Question

A load $F$ ( in $\text{Newtons}$) is applied, on a wire of length $2\text{m}$ having an area of cross-section $2\times {10}^{-6}{\text{m}}^2$. The extension $\left(\Delta L\right)$ versus load $\left(F\right)$ graph of wire is as shown in the figure below. Find the young's modulus of the material of wire.

• A. $2\times {10}^{11}{\text{N/m}}^2$
• B. $2\times {10}^{-11}{\text{N/m}}^2$
• C. $3\times {10}^{12}{\text{N/m}}^2$
• D. $2\times {10}^{13}{\text{N/m}}^2$

Answer 1

The correct option is A $2\times {10}^{11}{\text{N/m}}^2$

Given, Length of wire, $\left(L\right)=2\text{m}$
Area of cross section,$\left(A\right)=2\times {10}^{-6}{\text{m}}^2$
From the extension $\left(\Delta L\right)$ versus Load $\left(F\right)$ graph given in the question:


We know, Young's modulus,
$Y=\frac{F\times L}{A\times \Delta L}$
Since, the graph is linear in nature, we choose a point along the graph to find the young's modulus of elasticity.
Let us choose point B on graph, where $\left(\Delta L\right)=4\times {10}^{-4}\text{m}$ and Load applied $\left(F\right)=80\text{N}$
$\therefore Y=\frac{80\times 2}{2\times {10}^{-6}\times 4\times {10}^{-4}}$
$\Rightarrow Y=2\times {10}^{11}{\text{N/m}}^2$
Young's modulus of the material of wire is $2\times {10}^{11}{\text{N/m}}^2$.
Hence, option(a) is correct answer.