Explanation

(a) When strain is small, the ratio of the longitudinal stress to the corresponding longitudinal strain is called the Young’s modulus (Y) of the material of the body.
$Y = \frac{stress}{strain} = \frac{F/A}{l/L}$
Where F is force, A the area, $l$ the change in length and L the original length.
$\therefore\ $ $Y = \frac{FL}{{\pi r}^{2}l}$
$r\ \ $being radius of the wire.
Given $r_{2} = 2r_{1},\ \ L_{2} = {2L}_{1},\ F_{2\ } = {2F}_{1}$
Since, Young’s modulus is a property of material, we have
$Y_{1} = Y_{2}$
$\therefore\frac{F_{1}L_{1}}{\pi r_{1}^{2}l_{1}} = \frac{2F_{1} \times 2L_{1}}{\pi(2r_{1})^{2}l_{2}}$
${\ l}_{2} = l_{1} = l$
Hence, extension produced is same as that in the other wire.