Calculate the apprroximate change in density of water in a lake at a depth of `400 m ` below the surface. The density of water at the surface id `1030 kg//m^(3)` and bulk modulus of water is `2xx10^(9) N//m^(2)`.

Explanation

To calculate the approximate change in density of water in a lake at a depth of 400 m below the surface, we will follow these steps:

### Step 1: Identify the given values
- Density of water at the surface, \( \rho = 1030 \, \text{kg/m}^3 \)
- Bulk modulus of water, \( B = 2 \times 10^9 \, \text{N/m}^2 \)
- Depth, \( h = 400 \, \text{m} \)
- Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \)

### Step 2: Calculate the change in pressure (\( \Delta P \))
The change in pressure at a depth \( h \) is given by the formula:
\[ \Delta P = \rho g h \]
Substituting the values:
\[ \Delta P = 1030 \, \text{kg/m}^3 \times 9.8 \, \text{m/s}^2 \times 400 \, \text{m} \]
Calculating this:
\[ \Delta P = 1030 \times 9.8 \times 400 = 4,024,000 \, \text{N/m}^2 \]

### Step 3: Use the relationship between change in density and change in pressure
The change in density (\( \Delta \rho \)) can be calculated using the formula:
\[ \Delta \rho = \frac{\rho \Delta P}{B} \]
Substituting the values:
\[ \Delta \rho = \frac{1030 \, \text{kg/m}^3 \times 4,024,000 \, \text{N/m}^2}{2 \times 10^9 \, \text{N/m}^2} \]

### Step 4: Calculate \( \Delta \rho \)
Calculating the numerator:
\[ 1030 \times 4,024,000 = 4,139,920,000 \, \text{kg m}^{-2} \text{s}^{-2} \]
Now substituting into the equation:
\[ \Delta \rho = \frac{4,139,920,000}{2 \times 10^9} = 2.06996 \, \text{kg/m}^3 \]
Rounding off, we find:
\[ \Delta \rho \approx 2 \, \text{kg/m}^3 \]

### Step 5: Conclusion
The approximate change in density of water in the lake at a depth of 400 m is \( \Delta \rho \approx 2 \, \text{kg/m}^3 \).

---