A square brass plate of side `1.0 m` and thickness `0.0035m` is subjected to a force `F` on its smaller opposite edges, causing a displacement of `0.02 cm`. If the shear modulus o brass is `0.4 xx 10^(11) N//m^(2)`, the value on force `F` is

Options

A. `4 xx 10^(3)N`
B. `400 N`
C. `4 xx 10^(4)`
D. `1000 N`

Explanation

To find the force \( F \) applied to the square brass plate, we can use the relationship between stress, strain, and shear modulus. Here’s a step-by-step solution:

### Step 1: Identify the given values
- Side of the square plate, \( L = 1.0 \, \text{m} \)
- Thickness of the plate, \( h = 0.0035 \, \text{m} \)
- Displacement (shear), \( x = 0.02 \, \text{cm} = 0.0002 \, \text{m} \) (conversion from cm to m)
- Shear modulus of brass, \( \eta = 0.4 \times 10^{11} \, \text{N/m}^2 \)

### Step 2: Calculate the area of the plate
The area \( A \) on which the force is applied is given by:
\[ A = L \times h = 1.0 \, \text{m} \times 0.0035 \, \text{m} = 0.0035 \, \text{m}^2 \]

### Step 3: Calculate the shear strain
Shear strain \( \gamma \) is defined as the ratio of the displacement to the height:
\[ \gamma = \frac{x}{h} = \frac{0.0002 \, \text{m}}{1.0 \, \text{m}} = 0.0002 \]

### Step 4: Relate stress, strain, and shear modulus
The relationship between shear stress \( \tau \), shear modulus \( \eta \), and shear strain \( \gamma \) is given by:
\[ \tau = \eta \cdot \gamma \]

### Step 5: Calculate the shear stress
Substituting the values into the equation:
\[ \tau = 0.4 \times 10^{11} \, \text{N/m}^2 \times 0.0002 = 8 \times 10^{7} \, \text{N/m}^2 \]

### Step 6: Calculate the force \( F \)
Shear stress is also defined as the force per unit area:
\[ \tau = \frac{F}{A} \]
Rearranging this gives:
\[ F = \tau \cdot A \]
Substituting the values:
\[ F = 8 \times 10^{7} \, \text{N/m}^2 \times 0.0035 \, \text{m}^2 = 280000 \, \text{N} = 2.8 \times 10^{5} \, \text{N} \]

### Final Answer
The value of the force \( F \) is \( 2.8 \times 10^{5} \, \text{N} \).

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