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What will be the change in the resistance of a Eureka wire, when its radius is halved and length is reduced to one-fourth of its original length?

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Last updated date: 14th Jun 2024
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Answer
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Hint: As a first step, you could recall the expression of resistance in terms of conductor’s length and area of cross section. Or you could easily derive expression from the proportionality relation of resistance with length and area. You could first express resistance as per initial conditions and then express by making changes in its dimensions as per the question. Then you could compare the two and get the answer.

Formula used: Expression for resistance,
$R=\rho \dfrac{L}{A}$

Complete step by step answer:
Eureka wire is the wire made of copper-nickel alloy that consists of 55% copper and 45% nickel. The main specialty of this wire is its low thermal variation of its resistivity which is also constant over a range of temperatures. It is known to have a resistivity of $4.9\times {{10}^{-9}}\Omega m$ which is high enough in achieving suitable resistance values in even very small grids. Also, its temperature coefficient of resistance is fairly low.
Here, in the question we are given a Eureka wire and we are asked to find the change in resistance of the wire when its radius is halved and its length is reduced to one-fourth of its original length.
We know that, the resistance R is directly proportional to the length, that is,
$R\propto L$ ………………………. (1)
The resistance is also inversely proportional to the area of cross-section, that is,
$R\propto \dfrac{1}{A}$ ……………………… (2)
Combining equations (1) and (2), we have,
$R\propto \dfrac{L}{A}$
$\Rightarrow R=\rho \dfrac{L}{A}$ ………………………… (3)
Where, ρ is the constant of proportionality called resistivity. The resistivity of a material doesn’t depend on its dimensions but depends on its material.
Let L and A be the initial length and area of the wire respectively, then,
$A=\pi {{r}^{2}}$
The resistance of the wire initially is given by,
$\Rightarrow R=\rho \dfrac{L}{\pi {{r}^{2}}}$ ……………………………… (4)
When the length of the wire is made one-fourth of its initial length and radius is reduced to half, that is,
$\Rightarrow L'=\dfrac{L}{4}$
$\Rightarrow A'=\pi {{(r')}^{2}}=\pi {{\left( \dfrac{r}{2} \right)}^{2}}=\dfrac{\pi {{r}^{2}}}{4}$
Resistance now becomes,
$R'=\rho \dfrac{L'}{A'}=\rho \dfrac{\dfrac{L}{4}}{\dfrac{\pi {{r}^{2}}}{4}}$
$\Rightarrow R'=\rho \dfrac{L}{\pi {{r}^{2}}}$ …………………………. (5)
Comparing (4) and (5), we see that the resistance of the wire is the same even after changing the dimensions as per the given question.

Hence, when the radius is halved and length is reduced to one-fourth of its original length the resistance still remains constant.

Note: Resistance basically is the opposition offered to the flow of electric current. The SI unit of resistance is Ω. But, resistivity of a material quantifies how strongly a material resists or conducts. That is, low resistivity means that the material readily allows the electric current flow. SI unit of resistivity is Ωm.