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 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

Answer: (A)

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

Answer 2

:

Refer figure,

Increase in length,

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

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

$= 2L\left\lbrack 1 + \frac{1}{2}\frac{x^{2}}{L^{2}} \right\rbrack - 2L$ (By binomial theorem)

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

$\therefore Strain = \frac{\Delta L}{2L} = \frac{x^{2}}{2L^{2}}$