Question

When a wire of length 10m is subjected to a force of $100N$ along its length , the lateral strain produced is $0.01$ x $10$${}^{-3}$. The Poisson's ratio was found to be $0.4$. If the area of cross-section of wire is $0.025m$${}^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. $12.5\times {10}^{11}N/m^2$
• D. $16\times {10}^{10}N/m^2$

Answer 1

The correct option is A $1.6\times {10}^{8}N/m^2$

$\text{Poisson's ratio}=\frac{\Delta d/d}{\Delta d_0/d_0}$
$\frac{\Delta l}{l}=\frac{\Delta d/d}{0.4}$
$=\frac{0.01\times {10}^3}{0.4}$
$Y=\frac{F/A}{\Delta l/l}$
$=\frac{100/0.025}{0.01\times {10}^3/0.4}$
$=\frac{100\times 0.4}{0.025\times 0.01\times {10}^3}$
$=1.6\times {10}^{8}N/m^2$