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Asked by Maheshwari Last Modified
Answer 
Shivam Sharma
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.
For equations (i) and (ii), we get:
Therefore, the density of water at the given depth (h) is 1.034 × 103 kg m–3.
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