A potentiometer wire of length 1 m has a resistance of 10 Ω. It is connected to a 6 V battery in series with a resistance of 5 Ω. Determine the emf of the primary cell which gives a balance point at 40 cm. 

Given,

Total length of the potentiometer wire, L = 1m

Resistance of the wire, R = 10 Ω

Voltage of the battery = 6 V

Resistance of the battery = 5 Ω

Therefore, total resistance of the circuit, R = (R_AB + 5) Ω = 15 Ω

Using the figure given above, we have

Current in the circuit, I = V R = 6 x 15 A

Therefore, 

Voltage across AB, V_AB = i.R_AB = 4 V

Emf of the cell, e =                                 ... (1) 

Here,

Balance point is obtained at, l = 40 cm

Total length, AB = L = 1 m = 100 cm

Putting the values in equation (1), we have