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(A) The Young's modulus $(Y)$ is defined as the ratio of tensile stress to tensile strain: $Y = \frac{F/A}{\Delta l/l}$.Rearranging for elongation $(\

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Wire of length $50 \ cm$ and diameter $0.5 \ mm$

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(A) The Young's modulus $(Y)$ is defined as the ratio of tensile stress to tensile strain: $Y = \frac{F/A}{\Delta l/l}$.Rearranging for elongation $(\Delta l)$,we get: $\Delta l = \frac{F \cdot l}{A \cdot Y}$.Since the area $A = \pi r^2 = \pi (d/2)^2 = \frac{\pi d^2}{4}$,we substitute this into the equation:$\Delta l = \frac{4 F l}{\pi d^2 Y}$.Given that the tension $(F)$ and material $(Y)$ are the same for all wires,the elongation is proportional to the ratio $\frac{l}{d^2}$.Calculating the ratio $\frac{l}{d^2}$ for each option:$A: \frac{50}{(0.5)^2} = \frac{50}{0.25} = 200$.$B: \frac{200}{(2)^2} = \frac{200}{4} = 50$.$C: \frac{300}{(3)^2} = \frac{300}{9} \approx 33.33$.$D: \frac{100}{(1)^2} = \frac{100}{1} = 100$.Comparing the values,the ratio is maximum for option $A$.

Same tension is applied to the following four wires made of same material. The elongation is longest in

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