Question

A brass wire of diameter $1/mm$ and length $2/m$ is stretched by applying a force of $20/N$. If the increase in length is $0.51/mm$. Then

• A. Stress of the wire is $2.546\times {10}^{7}N/m^2$
• B. Strain in the wire is $2.546\times {10}^{-3}$
• C. Youngs modulus of the wire is $0.99\times {10}^{11}N/m^2$
• D. Stress of the wire is $4.646\times {10}^{6}N/m^2$

Answer 1

The correct option is A Stress of the wire is $2.546\times {10}^{7}N/m^2$

$\begin{array}{c}\text{diameter}=1mm\\ \text{Length}=2m\\ \text{Force}=20N\\ \text{elongation}=0.51mm\end{array}$

$\begin{array}{cc}\text{Stress}=& \frac{\text{Force}}{\text{Area}}=\frac{20}{\pi {\left(\frac{1}{2}\right)}^2\times {10}^{-6}}\\ & =\frac{20}{\pi }\times 4\times {10}^6\\ \text{Stress}& =2.546\times {10}^{7}N/m^2\end{array}$