Question

When a wire of length $10m$ is subjected to a force of $100N$ 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 wire is $0.025m^2$, its Young's modulus is

• A. $1.6\times 10^{8}N/m^2$
• B. $2.5\times 10^{10}N/m^2$
• C. $1.25\times 10^{11}N/m^2$
• D. $16\times 10^{9}N/m^2$

Answer 1

Answer: (A)

Poisson' s ratio $=\frac{\text{ lateral strain }}{\text{ longitudinal strain}}$
ie, $0.4=\frac{0.01\times 10^{-3}}{\Delta L/L}$
or $\frac{I}{\Delta L}=\frac{0.4}{0.01\times 10^{-3}}=4\times 10^4$
Young's modulus
$Y=\frac{FL}{A\Delta l}$
$=\frac{100}{0.025}\times 4\times 10^4$
$=1.6\times 10^{8}N/m^2$