How Far Does Mercury Rise in a U-Shaped Tube with Added Water?

  • Thread starter Thread starter bcjochim07
  • Start date Start date
  • Tags Tags
    Pressure Tube
Click For Summary

Homework Help Overview

The problem involves a U-shaped tube containing mercury, where water is added to one side, creating a pressure difference that affects the mercury levels in both arms of the tube. The objective is to determine how far the mercury rises in the right arm after the water is introduced.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between the height of the mercury and the height of the water added, with some attempting to derive equations based on pressure balance. Questions arise regarding the interpretation of variables and the correct reference points for measuring height changes.

Discussion Status

The discussion is ongoing, with participants providing different interpretations of the problem and questioning each other's reasoning. Some guidance has been offered regarding the setup of equations and the significance of measuring from the correct reference points, but no consensus has been reached on the final answer.

Contextual Notes

Participants note potential confusion regarding the initial positions of the mercury and the effects of adding water, as well as the implications of pressure changes in the system. There is also mention of the need to clarify what is being asked in the problem statement.

bcjochim07
Messages
366
Reaction score
0

Homework Statement


A U shaped tube, open to the air on both ends, contains mercury. Water is poured into the left arm until the water column is 10 cm deep. How far upward from its intitial position does the mercury in the right arm rise?


Homework Equations





The Attempt at a Solution



h is the height of mercury on the left side of the tube
x is the distance that the mercury moves up

101300 Pa + (1000kg/m^3)(9.80)(.10m) + (13600kg/m^3)(9.80)h= 101300 Pa + (13,600kg/m^3)(9.80)(h+x)

980 + 133280h = 133280h + 133280x
980=133280x
x= .00735m = .735cm I'm not sure if I did this correctly--this answer seems quite small
 
Physics news on Phys.org
Hi bcjochim07,

It looks to me like you are answering the wrong question. In your equation x is the distance the top of the mercury is above the bottom of the water on the other side. However, they are asking for how much the mercury rose up above where it was before the water was poured in.
 
101300 Pa + (1000kg/m^3)(9.80)(.10m) = 101300 Pa + (13,600kg/m^3)(9.80)(x)

finally,

result: x/2


hint: you don't need to write the same pressure both side. draw a horizontal line from top of the lowest height side of most bottom liquid. And calculate the pressures for only the upper side of line which you draw.
ejhgmh.jpg
 
Last edited:
What did you mean by x/2? I got the same number for x using your equation that I did mine. I really didn't need to put the h in.
 
x in your equation is the distance above the bottom of the water (where the purple line is in volcano's post). Pouring in the water pushes the mercury level down on the left side, and up on the right side. You found the difference in heights between the mercury levels on the left and right side; they want the difference in height between the new level on the right and the original level on the right.
 
If I try to visualize how it was before the water was added, I would just distributed the mercury evenly, so would I divide my answer by two then?
 
How much down(x/2), so much up(x/2) both sides. Think this, if one side down 10cm, how much rise other side? What is the distance difference between both sides?
 

Similar threads

  • · Replies 18 ·
Replies
18
Views
7K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
1
Views
5K
  • · Replies 1 ·
Replies
1
Views
15K
Replies
4
Views
4K
  • · Replies 8 ·
Replies
8
Views
5K