A cylindrical copper rod is reformed to twice its original length with no change in volume. The resistance between its ends before the change was R Now its resistance will be

R=ρl1πr12……….(1)Now the rod is reformed such that I2=2I1 and.πr12l1=πr22l2 (No change in volume)or r12/r22=l2/h……..(2)Now R2=ρl2πr22…………(3)From eqs. [1) and (3), we getRR2=l1l2×r22r22=l1l2×l1l2=12×12  [using eq. (2)]∴ R2=4R