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:
Answer & explanation
Correct answer: option 4
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,
$\frac{1.25}{ε_2}=\frac{0.35}{0.63}$
$ε_2=\frac{1.25×0.63}{0.35}$
$ε_2=1.25×\frac{9}{5}=2.25V$