The volume of an air bubble becomes three times as it rises from the bottom of a lake to its surface. Assuming atmospheric pressure to be 75 cm of Hg and the density of water to be `1//10` of the density of mercury, the depth of the lake is

Options

A. `5m`
B. `10 m`
C. `15 m`
D. `20 m `

Answer

To solve the problem of finding the depth of the lake based on the behavior of an air bubble rising to the surface, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Data:**
- The volume of the air bubble increases three times as it rises: \( V_2 = 3V_1 \).
- Atmospheric pressure at the surface: \( P_{atm} = 75 \, \text{cm of Hg} \).
- Density of water: \( \rho_{water} = \frac{1}{10} \rho_{Hg} \).

2. **Understand the Process:**
- The process is isothermal (constant temperature) since the bubble is rising slowly.
- According to Boyle's Law for isothermal processes:
\[ P_1 V_1 = P_2 V_2 \]

3. **Determine Initial and Final Pressures:**
- At the bottom of the lake, the pressure \( P_1 \) is the sum of atmospheric pressure and the pressure due to the water column above the bubble:
\[ P_1 = P_{atm} + \rho_{water} \cdot g \cdot h \]
- At the surface, the pressure \( P_2 \) is simply the atmospheric pressure:
\[ P_2 = P_{atm} \]

4. **Set Up the Equation Using Boyle's Law:**
- Substitute the known pressures into Boyle's Law:
\[ (P_{atm} + \rho_{water} \cdot g \cdot h) V_1 = P_{atm} \cdot (3V_1) \]
- Cancel \( V_1 \) from both sides (assuming \( V_1 \neq 0 \)):
\[ P_{atm} + \rho_{water} \cdot g \cdot h = 3P_{atm} \]

5. **Rearranging the Equation:**
- Rearranging gives:
\[ \rho_{water} \cdot g \cdot h = 3P_{atm} - P_{atm} = 2P_{atm} \]
- Thus:
\[ \rho_{water} \cdot g \cdot h = 2 \cdot 75 \, \text{cm of Hg} \]

6. **Convert Units:**
- Convert the pressure from cm of Hg to a consistent unit. The density of mercury \( \rho_{Hg} \) is approximately \( 13.6 \, \text{g/cm}^3 \).
- Therefore, \( P_{atm} = 75 \, \text{cm of Hg} = 75 \cdot 13.6 \, \text{g/cm}^2 = 1020 \, \text{g/cm}^2 \).

7. **Calculate the Depth:**
- Substitute \( \rho_{water} = \frac{1}{10} \rho_{Hg} = \frac{1}{10} \cdot 13.6 \, \text{g/cm}^3 = 1.36 \, \text{g/cm}^3 \):
\[ 1.36 \cdot g \cdot h = 2 \cdot 1020 \]
- Assuming \( g \approx 980 \, \text{cm/s}^2 \):
\[ 1.36 \cdot 980 \cdot h = 2040 \]
- Solving for \( h \):
\[ h = \frac{2040}{1.36 \cdot 980} \approx 1500 \, \text{cm} \]

8. **Convert to Meters:**
- Convert cm to meters:
\[ h = \frac{1500}{100} = 15 \, \text{m} \]

### Final Answer:
The depth of the lake is **15 meters**.