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Problem 23 Easy Difficulty

A barrel contains a 0.120 $\mathrm{m}$ layer of oil of density 600 $\mathrm{kg} / \mathrm{m}^{3}$ floating on water that is 0.250 $\mathrm{m}$ deep. (a) What is the gauge pressure at the oil-water interface? (b) What is the gauge pressure at the bottom of the barrel?

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Video Transcript

this problem starts by telling us that we have a container with layers of two materials. Inside. We have water on the bottom, we're told the height of that layer and the density of water, and we have oil on top. We know the height of that layer and the density of it as well. What our job is our job is to find the gauge pressure at two points. Our first point is going to be between the two layers That would be right here and at the bottom of the container. Now, this is pretty simple. The only thing we need to know for this equation is that gauge pressure. That's the difference in the pressure in the container and our atmospheric. So kind of. The relative pressure is equal to the density of the material times the acceleration due to gravity times the height of that material of that liquid, Um so to find it between layers were looking at Onley underneath the oil light layers. That's pretty simple. We know that our pressure is going to be equal to density times acceleration due to gravity times the height and again we're only caring about the material above us. That's on Lee the oil. The pressure is going to be equal to the density over oil which is 600 kilograms per cubic meter times acceleration due to gravity which is 9.8 meters per second squared times the height of that material 0.12 meters on what we can do Once we calculate that we will get our gauge pressure and our answer is going to be sorry. Our answer is going to be 705.6 and that unit is in Pascal's. So that is our gauge pressure between layers just underneath the oil and then at the bottom of the container. What we're going to dio is we're going to have the pressure from the oil which we already know is 705.6 past scales. We're going to need to add the pressure from the water which we do the exact same way. We're going to need the density of water 1000 kilograms per cubic meter, times acceleration due to gravity times the height of that water. We calculate that out. The pressure from the water is equal to 24 50 Pascal's, we need to add the pressure from the oil as well. So our total is 31 55.6. Past scales is our gauge pressure at the bottom of the container. If we wanted to, it does look like for these distances. We were given three sig figs in those distances. Weaken around these 236 figs. For the first, that would be 706 Pascal's, and for this one we would have 31 60 Pascal's.

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