An aluminum rod has a breaking strain of 0.2%. The minimum cross sectional area of the rod, in m² , in order to support a load of 10⁴ N is (Y = 7 x 10⁹ N / m²)

The correct answer is 7.1 x 10⁻⁴.

To find the minimum cross-sectional area of the aluminum rod, we use the fundamental relation of Young's modulus, which connects stress and strain:
$Y=\frac{\text{Stress}}{\text{Strain}}=\frac{F/A}{\epsilon }$
where:
- $Y$ is the Young's modulus of aluminum ($7\times {10}^9{\text{N/m}}^2$),
- $F$ is the applied load or force (${10}^4\text{N}$),
- $A$ is the cross-sectional area of the rod in ${\text{m}}^2$, and
- $\epsilon$ is the breaking strain.

The breaking strain is given as 0.2%. We convert this percentage into a decimal value:
$\epsilon =0.2\%=\frac{0.2}{100}=2\times {10}^{-3}$

We rearrange the formula to solve for the minimum cross-sectional area $A$:
$A=\frac{F}{Y\times \epsilon }$

Now, substitute the given values into the equation:
$A=\frac{{10}^4}{(7\times {10}^9)\times (2\times {10}^{-3})}$

Simplify the denominator:
$7\times 2=14$
${10}^9\times {10}^{-3}={10}^6$
Therefore, the denominator is $14\times {10}^6$.

Now calculate $A$:
$A=\frac{{10}^4}{14\times {10}^6}=\frac{1}{14}\times {10}^{4-6}=0.0714\times {10}^{-2}{\text{m}}^2$

Expressing the final value in scientific notation matching the options:
$A\approx 7.1\times {10}^{-4}{\text{m}}^2$