A vessel with vertical walls with a bottom area of 10 cm2 was poured with 100 ml of mercury

A vessel with vertical walls with a bottom area of 10 cm2 was poured with 100 ml of mercury, and on top of it, 100 ml of water. How many times did the pressure on the bottom of the vessel increase?

Given:

S = 10 cm ^ 2 = 0.001 m ^ 2 – the area of ​​the bottom of the vessel;

V1 = 100 ml = 0.1 liter = 0.0001 m ^ 3 – the volume of mercury poured into the vessel;

V2 = 100 ml = 0.1 liter = 0.0001 m ^ 3 – the volume of water poured into the vessel;

ro1 = 13500 kg / m ^ 3 is the density of mercury;

ro2 = 1000 kg / m ^ 3 – water density;

g = 10 m / s ^ 2 – acceleration of gravity.

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

Since the volume of the poured liquids is the same, the height of the column of both liquids will be equal to:

h = V1 / S = 0.0001 / 0.001 = 0.1 meter.

Then the pressure of mercury will be equal to:

P1 = ro1 * g * h = 13500 * 10 * 0.1 = 13500 Pascal.

The pressure of the water column will be equal to:

P2 = ro2 * g * h = 1000 * 10 * 0.1 = 1000 Pascal.

The total pressure will be equal to:

P = P1 + P2 = 13500 + 1000 = 14500 Pascal = 14.5 kPa.

Answer: the pressure on the bottom of the vessel will be equal to 14.5 kPa.