Determine the decrease in the cross-sectional area of a copper wire having an initial cross-sectional area of 0.01 cm ^2 that is subjected to a tension of 20 N . The Young modulus of copper is 1.1 10¹¹ N/m ^2 and its Poisson ratio is 0.32 . [Hint: A A = 2 r r ]

Step-by-step solution

The longitudinal strain in the wire is given by L L = F AY The lateral strain is given by r r = L L = F AY The fractional change in the cross-sectional area is A A = 2 r r = 2 F AY The decrease in the cross-sectional area is A = A ( 2 F AY ) = 2 F Y Substituting the given values F = 20 N , = 0.32 , and Y = 1.1 10¹¹ N/m ^2 : A = 2 0.32 20 1.1 10¹¹ m ^2 A = 12.8 1.1 10⁻¹¹ m ^2 = 11.636 10⁻¹¹ m ^2 Converting to cm ^2 : A = 11.636 10⁻¹¹ 10^4 cm ^2 1.164 10⁻⁶ cm ^2