Question

A copper wire of cross-sectional area 0.01 cm² is under a tension of 20N. Find the decrease in the cross-sectional area. Young modulus of copper = 1.1 × 10¹¹ N m⁻² and Poisson ratio = 0.32.

Answer 1

Given:
Cross-sectional area of copper wire A = 0.01 cm² = 10⁻⁶ m²
Applied tension T = 20 N
Young modulus of copper Y = 1.1 × 10¹¹ N/m²
Poisson ratio σ = 0.32

We know that:
$Y=\frac{FL}{A∆L}$

$$\begin{gathered} \Rightarrow \frac{∆L}{L}=\frac{F}{AY} \\ =\frac{20}{{10}^{-6}\times 1.1\times {10}^{11}}=18.18\times {10}^{-5} \end{gathered}$$

$$\begin{gathered} \mathrm{Poisson}\text{'}\mathrm{s}\mathrm{ratio},\sigma =\frac{{\displaystyle \frac{∆d}{d}}}{{\displaystyle \frac{∆L}{L}}}=0.32 \\ \mathrm{Where}d\mathrm{is}\mathrm{the}\mathrm{transverse}\mathrm{length} \\ \mathrm{So},\frac{∆d}{d}=\left(0.32\right)\times \frac{∆L}{L} \\ =0.32\times \left(18.18\right)\times {10}^{-5}=5.81\times {10}^{-5} \\ \mathrm{Again},\frac{∆A}{A}=\frac{2∆r}{r}=\frac{2∆d}{d} \\ \Rightarrow ∆A=\frac{2∆d}{d}A \\ \Rightarrow ∆A=2\times \left(5.8\times {10}^{-5}\right)\times \left(0.01\right) \\ =1.164\times {10}^{-6}{\mathrm{cm}}^2 \end{gathered}$$

Hence, the required decrease in the cross -sectional area is $1.164\times {10}^{-6}{\mathrm{cm}}^2$.