You come across an open container which is filled with two liquids. Since the tw

Question

You come across an open container which is filled with two liquids. Since the two liquids have different density there is a distinct separation between them. Water fills the lower portion of the container to a depth of 0.230 m which has a density of 1.00 times 10^3 kg/m^3. The fluid which is floating on top of the water is 0.335 m deep. If the absolute pressure on the bottom of the container is 1.049 times 10^5 Pa, what is the density of the unknown fluid? The acceleration due to gravity is g = 9.81 m/s^2 and atmospheric pressure is p_0 = 1.013 times^5 Pa.

Explanation / Answer

weight of the water = density * volume * g

volume = pi * radius^2 * depth

weight of the water = 1 * 10^3 * pi * radius^2 * 0.23 * 9.8

weight of the fluid = density * pi * radius^2 * 0.335 * 9.8

total weight = 1 * 10^3 * pi * radius^2 * 0.23 * 9.8 + density * pi * radius^2 * 0.335 * 9.8

pressure = weight / area

pressure = (1 * 10^3 * pi * radius^2 * 0.23 * 9.8 + density * pi * radius^2 * 0.335 * 9.8) / (pi * radius^2)

pressure = (1 * 10^3 * 0.23 * 9.8 + density * 0.335 * 9.8)

pressure at bottom = pressure + P0

1.049 * 10^5 = (1 * 10^3 * 0.23 * 9.8 + density * 0.335 * 9.8) + 1.013 * 10^5

density = 409.991 kg/m^3

density of the fluid = 409.991 kg/m^3

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