Question

A copper wire of cross-sectional area $0.01{\text{cm}}^2$ is under a tension of $20\text{N}$. Find the decrease in the cross-sectional area. Young's modulus and Poisson's ratio of copper are $1.1\times {10}^{11}{\text{N/m}}^2$ and $0.32$ respectively.

• A. $3.16\times {10}^{-6}{\text{cm}}^2$
• B. $1.16\times {10}^{-6}{\text{cm}}^2$
• C. $4.16\times {10}^{-6}{\text{cm}}^2$
• D. $2.16\times {10}^{-6}{\text{cm}}^2$

Answer 1

The correct option is B $1.16\times {10}^{-6}{\text{cm}}^2$

Given:
$F=20\text{N};A=0.01{\text{cm}}^2;Y=1.1\times {10}^{11}{\text{N/m}}^2$
We know that Young's modulus,
$Y=\frac{\text{stress}}{\text{strain}}$
$\Rightarrow Y=\frac{F/A}{\Delta L/L}$
$\Rightarrow \frac{\Delta L}{L}=\frac{F}{AY}$
$\Rightarrow \frac{\Delta L}{L}=\frac{20}{0.01\times {10}^{-4}\times 1.1\times {10}^{11}}=1.8181\times {10}^{-4}$

Now,
Poisson's ratio $=-\frac{\text{Transvere strain}}{\text{Longitudnal strain}}=\frac{-\frac{\Delta d}{d}}{\frac{\Delta L}{L}}$
$\Rightarrow \frac{\Delta d}{d}=-0.32\times 1.8181\times {10}^{-4}=-5.8179\times {10}^{-5}$

Also we know, fractional change in area
$\frac{\Delta A}{A}=2\frac{\Delta d}{d}$
$\Rightarrow \frac{\Delta A}{A}=2\times -5.8179\times {10}^{-5}$
$\Rightarrow \frac{\Delta A}{A}=-1.1636\times {10}^{-4}$
$\Rightarrow \Delta A=-1.1636\times {10}^{-4}\times 0.01$
$\Rightarrow \Delta A=-1.1636\times {10}^{-6}{\text{cm}}^2\approx -1.16\times {10}^{-6}{\text{cm}}^2$
Negative sign shows the decrease in area of cross -section when tension is applied in the wire.
Hence, $\left(b\right)$ is the correct option.