Answer
Verified
114.9k+ 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
JEE Main 2021 July 25 Shift 2 Question Paper with Answer Key
JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key
JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key
JEE Main 2021 July 20 Shift 2 Question Paper with Answer Key
Hybridization of Atomic Orbitals Important Concepts and Tips for JEE
Atomic Structure: Complete Explanation for JEE Main 2025
Trending doubts
JEE Main 2025: Application Form (Out), Exam Dates (Released), Eligibility & More
Class 11 JEE Main Physics Mock Test 2025
Learn About Angle Of Deviation In Prism: JEE Main Physics 2025
JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics
JEE Main Login 2045: Step-by-Step Instructions and Details
Degree of Dissociation and Its Formula With Solved Example for JEE
Other Pages
NCERT Solutions for Class 11 Physics Chapter 7 Gravitation
NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Fluids
Units and Measurements Class 11 Notes - CBSE Physics Chapter 1
NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements
NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs