Question

Calculate the apprroximate change in density of water in a lake at a depth of 400 m below the surface. The density of water at the surface id 1030 kg//m^(3) and bulk modulus of water is 2xx10^(9) N//m^(2).

Answer 1

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Explanation:

Answer 2

The approximate change in density of water is :

Given : Depth = 400m, density of water at the surface = 1030 kg/m^3, bulk modulus of water = 2x10^9 N/m^2

• Pressure from surface at a depth x m below is given as,

P =ρgx

• Also,

Bulk strain = P/B = ρgx/B

ΔV/V =ρgx/B

ΔV = (ρgx/B)×V

• Density at xm below the surface is given as,

ρ'= ρ×V/V'

where V' = compressed volume of V and V' = V-ΔV

ρ' = ρ×V/(V-ρgxV/B)

= ρB/(B-ρgx)

=1030×2x10⁹ / (2x10⁹ - 1030×10×400) =2.06x10¹² / 1.9958x10⁹

=1032 kg/m³

• The change in density, dρ =ρ'- ρ =1032-1030

• dρ = 2 kg/m³