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 𝜌w=1.00×10

Question

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 𝜌w=1.00×103 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 0.335 mm deep. If the absolute pressure on the bottom of the container is 1.049×105 Pa, what is the density, 𝜌l, of the unknown fluid? The acceleration due to gravity is 𝑔=9.81 m/s2 and atmospheric pressure is 𝑃o=1.013×105 Pa

Answer 1

This question can be solved by using the formula

Ptotal = Patm + P1 + P2

where,

P1 = pressure of fluid 1

P2 = pressure of fluid 2

When then substitute the formula for pressure of a column of fluid

P = density*g*h

where

density is the density of the fluid

g = gravitational acceleration

h = height of the column of fluid

Therefore

Ptotal = Patm + density1*g*h1 + density2*g*h2.

Now solving for the unknown density

density2 = (Ptotal - Patm - density1*g*h1)/g*h2.

Don't forget to convert mm to meters

1 m = 1000 mm