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

Answer: (A)

Given, $\sigma =0.32,F=20N$
$A=0.01c{m}^2=0.01\times 10^{-3}m$
and $Y=1-1\times 10^{11}N/m^2$
We know that
$\frac{\Delta l}{l}=\frac{F}{AY}=\frac{20}{0.01\times 10^{-3}\times 1.1\times 10^{11}}$
$=18.1\times 10^{-7}$
and we also known
$\sigma =\frac{-\Delta r/r}{\Delta l/l}$
$-\frac{\Delta r}{r}=0.32\times 18.1\times 10^{-7}=5.79\times 10^{-7}$
Hence, decrease in cross reactional area of wire is
$\Delta A=2\frac{\Delta r}{r}\times A=2\times 5.79\times 10^{-7}\times 0.01\times 10^{-3}$
$=0.158\times 10^{-10}{m}^2$
$=1.26\times 10^{-6}c{m}^2$