A cube of side 40 cm has its upper face displaced by 0.1 mm by a tangential force of 8 Kn. The shearing modulus of cube is :-

Options

A. `2xx10^(9)Nm^(-2)`
B. `4xx10^(9)Nm^(-2)`
C. `8xx10^(9)Nm^(-2)`
D. `16xx10^(9)Nm^(-2)`

Explanation

To find the shearing modulus of the cube, we will use the formula for shearing modulus (also known as the modulus of rigidity), which is given by:

\[ G = \frac{\text{Shearing Stress}}{\text{Shearing Strain}} \]

### Step 1: Calculate the Shearing Stress
Shearing stress (\( \tau \)) is defined as the tangential force (\( F \)) applied per unit area (\( A \)).

The formula for shearing stress is:

\[ \tau = \frac{F}{A} \]

Given:
- Force, \( F = 8 \text{ kN} = 8 \times 10^3 \text{ N} \)
- The area of the upper face of the cube, \( A = \text{side}^2 = (40 \text{ cm})^2 = (0.4 \text{ m})^2 = 0.16 \text{ m}^2 \)

Now, substituting the values into the formula:

\[ \tau = \frac{8 \times 10^3 \text{ N}}{0.16 \text{ m}^2} = 50000 \text{ N/m}^2 = 5 \times 10^4 \text{ N/m}^2 \]

### Step 2: Calculate the Shearing Strain
Shearing strain (\( \gamma \)) is defined as the displacement (\( \Delta L \)) divided by the original length (\( L \)).

The formula for shearing strain is:

\[ \gamma = \frac{\Delta L}{L} \]

Given:
- Displacement, \( \Delta L = 0.1 \text{ mm} = 0.1 \times 10^{-3} \text{ m} \)
- Original length (side of the cube), \( L = 40 \text{ cm} = 0.4 \text{ m} \)

Now, substituting the values into the formula:

\[ \gamma = \frac{0.1 \times 10^{-3} \text{ m}}{0.4 \text{ m}} = \frac{0.1 \times 10^{-3}}{0.4} = 0.25 \times 10^{-3} = 2.5 \times 10^{-4} \]

### Step 3: Calculate the Shearing Modulus
Now that we have both shearing stress and shearing strain, we can calculate the shearing modulus (\( G \)):

\[ G = \frac{\tau}{\gamma} \]

Substituting the values we calculated:

\[ G = \frac{5 \times 10^4 \text{ N/m}^2}{2.5 \times 10^{-4}} = 2 \times 10^8 \text{ N/m}^2 \]

### Final Answer
The shearing modulus of the cube is:

\[ G = 2 \times 10^8 \text{ N/m}^2 \]