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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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$1 \times 10^{- 2} m^{2}$

$1.4 \times 10^{- 3} m^{2}$

$3.5 \times 10^{- 3} m^{2}$

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

answer is D.

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Detailed Solution

$Y = \frac{F / A}{S t r a i n} \Rightarrow A = \frac{F}{Y \times S t r a i n} = \frac{10^{4}}{7 \times 10^{9} \times 0.002} = \frac{1}{14} \times 10^{- 2} = 7.1 \times 10^{- 4} m^{2}$

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