A wire of length L and diameter d is stretched by a force F. If the diameter is halved, what will be the new elongation for the same force? (2020)

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Q: A wire of length L and diameter d is stretched by a force F. If the diameter is halved, what will be the new elongation for the same force? (2020)

Solution: Elongation is inversely proportional to the area. Halving the diameter reduces the area to a quarter, thus elongation quadruples.

Steps: 8

Step 1: Understand that elongation (how much the wire stretches) depends on the area of the wire's cross-section.
Step 2: Recall that the area of a circle (the cross-section of the wire) is calculated using the formula A = π(d/2)², where d is the diameter.
Step 3: If the diameter d is halved, the new diameter becomes d/2.
Step 4: Calculate the new area with the halved diameter: A_new = π((d/2)/2)² = π(d/4)² = π(d²/16).
Step 5: Compare the new area to the original area: The original area A = π(d/2)² = π(d²/4).
Step 6: Notice that the new area is A_new = (1/4) * A (since π(d²/16) is one-fourth of π(d²/4)).
Step 7: Understand that elongation is inversely proportional to the area: If the area decreases, the elongation increases.
Step 8: Since the area is reduced to one-fourth, the elongation will increase by a factor of 4 (quadruples).