A potentiometer with a cell of 2.4 volt and internal resistance of 2 Ω maintains a potential drop across the resistance wire AB of length 2 meters and resistance 10 Ω. A standard cell which maintains a constant emf of 'V' volt with internal resistance 0.2 Ω gives a balance point at 1.6 m length of the wire. The value of emf of second (standard) cell (V) is:

Calculation:

In the given potentiometer experiment, the balancing length l is 1.6 m, and the total length of the wire AB is 2 m. The resistance per unit length of the wire AB is R = 10/2 = 5 Ω/m.

Let V be the emf of the standard cell to be determined. At the balancing point, the potential difference across the wire AB is equal to the emf of the standard cell, i.e., V = IR, where I is the current flowing through the wire AB.

The total resistance in the circuit is the sum of the resistance of the standard cell (0.2 Ω), the internal resistance of the cell in the potentiometer (2 Ω), and the resistance of the wire AB up to the balancing point (5 Ω/m x 1.6 m = 8 Ω).

The current I in the circuit can be calculated using the potential difference across the circuit, which is the sum of the emf of the cell in the potentiometer and the potential drop across the wire AB up to the balancing point. Thus, I = (2.4 V)/(2 Ω + 5 Ω/m x 1.6 m) = 0.24 A.

Substituting the value of I and R into the equation V = IR gives V = 0.24 A x (5 x 1.6 + 0.2)  = 1.968 V.

The correct answer is option (2) i.e., nearly equal to 1.9 V.