Question

Two soap bubbles of radii 3 mm and 4 mm confined in vacuum coalesce isothermally to form a new bubble. The radius of the bubble formed (in mm) is

• A. 3
• B. 3.5
• C. 4
• D. 5
• E. 7

Hint:

This "Pythagorean" relationship ($R^2 = R_1^2 + R_2^2$) for bubble coalescence only holds true in a vacuum. If performed in atmosphere, the calculation involves pressure and volume conservation ($PV$).

Answer 1

The Correct Option is D

Concept:
When soap bubbles coalesce isothermally in a vacuum, the total surface area is conserved because the internal pressure is balanced only by surface tension. [itemsep=8pt]
• Surface Energy/Area Conservation: In a vacuum, the work done by surface tension during coalescence is related to the surface area. For soap bubbles (which have two surfaces), the relation for isothermal coalescence is $R^2 = R_1^2 + R_2^2$.

Step 1: Apply the conservation of square of radii.
Given $R_1 = 3$ mm and $R_2 = 4$ mm. \[ R^2 = 3^2 + 4^2 \] \[ R^2 = 9 + 16 = 25 \]

Step 2: Calculate the final radius.
\[ R = \sqrt{25} = 5 \text{ mm} \]