Question

The graph shows the extension $\left(\Delta l\right)$ of a wire of length $1\text{m}$, suspended from the top of a roof at one end, with a load $W$ connected to the other end. If the cross-sectional area of the wire is ${10}^{-6}{\text{m}}^2$, Young's modulus of the material of the wire is

• 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 $2\times {10}^{11}{\text{N/m}}^2$

We know that, $Y=\frac{\frac{F}{A}}{\frac{\Delta l}{l}}$
where $Y$ = Young's modulus of elasticity
$\Rightarrow Y=\frac{F\times l}{A\times \Delta l}=\frac{W}{\Delta l}\times \frac{l}{A}$

At the point where $W=20\text{N}$, $\Delta l=1\times {10}^{-4}\text{m}$ [from the graph]
$\Rightarrow Y=\frac{20}{{10}^{-4}}\times \frac{1}{{10}^{-6}}$
$\Rightarrow Y=2\times {10}^{11}{\text{N/m}}^2$