An aluminum rod has a breaking strain 0.4%. The minimum cross-sectional area of the rod in m2 in order to support a load of 105N is if Young’s modulus is 7×109Nm−2 (Nearly)

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1.5×10−4

2.5×10−3

3.6×10−3

2.5×10−4

answer is C.

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Detailed Solution

y=FLAe Here eL×100=0.4 A=FyeL=1057×109×4×10−3 =3.6×10−3

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