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 follow the given parameters and formulas related to stress, strain, and elongation.

### Given Data:
- Mass (m) = 10 kg
- Length of wire (L) = 3 m
- Cross-sectional area (A) = 4 mm² = \(4 \times 10^{-6} \, m^2\) (conversion to standard units)
- Young's modulus (E) = \(2.0 \times 10^{11} \, N/m^2\)
- Acceleration due to gravity (g) = \(10 \, m/s^2\) (approximation)

### Step 1: Calculate the Load (Force)
The load (F) can be calculated using the formula:
\[ F = m \cdot g \]
Substituting the values:
\[ F = 10 \, kg \cdot 10 \, m/s^2 = 100 \, N \]

### Step 2: Calculate the Stress
Stress (\( \sigma \)) is defined as the load per unit area:
\[ \sigma = \frac{F}{A} \]
Substituting the values:
\[ \sigma = \frac{100 \, N}{4 \times 10^{-6} \, m^2} = 2.5 \times 10^7 \, N/m^2 \]

### Step 3: Calculate the Strain
Strain (\( \epsilon \)) is defined as the ratio of stress to Young's modulus:
\[ \epsilon = \frac{\sigma}{E} \]
Substituting the values:
\[ \epsilon = \frac{2.5 \times 10^7 \, N/m^2}{2.0 \times 10^{11} \, N/m^2} = 1.25 \times 10^{-4} \]

### Step 4: Calculate the Elongation
Elongation (\( \Delta L \)) can be calculated using the formula:
\[ \Delta L = \epsilon \cdot L \]
Substituting the values:
\[ \Delta L = 1.25 \times 10^{-4} \cdot 3 \, m = 3.75 \times 10^{-4} \, m \]

### Final Answers:
a. Stress = \( 2.5 \times 10^7 \, N/m^2 \)
b. Strain = \( 1.25 \times 10^{-4} \)
c. Elongation = \( 3.75 \times 10^{-4} \, m \)