$\Big[$Hint: $\frac{\triangle\text{A}}{\text{A}}=2\frac{\triangle\text{r}}{\text{r}}\Big]$
Given:
Cross-sectional area of copper wire A = 0.01cm2 = 10-6m2
Applied tension T = 20N
Young modulus of copper Y = 1.1 × 1011N/m2
Poisson ratio $\sigma=0.32$
We know that:
$\text{Y}=\frac{\text{FL}}{\text{A}\triangle\text{L}}$
$\Rightarrow\frac{\triangle\text{L}}{\text{L}}=\frac{\text{F}}{\text{AY}}$
$=\frac{20}{10^{-6}\times1.1\times10^{11}}=18.18\times10^{-5}$
Poisson's ratio, $\sigma =\frac{\frac{\triangle\text{d}}{\text{d}}}{\frac{\triangle\text{L}}{\text{L}}}=0.32$
Where d is the transverve length
So, $\frac{\triangle\text{d}}{\text{d}}=(0.32)\times\frac{\triangle\text{L}}{\text{L}}$
$=0.32\times (18.18)\times10^{-5}=5.81\times10^{-5}$
Again, $\frac{\triangle\text{A}}{\text{A}}=\frac{2\triangle \text{r}}{\text{r}}=\frac{2\triangle \text{d}}{\text{d}}$
$\Rightarrow\triangle\text{A}=\frac{2\triangle\text{d}}{\text{d}}\text{A}$
$\Rightarrow\triangle\text{A}=2\times(5.8\times10^{-5})\times(0.01)$
$=1.165\times10^{-6}\text{cm}^2$
Hence, the required decrease in the cross-sectional area is 1.164 × 10-6cm2.
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