Question

To apply a maximum stress of $1.2\times {10}^{8}N/m^2$ under the action of force $40N$ the minimum radius of wire required is

• A. $0.32$mm
• B. $0.66$mm
• C. $0.44$mm
• D. $0.22$mm

Answer 1

Answer: (A)

$0.32$mm

$\begin{array}{c}\text{Max stress}=1.2\times {10}^{8}N/m^2\\ \text{Force}=40N\\ \Rightarrow A=\frac{F}{\text{Stress}}=\frac{40}{1.2\times {10}^8}=\frac{400}{12\times {10}^8}\end{array}$

$\begin{array}{cc}& \Rightarrow \pi r^2=\frac{400}{12\times {10}^8}\\ & \Rightarrow r=\sqrt{\frac{400}{12\times {10}^8\times \pi }}\\ & \Rightarrow r=0.32mm\\ & \text{Ans :(A)}\end{array}$