An aluminum rod [Young's modulus $=7 \times 10^{9}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\( has a breaking strain of \)0.2 \%$ what is the minimum cross-sectional area of the rod in order to support a load of \(10^{4}\) Newtons? (A) \(1 \times 10^{-2} \mathrm{~m}^{2}\) (B) \(1.4 \times 10^{-3} \mathrm{~m}^{2}\) (C) \(3.5 \times 10^{-3} \mathrm{~m}^{2}\) (D) \(7.1 \times 10^{-4} \mathrm{~m}^{2}\)

The relationship between stress (σ), strain (ε), and Young's modulus (Y) is given by the formula: \(σ = Yε\) where σ is the stress in N/m² or Pa, ε is the strain (which is unitless), and Y is the Young's modulus in N/m² or Pa.

Given the breaking strain as \(0.2\%\), first, convert it to the decimal form: \(ε = 0.2 \% = 0.002\) Now using the Young's modulus, which is given as \(7 \times 10^{9} \mathrm{N/m}^{2}\), we can calculate the stress using the formula: \(σ = Yε\) \(σ = (7 \times 10^{9} \mathrm{N/m}^{2})(0.002)\) \(σ = 1.4 \times 10^{7} \mathrm{N/m}^{2}\)

We need to find the minimum cross-sectional area to support a load of \(10^{4}\) Newtons. We can use the formula for stress: \(σ = \cfrac{F}{A}\) Where F is the force in Newtons and A is the cross-sectional area in m². Rearranging the formula to find A: \(A = \cfrac{F}{σ}\) Now, substitute the values of F and σ: \(A = \cfrac{10^{4} \mathrm{N}}{1.4 \times 10^{7} \mathrm{N/m}^{2}}\) \(A = 7.14 \times 10^{-4} \mathrm{m}^{2}\) Comparing this value to the given options, we can round it up to the closest option: (A) \(1 \times 10^{-2} \mathrm{m}^{2}\) (B) \(1.4 \times 10^{-3} \mathrm{m}^{2}\) (C) \(3.5 \times 10^{-3} \mathrm{m}^{2}\) **(D) \(7.1 \times 10^{-4} \mathrm{m}^{2}\)** So, the minimum cross-sectional area of the rod in order to support a load of \(10^{4}\) Newtons is **(D) \(7.1 \times 10^{-4} \mathrm{m}^{2}\)**.