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(C) The formula for Young's modulus $(Y)$ is given by $Y = \frac{F/A}{\Delta \ell / \ell}$, where $F$ is the applied force, $A$ is the area of cross-s

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$20$

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(C) The formula for Young's modulus $(Y)$ is given by $Y = \frac{F/A}{\Delta \ell / \ell}$, where $F$ is the applied force, $A$ is the area of cross-section, $\ell$ is the original length, and $\Delta \ell$ is the change in length.Given:Force $F = 200 \, N$ (Note: In a wire pulled from both ends with $200 \, N$, the tension in the wire is $200 \, N$)Original length $\ell = 2 \, m$Area of cross-section $A = 2 \, cm^2 = 2 	imes 10^{-4} \, m^2$Young's modulus $Y = 1 	imes 10^{11} \, N \, m^{-2}$Rearranging the formula to find the extension $\Delta \ell$:$\Delta \ell = \frac{F \ell}{AY}$Substituting the values:$\Delta \ell = \frac{200 	imes 2}{(2 	imes 10^{-4}) 	imes (1 	imes 10^{11})}$$\Delta \ell = \frac{400}{2 	imes 10^7}$$\Delta \ell = 200 	imes 10^{-7} \, m$$\Delta \ell = 2 	imes 10^{-5} \, m$Converting to micrometers $(\mu m)$:$1 \, m = 10^6 \, \mu m$$\Delta \ell = 2 	imes 10^{-5} 	imes 10^6 \, \mu m = 20 \, \mu m$.

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