An air bubble starts rising from the bottom of a lake and its radius is doubled on reaching the surface. If the temperature is costant the depth of the lake is. (`1` atmospheric pressure `= 10m` height of water column)

Options

A. `7m`
B. `70m`
C. `10m`
D. `0.7m`

Explanation

To solve the problem of finding the depth of the lake where an air bubble rises and its radius doubles, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Initial and Final Conditions**:
- Let the initial radius of the bubble at the bottom of the lake be \( r_1 = r \).
- When the bubble reaches the surface, its radius becomes \( r_2 = 2r \).

2. **Determine the Volumes**:
- The volume of the bubble at the bottom (initial volume) is given by:
\[ V_1 = \frac{4}{3} \pi r_1^3 = \frac{4}{3} \pi r^3 \]
- The volume of the bubble at the surface (final volume) is:
\[ V_2 = \frac{4}{3} \pi r_2^3 = \frac{4}{3} \pi (2r)^3 = \frac{4}{3} \pi (8r^3) = 8 \times \frac{4}{3} \pi r^3 = 8V_1 \]

3. **Apply the Ideal Gas Law for Isothermal Process**:
- Since the temperature is constant, we can use the relation \( P_1 V_1 = P_2 V_2 \).
- Let \( P_1 \) be the pressure at the bottom of the lake and \( P_2 \) be the pressure at the surface.
- The pressure at the surface \( P_2 \) is equal to atmospheric pressure, which is \( 10 \, \text{m} \) of water column (or \( 1 \, \text{atm} \)).
- The pressure at the bottom \( P_1 \) is the sum of atmospheric pressure and the pressure due to the height of the water column \( h \):
\[ P_1 = 10 + h \]

4. **Set Up the Equation**:
- Substitute \( P_1 \), \( P_2 \), \( V_1 \), and \( V_2 \) into the equation:
\[ (10 + h) V_1 = 10 \times 8V_1 \]
- Since \( V_1 \) is common on both sides, we can cancel it out:
\[ 10 + h = 80 \]

5. **Solve for \( h \)**:
- Rearranging the equation gives:
\[ h = 80 - 10 = 70 \]

6. **Conclusion**:
- The depth of the lake is \( h = 70 \, \text{m} \).

### Final Answer:
The depth of the lake is **70 m**.