Water rises to the same level, no matter what the shape of the container is.

This demonstration is actually an illustration of what is known as the hydrostatic paradox, which is explained by Pascal’s principle, which is also called Pascal’s law, and which is usually stated: Pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and the walls of the containing vessel. If we imagine a vessel filled with liquid, either open to the atmosphere or closed with a piston pressing down on the top of the liquid, the pressure at any point within the liquid equals the sum of the external pressure (exerted either by the atmosphere or by the piston), which we can call p0, and the pressure exerted by the liquid between the top surface and that point, which is merely the weight of the liquid above unit area at that point, or ρgh, or p = p0 + ρgh, where ρ is the density of the liquid, g is the acceleration of gravity, and h is the height of the liquid above the point in question. If we now change the pressure on the surface, either by placing the vessel inside a chamber whose pressure we can make different from that of the atmosphere, or by changing the force on the piston, by Δp0, since liquids are essentially incompressible, the density does not change, and the new pressure at the same place in the liquid is just p + Δp0. Strictly speaking, when we change the pressure at one place in a liquid, there is an instantaneous density change, which propagates through the liquid as a wave at the speed of sound in that liquid. Once this wave has died away, the density throughout the liquid is again uniform, and Pascal’s principle holds. Pascal’s principle also holds for gases, except that it is necessary to take into account the changes in volume that occur when one changes the pressure on a confined gas.

We can see why the pressure in a fluid depends on the height of the fluid column, but not on the shape of the vessel in which it is contained, when we consider that wherever the fluid is in contact with the wall of the vessel, the force between it and the wall is perpendicular to the surface of the wall. It might be tempting to think that the greater volume, and thus greater weight, of water in the funnel at right in the photograph might force water into any of the other vessels that contained a smaller volume, and thus weight of water. The wall of the funnel, however, slopes, so that the forces it exerts on the liquid can be resolved into horizontal and vertical components. So outside the cylindrical column above the stem of the funnel, the vertical components support the weight of the liquid above the wall, and the pressure at the base of the funnel is due only to the weight of this cylindrical column. Whatever the shape of the vessel, the components of the forces of the vessel wall on the fluid add in such a way that at any point within the vessel, the pressure depends only on the surface pressure, gravity, density of the fluid, and the height of the column of fluid above that point.

Another way of posing the hydrostatic paradox is to consider three vessels, as shown in figure 17-11, for Question 4 in Chapter 17 of Physics, Part 1, by Robert Resnick and David Halliday (New York: John Wiley and Sons, 1977), p.379. Each vessel has the same area on the bottom. The first is shaped like a beaker with the wall sloping outward from the bottom, the second is shaped like an Erlenmeyer flask (wall sloping inward from the bottom), and the third is shaped like a regular beaker (vertical wall). All are filled to the same level with water. Since each is filled to the same level, the pressure at the bottom of each is the same, and since the area of the bottom of each is the same, then the bottom of each vessel must experience the same force (pressure multiplied by area). The question is why each of them has a different weight when you put them on a scale.

For the third vessel, the straight-walled beaker, all of the water is above the bottom, so the forces between the fluid and the wall are completely horizontal, and the forces against the bottom are purely vertical. The downward force of the water against the bottom of the beaker is just the weight of the cylindrical volume above it, so when you put it on a scale, the scale reads the weight of the beaker plus the weight of the water inside. The pressure is the weight of the water divided by the area of the bottom of the beaker.

The first vessel (with outward-sloping wall) is similar to our funnel. The water within the cylindrical volume above the bottom exerts a force on the bottom of the vessel, which is the same as the weight of the water in the straight-walled beaker. Outside this volume, the walls exert a force that has a vertical component that supports the weight of the water. Thus, the force on the bottom of the vessel is the same as that on the bottom of the straight-walled beaker, but the extra volume between the sloping wall and the cylindrical volume above the bottom, exerts an additional downward force on the vessel wall. So, assuming that the weights of the two vessels when empty are not significantly different, when they are filled with water to the same level, the one with outward-sloping walls weighs more than the straight-walled one. Even though the volume of the vessel with outward-sloping walls is greater than that of the straight-walled vessel, the pressure at the bottom of each is the same.

The second vessel, with inward-sloping wall, has a smaller volume than the cylindrical volume of the straight-walled vessel. The inward-sloping wall, however, in confining the water to this smaller volume, exerts forces on the water that now have downward vertical components. The cylindrical volume of water below the surface (which is now a smaller circle than the bottom of the vessel) is merely the weight of this volume of water, but outside this volume, the vertical components of force contributed by the vessel wall add to the weight of the water. The total force on the bottom of the beaker is thus the same as for the other two vessels, and so is the pressure. Since the volume of water is smaller than that in the straight-walled vessel, again assuming that the weights of the empty vessels are not significantly different, the vessel with the inward-sloping wall weighs less than the straight-walled vessel when you weigh them on a scale.

References:

1) Resnick, Robert and Halliday, David. Physics, Part One, Third Edition (New York: John Wiley and Sons, 1977), pp. 376, 379.
2) Sears, Francis Weston and Zemansky, Mark W. College Physics, Third Edition (Reading, Massachusetts: Addison-Wesley Publishing Company, Inc., 1960), pp. 240-242.