Question

h3 h2 A "U" shaped tube (with a constant radius) is filled with water and oil as shown. The water is a height h1 0.38 m above the bottom of the tube on the left side of the tube and a height h 0.12 m above the bottom of the tube on the right side of the tube The oil is a height h3-0.32 m above the water. Around the tube the atmospheric pressure is PA 101300 Pa. Water has a density of 103 kg/m3 1) What is the absolute pressure in the water at the bottom of the tube? Pa 2) What is the absolute pressure in the water right at the oil-water interface? Pa submt 3) What is the density of the oil? kg/m3 suimit 4) Now the oil is replaced with a height h3 = 0.32 m of glycerin which has a density of 1261 kg/m. 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? Pa subm. 5) How much higher is the top of the water in the tube compared to the glycerin? (labeled d in the diagram) 6) What is the absolute pressure in the water at the very bottom of the tube? Pa submit

Answer 1

3. A) We can take any side, leftor right, of the tube to find the absolute pressure. Let us take the left side where there is only water.

The absolute pressure is given by

P = $\rho$ gh1 + 101300 = (10^3*9.8*0.38 + 101300) Pa

= 105024 Pa

B) Let us now take the interface where the absolute pressure is given by

P1 = $\rho$ g(h1 - h2) + 101300

= [10^3*9.8*(0.38 - 0.12) + 101300] Pa

= 103848 Pa

C) Let $\rho'$ be the density of the oil. Now, pressure at the oil water interface due to the oil column

P2 = h3* $\rho'$ *g = 0.32* $\rho'$ *9.8 = 3.136 $\rho'$

Again this pressure is given by

P2 = $\rho$ g (h1 - h2) = 10^3*9.8*(0.38 - 0.12)

= 2548 Pa

So, 2548 = 3.136* $\rho'$

Or,    $\rho'$ = 2548/3.136 = 812.5 kg/m^3

Thus, density of the oil = 812.5 kg/m^3