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 :

\[\text{ Bulk modulus of water B }= 2 \times {10}^9 \text{ N/ m }^2 \]

Depth (d) = 400 m
Density of water at the surface (ρ₀) = 1030 kg/m³
We know that:

\[\text{ Density at surface } \rho_0 = \frac{\text{m}}{V_0}\]
\[\text{ Density at depth }\rho_d = \frac{\text{m}}{V_d}\]
\[ \Rightarrow \frac{\rho_d}{\rho_0} = \frac{V_0}{V_d} . . . \left( \text{i} \right)\]

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

\[\text{ Pressure at a depth d } = \rho_0 \text{ gd}\]
\[ \text{ Acceleration due to gravity g }= 10 {\text{ ms}}^2 \]
\[\text{ Volume strain } = \frac{V_0 - V_d}{V_0}\]
\[B = \frac{\text{ Pressure }}{\text {Volume strain}}\]
\[ \Rightarrow B = \frac{\rho_0 \text{ gd}}{\left( \frac{V_0 - V_d}{V_0} \right)}\]
\[ \Rightarrow 1 - \frac{V_d}{V_0} = \frac{\rho_0 \text{ gd}}{B}\]
\[ \Rightarrow \frac{V_d}{V_0} = \left( 1 - \frac{p_0 \text{ gd}}{B} \right) . . . \left(\text{ ii} \right)\]

Using equations (i) and (ii), we get:

\[\frac{\rho_d}{\rho_0} = \frac{1}{\left( 1 - \frac{\rho_0 \text{ gd }}{B} \right)}\]
\[ \Rightarrow \rho_d = \frac{1}{\left( 1 - \frac{\rho_0 \text{ gh}}{B} \right)} \rho_0 \]
\[ \Rightarrow \rho_d = \frac{1030}{\left( 1 - \frac{1030 \times 10 \times 400}{2 \times {10}^9} \right)} \approx 1032 \text{ kg/ m}^3 \]
\[\text{ Change in density }= \rho_d - \rho_0 \]
\[ = 1032 - 1030 = 2 \text{ kg/ m}^3\]

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