A shearing force of `10^(6)` N is applied on the top surface of a fixed cube of side 10 cm. If the top surface is displaced by 0.33 mm, find rigidity modulus of the material of the cube.

Answer

To find the rigidity modulus of the material of the cube, we can follow these steps:

### Step 1: Calculate the Area of the Top Surface
The top surface of the cube has a side length of 10 cm. To find the area (A), we use the formula:
\[ A = \text{side}^2 = (10 \, \text{cm})^2 = (0.1 \, \text{m})^2 = 0.01 \, \text{m}^2 \]

### Step 2: Calculate the Shear Stress
Shear stress (\(\tau\)) is defined as the force (F) applied per unit area (A). Given that the shearing force is \(10^6 \, \text{N}\):
\[ \tau = \frac{F}{A} = \frac{10^6 \, \text{N}}{0.01 \, \text{m}^2} = 10^8 \, \text{N/m}^2 \]

### Step 3: Convert Shear Displacement to Meters
The displacement (\(\delta x\)) is given as 0.33 mm. We convert this to meters:
\[ \delta x = 0.33 \, \text{mm} = 0.33 \times 10^{-3} \, \text{m} = 3.3 \times 10^{-4} \, \text{m} \]

### Step 4: Calculate the Rigidity Modulus
The rigidity modulus (G) can be calculated using the formula:
\[ G = \frac{\tau}{\theta} \]
where \(\theta\) is the shear strain, defined as:
\[ \theta = \frac{\delta x}{L} \]
Here, \(L\) is the original length of the cube (0.1 m). Thus:
\[ \theta = \frac{3.3 \times 10^{-4} \, \text{m}}{0.1 \, \text{m}} = 3.3 \times 10^{-3} \]

Now substituting the values into the rigidity modulus formula:
\[ G = \frac{10^8 \, \text{N/m}^2}{3.3 \times 10^{-3}} \approx 3.03 \times 10^{10} \, \text{N/m}^2 = 30.3 \, \text{GPa} \]

### Final Answer
The rigidity modulus of the material of the cube is approximately \(30.3 \, \text{GPa}\).

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