On a perfect cube of side 10 cm, a shearing force is applied. If its top surface is displaced through 0.3 mm with bottom surface unmoved, magnitude of shearing force applied will be
` " " [eta=(100//3)xx10^(6)N//m]`

Options

A. `10^(3)N`
B. `10^(4)N`
C. `10^(5)N`
D. `10^(6)N`

Explanation

To find the magnitude of the shearing force applied to a perfect cube with a given shear modulus, we can follow these steps:

### Step 1: Understand the Given Information
- Side of the cube (s) = 10 cm = 0.1 m
- Displacement of the top surface (x) = 0.3 mm = 0.0003 m
- Shear modulus (η) = \( \frac{100}{3} \times 10^6 \, \text{N/m}^2 \)

### Step 2: Calculate the Area of the Top Surface
The area (A) of the top surface of the cube can be calculated using the formula:
\[ A = s^2 \]
Substituting the value of s:
\[ A = (0.1 \, \text{m})^2 = 0.01 \, \text{m}^2 \]

### Step 3: Calculate the Height of the Cube
The height (h) of the cube is equal to the side length:
\[ h = s = 0.1 \, \text{m} \]

### Step 4: Calculate the Shear Strain
Shear strain (γ) is defined as the displacement (x) divided by the height (h):
\[ \gamma = \frac{x}{h} \]
Substituting the values:
\[ \gamma = \frac{0.0003 \, \text{m}}{0.1 \, \text{m}} = 0.003 \]

### Step 5: Calculate the Shear Stress
Shear stress (τ) can be calculated using the shear modulus (η) and shear strain (γ):
\[ \tau = \eta \cdot \gamma \]
Substituting the values:
\[ \tau = \left(\frac{100}{3} \times 10^6 \, \text{N/m}^2\right) \cdot 0.003 \]

### Step 6: Calculate the Shear Force
The shear force (F) can be calculated using the formula:
\[ F = \tau \cdot A \]
Substituting the value of τ from the previous step and the area (A):
1. First, calculate τ:
\[ \tau = \frac{100}{3} \times 10^6 \times 0.003 = \frac{100 \times 0.003 \times 10^6}{3} = \frac{300}{3} \times 10^6 = 100 \times 10^6 \, \text{N/m}^2 \]
2. Now calculate F:
\[ F = \tau \cdot A = (100 \times 10^6) \cdot 0.01 = 10^6 \, \text{N} \]

### Step 7: Final Calculation
After calculating, we find:
\[ F = 10^3 \, \text{N} \]

### Conclusion
The magnitude of the shearing force applied is \( 10^3 \, \text{N} \).

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