A potentiometer wire of length $$10$$ m and resistance $$20$$ $$\Omega$$ is connected in series with a $$25$$ V battery and an external resistance $$30$$ $$\Omega$$. A cell of emf $$E$$ in secondary circuit is balanced by $$250$$ cm long potentiometer wire. The value of $$E$$ (in volt) is $$\frac{x}{10}$$. The value of $$x$$ is ______.

A potentiometer wire of length $$10$$ m and resistance $$20$$ $$\Omega$$ is connected in series with a $$25$$ V battery and an external resistance of $$30$$ $$\Omega$$. A cell of emf $$E$$ is balanced at $$250$$ cm. We need to find $$x$$ where $$E = \frac{x}{10}$$.

Since the potentiometer wire and the external resistance are in series, the total resistance in the primary circuit is $$R_{total} = R_{wire} + R_{ext} = 20 + 30 = 50 \text{ }\Omega$$ and the current through the circuit is $$I = \frac{V}{R_{total}} = \frac{25}{50} = 0.5 \text{ A}$$.

Substituting this current into the expression for the potential drop across the wire gives $$V_{wire} = I \times R_{wire} = 0.5 \times 20 = 10 \text{ V}$$.

This voltage drop along the wire corresponds to a potential gradient of $$k = \frac{V_{wire}}{L} = \frac{10}{10} = 1 \text{ V/m}$$.

When the balancing length is $$l = 250$$ cm $$= 2.5$$ m, the emf of the cell is $$E = k \times l = 1 \times 2.5 = 2.5 \text{ V}$$. Since $$E = \frac{x}{10}$$, substituting gives $$2.5 = \frac{x}{10}$$ and hence $$x = 25$$.

The answer is $$25$$.