Question

A ductile metal wire has resistance R. What will be the resistance of this wire in terms of R if it is stretched to three times its original length, assuming that the density and resistivity of the material do not change when the wire is stretched? (Hint: The amount of metal does not change, so stretching out the wire will affect its cross-sectional area.)

Short answer

The resistance becomes 9R after the stretching.

Step-by-step solution

Step 1:

Given data:

The wire is three times its initial length, therefore if the initial length is L, the final length will be$L_f=3L$

The material's density and resistivity $\rho$do not change, but area A does change.

Step 2:

To ascertain the final resistance $R_f$upon stretching because the density does not vary as the area changes, the volume of the rod does not change.

Therefore, the initial and the final volume is:

$$\begin{gathered} V=V \\ LA=L_f{A}_f \\ LA=3L{A}_f \\ {A}_f=\frac{A}{3} \end{gathered}$$

Area decreases by three times.

Now, the relation between the resistance at the initial state, the area A and length L;

$R=\frac{\rho L}{A}$

Resistance $R_f$at the final state;

$$\begin{gathered} R=\frac{\rho L_f}{A_f} \\ =\frac{\rho \left(3L\right)}{\left(A/3\right)} \\ =9\left(\frac{\rho L}{A}\right)=9R \end{gathered}$$

To maintain the volume constant, the area shrinks three times as the length grows by the same factor. Simultaneously, resistance will increase by a factor of nine.

Hence, the resistance becomes 9R after the stretching.