
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
410.7k+ views
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.
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.
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

State and prove Bernoullis theorem class 11 physics CBSE

What organs are located on the left side of your body class 11 biology CBSE

How many valence electrons does nitrogen have class 11 chemistry CBSE
