Bulk modulus of water is `(2.3xx10^(9) N//m^(2))`.Taking average density of water` rho=10^(3)kg//m^(3)`,find increases in density at a depth of `1 km`. Take `g=10m//s^(2)`

Explanation

To find the increase in density of water at a depth of 1 km, we can follow these steps:

### Step 1: Calculate the change in pressure at a depth of 1 km
The change in pressure (\( \Delta P \)) at a depth \( h \) in a fluid can be calculated using the formula:
\[ \Delta P = \rho g h \]
Where:
- \( \rho \) = density of water = \( 10^3 \, \text{kg/m}^3 \)
- \( g \) = acceleration due to gravity = \( 10 \, \text{m/s}^2 \)
- \( h \) = depth = \( 1 \, \text{km} = 10^3 \, \text{m} \)

Substituting the values:
\[ \Delta P = (10^3 \, \text{kg/m}^3)(10 \, \text{m/s}^2)(10^3 \, \text{m}) = 10^7 \, \text{Pa} \]

### Step 2: Relate change in pressure to change in density using bulk modulus
The bulk modulus \( K \) is defined as:
\[ K = -\frac{\Delta P}{\frac{\Delta V}{V}} \]
Where \( \Delta V \) is the change in volume and \( V \) is the original volume. We can also express this in terms of density:
\[ K = \frac{\Delta P}{\frac{\Delta \rho}{\rho}} \]
Rearranging gives:
\[ \Delta \rho = \frac{\Delta P \cdot \rho}{K} \]

### Step 3: Substitute the known values into the equation
Given:
- \( K = 2.3 \times 10^9 \, \text{N/m}^2 \)
- \( \Delta P = 10^7 \, \text{Pa} \)
- \( \rho = 10^3 \, \text{kg/m}^3 \)

Substituting these values:
\[ \Delta \rho = \frac{(10^7 \, \text{Pa}) \cdot (10^3 \, \text{kg/m}^3)}{2.3 \times 10^9 \, \text{N/m}^2} \]

### Step 4: Calculate the change in density
Calculating the numerator:
\[ 10^7 \cdot 10^3 = 10^{10} \]
Now substituting into the equation:
\[ \Delta \rho = \frac{10^{10}}{2.3 \times 10^9} = \frac{10}{2.3} \approx 4.347 \, \text{kg/m}^3 \]

### Final Answer
The increase in density at a depth of 1 km is approximately:
\[ \Delta \rho \approx 4.347 \, \text{kg/m}^3 \]

---