There are two immiscible liquids in the vessel. The total height is 18 cm, and their masses are the same. What pressure do they exert on the bottom of the vessel (p = 800 kg / m3, p2 = 1000 kg / m3)?
h = 18 centimeters = 0.18 meters – the total height of the columns of two immiscible liquids;
ro1 = 800 kg / m3 (kilogram per cubic meter) – density of the first liquid;
ro2 = 1000 kg / m3 is the density of the second liquid;
g = 9.8 N / kg (Newton per kilogram) – acceleration of gravity;
m1 = m2 – the masses of liquids are the same.
It is required to determine P (Pascal) – what is the pressure of the liquid on the bottom of the vessel.
Find the ratio of the columns of liquids:
m1 = m2;
ro1 * V1 = ro2 * V2;
ro1 * s * h1 = ro2 * s * h2;
ro1 * h1 = ro2 * h2;
h1 = ro2 * h2 / ro1;
h1 = 1000 * h2 / 800 = 1.25 * h2.
Then we get:
h1 + h2 = h = 0.18;
1.25 * h2 + h2 = 0.18;
2.25 * h2 = 0.18;
h2 = 0.18 / 2.25 = 0.08, h1 = 0.1.
Then the pressure will be equal to:
P = P1 + P2 = ro1 * h1 * g + ro2 * h2 * g = g * (ro1 * h1 + ro2 * h2);
P = 9.8 * (800 * 0.1 + 1000 * 0.08) = 9.8 * (80 + 80) = 9.8 * 160 = 1568 Pascal.
Answer: the pressure of liquids at the bottom of the vessel is 1568 Pascal.