In an experiment to determine the Young's modulus, steel wires of...

We know that, Young's modulus is $Y = \dfrac{FL}{A ∆ L}$, where, $F$ is force or load acting on wire, $A$ is cross-section area, $L$ is length and $∆ L$ is extension in length.

Now, from given graph, the slope of the line is $= \dfrac{\dfrac{\DeltaL}{W}}{L} = \dfrac{\dfrac{\DeltaL}{L}}{W} = \dfrac{1}{YA}$

$⇒ Y = \dfrac{1}{\left(slope\right)A}$

Thus, Young's modulus of the steel wires is $Y = \dfrac{1}{2 \times {10}^{- 6}\left(0 . 25 \times {10}^{- 5}\right)}$

$⇒ Y = 2 \times {10}^{11}N{m}^{- 2}$.

Therefore, $x = 2$.