Question

A steel rod bas a radius of 20 mm and a length of 2.0 m. A force of 62.8 kN stretches it along its length. Young's modulus of steel is 2.0 × 10^{11} N/m^2 . The longitudinal strain produced in the wire is _______ × 10^{-5}

Answer 1

Correct Answer: 25

\(\begin{aligned} & Y=\frac{\text { stress }}{\text { strain }} \\ & \Rightarrow \text { strain }=\frac{\text { stress }}{Y}=\frac{F}{A Y}=\frac{62.8 \times 1000}{\pi r^2 \times 2 \times 10^{11}}=\frac{62.8 \times 1000}{3.14 \times 400 \times 10^{-6} \times 2 \times 10^{11}}=\frac{200}{8} \times 10^{-5}=25 \times 10^{-5} \end{aligned}\)

Answer 2

Young's Modulus and Strain Problem

Step 1: Young's Modulus and Strain

Young's modulus (\( Y \)) is defined as the ratio of stress to strain:

\( Y = \frac{\text{stress}}{\text{strain}} \)

Stress is defined as force (\( F \)) per unit area (\( A \)), and strain is the change in length (\( \Delta L \)) divided by the original length (\( L \)).

Step 2: Calculate Strain

We are given \( F = 62.8 \text{ kN} = 62.8 \times 10^3 \text{ N} \), \( r = 20 \text{ mm} = 20 \times 10^{-3} \text{ m} \), \( L = 2.0 \text{ m} \), and \( Y = 2.0 \times 10^{11} \text{ N/m}^2 \). The cross-sectional area of the rod is

\( A = \pi r^2 = \pi (20 \times 10^{-3} \text{ m})^2 = 400\pi \times 10^{-6} \text{ m}^2 \)

Strain is given by:

\( \text{strain} = \frac{\text{stress}}{Y} = \frac{F/A}{Y} = \frac{F}{AY} \)

\( \text{strain} = \frac{62.8 \times 10^3 \text{ N}}{(400\pi \times 10^{-6} \text{ m}^2)(2.0 \times 10^{11} \text{ N/m}^2)} = \frac{62.8 \times 10^3}{800\pi \times 10^5} = \frac{62.8}{800 \times 3.14} \times 10^{-2} \)

\( \text{strain} \approx \frac{62.8}{2512} \times 10^{-2} \approx 0.025 \times 10^{-2} = 25 \times 10^{-5} \)

Conclusion:

The longitudinal strain produced in the wire is \( \mathbf{25 \times 10^{-5}} \).