Question

A mild steel wire of length $2L$ and cross-sectional area $A$ is stretched, well within elastic limit, horizontally between two pillars. A mass $m$ is suspended from the mind point of the wire. Strain in the wire is

• A. $\frac{x^2}{2L^2}$
• B. $\frac{x}{L}$
• C. $\frac{x^2}{L}$
• D. $\frac{x^2}{2L}$

Answer 1

Answer: (A)

$\frac{x^2}{2L^2}$

$\begin{array}{c}\text{Increase in Length}=\Delta L=AB+BC-AC=2AB-AC\\ AB=\sqrt{x^2+L^2},AC=2L\\ \Rightarrow \Delta L=2\sqrt{x^2+L^2}-2L\\ \Rightarrow \Delta L=2L{(1+{\left(\frac{x}{L}\right)}^2)}^{1/2}-2L\\ \Rightarrow \Delta L=2L\{\sqrt{1+\frac{x^2}{L^2}}-1\}.\end{array}$

$\begin{array}{c}\text{Using Binomial theorem.}\\ {(1+\frac{x^2}{L^2})}^{\frac{1}{2}}=1+\frac{x^2}{2L^2}\\ \Delta L=2L\{1+\frac{x^2}{2L^2}-1\}=\frac{x^2}{L}\\ \text{strain}=\frac{\Delta l}{2L}=\frac{x^2}{2L^2}\end{array}$

$\text{Ans}\to A$