Question

A wire of length $10m$ is subjected to a force of $100N$ along its length and hence 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 will be

• 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 $=\frac{Lateralstrain}{Longitudinalstrain}$
$ie,0.4=\frac{0.01\times 10^{-3}}{\frac{l}{L}}$
or $\frac{L}{l}=\frac{0.4}{0.01\times 10^{-3}}=4\times 10^4$
Young's modulus
$Y=\frac{FL}{Al}$
$=\frac{100}{0.025}\times 4\times 10^4=1.6\times 10^{8}N{m}^{-2}$