A cylindrical steel wire of `3 m` length is to stretch no more than `0.2 cm` When a tensile force of `400 N` is applied to each end of the wire ? What minimum diameter is required for the wire ?? `Y_(steel) = 2.1xx10^(11) N//m^2`

Explanation

To find the minimum diameter required for the cylindrical steel wire, we will use the formula for elongation in terms of Young's modulus. Here are the steps to solve the problem:

### Step 1: Convert the elongation from centimeters to meters
The elongation given is `0.2 cm`. We need to convert this to meters:
\[ \Delta L = 0.2 \, \text{cm} = 0.2 \times 10^{-2} \, \text{m} = 0.002 \, \text{m} \]

### Step 2: Write down the formula for elongation
The formula for elongation (\(\Delta L\)) is given by:
\[ \Delta L = \frac{F \cdot L}{A \cdot Y} \]
where:
- \(F\) is the tensile force,
- \(L\) is the original length of the wire,
- \(A\) is the cross-sectional area,
- \(Y\) is Young's modulus.

### Step 3: Express the cross-sectional area in terms of diameter
The cross-sectional area \(A\) for a cylindrical wire is given by:
\[ A = \frac{\pi}{4} d^2 \]
where \(d\) is the diameter of the wire.

### Step 4: Substitute the area into the elongation formula
Substituting \(A\) into the elongation formula gives:
\[ \Delta L = \frac{F \cdot L}{\left(\frac{\pi}{4} d^2\right) \cdot Y} \]
Rearranging this to solve for \(d^2\):
\[ d^2 = \frac{4F \cdot L}{\pi \cdot Y \cdot \Delta L} \]

### Step 5: Substitute the known values into the equation
Now we substitute the values:
- \(F = 400 \, \text{N}\)
- \(L = 3 \, \text{m}\)
- \(Y = 2.1 \times 10^{11} \, \text{N/m}^2\)
- \(\Delta L = 0.002 \, \text{m}\)

\[ d^2 = \frac{4 \cdot 400 \cdot 3}{\pi \cdot (2.1 \times 10^{11}) \cdot 0.002} \]

### Step 6: Calculate \(d^2\)
Calculating the right-hand side:
\[ d^2 = \frac{4800}{\pi \cdot (2.1 \times 10^{11}) \cdot 0.002} \]

Calculating the denominator:
\[ \pi \cdot (2.1 \times 10^{11}) \cdot 0.002 \approx 1.31976 \times 10^{9} \]

Now substituting this back:
\[ d^2 = \frac{4800}{1.31976 \times 10^{9}} \approx 3.628 \times 10^{-6} \]

### Step 7: Take the square root to find \(d\)
\[ d = \sqrt{3.628 \times 10^{-6}} \approx 0.00191 \, \text{m} = 1.91 \, \text{mm} \]

### Final Answer
The minimum diameter required for the wire is approximately:
\[ \boxed{1.91 \, \text{mm}} \]