The balancing length for a cell is 560 cm in a potentiometer experiment. When an external resistance of 10 $$\Omega$$ is connected in parallel to the cell, the balancing length changes by 60 cm. If the internal resistance of the cell is $$\frac{n}{10}$$ $$\Omega$$, where n is an integer then value of n is

In a potentiometer experiment the potential gradient along the wire is constant, so the emf or terminal potential difference of a cell is directly proportional to its balancing length. We write this fact as

$$E = k\,L_1 \qquad\text{(1)}$$

where $$E$$ is the emf of the cell, $$k$$ is the potential gradient of the potentiometer wire and $$L_1$$ is the first balancing length.

We are told that without any external load the balancing length is

$$L_1 = 560\ \text{cm}.$$

Now a resistance $$R = 10\;\Omega$$ is connected externally (in parallel with the cell, i.e. as a load). A current now flows through the cell whose internal resistance is $$r$$. Under load the cell no longer supplies its full emf, it supplies a smaller terminal potential difference $$V$$ given by the well-known relation

$$V = E\;\frac{R}{R + r}.$$

This reduced potential difference again balances the potentiometer wire but at a different point. The question states that the balancing length “changes by 60 cm”. Because the terminal potential difference is smaller than the emf, the new balancing length must decrease, so

$$L_2 = L_1 - 60\ \text{cm} = 560\ \text{cm} - 60\ \text{cm} = 500\ \text{cm}.$$

Using the potentiometer proportionality for the second situation we have

$$V = k\,L_2 \qquad\text{(2)}.$$

Dividing equation (2) by equation (1) eliminates the unknown $$k$$:

$$\frac{V}{E} = \frac{L_2}{L_1}.$$

Substituting the numerical lengths,

$$\frac{V}{E} = \frac{500}{560} = \frac{50}{56} = \frac{25}{28}.$$

But we also have from the cell-load relation

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

Equating the two expressions for $$V/E$$ we get

$$\frac{R}{R + r} = \frac{25}{28}.$$

Now we substitute $$R = 10\;\Omega$$ and solve algebraically for $$r$$:

$$\frac{10}{10 + r} = \frac{25}{28}.$$

Cross-multiplying gives

$$28 \times 10 = 25 \,(10 + r).$$

So

$$280 = 250 + 25r.$$

Rearranging,

$$280 - 250 = 25r \quad\Longrightarrow\quad 30 = 25r.$$

Hence

$$r = \frac{30}{25} = 1.2\;\Omega.$$

The internal resistance is expressed in the question as

$$r = \frac{n}{10}\;\Omega.$$

Equating this to the value we just found,

$$\frac{n}{10} = 1.2 \quad\Longrightarrow\quad n = 1.2 \times 10 = 12.$$

Hence, the correct answer is Option 12.