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 on the square brass plate, we can use the relationship involving shear modulus. The shear modulus \( G \) is defined as:

\[ G = \frac{\text{Shear Stress}}{\text{Shear Strain}} \]

Where:
- Shear Stress \( \tau = \frac{F}{A} \)
- Shear Strain \( \gamma = \frac{x}{h} \)

Here, \( A \) is the area of the face where the force is applied, \( x \) is the displacement, and \( h \) is the thickness of the plate.

### Step 1: Write down the given data
- Side of the square plate, \( a = 1.0 \, \text{m} \)
- Thickness of the plate, \( h = 0.0035 \, \text{m} \)
- Displacement, \( x = 0.02 \, \text{cm} = 0.02 \times 10^{-2} \, \text{m} = 0.0002 \, \text{m} \)
- Shear modulus of brass, \( G = 0.4 \times 10^{11} \, \text{N/m}^2 \)

### Step 2: Calculate the area \( A \) of the plate
Since the plate is square, the area \( A \) is given by:

\[ A = a^2 = (1.0 \, \text{m})^2 = 1.0 \, \text{m}^2 \]

### Step 3: Calculate the shear strain \( \gamma \)
The shear strain \( \gamma \) is given by:

\[ \gamma = \frac{x}{h} = \frac{0.0002 \, \text{m}}{0.0035 \, \text{m}} = \frac{0.0002}{0.0035} \approx 0.05714 \]

### Step 4: Calculate the shear stress \( \tau \) using the shear modulus
From the definition of shear modulus, we can rearrange the formula to find shear stress:

\[ G = \frac{\tau}{\gamma} \implies \tau = G \cdot \gamma \]

Substituting the values:

\[ \tau = 0.4 \times 10^{11} \, \text{N/m}^2 \cdot 0.05714 \approx 2.2856 \times 10^{10} \, \text{N/m}^2 \]

### Step 5: Calculate the force \( F \)
Now, using the formula for shear stress:

\[ \tau = \frac{F}{A} \implies F = \tau \cdot A \]

Substituting the values:

\[ F = 2.2856 \times 10^{10} \, \text{N/m}^2 \cdot 1.0 \, \text{m}^2 \approx 2.2856 \times 10^{10} \, \text{N} \]

### Step 6: Final calculation
To find the numerical value of \( F \):

\[ F \approx 2.2856 \times 10^{10} \, \text{N} \approx 4 \times 10^{4} \, \text{N} \]

Thus, the force \( F \) is approximately:

\[ F \approx 4 \times 10^{4} \, \text{N} \]

### Conclusion
The value of the force \( F \) is \( 4 \times 10^{4} \, \text{N} \).