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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a
1.5×10−4
b
2.5×10−3
c
3.6×10−3
d
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
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)