Question

A copper wire of 3 m length and 1 mm diameter is subjected to a tension of 5 N. Calculate the elongation produced in the wire if the Young’s modulus of copper is 120 GPa.
​

Answer 1

Given :

The length of copper wire = 3m = 3000mm

Diameter of wire = 1mm

Tension / Force = 5N

Young's Modulus = 120GPa = 120000$N/mm^{2}$

To Find:

We have to find the elongation produced in the copper wire.

Solution:

The elastic properties of linear objects like wire, rods, etc. which are either stretched or compressed, a convenient parameter is the ratio of the stress to the strain, the parameter is called Young's modulus (E).

Young's Modulus, E =  $\frac{Stress}{Strain}$

Area of the copper wire, A = π×$r^{2}$ =π×$\frac{d^{2} }{4}$

                                        = $\frac{\pi }{4}$ $mm^{2}$

→   Stress, σ = $\frac{Force}{Area}$

  where; F is the tensile force

              A is the cross-sectional area of wire

            σ = $\frac{Force}{Area}$

               $\[\begin{array}{l} = \frac{{5 \times 4}}{\pi }\\\\ = \frac{{20}}{\pi }\,N/m{m^2}\end{array}\]$

 →  Strain,ε =  $\[\frac{{\Delta L}}{L}\]$

   where, $\[\Delta L\]$ is the elongation.

Combining the two equations we get,

              $\[\begin{array}{l}E = \frac{{{\raise0.7ex\hbox{ {20} } \!\mathord{\left/ {\vphantom {{20} \pi }}\right.\kern-\nulldelimiterspace}\!\lower0.7ex\hbox{ \pi }}}}{{{\raise0.7ex\hbox{ {\Delta L} } \!\mathord{\left/ {\vphantom {{\Delta L} L}}\right.\kern-\nulldelimiterspace}\!\lower0.7ex\hbox{ L }}}}\\\\\Delta L = \frac{{20 \times L}}{{\pi \times E}}\\\\\Delta L = \frac{{20 \times 3000}}{{\pi \times 120000}}\\\\\,\,\,\,\,\,\,\, = 0 \cdot 159\,mm\end{array}\]$

∴ The elongation of the copper wire = 0·159 mm ( As per the given data).