How to find the pressure at the bottom of a vessel with an area of 500 m2 in which there are 2 liters of liquid?

Given:

V = 2 liters = 0.002 cubic meters – the volume of liquid in the vessel;

S = 500 meters squared – the area of ​​the bottom of the vessel;

g = 10 m / s2 – acceleration of gravity.

It is required to determine P (Pascal) – the pressure of the liquid at the bottom of the vessel.

Since the condition of the problem is not specified, we assume that water is poured into the vessel as a liquid, and also that the vessel has a cylindrical shape.

Then, the height of the liquid column in the vessel is equal to:

h = V / S = 0.002 / 500 = 0.000004 meters.

The pressure will be equal to:

P = ro * g * h = 1000 * 10 * 0.000004 = 0.04 Pascal (where ro = 1000 kg / m3 is the density of water).

Answer: the pressure of the liquid at the bottom of the vessel is 0.04 Pascal.

Note: it is possible that an error was made in the problem statement when specifying the area of ​​the bottom of the vessel, since it is difficult to imagine a vessel with a bottom area of ​​500 square meters.