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प्रश्न
What is the density of water at a depth where the pressure is 80.0 atm, given that its density at the surface is 1.03 × 103 kg m–3?
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उत्तर १
Let the given depth be h.
Pressure at the given depth, p = 80.0 atm = 80 × 1.01 × 105 Pa
Density of water at the surface, ρ1 = 1.03 × 103 kg m–3
Let ρ2 be the density of water at the depth h.
Let V1 be the volume of water of mass m at the surface.
Let V2 be the volume of water of mass m at the depth h.
Let ΔV be the change in volume.
ΔV = V1 - V2
`=m(1/rho_1 - 1/rho_2)`
`:."Volumetric strain" = (triangle V)/V_1`
`= m(1/rho_1 - 1/rho_2) xx rho_1/m`
`:.triangleV/V_1 = 1 - rho_1/rho_2` ....(i)
Bulk modulus, `B = (rhoV_1)/(triangleV)`
`(triangleV)/V_1 = rho/B`
Compressibity of water = `1/B = 45.8 xx 10^(-11) Pa^(-1)`
`:.(triangleV)/V_1 = 80xx 1.013 xx 10^5 xx 45.8 xx 10^11 = 3.71 xx 10^(-3)` ...(ii)
For equation i and ii we get
`1 - rho_1/rho_2 = 3.71 xx 10^(-3)`
`rho_2 = (1.03 xx 10^3)/(1-(3.71xx10^(-3)))`
`=1.034 xx 10^3 kg m^(-3)`
Therefore, the density of water at the given depth (h) is 1.034 × 103 kg m–3.
उत्तर २
Compressibility of water `k = 1/B = 45.8 xx 10^11 Pa^(-1)`
Change in pressure, `trianglep = 80 atm - 1 atm`
`= 79 atm = 79 xx 1.013 xx 10^5 Pa`
Density of water at the surface
`rho = 1.03 xx 10^3 kg m^(-3)`
As `B= (trianglep.V)/(triangleV) or (triangleV)/V = (trianglep)/B = trianglep xx 1/B = trianglep xx k`
or `triangleV/V = 79 xx 1.013 xx 10^5 xx 45.8 xx 10^(-11) = 3.665 xx 10^(-5)`
Now `(triangleV)/V = ((M/rho)-(M/(rho')))/(M/rho) = 1 - rho/rho'`
or `rho/(rho') = 1 - (triangleV)/V`
or `rho' = rho/(1-(triangleV/V))`
or `rho' = (1.03 xx 10^3)/(1-3.665 xx 10^(-3)) = (1.03 xx 10^3)/0.996`
`= 1.034 xx 10^3 "kg/m"^3`
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