You come across an open container that is filled with two liquids. Since the two liquids have different densities, there is a distinct separation between them. Water, which has a density of rhow=1.00103 kg/m3, fills the lower portion of the container to a depth of 0.218 m. The fluid that is floating on top of the water is 0.328 m deep. If the absolute pressure on the bottom of the container is 1.049105 Pa, what is the density, rhol, of the unknown fluid? The acceleration due to gravity is g=9.81 m/s2 and atmospheric pressure is P0=1.013*105 Pa.

Question

Ezequiel

Physics

9 March, 22:56

You come across an open container that is filled with two liquids. Since the two liquids have different densities, there is a distinct separation between them. Water, which has a density of rhow=1.00103 kg/m3, fills the lower portion of the container to a depth of 0.218 m. The fluid that is floating on top of the water is 0.328 m deep. If the absolute pressure on the bottom of the container is 1.049105 Pa, what is the density, rhol, of the unknown fluid? The acceleration due to gravity is g=9.81 m/s2 and atmospheric pressure is P0=1.013*105 Pa.

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Answers (1)

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Travis Key · Answer 1

455.33 kg/m³

Explanation:

The pressure at the bottom of the container is

P = P' + P'' + P₀

P' is pressure due to water

P'' is pressure due to unknown liquid

P₀ is the atmospheric pressure

P = 1.049 * 10⁵ Pa

P' = ρgh

⇒P' = (1000) (9.8) (0.218) = 2136.4 Pa

P'' = P-P'-P₀

⇒P'' = (1.049 * 10⁵-2136.4-1.013 * 10⁵) Pa = 1463.6 Pa

P'' = ρ'gh'

ρ' = (1463.6) / (9.8*0.328) = 455.33 kg/m³