Question

3. A certain wire, 3 m long, stretches by 1.2 mm when under tension of 200 N. By how much does

Answer 1

Answer:

The extension of the second wire is   $e_2 = 0.0024 \ m = 2.4 mm$

Explanation:

From the question we are told that

    The length of the wire is $L = 3 \ m$

     The elongation of the wire is  $e = 1.2mm = \frac{1.2}{1000} = 0.0012 m$

        The tension is $F = 200 \ N$

       The length of the second wire is  $L_2 = 6 \ m$

     

Generally the Young's modulus(Y) of this material is  

        $Y = \frac{stress}{strain }$

Where $stress = \frac{F}{A}$

    Where A is the area which is evaluated as  

           $A = \pi r^2$

  and   $strain = \frac{extention}{length} = \frac{e}{L}$

   So

        $Y = \frac{\frac{F}{\pi r^2 } }{ \frac{e}{L} }$

Since the wire are of the same material Young's modulus(Y)  is constant

So we have  

              $\frac{F * L }{r^2 e} = \pi * Y = constant$

              $F * L = constant * r^2 e$

Now the ration between the first and the second wire is

         $\frac{F_1}{F_2} * \frac{L_1}{L_2} = \frac{r*2_1}{r^2} * \frac{e_1}{e_2}$

Since tension , radius are constant

   We have

           $\frac{L_1}{L_2} = \frac{e_1}{e_2}$

substituting values

          $\frac{3}{6} = \frac{0.0012}{e_2}$

          $0.5 e_2 = 0.0012$

         $e_2 = \frac{ 0.0012 }{0.5}$

          $e_2 = 0.0024 \ m = 2.4 mm$