Answer:
The radius is  [tex]r_{min} = 0.00226 \ m[/tex]
Explanation:
  From the question we are told that
   The  elastic limit(stress) is [tex]\sigma = 5.0*10^{8} \ N /m^2[/tex]
   The length is  [tex]L = 4.0 \ m[/tex]
   The weight of the commercial sign is   [tex]F_s = 8000 \ N[/tex]
    The maximum extension of the wire is  [tex]\Delta L = 5.0 \ cm = 0.05 \ m[/tex]
Generally the elastic limit of an alloy (stress) is is mathematically represented as
      [tex]\sigma = \frac{ F_s }{ A }[/tex]
Where A is the cross-sectional area of the wire which is mathematically represented as
     [tex]A = \pi r^2[/tex]
here [tex]r = r_{min}[/tex] which is the minimum radius of the wire that support the commercial sign
So
     [tex]\sigma = \frac{ F_s }{ \pi r_{min}^2 }[/tex]
=> Â Â Â [tex]r_{min} = \sqrt{\frac{F_s}{\sigma * \pi} }[/tex]
substituting values
       [tex]r_{min} = \sqrt{\frac{8000}{ 5.0* 10^8 * 3.142} }[/tex]
      [tex]r_{min} = 0.00226 \ m[/tex]
      Â
