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

Options

A. `3.18 xx 10^(6) N m^(-2)`
B. `3.18 xx 10^(7) Nm^(-2)`
C. `3.18 xx 10^(8) N m^(-2)`
D. `3.18 xx 10^(9) N m^(-2)`

Explanation

To solve the problem, we need to calculate the stress produced in the steel rod when a force is applied. Here are the steps to find the solution:

### Step 1: Identify the given values
- Length of the rod, \( L = 1 \, \text{m} \)
- Radius of the 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 of the rod
The cross-sectional area \( A \) of a circular rod can be calculated using the formula:
\[ A = \pi r^2 \]
Substituting the radius:
\[ A = \pi (10 \times 10^{-3})^2 \]
\[ A = \pi (100 \times 10^{-6}) \, \text{m}^2 \]
\[ A = \pi \times 10^{-4} \, \text{m}^2 \]

### Step 3: Calculate the stress
Stress \( \sigma \) is defined as the force \( F \) applied per unit area \( A \):
\[ \sigma = \frac{F}{A} \]
Substituting the values:
\[ \sigma = \frac{100 \times 10^{3}}{\pi \times 10^{-4}} \]
Calculating this gives:
\[ \sigma = \frac{100 \times 10^{3}}{3.14 \times 10^{-4}} \approx 318.31 \times 10^{6} \, \text{N/m}^2 \]
Converting to scientific notation:
\[ \sigma \approx 3.18 \times 10^{8} \, \text{N/m}^2 \]

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

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