Explanation

(d) Young’s modulus,$\ \ \ \ \ \ \ Y = \frac{Stress}{Strain} = \frac{\frac{Force}{Area}}{\frac{l}{L}}$
Where, $l$ is change in length and L the original length.
Force =$mg,$ Area $= A = {\pi r}^{2}$
$\therefore Y = \frac{FL}{{\pi r}^{2}l}$
$\therefore\ \ \frac{Y_{1}}{Y_{2}} = \frac{F_{1}L_{1}}{\pi r_{1}^{2}l_{1}} \times \frac{{\pi r}_{2}^{2}l_{2}}{F_{2}L_{2}}$
$_{⟹}$ $\frac{l_{1}}{l_{2}} = \frac{r_{2}^{2}}{r_{1}^{2}}$
(as all other quantities remain same for both the wires)
Given, $\ r_{2} = 2r_{1}$
$\therefore\ \frac{l_{1}}{l_{2}} = \frac{(2r_{1})^{2}}{r_{1}^{2}} = \frac{4}{1}$