The stress versus strain graphs for wires of two materials $A$ and $B$ are as shown in the figure. If $Y_A$ and $Y_B$ are the Young's modulii of the materials, then

We know that the slope of the line in a stress-strain curve represents the young's modulus for a wire. We also know that the slope of a line can be calculated as the tangent of the angle the line makes with the positive x -axis of the graph. So, for wire A, the stress-strain line of the material is at an angle of \(60^{\circ}\) from the positive \(x\)-axis. So the slope of the line \(\left(m_A\right)\) will be \(\begin{aligned} & m_A=\tan 60^{\circ} \\ & \Rightarrow m_A=\sqrt{3} \end{aligned}\) Hence the young's modulus of wire A will also be \(\sqrt{3}\). Similarly, for wire B , the stress-strain line of the material is at an angle of \(30^{\circ}\) from the positive \(x\) axis. So, the slope of the line \(\left(m_B\right)\) will be \(\begin{aligned} & m_B=\tan 30^{\circ} \\ & \Rightarrow m_B=\frac{1}{\sqrt{3}} \end{aligned}\) Hence the young's modulus of wire B will also be \(1 / \sqrt{3}\). Then taking the ratio of the young's modulus for wire A and B , we get \(\begin{aligned} & \frac{Y_A}{Y_B}=\frac{\sqrt{3}}{1 / \sqrt{3}} \\ & \therefore \frac{Y_A}{Y_B}=3 \end{aligned}\)