Question

If two soap bubbles each of radius \(2 \, \text{cm}\) combine in vacuum under isothermal conditions, then the radius of the new bubble formed is

• A. \(\sqrt{2} \, \text{cm}\)
• B. \(2\sqrt{2} \, \text{cm}\)
• C. \(0.5 \, \text{cm}\)
• D. \(2 \, \text{cm}\)

Hint:

When soap bubbles combine under isothermal conditions in vacuum, use the formula \( R = \sqrt{R_1^2 + R_2^2} \) to find the new radius.

Answer 1

The Correct Option is B

Step 1: In vacuum and under isothermal conditions, volume is conserved when two bubbles combine. \[ \text{Let the radius of each original bubble be } r = 2\, \text{cm} \] Step 2: Volume of each bubble = \(V = \frac{4}{3}\pi r^3\) Total volume of two bubbles: \[ V_{\text{total}} = 2 \times \frac{4}{3} \pi r^3 = \frac{8}{3} \pi r^3 \] Let \(R\) be the radius of the new bubble. Then: \[ \frac{4}{3} \pi R^3 = \frac{8}{3} \pi r^3 \Rightarrow R^3 = 2r^3 \Rightarrow R = r . \sqrt[3]{2} \] This is correct for general isothermal processes. However, in this particular question, **surface energy in vacuum** is also conserved (assuming ideal condition), and **pressure difference** inside bubble is considered. Hence, **pressure-volume** product is conserved due to **isothermal process**: \[ P_1V_1 + P_2V_2 = P_fV_f \] For small soap bubbles under vacuum, the relation reduces to: \[ R = \sqrt[3]{R_1^3 + R_2^3} \] Step 3: Since \(R_1 = R_2 = 2\, \text{cm}\), we get: \[ R = \sqrt[3]{2^3 + 2^3} = \sqrt[3]{16} = 2 \sqrt[3]{2} \] But in case of isothermal expansion of **gas** inside ideal soap bubbles (in vacuum), using pressure relation: \[ P \propto \frac{1}{R},
PV = \text{constant} \Rightarrow \frac{1}{R} . R^3 = \text{constant} \Rightarrow R^2 = \text{constant} \Rightarrow R = \sqrt{R_1^2 + R_2^2} \] Final Calculation: \[ R = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2} \] % Final Answer \[ \boxed{2\sqrt{2} \, \text{cm}} \]