Water rises in a capillary tube through a height h if the tube is inclined to the liquid surface at an angle the liquid will rise in the tube up to its length equal to
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The stretching of the air-liquid meniscus matches the pressure exerted by the liquid, not the mass of liquid in the tube.
In a closed tube of water the pressure at the top of the tube is P=P0−ρgh.
P=P0−ρgh
where h is the vertical distance (P0 is atmospheric pressure), So if we replace the closed end of the tubes by the air-water meniscus the curvature of the meniscus will be the same. It doesn't matter that there is a greater weight of water in the slanted tube. That's why the vertical height of the capillary rise is the same
In a closed tube of water the pressure at the top of the tube is P=P0−ρgh.
P=P0−ρgh
where h is the vertical distance (P0 is atmospheric pressure), So if we replace the closed end of the tubes by the air-water meniscus the curvature of the meniscus will be the same. It doesn't matter that there is a greater weight of water in the slanted tube. That's why the vertical height of the capillary rise is the same
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