A uniform metallic wire is elongated by 0 . 04m when subjected to a...

$F = Y . A \cdot \dfrac{\Deltal}{l}⇒ \Deltal = \dfrac{F}{Y . A .} \cdot l⇒ \Deltal = \dfrac{F . l}{Y . \pir^{2}}⇒ \Deltal ∝ \dfrac{l}{r2}$

$\dfrac{\Delta{l}_{2}}{\Delta{l}_{1}} = \left(\dfrac{l2}{l1}\right)\left(\dfrac{r1}{r2}\right)2= \left(2\right)\left(\dfrac{1}{2}\right)2⇒ \dfrac{\Delta{l}_{2}}{\Delta{l}_{1}} = \dfrac{1}{2}⇒ \Delta{l}_{2} = \dfrac{\Delta{l}_{1}}{2}$

$= \dfrac{0 . 04}{2}= 0 . 02m⇒ \Delta{l}_{2} = 2cm$.