The stress-strain graphs for materials $A$ and $B$ are shown in Figure
The graphs are drawn to the same scale.
$(a)$ Which of the materials has the greater Young’s modulus?
$(b)$ Which of the two is the stronger material?
(a) A ; (b) A
For a given strain, the stress for material $A$ is more than it is for material $B ,$ as shown in the two graphs.
Young's modulus $=\frac{\text { stress }}{\text { strain }}$
For a given strain, if the stress for a material is more, then Young's modulus is also greater for that material. Therefore, Young's modulus for material $A$ is greater than it is for material $B.$
The amount of stress required for fracturing a material, corresponding to its fracture point, gives the strength of that material. Fracture point is the extreme point in a stress-strain curve. It can be observed that material $A$ can withstand more strain than material $B$.
Hence, material $A$ is stronger than material $B$.
The diagram shows stress v/s strain curve for the materials $A$ and $B$. From the curves we infer that
The diagram shows the change $x$ in the length of a thin uniform wire caused by the application of stress $F$ at two different temperatures $T_1$ and $T_2$. The variations shown suggest that
In the below graph, point $D$ indicates
The graph shows the behaviour of a length of wire in the region for which the substance obeys Hook’s law. $P$ and $Q$ represent
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