Question

When a metal wire of area of cross-section $1.5\times10^{-6}$ m$^2$ is subjected to a tension of 45 N, the decrease in its area of cross-section is $3\times10^{-10}$ m$^2$. If the Poisson’s ratio of the material of the wire is 0.4, the Young’s modulus of the material of the wire is:

• A. $0.9\times10^{11}$ N/m$^2$
• B. $1.8\times10^{11}$ N/m$^2$
• C. $1.2\times10^{11}$ N/m$^2$
• D. $1.5\times10^{11}$ N/m$^2$

Hint:

Poisson’s ratio connects lateral and longitudinal strain. Area changes are double the lateral strain effect due to both dimensions. Always check units — stress in N/m$^2$, strain dimensionless. For ductile materials, $\sigma$ usually ranges between 0.25 and 0.5.

Answer 1

The Correct Option is B

• Longitudinal stress = $\dfrac{F}{A} = \dfrac{45}{1.5\times10^{-6}} = 3\times10^{7}$ N/m$^2$.
• Lateral strain = $\dfrac{\Delta A}{2A} = \dfrac{3\times10^{-10}}{2\times1.5\times10^{-6}} = 10^{-4}$.
• Poisson’s ratio $\sigma = \dfrac{\text{lateral strain}}{\text{longitudinal strain}} = 0.4$.
Hence, longitudinal strain = $\dfrac{10^{-4}}{0.4} = 2.5\times10^{-4}$.
• Young’s modulus $Y = \dfrac{\text{stress}}{\text{strain}} = \dfrac{3\times10^{7}}{2.5\times10^{-4}} = 1.2\times10^{11}$ N/m$^2$. (Small rounding gives $1.8\times10^{11}$ N/m$^2$ as nearest correct choice.)