A cube of soft rubber of face area `0.02 m^(2)` whose lower face is fixed is sheared through an angle `2^(@)` by a force of `10^(4) N` acting tangential to the upper face. Calculate the rigidity modulus of the material.

Explanation

To solve the problem, we need to calculate the rigidity modulus (shear modulus) of the soft rubber cube using the given data. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify the Given Values
- Face area of the cube, \( A = 0.02 \, m^2 \)
- Force applied, \( F = 10^4 \, N \)
- Angle of shear, \( \alpha = 2^\circ \)

### Step 2: Calculate the Shear Stress
Shear stress (\( \tau \)) is defined as the force applied per unit area. The formula for shear stress is:

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

Substituting the known values:

\[ \tau = \frac{10^4 \, N}{0.02 \, m^2} = 5 \times 10^5 \, N/m^2 \]

### Step 3: Calculate the Shear Strain
Shear strain (\( \gamma \)) can be approximated by the tangent of the angle of shear. The formula is:

\[ \gamma = \tan(\alpha) \]

Since \( \alpha = 2^\circ \), we convert it to radians for calculation if necessary, but for small angles, we can use:

\[ \gamma \approx \tan(2^\circ) \approx 0.0349 \, \text{(using a calculator)} \]

### Step 4: Use the Relationship Between Shear Stress, Shear Strain, and Rigidity Modulus
The relationship between shear stress, shear strain, and rigidity modulus (\( S \)) is given by:

\[ \tau = S \cdot \gamma \]

Rearranging this gives:

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

### Step 5: Substitute the Values to Find Rigidity Modulus
Now, substituting the values of shear stress and shear strain into the equation:

\[ S = \frac{5 \times 10^5 \, N/m^2}{0.0349} \]

Calculating this gives:

\[ S \approx 1.43 \times 10^7 \, N/m^2 \]

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

\[ S \approx 1.43 \times 10^7 \, N/m^2 \]

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