Question

An aluminium rod ($Y=7.0\times {10}^{9}N/m^2$) has a breaking strain $0.2$%. The minimum cross-sectional area of rod in order to support a load of ${10}^{4}N$ is

• A. $1.0\times {10}^{-2}{m}^2$
• B. $1.4\times {10}^{-2}{m}^2$
• C. $1.0\times {10}^{-3}{m}^2$
• D. $7.1\times {10}^{-4}{m}^2$

Answer 1

The correct option is D $7.1\times {10}^{-4}{m}^2$

$\begin{array}{c}\text{Given, young modulus of aluminium}=7\times {10}^9\frac{N}{m^2}\\ \text{Breaking strain}=2\times {10}^{-3}\\ \text{Load applied}={10}^{4}N\\ \text{Stress}=y\text{(Strain)}\\ \frac{F}{A}=Y\left(\frac{\Delta L}{L}\right)\end{array}$

$\begin{array}{cc}\Rightarrow \frac{{10}^4}{A}& =7\times {10}^9(2\times {10}^{-3})\\ A& =\frac{1}{7}\times {10}^{-2}{m}^2\\ & =\frac{7.1\times {10}^{-4}{m}^2}{A}\\ \text{Option}D& \text{is the correct answer.}\end{array}$