The shear modulus for a metal is `50000 Mpa`. Suppose that a shear force of `200 N` is applied on the upper surface of a cube of this metal that is `3.0 cm` on each edge. How far will the top surface be displaced?

Options

A. `2.67 xx 10^-7 m`
B. `4.5 xx 10^-7 m`
C. `6.0 xx 10^-7 m`
D. `1.33 xx 10^-7 m`

Explanation

To solve the problem, we need to determine how far the top surface of the cube will be displaced when a shear force is applied. We will use the relationship between shear stress, shear strain, and shear modulus.

### Step-by-Step Solution:

1. **Identify Given Values**:
- Shear modulus (G) = 50,000 MPa = \(50,000 \times 10^6 \, \text{Pa} = 5 \times 10^{10} \, \text{Pa}\)
- Shear force (F) = 200 N
- Edge length of the cube (L) = 3.0 cm = \(3.0 \times 10^{-2} \, \text{m}\)

2. **Calculate the Area of the Cube's Surface**:
- The area (A) of the top surface of the cube is given by:
\[ A = L^2 = (3.0 \times 10^{-2})^2 = 9.0 \times 10^{-4} \, \text{m}^2 \]

3. **Calculate Shear Stress**:
- Shear stress (\(\tau\)) is defined as the force per unit area:
\[ \tau = \frac{F}{A} = \frac{200 \, \text{N}}{9.0 \times 10^{-4} \, \text{m}^2} \approx 222222.22 \, \text{Pa} \]

4. **Calculate Shear Strain**:
- Shear strain (\(\gamma\)) is given by the formula:
\[ \gamma = \frac{\tau}{G} \]
- Substituting the values:
\[ \gamma = \frac{222222.22 \, \text{Pa}}{5 \times 10^{10} \, \text{Pa}} \approx 4.4444 \times 10^{-6} \]

5. **Relate Shear Strain to Displacement**:
- Shear strain is also defined as the change in length (displacement, \(\Delta L\)) divided by the original length (L):
\[ \gamma = \frac{\Delta L}{L} \]
- Rearranging gives:
\[ \Delta L = \gamma \cdot L \]
- Substituting the values:
\[ \Delta L = 4.4444 \times 10^{-6} \cdot (3.0 \times 10^{-2}) \approx 1.3333 \times 10^{-7} \, \text{m} \]

6. **Convert Displacement to Micrometers**:
- To express \(\Delta L\) in micrometers:
\[ \Delta L \approx 0.1333 \, \mu m \]

### Final Answer:
The top surface of the cube will be displaced approximately **0.1333 micrometers**.