Question

The excess pressure inside a soap bubble of volume \( V \) is three times the excess pressure inside a second soap bubble of volume \( V_1 \). The value of \( \left( \frac{V}{V_1} \right) \) is:

• A. 1:3
• B. 9:1
• C. 1:9
• D. 3:1

Hint:

The excess pressure inside a soap bubble is inversely proportional to its radius, which is related to the volume by \( r \propto V^{1/3} \).

Answer 1

The Correct Option is B

Step 1: Formula for Excess Pressure.
The excess pressure inside a soap bubble is given by: \[ P = \frac{4T}{r} \] where \( T \) is the surface tension and \( r \) is the radius of the bubble. The radius is inversely proportional to the volume \( V \), so we have: \[ P \propto \frac{1}{r} \propto \frac{1}{V^{1/3}} \] Since the excess pressure in the first bubble is three times that in the second bubble, we can equate the pressures and solve for the volume ratio: \[ \left( \frac{V}{V_1} \right) = 9:1 \] Step 2: Final Answer.
Thus, the value of \( \left( \frac{V}{V_1} \right) \) is 9:1.