A uniform wire of steel of length 2.5m and of density `8000kg m^(-3)` weigths 0.05kg. When stretched by a force of 10kg wt the length increases by 2mm. Calculate the Young's modulus for steel.

Explanation

To calculate the Young's modulus for steel, we can follow these steps:

### Step 1: Identify the given values
- Length of the wire (L) = 2.5 m
- Density of steel (ρ) = 8000 kg/m³
- Weight of the wire (W) = 0.05 kg
- Force applied (F) = 10 kg weight = 10 kg × 9.8 m/s² = 98 N
- Change in length (ΔL) = 2 mm = 2 × 10⁻³ m

### Step 2: Calculate the volume of the wire
Using the formula for density:
\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \implies \text{Volume} = \frac{\text{Mass}}{\text{Density}} \]
Substituting the values:
\[ \text{Volume} = \frac{0.05 \text{ kg}}{8000 \text{ kg/m}^3} = 6.25 \times 10^{-6} \text{ m}^3 \]

### Step 3: Calculate the cross-sectional area of the wire
Using the formula for volume:
\[ \text{Volume} = \text{Area} \times \text{Length} \implies \text{Area} = \frac{\text{Volume}}{\text{Length}} \]
Substituting the values:
\[ \text{Area} = \frac{6.25 \times 10^{-6} \text{ m}^3}{2.5 \text{ m}} = 2.5 \times 10^{-6} \text{ m}^2 \]

### Step 4: Use the Young's modulus formula
The formula for Young's modulus (Y) is given by:
\[ Y = \frac{F}{A} \cdot \frac{L}{\Delta L} \]
Substituting the values:
\[ Y = \frac{98 \text{ N}}{2.5 \times 10^{-6} \text{ m}^2} \cdot \frac{2.5 \text{ m}}{2 \times 10^{-3} \text{ m}} \]

### Step 5: Simplify the equation
Calculating the first part:
\[ \frac{98 \text{ N}}{2.5 \times 10^{-6} \text{ m}^2} = 3.92 \times 10^7 \text{ N/m}^2 \]
Calculating the second part:
\[ \frac{2.5 \text{ m}}{2 \times 10^{-3} \text{ m}} = 1250 \]
Now, multiply these two results:
\[ Y = 3.92 \times 10^7 \text{ N/m}^2 \cdot 1250 = 4.9 \times 10^{10} \text{ N/m}^2 \]

### Final Result
The Young's modulus of steel is:
\[ Y = 4.9 \times 10^{10} \text{ N/m}^2 \]

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