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

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The discussion centers on the relationship between the cross-sectional area and length of a wire during the process of wire drawing, where the volume of the wire remains constant. Reducing the cross-sectional area indeed increases the length of the wire, as the mass must remain constant according to the formula for mass (mass = density × volume). The confusion arises from the assumption that a reduction in area only affects the radius, while in reality, it also necessitates a proportional increase in length. The key takeaway is that as the cross-sectional area decreases, the length of the wire must increase to maintain a constant volume. Understanding this relationship clarifies the problem regarding resistivity and wire dimensions.
Ein Krieger
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Hello,

I was observing the solution of the following problem below:

765fa3055a66.png


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?
 
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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.
 
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.
 
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