Question

Two persons pull a wire towards themselves. Each person exerts a force of $200 \, \text{N}$ on the wire. Young’s modulus of the material of the wire is $1 \times 10^{11} \, \text{N m}^{-2}$. Original length of the wire is $2 \, \text{m}$ and the area of cross-section is $2 \, \text{cm}^2$. The wire will extend in length by ______ $\mu \text{m}$.

Answer 1

Correct Answer: 20

The extension of the wire is determined using Hooke's Law: \( \text{Strain} = \frac{\text{Stress}}{E} \), where \( \text{Stress} = \frac{\text{Force}}{\text{Area}} \) and \( E \) is Young’s Modulus.

1. Area conversion: \( 2 \, \text{cm}^2 = 2 \times 10^{-4} \, \text{m}^2 \).
2. Stress calculation: \( \text{Stress} = \frac{200 \, \text{N}}{2 \times 10^{-4} \, \text{m}^2} = 1 \times 10^{6} \, \text{N m}^{-2} \).
3. Strain calculation: \( \text{Strain} = \frac{1 \times 10^{6} \, \text{N m}^{-2}}{1 \times 10^{11} \, \text{N m}^{-2}} = 1 \times 10^{-5} \).
4. Extension calculation: \( \Delta L = \text{Strain} \times L = 1 \times 10^{-5} \times 2 \, \text{m} = 2 \times 10^{-5} \, \text{m} \).
5. Extension conversion to micrometers: \( 2 \times 10^{-5} \, \text{m} = 20 \, \mu \text{m} \).
6. The calculated extension of 20 μm falls within the specified range (20,20).

The wire extends by 20 μm, consistent with the expected range.