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: Understand the Problem
We are given:
- 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, \( x = 400 \, \text{m} \)
- Acceleration due to gravity, \( g \approx 10 \, \text{m/s}^2 \)

### Step 2: Calculate the Pressure at Depth
The pressure at a depth \( x \) in a fluid is given by:
\[ P = \rho g x \]
Substituting the values:
\[ P = 1030 \, \text{kg/m}^3 \times 10 \, \text{m/s}^2 \times 400 \, \text{m} = 4,120,000 \, \text{N/m}^2 \]

### Step 3: Calculate the Bulk Strain
The bulk strain \( \epsilon \) is given by:
\[ \epsilon = \frac{P}{B} \]
Substituting the pressure calculated in Step 2:
\[ \epsilon = \frac{4,120,000 \, \text{N/m}^2}{2 \times 10^9 \, \text{N/m}^2} = 0.00206 \]

### Step 4: Relate Bulk Strain to Change in Volume
The change in volume \( \Delta V \) can be expressed in terms of the original volume \( V \):
\[ \Delta V = \epsilon V \]

### Step 5: Calculate the Change in Density
Since density is inversely proportional to volume, the new density \( \rho' \) at depth \( x \) can be calculated as:
\[ \rho' = \frac{\rho V}{V - \Delta V} \]
Substituting \( \Delta V = \epsilon V \):
\[ \rho' = \frac{\rho V}{V - \epsilon V} = \frac{\rho}{1 - \epsilon} \]
Substituting \( \epsilon \):
\[ \rho' = \frac{1030 \, \text{kg/m}^3}{1 - 0.00206} \approx \frac{1030 \, \text{kg/m}^3}{0.99794} \approx 1032 \, \text{kg/m}^3 \]

### Step 6: Calculate the Change in Density
The change in density \( \Delta \rho \) is:
\[ \Delta \rho = \rho' - \rho = 1032 \, \text{kg/m}^3 - 1030 \, \text{kg/m}^3 = 2 \, \text{kg/m}^3 \]

### Final Result
The approximate change in density of water at a depth of 400 m below the surface is:
\[ \Delta \rho = 2 \, \text{kg/m}^3 \]

---