Hg in U-shape tube, water is added. final h Hg?

[locster314]

Homework Statement

A U shaped container containing Hg at rest. 11.2 cm of water is added to the right side of container. How much does the Hg rise?

Homework Equations

p = p + [tex]\rho[/tex]gh

The Attempt at a Solution

 

[berkeman]

Welcome to the PF, locster! An important PF rule is that you need to show some of your initial work in order for us to offer you some tutorial help. What are your initial thoughts on how to approach this problem? Hint -- the water will not diffuse through the water to any great extent, so the water is only significant for its weight down on one surface of the mercury. Think of it as a teeter-totter with weights on the ends -- what is the balance condition?

 

[m2003]

The Hg will rise until the hydrostatic pressure on both sides of the U-shaped tube are equal. This means that the height of the Hg on the right side will increase until the pressure from the water added is equal to the pressure from the Hg on the left side. This can be calculated using the equation p = p + \rhogh, where p is the atmospheric pressure, p is the density of the liquid (Hg), \rho is the density of water, g is the acceleration due to gravity, and h is the height of the Hg.

To solve for the final height of the Hg, we can rearrange the equation to h = (p - p)/(\rho g), where the final pressure will be equal to the atmospheric pressure plus the pressure from the added water, and the final density will be a combination of the densities of Hg and water.

Substituting in the values for p, p, \rho, and g, we get h = (101325 Pa + \rho g * 11.2 cm * 1000 kg/m^3 - 13600 Pa) / ((13546 kg/m^3 - 1000 kg/m^3) * 9.8 m/s^2) = 0.54 cm.

Therefore, the Hg will rise by 0.54 cm in the U-shaped tube when 11.2 cm of water is added to the right side.