1. Pressure inside a soap bubble

For a soap bubble in vacuum, the excess pressure inside is $\Delta P = \frac{4T}{r}$ where $T$ is the surface tension and $r$ is the radius.

So for the two initial bubbles: $P_1 = \frac{4T}{r_1}, \qquad P_2 = \frac{4T}{r_2}$ Since outside is vacuum, these are also their absolute pressures.

2. Use ideal gas law under isothermal conditions

At constant temperature, for each bubble: $PV = nRT$ So the number of moles is proportional to $PV$.

When the two bubbles combine, total gas is conserved: $P_1V_1 + P_2V_2 = PV$ where $P = \frac{4T}{R}$ and $R$ is the radius of the resulting bubble.

3. Substitute volumes

Volume of a sphere: $V = \frac{4}{3}\pi r^3$ Thus, $\frac{4T}{r_1}\cdot \frac{4}{3}\pi r_1^3 + \frac{4T}{r_2}\cdot \frac{4}{3}\pi r_2^3 = \frac{4T}{R}\cdot \frac{4}{3}\pi R^3$

4. Simplify

Cancel common factors $\frac{16}{3}\pi T$: $r_1^2 + r_2^2 = R^2$ Hence, $R = \sqrt{r_1^2 + r_2^2}$

5. Check options

• A: $\dfrac{r_1r_2}{r_1+r_2}$ — incorrect
• B: $\sqrt{r_1r_2}$ — incorrect
• C: $\sqrt{r_1^2+r_2^2}$ — correct
• D: $\dfrac{r_1+r_2}{2}$ — incorrect

Therefore, the correct option is C.