Answer:[tex]0.8\ mm[/tex]
Explanation:
Given
length of wire [tex]l=75\ cm[/tex]
change in length [tex]\Delta l=1.85\ mm[/tex]
mass of wire [tex]m=10\ kg[/tex]
Young's modulus for silver [tex]E=7.9\times 10^{10}\ N/m^2[/tex]
load on wire [tex]F=mg[/tex]
[tex]F=10\times 9.8=98\ kg[/tex]
change in length is given by
[tex]\Delta l=\dfrac{Pl}{AE}[/tex]
Where A=area of cross-section
[tex]A=\dfrac{Pl}{\Delta lE}[/tex]
[tex]A=\dfrac{98\times 0.75}{1.85\times 10^{-3}\times 7.9\times 10^{10}}[/tex]
[tex]A=\dfrac{73.5}{14.615\times 10^{7}}[/tex]
[tex]A=5.029\times 10^{-7}\ m^2[/tex]
also wire is the shape of cylinder so cross-section is given by
[tex]A=\dfrac{\pi d^2}{4}=5.029\times 10^{-7}\ m^2[/tex]
[tex]\Rightarrow d^2=\dfrac{5.029\times 10^{-7}\times 4}{\pi }[/tex]
[tex]\Rightarrow d^2=64.02\times 10^{-8}[/tex]
[tex]\Rightarrow d=8\times 10^{-4}\ m[/tex]
[tex]\Rightarrow d=0.8\ mm[/tex]
