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If there is a wire of resistance $R$ and we stretch it such that it becomes 2 times longer, then what should be the new resistance of the wire?

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    $\begingroup$ I guess you mean "stretch it to 2 times the lenght". $\endgroup$ – Steeven May 9 '15 at 16:38
  • $\begingroup$ ya....I mean to stretch it 2 times its length $\endgroup$ – Shubham Prateek May 9 '15 at 16:39
  • $\begingroup$ When you stretch a wire, its length and cross section will change. There are other effects as well but first investigate these. $\endgroup$ – Azad May 9 '15 at 17:10
  • $\begingroup$ i think ther eare several aspects to that problem, from simple ($volume=const$) to such considering strain and complex deformations. OP should provide context and some amount of work for this question to be OK for Physics.SE $\endgroup$ – aaaaa says reinstate Monica May 10 '15 at 13:27
  • $\begingroup$ I think in this case only the length and the cross sectional area is going to change. $\endgroup$ – Shubham Prateek May 11 '15 at 4:56
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Resistance is directly proportional to the length of the wire, and inversely proportional to the cross sectional area of the wire.

R = pl/A, where R is resistance, p is the material's resistance in ohms, l is the length, and A is the cross sectional area in m^2.

As a wire gets longer its resistance increases, and as it gets thinner its resistance also increases because its cross sectional area decreases.

Doubling the length will double the resistance, but the wire also must get thinner as it is stretched, because it will contain the same amount of metal in twice the length. The volume of a cylinder is length * cross sectional area, but in order to find the new cross section, you need to consider what the wire is made of. Most materials resist a change in volume more than they resist a change in shape, and because of that, they lose less volume than otherwise would be expected when stress is applied.

In the case of the stretched wire, its density will be less after stretching than it was before. This means that the volume of space occupied by the metal in the wire expands, due to the cross sectional area not shrinking in proportion to the stretch of its length. The amount of reduction in cross section is determined according to Poisson's ratio (http://silver.neep.wisc.edu/~lakes/PoissonIntro.html)

Assuming the wire is made of copper, Poisson's ratio is about .355, which is the ratio of strain in the cross sectional area to strain along the length of the stretched copper wire. Without considering Poisson's ratio, one would expect the area of the cross section to be halved, and the final volume to remain constant, but this would be wrong.

Another consideration is the effect on the wire's conductivity by cold-working the metal (assuming it is stretched without heating it). When the metal is put under strain, inter-atomic distances will change. This is called the piezo-electric effect, and it's used in strain gauges. I saw one study of this effect in which the resistance in a thin copper wire doubled.

As the cross sectional area of the stretched wire is less that of the original wire, then in addition to doubling the resistance by doubling the length, you further increase the resistance by reducing the cross section. The piezo-electric effect yet further increases the resistance. Thus, there are three ways resistance in the stretched wire increases: (1) Doubling the length, (2) Reducing the cross sectional area according to Poisson's Ratio, and (3) The piezo-electric effect.

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  • $\begingroup$ Cross section won't decrease by 1/2 read about poisson ratio $\endgroup$ – Azad May 9 '15 at 17:59
  • $\begingroup$ @Azad: Thank you for this correction. I'm editing the answer. $\endgroup$ – Ernie May 9 '15 at 18:27
  • $\begingroup$ Are you assuming that the resistivity of the copper does not change when the copper is worked by being drawn? $\endgroup$ – DJohnM May 9 '15 at 19:49
  • $\begingroup$ @User58220: When copper wire is stretched, the resistivity increases by a significant amount. One test of cold-worked wire resulted in the resistance doubling. This may be due to changes in the inter-atomic distances, which can be expected of a metal under strain. I'm editing the answer for this. Thank you for pointing it out. $\endgroup$ – Ernie May 9 '15 at 20:43
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    $\begingroup$ downvote as this is an answer to bad question $\endgroup$ – aaaaa says reinstate Monica May 10 '15 at 13:25

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