Step 1: Understanding the Question:
We need to calculate the tensile stress (\(\sigma\)) in a circular steel rod when a known tensile force (\(F\)) is applied to it. We are given the rod's radius (\(r\)).
Step 2: Key Formula or Approach:
Stress is defined as the force acting per unit of cross-sectional area. The formula is:
\[ \sigma = \frac{F}{A} \]
For a circular rod, the cross-sectional area \(A\) is given by:
\[ A = \pi r^2 \]
Step 3: Detailed Explanation:
First, we list the given values and convert them to consistent units (SI units: Newtons and meters).
Force, \( F = 100 \, \text{kN} = 100 \times 10^3 \, \text{N} \).
Radius, \( r = 10 \, \text{mm} = 10 \times 10^{-3} \, \text{m} = 0.01 \, \text{m} \).
Next, we calculate the cross-sectional area \(A\):
\[ A = \pi r^2 = \pi (0.01 \, \text{m})^2 = \pi (1 \times 10^{-4} \, \text{m}^2) \]
Now, we can calculate the stress \(\sigma\):
\[ \sigma = \frac{F}{A} = \frac{100 \times 10^3 \, \text{N}}{\pi \times 10^{-4} \, \text{m}^2} = \frac{10^5}{\pi \times 10^{-4}} \, \text{Pa} \]
\[ \sigma = \frac{10^9}{\pi} \, \text{Pa} \approx 318.3 \times 10^6 \, \text{Pa} \]
Since \(1 \, \text{MPa} = 10^6 \, \text{Pa}\), the stress is:
\[ \sigma \approx 318.3 \, \text{MPa} \]
Step 4: Final Answer:
The calculated stress is approximately 318 MPa, which matches option (B).