The load versus elongation graph for four wir | vedclass

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(C) The Young's modulus $Y$ is given by $Y = \frac{FL}{A \Delta l}$,where $F$ is the load,$L$ is the original length,$A$ is the cross-sectional area,a

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$OA$

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(C) The Young's modulus $Y$ is given by $Y = \frac{FL}{A \Delta l}$,where $F$ is the load,$L$ is the original length,$A$ is the cross-sectional area,and $\Delta l$ is the elongation.Since all wires are of the same material,$Y$ is constant. For wires of the same length $L$,we have $A = \frac{FL}{Y \Delta l}$.For a constant load $F$,the area $A$ is inversely proportional to the elongation $\Delta l$ $(A \propto \frac{1}{\Delta l})$.From the graph,for a given load $F$,the elongation $\Delta l$ is maximum for line $OA$ (i.e.,$\Delta l_A > \Delta l_B > \Delta l_C > \Delta l_D$).Since $A \propto \frac{1}{\Delta l}$,the wire with the maximum elongation will have the minimum cross-sectional area.Therefore,the line $OA$ represents the thinnest wire.

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

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