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 a load of ${10}^{4}N$ is if (Young's modulus is $7\times {10}^{9}N{m}^{-2}$)

• 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 $7.1\times {10}^{-4}$

We know

Young's modulus $\left(\gamma \right)=\frac{stress}{strain}$

For breaking (maximum) strain $0.2$% and load of ${10}^{4}N$, let the minimum supportive cross-sectional area be $A$ then,

$\gamma =\frac{\frac{L}{A}}{breakingstrain}$

$\Rightarrow$ $A=\frac{L}{r\times B.S}=\frac{{10}^4}{7\times {10}^9\times \left(\frac{2}{1000}\right)}$

$A=7.1\times {10}^{-4}{m}^2$