JEE Physics 2026 Question with Solution

Given: The cell is shunted with external resistances $R_1 = 4\,\Omega$ and $R_2 = 12\,\Omega$. The corresponding balance lengths are $l_1 = 120\,\text{cm}$ and $l_2 = 180\,\text{cm}$.

Find: The internal resistance $r$ of the cell.

Concept: In a potentiometer, balance length is proportional to terminal voltage.

$$\text{Balance length} \propto \text{terminal voltage}$$

For a cell of emf $E$ and internal resistance $r$ connected to an external resistance $R$,

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

Step 1: Write expressions for the two balance lengths. For $R_1 = 4\,\Omega$ and $l_1 = 120\,\text{cm}$,

$$l_1 \propto \frac{4}{4+r}$$

For $R_2 = 12\,\Omega$ and $l_2 = 180\,\text{cm}$,

$$l_2 \propto \frac{12}{12+r}$$

Step 2: Take the ratio of the balance lengths.

$$\frac{120}{180} = \frac{\frac{4}{4+r}}{\frac{12}{12+r}}$$ $$\frac{2}{3} = \frac{4(12+r)}{12(4+r)}$$

Step 3: Solve for $r$.

$$2 \times 12(4+r) = 3 \times 4(12+r)$$ $$24(4+r) = 12(12+r)$$ $$96 + 24r = 144 + 12r$$ $$12r = 48 \Rightarrow r = 4$$

Therefore, the internal resistance of the cell is $4\,\Omega$. The correct option is B.