Chapter 10, Problem 21P

To determine

To Find: The fraction by which the density of water differs from the density at the surface.

Explanation of Solution

Given information:

Depth of the water $=6.0km=6000m$

Density of the water = ${\rho }_{water}=1000kg/m^3=10^{3}kg/c{m}^3$

Bulk modulus = $2.2\times {10}^9\mathrm{N}/m^2$

Formula Used:

• The relationship between the change in volume and original volume is related as,

  $\frac{\Delta V}{V_0}=-\left(\frac{\Delta P}{B}\right)\mathrm{....}\left(\left(1\right)\right)$

Where, $\Delta V$ = change in volume

  $V_0$ = original volume

  $\Delta P$ = change in pressure and bulk modulus of elasticity.

• The change in pressure is also calculated as,

  $\Delta P={\rho }_{0}gh\mathrm{....}\left(2\right)$

Where, ${\rho }_0$ is density of sea water, $g$ is the acceleration due to gravity and $h$ is the depth.

Calculation:

Now, substituting the values in the formula which is written as above

  $\begin{array}{l}B=\frac{{\rho }_{0}gh}{\Delta \rho /\rho }\\ \frac{\Delta \rho }{\rho }=\frac{{\rho }_{0}gh}{B}\\ \frac{\Delta \rho }{\rho }=\frac{{10}^3\times 9.8\times 6.0}{2.2\times {10}^9}\\ \frac{\Delta \rho }{\rho }=26.7272\times {10}^{-6}\end{array}$

Conclusion:

It differs as per $\frac{\Delta \rho }{\rho }=26.7272\times {10}^{-6}$ from the density at the surface.