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

Explanation

To solve the problem step by step, we will use the formula for shear modulus and the relationship between shear stress, shear strain, and displacement.

### Step 1: Identify the given values
- Side of the square slab (a) = 50 cm = 0.5 m
- Thickness of the slab (t) = 10 cm = 0.1 m
- Shearing force (F) = \(9 \times 10^4\) N
- Shear modulus of lead (G) = \(5.6 \times 10^9\) N/m²

### Step 2: Calculate the area of the narrow face subjected to shear
The area (A) of the narrow face of the slab can be calculated as:
\[ A = \text{side} \times \text{thickness} = a \times t = 0.5 \, \text{m} \times 0.1 \, \text{m} = 0.05 \, \text{m}^2 \]

### Step 3: Calculate the shear stress
Shear stress (\(\tau\)) is given by the formula:
\[ \tau = \frac{F}{A} \]
Substituting the values:
\[ \tau = \frac{9 \times 10^4 \, \text{N}}{0.05 \, \text{m}^2} = 1.8 \times 10^6 \, \text{N/m}^2 \]

### Step 4: Relate shear stress to shear strain
Using the formula for shear modulus:
\[ G = \frac{\tau}{\phi} \]
Where \(\phi\) is the shear strain. Rearranging gives:
\[ \phi = \frac{\tau}{G} \]
Substituting the values:
\[ \phi = \frac{1.8 \times 10^6 \, \text{N/m}^2}{5.6 \times 10^9 \, \text{N/m}^2} = 3.21428571 \times 10^{-4} \]

### Step 5: Calculate the displacement
Shear strain (\(\phi\)) is also defined as:
\[ \phi = \frac{\Delta L}{L} \]
Where \(\Delta L\) is the displacement and \(L\) is the height of the slab (0.5 m). Rearranging gives:
\[ \Delta L = \phi \times L \]
Substituting the values:
\[ \Delta L = 3.21428571 \times 10^{-4} \times 0.5 \, \text{m} = 1.607142855 \times 10^{-4} \, \text{m} = 0.0001607142855 \, \text{m} \]
Converting to mm:
\[ \Delta L = 0.0001607142855 \, \text{m} \times 1000 \, \text{mm/m} = 0.1607142855 \, \text{mm} \approx 0.16 \, \text{mm} \]

### Final Answer
The upper edge of the slab will be displaced approximately **0.16 mm**.

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