Potentiometer wire `PQ` of `1 m` length is connected to a standard cell `E_(1)`. Another cell `E_(2)` of emf `1.02 V` is connected with a resistance `r` and a switch `S` as shown in the circuit diagram. With switch `S` open, the null position is obtained at a distance of `51 cm` from `P`.
(i) Calculate the potential gradient of the potentiometer wire.
(ii) Find the emf of cell `E_(1)`.
(iii) When switch `S` is closed, will the null point move toward `P` or toward `Q`? Give reason for your answer.

Explanation

(i) Potential gradient is
`k = (V)/(l) = (1.02)/(0.51) = 2 Vm^(-1)`
(ii) The emf of the cell is
`E_(1) = k_(l) = 2 xx 1= 2 V`
(iii) When the swich `S` is closed, there is no shift in the position of the null point because it depends on the potential gradient along the poteniometer wire (which depends on the emf of battery `E_(1)` and the resistance of the potentiometer wire circul and length of potentiometer) and the emf of cell `E_(2)`, which does not change when the swich `S` is closed.