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Question

Question: A cube of side \(40\,mm\) has its upper face displaced by \(0.1\,mm\) by a tangential force of \(8\,...

A cube of side 40mm40\,mm has its upper face displaced by 0.1mm0.1\,mm by a tangential force of 8kN8\,kN . The shearing modulus of cube is
A. 2×109N m22 \times {10^9}N{\text{ }}{m^{ - 2}}
B. 4×109N m24 \times {10^9}N{\text{ }}{m^{ - 2}}
C. 8×109N m28 \times {10^9}N{\text{ }}{m^{ - 2}}
D. 16×109N m216 \times {10^9}N{\text{ }}{m^{ - 2}}

Explanation

Solution

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:
η=Shearing stressShearing strain\eta = \dfrac{{{\text{Shearing stress}}}}{{{\text{Shearing strain}}}}
Where, η\eta is the shearing modulus.
Shearing stress = Shearing forceArea being sheared{\text{Shearing stress = }}\dfrac{{{\text{Shearing force}}}}{{{\text{Area being sheared}}}}
Shearing strain=ll{\text{Shearing strain}} = \dfrac{{\vartriangle l}}{l}
Where, l\vartriangle l is the total elongation and ll is the original length.

Complete step by step answer:
According to the question
Force applied is 8kN=8000N8kN = 8000\,N and
Area being sheared is 40×40=1600mm240 \times 40 = 1600\,m{m^2}
Shearing stress = 80001600{\text{Shearing stress = }}\dfrac{{8000}}{{1600}}
Shearing stress = 5×106Pa\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\vartriangle l = 0.1mm and original length is 40mm40\,mm .
Shearing strain=ll \because {\text{Shearing strain}} = \dfrac{{\vartriangle l}}{l} \\\
\Rightarrow {\text{Shearing strain}} = \dfrac{{0.1}}{{40}} \\\
\Rightarrow {\text{Shearing strain}} = 2.5 \times {10^{ - 3}} \\\ -----(2)
Now here we will calculate, the shearing modulus
η=Shearing stressShearing strain\eta = \dfrac{{{\text{Shearing stress}}}}{{{\text{Shearing strain}}}}
η=5×1062.5×103 η=5×1092.5 η=2×109 \Rightarrow \eta = \dfrac{{5 \times {{10}^6}}}{{2.5 \times {{10}^{ - 3}}}} \\\ \Rightarrow \eta = \dfrac{{5 \times {{10}^9}}}{{2.5}} \\\ \Rightarrow \eta = 2 \times {10^9} \\\
η=2×109N m2\therefore \eta = 2 \times {10^9}N{\text{ }}{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.