# Fluid statics – problems and solutions

**Liquid pressure
**

1. What is the difference between the hydrostatic pressure of blood between the brain and the soles of the feet of a person whose height 165 cm (suppose the density of blood = 1.0 × 10^{3} kg/m^{3}, acceleration due to gravity = 10 m/s^{2})

__Known :__

Height (h) = 165 cm = 165/100 m = 1.65 meters

Density of bloods (ρ) = 1.0 × 10^{3} kg/m^{3}

Acceleration due to gravity (g) = 10 m/s^{2}

__Wanted:__ liquid pressure

__Solution :__

P = ρ g h

P = (1.0 × 10^{3})(10)(1.65)

P = (1.0 × 10^{4})(1.65)

P = 1.65 x 10^{4 }N/m^{2}

**Pipe U**

2. A U pipe is initially filled with water than on one pipe filled with oil, as shown in the figure below. The density of water is 1000 kg/m^{3}. If the height of oil is 8 cm and the height of the water is 5 cm, what is the density of oil?

__Known :__

Density of water = 1000 kg.m^{-3}

The height of water (h_{2}) = 5 cm

The height of oil (h_{1}) = 8 cm

__Wanted :__ density of oil

__Solution :__

ρ_{1} g h_{1} =ρ_{2} g h_{2}

ρ_{1} h_{1} =ρ_{2} h_{2}

(1000)(5) = (ρ_{2})(8)

5000 = (ρ_{2})(8)

ρ_{2} = 625 kg.m^{-3}

3. A U pipe was first filled with kerosene then added water. If the mass of kerosene is 0.8 grams/cm^{3} and the density of water is 1 gram/cm^{3} and the cross sectional area is 1.25 cm^{2}. Determine how much water should be added so that the height difference of the kerosene surface is 15 cm

A. 9 ml

B. 12 ml

C. 15 ml

D. 18 ml

__Known :__

Density of kerosene (ρ_{1}) = 0.8 gram/cm^{3}

Density of water (ρ_{2}) = 1 gram/cm^{3}

Sectional area of the pipe = 1.25 cm^{2}

The height difference of the surface of kerosene (h_{1}) = 15 cm

__Wanted :__ Volume of water

__Solution :__

The height of water (h_{2}) :

ρ_{1 }g h_{1} = ρ_{2 }g h_{2 }

(0,8)(15)(1)(h_{2})

h_{2} = 12 cm

Volume of water :

V = (Sectional area of the pipe)(height of water)

V = (1.25 cm^{2})(12 cm)

V = 15 cm^{3}

*1 liter = 1 dm*^{3}* = 10*^{3}* cm*^{3}

*1 mililiter = 10*^{-3}* liter**s** = (10*^{-3}*)(10*^{3}*) cm*^{3}* = 1 cm*^{3}

Volume of water is 15 cm^{3} = 15 mililiters

The correct answer is C.

4. A pipe U filled with water with density of 1000 kg/m^{3}. One column of pipe U filled with glyserin with density of 1200 kg/m^{3}. If the height of glyserin is 4 cm, determine the height difference of both columns of the pipe.

A. 0.8 cm

B. 4 cm

C. 8 cm

D. 12 cm

__Known :__

Density of water (ρ_{1}) = 1000 kg/m^{3}

Density of glycerin (ρ_{2}) = 1200 kg/m^{3}

Height of glycerin (h_{2}) = 4 cm

__Wanted:__ The height difference of both columns of the pipe.

__Solution :__

The height of the column of the pipe (h_{1}) :

ρ_{1 }h_{1} = ρ_{2 }h_{2 }

(1000)(h_{1}) = (1200)(4)

(1000)(h_{1}) = 4800

h_{1} = 4.8 cm

The height difference of both columns of the pipe U = h_{1} – h_{2} = 4.8 cm – 4 cm = 0.8 cm

The correct answer is A.

5. A pipe U has two ends are open filled with water with a mass of 1 g/cm^{3}. The sectional area along the pipe is the same, that is 1 cm^{2}. Someone blows on one end of the foot of the pipe so that the surface of the water at the other foot rises 10 cm from its original position. If the acceleration due to gravity is 10 m/s2 then determine the force acted by that person.

A. 20 kilodyne

B. 10 kilodyne

C. 2 kilodyne

D. 1 kilodyne

__Known :__

*Change all units to the International system.*

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

Cross sectional area of pipe (A) = 1 cm^{2 }= 10^{-4} m^{2}

The change of column of pipe (h) = 10 cm = 1 dm = 10^{-1} m

Acceleration due to gravity (g) = 10 m.s^{-2 }= 10^{1} m.s^{-2}

Volume of moved water (V) = (A)(h) = (1 cm^{2})(10 cm) = 10 cm^{3 }= (10^{1})(10^{-6} m^{3}) = 10^{-5} m^{3}

__Wanted :__ Force (F) acted by the person.

__Solution :__

The force that acted by that person = weight of water with a height of 10 cm

F = w

F = m g —–> Equation of density :* *m = ρ V

F = ρ V g

F = (10^{3})(10^{-5})(10^{1})

F = (10^{4})(10^{-5})

F = 10^{-1} Newton —–> *1 Newton = 10*^{5}* dyne*

F = (10^{-1})(10^{5} dyne)

F = 10^{4 }dyne

F = 10 kilodyne

The correct answer is B.

6. A Y-shaped tube is inserted upside down so that the left foot and right foot are immersed in two kinds of liquid. After both feet are immersed in the liquid, then the top of the Y pipe is closed with the finger and pulled upwards, so that the two legs of the Y pipe are filled with a column of different high-density liquids. If the density of the first liquid is 0.80 gram.cm^{-3 }and the second density is 0.75 gram.cm^{-3}, and the lower liquid column is 8 cm, then determine the height difference between the two liquid columns on U pipe.

A. 1.0666 cm

B. 0.9375 cm

C. 0.3533 cm

D. 0.5333 cm

__Known :__

Density of first liquid (ρ_{1}) = 0,80 gram.cm^{-3}

Density of second liquid (ρ_{2}) = 0,75 gram.cm^{-3}

The height of the lower liquid (h_{1}) = 8 cm

__Wanted :__ The height difference between the two liquid columns on U pipe

__Solution :__

The height of the higher liquids (h_{2}) :

ρ_{1 }h_{1} = ρ_{2 }h_{2 }

(0.80)(8) = (0.75)(h_{2})

6.4 = 0.75 (h_{2})

h_{2 }= 6.4 / 0.75

h_{2 }= 8.5 cm

The height difference of liquids = h_{2} – h_{1} = 8.5333 cm – 8 cm = 0.5333 cm

The correct answer is D.

7. A stone with the volume of 0.5 m^{3} placed in a liquid with the density of 1.5 gr cm^{–3}. Acceleration due to gravity is 10 m s^{-2}. What is the buoyant force?

__Known :__

Volume of stone (V) = 0.5 m^{3}

Density of water (ρ) = 1.5 gr cm^{–3} = 1500 kg m^{-3}

Acceleration due to gravity (g) = 10 m s^{-2}

__Wanted:__ buoyant force (F_{A})

__Solution :__

The equation of the buoyant force :

F_{A} = ρ g V = (1500 kg m^{-3})(10 m s^{-2})(0.5 m^{3}) = 7500 kg m/s^{2} = 7500 Newton

**Float**

8. A block of ice float in the sea as shown in the figure below. The density of sea is 1.2 gr cm^{–3 }and density of ice is 0.9 gr c^{–3}. The volume of ice in sea water = ……. x the volume of ice in the air.

__Known :__

Density of sea (ρ_{sea}) = 1.2 gr cm^{–3}

Density of ice (ρ_{ice}) = 0.9 gr c^{–3}

__Wanted:__ The volume of ice in sea water = ……. x the volume of ice in the air.

__Solution :__

The volume of ice in sea = 0.75

The volume of ice in air = 0.25

The volume of ice in sea water = 3 x the volume of ice in air (3 x 0.25 = 0.75).

9. An object float in a liquid where 2/3 of the object in the liquid. If the density of the object is 0.6 gr cm^{3}, then what is the density of water.

__Known :__

The part of the object in liquid = 2/3

Density of object = 0.6 gr cm^{3} = 600 kg m^{3}

__Wanted:__ the density of the liquid (x)

__Solution :__

The density of the liquid is 900 kg m^{3}

10. A wood float in water, where 3/5 part of wood in the water. If the density of water is 1 × 10^{3} kg/m^{3}, what is the density of wood?

__Known :__

Part of object in water = 3/5

Density of water = 1×10^{3} kg/m^{3 } = 1000 kg/m^{3 }

__Wanted :__ The density of wood (x)

__Solution :__

The density of wood is 600 kg/m^{3} = 6 x 10^{2 }kg/m^{3}