In an experiment to find emf of a cell using potentiometer, the length of null point for a cell of emf 1.5 V is found to be 60 cm. If this cell is replaced by another cell of emf $$E$$, the length-of null point increases by 40 cm. The value of $$E$$ is $$\frac{x}{10}$$ V. The value of $$x$$ is _____.

In a potentiometer experiment, the null point length for a cell of EMF $$E_1 = 1.5$$ V is $$l_1 = 60$$ cm. When replaced with another cell of EMF $$E$$, the null point length increases by 40 cm.

So the new null point length is:

$$l_2 = 60 + 40 = 100 \text{ cm}$$

Now, in a potentiometer the EMF is directly proportional to the length of the null point:

$$\frac{E}{E_1} = \frac{l_2}{l_1}$$

$$E = E_1 \times \frac{l_2}{l_1} = 1.5 \times \frac{100}{60} = \frac{150}{60} = 2.5 \text{ V}$$

We are given that $$E = \frac{x}{10}$$ V, so:

$$2.5 = \frac{x}{10} \implies x = 25$$

Hence, the answer is $$25$$.