Question

Two soap bubbles each with radius \(r_1\) and \(r_2\) coalesce in vacuum under isothermal conditions to form a bigger bubble of radius \(R\). Then \(R\) is equal to

• A. \(\sqrt{r_1^2 + r_2^2}\)
• B. \(\sqrt{r_1^2 - r_2^2}\)
• C. \(r_1 + r_2\)
• D. \(\frac{\sqrt{r_1^2 + r_2^2}}{2}\)
• E. \(2\sqrt{r_1^2 + r_2^2}\)

Hint:

In a vacuum under isothermal conditions, simply remember that the square of the final radius is the sum of the squares of the individual radii: \(R^2 = \sum r_i^2\).

Answer 1

The Correct Option is A

Concept: When soap bubbles coalesce in a vacuum under isothermal (constant temperature) conditions, the total surface energy remains conserved if no external work is done, or more fundamentally, the pressure-volume relationship and surface area properties are used.
• Surface Area of a sphere: \(4\pi r^2\)
• A soap bubble has two surfaces (inner and outer), so total surface area is \(8\pi r^2\).
• Under isothermal conditions in a vacuum, the final surface area is the sum of the initial surface areas: \(S_{total} = S_1 + S_2\).

Step 1: Equate the surface areas.
The energy of a soap bubble is given by \(E = T \times \text{Area}\), where \(T\) is surface tension. For two bubbles coalescing: \[ 8\pi R^2 T = 8\pi r_1^2 T + 8\pi r_2^2 T \]

Step 2: Solve for \(R\).
Dividing the entire equation by \(8\pi T\): \[ R^2 = r_1^2 + r_2^2 \] \[ R = \sqrt{r_1^2 + r_2^2} \]