Question

A student finds an unlabeled liquid container in his lab. He notices that the container has two liquids. Since the two liquids have different density, there is a distinct separation between them. Water, which has a density of 1.00g/cm^3, fills the lower portion of the container to a depth of 20.0 cm.The fluid which is floating on top of the water is 30.7 cm deep. If the absolute pressure on the bottom of the container is 104900 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 P0=101300 Pa.

Answer 1

`\rho = 1848.03 kg m^(-3)`

Step-by-step explanation:

given data:

density of water \rho = 1 gm/cm^3 = 1000 kg/m^3

height of water = 20 cm =0.2 m

Pressure p = 1.01300*10^5 Pa

pressure at bottom

`P =  P_(fluid) + P_(h_2 o)`

`P   = P_(fluid)  + \rho g h`

`P_(fluid)  = P - \rho g h`

= 1.01300*10^5 - 1000*0.2*9.8

= 99340 Pa

`p_(fluid)  = P_(atm) + \rho g h_(fluid)` h_[fluid} = 0.307m

`99340 = 104900 + \rho *9.8*0.307`

`\rho = 1848.03 kg m^(-3)`