A tension of 22 ~N is applied to a copper wire of cross-sectional area 0.02 ~cm ^2 Young's modulus of copper is 1.1 10¹¹ ~N / m ^2 and Poisson's ratio 0.32 . The decrease in cross-sectional area will be

Step-by-step solution

Young's modulus of materials Y= F l A l , l l = F A Y = 22 0.02 10⁻⁴ 1.1 10¹¹ =10⁻⁴ Poisson's ratio, = l l r r aligned l l & = r r =0.32 r r & =0.32 10⁻⁴=32 10⁻⁶ aligned The decrease in cross-sectional area, aligned & A A = A 0.02 =0.32 l l & A=0.64 10⁻⁶ ~cm ^2 aligned