A weight of 110N is suspended from the cceiling by an aluminium wire of lenth 2m and radius 1 mm. Calculate (i) the extension produced and (ii) the elastic energy stored in the wire. [The Young's modulus of aluminium is `7 xx 10^(10) Nm^(-2)`

Explanation

To solve the problem step by step, we will calculate the extension produced in the aluminum wire and then find the elastic energy stored in it.

### Given Data:
- Weight (Force, F) = 110 N
- Length of the wire (L) = 2 m
- Radius of the wire (r) = 1 mm = 1 × 10^(-3) m
- Young's modulus of aluminum (Y) = 7 × 10^(10) N/m²

### Step 1: Calculate the Cross-Sectional Area (A) of the Wire
The cross-sectional area of the wire can be calculated using the formula for the area of a circle:
\[ A = \pi r^2 \]
Substituting the radius:
\[ A = \pi (1 \times 10^{-3})^2 \]
\[ A = \pi (1 \times 10^{-6}) \]
\[ A \approx 3.14 \times 10^{-6} \, \text{m}^2 \]

### Step 2: Use Young's Modulus to Calculate the Extension (ΔL)
Young's modulus (Y) is defined as:
\[ Y = \frac{\text{Stress}}{\text{Strain}} \]
Where:
- Stress = \(\frac{F}{A}\)
- Strain = \(\frac{\Delta L}{L}\)

Rearranging the formula gives:
\[ \Delta L = \frac{F \cdot L}{A \cdot Y} \]

Substituting the known values:
\[ \Delta L = \frac{110 \, \text{N} \cdot 2 \, \text{m}}{(3.14 \times 10^{-6} \, \text{m}^2) \cdot (7 \times 10^{10} \, \text{N/m}^2)} \]

Calculating the denominator:
\[ A \cdot Y = (3.14 \times 10^{-6}) \cdot (7 \times 10^{10}) \approx 2.198 \times 10^5 \]

Now substituting back:
\[ \Delta L = \frac{220}{2.198 \times 10^5} \]
\[ \Delta L \approx 1.00 \times 10^{-3} \, \text{m} \]

### Step 3: Calculate the Elastic Energy Stored in the Wire
The elastic energy (U) stored in the wire can be calculated using the formula:
\[ U = \frac{1}{2} F \Delta L \]

Substituting the values:
\[ U = \frac{1}{2} \cdot 110 \, \text{N} \cdot (1.00 \times 10^{-3} \, \text{m}) \]
\[ U = \frac{110 \cdot 1.00 \times 10^{-3}}{2} \]
\[ U = 0.055 \, \text{J} \]

### Final Answers:
(i) The extension produced in the wire is approximately \( \Delta L \approx 1.00 \, \text{mm} \) or \( 1.00 \times 10^{-3} \, \text{m} \).

(ii) The elastic energy stored in the wire is \( U \approx 0.055 \, \text{J} \).