Overview

Resistance of whole wire is

$R=r^{\text{'}}*l$

$20=1*l$

$l=20m$

Let length of the shorter section = a

$\frac{1}{R}=\frac{1}{a}+\frac{1}{\left(20-a\right)}$

$\frac{1}{1.8}=\frac{20-a+a}{a\left(20-a\right)}?20a-a^2=36$

$a=2and18$

$?a=2m$

(Explanation can be elaborate)

$l=\frac{25}{R_{eq}}$

$R_{eq}=25\mathrm{\Omega }$

$l=\frac{25}{25}1A$

Please consider the following Image  

$\Rightarrow \frac{R}{100}=\frac{40}{60}$

$\Rightarrow \Delta R=\left(\frac{\Delta l}{l}+\frac{\Delta l}{\left(100-l\right)}\right)R$

$=\left(\frac{0.1}{40}+\frac{0.1}{60}\right)*66.7=0.27=0.3$

When wire is stretched uniformly

$R\propto \mathcal{R}$

$\frac{R}{R^{\text{'}}}={\left(\frac{l}{nl}\right)}^2=\frac{1}{n^2}$

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

When this elongated wire is cut into 5 parts, resistance of each part $=\frac{n^{2}R}{5}$  

Effective resistance between A and C $=\frac{n^{2}R}{5}$  

$01*10^4\Omega \pm 5\mathrm{\%}=10*10^3\Omega \pm 5\mathrm{\%}$

Resistance of whole wire is

$R=r^{\text{'}}*l$

$20=1*l$

$l=20m$

Let length of the shorter section = a

$\frac{1}{R}=\frac{1}{a}+\frac{1}{\left(20-a\right)}$

$\frac{1}{1.8}=\frac{20-a+a}{a\left(20-a\right)}?20a-a^2=36$

$a=2and18$

$?a=2m$

  $R_i=\frac{\rho \mathcal{l}}{A}$
$R_f=\frac{\rho \left(1.25\mathcal{l}\right)}{\left(A/1.25\right)}={\left(1.25\right)}^2*\frac{\rho \mathcal{l}}{A}$

$\Rightarrow R_f=1.5625*R_i$

$\Rightarrow \frac{R_f-R_l}{R_l}*100=56.25\mathrm{\%}$