What is the percentage increase in length of a wire of diameter 2.5 mm, stretched by a force of 100 kg wt? Young's modulus of elasticity of the wire is

Percentage Increase in the Length of a Wire

Question: What is the percentage increase in length of a wire of diameter 2.5 mm, stretched by a force of 100 kg wt? Young's modulus of elasticity of the wire is \( 12.5 \times 10^{11} \, \text{dyne/cm}^2 \).

Solution:

We use the formula for elongation under a stretching force:

\[ \Delta L = \frac{F L}{A Y} \]

The percentage elongation is given by:

\[ \text{Percentage elongation} = \frac{\Delta L}{L} \times 100 = \frac{F}{A Y} \times 100 \]

Step 1: Convert Force to Dynes

\[ F = 100 \, \text{kg wt} = 100 \times 980 \times 10^5 = 9.8 \times 10^7 \, \text{dynes}. \]

Step 2: Calculate the Cross-sectional Area

The diameter of the wire is \( 2.5 \, \text{mm} = 0.25 \, \text{cm} \), so the radius is:

\[ r = \frac{\text{Diameter}}{2} = \frac{0.25}{2} = 0.125 \, \text{cm}. \]

The cross-sectional area \( A \) is:

\[ A = \pi r^2 = \pi (0.125)^2 = \pi \times 0.015625 = 0.0491 \, \text{cm}^2. \]

Step 3: Substitute into the Formula

\[ \text{Percentage elongation} = \frac{9.8 \times 10^7}{0.0491 \times 12.5 \times 10^{11}} \times 100. \]

Simplifying the denominator:

\[ 0.0491 \times 12.5 \times 10^{11} = 6.1375 \times 10^{10}. \]

Simplifying the fraction:

\[ \frac{9.8 \times 10^7}{6.1375 \times 10^{10}} = 1.596 \times 10^{-3}. \]

Convert to a percentage:

\[ \text{Percentage elongation} = 1.596 \times 10^{-3} \times 100 = 0.1596\%. \]

Final Answer: The percentage increase in the length of the wire is approximately \( 0.16\% \).