Pressure inside two soap bubbles are `1.01` and `1.02` atmospheres. Ratio between their volumes is

Options

A. `102 : 101`
B. `(102)^3 : (101)^3`
C. `8:1`
D. `2:1`

Answer

To solve the problem of finding the ratio of the volumes of two soap bubbles given their internal pressures, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Pressures:**
- Let the pressure inside the first bubble (P1) be 1.01 atmospheres.
- Let the pressure inside the second bubble (P2) be 1.02 atmospheres.

2. **Calculate the Excess Pressures:**
- The excess pressure (ΔP) inside a soap bubble is given by:
\[ \Delta P = P - P_{\text{atm}} \]
- For the first bubble:
\[ \Delta P_1 = P_1 - P_{\text{atm}} = 1.01 - 1 = 0.01 \text{ atmospheres} \]
- For the second bubble:
\[ \Delta P_2 = P_2 - P_{\text{atm}} = 1.02 - 1 = 0.02 \text{ atmospheres} \]

3. **Relate Excess Pressure to Radius:**
- The excess pressure inside a soap bubble is related to its radius (r) by the formula:
\[ \Delta P = \frac{4\sigma}{r} \]
- From this, we can say that:
\[ \Delta P \propto \frac{1}{r} \]
- Therefore, the ratio of the excess pressures can be expressed as:
\[ \frac{\Delta P_1}{\Delta P_2} = \frac{r_2}{r_1} \]

4. **Substitute the Values:**
- Substitute the values of ΔP:
\[ \frac{0.01}{0.02} = \frac{r_2}{r_1} \]
- Simplifying this gives:
\[ \frac{1}{2} = \frac{r_2}{r_1} \]
- This implies:
\[ r_2 = \frac{1}{2} r_1 \]

5. **Calculate the Volume Ratio:**
- The volume (V) of a sphere is given by:
\[ V = \frac{4}{3} \pi r^3 \]
- Therefore, the volumes of the two bubbles are:
\[ V_1 = \frac{4}{3} \pi r_1^3 \]
\[ V_2 = \frac{4}{3} \pi r_2^3 \]
- The ratio of the volumes is:
\[ \frac{V_1}{V_2} = \frac{r_1^3}{r_2^3} \]
- Substituting \( r_2 = \frac{1}{2} r_1 \):
\[ \frac{V_1}{V_2} = \frac{r_1^3}{\left(\frac{1}{2} r_1\right)^3} = \frac{r_1^3}{\frac{1}{8} r_1^3} = 8 \]

6. **Final Result:**
- Thus, the ratio of the volumes \( V_1 : V_2 \) is:
\[ V_1 : V_2 = 8 : 1 \]

### Conclusion:
The ratio of the volumes of the two soap bubbles is 8:1.