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 ______ $\mu \text{m}$.
Correct Answer: 20
Approach Solution - 1
To determine the extension in length of the wire, we apply the formula from Hooke's Law relating stress, strain, and Young’s modulus: \( \text{Strain} = \frac{\text{Stress}}{E} \), where \( \text{Stress} = \frac{\text{Force}}{\text{Area}} \) and \( E \) is Young’s Modulus.
- Convert the area from \( \text{cm}^2 \) to \( \text{m}^2 \):
\( 2 \, \text{cm}^2 = 2 \times 10^{-4} \, \text{m}^2 \). - Calculate the stress on the wire:
\( \text{Stress} = \frac{200 \, \text{N}}{2 \times 10^{-4} \, \text{m}^2} = 1 \times 10^{6} \, \text{N m}^{-2} \). - The strain in the wire is given by:
\( \text{Strain} = \frac{1 \times 10^{6} \, \text{N m}^{-2}}{1 \times 10^{11} \, \text{N m}^{-2}} = 1 \times 10^{-5} \). - Calculate the extension \( \Delta L \) using \( \text{Strain} = \frac{\Delta L}{L} \):
\( \Delta L = \text{Strain} \times L = 1 \times 10^{-5} \times 2 \, \text{m} = 2 \times 10^{-5} \, \text{m} \). - Convert the extension into micrometers (\(\mu \text{m}\)):
\( 2 \times 10^{-5} \, \text{m} = 20 \, \mu \text{m} \). - Verify the calculated extension is within the specified range (20,20), confirming the accuracy of our computation.
Therefore, the wire will extend in length by 20 μm, fitting perfectly within the expected range.
Approach Solution -2
Step 1: Relation between stress and strain Young’s modulus is given by:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{\frac{F}{A}}{\frac{\Delta \ell}{\ell}}. \]
Rearranging for \( \Delta \ell \):
\[ \Delta \ell = \frac{F \ell}{A Y}. \]
Step 2: Substitute given values
- Force, \( F = 200 \, \text{N} \),
- Original length, \( \ell = 2 \, \text{m} \),
- Area of cross-section, \( A = 2 \, \text{cm}^2 = 2 \times 10^{-4} \, \text{m}^2 \),
- Young’s modulus, \( Y = 1 \times 10^{11} \, \text{N/m}^2 \).
Substitute into the formula:
\[ \Delta \ell = \frac{200 \cdot 2}{2 \cdot 10^{-4} \cdot 10^{11}}. \]
Step 3: Simplify the expression
\[ \Delta \ell = \frac{400}{2 \times 10^7} = 2 \times 10^{-5} \, \text{m}. \]
Convert to micrometers (\( \mu \text{m} \)):
\[ \Delta \ell = 20 \, \mu \text{m}. \]
Final Answer: 20 \( \mu \text{m} \).
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