Question

The pressure at the bottom of a cylindrical container with a cross-sectional area of 45.5 cm2 and holding a fluid of density 420 kg/m3 is 115 kPa. (a) Determine the depth of the fluid. How is the pressure on the bottom of the container related to atmospheric pressure and the pressure due to the depth of the fluid

Answer 1

Answer:

3.33 m

Explanation:

Pressure is the distributed force applied to the surface of an object per unit area. The force is applied perpendicular to the surface of the object. The SI unit of pressure is N/m² or Pa.

Hydrostatic pressure is the pressure that a fluid exerts at a point due to the force of gravity.

The relationship between  pressure on the bottom of the container,  atmospheric pressure and the pressure due to the depth of the fluid is given by:

$P_{bottom}-P_{atm}=P_{depth}\\\\where\ P_{bottom}=pressure\ at\ the \ fluid\ bottom,\ P_{atm}=atmospheric\ pressure\\P_{depth}=pressure\ due\ to\ fluid\ depth=\rho gh. \ Hence:\\\\P_{bottom}-P_{atm}=\rho gh\\\\Given \ that\ P_{bottom}=115\ kPa=115*10^3\ Pa, let\ us\ assume\ P_{atm}=101\ kPa=101*10^3\ Pa,\rho=420\ kg/m^3,g=acceleration\ due\ to \ gravity=10\ m/s^2.\\\\Therefore:\\\\115*10^3-101*10^3=420*10*h\\\\14*10^3=4200h\\\\h=3.33\ m$