Can someone explain this physics passage and picture?

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SUMMARY

The discussion focuses on the behavior of liquids in conical capillaries, specifically addressing how the radius of the capillary affects the liquid's movement. It is established that the force drawing water up is proportional to the circumference of the capillary (F ~ 2πR), while the weight of the water is proportional to its volume (V = hπR²). As the radius decreases, the height of the water column increases, leading to a larger concave surface. The pressure difference at each end of the tube is explained through Laplace's law, which relates pressure jump to surface curvature, resulting in different behaviors for water and mercury in capillaries.

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On page 319 of this book, with regards to the conical capillaries: http://books.google.com/books?id=Eo...det&sig=Xr8x61FArV8fK4r0vpR_McJlQIk#PPA319,M1

One particular point that I feel needs further clarifying is that "...and the liquid is drawn in the direction of the of the greatest force per unit of area."

It leaves me with a lot of questions. I know water moves up the sides of the tube and pulls the water with it, leaving a concave surface. But why is there a larger concave surface when the radius gets smaller?

Why is there a pressure difference at each end of the tubes?

Thanks for any help.
 
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The force to draw water up is proportional to the circumstance of the capillary (F ~2.pi.R), while the weight of the water drawn is propotional to the volume V: V = h.pi.R^2. Then you can see h depends on R reciprocally: smaller the radius, the higher the water column drawn up.
 
Thanks but did you look at the picture in the link?

Maybe I should ask in steps: why does a smaller radius pull water more than a larger radius? If you look at the pictures, you'll see the drop of water is pulled toward the smaller end.

EDIT - in addition, the capillary is turned sideways so that gravity and weight of water is not a factor.
 
I'm sorry, I can not retrieve the picture.
 
There is an unbalanced force. There's two semi-independent effects; one of wetting and one of the pressure jump across the surface.

For the Hg in glass, since mercury does not wet the glass, it tries to minimize the amount of Hg in contact with the glass. In the picture, the water wets the glass, and so the water tries to maximize the contact area.

Now, becasue both ends of the fluid column contain (roughly) spherical caps of different diameters, there is an unbalanced pressure: Laplace's law equates the pressure jump (i.e. force per unit area) to the surface curvature.

For Hg, the interior pressure is higher than the exterior pressure, and so the smaller cap exists at a higher pressure and the fluid moves to the larger diameter region of the capillary.

For water, the interior pressure is lower than the exterior pressure, and so the smaller cap is at a lower pressure than the larger cap, and so the fluid is pulled into the smaller region of the capillary.

Clever idea, actually...
 

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