A steel rod has a radius 10 mm and a length of 1.0 m. A force stretches it along its length and produces a strain of `0.32%`. Young's modulus of the steel is `2.0xx10^(11Nm^(-2)`. What is the magnitude of the force stretching the rod?

Options

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

Answer

To find the magnitude of the force stretching the steel rod, we can follow these steps:

### Step 1: Understand the relationship between Young's modulus, stress, and strain.
Young's modulus (E) is defined as the ratio of stress (σ) to strain (ε):
\[ E = \frac{\sigma}{\epsilon} \]
where:
- Stress (σ) is defined as force (F) per unit area (A):
\[ \sigma = \frac{F}{A} \]
- Strain (ε) is defined as the change in length (ΔL) divided by the original length (L):
\[ \epsilon = \frac{\Delta L}{L} \]

### Step 2: Rearrange the formula to find the force (F).
From the definition of Young's modulus, we can rearrange the equation to find the force:
\[ F = E \cdot A \cdot \epsilon \]

### Step 3: Calculate the area (A) of the rod.
The area of a circular cross-section is given by:
\[ A = \pi r^2 \]
Given the radius \( r = 10 \, \text{mm} = 10 \times 10^{-3} \, \text{m} = 0.01 \, \text{m} \):
\[ A = \pi (0.01)^2 = \pi \times 0.0001 \, \text{m}^2 = 3.14 \times 10^{-4} \, \text{m}^2 \]

### Step 4: Convert the strain (ε) into decimal form.
The strain given is \( 0.32\% \):
\[ \epsilon = \frac{0.32}{100} = 0.0032 \]

### Step 5: Substitute the values into the force equation.
Now we can substitute the values into the equation for force:
\[ F = E \cdot A \cdot \epsilon \]
Substituting the known values:
- \( E = 2.0 \times 10^{11} \, \text{N/m}^2 \)
- \( A = 3.14 \times 10^{-4} \, \text{m}^2 \)
- \( \epsilon = 0.0032 \)

\[ F = (2.0 \times 10^{11}) \cdot (3.14 \times 10^{-4}) \cdot (0.0032) \]

### Step 6: Calculate the force (F).
Calculating the above expression:
\[ F = 2.0 \times 10^{11} \times 3.14 \times 10^{-4} \times 0.0032 \]
\[ F \approx 2.0 \times 3.14 \times 0.0032 \times 10^{11 - 4} = 2.0 \times 3.14 \times 0.0032 \times 10^{7} \]
\[ F \approx 0.020096 \times 10^{7} = 2.0096 \times 10^{5} \, \text{N} \]

### Step 7: Convert the force to kilonewtons (kN).
To convert Newtons to kilonewtons:
\[ F \approx 2.0096 \times 10^{5} \, \text{N} = 200.96 \, \text{kN} \approx 201 \, \text{kN} \]

### Final Answer:
The magnitude of the force stretching the rod is approximately **201 kN**.

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