Question

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 moduli of the materials, then ​
A) $Y_B=2Y_A$
B) $Y_A=Y_B$
C) $Y_B=3Y_A$
D) $Y_A=3Y_B$

Answer 1

Hint : The young’s modulus of wire is the slope of the line of the wire in the graph of stress versus strain. The slope of a line can be determined using the angle the line makes with the $x$ -axis.

Complete step by step answer
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 $(m_A)$ will be
 $m_A=\tan {60}^{\circ }$
 $\Rightarrow m_A=\sqrt{3}$
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 $(m_B)$ will be
 $m_B=\tan {30}^{\circ }$
 $\Rightarrow m_B={\displaystyle \frac{1}{\sqrt{3}}}$
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
 ${\displaystyle \frac{Y_A}{Y_B}}={\displaystyle \frac{\sqrt{3}}{1/\sqrt{3}}}$
 $\therefore {\displaystyle \frac{Y_A}{Y_B}}=3$
Hence, we can write
$Y_A=3Y_B$ which corresponds to option (D) which is the correct choice.

Note
We can only calculate the slope of the line in such a way if the stress-strain curve for a wire is a straight line. For practical wires, the stress-strain curve is linear only for a range of values of stress applied on the wire. While calculating the slope of the wire, we must calculate the tangent of the line made with the $x$ -axis and not the $y$ -axis if the strain is represented on the $x$ -axis of the graph.