Question

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 wire is $0.025\, m ^{2}$, 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
$Y =\frac{F L}{A \Delta l}$
$=\frac{100}{0.025} \times 4 \times 10^{4}$
$=1.6 \times 10^{8}\, N / m ^{2}$