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

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 \( \mu \text{m} \).