Question

A metal wire of resistance $6\Omega$is stretched so that its length is increased to twice its original length. Calculate its new resistance.

Answer 1

Step-1 Given data

Resistance of wire $R_1=6\Omega$

Let, the original length of wire $l_1=l$

Length of wire after stretched $l_2=2l$

Step-2 Write the relation between resistance and length of wire

We know that $R=\frac{\rho l}{A}$ Where $\rho$ is the resistivity of the material, $l$ is the length of wire and $A$is the area cross-sectional.

Let $V$is the volume of material.

$$\begin{gathered} V=Al \\ A=\frac{V}{l} \end{gathered}$$

Then, $R=\frac{\rho l^2}{V}$

Since the resistivity and the volume of the wire is a constant.

The resistance of a wire is directly proportional to the square of the length.

$R\propto l^2$

SO, $\frac{R_1}{R_2}={\left(\frac{l_1}{l_2}\right)}^2$

Step-3 Finding new resistance of the wire

$\frac{6}{R_2}={\left(\frac{l}{2l}\right)}^2$

$R_2=6\times 4$

$R_2=24\Omega$

Hence, the new resistance of the wire will be $24\Omega$.