A wire of length L and diameter d is stretched by a force F. If the diameter is halved while keeping the length constant, what happens to the stress? (2020)
Practice Questions
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Q1
A wire of length L and diameter d is stretched by a force F. If the diameter is halved while keeping the length constant, what happens to the stress? (2020)
It doubles
It quadruples
It halves
It remains the same
Stress = Force / Area. Halving the diameter increases the area by a factor of 1/4, thus stress quadruples.
Questions & Step-by-step Solutions
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Q
Q: A wire of length L and diameter d is stretched by a force F. If the diameter is halved while keeping the length constant, what happens to the stress? (2020)
Solution: Stress = Force / Area. Halving the diameter increases the area by a factor of 1/4, thus stress quadruples.
Steps: 8
Step 1: Understand what stress is. Stress is defined as the force applied to an object divided by the area over which the force is applied.
Step 2: Write the formula for stress: Stress = Force / Area.
Step 3: Identify the area of a wire with diameter d. The area (A) of a circle is calculated using the formula A = π(d/2)².
Step 4: If the diameter is halved, the new diameter becomes d/2. Calculate the new area with the new diameter: A' = π((d/2)/2)² = π(d/4)² = π(d²/16).
Step 5: Compare the new area (A') to the original area (A). The original area is A = π(d/2)² = π(d²/4).
Step 6: Find the ratio of the new area to the original area: A' / A = (π(d²/16)) / (π(d²/4)) = (d²/16) / (d²/4) = 1/4.
Step 7: Since the area decreases by a factor of 1/4, the stress increases by a factor of 4 (because stress is inversely proportional to area).
Step 8: Conclude that halving the diameter while keeping the length constant causes the stress to quadruple.
