Key idea: axes are interchanged

Normally, the graph is plotted as:

• stress on the $y$-axis
• strain on the $x$-axis

But here, by mistake, the student has plotted:

• stress on the $x$-axis
• strain on the $y$-axis

So we must interpret the graph carefully.

Young's modulus from the graph

Young's modulus is $Y = \frac{\text{stress}}{\text{strain}}$

In the usual stress-strain graph, slope in the elastic region is $\frac{\text{stress}}{\text{strain}} = Y$

But in the mistaken graph, slope is $\frac{\text{strain}}{\text{stress}} = \frac{1}{Y}$

Hence:

• larger slope in the mistaken graph $\Rightarrow$ smaller Young's modulus
• smaller slope in the mistaken graph $\Rightarrow$ larger Young's modulus

From the figure, the initial slope of curve $P$ is greater than that of $Q$. Therefore, $\frac{1}{Y_P} > \frac{1}{Y_Q} \implies Y_P < Y_Q$

So statement D is false.

Tensile strength

Tensile strength corresponds to the maximum stress sustained by the material.

Since stress is on the $x$-axis here, tensile strength is determined by the maximum horizontal extent of the curve.

From the graph, curve $P$ extends farther to the right than curve $Q$. Therefore, the maximum stress for $P$ is greater than for $Q$.

So, P has more tensile strength than Q.

Hence statement A is true.

Ductility / brittleness

Ductility is judged by the strain at fracture.

Since strain is on the $y$-axis here, ductility is determined by the vertical extent (strain value at breaking).

From the graph, curve $P$ reaches a larger strain before fracture than curve $Q$. Therefore, $P$ is more ductile than $Q$.

Hence statement B is true.

A more brittle material has smaller strain before breaking, so $P$ cannot be more brittle than $Q$. Therefore statement C is false.

Final evaluation of options

• A: True
• B: True
• C: False
• D: False

Derived answer

The correct options are: $\boxed{A,\ B}$

Comparison with stored correct answer

Stored correct answer: A, B

This matches the derived answer.