Balance point of the potentiometer, $l_{1}=35 \,cm$

The cell is replaced by another cell of $emf$ $E_{2}$

New balance point of the potentiometer, $l_{2}=63 \,cm$

The balance condition is given by the relation,

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

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

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

Therefore, $emf$ of the second cell is $2.25\, V$