A metal plate has the dimensions `10cm xx 10cm xx 1mm`. One of its faces having larger area is fixed and a tangential force is applied to the opposite larger face. If the lateral displacement between the two surfaces is `1.2 xx 10^(-3) mm`, and the modulus of rigidity of the metal is `5 xx 10^(10) N//m^(2)`, then the tangential force is

Options

A. `2xx10^(5) N`
B. `4xx10^(5) N`
C. `6xx10^(5) N`
D. `8xx10^(5) N`

Explanation

To solve the problem step by step, we will use the formula for shear stress and shear strain, as well as the relationship between them through the modulus of rigidity.

### Step 1: Understand the dimensions of the plate
The dimensions of the plate are given as:
- Length (L) = 10 cm = 0.1 m
- Width (W) = 10 cm = 0.1 m
- Thickness (t) = 1 mm = 0.001 m

The area (A) of the face where the tangential force is applied is:
\[ A = L \times W = 0.1 \, \text{m} \times 0.1 \, \text{m} = 0.01 \, \text{m}^2 \]

### Step 2: Identify the given values
- Lateral displacement (x) = \( 1.2 \times 10^{-3} \, \text{mm} = 1.2 \times 10^{-6} \, \text{m} \)
- Modulus of rigidity (η) = \( 5 \times 10^{10} \, \text{N/m}^2 \)

### Step 3: Calculate shear strain (φ)
Shear strain (φ) is defined as the ratio of lateral displacement (x) to the original length (L):
\[ \phi = \frac{x}{L} \]
Here, the original length (L) is the thickness of the plate:
\[ L = 0.001 \, \text{m} \]
Thus,
\[ \phi = \frac{1.2 \times 10^{-6} \, \text{m}}{0.001 \, \text{m}} = 1.2 \times 10^{-3} \]

### Step 4: Use the relationship between shear stress, shear strain, and modulus of rigidity
The relationship is given by:
\[ \eta = \frac{\text{Shear Stress}}{\text{Shear Strain}} \]
Shear stress (τ) can be expressed as:
\[ \tau = \frac{F}{A} \]
Where F is the tangential force and A is the area. Therefore:
\[ \eta = \frac{F/A}{\phi} \]
Rearranging gives:
\[ F = \eta \cdot \phi \cdot A \]

### Step 5: Substitute the known values into the equation
Now we can substitute the values we have:
\[ F = (5 \times 10^{10} \, \text{N/m}^2) \cdot (1.2 \times 10^{-3}) \cdot (0.01 \, \text{m}^2) \]

### Step 6: Calculate the force
Calculating:
\[ F = 5 \times 10^{10} \cdot 1.2 \times 10^{-3} \cdot 0.01 \]
\[ F = 5 \times 1.2 \times 0.01 \times 10^{10} \times 10^{-3} \]
\[ F = 6 \times 10^{5} \, \text{N} \]

### Final Answer
The tangential force is:
\[ F = 6 \times 10^{5} \, \text{N} \]