Question

The excess pressure inside the first soap bubble of radius \(R_1\) is three times that inside the second soap bubble of radius \(R_2\). The ratio of volumes of the first to second bubble is

• A. \(1:27\)
• B. \(1:18\)
• C. \(1:9\)
• D. \(1:3\)

Hint:

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

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
For a soap bubble with two surfaces, excess pressure \(P = \dfrac{4T}{R}\). So \(R \propto \dfrac1P\).

Step 2: Radius ratio
\(P_1 = 3P_2\), so \(R_1 = \dfrac{R_2}{3}\), that is \(R_1 : R_2 = 1 : 3\).

Step 3: Volume ratio
Volume goes as \(R^3\):
\[ \frac{V_1}{V_2} = \left(\frac{1}{3}\right)^3 = \frac{1}{27} \]
So the ratio is 1:27. Option (D), 1:3, is the radius ratio.

Final Answer:
The volume ratio is 1:27, option (A). \[ \boxed{1:27} \]