Question

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

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

Answer 1

The correct option is A $\frac{\rho B}{B-(n-1)P_0}$

Increase in pressure $\left(\Delta P\right)=n{P}_0-P_0=(n-1)P_0\dots \left(1\right)$
Let $V$ be the volume of a certain mass $M$ of water at the surface.
Then, $M=\rho V\dots \left(2\right)$
Decrease in volume under pressure $\Delta P$ is
$\Delta V=\frac{V\Delta P}{B}\dots \left(3\right)$
Volume of the same mass $M$ of water at given depth is
$V^{{}^{\prime }}=V-\Delta V=V-\frac{V\Delta P}{B}$
$=V(1-\frac{\Delta P}{B})=\frac{V}{B}(B-\Delta P)$
[using (3)]
Density of water at that depth is
${\rho }^{{}^{\prime }}=\frac{M}{V^{{}^{\prime }}}=\frac{\rho V}{V^{{}^{\prime }}}=\frac{\rho V}{\frac{V}{B}(B-\Delta P)}$ [using (2)]
$\Rightarrow {\rho }^{{}^{\prime }}=\frac{\rho B}{B-\Delta P}=\frac{\rho B}{B-(n-1)P_0}$ [using (1)]