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
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
- A
$1 \times {10^{ - 2}}\,{m^2}$
- B
$1.4 \times {10^{ - 3}}\,{m^2}$
- C
$3.5 \times {10^{ - 3}}\,{m^2}$
- D
$7.1 \times {10^{ - 4}}\,{m^2}$
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