Determine the pressure of 0.6 m in water, kerosene, mercury?

Given:

H = 0.6 meters – the height of the liquid column;

ro1 = 1000 kg / m ^ 3 – water density;

ro2 = 800 kg / m ^ 3 is the density of kerosene;

ro3 = 13500 kg / m ^ 3 is the density of mercury;

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

It is required to determine the pressure P1, P2, P3 (Pascal) of different liquids.

The pressure of the water column is:

P1 = ro1 * g * H = 1000 * 0.6 * 10 = 1000 * 6 = 6000 Pascal = 6 kPa.

The pressure of the kerosene column is:

P2 = ro2 * g * H = 800 * 10 * 0.6 = 800 * 6 = 4800 Pascal = 4.8 kPa.

The pressure of the column of mercury is:

P3 = ro3 * g * H = 13500 * 10 * 0.6 = 13500 * 6 = 81000 Pascal = 81 kPa.

Answer: the pressure of a column of water is 6 kPa, a column of kerosene is 4.8 kPa, and a column of mercury is 81 kPa.