Question

A mild steel wire of length 2L and cross-sectional area A is stretched, well within elastic limit, horizontally between two pillars (figure). A mass m is suspended from the mid-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

The correct option is A $\frac{x^2}{2L^2}$

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

$\begin{array}{c}\text{using Binomial theorom.}\\ \begin{array}{c}{(1+\frac{x^2}{L^2})}^{\frac{1}{2}}=1+\frac{x^2}{2L^2}\\ \Rightarrow \Delta L=2L\{1+\frac{x^2}{2L^2}-1\}=\frac{x^2}{L}\end{array}\\ \text{strain}=\frac{\Delta l}{2L}=\frac{x^2}{2L^2}\\ \text{Ans}\to A\end{array}$