A copper wire of cross-sectional area 0.01 cm2 is under a tension of 20 N. Find the decrease in the cross-sectional area. Young modulus of copper = 1.1 × 1011 N m-2 and Poisson ratio = 0.32. [Hint: ∆A/A=2 ∆r/r]

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Deepak Patra · Answer

Sol. Y = FL/A∆L ⇒ ∆L/L = F/Ay σ = ∆D/D/∆L/L ⇒ ∆D again, ∆A/A = 2∆r/r ⇒ ∆A = 2∆r/r

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A copper wire of cross-sectional area 0.01 cm2 is under a tension of 20 N. Find the decrease in the cross-sectional area. Young modulus of copper = 1.1 × 1011 N m-2 and Poisson ratio = 0.32. [Hint: ∆A/A=2 ∆r/r]

A copper wire of cross-sectional area 0.01 cm2 is under a tension of 20 N. Find the decrease in the cross-sectional area. Young modulus of copper = 1.1 × 1011 N m-2 and Poisson ratio = 0.32. [Hint: ∆A/A=2 ∆r/r]

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A copper wire of cross-sectional area 0.01 cm2 is under a tension of 20 N. Find the decrease in the cross-sectional area. Young modulus of copper = 1.1 × 1011 N m-2 and Poisson ratio = 0.32. [Hint: ∆A/A=2 ∆r/r]

A copper wire of cross-sectional area 0.01 cm2 is under a tension of 20 N. Find the decrease in the cross-sectional area. Young modulus of copper = 1.1 × 1011 N m-2 and Poisson ratio = 0.32. [Hint: ∆A/A=2 ∆r/r]

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