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The balancing length of a potentiometer wire is 220cm, when a cell of emf 1.1V is used in the secondary circuit. If it is replaced by a cell of emf 2.5V, the balancing length becomes

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300 cm

500 cm

250 cm

400 cm

answer is D.

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Detailed Solution

$\frac{E_{1}}{E_{2}} = \frac{l_{1}}{l_{2}}; \frac{1 .1}{2 .5} = \frac{220}{l_{2}} \Rightarrow \frac{11}{25} = \frac{220}{l_{2}} \Rightarrow l_{2} = 500 cm$

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