# Homework Help: Torricellian tube inclined

1. Sep 7, 2014

### Rugile

1. The problem statement, all variables and given/known data

A Torricellian tube (one end open, one end closed) of 1m length is filled completely with mercury and situated vertically the closed end down. Then the tube is flipped around 180 degrees and there is 760mm of mercury left. If we incline the tube at an angle $\phi$, what is the height of column of mercury in the tube? Find the dependency h($\phi$), if when phi = 0 the tube is vertical, and when phi = 90 it is horizontal.

2. Relevant equations

$p = \rho g h$

3. The attempt at a solution

It is not said in the problem statement that the tube is emerged into a bowl of mercury, thus the height of the column when inclined can't stay 760mm. So I'm not really sure what happens when we rotate the tube - my guess is that some air goes through the mercury up to where Torricellian vacuum is supposed to be? If so, I have an equation (which is based on intuition only, can't really prove it) of equilibrium, but I don't really know if it is right: $p_a = p_{\phi} + \rho g h_{\phi}; p_a = p_a sin\phi + \rho g h_{\phi}$, pa is atmospheric pressure. It does make sense in a way, that when phi = 0, sin(phi) = 0 and pa = ρgh, and when phi = 90, then h = 0. But I'm unable to prove it with physics... Any hints?

2. Sep 7, 2014

### BvU

You mean: It is not said in the problem statement that the bottom end of the tube is immersed into a bowl of mercury. But the exercise really wants you to make that assumption.

No air sneaking in either.

There is an analogy with communicating vessels (not identical, but Δp = ρgh in both cases). What happens when the third one from the left is tilted more and more (while remaining of the same length) ?

This Δp = ρgh is something you should take under thorough consideration: what does it mean, precisely. (note I already added a Δ).

Can't distill sense out of your equations. No idea what $p_{\phi}$ stands for.

(the way you tackle this made me think of an Einstein joke where the blackboard shows E = ma2 crossed out, mb2 crossed out too, and mc2 marked with a big eureka exclamation point. Sorry).

Imagine the tube is upright and has a huge diameter. Then think of a thin cylinder at an angle $\phi$, well within the big tube. Thin cylinder is long enough to go from vacuum above to Hg bath at foot. Close top end of thin cylinder and let big tube drain. Would anything change for the thin tube ?

By the way, I do find the exercise wording somewhat misleading: do they want the (average) length of the mercury column ? or the difference in height ?

3. Sep 10, 2014

### Rugile

Thank you for the answer. Well if you say it is safe to assume that the bottom of the tube is emerged in mercury, then I do know how to solve it :)

My equation had the idea that p(phi) is the pressure of air in the tube above mercury, if there is any air sneaking. As I said, it was completely unmotivated, just intuition (no hard feelings for the joke, it is funny!).

They want the length of mercury column in the tube. Not difference, just dependence of how high the column is depending on the inclined angle.

4. Sep 10, 2014

### BvU

Problem is height is in the vertical direction, length can be along the tube. Up to you. As soon as the top of the tube is less than 760 mm above the liquid level, there is no more vacuum.