Question

A copper wire of cross- sectional area 0.01 $c{m}^2$ is under a tension of 20 N. Find the decrease in the cross-sectional area. Young modulus of copper = $1.11\times N{m}^{-2}$ and Poisson ratio = 0.32 \)

Answer 1

A = $0.01c{m}^2$

= ${10}^{-6}{m}^2$

T = 20 N,

Y = $1.1\times {10}^{11}N{m}^2$

We know that, Y =$\frac{FL}{A\Delta L}$

$\frac{\Delta }{L}=\frac{F}{AY}$

= $\frac{20}{{10}^{-6}\times 1.1\times {10}^{11}}\times 18.18\times {10}^{-5}$

$d=\frac{\frac{\Delta D}{D}}{\frac{\Delta L}{L}}=0.32$

So, $\frac{\Delta D}{D}=\left(0.32\right)\times \frac{KL}{L}$

= $32\times \left(18.18\right)\times {10}^{-5}=5.81\times {10}^{-5}$

Again, $\frac{\Delta A}{A}=\frac{2\Delta r}{r}$

$\Rightarrow \Delta A=\frac{2\Delta r}{r}$

$DeltaA=2\times (5.8\times {10}^{-5})\times \left(0.01\right)$

= $1.16\times {10}^{-5}c{m}^2$