Question

The density of water at the surface of the ocean is $\rho$ and atmospheric pressure is $P_0$. If the bulk modulus of water is $K$, what is the density of ocean water at a depth where the pressure is then$n{P}_0$?

• A. $\frac{\rho K}{K-n{P}_0}$
• B. $\frac{\rho K}{K+n{P}_0}$
• C. $\frac{\rho K}{K-(n-1)P_0}$
• D. $\frac{\rho K}{K+(n-1)P_0}$.

Answer 1

Answer: (C)

$\frac{\rho K}{K-(n-1)P_0}$

Let the volume of certain mass $\left(M\right)$ of water at surface be $V$

Density of water, $\rho =\frac{M}{V}$

Pressure at a depth, $P^{\prime }=n{P}_o$

Thus, change in pressure, $\Delta P=n{P}_o-P_o=(n-1)P_o$

Given Bulk Modulus $=K=\frac{\Delta P}{\frac{\Delta V}{V}}⟹\Delta V=\frac{(n-1)P_{o}V}{K}$

As at a depth, the pressure increases. Thus the volume of water will decrease at that depth.

$\therefore$ Volume of water of mass $\left(M\right)$ at that depth, $V^{\prime }=V-\Delta V$

$\therefore V^{\prime }=V-\frac{(n-1)P_{o}V}{K}=(K-(n-1\left)P_o\right)\frac{V}{K}$

$\therefore$ Density at that depth, ${\rho }^{\prime }=\frac{M}{V^{\prime }}=\rho \frac{V}{V^{\prime }}$

$\therefore {\rho }^{\prime }=\frac{\rho K}{K-(n-1)P_o}$