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EEJ MAIN Physics
QUESTION #7085
Question 1
A wire stretches by $0.04\text{ m}$ under force $F$. What is the elongation if length and diameter are doubled under the same force?
Correct Answer Explanation
Elongation $\Delta L = \frac{FL}{AY} = \frac{FL}{\pi(d/2)^2 Y} \propto \frac{L}{d^2}$.
- New elongation $\Delta L' \propto \frac{2L}{(2d)^2} = \frac{2L}{4d^2} = \frac{1}{2} \Delta L$
- $\Delta L' = \frac{1}{2} \times 0.04 = 0.02\text{ m}$.
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