Question

The radii of two soap bubbles are $r_{1}$ and $r_{2}$. In isothermal condition they combine with each other to form a single bubble. The radius of resultant bubble is

• A. $R=\frac{r_{1}+r_{2{2}$
• B. $R=r_{1}(r_{1}r_{2}+r_{2})$
• C. $R=\sqrt{r_{1}^{2}+r_{2}^{2$
• D. $R=r_{1}+r_{2}$

Hint:

Logic Tip: When bubbles coalesce isothermally in a vacuum, the sum of their surface areas remains constant.

Answer 1

The Correct Option is C

Concept: When two soap bubbles combine under isothermal conditions, the gas inside them merges. Since temperature is constant, the process follows Boyle’s law, and effectively the total volume is conserved.
Step 1: Write volume relation Volume of a spherical bubble: \[ V = \frac{4}{3}\pi r^3 \] For two bubbles of radii $r_1$ and $r_2$: \[ V_1 + V_2 = V_{\text{final \] \[ \frac{4}{3}\pi r_1^3 + \frac{4}{3}\pi r_2^3 = \frac{4}{3}\pi R^3 \]
Step 2: Simplify Cancel $\frac{4}{3}\pi$: \[ r_1^3 + r_2^3 = R^3 \]
Step 3: Final expression \[ R = \left(r_1^3 + r_2^3\right)^{1/3} \] Conclusion: \[ \boxed{R = \left(r_1^3 + r_2^3\right)^{1/3 \]