Physics - Fluid Question with Solution | TestHub

Let $P_1$ and $P_2$ be the pressures inside the two soap bubbles of radii $r_1$ and $r_2$ respectively.

Since the bubbles are in vacuum, the external pressure is zero. The pressure inside a soap bubble of radius $r$ is given by $P=\frac{4T}{r}$, where $T$ is the surface tension.

The volumes of the bubbles are $V_1=\frac{4}{3}\pi r_1^3$ and $V_2=\frac{4}{3}\pi r_2^3$.

Under isothermal conditions, the temperature remains constant. The total number of moles of gas is conserved when the bubbles coalesce:

$n_1+n_2=n$

Using the ideal gas law $PV=nRT$, we get:

$P_1{V}_1+P_2{V}_2=PV$

Substituting the expressions for pressure and volume:

$\left(\frac{4T}{r_1}\right)\left(\frac{4}{3}\pi r_1^3\right)+\left(\frac{4T}{r_2}\right)\left(\frac{4}{3}\pi r_2^3\right)=\left(\frac{4T}{r}\right)\left(\frac{4}{3}\pi r^3\right)$

$\frac{16\pi T}{3}{r}_1^2+\frac{16\pi T}{3}{r}_2^2=\frac{16\pi T}{3}{r}^2$

$r_1^2+r_2^2=r^2$

$r=\sqrt{r_1^2+r_2^2}$

Answer: $\sqrt{r_1^2+r_2^2}$