A current of 2 A flows through a 2 Ω resistor when connected with a cell. When the resistance is replaced by 10 Ω, a current of 0.5 A flows. The internal resistance of the cell is

2/3 Ω

The correct answer is Option (4) → 2/3 Ω

Given:

Case 1: $I_1 = 2\ \text{A}$, $R_1 = 2\ \Omega$

Case 2: $I_2 = 0.5\ \text{A}$, $R_2 = 10\ \Omega$

Let the emf of the cell be $E$ and internal resistance be $r$.

From Ohm’s law for both cases:

$E = I_1(R_1 + r) = I_2(R_2 + r)$

Substitute values:

$2(2 + r) = 0.5(10 + r)$

$4 + 2r = 5 + 0.5r$

$2r - 0.5r = 5 - 4$

$1.5r = 1$

$r = \frac{1}{1.5} = \frac{2}{3}\ \Omega$

Therefore, internal resistance of the cell is $\frac{2}{3}\ \Omega$.