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When 6 identical cells of no internal resistance are connected in series in the secondary circuit of a potentiometer, the balancing length is l. If two of them are wrongly connected the balancing length becomes

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$\frac{l}{3}$

$\frac{2 l}{3}$

$\frac{l}{4}$

answer is B.

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Detailed Solution

$\frac{nE}{(n - 2 m) E} = \frac{l}{l'}$

$\Rightarrow \frac{6 E}{(6 - 2 \times 2) E} = \frac{l}{l'}$

$\Rightarrow l' = l / 3$

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