The load versus elongation graph for four wir | vedclass

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(C) From the formula for Young's modulus,$Y = \frac{F L}{A \Delta L}$,we have $\Delta L = \frac{F L}{A Y}$.Since the material is the same,$Y$ is const

Vedclass · Accepted Answer

$OD$

Vedclass · Answer

(C) From the formula for Young's modulus,$Y = \frac{F L}{A \Delta L}$,we have $\Delta L = \frac{F L}{A Y}$.Since the material is the same,$Y$ is constant. For wires of the same length $L$,we have $\Delta L \propto \frac{1}{A}$.Since $A = \pi r^2$,we have $\Delta L \propto \frac{1}{r^2}$.This means that for a given load $F$,the wire with the largest radius (thickest wire) will show the minimum elongation $\Delta L$.In the load versus elongation graph,the slope is given by $\frac{	ext{load}}{	ext{elongation}} = \frac{F}{\Delta L} = \frac{A Y}{L}$.$A$ steeper slope indicates a larger area $A$,which corresponds to a thicker wire.Looking at the graph,the line $OD$ has the steepest slope,representing the minimum elongation for a given load.Therefore,the thickest wire is represented by the line $OD$.

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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