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

In a potentiometer, the balancing length is proportional to the EMF of the cell:

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

Given: $$E_1 = 1.20 \text{ V}$$, $$l_1 = 36 \text{ cm}$$, and $$E_2 = 1.80 \text{ V}$$.

Finding $$l_2$$:

$$l_2 = l_1 \times \dfrac{E_2}{E_1} = 36 \times \dfrac{1.80}{1.20} = 36 \times 1.5 = 54 \text{ cm}$$

The difference in balancing lengths:

$$\Delta l = l_2 - l_1 = 54 - 36 = 18 \text{ cm}$$

Therefore, the difference is $$\boxed{18}$$ cm.