Treat water as slightly compressible with surface density 1000 kg/m³ and bulk modulus 2 GPa. At a depth of 2000 m, take g = 10 m/s² and use the first-order approximation in which the pressure increase is calculated using the surface density. If the density at this depth is N kg/m³, find N.
Physics · Properties of solid and fluid · properties of solid stress ,strain and modulus of elasticity
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Question

Treat water as slightly compressible with surface density 1000 kg/m³ and bulk modulus 2 GPa. At a depth of 2000 m, take g = 10 m/s² and use the first-order approximation in which the pressure increase is calculated using the surface density. If the density at this depth is N kg/m³, find N.

Physics · Properties of solid and fluid · properties of solid stress ,strain and modulus of elasticity

G1 Physics · Accepted Answer

Using the surface density for the first-order pressure estimate, ΔP = ρgh. Therefore, ΔP = 1000×10×2000 = 2×10⁷ Pa = 20 MPa. For small compression, Δρ/ρ ≈ ΔP/K. Thus, Δρ/ρ = 20 MPa / 2000 MPa = 0.01. Hence, ρ' = ρ(1+0.01) = 1000×1.01 = 1010 kg/m³. Therefore, N = 1010.