Question

Two spherical soap bubbles of radii '$a$' and '$b$' in vacuum coalesce under isothermal conditions. The resulting bubble has a radius equal to}

• A. a + b
• B. \frac{a+b}{2}
• C. \sqrt{a^2 + b^2}
• D. \frac{a+b}{ab}

Hint:

When two soap bubbles coalesce isothermally in a vacuum, the total surface area is conserved, leading to the relation $R^2 = r_1^2 + r_2^2$.

Answer 1

The Correct Option is A

Step 1: In a vacuum, the excess pressure inside a soap bubble of radius $r$ is $P = \frac{4T}{r}$.
Step 2: Under isothermal conditions, the product of pressure and volume ($PV$) is conserved when the bubbles coalesce ($P_1 V_1 + P_2 V_2 = P_3 V_3$).
Step 3: Substitute the expressions for pressure and volume ($V = \frac{4}{3}\pi r^3$): \[ \left(\frac{4T}{a}\right)\left(\frac{4}{3}\pi a^3\right) + \left(\frac{4T}{b}\right)\left(\frac{4}{3}\pi b^3\right) = \left(\frac{4T}{c}\right)\left(\frac{4}{3}\pi c^3\right) \]
Step 4: Simplify the equation by canceling out the constant terms $4T$ and $\frac{4}{3}\pi$: \[ a^2 + b^2 = c^2 \Rightarrow c = \sqrt{a^2 + b^2} \]