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\text{Hint: Bulk modulus } = -\frac{\Delta P}{\frac{\Delta V}{V}} \text{Step 1: Calculate the change in volume at the depth for the same mass.} \text{Let the given depth be h.} \text{Pressure at the given depth, } P = 80.0 \text{ atm} = 80 \times 1.01 \times 10^5 \text{ Pa} \text{The density of water at the surface, } \rho_1 = 1.03 \times 10^3 \text{ kg m}^{-3} \text{Let } \rho_2 \text{ be the density of water at the depth.} \text{Let } V_1 \text{ be the volume of water of mass m at the surface.} \text{Let } V_2 \text{ be the volume of water of mass m at the depth h.} \text{Let } \Delta V \text{ be the change in volume.} \Delta V = V_1 - V_2 = m \left[\left(\frac{1}{\rho_1}\right) - \left(\frac{1}{\rho_2}\right)\right] \text{Step 2: Calculate the Compressibility of water } = \left(\frac{1}{B}\right) \text{As} \text{Volumetric strain } = \frac{\Delta V}{V_1} = m \left[\left(\frac{1}{\rho_1}\right) - \left(\frac{1}{\rho_2}\right)\right] \times \left(\frac{\rho_1}{m}\right) \frac{\Delta V}{V_1} = 1 - \left(\frac{\rho_1}{\rho_2}\right) \text{ .....(i)} \text{Bulk modulus, } B = \frac{P V_1}{\Delta V} \Rightarrow \frac{\Delta V}{V_1} = \frac{P}{B} \text{Compressibility of water } = \left(\frac{1}{B}\right) = 45.8 \times 10^{-11} \text{ Pa}^{-1} \text{Step 3: Calculate the density of water at depth h.} \frac{\Delta V}{V_1} = 80 \times 1.013 \times 10^5 \times 45.8 \times 10^{-11} = 3.71 \times 10^{-3} \text{ .....(ii)} \text{From equations (i) and (ii), we get:} 1 - \left(\frac{\rho_1}{\rho_2}\right) = \frac{\Delta V}{V_1} = 3.71 \times 10^{-3} \Rightarrow \rho_2 = \frac{1.03 \times 10^3}{\left[1 - \left(3.71 \times 10^{-3}\right)\right]} = 1.034 \times 10^3 \text{ kg m}^{-3} \text{Therefore the density of water at depth (h) is } 1.034 \times 10^3 \text{ kg m}^{-3}