# Hollow aluminum cylinder

1. Jul 19, 2010

### delve

1. The problem statement, all variables and given/known data

A hollow aluminum cylinder is 2.5 m long and has an inner radius of 3.20 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 the inner and outer surfaces?

2. Relevant equations

R=[(resistivity of hollow cylinder)*(length of hollow cylinder)]/Area

V=IR

3. The attempt at a solution

I was given the resisitivity of the aluminum cylinder and the length, but I am at a loss: the area inner part of the cylinder is less than the area of the outer part of the cylinder. My question is this: I don't know what area to use: do I use the area of the outer cylinder or the inner cylinder. Thanks.

2. Jul 20, 2010

### Onamor

A is the cross-sectional area of the current flow.
Think about why both the length and A are in the equation for resistance. Imagine their effects on a microscopic level.

Can you do it now?