What is the Resistance of a Cylinder Conductor Made of Aluminum?

[JSGandora]

Homework Statement

A cylindrical conductor of radius 2 mm and length 2.33 m is made of aluminum (p=2.75E-8$$\Omega$$m). Calculate the Resistance of the conductor.

Homework Equations

Unknown (not given).

The Attempt at a Solution

I purely do not know how to do it. I'm thinking that we find the volume of the conductor then multiply it by the 2.75 but I don't think that works.

 

[gneill]

Look up "resistivity".

 

[tiny-tim]

Welcome to PF!

Hi JSGandora! Welcome to PF!

(have a rho: ρ and an omega: Ω )

Look at the units … Ω.m, not Ω.m³

(actually, that m is really m²/m, ie area/length )

 

[JSGandora]

Thank you! :)

I got the following equation:
$$R=\frac{\rho L}{A}$$
where R is the resistance, p is the resistivity (by the way, what is the symbol for the "p"-like symbol called?), and A is the cross sectional area. Therefore, we have
$$R=\frac{2.75\Omega m\times10^{-8}(2.33m)}{(0.0002m)^2\pi}\approx0.51\Omega$$?

The answer key says the answer is $$0.0051\Omega$$. It looks like I am wrong by a factor of 100. I can't see why though...hmm...

 

[tiny-tim]

erm … too many 0s in your 2mm !

(and of course, it's called "rho")

 

[JSGandora]

OHHH, failure on my part. Thanks for pointing that out, I thought there were 100 millimeters in one centimeter when I did that. ><