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A cell of emf 10 volt and internal resistance $2 Ω$ gives a null point in a potentiometer and the balancing length is found to be 55 cm. Now the cell is short circuited through an external resistance. R and the balancing length is found to be 50 cm then the value of R is

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20 $Ω$

30 $Ω$

40 $Ω$

25 $Ω$

answer is B.

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Detailed Solution

$r = R (\frac{l_{1}}{l_{2}} - 1)$

$∴ R = \frac{2}{(\frac{55}{50} - 1)} Ω = 20 Ω$

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