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I've had contradictory experimental results from my calculations of displaced fluid height after adding a floating object.

Consider the following scenario: Using a test tube of diameter $D=1.70$ cm and height $H=17.50$ cm, we add sea salt completely saturated water ($ρ_f = 1.202$ g/cm^3) to it until the initial fluid height $H_1$ is 10 cm. After adding an ice cylinder made of purifed water ($ρ_I=0.915$ g/cm^3) with dimensions $d=0.85$ cm and $h=9.00$ cm, the task is to determine the displaced fluid height.

Using Archimedes' principle, we can solve for $h_d$:

$$F_b = W_I\\ ρ_f V_d = ρ_I V_I$$

Consider that $V_d$ has the shape of a cylindrical ring, therefore:

$$ρ_f\frac{\pi}{4}(D-d)^2h_d = ρ_I\frac{\pi}{4}d^2h$$

Solving for $h_d$:

$$h_d =\frac{ρ_I}{ρ_f}\left(\frac{d}{D-d}\right)^2 h = 6.85~\text{cm}$$

Yet, experimentally I have obtained multiple times a $h_d$ of around 2 cm. What have I done wrong? I have provided two images of one of my many tests below.

Some further notes: I have tried minimizing the amount of bubbles in the ice cylinder by directional freezing. I have also measured the density of the salt saturated solution always before starting the test, and dimensions of the ice cylinder are as close to $9.00$ cm as I could get them to be.

I also apologize for may lack of formatting. I'm new to this forum and do not know how to use the proper format.

Test tube with initial fluid height

Test tube with the ice cylinder displacing the fluid

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  • $\begingroup$ Consider using MathJax to format the equations in the question, as it's very hard to read it in it's current state. $\endgroup$
    – Cross
    Commented Jun 18, 2022 at 16:32
  • $\begingroup$ Its hard to tell from the photo but it appears as if your ice cylinder is hitting the bottom of your test tube before the ice freely floats. This will affect your results. $\endgroup$
    – Stevan V. Saban
    Commented Jun 18, 2022 at 16:45
  • $\begingroup$ @Cross Thank you for your advice! $\endgroup$
    – Rodrigo Alfaro
    Commented Jun 18, 2022 at 16:55
  • $\begingroup$ @SteveSaban I have changed the image to a clearer one. I can assure you that the ice cylinder does not touch the bottom of the test tube. Although it does touch the left side, but I presume any effect that may contibute might be negligible, as would the air bubbles inside the ice cylinder which i tried to minimize with directional freezing. $\endgroup$
    – Rodrigo Alfaro
    Commented Jun 18, 2022 at 16:58

1 Answer 1

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Your mistake is assuming $V_d$ is a cylindrical ring. $V_d$ is just a cylinder $$\frac {\pi D^2h_d}{4}$$

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  • $\begingroup$ Thank you for your answer, now the numbers seem to agree with experimental results! Although I have a hard time understanding why the shape of the displaced fluid volume would be equivalent to the volume of a cylinder. Isn't the ice cylinder only displacing the fluid around it? $\endgroup$
    – Rodrigo Alfaro
    Commented Jun 18, 2022 at 17:23
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    $\begingroup$ Thinking in terms of shapes is unwise. A given volume of liquid can have any shape. In the simplest of terms, two volumes have been added to create a new volume: $V_f = V_0 +V_d$ The shape for all volumes is normalized to that of a cylinder with a cross-sectional area $\pi D^2$/4. $\endgroup$
    – Stevan V. Saban
    Commented Jun 18, 2022 at 17:32
  • $\begingroup$ Interesting idea, but I think my mistake was just not realizing my initial formula for a cylindrical ring was wrong all along. I used $$(D-d)^2$$ insted of $$(D^2-d^2)$$. I've calculated a resultant height of 2.28 cm. Do you think this result is wrong? It does seem to agree within a reasonable amount of error $\endgroup$
    – Rodrigo Alfaro
    Commented Jun 18, 2022 at 18:01

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