Question

The density $\rho$ of water of bulk modulus$B$ at a depth $y$ in the ocean is related to the density at surface ${\rho }_0$ by the relation:

• A. $\rho ={\rho }_0(1-\frac{{\rho }_{0}gy}{B})$
• B. $\rho ={\rho }_0(1+\frac{{\rho }_{0}gy}{B})$
• C. $\rho ={\rho }_0(1+\frac{B}{{\rho }_{0}gy})$
• D. $\rho ={\rho }_0(1-\frac{B}{{\rho }_{0}gy})$

Answer 1

The correct option is B $\rho ={\rho }_0(1+\frac{{\rho }_{0}gy}{B})$

The Bulk modulus is given as, $B=\rho dP/d\rho -\left(1\right)$

As fluid is compressible, the density changes.

Thus we cannot write $dP={\rho }_{0}gy$

But we have to use, $dP=\rho gdy$

Using this relation in the equation $\left(1\right)$ we get,

$\frac{d\rho }{{\rho }^2}=gdy/B$

Integrating above,

${\int }_{{\rho }_o}^{\rho }\frac{d\rho }{{\rho }^2}={\int }_0^{y}gdy/B$

$\Rightarrow \frac{1}{{\rho }_o}-\frac{1}{\rho }=gy/B$

$\Rightarrow \rho ={\rho }_o/(1-{\rho }_{o}gy/B)$

As, ${\rho }_{o}gy/B<<1$,

$\rho ={\rho }_o(1+{\rho }_{o}gy/B)$