A cylindrical wire of radius 1, mm,length 1, | vedclass

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(D) Given: Radius $r = 1\, mm = 10^{-3}\, m$,Length $L = 1\, m$,Force $F = 100\, N$,Young's modulus $Y = 2 	imes 10^{11}\, N/m^2$,Poisson's ratio $\m

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$0.99995$

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(D) Given: Radius $r = 1\, mm = 10^{-3}\, m$,Length $L = 1\, m$,Force $F = 100\, N$,Young's modulus $Y = 2 	imes 10^{11}\, N/m^2$,Poisson's ratio $\mu = \pi / 10$,Area $A = \pi r^2 = \pi 	imes (10^{-3})^2 = \pi 	imes 10^{-6}\, m^2$.$1$. Calculate longitudinal strain $(\epsilon_L)$: $\epsilon_L = \frac{\Delta L}{L} = \frac{F}{AY} = \frac{100}{(\pi 	imes 10^{-6}) 	imes (2 	imes 10^{11})} = \frac{100}{2\pi 	imes 10^5} = \frac{1}{2\pi} 	imes 10^{-3}$.$2$. Calculate lateral strain $(\epsilon_d)$: Poisson's ratio $\mu = - \frac{	ext{lateral strain}}{	ext{longitudinal strain}} = - \frac{\Delta r / r}{\Delta L / L}$.$\frac{\Delta r}{r} = - \mu 	imes \epsilon_L = - (\frac{\pi}{10}) 	imes (\frac{1}{2\pi} 	imes 10^{-3}) = - \frac{1}{20} 	imes 10^{-3} = - 0.05 	imes 10^{-3} = - 5 	imes 10^{-5}$.$3$. Calculate change in radius $(\Delta r)$: $\Delta r = - 5 	imes 10^{-5} 	imes r = - 5 	imes 10^{-5} 	imes 1\, mm = - 0.00005\, mm$.$4$. New radius $(r')$: $r' = r + \Delta r = 1\, mm - 0.00005\, mm = 0.99995\, mm$.

$A$ cylindrical wire of radius $1\, mm$,length $1\, m$,Young's modulus $Y = 2 \times 10^{11}\, N/m^2$,and Poisson's ratio $\mu = \pi / 10$ is stretched by a force of $100\, N$. What will be its new radius (in $, mm$)?

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