The load versus elongation graph for four wir | vedclass

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Question

(A) The Young's modulus $Y$ is given by $Y = \frac{	ext{stress}}{	ext{strain}} = \frac{F/A}{\Delta L/L} = \frac{F}{\Delta L} \cdot \frac{L}{A}$.Sinc

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$OA$

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(A) The Young's modulus $Y$ is given by $Y = \frac{	ext{stress}}{	ext{strain}} = \frac{F/A}{\Delta L/L} = \frac{F}{\Delta L} \cdot \frac{L}{A}$.Since the material and length $L$ are the same for all wires,$Y$ is constant. Thus,the slope of the load-elongation graph,which is $\frac{F}{\Delta L}$,is proportional to the cross-sectional area $A$ $(slope \propto A)$.The wire with the smallest slope will have the smallest cross-sectional area $A$,meaning it is the thinnest wire.Looking at the graph,the line $OA$ has the smallest slope.Therefore,$OA$ represents the thinnest wire.

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

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