An accumulator with a steady emf of 2 V is connected across a potentiometer wire at 6.732 m If a resistance of 2.5 `Omega` is put in series with the wire find the new position of the null point.
Correct Answer - Put `R_(1)` =1018 `Omega`(1000 times emf of Cd-cell),adust `R_(2)` till no deflection ocurs in the galvanometer , `R_(2)`=972 `Omega`;5.1 m