In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35.0 cm length of the wire.
Question:

In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35.0 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63.0 cm, what is the emf of the second cell?

Solution:

Emf of the cell, E1 = 1.25 V

Balance point of the potentiometer, l1= 35 cm

The cell is replaced by another cell of emf E2.

New balance point of the potentiometer, l2 = 63 cm

The balance condition is given by the relation,

$\frac{E_{1}}{E_{2}}=\frac{l_{1}}{l_{2}}$

$E_{2}=E_{1} \times \frac{l_{2}}{l_{1}}$

$=1.25 \times \frac{63}{35}=2.25 \mathrm{~V}$

Therefore, emf of the second cell is $2.25 \mathrm{~V}$.