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 is1.5 , the difference in the balancing length of the potentiometer wire in the two cases is

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15 cm

12.5 cm

10 cm

5.0 cm

answer is C.

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Detailed Solution

Balancing length l1=30 cm Ratio of emfs of two sells E1E2=1.5 E1E2=l1l2 ⇒1.5=30l2 l2=20 cm Difference in the balancing length = l1−l2=30−20=10 cm .

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