Question

Pressure inside two soap bubbles are 1.01 and 1.02 atmosphere, respectively. The ratio of their volumes is:

• A. 4:1
• B. 0.8:1
• C. 8:1
• D. 2:1

Hint:

Smaller soap bubbles have higher internal pressure due to surface tension effects.

Answer 1

The Correct Option is C

Step 1: {Use Laplace’s pressure equation}
\[ \Delta P = \frac{4T}{R} \] Step 2: {Compare pressures and radii}
\[ \frac{\Delta P_1}{\Delta P_2} = \frac{R_2}{R_1} \Rightarrow R_1 = 2 R_2 \] Step 3: {Find volume ratio}
\[ V_1 : V_2 = R_1^3 : R_2^3 = 8:1 \] Thus, the correct answer is 8:1.

Answer 2

Step 1: For a soap bubble, the excess pressure inside is given by:
\( \Delta P = \frac{4T}{r} \), where:
- \( T \) is surface tension
- \( r \) is the radius of the bubble

Step 2: Since surface tension \( T \) is the same for both bubbles, we can write:
\( \frac{\Delta P_1}{\Delta P_2} = \frac{r_2}{r_1} \)

Step 3: Now relate volumes:
Volume of a sphere \( V = \frac{4}{3}\pi r^3 \), so:
\( \frac{V_1}{V_2} = \left( \frac{r_1}{r_2} \right)^3 \)

Step 4: Use the pressure values:
Let external pressure be atmospheric (1 atm), so:
Excess pressure in bubble 1: \( \Delta P_1 = 1.01 - 1.00 = 0.01 \, atm \)
Excess pressure in bubble 2: \( \Delta P_2 = 1.02 - 1.00 = 0.02 \, atm \)

So, \( \frac{\Delta P_1}{\Delta P_2} = \frac{0.01}{0.02} = \frac{1}{2} = \frac{r_2}{r_1} \Rightarrow \frac{r_1}{r_2} = 2 \)

Step 5: Now calculate volume ratio:
\( \frac{V_1}{V_2} = \left( \frac{r_1}{r_2} \right)^3 = 2^3 = 8 \)
So, \( \frac{V_1}{V_2} = 8:1 \)

Final Answer: \( 8:1 \)