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NCERT Exemplar · Q6

Q.A mild steel wire of length 2L2L and cross-sectional area AA is stretched, well within the elastic limit, horizontally between two rigid pillars that are a distance 2L2L apart, so that it initially lies along a straight horizontal line. A mass mm is then suspended from the mid-point of the wire, and the mid-point is pulled down (sags) through a small vertical distance xx (x≪Lx \ll L) below the original straight line. The strain produced in the wire is

(a) x22L2\dfrac{x^{2}}{2L^{2}}
(b) xL\dfrac{x}{L}
(c) x2L\dfrac{x^{2}}{L}
(d) x22L\dfrac{x^{2}}{2L}
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The wire is fixed at both pillars, so each half keeps its own fixed length LL. When the midpoint sags by xx, each half stretches from LL to L2+x2\sqrt{L^{2}+x^{2}}. For a small sag this extension is ≈x2/2L\approx x^{2}/2L, and dividing by the original length LL gives a strain of x2/2L2x^{2}/2L^{2}.

Concept

Strain is the fractional change in length, strain=ΔLL0\text{strain}=\dfrac{\Delta L}{L_0}, and it must be dimensionless. Here the wire is anchored at the two pillars, so its half on each side of the mass has a fixed unstretched length LL (half of 2L2L).

Why this geometry

Before loading, each half lies horizontally with length LL. After the mass sags the midpoint down by xx, each half is the hypotenuse of a right triangle whose horizontal leg is LL and vertical leg is xx.

Steps

  1. New length of each half:

L′=L2+x2=L1+x2L2.L' = \sqrt{L^{2}+x^{2}} = L\sqrt{1+\frac{x^{2}}{L^{2}}}.

  1. For a small sag, x≪Lx \ll L, use the binomial approximation 1+u≈1+12u\sqrt{1+u}\approx 1+\tfrac{1}{2}u:

L′≈L(1+x22L2)=L+x22L.L' \approx L\left(1+\frac{x^{2}}{2L^{2}}\right)=L+\frac{x^{2}}{2L}.

  1. Extension of each half: …

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