The following four wires are made of the same | vedclass

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(A) Since the material is the same,Young's modulus $(Y)$ is constant. Given that the tension $(F)$ is also constant,the extension $(\Delta L)$ is give

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length $0.5 \ m$,diameter $0.5 \ mm$.

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(A) Since the material is the same,Young's modulus $(Y)$ is constant. Given that the tension $(F)$ is also constant,the extension $(\Delta L)$ is given by the formula:$\Delta L = \frac{F L}{Y A}$Substituting the area $A = \frac{\pi D^2}{4}$,we get:$\Delta L = \frac{4 F L}{Y \pi D^2} \Rightarrow \Delta L \propto \frac{L}{D^2}$Now,we calculate the ratio $\frac{L}{D^2}$ for each option:$(A)$ $\frac{0.5}{(0.5 	imes 10^{-3})^2} = \frac{0.5}{0.25 	imes 10^{-6}} = 2 	imes 10^6 \ m^{-1}$$(B)$ $\frac{1}{(1 	imes 10^{-3})^2} = \frac{1}{1 	imes 10^{-6}} = 1 	imes 10^6 \ m^{-1}$$(C)$ $\frac{2}{(2 	imes 10^{-3})^2} = \frac{2}{4 	imes 10^{-6}} = 0.5 	imes 10^6 \ m^{-1}$$(D)$ $\frac{3}{(3 	imes 10^{-3})^2} = \frac{3}{9 	imes 10^{-6}} = 0.33 	imes 10^6 \ m^{-1}$Comparing the values,the ratio $\frac{L}{D^2}$ is largest for option $(A)$. Therefore,the wire in option $(A)$ has the largest extension.

The following four wires are made of the same material. If the same tension is applied to each,the wire having the largest extension is

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