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

Density comes $2 kg/m^3$

Explanation:

Bulk Modulus of Water $B = 2 \times 10^9 N/m^2$

Depth (d) = 400 m

The Density of Water at the surface $(\rho_0) = 1030\ kg/m^3$

We know that -

Density at surface $\rho_0 = \frac {m}{V_0}$

Density at Depth $\rho_d = \frac {m}{V_d}$

$\frac {\rho_d}{\rho_0} = \frac {V_0}{V_d}$ ''''''(i)

Here, $\rho_d$ = density of water at a depth

m = mass

$V_0$ = Volume at a surface

$V_d$ = Volume at a depth

Pressure at a depth $d = \rho_0 gd$

Acceleration due to gravity $g = 10\ m/s^2$ (Approx)

Volume Strain = $\frac {V_0 - V_d}{V_0}$

B = $\frac {Pressure}{Volume\ strain}$

or B = $\frac {\rho_0 gd}{(V_0 - V_d)/V_0}$

$1 - \frac {V_d}{V_0} = \frac {\rho_0 gd}{B}$

$\frac {V_d}{V_0} = (1 - \frac {\rho_0 gd}{B})$ """"(ii)

using equations (i) and (ii) -  

$\frac {\rho_d}{\rho_0} = \frac {1}{1 - \frac {\rho_0 gd}{B}}$

$\rho_d = \frac {\rho_0}{1} \times ({1 - \frac {\rho_0 gd}{B})^-^1$

$\rho_d = 1032\ kg/m^3$

Change in density = $\rho_d - \rho_0$

= 1032 - 1030  

= $2 kg/m^3$

So, the required density at 400 m depth below surface is 2 kg/m^3

Answer 2

Explanation:

Bulk Modulus of Water B = 2 \times 10^9 N/m^2B=2×109N/m2

Depth (d) = 400 m

The Density of Water at the surface (\rho_0) = 1030\ kg/m^3(ρ0)=1030 kg/m3

We know that -

Density at surface \rho_0 = \frac {m}{V_0}ρ0=V0m

Density at Depth \rho_d = \frac {m}{V_d}ρd=Vdm

\frac {\rho_d}{\rho_0} = \frac {V_0}{V_d}ρ0ρd=VdV0 ''''''(i)

Here, \rho_dρd = density of water at a depth

m = mass

V_0V0 = Volume at a surface

V_dVd = Volume at a depth

Pressure at a depth d = \rho_0 gdd=ρ0gd