In a potentiometric arrangement ,a cell is connected to the potentiometer wire of 60 cm in length to make the deflection zero in galvanometer . Now if the cell is shunted with a `6Omega` resistor , a null point is found in 50 cm length of the wire . What is the internal resistance of the cell ?

Explanation

Here , shunt `R=6Omega,l_(1)=60"cm" ,l_(2)=50"cm".`
`therefore` Internal resistance of the cell ,
`r=((l_(1))/(l_(2))-1)R=((60)/(50)-1)xx6=0.2xx6=1.2Omega`