Question

A copper wire of length 2.2 m and a steel wire of length 1.6 m, both of diameter 3.0 mm, are connected end to end. When stretched by a force, the elongation in length 0.50 m is produced in the copper wire. The stretching force is ( \( Y_{\text{cu}} = 1.1 \times 10^{11} \, \text{N/m}^2, Y_{\text{steel}} = 2.0 \times 10^{11} \, \text{N/m}^2 \))

• A. \( 5.4 \times 10^2 \, \text{N} \)
• B. \( 3.6 \times 10^2 \, \text{N} \)
• C. \( 2.4 \times 10^2 \, \text{N} \)
• D. \( 1.8 \times 10^2 \, \text{N} \)

Hint:

The elongation is inversely proportional to the Young's modulus and cross-sectional area of the material.

Answer 1

The Correct Option is D

Step 1: Elongation formula.
The elongation in a material under a force is given by: \[ \Delta L = \frac{F L}{A Y} \] where \( F \) is the force, \( L \) is the length, \( A \) is the cross-sectional area, and \( Y \) is Young's modulus. We can solve for \( F \) using the elongation in copper.

Step 2: Conclusion.
The required stretching force is \( 1.8 \times 10^2 \, \text{N} \).