When $6$ identical cells of no internal resistance are connected in series in the secondary circuit of a potentiometer, the balancing length is $l$ , balancing length becomes $\frac{l}{3}$ when some cells are connected wrongly are

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Solution

The correct option is A $2$
Given, $l_{1} = l; l_{2} = \frac{l}{3}; n = 6$

$E_{1} = 6 E; E_{2} = E (6 - 2 m)$

Comparision of emf's of two cell is,

$\frac{E_{2}}{E_{1}} = \frac{l_{2}}{l_{1}} \Rightarrow \frac{E (6 - 2 m)}{6 E} = \frac{l}{3 l}$

$\Rightarrow 6 - 2 m = 2$

$\Rightarrow m = 2$

 Hence, $(A)$ is the correct answer.

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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, balancing length becomes l3 when some cells are connected wrongly are

The correct option is A 2Given, l1=l ; l2=l3 ; n=6 E1=6E ; E2=E(6−2m) Comparision of emf's of two cell is, E2E1=l2l1⇒E(6−2m)6E=l3l ⇒6−2m=2 ⇒m=2  Hence, (A) is the correct answer.

When $6$ identical cells of no internal resistance are connected in series in the secondary circuit of a potentiometer, the balancing length is $l$ , balancing length becomes $\frac{l}{3}$ when some cells are connected wrongly are