Question

A tension of $20N$ is applied to a copper wire of cross sectional area $0.01c{m}^2$, Young's Modulus of copper is $1.1\times {10}^{11}N/m^2$ and Poisson's ratio is 0.32. The decrease in cross sectional area of the wire is:

• A. $1.16\times {10}^{-6}c{m}^2$
• B. $1.16\times {10}^{-5}{m}^2$
• C. $1.16\times {10}^{-4}{m}^2$
• D. $1.16\times {10}^{-3}c{m}^2$

Answer 1

The correct option is A $1.16\times {10}^{-6}c{m}^2$

In case Poisson ratio ($\nu$) is given, the relation between stress and strain contains factor of 2,

Hence, $\frac{\Delta A}{A}=\frac{2}{Y}\frac{F}{A}\nu$

$\Delta A=\frac{2F}{Y}\nu =2\frac{20N}{1.1\times {10}^{11}N/m^2}\left(0.32\right)=1.16\times {10}^{-6}{cm}^2$