physics promblem kind of wierd question?

Question

Two identical containers are open at the top and are connected at the bottom via a tube of negligible volume and a valve which is closed. Both containers are filled initially to the same height of 1.00 m, one with water, the other with mercury, as the drawing indicates. The valve is then opened. Water and mercury are immiscible. Determine the fluid level in the left container when equilibrium is reestablished.
distance from the bottom

Keep getting the wrong solution even though I did it using the formula my teacher told me to use. Can you help me find my error?

PwgH=PHggh
Pwh=PHgh
(1.0043)(9.8)=13600h(9.8)
x=1-h
got the answer to be 1.9265<

Answer 1

H1 = height of mercury in the mercury + water tank
H2 = height of the mercury in the mercury alone tank

The '1' comes from density of water x's 1 meter height:
13.534xH2 = 1 + 13.534xH1
H1 + H2 = 1 liter
These lead to
1 - 2H1 = 1/13.534
so H1 = .463m, H2=.537m
so the height of fluid in the water + mercury tank is 1.463m

Answer 2

Remember that any mercury that exits one container is added to the bottom of the other container.

If you have (x) meters of mercury in one container, you have (1 - x) meters of mercury on the other side. The total amount of mercury must equal 1 meter.

You need to solve the equation:
(H20 density)*(H2O depth) + (Hg density)*(1-x meters) = (Hg density)*(x meters)

You know the densities of each liquid, and you know the H2O depth is 1 meter. Now simply solve for x and figure out the depth of the mercury on each side. Then you can add the depth of the water to the proper side to obtain the right answer.

Answer 3

Remember that any mercury that exits one container is added to the bottom of the other container.

If you have (x) meters of mercury in one container, you have (1 - x) meters of mercury on the other side. The total amount of mercury must equal 1 meter.

You need to solve the equation:
(H20 density)*(H2O depth) + (Hg density)*(1-x meters) = (Hg density)*(x meters)

You know the densities of each liquid, and you know the H2O depth is 1 meter. Now simply solve for x and figure out the depth of the mercury on each side. Then you can add the depth of the water to the proper side to obtain the right answer.

Answer 4

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