Question

A telephone wire $125\,m$ long and $1\,mm$ in radius is stretched to a length $125.25\,m$ when a force of $800\,N$ is applied. What is the value of Young’s modulus for material wire?

Answer 1

Hint: Young’s modulus describes the relationship between stress (force per unit area) and strain (proportional deformation in an object. The Young’s modulus is named after the British scientist Thomas Young. A solid object deforms when a particular load is applied to it.

Complete step by step answer:
Given, Original length of telephone wire, $l = 125\,m$
Length after stretched, ${l_f} = 125.25\,m$
Radius of telephone wire, $r = 1\,mm$
Radius of telephone wire, $r = 0.001\,m$
Cross section area of telephone wire,
$\therefore A = \pi {r^2}$
Put the value
\[ A = \dfrac{{22}}{7} \times {(0.001)^2}\]
$ \Rightarrow 3.14 \times {10^{ - 6}}$
Change in length, $\Delta l = ?$
$\therefore \Delta l = {l_f} - l$
Put the value
$ \Delta l = 125.25 - 125$
$ \Rightarrow \Delta l = 0.25\,m$
Strain in telephone wire, $\varepsilon = ?$
As we know that
$\varepsilon = \dfrac{{\Delta l}}{l}$
Put the value
$ \varepsilon = \dfrac{{0.25}}{{125}} \\$
$ \Rightarrow \varepsilon = 0.002$
Now
Young’s modulus,
$Y = \dfrac{{\dfrac{F}{A}}}{{\dfrac{{\Delta l}}{l}}}$
Put the value
$Y = \dfrac{{\dfrac{{800}}{{3.14 \times {{10}^6}}}}}{{\dfrac{{0.25}}{{125}}}} \\$
Simplify
$\Rightarrow Y = \dfrac{{800 \times 125}}{{3.14 \times {{10}^{ - 6}} \times 0.25}} \\$
$\therefore Y = 1.27 \times {10^{11}}Pa$

Hence, the value of Young’s modulus for material wire is $1.27 \times {10^{11}}\,Pa$.

Note: The young’s modulus of a material is a fundamental property of every material that cannot be changed. It is dependent upon temperature and pressure however. The young’s modulus is the essence, the stiffness of a material. In other words, it is how easily it is bended or stretched.