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 0.221 m which has a density of 1.00 × 10^3 kg/m3. The fluid which is floating on top of the water is 0.349 m deep. If the absolute pressure on the bottom of the container is 1.049 × 10^5 Pa, what is the density of the unknown fluid? The acceleration due to gravity is g = 9.81 m/s2 and atmospheric pressure is P0 = 1.013 × 10^5 Pa.

Explanation / Answer

Given

depth of the fluid is h1 = 0.349 m , density rho1 =?

depth of the water is h2 = 0.221 m , density rho2 = 1000 kg/m^3

the absolute pressure on the bottom of the container is Pw = 1.049 × 10^5 Pa

atmospheric pressure is P0 = 1.013 × 10^5 Pa

we know that the pressure of the fluid is Pl =P0+rho_l*g*h1

the pressure on the water due to fluid on it is Pw = Pl+ rho_w*g*h2

here Pw = Pl+ rho_w*g*h2

Pw = P0+rho_l*g*h1 + rho_w*g*h2

rho_l*g*h1 = Pw - rho_w*g*h2 - P0

rho_1 = (Pw - rho_w*g*h2 - P0)/(g*h1)

= (1.049 × 10^5 -1000*9.8*0.221 - 1.013 × 10^5)/(9.8*0.349)

= 419.3322 kg/m^3

the density of the unknown liquid is rho_1 = 419.3322 kg/m^3

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