b
\(Emf\) of the cell, \(E_{1}=1.25\, V\)

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 \(2.25\, V\)