Question

A square brass plate of side $1.0m$ and thickness $0.005m$ is subjected to a force F on its smaller opposite edges causing a displacement of $0.02\text{ }cm$. If the shear modulus of brass is $0.4\times {{10}^{11}}N/{{m}^{2}}$ the value of the force $F$ is
(a) $0.4\times {{10}^{3}}N$
(b) $400N$
(c) $0.4\times {{10}^{4}}N$
(d) $1000N$

Answer 1

In order to solve this question, we need to use the formula for shear modulus. Shear modulus is calculated by as follows,
$Shear\text{ }modulus=Shearing\text{ }stress/shearing\text{ }stress$

Complete step-by-step answer:
From the question, given is,
$A=1{{m}^{2}}$
$h=0.005m$
$\Delta l=0.002m$
$\eta =0.4\times {{10}^{11}}$
Now,
$\eta =\dfrac{\dfrac{F}{A}}{\dfrac{\Delta l}{l}}$
$=\eta \times \dfrac{\Delta l}{l}\times A$
Here,
$\eta =0.4\times {{10}^{11}}N/{{m}^{2}}$
$\Delta l=0.02cm=\dfrac{0.02}{100}m$
$l=1.0m$
$A=l\times t$
$=(1.0\times 0.005){{m}^{2}}$
Putting these values, we get,
$F=0.4\times {{10}^{11}}\times \dfrac{0.02}{100\times 1}\times (1.0\times 0.005)$
$=0.4\times {{10}^{4}}N$

Thus, the answer to this question is option (c).

Additional Information: A lateral deformation is observed in the object when a shear force is applied to it. The elastic coefficient is known as shear modulus of rigidity. Shear modulus rigidity is the measurement of the rigidity of the object and it is obtained by measuring the ratio of shear stress of the object to the shear strain of the object.In materials science, shear modulus or modulus of rigidity, denoted by G, or sometimes S or μ, is defined as the ratio of shear stress to the shear strain: where = shear stress is the force which acts is the area on which the force acts = shear strain.

Note: While solving this question, we should be aware of the different types of formula used here. Especially shear modulus and how the different values of the variable of the formula is used from the question. The shear modulus formula is modified and used here to take out the required solution for the problem given here.