Answer:0.143 mm
Explanation:
Given
Length of wire [tex]L=2.8\ m[/tex]
Cross-section of wire [tex]A=1.2\times 1.2\ mm^2[/tex]
Extension in wire on application of load [tex]\delta L=1.43\ mm[/tex]
Mass of support [tex]m=17\ kg[/tex]
Weight of support [tex]W=mg=17\times 9.8=166.6\ N[/tex]
Extension in the wire is given by
[tex]\delta L=\frac{PL}{AE}[/tex]
where E=Young's Modulus
substitute values
[tex]1.43=\frac{166.6\times 2.8}{1.44\times E}----1[/tex]
When wire is cut for length of [tex]L'=0.28\ m[/tex]
extension is given by
[tex]\delta L'=\frac{166.66\times 0.28}{1.44\times E}---2[/tex]
divide 1 and 2 we get
[tex]\frac{1.43}{\delta L'}=\frac{2.8}{0.28}[/tex]
[tex]\delta L'=0.143\ mm[/tex]
