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(B) The bulk modulus $B$ is defined as $B = -V_0 \frac{\Delta p}{\Delta V}$,which implies $\Delta V = -V_0 \frac{\Delta p}{B}$.Since $V = V_0 + \Delta

Vedclass · Accepted Answer

$\rho = \rho_0 \left[ 1 + \frac{\rho_0 gy}{B} \right]$

Vedclass · Answer

(B) The bulk modulus $B$ is defined as $B = -V_0 \frac{\Delta p}{\Delta V}$,which implies $\Delta V = -V_0 \frac{\Delta p}{B}$.Since $V = V_0 + \Delta V$,we have $V = V_0 \left[ 1 - \frac{\Delta p}{B} \right]$.Density $\rho = \frac{m}{V} = \frac{m}{V_0 [1 - \frac{\Delta p}{B}]} = \rho_0 [1 - \frac{\Delta p}{B}]^{-1}$.Using the binomial approximation $(1 - x)^{-1} \approx 1 + x$ for small $x$,we get $\rho = \rho_0 [1 + \frac{\Delta p}{B}]$.At depth $y$,the pressure difference is $\Delta p = \rho_0 gy$.Substituting this,we get $\rho = \rho_0 [1 + \frac{\rho_0 gy}{B}]$.

The density $\rho$ of water of bulk modulus $B$ at a depth $y$ in the ocean is related to the density at the surface $\rho_0$ by the relation:

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