Question

Two spherical soap bubbles of radii $r_{1}$ and $r_{2}$ in vacuum combine under isothermal conditions. The resulting bubble has radius equal to

• A. $\frac{r_{1} + r_{2}}{2}$
• B. $\frac{r_{1} \, r_{2}}{r_{1} + r_{2}}$
• C. $\sqrt{r_{1} r_{2}}$
• D. $\sqrt{r_{1}^{2} + r_{2}^{2}}$

Answer 1

Answer: (D)

Excess of pressure, inside the first bubble $p_{1}=\frac{4 T}{r_{1 \, \, }}$
$Similarly, \, \, p_{2}=\frac{4 T}{r_{2 \, }}$
Let the radius of the large bubble be R. then, excess of pressure inside the large bubble $p=\frac{4 T}{R}$
Under isothermal condition, temperature remains constant.
$So, \, \, pV=p_{1} \, V_{1}+p_{2} \, V_{2}$
$\frac{4 T}{R}\left(\frac{4}{3} \pi r^{3}\right)=\frac{4 T}{r_{1}}\left(\frac{4}{3} \pi r_{1}^{3}\right)+\frac{4 T}{r_{2 \, }}\left(\frac{4}{3} \pi r_{2}^{3}\right)$
$R^{2}=r_{1}^{2}+r_{2}^{2}$
$\Longrightarrow \, R=\sqrt{r_{1}^{2} + r_{2}^{2}}$