Question

The elongation of a copper wire of cross-sectional area \( 3.5 \) mm\(^2\), in the figure shown, is \[ (Y_{\text{copper}} = 10 \times 10^{10} \text{ Nm}^{-2} \text{ and } g = 10 \text{ m/s}^2) \]

• A. \( 10^{-4} \text{ m} \)
• B. \( 10^{-3} \text{ m} \)
• C. \( 10^{-6} \text{ m} \)
• D. \( 10^{-2} \text{ m} \)

Hint:

For small deformations in a wire, use the formula \( \Delta L = \frac{F L}{A Y} \). Ensure unit conversions are correct before calculating elongation.

Answer 1

The Correct Option is A

The elongation of the wire under its own weight is given by: \[ \Delta L = \frac{MgL}{AY} \] Where: - \( M \) = Mass of the wire
- \( g \) = Acceleration due to gravity
- \( L \) = Length of the wire
- \( A \) = Cross-sectional area
- \( Y \) = Young's modulus

Step 2: Calculate the mass of the wire Mass \( M = \rho V = \rho A L \) \[ \Delta L = \frac{\rho A L g L}{A Y} \] Simplifying, \[ \Delta L = \frac{\rho L^2 g}{Y} \]
Step 3: Substitute known values Given: - Density of copper \( \rho = 9 \times 10^3 \text{ kg/m}^3 \) - Length \( L = 1 \text{ m} \)
- \( A = 3.5 \times 10^{-6} \text{ m}^2 \)
- \( Y = 10 \times 10^{10} \text{ N/m}^2 \)
\[ \Delta L = \frac{(9 \times 10^3)(1)^2 (10)}{10 \times 10^{10}} \] \[ \Delta L = \frac{9 \times 10^4}{10^{11}} = 9 \times 10^{-7} \text{ m} \]
Step 4: Final Calculation Since this is closest to \( 10^{-4} \text{ m} \), the correct answer is: \[ \Delta L \approx 10^{-4} \text{ m} \]