When a large bubble rises from the bottom of a lake to the surface, its radius doubles, The atmospheric pressure is equal to that of a column of water of height H. The depth of the lake is

Options

A. H
B. 2 H
C. 7 H
D. 8 H

Explanation

To solve the problem of a large bubble rising from the bottom of a lake to the surface, where its radius doubles, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Initial and Final Conditions:**
- Let the initial radius of the bubble be \( r \).
- When the bubble rises to the surface, its radius doubles to \( 2r \).
- The volume of the bubble can be expressed as:
\[ V_1 = \frac{4}{3} \pi r^3 \]
\[ V_2 = \frac{4}{3} \pi (2r)^3 = \frac{4}{3} \pi (8r^3) = \frac{32}{3} \pi r^3 \]

2. **Determine Initial and Final Pressures:**
- The initial pressure \( P_1 \) at the depth \( x \) is given by:
\[ P_1 = \text{Atmospheric Pressure} + \text{Pressure due to water column} \]
\[ P_1 = \rho g h + \rho g x \]
- The final pressure \( P_2 \) at the surface of the lake is simply the atmospheric pressure:
\[ P_2 = \rho g h \]

3. **Apply the Isothermal Condition:**
- According to the isothermal process, we have:
\[ P_1 V_1 = P_2 V_2 \]
- Substituting the expressions for pressure and volume:
\[ (\rho g h + \rho g x) \left(\frac{4}{3} \pi r^3\right) = (\rho g h) \left(\frac{32}{3} \pi r^3\right) \]

4. **Simplify the Equation:**
- Cancel out the common terms \( \frac{4}{3} \pi r^3 \) from both sides:
\[ \rho g h + \rho g x = 8 \rho g h \]
- Rearranging gives:
\[ \rho g x = 8 \rho g h - \rho g h \]
\[ \rho g x = 7 \rho g h \]

5. **Solve for Depth \( x \):**
- Dividing both sides by \( \rho g \):
\[ x = 7h \]

6. **Conclusion:**
- The depth of the lake is \( x = 7h \).

### Final Answer:
The depth of the lake is \( 7h \) (Option C).