The length of a potentiometer wire is 1200 cm and it carries a current of 60 mA. For a cell of emf 5 V and internal resistance of 20 $$\Omega$$ the null point on it is found to be at 1000 cm. The resistance of whole wire is:

We have a potentiometer wire of total length $$L = 1200\ \text{cm}$$ through which a steady current $$I = 60\ \text{mA} = 0.06\ \text{A}$$ is flowing.

If the resistance of the entire potentiometer wire is $$R_{\text{w}}$$, then, by Ohm’s law (stated as $$V = IR$$), the total potential difference across the whole wire is

$$V_{\text{total}} = I\,R_{\text{w}}.$$

The potential gradient (potential drop per unit length) along the wire is therefore

$$k = \frac{V_{\text{total}}}{L} = \frac{I\,R_{\text{w}}}{1200\ \text{cm}}.$$

During the experiment the auxiliary cell of emf $$E = 5\ \text{V}$$ is connected, and a null point is obtained at a length $$l = 1000\ \text{cm}$$. At the null point, no current flows through the cell, so its internal resistance does not play any role, and the emf is balanced exactly by the IR drop along that length of wire:

$$E = k\,l.$$

Substituting the expressions for $$k$$ and the given values, we write

$$5\ \text{V} = \left(\frac{I\,R_{\text{w}}}{1200}\right) (1000).$$

Now substituting $$I = 0.06\ \text{A}$$, we get

$$5 = 0.06\,R_{\text{w}}\left(\frac{1000}{1200}\right).$$

Simplifying the numerical fraction, $$\dfrac{1000}{1200} = \dfrac{5}{6},$$ so

$$5 = 0.06\,R_{\text{w}}\left(\frac{5}{6}\right).$$

Multiplying the numbers inside the brackets first:

$$0.06 \times \frac{5}{6} = 0.06 \times 0.8333\ldots = 0.05.$$ Hence

$$5 = 0.05\,R_{\text{w}}.$$

Finally, solving for $$R_{\text{w}}$$,

$$R_{\text{w}} = \frac{5}{0.05} = 100\ \Omega.$$

Hence, the correct answer is Option D.