In a potentiometer experiment, it is found that no current passes through the galvanometer when the terminals of the cell are connected across 52 cm of the potentiometer wire. If the cell is shunted by a resistance of 5 $$\Omega$$, a balance is found when the cell is connected across 40 cm of the wire. Find the internal resistance of the cell.

First, recall that in a potentiometer, the potential drop along a uniform wire is proportional to its length. If the potential gradient (potential per unit length) is $$k$$, then

$$\text{potential drop} = k \times \text{length}.$$

We have a cell of emf $$E$$ and unknown internal resistance $$r$$. When this cell is connected to the potentiometer without any external load, the galvanometer shows null deflection at a length of 52 cm. Hence

$$E = k \times 52.$$

Now the cell is “shunted”, that is, a resistor of $$5 \,\Omega$$ is connected directly across its terminals. Under null conditions the potentiometer draws no current, so the only current leaving the cell is the current through this 5 $$\Omega$$ shunt. Let the terminal voltage across the cell in this loaded condition be $$V$$. Balance is now obtained at 40 cm, giving

$$V = k \times 40.$$

Dividing the two balance-point equations eliminates $$k$$:

$$\frac{E}{V} = \frac{k \times 52}{k \times 40} = \frac{52}{40} = \frac{13}{10}.$$

Next, relate $$V$$ to $$E$$, $$r$$ and the external resistance. With the 5 $$\Omega$$ resistor connected across the cell, the circuit inside the cell consists of the internal resistor $$r$$ in series with the emf $$E$$, while outside we merely have the 5 $$\Omega$$ resistance. The current through the cell is therefore

$$I = \frac{E}{r + 5}.$$

The terminal voltage $$V$$ is the drop across the external 5 $$\Omega$$ resistor:

$$V = I \times 5 = \frac{E}{r + 5}\times 5 = E\,\frac{5}{\,r + 5\,}.$$

Taking the ratio $$E/V$$ from this expression, we obtain the theoretical relation

$$\frac{E}{V} = \frac{r + 5}{5}.$$

But from the potentiometer readings we already have $$E/V = 13/10$$. Equate the two results:

$$\frac{r + 5}{5} = \frac{13}{10}.$$

Cross-multiply to solve for $$r$$:

$$10\,(r + 5) = 5 \times 13,$$

$$10r + 50 = 65,$$

$$10r = 65 - 50 = 15,$$

$$r = \frac{15}{10} = 1.5 \,\Omega.$$

Hence, the correct answer is Option C.