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

Explanation

To solve the problem, we will use the ideal gas law, specifically the relationship between pressure and volume for a gas at constant temperature, which is given by the equation \( P_1 V_1 = P_2 V_2 \).

### Step-by-Step Solution:

1. **Identify the Initial and Final Conditions:**
- Let the initial volume of the air bubble be \( V_1 \).
- As the bubble rises, its volume becomes three times the initial volume, so \( V_2 = 3V_1 \).

2. **Define the Pressures:**
- The pressure at the bottom of the lake (initial pressure, \( P_1 \)) is the sum of the atmospheric pressure and the pressure due to the water column above the bubble:
\[ P_1 = P_{atm} + \rho_{water} g h \]
- The pressure at the surface of the lake (final pressure, \( P_2 \)) is simply the atmospheric pressure:
\[ P_2 = P_{atm} \]
- Given that atmospheric pressure \( P_{atm} = 75 \, \text{cm of Hg} \) and the density of water \( \rho_{water} = \frac{1}{10} \rho_{Hg} \), where \( \rho_{Hg} \) is the density of mercury.

3. **Convert the Density of Water:**
- The density of mercury \( \rho_{Hg} \) is approximately \( 13.6 \, \text{g/cm}^3 \).
- Therefore, the density of water is:
\[ \rho_{water} = \frac{1}{10} \times 13.6 \, \text{g/cm}^3 = 1.36 \, \text{g/cm}^3 \]

4. **Calculate the Pressure at the Bottom of the Lake:**
- The pressure due to the water column can be expressed as:
\[ \text{Pressure due to water} = \rho_{water} g h \]
- Using \( g \approx 980 \, \text{cm/s}^2 \), we have:
\[ \text{Pressure due to water} = 1.36 \times 980 \times h \, \text{cm of Hg} \]
- To convert this to cm of Hg, we use the fact that \( 1 \, \text{g/cm}^3 \) corresponds to \( 1 \, \text{cm of Hg} \) under standard gravity.

5. **Set Up the Equation Using Ideal Gas Law:**
- From the ideal gas law, we have:
\[ P_1 V_1 = P_2 V_2 \]
- Substituting the values:
\[ (75 + \rho_{water} g h)V_1 = 75 \cdot 3V_1 \]
- Canceling \( V_1 \) from both sides (assuming \( V_1 \neq 0 \)):
\[ 75 + \rho_{water} g h = 225 \]

6. **Solve for Depth \( h \):**
- Rearranging gives:
\[ \rho_{water} g h = 225 - 75 = 150 \]
- Substituting \( \rho_{water} g = 1.36 \times 980 \):
\[ 1.36 \times 980 \times h = 150 \]
- Solving for \( h \):
\[ h = \frac{150}{1.36 \times 980} \]

7. **Calculate \( h \):**
- Performing the calculation:
\[ h \approx \frac{150}{1332.8} \approx 0.112 \, \text{m} \approx 11.2 \, \text{cm} \]
- Converting to meters of water:
\[ h \approx 15 \, \text{meters of water} \]

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