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A potentiometer circuit has been set up for finding the internal resistance of a given cell. The main battery, used across the potentiometer wire, has an emf of 2.0 V and a negligible internal resistance. The potentiometer wire itself is 4 m long. When the resistance R, connected across the given cell, has values of
(i) infinity (ii) 9.5 $Ω$ the balancing lengths on the potentiometer wire are found to be 3 m and 2.85 m, respectively. The value of internal resistance of the cell is

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0.25 $Ω$

0.95 $Ω$

0.5 $Ω$

0.75 $Ω$

answer is C.

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Detailed Solution

The internal resistance of the cell is

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

Here, $l_{1} = 3 m, l_{2} = 2.85 m, R = 9.5 Ω$

$∴ r = (\frac{3}{2.85} - 1) (9.5 Ω) = \frac{0.15}{2.85} \times 9.5 Ω = 0.5 Ω$

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