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 in the wire? (2022)
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 in the wire? (2022)
It doubles
It quadruples
It remains the same
It halves
Stress (σ) = Force (F) / Area (A). Halving the diameter increases the area by a factor of 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 in the wire? (2022)
Solution: Stress (σ) = Force (F) / Area (A). Halving the diameter increases the area by a factor of 4, thus stress quadruples.
Steps: 8
Step 1: Understand that stress (σ) is calculated using the formula σ = F / A, where F is the force applied and A is the cross-sectional area of the wire.
Step 2: Recognize that the area (A) of a wire with diameter (d) is calculated using the formula A = π(d/2)², which simplifies to A = πd²/4.
Step 3: If the diameter (d) is halved, the new diameter becomes d/2.
Step 4: Calculate the new area (A') using the new diameter: A' = π((d/2)/2)² = π(d/4)² = πd²/16.
Step 5: Compare the new area (A') to the original area (A): A' = (1/4) * A, meaning the new area is one-fourth of the original area.
Step 6: Since stress (σ) is inversely proportional to area (A), if the area decreases, the stress increases.
Step 7: Calculate the new stress (σ') using the new area: σ' = F / A' = F / (A/4) = 4F / A = 4σ.
Step 8: Conclude that halving the diameter of the wire while keeping the length constant causes the stress in the wire to quadruple.
