Question

A tension of $20\, N$ is applied to a copper wire of cross sectional area $0.01\, cm ^{2}$, 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. $1.16 \times 10^{-6} cm ^{2}$
• B. $1.16 \times 10^{-5} m ^{2}$
• C. $1.16 \times 10^{-4} m ^{2}$
• D. $1.16 \times 10^{-3} cm ^{2}$

Answer 1

Answer: (A)

Given, $\sigma=0.32,\, F=20\, N$
$A=0.01\, cm ^{2}=0.01 \times 10^{-3}\, m$
and $Y=1-1 \times 10^{11}\, N / m ^{2}$
We know that
$\frac{\Delta l}{l}=\frac{F}{A Y}=\frac{20}{0.01 \times 10^{-3} \times 1.1 \times 10^{11}}$
$=18.1 \times 10^{-7}$
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
$\Delta A=2 \frac{\Delta r}{r} \times A=2 \times 5.79 \times 10^{-7} \times 0.01 \times 10^{-3}$
$=0.158 \times 10^{-10} m ^{2}$
$=1.26 \times 10^{-6} cm ^{2}$