Question

A piece of wire of resistance $20\Omega$ is drawn out so that its length increases to twice its original length. Calculate the resistance of the wire in the new situation.

Answer 1

When length of wire is doubled i.e. $2l$

Also it's area of cross section becomes half i.e $\frac{A}{2}$

Resistance $R=\frac{\rho l}{A}=20\Omega$

New Resistance ,

$R^{\prime }=\frac{\rho l^{\prime }}{A^{\prime }}$

$=\frac{\rho \left(2l\right)\times 2}{A}$

$=4\frac{\rho l}{A}=4\left(\frac{\rho l}{A}\right)$

$=4\times 20$

$=80\Omega$