A potentiometer circuit has been set up for finding the internal resistance of a given cell. The main battery, used across the potentiometer wire, has an emf of 2.0 V and a negligible internal resistance. The potentiometer wire itself is 4 m long. When the resistance R, connected across the given cell, has values of (i) infinity (ii) 9.5 Ω the balancing lengths on the potentiometer wire are found to be 3 m and 2.85 m, respectively. The value of internal resistance of the cell is

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0.25 Ω

0.95 Ω

0.5 Ω

0.75 Ω

answer is C.

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Detailed Solution

The internal resistance of the cell is r=l1l2-1RHere, l1=3 m,l2=2.85 m,R=9.5Ω∴ r=32.85-1(9.5Ω)=0.152.85×9.5Ω=0.5Ω

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