The depth of a lake at which density of water is 2% greater than on surface. [Compressibility of water = `50xx10^(-6)`/atm , 1 atm = `10^5N//m^2`]

Options

A. 2 km
B. 4 km
C. 8 km
D. 40 km

Explanation

To solve the problem of finding the depth of a lake at which the density of water is 2% greater than at the surface, we will use the concept of compressibility and the relationship between pressure, density, and volume.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We need to find the depth \( h \) at which the density of water increases by 2% compared to its density at the surface.

2. **Given Data**:
- Compressibility of water, \( C = 50 \times 10^{-6} \, \text{atm}^{-1} \)
- \( 1 \, \text{atm} = 10^5 \, \text{N/m}^2 \)
- Change in density, \( \Delta \rho = 0.02 \rho_0 \) (where \( \rho_0 \) is the density at the surface, approximately \( 1000 \, \text{kg/m}^3 \))

3. **Convert Compressibility**:
Convert the compressibility from atm to \( \text{N/m}^2 \):
\[ C = 50 \times 10^{-6} \, \text{atm}^{-1} \times \frac{1 \, \text{atm}}{10^5 \, \text{N/m}^2} = 50 \times 10^{-6} \times 10^{-5} \, \text{N/m}^2 = 5 \times 10^{-10} \, \text{N/m}^2 \]

4. **Using the Bulk Modulus Relation**:
The change in pressure \( \Delta P \) at depth \( h \) is given by:
\[ \Delta P = \rho g h \]
where \( \rho \) is the density, \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)), and \( h \) is the depth.

5. **Relating Change in Density to Change in Pressure**:
From the definition of compressibility:
\[ \Delta \rho = C \Delta P \]
Substituting \( \Delta P \) into this equation gives:
\[ \Delta \rho = C (\rho g h) \]

6. **Substituting the Change in Density**:
Since \( \Delta \rho = 0.02 \rho_0 \):
\[ 0.02 \rho_0 = C (\rho_0 g h) \]
We can cancel \( \rho_0 \) from both sides (assuming \( \rho_0 \neq 0 \)):
\[ 0.02 = C g h \]

7. **Solving for Depth \( h \)**:
Rearranging the equation gives:
\[ h = \frac{0.02}{C g} \]
Substituting \( C = 5 \times 10^{-10} \, \text{N/m}^2 \) and \( g = 10 \, \text{m/s}^2 \):
\[ h = \frac{0.02}{5 \times 10^{-10} \times 10} = \frac{0.02}{5 \times 10^{-9}} = \frac{2 \times 10^{-2}}{5 \times 10^{-9}} = 4 \times 10^{6} \, \text{m} \]

8. **Final Calculation**:
Converting \( 4 \times 10^{6} \, \text{m} \) to kilometers:
\[ h = 4 \times 10^{3} \, \text{km} = 4 \, \text{km} \]

### Conclusion:
The depth of the lake at which the density of water is 2% greater than at the surface is **4 kilometers**.