A load of 10 kg is suspended by a metal wire 3 m long and having a cross sectional area `4mm^2`. Find a. the stress b. the strain and c. the elongation. Young modulus of the metal is`2.0xx10^11Nm^-2`

Explanation

To solve the problem step by step, we will calculate the stress, strain, and elongation of the metal wire under the given load.

### Given Data:
- Mass of the load (m) = 10 kg
- Length of the wire (L) = 3 m
- Cross-sectional area (A) = 4 mm² = 4 × 10⁻⁶ m² (conversion from mm² to m²)
- Young's modulus (Y) = 2.0 × 10¹¹ N/m²
- Acceleration due to gravity (g) = 9.8 m/s²

### Step 1: Calculate the Force (F)
The force exerted by the load can be calculated using the formula:
\[ F = m \cdot g \]
Substituting the values:
\[ F = 10 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 98 \, \text{N} \]

### Step 2: Calculate the Stress (σ)
Stress is defined as the force applied per unit area. The formula for stress is:
\[ \sigma = \frac{F}{A} \]
Substituting the values:
\[ \sigma = \frac{98 \, \text{N}}{4 \times 10^{-6} \, \text{m}^2} \]
Calculating this gives:
\[ \sigma = 2.45 \times 10^7 \, \text{N/m}^2 \]

### Step 3: Calculate the Strain (ε)
Strain is defined as the ratio of change in length (ΔL) to the original length (L). We can use Young's modulus to find the strain:
\[ Y = \frac{F \cdot L}{A \cdot \Delta L} \]
Rearranging gives:
\[ \frac{\Delta L}{L} = \frac{F}{A \cdot Y} \]
Substituting the known values:
\[ \frac{\Delta L}{3 \, \text{m}} = \frac{98 \, \text{N}}{4 \times 10^{-6} \, \text{m}^2 \times 2.0 \times 10^{11} \, \text{N/m}^2} \]
Calculating this gives:
\[ \frac{\Delta L}{3} = 1.225 \times 10^{-4} \]
Thus,
\[ \Delta L = 3 \times 1.225 \times 10^{-4} = 3.675 \times 10^{-4} \, \text{m} \]

### Step 4: Final Results
- **Stress (σ)**: \( 2.45 \times 10^7 \, \text{N/m}^2 \)
- **Strain (ε)**: \( 1.225 \times 10^{-4} \) (dimensionless)
- **Elongation (ΔL)**: \( 3.675 \times 10^{-4} \, \text{m} \) or \( 0.3675 \, \text{mm} \)