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 solve the problem of finding the increase in density of water at a depth of 1 km, we will use the relationship between bulk modulus, pressure change, and density change. Here’s a step-by-step solution:

### Step 1: Write down the given data
- Bulk modulus of water, \( B = 2.3 \times 10^9 \, \text{N/m}^2 \)
- Average density of water, \( \rho = 10^3 \, \text{kg/m}^3 \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)
- Depth, \( h = 1 \, \text{km} = 1000 \, \text{m} \)

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