r _{2}= r$
$T _{\text {soap }}=\frac{1}{2} T _{\text {water }}$
$P _{\text {excess}}$ of drop $=\frac{2 T _{\text {water }}}{ r }$
$P_{\text {excess }}$ of soap solution $=\frac{4 T_{\text {soap }}}{r}$
$\frac{P_{\text {drop }}}{P_{\text {bubble }}}=\frac{\frac{2 T_{\text {water }}}{r}}{\frac{4 T_{\text {soap }}}{r}}=\frac{T_{\text {water }}}{2 T_{\text {soap }}}$
$\frac{ P _{\text {drop }}}{ P _{\text {bubble }}}=\frac{2}{2}=1 ; \quad \frac{ P _{\text {drop }}}{ P _{\text {bubble }}}=1: 1$