Question

An aluminium rod has a breaking strain $0.2\%$. The minimum cross-sectional area of the rod in $m^2$ in order to support of load of ${10}^{4}N$ is

• A. $1.7\times {10}^{-4}$
• B. $1.7\times {10}^{-3}$
• C. $7.1\times {10}^{-4}$
• D. $1.4\times {10}^{-4}$

Answer 1

The correct option is C $7.1\times {10}^{-4}$

$\begin{array}{c}\text{Given breaking strain}=0.2\%\\ \text{load to be supported}={10}^{4}N\\ \text{Young's modulus of alluminium}=7\times {10}^9{\text{Nm}}^{-2}\end{array}$

$\begin{array}{c}\text{Young's modulus}=\frac{\text{Stress}}{\text{Stratn}}\\ \text{we know that stress=}\frac{\text{Force}}{\text{Area}}\\ \therefore 7\times {10}^9=\frac{\frac{\text{Force}}{\text{Area}}}{\text{strain.}}\end{array}$

$\begin{array}{cc}\text{Area}& =\frac{\text{Force}}{(7\times {10}^9)\text{strain}}\\ & =\frac{{10}^4}{7\times {10}^9\times \left(0.2\right)\times {10}^{-2}}\\ \text{Area}& =7.1\times {10}^{-4}{m}^2\end{array}$