Question

Estimate the change in the density of water in ocean at a depth of 400 m below the surface. The density of water at the surface = 1030 kg m⁻³ and the bulk modulus of water = 2 × 10⁹ N m⁻².

Answer 1

Given:
$\mathrm{Bulk}\mathrm{modulus}\mathrm{of}waterB=2\times {10}^9\mathrm{N}/{\mathrm{m}}^2$
Depth (d) = 400 m
Density of water at the surface (ρ₀) = 1030 kg/m³
We know that:
$$\begin{gathered} \mathrm{Density}\mathrm{at}\mathrm{surface}{\rho }_0=\frac{m}{{\mathrm{V}}_0} \\ \mathrm{Density}\mathrm{at}\mathrm{depth}{\rho }_{\mathrm{d}}=\frac{m}{{\mathrm{V}}_{\mathrm{d}}} \\ \Rightarrow \frac{{\rho }_{\mathrm{d}}}{{\rho }_0}=\frac{V_0}{V_d}...\left(\mathrm{i}\right) \end{gathered}$$

Here: ρ_d = density of water at a depth
m = mass
V₀ = volume at the surface
V_d = volume at a depth

$$\begin{gathered} \mathrm{Pressure}\mathrm{at}\mathrm{a}\mathrm{depth}d={\mathrm{\rho }}_{0}gd \\ \mathrm{Acceleration}\mathrm{due}\mathrm{to}\mathrm{gravity}g=10{\mathrm{ms}}^2 \\ \mathrm{Volume}\mathrm{strain}=\frac{V_0-V_d}{V_0} \\ B=\frac{\mathrm{Pressure}}{\mathrm{Volume}\mathrm{strain}} \\ \Rightarrow B=\frac{{\mathrm{\rho }}_{0}gd}{\left(\frac{V_0-V_d}{V_0}\right)} \\ \Rightarrow 1-\frac{V_d}{V_0}=\frac{{\mathrm{\rho }}_{0}gd}{B} \\ \Rightarrow \frac{V_d}{V_0}=\left(1-\frac{p_{0}gd}{B}\right)...\left(\mathrm{ii}\right) \end{gathered}$$
Using equations (i) and (ii), we get:

$$\begin{gathered} \frac{{\rho }_{\mathrm{d}}}{{\rho }_0}=\frac{1}{\left(1-{\displaystyle \frac{{\rho }_{0}gd}{B}}\right)} \\ \Rightarrow {\rho }_d=\frac{1}{\left(1-{\displaystyle \frac{{\rho }_{0}gh}{B}}\right)}{\rho }_0 \\ \Rightarrow {\rho }_d=\frac{1030}{\left(1-{\displaystyle \frac{1030\times 10\times 400}{2\times {10}^9}}\right)}\approx 1032\mathrm{kg}/{\mathrm{m}}^3 \\ \mathrm{Change}\mathrm{in}\mathrm{density}={\rho }_{\mathrm{d}}-{\rho }_0 \\ =1032-1030=2\mathrm{kg}/{\mathrm{m}}^3 \end{gathered}$$

Hence, the required density at a depth of 400 m below the surface is 2 kg/m³.​