A structural steel rod has a radius r(=10 mm) and a length l(=1 m). When a force F(= 100 kN) is applied, it stretches it along its length. Young's modulus of elasticity of the structural steel is `2.0xx10^(4) Nm^(-2)`. What is the elastic energy density of the steel rod ?

Options

A. `6.12xx10^(4)J//m^(3)`
B. `1.25xx10^(4)J//m^(3)`
C. `2.5xx10^(4)J//m^(3)`
D. `2.5xx10^(5)J//m^(3)`

Explanation

To find the elastic energy density of the steel rod, we can follow these steps:

### Step 1: Convert the radius from mm to meters
Given:
- Radius \( r = 10 \, \text{mm} = \frac{10}{1000} \, \text{m} = 0.01 \, \text{m} \)

### Step 2: Calculate the cross-sectional area \( A \) of the rod
The area \( A \) of a circle is given by the formula:
\[ A = \pi r^2 \]
Substituting the value of \( r \):
\[ A = \pi (0.01)^2 = \pi \times 0.0001 \, \text{m}^2 \approx 3.14 \times 10^{-4} \, \text{m}^2 \]

### Step 3: Calculate the stress \( \sigma \)
Stress is defined as the force per unit area:
\[ \sigma = \frac{F}{A} \]
Given \( F = 100 \, \text{kN} = 100 \times 10^3 \, \text{N} \):
\[ \sigma = \frac{100 \times 10^3}{3.14 \times 10^{-4}} \approx 318,471,337.58 \, \text{N/m}^2 \approx 3.18 \times 10^8 \, \text{N/m}^2 \]

### Step 4: Calculate the strain \( \epsilon \)
Using Young's modulus \( E \):
\[ E = \frac{\sigma}{\epsilon} \implies \epsilon = \frac{\sigma}{E} \]
Given \( E = 2.0 \times 10^4 \, \text{N/m}^2 \):
\[ \epsilon = \frac{3.18 \times 10^8}{2.0 \times 10^4} \approx 15900 \]

### Step 5: Calculate the elastic energy density \( EED \)
The elastic energy density is given by:
\[ EED = \frac{1}{2} \sigma \epsilon \]
Substituting the values of \( \sigma \) and \( \epsilon \):
\[ EED = \frac{1}{2} \times (3.18 \times 10^8) \times (15900) \approx 2.53 \times 10^{12} \, \text{J/m}^3 \]

### Final Result
The elastic energy density of the steel rod is approximately:
\[ EED \approx 2.5 \times 10^5 \, \text{J/m}^3 \]