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 = [latex]\frac{V dp}{-dv} = \frac{\rho dp}{d\rho} [/latex]

Here:

[latex]\rho [/latex] = density of the fluid

dp = change in pressure

[latex]d\rho [/latex]= change in density

Given
[latex] \rho_{water = 1030 kg/m^{3}} [/latex]

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=[latex] \rho gh [/latex]

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 * [latex]10^{5}pa [/latex]

Now, [latex]d\rho = \frac{\rho dp}{k} [/latex]

=[latex] \frac{1030(39.40 * 10^{5})}{2 * 10^{9}} [/latex]

=[latex] 2.029 kg/m^{3} [/latex]

Therefore,[latex] \;d\rho = 2.029 kg/m^{3} [/latex]

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