Answer
The minimum diameter of the wire is 1.89 mm.
Work Step by Step
The Young's Modulus for steel is $20\times 10^{10}~N/m^2$.
We can find the required area of the wire.
$Y = \frac{F/A}{\Delta L/L_0}$
$A = \frac{F~L_0}{Y~\Delta L}$
$A = \frac{(700~N)(2.00~m)}{(20\times 10^{10}~N/m^2)(0.0025~m)}$
$A = 2.80\times 10^{-6}~m^2$
We can find the radius of the wire with this area.
$A = 2.80\times 10^{-6}~m^2$
$\pi~R^2 = 2.80\times 10^{-6}~m^2$
$R^2 = \frac{2.80\times 10^{-6}~m^2}{\pi}$
$R = \sqrt{\frac{2.80\times 10^{-6}~m^2}{\pi}}$
$R = 0.944\times 10^{-3}~m$
$R = 0.944~mm$
Since the diameter is twice the radius, the minimum diameter of the wire is 1.89 mm.
