Question

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 $=1030kg/m^3$ and The Bulk Modulus Of Water $=2\times {10}^{9}n{m}^{-2}$.

Answer 1

Step 1: Given

Depth = $400m$

The Density Of Water At The Surface $=1030kg/m^3$

Bulk Modulus Of Water $=2\times {10}^{9}n{m}^{-2}$.

Step 2: Bulk modulus

We know bulk modulus $k=$

$\frac{vdp}{-dv}=\frac{\rho dp}{d\rho }$

Here:

$\rho =$density of the fluid

$dp=$change in pressure

$d\rho =$change in density

Step 3: Change in density

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

Pressure at a height of 400 m below the surface of water $P_2=\rho gh$

Here

$g=$acceleration due to gravity

$h=$height $=400m$

By substituting the above values, we get,

$$\begin{gathered} P_2=1030\times 9.81\times 400 \\ =4041720Pa \\ \therefore dp=4041720-101325 \\ =3940395 \\ =39.40395\times {10}^{5}Pa \end{gathered}$$

Now,

$$\begin{gathered} d\rho =\frac{\rho dp}{k} \\ =\frac{1030\left(39.40\times {10}^5\right)}{2\times {10}^9} \\ =\frac{40582\times {10}^5}{2\times {10}^9} \\ =20291\times {10}^{5-9} \\ =2.09\times {10}^{9-9} \\ =2.09kg/m^3 \end{gathered}$$

$\therefore d\rho =2.029kg/m^3$