If the normal density of sea water is `1.00 g//cm^(3)`, what will be its density at a depth of 4km? Given compressibility of water = 0.00005 per atmosphere. 1 atmospheric pressure `=10^(6) dyn e//cm^(2), g=980 cm//s^(2)`.

Explanation

To find the density of seawater at a depth of 4 km, we can follow these steps:

### Step 1: Gather the given data
- Normal density of seawater, \( \rho = 1.00 \, \text{g/cm}^3 \)
- Compressibility of water, \( C = 0.00005 \, \text{per atmosphere} \)
- Depth, \( h = 4 \, \text{km} = 4 \times 10^5 \, \text{cm} \)
- Atmospheric pressure, \( P_0 = 10^6 \, \text{dyn/cm}^2 \)
- Acceleration due to gravity, \( g = 980 \, \text{cm/s}^2 \)
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