Question

A mild steel wire of length $2L$ and cross sectional area $A$ is stretched, well within elastic limit, horizontally between two pillars as shown in the figure.

• 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 $\frac{x^2}{{2L}^2}$

Increase in length $\Delta L=(PR+RQ)-PQ=2PR-PQ$

$\Delta L=2{(L^2-x^2)}^{1/2}-2L=2L(1+\frac{x^2}{L^2})-2L=2L[1+\frac{1}{2}\frac{x^2}{L^2}]-2L=\frac{x^2}{L^{}}$ (BY binomial theorem)
$\therefore Strain=\frac{\Delta L}{2L}=\frac{x^2}{{2L}^2}$