A 4m long aluminium wire whose diameter is 3 mm is used to support a mass of 50 kg. What will be elongation of the wire ? Y for aluminium is `7xx10^(10) Nm6(-2)`

Answer

To find the elongation of the aluminium wire, we will use the formula for elongation derived from Young's modulus:

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

Where:
- \(\Delta L\) = elongation of the wire
- \(F\) = force applied (weight of the mass)
- \(L\) = original length of the wire
- \(A\) = cross-sectional area of the wire
- \(Y\) = Young's modulus of the material

### Step 1: Calculate the force \(F\)
The force \(F\) can be calculated using the formula:

\[ F = m \cdot g \]

Where:
- \(m\) = mass supported by the wire = 50 kg
- \(g\) = acceleration due to gravity = 9.8 m/s²

\[ F = 50 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 490 \, \text{N} \]

### Step 2: Determine the original length \(L\)
The original length \(L\) of the wire is given as:

\[ L = 4 \, \text{m} \]

### Step 3: Calculate the cross-sectional area \(A\)
The diameter of the wire is given as 3 mm, so the radius \(r\) is:

\[ r = \frac{d}{2} = \frac{3 \, \text{mm}}{2} = 1.5 \, \text{mm} = 1.5 \times 10^{-3} \, \text{m} \]

The cross-sectional area \(A\) of the wire can be calculated using the formula for the area of a circle:

\[ A = \pi r^2 \]

Substituting the radius:

\[ A = \pi (1.5 \times 10^{-3} \, \text{m})^2 = \pi (2.25 \times 10^{-6} \, \text{m}^2) \approx 7.06858 \times 10^{-6} \, \text{m}^2 \]

### Step 4: Use Young's modulus \(Y\)
The Young's modulus for aluminium is given as:

\[ Y = 7 \times 10^{10} \, \text{N/m}^2 \]

### Step 5: Substitute values into the elongation formula
Now we can substitute all the values into the elongation formula:

\[ \Delta L = \frac{F \cdot L}{A \cdot Y} = \frac{490 \, \text{N} \cdot 4 \, \text{m}}{7.06858 \times 10^{-6} \, \text{m}^2 \cdot 7 \times 10^{10} \, \text{N/m}^2} \]

Calculating the denominator:

\[ A \cdot Y = 7.06858 \times 10^{-6} \, \text{m}^2 \cdot 7 \times 10^{10} \, \text{N/m}^2 \approx 4.958006 \times 10^5 \, \text{N} \]

Now substituting back into the elongation formula:

\[ \Delta L = \frac{1960 \, \text{N m}}{4.958006 \times 10^5 \, \text{N}} \approx 3.95 \times 10^{-3} \, \text{m} \]

### Step 6: Convert to mm
To convert meters to millimeters:

\[ \Delta L \approx 3.95 \, \text{mm} \]

### Final Answer
The elongation of the wire is approximately \(3.95 \, \text{mm}\).

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