The force required to punch a hole of diameter 2 mm, will be (If a shearing stress `4xx10^(8)N//m^(2)`.)

Options

A. `400pi N`
B. `1600 pi N`
C. `1800pi N `
D. `1200pi N`

Explanation

To solve the problem of finding the force required to punch a hole of diameter 2 mm under a shearing stress of \(4 \times 10^{8} \, \text{N/m}^{2}\), we can follow these steps:

### Step 1: Identify the given values
- Diameter of the hole, \(D = 2 \, \text{mm} = 2 \times 10^{-3} \, \text{m}\)
- Shearing stress, \(S = 4 \times 10^{8} \, \text{N/m}^{2}\)

### Step 2: Calculate the radius of the hole
The radius \(r\) is half of the diameter:
\[ r = \frac{D}{2} = \frac{2 \times 10^{-3}}{2} = 1 \times 10^{-3} \, \text{m} \]

### Step 3: Calculate the area of the hole
The area \(A\) of the circular hole can be calculated using the formula for the area of a circle:
\[ A = \pi r^2 \]
Substituting the value of \(r\):
\[ A = \pi (1 \times 10^{-3})^2 = \pi (1 \times 10^{-6}) \, \text{m}^2 = \pi \times 10^{-6} \, \text{m}^2 \]

### Step 4: Use the relationship between stress, force, and area
The relationship between stress (\(S\)), force (\(F\)), and area (\(A\)) is given by:
\[ S = \frac{F}{A} \]
Rearranging this formula to find the force:
\[ F = S \times A \]

### Step 5: Substitute the values to find the force
Now substituting the values of \(S\) and \(A\):
\[ F = (4 \times 10^{8} \, \text{N/m}^{2}) \times (\pi \times 10^{-6} \, \text{m}^2) \]
\[ F = 4\pi \times 10^{2} \, \text{N} \]

### Step 6: Calculate the numerical value
Using the approximate value of \(\pi \approx 3.14\):
\[ F \approx 4 \times 3.14 \times 10^{2} \approx 12.56 \times 10^{2} \, \text{N} \approx 1256 \, \text{N} \]

### Final Answer
The force required to punch a hole of diameter 2 mm is approximately \(1256 \, \text{N}\).

---