A wire of 1   Ω has a length of 1   m . It is stretched till its length increases by 25 % .…

$R_0=1 \mathrm{\Omega } R_1=$

${\ell }_0=1 \mathrm{m} {\ell }_1=1.25 \mathrm{m}$

$A_0=A$

As volume of wire remains constant so

$A_0{\ell }_0=A_1{\ell }_1\Rightarrow A_1=\frac{{\ell }_0 A_0}{{\ell }_1}$

Now, Resistance $\left(R\right)=\frac{\rho \ell }{A}$

$\frac{R_0}{R_1}=\frac{{\ell }_0/A_0}{\rho {\ell }_1/A_1}$

$\frac{1}{R_1}=\frac{{\ell }_0}{ A_0}\left(\frac{{\ell }_0 A_0}{{\ell }_1\times {\ell }_1}\right)R_1=\frac{{\ell }_1^2}{{\ell }_0^2}=1.5625 \mathrm{\Omega }$

So $\%$ change in resistance

$=\frac{R_1-R_0}{R_0}\times 100\%$

$=\frac{1.5625-1}{1}\times 100\%$

$=56.25\%$