When an air bubble of radius `'r'` rises from the bottom to the surface of a lake, its radius becomes `5r//4` (the pressure of the atmosphere is equal to the `10 m` height of water column). If the temperature is constant and the surface tension is neglected, the depth of the lake is

Options

A. `3.53 m`
B. `6.53 m`
C. `9.53 m`
D. `12.53 m`

Explanation

To solve the problem, we need to analyze the situation of the air bubble rising from the bottom of the lake to the surface. We will use the principles of the ideal gas law and the relationship between pressure, volume, and temperature, given that the temperature remains constant.

### Step-by-Step Solution:

1. **Identify the Initial and Final Conditions**:
- Initial radius of the bubble at the bottom: \( r \)
- Final radius of the bubble at the surface: \( \frac{5r}{4} \)

2. **Calculate the Volume of the Bubble**:
- Volume at the bottom (\( V_1 \)):
\[ V_1 = \frac{4}{3} \pi r^3 \]
- Volume at the surface (\( V_2 \)):
\[ V_2 = \frac{4}{3} \pi \left(\frac{5r}{4}\right)^3 = \frac{4}{3} \pi \cdot \frac{125r^3}{64} = \frac{125}{64} \cdot \frac{4}{3} \pi r^3 \]

3. **Set Up the Pressure-Volume Relationship**:
- According to Boyle's Law (since temperature is constant):
\[ P_1 V_1 = P_2 V_2 \]
- Where:
- \( P_1 \) is the pressure at the bottom of the lake.
- \( P_2 \) is the pressure at the surface (atmospheric pressure).

4. **Define the Pressures**:
- Atmospheric pressure at the surface (\( P_2 \)):
\[ P_2 = 10 \text{ m of water} \]
- Pressure at the bottom (\( P_1 \)):
\[ P_1 = P_2 + \text{pressure due to water column} = 10 + h \text{ (in meters)} \]

5. **Substitute into the Pressure-Volume Equation**:
\[ (10 + h) \cdot \left(\frac{4}{3} \pi r^3\right) = 10 \cdot \left(\frac{125}{64} \cdot \frac{4}{3} \pi r^3\right) \]
- Cancel \( \frac{4}{3} \pi r^3 \) from both sides:
\[ 10 + h = 10 \cdot \frac{125}{64} \]

6. **Solve for \( h \)**:
\[ 10 + h = \frac{1250}{64} \]
\[ h = \frac{1250}{64} - 10 \]
- Convert 10 to have a common denominator:
\[ 10 = \frac{640}{64} \]
\[ h = \frac{1250 - 640}{64} = \frac{610}{64} = 9.53125 \text{ m} \]

7. **Final Answer**:
The depth of the lake is approximately:
\[ h \approx 9.53 \text{ m} \]