Question

Consider a U-tube whose arms are open to the atmosphere. Now water is poured into the U-tube from one arm, and light oil (r 5 790 kg/m3) from the other. One arm contains 70-cm-high water, while the other arm contains both fluids with an oil-to-water height ratio of 4. Determine the height of each fluid in that arm.

Answer 1

Answer:

water height: 3cm

oil height: 12 cm

Explanation:

70 cm = 0.7 m

Suppose both arms have the same cross-section area S:

Let water density be $\rho_w = 1000 kg/m^3$

Let x be the height of water in the 2nd arm, then 4x is the heigh of oil in the 2nd arm.

For the system to be balanced, then the fluid mass of both arms must be the same

$m_1 = m_2$

$V_1\rho_w = V_w\rho_w + V_o\rho_o$

$Sh_1\rho_w = Sx\rho_w + S4x\rho_o$

Where V1 is the fluid volume  is the 1st arm, Vw is the water volume in the 2nd arm, Vo is the oil volume in the 2nd arm, $h_1 = 0.7 m$ is the fluid height in the 1st arm, $\rho_o = 5790kg/m^3$ is the oil density.

We can divide both sides by S

$h_1\rho_w = x\rho_w + 4x\rho_o$

$h_1\rho_w = x(\rho_w + 4\rho_o)$

$x = \frac{ h_1\rho_w }{(\rho_w + 4\rho_o } = \frac{0.7*1000}{1000 + 4*5790} = 0.03m = 3 cm$

So the water height in the 2nd arm is 3 cm and the oil height is 3*4 = 12 cm