I have a couple of brain blocks.

[falcon0311]

I've got a U-tube filled with Mercury and water. If I add 11.2 cm of water to the left side, how much does the mercury rise on the right side?

For simplicity's sake, D will stand for density, p0 for initial pressure, w = water, m = mercury, h = 11.2 cm for the water and d = the distance the mercury rises, and g is gravity.

The equation I think I'm supposed to use is

p0 + Dw(g)(1/2 * h) = p0 + Dm(g)(d)

--> Dw(1/2 * h) = Dm(d)

d = [ Dw(1/2 * h) ] / Dm = [ 1,000 kg/m^3 ( 1/2 * 11.2 cm ) ] / 13.6*10^3 kg/m^3

d = .412 cm which is what the book says, but my question is why did I use (1/2 * h) instead of h? Is there another way to do this problem?

 

[genxhis]

I believe the correct answer depends on whether it's mercury or water that occupies the lower, bent portion of the U-tube. Let's assume however that mercury fills the bottom. Then the water-mercury interface is on the left side (water side) at a point where the tube remains vertical. The importance of the interface position is that for a height change [tex]\inline{dx}[/tex] of the mercury surface level, the distance change of the mercury surface to water-mercury interface is [tex]\inline{2dx}[/tex]. Thus, if we add 11.2 cm of water to the left-hand side, then [tex]\inline{ \rho_w g [11.2 \text{ cm}]= \rho_m g [2dx] }[/tex].

 

[Leong]

If the mercury level falls by x cm on the left side. then the mercury on the right side rises by x cm too relative to the initial level.

See the attached file. Do you know why now?

The density of mercury is greater than water, so it is always beneath water.

 

[falcon0311]

Thanks for the help; you cleared that up pretty well.