Question

The radii of two soap bubbles are r₁ and r₂. In isothermal conditions, two meet together in vacuum. Then the radius of the resultant bubble is given by

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

Answer 1

Answer: (C)

$R^{2} = r_{1}^{2} + r_{2}^{2}$

Answer 2

Under isothermal condition surface energy remain constant

∴ $8\pi r_{1}^{2}T + 8\pi r_{2}^{2}T = 8\pi R^{2}T$ ⇒ $R^{2} = r_{1}^{2} + r_{2}^{2}$