Question

The pressures inside two soap bubbles A and B are \(1.02\) atmosphere and \(1.04\) atmosphere respectively. The ratio of volume of bubble A to that of bubble B is (outside pressure \(= 1\) atmosphere)

• A. \(1:8\)
• B. \(8:1\)
• C. \(4:1\)
• D. \(1:4\)

Hint:

Excess pressure in a soap bubble is 4T/r, so volume goes as 1/(excess pressure)^3.

Answer 1

The Correct Option is B

Step 1: Excess Pressure:
For a soap bubble, \(\Delta P=\dfrac{4T}r\), so \(r\propto\dfrac1{\Delta P}\).

Step 2: Compute:
\(\Delta P_A=1.02-1=0.02\) atm and \(\Delta P_B=1.04-1=0.04\) atm. So \(\dfrac{r_A}{r_B}=\dfrac{0.04}{0.02}=2\).

Step 3: Volume Ratio:
\[ \frac{V_A}{V_B}=\left(\frac{r_A}{r_B}\right)^3=2^3=8 \]

Step 4: Check the Options:
Option (A) \(1:8\) is the inverse. Options (C) and (D) correspond to ratios of 4 and 1/4, which would need a radius ratio of \(4^{1/3}\), not 2. So (B) is correct.

Final Answer:
The volume ratio is \(8:1\), option (B). \[ \boxed{\text{(B) } 8:1} \]