Q. 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) $\rho = \rho_{0}\left\lbrack 1 - \frac{\rho_{0}gy}{B} \right\rbrack$
B) $\rho = \rho_{0}\left\lbrack 1 + \frac{\rho_{0}gy}{B} \right\rbrack$
C) $\rho = \rho_{0}\left\lbrack 1 + \frac{B}{\rho_{0}hgy} \right\rbrack$
D) $\rho = \rho_{0}\left\lbrack 1 - \frac{B}{\rho_{0}gy} \right\rbrack$
Explanation
(b) Bulk modulus, $B = - V_{0}\frac{\mathrm{\Delta}p}{\mathrm{\Delta}V} \Rightarrow \mathrm{\Delta}V = - V_{0}\frac{\mathrm{\Delta}p}{B}$
$\Rightarrow V = V_{0}\left\lbrack 1 - \frac{\mathrm{\Delta}p}{B} \right\rbrack$
$\therefore$ Density, $\rho = \rho_{0}\left\lbrack 1 - \frac{\mathrm{\Delta}p}{B} \right\rbrack^{- 1} = \rho_{0}\left\lbrack 1 + \frac{\mathrm{\Delta}p}{B} \right\rbrack$
Where, $\mathrm{\Delta}p = p - p_{0} = h\rho_{0}g$
= pressure difference between depth and surface of ocean
$\therefore\rho = \rho_{0}\left\lbrack 1 + \frac{\rho_{0}gy}{B} \right\rbrack$ (As $h = y$)
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.