Question

71. A long, round wire has resistance$R$. What will the wire's resistance be if you stretch it to twice its initial length?

Short answer

The wire's resistance, if stretch to twice its initial length, is $4R$.

Step-by-step solution

Given information

A long, round wire has resistance$\mathrm{R}$. Resistance of the wire is $R=\rho \frac{L}{A},$where p indicates the resistivity, L is the length and A is the cross sectional area.

Explanation

In the initial state of the wire will be,
$R=\rho \frac{L}{A}$
In elongation, the volume is the quantity that doesn't switch, but the density accomplishes. This indicates that the product of the length with the cross-section surface area is stable. To discover the area after the stretching as,
$LA=L^{\text{'}}{A}^{\text{'}}\Rightarrow A^{\text{'}}=\frac{L}{L^{\text{'}}}A$
The resistance of the elongated wire evolves:
$R^{\text{'}}=\rho \frac{L^{\text{'}}}{A^{\text{'}}}=\frac{A}{A^{\text{'}}}\frac{L^{\text{'}}}{L}R$

Find the ratio from previous term,
$\frac{A}{A^{\text{'}}}=\frac{L^{\text{'}}}{L}$

Applying,

$R^{\text{'}}={\left(\frac{L^{\text{'}}}{L}\right)}^{2}R$

$R^{\text{'}}=2^{2}R$

$=4R.$

Therefore, The wire's resistance, if stretch to twice its initial length, is$4R$