Q. 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) $\sqrt{r_{1}^{2} + r_{2}^{2}}$
Explanation
(d) Excess of pressure, inside the first bubble $p_{1} = \frac{4T}{r_{1\ \ }}$
$Similarly,\ \ p_{2} = \frac{4T}{r_{2\ }}$
Let the radius of the large bubble be R. then, excess of pressure inside the large bubble $p = \frac{4T}{R}$
Under isothermal condition, temperature remains constant.
$So,\ \ \ \ \ \ \ \ pV = p_{1}\ V_{1} + p_{2}\ V_{2}$
$\frac{4T}{R}\left( \frac{4}{3}\pi r^{3} \right) = \frac{4T}{r_{1}}\left( \frac{4}{3}\pi r_{1}^{3} \right) + \frac{4T}{r_{2\ }}\left( \frac{4}{3}\pi r_{2}^{3} \right)$
$R^{2} = r_{1}^{2} + r_{2}^{2}$
$\Longrightarrow \ R = \sqrt{r_{1}^{2} + r_{2}^{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.