A tension of $22 \mathrm{~N}$ is applied to a copper wire of cross-sectional area $0.02 \mathrm{~cm}^2$…

A tension of $22 \mathrm{~N}$ is applied to a copper wire of cross-sectional area $0.02 \mathrm{~cm}^2$ Young's modulus of copper is $1.1 \times 10^{11} \mathrm{~N} / \mathrm{m}^2$ and Poisson's ratio 0.32 . The decrease in cross-sectional area will be
  1. $1.28 \times 10^{-6} \mathrm{~cm}^2$
  2. $1.6 \times 10^{-6} \mathrm{~cm}^2$
  3. $2.56 \times 10^{-6} \mathrm{~cm}^2$
  4. $0.64 \times 10^{-6} \mathrm{~cm}^2$

Solution

Young's modulus of materials $Y=\frac{F \times l}{A \Delta l}, \frac{\Delta l}{l}=\frac{F}{A Y}$ $=\frac{22}{0.02 \times 10^{-4} \times 1.1 \times 10^{11}}=10^{-4}$ Poisson's ratio, $\sigma=\frac{\frac{\Delta l}{l}}{\frac{\Delta r}{r}}$ $\begin{aligned} \frac{\Delta l}{l} & =\sigma \frac{\Delta r}{r}=0.32 \times \frac{\Delta r}{r} \\ & =0.32 \times 10^{-4}=32 \times 10^{-6}\end{aligned}$ The decrease in cross-sectional area, $\begin{aligned} & \frac{\Delta A}{A}=\frac{\Delta A}{0.02}=0.32 \times \frac{\Delta l}{l} \\ & \Delta A=0.64 \times 10^{-6} \mathrm{~cm}^2\end{aligned}$

Asked in: AP EAMCET 2012

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