Calculate Young's Modulus for Steel

A uniform wire of steel of length \( 2.5 \, \text{m} \) and density \( 8000 \, \text{kg/m}^3 \) weighs \( 0.05 \, \text{kg} \). When stretched by a force of \( 10 \, \text{kgf} \), the length increases by \( 2 \, \text{mm} \). Calculate the Young's modulus for steel.

Step 1: Identify the given values

• Original length of the wire: \( L = 2.5 \, \text{m} \)
• Density of steel: \( \rho = 8000 \, \text{kg/m}^3 \)
• Weight of the wire: \( m = 0.05 \, \text{kg} \)
• Force applied: \( F = 10 \, \text{kgf} = 10 \times 9.8 = 98 \, \text{N} \)
• Extension in length: \( \Delta L = 2 \, \text{mm} = 2 \times 10^{-3} \, \text{m} \)

Step 2: Calculate the volume of the wire

\[ V = \frac{m}{\rho} = \frac{0.05}{8000} = 6.25 \times 10^{-6} \, \text{m}^3 \]

Step 3: Calculate the cross-sectional area \( A \)

\[ A = \frac{V}{L} = \frac{6.25 \times 10^{-6}}{2.5} = 2.5 \times 10^{-6} \, \text{m}^2 \]

Step 4: Substitute values into the Young's modulus formula

\[ Y = \frac{F \cdot L}{A \cdot \Delta L} = \frac{98 \cdot 2.5}{2.5 \times 10^{-6} \cdot 2 \times 10^{-3}} \]

Step 5: Simplify the expression

\[ Y = \frac{245}{5 \times 10^{-9}} = 49 \times 10^{9} \, \text{Pa} \]

Final Answer:

The Young's modulus for steel is:

\[ Y = 4.9 \times 10^{10} \, \text{Pa} \]