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Q.

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
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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)