A steel rod of length 1m and radius 10 mm is stretched by a force 100 KN along its length. The stress produced in the rod `Y_("Steel")=2xx10^(11)" N m"^(-2)`

Options

A. `3.18xx10^(6)" N m"^(-2)`
B. `3.18xx10^(7)" N m"^(-2)`
C. `3.18xx10^(8)" N m"^(-2)`
D. `3.18xx10^(9)" N m"^(-2)`

Explanation

To solve the problem of finding the stress produced in a steel rod when a force is applied, we can follow these steps:

### Step 1: Identify the Given Values
- Length of the steel rod, \( L = 1 \, \text{m} \)
- Radius of the steel rod, \( r = 10 \, \text{mm} = 10 \times 10^{-3} \, \text{m} \)
- Force applied, \( F = 100 \, \text{kN} = 100 \times 10^3 \, \text{N} \)

### Step 2: Calculate the Cross-Sectional Area (A)
The cross-sectional area \( A \) of the rod can be calculated using the formula for the area of a circle:
\[ A = \pi r^2 \]
Substituting the radius:
\[ A = \pi (10 \times 10^{-3})^2 \]
Calculating \( r^2 \):
\[ r^2 = (10 \times 10^{-3})^2 = 100 \times 10^{-6} = 10^{-4} \, \text{m}^2 \]
Now substituting back into the area formula:
\[ A = \pi \times 10^{-4} \approx 3.14 \times 10^{-4} \, \text{m}^2 \]

### Step 3: Calculate the Stress (σ)
Stress is defined as the force per unit area:
\[ \sigma = \frac{F}{A} \]
Substituting the values of \( F \) and \( A \):
\[ \sigma = \frac{100 \times 10^3}{3.14 \times 10^{-4}} \]
Calculating the stress:
\[ \sigma = \frac{100000}{3.14 \times 10^{-4}} = \frac{100000 \times 10^4}{3.14} = \frac{10^9}{3.14} \]
Now, performing the division:
\[ \sigma \approx 318471.34 \, \text{N/m}^2 \approx 3.18 \times 10^8 \, \text{N/m}^2 \]

### Conclusion
The stress produced in the steel rod is approximately \( 3.18 \times 10^8 \, \text{N/m}^2 \).

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