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

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 $2.25 \mathrm{~V}$.