A U-tube is filled with water, and the two arms are capped. The tube is cylindrical, and the right arm has twice the radius of the left arm. The caps have negligible mass, are watertight, and can freely slide up and down the tube. A one-inch depth of sand is poured onto the cap on each arm. After the caps have moved (if necessary) to reestablish equilibrium, is the right cap higher, lower, or the same height as the left cap?
p = p_0 + pgd
The Attempt at a Solution
I'm slightly confused by this problem because my text doesn't really discuss closed liquid systems.
It seems that if the left cap is wider, yet they both have a 1" depth of sand poured in, that there would be more sand (more mass) on the left cap. Would that result in a greater surface pressure (p_0)?
I know that if the water is in hydrostatic equilibrium and if the container is open, then the pressure under both caps is the same and they would rise to the same height.
But since the u-tube is capped/closed would the left cap (larger diameter) sink lower than the right cap in order to reestablish equilibrium?