Question

Two wires of different materials have same length \( L \) and same diameter \( d \). The second wire is connected at the end of the first wire and forms one single wire of double the length. This wire is subjected to stretching force \( F \) to produce the elongation \( \Delta L \). The two wires have

• A. Same stress and same strain.
• B. Different stress but same strain.
• C. Different stress and different strain.
• D. Same stress but different strain.

Hint:

When two wires are subjected to the same force, their strain depends on the material's Young's modulus, while the stress depends on the cross-sectional area.

Answer 1

The Correct Option is D

Step 1: Understanding stress and strain.
Stress \( \sigma \) is given by \( \sigma = \frac{F}{A} \), where \( A \) is the cross-sectional area of the wire. Since both wires have the same diameter, the stress in both will be the same.
Step 2: Strain relationship.
Strain \( \epsilon \) is given by \( \epsilon = \frac{\Delta L}{L} \). The strain will be different for the two wires as their materials differ. The elongation produced by the same force will vary depending on the Young’s modulus of the material.
Step 3: Conclusion.
The stress will be the same because the diameter is the same, but the strain will be different due to the different materials. Thus, the correct answer is (D).