A copper wire of cross-sectional area 0.01 ~cm ^2 is under a tension of 22 ~N . The decrease in the cross-sectional area is (Young modulus =1.1 10¹¹ Nm ⁻² , Poisson's ratio =0.32 )

Step-by-step solution

Young's modulus, aligned Y & = F / A l / l l l & = F Y A aligned where, l l = longitudinal strain Given, F=22 ~N , Y=1.1 10¹¹ ~N - m ^2 , aligned A & =0.01 ~cm ^2=10⁻⁶ ~m ^2 l l & = 22 1.1 10¹¹ 10⁻⁶ =2 10⁻⁴ aligned Now, Poisson ratio aligned & = Lateral strain Longitudinal strain = d / d l / l d d & = l l =0.32 2 10⁻⁴ aligned Change in diameter, d d =6.4 10⁻⁵ or change (decrease) in radius, r r =6.4 10⁻⁵ Area, A= r^2 Fractional change in area, A A =2 r r aligned & A A =2 6.4 10⁻⁵ & A= (12.8 10⁻⁵ ) A aligned Decreas