When $6$ identical cells of negligible internal resistance are connected in series in the secondary circuit of a potentiometer the balancing length is 'I', balancing length becomes 1/3 when some cells are connected wrongly , the number of cells connected wrongly are :

1

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3

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2

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4

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Solution

The correct option is C $2$
According to formula
$\frac{E_{1}}{E_{2}} = \frac{l_{1}}{l_{2}}$
given
$E_{1} = 6$
$E_{2} = 6 - 2 m$
$l_{1} = l$
$l_{2} = \frac{l}{3}$
$\frac{6}{6 - 2 m} = \frac{l}{\frac{l}{3}} = \frac{3 l}{l} = 3$
$6 = 18 - 6 m$
$- 12 = - 6 m$
$m = 2$
Answer: c

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When 6 identical cells of negligible internal resistance are connected in series in the secondary circuit of a potentiometer the balancing length is 'I', balancing length becomes 1/3 when some cells are connected wrongly , the number of cells connected wrongly are :

The correct option is C 2According to formulaE1E2=l1l2givenE1=6 E2=6−2m l1=l l2=l366−2m=ll3=3ll=36=18−6m−12=−6mm=2Answer: c

When $6$ identical cells of negligible internal resistance are connected in series in the secondary circuit of a potentiometer the balancing length is 'I', balancing length becomes 1/3 when some cells are connected wrongly , the number of cells connected wrongly are :

The correct option is C $2$
According to formula
$\frac{E_{1}}{E_{2}} = \frac{l_{1}}{l_{2}}$
given
$E_{1} = 6$
$E_{2} = 6 - 2 m$
$l_{1} = l$
$l_{2} = \frac{l}{3}$
$\frac{6}{6 - 2 m} = \frac{l}{\frac{l}{3}} = \frac{3 l}{l} = 3$
$6 = 18 - 6 m$
$- 12 = - 6 m$
$m = 2$
Answer: c