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Question

A tension of 22 N is applied to a copper wire of cross-sectional area 0.02 cm2. Young's modulus of copper is 1.1×1011N/m2 and Poisson's ratio 0.32. The decrease in cross sectional area will be:

A
1.28×106cm2
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B
1.6×106cm2
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C
2.56×106cm2
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D
0.64×106cm2
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Solution

The correct option is A 1.28×106cm2
Young's modulus, Y=FlAΔl and Poisson's ratio , σ=Δr/rΔl/l

From these we get, Δr/r=FσAY=22×0.320.02×104×1.1×1011=32×106

Cross sectional area , A=πr2 and ΔA=2πrΔr

Thus, ΔAA=2Δrr=64×106

ΔA=64×106×0.02=1.28×106cm2

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