In plotting stress versus strain curves for t | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The Young's modulus $Y$ is defined as $Y = \frac{	ext{stress}}{	ext{strain}}$.In the given graph,strain is on the $y$-axis and stress is on the

Vedclass · Accepted Answer

$(B, C)$

Vedclass · Answer

(C) The Young's modulus $Y$ is defined as $Y = \frac{	ext{stress}}{	ext{strain}}$.In the given graph,strain is on the $y$-axis and stress is on the $x$-axis. Therefore,the slope of the curve is $\frac{	ext{strain}}{	ext{stress}} = \frac{1}{Y}$.Since the slope of material $P$ is greater than the slope of material $Q$,we have $\frac{1}{Y_P} > \frac{1}{Y_Q}$,which implies $Y_P Material $P$ shows a larger strain for a given stress compared to $Q$,indicating that $P$ is more ductile than $Q$. Thus,statement $(B)$ is correct.Since $P$ is more ductile,it is less brittle than $Q$. Thus,statement $(C)$ is incorrect.Tensile strength is determined by the maximum stress a material can withstand before breaking. From the graph,$Q$ can withstand more stress than $P$ before the curve ends,so $Q$ has more tensile strength than $P$. Thus,statement $(A)$ is incorrect.Therefore,only statement $(B)$ is correct. However,based on the provided options,the question implies multiple correct statements. Re-evaluating: $P$ is more ductile ($B$ is correct). Since $P$ is more ductile,it is less brittle. The options provided are combinations. Given the standard interpretation of such problems,$(B)$ is the only physically correct statement.

Similar Questions

Assertion $(A)$: Ductile metals are used to prepare thin wires.
Reason $(R)$: In the stress-strain curve of ductile metals,the length between the points representing elastic limit and breaking point is very small.

Plot a stress $\to$ strain curve for a metal and explain.

The stress-strain curves for brass,steel,and rubber are shown in the figure. The lines $A, B,$ and $C$ correspond to:

The diagram shows the stress versus strain curve for materials $A$ and $B$. From the curves,we can infer that

The graph is drawn between the applied force $F$ and the strain $(x)$ for a thin uniform wire. The wire behaves as a liquid in the part:

Vedclass Test Series

Exam Paper Generator

Online Exam Module