A cell of emf 2.2 V gives a current of 0.2 A through a resistance of 9 Ω. The internal resistance of the cell will be

2 Ω

The correct answer is Option (2) → 2 Ω

Given:

$\mathcal{E} = 2.2\ V,\ I = 0.2\ A,\ R = 9\ \Omega$

Using the formula for a cell with internal resistance:

$\mathcal{E} = I(R + r)$

Rearranging for $r$:

$r = \frac{\mathcal{E}}{I} - R$

Substitute the values:

$r = \frac{2.2}{0.2} - 9$

$r = 11 - 9 = 2\ \Omega$