An aluminum rod (Young's modulus $ = 7 \times {10^9}\,N/{m^2})$ has a breaking strain of $0.2\%$. The minimum cross-sectional area of the rod in order to support a load of ${10^4}$Newton's is
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(d) $Y = \frac{{F/A}}{{{\rm{strain}}}} \Rightarrow A = \frac{F}{{Y \times {\rm{strain}}}}$= $\frac{{{{10}^4}}}{{7 \times {{10}^9} \times 0.002}}$
= $\frac{1}{{14}} \times {10^{ - 2}}$$ = 7.1 \times {10^{ - 4}}{m^2}$
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