Answer:
The extension of the second wire is [tex]e_2 = 0.0024 \ m = 2.4 mm[/tex]
Explanation:
From the question we are told that
The length of the wire is [tex]L = 3 \ m[/tex]
The elongation of the wire is [tex]e = 1.2mm = \frac{1.2}{1000} = 0.0012 m[/tex]
The tension is [tex]F = 200 \ N[/tex]
The length of the second wire is [tex]L_2 = 6 \ m[/tex]
Generally the Young's modulus(Y) of this material is
[tex]Y = \frac{stress}{strain }[/tex]
Where [tex]stress = \frac{F}{A}[/tex]
Where A is the area which is evaluated as
[tex]A = \pi r^2[/tex]
and [tex]strain = \frac{extention}{length} = \frac{e}{L}[/tex]
So
[tex]Y = \frac{\frac{F}{\pi r^2 } }{ \frac{e}{L} }[/tex]
Since the wire are of the same material Young's modulus(Y) is constant
So we have
[tex]\frac{F * L }{r^2 e} = \pi * Y = constant[/tex]
[tex]F * L = constant * r^2 e[/tex]
Now the ration between the first and the second wire is
[tex]\frac{F_1}{F_2} * \frac{L_1}{L_2} = \frac{r*2_1}{r^2} * \frac{e_1}{e_2}[/tex]
Since tension , radius are constant
We have
[tex]\frac{L_1}{L_2} = \frac{e_1}{e_2}[/tex]
substituting values
[tex]\frac{3}{6} = \frac{0.0012}{e_2}[/tex]
[tex]0.5 e_2 = 0.0012[/tex]
[tex]e_2 = \frac{ 0.0012 }{0.5}[/tex]
[tex]e_2 = 0.0024 \ m = 2.4 mm[/tex]
