A potentiometer wire of length $$300 \text{ cm}$$ is connected in series with a resistance $$780 \Omega$$ and a standard cell of emf $$4 \text{ V}$$. A constant current flows through potentiometer wire. The length of the null point for cell of emf $$20 \text{ mV}$$ is found to be $$60 \text{ cm}$$. The resistance of the potentiometer wire is ______ $$\Omega$$.

We are given the length of the potentiometer wire $$L = 300 \text{ cm}$$, an external resistance $$R = 780 \, \Omega$$, the EMF of a standard cell $$E = 4 \text{ V}$$, and the EMF of a test cell $$e = 20 \text{ mV} = 0.02 \text{ V}$$, with the null point occurring at a length $$l = 60 \text{ cm}$$. Let the resistance of the potentiometer wire be $$R_w \, \Omega$$.

Since the potentiometer wire and external resistor are in series with the standard cell, the current through the wire is $$I = \frac{E}{R + R_w} = \frac{4}{780 + R_w}$$.

Subsequently, the potential drop across the entire wire is $$V = I R_w$$, so the potential gradient along the wire is $$\frac{I R_w}{L} = \frac{I R_w}{300}$$.

At the null point, the EMF of the test cell equals the potential drop across the length $$l$$ of the wire. Therefore, $$e = \frac{I R_w}{300} \times l$$. Substituting $$l = 60$$ yields $$0.02 = \frac{I R_w \times 60}{300} = \frac{I R_w}{5}$$, which gives $$I R_w = 0.1 \text{ V}$$.

Substituting $$I = \frac{4}{780 + R_w}$$ into $$I R_w = 0.1$$ leads to $$\frac{4 R_w}{780 + R_w} = 0.1$$. This gives $$4R_w = 0.1(780 + R_w) = 78 + 0.1R_w$$, so $$3.9R_w = 78$$ and hence $$R_w = \frac{78}{3.9} = 20 \, \Omega$$.

Therefore, the resistance of the potentiometer wire is $$\textbf{20} \, \Omega$$.