A cube of sponge rubber with edge length 5cm has a force of 2N applied horizontally to the top face (parallel to an edge) while the bottom face is held fixed. If the top face is displaced horizontally through a distance of 1mm, find the shear modulus for the sponge rubber.
$\eqalign{
  & {\text{A}}{\text{. 7}} \times {\text{1}}{{\text{0}}^4}N{m^{ - 2}} \cr
  & {\text{B}}{\text{. 6}} \times {\text{1}}{{\text{0}}^4}N{m^{ - 2}} \cr
  & {\text{C}}{\text{. 4}} \times {\text{1}}{{\text{0}}^4}N{m^{ - 2}} \cr
  & {\text{D}}{\text{. 5}} \times {\text{1}}{{\text{0}}^4}N{m^{ - 2}} \cr} $

Hint: In the question we are given that top faced is displaced by 1 mm means elongation is of 1 mm. So we know that shear modulus is defined as stress divided by strain. We will simply put values in the formula of shear modulus and obtain the answer.

Formula used:
$Stress = \dfrac{F}{A}$
$\eqalign{
  & {\text{shear modulus}}\left( G \right){\text{ }} = \dfrac{{{\text{shear stress}}}}{{{\text{shear strain}}}} \cr
  & \dfrac{{{\text{shear stress}}}}{{{\text{shear strain}}}} = \dfrac{{\dfrac{F}{A}}}{{\tan \theta }} \cr} $

Complete step by step answer:
We know that stress is defined as:
$Stress = \dfrac{F}{A}$
where A represents the cross sectional area of top faced.
And F represents the Force acting on that area.

The word shear in the formula represents that force acting is the shear force i.e. the force is pulling one end of the body in a specific direction and the other part is being subjected to the force this force in another direction. So we can equate the expression for stress with the expression of shear stress. Thus we get:
$G\left( {shear{\text{ }}modulus} \right){\text{ }} = \dfrac{{{\text{shear stress}}}}{{{\text{shear strain}}}} = \dfrac{{\dfrac{F}{A}}}{{\tan \theta }} \cdots \cdots \cdots \left( 1 \right)$
Given:
Edge length, $l = 5 \times {10^{ - 2}}m$
Force, $F = 2N$
Horizontal displacement, $\Delta l = 1 \times {10^{ - 3}}m$
For finding the shear modulus we need to find the value of $\tan \theta $ as it is unknown. This quantity is nothing but the ratio of the horizontal displacement over the edge length of the cube, so we have:
$\eqalign{
  & \tan \theta = \dfrac{{\Delta l}}{l} \cr
  & \Rightarrow \tan \theta = \dfrac{{1 \times {{10}^{ - 3}}}}{{5 \times {{10}^{ - 2}}}} \cr
  & \Rightarrow \tan \theta = 0.2 \times {10^{ - 1}} \cr
  & \therefore \tan \theta = 0.02N{m^{ - 2}} \cr} $
Now as all the quantities are known so we will find the shear modulus by substituting all the values in the equation (1):
$\eqalign{
  & G = \dfrac{{\dfrac{F}{A}}}{{\tan \theta }} \cr
  & \Rightarrow G = \dfrac{{\dfrac{2}{{{{\left( {5 \times {{10}^{ - 2}}} \right)}^2}}}}}{{0.02}}{\text{ }}\left[ {\because A = {l^2}} \right] \cr
  & \Rightarrow G = \dfrac{2}{{25 \times {{10}^{ - 4}} \times 0.02}} \cr
  & \therefore G = 4 \times {10^{ - 4}}N{m^{ - 2}} \cr} $
The shear modulus for the sponge rubber is $4 \times {10^{ - 4}}N{m^{ - 2}}$.

Therefore the correct option is C.

Note:
We can define Shear modulus as the measure of the ability of a material to resist transverse deformations. It is used for measuring the stiffness of a given material. The shear modulus of metals is usually observed to decrease as temperature increases. Additionally, at high pressure also the shear modulus tends to increase with the pressure being applied.