In a potentiometer arrangement, a cell of emf $1.20 V$ gives a balance point at $30 cm$ length of the wire. This cell is now replaced by another cell of unknown emf. It the ratio of the emf’s of two cells is $1.5$ , the difference in the balancing length of the potentiometer wire in the two cases is

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$5.0 cm$

$12.5 cm$

$10 cm$

$15 cm$

answer is C.

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Detailed Solution

Balancing length $l_{1} = 30 c m$

Ratio of emfs of two sells $\frac{E_{1}}{E_{2}} = 1.5$

$\frac{E_{1}}{E_{2}} = \frac{l_{1}}{l_{2}}$

$\Rightarrow 1.5 = \frac{30}{l_{2}}$

$l_{2} = 20 cm$

Difference in the balancing length $= l_{1} - l_{2} = 30 - 20 = 10 cm$ .

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