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Physics Question on mechanical properties of solids

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

x22L2\frac{x^{2}}{2L^{2}}

B

xL\frac{x}{L}

C

xL2\frac{x}{L^2}

D

x2L\frac{x}{2L}

Answer

x22L2\frac{x^{2}}{2L^{2}}

Explanation

Solution

A mild steel wire of length 2L and cross-sectional area A is stretched, well within elastic limit

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

ΔL=2(L2−x2)1/2−2L\Delta L = 2 \left( L^2 - x^2 \right)^{1/2} - 2L

ΔL=2L(1+x22L2)−2L\Delta L = 2L \left( 1 + \frac{x^2}{2L^2} \right) - 2L

Use the binomial theorem to approximate ΔL\Delta L.
To find the strain ϵ\epsilon:
ϵ=ΔL2L\epsilon = \frac{\Delta L}{2L}

Substituting the expression for ΔL\Delta L:
ϵ=2L(1+x22L2)−2L2L\epsilon = \frac{2L \left( 1 + \frac{x^2}{2L^2} \right) - 2L}{2L}
ϵ=1+x22L2−1\epsilon = 1 + \frac{x^2}{2L^2} - 1

ϵ=x22L2\epsilon = \frac{x^2}{2L^2}

So, the correct option is (A): x22L2\frac{x^2}{2L^2}