A metal of fixed volume $V$ is drawn into a wire of uniform cross-sectional area $A$. When this wire

Question

A metal of fixed volume $V$ is drawn into a wire of uniform cross-sectional area $A$. When this wire is subjected to a constant stretching force $F$, the extension produced is $\Delta l$. Which of the following graphs will be a straight line?

Professor PC Thomas Classes · Answer

(a) : The volume of the wire is $V = A 	imes l$, which means the length of the wire is $l = \frac{V}{A}$. Young's modulus is $Y = \frac{F/A}{\Delta l/l} = \frac{Fl}{A \Delta l}$. Substituting $l = \frac{V}{A}$ into the equation gives $\Delta l = \frac{F(V/A)}{Y A} = \frac{FV}{Y A^{2}}$. Since $F, V,$ and $Y$ are constants, $\Delta l \propto \frac{1}{A^{2}}$. Therefore, the graph of $\Delta l$ versus $1/A^{2}$ is a straight line.