Determine the elongation of the steel bar `1m` long and `1.5 cm^(2)` cross-sectional area when subjected to a pull of `1.5xx10^(4) N.`
`(Take Y=2.0xx10^(11 )N//m^(2))`.

Explanation

To determine the elongation of the steel bar, we will use the formula for elongation (ΔL) given by:

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

where:
- \( F \) = applied force (in Newtons)
- \( L \) = original length of the bar (in meters)
- \( A \) = cross-sectional area (in square meters)
- \( Y \) = Young's modulus (in Newtons per square meter)

### Step-by-step Solution:

1. **Identify the given values**:
- Length of the steel bar, \( L = 1 \, \text{m} \)
- Cross-sectional area, \( A = 1.5 \, \text{cm}^2 = 1.5 \times 10^{-4} \, \text{m}^2 \) (conversion from cm² to m²)
- Applied force, \( F = 1.5 \times 10^4 \, \text{N} \)
- Young's modulus, \( Y = 2.0 \times 10^{11} \, \text{N/m}^2 \)

2. **Substitute the values into the elongation formula**:
\[ \Delta L = \frac{F \cdot L}{A \cdot Y} \]
\[ \Delta L = \frac{(1.5 \times 10^4) \cdot (1)}{(1.5 \times 10^{-4}) \cdot (2.0 \times 10^{11})} \]

3. **Calculate the denominator**:
\[ A \cdot Y = (1.5 \times 10^{-4}) \cdot (2.0 \times 10^{11}) = 3.0 \times 10^{7} \, \text{N} \]

4. **Calculate the elongation**:
\[ \Delta L = \frac{1.5 \times 10^4}{3.0 \times 10^7} \]
\[ \Delta L = 0.5 \times 10^{-3} \, \text{m} \]

5. **Convert the elongation to millimeters**:
\[ \Delta L = 0.5 \, \text{mm} \]

### Final Answer:
The elongation of the steel bar is \( 0.5 \, \text{mm} \).

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