In a potentiometer circuit a cell of EMF 1.5 V gives balance point at 36 cm length of wire. If another cell of EMF 2.5 V replaces the first cell, then at what length of the wire, the balance point occurs ?

The correct answer is 60 cm.

Underlying Principle:
In a potentiometer circuit, the potential gradient along the wire is constant when a steady current flows through it. This means the electromotive force (EMF) $E$ of a cell is directly proportional to the balancing length $l$ of the wire:

$E\propto l$

For two different cells with EMFs $E_1$ and $E_2$ obtaining balance points at lengths $l_1$ and $l_2$ respectively, the relationship is given by:

$\frac{E_1}{E_2}=\frac{l_1}{l_2}$

Given Data:
EMF of the first cell, $E_1=1.5\text{V}$
Balancing length for the first cell, $l_1=36\text{cm}$
EMF of the second cell, $E_2=2.5\text{V}$

Step-by-Step Calculation:
We need to find the new balancing length $l_2$. Rearranging the formula to solve for $l_2$:

$l_2=l_1\times \frac{E_2}{E_1}$

Substituting the given values into the equation:

$l_2=36\text{cm}\times \frac{2.5\text{V}}{1.5\text{V}}$

Simplify the ratio of the EMFs:

$l_2=36\text{cm}\times \frac{5}{3}$

Calculate the final balancing length:

$l_2=12\times 5\text{cm}$

$l_2=60\text{cm}$

Therefore, the balance point occurs at a wire length of 60 cm.