The load versus elongation graphs for four wi | vedclass

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(A) Young's modulus $Y$ is given by $Y = \frac{FL}{A\Delta L}$,where $F$ is the load,$L$ is the length,$A$ is the area of cross-section,and $\Delta L$

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$OA$

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(A) Young's modulus $Y$ is given by $Y = \frac{FL}{A\Delta L}$,where $F$ is the load,$L$ is the length,$A$ is the area of cross-section,and $\Delta L$ is the elongation.Rearranging for elongation,we get $\Delta L = \frac{F}{A} \cdot \frac{L}{Y}$.Since all wires have the same length $L$ and are made of the same material (same $Y$),the elongation $\Delta L$ is inversely proportional to the area of cross-section $A$ for a constant load $F$ (i.e.,$\Delta L \propto \frac{1}{A}$).The thinnest wire has the smallest area of cross-section $A$,which means it will have the maximum elongation for a given load.From the graph,for a constant load,the elongation is maximum for the line $OA$.Therefore,$OA$ represents the thinnest wire.

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

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