Question

A metal cube of side $10\,cm$ is subjected to a shearing stress of ${10^4}\,N{\text{ }}{m^{ - 2}}$ . The modulus of rigidity if the top of the cube is displaced by $0.05cm$ with respect to its bottom is
A. $2 \times {10^6}N{\text{ }}{m^{ - 2}}$
B. ${10^5}N{\text{ }}{m^{ - 2}}$
C. $1 \times {10^7}N{\text{ }}{m^{ - 2}}$
D. $4 \times {10^5}N{\text{ }}{m^{ - 2}}$

Answer 1

To solve this question, we must know about rigidity modulus, stress and strain and we are ready to solve this question after that just put the values in the formula and we will get the answer. Firstly, we will calculate shearing strain and then modulus of rigidity. Shear modulus also known as Modulus of rigidity is the measure of the rigidity of the body, given by the ratio of shear stress to shear strain.

Formula used:
${\text{Shearing strain}} = \dfrac{{\vartriangle l}}{l}$
Where, $\vartriangle l$ is the total elongation and $l$ is the original length.
$\eta = \dfrac{{{\text{Shearing stress}}}}{{{\text{Shearing strain}}}}$
Where, $\eta$ is the rigidity modulus.

Complete step by step answer:
According to the question,
Length of cube is $l = 10\,cm$ ,
Shearing displacement is $\vartriangle l = 0.05\,cm$ and
Shearing stress is ${10^4}N{\text{ }}{m^{ - 2}}$
Now, putting the values, here we will calculate shearing strain,
$\because {\text{Shearing strain}} = \dfrac{{\vartriangle l}}{l} \\\ \Rightarrow {\text{Shearing strain}} = \dfrac{{0.05}}{{10}} \\\ \Rightarrow {\text{Shearing strain}} = 0.005m \\\$
Now, putting the values to calculate modulus of rigidity,
$\because \eta = \dfrac{{{\text{Shearing stress}}}}{{{\text{Shearing strain}}}} \\\ \Rightarrow \eta = \dfrac{{{{10}^4}}}{{0.005}} \\\ \Rightarrow \eta = \dfrac{{10000 \times 1000}}{5} \\\ \Rightarrow \eta = 2000000 \\\ \therefore \eta = 2 \times {10^6}N{\text{ }}{m^{ - 2}}$
Hence the modulus of rigidity is $2 \times {10^6}N{\text{ }}{m^{ - 2}}$.

Therefore, the correct option is A.

Note: Shearing stress and modulus of rigidity are two different things. Shear Modulus of elasticity is one of the measures of mechanical properties of solids. Other elastic moduli are young’s modulus and bulk modulus. The shear modulus of material gives us the ratio of shear stress to shear strain in a body. It can be used to explain how a material resists transverse deformations but this is practical for small deformations only, following which they are able to return to the original state. This is because large shearing forces lead to permanent deformations (no longer elastic body).