An ideal battery of emf 4V and resistance R are connected in series in the primary circuit of a potentiometer of length 1 m and resistance $$5\Omega$$. The value of R, to give a potential difference of 5 mV across 10 cm of potentiometer wire, is:

We have an ideal battery whose emf is $$4\text{ V}$$. This battery is connected in series with an unknown resistance $$R$$ and the potentiometer wire of resistance $$5\ \Omega$$. Because the battery is ideal, its internal resistance is zero, so the total resistance in the primary circuit is simply $$R + 5\ \Omega$$.

The current flowing through the potentiometer wire is therefore obtained from Ohm’s law, which states $$I = \dfrac{V}{\text{Total resistance}}$$. Stating the formula first,

$$I = \frac{\text{emf of battery}}{R + 5} = \frac{4}{R + 5}\ \text{ampere}.$$

Next, we consider the potentiometer wire itself. Its length is $$1\text{ m}=100\text{ cm}$$ and its resistance is $$5\ \Omega$$. Hence, the resistance per centimetre of the wire is

$$\text{Resistance per cm} = \frac{5\ \Omega}{100\ \text{cm}} = 0.05\ \Omega/\text{cm}.$$

Now we focus on a segment of length $$10\text{ cm}$$. The resistance of this segment is obtained by multiplying the resistance per centimetre by the length in centimetres, that is,

$$R_{10\text{ cm}} = 0.05\ \Omega/\text{cm} \times 10\ \text{cm} = 0.5\ \Omega.$$

According to the requirement of the question, the potential difference across this $$10\text{ cm}$$ portion must be $$5\text{ mV}=0.005\text{ V}$$. Using Ohm’s law again for this small section, we state the relation $$V = IR$$ and write

$$0.005 = I \times 0.5.$$

Solving for $$I$$, we divide both sides by $$0.5$$:

$$I = \frac{0.005}{0.5} = 0.01\ \text{A}.$$

So, the current that must flow through the entire primary circuit is $$0.01\ \text{A}$$. Substituting this value of $$I$$ back into the earlier expression $$I = \dfrac{4}{R + 5}$$, we get

$$0.01 = \frac{4}{R + 5}.$$

Now we cross-multiply to isolate $$R + 5$$:

$$0.01\,(R + 5) = 4.$$

Dividing both sides by $$0.01$$ gives

$$(R + 5) = \frac{4}{0.01} = 400.$$

Finally, subtracting $$5$$ from both sides yields the required resistance $$R$$:

$$R = 400 - 5 = 395\ \Omega.$$

Hence, the correct answer is Option C.