# Fluids and pressures

1. Nov 1, 2005

### thenewbosco

Mercury is poured into a u-tube. The left arm has cross sectional area A1 = 10 cm^2, the right has area A2, 5cm^2.
100 g of water are then poured into the right arm.
determine the length of the water in the right arm and given the density of Hg = 13.6 g/cm^3, what distance h, does the mercury rise in the left arm.

i have a diagram here and have calculated the length of the water in the right arm to be 0.2 m. i dont know now how to go about calculating the height that the mercury rises though. any help?

2. Nov 1, 2005

### Pyrrhus

Remember the total volume of mercury has not changed, therefore the volume displaced on the right must be the same volume that rose a distance h.

3. Nov 1, 2005

### lightgrav

At the bottom of the water column, it touches the mercury.
On the other side of the U-tube "balance", at that height,
there's an equal *Pressure* caused by mercury above it.

4. Nov 1, 2005

### thenewbosco

ok so i know at that depth on the right side the pressure = external pressure + densityofwater*g*0.2
so i equate this and get $$\rho_{Hg}gh=\rho_{water}g(0.2)$$
but the h here is not the h i require, it is height i require plus some other height...
where that other height is the length displaced on the right i think??

5. Nov 1, 2005

### lightgrav

If you would replace the water height
with half of the new excess mercury height,
the two sides would be the same height.
just like they started.
(see cyclovenom's post, above)

6. Nov 1, 2005

### thenewbosco

quite a tricky little problem, but thanks for all your help i got the correct answer now