If a rod has resistance 4 Ω and if rod is turned as half cycle then the resistance along diameter

Suppose the length of the rod = /. Given that its resistance is 4 Ω∴ Resistance per unit length =(4/l)ΩThe half circle of turned rod is shown in fig.From figure πr+r+r=lor r=l(2+π)Its diameter =2r=2l(2+π)Now, resistance along diameter =2l(2+π)×4l=8(2+3⋅14)=1⋅56ohm