Calculating Steel Wire Length Change

This problem requires calculating the change in length ($\Delta L$) of a steel wire when a weight is suspended from it. We can use the formula derived from Young's modulus ($Y$).

Physics Principle: Young's Modulus

Young's modulus relates stress and strain in a material. The formula is:

$ Y = \frac{\text{Stress}}{\text{Strain}} $

Where:

• Stress = Force per unit area ($ \frac{F}{A} $)
• Strain = Change in length over original length ($ \frac{\Delta L}{L} $)

Substituting these into the formula gives:

$ Y = \frac{F/A}{\Delta L/L} = \frac{F \cdot L}{A \cdot \Delta L} $

Deriving Formula for Change in Length

Rearranging the formula to solve for the change in length ($\Delta L$):

$ \Delta L = \frac{F \cdot L}{A \cdot Y} $

Calculations

First, calculate the force ($F$) due to the suspended mass ($m=5\text{ kg}$). We use the acceleration due to gravity, $g \approx 9.8\text{ m/s}^2$.

$ F = m \cdot g = 5\text{ kg} \times 9.8\text{ m/s}^2 = 49\text{ N} $

Now, substitute all the given values into the formula for $\Delta L$:

• Force ($F$): $49\text{ N}$
• Original Length ($L$): $2.5\text{ m}$
• Cross-sectional Area ($A$): $2.5 \times 10^{-6}\text{ m}^2$
• Young's Modulus ($Y$): $2 \times 10^{11}\text{ N/m}^2$

$ \Delta L = \frac{(49\text{ N}) \times (2.5\text{ m})}{(2.5 \times 10^{-6}\text{ m}^2) \times (2 \times 10^{11}\text{ N/m}^2)} $

Simplify the expression:

$ \Delta L = \frac{49 \times 2.5}{2.5 \times 2 \times 10^{-6 + 11}} $

$ \Delta L = \frac{49}{2 \times 10^{5}} $

$ \Delta L = \frac{49}{2} \times 10^{-5}\text{ m} $

$ \Delta L = 24.5 \times 10^{-5}\text{ m} $

Convert to the required format:

$ \Delta L = 2.45 \times 10^{-4}\text{ m} $

Final Result

The change in length of the wire is $2.45 \times 10^{-4}\text{ m}$.