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In the given figure, battery E is balanced on 55 cm length of potentiometer wire but when a resistance of 10 $Ω$ is connected in parallel with the battery then it balances on 50 cm length of the potentiometer wire. Then internal resistance of r the battery is

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1 $Ω$

3 $Ω$

10 $Ω$

5 $Ω$

answer is A.

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Detailed Solution

$\begin{array}{l} r = (\frac{l_{1} - l_{2}}{l_{2}}) \times R \\ r = (\frac{55 - 50}{50}) \times 10 = 1 Ω \end{array}$

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