Question

A cylindrical wire of radius 1mm, length 1 m, Young's modulus $=2\times {10}^{11}N/m^2,$ poisson's ratio $\mu =\pi /10$ is stretched by a force of 100 N. Its radius will become

• A. $0.99998mm$
• B. $0.99999mm$
• C. $0.99997mm$
• D. $0.99995mm$

Answer 1

The correct option is D $0.99995mm$

$\begin{array}{c}\text{Given,}\\ \text{radius of wire}=1mm=1\times {10}^{-3}m\\ \text{Lenght of wire}=1m\\ \text{young's mo dulus}=2\times {10}^{11}N/m2\\ \text{Force}=100N\\ \text{On applying streching force length as well as}\\ \text{diameter will change.}\end{array}$

$\begin{array}{cc}Y& =\frac{\text{stress}}{\text{strain}}\\ Y& =\frac{\left(\frac{F}{A}\right)}{\left(\frac{\Delta L}{L}\right)}\\ \Delta L& =\frac{FL}{AY}\end{array}$

$\begin{array}{c}\text{Now Poisson's ratio}\left(\mu \right)=\frac{\left(\frac{\Delta r}{r}\right)}{\left(\frac{\Delta L}{L}\right)}\\ \text{Putting Know value,}\\ \begin{array}{cc}\Delta r& =\frac{\mu \Delta L\cdot r}{L}\\ \Delta r& =\frac{\mu \cdot \left(\frac{FL}{AY}\right)\cdot r}{L}\end{array}\end{array}$

$\begin{array}{c}\Delta r=\frac{\frac{\pi }{10}\times 100\times 1\times 1\times {10}^{-3}}{\pi {(1\times {10}^{-3})}^2\times 1\times 2\times {10}^{11}}\\ \Delta r=-0.5\times {10}^{-4}\\ r_{\text{final}}-r_{\text{initial}}=-0.5\times {1.0}^{-4}\end{array}$

$=0.99995mm$