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(D) The elongation $\Delta \ell$ of a wire is given by the formula: $\Delta \ell = \frac{F \ell}{A Y} = \frac{F \ell}{\pi r^2 Y}$.Since the material (

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radius $3 \ mm$,length $2 \ m$

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(D) The elongation $\Delta \ell$ of a wire is given by the formula: $\Delta \ell = \frac{F \ell}{A Y} = \frac{F \ell}{\pi r^2 Y}$.Since the material (Young's modulus $Y$) and the force $F$ are the same for all wires,the elongation is proportional to $\frac{\ell}{r^2}$.We calculate the ratio $k = \frac{\ell}{r^2}$ for each option:$A$: $k = \frac{3}{3^2} = \frac{3}{9} = 0.333 \ m/mm^2$.$B$: $k = \frac{0.5}{0.5^2} = \frac{0.5}{0.25} = 2.0 \ m/mm^2$.$C$: $k = \frac{2}{2^2} = \frac{2}{4} = 0.5 \ m/mm^2$.$D$: $k = \frac{2}{3^2} = \frac{2}{9} = 0.222 \ m/mm^2$.Comparing the values,option $D$ has the minimum ratio,and therefore the minimum elongation.

Four uniform wires of the same material are stretched by the same force. The dimensions of the wires are given below. Which one has the minimum elongation?

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