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.

In a potentiometer circuit, the principle of operation states that the electromotive force (EMF) of a cell is directly proportional to the balancing length of the potentiometer wire, provided the potential gradient along the wire remains constant.

Mathematically, this relationship is given by the formula:

$\frac{E_1}{E_2}=\frac{L_1}{L_2}$

From the problem statement, we are given the following values:

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, $L_2$.

Rearranging the formula to solve for $L_2$, we get:

$L_2=\frac{E_2}{E_1}\times L_1$

Substitute the given values into the equation:

$L_2=\frac{2.5}{1.5}\times 36$

Simplify the fraction by multiplying the numerator and denominator by 10, then reducing:

$L_2=\frac{25}{15}\times 36$

$L_2=\frac{5}{3}\times 36$

Calculate the final value:

$L_2=5\times 12$

$L_2=60\text{cm}$

Therefore, when the 1.5 V cell is replaced by a 2.5 V cell, the balance point occurs at a wire length of 60 cm.