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 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 - AO
$=[L^2+x^2{]}^{\frac{1}{2}}-L=L[1+\frac{1}{2}\frac{x^2}{L^2}]-L=\frac{x^2}{2L}$
Longitudinal strain $=\frac{\Delta L}{L}=\frac{x^2/2L}{L}=\frac{x^2}{2L^2}$