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(D) The formula for elongation $\Delta L$ is given by $\Delta L = \frac{FL}{AY}$,where $F$ is the tension,$L$ is the length,$A$ is the cross-sectional

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Length $400 \, cm$,diameter $0.01 \, mm$

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(D) The formula for elongation $\Delta L$ is given by $\Delta L = \frac{FL}{AY}$,where $F$ is the tension,$L$ is the length,$A$ is the cross-sectional area,and $Y$ is Young's modulus.Since all wires are made of the same material,$Y$ is constant. For the same tension $F$,the elongation $\Delta L \propto \frac{L}{A}$.Since $A = \pi r^2 = \pi (d/2)^2 = \frac{\pi d^2}{4}$,we have $\Delta L \propto \frac{L}{d^2}$.Calculating the ratio $L/d^2$ for each option:$A: \frac{500}{(0.05)^2} = \frac{500}{0.0025} = 200,000$$B: \frac{200}{(0.02)^2} = \frac{200}{0.0004} = 500,000$$C: \frac{300}{(0.03)^2} = \frac{300}{0.0009} \approx 333,333$$D: \frac{400}{(0.01)^2} = \frac{400}{0.0001} = 4,000,000$Comparing the values,option $D$ has the largest ratio,hence it will have the largest elongation.

The following four wires are made of the same material. Which one will have the largest elongation when subjected to the same tension?

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