The elastic limit of steel is `8xx10^8Nm^-2` and its Young modulus `2xx10^11Nm^-2`. Find the maximum elongation of a half meter steel wire that can be given without exceeding the elastic limit.

Explanation

To find the maximum elongation of a half-meter steel wire without exceeding the elastic limit, we can follow these steps:

### Step 1: Identify the given values
- Elastic limit of steel (maximum stress, σ) = \(8 \times 10^8 \, \text{N/m}^2\)
- Young's modulus (E) = \(2 \times 10^{11} \, \text{N/m}^2\)
- Original length of the wire (L) = 0.5 m

### Step 2: Calculate the maximum stress
The maximum stress is already given as:
\[ \sigma = 8 \times 10^8 \, \text{N/m}^2 \]

### Step 3: Use Young's modulus to find strain
Young's modulus relates stress and strain as follows:
\[ E = \frac{\sigma}{\text{strain}} \]
From this, we can express strain as:
\[ \text{strain} = \frac{\sigma}{E} \]
Substituting the values:
\[ \text{strain} = \frac{8 \times 10^8}{2 \times 10^{11}} = 4 \times 10^{-3} \]

### Step 4: Relate strain to elongation
Strain is also defined as the ratio of elongation (ΔL) to the original length (L):
\[ \text{strain} = \frac{\Delta L}{L} \]
Rearranging gives us:
\[ \Delta L = \text{strain} \times L \]
Substituting the values:
\[ \Delta L = 4 \times 10^{-3} \times 0.5 = 2 \times 10^{-3} \, \text{m} \]

### Step 5: Convert elongation to millimeters
To express the elongation in millimeters:
\[ \Delta L = 2 \times 10^{-3} \, \text{m} = 2 \, \text{mm} \]

### Final Answer
The maximum elongation of the half-meter steel wire that can be given without exceeding the elastic limit is **2 mm**.

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