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Q.

Balancing length of a potentiometer wire of a cell of emf 20 volt is 80 cm. What will be the balancing length of a cell of emf 15 volt in the same potentiometer wire?

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a

65 cm

b

55 cm

c

60 cm

d

70 cm

answer is C.

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Detailed Solution

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$\frac{E_{2}}{E_{1}} = \frac{l_{2}}{l_{1}} \Rightarrow l_{2} = 80 \times \frac{15}{20} c m = 60 c m$

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