Given,
The length of the string, L=0.85 m
The diameter of the string, d=6×10⁻⁴ m
The radius of the string, r=3×10⁻⁴ m
The Young's Modulus of the string, Y=190 GPa
The tension, F=66 N
a)
The Young's Modulus is given by,
[tex]Y=\frac{FL}{Al}[/tex]Where A is the area of cross-section of the string and l is the extension produced in the string.
On rearranging the above equation,
[tex]l=\frac{FL}{YA}[/tex]On substituting the known values,
[tex]\begin{gathered} l=\frac{66\times0.85}{190\times10^9\times\pi(3\times10^{-4})^2} \\ =1.04\times10^{-3}\text{ m} \end{gathered}[/tex]Thus the extension of the wire that is needed to be produced is 1.04×10⁻³ m
b)
The work done in producing the extension is given by,
[tex]W=\frac{1}{2}Fl[/tex]On substituting the known values,
[tex]\begin{gathered} W=\frac{1}{2}\times66\times1.04\times10^{-3} \\ =0.034\text{ J} \end{gathered}[/tex]Thus the work done in producing the extension is 0.034 J