A tall cylindrical beaker 10 cm in radius is placed on a picnic table outside. You pour 5 L of an 8,000 kg/m 3 liquid and 10 L of a 6,000 kg/m 3 liquid into. Calculate the total pressure at the bottom of the beaker. (Use g =10 m/s 2 .)

Everyone. So in this problem, we're gonna calculate the total pressure when we have 2 liquids, but there's something a little bit different about this problem. Hopefully, you guys try it on your own. Let's check it out. So we have a tall cylindrical beaker that's 10 centimeters in radius. So in other words, the radius of this little cylinder over here, I'm gonna say this is r is equal to 0.1. We're gonna mix 2 different liquids inside, right? So So we're gonna pour 5 liters of an 8,000, density, liquid and then 10 liters of a 6,000. Now remember, when you have 2 different densities, the higher one's going to settle towards the bottom. So in other words, this density over here is going to be 8,000. This volume here is going to be 5 liters and this one is gonna be 6,000 and this one's gonna be 10 liters. Right? Notice how it's kind of scaled. So what I'm just gonna do is I'm gonna call this 1 1 and this one 2 just to make it a little simple. Alright, so we're gonna calculate the total pressure at the bottom of the beaker. So in other words, the the place where we're trying to figure out the pressure is right here. I'm gonna call this Pbottom. Alright. Now in order to figure out Pbottom, I have to use my pressure equation. So we're gonna start off with a Ptop is equal to I'm sorry, Pbottom is equal to Ptopplusrowgh. Now there's 2 places where you could pick to be your ptop. Right? So basically the the very top of both of the liquids, or you could pick the red green boundary. Either one of those would work, except what happens is that we don't know what the pressure is at the red green liquid boundary. And so if we wanna calculate the other pressure, remember, we need to compare this to a pressure pressure that we know. So we're actually not going to use the boundary and instead what we're going to do is we're basically just going to start all the way from top and go through both liquids. Okay, so this is really where you're going to measure your P top from. And the reason this is advantageous, the reason you want to do this is because you know something about this P top, which is that it touches air and so therefore, this P top is really just equal to the standard air pressure around it. Alright. So this P top here is just the air pressure, so we know that. And so what happens here is if we're going from the top all the way down, there's actually 2 liquids that we're going through. We're gonna go all the way down the red liquid and then we're gonna have to cross the boundary and then go all the way down the green liquid. So we have h one and then we have h two. Because I'm going through 2 different liquids and their different densities, we actually can't just use one rho g h term. We have to use both of them. Alright? Remember that you can go from the top of the liquid all the way down to the bottom. You just have to have a rho g h term for every single liquid that you encounter. That's a different density. Alright? So just make this super simple. We're gonna have to do, super clear. We're We're gonna have to go through the first liquid, which is row 1gh1, plus and then row 2gh2. Alright? So we have that p bottom here. If you take a look at your variables, we know this p top here is gonna be. We know what the densities of both of the liquids are. The only thing that we don't have in this problem is we don't have the heights. Normally, in these problems, you do have the heights, but in this one, we actually don't. The only thing we know about these problems or about these liquids is their volumes. So somehow we have to go from height and we have to get sorry, somehow we have to go from volume and then get this basically to height. Now you might be thinking that you could use, the density equation. You might be thinking that you could use row equals mass over volume because we have volume. But this is only gonna help you solve for the mass, not the height. So that's not really gonna be the equation that you should use. Alright, so we're not gonna use rho, mass over volume. Instead, we're gonna use the volume equation. Remember that the volume of a cylinder, is really just the cross sectional area of the cylinders. In other words, the area of the circle thing like this. This is gonna be the area times the height, right? So that's the volume of a cylinder and the area of a circle is really just gonna be area circle equals Pi R squared. So I'm going to use Pi big R squared. So in other words, it's just Pi big R squared times the height over here. So this is how we're gonna get to the height here. Notice how we have an h term. So basically, here's what's going on here. So h one is going to equal the volume divided by Pi R squared. Just going right from this equation over here, right? If V cylinder equals Pi R squared times h, you can rearrange and solve for the heights. So basically what you can do is you can model both of these liquids as if they're just cylinders that are sort of stacked on top of each other. So h one is gonna be a v over pi r squared. You're gonna get, 4 you're gonna get 10 liters divided by Pi. This is gonna be 0.1 squared. Now it's important to sort of keep track of the units here. I remember, so you always want to keep track of units. This is going to be meters. This is going to be meters squared and this is going to be liters. Alright. Now what happens here is that I can't just go ahead and cancel these things because I have a liter over a meter cubed, and the liters is an SI. Right? So what I'm gonna have to do is I'm gonna have to, sort of rewrite this using a conversion factor that I know. One of the conversion factors that is really, really useful to know is that, 1 meter cubed is equal to I've got 10,000 liters. So So I wanna cancel out the liters and then basically, what you're ending up with is you're ending up with meters on top of meters that you can cancel. Alright? So when you work this out, you're gonna get here is you're gonna get a height of 0.32, and this is gonna be in meters. Alright. So that's the height of basically h one. Now you're gonna do the exact same thing. I'm gonna do a little bit faster with h 2. This cylinder here, we can model this as a cylinder. So in other words, it's gonna be v 2 divided by pi r squared. So, in other words, this is just gonna be 5 divided by pi times 0.1 squared. And rather than go through all the math again, basically, you can just do one over 1,000. You plug in all the units, you'll see that they cancel. One thing you can see here is that the numerator the denominators of both of these fractions are gonna be the same. So the only thing that's different about this is that this one is half of the other one that we calculated. So instead of getting 0.32, you're gonna get 0.16 meters, and that's gonna be your 2nd height. Okay, so just kinda like using a little shortcut there, but that's your h's. You got h one and h two. So now I'm just gonna put it all together into my pressure equation. So p p bottom is gonna equal, p top, which is air, which is gonna be a 101,000 plus, and then we've got, this is gonna be 6,000 times 9.8 times, and then we're gonna use, 0.32 plus and now we've got the other density, the other liquid at the bottom, which is gonna be your 8,000 times 9.8 times your 0.16. Now when you pull this all together in your calculator, you plug this whole mess in, what you're gonna get for the pressure at the bottom of this cup is you're gonna get a pressure of 1.33 times 10 to the 5th, Pascals. Alright? So we can actually see that even in just a small little container like this, right, with 10 liters and 5 liters, it's not a whole lot of liquid. The pressure actually changes significantly here, and that's because the densities of these materials, 6,008,000, are actually relatively dense. Right? They're like 6 and 8 times denser than fresh water. So it doesn't take a whole lot of liquid actually to change the pressure very drastically and only a certain amount of heights. Alright? So that's it for this one folks. Let me know if you have any questions.

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19. Fluid Mechanics

Intro to Pressure

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Intro to Pressure

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Multiple Choice

A tall cylindrical beaker 10 cm in radius is placed on a picnic table outside. You pour 5 L of an 8,000 kg/m³ liquid and 10 L of a 6,000 kg/m³ liquid into. Calculate the total pressure at the bottom of the beaker. (Use g=10 m/s².)

3.20×10⁴

3.52×10⁴

1.33×10⁵

1.36×10⁵

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Verified step by step guidance

First, calculate the volume of each liquid in cubic meters. Since 1 L = 0.001 m³, the volume of the first liquid is 5 L = 0.005 m³ and the volume of the second liquid is 10 L = 0.010 m³.

Next, calculate the mass of each liquid using the formula: mass = density × volume. For the first liquid, mass = 8000 kg/m³ × 0.005 m³. For the second liquid, mass = 6000 kg/m³ × 0.010 m³.

Determine the total height of the liquid column in the beaker. The beaker has a radius of 10 cm, which is 0.1 m. Use the formula for the volume of a cylinder: volume = π × radius² × height. Rearrange to find height: height = volume / (π × radius²). Calculate the height for each liquid separately and add them together.

Calculate the pressure at the bottom of the beaker using the formula: pressure = (density × g × height). Calculate the pressure contribution from each liquid separately and then add them together to find the total pressure at the bottom of the beaker.

Finally, add the atmospheric pressure to the calculated pressure at the bottom of the beaker, if required. In this problem, we assume the beaker is open to the atmosphere, so atmospheric pressure should be considered if not already included in the options.