A rod 100 cm long and of `2cmxx2cm` cross-section is subjected to a pull of 1000 kg force. If the modulus of elasticity of the materials is `2.0xx10^(6) kg//cm^(2)`, determine the elongation of the rod.

Explanation

To determine the elongation of a rod subjected to a pull, we can use the formula for elongation derived from Young's modulus:

\[ \Delta L = \frac{F \cdot L}{A \cdot E} \]

Where:
- \(\Delta L\) = elongation of the rod
- \(F\) = force applied (in kgf)
- \(L\) = original length of the rod (in cm)
- \(A\) = area of cross-section (in cm²)
- \(E\) = modulus of elasticity (in kg/cm²)

### Step-by-Step Solution:

1. **Identify the given values:**
- Length of the rod, \(L = 100 \, \text{cm}\)
- Cross-sectional area, \(A = 2 \, \text{cm} \times 2 \, \text{cm} = 4 \, \text{cm}^2\)
- Force applied, \(F = 1000 \, \text{kgf}\)
- Modulus of elasticity, \(E = 2.0 \times 10^6 \, \text{kg/cm}^2\)

2. **Substitute the values into the elongation formula:**
\[ \Delta L = \frac{F \cdot L}{A \cdot E} = \frac{1000 \, \text{kgf} \cdot 100 \, \text{cm}}{4 \, \text{cm}^2 \cdot 2.0 \times 10^6 \, \text{kg/cm}^2} \]

3. **Calculate the numerator:**
\[ F \cdot L = 1000 \cdot 100 = 100000 \, \text{kgf} \cdot \text{cm} \]

4. **Calculate the denominator:**
\[ A \cdot E = 4 \cdot 2.0 \times 10^6 = 8.0 \times 10^6 \, \text{kg/cm}^2 \]

5. **Now substitute the values into the elongation formula:**
\[ \Delta L = \frac{100000}{8.0 \times 10^6} \]

6. **Perform the division:**
\[ \Delta L = \frac{100000}{8000000} = \frac{1}{80} \, \text{cm} \]

7. **Convert to decimal form:**
\[ \Delta L = 0.0125 \, \text{cm} \]

### Final Result:
The elongation of the rod is \(0.0125 \, \text{cm}\).