A metal cube of side length `8.0cm` has its upper surface displacement with respect to the bottom by `0.10mm` when a tangential force of `4xx10^(9)N` is applied at the top with bottom surface fixed. The rigidity modulus of the material of the cube is

Options

A. `4xx10^(14) N//m^(2)`
B. `5xx10^(14) N//m^(2)`
C. `8xx10^(14) N//m^(2)`
D. `1xx10^(14) N//m^(2)`

Explanation

To find the rigidity modulus (also known as the shear modulus) of the material of the cube, we can use the formula:

\[ \text{Rigidity Modulus} (Y) = \frac{F/A}{\Delta L/L} \]

Where:
- \( F \) is the applied tangential force,
- \( A \) is the area of the surface,
- \( \Delta L \) is the displacement (shear),
- \( L \) is the original length of the side of the cube.

### Step 1: Identify the given values
- Side length of the cube, \( L = 8.0 \, \text{cm} = 8.0 \times 10^{-2} \, \text{m} \)
- Displacement, \( \Delta L = 0.10 \, \text{mm} = 0.10 \times 10^{-3} \, \text{m} \)
- Tangential force, \( F = 4 \times 10^{9} \, \text{N} \)

### Step 2: Calculate the area \( A \)
The area \( A \) of the upper surface of the cube is given by:

\[ A = L^2 = (8.0 \times 10^{-2} \, \text{m})^2 = 6.4 \times 10^{-3} \, \text{m}^2 \]

### Step 3: Calculate the ratio \( \Delta L/L \)
Now, we can calculate the ratio of the displacement to the original length:

\[ \frac{\Delta L}{L} = \frac{0.10 \times 10^{-3} \, \text{m}}{8.0 \times 10^{-2} \, \text{m}} = \frac{0.10 \times 10^{-3}}{8.0 \times 10^{-2}} = 1.25 \times 10^{-3} \]

### Step 4: Calculate the rigidity modulus \( Y \)
Now we can substitute the values into the rigidity modulus formula:

\[ Y = \frac{F/A}{\Delta L/L} = \frac{4 \times 10^{9} \, \text{N} / (6.4 \times 10^{-3} \, \text{m}^2)}{1.25 \times 10^{-3}} \]

Calculating the force per unit area:

\[ \frac{F}{A} = \frac{4 \times 10^{9}}{6.4 \times 10^{-3}} = 6.25 \times 10^{11} \, \text{N/m}^2 \]

Now substituting this back into the equation for \( Y \):

\[ Y = \frac{6.25 \times 10^{11}}{1.25 \times 10^{-3}} = 5.0 \times 10^{14} \, \text{N/m}^2 \]

### Final Answer
The rigidity modulus of the material of the cube is:

\[ Y = 5.0 \times 10^{14} \, \text{N/m}^2 \]