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. $\frac{x}{L^2}$
• D. $\frac{x}{2L}$

Answer 1

The Correct Option is A

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

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

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

Use the binomial theorem to approximate \(\Delta L\).
To find the strain \(\epsilon\):
\(\epsilon = \frac{\Delta L}{2L}\)

Substituting the expression for \(\Delta L\):
\(\epsilon = \frac{2L \left( 1 + \frac{x^2}{2L^2} \right) - 2L}{2L}\)
\(\epsilon = 1 + \frac{x^2}{2L^2} - 1\)

\(\epsilon = \frac{x^2}{2L^2}\)

So, the correct option is (A): \(\frac{x^2}{2L^2}\)