The load versus elongation graph for four wir | vedclass

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(A) The formula for elongation $l$ is given by $l = \frac{FL}{AY}$,where $F$ is the load,$L$ is the original length,$A$ is the cross-sectional area,an

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$OD$

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(A) The formula for elongation $l$ is given by $l = \frac{FL}{AY}$,where $F$ is the load,$L$ is the original length,$A$ is the cross-sectional area,and $Y$ is Young's modulus.Since the wires are of the same material ($Y$ is constant) and have the same length ($L$ is constant),for a given load ($F$ is constant),the elongation $l$ is inversely proportional to the cross-sectional area $A$ $(l \propto \frac{1}{A})$.Since $A = \pi r^2$,we have $l \propto \frac{1}{r^2}$,where $r$ is the radius of the wire.This means that for the same load,the thickest wire (largest $r$) will show the minimum elongation.In the given graph,for a constant load on the $y$-axis,the line $OD$ corresponds to the minimum elongation on the $x$-axis.Therefore,the line $OD$ represents the thickest wire.

The load versus elongation graph for four wires of the same material is shown in the figure. The thickest wire is represented by the line

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