Question

If a wire of resistance 1$\Omega$ is stretched to double its length, then what will be the resistance of the new wire?

Answer 1

The expression for the resistance of a wire is given as,
$R = \rho \dfrac{l}{A}$
Where,
$\rho$ is the resistivity of a material.
l is the length of wire.
A is the cross-sectional area.
Upon stretching the length of the wire will be doubled while the area will be halved.
The new length will be: $l\prime {\text{ }} = {\text{ }}2l$, put this in the above formula and find the new resistance.

Complete step by step answer:
Here, on stretching the length of the wire:
the volume of the wire will remain constant.
the length of the wire will increase.
Area of the cross section will decrease.
Resistivity of wire will remain constant.
And the area of the cross section of the wire will decrease such that the volume of the wire will remain the same.
Using this fact, the formula for resistance of a wire can also be written as:
$R = \rho \dfrac{l}{A} \times \dfrac{l}{l} = \rho \dfrac{{{l^2}}}{V}$ ($\because$area $\times$ length = volume i.e. A x l = V) --(i)
Now, if the length of the wire gets doubled.
The new length of wire will be,
l′ = 2l

Now, the new resistance of the wire is given as:
${R_{new}} = \rho \dfrac{{{{\left( {{l'}} \right)}^2}}}{V} = \rho \dfrac{{{{\left( {2l} \right)}^2}}}{V} = \rho \dfrac{{4{l^2}}}{V}$
${R_{new}} = 4\rho \dfrac{{{l^2}}}{V} = 4R$ ($\because$ from (i) $R = \rho \dfrac{{{l^2}}}{V}$ )
$\therefore$ The new resistance is four times the previous resistance.
Hence, the resistance of the new wire is 4 $\times$1$\Omega$ = 4$\Omega$.

Note:
Resistance: Resistance is defined as the property of a material to resist the flow of the current flowing in the conductor. The SI unit used for the measurement of resistance is ohms$\left( \Omega \right)$.
By ohm’s law resistance can be calculated as:
$R = \dfrac{V}{I}$, where V is a potential difference across the wire and I is currently flowing through it.