Question

A brass rod is to support a load of $400\text{N}$. If its elastic limit is $4.0\times {10}^8{\text{N/m}}^2$, its minimum diameter must be :

• A. $\frac{{10}^{-3}}{\sqrt{\pi }}\text{mm}$
• B. $\frac{2\times {10}^{-3}}{\sqrt{\pi }}\text{mm}$
• C. $\frac{4\times {10}^{-3}}{\sqrt{\pi }}\text{mm}$
• D. $\frac{2}{\sqrt{\pi }}\text{mm}$

Answer 1

The correct option is D $\frac{2}{\sqrt{\pi }}\text{mm}$

Limiting stress $=4.0\times {10}^8{\text{N/m}}^2$
Limiting stress =$\frac{F}{A_{min}}=\frac{400}{A_{min}}=4.0\times {10}^8$

$\Rightarrow A_{min}={10}^{-6}{\text{m}}^2=\frac{\pi }{4}{D}_{min}^2$
$\therefore D_{min}={\left(\frac{4A_{min}}{\pi }\right)}^{1/2}={\left(\frac{4\times {10}^{-6}}{\pi }\right)}^{1/2}$
$=\frac{2\times {10}^{-3}}{\sqrt{\pi }}\text{m}==\frac{2}{\sqrt{\pi }}\text{mm}$