In a potentiometer arrangement for comparing emf of two cells, it was found that a cell of emf 1.25 V gives a balance point at 0.35 m length of the wire. On replacing the cell by another cell, the balance point shifted to 0.63 m. The emf of the second cell is:

2.25 V

The correct answer is Option (4) → 2.25 V

Given,

Emf of the first cell, $ε_1=1.25 V$

Balance length for the first cell, $l_1=0.35 m$

Balance length for the second cell, $l_2=0.63 m$

Let Emf of the second cell be $ε_2$

from the principle of potentiometer,

$\frac{ε_1}{ε_2}=\frac{l_1}{l_2}$

On substituting the value,

$\frac{1.25}{ε_2}=\frac{0.35}{0.63}$

$ε_2=\frac{1.25×0.63}{0.35}$

$ε_2=1.25×\frac{9}{5}=2.25V$