Question

A wire of length 2 m, radius of cross-section 20 mm and Young's Modulus 2×10¹¹ N/m is subjected to a force of 62.8 kN.The change in length of the wire is p×10^–5 (in m). Find P. 

Answer 1

The change in length (\( \Delta L \)) of a wire subjected to a force is given by the formula: \[ \Delta L = \frac{F \cdot L}{A \cdot Y} \] where: 
- \( F \) is the applied force, 
- \( L \) is the original length of the wire, 
- \( A \) is the cross-sectional area of the wire, 
- \( Y \) is Young's Modulus of the material. 
Step 1: Substitute the given values.
Given: - Force \( F = 62.8 \text{ kN} = 62.8 \times 10^3 \text{ N} \), 
- Length \( L = 2 \text{ m} \), 
- Radius of cross-section \( r = 20 \text{ mm} = 20 \times 10^{-3} \text{ m} \), 
- Young's Modulus \( Y = 2 \times 10^{11} \text{ N/m}^2 \). 
The cross-sectional area \( A \) of the wire is: \[ A = \pi r^2 = \pi (20 \times 10^{-3})^2 = \pi \times 400 \times 10^{-6} = 1.2566 \times 10^{-3} \text{ m}^2 \] Step 2: Calculate the change in length.
Now, substituting all the values into the formula for \( \Delta L \): \[ \Delta L = \frac{(62.8 \times 10^3) \cdot 2}{(1.2566 \times 10^{-3}) \cdot (2 \times 10^{11})} \] \[ \Delta L = \frac{125.6 \times 10^3}{2.5132 \times 10^8} \] \[ \Delta L = 50 \times 10^{-5} \text{ m} \] Final Answer:
The change in length of the wire is \( 50 \times 10^{-5} \text{ m} \), so \( P = 50 \).