A potentiometer wire AB has a length 0.5 m and a resistance `0.5 Omega`. As shown in the diagram, it is connected with a cell of emf 3V and a combination of resistance `R=S=5 Omega.` . Find the potential drop per unit length of the wire.

Explanation

Total resistance in circuit ` = 0.5 ((5xx5)/(5+5)) = 3 Omega`
Current in potentiometer wire
Potential drop across potentiometer wire = IR
` = 1 xx 0.5 = 0.5 V`
Potential drop per unit lenght of wire .
` = (V)/(l ) = (0.5)/(0.5) = 1 V//m`