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 can use the formula for Young's modulus (Y), which relates stress and strain:

\[ Y = \frac{\text{Stress}}{\text{Strain}} \]

Where:
- Stress = \(\frac{F}{A}\)
- Strain = \(\frac{\Delta L}{L}\)

Here, \(F\) is the force applied, \(A\) is the cross-sectional area, \(\Delta L\) is the elongation, and \(L\) is the original length of the bar.

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

### Step 2: Calculate the stress
Using the formula for stress:

\[ \text{Stress} = \frac{F}{A} = \frac{1.5 \times 10^4 \, \text{N}}{1.5 \times 10^{-4} \, \text{m}^2} \]

Calculating this gives:

\[ \text{Stress} = 1.0 \times 10^8 \, \text{N/m}^2 \]

### Step 3: Calculate the strain
Using the relationship of Young's modulus:

\[ Y = \frac{\text{Stress}}{\text{Strain}} \implies \text{Strain} = \frac{\text{Stress}}{Y} \]

Substituting the values:

\[ \text{Strain} = \frac{1.0 \times 10^8 \, \text{N/m}^2}{2.0 \times 10^{11} \, \text{N/m}^2} \]

Calculating this gives:

\[ \text{Strain} = 0.5 \times 10^{-3} \]

### Step 4: Calculate the elongation
Now, using the definition of strain:

\[ \text{Strain} = \frac{\Delta L}{L} \implies \Delta L = \text{Strain} \times L \]

Substituting the values:

\[ \Delta L = 0.5 \times 10^{-3} \times 1 \, \text{m} \]

Calculating this gives:

\[ \Delta L = 0.5 \times 10^{-3} \, \text{m} = 0.5 \, \text{mm} \]

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

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