Question

A copper wire of cross-sectional area $0.01c{m}^2$ is under a tension of $22N$. The decrease in the cross-sectional area is (Young modulus $=1.1\times 10^{11}N{m}^{-2}$, Poisson's ratio $=0.32$ )

• A. $0.128\times 10^{-6}c{m}^2$
• B. $128\times 10^{-6}c{m}^2$
• C. $12.8\times 10^{-6}c{m}^2$
• D. $1.28\times 10^{-6}c{m}^2$

Answer 1

Answer: (D)

Young's modulus,
$Y=\frac{F/A}{\Delta l/l}$
$\frac{\Delta l}{l}=\frac{F}{YA}$
where, $\frac{\Delta l}{l}=$ longitudinal strain
Given, $F=22N,Y=1.1\times 10^{11}N-m^2$
$A=0.01c{m}^2=10^{-6}{m}^2$
$\frac{\Delta l}{l}=\frac{22}{1.1\times 10^{11}\times 10^{-6}}$
$=2\times 10^{-4}$
Now, Poisson ratio
$\sigma =\frac{\text{ Lateral strain }}{\text{ Longitudinal strain }}=\frac{\Delta d/d}{\Delta l/l}$
$\frac{\Delta d}{d}=\sigma \cdot \frac{\Delta l}{l}=0.32\times 2\times 10^{-4}$
Change in diameter, $\frac{\Delta d}{d}=6.4\times 10^{-5}$
or change ( decrease) in radius,
$\frac{\Delta r}{r}=6.4\times 10^{-5}$
Area, $A=\pi r^2$
Fractional change in area,
$\frac{\Delta A}{A}=2\cdot \frac{\Delta r}{r}$
$\frac{\Delta A}{A}=2\times 6.4\times 10^{-5}$
$\Delta A=\left(12.8\times 10^{-5}\right)A$
Decrease in area,
$\Delta A=\left(12.8\times 10^{-5}\right)\times (0.01)c{m}^2$
$\Delta A=1.28\times 10^{-6}c{m}^2$