Question

Under isothermal conditions, two soap bubbles of radii $r_{1}$ and $r_{2}$ coalesce to form a big bubble. The radius of the big bubble is}

• A. $(r_{1}+r_{2})^{1/2}$
• B. $(r_{1}+r_{2})^{2}$
• C. $(r_{1}^{2}+r_{2}^{2})^{1/2}$
• D. $(r_{1}+r_{2})^{3}$

Hint:

Coalescence in vacuum: $R^2 = r_1^2 + r_2^2$. Coalescence under atmospheric pressure: $R^3 = r_1^3 + r_2^3$.

Answer 1

The Correct Option is C

Step 1: Concept
In isothermal coalescence in a vacuum (or assuming constant external pressure), the surface energy is conserved if no external work is done.

Step 2: Analysis
Surface energy $E = T \times \text{Area}$. For a soap bubble (two surfaces), $E = 2 \times T \times (4\pi r^2)$.
$8\pi T r_1^2 + 8\pi T r_2^2 = 8\pi T R^2$.

Step 3: Calculation
$r_1^2 + r_2^2 = R^2 \implies R = \sqrt{r_1^2 + r_2^2}$.

Step 4: Conclusion
Hence, the radius of the big bubble is $(r_{1}^{2}+r_{2}^{2})^{1/2}$. Final Answer: (C)