A square lead slab of side 50 cm and thickness 10 cm is subjected to a shearing force (on its narrow face) of `9xx10^4N`. The lower edge is riveted to the floor. How much will the upper edge be displaced? (Shear modulus of lead`=5.6xx10^9Nm^-2`)

Options

A. 0.16mm
B. 1.6mm
C. 0.16 cm
D. 1.6 cm

Explanation

To solve the problem, we need to find the displacement of the upper edge of a square lead slab subjected to a shearing force. We will use the formula relating shear stress, shear strain, and shear modulus.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Side of the square slab, \( a = 50 \, \text{cm} = 0.5 \, \text{m} \)
- Thickness of the slab, \( t = 10 \, \text{cm} = 0.1 \, \text{m} \)
- Shearing force, \( F = 9 \times 10^4 \, \text{N} \)
- Shear modulus of lead, \( G = 5.6 \times 10^9 \, \text{N/m}^2 \)

2. **Calculate the Area of the Narrow Face:**
The area \( A \) of the narrow face subjected to the shearing force is given by:
\[ A = \text{side} \times \text{thickness} = a \times t = 0.5 \, \text{m} \times 0.1 \, \text{m} = 0.05 \, \text{m}^2 \]

3. **Use the Shear Modulus Formula:**
The shear modulus \( G \) is defined as:
\[ G = \frac{\text{Shear Stress}}{\text{Shear Strain}} \]
where:
- Shear Stress \( \tau = \frac{F}{A} \)
- Shear Strain \( \gamma = \frac{\Delta L}{L} \)

Rearranging gives:
\[ G = \frac{F/A}{\Delta L / L} \implies \Delta L = \frac{F \cdot L}{A \cdot G} \]

4. **Calculate the Original Length \( L \):**
The original length \( L \) is equal to the side of the slab:
\[ L = 0.5 \, \text{m} \]

5. **Substitute the Values into the Formula:**
Now we can substitute the values into the equation for \( \Delta L \):
\[ \Delta L = \frac{(9 \times 10^4 \, \text{N}) \cdot (0.5 \, \text{m})}{(0.05 \, \text{m}^2) \cdot (5.6 \times 10^9 \, \text{N/m}^2)} \]

6. **Calculate \( \Delta L \):**
\[ \Delta L = \frac{4.5 \times 10^4}{0.28 \times 10^9} = \frac{4.5 \times 10^4}{2.8 \times 10^8} \]
\[ \Delta L = 1.607 \times 10^{-4} \, \text{m} = 0.0001607 \, \text{m} = 0.1607 \, \text{mm} \]

7. **Final Result:**
The displacement of the upper edge of the slab is approximately:
\[ \Delta L \approx 0.16 \, \text{mm} \]