Question

An absolute pressure gauge is submerged 50 m (1979 in) below the surface of the ocean and reads P 6.064 bar. This is P 2434 in of H2O), according to the unit conversions built into a particular calculator. Estimate the density of seawater. The density of seawater is kg/m,2 What is the reason for the apparent discrepancy between the pressure measurement and the actual depth of submersion The apparent discrepancy between the pressure measurement and the actual depth of submersion is due to the fact that the density of sea water is is higher higher than the density of fresh water.

Answer 1

Pressure at submerged point

= atmospheric pressure + density x g x h

6.064 bar x 10^5 Pa/bar = 101325 Pa + density x 9.81 m/s2 x 50 m

Density of sea water = 1029.71 kg/m3

Discrepancy is due to higher density of sea water

Answer 2

P=F/A P=mg/A-----(1) Density=m/v m=density ×v Volume = A*h Put in 1 P=density*g*A*h/A P=density*g*h P=1025*9.8*50 P=512500