When a resistor of 11 ohm is connected in series with an electric cell the current flowing in it is 0.5 A. Instead, when a resistor of 5 $Ω$ is connected to the same electric cell in series, the current increases by 0.4A. The internal resistance of cell is

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1.5 $Ω$

2 $Ω$

2.5 $Ω$

3.5 $Ω$

answer is B.

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Detailed Solution

Current taken from the cell.
$\begin{array}{l} i = \frac{E}{r + R}; 0.5 = \frac{E}{11 + r} \to (1) \\ 0.5 + 0.4 = \frac{E}{5 + r} \to (2); \frac{(1)}{(2)}; \frac{5}{9} = \frac{5 + r}{11 + r} \\ Solving r = 2.5 Ω \end{array}$

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