Question

In: Physics

A “U” shaped tube (with a constant radius) is filled with water and oil as shown....

A “U” shaped tube (with a constant radius) is filled with water and oil as shown. The water is a height h1 = 0.36 m above the bottom of the tube on the left side of the tube and a height h2 = 0.1 m above the bottom of the tube on the right side of the tube. The oil is a height h3 = 0.33 m above the water. Around the tube the atmospheric pressure is PA = 101300 Pa. Water has a density of 103 kg/m3.

A)What is the absolute pressure in the water at the bottom of the tube?
B)What is the absolute pressure in the water right at the oil-water interface?
C)What is the density of the oil?
D)Now the oil is replaced with a height h3 = 0.33 m of glycerin which has a density of 1261 kg/m3. Assume the glycerin does not mix with the water and the total volume of water is the same as before.

Now what is the absolute pressure in the water at the interface between the water and glycerin?
E)How much higher is the top of the water in the tube compared to the glycerin? (labeled d in the diagram)
F)What is the absolute pressure in the water at the very bottom of the tube?

Solutions

Expert Solution

a) pressure at the bottom is found as

P = Patm + gh

P = 101300 + 1000 * 9.8 * 0.36

P = 104828 Pa

-------------------------

b)

pressure at the interface

P = 101300 + 1000 * 9.8 * ( 0.36 - 0.1)

P = 103848 Pa

-----------------------------

c)

pressure in left column = pressure in right column

water g h1 = oil g h3 + water g h2

1000 * 9.8 * 0.36 =  oil * 9.8 * 0.33 + 1000 * 9.8 * 0.1

oil = 787.87 Kg/m3

-----------------------------------

d)

now , oil is replaced with glycerin

so,

P ( interface) = 101300 + 1261 * 9.8 * 0.33

P ( interface) = 105378 Pa

--------------------------------------

(e)

p ( left) = p ( right)

101300 + 1000 * 9.8 * h = 105378

h = 0.416 m

so,

top of water = 0.416 - 0.33 = 0.08613 m

-------------------------------

(f)

height of water in right tube = 0.1 - 0.08613 = 0.01387 m

so,

P = 101300 + 1261 * 9.8 * 0.33 + 1000 * 9.8 * 0.01387

P = 105514 Pa


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