In the potentiometer, when the cell in the secondary circuit is shunted with 4Ω resistance, the balance is obtained at the length 120 cm of wire. Now when the same cell is shunted with 12Ω resistance, the balance is shifted to a length of 180 cm. The internal resistance of cell is_________Ω

We need to find the internal resistance of a cell using potentiometer readings with two different shunt resistances.

Formula: When cell is shunted with resistance S, the balance length is:

$$l = \frac{E \cdot S}{S + r} \times k$$ (where k is a constant depending on the potentiometer)

For shunt S₁ = 4Ω: $$l_1 = 120$$ cm → $$120 = \frac{E \times 4}{4+r} \times k$$

For shunt S₂ = 12Ω: $$l_2 = 180$$ cm → $$180 = \frac{E \times 12}{12+r} \times k$$

Taking the ratio:

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

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

$$2 \times 3(4+r) = 3(12+r)$$

$$6(4+r) = 3(12+r)$$

$$24 + 6r = 36 + 3r$$

$$3r = 12$$

$$r = 4 \, \Omega$$

Therefore, the internal resistance is Option 3: 4 Ω.