Question

Two spherical soap bubbles of radii $r_1$ and $r_2$ in vacuum combine under isothermal conditions. The resulting bubble has a radius equal to

• A. $\sqrt{r_1^2+r_2^2}$
• B. $\frac{r_1+r_2}{2}$
• C. $\frac{r_1{r}_2}{r_1+r_2}$
• D. $\sqrt{r_1{r}_2}$

Answer 1

The correct option is A $\sqrt{r_1^2+r_2^2}$


No. of moles is conserved when two bubbles combine to form a single bubble.

$n_1+n_2=n_3$

Using ideal gas equation,

$\frac{P_1{V}_1}{R{T}_1}+\frac{P_2{V}_2}{R{T}_2}=\frac{P_3{V}_3}{R{T}_3}$

As the process takes place in isothermal condition,

$\Rightarrow P_1{V}_1+P_2{V}_2=P_3{V}_3$

Using excess pressure inside the bubble formula,

$\Rightarrow \frac{4S}{r_1}\left(\frac{4}{3}\pi r_1^3\right)+\frac{4S}{r_2}\left(\frac{4}{3}\pi r_2^3\right)=\frac{4S}{r_3}\left(\frac{4}{3}\pi r_3^3\right)$

$\Rightarrow r_1^2+r_2^2=r_3^2$

$\Rightarrow r_3=\sqrt{r_1^2+r_2^2}$

Hence, option $\left(C\right)$ is correct.