A hollow aluminum sphere with outer diameter 10.0 cm has a mass of 690 g. What is the sphere's inner diameter?

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Hey, everyone, here's a problem that asks about a spherical copper shell. Uh The problem mentions that it's hollow. So you could imagine a cross section of the shell kind of looking like this where what I just drew, the larger circle is the outer diameter of the shell, but it's hollow. So there's also a bunch of emptiness inside. So this area that I'm shading here is kind of where the actual mass is. And the problem mentions that there is an inner diameter of 14 centimeters. And you're doing this using a mass of 1. kg. So M equals 1.4 K G. And we were asked to find the expected outer diameter of the shell. So we're looking for the out which is unknown to us. And lastly, the problem gives us the density of copper which is about kilograms per cubic meter. Now, in case it's a little unclear to you what exactly is happening in this problem or what it means when it talks about outer diameters and inner diameters. Well, as I mentioned earlier, this is a hollow shell. So that means there are two different radii that we have to work with, on the one hand, there's the distance from the center of the shell all the way to where you first hit material. And this would be called the inner radius. But there's also the length from the center of the sphere all the way to the outermost point of material. The point that touches the outside, we would call this the outer radius or are out. So that's where the two different types of radii come from. So what we generally want to do is relate these two radii to the volume of the sphere. So first recall that the volume of the sphere is equal to four thirds pi multiplied by the cube of the sphere's radius. And the thing is in this problem, we have two different radii to work with. So what we can do to represent the volume of the shell is use this sphere formula on the outer radius and then subtract what we'd get from the same formula. But with the inner radius. And the reason why this helps us is because we'll think about what this would do for the outer radius. This term will basically give us what the volume of the sphere would be if it was completely solid. But by then subtracting the volume we'd get from the inner radius that essentially takes away all of the volume from the empty space of the hollow shell. So this is an expression that will accurately give us the volume of just the shell itself. So let's expand it out a bit. So it's four thirds pi are out cubed minus four thirds pi R N cubed. And I'm gonna factor out the four thirds pi since they're common terms. So four thirds pi and then in parentheses are out cubed minus R in cubed. Let me fix that up a little bit. There we go. Now, as a side note, recall that the diameter of a sphere or a circle is equal to twice the radius. So two multiplied by the radius of the sphere. Now we're looking for the diameter, the outer diameter of the sphere. So the first thing we'll need to find then is the outer radius of the sphere. And the outer radius of the shell is one of our terms in this equation we have. So the next thing I'm gonna wanna do is take this equation and solve it for R sub out. We can do this using some basic algebra steps. We'll want to divide both sides of the equation by four thirds pi. So the fraction becomes flipped and then this V gets divided by pi and then two of that, we want to add R in cubed. So doing that, we end up with a formula saying that R out cubed is equal to three V over four pi plus are in cubed. And then lastly, we're going to want to take the cubed root of both sides of the equation so that we can get the R out term on its own without that cubed. Now note that if your calculator doesn't have a cubed root term, this is the exact same thing as putting something in parentheses and raising it to the power of one third. So that might come in handy for some of you knowing that little trick. But what we find is that R out is equal to the cube root of three V over four pi plus R N cubed. Now note that we technically have R M because based on what I've said earlier about the densities in the radius and we're in the, and the fact that we're given the inner diameter, we know that the radius is equal to the diameter divided by two. So we can take the inner diameter that we were given and divide it by two to get the radius that we're supposed to use here for R M V might be a little trickier though because we're not actually given the specific volume of the shell. We did find an expression for it over here with the V out minus A V N. But we're still gonna want to find a numerical value for it. If we want to plug it into this equation here, we can find it though with the other formula that relates volume with some other variables we're given because the general density formula tells us that density is equal to mass divided by volume. And if we algebraically solve this for volume, then we find that the volume is equal to the mass divided by the density. And if we look at the uh the problem statement, we're given both the mass and the density of the copper. So into a calculator, we'll plug in 1.4 kg for mass and then 8900 kg per cubic meter in for density. And if you put that into a calculator, then we find a volume of about 1.57 times 10 to the power of negative four cubic meters. So this is the actual volume of the shell which we can put in for our V right here. So now it's time to actually put that into our calculator. So for V we're gonna plug in this value we were given and then for our sub in, remember I mentioned, we just had to take the inner diameter we were given and then divide it by two. So 14 centimeters divided by two, that's just equal to seven centimeters. And then because we're looking for an answer in meters, judging by our multiple choice option, we're also gonna want to convert that into meters. So seven centimeters converted into meters is just 0.7 m. So that's the value we're going to plug in for our sub in. And then if we put that into our calculator, we're plugging in the cube root of three times 1.57 times 10 to the power of negative four divided by four pi plus the cube route of 0.7 m. So if we put that into a calculator, then what we find is an outer radius of about 0.725 m. Now this is the outer radius, but we're not done yet because remember the problem is asking for the outer diameter. So the last thing we need to do is make use of this formula right here and multiply the radius we got by two to get it as a diameter. So if we multiply this by 22 times 0. to five m, then we find an outer diameter of 0.145 m. And if we look at our multiple choice options, then you'll notice that we do indeed have that as an option. It's option B. So option B is the answer to this problem and that's it. I hope this video helped you out if you need help with this and you want to learn more, check out some of our other videos and find other practice problems that will help you. But that's it for now. And I hope you all have a lovely day. Bye-bye.

A hollow aluminum sphere with outer diameter 10.0 cm has a mass of 690 g. What is the sphere's inner diameter?

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Key Concepts

Density

Volume of a Sphere

Mass and Volume Relationship