# Pressure of fluids – problems and solutions

1.

Acceleration due to gravity is 10 N/kg. The surface area of fish pressed by the water above it is 6 cm^{2}. Determine the force of water above fish that acts on fish.

__Known :__

Surface area of fish (A) = 6 cm^{2} = 6 x 10^{-4 }m^{2}

Density of water = 1 gram/cm^{3 }= 1 (10^{-3 }kg) / 10^{-6} m^{3} = 10^{3} kg/m^{3} = 1000 kg/m^{3}

Height of water = 100 cm – 15 cm = 85 cm = 85 x 10^{-2} meters

__Wanted :__ Force of the water above fish that acts on fish

__Solution :__

Equation of pressure:

P = F / A

F = P A

F = ρ g h A

F = (1000)(10)(85 x 10^{-2})(6 x 10^{-4 }m^{2})

F = (10^{4})(510 x 10^{-6})

F = 510 x 10^{-2}

F = 5.1 Newton

2. The normal pressure of blood is 80 mm hg to 120 mm hg. This value is equal to…

A. 80 Pascal to 120 Pascal

B. 800 Pascal to 1200 Pascal

C. 10256 Pascal to 17589 Pascal

D. 10526 Pascal to 15789 Pascal

__Known :__

The normal pressure of blood = 80 mm Hg – 120 mm Hg

__Wanted:__ This value is equal to… Pascal

__Solution :__

The height of mercury in barometer (above the surface of the sea) = 760 mm Hg

Pressure of air (above the surface of sea) = 1 atm = 1.0 x 10^{5} Pascal

1 mm Hg = ……. Pascal?

80 mm Hg = 80 x 131.58 Pascal = 10,526 Pascal

120 mm Hg = 120 x 131.58 Pascal = 15,789 Pascal

The correct answer is D.

3. Above the surface of the sea, the height of mercury in a barometer is 760 mm. If at a place, the height of mercury in a barometer is 700 mm. Determine the pressure of air at that place.

__Known :__

The height of mercury in barometer (above the surface of the sea) = 760 mm Hg

Pressure of air (above the surface of sea) = 1 atm = 1,013 x 10^{5} Pascal

__Wanted:__ Determine the pressure of air if the height of mercury is 700 mm

__Solution :__

760 mm Hg = 1.013 x 10^{5} Pascal

700 mm Hg = ……… Pascal?