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(A) Poisson's ratio $(\sigma) = \frac{	ext{lateral strain}}{	ext{longitudinal strain}}$.Given: lateral strain $= 0.01 	imes 10^{-3}$,$\sigma = 0.4$

Vedclass · Accepted Answer

$1.6 	imes 10^8 \ N/m^2$

Vedclass · Answer

(A) Poisson's ratio $(\sigma) = \frac{	ext{lateral strain}}{	ext{longitudinal strain}}$.Given: lateral strain $= 0.01 	imes 10^{-3}$,$\sigma = 0.4$.Longitudinal strain $(\frac{\Delta L}{L}) = \frac{	ext{lateral strain}}{\sigma} = \frac{0.01 	imes 10^{-3}}{0.4} = 0.025 	imes 10^{-3} = 2.5 	imes 10^{-5}$.Young's modulus $(Y) = \frac{	ext{Stress}}{	ext{Strain}} = \frac{F/A}{\Delta L/L} = \frac{F}{A} 	imes \frac{L}{\Delta L}$.Given: $F = 100 \ N$,$A = 0.025 \ m^2$,$\frac{\Delta L}{L} = 2.5 	imes 10^{-5}$.$Y = \frac{100}{0.025 	imes 2.5 	imes 10^{-5}} = \frac{100}{0.0625 	imes 10^{-5}} = \frac{100}{6.25 	imes 10^{-7}} = 16 	imes 10^7 \ N/m^2 = 1.6 	imes 10^8 \ N/m^2$.

When a wire of length $10 \ m$ is subjected to a force of $100 \ N$ along its length,the lateral strain produced is $0.01 \times 10^{-3} \ m$. The Poisson's ratio was found to be $0.4$. If the area of cross-section of the wire is $0.025 \ m^2$,its Young's modulus is

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