In a potentiometer arrangement, a cell of emf $1.20 \mathrm{V}$ gives a balance point at $36 \mathrm{cm}$ length of wire. This cell is now replaced by another cell of emf $1.80 \mathrm{V}$. The difference in balancing length of potentiometer wire in above conditions will be _____ $\mathrm{cm}$.

In a potentiometer, null point or balance point is defined as the point where the deflection of the galvanometer is zero. If emf of the unknown battery is $\epsilon$ then,

$\epsilon =$(Potential Gradient)$\times$(Balancing length $l$). Therefore,

$\frac{{\epsilon }_1}{{\epsilon }_2}=\frac{l_1}{l_2}$

$\Rightarrow \frac{1.20}{1.80}=\frac{36}{l_2}$

$\Rightarrow l_2=18\times 3=54 \mathrm{cm}$

Hence, difference in balancing length is $l_2-l_1=54-36=18 \mathrm{cm}$