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A glass capillary tube is of the shape of a truncated cone with an apex angle alpha so that it's two ends have cross sections of different radius. When dipped in water vertically, water rises to a height h , where the radius of cross section is b. If the surface tension of water is S, its density $\rho$ , and it's contact angle with glass is $\theta$, the value of $h$ will be..?

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Source

In the very first line of the solution to the problem, it is taken that $P_0 -P_1 = \rho gh$

Where $P_0$ is the pressure just outside and $P_1$ is the pressure inside

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But I don't get it, $\rho g h$ is the pressure difference between the top most point of the beaker open and the bottom most point near which the edges converge, how does it also give the pressure difference between right outside and inside the water surface open to atmosphere?

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  • $\begingroup$ Would you edit and finish off the last line at the top section of your question please. Also maybe check the positions of the places where $P_0$ and $P_1$ are... $\endgroup$ Jun 27, 2021 at 19:41
  • $\begingroup$ Sorry, I didn't know what I was thinking when I wrote the title. I went a bit too fast. I have corrected it now, thank you for checking out my question carefully. Hmm I don't quite get your point about position of $P_0$ and $P_1$ $\endgroup$ Jun 27, 2021 at 20:26
  • $\begingroup$ Well it's probably the same confusion that you have, but it seemed strange that they were put where they were. The edit referred to was "the value of * will be", please insert something instead of * $\endgroup$ Jun 27, 2021 at 20:30

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As well as being the pressure in the air just above the meniscus, $P_\text 0$ is also the pressure at the water surface in the bowl (neglecting pressure changes in the air over height $h$). So the equation that you quote simply gives the hydrostatic pressure difference in the liquid column – and that is the same as the pressure difference across the meniscus!

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  • $\begingroup$ Hmm I am not sure I understand, how does we want to get the pressure difference between right outside and right inside.. why do we need it till the very bottom of the tube? $\endgroup$ Jun 27, 2021 at 20:52
  • $\begingroup$ I'm afraid I can't understand your comment. I've added to my answer, hoping that it will now be clear to you. $\endgroup$ Jun 27, 2021 at 21:32
  • $\begingroup$ Hi, it helps a bit, but what I don't get is why is it that the pressure difference across the meniscus is same as the hydrostatic pressure difference across the liquid column. I understand the situation but not why the situation is as it is. $\endgroup$ Jun 27, 2021 at 21:38
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    $\begingroup$ Pressure above meniscus – Pressure just below meniscus = $p_0 - p_1$. Pressure at bottom of column – Pressure just below meniscus = $p_0 - p_1$ =$h\rho g$. Therefore Pressure above meniscus – Pressure just below meniscus =$p_0 - p_1$ =$h\rho g$. If you still don't understand, read the first sentence of my answer again, and think about it! $\endgroup$ Jun 27, 2021 at 21:50
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    $\begingroup$ "The result feels really unbelievable though... " But water really does rise up a capillary tube! And the height rise agrees with the equation we've just derived! $\endgroup$ Jun 28, 2021 at 9:43
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There is a formula here

https://thefactfactor.com/facts/pure_science/physics/numerical-problems-on-capillary-action/5329/

That gives the height of the capillary column

$$h = \frac{2T\cos\theta}{r \rho g}\tag1$$

The surface tension will have to support the weight of the capillary column, that's where the pressure difference comes in, it's $h \rho g \times \pi a^2$ where $a$ is the radius at the bottom.

This is just an idea, but perhaps formula 1) came from this derivation, putting the up and downward forces equal $$\pi r^2 h \rho g = Tcos \theta 2 \pi r \tag 2$$

where the left $r$ will later be replaced with $a$ and the right $r$ replaced with $b$, although $a$ might need to be found in terms of $h$ and $\theta$...best of luck.

P.S. Think your question was on the internet, here ( https://www.concepts-of-physics.com/mechanics/capillary-rise.php ), still understood best by putting upward forces equal to downward forces, (their pressure method does seem confusing).

You'll also need to change $\theta$, (as it should be the angle to the vertical), to $\theta + \frac{\alpha}{2}$, then the answer will match, if the first $r$ is left as $b$ (although $a$ seems better, as the sides of the glass support the weight of the extra area), but at least that way you'll get a match to the given answer.

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  • $\begingroup$ Hi John, the site seems helpful for my studies but my question was related to how the equality of forces was achieved, why is the surface tension force exactly equal to weight of the whole column? $\endgroup$ Jun 27, 2021 at 20:53
  • $\begingroup$ The liquid column has a weight that needs supporting, that comes from the upwards force of the surface tension , see the top diagram of concepts-of-physics.com/mechanics/capillary-rise.php $\endgroup$ Jun 27, 2021 at 21:44

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