A hollow aluminum cylinder is 2.50 m long and has an inner...

A hollow aluminum cylinder is 2.50 m long and has an inner radius of $2.75 \mathrm{~cm}$ and an outer radius of $4.60 \mathrm{~cm} .$ 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?

Transcript

00:01 Okay, so we're given a hollow aluminum cylinder of length 2 .5 meters, inner radius of 2 .75 centimeters or 2 .75 times 10 to the minus 2 meters, an outer radius of 4 .6 centimeters or 4 .6 times than 10 minus 2 meters.

00:21 And we're told to treat each of the surfaces as an equipotential surface.

00:27 And in part a, you're asked to find the resistance between the two faces of the senator.

00:39 So the faces are this area right here, the circle.

00:47 And we're also told that an o meter is connected between the two faces.

00:52 So you have to find the resistance.

00:55 And to do this, you can use equation 2510.

01:01 Write down equation 2510, which tells us that the resistance is the resistivity times the length over the cross -sectional area.

01:18 We're already given the length.

01:22 We're told it's aluminum so we can find a resistivity, but we don't know the cross -sectional area.

01:33 However, we can easily find the cross -sectional area by keeping in mind that the part where there is no aluminum is in the inner radius.

01:48 So aluminum exists between the outer radius and the inner radius.

01:55 So kind of in this shaded region here, this is where the aluminum is.

02:03 So let's say rd, which is the difference between the outer and inner radius, that is the radius in which the aluminum lives.

02:22 So then we know the area is pi r d squared, which is pi times r o minus r i squared.

02:40 We're given the outer and inner radii in the problem already.

02:48 So then we can find the resistance.

02:51 It's a resistivity of aluminum, which is 2 .75 times 10 to minus 8 o meter times the length of the cylinder, which is 2 .5 meters, divided by the length of the slender, which is 2 .5 meters, divided by the length of the slender, by the cross -sectional area, which is pi times 4 .6 times 10 to minus 3 minus 2 .75.

03:39 That's meter times tens of the minus 3 meters square.

03:47 And when you plug in these numbers, you find that the resistance is 6 .39 times 10th minus three oms.

04:02 So that's it for part a.

04:04 Now we move on to part b...

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