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(A) Young's modulus is given by $Y = \frac{FL}{A \Delta L} = \frac{4FL}{\pi D^2 \Delta L}$.Rearranging for extension,we get $\Delta L = \frac{4FL}{\pi

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length $= 50 \; cm$,diameter $= 0.5 \; mm$

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(A) Young's modulus is given by $Y = \frac{FL}{A \Delta L} = \frac{4FL}{\pi D^2 \Delta L}$.Rearranging for extension,we get $\Delta L = \frac{4FL}{\pi D^2 Y}$.Since all wires are made of the same material and subjected to the same tension $F$,$Y$ and $F$ are constant.Therefore,$\Delta L \propto \frac{L}{D^2}$.Calculating the ratio $\frac{L}{D^2}$ for each case:$(a)$ $\frac{50}{(0.5)^2} = \frac{50}{0.25} = 200 \; cm^{-1}$$(b)$ $\frac{100}{(1)^2} = 100 \; cm^{-1}$$(c)$ $\frac{200}{(2)^2} = \frac{200}{4} = 50 \; cm^{-1}$$(d)$ $\frac{300}{(3)^2} = \frac{300}{9} \approx 33.3 \; cm^{-1}$Comparing the values,the ratio is largest for option $(a)$.

The following four wires are made of the same material. Which of these will have the largest extension when the same tension is applied?

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