A bubble of gas released at the bottom of a lake increases to eight times its original volume when it reaches the surface. Assuming that atmospheric pressure is equivalent to the pressure exerted by a column of water 10 m height, the depth of the lake is

Options

A. 80 m
B. 90 m
C. 40 m
D. 70 m

Explanation

To solve the problem, we will use Boyle's Law, which states that for a given mass of gas at constant temperature, the product of pressure and volume is constant (P1V1 = P2V2). We will also consider the pressure exerted by the column of water and atmospheric pressure.

### Step-by-step Solution:

1. **Identify Initial and Final Conditions**:
- Let the initial volume of the bubble at the bottom of the lake be V1.
- When the bubble reaches the surface, its volume increases to V2 = 8V1.

2. **Define Pressures**:
- At the bottom of the lake (initial state), the pressure (P1) is the sum of atmospheric pressure (P0) and the pressure due to the water column.
- The pressure due to the water column can be expressed as:
\[ \text{Pressure due to water} = h \cdot \rho \cdot g \]
- Therefore, the total pressure at the bottom is:
\[ P1 = P0 + h \cdot \rho \cdot g \]

3. **Pressure at the Surface**:
- At the surface of the lake (final state), the pressure (P2) is simply the atmospheric pressure:
\[ P2 = P0 \]

4. **Apply Boyle's Law**:
- According to Boyle's Law:
\[ P1V1 = P2V2 \]
- Substituting the values of P1, P2, V1, and V2:
\[ (P0 + h \cdot \rho \cdot g)V1 = P0 \cdot (8V1) \]

5. **Cancel V1**:
- Since V1 is common on both sides, we can cancel it out:
\[ P0 + h \cdot \rho \cdot g = 8P0 \]

6. **Rearrange the Equation**:
- Rearranging gives:
\[ h \cdot \rho \cdot g = 8P0 - P0 \]
- This simplifies to:
\[ h \cdot \rho \cdot g = 7P0 \]

7. **Substitute for P0**:
- We know that the atmospheric pressure (P0) is equivalent to the pressure exerted by a column of water 10 m high:
\[ P0 = 10 \cdot \rho \cdot g \]

8. **Substituting P0 into the Equation**:
- Substitute P0 into the equation:
\[ h \cdot \rho \cdot g = 7 \cdot (10 \cdot \rho \cdot g) \]
- This simplifies to:
\[ h \cdot \rho \cdot g = 70 \cdot \rho \cdot g \]

9. **Cancel ρ and g**:
- Since ρ and g are common on both sides, we can cancel them:
\[ h = 70 \text{ m} \]

10. **Conclusion**:
- The depth of the lake is therefore:
\[ \text{Depth of the lake} = 70 \text{ m} \]