# Estimate The Change In The Density Of Water In The Ocean At A Depth Of 400m Below The Surface. The Density Of Water At The Surface =1030 Kg M−3and The Bulk Modulus Of Water = 2 × 109n M−2.

Sol:

We know bulk modulus k = $\frac{V dp}{-dv} = \frac{\rho dp}{d\rho}$

Here:

$\rho$ = density of the fluid

dp = change in pressure

$d\rho$= change in density

Given
$\rho_{water = 1030 kg/m^{3}}$

Now:

Pressure at surface of the ocean P1 = 1 atm = 101325 Pa

Pressure at a height of 400 m below the surface of water P2=$\rho gh$

Here

g = acceleration due to gravity

h = height = 400 m

By substituting the above values, we get,

P2 = 1030 * 9.81 * 400 = 4041720 Pa.

Therefore, dp = 4041720 – 101325 = 39.40395 * $10^{5}pa$

Now, $d\rho = \frac{\rho dp}{k}$

=$\frac{1030(39.40 * 10^{5})}{2 * 10^{9}}$

=$2.029 kg/m^{3}$

Therefore,$\;d\rho = 2.029 kg/m^{3}$

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