A steel rod has a radius `10mm` and a length of `1m`. A force stretches it along its length and produces a strain of `0.32%`. Younng's modulus of steel is `2xx10^(11)Nm^(-2)`, the magnitude of force stretching the rod is

Options

A. `100.5 kN`
B. `201 kN`
C. `78 kN`
D. `150 kN`

Explanation

To find the magnitude of the force stretching the steel rod, we can use the relationship defined by Young's modulus. Here are the steps to solve the problem:

### Step-by-Step Solution:

1. **Understand the given values:**
- Radius of the rod, \( r = 10 \text{ mm} = 10 \times 10^{-3} \text{ m} \)
- Length of the rod, \( L = 1 \text{ m} \)
- Strain, \( \text{strain} = 0.32\% = \frac{0.32}{100} = 0.0032 \)
- Young's modulus of steel, \( Y = 2 \times 10^{11} \text{ N/m}^2 \)

2. **Calculate the change in length (\( \Delta L \)):**
\[ \Delta L = \text{strain} \times L = 0.0032 \times 1 \text{ m} = 0.0032 \text{ m} \]

3. **Calculate the cross-sectional area (\( A \)) of the rod:**
\[ A = \pi r^2 = \pi (10 \times 10^{-3})^2 = \pi (100 \times 10^{-6}) = 100\pi \times 10^{-6} \text{ m}^2 \]

4. **Use Young's modulus formula:**
Young's modulus is defined as:
\[ Y = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{\Delta L/L} \]
Rearranging for force (\( F \)):
\[ F = Y \cdot A \cdot \frac{\Delta L}{L} \]

5. **Substituting the values into the formula:**
\[ F = 2 \times 10^{11} \cdot (100\pi \times 10^{-6}) \cdot \frac{0.0032}{1} \]

6. **Calculate the force:**
\[ F = 2 \times 10^{11} \cdot 100\pi \times 10^{-6} \cdot 0.0032 \]
\[ = 2 \times 100 \times 0.0032 \times \pi \times 10^{5} \]
\[ = 0.64\pi \times 10^{5} \text{ N} \]
\[ \approx 2.01 \times 10^{5} \text{ N} \quad (\text{using } \pi \approx 3.14) \]
\[ \approx 2.01 \text{ kN} \]

### Final Answer:
The magnitude of the force stretching the rod is approximately **2.01 kN**.