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An aluminum 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 $7 \times 10^{9} {Nm}^{- 2}$

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$1.7 \times 10^{- 4}$

$1.7 \times 10^{- 3}$

$7.1 \times 10^{- 4}$

$1.4 \times 10^{- 4}$

answer is C.

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Detailed Solution

$7 \times 10^{9} = \frac{\frac{10^{4}}{A}}{0.002}$

$A = \frac{10^{4}}{0.002 \times 7 \times 10^{9}} = \frac{10^{- 2}}{14} = 0.71 \times 10^{- 3}$

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