Question

Modulus of Rigidity of diamond is
A. Too Less
B. Greater than all matters
C. Less than all matters
D. Zero

Answer 1

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.