Answer:
7.65 mm
Explanation:
Stress, [tex]\sigma=\frac {F}{A}[/tex] where F is the force and A is the area
Also, [tex]\sigma=E\times \frac {\triangle L}{L}[/tex]
Where E is Young’s modulus, L is the length and [tex]\triangle L[/tex] is the elongation
Therefore,
[tex]\frac {F}{A}= E\times \frac {\triangle L}{L}[/tex]
Making A the subject of the formula then
[tex]A=\frac {FL}{E\triangle L}=\frac {6660\times 380}{110\times 10^{9}\times 0.5}=4.60145\times 10^{-5} m^{2}[/tex]
Since [tex]A=\frac {\pi d^{2}}{4}[/tex] then
[tex]d=\sqrt {\frac {4A}{\pi}}=\sqrt {\frac {4\times 4.60145\times 10^{-5}}{\pi}}= 0.00765425m= 7.654249728 mm\approx 7.65 mm[/tex]
