Question

A tension of $22$ N is applied to a copper wire of cross-sectional area $0.02c{m}^2$. Young's modulus of copper is $1.1\times {10}^{11}N/m^2$ and Poisson's ratio $0.32$. The decrease in cross sectional area will be:

• A. $1.28\times {10}^{-6}c{m}^2$
• B. $1.6\times {10}^{-6}c{m}^2$
• C. $2.56\times {10}^{-6}c{m}^2$
• D. $0.64\times {10}^{-6}c{m}^2$

Answer 1

The correct option is A $1.28\times {10}^{-6}c{m}^2$

Young's modulus, $Y=\frac{Fl}{A\Delta l}$ and Poisson's ratio , $\sigma =\frac{\Delta r/r}{\Delta l/l}$

From these we get, $\Delta r/r=\frac{F\sigma }{AY}=\frac{22\times 0.32}{0.02\times {10}^{-4}\times 1.1\times {10}^{11}}=32\times {10}^{-6}$

Cross sectional area , $A=\pi r^2$ and $\Delta A=2\pi r\Delta r$

Thus, $\frac{\Delta A}{A}=\frac{2\Delta r}{r}=64\times {10}^{-6}$

$\therefore \Delta A=64\times {10}^{-6}\times 0.02=1.28\times {10}^{-6}c{m}^2$