The vessels are filled with three layers of immiscible liquids – water, oil and mercury. The height of each layer is h = 10.

The vessels are filled with three layers of immiscible liquids – water, oil and mercury. The height of each layer is h = 10. Determine the total pressure p of the liquids at the bottom of the vessel and the pressure p1 at a depth of h1 = 15 cm.

Three layers of immiscible liquids are poured into the vessel – mercury, water and oil, having different densities ρ₁, ρ₂ and ρ₃, respectively. There will be a column of mercury in the lower part of the vessel, a column of water will be located above, and then a column of oil, the height of each layer is h = 10 cm = 0.1 m.The total pressure p of the liquids at the bottom of the vessel will be the sum of the weight pressures of all three liquids:
p = g ρ₁ h + g ρ₂ h + g ρ₃ h or p = g h (ρ₁ + ρ₂ + ρ₃), where the coefficient g = 9.8 N / kg, and the values ​​of the density of liquids are in the reference tables: ρ₁ = 13600 kg / m³, ρ₂ = 1000 kg / m³ and ρ₃ = 900 kg / m³. Substitute the values ​​of physical quantities into the calculation formula and find the total pressure p of the liquids at the bottom of the vessel:
p = 9.8 N / kg · 0.1 m · (13600 kg / m³ + 1000 kg / m³ + 900 kg / m³);
p = 15190 Pa.
To determine the pressure p₁ at a depth of h₁ = 15 cm = 0.15 m, it is necessary to take into account that the pressure p₁ will be created by the weight pressures of the mercury column and half of the water column: р₁ = g ρ₁ h + g ρ₂ h / 2 or р₁ = g h (ρ₁ + ρ₂ / 2). We get:
p₁ = 9.8 N / kg · 0.1 m · (13600 kg / m³ + 1000 kg / m³ / 2);
p₁ = 14308 Pa.
Answer: the total pressure p of liquids at the bottom of the vessel is ≈ 15.2 kPa, at a depth of h₁ the pressure will be p₁ ≈ 14.3 kPa.