A 6 V battery is connected to the terminals of a three metre long wire of uniform thickness and resistance of 100  Ω. The difference of potential between two points on the wire separated by a distance or 50 cm, will be

Total current drawn from the batteryi= ER+r =6100+0=0.06 AResistance of 5o cm wire is R'=ρl'A =  ρAl'=Rl l'    ∵   R=ρlA =100300 × 50so,   R'=5030 ΩHence, the potential difference between two points on the wire separated by a distance l isV= iR'=0.06  × 503   =  1 V