A hollow aluminum cylinder is 2.50 m long and has an inner radius of 2.75 cm and an outer radius of 4.60 cm. Treat each surface (inner, outer, and the two end faces) as an equipotential surface. At room temperature, what will an ohmmeter read if it is connected between (a) the opposite faces and (b) the inner and outer surfaces?

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Hi everyone today, we are going to measure the resistance of the pipe when first measure along its length and second measure along its wall thickness. We're going to do this by assuming that each area will act like an every potential service. And it is known from the problem that the length The L of the pipe is going to be five m and the outer diameter is going to be 5. inch. We do not like inches. So we're going to deal with that after this. And then the inner diameter is 4.805 inches. Okay, so first I'm gonna first divide this into radius because that's what we're going to be dealing with. So I'm just going to divide this by two to make it, it's I'm gonna put our to ask the other radius and then I'm going to define this by two as well to put our one R two is then going to be 2.56- inches and our one is then going to be 2.40 to 5 inches like. So I am just gonna remove this over to just so that you guys are not confused with that value being the diameter instead. So now that we have the radius here or I'm just gonna do just divided or multiplied by half like so to get the radius and then now we want to convert this into a meter. So I'm just multiplying with 2.54 times 10 to the power of minus two m over one inch. Just like so and this will result to be 6.51 times 10 to the power of -2 - meter. And then we want to do the same thing with the inner diameter here to multiply a nearby 2.54 times 10 to the power of minus two divided by one inch. And this will be 6.1 times 10 to the power of minus two m. Just like. So okay, now that we pretty much have our inner and outer diameter. So what we want to do next is to actually start solving. So we know that the current that flows along the copper pipe is measured by connecting an zero m between the opposite faces. On the other hand, the resistance along the pipes, wall thickness is going to be measured by connecting the zero m along the wall thickness, radial e outward. So first we want to start with the first case. So in the case of a hollow cylinder, the resistance is going to be given by this formula. Right here, our will be equals to rho L over a right where in this case the row is going to be the receptivity. The L. Is going to be the length of the copper pipe. The A. Is going to be the pipes cross sectional area and in this case the A. Is going to be pi r squared where r is the radius of the pipe. However, for a hollow cylinder, which is the one that we have here. The A. Is then going to be pie the outer the honor the outer radius squared minus the inner radius squared just like so where are two is the outer radius? The R one is the inner radius. And now we can actually start plugging in all the formula or all the known values that we have into our formula of resistance resistance here. So for for a copper, the rest activity is going to be a constant of 1.72 times 10 to the power of minus eight. Multiplied by meter. Just like so so we can start with for part one are real equals to rho L over A. And Rio is going to be 1.72 times 10 to the power of minus eight row own multiplied by meter. And then the L. A. Is going to be five m which is known in this problem. And then the A. We're gonna substitute with this formula right here, A hollow cylinder is going to be pi R two square is going to be this first one, the outer diameter. So it's gonna be 6.51. Then stand to the power of -2 m squared minus 6.1 times 10 to the power of minus two m squared just like so and this will actually come out to be 5. times 10 to the power of minus five. Oh. So from this first part we know that the resistance of the pipe where measured first along its length is going to be 5.29 times 10 to the power of -50. So from here we can actually kind of like uh know that the answer is going to be B. But we want to check the second part first. Okay, so now let's consider a thin cylindrical shell of the copper pipe. So we know that the inner radius is gonna be our and the thickness is going to be D. R. So in this case I'm gonna write it down. So the thickness for a thin cylindrical shell of copper pipe thickness D. R. And in the radius or and then in this case because it is a cylindrical shell of the copper pipe and we want to measure it along its wall thickness. The A. Is going to be two pi R. L. Which is essentially the area of the cylindrical wall itself. Therefore we can manipulate our previous um resistance formula. So that for this case we want to start with using this formula right here, Over two pi r. l. Or it's pretty much just going to be row D. R. Over area, like so okay, so for the current to flow from R. One to R two, we wanna solve this by integrating each side going from R one to R. Two. So we want to integrate both sides of our equation. So doing that, we will pretty much get, I'm gonna write it down here integrating both sides, we will get uh integral of the R equals the integral of rho D R over two pi R L. And then I'm gonna remove all constants to one side outside of the inter girl. So this is going to be two pi l integral of one over R. Multiplied by the are just like so so the integral of D. R is just going to be our and then on the right side all the constants will remain the same just like so and then recall that integral of one over X D X is going to be the natural lock. So it's going to be the Ln of are in this case or the Ln of X. Don't forget that in this particular integral for the right side we are integral ng it from R one to R two, just like so so in this case I'm just gonna write it down that, recall that the integral of one over X the X is going to be the Ln of X. And in this case we're integrating it from R two to R one. So this part is going to be the Ln of R two -301, just like so okay, and then we can actually simplify this even further. The R is just going to be row two by L multiplied by the L N, combining R two minus R. One. Recall that L. N. B minus Ln A. Will equals two L. N. B over A. So this can be combined into L. N. R. Two over R. One just like so okay, so now that we get our formula that we want to use, we can just start plugging everything in just like so so the R. Is then going to be, There was activity which is going to be 1.72 times 10 to the part of -8. 1.72 times 10 to the power of minus 80. Multiplied by meter. And now we have two pi L which is two pi times five m. And then we have the L. N. Here and then of our to recall earlier we found our two to be 6.51 times 10 to the power minus two and R one to be 6.1 times 10 to the power of minus two Dumps them to the power of -2 m and then R one 6.1 them standard apart of -2 m also. Okay, so we have everything here and now the R can actually be found to be 3.56 times 10 to the power of -11. And that will be the answer for the second part. So the resistance of the pipe here when measured along its wall thickness is going to be 3.56 times that stand to the part of -110. So we can determine that the answer to our problem is going to be be with the resistance of the pipe measure along its line to be 5.29 times 10 to the power of minus five. While the resistance measure along its wall thickness is 3. times 10 to the power of minus 11 oh. So the answer is gonna be part B or option B. So if you guys have any more questions on this or any sort of confusion, make sure to check out our other videos regarding this topic and that will be all for this particular practice problem. Thank you.

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Key Concepts

Ohm's Law

Resistance of a Cylinder

Equipotential Surfaces