Question

A mild-steel wire of length $2L$ and cross-sectional area $A$ is stretched, well within the elastic limit, horizontally between two pillars. A mass $m$ is suspended from the midpoint 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 $\frac{x^2}{2L^2}$

Change in length, $\Delta L=(AC+BC)-AB$

$\Delta L=2AO-2L=2[AO-L]$

$\Delta L=2\left[\right(L^2+x^2{)}^{\frac{1}{2}}-L]$

$\Delta L=2\left[L\right(1^2+\frac{x^2}{L^2}{)}^{\frac{1}{2}}-L]$

For small values of $x$ applying Binomial theorem,
$\Rightarrow \Delta L=2L[1+\frac{1}{2}\frac{x^2}{L^2}]-2L=\frac{x^2}{L}$
Longitudinal strain, $\frac{\Delta L}{2L}=\frac{x^2/L}{2L}=\frac{x^2}{2L^2}$.