a
Let $'L'$ and $A$ be lenght and area of cross section of each wire. In order to have the lower ends of the wires to be at the same level $(i.e.,\,same\,elongation\,is\,produced\,in\,both\,wiers)$ let weights $W_s$ and $W_b$ are added to steel and brass wires respectively. Then, By definition of $Young's$ modulus, the elongation produced in the steel wire is
$\Delta {L_s} = \frac{{{W_s}L}}{{{Y_s}A}}$ $\left( {asY = \frac{{W/A}}{{\Delta L/L}}} \right)$
and that in the brass wire is $\Delta {L_b} = \frac{{{W_b}L}}{{{Y_b}A}}$
But $\Delta {L_s} = \Delta {L_b}$ $(given)$
$\therefore \frac{{{W_s}L}}{{{Y_s}A}} = \frac{{{W_b}L}}{{{Y_b}A}}\,\,or\,\,\frac{{{W_s}}}{{{W_b}}} = \frac{{{Y_s}}}{{{Y_b}}}$
$As\,\frac{{{Y_s}}}{{{Y_b}}} = 2$ $(given)$
$\therefore \frac{{{W_s}}}{{{W_b}}} = \frac{2}{1}$
