Question

A cube of side $40mm$ has its upper face displaced by $0.1mm$ by a tangential force of $8kN$ . The shearing modulus of cube is
A. $2\times {10}^{9}N{m}^{-2}$
B. $4\times {10}^{9}N{m}^{-2}$
C. $8\times {10}^{9}N{m}^{-2}$
D. $16\times {10}^{9}N{m}^{-2}$

Answer 1

Hint:To solve this question, we must know about shearing 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:
$\eta ={\displaystyle \frac{\text{Shearing stress}}{\text{Shearing strain}}}$
Where, $\eta$ is the shearing modulus.
$\text{Shearing stress = }{\displaystyle \frac{\text{Shearing force}}{\text{Area being sheared}}}$
$\text{Shearing strain}={\displaystyle \frac{△l}{l}}$
Where, $△l$ is the total elongation and $l$ is the original length.

Complete step by step answer:
According to the question
Force applied is $8kN=8000N$ and
Area being sheared is $40\times 40=1600m{m}^2$
$\text{Shearing stress = }{\displaystyle \frac{8000}{1600}}$
$\Rightarrow \text{Shearing stress = }5\times {10}^{6}Pa$ -----(1)
Now, we have to calculate Shear strain according to the question, total elongation is $△l=0.1mm$ and original length is $40mm$ .
$\because \text{Shearing strain}={\displaystyle \frac{△l}{l}}$
$\Rightarrow \text{Shearing strain}={\displaystyle \frac{0.1}{40}}$
$\Rightarrow \text{Shearing strain}=2.5\times {10}^{-3}$ -----(2)
Now here we will calculate, the shearing modulus
$\eta ={\displaystyle \frac{\text{Shearing stress}}{\text{Shearing strain}}}$
$$\begin{gathered} \Rightarrow \eta ={\displaystyle \frac{5\times {10}^6}{2.5\times {10}^{-3}}} \\ \Rightarrow \eta ={\displaystyle \frac{5\times {10}^9}{2.5}} \\ \Rightarrow \eta =2\times {10}^9 \end{gathered}$$
$\therefore \eta =2\times {10}^{9}N{m}^{-2}$

Hence, the correct option is A.

Note:Shear strain is measured in radians and hence has no units.Shear stress arises due to shear forces. They are the pair of forces acting on opposite sides of a body with the same magnitude and opposite direction. Shear stress is a vector quantity. Which means, here the direction is also involved along with magnitude.