Question

A wire has a resistance of $10\Omega$. It is stretched by $10$% of its original length, what will be the new resistance?

• A. $10\Omega$
• B. $11\Omega$
• C. $9\Omega$
• D. $12.1\Omega$

Answer 1

Answer: (D)

$12.1\Omega$

Let resistance of a conductor be R, which is given by:
$R=\frac{\rho L}{A}$

Where $\rho$ is the resistivity of the material, $L$ is the length and $A$ is the cross-sectional area.

Assume the wire is cylindrical. then

Volume of a cylinder, $V=\pi r^{2}h$

Where $r$ is the radius and $h$ is the height.

Here, $A=\pi r^2$ and $h=L$

So, $V=AL$

While stretching the wire, the volume will remain the same.

$V=A^{\prime }{L}^{\prime }=A^{\prime }\left(1.1\right)L$ (stretching by 10%= 1 + 0.1 = 1.1)

Thus $A^{\prime }=\frac{V}{\left(1.1L\right)}=0.909A$

So new resistance, $R^{\prime }=\frac{\rho L^{\prime }}{A^{\prime }}=\frac{\rho \left(1.1L\right)}{\left(0.909xA\right)}=1.21R=12.1\Omega$