In a potentiometer arrangement, a cell gives a balancing point at $$75$$ cm length of wire. This cell is now replaced by another cell of unknown emf. If the ratio of the emf's of two cells respectively is $$3 : 2$$, the difference in the balancing length of the potentiometer wire in above two cases will be ______ cm.

In a potentiometer, the balancing length is proportional to the emf of the cell: $$E \propto l$$. Let the emf of the first cell be $$E_1$$ and of the second be $$E_2$$ with $$\frac{E_1}{E_2} = \frac{3}{2}$$. The first cell gives a balancing length of $$l_1 = 75$$ cm, and since emf is proportional to balancing length, we have $$\frac{E_1}{E_2} = \frac{l_1}{l_2}$$, which leads to $$\frac{3}{2} = \frac{75}{l_2}$$. Solving for $$l_2$$ yields $$l_2 = \frac{75 \times 2}{3} = 50 \text{ cm}$$. The difference in balancing lengths is $$\Delta l = l_1 - l_2 = 75 - 50 = 25 \text{ cm}$$. The answer is 25 cm.