
Modulus of Rigidity of diamond is
A. Too Less
B. Greater than all matters
C. Less than all matters
D. Zero
Answer
163.8k+ views
Hint:In the case, when a problem is based on Mechanical properties of solids, we know that the Modulus of Rigidity plays a significant role in establishing a relationship between shear stress and shear strain hence, use this formula of modulus of rigidity in the given problem in order to select a most appropriate option.
Complete answer:
We know that $Modulus{\text{ }}of{\text{ }}Rigidity(\eta ) = \dfrac{{ShearStress({\sigma _s})}}{{ShearStrain(\theta )}}$ … (1)
But $ShearStress({\sigma _s}) = \dfrac{{Force}}{{Area}} = \dfrac{F}{A}$ and $ShearStrain = \theta = \dfrac{{\Delta x}}{l}$
From eq. (1), we get
$ \Rightarrow \eta = \dfrac{{\dfrac{F}{A}}}{\theta } = \dfrac{F}{{A\theta }}$
Now, we know that “The hardest substance known in existence is Diamond”. Therefore, to produce even a very little amount of strain in diamond, we have to exert very high stress.
i.e., For $Shear{\text{ }}Strain{\text{ }}\left( \downarrow \right)$ ; $Shear{\text{ }}Stress{\text{ }}\left( \uparrow \right)$ and therefore, modulus of rigidity will be high for diamond $\eta {\text{ }}\left( \uparrow \right)$ .
$\therefore {\eta _{Diamond}} > {\eta _{all{\text{ }}matters}}$
Thus, Modulus of Rigidity of diamond is greater than all matters.
Hence, the correct option is (B) Greater than all matters
Therefore, the answer is option (B)
Note: Since this is a conceptual problem related to stress-strain analysis in Mechanical properties of solids hence, given conditions must be analyzed very carefully to give a precise explanation. While writing an explanation for this kind of conceptual problem, always keep in mind to provide the exact reasons in support of your explanation.
Complete answer:
We know that $Modulus{\text{ }}of{\text{ }}Rigidity(\eta ) = \dfrac{{ShearStress({\sigma _s})}}{{ShearStrain(\theta )}}$ … (1)
But $ShearStress({\sigma _s}) = \dfrac{{Force}}{{Area}} = \dfrac{F}{A}$ and $ShearStrain = \theta = \dfrac{{\Delta x}}{l}$
From eq. (1), we get
$ \Rightarrow \eta = \dfrac{{\dfrac{F}{A}}}{\theta } = \dfrac{F}{{A\theta }}$
Now, we know that “The hardest substance known in existence is Diamond”. Therefore, to produce even a very little amount of strain in diamond, we have to exert very high stress.
i.e., For $Shear{\text{ }}Strain{\text{ }}\left( \downarrow \right)$ ; $Shear{\text{ }}Stress{\text{ }}\left( \uparrow \right)$ and therefore, modulus of rigidity will be high for diamond $\eta {\text{ }}\left( \uparrow \right)$ .
$\therefore {\eta _{Diamond}} > {\eta _{all{\text{ }}matters}}$
Thus, Modulus of Rigidity of diamond is greater than all matters.
Hence, the correct option is (B) Greater than all matters
Therefore, the answer is option (B)
Note: Since this is a conceptual problem related to stress-strain analysis in Mechanical properties of solids hence, given conditions must be analyzed very carefully to give a precise explanation. While writing an explanation for this kind of conceptual problem, always keep in mind to provide the exact reasons in support of your explanation.
Recently Updated Pages
Uniform Acceleration - Definition, Equation, Examples, and FAQs

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 Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Degree of Dissociation and Its Formula With Solved Example for JEE

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

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

Motion in a Straight Line Class 11 Notes: CBSE Physics Chapter 2

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line

Important Questions for CBSE Class 11 Physics Chapter 1 - Units and Measurement
