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(D) Consider a volume $V_1$ of liquid at the top surface. Due to pressure at depth $H$, the same mass of liquid occupies a volume $V_2$.From the conse

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$\frac{-1}{B_0}$

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(D) Consider a volume $V_1$ of liquid at the top surface. Due to pressure at depth $H$, the same mass of liquid occupies a volume $V_2$.From the conservation of mass, we have $\rho_0 V_1 = \rho V_2$, which implies $\frac{V_1}{V_2} = \frac{\rho}{\rho_0}$.At depth $H$, the gauge pressure is $P = \rho g H$.From the definition of bulk modulus $B_0$, we have $B_0 = -\frac{\Delta P}{\Delta V / V_1} = -\frac{P}{(V_2 - V_1) / V_1}$.Rearranging this, we get $\frac{V_2 - V_1}{V_1} = -\frac{P}{B_0} = -\frac{\rho g H}{B_0}$.Thus, $\frac{V_2}{V_1} = 1 - \frac{\rho g H}{B_0}$.Taking the reciprocal, $\frac{V_1}{V_2} = \frac{1}{1 - \frac{\rho g H}{B_0}}$.Substituting this into the mass conservation equation $\rho = \rho_0 \left( \frac{V_1}{V_2} \right)$, we get $\rho = \frac{\rho_0}{1 - \frac{\rho g H}{B_0}}$.Comparing this with the given expression $\rho = \frac{\rho_0}{1 + \alpha \rho g H}$, we identify $\alpha = -\frac{1}{B_0}$.

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