An aluminium rod has a breaking strain 0.2% . The minimum cross-sectional area of the rod in m2 in order to support a load of 104N is, if Young's modulus is 7×109Nm−2 ,

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1.7×10−4

1.7×10−3

7.1×10−4

1.4×10−4

answer is C.

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Detailed Solution

Δℓℓ=0.2100 A=Fy(ΔLL)=1047×1090.2100=7.1×10−4m2

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