Question

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

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

Answer 1

The correct option is D $0.99995mm$

$\begin{array}{c}r=1mm,Y=2\times {10}^{11}Pa,L=1m,\mu =\frac{\pi }{10},F=100N\\ \text{strain}=\frac{\text{stress}}{Y}=\frac{100/\pi \times {10}^{-6}}{2\times {10}^{11}}=\frac{100}{2\pi \times {10}^5}=\frac{1}{2000\pi }\end{array}$

$\begin{array}{cc}\mu & =\frac{\Delta r/r}{strain}\\ \Rightarrow \frac{\pi }{10}& =\frac{\Delta r\times 2000\pi }{r\times 1}\\ \Rightarrow & \Delta r=\frac{r}{2\times {10}^4}\\ \Rightarrow & \Delta r=0.00005\end{array}$

$\begin{array}{c}\text{new radius}=r-\Delta r=0.99995mm\\ \text{Answer}:D\end{array}$