Question

The excess pressure inside a soap bubble is 1.5 times the excess pressure inside a second soap bubble. The volume of the second bubble is ' \( x \) ' times the volume of the first bubble. The value of ' \( x \) ' is

• A. \( \frac{3}{2} \)
• B. \( \frac{9}{4} \)
• C. \( \frac{8}{27} \)
• D. \( \frac{27}{8} \)

Hint:

Pressure $\propto \frac{1}{R}$; Volume $\propto R^3$. Therefore, Volume $\propto \frac{1}{P^3}$.

Answer 1

The Correct Option is D

Step 1: Pressure-Radius Relation
Excess pressure in soap bubble: $P = \frac{4T}{R} \implies P \propto \frac{1}{R}$.
Given $P_1 = 1.5 P_2 \implies \frac{P_1}{P_2} = \frac{3}{2} = \frac{R_2}{R_1} \implies R_2 = 1.5 R_1$.
Step 2: Volume-Radius Relation
Volume $V = \frac{4}{3}\pi R^3 \implies V \propto R^3$.
$\frac{V_2}{V_1} = \left( \frac{R_2}{R_1} \right)^3$.
Step 3: Calculation
$x = \frac{V_2}{V_1} = (1.5)^3 = \left( \frac{3}{2} \right)^3 = \frac{27}{8}$.
Final Answer: (D)