To measure the internal resistance of a battery, potentiometer is used. For R = 10Ω, the balance point is observed at l = 500 cm and for R = 1Ω the balance point is observed at l = 400 cm. The internal resistance of the battery is approximately:

In the potentiometer method for measuring the internal resistance of a cell, the relationship is given by $$r = R\left(\frac{l_1 - l_2}{l_2}\right)$$ where $$l_1$$ is the balance length when the cell is in open circuit and $$l_2$$ is the balance length when an external resistance $$R$$ is connected.

Since the balance length is proportional to the terminal voltage across the cell, when the external resistance $$R$$ is connected the terminal voltage is $$V = \frac{E\,R}{R + r}$$ and hence the corresponding balance length $$l$$ satisfies $$l \propto \frac{E\,R}{R + r}\,.$$

For $$R_1 = 10\ \Omega$$ and balance length $$l_1 = 500\ \text{cm}$$, it follows that $$500 \propto \frac{10E}{10 + r}\quad\text{...(1)}$$ and for $$R_2 = 1\ \Omega$$ with balance length $$l_2 = 400\ \text{cm}$$, $$400 \propto \frac{E}{1 + r}\quad\text{...(2)}\,. $$

Dividing equation (1) by equation (2) gives $$\frac{500}{400} = \frac{10(1 + r)}{10 + r}\,,$$ so $$\frac{5}{4} = \frac{10 + 10r}{10 + r}\,.$$ Cross‐multiplying yields $$5(10 + r) = 4(10 + 10r),$$ $$50 + 5r = 40 + 40r,$$ $$10 = 35r,$$ $$r = \frac{10}{35} = \frac{2}{7} \approx 0.286 \approx 0.3\ \Omega\,. $$

The correct answer is Option 2: 0.3 Ω.