In a potentiometer arrangement a cell of 1.5 V gives a balance point at 45.0 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 75.0 cm, what is the emf of the second cell?

2.5 V

The correct answer is Option (1) → 2.5 V

The relationship between the emf of the cell and the balance point is,

$\frac{E_1}{L_1}=\frac{E_2}{L_2}$

$⇒\frac{1.5}{45.0}=\frac{E_2}{75.0}$

$⇒E_2=\frac{1.5×75}{45}=2.5V$