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 option is 60 cm.

Principle of a Potentiometer:
For a potentiometer wire carrying a constant current, the potential difference (or electromotive force, EMF) across any length of the wire is directly proportional to that balancing length.

Mathematically, we can write:
$E\propto l$
where $E$ is the EMF of the cell and $l$ is the balancing length of the wire.

For two different cells with EMF values $E_1$ and $E_2$ balanced at lengths $l_1$ and $l_2$ respectively, we have the relation:
$\frac{E_1}{E_2}=\frac{l_1}{l_2}$

Given Parameters:
Initial EMF, $E_1=1.5\text{V}$
Initial balancing length, $l_1=36\text{cm}$
New EMF, $E_2=2.5\text{V}$

We need to find the new balancing length, $l_2$.

Step-by-Step Calculation:
Substituting the given values into our formula:
$\frac{1.5}{2.5}=\frac{36}{l_2}$

Simplify the ratio on the left-hand side:
$\frac{3}{5}=\frac{36}{l_2}$

Solving for $l_2$:
$l_2=36\times \frac{5}{3}$
$l_2=12\times 5$
$l_2=60\text{cm}$

Thus, the balance point occurs at a length of 60 cm when the 2.5 V cell is connected.