Answer
The midpoint moves a distance of $0.80~mm$
Work Step by Step
$Y = \frac{F/A}{\Delta L/L}$
$Y$ is Young's modulus
$F$ is the force
$A$ is the cross-sectional area
$\Delta L$ is the change in length
$L$ is the original length
We can find an expression for the change in length of the thinner wire:
$\Delta L_1 = \frac{F~L}{Y~A_1}$
Since the thicker wire has twice the radius, the area $A_2$ of the thicker wire is $4 A_1$. We can find an expression for the change in length of the thicker wire:
$\Delta L_2 = \frac{F~L}{Y~A_2}$
$\Delta L_2 = \frac{F~L}{Y~(4A_1)}$
$\Delta L_2 = \frac{1}{4}\times \frac{F~L}{Y~A_1}$
$\Delta L_2 = \frac{1}{4}\times \Delta L_1$
The total change in length of both wires is $\frac{5}{4}~\Delta L_1$. We can find $\Delta L_1$:
$\frac{5}{4}~\Delta L_1 = 1.0~~mm$
$\Delta L_1 = 0.80~~mm$
Therefore, the midpoint moves a distance of $0.80~mm$.
