A wire of 1Ω has a length of 1m . It is stretched till its length...

{R}_{0} = 1Ω{R}_{1} =$</p><p>${ℓ}_{0} = 1m{ℓ}_{1} = 1 . 25m$</p><p>${A}_{0} = A$</p><p>As volume of wire remains constant so</p><p>${A}_{0}{ℓ}_{0} = {A}_{1}{ℓ}_{1} ⇒ {A}_{1} = \dfrac{{ℓ}_{0}{A}_{0}}{{ℓ}_{1}}$</p><p>Now, Resistance \left(R\right) = \dfrac{ρℓ}{A}$

$\dfrac{{R}_{0}}{{R}_{1}} = \dfrac{{ℓ}_{0} / {A}_{0}}{ρ{ℓ}_{1} / {A}_{1}}$

$\dfrac{1}{{R}_{1}} = \dfrac{{ℓ}_{0}}{{A}_{0}}\left(\dfrac{{ℓ}_{0}{A}_{0}}{{ℓ}_{1} \times {ℓ}_{1}}\right){R}_{1} = {ℓ}_{1}^{2}{ℓ}_{0}^{2} = 1 . 5625Ω$

So $$ change in resistance

$= \dfrac{{R}_{1} - {R}_{0}}{{R}_{0}} \times 100$

$= \dfrac{1 . 5625 - 1}{1} \times 100$

$= 56 . 25$