Question

If the length of a copper wire is increased by $5\%$ due to stretching. What will be the percentage increase in its resistance?

• A. $12\%$
• B. $11.5\%$
• C. $20.25\%$
• D. $10.25\%$

Answer 1

The correct option is D

$10.25\%$

Resistance, $R=\rho \frac{L}{A}$
where:
$\rho$: Resistivity,
$L$: Length of the wire
$A$: Cross-sectional area of the wire.

The length of the copper wire is increased by $5\%$.

So, new length is:
$L^{\prime }=L+\frac{5}{100}L=1.05L$

Let the new cross-sectional area be $A^{\prime }$.
On stretching the wire, its volume will remain constant.
$⟹LA=L^{\prime }{A}^{\prime }$
$A^{\prime }=\frac{A}{1.05}$

New resistance will become:
$R^{\prime }=\rho \frac{L^{\prime }}{A^{\prime }}$
$R^{\prime }=\rho \left(\frac{1.05L}{A/1.05}\right)$
$R^{\prime }={1.05}^{2}R=1.1025R$

Percentage increase in its resistance is,
$p=\frac{R^{\prime }-R}{R}\times 100$
$p=\frac{(1.1025-1)R}{R}\times 100$
$p=10.25\%$