A wire with a resistance of 6.0$\mathbf{\Omega }$is drawn out through a die so that its new length is three times its original length. Find the resistance of the longer wire, assuming that the resistivity and density of the material are unchanged.

1. Resistance of the wire before the change,${\mathrm{R}}_0=6.0\mathrm{\Omega }$
2. New length in relation to the original length,$\mathrm{L}=3{\mathrm{L}}_0$

The resistance of the wire is directly proportional to its length and inversely proportional to the area of the cross-section. The constant of proportionality is a characteristic of that material called resistivity.

We have to use the given relation between the original wire and the longer wire and the formula of resistance to find the resistance of the longer wire.

Formulae:

The resistance of the wire material, R = pL/A …(i)

Here, R is resistance, p is resistivity, L is the length, and A is the area of the cross-section of the wire.

Resistance is given as,

$R=\frac{pL}{A}$

Since the mass and density of the material do not change, the volume remains the same. If $L_0$and $A_0$are the original length area and if L and A are the new length and area respectively, then, considering the volume relation we get

$$\begin{gathered} L_0{A}_0=LA \\ A=\frac{L_0{A}_0}{L} \\ =\frac{L_0{A}_0}{3L_0} \\ =\frac{A_0}{3} \end{gathered}$$

Thus, the resistance of the longer wire using equation (i) is given as follows:

$$\begin{gathered} \mathrm{R}=\frac{\mathrm{p}\left(3{\mathrm{L}}_0\right)}{{\displaystyle \frac{{\mathrm{A}}_0}{3}}} \\ =\frac{9{\mathrm{pL}}_0}{{\mathrm{A}}_0} \\ =9{\mathrm{R}}_0 \\ =9\left(6.0\mathrm{\Omega }\right) \\ =54\mathrm{\Omega } \end{gathered}$$

Therefore, the resistance of the longer wire is$54\mathrm{\Omega }$ .