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

Answer: 1.00

$r_1=r_2=r$
$T_{\text{soap }}=\frac{1}{2}{T}_{\text{water }}$
$P_{\text{excess}}$ of drop $=\frac{2T_{\text{water }}}{r}$
$P_{\text{excess }}$ of soap solution $=\frac{4T_{\text{soap }}}{r}$
$\frac{P_{\text{drop }}}{P_{\text{bubble }}}=\frac{\frac{2T_{\text{water }}}{r}}{\frac{4T_{\text{soap }}}{r}}=\frac{T_{\text{water }}}{2T_{\text{soap }}}$
$\frac{P_{\text{drop }}}{P_{\text{bubble }}}=\frac{2}{2}=1;\frac{P_{\text{drop }}}{P_{\text{bubble }}}=1:1$