When an external resistance of \(5\ \Omega\) is connected across terminals of a cell, a current of \(0.25\text{ A}\) flows through it. When the \(5\ \Omega\) resistor is replaced by a \(2\ \Omega\) resistor, a current of \(0.5\text{ A}\) flows through it. The internal resistance of the cell is _______ \(\Omega\).

Let the EMF of the cell be \(E\) and its internal resistance be \(r\).

The current \(I\) in the circuit is given by \(I = \dfrac{E}{R + r}\).

For the first case, \(R_1 = 5\ \Omega\) and \(I_1 = 0.25\text{ A}\):

\(0.25 = \dfrac{E}{5 + r} \Rightarrow E = 0.25(5 + r)\)

For the second case, \(R_2 = 2\ \Omega\) and \(I_2 = 0.5\text{ A}\):

\(0.5 = \dfrac{E}{2 + r} \Rightarrow E = 0.5(2 + r)\)

Equating the expressions for \(E\):

\(0.25(5 + r) = 0.5(2 + r)\)

\(5 + r = 2(2 + r)\)

\(5 + r = 4 + 2r\)

\(r = 1\ \Omega\)

Answer: \(1\)