
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 Young's moduli of the materials, then:

(A) ${Y_B} = 2{Y_A}$
(B) ${Y_B} = {Y_A}$
(C) ${Y_B} = 3{Y_A}$
(D) $3{Y_B} = {Y_A}$
Answer
242.4k+ views
.Hint. We solve this question by finding the values of Young’s modulus for both the materials. We find Young’s modulus by finding the slope of the line in the graph for each material. We find the slope by using trigonometric properties. The angle made by the line in the stress vs strain graph is already given. By using these angles in trigonometry properties to find Young’s modulus we get a relation between ${Y_A}$and ${Y_B}$
Complete step by step answer. The slope of stress vs strain graph gives us Young’s modulus.
In the graph, we are given the angle made by the line in the stress vs strain graph in both cases.
We know that slope is equal to $\tan \theta $
In the case of material (A)
The slope is equal to $\tan 60^\circ $
Material (B)
The slope is equal to $\tan 30^\circ $
We know that slope is equal to Young’s modulus hence
${Y_A} = \tan 60^\circ = \sqrt 3 $
${Y_B} = \tan 30^\circ = \dfrac{1}{{\sqrt 3 }}$
Diving Young’s modulus of (A) with Young’s modulus of (B) we get $\dfrac{{{Y_A}}}{{{Y_B}}} = \dfrac{{\sqrt 3 }}{{\dfrac{1}{{\sqrt 3 }}}} = 3$
Hence $3{Y_B} = {Y_A}$
Option (D) $3{Y_B} = {Y_A}$ is the correct answer.
Additional information Young’s modulus is defined as the measure of ability of a material to withstand changes in length when it is under a lengthwise tension of compression. It is also referred to as modulus of elasticity.
Note We find the slope in the graph by using tanθ because we are assuming the graph as a right-angled triangle. And $\tan \theta $is equal to the opposite by the adjacent side of the assumed triangle. In the case of stress vs strain graph opposite is equal to stress and adjacent is equal to the strain. Hence $\tan \theta $ is equal to Young’s modulus. Therefore, we use $\tan \theta $ to find the slope.
Complete step by step answer. The slope of stress vs strain graph gives us Young’s modulus.
In the graph, we are given the angle made by the line in the stress vs strain graph in both cases.
We know that slope is equal to $\tan \theta $
In the case of material (A)
The slope is equal to $\tan 60^\circ $
Material (B)
The slope is equal to $\tan 30^\circ $
We know that slope is equal to Young’s modulus hence
${Y_A} = \tan 60^\circ = \sqrt 3 $
${Y_B} = \tan 30^\circ = \dfrac{1}{{\sqrt 3 }}$
Diving Young’s modulus of (A) with Young’s modulus of (B) we get $\dfrac{{{Y_A}}}{{{Y_B}}} = \dfrac{{\sqrt 3 }}{{\dfrac{1}{{\sqrt 3 }}}} = 3$
Hence $3{Y_B} = {Y_A}$
Option (D) $3{Y_B} = {Y_A}$ is the correct answer.
Additional information Young’s modulus is defined as the measure of ability of a material to withstand changes in length when it is under a lengthwise tension of compression. It is also referred to as modulus of elasticity.
Note We find the slope in the graph by using tanθ because we are assuming the graph as a right-angled triangle. And $\tan \theta $is equal to the opposite by the adjacent side of the assumed triangle. In the case of stress vs strain graph opposite is equal to stress and adjacent is equal to the strain. Hence $\tan \theta $ is equal to Young’s modulus. Therefore, we use $\tan \theta $ to find the slope.
Recently Updated Pages
WBJEE 2026 Registration Started: Important Dates Eligibility Syllabus Exam Pattern

Dimensions of Charge: Dimensional Formula, Derivation, SI Units & Examples

How to Calculate Moment of Inertia: Step-by-Step Guide & Formulas

Circuit Switching vs Packet Switching: Key Differences Explained

Dimensions of Pressure in Physics: Formula, Derivation & SI Unit

JEE Extractive Metallurgy Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2026: Session 1 Results Out and Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

JEE Main Participating Colleges 2026 - A Complete List of Top Colleges

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Electric Field of a Uniformly Charged Ring

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Important Questions For Class 11 Physics Chapter 1 Units and Measurement - 2025-26

CBSE Notes Class 11 Physics Chapter 4 - Laws of Motion - 2025-26

CBSE Notes Class 11 Physics Chapter 14 - Waves - 2025-26

JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

CBSE Notes Class 11 Physics Chapter 9 - Mechanical Properties of Fluids - 2025-26

