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(C) Given: Length of the wire,$L = 10 \ m$,diameter,$D = 0.6 	imes 10^{-3} \ m$,Poisson's ratio,$\sigma = 0.3$,and change in wire length,$\Delta L =

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$10.8 	imes 10^{-8} \ m$

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(C) Given: Length of the wire,$L = 10 \ m$,diameter,$D = 0.6 	imes 10^{-3} \ m$,Poisson's ratio,$\sigma = 0.3$,and change in wire length,$\Delta L = 6 	imes 10^{-3} \ m$.The Poisson's ratio is defined as the ratio of lateral strain to longitudinal strain:$\sigma = \frac{	ext{lateral strain}}{	ext{longitudinal strain}} = \frac{\Delta D / D}{\Delta L / L}$Rearranging the formula to solve for the change in diameter $\Delta D$:$\Delta D = \sigma 	imes D 	imes \frac{\Delta L}{L}$Substituting the given values:$\Delta D = 0.3 	imes (0.6 	imes 10^{-3} \ m) 	imes \frac{6 	imes 10^{-3} \ m}{10 \ m}$$\Delta D = 0.3 	imes 0.6 	imes 10^{-3} 	imes 6 	imes 10^{-4} \ m$$\Delta D = 1.08 	imes 10^{-7} \ m = 10.8 	imes 10^{-8} \ m$Therefore,the change in the diameter of the wire is $10.8 	imes 10^{-8} \ m$. The correct option is $C$.

$A$ uniform wire of length $10 \ m$ and diameter $0.6 \ mm$ is stretched by $6 \ mm$ with a certain force. If the Poisson's ratio of the material of the wire is $0.3$,then the change in diameter of the wire is

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