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

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

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(A) Poisson's ratio $(\sigma) = \frac{	ext{lateral strain}}{	ext{longitudinal strain}}$.Given lateral strain $= 0.01 	imes 10^{-3} \ m$ (Note: Strain is dimensionless, assuming the value represents the change in diameter $\Delta D$ or similar, but here we use it directly as the numerator for the ratio).$\sigma = \frac{	ext{lateral strain}}{\Delta L / L} = 0.4$.$\frac{\Delta L}{L} = \frac{0.01 	imes 10^{-3}}{0.4} = 0.025 	imes 10^{-3} = 2.5 	imes 10^{-5}$.Young's modulus $Y = \frac{F/A}{\Delta L/L} = \frac{F \cdot L}{A \cdot \Delta L}$.$Y = \frac{100 \ N}{0.025 \ m^2 	imes (2.5 	imes 10^{-5})} = \frac{100}{0.0625 	imes 10^{-5}} = \frac{100}{6.25 	imes 10^{-7}} = 16 	imes 10^7 = 1.6 	imes 10^8 \ N/m^2$.

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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