Question

Two soap bubbles of radii $3\, cm$ and $4\, cm$ coalesce under isothermal conditions in vacuum to form a single bubble of radius $r _{1}$. If the two same soap bubbles coalesce in atmosphere, they form a single bubble with radius of interface $r_{2}$. Find $\frac{r_{2}}{r_{1}}$.

Answer 1

In vacuum under isothermal conditions, radius of drop formed, $r_{1}=\sqrt{r_{A}^{2}+r_{B}^{2}}=\sqrt{3^{2}+4^{2}}=5\, cm$
Under atmospheric condition, radius of interface of drop formed
$r_{2}=\frac{r_{A} r_{B}}{r_{B}-r_{A}}=\frac{3 \times 4}{4-3}$
$=12\, cm$
$\therefore \frac{r_{2}}{r_{1}}=\frac{12}{5}=2.4$