# Couldn't understand the way the problem is solved regarding resistivit

1. Dec 8, 2013

### Ein Krieger

Hello,

I was observing the solution of the following problem below:

Below it says that reducing the cross-sectional area of the wire triples its length. However, according to cross-sectional area equation A=pi*r^2 we observe that reduction of area only leads to reduction of radius .

Is there a mistake or I have some gaps in knowledge regarding above mentioned relationship?

2. Dec 8, 2013

### AlephZero

It says the wire is drawn out to reduce it cross section area. See http://en.wikipedia.org/wiki/Wire_drawing if you don't know what "wire drawing" means.

Assume the volume of the wire is constant. As you reduce the cross section area, you increase the length.

3. Dec 8, 2013

### Integral

Staff Emeritus
In the original configuration there is some fixed mass of wire. After stretching will there not be the same mass of wire? You need to make the wire longer to compensate for the loss in radius.

The wires mass before+
mass= ρ*Volume. the volume = A*L
so
Mass = ρAL

I will now let you find the length after stretching.