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 potential gradient $$k$$ (in V/cm) is constant along the wire.

For the first cell of emf $$1.5\text{ V}$$ and null point length $$l_1 = 60\text{ cm}$$, we use the definition of potential gradient:
$$k = \frac{\text{emf}}{\text{length}}$$ Hence, $$k = \frac{1.5}{60}$$ $$-(1)$$

Evaluating equation $$(1)$$:
$$k = \frac{1.5}{60} = \frac{15}{600} = \frac{1}{40}\;\text{V/cm}$$

The cell is now replaced by another cell of emf $$E$$, and the new null point length is $$l_2 = 60 + 40 = 100\text{ cm}$$. Using the same gradient:
$$k = \frac{E}{l_2}$$ $$-(2)$$

Equate equations $$(1)$$ and $$(2)$$:
$$\frac{1}{40} = \frac{E}{100}$$

Solving for $$E$$:
$$E = \frac{100}{40} = \frac{10}{4} = 2.5\text{ V}$$

Given that $$E = \frac{x}{10}\text{ V}$$, we have:
$$\frac{x}{10} = 2.5$$
Multiplying both sides by $$10$$:
$$x = 25$$

Therefore, the required value of $$x$$ is $$25$$.