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**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.

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