Question

Two soap bubbles, A and B, have radii in the ratio \(3:2\). The excess pressure inside the bubble A to that of B is in the ratio

• A. \(2:3\)
• B. \(3:2\)
• C. \(1:5\)
• D. \(4:9\)

Hint:

Excess pressure inside a soap bubble is 4T/R, so radius is inversely proportional to pressure.

Answer 1

The Correct Option is A

Step 1: Solve for radii:
Write $R_2 = \frac{4T}{\Delta P_2}$. The first bubble has $\Delta P_1 = 3\Delta P_2$, so $R_1 = \frac{4T}{3\Delta P_2} = \frac{R_2}{3}$.

Step 2: Cube it:
$\frac{V_1}{V_2} = \frac{\frac43\pi R_1^3}{\frac43\pi R_2^3} = \frac{1}{27}$ (A).

Final Answer:
1:27. \[ \boxed{1:27} \]