One of the square faces of a metal slab of side 50 cm and thickness 20 cm is rigidly fixed on a horizontal surface. If a tangential force of 30 N is applied to the top face and it is known that the shear modulus of the material is `4 xx 10^(10) N//m^(2)`, then the displacement (in m) of the top face is

Options

A. `4 xx 10^(-12)`m
B. `4 xx 10^(-10)`m
C. `6 xx 10^(-10)`m
D. `8 xx 10^(-10)`m

Answer

To solve the problem, we need to determine the displacement (ΔL) of the top face of the metal slab when a tangential force is applied. We can use the formula related to shear modulus (η) to find this displacement.

### Step-by-step Solution:

1. **Identify the given values:**
- Side of the square face of the slab (s) = 50 cm = 0.5 m
- Thickness of the slab (L) = 20 cm = 0.2 m
- Tangential force (F) = 30 N
- Shear modulus (η) = \(4 \times 10^{10} \, \text{N/m}^2\)

2. **Calculate the area (A) of the square face:**
\[ A = s^2 = (0.5 \, \text{m})^2 = 0.25 \, \text{m}^2 \]

3. **Use the formula for shear modulus:**
The relationship between shear modulus, force, area, and displacement is given by:
\[ \eta = \frac{F \cdot L}{A \cdot \Delta L} \]
Rearranging this formula to solve for displacement (ΔL):
\[ \Delta L = \frac{F \cdot L}{A \cdot \eta} \]

4. **Substitute the known values into the formula:**
\[ \Delta L = \frac{30 \, \text{N} \cdot 0.2 \, \text{m}}{0.25 \, \text{m}^2 \cdot 4 \times 10^{10} \, \text{N/m}^2} \]

5. **Calculate the numerator:**
\[ 30 \, \text{N} \cdot 0.2 \, \text{m} = 6 \, \text{N} \cdot \text{m} \]

6. **Calculate the denominator:**
\[ 0.25 \, \text{m}^2 \cdot 4 \times 10^{10} \, \text{N/m}^2 = 1 \times 10^{10} \, \text{N} \]

7. **Now substitute these values back into the equation for ΔL:**
\[ \Delta L = \frac{6 \, \text{N} \cdot \text{m}}{1 \times 10^{10} \, \text{N}} = 6 \times 10^{-10} \, \text{m} \]

8. **Final answer:**
The displacement of the top face is:
\[ \Delta L = 6 \times 10^{-10} \, \text{m} \]