U-tube filled with water and inmiscible liquid

[xannaxiero]

Homework Statement

A U-shaped tube is partly filled with water and partly filled with a liquid that does not mix with water. Both sides of the tube are open to the atmosphere. If h1 = 0.52 m and h2 = 0.16 m, what is the density of the liquid?

Homework Equations

ρ$_{u}$=h$_{1}$/h$_{2}$ * ρ$_{k}$

as in: unknown density = ratio of heights times known density

ρ$_{k}$= density of water = 1 cm$^{3}$/mL

The Attempt at a Solution

so I just plugged the info in, seems relatively simple...
ended up with

ρ$_{u}$= 52 cm/16 cm * 1 cm$^{3}$/mL = 3.25 cm$^{3}$/mL

but for some reason this is wrong.

just for kicks I also tried
ρ$_{u}$= 16 cm/52 cm * 1 cm$^{3}$/mL = .31 cm$^{3}$/mL

but this is also wrong.

any ideas?

 

[I like Serena]

Homework Statement

A U-shaped tube is partly filled with water and partly filled with a liquid that does not mix with water. Both sides of the tube are open to the atmosphere. If h1 = 0.52 m and h2 = 0.16 m, what is the density of the liquid?

Homework Equations

ρ$_{u}$=h$_{1}$/h$_{2}$ * ρ$_{k}$

as in: unknown density = ratio of heights times known density

ρ$_{k}$= density of water = 1 cm$^{3}$/mL

The Attempt at a Solution

so I just plugged the info in, seems relatively simple...
ended up with

ρ$_{u}$= 52 cm/16 cm * 1 cm$^{3}$/mL = 3.25 cm$^{3}$/mL

but for some reason this is wrong.

just for kicks I also tried
ρ$_{u}$= 16 cm/52 cm * 1 cm$^{3}$/mL = .31 cm$^{3}$/mL

but this is also wrong.

any ideas?

Hi xannaxiero, welcome to PF!

Your unit for density is wrong.
It's not cm³/mL.

Btw, your formula assumes that the separation of the fluids is at the bottom of the U-tube.
I guess you have to make that assumption, because otherwise you do not have enough data.

 

[xannaxiero]

Oops you're right, I'm sleepy >.< Its g/mL. My bad. That doesn't make any difference in the equation though does it? And yes, I'm assuming separation is in the bottom...this is supposed to be a pretty simple, standard u-tube problem, no tricks...

 

[I like Serena]

Beyond that, your answers are right.
Why do you think they are wrong?

 

[xannaxiero]

Online homework...they're being graded as incorrect :'(

 

[I like Serena]

Oh those!
That usually means you did not follow the format that they expected.

Try rounding to 2 digits (since your input data is 2 digits each).
And try the unit kg/L or perhaps kg/m3 (adjusting of course the result to match).

 

[xannaxiero]

The answer ended up being .563 g/mL. Any idea why this is? There's no explanation with the homework :(

 

[I like Serena]

Nope. No idea.

I can only guess that there is a typo in the problem statement, or that there is more information that is not given.

 

[xannaxiero]

Do you think it could be because the tube is open to the atmosphere? Would that impact my formula at all?

 

[I like Serena]

No, the atmosphere gives the same pressure of 1 atmosphere on both tubes.
It cancels out.

It would matter if the tubes had different diameters, or if the separation is not at the bottom.