A lead cube of side 20 cm is subjected under a shearing force of magnitude `5.6 x 10^6 N`. If the lower face of the cube is fixed, then the displacement in the upper face is (If shear modulus of lead is `5.6 * 10^6 N/m^2)`

Options

A. 15 mm
B. 10 mm
C. 20 mm
D. 5 m

Answer

To solve the problem, we need to find the displacement in the upper face of a lead cube subjected to a shearing force. We will use the formula for shear modulus, which relates shear stress, shear strain, and shear modulus.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Side of the cube, \( a = 20 \, \text{cm} = 0.2 \, \text{m} \)
- Shearing force, \( F = 5.6 \times 10^6 \, \text{N} \)
- Shear modulus of lead, \( G = 5.6 \times 10^6 \, \text{N/m}^2 \)

2. **Calculate the Area of the Upper Face:**
\[ \text{Area} (A) = a^2 = (0.2 \, \text{m})^2 = 0.04 \, \text{m}^2 \]

3. **Calculate Shear Stress (\( \tau \)):**
Shear stress is given by the formula:
\[ \tau = \frac{F}{A} \]
Substituting the values:
\[ \tau = \frac{5.6 \times 10^6 \, \text{N}}{0.04 \, \text{m}^2} = 1.4 \times 10^8 \, \text{N/m}^2 \]

4. **Relate Shear Stress, Shear Modulus, and Shear Strain:**
The relationship is given by:
\[ G = \frac{\tau}{\phi} \]
Rearranging gives:
\[ \phi = \frac{\tau}{G} \]

5. **Calculate Shear Strain (\( \phi \)):**
Substituting the values:
\[ \phi = \frac{1.4 \times 10^8 \, \text{N/m}^2}{5.6 \times 10^6 \, \text{N/m}^2} = 25 \]

6. **Calculate Displacement (\( x \)):**
Shear strain is defined as:
\[ \phi = \frac{x}{L} \]
where \( L \) is the original length (or height) of the cube. Here, \( L = a = 0.2 \, \text{m} \). Rearranging gives:
\[ x = \phi \times L = 25 \times 0.2 \, \text{m} = 5 \, \text{m} \]

### Final Answer:
The displacement in the upper face of the cube is \( 5 \, \text{m} \).

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