Answer :

Given data,


Pressure at the given depth, p = 80.0 atm


Pressure = 80× 1.01 × 105 Pa.


Density of water at the surface, ρ1 = 1.03 × 103 kgm-3


Let the given depth be h.


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


V = m[(1/ρ1)-(1/ρ2)]


Volumetric strain = ΔV/V1


Volumteric Strain = m [(1/ρ1)-(1/ρ2)] × (ρ1/m)


ΔV/V1 = 1-(ρ12) ………… (1)


Bulk modulus, B = pV1/ ΔV


ΔV/V1 = p/B


Compressibility of water = (1/B) = 45.8× 10-11 Pa-1


Compressibility is the fractional change in volume per unit increase in pressure. For each atmosphere increase in pressure, the volume of water would decrease 46.4 parts per million.


ΔV/V1 = 80× 1.013× 105× 45.8× 10-11 = 3.71 × 10-3 ……. (2)


From equations 1 & 2, we get:


1-(ρ12) = 3.71 × 10-3


ρ2 = 1.03× 103 / [1-(3.71 × 10-3)]


ρ2 = 1.034× 103 kgm-3


Therefore, the density of water at the given depth (h) is 1.034× 103 kgm-3.


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