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

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SUMMARY

The discussion centers on the relationship between the cross-sectional area and length of a wire during the process of wire drawing. It is established that reducing the cross-sectional area of a wire results in an increase in its length, assuming constant volume. The formula for mass, expressed as Mass = ρAL, where ρ is density, A is cross-sectional area, and L is length, supports this conclusion. The confusion arises from the interpretation of the area reduction and its direct impact on the wire's dimensions.

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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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