Question

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

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

Answer 1

Answer: (D)

Young's modulus,
$Y =\frac{F / A}{\Delta l / l} $
$\frac{\Delta l}{l} =\frac{F}{Y A}$
where, $\frac{\Delta l}{l}=$ longitudinal strain
Given, $F=22 \,N ,\, Y=1.1 \times 10^{11}\, N - m ^{2}$
$A =0.01\, cm ^{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) \,cm ^{2} $
$\Delta A=1.28 \times 10^{-6} \,cm ^{2}$